2.2.1 Dead loads

The dead loads are the self-weight of the structure and the weight of the permanent fixtures on the structure. The self-weight includes weight of the individual structural members and depends on their overall dimensions and the material of construction. Since the section of the member is derived based on the overall load application which itself is a function of the sizes being adopted, the process of initial determination of dead weight of structural members is somewhat iterative. For this, suitable assumptions about the components sizes are made in the beginning and are verified at the end, with the derived sections. Any correction can be incorporated at that stage.

Besides weight of the bare members, the dead weight takes into account permanent finishing like plaster, floorings, partitions, etc. The weight of all permanent installations, too, is considered within the dead load category. The data pertaining to the unit weight of the construction materials or the weight of the installations are obtained from the design codes and the manufacture’s specifications.

2.2.2 Live loads

The live loads, or the imposed loads, are the weight of its habitants, stored materials, temporary fixtures or movable equipment. The prominent difference between dead load and live load is that the former remains fixed but the latter is variable during the use of the structure. The live loads may also change their point of application that needs to be accounted while in the process of design.

The live loads on a structure can be classified into the following main categories.

• Weight of movable furniture and partitions
• Weight of occupants of the building
• Loads due to impact and vibrations
• Weight of movable equipments and fixtures

Most of the counties across the world have adopted certain standard values of live loads applicable on floors, roofs and other elements of the buildings. These standard values are related to factors like nature of building, occupancy, floor location, accessibility, etc. These live loads are expressed as uniformly distributed loads (UDL) over the area of application. Any dynamic or concentrated load is considered over and above this. Besides individual countries, the International Organization of Standardization (ISO) has published the codes on imposed floor loads for various types of buildings.

2.2.3 Lateral loads

Lateral loads are mostly imposed by nature in the form of wind and earthquakes. However, in the case of retaining structures like tanks and soil retaining walls, lateral loads also occur due to the pressure imposed by the material retained. Such load can be calculated with the help of established scientific theories.

The wind and earthquake-induced loads do affect the taller structure more than low-rise structures and, therefore, are of extreme importance in the analysis of high-rise structures and buildings. Both of these loads are transient loads.

Wind load:

In the case of wind loads, the applicability depends on the geographical region, geometry of the structure, its proximity with other structures, etc. Relative to the surface of earth, the natural air remains in motion. The difference in solar and terrestrial radiations results in differences in temperatures, which gives rise to convection, both upwards or downwards. The high-speed wind usually blows horizontal to the ground. And, therefore, the term wind denotes almost always horizontal wind. Vertical winds are specifically identified as such. Weak winds (i.e., winds at low speeds in the range of 5 kmph to 15 kmph) are termed as breeze and strong winds (i.e., winds at a speed of about 60 kmph) are termed storms. It is the storm the effect of which needs to be factored while designing.

In addition to normal wind, there are sudden blasts of wind that may last for few seconds. These are called gusts and may also require consideration. These gusts cause an increase in air pressure, but their effect on the stability of the building may not be significant. Often, the gusts influence only parts of the building and the increased local pressures may be more than balanced by a momentary reduction in pressure elsewhere.

As the wind velocity increases with height, a gradient of wind pressure is developed over the structure. Other important factors that come into play are the effective frontal area, depth with respect to wind direction, angle of attack, etc. The potential effect due to wind is analyzed and derived after giving consideration to all these factors.

Earthquake load:

The earthquake or seismic loads (since they occur due to movement within sub-terrain strata) are important to consider if the location of the structure falls in the active seismic zone. Seismic areas are regions, which are geologically young and unstable and, therefore, have a likelihood of experiencing earthquakes. The earthquake shock causes the earth to move and the structure too vibrates consequently. The resultant effect can be resolved in three mutually perpendicular directions. The horizontal effect on structure is generally considerable compared to the vertical one. The earthquake causes effective acceleration. The ratio of this acceleration to the acceleration due to gravity (g) is known as the seismic coefficient, which is applied during the design process as additional load over the structure.

Based on the type and importance of the structure, that may be designed either for horizontal seismic forces or both horizontal seismic forces and vertical seismic forces. The seismic accelerations for the design may be arrived at, from seismic coefficient, which is defined as the ratio, of acceleration, due to earthquakes, and acceleration, due to gravity.

Hydrostatic load:

The hydrostatic or water pressure is crucial for those structures which either retain aqueous liquids or are in contact with the underground water table. The earth or soil pressure again is imposed on underground structures or the soil retaining arrangements.

The hydrostatic load acts on a horizontal surface in accordance with the distribution shown in Figure 2.8. As is obvious, the intensity of the pressure is highest at the bottommost point of the surface. Therefore, the deep water structures are exposed to high intensity of pressure and overall load.

Earth or Soil load:

Like hydrostatic pressure, the earth pressure is the lateral force that acts on the structures those retain soil or has been exposed to soil. However, unlike hydrostatic pressure, the extent of soil pressure depends on the intrinsic physical properties of the soil like its unit weight, cohesion, angle of internal friction, etc—besides factors like extent of overburden, angle of overburden, etc.

The earth pressure may be calculated with the help of established theories of soil mechanics. For instance, as illustrated for a granular sort of soil like sand, the distribution of pressure on the retaining surface would be distributed in a triangular way. In the case of a more cohesive soil like clay, the pressure as well as the distribution pattern changes considerably. In this example, the active pressure at the depth H would be

It must be noted that for the same system, the value of the earth pressure would go up substantially if the earth pressure turns passive from the active earlier.
In addition to the normal pressure due to soil, the retaining structure may be exposed to more loads under special circumstances. The structures that retain expansive soils like clay are subject to further lateral pressure when the soils expand under moisture movement.

2.2.4 Snow load

In cold zones, the weight of snow on the structure can be predominant. The load due to snow acts as a dead load over the structure. Its occurrence and magnitude depend upon climatic conditions of the area and the exposure of the structure, its shape and geometry of its components.

The snow load depends upon the latitude of the place and the atmospheric humidity. The snow load acts vertically and it is expressed in kN/M² of plan area of the building or structure. The actual load due to snow depends upon the shape of the roof and its capacity to retain the snow. When actual data for snow load are difficult to obtain, load due to snow may be assumed to be 25 N/M² per mm depth of snow over the plan area of the structure.

During analysis, it is a usual practice to assume that snow load and maximum wind load will not be acting simultaneously on the structure.

2.2.5 Thermal loads

Thermal loads (or stresses) are generated because of inbuilt constraints that restrict the free expansion or contraction of structural members that otherwise would have taken place due to temperature variation. These loads are a function of the thermal property of the construction material used.

Variations in temperatures result in expansion and contraction of the structural members, while they are contained in the overall arrangement. This induces stresses within the members, which can be quite substantial at times. The range of variation in temperature varies from location to location and also from season to season. The temperature effects thus need to be accounted for properly and adequately. The combined value of stress developed because of design loads and due to temperature effects should not exceed the allowable limit. It is very important to take temperature stresses into account when the member is likely to face large variation in temperatures. The magnitude of thermal load is calculated as

The coefficient of expansion is a physical property of the material and has different values for different materials.

2.2.6 Impact and vibrations

Dynamic or moving loads cause the impacts and vibrations on structures. The dynamic effects resulting from moving loads are accounted for by adopting an impact factor. The live load on the structure is computed in the normal way and the magnitude of live load is then increased by the impact factor. This takes care of development of additional stresses, due to the dynamic nature of the loads.

Vibrations can be an important factor while designing an industrial structure in which continuous vibrations of machines can cause severe stress in the supporting structures.

2.2.7 Erection and handling loads

Erection loads include all forces, to which a structure, or a part of a structure, is subjected during its transportation and/or erection. Erection loads also take into account the placement and storage of materials during the construction phase. Proper provisions should be made, for example, by providing additional reinforcement in the RCC, or temporary bracings in steel structures, to take care of all stresses caused during erection. The stresses developed because of handling and erection should never exceed the allowable values.

2.5.1 Free-body diagram

When a body is in the state of static equilibrium, every part of it must also be in the same state. Therefore, if a cut is made at any place in the body and a part is isolated, that part must also be in equilibrium under the forces acting on it and the internal reactions at the section of cut.

A free-body diagram is a graphic representation of the body (i.e., a structure, an element of a structure or a piece of an element) with all of its connecting parts removed. Each one of these removed parts is replaced with external forces, which correspond to the internal forces present where that part connects with the body, such that the original state of equilibrium of the body is maintained. All the original loads on the body are also included in the force system.

All the required information to analyze and solve the structure under a force system is included in the ‘free-body diagram’. A ‘free-body diagram’ is a very useful tool to model the structure, structural element or sub-element that is under scrutiny. These diagrams may not be really necessary in relatively simple structural situations but have great utility when the structures become complex.

2.5.2 Preparation of a free-body diagram

An example of how to prepare a free-body diagram has been illustrated below. We take the case of a beam AB of length l, supported at both ends on pin supports (which allow for rotation of the joint). The beam has been loaded at its center C with load W. The arrangement is shown in Figure 2.12.

At supports A and B, upward reaction forces R_A and R_B would act. It is clear that the sum of both these reactions is equal to the total load W.
Putting mathematically,

R_A + R_B     = W

Taking the moment of all forces about support A,

W*(l/2) – R_B*l     = 0

From this, R_B = W/2

Similarly, R_A = W/2

We have decided to prepare a free-body diagram of the portion AD of the beam. For this, a cut is made at location D, which is at a distance x from the center, and the portion AD is separated from the system. Now, the forces acting on the isolated portion are as follows.

• Vertical external load W at point C
• Vertical reaction at support A, equivalent to 

In addition, we have a moment at the end D of the free body M_D and a shear force V_D acting at D.

Applying the equations of equilibrium for the case:

This demonstrates how useful a free-body diagram can be in the process of analysis of a structure.
To make it clearer, the process for determining the reactions at the support of a cantilever beam with a concentrated load W acting at its free end is explained. The fixture at the wall is a rigid support for the beam, as it allows neither any rotation nor any displacement of the member. The arrangement is shown in Figure 2.13.

The steps involved are as follows:

• First of all, one needs to draw a single-line diagram describing the system.
• Next, cut the beam free from the wall and replace the wall with the forces that were supporting the beam at the wall before it was cut free. The exact magnitude of these forces is not known at this stage but the nature is known, as they are acting in a way that keeps the beam in equilibrium.
• The internal forces in the beam before it was cut free from its support are determined when the forces that will keep the free-body diagram in equilibrium are found. The magnitude of the forces at the section cut will be identical to the internal forces in the beam at that point before it was cut.

A fixed support will resist translation in all directions as well as the rotation of the member. This will lead to the generation of corresponding restraining reactions that are consistent with the load system. In our situation, no lateral or horizontal force has been acting on the beam; therefore, there will be no horizontal reaction mobilized at the support. By applying the principles of equilibrium in the free-body figure above,

Σ Fx = 0
And, Σ Fy = W + V = 0;      hence, V = (–) W

This indicates that the vertical reaction at the wall end will be equal in value to the load W but will act in the opposite direction. The load in combination with vertical reaction force causes a clockwise couple of value W×l that must be resisted by a moment at the end of the cut section acting in a counter-clockwise direction.
Again, applying the principle of equilibrium to obtain the value of this balancing moment,

Σ M = 0

Hence, balancing moment = (–) W×l

This explains the procedure of determining the internal forces within a structural element, with the help of a free-body diagram. To solve complex structures, a free-body diagram is prepared for each segment one after the other and the internal force determination is done in stages.
The same has been illustrated as below.

Illustration – 2.1: The beam shown in Figure 2.14 is simply supported at both the ends. We will draw the free-body diagram to determine reactions at both the supports.

In order to determine support reaction forces, the free-body diagram of the entire structural system has been drawn in Figure 2.8. For this, the supports at A and B have replaced with the upward reaction R_A and R_B respectively, balancing the downward load of 2T.
We know that moment at A is equivalent to zero. Hence,

M_A = 0

On the other hand, from the free-body diagram,

M_A = R_B x 4 – 2 x 3

Therefore,

R_B x 4 – 6 = 0

R_B = 1.5kN

From the equilibrium condition Σ Fx = 0,

R_A + R_B – 2    =    0

R_A + 1.5 – 2    =    0

R_A    =        0.5 kN

These derived values have been indicated in the free-body diagram, which has been drawn for the system.

2.7.1 Statically determinate structures

Structures which can be completely analyzed with the help of statics alone are known as “statically determinate structures.” The process of finding the internal forces and reactions in these structures involves only the application of the equations of equilibrium. The process needs information on the overall arrangement of the members in the structure and the types of support constraints. At the same time the process is independent of all other geometric considerations, which means that the determination of internal and external forces and reactions can be done without giving any attention to the deformed shape of the structure.

To explain further, a determinate structure will have only as many supports as are absolutely essential for its stability under imposed loading conditions. Thus, a statically determinate structure can also be defined as the one in which the number of external forces and reactions equals the number of the equations of statics which are applicable for the particular situation.

2.7.2 Statically indeterminate structures

Those structures which cannot be fully analyzed with the help of statics alone are known as statically indeterminate structures. In these cases, the number of unknown reactions exceeds the number of applicable equilibrium equations.

Let us look at an example. Consider the case of a beam AB in Figure 2.19, which is fixed at end A and supported at end B. Structurally, this arrangement is known as a propped cantilever. An inclined load W has been imposed on the beam.

As is obvious from the diagram, there are four unknown support reactions: three at support A and one at support B. However, there are only three equations of statics (ΣH, ΣV and ΣM = 0) available; therefore, beam AB is a statically indeterminate structure.

As we are aware by now, in the case of plane structures there can only be three possible equations of equilibrium. Does that mean that all structures with more than three unknown support reactions are statically indeterminate? Not necessarily. Consider a modification in the beam in Figure 2.19. Between A and B, a hinge has been installed at C. Clearly, the condition that the moment at C should be equal to zero provides an additional equation because the vertical reaction at B can now be found by considering the moment equilibrium of CB about C. With this additional equation, it has become possible to analyze all the four unknown reactions and, therefore, the structure has become statically determinate.

Let us consider another situation for beam AB. We decide to remove the support at B. The structure will still be stable under the influence of the load W. Simultaneously, it has become statically determinate even without the introduction of any hinge (in fact, in this situation, the introduction of a hinge would render it unstable). It can be concluded that the support B was redundant. We can therefore say that in a statically indeterminate structure, the number of support reactions is more than those necessary for the stability of the structure. This difference, or alternatively the number of redundant members in the structure, is known as the degree of indeterminacy of the structure.

2.7.3 Determination of degree of indeterminacy

As stated earlier, the degree of indeterminacy is the number of redundant support reactions in a structure, which, on removal, would make the structure statically determinate. Two popular methods to determine the degree of indeterminacy have been explained below. The scope has been confined to plane or two-dimensional structures only.

Cantilever tree method:

This method is useful for rigid jointed frames.
To work out the degree of determinacy for frames by this method, the first step is to ensure that all the joints be rigid ones. This is achieved by providing extra restraints at the non-rigid joints (to convert them into rigid ones). The number of such restraints is counted (R).
Now, the closed frame structure is cut in such a way that each isolated portion resembles a tree with branches on one or both sides. The numbers of cuts in the total structure are also counted (C).

Degree of Indeterminacy = 3×C – R

We will consider the example of the frame shown in Figure 2.20.

The frame in the example has two hinged supports. To convert them to rigid, each has had a single restraint applied.

The underlying principle can be applied to any structure to find its indeterminacy. In the case of indeterminate beams, what is required is to make any one end of the beam fixed (if it is not already) by introducing restraints. Once this has been done, remove all the redundant supports. The degree of indeterminacy can be worked out with the following relationship.

Degree of Indeterminacy     = Supports Removed – Constraints Added

Method of joints:

This method is suitable for pin-jointed structures with a relatively large number of members. A truss is one example of a structure of this type. The principle applied is that each member of the structure has an unknown axial force, the number of which is added to the total number of support reactions. Now, each of the pin-joints contributes two of equilibrium equations. Therefore in the case of a plane structure:

Let us take the truss shown in Figure 2.21 as an example for illustration. The truss has a total of 13 members and 8 joints. There are two supports: a hinge and a roller. Total support reactions will be three in number (two for a hinge and one for a roller support).

Degree of Indeterminacy = 13 + 3 – 2×8 = 0

The zero degree of indeterminacy indicates that the truss is statically determinate. It may be noted here that if the roller support is converted into a hinged one, the total number of support reactions will increase by one; hence, the degree of indeterminacy will become one and the truss will no longer remain statically determinate.

2.8.1 Sign for bending moment and shear force

Normally, the bending moment which tends to produce the bending of the beam concave upwards is termed a positive bending moment and opposite to it, i.e. concave downwards, is denoted as negative. In the case of shear force, the force that tends to move the left segment upwards with respect to the right is considered a positive shear force. Similarly, the force that tends to do the opposite is termed a negative shear force. These sign conventions are shown in Figure 2.23.

2.8.2 Bending moment – influencing factors

In common structural problems, the quantum of the bending moment (BM) generated, including its nature, depends upon the following.

• Nature of loading
• Nature of supports
• Geometry of the structure

The nature of traverse loading is crucial for the distribution of the bending moment along the overall span of the beam. A load application can be of several types. A concentrated load or point load is the one that acts on a very small area compared to its intensity, and can be considered to be acting at a point. On the other hand, distributed loads have a larger area of application and are considered to be acting on a considerable length of the structure. Distributed loads can have several patterns of distribution. A uniformly distributed load is one that has even distribution of load along the length of the application. The uniformly varying or triangularly distributed load is the pattern where the application of the load increases at a constant rate. There can also be a non-uniformly distributed load. All these loading conditions are shown in Figure 2.24. It may be worth mentioning that in a large number of cases, the actual patterns encountered are non-uniformly distributed but from the point of view of ease in analysis, they are considered to be either uniformly distributed or uniformly varying patterns of loading.

The loading pattern on the structure affects the development of bending moment in a major way.

Loads can be further classified as stationary or moving, as we will see later on.

The type of supports and overall geometry of the structural system also have a major influence upon the amount and distribution pattern of the bending moment and shear force. Two situations of beams having different spans (hence geometry), loading situations and support locations have been analyzed below.

To understand the concept of bending moment and shear force development clearly, we will consider a couple of situations and observe the process of determination.

Illustration – 2.2: A 4 m span cantilever beam has a single concentrated load of 4 kN working at a point 3 m away from the fixed end. There has been no other load over the beam. Find the value of the bending moment at the support.

As we know, a cantilever is a beam that has one end fixed and other free. The fixed end is rigid and prevents any rotation.
If we consider the forces for equilibrium, the only external load is 4 kN that is acting downwards. The only reaction possible is at support and, for the sake of equilibrium, that would be equivalent to 4 kN, acting upwards.

Considering section at the support, since the reaction does pass through it, the moment would be equivalent to the product of only load and the distance from the support.

M = 4 kN x 3m
=       12 kNm

Therefore, the value of the bending moment at support will be 12 kNm. As it is obvious that the beam will curve concave downwards, the sign of bending moment will be negative.

Illustration – 2.3: A simply supported beam, shown in Figure 2.25, is subjected to single-point load as indicated. We will draw the bending moment and shear force diagram for the entire span.

The first step is to determine the reactions at both the supports.
For this, let us use the equilibrium condition Σ M_A = 0.

Now, in order to determine the reaction force at A, we will use the equilibrium condition Σ F_Y = 0:

With R_A and R_B known, we can proceed to draw the bending moment and shear force diagrams in the following way. Since the bending moment at a section has been defined as the algebraic sum of the moment due to all the load on either side of the section and the shear force has been defined as the net imbalance in the lateral load on either side of the section, we have to use these definitions by converting them into mathematical expressions, to work out the value of the bending moment and shear force.

First of all, consider a section C on the span, located between point A and the 10kN load, spaced x from A. As per definition, moment at this section,

M_C =   R_A*x
M_C =   4x

As is obvious from the expression, the relationship between bending moment and the distance from the support is linear.

Now, consider another section D on the span, located between B and the 10kN load, spaced y from B. As per definition, moment at this section,

M_D =   R_B*y

M_D =   6y

As is obvious from the expression, the relationship between bending moment and the distance from the support is linear.

Resultant bending moment diagram is shown in Figure 2.19.
To draw shear force diagram, we will treat the entire span as two sections: first one, A to the 10 kN load and the second one as B to 10 kN load.

In the first section, the single vertical force acting between A and just left of the load is the reaction measuring 4 kN, irrespective of the location. Therefore, the shear force all along this segment would be 4 kN. Since the tendency would be to move the left segment upward with respect to right one, the sign of shear force is positive.

In the second section, the single vertical force acting between B and just right of the load is the reaction measuring 6 kN, again irrespective of the location. Therefore, the shear force all along this segment would be 6 kN. As the tendency now would be to move the right segment upward with respect to the left one, the sign of shear force is negative.

The section of the span subjected to downward load of 10 kN is the one where shear force changes its sign.

The shear force diagram is shown in Figure 2.25.

Illustration – 2.4: The objective is to draw the bending moment and shear force diagram for the beam shown in Figure 2.26.

The beam in the example is simply supported at both ends. We will start by computing the reactions at the support. For this, use the equilibrium equation Σ M_A = 0:

50×5 + 5×10(15 +2.5) – R_B × 20 = 0

Hence,

R_B = 56.25 kN

Now, by applying the equation Σ F_Y= 0,

R_A – 50 – 5×10 + 56.25 = 0

Hence,

R_A = 43.75 kN

Now, with R_A and R_B known, we will consider three sections along the span of the beam AB to derive the values of the bending moment and shear force along the entire span. The first one is at location X at a distance of x m from B, the second one at Y at a distance of y m from C and the third one at Z at a distance of z m from A.

Consider the section at X first. Again, as the bending moment at a section has been defined as the algebraic sum of the moment due to all the load on either side of the section and the shear force has been defined as the net imbalance in the lateral load on either side of the section, we will use these definitions to work out the value of the bending moment and shear force.

Consider segment AZ first. The loads acting on this segment have are in the diagram.

M_Z = R_Az = 43.75z

It is clear from the equation that the relationship between M_z and z is a linear one.

For z = 0 (Section A), M_A = 0
For z = 5 (Section C), M_C = 218.75 kN.m

Hence, the bending moment varies from zero at A to 218.75 kN.m at C in a linear fashion.

Similarly, V_Z = R_A = 43.75 kN

Therefore, the value of the shear force along AC is constant independent of the location:

Consider segment CY next and refer to the loads shown in this segment on the diagram:

M_y   = R_A (5 + y) – 50y
M_y   = 43.75 (5 + y) – 50y = 218.75 – 6.25y

Again, the relationship between M_y and y has been found to be linear.

Note that the values of M_c worked out previously and now are the same.

The value of shear force has been derived as a constant for the CD segment as well. However, notice the change in its sign. Also, the value of V_c derived earlier was 43.75 kN. The explanation for this is that due to the introduction of a point load at C, there has been an abrupt change in the value of the shear force. The value derived earlier is applicable to the segment on the left of C and the recent one is for the segment right of C.

Finally, we will consider segment BX, and watch the behavior of the bending moment and shear force along this portion of the beam. In the case of a uniformly distributed load, the total force from the block of the load is assumed to be acting at the centroid of the load block:

M_x = 56.25 x – 5x²

The above relationship between M_x and x is the equation of a parabola. Therefore, the bending moment diagram of a beam under the influence of uniformly distributed load will assume the shape of a parabola.

For x = 0 (Section B), M_B = 0
For x = 5 (section D), M_D = 156.25 kN.m

In the case of shear force,

V_x = 10x – R_B
V_x = 10x – 56.25

i.e., under the influence of uniformly distributed load, the variation in the shear force along the span is a linear one.

For x = 0 (Section B), V_x = (–) 56.25 kN
For x = 5 (Section D), V_x = (–) 6.25 kN

The graphical representation of the distribution of the bending moment and shear force is shown in Figure 2.26.

Illustration – 2.5: A similar exercise of drawing the bending moment and shear force diagram for the beam is illustrated in Figure 2.27.

The beam is supported at A and B, with A being a hinged support and B a roller support. BC is a cantilever span of 3 m. The force at point C is acting at an angle of 45 degrees to the axis of the beam.

To determine the bending moment and shear force on the different sections, we will only visualize the states of the beam segments, instead of drawing them separately. This will speed up the process.

First of all, we need to determine the support reactions. In this case, the load at point C is inclined at 45 degrees to the axis of the beam. This has to be resolved into two components, along and perpendicular to the axis respectively.

The horizontal component = 20cos 45º = 14.14 kN
The vertical component      = 20sin 45º = 14.14 kN

From the moment equilibrium about A,

30×6 – R_b9 + 14.14×12 = 0
R_b = 38.85 kN

From the vertical equilibrium,

R_a = 30 – 38.85 + 14.14 = 5.29 kN

The load at C has a horizontal component. Since only support A can offer resistance to the horizontal force, the entire horizontal reaction will be mobilized at A and will act along the centroidal axis of the beam.

The bending moment and shear force diagrams are shown in Figure 2.25.

Note that the bending moment was defined as the algebraic sum of the moments about the lateral axis at a section, due to loads on either side of the section. Since the horizontal force and reaction are acting along the centroidal axis, there has been no bending effect due to the action of these forces.

2.8.3 Bending moment and shear force diagrams – notable features

We have drawn the bending moment and shear force diagrams for the two examples above. It is worth examining the profile of the bending moment and shear force diagrams and their interrelation. In the case of the shear force diagram, the abrupt change in the shear force at a concentrated vertical load is represented by a vertical line drawn to join the two values on either side of the loading.

The second observation is that in both examples, the bending moment assumes the highest positive or negative value at the section where either the shear force is zero or reversal in its sign takes place.

We will now observe its mathematical derivation. We have seen that the relation between the moment at a section M_x and its location along the X-axis, designated by x, has a mathematical relation.

Hence, M_x = f (x)

To find the value/values of x that maximize the bending moment, we know from calculus that M_x must be differentiated with respect to x and the result equated to zero. Solving this equation for x gives the points where the bending moment is maximized. However, we have seen from the example that the shear force is zero at this section. Therefore, dMx/dx must be the function for the shear force, V:

i.e., dM_x = 0

dx
This relationship will be proved in the section on Strength of Materials. Therefore, it can be concluded that if the bending moment has a maximum value at a section along the span, the shear force at the section will become zero, and vice versa.

The third interesting feature, which is noticeable in the bending moment diagram of the second example, is the change in the sign of the bending moment at a certain section in the span. In this case, the section where reversal takes place is designated as I on the diagram. According to the sign convention for bending moment, a positive moment tries to deform the beam with concavity upward and the negative one with concavity downward. The reversal of the moment indicates that the shape of the curve is concave upwards in the segment AI and concave downwards in the segment IC. The point I, where the bending moment is zero, is called the point of inflexion (meaning no flexure) or point of contraflexure (meaning the direction of flexure changes)

2.9.1 Bending moment under moving loads

In the case of moving loads, it is critical to determine the specific location of the load set that causes the generation of the maximum bending moment at any section along the span.

For a set of concentrated loads, the maximum bending moment caused by the whole set occurs directly under one of the loads. Therefore, it is considered necessary to determine the values of the bending moments under each load, when the position of each load is positioned to cause maximum moment under it. The largest of these determined values will be the one governing the design of the beam.

In Figure 2.28, the case of a simply supported beam AB of span l is shown. The beam is subjected to a set of three loads W₁, W₂ and W₃ acting at a distance of a and b from each other. These loads move as a set on the entire span of the beam and are free to occupy any location.

We will now show the example of locating position of the central load W₂ when the bending moment under this load is the largest. Let us denote the resultant of this force system as W, acting at a distance of p from W₂.

Applying the equilibrium equation:

Σ M_b = 0 = R_al – W(l – x – p)

Hence, 

Now, the bending moment under W₂ is

To obtain the value of x for which M_w2 is maximized, we differentiate the equation derived for M_w2 with respect to x and equate the value to zero:

This relationship determines the location of the load at which it would cause the maximum bending moment. It can be expressed in general form as follows.

In the case of a set of loads acting on a beam, the bending moment under a particular load will assume maximum value if the load occupies a location such that the center point of the beam is at equal distance from that individual load and the resultant of the entire load system.

2.9.2 Shear force under moving loads

Under the influence of moving loads, the maximum shear force will occur at one of the supports. There can be two situations of a load set for which shear force can be the maximum:

• When the extreme left load of the set is over the left support, the shear force on the left support will be the maximum.
• When the extreme right load of the set is over the right support, the shear force on the right support will be the maximum.

In the case of moving loads, the shear force is determined for both the conditions and the most severe value out of these is adopted for the design exercise.

2.10.1 Statically determinate trusses

A truss for which the determination of all forces in the member, as well as the support reactions, can be done by applying the principles of statics is termed a statically determinate truss.

As we have seen earlier, in the case of a plane truss,

Degree of Indeterminacy = Total Members + Total Support Reactions – 2×Number of Joints

If a truss has a degree of indeterminacy greater than zero, the implication of this is that the truss either has (a) redundant member(s) or has more restraints at the supports than those absolutely essential from the point of view of stability.

If the degree of indeterminacy is less than one, the frame will probably be unstable under loads. There can be a situation where the configuration of the load system itself may provide stability to this type of frame, but inherently such a system is unacceptable. A system of this nature is called a mechanism.

In this section, we will observe the methods for force determination of members of statically determinate trusses. Two widely used analytical methods for this are:

• Method of joints
• Method of sections

Both use the principles of static equilibrium to find member forces. The method of joints is more appropriate when all member forces are to be determined and the method of sections is more suitable when only selected member forces are to be determined.

In this section, we will describe the method of joints.

2.10.2 Method of joints

In this case, each of the joints is analyzed separately by considering its equilibrium.

To explain the concept, consider the triangular pin-jointed truss shown in Figure 2.31. The load W has been applied at the crown C of the triangle. Node A has a hinge support and node B a roller support. For this truss,

Therefore, the truss is statically determinate. To derive the vertical support reaction at B, we consider the moment equilibrium of the whole truss about A:

From the vertical equilibrium:

R_a + R_b = W. Hence, R_a = 

Now, for the next step, we will consider node B in isolation and examine its stability under the influence of the forces acting on it. These forces are those in line with the longitudinal axes of the members AB and CB and the vertical reaction R_B. We initially assume that the forces in the members are tensile. That is, they act on the body of joint B as shown in Figure 2.31.

From the vertical equilibrium of joint B:

The negative sign coming from this equilibrium equation indicates that the assumed direction of force P_cb was incorrect and it is actually a compressive force.

Members in tension tend to pull their ends axially inwards and members in compression tend to push their ends axially outwards.

In the same way, if we consider the equilibrium of horizontal forces at node B, we get:

This is a tensile force.

The internal force in the member meeting at A can be determined in the same way. Considering the vertical equilibrium of A:

This force again is compressive in nature.

Similarly, considering the horizontal equilibrium at A:

Therefore, the horizontal component of the reaction at A is zero. Although the hinged support A has potential to mobilize horizontal reaction, in this case it has not happened due to symmetrical nature of the loading.

Illustration 2.6 – Using the method of joints, we will undertake the analysis of the truss shown in Figure 2.32, which has loads acting perpendicular to its top chord.

Also, Length l₁₂ = l₂₃ = l₃₅ = l₅₆ = l₂₅ = l₄₅ = 2.1213 m

In order to derive the support reactions, we will consider the moment equilibrium of node 1 and node 6, one by one:

To derive the horizontal reaction at node 1, equate the horizontal components of the external forces with the R₁h,

R1H = 20cos 45º + 10cos 45º = 21.2132 kN

Now, to determine the forces in the members, we will consider the horizontal and vertical equilibrium of joints separately.

For Joint 6,

Σ Fx = 0, F65(sin 45º) = 7.071, F65 = 10 kN (C)

Σ Fy = 0, F 65(cos 45º) = F65, F64 = 7.071 kN (T)

For Joint 5,

There are three members that meet at this node. Out of these, two members – 56 and 53 – are in the same axis and the member 45 has its axis perpendicular to them. Now, it is obvious that the members 56 and 53 cannot have any force components perpendicular to their own axis. Hence there cannot be induction of force in the member 45 as otherwise joint 5 would not meet the equilibrium criteria. Therefore,

For Joint 3,

Σ Fx = 0, F32(cos 45º) = F35(cos 45º),        F32 = 10 kN    (C)

Σ Fy = 0, F34     = F32(sin 45º) + F35(sin 45º), F34     = 14.1421 kN (T)

For Joint 2,

Joint 2 has member configuration similar to Joint 5. The difference is in the presence of a 20 kN load, which has been acting along the axis of the member 24.

For Joint 1,

All these member forces have been tabulated below. This table is useful to determine the required sectional properties of the member.

