2. action of beams involves combination of bending and shear Beams 2/39 devices for transferring vertical loads horizontally

12. What the Loads Do (cont.) 12/39 e Bending e e e C T Shear

16. Take section anywhere at distance, x from end Shear V = W constant along length (X = 0 -> L) Add in forces, V = W and moment M = - Wx Bending Moment BM = W.x when x = L BM = WL when x = 0 BM = 0 Example 1 - Cantilever Beam Point Load at End (cont1.) 16/39 V = W Shear Force Diagram Bending Moment Diagram BM = WL BM = Wx x W V = W M = -Wx

17. For maximum shear V and bending moment BM vertical reaction, R = W = wL and moment reaction M R = - WL/2 = - wL 2 /2 Example 2 - Cantilever Beam Uniformly Distributed Load 17/39 L w /unit length Total Load W = w.L L/2 L/2 R = W = wL M R = - WL/2 = - wL 2 /2

18. Take section anywhere at distance, x from end Shear V = wx when x = L V = W = wL when x = 0 V = 0 Add in forces, V = w.x and moment M = - wx.x/2 Bending Moment BM = w.x 2 /2 when x = L BM = wL 2 /2 = WL/2 when x = 0 BM = 0 (parabolic) For distributed V and BM Example 2 - Cantilever Beam Uniformly Distributed Load (cont.) 18/39 V = wL = W Shear Force Diagram X/2 wx X/2 V = wx M = -wx 2 /2 Bending Moment Diagram BM = wL 2 /2 = WL/2 BM = wx /2 2

24. Sign Conventions Bending Moment Diagrams (cont.) 24/39 Sagging bending moment is POSITIVE (happy) + Hogging bending moment is NEGATIVE (sad) -

33. Standard BM Coefficients Cantilevers 33/39 End point load Max bending moment = -WL Uniformly Distributed Load Max bending moment = -WL/2 or - w L 2 /2 where W = w L W L ( w per metre length) Total load = W L

34. Standard BM Coefficients Simple Beams Beam Cable BMD 34/39 W W W W W W W W W W W W W /m W /m

35. SFD & BMD Simply Supported Beams 35/39 V = +W M max = -WL L W W L V = +W/2 V = -W/2 M max = WL/4 L W = w L V = +W/2 V = -W/2 M max = WL/8 = w L 2 /8 L W = w L V = +W M max = -WL/2 = - w L 2 /2

39. next lecture Finding Section Properties 39/39 we need to find the Section Properties