Chapter 8. Shear Force and Bending Moment Diagrams for Uniformly Distributed Loads.

Transcription

1 hapter 8 Shear Force and ending Moment Diagrams for Uniformly Distributed Loads. 8.1 Introduction In Unit 4 we saw how to calculate moments for uniformly distributed loads. You might find it worthwhile to revisit that unit to refresh your memory. The rule for calculating bending moments for uniformly distributed loads is shown in Fig. 1. With reference to this figure, the moment of the uniformly distributed load about is the total load multiplied by the distance from the centre line of the UDL, to the point about which we re taking moments. The total UDL is, the distance concerned is a, so: Moment of UDL about = pply this principle whenever you re working with uniformly distributed loads. w kn/m x a entre line of Loaded length Figure 1: ending moment calculation for uniformly distributed load (UDL): general case

2 Practical Example 1 eam G, shown in figure 2, spans 6 metres. It supports a uniformly distributed load of 4kN/m along its entire length. Draw the shear force and bending moment diagrams. 4 kn/m D E F G 6 m R R G Figure 2: Example 1: eam Diagram First of all, calculate the reactions. This is easy in this case because of the symmetry of both the beam itself and its loading. Each end reaction will be half the total load on the beam. So ( ) We will now try the metre-by-metre approach as pioneered in chapter 7 - to drawing the shear force and bending moment diagrams. So, we are going to calculate the shear force and bending moment values at points,,, D, E, F and G. SHER FORES (Remember always look at what s going on to the left of the point at which you re trying to calculate shear force.) s before, draw a horizontal straight line representing zero shear force. This will be the base line from which the shear force diagram is drawn. There is nothing to the left of point, so the shear force at point is zero. If we go a very small distance (say 2 millimetres)to the right of, there is now a 12 kn upward force (the reaction at ) to the left of the point we re considering. So the shear force at this point is 12 kn. We can represent this effect by a vertical straight line at point, starting at the zero force base line and going up to a point representing 12 kn.

3 Each of points,, D, E, F and G has this 12 kn upward force to the left of it (i.e. the reaction at point ), but they also have downward forces to the left. Let s consider each of these points in turn. Point Upward Force Downward Force Shear force 12kN (4kN/m x 1m) =4kN 12 4 = 8kN 12kN (4kN/m x 2m) = 8kN 12 8 = 4kN D 12kN (4kN/m x 3m) = 12kN = kn E 12kN (4kN/m x 4m) = 16kN = 4kN F 12kN (4kN/m x 5m) = 2kN 12 2 = 8kN G 12kN (4kN/m x 6m) = 24kN = 12kN t point G, there is an upward reaction of 12kN. So the net shear force at G will be = kn. These values can be plotted on our shear force diagram in Fig. 3 (b). ending Moments Once more, we will be looking solely at forces and moments to the left of the point we re considering. s in earlier examples, we will calculate the momnet at each point, remembering that: lockwise moments are positive, and anticockwise moments are neagtive; Distances are measured from the force concerned to the point considered. Point ending Moment alculation ending Moment (kn.m) +(12kN x m) +(12kN x 1m) (4kN/m x 1m x.5m) 12 2 = 1 +(12kN x 2m) (4kN/m x 2m x 1m) = 24 8 = 16 D +(12kN x 3m) (4kN/m x 3m x 1.5m) = = 18 E +(12kN x 4m) (4kN/m x 4m x 2m) = = 16 F +(12kN x 5m) (4kN/m x 5m x 2.5m) = 5 = 1 G +(12kN x 6m) (4kN /m x 6m x 3m) = = The bending moment diagram is drawn in Fig. 3 (c).

4 4 kn/m D E F G 6 m R = 12kN (a) eam Diagram R G = 12kN (b) Shear Force Diagram (c) ending Moment Diagram Figure 3: Practical Example 1 Graphical interpretation

5 8.2 The Shape of Shear Force and ending Moment Diagrams where Uniformly Distributed Loads are Present If you examine the shape of the shear force and bending moment diagrams in Fig. 3 you will notice the following features: The shear force diagram comprises sloping straight lines. The bending moment diagram is curved (parabolic). In general, where a beam is loaded with uniformly distributed loads along all or part of its length, the shear force and bending moment diagrams along the part of the beam concerned have the above features. To summarise: where a beam experiences uniformly distributed loads, the shear force diagram will compromise sloping straight lines and the bending moment diagram will be curved. Shear Force and ending Moment Diagrams for Standard ases There are three standard cases of beam loading that are so common that the reader would be well advised to commit the results to memory. These are: eam with a central point load; eam with a non-central point load; eam carrying a uniformly distributed load over its entire length. These cases, along with their respective shear force and bending moment diagrams, are shown in Figs Using the techniques discussed above, you should be able to obtain these reactions and shear force and bending moment values for yourself.

6 P L R = P/2 (a) eam Diagram R = P/2 P/2 P/2 P/2 P/2 (b) Shear Force Diagram PL/4 (c) ending Moment Diagram Figure 4: Standard case 1: Shear force and bending moment diagrams for a beam carrying a central point load

7 a P b L R = Pb/L (a) eam Diagram R = Pa/L Pb/L Pb/L Pa/L Pa/L (b) Shear Force Diagram Pab/L (c) ending Moment Diagram Figure 5: Standard case 2: Shear force and ending moment diagrams for a beam carrying a non-central point load

8 Note that the result for the maximum bending moment in a beam with a uniformly distributed load over its entire length ( ) is particularly commonly used in practise. w kn/m L m R =(wl ) (a) eam Diagram R = (wl ) (wl ) (wl ) (b) Shear Force Diagram (wl ) (c) ending Moment Diagram Figure 6: Standard case 3: Shear force and bending moment diagrams for a beam carrying distributed load over its entire length

9 More Examples Involving Uniformly Distributed Loads Draw the shear force and bending moment diagrams for each of the beams shown in Fig. 7. The solutions are given during the class. 6 kn/m 3 m 3 m R (a) R 5 kn 5 kn 1 kn/m 3 kn/m 4 m 6 m 5 m R (b) R 12 kn/m 5 kn 5 kn D E 3 m 1 m 1 m 1 m R (c) R E Figure 7: Further shear force and bending moment diagram examples