2.11.1 Influence lines for statically determinate beams

Consider the beam AB of span L, supported at both ends. The end A has a hinge support and B a roller support. It is subjected to a traverse load of unit value, which is moving along the span. At any instant, the load is at a distance cL from the end A. We will observe the variation in the bending moment and shear force at a section X along the span, which is located at a distance x from the support A. In addition, we will draw the influence lines for the support reactions. The entire structural arrangement is shown in Figure 2.33.

On the basis of equilibrium conditions,

The value of ‘c’ varies from 0 to 1. As this variation in the value takes place, the resultant influence line diagrams for R_a, R_b, V_x and M_x are indicated in Figure 2.33. In a similar way, influence line diagrams for any statically determinate structure can be determined.

We will now observe the application of the influence line diagram for structural analysis. With the influence line diagram, it is possible to determine the value of the stress function for a given set of loading. We will take the case of a load system consisting of a number of concentrated loads P1, P2, P3, P4,……. working at points 1,2,3,4, … along the span. If the values of ordinates of the influence line at those sections are a1, a2, a3, a4,……. respectively, the total value of this stress function can be derived by just algebraically summing the products of the magnitude of forces with their respective ordinates. Hence, the overall value will be

= P1(a1) + P2(a2) + P3(a3) + P4(a4) + ……….

In case the load is uniformly distributed instead of being concentrated, the influence line still can be used to determine the stress function. Suppose a load of intensity w per unit length is working on the span of the beam covering a length x from points X to X’ as shown in Figure 2.34. The influence line under consideration is for the support reaction R_a.

As shown in the diagram, the application of the load starts at X at a distance ‘a’ from end A and ends at X’ at a distance ‘b’ from A.

Hence, x = b – a

As shown in the diagram, the load on a small element dx is w*dx. If the coordinate at the location is a1, the value of the support reaction at A,

= a1 (wdx)

The total reaction at support A due to the entire uniformly distributed load

Illustration 2.7: The problem below will illustrate the procedure of drawing the influence line diagram and its application.

For the beam shown in Figure 2.35, we will draw the influence lines for the following:

• Support reaction at A
• Bending Moment at Section X, located 4 m from the end A
• Shear Force at section X

We will also work out the values of above stress functions, under the influence of a set of four concentrated loads, as shown in the diagram

3.2.1 Stress and strain

When a prismatic member has a load applied along its axis, the load is uniformly distributed along its entire cross-sectional area.

Again refer to Figure 3.1. Due to the application of the tensile force F on the bar, an increase in the overall length of the bar takes place. Since the end B of the bar is fixed, its other end B’ will shift to B” to accommodate this change. The change in the length of the bar B’B” is ΔL.

As shown in Figure 3.1, the prismatic bar BB’ of length L has had a tensile force of magnitude F applied along its axis. The cross-sectional area of the bar is A. On section XX’ of the bar,

Force per unit area = F/A = σ

This force per unit of sectional area is known as the stress. It is the internal force developed within the bar to resist its structural failure. As is obvious from the derivation, the unit of stress is kgfcm² (N/mm² in SI units).

Elongation per unit length of the bar .

This elongation per unit length (contraction in the case of compressive force) is known as the strain. As the strain is the ratio of two quantities having similar units, it is just a number without any unit.

3.2.2 Hooke’s law and modulus of elasticity

‘Within elastic limits, stress is directly proportional to the strain.’

The statement above is known as Hooke’s law. It is a very important relationship between stress and strain and has wide application in the design of several mechanisms, apart from the design of structural members.

Hooke’s law, presented mathematically, can be stated as follows.

Within elastic limits, variation in stress is directly proportional to strain:

This constant E is known as the modulus of elasticity or Young’s modulus. Theoretically, it is the value of the magnitude of the stress required to be induced in a specimen of the material to increase its length to double the original length, provided the material remains in the elastic state throughout. The modulus of elasticity has different values for different materials, which invariably are very large. Since it is the ratio of stress to strain with strain being a unitless quantity, the unit of the modulus of elasticity is the same as that for stress, i.e., kg/cm² (N/mm² in SI units).

If we plot the strain of an elastic member against the corresponding value of stress within it, the resultant graph would be a straight line. The same is shown in Figure 3.2. The strain has been plotted on the X axis and the stress on the Y axis. It may be noted here that, by convention, strain is always plotted on X axis in a stress strain figure.

The slope of the line OP in Figure 3.2 indicates the modulus of elasticity for the material.

3.2.3 Lateral strain and Poisson’s ratio

As is clear now, when a material is placed under tensile stress, a corresponding strain is generated along the direction of the stress, which increases the length of the specimen. It has been observed during studies that this elongation will always be accompanied by a contraction in the directions lateral to the stress. Within elastic limits, the ratio of these strains is constant for a material.

Lateral strain/axial strain = μ (constant)

This ratio μ is known as Poisson’s ratio for the material.

Theoretically, it has been calculated that the value for Poisson’s ratio for most materials should range between 0.25 and 0.33, while it should be 0.25 for an isotropic material. Isotropic materials are those having physical properties that do not vary with direction. Let’s try to examine the limiting value of Poisson’s ratio.

We will consider the case of a prismatic specimen of a material, for which the values of E and μ are known. Suppose the original length of the specimen is l_o and the original cross-sectional area is A_o.

Hence, the original volume of the specimen V_o = A_o × l_o.

The tensile stress of value T per unit area is applied on the specimen such that the material remains within the elastic limit and also follows Hooke’s law. The new length of the specimen is then

Corresponding to this increase in length, there will be some contraction in lateral directions. The effect of these contractions will be a reduced cross-sectional area for the specimen. With Poisson’s ratio for the material being μ, the new cross-sectional area will be

A_n = A_o (1– μ×ɛ)²

Hence, the new volume of the specimen will be

V_n = A_o(1– μ×ɛ)² l_o (1+ ɛ)

Since ɛ would be a very small value normally, we will ignore the value of its square from the expression.

It results in,

V_n= A_ol_o(1 + ɛ – 2 μɛ)

The change in volume per unit of volume is

We can safely assume that the volume of any material will not reduce while under tension the value of the above expression will not become negative. To meet this condition, the limiting value of Poisson’s ratio can be 0.5, which denotes the unchanged volume during the process of deformation. In the case of metals, most of them exhibit a value of Poisson’s ratio between 0.25 and 0.35, which indicates some increase in the overall volume. Concrete, on the other hand, generally exhibits a value around 0.1.

With the knowledge of Young’s modulus and Poisson’s ratio, we understand two very important properties of matter from the point of view of structural engineering. The knowledge about Young’s modulus for materials helps in deriving the values of deformations in the structures under stress. Poisson’s ratio, on the other hand, helps in understanding the failure mechanism of various structural materials. Both these aspects will be elaborated on further as we move forward in the chapter.

3.4.1 Values of the moment of inertia for common sections

While discussing the bending stresses and deriving their expressions, we have seen the importance of the value of the moment of inertia for the section of the member. As defined previously, the moment of inertia for a section is the sum of the multiplication of all the elements of the area with the square of its distance from the centroidal axis of the area. Since the moment of inertia is the product of multiplication of the area with the square of the distance, its unit is the (length).⁴

It is always possible to derive the value of the moment of inertia for a section from the first principle, by applying the process of mathematical integration. However, while designing a section for a beam or any other member, we normally adopt the regular geometrical sections symmetrically distributed about one or more of its axes. Hence, it may be useful here to provide the expressions for the moments of inertia for some common plane shapes.

Shape	Moment of Inertia (I)	Section Modulus (Z)
Rectangle – Breadth = b Height = h	– About axis X-X – About axis Y-Y
Circle – Diameter = d	– About axis X-X & Y-Y both
Triangle – Base = b Height = h	– About axis X-X

Parallel axis theorem

This theorem is very useful to work out the moment of inertia of an area with respect to any axis, as long as the value of the moment of inertia about an axis parallel to it is known.

As per this theorem, if the moment of inertia with respect to axis X-X is I_x, the total area of the shape is A and the distance of the new axis from X-X is d, the moment of inertia about the new axis is

I_new = I_x + Ad²

Illustration 3.1: We will determine the moment of inertia of the section shown in Figure 3.6, about the X axis passing through the top of the section.

As is obvious, the section shown consists of two rectangles. The axis A-A’ is the one about which we have to determine the moment of inertia.

The moment of inertia of the top rectangle about its centroidal axis is

Similarly, the moment of inertia of the bottom rectangle about its centroidal axis is

The area of the top rectangle is

A_top  =  40 x 20 = 800 cm²

And the distance of its centroidal axis from A-A is

d_t  =  10 cm

As per parallel axis theorem, the moment of inertia of this segment about A-A is

The area of the bottom rectangle is

A_bottom  =  30 x 10  =  300 cm²

And the distance of its centroidal axis from A-A’ is

d_b = 15+20 = 35 cm

As per parallel axis theorem, the moment of inertia of this segment about A-A’ is

Therefore, the gross moment of inertia about A-A’ is

3.4.2 The application of the principles of bending stress

We have seen in the expression for bending stress that the value of stress increases when the distance of the fibre from the neutral axis increases. In other words, the fibres located near the neutral axis would be less stressed in comparison to the fibres located at the edge of the section.

The other observation is that for a fixed maximum stress value, the moment-resisting capacity of a section is directly proportional to its section modulus, which is a function of the moment of inertia and overall depth of the beam. Therefore, it would prove economical if the moment of inertia of the section could be increased without altering the overall area substantially.

To see the achievement of these objectives, let’s start with the case of a beam with a rectangular section of b×h. The same is shown in Figure 3.7 (a).

As the stress generation is only due to the bending moment, the area adjacent to the neutral axis (which passes through the center of the area) is subjected to less severe stresses compared to the fibres at the extremity. As the material is assumed to be homogeneous, the stress on fibres at the extremity would govern the design, and fibres near the neutral axis would remain under-stressed. This is an example of inefficient use of the material in the construction.

It is obvious from the expression of the section modulus (for a rectangular section) that the value (of the section modulus) will be higher for a section with a higher depth-to-width ratio than for a section with a lower depth-to-width ratio, even if the area is the same in both cases. Therefore, the deeper section is more economical in comparison to a wider section of the same area. However, there is a limit beyond which the section will become laterally unstable and would tend to buckle.

We will try to develop a more efficient section now. Since the moment of inertia will increase if the area distribution is such that more of it is distributed away from the neutral axis, it may be helpful to provide the sides a slope away from the neutral axis. This arrangement is shown in Figure 3.7 (b).

With this, the value of the section modulus has increased and the section has become more efficient. However, it may not be convenient to form sloping sides during construction or fabrication. Therefore, another improvement can be taken up here. It may be decided to form wide flanges at equal distance on both sides of the neutral axis. As we can see from the parallel axis theorem, the value of the moment of inertia of this new arrangement about axis X-X would be substantially higher. Since two flanges alone cannot remain stable, a web has been provided to connect both with each other. This arrangement is shown in Figure 3.7 (c). The web also resists most of the vertical shear, as we will see later on.

This development of ‘I’- or ‘H’-shaped sections is one of the examples where the understanding of the development of bending stresses helps in more efficient use of the material in design.

Everything described above is true for a material that has similar strength in both tension and compression. However, there are materials that have lower tensile strength in comparison to compressive strength. Cast iron is one example of such materials. Now, if such a material is exclusively used for the construction of a beam, a section symmetrical about its X-X axis would prove uneconomical. This is because the tensile stress in the outermost fibre would reach its limit even when the stress on the compression face fibre has still not reached its limit. In this case, it would be more economical to design an asymmetrical section. Such a section would have its centroid nearer to the tensile face of the section. In such case, Z_c would be less than Z_t in magnitude. Since the magnitude of stress generated is inversely proportional to the value of Z, the value of tensile stress in the outermost fibre would always be less than the value of the compressive stress for the corresponding fibre. A ‘T’-shaped section is a classic example of sections useful under such situations.

We will now illustrate the process of bending stress determination with the help of an example.

Illustration 3.2: Figure 3.8 shows a simply supported beam of ‘I’-shaped cross section. The loading on the beam and the cross-sectional dimensions are described in the sketch. The objective is to find out the value of the maximum stress developed in the material at any section.

First of all, we will determine the maximum value of the bending moment, which is being generated due to this loading.

As the load system is only vertical, the reaction at A and B will only be vertical.

Taking the moment about B and equating the same to zero:

It is clear from the loading pattern that the maximum value of the bending moment would be generated at C, right under the concentrated load of 25 kN:

The resultant bending moment distribution is shown in Figure 3.8. The curved shape between A and C is the effect of the uniformly distributed load.

The next step is the determination of the moment of inertia of the beam section, with respect to the axis of bending. As we have seen, the neutral axis for a section passes through the centroid of the area. The ‘I’ section under consideration is symmetrical about both X and Y axes, respectively. Hence, the neutral axis would coincide with this axis of symmetry. The axis X-X is shown in the figure.

To calculate the moment of inertia about axis X-X, the three elements of the section have to be considered separately. These elements are:

• A vertical web of size 560 mm × 10 mm
• Two horizontal flanges of size 400 mm × 20 mm each

Both are rectangular sections and we know that the moment of inertia for a rectangular section about axis 

For the sake of convenience in calculations, we will adopt the cm unit of measurement of length.
Hence, I_web = 1 × 56³/12 = 14634 cm⁴
And, I_flange = 40 × 2³/12 = 26.67 cm⁴ (for each flange)

It may be noted that the value derived for the moment of inertia of the flange is about the X-X axis of the flange itself and not of the overall section. For this conversion, we will be required to apply the theorem of parallel axis.

As per the theorem of parallel axis,

It may be worth noticing here the amount by which the effective moment of inertia has increased per flange is due to a more efficient distribution of material with respect to the axis.

Therefore,

We also know that section modulus , where y is the distance of the outermost fibre from the neutral axis. In this case, this distance is 30 cm.

Therefore, the maximum value of the stress encountered by a fibre of the material would be 1884.45 N/cm².

3.8.1 Rectangular section – shear stress distribution

We have already derived the expression for the distribution of shear stress for any section. We will now see the distribution of shear stress along a rectangular section of height h and width b, as shown in Figure 3.15. The distance of the outermost fibres from the neutral axis is h/2, for both top and bottom.

We know that the value of shear stress at any layer is

Now for a rectangular section, the area of the strip (dA) will be represented by b×dy. The upper limit of integration will also be h/2 in this case:

Substituting this value of Q into the expression,

In this expression, V, I and H are constants for the particular section. Hence the distribution of shear stress across the section is parabolic (where the value of y is variable).

To determine the values at the center and at the edges, we will put the respective values of y in the expression.

Therefore, the value of shear stress at the edges is zero. We have studied the reason for this.

For the center, y = 0

Here, V/A is the value of the average shear stress on the section. It means that the maximum value of the shear stress in a rectangular section is 1.5 times the value of the average shear stress on the section.

Illustration 3.3: We will see the application of shear stress distribution in the illustration below.

The beam AB, shown in Figure 3.16, is a simply supported structure of 8 m span with load W-N acting at the center of the supports. The width of the section is 15 cm and depth is 40 cm. The permissible stresses in the materials are as follows:

The objective is to determine the maximum value of W, which meets the criteria of maximum stresses.

For this case,

support reactions 

The maximum value of the bending moment in this case will be at the center of the span, as the concentrated load W has been acting on this section.

The bending moment at the center is  = 2 W N-m = 200 W N-cm.

The maximum value of the shear force will be at any of the supports and would be equal to W/2.

Hence, V = W/2 Newton.

The dimensions of the rectangular cross section are also known. The area and section modulus can be worked out on its basis.

Area is A = b×h = 15×40 = 600 cm².

Section modulus is .

We know from the bending stress formula that

and from shear stress for a rectangular section that

The permissible value of the load derived from the criteria of bending stresses is much smaller compared to a similar value derived from the criteria of shear stress. Therefore, it is clear that maximum bending stress will govern the design in the case and the maximum permissible value of W will be 18,000 N.

3.9.1 Elastic curve

The projection of the neutral surface of the beam in a vertical plane, when the beam is in the state of deflection, is known as the elastic curve of the beam. If the equation of the elastic curve is determined by some method, it would give the relationship between the x and y coordinates of the beam and can be used to determine the value of deflection (y) at any section of the span (x).

To determine such an equation, we refer to Figure 3.17. The elastic curve OB of a prismatic beam has been shown. We consider a small element pq of the elastic curve. The length of ‘pq’ is d_s and its projection on the X axis is dx. The tangents drawn at points p and q form an angle of value d_θ between them.

The curve pq has its center at C and the radius of curvature is r. We know from geometry that the normals drawn from C to the points p and q also form an angle d_θ between them. The deflections in this case are assumed to be very small compared to the overall dimensions of the beams. This is actually true for almost all real-life situations. Due to this, the difference between the length of the deflected shape of pq and its projection on X axis is very small. Therefore, d_s = dx.

The value of the slope between any two points of the curve is

Again since the deformations are very small, θ would be a very small value. In such a case,

Differentiating the expression further with respect to x,

Also, if we consider C_pq as a triangle with its arm ‘pq’ equal to dx,

In Section 3.4, we have seen the establishment of a relationship for r. As per Section 3.4,

This, along with the previous expression for 1/r, establishes the relationship

This expression is the differential equation of the elastic curve for the beam, which is subjected to bending. When it is solved to bring to the form y = f (x), it would give the true equation of the elastic curve. For this, we would twice integrate the equation with respect to x. On first integration,

where C₁ is the constant of integration, the value of which is required to be obtained for a given condition of loading.

In the above expression,  is the value of the slope of the elastic curve at that point. It is the angle which the tangent drawn at the point forms with the original axis of the beam. Since the beam generally has a horizontal axis, the slope at any point will become the angle between the tangent drawn at the point and the X axis.

On second integration,

This is the desired equation for the elastic curve. The value of the second constant of integration C₂ also needs to be determined for the given loading condition.

This method of determination (of the equation of an elastic curve) is known as the method of double integration. As stated, values of constants C₁ and C₂ are dependent on a given loading condition. It implies that the equation of the beam has to be written all over again whenever the loading condition changes. This limitation can be overcome by writing the equation in such a form that its applicability remains for the entire span. Carefully selecting and ignoring the applicable portions of the equation further achieves this.

Illustration 3.4: We will take up the following example to determine the equation of the elastic curve and the maximum deflection for a simply supported beam, if the beam is loaded with a uniformly distributed load of w intensity per unit length, throughout the span.

As shown in Figure 3.18, the beam OB is a simply supported one with span L. The loading intensity, as described above, is w per unit length. The beam is prismatic so the value of the moment of inertia is constant throughout the span.

We know that

R_o = R_b = 

To determine the equation of the elastic curve, we assume that point O at one end of the beam is the intersection of X and Y axes. Consider a point A along the curved shape of the beam, the abscissa of which is x:

Replacing M with the value derived in Section 3.91,

Integrating both sides with respect to x,

Here, C₁ is the constant of integration. To find out its value, we will first have to find a defined situation for the slope, since represents the slope of the elastic curve. In this case, since the loading is symmetrical about the center of the span, the value of the deflection will be the largest at the center of the span. Therefore, the tangent at this point of the elastic curve will be parallel to the X axis, with its slope with respect to the X axis being zero. Mathematically,

Substituting these values in the expression above,

Putting this value of C₁ back into the equation,

Taking the integral of both sides once again,

To find the value of C₂ of this expression, we would look for a defined value of deflection y, which would be consistent with the arrangement. In the case, since both the ends of the beam are supported, the value of deflection at any of the support is nil.

Mathematically,

for x = 0, y = 0

Substituting these values into the above equation yields that the value of C₂ is zero.

Just from the academic point of view, if we put the values x = L and y = 0 in the expression, which is the case for the end B, the result obtained would be similar.

Hence, the equation of the elastic curve is

After deriving the equation for the elastic curve, the next objective is to find the value of the maximum deflection. As explained earlier, the value of the deflection will be the maximum for the center of the span, i.e., at x = .

With this,

The negative sign indicates the direction of the deflection with respect to the X axis.

3.9.2 Moment area method

Moment area method is another popular method of determining the values of deflection and slope for a structural element, which is subjected to bending. The principle of the method is the same as for the earlier explained method of double integration.

Again refer to Figure 3.19. LN is the elastic curve under consideration. The bending moment diagram for the same due to loading is L’N’. The tangents drawn at L and N are LL₁ and NN₁ respectively. The angle between the two tangents is θ. A very small portion of the elastic curve of length ds subtends an angle dθ center C of the arc.

As we have seen in Section 3.9.1, for similar elastic deformed body, which follows Hooke’s law,

Since d_s is very small and its curvature is not very steep, we replace d_s with dx – the projection of the arc on the X axis:

Graphically, the value of Mdx is equal to the area of the bending moment diagram for portion d_s. Since dx is a very small length, the value of the bending moment along this length can be assumed to be uniform, although it might be varying from section to section.

Now, to determine the value of the angle θ, both sides of the expression will be integrated between the limits L and N:

The expression on the right-hand side is the total area of the bending moment diagram for the entire span LN of the beam, divided by EI.

In the same Figure 3.19, dt is the intercept which the tangents drawn on both the ends of the curve element ds, cuts on an axis along N that is perpendicular to the original axis of the beam. Since the distance of the segment from point N is x and the angle between two tangents is dθ, the intercept dt may be considered equal to the arc of the circle with radius x subtending angle dθ at the center. From this relationship,

d_t = x dθ

Putting the value of dθ from the earlier derived expression,

We can see that the distance measured perpendicular to the original axis of the beam from N that will intersect a tangent drawn from end L of the elastic curve will be equal to the sum of such intercepts for all segments of the curves between the two extreme points. From the above expression this can be derived by integrating both sides of the expression within limits L and N:

The expression ∫x Mdx is the ‘moment of the bending moment diagram’ with respect to N. It may be noted here that the length of the intersection is required to be measured from the point where the perpendicular cuts the elastic curve. Since we have considered the case pertaining to the extreme points of the elastic curve (which do not yield during deformation), the original distance remains unchanged after the deformation.

Another aspect that needs attention is that the expression (above) provides the value of intersection at the line through N perpendicular to the original beam axis. Consider the case when we need to work out the corresponding value of intersection at the line through L when the tangent has been drawn at N. In this situation, the value of ∫ x Mdx is the moment of bending moment diagram with respect to L. Although the bending moment diagram is similar in both the cases, the length of the moment arm (x) will be different except for the case when bending moment diagram is symmetrical about both the points. In normal situations,

t (n-l) =/= t (l-n)

3.9.3 Moment area theorems

On the basis of what has been derived in this section, we define the two moment area theorems as below.

Theorem – I : For an elastic curve, the angle between tangents at any two points on the curve is equivalent to the total area of the bending moment diagram between these two points, divided by EI.

Theorem – II : For an elastic curve, the deviation of any point perpendicular to the original beam axis, relative to the tangent drawn on the curve at any other point, is equivalent to the moment of the area of the bending moment about the first point, divided by EI.

In both the above theorems, it is assumed that EI, the product of the Young’s modulus for the material and the moment of inertia of the section, is a constant. This product EI is known as the flexural rigidity of the section. For it to be a constant, both E and I are required to be constants. For it to be true, the material of the beam should be homogeneous so the value of E remains unchanged throughout. Also, the section of the beam is uniform (please remember, we have assumed the member to be prismatic), so the sectional properties remain unchanged.

These conditions are to be kept in mind while applying the theorems. In the case that the value of either E or I or both are not constant throughout the span of the beam, the pattern of their variation with respect to x will be needed for the application of the theorem. To simplify such situations, the ordinates of the bending moment diagrams itself are drawn to directly give the value of M/EI instead of just M, and the same M/EI diagram is used for the application.

3.9.4 Moment area theorems – applications

With the application of the moment area theorems, it is possible to determine the values of the slope of an elastic curve or the deflection of a point on it, in a semi-graphical way.

To obtain the desired values of the slope at any point, note the following procedure. As a first step, draw the bending moment diagram due to the load systems on the beam. We have learnt the procedure of drawing a bending moment diagram and therefore it should not be difficult.

However, we eventually would be required to accurately work out the area of the bending moment diagram and, in view of this, we may be required to follow a number of additional steps.

For this purpose, it becomes necessary in the case of multiple loadings to draw separate bending moment diagrams for each type of load. We know that the bending moment diagram for a concentrated load does form a triangular shape. The calculation of the area of this triangular diagram and determination of the centroid of this area are not difficult. If the load is uniformly distributed or gradually varying in nature, the shape of the bending moment diagram will be a curve.

As stated earlier, to solve the problem of the determination of the area of bending moment diagrams, it will be helpful if we break it into a number of simple structural situations.

For a loaded cantilever beam, the general expressions for the area and the centroid of such shapes are as follows:

Here, b is the length of the base of the curve and h is the maximum ordinate or height of the geometric shape. The value of n would be dependent on the nature of the loading.

The cases representing various types of loading on a cantilever beam and the corresponding bending moment diagrams are shown in Figure 3.20. The areas of such diagrams and their centroid locations have been tabulated below.

While determining the area of the bending moment diagram and the distance of its centroid, the effort lies in reconfiguring the loading on the beam such that each part of it is similar to one or the other standard loading condition.

The illustration of the procedure to determine the area and centroid by solving the case ‘by parts’ has been given below.

Illustration 3.4: We have taken the case of the simply supported beam AB of 4 m span, with a concentrated load of 5000 N acting at the center of the span and a uniformly distributed load of intensity 400 N/m acting on the right hand half of the beam. The arrangement is shown in Figure 3.21.

From statics,

Ra =  2700 N
Rb =  3100 N

As a first step towards the solution, assume that the end B has the capacity to withstand moment. For this purpose, we assume the end B is a fixed end. Now with B as fixed, we consider the effect moment generating effect of Ra alone, by removing the other loads from the system. Note that the removal of these loads will not change the value of Ra.

Hence, M_b1 = 2700 × 4 = 10800 N.m.

This moment is positive in nature and the distribution along the span is shown in Figure 3.20.

Similarly, considering the presence of the concentrated load alone at the mid span,

M_b2 = 5000 × 2 = 10000 N.m

This is a negative movement and the distribution is shown again in Figure 3.20.

Finally, considering the presence of the uniformly distributed load alone at the right half of the span,

M_b3 = 400 × 2 × 2/2 = 800 N.m

The sign of this moment is also negative.

Before moving further, we observe that the value of the overall moment at B has been arrived as zero (10800 – 10000 – 800), which is consistent with the support condition at B. We have analytically and graphically shown the breakup of the moment generation due to load components.

If we plot these three moment diagrams on a common baseline, we obtain the actual bending moment diagram due to loading.

Now, we want to determine the area of the bending moment diagram due to the cumulative effect of loads. This area A will be equal to

A =A₁ + A₂ + A₃

Here, A₁, A₂ and A₃ are the areas of the respective bending moment diagrams due to individual loading. By substituting their values in the equation

As per theorem 1 of the moment area, the angle between the tangents at A and B of the elastic curve will be equal to the area of the bending moment diagram between A and B, divided by EI.

Again, the moment of the overall area about B can be determined by applying the principle of the parts:

By applying theorem II of the moment area, the length of the intercept from B on a line drawn perpendicular to the original axis, by the tangent drawn from A,

We have seen the procedure to determine the area of the bending moment diagram between two points on the span of the beam. We have also seen the procedure to determine the moment of the bending moment diagram about a point by determining the centroids of the part bending moment diagrams. When the results of this exercise are used in conjunction with the moment area theorems, we can derive the following deformations for an elastic curve:

• Slope between the tangents drawn at the two points
• Length of the vertical intercept from one point of the elastic curve by the tangent drawn at another point along the curve.

Although the second information is useful, our main objective is to find out the deflection of a particular point on the elastic curve from its original position. This information will not be directly available from the expressions we have seen so far, except for the condition when the beam under consideration is a cantilever. This is because, with a cantilever, due to a no-rotation condition of the fixed end, the tangent drawn on this point of the elastic curve coincides with the original axis of the beam. Therefore the intercept between the tangent and any point on the elastic curve would be equal to the deflection of that point with respect to its original position on the beam axis.

This point can be further understood with the help of Figure 3.22. Beam AB is a cantilever with load W acting at its free end B. Due to the action of the load, the beam will deform to the shape shown in the diagram.

Consider point C in the original beam, which takes position C’ in the deformed beam. The tangent drawn at A clearly coincides with the original beam axis. Now we draw a line through C’, perpendicular to the original beam axis. The perpendicular distance between C’ and the original axis of the beam is the deflection at point C. At the same time, this distance is equal to the intercept between point C’ and the tangent drawn from A. As per the second moment area theorem, the value of this intercept {t _(c-a)} will be equal to the moment of the bending moment diagram between A and C, with respect to C, divided by EI.

Illustration 3.5: We will consider the case of a cantilever beam AB of 4M span, with a single concentrated load of 2000 N working at the middle of the span at point C. We want to determine the deflection at the points:

• Right below the application of the load
• At the free end of the cantilever.

The structural arrangement and the bending moment diagram have been shown in Figure 3.23. The vertical reaction at B is equal to the total vertical load 2000 N.

As is clear, there has been no force acting in the portion CB of the beam and therefore there is no bending moment in the region between C and B. In the region AC, the bending moment distribution along the span has been shown in the bending moment diagram. The distribution is triangular with a maximum value of 4000 N.m at A and reduces to zero at C in a linear fashion.

Note the shape of the deflected beam or the elastic curve. The portion of the elastic curve between A and C’ has curvature of some degree. This curvature is the function of the shape and area of the bending moment diagram between the points lying in section AC’. At the same time, the shape of the elastic curve between C’ and B’ has been a straight line. The explanation for it from a physical point of view is that there has been no load acting at this section, which can intervene to alter the slope of the elastic curve; hence, the slope between C’ and B’ remains unchanged. Alternatively, as there has been no bending moment between C’ and B’, the slope between any two points taken between C’ and B’ would be zero (as per the first moment area theorem).

To determine the value of deflection at midpoint C, we will use the second moment area theorem. According to the theorem,

Note that the moment is being taken about point C of the bending moment diagram and hence the length of the moment arm is (2 – 2/3):

Similarly, to determine the deflection of the point B at the free end,

The area of the bending moment diagram  will remain unchanged in the case. What may be noticed is the change in the length of the moment arm:

In this illustration, we have seen the method to derive the deflection at a point for a cantilever. When the beam under consideration is not a cantilever but a simply supported beam, we can still use the moment area theorem to derive the value of the deflection at a point of it. In this situation, however, there will be some additional steps required in the procedure.

To explain this, we take the simply supported beam AB of span L, as shown in Figure 3.24. The elastic curve of the beam has also been shown in the diagram. It is clear that the points at ends A and B are unyielding due to the support’s constraints and, therefore, they continue to remain at their original positions even after the deformation.

Consider a point C (in the original beam axis), whose deflection we want to determine. Point C is located at a distance l from point A. After the deformation, the point deflects by δ and assumes its new position at C’.

A tangent has been drawn at A, which cuts the line through B and perpendicular to the original axis at D. The intercept BD is t _(b-a), the value of which can be determined from the second moment area theorem.

In the same way, the value of the intercept C’C’’, which is t _(c-a), can also be determined. Once the values of t_(b-a) and t_(b-a) are known, we move forward by applying the principle of geometry. In Figure 3.24, the lines CC’’ and BD are parallel to each other. Therefore, ΔABD and ΔACC’’ are similar triangles.

From the properties of similar triangles,

Here, CC’’ is the sum of the deflection of point C (δ) and intercept at C’ [t_(c-a)]. Putting these values in the equation above,

In the above expression, we know the values of all except δ. Hence, by putting all respective values in the expression, it is possible to determine the deflection δ for the point.

We will see the illustration of this method, too, with the help of the following example.

Illustration 3.6: Take the case of a simply supported beam XX’ of span 5 m. A concentrated load of 2000 N has been applied at point A, 3 m away from end X. We will see the determination of deflection at the point of application of load.

Figure 3.25 illustrates the structural arrangement and the bending moment diagram. First of all, the reactions at both the supports X and X’,

R_x = 800 N
R_x’ = 1200 N

To draw the simplified bending moment diagram, we will assume end X’ as fixed one and consider one by one the moment generation effect of reaction Rx and vertical load.

Due to R_x, moment at X’ = 800 × 5 = 4000 N.m (+ve).
Due to 2000 N load, moment at X’= 2000 × 2 = 4000 N.m (–ve).

As shown in the representation of elastic curve in Figure 3.25, t _(a-x) and t _(x’-x) are the two intercepts of our interest. As per the second moment area theorem,

The triangular bending moment diagram between A and X would have a height equal to 2400 N.m at A. This results from the characteristics of the similar triangles.

From the elastic curve diagram of the beam,

Here, δ_a is the deflection of point A with respect to its original position. Putting values in the expression,

Therefore, the deflection at A would be equal to .

The estimation of the deformation of a structure, or its element, is important from the functional point of view of the structure. On occasions, despite the stresses remaining within limit, the deflection of the structure may become excessive. Excessive deformation of a structure can lead to two problems.

