# Shear Force and Bending Moment

**Contents**

## Introduction

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**Shear**Forces occurs when two parallel forces act out of alignment with each other. For example, in a large boiler made from sections of sheet metal plate riveted together, there is an equal and opposite force exerted on the rivets, owing to the expansion and contraction of the plates.

**Bending**Moments are rotational forces within the beam that cause bending. At any point within a beam, the Bending Moment is the sum of: each external force multiplied by the distance that is perpendicular to the direction of the force.

## Shearing Force

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**Shearing Force**at the section AA. It may be defined as follows:-

*The shearing force at any section of a beam is the algebraic sum of the lateral components of the forces acting on either side of the section.*Where forces are neither in the lateral or axial direction they must be resolved in the usual way and only the lateral components are used to calculate the shear force.

## Bending Moments

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*Bending Moment at AA is defined as the algebraic sum of the moments about the section of all forces acting on either side of the section*.

*Bending moments are considered positive when the moment on the left portion is clockwise and on the right anticlockwise*. This is referred to as a

**sagging**bending moment as it tends to make the beam concave upwards at AA. A negative bending moment is termed

**hogging**.

## Types Of Load

A beam is normally horizontal and the loads vertical. Other cases which occur are considered to be exceptions. A**Concentrated load**is one which can be considered to act at a point, although in practice it must be distributed over a small area. A

**Distributed load**is one which is spread in some manner over the length, or a significant length, of the beam. It is usually quoted at a weight per unit length of beam. It may either be uniform or vary from point to point.

## Types Of Support

A**Simple**or

**free**support is one on which the beam is rested and which exerts a reaction on the beam. It is normal to assume that the reaction acts at a point, although it may in fact act act over a short length of beam.

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**Built-in**or

**encastre'**support is frequently met . The effect is to fix the direction of the beam at the support. In order to do this the support must exert a "fixing" moment M and a reaction R on the beam. A beam which is fixed at one end in this way is called a

**Cantilever**. If both ends are fixed in this way the reactions are not statically determinate. In practice, it is not usually possible to obtain perfect fixing and the fixing moment applied will be related to the angular movement of the support. When in doubt about the rigidity, it is safer to assume that the beam is freely supported.

## The Relationship Between W, F, M.

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*inflection*or

*contraflexure*. By integrating equation (2) between the x = a and x = b then: Which shows that the increase in bending moment between two sections is the area under the shearing force diagram. Similarly integrating equation (4) equals the area under the load distribution diagram. Integrating equation (5) gives: These relations can be very valuable when the rate of loading cannot be expressed in an algebraic form as they provide a means of graphical solution.

## Concentrated Loads

##### Example - Example 1

*l*carries a concentrated load

*W*at its free end. Draw the Shear Force (SF) and Bending Moment (BM) diagrams.

*x*from the free end. Then F = - W and is constant along the whole cantilever i.e. for all values of

*x*Taking Moments about the section gives M = - W

*x*so that the maximum Bending Moment occurs when

*x*=

*l*i.e. at the fixed end. From equilibrium considerations it can be seen that the fixing moment applied at the built in end is WL and the reaction is W. Hence the

**SF**and

**BM**diagrams are as follows:

**The shearing force suffers sudden changes when passing through a load point. The change is equal to the load.****The bending Moment diagram is a series of straight lines between loads. The slope of the lines is equal to the shearing force between the loading points**.

## Uniformly Distributed Loads

##### Example - Example 3

*l*carrying a uniformly distributed load

*w*per unit length which occurs across the whole Beam.

*wl*and by symmetry the reactions at both end supports are each wl/2 If

*x*is the distance of the section measured from the left-hand support then: This give a straight line graph equal to the rate of loading. The end values of Shearing Force are The Bending Moment at the section is found by assuming that the distributed load acts through its center of gravity which is

*x/2*from the section. This is a parabolic curve having a value of zero at each end. The maximum is at the center and corresponds to zero shear force. From Equation (2) Putting

*x*=

*l/2*

## Combined Loads.

##### Example - Example 4

**The Shearing Force**Starting at A F = 7.25. As the section moves away from A F decreases at a uniform rate of w per unit length ( i.e. f = 7.25 - wx) and reaches a value of - 2.75 at E. Between E and D, F is constant ( There is no load on Ed) and at D it suffers a sudden decrease of 2 tons ( the load at D) . Similarly there is an increase at B of 7.75 tons ( the reaction at B). This results in a value of F = 3 tons at B which remains constant between B and C. Note this value agrees with the load at C. Bending Moment From A to E: This is a parabola which can be sketched by taking several values of x. Beyond E the value of x for the distributed load remains constant at 5 ft. from A Between E and D This produces a straight line between E and D. Similar equations apply for sections DB and BC. However it is only necessary to evaluate M at the points D and B since M is zero at C. The diagram consists in straight lines between these values. At D At B This last value was calculated for the portion BC We were required to find the position and magnitude of the maximum BM. This occurs where the shearing force is zero. i.e.at 7.25 ft. from A The point of contraflexure occurs when the bending moment is zero and this is between D and B at:

## Varying Distributed Loads.

##### Example - Example 7

**Note**. These are the centroids of the triangles which represent the load distribution) Taking Moments about B At a distance

*x*(<18)from A the loading is x/18 tons/ft.. The Total distributed load on this length is: The centre of gravity of this load is from A.

**For 0<x<6**At x = 6 ft. At x = 6 ft.

**6< x <18**The maximum Bending Moment occurs at zero s earing force 1.e.x = 7.58 ft. The section BC can be more easily calculated by using a variable X measured from C. Then by a similar argument:-

The complete diagrams are shown. It can e seen that for a uniformly varying distributed load, the Shearing Force diagram consists of a series of parabolic curves and the Bending Moment diagram is made up of "cubic" discontinuities occurring at concentrated loads or reactions. It has been shown that Shearing Forces can be obtained by integrating the loading function and Bending Moment by integrating the Shearing Force, from which it follows that the curves produced will be of a successively "higher order" in x ( See equations (6) and(7))

## Graphical Solutions

**Note**This method may appear complicated but whilst the proof and explanation is fairly detailed, the application is simple and straight forward. Earlier it was shown that the change of Bending Moment is given by the double Integral of the rate of loading. This integration can be carried out by means of a

**funicular polygon**. See diagram. Suppose that the loads carried on a simply supported beam are are the reactions at the supports. Letter the spaces between the loads and reactions A, B, C, D, E, and F. Draw to scale a vertical line such that. Now take any point "O" to the left of the line and join O to a, b, c, d, and e. This is called

**The Polar Diagram**Commencing at any point p on the line of action of draw pq parallel to Oa in the space "A" , qr parallel to Ob in the space "B" and similarly rs, st, and tu. Draw Of parallel to pu. It will now be shown that fa represents . Also, pqrstu is the Bending Moment diagram drawn on a base pu, M being proportional to the vertical ordinates. is represented by ab and acts through the point q; it can be replaced by forces aO along qp and Ob along qr. Similarly, can be replaced by forces represented by bO along rq and Oc along rs, by cO along sr and Od along st etc. All of these forces cancel each other out except aO along qp and Od along te, and these two forces must be in equilibrium with . This can only be so if is equivalent to a force Oa along pq and fO along up, being equivalent to eO along ut and Of along pu. Hence, is represented by fa and by ef.

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