Shear Force and Bending Moment
IntroductionShear 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 ForceThe shearing force (SF) at any section of a beam represents the tendency for the portion of the beam on one side of the section to slide or shear laterally relative to the other portion. The diagram shows a beam carrying loads . It is simply supported at two points where the reactions are . Assume that the beam is divided into two parts by a section XX. The resultant of the loads and reaction acting on the left of AA is F vertically upwards, and since the whole beam is in equilibrium, the resultant force to the right of AA must be F downwards. F is called the 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 MomentsIn a similar manner it can be seen that if the Bending moments (BM) of the forces to the left of AA are clockwise, then the bending moment of the forces to the right of AA must be anticlockwise. 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 LoadA 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 SupportA 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. A 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.In the following diagram is the length of a small slice of a loaded beam at a distance x from the origin O Let the shearing force at the section x be F and at . Similarly, the bending moment is M at x, and . If w is the mean rate of loading of the length , then the total load is , acting approximately (exactly if uniformly distributed) through the centre C. The element must be in equilibrium under the action of these forces and couples and the following equations can be obtained:- Taking Moments about C:
Example - Example 1
- 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
Example - Example 4
Varying Distributed Loads.
Example - Example 7
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 SolutionsNote 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. triangles pqv and Oaf are similar and hence:
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