• The finishing over the main structure members (for instance plastering over concrete beams and paint-coating over steel beams) may not be able to withstand large deformations and may loose its integrity.
• The second reason is psychological. Excessive deflection can affect the aesthetics of the structure and can also make users fear the safety of the structure.

To overcome this, structural engineering codes prescribe means to limit the deformation of structural members. As we can see from the expression of deflection for a particular structural arrangement and loading, the amount of deflection varies only with the variation in the modulus of elasticity (E) of the material and moment of inertia (I) of the section. In fact, the value of deflection is inversely proportional to the product of the two. Here, we may consider even the modulus of elasticity as non-variable as the material of construction is finalized beforehand, the selection being based on several factors. Hence the only variable remaining in the expression is the moment of inertia. We know that the value of I depends on the geometrical properties of the section and can be controlled by altering the sectional dimensions or the pattern of distribution of the material about the neutral axis.

To overcome the adverse effect on aesthetics, the practice has been to introduce the deformation of reverse order in the structure at the stage of construction. This is done by accurately assessing the value of future deformation under load and introducing the deformation of similar magnitude in the reverse direction at the time of construction or fabrication of the structure. This takes care to neutralize the deformation, which takes place when the intended loads are applied on the structure during its life.

3.10.1 Combined stresses in columns

A column or strut is a structural member, which is predominantly subjected to axial compressive loads. Strut is a general term to define the members under compression, irrespective of the position of their axes with respect to the horizon. A column is a strut, which has its axis vertical.

We are quite aware of the effect of direct loads and also know the pattern of stress generation. We know that axial loads generate even stress along the cross section of the member. At the same time, this situation is valid under the assumption that the load is acting along the axis of the member. Any variation from this assumption would lead to a different stress development pattern across the section. We will observe the effect if the load application is off the axis.

Let’s take the example of the column AA’. For simplicity, consider the cross section of the column to be rectangular with dimensions b × d. As shown in Figure 3.27, the column is subjected to a vertical compressive load of magnitude W. The application of load is at a distance e from the axis of the column. To analyze this situation, at A we introduce forces W_u and W_d acting vertically upward and downward respectively, along the axis of the beam. The magnitude of each force is equal to W. Since W_u and W_d do neutralize each other, the overall structural arrangement remains unaltered.

This combination of forces W, W_u and W_d can be considered to be acting in the following ways:

• The force W_u acts along the axis of the column.
• The forces W and W_d, being equal and opposite, form a couple with force W and arm e.

The couple induces a moment along the entire length of the column. The magnitude of the moment is

M = W×e

It is clear to us from the previous sub-section that due to the combined effect of the moment and direct force, the stresses on the outermost fibres of the column section would be

As  is the compressive stress, the use of the positive sign will give the value of the maximum compressive stress and the use of the negative sign will give the value of the maximum tensile stress on the section.

As is evident from Figure 3.26, the position of the neutral axis of the section is now not along the principal axis but is at a distance x from it. By definition, neutral axis is the location of zero stress. To find its location, we equate the value of the stress to zero:

The above expression defines the shift in the location of the neutral axis if combined bending and direct forces are acting on it. It is a general expression for all the cases. In this case, when the section is rectangular with known dimensions,

Therefore, for a given section and load, the shift in neutral axis due to eccentric application of direct load is inversely proportional to the amount of eccentricity.

It is obvious that the shifting of the neutral axis is accompanied by a reduction in tensile stress on one extreme of the cross section. The height of the triangular stress diagram will keep on decreasing as the neutral axis shifts towards the extreme edge of the section. It will become zero when the neutral axis coincides with the edge. This is the limiting condition when there is no tensile stress on any fibre of the section. In this situation, the shift in the neutral axis (x) is equivalent to one half of the overall depth of the section. Figure 3.28 can be referred to for clarity.

Putting this value of x in the expression above,

Therefore, in the case of a rectangular section, the value of extreme stress would remain compressive as long as the application of compressive load is within the inner one-third depth of the section.

3.11.1 Critical load – Euler’s column formula

Consider the case of a long column AB, the bottom end A of which is built-in and the top end B is free. The column is shown in Figure 3.30. We assume that the column has a perfectly vertical axis, homogeneous material properties and constant section throughout its length. The column has first had a horizontal force H applied at A, normal to its axis. Due to this, the end A deflects and assumes a new position A’, as shown in the diagram. At this stage, a compressive force P is applied along the axis of the column. Keeping P in position, the force H is removed. The free end of the column comes back to its original position at A.

This exercise is repeated with a gradually increasing magnitude of vertical force P. The end A will continue to come back to its original position on the removal of H, until the force P reaches a certain magnitude. At that magnitude of vertical force, the end A of the column would continue to stay at its new position A’, despite the removal of the horizontal force H. At this stage, the column is treated as failed due to lateral buckling. The minimum load, on the application of which a column fails due to lateral buckling, is known as the critical load for the column. The notation for it is P_{cr .}

Euler, a Swiss mathematician, theoretically analyzed the failure mechanism of long columns and from there derived the expression for critical load. He based his analysis on the solution of the differential equation for the elastic curve of the column. Euler solved the expression for moment, , for the condition of buckling. Based on his analysis, the expression for critical load is as follows:

The above expression for P_cr is for the column of length L, both the ends of which are hinged. The value of n in the expression depends upon the shape of the column axis after the buckling.

The cases for values of n = 1, 2 and 3 are shown in Figure 3.31. Out of this, the shape representing n = 1 is the most common and is encountered frequently. For this situation, the value of critical load,

The buckling conditions for which n = 2, 3, etc., are not very common. These are possible if the column is provided with bracing. It may be noted here that for the same length of column, the bracing at midpoint would increase the value of critical load (and therefore the ultimate load carrying capacity of the column) to four times. Similarly, the bracing at one-third and two-third length of the column would increase the value of the critical load to nine times.

On analyzing the expression for critical load of a long column, we observe the following:

• The value of the critical load for a column depends on the length of the column (effective length, as we will see under the next observation) and the values of modulus of elasticity of the material and the moment of inertia of the section. It is independent of the strength, or the stress carrying capacity, of the material of construction. Therefore, the critical load for the columns made of different grades of steels will be the same, as long as the effective length and the sectional properties are similar. This is due to the fact that despite the large variation in the strength of the various grades of steels, the values of the moment of elasticity for them remain in a very close range.
• The length L in the equation is the effective length of the column arrangement. In the fundamental case, when both ends of the column are hinged, the effective length is equal to the overall length of the column. For any other end position, the effective length of the column would vary. In general, the effective length is considered as the length between the points of inflexion in the column. Let’s consider the situation where both ends of the column are fixed. If we draw the bending moment diagram, the points of inflexion fall at distance L/4 from both ends. Therefore, the distance between the points of inflexion is L/2, which has to be treated as the effective length of the column. Putting this value of effective length in the expression, we discover that the value of critical load has increased to four times compared to the case when both the column ends were hinged.

The different end conditions for column and their effect on the value of critical load are tabulated below.

A column will always tend to buckle in the direction where the moment of inertia is the lowest. Due to this, it is necessary to put the value of the least moment of inertia of the column section in the expression while determining the critical load.

3.11.2 Use of Euler’s formula and its limitations

Euler’s formula provides the value of the critical load on a long column by the expression

If we divide both sides by the area of the cross section of the column A:

The value of expression√(I/A) is called the radius of gyration of the section and the notation for it is r. Also, P_cr/A is the stress on the section during critical load and is denoted as σ_cr. Putting σ_cr and r in the above relationship,

This ratio L/r is known as the slenderness ratio of the column. If for a given material, a graph is plotted between values of the critical compressive stress (σ_cr) and slenderness ratio (L/r), its shape would be as shown in Figure 3.32.

It is obvious to us that Euler’s equation to determine compressive stress is valid only if the value derived by it is less than the permissible compressive stress as per Hooke’s law. Since Euler’s formula does not recognized the concept of the proportionate limit of elasticity, the value of critical stress derived for shorter columns (of lower slenderness ratio) may exceed this stress limit. This is one of the limitations of the formula and is required to be kept in mind during application.

By putting the value of the proportionate limit stress (σ_pr) in the expression above, we can find the value of slenderness ratio, for which σ_cr is equivalent to σ_pr.

The graph in Figure 3.32 can only be used when slenderness ratio values are above this limit. The portion of the graph, where σ_cr is more than the value of σ_pr, has been shown with broken lines and should be ignored during analysis.

The other important aspect is that the value of critical stress determined by Euler’s formula is the ultimate or failure stress. One can notice that at any stage, no factor of safety has been introduced in the expression. Therefore, it is important that once critical stress has been determined, it be suitably factored to derive permissible stress.

We will demonstrate the steps involved in the analysis of columns, with the help of an illustration below. In this illustration, instead of providing a specific solution to a design related problem, how one should proceed and use the various design inputs is shown. The reason for this is that to derive the permissible value of critical stress for a given slenderness ratio, we need the related stress graphs or tables for the material under consideration. It would not be convenient to reproduce them on these pages and once the concept is understood, applicable codes can be referred for them.

Illustration 3.7: Take the case of a rectangular section of the size 4 cm X 6 cm. It has been made of steel, which has the modulus of elasticity of 2 × 10⁶ kg/cm². The value of stress at proportionate limit of elasticity is 2000 kg/cm². If the column is fixed at one end and free at the other, we need to determine at what length of the column critical stress is equal to proportionate limit stress.

First of all, we will determine the value of the moment of inertia of the section. As discussed earlier, we need to take the lowest value of the moment of inertia for this purpose:

We know the relation,

To derive the value of L when critical stress equals proportionate stress, we would put the value of the proportionate stress in the expression

Therefore, till the effective length of the column is 114.74 cm or below, it is not desirable to use the value of critical stress determined from Euler’s formula.

Here we have determined the value of the effective length. It is stated in the problem that the column is fixed at one end and is free at the other. We know that for this condition, the effective column length becomes double of the actual length.

Hence, actual column length for the limiting condition

= 114.74 / 2

= 57.37 cm.

In a way, this reduction in actual length (we can also call it the permissible length for this condition) demonstrates the influence of the end condition on the column.

For this case, we have enough information available to determine the critical stress for any length and end condition. The simplicity of the adopted section has made things very easy for us. However in actual practice, we will come across sections for which it is not very simple to calculate the values of I and A. In such a scenario, it becomes imperative to refer to the manufacture’s data table or the structural engineering codes, to determine this information.

4.1 Structural classification based on the degree of indeterminacy

We have learnt that structures can be classified as structurally determinate or non-determinate, based on the procedure of their analysis. We will further observe the characteristics that lead to this classification.

4.1.2 Statically indeterminate structures (from chapter 2)

Those structures which cannot be fully analyzed with the help of statics alone are known as statically indeterminate structures. In these cases, the number of unknown reactions exceeds the number of applicable equilibrium equations.

We will consider an example here. Consider the case of a beam AB in Figure 4.1, which is fixed at end A and supported at end B. Structurally, this arrangement is known as the propped cantilever. An inclined load W has been imposed on the beam.

As is obvious from the diagram, there are four unknown support reactions: three at the support A and one at the support B. Only three equations of statics are available (ΣH, ΣV and ΣM = 0). Therefore, the beam AB is a statically indeterminate structure.

As we are aware by now, there can only be three possible equations of equilibrium in the case of plane structures. Does it mean that all structures having more than three unknown support reactions are statically indeterminate? Not necessarily. Now consider a modification in the beam of Figure 4.2. Between A and B, a hinge has been installed at C. Clearly, the condition at that moment at C should be equal to zero and provide an additional equation. With this additional equation, it has become possible to analyze all of the four unknown reactions and, therefore, the structure has become statically determinate.

Let us consider another situation for the beam AB. We decide to remove the support at B. The structure would still be stable under the influence of the load W. Simultaneously, it has become statically determinate even without the introduction of any hinge (in fact, in this situation the introduction of a hinge would render it unstable). It can be concluded that the support B was redundant. Therefore, we can say that in a statically indeterminate structure, the number of support reactions is more than those necessary for the stability of the structure. This difference, or alternatively, the number of redundants in the structure, is known as the degree of indeterminacy of the structure.

4.1.3 Determination of degree of indeterminacy

As stated earlier, the degree of indeterminacy is the number of redundant support reactions in a structure, which, on removal, would make the structure statically determinate. Two popular methods to determine the degree of indeterminacy have been explained below. The scope has been confined to plane or two-dimensional structures only.

Cantilever tree method: This method is very useful for rigid jointed frames. Nature has generated examples of some very efficient structural systems, a tree being one such case.

To work out the degree of determinacy for frames by this method, the first step is to ensure that all of the joints should be rigid. This is achieved by providing extra restraints at non-rigid joints (converting them into rigid ones). The number of such restraints is counted (R).

Now, the closed frame structure is cut in such a way that each isolated portion resembles a tree with branches on one or both sides. The numbers of cuts in the total structure are also counted (C).

Degree of indeterminacy = 3×C – R.

We will consider an example of the frame shown in Figure 4.2.

The frame in the example has two hinged supports. To convert them to rigid, each one has had a single restraint applied:

Total number of restraints introduced ‘R’	= 1 + 1 = 2
The number of cuts required to convert into tree ‘C’	= 6
Hence, degree of indeterminacy	= 3×6 – 2 = 16

The underlying principle can be applied to any structure to find its indeterminacy. In the case of indeterminate beams, what is required is to convert any one end of the beam as fixed (if it is not already so) by introducing restraints. Once done, remove all the redundant supports. The degree of indeterminacy can be worked out with the following relationship:

Degree of Indeterminacy = Supports Removed – Constraints Added

Method of joints: This method is more suitable for pin-jointed structures with relatively large number of members. A truss is one example of this type of structure. The principle applied is that each member of the structure has an unknown axial force contributing to the indeterminacy, which is added to the total number of support reactions. Also, each one of the pin joints contributes two numbers of equilibrium equations. Therefore, in the case of a plane structure,

Degree of Indeterminacy	= Total Members + Total Support Reactions
	(– 2) × Number of Joints

Let us take the truss shown in Figure 4.3 as an example for illustration. The truss has a total of 13 members and eight joints. There are two supports: a hinged and a roller. Total support reactions shall be three (two for hinged and one for roller support):

Degree of Indeterminacy = 13 + 3 – 2×8 = 0

The zero degree of indeterminacy indicates that the truss is statically determinate. It may be noted here that if the roller support is converted into the hinged one, the total number of support reactions will increase by one, resultant to which the degree of indeterminacy will become one and the truss will no longer remain statically determinate. The illustrations below would be of help to further understand the procedure.

Illustration 4.1: The frame shown in Figure 4.4 has all the joints rigid. It is needed to determine if the frame is statically indeterminate.

As we see, the two intermediate supports for the frame are of hinged type. In order to convert them into fixed supports, they need to be provided with one restraint each.

Therefore, Number of Restraints ‘R’ = 2.

Once that done, the frame is a fully rigid one with fixed supports. Now, to convert it into a set of independent cantilever trees, a number of cuts have been introduced in the horizontal members. To find out the number of such cuts required, the diagram of ‘equivalent frame’ is shown in Figure 4.4.

According to this, Number of Cuts ‘C’ = 4.

We know that

Degree of Indeterminacy = 3×C – R

Substituting relevant values,

Degree of Indeterminacy  = 3×4 – 2 = 10

Therefore, the frame has the degree of indeterminacy equivalent to 10.

Illustration 4.2: A pin-jointed truss is shown in Figure 4.5. It is needed to determine whether the frame is statically indeterminate.

As we know that for pin-jointed frames, the degree of indeterminacy can be determined from the following expression:

Degree of Indeterminacy =	Total Members + Total Support Reactions
	(– 2) × Number of Joints

Total Members	=	17
Total Support Reactions	=	3 (one at roller & 2 at hinged support)
Total Number of Joints	=	10

Therefore, Degree of Indeterminacy	=	17 + 3 – 2 × 10
	=	20 – 20 = 0
This establishes that the truss is statically determinate.

4.2 Principle of superposition

The superposition is one of the widely used principles in structural analysis. It states that if a structure is under application of multiple loads and the stress–strain relationship follows Hooke’s law, the overall deformation caused by the load system at a particular point shall be equal to the algebraic sum of the deformation caused by the application of each load of the system acting one at a time.

This principle is derived from Hooke’s law. We know that the relationship between the applied force and deformation shall be linear, as long as the material follows Hooke’s law. Let’s assume a force system, consisting of forces F1, F2 and F3, which has been acting on a structure. As shown in the stress–strain relationship in Figure 4.6, the deformations caused by these individual forces are δ1, δ2 and δ3, respectively.

As is obvious from the figure, when we represent the overall force F, as the sum of the individual forces F1, F2 and F3, the corresponding deformation δ shall be equal to the sum of the deformations δ1, δ2 and δ3.

It is clear that the principle of superposition would not be valid for those cases in which the stress–strain relationship is non-linear. Other circumstances in which it would not be valid are when the application of one force tends to affect the magnitude of another. A case in point is the column subjected to both axial and lateral loads. Here, when lateral load is applied to the column, the resultant deflection of the axis tends to induce an additional moment on the section of deflection. The magnitude of this moment is equal to the product of the axial force and deflection at that point. The value of this moment keeps on increasing with every incremental increase in the lateral force, though the value of the longitudinal force may remain unchanged throughout. When both forces are applied individually, this moment shall not be present in any of the situations.

Here, the cumulative effect of the application of both the forces is not equal to the algebraic sum of their individual applications.

4.3 Analysis of statically indeterminate beams

In addition to the equations of statics, statically indeterminate structural cases require equations that show relationships, between the elastic deformations, in order to derive all the variables necessary for analysis.

For the analysis of statically determinate beams, we will examine two popular techniques. One is the ‘method of superposition’ in which superposition is used for analysis. The other is the ‘method of area moment’.

4.3.1 Method of superposition

This technique of analysis of structures is based on the ‘principle of superposition’. We know that in the case of a statically indeterminate beam, one or more of the support reactions are redundant. Therefore, if the constraints, which are responsible for those reactions, are removed, the beam becomes statically determinate and can be analyzed for those support reactions.

Once that is done, geometrical constraints can then be imposed on the system. These constraints, which relate to either deflection or slope of the elastic curve of the beam, can then be expressed in terms of loads. Such relationships form the additional equations, which are used together with the equations of statics, to determine the values of the support reactions.

To understand it fully, take the case of the beam AB, supported on rollers at A and a hinge at B, as shown in Figure 4.7. The beam has been provided an additional support at C located between A and B. Two concentrated loads, each one in the spans AC and AB, have been acting on the beam. We have two conditions of equilibrium (ΣH = 0 is not relevant in the absence of any horizontal load component). Clearly, the introduction of support C has made one of the support reactions redundant. For the sake of explanation, it is possible to treat either support A or C as redundant; as the beam remains stable if either one is removed.

In this case, we decide to remove support C. The structural arrangement now is statically determinate. The beam under the influence of loads and support constraints would deform as shown in Figure 4.7. Table 4.1 describes expressions for slope and deflection for some standard loading conditions.

If the loading system is different from any standard condition, it is always possible to determine these values with the help of the analytical methods already studied. Let’s say that deflection at point C of the beam in the absence of support is δ1.

In the second step, we remove the imposition of the concentrated loads and place support C back to its original position. The reaction at support C ensures that the beam has no deflection at that point. In other words, the reverse deflection at C due to reaction is equal to the deflection due to imposition of the earlier concentrated loads. Let’s say this deflection is δ2.

It is obvious that δ1 is expressed in terms of loads imposed on the beam, whereas δ2 is expressed in terms of reaction at C.

The relation

δ1 = δ2

would provide the magnitude of reaction at C. Once it is known, the reactions at A and B can be determined with the help of relations Σ M = 0 and Σ V = 0.

Table 4.1
Expressions for slope and deflection in some standard cases

BEAM DESCRIPTION	SLOPE		MAXIMUM DEFLECTION
	At End ‘A’	At End ‘B’
Cantilever Beam – Span ‘l’ End ‘A’ Fixed and ‘B’ Free Point Load ‘P’ applied at B	Nil	Pl2/2EI	Pl3/3EI
Cantilever Beam – Span ‘l’ End ‘A’ Fixed and ‘B’ Free Point Load ‘P’ applied at distance ‘a’ from A and ‘b’ from B	Nil	Pa2/2EI	Pa2(3l-a)/6EI
Cantilever Beam – Span ‘l’ End ‘A’ Fixed and ‘B’ Free Uniformly Distributed Load ‘w’ per length applied throughout the span	Nil	wl3/6EI	wl4/8EI
Cantilever Beam – Span ‘l’ End ‘A’ Fixed and ‘B’ Free Uniformly Varying Load ‘w’ per length at A to zero at B applied on the span	Nil	wl3/24EI	wl4/30EI
Cantilever Beam – Span ‘l’ End ‘A’ Fixed and ‘B’ Free Moment of magnitude ‘M’ applied at B	Nil	Ml/EI	Ml2/2EI
Simply Supported Beam – Span ‘l’ Point Load ‘P’ applied at the centre of the span	Pl2/16EI	Pl2/16EI	Pl3/48EI
Simply Supported Beam – Span ‘l’ Point Load ‘P’ applied at distance ‘a’ from A and ‘b’ from B	Pb(l2-b2) /6lEI	Pab(2l-b) /6lEI	Pb(l2-b2)3/2/9√3lEI at √{(l2-b2)/3} from end A
Simply Supported Beam – Span ‘l’ Uniformly Distributed Load ‘w’ per length applied throughout the span	wl3/24EI	wl3/24EI	5wl4/384EI
Simply Supported Beam – Span ‘l’ Uniformly Varying Load ‘w’ per length at A to zero at B applied on the span	7wl3/360EI	wl3/45EI	0.00652wl4/EI at 0.519l from A
Simply Supported Beam – Span ‘l’ Moment of magnitude ‘M’ applied at B	Ml/6EI	Ml/3EI	Ml2/9√3EI at l/√3 from A

To illustrate the technique, the analysis of a propped cantilever beam has been demonstrated below.

Illustration 4.3: The propped cantilever AB has been fixed at A and supported at B. The beam has a span of 6 m and a uniformly distributed load of 600 N/m is acting on the entire span of the beam. The arrangement is shown in Figure 4.8.

Firstly, we use the equilibrium equations applicable to the situation. As per these,

Σ V = 0 and Σ M = 0.

Hence,

If we assume the case when support B is not present and the beam AB is a normal cantilever, the downward deflection at end B due to the imposed load shall be expressed as follows:

Also, if we consider the situation when reaction at B is the only applicable load on the beam, the upward deflection in the beam due to reaction at B shall be expressed as follows:

As there has been no net deflection at point B, the following relation emerges from the principle of superposition:

By substituting this value of Vb into equation (1),

	Va + 1350 = 3600
Hence,	Va = 2250 N

Similarly, putting the value of Vb into equation (2),

Ma = 10800 – 1350×6

Ma = 2700 N.m

By this way, magnitudes of all the support reactions have been determined.

It may be worth noticing the effect of providing a prop at the free end of the cantilever. The moment at fixed end, which would have been 10800 N.m in the absence of the prop, has come down in the case by 75% to 2700 N.m.

4.3.2 Method of area moment

We already know that in the case of statically indeterminate beams, it is always possible to define some constraints of slope or deflection somewhere along the span. These constraints lead to certain conditions, which can be expressed mathematically with the help of the moment area theorems, described in Section 4.6. These expressions, along with the equations derived from statics, can be solved simultaneously to get the values of support reactions.

To further clarify, take the example of a beam AB of span l, built-up in the supports at both the ends. The beam has had two concentrated loads applied, W1 and W2, which are acting vertically downward. The arrangement is shown in Figure 4.9.

Under the influence of this load system, the support reactions generated are Va, Ma, Vb and Mb.

From the condition of equilibrium, we derive the following two equations:

	Va + Vb = W1 + W2
and	Ma= Mb – Vb×l + W1×l1 + W2×l2

It is clear that four unknowns cannot be determined with the help of the two equations. Hence, we would go further and identify some critical constraints of the system. There can be three such constraints.

The ends A and B would not allow any rotation, due to which the tangents drawn at both the points in the elastic curve would be parallel (in fact, tangents would coincide as A and B are at the same level with respect to the horizon). Therefore, the angle between A and B is nil. We know from the first moment area theorem that the angle between two points is equal to the area of the bending moment diagram between these points, divided by EI.

As the tangents on points A and B of the elastic curve are parallel to each other, the deviation at A from tangent drawn at B would be equal to zero. Similarly, the deviation at B from the tangent drawn at A would also be equal to zero. From the second moment area theorem, such deviation shall be equal to the moment of the bending moment diagram about A and B respectively, divided by EI:

In all these expressions, the values shall be expressed in the form of imposed loads, span and reactions. With the help of one of these two equations, along with the two equations earlier derived from statics, it is possible to determine the value of all the four unknowns.

We will demonstrate the application of area–moment method with the help of the following illustration.

Illustration 4.4: The beam AB has both the ends A and B rigidly fixed. The beam has a span of 9 m and a concentrated load of 5000 N is acting at point C, located 3 m away from B. The arrangement is shown in Figure 4.10.

There are four unknowns in the arrangement, which are Va, Ma, Vb and Mb.

Firstly, we use the equilibrium equations applicable to the situation. As per these,

Further to apply the area–moment method, we would draw the bending moment diagram ‘by parts’ as explained previously in the application of moment–area theorems. Considering B as the fixed end, the bending moment diagram is shown in Figure 4.10.

Due to the rigidity of the fixed ends, there would be no rotation permitted and the tangents at the elastic curve at these two points shall be parallel. In this situation, the angle between these two points would be zero. According to the first moment–area theorem, the area of the bending moment diagram between A and B also shall be equal to zero:

Also, since the tangents are parallel and at the same level at A and B, the deviation at B with the tangent drawn at A would be zero. From the second moment area theorem, the bending moment diagram between A and B with respect to B shall be equal to zero:

t_(a-b) = M_area-ab×x_a = 0

Solving equations (3) and (4) simultaneously, we get

Va = 1296.3 N and Ma = 3333.35 N.m

Putting the value of Va into equation (1),

Vb = 3703.7 N

Putting the values of Ma and Vb into equation (2),

Mb = 6666.65 N.m

4.4 Multispan or continuous beams

Till now, we have observed the analysis of beams, which had one or at the most two spans. If the beam has a larger number of spans, it is still possible to use any of the above-described methods for analysis. However, the number of simultaneous equations increases with the number of redundants in the beam. This increases the complexity of the procedure. A more convenient way would be to first determine the magnitude of the moments at the supports. Determining the support reactions would be a subsequent step. We will examine one such method in the section.

4.4.1 Three-moment equation

In the three-moment equation, a general relation between bending moments at any three sections of a beam is established. It is a relationship between the moments at those sections, distances between them and the deflection of the elastic curve at those points. Since the reaction moments are more critical in the case of a multispan beam, the sections chosen for application of the three-moment equation can be three consecutive supports.

The expression of three-moment theorem is as follows:

Here,

M1, M2 and M3	=	Moment at Section – 1, 2 and 3 respectively
L1	=	Distance between Section – 1 and 2
L2	=	Distance between Section – 2 and 3
A1	=	Area of bending moment diagram between Section – 1 and 2
a1	=	Moment arm of A1 with respect to Section – 1
A2	=	Area of bending moment diagram between Section – 1 and 2
b2	=	Moment arm of A2 with respect to Section – 3
h1	=	Height of point 1 of the elastic curve with respect to point 2
h3	=	Height of point 3 of the elastic curve with respect to point 2

Normally when we consider a horizontal beam and apply the three-moment theorem to any three non-yielding supports, those that are not subjected to settlement, the values of h₁ and h₃ become zero. In such case, the three-moment equation assumes a relatively simple form, which is

To simplify the procedure further, it is helpful to always select at least one support for which the value of the moment is determinable. For instance, if the end supports are not fixed, the value of the moment there shall be zero. Also if there has been an overhang beyond one of the end supports, the value of the moment at this support due to loading at the overhang portion can be determined easily.

4.4.2 Determination of shear force from moment diagram

We have observed the procedure of direct determination of moment at the beam sections. The application of the three-moment equation can provide the value of moments at the sections or supports. However, it does not directly provide the magnitude of the support reactions. It is crucial to know the support reactions in order to determine the shear force at different sections of the beam.

To determine the support reactions, the below procedure is required to be followed. Figure 4.11 should be referred in order to understand the steps involved.

M = M₁ – M₂

• Isolate the beam span and represent it as a free body with all the imposed loads. Suppose the length of the span is L. The moments as determined by the analysis should be indicated as acting on both the ends.
• The support reactions, which are unknown, would also be shown as acting on both the supports.
• To determine the magnitude of support reactions, the diagram would be further resolved into two independent ones. The first one would show the external loads imposed sans support moments and the reactions as applicable in the case of a simply supported beam.
• The second diagram would show the imposition of moments M₁ and M₂ at both the ends. Assuming that both the moments would not be equal (and they actually would not be on most of the occasions), the net moment acting on the span would be equal to

The moments M₁ and M₂ would be given the proper signs as applicable in a particular case. Here it is assumed that both are positive and M₁ > M₂.

Since M₁ is assumed to have larger magnitude, its overall effect would be to rotate the beam in a clockwise direction. For equilibrium, the reactions at both the ends would form a moment couple of equal magnitude, acting anticlockwise.

If the force of the couple is R, the magnitude of the couple would be R×L, as L is the arm of the couple.

In the state of equilibrium,

R×L = M₁ – M₂

Hence, 

Therefore, the magnitude of the reaction would be similar for both ends. The only difference being that it would act downward at end 1 and upward at end 2.

• The net reaction at ends 1 and 2 would be the algebraic sum of the reaction due to applied load and the reaction due to moment’s application.
• The shear force diagram would now be drawn in the usual way by starting from an end.

This procedure of drawing the shear force diagram is very useful for the statically indeterminate structures with a high degree of indeterminacy. Besides beams, the method can be applied to any type of structure. The same procedure would be used to draw shear force diagrams for frames, as we would see in the further sections.

4.4.3 Determination of shear force

As we have seen earlier, the reaction at a particular support shall be the algebraic sum of the reaction due to loads imposed and the reaction due to moment differential at the two supports. This would be equal and opposite at both the supports to form a resisting couple. We also know that the value of this reaction would be

To make it easy to calculate, the values of both these components for each support reaction are shown in Figure 4.12 and, thus, the net reaction for each support has been worked out. These values have been used to draw the shear force diagram, which is also shown in Figure 4.12.

To illustrate the application of the three-moment theorem to determine the support moments and utilization of the results to determine the shear force diagram, we will look at the following situations.

Illustration 4.5: The beam ABC has ends A and C simply supported and is continuous over the support B. The spans and the loads imposed are shown in Figure 4.12. We have to determine the bending moment at critical locations and draw the moment diagram.

To start with, we would apply the three-moment theorem to the set of supports A, B and C. As the supports are non-yielding, h₁ and h₃ would be equal to zero. As is obvious, the magnitude of the moment at the end supports A and C would be equal to zero. Also, in view of second span having no load on it, the moment in bending moment diagram would also be zero. The applicable relationship equation would be

To determine the value of the expression A₁a₁, we need to draw the bending moment diagram for the load on the span AB. Refer to Figure 4.12, where bending moment diagram in parts for span AB has been drawn. It has been drawn in this way to make it easy to calculate the moment of the area about A, the first section. The ordinates to calculate the area have also been indicated in the diagram.

Therefore,

Substituting this value along with others (M₁, M₃ = 0) into the three-moment equation,

Therefore, the value of moment at support B is (–) 3375 Nm. In order to draw the bending moment diagram, we also need to derive the bending moment due to load applied.

Considering span AB, the reaction due to load would be 2500 N on each support.

Mid span moment = 2500 x 3 = 7500 NM.

Using these values of moment, the overall bending moment diagram is shown in Figure 4.12.

Now, in order to determine the shear force, let’s consider both the spans separately.

Span AB
The support moment at A is zero and at B is 3375 NM.
Hence, net moment = 3375 NM.

The moment has tendency to rotate the span in a clockwise direction. The anti-clockwise couple of reaction R₁ would resist this.

From this, the value of reaction R₁ = 3375/6 = 562.5 N.

Span BC
Similarly for span BC, the support moment at C is zero and at B is 3375 NM.

Hence, net moment = 3375 N-m.

The moment has tendency to rotate the span in an anti-clockwise direction. The clockwise couple of reaction R₂ would resist this.

From this, the value of reaction R₂ = 3375/4 = 843.75 N.

The combined effect of the reactions due to vertical load and moments is shown in Figure 4.12, along with the resultant shear force diagram.

Illustration 4.6: The beam ABCD has both the ends A and D simply supported and is continuous over the supports B and C. All the supports are non-yielding types. The spans and the loads imposed are shown in Figure 4.13. The objective is to draw the bending moment and shear force diagram for the beam system.

First of all, we would apply the three-moment theorem to the set of supports A, B and C. Since the supports are non-yielding, the values of h₁ and h₃ would be zero. Also, the magnitude of the moment at the end support A would be equal to zero and since the first span has no load on it, the moment of bending moment diagram would also be zero. The applicable relationship equation would be

To determine the value of the expression A₂b₂, we would draw the bending moment diagram for the load on the span BC. Since moving further we need the values of similar expression for the next span also, we would draw the bending moment diagram for span CD, too. Refer to Figure 4.10(a), in which bending moment diagram in parts for span BC has been drawn. It has been drawn in this way to make it easy to calculate the moment of the area about C, the third section. The ordinates to calculate the area have also been indicated in the diagram.

Hence,

Substituting this value along with others into the three-moment equation,

Similarly, when we consider sections B, C and D for the application of the three-moment theorem, the applicable expression shall be as follows:

To calculate the values of A₂a₂ and A₃b₄, we will refer back to Figures 4.13. It may be noted that the bending moment diagram for span BC in 4.13 is essentially similar to the one drawn below it, but has been drawn in such a way to make it easier to calculate its moment about B, which is the requirement this time.

Substituting all the values into the expression,

Solving equations (1) and (2) simultaneously,

	M2	=	(–) 1850 NM
and	M3	=	(–) 2525 NM

The resultant bending moment diagram is shown in Figure 4.14. Also have been derived the shear force diagram, adopting the procedure explained under illustration 4.5.

4.5 Slope-deflection method

Once we move to cases of indeterminate structures, where instances like yielding of supports, rotation and deflection of joints, geometrical arrangements other than a linear beam, etc. are involved, it becomes quite complex to use any of the earlier described method. For such situations, there are a few more good techniques available.

The slope-deflection method is one such technique. Here, the main assumptions used to write the equations are as follows.

• First of all, the joints of structural members are assumed rigid. Therefore, if the joints were not restrained against rotation, they would rotate in such a way that the angle between the members emanating from a joint would remain unchanged after the rotation.
• In a structural system, the algebraic sum of the end moments of the members, meeting at a joint, which is free to rotate, would always be equal to zero.

In the slope-deflection method, the end moments are expressed in terms of slope and deflection at the ends of the members.

4.5.1 Derivation of the slope-deflection equation

While writing an equation the following sign conventions are followed:

• Clockwise moments are designated as positive and anticlockwise as negative.
• Clockwise angles of rotation are expressed as positive and anticlockwise as negative.
• The deflection responsible to rotate the member clockwise is treated as positive and the one, which rotates anticlockwise, is treated as negative.

To derive the slope-deflection equation, consider the case of the beam AB of length L, as shown in Figure 4.15(a). Suppose the ends are free to rotate for the beam. A set of traverse loads has been applied to the beam, which has not been shown in the diagram for the sake of simplicity.

Due to influence of these loads, the net values of moment at A and B are M_AB and M_BA respectively. The angles of rotation of both the ends are θ_A and θ_B respectively. In addition, the end B has undergone a lateral deflection of magnitude δ_BA with respect to end A.

It is clear that this entire arrangement can be divided into following components.

• A beam with both the ends fixed has all the transverse loads applied to it. The same is shown in Figure 4.15(b). The beam has been subjected to the fixed end moments of magnitudes M_FAB and M_FBA.

• A beam has end B fixed and A simply supported with no lateral loads, but subject to moment M’_A at end A. This is shown in Figure 4.15(c). The moment has induced a rotation of θ_A at A, whereas B is unaffected_. The resultant moment at B would be M’_B.
• A beam has end A fixed and B simply supported with no lateral loads, but subject to moment M’’_B at end B. This is shown in Figure 4.15(d). The moment has induced a rotation of θ_B at B, whereas A is unaffected_. The resultant moment at B would be M’’_A.
• Fixed beam with no lateral loads, but subject to a deflection of magnitude δ_BA of end B with respect to end A. This is shown in Figure 4.15(e). The induced moments due to it are M’’’_A and M’’’_B respectively.

The sum of these four would geometrically and statically be equivalent to the original beam system.

Hence,	MAB	=	MFAB + M’A + M’’A + M’’’A.
And	MBA	=	MFBA + M’B + M’’B + M’’’A.

Let us proceed to determine and express the values of end moments in the form of deformations.

The fixed end moments M_FAB and M_FBA are the function of the loads imposed on the member and its length.

For cases 2 and 3, the value of the moment at the end of rotation would be equal to 4Eiθ/L, whereas for the fixed end the value of the moment would be equal to 2Eiθ/L.

Therefore, in case 2,

For the fourth case, where deflection has been inducing the moments at both the ends, the magnitude of such moment would be equal to 6Eiδ/L².

It may be noted here that the direction of the moment at both the ends would be anti-clockwise (hence negative sign), whereas the rotation of the member itself has been positive. Substituting these values into the moment relationships,

In the same way,

Equations (A) and (B) are slope-deflection equations. The expression δ/L in them is sometimes referred to as the rotation of the member.

4.5.3 Application of slope-deflection equations

The stepwise application of the slope-deflection equation to analyze a structure, in its general form, is as follows:

• First of all, consider all the joints as fixed and calculate the fixed end moments for the nodes.
• For each span of the structure, express both the end moments in the form of deformations using slope-deflection equations.
• Establish the equation for moment equilibrium of each joint, which is free to rotate. For equilibrium conditions, the algebraic sum of the end moments of all the members of the joint would be equal to zero.
• In the case of any other deformation like side sway or settlement, the additional equilibrium equation for stability would also be arrived at.
• Calculate the values of the deformations by solving the above equations simultaneously.
• All such calculated values of deformations would be substituted into the equations to determine the values of the moments.

The illustration below will be useful to understand the procedure better.

Illustration 4.7: The beam ABC has ends A and C simply supported and is continuous over support B, which is non-yielding. The spans and the loads imposed are shown in Figure 4.16.

First of all, we shall consider all the joints as fixed and calculate the end moment for each span, resultant to the imposed loading:

MFAB	=	MFBA =	0 (Since no external load)
MFCB	=	(-)MFBC =	wL2/12 = 1000×62/12 = 3000 NM

In the next step, we would write the slope-deflection equation for each joint. The basic equation is

For joint A, the values of M_FAB, and Δ_BA would be zero.

In the above equations for moment, it is clear that the second expression on right-hand side is the multiplication of the angles of rotation of joints, with the value equivalent to 2EI/L. Since we are not interested in determining the value of the individual angles of rotation, it is okay if we assign some arbitrary value to EI, which makes it convenient for our further calculations. The ultimate values derived for the angles would be perfectly okay when used in combination with this value of EI.

In this particular case, let us assign a value of 6 to EI.

Now,	MAB	=	6θA + 3θB
	MBA	=	3θA + 6θB
	MBC	=	– 3000 + 4θB + 2θC
	MCB	=	3000 + 4θC + 3θB

The next step is to write the equilibrium equations for the joints, where a clear relationship emerges.

These equations are

	MAB	= 0
	MBA + MBC	= 0
And,	MCB	= 0.

Substituting derived values of the moments in these condition equations, we get

We now need to solve equation (1), (2) and (3) simultaneously to determine the values of three unknown angles.

From equation (1), θA	=	(-)θB / 2
From equation (3), θC	=	-750 – 3θB / 4

Substituting these values into equation (2),

-3θB / 2 + 10θB – 1500 – 3θB / 2	=	3000
θB	=	642.85
From this, θA	=	(−)321.43
θC	=	(−)1232.14

Substituting these values of slope to derive the fixed end moments,

MAB	=	0
MBA	=	3θA + 6θB
	=	3 x ( −321.43) + 6 x 642.85
	=	2892.8 NM
MBC	=	– 2892.8 NM
MCB	=	0

The resultant bending moment diagram is shown in Figure 4.15.

In order to bring further clarity, we once again shall solve the example of continuous beam of Illustration 4.6, which previously has been solved by using three-moment theorem.

Illustration 4.8: The beam ABCD has both the ends A and D simply supported and is continuous over the supports B and C. All the supports are non-yielding type. The spans and the loads imposed are shown in Figure 4.16.

In the first step, we shall consider all the joints as fixed and calculate the fixed end moment at the end of each span, due to the imposed loading:

M_FAB = M_FBA = 0

MFCB	=	(−)MFBC	=	WL/8	=	5000×4/8	=	2500 NM
MFDC	=	(−)MFCD	=	wL2/12	=	1000×32/12	=	750 NM

In the next step, we would write the slope-deflection equation for each joint. The basic equation is

For joint A, the values of M_FAB and Δ_BA would be zero.

For this case, we assign a value of 6 to EI.

Now,	MAB	=	8θA + 4θB
	MBA	=	4θA + 8θB
	MBC	=	– 2500 + 6θB + 3θC
	MCB	=	2500 + 6θC + 3θB
	MCD	=	–750 + 8θC + 4θD
	MDC	=	750 + 8θD+ 4θC

The next step is to write equilibrium equations for the joints where the relationship is clear.

These equations are

Substituting the derived values of the moments into these condition equations, we get

We are now required to solve equations (1), (2), (3) and (4) simultaneously to find the values of all the four unknown angles. For this, we would need to perform the Gauss–Seidel iteration. This would require us to rewrite the above equations in the following forms:

Setting θ_B = 0 in equation (1),

θ_A = 0

Setting θ_A = 0 and θ_C = 0 in equation (2),

θ_B = 178.57

Setting θ_B = 178.57 and θ_D = 0 in equation (3),

θ_C = –163.21

Setting θ_C = –163.21 in equation (4),

θ_D = –12.15

A number of such cycles of iteration would require to be performed, by setting the previously obtained values, till we obtain two consecutive sets of values, which are almost similar to each other. The results of this exercise are shown in Table 4.3.

Table 4.3
Iteration table

Cycle	θA	θB	θC	θD
1	0	178.57	–163.21	–12.15
2	–89.3	239.7	–172.8	–7.35
3	–119.9	249.8	–176.4	–5.55
4	–124.9	252.0	–177.3	–5.1
5	–126.0	252.5	–177.6	–5.0

The set of simultaneous equations can be solved by any other technique also. The Gauss–Seidel iteration is one of the popular techniques for the solution of simultaneous equations in case the number of unknown exceeds three.

Since the differences in the values obtained in the fourth and fifth cycles are nominal, we would adopt the values of the fifth cycle as a solution to these equations.

By substituting these values of slopes, the values of end moments are as follows:

MAB	=	0
MBA	=	1517 NM
MBC	=	– 1517 NM
MCB	=	2192 NM
MCD	=	– 2192 NM
MDC	=	0

The bending moment diagram, shear force diagram and the deformed shape (dotted lines) of the beam are shown in Figure 4.16.

4.5.4 Application of slope-deflection equations for frame analysis

The slope-deflection method is the major application in the analysis of frames. The frames are the structure, which is formed of vertical and horizontal (sometimes inclined too) members. A typical frame in RCC (reinforced cement concrete) is shown in Figure 4.17.

In our case, we are only considering frames with rigid joints. These joints are capable of rotation as a whole, but the angles between the members at a joint remain unaltered.

In the case of frames, unlike beam junctions, the joints can be unrestrained and therefore can shift from their original positions under loading. Since we are considering two-dimensional frames with load application in the same plane, the shift or the movement of joints would remain inter-plane. This shift is termed as sway.

Coming back to frames, these are of two types. One has symmetry about some vertical axis. Apart from similar geometry on both sides of the axis, the corresponding members would also have the same sectional properties and end conditions. Frames meeting this criterion are known as symmetric frames. All others are known as the anti-symmetric frames.

Similarly, load applications can also be symmetric or anti-symmetric in nature.

Some examples of symmetric and anti-symmetric frames, as well as symmetric and anti-symmetric loadings are shown in Figure 4.18.

While analyzing the behavior of the frame under loading, we need to remember that the frames may or may not be subjected to sway. The sway-related behavior of the beam for different combinations of frames, loadings and restrains has been described in Table 4.4.

Table 4.4
Say / No-sway Conditions for Frames

Type of Frame	Type of Loading	Joint Condition	Result
Symmetric	Symmetric	Unrestrained	No-sway
Symmetric	Symmetric	Restrained	No-sway
Symmetric	Non-symmetric	Unrestrained	Sway
Symmetric	Non-symmetric	Restrained	No-sway
Non- symmetric	Symmetric	Unrestrained	Sway
Non- symmetric	Symmetric	Restrained	No-sway
Non- symmetric	Non -symmetric	Unrestrained	Sway
Non- symmetric	Non -symmetric	Restrained	No-sway

It is clear that the condition of non-symmetry either of frame or of loading is sufficient to make the frame sway, provided there has been no physical restraint against it.

If it can be recollected, we already have discussed the effect of restraint on a frame in Section – 2. It is obvious that the flexural behavior of the members of the same frame can be very different in the sway and no-sway conditions.

Analysis of frames meeting ‘no-sway’ condition

For frames with no-sway, the procedure of analysis is similar to the one we have followed in the analysis of the continuous beam. It involves steps like calculations of fixed end moment, writing slope-deflection equations for each span, defining conditions of constraints and solution of resultant equations simultaneously.

The following illustration should make the procedure amply clear.

Illustration 4.9: The frame ABCD has both the ends A and B fixed, and C and D free. The spans, loads imposed and relative moment of inertia for each member are shown in Figure 4.19.

In the first step, we shall consider all the joints as fixed and calculate the fixed end moment at the end of each span, due to the imposed loading:

M_FAC = M_FcA = 0

M_FDC = (–)M_FCD = WL/8 = 10000×6/8 = 7500 NM

M_FDB = (–)M_FBD = 0

In the next step, we would write the slope-deflection equation for each joint. The basic equation is

Here, the values of deflection at C and rotation at A are zero.

Let us assign a value of 6 to EI:

The equations of constraints are as follows:

M_CA + M_CD = 0

And, M_DC + M_DB = 0

Substituting the values of moments into these relation equations, we get

On solving equations (1) and (2) simultaneously, we get the following values:

θ_C = 750

And, θ_D = (–)750

Substituting these values of θ_C and θ_D, we derive following values of end moments:

M_AC = 2250 NM

M_CA = 4500 NM

M_CD = (−) 4500 NM

M_DC = 4500 NM

M_DB  = (−) 4500 NM

M_BD = (−) 2250 NM

The resultant bending moment diagram is shown in Figure 4.19.

To draw the shear force diagram, the previously described method of drawing free-body diagrams for individual members shall be used to determine the reactions.

For member AC,

Net Moment	=	2250 +4500	= 6750 NM (Clockwise)
Shear Force	=	6750 / 4	= 1687.5 N

Span AC has no other force acting on it; hence, the values determined above shall be the net reaction force acting at the nodes.

For member CD,

Net Moment	=	–4500 +4500	= 0
Shear Force			= 0

A concentrated load of 10000 N has been acting on the centre of the span CD. Due to it, each support shall offer a reaction of 5000 N.

For member CB,

Net Moment	=	− 2250 − 4500	= 6750 NM (Anticlockwise)
Shear Force	=	6750 / 4	= 1687.5 N

Span CB has no other force acting on it; hence, the values determined above shall be the net reaction force acting at the nodes.

The resultant shear force diagram is shown in Figure 4.19.

Analysis of frames subjected to ‘sway’

For frames, which are subjected to side sway either due to anti-symmetric configuration or loading or both, the process of analysis includes the accounting for the sway in the slope-deflection equations.

For this, the direction of sway and its relative magnitude are assessed for each member. Mostly, it is easy to determine the direction of the sway based on the load configuration. Even if the direction were assessed opposite to the actual one, the ultimate sign of the value derived would help in making the required correction.

The other important addition in the procedure is the introduction of the shear equation, as one of the condition equations. This is derived by applying the condition of horizontal stability to the entire frame. According to this condition,

ΣH = 0

For this, the net horizontal reactions on each member are expressed in the form of the sum of the end moments, divided by member length. The cumulative reaction of these net horizontal reactions added with direct horizontal force is expressed equal to zero.

So, ΣH = H1 + H₂ +…….. + (M_AB + M_BA)/L_AB + (M_CD + M_DC)/L_CD + ……… = 0

To clarify the procedure, we solve a simple case in the illustration below.

Illustration 4.10: The frame ABCD has both the ends A and B fixed, and C and D free. The spans, load imposed and relative moment of inertia for each member are shown in Figure 4.20.

It is very clear that due to the load imposed on joint C, there would be a lateral sway in the frame. The location of the joint C and D would shift to the right, consistent with the direction of the application of load at C. The new positions assumed by C and D would be C’ and D’ respectively. Let us assume that the magnitude of sway of the joint C to its new position C’ is δ. Since the lengths of the member can be assumed to remain unchanged due to loading, the corresponding sway at D would also be equal to δ in magnitude.

All the three members of the frame have no load applied between the nodes; hence, the fixed end moments for all the cases would be zero.

Therefore, M_FAC = M_FcA = 0

M_FDC = M_FCD = 0

M_FDB = M_FBD = 0

In the next step, we would write the slope-deflection equation for each joint. The basic equation for member AB is

In this case, the values of M_FAC and θ_A are zero. Also, the relative deformation of the joints is such that the member rotates in a clockwise direction. Hence, the magnitude of δ_CA would also be considered positive and the expression 3δ/L would retain its original negative sign.

By substituting an arbitrary value of 2 for the expression EI, we get

In the same way,

The conditional equations for this case are

M_CA + M_CD  = 0

M_DC + M_DB  = 0

And the shear equation is

5000 + (M_AC + M_CA)/L_AC + (M_DB + M_BA)/L_DB  = 0

Substituting the values,

By solving equations (1), (2) and (3) simultaneously, we obtain the following values:

θ_C = 74.6, θ_D = 559.68 and δ = 1144.23

Substituting these values into the end moment equations,

MAC	=	(–) 783.57 NM, MCA	=	(–) 708.9 NM,
MCD	=	708. 9 NM, MDc	=	1193.96 NM,
MDB	=	(–) 1193.96 NM, MBD	=	(–) 2313.33 NM

To draw the shear force diagram, we need to determine the end reaction of each member. In this case, apart from the load at node C, no external load has been acting on the span of any member. Any load at the node does not induce shear force; hence, the shear force generation would only be due to the end moments.

For span AC,

Rac = Rca	=	− (783.57 + 708.9)/ 4	= 373.1 N (clockwise couple)
Rcd = Rdc	=	(708.9 + 1193.96)/ 4	= 475.7 N (anticlockwise couple)
Rdb = Rbd	=	− (1193.96 – 2313.33)/ 2	= 1753.64 N (clockwise couple)

The bending moment and shear force diagrams are shown in Figure 4.20.

The above illustration makes clear the method of application of shear equation to solve structural situations, where sway of the frame is involved.

The slope-deflection method can similarly be used to solve structural problems pertaining to multibay frames (frames with more than one span), multistoried frames or frames with inclined members. The advantages of the slope-deflection method are given below:

The method has been derived from the basics and in turn can be used as a basis for other techniques.

It can be used to solve any type of structural problem.

The results derived are quite accurate.

The technique however, has a disadvantage. The number of linear simultaneous equations required to derive a complete solution is equal to the number of unknowns. When increasing the number of structural nodes, the number of the equations can become quite large. This does make the solution complex and cumbersome. In such a situation, it becomes imperative to use some ‘stand-alone’ iteration technique to solve the large number of simultaneous equations. This disadvantage of the technique has been overcome in the ‘moment-distribution method’.

4.6 Moment-distribution method

In the moment-distribution method of structural analysis, the joint equilibrium equations, already described in the slope-deflection method, are solved, without writing them explicitly. In this method, each of the joints is released one at a time instead of releasing all of them together as in the slope-deflection method. The iteration of values is done at each such step. This technique involves a tabular form of calculations, which makes it much simpler compared to the solution of slope-deflection equations.

The concepts and fundamentals of the technique have been described below.

4.6.1 Rotational stiffness of the member

Let’s consider the case of a member AB of length L. The member has a prismatic section and its moment of inertia is I. A moment of magnitude M_AB has been applied at the joint A. The joint A is free to rotate and the rotation of the joint due to the application of the moment is θ_A. On the other end, the joint B can have either of the following two conditions.

First Situation: End B is fixed and unable to rotate

In this situation, the introduction of moment M_AB at A would induce a moment M_BA at B such that the rotation at B becomes zero. This situation can be described as follows.

• End B is on hinge and subjected to a rotation equivalent to θ’_B due to the application of moment at A. In this situation, the rotation at A itself is equivalent to θ’_A.
• End B is subjected to a moment M_BA due to which the rotation at B becomes θ’’_B. This moment does induce θ’’_A rotation at A.

These situations are shown in Figure 4.21.

This way, the net rotation at B would be zero. At A, the net rotation is the algebraic sum of the rotations induced in both the situations.

Hence, θ_A = θ’_A – θ’’_A^.

And θ_B = 0 = θ’_B – θ’’_B^.

In the first case, when A is subjected to moment M_AB, the values of rotations at A and B would be the following:

Similarly, due to induction of moment M_BA at B the values of rotations at A and B would be the following:

Substituting these values into the relationships of the rotations,

From above, M_AB = 4EI θ_A / L.

If we equate θ_A to one radian, the value of moment becomes

M_AB = 4EI / L

The above expression, which provides the value of moment required at one end to induce unit rotation there, is known as the rotational stiffness of the member. The notation for this is K_AB. It may be noted that this expression of rotational stiffness is for the condition when the other end of the member is fixed.

Second Situation: End B is on hinge and is free to rotate

In this situation, when a moment M_AB is applied at A, the moment induced at the other end would be zero due to the hinge effect. The rotation at A,

Hence, M_AB = 3EI θ_A / L.

As per the definition of rotational stiffness, the value of θ_A would be equal to one unit:

K_AB =  M_AB  = 3EI / L

Carry-over factor

While deriving the rotational stiffness of a member for the above two situations, we have seen that in the case of end B fixed, the moment induced at B would be half of the moment introduced at A. This means that one half of the moment at A has been carried over to B. This factor of ½ is known as the carry-over factor.

In the case when the end B is hinged, the carry-over factor is zero.

These values of carry-over factors greatly help in making an assessment of the moment values at the other end during process of iteration.

Lateral stiffness of members

Lateral stiffness of the member is the magnitude of moment, which gets induced at the end A of the member AB when a relative displacement of unit magnitude takes place between the ends A and B of the member.

Similar to the conditions for rotational stiffness, the value of lateral stiffness is also dependent on the support condition of the other end.

When the end B is fixed

In the case when both ends of the beam are fixed, the moment at one end is due to relative deflection ‘d’ with the other end:

M = 6EId / L²

The value of lateral stiffness in this case (value of moment when d = 1),

LS = 6EI / L²

When the end B is on hinge and is free to rotate

The value of lateral stiffness in this case would be equal to

LS = 3EI / L²

Distribution factor

In the case of frames or any other structural system, where more than one member joins at a node, that joint is assumed as rigid. It means that even if the joint as a whole rotates due to load application, the individual members still maintain the original slopes between them. This implies that all the members undergo the same amount of rotation.

If a number of members are meeting at a joint which undergoes rotation due to moment application of magnitude M, then this moment shall be shared by each member in proportion to its rotational stiffness with respect to the cumulative value of the rotational stiffness of all members meeting at the joint.

Therefore, if the cumulative value of rotational stiffness at the joint is Kcum and the value for the member under consideration is K’, the moment carried by the member shall be

M’ = (K’ / Kcum) M

This factor K’/Kcum is known as the distribution factor for an individual member.

4.6.2 Application of moment distribution method

The step-wise description of the application of the moment distribution method is as follows:

• First of all, it is required to calculate the fixed end moment for each member under the applied load, in a manner similar to that described in the slope-deflection technique. While doing this, both the ends of the members shall be considered as fixed, irrespective of their actual structural status.
• The next step in the process is to work out the rotational stiffness of members at each joint. The expression K = 4EI/L would be used to determine it. It may be noted here that even if the other end of the member is hinged, it is procedurally correct to use the value of stiffness considering the other end as fixed and then release the fixity of the end in every cycle in the way described under (5). Otherwise, for hinged ends, the expression K = 3EI/L can also be used and it would obviate the need of repeating the release of the joint in each subsequent cycle.
• With the help of the values of individual K’s for each member, the values of the distribution factor for each member converging at the joint would be determined.
• This entire information about the fixed end moment and the distribution factors shall be recorded in a tabular form. The format of the table would be described in the illustration below.
• The further step would involve the releasing of the joints, which are actually hinged but were considered fixed. To do that, a moment, which is equal to the fixed end moment in magnitude, but of opposite sign, shall be applied to the joint.
• Together with this, the joints, which have more than one member meeting, would need balancing of the moment. In the act of balancing, the net moment at the joints, due to the applicable fixed end moments, would be determined first. This would be the algebraic sum of all the moments. Once determined, this unbalanced value of the moment shall be balanced by applying a moment at the joint, equal in magnitude to the unbalanced moment but opposite in direction.
• The distribution of this balancing moment shall be in proportion to the distribution factor of the members meeting at the joint.
• Once the joints have been released/balanced in the way described above, the carry-over of the moments applied would need to be considered. As we have observed in concepts, half of the moment applied at one end of the member gets carried-over to the other end.
• Once the carry-over (CO) cycle is completed, the joints will again turn non-balanced. This would need further cycles of joint releasing/balancing activities, accompanied with the carry-over of the moments.
• After a number of balancing cycles, the carried-over moments would become too small. The procedure would be stopped at that point. The last cycle would consist of the joint balancing operation but no carried over to the junctions. However, carry-over to an end joint can be performed.
• The algebraic sum of the value of original fixed end moment and the subsequent values of all the balancing and carried-over moments for any end (all the moments figuring in single column) shall be the value of the final end moment.

The following illustrations would further clarify the procedure.

Illustration 4.11: The continuous beam ABC has ends A and C supported over a hinge. The point B is continuous over the supports. The spans, load imposed and relative moment of inertia for each member are shown in Figure 4.22. We have to draw the bending moment diagram.

First of all, the fixed end moments for each span would be calculated as follows:

Also, the values of rotational stiffness of each member shall be determined by considering both the ends as fixed:

K_AB = 4EI/4 = EI,

K_BC = 4EI/6 = 0.67 EI

The values of distribution factor for the nods B are as follows:

D_BA  = EI/ (1+0.67) EI = 0.60

D_BC =  0.67EI/ (1+0.67) EI = 0.40

The fixed end moments and the distribution factors have been tabulated as below.

The next step is the release of the joint C from an assumed position of fixity. It may be noted that due to non-imposition of load on span AB, the fixed end moment anyway is equivalent to zero. Thus, there is no need to apply the procedure on joint A.

To achieve release of joint C, moment of magnitude –1500 NM would be applied at C. Half of it would get transferred to the other end (B) of the span.

The next step would be the balancing of the joint B.

At B,

Unbalance Moment	= − 1500 − 750=	− 2250
Balancing Moment	= 2250

The balancing moment at B would be distributed in proportion of the distribution factors:

Towards A	=	0.6 X 2250	=	1350
Towards C	=	0.4 X 2250	=	900

Half of the magnitude of these moments would be carried-over to the other respective ends, A and C.

In the next step, the ends A and C would be released from the carried-over moments by introduction of moments of similar magnitude but

The resultant bending moment diagram is shown in Figure 4.22.

Illustration 4.12: The continuous beam ABCD has ends A fixed and D supported over a hinge. B and C are continuous over the supports. The spans, load imposed and relative moment of inertia for each member are shown in Figure 4.23.

First of all, the fixed end moments for each span would be calculated as follows:

Also, the values of rotational stiffness of each member shall be determined by considering all the ends fixed:

KAB	=	4EI/5 = 0.8 EI,
KBC	=	4EI/4 = EI,
KCD	=	4EI/4 = EI

It should be noted here that the value of K_CD has been determined by assuming end D as fixed and, therefore, it would be necessary to release the joint D in each cycle.

The values of distribution factors for the nodes B and C are as follows:

These values of fixed end moments and distribution factors have been joint-wise tabulated on the next page.

The next step is the release of the joint D from an assumed position of fixity. It may be noted that since joint A is fixed originally, there would be no need to release this joint. To release D, the moment of magnitude –5000 NM would be applied.

The next step would be the balancing of the joints B and C.

At B,

Unbalance Moment	= − 5000 + 2084	=	− 2916
Balancing Moment	= 2916

The balancing moment at B would be distributed in proportion of the distribution factors:

Towards A	=	0.444 X 2916	=	1295
Towards C	=	0.556 X 2916	=	1621

Once these two moments are applied at junction B, half of these would carry forward to the other respective ends.

Moment Distribution Table:

The joint C was originally in a balanced state but after release of joint D and the moment carry-over from B it has turned unbalanced:

Unbalance Moment	= −2500 + 810 =	−1690
Balancing Moment	= 1690

The balancing moment at C would be distributed in proportion to the distribution factors:

Towards B	=	0.5 × (1690)	=	845
Towards D	=	0.5 × (1690)	=	845

Once these two moments are applied at junction C, half of these would carry forward to the other respective ends.

After this carried over operation, the end D and joint B have again turned unbalanced. Therefore, we need to repeat the cycle of releasing, balancing and carry-over once again for these two.

Again, with the carry-over performed from B and D, the joint C has unbalanced moment. We need to balance the joint C once again. This, along with resultant carry-over, is shown in the table.

In the next step, the balancing of B and D has been done. Since the values of carry-over moments have become quite small now, we have decided to stop the process here itself. The values of end moments have been worked by algebraically summing all the moment values into a single column.

It is noticeable that the algebraic sum of all the moments at mid joints B and C has been zero, which is a conceptually correct situation.

Once the end moment values are derived, the further procedure of drawing the bending moment diagram and shear force diagram is similar to the one adopted in the slope-deflection method.

The resultant bending moment diagram is shown in Figure 4.23.

Illustration 4.13: The frame ABCD has ends A and D supported on a hinge. The nodes B and C are rigid, but unsupported. The spans, load imposed and relative moment of inertia for each member are shown in Figure 4.24.

Both frame and loading are symmetric in this case and, therefore, there would be no-sway in the frame.

First of all, the fixed end moments for each span would be calculated as follows.

MFAB	=	MFAB	= 0
MFBC	=	(−) MFCB	= (−) 1000×62 / 12  = (−) 3000 NM
MFCD	=	MFDC	= 0

Next, the values of rotational stiffness of each member shall be determined in accordance with the actual end conditions:

KAB	=	3EI/4	= 0.75EI, (End A is hinged)
KBC	=	4E (2.25I)/6	= 1.5EI,
KCD	=	3EI/4	= 0.75EI (End A is hinged)

The values of distribution factors for the nodes B and C are as follows:

For B, D_BA =  0.75 EI/ (0.75 + 1.5) EI = 0.333

D_BC = 1.5 EI/ (0.75 + 1.5) EI = 0.667

For C, D_CB = 1.5 EI/ (0.75 + 1.5) EI = 0.667

D_CD = 0.75 EI/ (0.75 + 1.5) EI = 0.333

These values of fixed end moments and distribution factors have been joint-wise tabulated as described in the previous illustration.

The procedure adopted in moment writing in the table is self-explanatory. The only noticeable point is that there has been no moment carry-over shown toward ends A and D. The reason for this is the hinge supports at both the ends. Due consideration has been given while calculating the rotational stiffness of the members AB and CD. Incidentally, in this case, the values of fixed end moments at A and D are zero, due to the absence of any loads on the spans. Had there been any fixed end moments present for these ends, it would have necessitated the inclusion of a step, for release of these two joints to make their fixed end moment zero.

Moment Distribution Table:

The slight variation observed in the values of the moments at joints B and C is due to the inherent characteristics of the procedure. The accuracy level of the results increases, with the number of cycles performed. We will rationalize the absolute moment values to 1496 NM.

With the fixed end moments known, the bending moment diagram for the frame ABCD is shown in Figure 4.20.

To draw the shear force diagram, the end shear force due to moments has been calculated as follows:

The shear force diagram and the elastic curve for the frame are shown in Figure 4.20.

Analysis of frames subjected to ‘sway’

If the frame is subjected to sway, the analysis by moment distribution method involves preparation of more than one moment, balancing tables. The first table will derive a usual ‘no-sway’ solution, by assuming that the sway is prevented by introduction of some restraint.

The next step would be to assign an arbitrary value of sway d, in the direction consistent with the load arrangement, preferably in terms of some multiple of EI. With the help of a lateral stiffness expression of each member, the end moment values for the joints will be derived exclusively, due to the sway. At this stage, another set of moment distribution shall be performed in order to derive the set of end moments, due to sway.

Since the sway value was assumed arbitrarily, the final step would be to apply the correction factor to the values of end moments due to sway. For this, the net horizontal shear in the first case would be determined. This would necessarily be =/= 0; otherwise, the sway would not have taken place. Suppose the value of this net shear force is V. For the overall stability, the net value of shear in the sway case should have been equal to (–) V. If it is different from (–) V, the factor likely to make it (–) V is called the correction factor.

The ultimate end moment values are the algebraic sum of the end moment due to normal loading and the factored end moment, due to sway.

The following illustration should make the procedure clear.

Illustration 4.14: The frame ABCDEF has ends A and D as fixed supports and F is supported on a hinge. The nodes B and C are rigid, but unsupported. The spans, load imposed and relative moment of inertia for each member are shown in Figure 4.25.

In this case the frame itself is symmetric but is subjected to asymmetric loading and, therefore, there would be sway in the frame.

In the first step, the frame would be solved for loads applied, assuming it to be a non-sway case.

Step – I: No-sway solution

First of all, the fixed end moments for each span would be calculated as follows:

Next, the values of rotational stiffness of each member shall be determined in accordance with the actual end conditions:

KAB	=	KCD	=	4EI/5	= 0.8EI,
KBE	=	KED	=	4E (2I)/4	= 2EI,
KEF	=			3EI/5	= 0.6EI (End F is hinged)

The values of distribution factors for the nodes B and C are as follows:

These values of fixed end moments and distribution factors have been joint-wise tabulated as follows.

Moment Distribution Table:

Step – II: Sway solution

For this part, let’s assign an arbitrary value of 2500/EI to the sway d. The direction of sway is assumed such that the members AB, FE and DC rotate clockwise.

In this situation,

M_FAB = M_FBA = 6EId / L² = (6EI×2500/EI) / 5² = 600

M_FCD = M_FDC = 6EId / L² = (6EI×2500/EI) / 5² = 600

But since the end F is hinged,

M_FEF = 3EId / L² = (3EI×2500/EI) / 5² = 300

And M_FFE = 0.

Moment Distribution Table:

Step – III: Determination of correction – factor

As the values of end moment determined for sway are based on an arbitrarily assumed magnitude of sway, we are required to determine its correction factor.

For this, the value of net horizontal shear force in the ‘no-sway’ condition is

And the value of net horizontal force in the ‘sway’ condition is

The correction factor k would be determined from the relationship

81.8 + k×440 = 0

Hence, k = (−) 0.186.

With this value of correction factor, the net composite values of end moments are as follows.

Designation	End Moment : No-sway (A)	End Moment: Sway (B)	Correction Factor X End Moment: Sway (C)	Net End Moment in NM (A + C)
MFAB	378	514	–96	282
MFBA	758	428	–80	678
MFBE	–758	–428	80	–678
MFEB	2566	–158	29	2595
MFEF	–104	316	–59	–163
MFFE	0	0	0	0
MFEC	–2462	–158	29	–2433
MFCE	415	–428	80	495
MFCD	–415	428	–80	–495
MFDC	–208	514	–96	–304

The bending moment diagram and the elastic curve (deflected shape) of the structure are shown in Figure 4.21. The procedure to draw the shear force diagram is standard for this situation, too.

After understanding both slope-deflection and moment distribution techniques, we can appreciate the obvious simplicity of the moment distribution technique. The technique obviates the need to write equations for each fixed end moment, as well as the solution of a set of simultaneous conditional equations, which emerge from these end moment equations. Although the results derived from moment distribution techniques are not hundred percent accurate, it is possible to reach to a very high degree of accuracy by increasing the number of iteration cycles. For most of the problem though, an acceptable degree of accuracy is reached after just one or two such cycles.

4.7 Influence line diagram for statically indeterminate structures

We have studied the concept and usefulness of influence lines in Section 2.11. In that section, we have seen how the influence line can be useful for determination of various stress functions, in the case of statically determinate structures. In this section, we will study the technique of drawing influence lines for statically indeterminate structures and their use in analyzing such structures.

To reiterate, the influence line is a graphical representation of the magnitude of a reaction or stress function on a section of the span, when a unit concentrated load moves along the span. The influence line can be quite helpful for the situation, when the load system on the structure is not stationary. Therefore, this fact needs consideration in the analysis.

In the case of statically determinate structures, the analytical method of drawing an influence line diagram is quite simple, as we have seen earlier. It is based on the principles of statics. In the case of statically determinate structures, the procedure of drawing an influence line diagram for a stress function is obviously not that simple. There are a number of techniques to draw influence lines. A good example is the application of ‘the principle of virtual work’. The most widely used method for this category of structures is the application of the ‘Muller–Breslau’s principle’, which can be backed by any other quantitative technique such as moment distribution to determine the ordinates.

4.7.2 Influence line using Muller–Breslau’s principle

We have understood the principle in the previous section. Here, we would observe the application of the principle to draw the influence line diagram for different stress functions.

Influence line for bending moment

We consider the case of the statically indeterminate beam ABCD, with both the supports on hinges and the beam continuous over both the mid supports. A unit load is moving along the span and has presently occupied a position in the span CD. Due to its present location, the elastic curve for the beam has adopted the shape shown in Figure 4.26. Although this step is not essential, it provides the idea about the shape the elastic curve would take under restraints imposed.

Influence line for shear force

We will again consider the case of the statically indeterminate beam ABCD of the previous sub-section.

To draw the influence line diagram for shear force, we again select the same section I-I on span BC. The next step has been to introduce a cut at the section and apply a force in such a way that results in a unit relative displacement, between the points on either side of the cut. At the same time, the slopes on either side of the cut would be similar. The deformed shape of the beam in this situation is shown in the same diagram. Since the reverse of it would be the influence line at some scale, the shape of the influence line is shown in Figure 4.27.

Influence line for support reaction

For this too, we consider the case of the statically indeterminate beam ABCD of the previous sub-section.

To draw the influence line diagram for support reaction, we select the support C. The next step is to remove the constraint of ‘no-settlement’ at C and apply a force in such a way that results in a unit magnitude of settlement at C. The deformed shape of the beam in this situation is shown in the same diagram. Since the reverse of it would be the influence line at some scale, the shape of the influence line is shown in Figure 4.28.

Influence line for support moment

For support moment, we need to consider one of the internal supports only, since for end supports the moment would always remain zero. For this too, we consider the case of the statically indeterminate beam ABCD as in the earlier examples.

To draw the influence line diagram for support moment, we select the support C. The next step is to insert a ‘hinge’ at C and apply a force in such a way that results in a unit angle of rotation at C. The deformed shape of the beam in this situation is shown in the same diagram. Since the reverse of it would be the influence line at some scale, the shape of the influence line is shown in Figure 4.29.

4.8 Conclusion

In this chapter, we have studied the techniques dedicated to the analysis of the continuous beams, as they are one of the most commonly encountered statically indeterminable systems. We can now also analyze other beam systems such as fixed beams, propped cantilevers, etc.

We have also studied the analysis techniques, which are useful for the analysis of frames and other structures, in addition to beams. For this, the techniques of slope-deflection and moment distribution were described. These are the two most commonly used techniques for this purpose. With these two known, it is possible to analyze almost any kind of structural arrangement with ease.

We ended the section on the introduction by explaining the method of drawing influence line diagrams for statically determinate structures. We have already studied influence lines and their use in the case of statically determinate sections in Chapter 2. The information on the Muller–Breslau principle supplements the earlier material studied on influence lines.

In the previous three chapters we have observed the analysis of statically determinate and indeterminate structures separately, as well as studied the principles and applications of the strength of materials. With this, we are ready to move to practical applications of these principles, for actual design exercises. We will study the types of loading, failure mechanisms and major design theories in the next chapter.

5.1.1 Behavior of steel beyond the yield point

When a mild steel specimen is stretched beyond its yield point, the material hardens and the value of stress imposition as value of stress imposition goes up the bar elongates. At this stage, elongation of the specimen is accompanied with uniform reduction of the cross-sectional area so that overall volume of the specimen remains practically unchanged. The major part of the mechanical energy given at this stage is converted into heat, while a smaller part of it remains in the specimen in the form of strain energy.

5.2.1 Working stress theory

In the elastic design process, the requirement is to simultaneously fulfill the equilibrium condition, the compatibility condition and the limiting stress condition. According to the first condition, the structure subjected to loading should remain in equilibrium. According to the second condition, the deformation of the various fibres should be compatible with those of the adjacent fibres. According to the third condition, the maximum stress at any section in any fibre should be less than the yield stresses; in other words, the bending moment at the section should be less than the yielding moment.

A stress–strain diagram gives valuable information on the mechanical properties of a structural material. With information on the limit of proportionality, yield point stress and the ultimate failure stress of the material, it is possible to establish the magnitude of stress, specific to any engineering use, which may be considered a safe stress. This stress is typically known as the working stress, and is applied as a limiting value for design of the structural members in the ‘working stress method’ of design.

As a principle, working stress is determined by the following relationship:

Working Stress = Failure Stress / Factor of Safety

Different factors govern the selection of the basis for working stress, for different materials. For instance, in choosing the magnitude of the working stress for steel, it is taken into consideration that, at stress level below the yield point, the material may be considered as perfectly elastic. Beyond this limit, a part of the strain usually remains after unloading the structure, which results in permanent deformation or set. In order to ensure that the structural components remain in the elastic condition throughout, as well as to eliminate the possibility of a permanent set in the components, it is sensible to keep the working stress within the yield point limit. As the yield point stress is quite easy to determine with the help of the stress–strain curve, the working stress value can easily be correlated to the yield point for structural steel.

In the case of materials like concrete, the value of crushing strength of the specimen is the most important, since concrete is essentially designed to withstand compression. For other brittle materials such as cast iron, various kinds of stone and also for such material as wood, the ultimate strength is usually taken as a basis for determining the working stresses.

5.2.2 Estimation of ‘Factor of Safety’

It would be worthwhile to develop a mathematical model, which can help in determining the factor of safety, for a given condition. Such relationships for the factor of safety may be decided by defining it in a rational method. As is obvious, the factor of safety may be defined as the ratio of computed strength, T, of the structure or the structural element to the respective computed internal force, F.

Therefore,

Factor of Safety M = T / F

The magnitude of lowest probable strength, T, is influenced by the uncertainties in the mechanism failure, properties of material and the workmanship. The magnitude of highest probable internal load, F, is influenced by the uncertainties in loading conditions, structural behavior, method of analysis, etc. These uncertainties make the probable values of T and F deviate from the computed values.

The minimum probable value of strength is expressed as follows:

T_min = (T – ΔT)

where ΔT = maximum probable deviation of actual value from the computed value of strength.

Similarly, the maximum probable value of the internal force is expressed as follows:

F_min = (F + ΔF)

where ΔF = maximum probable deviation of actual value from the computed value of internal load.

We now will define the limiting condition to prevent structural failure. If the minimum probable value of strength is just equal to the maximum probable value of the internal load, then the failure of a structure is just prevented under most severe conditions.

Expressed mathematically, it would be

As we have seen earlier, T/F = M, the factor of safety.

With this, we can now determine the value of factor of safety. In the above expressions, the values of subexpressions ΔF/F and ΔT/T can be assigned values as percentages, depending on the assessment of the situation.

Again as a limiting case, let us assume that the values of deviations ΔF and ΔT in terms of percentage are 25% of the respective calculated values. In that case,

or M = 1.25 / 0.75 = 1.67

This value of 1.67 for factor of safety, though just indicative, can be considered as its minimum value.

It may be understood here that in the case of a statically determinate structural system, the stresses in the structural components will vary in proportion to the applied loads, even if the proportionate limit is exceeded. Thus, in such systems, the application of the working stress method fulfills the desired objective. However, in the case of a statically indeterminate structural system, the stresses in the members are the functions of their deformations. Here, the methods of analysis based on Hooke’s law do not apply beyond the proportionate limit.

5.2.3 Limit-state theory

The previously described working stress design procedure incorporates a single all-embracing safety factor. This safety factor would be appropriate for some aspects of the design, but may not necessarily be appropriate for all. This can lead to inefficiencies in design or even an unacceptable risk level.

Limit state is a way of designing an object by separately ensuring that it can perform each of the required functions. The method takes account of the risk involved, the factors that affect the object’s ability and the factors that affect the anticipated duty or loading. Limit-state design of structure enables one to predict loading, under which the structure will collapse due to simultaneous failure of some or all of its members. According to this theory, we multiply the actual service load with a load factor n and undertake the design treating this inflated load as the collapse load. This way, a true factor of safety n is realized against complete failure of the structure. This is the philosophy of limit design, which is gaining increasing favor among design engineers because its application normally results in more efficient and economical designs.

Serviceability and strength are the two limit states used in the design of structures. Limit state of serviceability is the level at which the structure remains fully functional, with no permanent distortion of the components and also keeping the magnitude of deflection within an acceptable limit. Strength limit state is the point beyond which the structure would no longer stay in place or no longer sustain the imposed loads.

To start with, it is convenient to represent the portion within curve OYMN of the stress–strain diagram for steel, as shown in Figure 5.3. The assumption made here is that the yield point is the limit of proportionality and the material deforms indefinitely beyond it. Therefore, the material is said to be either perfectly elastic or perfectly plastic, depending on whether the stress is below or above the yield point value. It will be further assumed here that the yield point has the same values in both tension and compression. The value of yield stress (fy) for ordinary structural steel may be taken as 280 Newton/mm².

The strength of steel is expressed in terms of its yield stress. The strength of materials, which do not have a distinct yield point, is specified in terms of 0.2% proof stress. This indicates a stress level, at which a residual strain of 0.2% is left over after unloading. High tensile steel, aluminum, etc., fall in this category.

In order to explain the principle of the limit-state design, let us consider the statically indeterminate system, consisting of three members, as shown in Figure 5.4.

We assume here that all three members are made of similar steel and have the same cross-sectional area (say, A). As is clear, all the members of the system are subjected to direct stresses only. Consider the situation when the magnitude of the load F keeps on increasing. As the magnitude of the load F increases, the values of the axial forces P and Q in the members also increase, but the ratio of P and Q would remain constant during this phase. This situation remains until the load reaches a certain value Pi, at which point, the force in the vertical bar attains the magnitude Fi = fy×A, i.e., the stress in this bar reaches the yield point. At this moment, the material of the vertical member becomes plastic. On further increase in the load F, the vertical member would stretch without further increase in tensile stress within it. In other words, the member will not contribute towards sharing of the incremental load but the inclined bars, subjected to the smaller value of stress would still be elastic and capable of carrying further load. Therefore, as the load F increases beyond the value Fi, the tensile force P in the vertical bar remains constant but the tensile forces Q in the inclined bars continue to increase. At this stage, the earlier relationship describing the ratio of P and Q would not be valid. The only condition now would be to satisfy the equilibrium equation

F = P + 2Q cosθ

On further increase in the load F, finally, the force in the inclined members would also reach the yield point. At this stage,

P = Q = fy×A

Putting these values in the equilibrium equation gives

F_ultimate = fy×A (1 + 2 cosθ)

This value of F_ultimate is termed as the collapse load or limit load of the structure. A working load F = F_ultimate /N will have a true factor of safety N against complete collapse of the entire structural system.

If we evaluate working load as per limit state of strength vis-à-vis the similar value permitted as per working stress principle, for same value of factor of safety, the value permitted by limit state would be higher than the working stress. The reason is that in the case of working stress, no plastic deformations are permitted and the analysis would have restricted the load to the value, which induces yield stress in the vertical member. Therefore, the limit-state design offers a more efficient solution in structural design.

The design principles and their application under limit-state design method have further been elaborated in Chapter 8.

6.3.1 Chemical composition

On the basis of chemical composition, structural steel is generally classified as mild steel, high-carbon steel and low-alloy steel. This classification depends upon the content of carbon and other elements in the steel. Steel of other classifications, i.e., high-alloy steel, is normally used for special purposes and not so much for the manufacturing of structures.

The range of constituents for different classifications of structural steels is as follows:

Mild steel

High-carbon steel

Low-alloy steel

Low-alloy steel contains a larger number of elements in comparison with carbon steels. There are different grades of low-alloy structural steel, with varying physical and chemical properties. A sample composition for low-alloy steel is given below.

6.3.2 Physical properties

The physical properties, the magnitude of which can be considered similar to all grades of steels, are as follows:

Apart from these, other important physical characteristics for structural steel are strength, ductility and hardness. The magnitudes of all these characteristics vary, from one type of steel to the other. Out of these, strength is the most important characteristic. It determines the actual section of the structural element. Ductility is important from the point of view of the allowable deformation and is measured as the percentage elongation of the tensile test specimen for steel at the moment of rupture.

The engineering codes of different countries classify structural steel in different grades. The indicative range of yield strength and ultimate tensile strength for different grades of steel is given below.

The classifications indicated above are arbitrary and for the purpose of providing an indication of the stress range for steel. However, during actual design process, it is absolutely vital to use the permissible stress values provided by the manufacturer or the relevant code.

Simple design

This method is based on the elastic theory and applies to structures, in which the end connections between members are such that they will not develop restraint moments, affecting the members and the structures as a whole and in consequence the structure may, for the purpose of design, be assumed to be pin jointed. The method of simple design involves the following assumptions.

• The stress–strain relationship for steel is linear.
• The plane sections normal to the axis do remain plane after bending.
• All the beams are simply supported.
• All connections of beams, girders, or trusses are flexible and are proportioned for the reaction shears applied at the appropriate eccentricity.
• The members in compression are subjected to forces applied at the appropriate eccentricities.
• The members in tension are subjected to longitudinal forces applied over the net area of the section.

Semi-rigid design

This method as compared with the simple design method permits a reduction in the maximum bending moment in beams suitably connected to their supports, so as to provide a degree of direction fixity. In the case of triangulated frames, it permits rotation account being taken of the rigidity of the connections and the moment of interaction of members. In cases where this method of design is employed, it is ensured that the assumed partial fixity is available and calculations based on general or particular experimental evidence shall be made to show that the stresses in any part of the structure are not in excess of those laid down in the relevant codes.

Rigid design

This method as compared to the methods of simple and semi-rigid design assumes that the end connections are fully rigid and are capable of transmitting moments and shears. It is also assumed that the angle between the members at the joint does not change when it is subjected to loading. This method gives economy in the weight of steel used when applied in appropriate cases. The end connections of members of the frame shall have sufficient rigidity to hold virtually unchanged, original angles between such members and the members they connect. The design should be based on accurate methods of elastic analysis, and calculated stresses shall not exceed permissible stresses.

Plastic design

The method of plastic analysis and design was developed recently. The method was thought of around year 1935. In this method, the structural usefulness of the material is extended up to ultimate load. This method has its main application in the analysis and design of statically indeterminate frame structures. This method provides striking economy, as far as the weight of the steel structure is concerned. This method provides a margin of safety in terms of the load factor instead of factor of safety. A load factor of 1.85 is adopted for dead load and live load, and a load factor of 1.40 is adopted for wind or earthquake forces. The deflections under working loads should not exceed the limits prescribed in the design codes.

6.6.1 Design of riveted joints

Rivets are the rounded pieces of steel, which are inserted through holes drilled in the interconnecting members and then forged in place. This involves heating the rivet shaft to a red-hot condition and then placing it in the hole. The ends of the rivets are then beaten (driven) into heads of a predetermined shape. The material of the rivet fills the hole entirely. This body of the rivet is known as the shank. Rivets are permanent fasteners and opening of such joints can be achieved only by destroying the rivets.

Besides the material of manufacturing, rivets can be classified based on the shape of their heads or on the method of driving. From the design perspective, the method of driving is important. There can be two types of driving-based classifications.

• Power-Driven Rivets – Normally a power-driven pneumatic tool is used to drive such rivets.
• Hand-Driven Rivets – In the case of hand-driven rivets, the application of drive (blow) is manual and the result can be less satisfactory than the first category.

Riveting terminology

Important terms used in riveting are described below:

Nominal diameter – The nominal diameter of the rivet is the shaft diameter before driving.

Gross diameter – The diameter of the rivet post driving, which is effectively equal to the diameter of the hole, is termed as the gross diameter of the rivet.

Pitch of rivet (p) – The pitch of rivets is the centre-to-centre distance between consecutive rivets, which lie on the same rivet line. The pitch is measured parallel to the direction of the force in the structural member. Normally, a pitch of less than two and a half times the gross diameter of the rivet is not recommended. On the upper side, the pitch is normally restricted to 12–16 times the thickness of the member. While designing, an effort is made to keep the pitch of the rivet constant over, as large a length as possible. This facilitates the easy and accurate marking and inspection of the job.

Staggered pitch – The staggered pitch or alternate pitch is the distance measured along one rivet line, from the centre of a rivet on it, to the centre of the adjoining rivet on the adjacent parallel rivet line. The staggered pitch occurs between the double rivet lines.

Edge distance – The distance between the edges of the member to the centre of the nearest rivet hole is known as the edge distance. Normally, a minimum edge distance of one and a half times the gross diameter of the rivet is provided to protect it from the shearing, splitting or bearing failure of the member.

Gauge distance of rivets (g) – The gauge distance is the transverse distance between two rivets of adjacent chains (parallel rivet lines) and is measured at right angles to the direction of the force in the structural member.

Rivet line – The rivet line, also termed as the gauge line, is the line along which rivets are placed. The rolled steel section (e.g., I-sections, channel sections, T-sections) has been assigned standard positions of the rivet lines. Standard positions of the rivet lines depend upon the flange widths in the case of I-sections, channel sections and T-sections, respectively. The standard positions of the rivet lines for the angle sections depend upon the length of the legs. While designing riveted connections, an effort should be made to place the rivets along these standard rivet lines to the possible extent.

The rivet heads can be of several shapes. Some of these are shown in Figure 6.4. All dimensions of the rivets are expressed in terms of ‘d’, the diameter of the shank. Out of all these head shapes, snap and countersunk are the most widely used.

Allowable stresses and failures of rivets

A riveted joint can fail due to one of the following reasons:

• Shear failure of rivets – In this kind of failure, the tensile force between the connected member’s results in shearing of the rivet shank along its cross section. There can be single or double shear of the rivet, depending upon the number of sections sheared.
• Bearing failure of rivets – The area of the rivets directly subjected to the force may get crushed due to excessive load. This failure is termed as the bearing failure of the rivet.
• Tear failure of member – The member connected through riveted joints may fail in this manner as the limiting case if its strength (i.e., strength of the member) is less than that of the rivets and it is subjected to tensile force.
• A rivet joint may fail because of inadequate edge distance.
• It may also fail because of cracking of the plate at the back of the rivet.
• Tearing can take place along the diagonal or zig-zag direction if the net area is less.

These failure conditions are shown in Figure 6.5.

In addition, there can be failure situation like shearing of the member, splitting of the member, etc., which can be avoided by providing sufficient edge distance and sufficient net area for the rivets.

The exact development of stress in the driven rivet is quite complex. Due to this, normally a higher factor of safety is provided to determine the working stresses in rivets. These stresses are allowed over the gross area of the rivet, which is the area of the holes in which it is driven. To account for a possible variation in the cross-sectional distribution along the longitudinal axis of the rivet, a higher factor of safety is allowed for direct stress in the rivet. The working stress value for tension and shear is taken as similar, though steel is known to be stronger in tension compared to shear.

The working stress allowed for rivets also depends upon the composition of the rivet material. In the case of mild steel, the indicative range of the working stresses is as given in Table 6.1.

Table 6.1
Permissible Stress Values in Steel

S No	Type of Stress	Range of Permissible Stress (N/mm2)
1	Tensile (σt) Power-Driven Rivet Hand-Driven Rivet	90–100 75–80
2	Shear (τ) Power-Driven Rivet Hand-Driven Rivet	90–100 75–80
3	Bearing (σb) Power-Driven Rivet Hand-Driven Rivet	280–300 240–250

These values of stresses are applicable for the shop-driven bolts. Sometimes, the rivets are driven at the site itself instead of at the shop. In such cases, the values of the permissible working stresses are normally taken as 10 percent less than the corresponding value for the shop-driven bolts.

There can also be situations when the rivets are subjected to both shear and direct tensile forces. In such cases, apart from ensuring that the actual values of these stresses do not exceed the respective permissible values, it is also necessary to ensure that they are so proportioned that the value of the expression (σ_t.actual / σ_t + τ_actual / τ) should not exceed n. The limiting value of n is provided by the design codes and has been adopted as 1.4 for the purpose of this manual.

Efficiency of the riveted joint

The efficiency of a joint is defined as the ratio of the minimum strength of a riveted joint to the strength of solid plate. This is also known as percentage strength of riveted joint and is expressed in percentage:

The following illustration will be helpful to understand the design procedure with clarity.

Illustration 6.1: A 150 mm wide and 8 mm thick plate needs to be connected to the gusset plate through rivets, as shown in Figure 6.6. We need to determine if 2 rivets, 20 mm in diameter, are safe to provide.

For the safety of the joint, the requirement is that the capacity of the rivets in shear should not be less than the tensile strength of the plate.

The weakest section for plate in tension is along the rivet holes. For 20 mm diameter rivets, the diameter of each rivet hole shall be 21.5 mm. As the permissible tensile stress for mild steel is 125 N/mm², the tensile strength of the plate,

Tp	=	(150 – 2 x 21.5) x 8 x 125
	=	107000 N

On the other hand, the strength of two rivets in shear,

Sr	=	2 x π/4×d2 ×τ
	=	2 x π/4 × 202 × 100
	=	62800 N

Therefore, the withholding capacity of rivets is less than the strength of plate in tension and they are not enough to provide.

Illustration 6.2: Two plates, each 12 mm thick, are to be joined by double riveted/double cover butt joint. It is required to design the joint using power-driven shop rivets. The permissible tensile strength on the plate is 130 N/mm².

First of all, we will estimate the optimum rivet diameter for the situation. Using Urwin’s formula, the diameter of the rivet for 12 mm thick plates should be

   d = 6.04×√12 = 20.9

Let us select the nominal rivet diameter of 20 mm. The gross diameter for the rivets would be 21.5 mm (nominal diameter + 1.5 mm).

Since the rivets are power-driven shop rivets, the permissible values of stress would be as follows:
Shear Stress in the rivet (τ) = 100 N/mm²
Bearing Stress in the rivet (σ_b) = 300 N/mm²

Determination of rivet value

As the design envisages a butt joint with two cover plates, the rivets shall be subjected to double shear.

Since the least of the above two would be the rivet value, the rivet value in this case is 72610 N.

For designing an efficient fastening, it would be the best situation if the post riveting strength of the plate equals the rivet strength. Since it is a continuous joint, covering a large width of the plate, we would compare the strength per pitch length, which would determine the value of the pitch (p). For this situation, the strength of plate per pitch length should be equal to two times the rivet value.

This is the upper limit of the pitch. Since the lower limit for pitch is 2.5 d (53.75 mm in this case), we would adopt the value of 110 mm for pitch.

Also, the edge distance should be minimum 1.5 d (32.25 mm); we would maintain the edge distance of 35 mm in this case.

The details of the joint are shown in Figure 6.7.

6.6.2 Design of bolted joints

As stated earlier, various types of fasteners are used for connecting the steel structural members. Different types of fasteners essentially serve the same function of transferring loads from one structural component to another. The choice of type of fastener is normally governed by the rigidity requirement, availability of technical inputs to fabricate and erect the structures, and service conditions for structure. In view of the ease in providing, bolted connections are preferred to riveted connections. In this case, the bolts along with nuts are used to fasten the components. In direct tension or compression members bolts act just like the rivets. For all other uses, unlike rivets, bolts form a semi-rigid connection. The failure conditions for bolts and the requirements related to pitch and minimum edge distance are the same as those for the rivets. The photograph of a typical bolted connection is shown in Figure 6.8.

Quite similar to bolts are pinned connections, where pins are used instead of bolts. Normally if structures are subjected to shock or vibrations during service, for instance when they support machinery or dynamic loads, bolted connection is best avoided for critical structural connections. In such cases, pinned connections are preferred. They form flexible connections as far as rotation of the member at the joint is concerned. The pins are also used for hinged arches and similar other structures.

The advantages of bolted connections are as follows:

• Their application does not pose fire, heat or noise hazards.
• Bolted connections can be done much faster than any other connection.
• Bolted connections are economical to use due to requirement of lesser manpower, with low or moderate skills.
• Being a factory-finished product, the quality of the product is more assured.

At the same time, drawbacks in the case of bolted connections are as follows:

• The nut-bolt connection, being threaded, is suceptible to get loose. This can considerably reduce the load-transmitting capacity.
• The presence of threads results in reduced cross-sectional area and therefore reduced resistance to direct tension.
• The fixing requirement of holes with dimensional clearance causes vibration of structure.

Types of bolts

A bolt is a metal pin with a head formed at one end to retain it in position and the shank threaded at the other in order to receive a nut. The bolts used for structural purposes are manufactured from mild steel and have high strength steel. As described above, a bolt consists of a head and a body (or shank). Bolts are manufactured out of forgings in different lengths.

The size of a bolt is expressed by the diameter of the shank. The shank is provided with threads at the free end, where one or more nut can be fixed. To assure full grip between nut and bolt, the threaded portion of each bolt is dimensioned to project through the nut by at least one thread. The assembly is also provided with a washer of adequate thickness under the nut. A typical nut-bolt assembly is shown in Figure 6.9.

Structural bolts can be classified as follows:

• Unfinished bolts – Unfinished bolts, also termed as rough bolts or black bolts, are manufactured from low-carbon steel. The shanks of unfinished bolts are not uniform in diameter. Unfinished bolts are designed for a lower permissible stress in shear and bending in comparison to rivets and turned bolts. Unfinished bolts are not used for structural connections of members that are subjected to shock or vibrations. Unfinished bolts, generally, are used for temporary work or smaller loads.
• Turned bolts – Turned bolts are also termed fitted bolts, as these are forced fitted. Turned bolts are manufactured from forgings with hexagonal section. The bars are turned to required diameters. The threads in turned bolts are usually cut with a die. If the holes for the bolts are reamed, then for many situations the turned bolts can be treated, so as to have the same strength as rivets.
• High-strength friction grip bolts – The use of high-strength friction grip bolts or HSFG bolts is a relatively recent development in the field of steel structure fasteners. To act in the desired way, HSFG bolts are tightened to a predetermined shank tension that is equal to the proof load. Resultant to this, the structural elements connected together are clamped between the nuts and the head of the bolt. The load is transferred from one structural element to the other by friction between the parts. When the HSFG bolt is tightened, longitudinal tensile stress and torsional stress are developed in the shank of the bolt. The shank tension reduces slightly after completion of tightening. The torsion drops appreciably. Later on, the shank tension remains more or less constant for the whole of its working life, irrespective of whether the connections remain loaded or unloaded.

The bolts do not allow any relative movement between the connected elements. For HSFG bolts to act effectively, all the contact surfaces should be ensured free of scale, oils and other foreign material that could prevent perfect contact between the metals.

High-strength friction grip bolts have advantages over rivets and traditional bolts. Unlike these, the HSFG bolts are only subjected to uniform tensile stresses and no shear or bearing stress. These bolts have higher resistance against fatigue. Stress concentration in the bolthole remains absent. Also as HSFG bolts do not bear against the plates, the uneven distribution of stress does not occur.

Common terms in bolting

Some of the common terms related to bolts and their meanings are as follows:
Nominal diameter – The nominal diameter of a bolt is defined as the diameter of the shank prior to threading.

Gross area of the bolt –The gross area of a bolt is the cross-sectional area of the bolt, calculated from the gross diameter of the bolt.

Net area of the bolt – The net area of a bolt is the area at the root of the threaded part of the cross-section. Normally, it is taken as 75% to 80% of the gross area.

Diameter of the bolthole – The diameter of a bolthole is the nominal diameter of the bolt plus the clearance specified.

Bolt grip – The grip of a bolt is equal to the total thickness of the structural components to be held together. It is the distance between inside of the bolt head and inside of the nut, excluding the thickness of the washer, if any, in the tightened state of the bolt.

Length of the bolt – The length of the bolt is defined as the distance from the inside of the bolt head to the extreme end of the shank, including any camber or radius at the free end.

Effective interface – The effective interface is the common contact surface between the bolt and the connected members.

6.6.3 Design of welded joints

Welding is one of the effective methods of connecting two members of similar or different materials. In welding, a joint is developed between two pieces, by rendering their interface either plastic or liquid, by application of heat or pressure. This results into a permanent bond at the interface. A typical in situ welding process of steel structures is illustrated in the photograph in Figure 6.10.

For structural steel fastening, welding is a widely used means. A typical weld for steel is shown in Figure 6.11. This type of weld is known as the groove or butt weld. The interface has been sloped or beveled to receive the filler metal. The zone in which the material has melted is termed as the fusion zone. The zone next to the fusion zone is the heat-affected zone. Although the metal in the heat-affected zone does not melt, its metallurgy undergoes a change due to the effect of heat received. The entire area consisting of filler, fusion zone and heat-affected zone together is known as the weld zone.

The welding process may be used for joining metals other than steel too, provided it is economically feasible to do so. Mild steel, cast iron, copper, brass and aluminum can be welded by application of heat with or without pressure.

Design of bull welded connections

The strength of a butt weld or groove weld is taken as equal to the strength of the parent metal, provided proper penetration of the weld metal is ensured. Out of the various types and shapes of butt welds, it is possible to ensure the penetration of the weld metal in the case of double-V, U, J and double-bevel joints, as the shape of the joints allows weld metal deposition from both faces of the joints. The permissible stress for the weld is taken to be the same as for the parent metal.

However, in the case of single – V, U, J and single-bevel joints, the weld penetrations generally remain incomplete. Due to this, the effective throat thickness for these types of joints is taken as 5/8 times the thickness member. In case the thickness of the two joining parts is different, the thickness of the thinner part is considered for the purpose of calculation of strength.

In the case of joining of two parts of unequal thickness, it is crucial to properly design the weld joint. In case the difference in thickness is more than 25% of thickness of thinner part or 3 mm, it is necessary to provide a taper, usually not more than 1 in 5, to ensure gradual change in the thickness.

Another important aspect is the selection of a proper sized electrode for welding. The diameter selected should be such that the resultant arc should be able to reach the root or the neck of the butt joint.

Design of fillet weld joints

As described earlier, the fillet weld is the deposit of weld metal outside the projected planes of the members. This is the single most important factor, which differentiates fillet welds from butt welds. The cross section of the fillet weld is triangular, as shown in Figure 6.13.

Some of the important terms, used while designing a fillet weld, have been described as follows.

Size of the fillet weld: The length of the sides of the largest right-angled triangle in the cross section of the fillet weld is denoted as the size of the fillet weld. Usually, the perpendicular sides of such a right-angled triangle are equal and the size of a fillet weld may be specified in one dimension only.

The minimum size of the fillet weld is usually specified on the basis of the thickness of the parts to be joined. The indicative sizes for different thickness ranges are specified, in the table below:

Thickness of Thicker Part	Minimum Fillet Weld Size
Up to and including 10 mm	3 mm
Over 10 mm up to and including 20 mm	5 mm
Over 20 mm up to and including 30 mm	6 mm
Over 30 mm up to and including 50 mm	8 mm first run 10 mm minimum

The maximum size of the fillet weld is also specified. The maximum size of the fillet weld applied to the square edge of a plate or member should be 1.5 mm less than the nominal thickness of the member or edge. The size of the fillet weld used, along the toe of an angle or the rounded edge of a flange, should not exceed three-fourths the nominal thickness of an angle or flange leg.

Throat of fillet weld: The throat of a fillet is the length of the perpendicular, from the right-angle corner to the hypotenuse, as shown in Figure 6.11. The effective thickness of throat is dependent on the size of the fillet, as well as the angle between the edges to be joined. The throat can be calculated with the following expression:

Throat Thickness = K × Fillet Size

In most of the cases, a right-angled fillet is formed during welding. For this case, the size of throat is equal to the 1/√2 times the size of the fillet. For this case,

K = 1/√2 = 0.7

Effective length of fillet weld: The effective length of a fillet weld is equal to its overall length minus twice the weld size. This is considered as the length that is truly effective in withstanding the load. This reduction from actual length takes care of the weld imperfections, at the start and at the end of the weld. The effective length of a fillet weld, designed to transmit loading, should never be less than four times the weld size. The length shown by the designer on the drawing is the effective length for the weld. The welder is supposed to provide the additional length (i.e. 2 x weld size) of the weld.

End return: The end return is the precaution about termination of weld at the end of the member. The fillet weld terminating, at the end or side of a member, should be returned around the corner, whenever practicable, for a distance not less than twice the weld size. This provision applies, in particular, to fillet welds in tension connecting beam seating, brackets, etc.

Side fillet: Side fillets are separate rows of fillet welds for the same member. These are designed to resist the force in combination with each other. For a lap joint made by a side or longitudinal fillet weld, the length of each fillet weld should not be less than the perpendicular distance between them. The perpendicular distance between the side fillets should also not exceed sixteen times the thickness of the thinner part connected. In case this limit cannot be adhered to, an additional end fillet plug or slot weld is provided to prevent buckling or separation of the parts.

Intermittent fillet weld: Intermittent or discontinuous fillet welds may be used when the length of the smallest-size fillet weld required to transmit stress is less than the continuous length of the joint. As stated earlier, any section of an intermittent fillet weld should have an effective length of not less than four times the weld size. The clear spacing between the ends of the effective lengths of intermittent fillet-weld carrying stresses should not exceed 12 t for compression and 16 t for tension, where t = thickness of thinner part joined, and in no case should be more than 20 cm.

Chain intermittent welding, where the weld length and gaps have a regular pattern, is to be preferred to irregular intermittent welding. When staggered intermittent welding is used, the ends of the component parts should preferably be welded on both the ends.

Single fillet weld: The single fillet weld is a single line of fillet welds. When single fillet weld is used, it should not be subjected to bending about its longitudinal axis.

Permissible stress and strength of fillet weld

As we can visualize, the fillet weld may be subjected to two types of loads. In the first case, which is the most frequent one, the members can be subjected to tension where the load would act along the longitudinal axis of the weld. In such cases, the weld metal would be subjected to direct shear force. Both the perpendicular edges being fused with the different members would tend to move in the different directions. This would cause the weldment to slide longitudinally along its weakest section.

In the other case, the direction of load can be in transverse direction to the weld axis. In such cases, the weld would be termed as ‘in tension’ and the rupture again would be along the smallest length of the cross section.

Fillet welds are designed for shear stress at its minimum section, i.e. the throat of the weld. The permissible stress in the fillet weld is 105–115 N/mm². The shear strength of a fillet weld is given by the following equation:

Since for most of the cases the angle between the welded surfaces for fillet weld is either 90º or very close to it, the value of K can be adopted as 0.70.
Hence, t = 0.7s

The values of permissible stresses, in shear and tension, are reduced to 80% for field welds made during erection.

To explain the process of fillet weld design in detail, following illustration has been used.

Illustration 6.4: Two 200 mm wide steel plates of 10 mm thickness each are to be fastened together with a 6 mm sized fillet weld. If the total weld length is 200 mm and the permissible shear stress in the weld metal is 115 N/mm², we need to determine how much load can be safely transferred through the arrangement.

As is obvious, the total load transferred can be lower than the capacity of the plates under tension and the weld capacity under shear. Since the plates have been as wide as the total length of weld, substantially thicker than the weld size, and steel’s capacity to withstand tension is higher than its capacity to withstand shear, the weld’s capability to transfer load would be the limiting capacity of the overall arrangement.

For a 90⁰ fillet weld, the capacity to withstand force is given by the following relationship:

Illustration 6.5: The objective is to design a lap joint for two plates of 8 mm and 12 mm thickness respectively. The width of each plate is 150 mm. The values of permissible stresses are as follows:

for plate in tension        – 145 N/mm²
for weld metal in shear – 110 N/mm²

The arrangement of the joint is shown in Figure 6.14.

Since the plates have equal width, the smaller of the two thicknesses, i.e. 8 mm, would determine the design of the joint.

As per norms, the overlap between the two plates should not be less than five times the thickness of the thinner plate. Hence the length of overlap

= 5 × 8 = 40 mm

As already stated, the strength of the thinner plate would govern the design of the weld joint. The tensile force carrying capacity of this plate is as follows:

This force of 174000 N would be shared by the two lengths of fillet welds, each of 150 mm. If the forces shared between them are T1 and T2 respectively, then

Since both the plates would elongate together, the value of elongation in both should be same for the overlap length. To satisfy this condition,

Since both the welds are 90º fillet welds, the size of each can be determined by the following relationships:

Since the maximum weld size should be smaller by at least 1.5 mm from the member thickness or the edge size, we can adopt fillet weld of size 6.5 mm and 10 mm respectively for the 8 mm thick and 12 mm thick plate, respectively.

Design of eccentrically loaded fillet weld joint

In day-to-day work, designers come across problems when the welded joint is subject to eccentric loading and, consequently, the moment. For instance, consider the case of the bracket shown in Figure 6.15. The bracket has been welded with the column flange through fillet weld of size s. The depth of the bracket, as well as the length of each run, of weld is equal to d.
As shown in the diagram, a concentrated load of magnitude W has been acting on the bracket. The load is eccentric with respect to weld axes by distance e. Under this situation,
Total length of weld = 2d

Vertical shear stress in the weld (pv) = 

The eccentric loading would also cause bending moment in the fillet weld. The value of this moment M is equal to W×e.

Magnitude of the horizontal shear stress due to bending in the extreme fibre,

With the help of these determined values of pv and ph, the value of resultant stress can be determined by the following relationship:
Resultant Stress (pr) = √ (pv² + ph²)

The value of the resultant stress should not exceed the permissible value of shear stress in the weld material.
This has been explained with the help of an illustration below.

Illustration 6.6: The objective is to determine the size of the fillet weld to fix a bracket plate of 12 mm thickness with the column. The size of the bracket, load intensity and its eccentricity are shown in Figure 6.16. The permissible stress for weld metal in shear has been 110 N/mm².

Let’s begin with the assumption that the weld size is s mm.
Vertical Shear Stress, pv = 75000 / (600×0.7s) = 178.57/s.
Total Moment due to eccentricity, M = 75000 × 300 = 22500000 N-mm.
Horizontal shear stress at extreme fibre due to bending moment

Therefore, we can provide 10 mm sized fillet welds. Since the plate thickness is 12 mm, this weld size would be fully acceptable.

6.7.1 Types of tension members

A tension member is defined as the structural component subjected to tensile force. Steel is one of the best construction materials, which has a good capacity to bear tensile stress. The nature of the structure, magnitude of stress and the method of connection govern the selection of tension member. To the possible extent, efforts are made in the case of tension members to connect them directly to gusset plates or splice plates. This helps in minimizing the possibility of stress concentration in the members.

The various types of tension members used may be grouped as follows:

• Wires and cables
• Rods or bars
• Rolled structural sections
• Built-up members

Wires and cables: Wire ropes and cables are used as tie members in the case of low-stressed or temporary applications such as hoists, derricks, rigging slings, guy wires, etc. Untwisted parallel wires are used in the manufacture of main cables of suspension bridges. Wire ropes have the advantages of flexibility and strength. At the same time, they require specially designed fittings for proper end connections as traditional structural fasteners can be of little use. Wire ropes are initially stressed when used as guy ropes for steel towers. This initial tension prestresses the frame due to which the effectiveness of the frame in resisting the external load is increased.

Rods or bars: Square and round bars are frequently used as tension members. To facilitate jointing, round bars are threaded at the ends and are used with nuts. Often, instead of providing threads, the bars are bent to form loops at their ends. Round bars with clevises as well as loop rods are made in two portions. These portions are joined and tightened with the help of turnbuckles. Such members are commonly used for lateral bracing. In the case of rectangular section bars and small width flats, the ends are enlarged by forging and holes are driven in them to form eye bars. These eye bars are used with pin connections. Inadequate stiffness of rods and bars results in noticeable sag under self-weight.

Rolled structural sections: A variety of rolled structural sections, viz. angle sections, tee sections, channel sections, etc., are used as tension members in structures. Angle sections have considerably more rigidity as compared to wire ropes and bars. Angle sections are also widely used for the purpose of bracing. A single-angle section develops bending stress due to the eccentricity between the end connections and the centre of gravity of the angle section. Unlike ropes and bars, these members can also resist some compression if reversal of stress takes place. Channel sections can also be effectively employed as tension members. Occasionally, I-sections are also used as tension members. These sections have the advantage of a high level of rigidity.

Built-up members: Built up members are the combination of either more than one rolled sections fastened together or a combination of one rolled section along with plates, or a section developed only with the plates. When single rolled steel sections cannot fulfill the requirement of cross-sectional area, built-up sections are used. Obviously, members constructed with built-up sections are used in the case of heavy tensile force. Built-up sections are normally designed as more rigid and stiffer than single rolled structural sections. As built-up sections are custom made, the sectional properties, such as total area, moment of inertia, etc., are properly controlled. These sections can resist compression force if the reversal of stresses takes place in the structure.

Some commonly used built-up sections are shown in Figure 6.17. Double-angle sections, placed back-to-back, with equal or unequal legs are extensively used as tension members in the roof trusses. Two angle sections may be arranged in the star shape, with the diagonals opposite each other and with legs on the outer sides. The star arrangement provides a symmetrical and concentric connection. Two angle sections have the advantage of adjusting the distance between them. Four angle sections are also used in the large biplanar trusses. A built-up section may be made of two channels, placed back-to-back, with a gusset plate between them. Such sections are used for medium force in a single-plane truss. In bi-planar trusses, two channels are arranged at a distance with their flange turned inward. It simplifies the transverse connections and also minimizes lacing. The flanges of two channels are occasionally turned outward in order to have greater lateral rigidity. The heavy built-up tension members in the bridge truss girders are made of angles and plates.

Determination of net cross-sectional area for tension members

As is quite obvious, if a steel section has been fastened by drilling holes as is the case with rivetting and bolting, it would not be possible for the member to transfer the tensile force to those fasteners. In such cases therefore, the net or effective area resisting tensile force would be less than the gross cross-sectional area. The net cross-sectional area of a tension member is the gross cross-sectional area of the member less the deductions made for holes. The projected area for each hole would be a rectangle: one side of which would equal the thickness of the member and the other side would equal the diameter of the hole.

Let us examine the case of some specific sections and the ways to apply deductions to them.

Plates: In the case of plates, the pattern of holes for fasteners determines the amount of deductions.

• For chain riveting or bolting, the net area of the member at cross section is the gross cross-sectional area less the area to be obtained from the product of thickness of the plate and the sum of the diameters of all rivet and bolt holes in that section.

If the plate has a thickness t and width b, the gross diameter of holes is d and the section has n such holes,
A_net = t × (b – n×d)

• If fasteners are placed in a zig-zag fashion, the failure of a plate may occur along this zig-zag section and would pass through these fasteners instead of a straight section, passing through the section, with higher numbers of fasteners. When one or more than one hole falls off the line, the failure of the plate depends upon the staggered pitch (p), gauge distance (g), and the diameter of the hole (d). When the gauge distance is larger than the staggered pitch, the failure may occur in zig-zag line. When the gauge distance is small, as compared to the staggered pitch, the failure may occur along a straight right angle section, passing through the largest number of holes. In case both the staggered pitch and the gauge are similar, the probability of the zig-zag failure occurring increases considerably as the diameter of the hole increases.

From theoretical investigations and laboratory tests, it is found that the net area of the plate for such a section would be the actual net area calculated along the zig-zag section plus additional area calculated from the following expression:

Additional Area = p²×t/4g

In case the rivets are connected in different staggered pitches p₁ and p₂ and the gauge distances g₁ and g₂, as shown in Figure 6.18, the net area along the section ABCDE shall be

A_net = t × {b – n×d + (p₁²×t / 4g₁ + p₂²×t / 4g₂)}

Angles and tee-sections: When an angle section is used as a tension member, it is connected with the gusset plate by one leg. When a tee section is used as a tension member, it is generally connected with the gusset plate by its flange. In such cases, the rivets connecting the tension member and the gusset plate do not lie on the line of action of the load, the result of which is an eccentricity moment on the connection. The distribution of stress becomes non-uniform because of eccentric connection. The net cross-sectional area of an angle or a tee member is reduced to account for this non-uniform distribution of stress. The reduced net cross-sectional area of such members is called net effective area.

Some commonly encountered situations are shown in Figure 6.19. The value of net effective areas for different situations is as follows.

• For a single angle connected by one leg only – The net effective area for this situation would be as per the following expression:

A_net = A₁ + A₂ × k₁
Here,  A₁ = Net cross-sectional area of the connecting leg
A₂ = Gross cross-sectional area of the other leg
k₁ = 3× A₁ / (3× A₁ + A₂)

• For two angles connected back-to-back or single tee – The net effective area for this situation would be as per the following expression:

A_net = A₁ + A₂ × k₁
Here,  A₁ = Net cross-sectional area of the connecting legs of angle or of flange of tee
A₂ = Gross cross-sectional area of the other legs or of web of tee
k₁ = 5× A₁ / (5× A₁ + A₂)

• For two angles connected back-to-back or two tees connected to each side of the gusset – The net effective area is calculated by reducing the area of the holes from the gross sectional area, provided the members are tack-fastened along their length. In case they are not tacked together, each would be considered as an individual section on one side of the gusset and the area would be determined accordingly.

However, if staggered fasteners are used in two legs of an angle section, then the net area would also be determined by the provisions for zig-zag fastening.

Steps in design of axially loaded tension member

The below steps are followed while designing an axially loaded tension member:

• Determine the total load for which the member has to be designed.
• Calculate the net sectional area required as –

Required A_net = Total force / Permissible tensile stress

• Conduct a trial with a suitable section, with a gross sectional area 20% to 30% larger than the A_net calculated. If a selection has to be made out of rolled sections, the sectional area and other properties of these sections are obtained from the relevant design codes and tables.
• The next step would be to calculate the A_net available in this trial section. At this stage, the following deductions for fastening holes may be assumed as a thumb rule:
  1. For plates – one rivet hole for each 150 mm width
  2. For single angle or double angle – one rivet hole from each angle
  3. Double channel – two rivet holes from each channel web or one rivet hole from each flange
• If A_net for this section is equal to or more than the required net area, the section can be adopted.
• As a special case, if the nature of loading on the member is such that the reversal of stress can occur during the life span, it would be necessary to calculate the slenderness ratio and verify its meeting norms.
• The final step in the process is to design the end connections.

The process of determination of net effective area for the tension members and the influence of different factors on it are shown in the illustration below.

Illustration 6.7: We need to assess the tension-carrying capacity of the member. As shown in Figure 6.17, the flat of size 175 mm × 12 mm has been used as a tension member. Its connection to another member is through a gusset plate of adequate size and strength. The fastening is through 20 mm nominal diameter rivets. The proposed arrangement of rivets is as per the diagram. The yield strength of steel is 250 N/mm².

 

For this situation,

Diameter of holes = Gross diameter of rivets = 20 + 1.5 = 21.5 mm

Permissible tensile stress = 60% of 250 N/mm² =150 N/mm²

As per the rivet arrangement, this is the area number of potential failure sections. We need to separately examine each to determine which section offers the least area and to resist the tensile force. We will convert all the dimensions into cm.

Section 6-5-7:
This section has one rivet hole. As the gauge is smaller than the staggered pitch in this case, the influence of the nearby holes will not be felt:

A_net = 1.2×(17.5 – 1×2.15) = 18.42 cm²

Section 1-2-3-4:
The section has two rivet holes. At the same time, the strength is additional due to existence of rivet ‘5’. In this case, the net effective area would be
A_net = t × {b – n×d + p²/ 4g}
A_net = 1.2× {17.5 – 2×2.15 + 5² / 4×4} = 17.71 cm²

Section 1-2-5-7:
The section has two rivet holes. Apart from the reduction in effective area due to these, there will also be strength provided by rivet ‘3’. In this case, the net effective area would be
A_net = t × {b – n×d + p²/ 4g}
A_net = 1.2× {17.5 – 2×2.15 + 5² / 4×4} = 17.71 cm²

Section 1-2-5-3-4:
This section has three rivet holes within it. Apart from the reduction in effective area due to these, there would also be strength provided by two rivets. In this case, the net effective area would be
A_net = t × {b – n×d + 2×p² / 4g}
A_net = 1.2× {17.5 – 3×2.15 + 2×5² / 4×4} = 17.01 cm²

Of the entire scenario examined, the least effective area emerges as 17.01 cm² or 1701 mm². This is the critical area for the arrangement.

Since the permitted tensile stress for the section is 250 N/ mm², the maximum tensile load-carrying capacity would be as follows:
T = 1701 × 150 = 255150 N or 255.15 kN

Illustration 6.8: An angle section of size 75 X 75 X 8 mm has been used as a tension member. One leg of the member has been bolted to the gusset plate with 16 mm bolts in a single row. The objective is to determine the tensile force-carrying capacity of the member. We also need to determine the maximum tensile load that can be carried through this section. The yield strength of steel is 250 N/mm².

For this situation,
Diameter of bolt holes = 16 + 1.5 = 17.5 mm
Permissible tensile stress = 60% of 250 N/mm² =150 N/mm²

Though the angle section will have an actual thickness of varying degree, we would consider it as a section with parallel and perpendicular edges, without any curvatures. The precise geometry of the section and its area statement can be obtained from the relevant codes. The sketch of the angle section and arrangement is shown in Figure 6.21.

We already know that in the case of a single-angle tensile member with holes driven into it, the net effective area would be given by the following expression:

A_net = A1 + A₂ × k₁

Here,  A₁ = Net cross-sectional area of the connecting leg
A₂ = Gross cross-sectional area of the other leg
k₁ = 3× A₁ / (3× A₁ + A₂)

Hence, area of each leg of the angle,

And, A_net = 428 + 568 × 0.693 = 821.6 mm²

Based on this net area, the maximum tensile load-carrying capacity of the member,

T = 821.6 × 150 = 123240 N or 123.24 kN

To determine the maximum possible tensile load-carrying capacity of the member, the holes for any fasteners would require to be eliminated. The member would be fixed to the gusset through welding and, therefore, no loss in the gross cross-sectional area would take place. We also assume that the weld design has been done to take care of the eccentric center of gravity of the angle section, and no moment is induced in the section.

In this case, overall cross-sectional area = 2× A₂ = 2× 568 = 1136 mm²

Overall tensile load-carrying capacity in this case,

T = 1136 × 150 = 170400 N or 170.4 kN

It may be observed here that the reduction in load-carrying capacity of a member is due to holes. In this case, it is about 27.6%.

Axially loaded columns

In an axially loaded column, the load is applied at the centroid of the cross section and it acts in the direction parallel to the longitudinal axis of the column. The term concentrically loaded is also used to specify axially loaded columns.

An axially loaded column can be defined as the one which transmits only compressive force, with no explicit design requirement to withstand either lateral loads or end moments. However, a perfectly straight column with central application of load is an ideal condition. During fabrication and erection of structural columns, it is difficult to avoid accidental eccentric load application, crookedness and curvature, and the presence of residual stresses. The effect of these is to reduce the overall load bearing capacity that can be addressed by making appropriate modifications in the analytical expressions and using suitable factors of safety in the design formula.

At this stage, we should review our earlier understanding about the phenomenon of critical loads for a column, which we had studied in the section on column analysis. Consider the case of an ideal column, which is vertically fixed at the base and free at the upper end and subjected to an axial load F. The column is assumed to be perfectly elastic. When the value of load F is less than the critical load and stress is within the limit of proportionality, the column remains straight. In this stage, the column is said to be in stable equilibrium. This means that if a small lateral load is applied at the free end the column deflects and the free end moves laterally. However on withdrawal of the lateral load, the column resumes its vertical position and deflection vanishes. When the axial load F is gradually increased, a stage would eventually be reached when the column would attain the position of unstable equilibrium. At this stage if a small lateral load is applied, the resultant lateral deflection will not disappear on withdrawal of this lateral load. The magnitude of minimum axial load required to keep the column in such a deflected shape is termed as critical load. Critical load is also called buckling load or crippling load.

Buckling is the sudden bending or warping of the members under compressive load. Buckling load is the load at which a member buckles and, consequently, fails functionally. The direction of buckling of a column depends upon flexural rigidity, El, of the column section. The compression member buckles in a direction perpendicular to the axis, about which the moment of inertia of the section is minimum.

Euler, a Swiss mathematician, had derived one of the most widely used expressions for the critical load for columns. This is known as Euler’s column formula:

Critical Load, F_cr = 

Here,  E  = Modulus of elasticity of the material
I   = Least moment of inertia of the column cross section
L_eff = Effective Length of the column

Effective length of the compression members

The effective length of a compression member depends upon the end restraint conditions. These end restraints are of two types:

• Restraints for position : In position restraint the end of the column is not allowed to shift from its position, but is permitted to rotate. The example of this is the hinged connection.
• Restraints for rotation: In rotation restraint the end of the column is free to change its position, but is not allowed to rotate.

A column end can be free or can have either one of the restraints. It can also be subjected to both the restraints. The end conditions of the column determine its effective length, which has a bearing on the value of critical load for the column.

There are various possible combinations of restraints about either or both axes. The restraint conditions at the two ends of a column can remain same or can differ. Given below are some standard cases of end conditions:

• Both ends of column hinged
• Both ends of column fixed
• One end of column fixed and the other end hinged
• One end of column fixed and the other end free

The representative diagrams, for a number of such end restraint conditions, are shown in Figure 6.22.

In all the cases, the actual length of the column is L. Case 1, i.e., both ends of the column hinged, has been considered as a standard case. The effective length (L_eff) of a column is expressed in terms of equivalent length of compression members hinged at both the ends. The equivalent length can be defined as the length of a column between two adjacent points of zero moments, and is represented by L. It is also termed as unsupported length.

The effective length of a compression member can be adopted as per the table given below. Although the perfect end conditions, assumed in the analysis, are nearly impossible to get in practice, the values of effective lengths indicated in the table will be accurate enough for most cases. These values of effective length can also be adopted if the columns form part of the framed structures.

Important sectional properties for compression members

Effective cross-sectional area for compression member: In the case of compressive load, a member can also transfer the load to the fasteners. Therefore, the effective cross-sectional area of a compression member is the gross cross-sectional area of the member. No deduction is made for holes meant for rivets, bolts and pins, as long as the fasteners fill them. However, the plate element or outstand of a section may buckle under the compressive force before the failure of the entire member. To avoid this, the effective area is modified when the ratio of the outstand to thickness exceeds the limits specified. The outstand in the case of channels and angles is measured as nominal width of the sections. In the case of flanges of beams and tee section, it is measured as half the nominal width. The outstand for a number of sections is shown in Figure 6.23.

The thickness is taken as 1/16 of outstand of any member in compression for unstiffened outstand and 1/20 of outstand of any member in compression for stiffened outstand. The thickness is taken as 1/90 of outstand for unsupported width of unstiffened plate forming any part of a compression member, measured between an adjacent line of rivets. But for the purpose of calculating effective cross-sectional area and radius of gyration, only 50 times the thickness of plate will be taken, instead of 90 times the thickness, in case a single plate is used. However, it is the total thickness of two or more plates effectively tacked together.

A deduction is also made from the gross cross-sectional area for excessive effective plate width (if any) to determine the effective sectional area.

Radius of gyration: The radius of gyration of a section is a geometrical property of the section and it is denoted by r. The expression for this has been
r = √ (I/A)
where I = Moment of inertia of the section about the axis
A = Effective sectional area of the section

Slenderness ratio: The slenderness ratio of a compression member is defined as a ratio of effective length of compression member (L_eff) to appropriate radius of gyration (r).

Slenderness ratio, λ = L_eff lr.

The slenderness ratio of a compression member should be as small as possible, so that the material may be stressed to its largest permissible limit. In the case of rolled steel sections, the values of I, A and r are obtained from the design tables, which provide details of sectional properties. The radii of gyration about various axes of rolled steel sections can be obtained from structural steel section tables. The minimum value of radius of gyration gives the maximum value of slenderness ratio. For built-up compression members, the value of radius of gyration is calculated independently.

The end restraints of columns are often different in the two principal planes. The different values of moments of inertia of the column cross section in these planes are sometimes beneficial to achieve approximately equal slenderness ratios in both the planes of bending.

Compression members composed of two sections

As we have seen in the case of Euler’s formula, the value of critical load for a section depends upon the moment of inertia of the section. Incidentally, as per Euler’s formula, the value of critical load is independent of the sectional area. In the case of other column formulae too, the overall sectional properties determine the value of critical load. Hence, it sometimes becomes important to enhance the sectional properties of a rolled section by using it in combination with plates or other sections. One of the most effective and efficient ways would be to form a new section by placing two similar rolled sections back-to-back. This process necessitates certain norms which should be followed to ensure that the two different sections work as one.

• A compression member with two similar sections back-to-back, either in contact or separated by a small distance, should be joined together by riveting, bolting or welding so that the slenderness ratio of each member between the connections is neither greater than 40 nor greater than 0.6 times the most unfavorable slenderness ratio of the whole strut. The spacing of tacking rivets in a line should not exceed 600 mm for such members.
• For other types of built-up compression members, where, say, cover plates are used, the pitch of tacking rivets should not exceed 32 t or 300 mm, whichever is less (where t is the thickness of the thinner outside plate).
• The rivets, bolts or welds in these connections should be sufficient to carry the shear force and moments, if any specified for battened struts. The diameter of the connecting rivets should not be less than the minimum diameter given below.

• The ends of the struts should be connected together with not less than two rivets or bolts or their equivalent in welding and there should not be less than two additional connections spaced equidistant in the length of the strut.
• A minimum of two rivets or bolts should be used in each connection: one on each line gauge mark, where the legs of the connected angles or the tables of the connected tees are 125 mm wide or over or where the webs of channels are 150 mm wide or over.

Load bearing capacity of compression member

The strength of a compression member is defined as its load-carrying capacity. The strength of a centrally loaded straight steel column depends on diverse factors like effective cross-sectional area, radius of gyration (viz., shape of the cross section), the effective length, the magnitude and distribution of residual stresses due to heat treatment and cold straightening, etc. The importance of all these factors is of varying degree.

The effective cross-sectional area and the slendemess ratio have high significance for a compression member. The permissible stress for the member is dependent on the slenderness ratio. Once the value of slenderness ratio is known, the permissible stress for a particular steel grade can be determined either from design charts or from graphs. A typical graph for the purpose is shown in Figure 6.24. Other factors described above are of relatively lesser importance.

The steps involved in determining the safe bearing capacity of a compression member are as follows:

• First of all, one needs to determine the effective length from the actual length of the compression member and the end support conditions.
• With the section of the compression member known, the radius of gyration about various axes of the section either has to be calculated or to be directly read from steel section tables. The least value of the radius of gyration is adopted for the analysis.
• With effective length and radius of gyration known, the maximum slendemess ratio (L/r) is determined for the compression member.
• The allowable working stress (σ_cd), in the direction of compression, is seen from the table, corresponding to the maximum slendemess ratio of the column and the grade of the structural steel.
• The effective sectional area (A) of the member is noted from structural steel section tables. For the built-up members, it can be calculated.
• With the values of effective area and permissible stress known, the safe load-carrying capacity of a member is determined from the relation

F = A × σ_cd

In case the objective is to design a section, for the compression member, capable of withstanding a particular magnitude of load safely, the steps would be as follows:

• As the actual length of member and end conditions for the supports are known, the effective length of the member would be computed first.
• It would be required to assume a value for safe stress in the steel. As a thumb rule, for a low-load case where a single rolled section is going to be adopted, a stress value corresponding to slenderness ratio of around 100 can be assumed. For a high-load cum built-up section scenario, the value of stress corresponding to a slenderness ratio of 60 -70 may be assumed at that stage.
• With the value of permissible stress assumed, the next step is to work out the cross-sectional area requirement for the given load.
• With the target cross-sectional area known, a trial section (single or built-up) would be developed at this stage. For this section, the radius of gyration and actual slenderness ratio would need to be determined.
• The permissible stress for this permissible ratio would be read from the chart /table.
• The safe load for the trial section would be determined and compared with the actual load value. In case the actual exceeds the computed load, a fresh trial with a further enhanced section would need to be made.

Built-up compression members

We already know that the load-carrying capacity of a compression member is dependent on its slenderness ratio, which in turn is dependent on the member’s effective length as well as the moment of inertia of its cross section. To put it in another way, the load-carrying capacity of a compression member of a given material and fixed effective length is the function of the moment of inertia of its cross section. Therefore, the distribution of the material in the cross section is quite important. In the cases of heavy compression loads, built-up sections (sections made of two or more sectional members/plates) are provided. While designing such a section, an effort is made to place the material uniformly with respect to both the major axes of the cross section and also to space it as far from the centre as possible.

To ensure that the different components of the built-up section act as a single column, it is necessary to connect them. These connections are designed to withstand all the forces, which try to separate the components. This connection is achieved either by lacing or battening. In lacing, the members with axis inclined with horizontal are provided to connect the main members. Between them, the lacing members form a lattice-like structure. In the case of battening, horizontal plates are used to connect the main members.

A detailed design procedure for lacing has been explained in the following section.

Design of lacing

Lacing is the process of connecting those sections with members inclined in the elevation. The sloping members would be made of plates in most of the cases, but can also be of rolled sections like angles, small channels, etc., if the loads are heavy. As a general guideline, the moment of inertia of the section perpendicular to the plane of lacing should not be less than the moment of inertia parallel to it. The lacing system on two faces should preferably be mirror images of each other.

Typical lacing systems are shown in Figure 6.25. The points to be considered in a design process are as follows:

• The slope of the lacing members with respect to the column axis should be kept between 40° and 70°.
• The slenderness ratio of the lacing member should not be higher than 1.45. The cross section of the lacing member shall be determined, based on this criterion. The effective length of the member for this purpose is determined as follows:

• The thickness of the lacing bar shall not be less than l/40 for single-laced system and l/60 for double-laced system.
• The following relation shall determine the spacing between parallel lacing members:
• Spacing L ≤ 50× r_min
• or L ≤ 0.7× Maximum Slenderness Ratio of the member × r_min, whichever is less.
• The lacing member shall be designed to withstand a transverse shear equivalent to 2.5% of the axial load in the compression member divided equally among all lacing members.
• The slenderness ratio of the lacing member should not be allowed to exceed 145. In the case of a flat lacing member with width d and thickness t, the value of the slenderness ratio can be expressed as ‘l_eff×√12 / t’.

The design procedure has been further explained with the help of the illustration below.

Illustration 6.9: A built-up steel section, shown in Figure 2.26, is restrained at both the ends against rotation. The length of the column is 3 m. The modulus of elasticity for steel is 200,000 N/mm².

As per Euler’s formula, the critical load for the column is determined by the expression

For this section, the distribution of material is such that the moment of inertia about the XX axis is obviously greater than the moment of inertia about the YY axis.
The value of moment of inertia about YY,

With the constrained defined, the effective length of the column shall remain equal to its actual length.

Hence, L_eff = 3000 mm
Therefore, critical load,

Therefore, the safe load-carrying capacity of the column is 5848.2 kN.

Illustration 6.10: Two channel sections, back-to-back, have been used to withstand a compressive load of 1100 kN. The overall arrangement of this section, along with sectional properties, is shown in Figure 6.27. The objective now is to design a lacing system for the built-up section. The value of yield stress for the steel is 250 N/mm.

We now undertake the design of lacing fastened with rivets. As the back-to-back distance for the channels is 22 cm and the rivet line is at 5 cm from the back of the channel, the horizontal distance between the fasteners would be 35 cm.

As the first step, the lacing member would be kept at 45° with the horizontal. Also, the fastening with the main member is through a single rivet. In this case,

Effective length of the lacing bar l = 32×√2 = 45.3 cm

Spacing between the lacing bars L = 2 × 32 × tan 45° = 64 cm

We know that for the independent channel sections, r_min = 2.8 mm. Also, the overall slenderness ratio for the built-up section is 4.5.8.

Hence, 50× r_min = 50 × 2.8 = 140.
And 0.7× Maximum Slenderness Ratio × r_min = 0.7×45.8×2.8 = 89.76.
Since L is less than both of these values, it is acceptable to adopt.

The next step would be to determine the appropriate section for the lacing bar.

For this, we know that the minimum thickness of a lacing bar in single lacing should not be less than l/40. In this case,
l/40 = 45.3 /40 = 1.13 cm.

Hence, the design would adopt the flat, with a 12 mm thickness. Since the battens would be fastened with rivets, we take the width of the trial section as 65 mm.

Hence, the slenderness ratio for batten plate = l_eff×√12 / t = 45.3×√12 / 1.2 = 130.77.

Since the value of the slenderness ratio is less than 145, this section can be adopted. Also, for the slenderness ratio of 130.76, the permissible direct compressive stress for steel (yield stress = 250 N/mm²) obtained from the table is about 56.6 N/mm².

The column is subjected to direct compressive load of 1100 kN. As per the design requirement, the lacing member should withstand a shear force equivalent to 2.5% of the direct load.

Transverse shear force on lacing = 2.5% × 1100 = 27.5 kN.

Since lacing is being provided on both the faces of the built-up section, the corresponding bars on both faces would share the load equally.

Hence, shear for one lacing face, T = 27.5 / 2 sin 45 = 13.75 kN.

The direct force induced in the lacing bar due to this shear force,

F = T/sin 45º = 13.75/sin 45º = 19.45 kN.

Each lacing member would be subjected to direct tensile or compressive force of this magnitude. Let’s check if the section of the lacing bars is adequate for it.

Net compressive stress = 19.45×1000 / 12×65 = 24.93 N/mm².

This is less than the observed maximum permissible value of 56.6 N/mm².

If the member is fastened with 20 mm nominal diameter rivets,

Net tensile stress = 19.45×1000 / 12×(65–21.5) = 37.26 N/mm².

In this case too, the actual stress is much less than the maximum allowed stress of 60% of the yield value. The design can safely be adopted.

Splicing in columns

Rolled steel sections come in standard lengths and it is not always possible to obtain the desired length. In some situations, the section of column itself may undergo a change at a particular location. In all such situations, it becomes necessary at times to join the separate pieces of rolled section. The method of joining those pieces with the help of additional plates is known as splicing.

Splice plates are designed to transmit all the forces that are in existence at the joint. It includes direct compression as well as the bending moment and shear force, if present. It is clear that if the sectional areas of the plates for web and flanges were at least equal to the original area of the individual web or flange, they would serve their purpose. The further design process involves determination of total load on each splice and accordingly decides the fastening for the splice plates.

The following example explains the design process of column splicing.

Illustration 6.11: An I-section of 200 mm depth and 125 mm flange width has a compressive load of 350 kN acting axially. In addition, there is moment acting at the splice section of 10 kNM.

Since the section under consideration is symmetrical about both X and Y axes, the splice system would also be symmetrical about these axes. In other words, the size and location of both the flange plates would be similar. The same applies to the web plates.

Total axial load = 350 kN.
Load on each splice = 350/2 = 175 kN.
The moment at the splice plate would add to the load on the splice. The magnitude of the load due to moment would be equal to
Moment / Section Depth = 10 / (200/1000) = 50 kN.

This load due to moment would be tensile for one splice and compressive for the other. Since we require both the plates to be similar, we would design them for the worst combination of loads.

The largest magnitude of load on each splice = 175 + 50 = 225 kN.

The average thickness of flange and web are 14 mm and 9 mm respectively. Hence, 16 mm thick and 200 mm wide plates on each of the flange would serve the purpose. For the web, a plate of 8 mm thickness on both the sides of the web has been provided.

Design of fasteners

It has been decided to provide fastening with 20 mm nominal diameter rivets. For these rivets,
Allowable stress in shear = 90 N/mm²
Allowable stress in bearing = 270 N/mm²
Strength of rivet in single shear = π/4 ×(21.5) ² × 90 = 32674 N = 32.67 kN
Strength of rivet in bearing = 21.5 × 14 × 270 = 81270 N = 81.27 kN
Therefore the rivet value is 32.67 kN.

Number of rivets required for each splice,
N = 225/32.67 = 6.88
Hence each flange plate shall be provided with eight rivets on both sides of the joint.
The rivet arrangement is shown in Figure 6.28.

Classifications of beams

In the case of a grid of beams, those beams that transfer the load directly to the vertical structural members, i.e., columns and walls, are termed as main beams. The beams, which directly support the floor construction and in turn are supported on main beams, are termed as secondary beams. Main beams receive the load from secondary beams. The term joist is used for beams made of light structural sections. A heavy-sectioned major beam in a structure is known as a girder. In the case of excessively heavy loads, which may be encountered in industrial structures, bridges, etc., it is common practice to form a suitable section by fastening plates of different sections together.

Such beams are known as plate girders. The photograph of such a plate girder is shown in Figure 6.29.

Beams are also classified, based on the function they perform. In the case of sheds made of columns and roof trusses, horizontal beams spanning two adjacent trusses are known as purlins. Those beams that rest on the purlins are known as rafters. In the case of buildings, the beams that span doors, windows and other openings in the walls and support the masonry above these openings are known as lintels. The beams provided along the peripheral wall of a building, that support part of the floor, as well as the wall up to the floor above, are known as spandrel beams. The beams that support the stairs are called stringers.

Besides, based on the supporting systems, beams are classified as simply supported, cantilever, fixed, continuous, etc.

Beams of varying strength

As we studied in the previous section, the extreme fibre faces the maximum amount of stress in the beams. To make the beam economical, the strength in extreme fibre is designed to bear the maximum stress, while the central fibres, where the stress is less, are designed accordingly. This is the reason that most of the beams are made up of “I”- or “C”-shaped channels.

Rolled steel sections and their behavior under lateral loading

We are already familiar with rolled steel sections and their various shapes available. Among the rolled steel sections, I-sections, channel sections, tee sections, etc., are the regular sections, which are most commonly used as beams. In all these sections, most of the material is located near the top and bottom faces, i.e., in the flanges as compared to the web portion. Such rolled steel sections provide a large moment of inertia about the X axis. As a result, these sections provide a larger moment of resistance as compared to solid rectangular, square and circular sections of the same cross-sectional area. These sections, as such, are considered most efficient and economical. In the case of I-sections, the large flange width of these sections provides greater lateral stability. The web-flange construction of the section also facilitates easy connection with other structural members. Channel sections have reasonably good lateral strength but poor lateral stability. They are also unsymmetrical about the Y axis and are subjected to unsymmetrical bending on loading. Angle sections and tee sections are used as beams for light loads. There sections are weak in resisting bending.

In most situations, beams are loaded in the direction perpendicular to the X axis. This is done to ensure that the bending of beams occurs about X axis, the moment of inertia of the section about which is actually the largest. Very rarely is the application of load allowed in the direction perpendicular to the Y axis.

In the case of loading the beam, the application is considered to be through the shear centre of the beam sections. The shear centre for the beam section is the centroid of the sectional area. Such load application ensures that the loading produces just plain bending about either one or both of the axes. In such cases, the section is subjected to simple bending and shear. When the load does not pass through the shear center, a torsional moment is produced in addition, along with the bending moment, which attempts to twist the section. It becomes imperative then to design the sections by considering stresses induced by torsion, in addition to normal flexural and shear stresses. Sections like angles and channels are prone to inducement of torsional stresses. For such sections, a special load device may be used so that the load passes through the shear center of the section and the torsional moment may be avoided.

A number of times the sectional properties of a rolled steel section are enhanced by cutting the beam in a pre-determined pattern along the web and then rejoining both the separated parts in a fresh pattern. This way, the section modulus of the beam increases, without any increase in the weight of the beam. These sections are known as castellated or open web sections. The reason for calling them ‘open web’ is that cutting and rejoining of the web leaves a number of similar and regular openings in the web. Examples of a few such sections are shown in Figure 6.30 and a photograph in Figure 6.31. Castellated sections are economical for light loading conditions.

Moment resisting properties of the section

Here, it would be pertinent to recall the important expression pertaining to bending moment, the derivation of which was observed in Chapter 3. The expression has been

Here, M is the value of bending moment at a section, I is the moment of inertia of the section, y is the distance of fibre from the neutral axis and f is the magnitude of stress in the fibre.

Or M = f × (I/y).

The above equation holds well when the plane of bending coincides with the plane of symmetry of the beam section. The bending of the beam occurs in the principal plane of the beam section. In this expression, it is possible to treat any component as variable. However, while undertaking the actual design of a beam, normally the construction material is frozen beforehand based on a number of factors. Along with it, the value of f also gets decided, though there always remains some degree of flexibility regarding the ultimate grade of the construction material.

The magnitude of maximum bending moment (M) in the beam is a function of structural geometry and the loading system. Hence, it is also known at the time of actual determination of the beam section. The self-load of the beam is factored in the dead load of the structure at the beginning of the analysis. A small variation in it will not alter the design results in any significant way.

The last component of the expression is I/y. Though constituted of two elements, it is reasonable to treat it as one. The reason is that both I and y are functions of the size and shape of the cross section. As is clear from the expression, other thing being constant, the moment-carrying capacity of a beam is directly proportional to the moment of inertia of the section and inversely proportional to the distance of the outermost fibre from a neutral axis. This expression, I/y, is termed as the section modulus and denoted with Z. In most beam design problems, the objective is to provide a section with the required value of Z in combination with the lowest possible value of sectional area.

Mathematically stated, Z ≥ M/(f_tan or f_ten).

When the section of beam is symmetrical about the neutral axis, the value of y is equal to half the depth of the section. In addition, the maximum bending stress in compression and in the tension at the extreme fibres would be equal. In case the beam section is not symmetrical about the neutral axis, then the distances of extreme compression fibre and extreme tension fibre, from the neutral axis, would be different.

As shown in Figure 6.32, if y₁ and y₂ are the distances of two extreme fibres from the neutral axis, the values of extreme bending stresses would be different. It is also to be ensured that maximum bending stress in the beam section is less than the allowable bending stresses, whether tensile or compressive. The stress triangles for the section above and below the neutral axis are shown in the diagram. For the state of equilibrium at the section, the total compressive force ‘C’ above the neutral axis would be equal to the tensile force ‘T’. These two forces act in opposite directions and form a couple. The arm of the couple is equal to the distance between the centroids of the two stress triangles.

This couple resists the bending moment due to external loads and, therefore, its magnitude is known as the moment of resistance (M_r):

M_r = (C × Arm ‘L’) or (T × Arm ‘L’)

6.9.1 Beam design process

We do by now understand the theoretical aspects of beam design. Now we observe the process through which a design is actually undertaken and the factors that influence the process.

Checking for web crippling

The section is further checked for web crippling. As shown in Figure 6.33, the web may cripple locally due to the effect of concentrated load. The potential spot for this is the location where the flange joins with the web. As shown in the diagram, the load is supposed to disperse at an angle of 30º. If the magnitude of load/reaction is W, the width of bearing for load is a and the depth of the critical section is h_f, the value of bearing stress at the section would be

And 

The maximum permissible value of σ_p is normally not allowed to exceed 0.75×fy.

Checking for web buckling
Besides the potential of failure in bearing, the web may buckle under the influence of a concentrated load. This again is a local phenomenon. The load is supposed to disperse at an angle of 45º. The effective web length for this purpose is taken at the level of the neutral axis. If W is the magnitude of load/reaction, hw is the clear height of web between the flanges and B is the effective web length offering resistance, the value of direct compressive stress in the web would be as follows:

The value of σc thus calculated should be less than the permissible compressive stress for a member with the slenderness ratio hw√3 / tw.

In case this condition fails to get fulfilled, it would be necessary to provide stiffeners below the location of concentrated load.

Checking for deflection

The maximum permissible deflection in a beam is normally expressed in terms of the span of the beam. As per general norms, it should not exceed 1/325 of the total span, though there can be higher deflections permitted in certain cases. Since deflection is a functional aspect, a certain degree of flexibility is allowed based on the material of construction and finishing, which are being supported.

The expression for maximum deflection in any beam is as follows:

Here all the values, except K, are quite familiar. K is the constant, which depends on the load configuration and supporting arrangement for the beam. The values of K for different conditions are given in Figure 6.34.

The following illustration further explains the steps involved in the beam design process.

Illustration 6.12: A steel beam of 6 m span is resting on 250 mm thick brick walls at both its ends. The beam has been made of a 450 mm deep I-section. Both the flanges have been further strengthened with flange plates of 16 mm × 200 mm section. We are now required to determine the maximum amount of uniformly distributed load, which can be allowed on the beam, if the compression flange is laterally restrained. The yield stress for steel can be taken as 250 N/mm².

The sectional properties for a 450 mm deep I-section are as follows:

First of all, we determine the permissible bending stress in tension and compression. Since the compression flange is effectively restrained,

Permissible value of bending stresses = 0.66 × fy =0.66 × 250 = 165 N/mm².

The next step would be to determine the moment of inertia of the composite section. Since the composite section is also symmetrical about the X axis of rolled section, the neutral axis for the composite section would continue to remain the same. The arrangement is shown in Figure 6.35.

The moment of inertia of the combined section with respect to the X axis,

The value of section modulus for this section,

Z = I_x–com / (d/2) = 65144.96 / (22.5 + 1.6) = 2703.1 cm³

The maximum moment carrying capacity of the section,
M = Z × σ_bc or σ_bt (σ_bc and σ_bt do have similar values in this case)
Hence, M = 2703.1 × 165 × 100 = 44601150 N-cm = 44601 kN–cm.

Since the beam is simply supported with a clear span of 4 m and an end bearing of 250 mm, the effective span would be equal to the center-to-center distance between the supports.
Effective Span ‘L’ = 6 + 0.125 + 0.125 = 6.25 m.

 

In the structural terms, the beam is a simply supported one. If the amount of uniformly distributed load on the beam is w kN/m, the maximum bending moment (at mid span) shall be expressed by the following equation:
M_max = w×L²/8
Equating it to the maximum load-carrying capacity of the section,
44601 = w× 100 ×(6.25 × 100)²/8.
Hence, w = 91.3 kN.

From the parameter of flexural stress, the beam can be subjected to a UDL of 91.3 kN/M. Here, for the sake of simplicity, we are ignoring the self-weight of the beam.

However, it is still required to check the section against shear, deflection and web buckling.

Design of truss members

As we already know, truss members are only subjected to direct forces, either compressive or tensile. Though there can be exceptions, the members are not allowed to be subjected to bending and shear force. Hence, once the analysis for the truss has been conducted and the load for each member is known, further process is only to design the section of these tension or compression members, as per the procedure applicable.

Since the design of truss involves the analysis of forces in its members and determination of appropriate sections, for these members, which already has been explained in the earlier part of this manual, we will not repeat the same. At the same time, certain important points related to truss design have been described below:

• The pitch of the truss (its height to span ratio) is maintained in the range of 1/4 to 1/6.
• The purlins, with their axis inclined with respect to the vertical, would be subjected to bending about both its axes.

Trusses may be subjected to either moving load if they are supporting a bridge deck, or wind load, if they are supporting a roof. Due to the dynamic nature of these loads, the magnitude of induced forces in individual members keeps varying. In fact, they can even change in nature from tensile to compressive and vice versa. Hence each of the members is designed for the highest force it can subject to, with due consideration to possible reversal of forces.

While designing the rafter bracing, the stiffness provided by purlins is totally ignored.

7.1.1 Cement

A product with binding characteristics is called cement. For concrete, the most commonly used cement is Portland cement. Portland cement is the name given to cement obtained by intimately mixing together calcareous and argillaceous materials, alumina, and iron oxide-bearing materials; burning them at a high temperature (called clinkering temperature) and finally grinding the resulting clinker. Normally no material other than gypsum, water and grinding aids may be added after burning.

From the description of Portland cement given above, it can be seen that it is made primarily from a combination of a calcareous material, such as limestone or chalk, and of silica and alumina found as clay. The manufacturing process, described briefly below, consists of grinding the raw materials into a very fine powder, mixing them intimately in predetermined proportions and then burning the mix in a large rotary kiln at a temperature of about 1350 °C. In the kiln, the material fuses partially and forms a clinker. The clinker is cooled and ground to a fine powder with some gypsum added to produce commercial Portland cement.

A typical composition limit for Portland cement is shown in the table below:

The mixing and grinding of the raw materials can be performed either in water or in a dry condition and the process classifications, wet and dry, arrive from here. The mixture is fed into a rotary kiln, which can be as large as 7 m in diameter and 225 m in length. The longitudinal axis of the kiln remains slightly inclined. The mixture is fed at the upper end, while pulverized coal is blown in by an air blast at the lower end of the kiln. As the mixture of raw materials moves down the kiln, it encounters a progressively higher temperature so that various chemical changes take place along the kiln. First, water, if any, is driven off and CO₂ is liberated from calcium carbonate. Further on, the dry material undergoes a series of chemical reactions and in the hottest part of the kiln about 30 percent of the material turns liquid. All the raw materials—lime, silica and alumina—recombine at this stage. The mass then converts into balls of 3 to 25 mm diameter, known as clinkers. These clinkers are cooled through heat exchangers. The cool clinkers, which are very hard, are ground with gypsum to avoid flash or very rapid setting of the cement during hydration.

Chemistry of cement

We have seen that the raw materials used in the manufacture of Portland cement consist mainly of lime, silica, alumina and iron oxide. In the kiln, these compounds react with each other to form a series of more complex products and a state of chemical equilibrium is reached. However, equilibrium is not maintained during cooling, and the rate of cooling will affect the degree of crystallization and the amount of amorphous material present in the cooled clinker. Nevertheless, cement can be considered as being in frozen equilibrium, i.e., the cooled products are assumed to reproduce the equilibrium.

Four compounds are regarded as the major constituents of cement. They have been listed in the table below:

The silicates, C₃S and C₂S, are the most important compounds. These are responsible for the strength of the hydrated cement paste. In reality, the silicates in cement are not pure compounds but contain minor oxides. These oxides have significant effects on the atomic arrangements, crystal form and hydraulic properties of the silicates.

The presence of C₃A in cement is undesirable as it contributes little or nothing to the strength of cement, except at the early stage of hydration. However, C₃A is beneficial in the manufacture of cement since it facilitates the combination of lime and silica.

C₄AF’s presence in cement is small in proportion and its effect on the cement characteristics is relatively insignificant. However, it reacts with gypsum to form calcium sulphoferrite and its presence may accelerate the hydration of silicates.

The amount of gypsum added to the clinker is crucial and depends upon the C₃A content and the alkali content of cement. Increasing the fineness of cement has the effect of increasing the quantity of C₃A available at early ages, which raises the gypsum requirement. An excess of gypsum however leads to expansion and consequent disruption of the set cement paste. The optimum gypsum content is determined on the basis of the overall generation of the heat of hydration so that a desirable rate of early reaction occurs. This ensures that there is little C₃A available for reaction after all the gypsum has combined.

In addition to the main compounds described, there are minor compounds such as MgO, TiO₂, Na₂O, K₂O, Mn₂O₃, etc. They usually do not exceed a very small percentage of the mass of cement. Two of the minor compounds, the oxides of sodium and potassium (Na₂O and K₂O), are termed as alkalis. They have been found to react with some aggregates. The products of the alkali aggregate reaction can cause disintegration of concrete. This reaction also affects the rate of the gain of strength of cement. We, therefore, should understand that the use of the term minor refers primarily to the quantity of the compounds and not necessarily to their importance.

Hydration of cement

The process of chemical reaction of cement with water is known as hydration. Till now, we have discussed the properties of cement, but the material of actual interest to a structural engineer in practice is the set cement paste. This is the product of the reaction of cement with water. In the presence of water, the silicates and aluminates of Portland cement form hydrates. These produce the hardened cement paste in the course of time. As stated earlier, the two calcium silicates (C₂S and C₃S) are the main cementitious compounds, with the former hydrating much more rapidly than the latter. In normal cement, the calcium silicates contain small impurities from some of the oxides present in the clinker. These can have a strong effect on the properties of the hydrated silicates. The product of hydration of C₃S is the microcrystalline hydrate, with some lime separating out as crystalline Ca (OH) ₂. C₂S behaves similarly though it takes less lime. The reaction of pure C₃S with water is very rapid and would lead to a flash set, which is prevented by the addition of gypsum to the cement clinker. Even so, the rate of reaction of C₃S is quicker than that of the calcium silicates.

The chemical reaction involved in the hydration of cement is exothermic. The resultant heat generated is known as the heat of hydration and is expressed per gram of cement. The temperature at which hydration occurs has a bearing on the amount of heat produced. In normal Portland cement, about one-half of the total heat is liberated between the first and third day and about three-quarters of the total heat is liberated in the first seven days of reaction.

Types of cement

There are various types of cement that differ in properties, like setting time, strength, etc. Some of them are used only under adverse situations. The common types of cement have been described below:

• Portland pozzolana cement – The term Pozzolana is used for a siliceous material which possesses little or no cementitious properties, but in the presence of water it reacts with calcium hydroxide (liberated during the hydration of cement) to form compounds possessing cementitious properties. The commonly used pozzolanas in the manufacture of this type of cement are burnt clay, shale or fly ash. The proportion of Pozzolana in cement may vary from 10% to 25% of the overall weight. Portland Pozzolana cement produces lower heat of hydration and has greater resistance to attack of chemical agencies than ordinary Portland cement.
• Portland blast furnace slag cement – In this case, the normal cement clinkers are mixed with up to 65% of blast furnace slag for the final grinding. The resulting cement is quite cheap and has almost the same characteristics as Portland cement. In its early stage, the strength development remains slower than that of normal cement but the eventual strength remains the same. In addition, it offers very high resistance to dilute acids. In view of its having a low heat of hydration and better resistance to chemicals, this cement can be used in mass concrete works such as dams, heavy foundations, retaining walls, etc., as well as for the construction within a sea.
• High alumina cement – This cement is manufactured by using 40% each of bauxite and lime, 15% of iron oxide and 5% of silica, magnesia, etc., as raw materials and grinding the clinkers to a very fine powder. High alumina cement has an initial setting time of about three to four hours. Once set, it hardens quite rapidly and the 24 hours compressive strength becomes as high as can be obtained from other types of cement after a minimum period of seven days. It also has excellent resistance to action of sulphates.
• Rapid hardening cement – Rapid hardening cement attains high strength at an early stage. The property of high early strength is achieved due to a high degree of fineness, burning at relatively high temperature and high lime content. The strength development of this cement at the age of three days and seven days is almost the same as the respective seven days and 28 days strength of ordinary Portland cement with the same water cement ratio. The main advantage of using rapid hardening cement is that the formwork/moulds for the members can be removed faster.
• Low heat cement – In the case of mass concrete work such as dams, massive retaining walls, etc., the rate of loss of heat from the surface is much lower than the buildup of heat of hydration inside the concrete mass. This results in setting up tensile stresses during curing and finally leads to cracking of concrete. In low heat Portland cement, the amounts of tri-calcium aluminate and tri-calcium silicate, which contribute the maximum towards heat generation, are kept to the minimum.
• Quick setting cement – If concrete is to be laid under water, quick setting cement is used. The setting action of such a cement starts within five minutes and it becomes stone hard in less than an hour. Quick setting cement is mixed with a small percentage of aluminum sulphate and is ground much finer than ordinary Portland cement.
• White cement – White-colored cement is useful for terrazzo flooring, face plaster to walls and other ornamental works. It is also used for putting markings on grey cement surfaces like roads or airstrips. To manufacture white cement, the raw materials used are china clay and pure limestone while completely excluding iron oxide. For burning the rotary kiln, oil is used as fuel instead of coal. The strength of white cement is less than that of ordinary cement.
• Sulphate-resisting cement – This type of cement is used in the construction of foundations in soils where the sub-soil water contains a very high proportion of sulphates. The constituents of sulphate-resisting Portland cement are such adjusted that the tri-calcium aluminate (C₃A) and tetra calcium aluminate-ferrite (C₄AF) remain very small. Cement with such composition has excellent resistance to sulphate attack.

7.1.2 Aggregates

Aggregates are materials bonded by the cement paste in concrete. Aggregates are largely viewed as inert, inexpensive materials dispersed throughout the cement paste so as to produce a large volume of concrete. But strictly speaking, aggregates are not truly inert because their physical, thermal and chemical characteristics influence the performance of concrete. From the economic viewpoint, it is advantageous to use a mix with as much aggregate and as little cement as possible. The cost benefit has off course to be balanced against the desired properties of concrete in its fresh and hardened state. As aggregate form one of the phases of the two-phased composition of concrete, the properties of aggregate directly affect the properties of the concrete. The aggregate for a concrete mix should be hard, strong, dense, durable and free from foreign materials like clay, loam, vegetable and other injurious matter. The presence of these deleterious substances prevents proper adhesion of cement, on the surface of aggregates, and thus adversely affects the properties of the concrete.

Aggregates are derived from natural rocks or alternatively manufactured from products like slag, etc. Natural aggregates are formed by the process of weathering and abrasion or by crushing of a larger parent mass.

Many properties of the aggregate depend on the properties of the parent material, which include chemical and mineral composition, specific gravity, hardness, strength, pore structure, colour, etc. In addition, there are certain properties of the aggregate, which are not dependent on the parent rock. These are particle shape and size, surface texture, etc. All of these properties do have a considerable influence on the quality of fresh or hardened concrete.

Even in the case if all the aggregate properties are known, it is not quite easy to define a good aggregate for concrete. While aggregate with satisfactory properties would make good concrete, aggregates, which seem to have some drawback, may also produce good concrete. For instance, a material in rock form may disrupt on freezing but may not do so when embedded in concrete in aggregate form. In general though, aggregate performing poorly under one or more than one criteria is unlikely to make satisfactory concrete.

Size and origin of aggregates

Concrete is made with aggregate particles, covering a range of sizes with maximum particle size normally between 10 mm and 75 mm, the selection depending on a host of factors. Particle size of about 20 mm for aggregate is widely adopted for concrete production. At times, low-strength concretes are made with a single aggregate containing all the sizes and known as all in aggregate. However, good-quality concretes are always produced with multiple aggregate sizes, termed as fine and course. The size limit of 5 mm divides the fine aggregate (also called sand, if obtained naturally) from coarse aggregate. Fine aggregates or sand, normally, have a lower size limit of about 0.075 mm (75 μ) or a little less. Material with particle size between 0.0 6mm and 0.02 mm is classified as silt and still smaller particles are termed as clay.

Fine aggregate – As stated earlier, the material below 5 mm in size is termed as fine aggregate. Natural sand or crushed stone dust is mostly used in concrete mix. Sand used from any source, when used in a concrete mix, needs to be properly tested to ensure that the total percentage of clay, silt, salts and other such organic matter does not exceed specified limits. Sand grains may be round or angular. Angular grained sand has good interlocking property that results in a strong mortar. Sea sand should not be used in its natural state for RCC work, as the chloride salts present in the seawater are liable to attack the steel reinforcement.

Coarse aggregate – The material with particle size in excess of 5 mm is termed as coarse aggregate. The size of the coarse aggregate used depends upon the nature of work, size of the component, density of reinforcement, etc. The maximum size may be up to 200 mm for mass concrete applications such as dams. Crushed hard stone and gravel are the most common materials used as coarse aggregates for structural concrete. Clinker slag and coke breeze are also used as aggregates for lightweight and insulating concrete where strength requirement is not really high.

The photograph of a typical coarse aggregate is shown in Figure 7.1.

Grading of aggregates

To produce concrete of a desired strength, density, durability and imperviousness, the proportion of all its ingredients has to be such that the cement paste is sufficient to fill the voids in the fine aggregates and the mortar thus produced is sufficient to fill the voids in the coarse aggregates. To fulfill this objective, the fine as well as the coarse aggregate is graded suitably. It means that the proportion of different particle sizes is so maintained that the smaller particles fill the voids between the larger particles to the maximum extent. This results in a low percentage of voids in the mix. The reduction in percentage of voids results in a corresponding reduction in the quantity of mortar required to fill the voids of the coarse aggregates and, consequently, there is a reduction in the quantity of cement required to make the concrete. Thus, grading of aggregates is necessary to effect economy in the production of concrete.

The grading limits for aggregates are maintained as per the provision of relevant concrete mix design codes. The codes provide tables and graphs for the purpose, applicable for all categories of aggregate.

7.1.3 Water

Water is another important ingredient of concrete. Water chemically reacts with cement and the resultant hydration produces the gel of requisite bonding characteristics. Therefore, it is not possible to produce concrete without water.

The water for concreting should meet the following norms:

• It should be free from oils, fats, etc.
• It should be free from injurious amounts of acids or alkalis.
• It should be free from iron content, organic matter and any other substance, that is likely to have an adverse effect on concrete or reinforcement.
• It should be fit for drinking purposes

The functions of water in the concrete mix can be described as below:

• It acts chemically with cement to form the binding paste for the aggregates as well as the steel reinforcement.
• It acts as a lubricant for the aggregates in the mix and facilitates proper mixing.
• It provides moisture to the aggregate in order to prevent them from absorbing water that is vital for chemical action.

As true for any chemical process, a minimum quantity of water is essential to accomplish the complete hydration process. This is expressed in terms of percentage weight of cement and the range varies from 20% to 25%, depending on the type of cement. However, the fact is that if the water proportion in the mix were maintained just enough to complete the hydration, the mix would hardly be workable and would have low moulding ability. Hence, for a well-prepared plastic concrete, the requirement of water is higher than that derived from chemical consideration.

Water–cement ratio

The ratio of the weight of water to the weight of cement within a concrete mix is known as the water–cement ratio for that mix. Research has established that for a given proportion of ingredients in a concrete mix, there is almost a fixed amount of water (optimum, which gives maximum strength). A small variation from the optimum quantity of water causes a large variation in the strength of concrete. If the amount of water used is less than the optimum quantity, the resultant concrete will be difficult to place in the formwork and can pose problems in compaction. It is difficult to ensure complete setting of the cement in these circumstances and the strength of concrete gets reduced appreciably. On the other hand, if the quantity of water used is more than optimum, the result would be the formation of excessive voids in the hardened concrete, which can reduce the strength and density of the concrete and can also affect its durability. Thus, the water–cement ratio serves as a yardstick for obtaining concrete of the desired strength. The lower the ratio, the greater would be the strength of the mix. Since the water requirement for chemical action is much lower than the one required for workability of the concrete, the water proportion in a concrete mix can always be brought down with appreciation in its other properties. A mix rich in cement would thus have a lower water–-cement ratio and, therefore, higher strength compared to a mix lean in cement.

7.1.4 Admixtures

Admixtures in the concrete are external compounds, which are added to concrete in order to enhance one or more of the physical properties of the concrete. The addition of admixtures is purely optional in the manufacturing of concrete. At times, it is possible to change some of the properties of more commonly used cement types by incorporating a suitable additive or an admixture. In other cases, such addition of admixtures serves the sole purpose of achieving the desired effect. A large number of products are available in the market with their desirable effects described. But such products may also have some undesirable effects, so a cautious approach is warranted while using the admixtures. In normal practice, the terms ‘additive’ and ‘admixture’ are used interchangeably, although, strictly speaking, additive refers to a substance added at the cement manufacturing stage, while admixture is added at the mixing stage.

In most cases, the admixture acts chemically with the cement to produce the desired result. The examples of chemical admixtures are plasticizers, retarders and accelerators. In addition, there are compounds like air-entraining agents whose main purpose is to protect concrete from the deleterious effects of freezing and thawing.

Accelerators

Accelerators are generally referred as those admixtures that accelerate the process of development of the early strength of the concrete. The strength accelerator may not affect the setting time but necessarily influences the hardening pattern of the concrete. It may be worth mentioning here that quick-setting admixtures also exist, which specifically reduce the setting time. Sodium carbonate is one example of a quick-setting admixture that promotes a flash set in the cement.

Retarders

Retarders are those admixtures, which delay the setting of concrete. In other words, retarders increase the setting time for cement. Retarders are useful when concrete is being poured in hot weather and higher temperature reduces the setting time. Sugar is one of the most common examples of retarder. Apart from this, carbohydrate derivatives, such as soluble zinc salts, etc., are the others in this category. When used in a carefully controlled manner, about 0.05 per cent of sugar by mass of cement will delay the setting time by about 4 hours.

The use of retarding admixtures slows down the development of the early strength of concrete, but the rate of strength development goes up later, so the ultimate strength development remains practically unaltered. Retarders also tend to increase the plastic shrinkage because the plastic stage is extended, but the drying shrinkage remains unaffected.

Plasticizers

Plasticizers are essentially water-reducing admixtures. In concrete, these are used to achieve the following objectives:

• To increase the workability of the mix, so as to ease its placement in hard to access locations.
• To achieve a higher strength by decreasing the water/cement ratio at the same workability as an admixture-free mix.

The principal active components of water-reducing admixtures are surface-active agents, which are concentrated at the interface between two immiscible phases and which alter the physico-chemical forces at this interface. With the influence of admixtures, water freed from the system becomes available to lubricate the mix, so that its workability goes up. The reduction in the quantity of mixing water, which is possible owing to the use of admixtures, varies between 5 percent and 15 percent. A part of this reduction is, in many cases, due to the entrained air introduced by the admixture. The actual decrease in mixing water depends on the cement content, aggregate type and the air-entraining agent, if present. Trial mixes are therefore essential to achieve optimum properties as well as to ascertain any potential undesirable side effects like segregation, bleeding and loss of workability with time.

The plasticity of wet concrete is a very important characteristic, especially when the structural member is slender or the concentration of reinforcement within it is high. This can be appreciated by observing the case of concrete wall construction, the shuttering of which prior to concrete pouring and its finish post concreting are shown in Figure 7.2.

7.1.5 Properties of concrete

The most important properties of concrete are the following:

• Strength
• Durability
• Workability

Strength of concrete – Strength is the most important property of concrete. The hardened concrete must be strong enough to withstand all the imposed stresses, taking into account the factor of safety. Factors like strength of the cement, water–cement ratio, aggregate properties, etc. have basic influence on the strength of concrete. Proper placement, compaction and curing are the other factors governing the strength of concrete. Curing is important to supplement the moisture to the green concrete, the loss of which can retard the process of hydration.

The strength of the concrete is measured by the crushing test on the cubical or cylindrical specimen of it. As is obvious, the test provides the compressive strength of the concrete. All other types of strength for concrete are the function of its compressive strength. The structural design of concrete members is based on the characteristic compressive strength of a 15 cm cube specimen of age of 28 days. It is, however, important to note that the process of hydration of cement in concrete continues even after 28 days and, consequently, concrete continues to harden with age, although at a slower rate.

The strength of concrete is expressed in the form of its 28-day characteristic strength. ‘28 days’ indicates the age of the concrete. The age is worked out from the instant when water is mixed with other ingredients, cement and aggregates of the concrete. The characteristic strength of concrete is the value of compressive stress at which not more than 5% of the specimens are expected to fail. In other words, 95% or more specimens of the concrete are certain to attain that strength.

A factor of safety is applied to the strength of concrete, the value of which depends on the nature of stress and circumstances of use. In the case of bending compression, a factor of safety of 3 is normally used. A higher factor of safety is adopted for the direct compressive force.

Durability of concrete – Durability is that property of concrete by virtue of which it resists the process of disintegration and decay. Disintegration of concrete may be the result of:

• Use of unsound cement or less durable aggregates, coarse or fine, in the preparation of concrete.
• Its exposure to harmful chemicals.
• Freezing and thawing of water within it, sucked through the cracks by capillary action.
• Expansion and contraction resulting from temperature changes.

The resistance of concrete to chemical attack, weathering, etc., to a very large extent, depends upon the conditions of exposure, grade of concrete used, quality of its materials and the extent of voids and pores present in the concrete mass. The magnitude of cover provided over reinforcement and the degree of imperviousness of a concrete mix also influence the durability of the concrete. Concrete can be made durable:

• By using good-quality materials.
• By using adequate quantity of cement.
• By maintaining a low water–cement ratio.
• By reducing the extent of voids by suitable mix design.
• By adequate compaction and curing of the concrete.

Workability of concrete – Workability is the property of wet concrete due to which the concrete can easily be mixed, transported, placed in formworks and compacted. Despite its importance, workability is a somewhat elusive property of concrete and is difficult to define and measure. A workable concrete mix must be fluid enough so that it can be compacted with the minimum labor. A workable concrete does not result in bleeding of slurry or segregation of aggregate from the mix.

The Slump Test measures the workability of freshly prepared concrete, where concrete is poured into a container of the shape of frustum of a cone. The cone is removed post compaction and the slump is measured. This is indicated in the photograph in Figure 7.3.

Placement, compaction and curing of concrete

For best results, concrete should be placed and compacted immediately after mixing. The arrangement for the conveyance of the concrete mix should be so planned that the mixed mass is used before its initial set, which normally is a period between thirty to sixty minutes. Hence, it should be possible to place the quantity of concrete prepared in one batch within this period. In order to prevent the shuttering from absorbing the water from the concrete or getting stuck to it, a coat of oil or grease is usually applied to it (the shuttering).

The fresh concrete is prepared by mixing all the ingredients either in a portable mixer (photograph of shown in Figure 7.4) or in large capacity batching plants (shown in Figure 7.5).

While placing this freshly prepared concrete, it should be laid in layers 100 to 300 mm thick based on the shape and size of the overall component, with each layer properly compacted before laying the next one. Compaction of concrete should proceed immediately after placing. The function of concrete compaction is to expel the air bubbles in the mass and make it impermeable enough to attain the desired strength. Compaction is normally done by the use of mechanical vibrators. The resultant concrete is more impermeable, dense and free from honeycombing. The process of concrete placement around reinforcement and its compaction by vibrators is shown in Figure 7.6.

Curing of concrete is one of the essential requirements after placement. Curing is the process of keeping the set concrete continuously damp for some days, in order to enable the concrete to gain more strength. It has been established that the strength of concrete increases with age provided it is kept moist. The period for which curing should be continued depends upon atmospheric conditions, such as temperature, humidity and wind velocity. In general, the process of curing should be continued for seven to 10 days. In cold weather, however, the concrete ought to be cured for longer, as the rate of hardening of cement is low. Efficient curing increases both impermeability and durability of the concrete. Both covering the exposed surface with moist cloth and sprinkling water at intervals to keep the covering wet or by impounding water within bunds over the entire flat area may do curing. The alternate process of curing with steam enables the concrete to gain the required strength in lesser time.

7.2.1 Advantages of reinforced concrete

RCC has the following main advantages when compared with other alternatives like steel:

• RCC normally proves more cost-effective compared to steel.
• Different components of the structure are cast monolithically, which gives high rigidity to the structure.
• Concrete is durable, fire resisting and free form corrosion. These properties get extended to the RCC also.
• Concrete, and therefore RCC, is almost impermeable to moisture.
• The flexibility of reinforcement and the plastic properties of concrete make the moulding of reinforced concrete members in any desired shape possible. Hence, structures of any size and sectional shapes can be constructed with the RCC.

7.2.2 Reinforcement and its bond with concrete

For RCC construction, concrete is reinforced with steel and this material acts as a composite one. Since there can be no fastening to join the two materials, their composite behavior is completely dependent on the development of the bond between the two.

The bondage strength between steel and concrete depends upon the two factors:

• Concrete quality
• Surface characteristics of steel

The quality of concrete, measured as strength and denseness, contributes to its capacity to hold the steel firmly. The more the strength and compactness of concrete, the more will be the strength in bonding. As the strength of the bond depends upon the surface of steel also, the bondage will be better if the steel bar has small ribs on its surface. Such steel is termed as deformed steel. Design codes provide the values of the bond strength, which normally range from 0.6 N/mm² to 1.2 N/mm² for concrete with a characteristic strength from 15 N/mm² to 40 N/mm². The bond strength is normally increased for deformed bars by a factor of 1.4.

When a reinforcement bar is embedded within the concrete, the effort to pull it out would certainly be met with some resistance. This resistance would be equal to the permissible bond stress for the combination of steel and concrete, multiplied by the surface area on which it is working. If the reinforcement steel is of circular section of diameter d and the permissible bond stress is τ_b, the resistance force would be

= τ_b ×π×d×L

Here, L is the length of embedment of steel. The above expression gives the value of the force required to break the bondage. It is obvious that the steel bar would be imposed to this much force before it comes out of concrete.

If the length of embedment is increased gradually, a stage will arrive when the tensile strength in the steel would equal to its limiting value. If the permissible tensile strength for steel were σ_s, the tensile force in the bar would be as follows:

T = π×d²×σ_s,

At this stage, the load developed in the bar is the ultimate allowed and the length of embedment required is known as the development length of reinforcement for its full strength. If the length were L_d, the bond strength developed would be equal to

= τ_b ×π×d×L_d

Equating both the expressions above,

Hence, .

The above relationship informs about the minimum length of reinforcement required beyond the section to develop appropriate amount of strength in it. It may be noted that we did consider the pulling out of reinforcement in the case, which induces tensile strength in the bar. If instead of the pulling out a rod we try pushing it within the concrete, it can be visualized that the resistance will be higher. Hence for compression, the development length can be calculated by considering 25% higher value of bond stress.

In the compression, .

7.2.3 Theory of RCC section design

To design an RCC structural section, the following assumptions are made:

• A transverse section, which is plane before bending, remains plain after bending.
• The modulus of elasticity of steel and concrete are constant and not stress dependent.
• All tensile stresses are taken up by the steel reinforcement only and the tensile load on concrete is nil.
• The bond between concrete and steel is sufficient to make them act jointly to resist tensile or compressive forces. No slippage takes place between the two materials.

For any composite material, the relationship between the modulus of elasticity of its components is extremely important. This is helpful in deriving the distribution of the load between its components. In the case of RCC section design, the ratio between the modulus of elasticity of steel and concrete is known as the modular ratio and is denoted by m. Hence, if E_s and E_c are the modulus of elasticity for steel and concrete respectively, the modular ratio can be described as

m = E_s / E_c

Now, we would derive the actual distribution of the load. As per assumptions made above, it is evident that if a beam were subjected to bending that causes compression at the top face and tension at the bottom, the strain in the steel and concrete would be equal at the interface where both meet.

For steel,

Strain = σ_st / E_s

And for concrete,

Strain = σ_cbc / E_c

where

σ_st = Tensile stress in steel

σ_cbc = Compressive stress in concrete

Since both the strains are equal,

σ_st / E_s = σ_cbc / E_c

σ_st / σ_cbc = E_s / E_c

σ_st / σ_cbc = m

σ_st = σ_cbc × m

Therefore, the stress in the steel would be m times the value of the stress in the surrounding concrete.

In the next step, we need to determine the overall stress bearing capacity of an RCC section. For this, consider the section shown in Figure 7.7. The overall area of the section is A, with the area of concrete and steel being A_c and A_t respectively. If the permissible compressive stress in steel and concrete were σ_sc and σ_cc respectively, the overall capacity of the section to withstand the compressive load would be

We know that A_c + A_t = A. Hence,

The above is the relationship between the total compressive force and the compressive stress in the concrete. The expression ‘A + A_t(m – 1)’ is termed as the equivalent area of the section, represented in the form of concrete area.

During bending, the transverse section, which is plane before bending, remains plane after bending. It is, therefore, obvious that when bending takes place in a beam, the strain in any layer remains proportional to its distance from the neutral axis. Thus, the strain diagram will be a straight-line diagram as shown in Figure 7.7. Also drawn is the stress diagram. Since it is assumed that the overall section is transformed into equivalent concrete section, it becomes essential to write the stress in steel in terms of equivalent stress in concrete at the level of steel. In the stress diagram, the stress in steel is replaced by equivalent stress in the concrete surrounding the steel.

Unlike steel sections, the location of the neutral axis in the RCC beam of given dimensions is not fixed and varies with the amount of steel provided in the section. We will observe the reason for it. It is clear from Figure 7.7 that the section of the beam above the neutral axis is subjected to compression, while that below the neutral axis is subjected to tension. As is assumed that no tensile force is resisted by concrete, the whole of the tension below the neutral axis is supposed to be resisted by the steel reinforcement. Thus, the effective depth of the beam would be equal to the distance measured from the top face of the beam to the center of the tensile reinforcement.

We will consider the case of a singly reinforced section shown in Figure 7.7. In the singly reinforced section, the steel bars are provided to resist only the tensile force and the concrete resists the entire compression. Assuming that for the case the steel and concrete have been stressed to their ultimate permissible values, the neutral axis of a beam can be determined as below.

Referring to the stress diagram in Figure 7.7,

From this,

If we represent the stress ratio σ_st/σ_cc as r, the expression would become

The above expression gives the location of the neutral axis for a reinforced concrete section. It may be noted here that both concrete and steel have been assumed to be stressed to their highest permissible limits. Such a section is known as the balanced section and the value of k would be dependent on the stress ratio and the modular ratio.

Next, we would determine the capacity of the balanced section to withstand moment. In the case of RCC section, the size or diameter of steel reinforcement is quite small compared to the overall size of the section. The variation of bending stress along this small dimension would be quite small. Hence, the tensile stress in the steel is assumed to be uniform over the total area of tensile steel and is considered to act at the centre of the steel. For the purposes of deriving design expressions, the rectangular section of the beam is treated to consist of a block of concrete above the neutral axis for resisting the whole of the compression force and the area of steel below the neutral axis is assumed to resist all the tensile forces. The concrete below neutral axis is completely ignored as far as resisting the load is concerned.

Total compressive force in the concrete above neutral axis = b×k×d×(σ_cc /2)

Total tensile force in the steel below neutral axis = A_s × σ_st

As is evident from the stress diagram, the compressive force would act at a distance of n/3 or k×d/3 from the top edge of the section. The tensile force would work at the bottommost edge of the section.

For a section to be in equilibrium, both the above forces should be equal in magnitude:

It is obvious that these forces would form a couple. The length of arm of the couple would be equal to (d – n/3).
Hence, moment of resistance of the section,

The lever arm (d – n/3) can be expressed as equivalent to j×d, where j is known as the lever arm factor:

Using this value of j in the expression for the moment resistance of the section,

Here again, the expression (j×k×ς_cc/2) is denoted as R, the moment of resistance factor.

From the above relationships, for a particular value of moment, the requirement of sectional size as well as steel area for a balanced singly reinforced RCC section can be determined.

Under and over reinforced RCC sections

The RCC section where the area of steel provided is less than that required as per the above-derived expression is known as the under reinforced section. In this case, the steel being deficient in quantity it attains the maximum stress value ahead of concrete and the concrete, in turn, never reaches its maximum stress value. In such cases, the location of the neutral axis remains above the neutral axis for the balanced section, as shown in Figure 7.8.

For the under reinforced section, the moment of resistance is derived by the relation

M = A_s×σ_st×(d – n/3)

The RCC section where the area of steel provided is larger than that required as per the expression valid for the balanced section, the section is known as the over reinforced section. In this case, the concrete attains its maximum stress value ahead of steel. In this situation, steel never reaches its maximum stress value. In such cases, the location of the neutral axis remains below the neutral axis for the balanced section, as shown in Figure 7.8

For the under reinforced section, the moment of resistance is derived by the relation

M = b×k×d×(σ_cc/2)×(d – n/3)

Process for design of singly reinforced section

The objective of design of a moment resistant section in RCC is to derive the most economical sectional size for the given value of moment. Accordingly, the area of steel required is calculated.

Since the values of permissible stresses and modular ratio are known, the first step would be to calculate the value of k from the relevant formula. With this known, the lever arm factor j can be derived.

Next would be the derivation of the moment of resistance factor, R, with the formula given below:

R = j×k×σ_cc/2

The moment of resistance of the section is the function of R and the width as well as the depth of the section. Since the size of the section has to be determined yet, we move ahead by assuming a suitable width for the section. With the help of this, the required depth of the section can be determined by using the formula

M = Rbd²

As a final step, the area of steel reinforcement can be determined by the expression

The illustration below would explain the procedure more clearly.

Illustration 7.1: An RCC beam is subjected to bending moment of 150000 N-m, with the moment due to self-load factored in. It is desirable to determine a section for it. Since the wall size right below the beam is 250 mm, it is desirable to restrict the beam width to 250 mm only.

The allowable stresses are as follows:

First of all, we need to determine the value of k for the section. Using expression,

Moment arm factor,

For the section, the value of coefficient of the resisting moment,

We know that the moment of resistance of the section,

M = Rbd²

The value of M is known and the value of R for the case has been derived. Also, the width b of the section has been restricted to 250 mm. By placing these values into the expression above,

The effective depth of the beam should be 667.2 mm. Giving consideration to the cover for the bottom reinforcement, the overall depth of 700 mm can be provided.

With the section being designed as balanced, the area of steel required would be derived from the following expression:

To fulfill this requirement, four steel bars of 25 mm diameter each (total area = 1963 mm²) can be provided as reinforcement.

Design of doubly reinforced RCC sections

In the previous section, we have studied the design principles and procedure for singly reinforced RCC sections. It may be noted here that the singly reinforced RCC beams of given dimensions have a limiting value of moment which it can resist. To enhance this value what is required is to alter the composition or grade of its components. This may not be feasible or economical at times.

Also while analyzing continuous beams and frames, we have seen that the sign of the bending moments can be different for different portions of a beam. It may be positive at the centre of the span but can turn negative over the supports. This means that the tension generation may be at different faces within a single span of the beam.

In situations as above, designing a doubly reinforced section may be a solution. In doubly reinforced beams, steel is provided in tension as well as in the compression zone. Unlike singly reinforced sections, steel bears both tensile and compressive loads in a doubly reinforced section. Since the compressive reinforcement is restrained from all sides by the concrete, there is no scope for compression reinforcement to buckle and, therefore, the value of permissible stress is independent of the slenderness of the steel bars.

When subjected to compressive stress, concrete undergoes elastic deformation. Apart from this, when concrete is consistently subjected to compressive loads, it undergoes a slow plastic deformation, also known as the creep. Therefore, for the same amount of stress, the deformation of concrete is more in the case of compression in comparison to tension. As a result, the modulus of elasticity for concrete is more in tension and somewhat lower in compression. In other words, the modular ratio for compression would be higher in the case of a doubly reinforced section. If the modular ratio in tension is m, a corresponding value of 1.5m is adopted for compression. In view of this, the equivalent area of steel in compression would be equivalent to

The stress diagram for a doubly reinforced section is similar to a singly reinforced section. Hence, the following relationship, for the location of neutral axis, would be valid in the case of a doubly reinforced section also:

As was in the case of a singly reinforced section, the above expression gives the location of neutral axis for the balanced section. If the actual neutral axis does not coincide with it, the exact location of the neutral axis can be derived by taking moment of all the forces on the section with respect to the neutral axis.

Referring to Figure 7.9, by equating the moment of compressive force with that of tensile force, we will have

bn² / 2 + (m_c– 1)×A_sc×(n – d_c) = m×A_st×(d – n)

The moment of resistance of a doubly reinforced beam would be equal to two components. First is the moment of resistance, ignoring the compressive reinforcement and treating the beam as singly reinforced (M₁). The other one would be the moment of compressive force in the steel reinforcement, with the distance between the center of tensile and compressive reinforcement as the lever arm (M₂). The total moment of resistance would be equal to the sum of the two.

Hence, M = M₁ + M₂.

We already know the expression for the moment of resistance, for a singly reinforced section, and the procedure to derive the area of reinforcement. Here, we would see the expression for moment due to compressive reinforcement. Again referring to Figure 7.9 and taking moment about the centre of the tensile reinforcement,

To balance the compressive reinforcement, there would be need of additional reinforcement in the tensile zone. Its area can be calculated from the following expression:

A_st2 = M₂/ ς_st× (d – d_c)

Shear stress treatment in RCC section

We understand that in a beam section, the bending moment is generally accompanied with shear force. This is due to the fact that load and support systems, which induce bending moment, are also conducive to the generation of shear force.

If a concrete section, with width b and depth d, is subjected to shear force V, the value of shear stress in the section should be equal to

τ = V / b×d

We also know that the shear stress distribution along the section is not uniform but parabolic. Hence, the above expression gives only the average shear stress on the section. The maximum value of shear stress would be along the neutral axis and also be equal to 1.5 times the average value.

In the case of RCC sections, the treatment of distribution of shear stress is still further different from a pure concrete section. In the case of RCC sections, the concrete in the tensile zone is neglected and the shear force is assumed to be resisted by the bond between steel and concrete. The distribution of the shear stress across the section is shown in Figure 7.10.

The area of the shear force diagram in this case is the sum of the area of the parabolic portion above neutral axis and the rectangular portion below neutral axis.

We know that (d – n/3) is equal to the lever arm, j×d.

When the shear stress acts on an RCC section, as shown in Figure 7.11, a set of complementary stresses develops in a plane perpendicular to the shear force. The resultants of these do act along the diagonal of the block under shear force. As is obvious from the diagram, the force along one diagonal plane would be tensile and compressive along other.

Due to the development of these forces, there would be a tendency of either crushing of the material along one diagonal or the rupture of the material along the other. The actual failure would be dependent on the tensile and compressive strength of the material, as the lower value of the two would govern the type of failure. We know that concrete has a much lower strength in tension than in compression. Hence, the concrete section would tend to split along the diagonal representing tensile force.

When a concrete section is subjected to shear force, it resists a certain amount of it due to a combination of frictional resistance in the compression zone (tension zone is ignored as concrete has a tendency to separate here) and the vertical component of the aggregate interlock force. In the RCC section, the shear force across the reinforcement adds to this resistance. Hence, a certain amount of shear stress can safely be allowed to develop in the RCC section, the value of which is decided on the basis of the characteristic strength of the concrete in combination with the percentage area of reinforcement in the overall RCC section. If the actual shear stress exceeds this value, shear reinforcement is provided in the shape of vertical stirrups. However, it may be noted that the value of shear stress in a section, for a particular grade of concrete, is never allowed to exceed the values provided in the design code. If the value of shear stress is found to be more than this value, the section itself is required to be redesigned. These stirrups act as direct tension members and prevent the separation of concrete along the plane of shear failure.

For these vertical stirrups, the design shear force is taken as the difference between the total shear force at the section and the shear resistance of the section. If a section has width b and depth d, the designs shear force V_d shall be derived by the following expression:

V_d = V – τ_p×b×d

Here, τ_p is the permissible shear stress for the section.

The vertical stirrups are made of plain steel bars of diameter ranging from 5 mm to 12 mm. The stirrups can be single or multi-legged, depending upon the shear force they are required to resist. The typical shapes of the stirrups are shown in Figure 7.12. They are either anchored with the compression reinforcement or taken around both the tension and compression reinforcement to form a closed loop.

While designing a stirrup system, the diameter and number of legs are decided by judgment and based on these, the spacing between these stirrups is derived.

We would refer back to Figure 7.10, where the elevation of a beam near its support has been shown. We know that for a simply supported beam, the amount of shear force becomes largest at the supports. Hence, the area around support would be prone to the maximum shear stress. The shear crack development is at the 45º to the axis of the beam; hence, longitudinally, it would extend to the distance equal to beam depth from the edge of the support.

If, d  = Effective depth of the beam
σ_st = Permissible tensile stress in the stirrup
A_s = Cross-sectional area of the stirrup
s   = Spacing of the stirrups

then
Tensile force resistance capacity of each stirrup = A_s × σ_st

Within the critical longitudinal length of d, the number of stirrups would be equal to d/s.

Hence, tensile force resistance capacity of all stirrups 

This tensile force bearing capacity has to be equal to or more than the value of shear force V_d, derived earlier.

This expression gives the design value of stirrup spacing in the design of vertical shear stirrups.

By now, we are quite familiar with the characteristics of concrete and design of RCC as a composite material. Let us examine the application of this knowledge in the design of some structures.

7.3.1 Effective span

Simply supported beam – The effective span of a simply supported beam or slab is taken as the distance between the center-to-center of support or the clear distance between the supports, plus the effective depth of the beam, whichever is smaller.

Continuous beam – In the case of a continuous beam or slab where the width of the support is less than 1/12 of the clear span, the effective span shall be calculated as in the case of a simply supported beam.

In case the supports are wider than 1/12 of the clear span or 600 mm, whichever is less, the effective span shall be calculated as below:

• For end span with one end fixed and the other continuous or for intermediate spans, the effective span shall be the clear span between supports.
• For end span with one end free and the other continuous, the effective span shall be equal to the clear span plus half the effective depth of the beam or slab or the clear span plus half the width of the discontinuous support, whichever is less.

Beams part of frames – In the analysis of a continuous beam, which is part of a frame, the effective span shall be taken as the center-to-center distance between the supports.

Depth of beams from the viewpoint of deflection

It is necessary to impose a limit on the magnitude of deflection in a structural member with a view to ensure that the extent of deflection does not adversely affect the appearance or efficiency of the structure. Also, the structural behavior of the material may be different from the assumed one if the deformation is excessive. We can theoretically calculate the deflection for a specified case of loading on the structure. However, till 10 M span of the beam, the requirement related to deflection is very likely to be met if span to depth ratios for different beams are within the following limits:

• For Simply Supported Beam – 20
• For Cantilever Beam / Overhang – 7
• For Continuous Beams – 26

Proportioning against slenderness

As in the case of steel beams, the compression zone of the RCC beam would also have a tendency to buckle. The required stability is maintained, by proportioning the cross section of the beam. For non-cantilever beams, the clear distance between the lateral supports does not exceed 60b or 250b²/d, whichever is less. Here, d is the effective depth of the RCC section and b is the breadth of the compression face of the beam, with a location at the middle of the supports. In the case of cantilever beams, the clear distance from the free end to the lateral restraint should not exceed 25b or 100 b²/, whichever is less.

Limits related to reinforcement steel

For an RCC section, the provision of reinforcement steel is governed by certain norms, which ensure optimum performance.
Some of these norms are as follows:

Tension reinforcement

• The minimum area of the tension reinforcement in beams shall not be less than that derived by the following expression:A_s / bd = 0.35 / fy

Here, A_s is the area of steel as tensile reinforcement and fy is the yield strength of the steel.

• The maximum area of tension reinforcement in a beam should not exceed 0.04 bD, where D is the overall depth of the beam.

Compression reinforcement

There has been no lower limit for compression reinforcement, as it can very well remain absent in the section. However, the maximum area of compression reinforcement in a beam shall not exceed 0.04 bD, where D is the overall depth of the beam.

Curtailment of reinforcement

It is known that the value of bending moment varies across the length of the beam. Reinforcement in a beam section is decided according to the highest value of the bending moment, and, therefore, some of it can become redundant at sections away from the high moment locations. In view of this, the main reinforcement in beams and slabs may be discontinued or curtailed at the location where it is no longer required to resist bending. However, it is necessary that the curtailed reinforcement should extend beyond the section for a distance equal to the effective depth of the member or 12 times the bar diameter, whichever is greater.

Side face reinforcement

Where the depth of the RCC beam exceeds 750 mm, it becomes necessary to provide side-face reinforcement along both the faces of the beam. The total area of such reinforcement shall be not less than 0.1 per cent of the sectional area of beam web and they shall be distributed equally on the two faces, at a spacing not exceeding 300 mm or web thickness, whichever is lower.

Shear reinforcement

In RCC beams, minimum shear reinforcement in the form of stirrups should be provided as derived by using the following expression, irrespective of the requirement as per stress level:

The maximum spacing of shear reinforcement along the axis of the beam should be maintained as 0.75d or 450 mm, whichever is less.

Concrete cover to reinforcement

The thickness of the concrete portion between the outermost edge of reinforcement steel and the outermost edge of the concrete section is known as the clear cover for the reinforcement. The clear cover of concrete to reinforcement in different structural members should be as follows:

• The clear cover of longitudinal reinforcing bar in a beam should not be less than 25 mm or the diameter of the reinforcing bar, whichever is higher.
• The clear cover at the each end of reinforcing bar in a beam should be equal to or more than 25 mm or twice the diameter of such bar, whichever is higher.
• If the surface of concrete of a structural member is exposed to the action of harmful chemicals, acids, saline atmosphere, etc., or is in direct contact with soil containing such chemicals, it is suggested to provide additional cover ranging from 15 mm to 50 mm over and above the normal cover.
• For reinforced concrete members permanently or periodically immersed in seawater, the cover of concrete should be 50 mm more than the normal cover.

Spacing of reinforcement

The gap between the reinforcements in RCC sections is governed by factors like diameter of the bars, aggregate size of the concrete, etc. The norms for spacing of reinforcement are within the section are as follows:

• The minimum clear gap between two parallel main reinforcing bars should not be less than the diameter of the bar (or the diameter of the larger bar, if unequal) or 5 mm more than the nominal maximum size of coarse aggregate used in the concrete, whichever is more.
• If main reinforcement bars are provided in two or more layers one over the other, the minimum clear gap between any two layers of the bars should normally be the largest of the three,
  1. 15 mm,
  2. two-thirds the nominal maximum size of aggregate, or
  3. the maximum size of the bar.

The following illustration would make clear how these provisions are followed in the design exercise.

Illustration 7.2: A simply supported RCC beam is subjected to a live load of 13500 N/m, apart from its self-weight. The beam is supported on columns of 250 mm × 250 mm section. The clear span for the beam is 4.5 m. The objective is to design an appropriate section.

To design a section, we need to determine the highest bending moment within the span of the beam. Since the sectional size of the beam is not known, we assume an overall section of 250 mm × 400 mm for it to determine the bending moment. Since the values of 60b and 250d²/b both are much higher than the clear span of 4 m, the section meets the safety norms related to slenderness.

For each meter length of beam, the self-weight would be

Total load on beam = Live Load + Self-Load

Since the center-to-center distance in the beam (4.5 + 0.25 = 4.75 m) is less than the sum of the clear span and beam depth (4.5 + 0.4 = 4.9 m), 4.9 m would be considered as the span for the beam.

Now, from the data about material characteristics, we will determine the value of k for the section. Using expression,

Moment arm factor,

For the section, the value of coefficient of the resisting moment,

We know that the moment of resistance of the section,

The value of M is known and the value of R for the case has been derived. Also, the width b of the section has been kept 250 mm, according to the width of the supporting columns. By substituting these values into the above expression,

The effective depth of the beam should be 372.4 mm. We have assumed an overall depth of 400 mm, which should be fine giving consideration to the clear cover requirement of 25 mm for the main tensile reinforcement.

As we are designing a balanced section, the area of steel required would be derived from the following:

To provide reinforcement as per this requirement, four steel bars of 20 mm diameter each (total area = 1256 mm²) are proposed.

This is very close to the minimum norm of 25 mm, and, hence, is acceptable.

Check for minimum area requirement for tension reinforcement

The minimum area for tension reinforcement can be calculated from the following expression:

Since the area of actual steel provided is much higher (1256 mm²), the section meets the relevant norm completely.

Design of shear reinforcement

We know that the maximum shear force would be at either one of the supports.

For this percentage of steel and concrete with characteristic strength of 20 N/mm², the permissible value of shear stress in the section is 0.435 N/mm².

Hence, the shear force withstanding capacity of the section = 0.435 × 250 × 366.7
= 39878.62

This capacity of section exceeds the maximum value of shear force at any section of the beam. Hence, there has theoretically been no need to provide shear reinforcement. However, we know that RCC sections need to be provided with a minimum quantity of shear reinforcement under all circumstances. This is derived as per the following expression:

We decide to provide two-legged stirrups of 6 mm diameter as shear reinforcement. The cross-sectional area of this stirrup would be 56 mm².

Substituting all the relevant values into the expression,

Hence, the spacing for stirrup has been decided as 125mm c/c (center to center).

The section with all reinforcements is shown in Figure 7.13:

7.4.1 Design of slabs

Slabs are designed by using the principle of flexure and shear as are used for beams. The following methods of slab analysis and design are available for adoption:

• Elastic method – by idealizing the slabs as strips of beams
• Limit state method – based on semi-empirical coefficients provided in the design codes
• Yield line theory