278 Mechanics of Materials the mass falls)it can,for all practical purposes,be ignored in the above calculation: deflection produced (3) 200×109 i.e. elongation of bar =1.26 mm (b)Consider the loading system shown in Fig.11.8.Let WE be the equivalent force that produces the same deflection and stress when gradually applied as that produced by the falling mass. Then WE Ws maxδ where W is a known load,gradually applied to the beam at mid-span,producing deflection and stress os Then _Wed,_WE×5×10-3 W 1×103 5 dmn=10s形e Now wh+δ)=2 1[品+]- L10+10] 12 500WE_2.5W星 106 106 经 500WE1.2×10 2.5 =0 2.5 and W2-200WE-0.48×106=0 By factors, WE=800N or -600N WE=800N By proportion Os G max W WE and the maximum stress is given by 158×10×800 W 1×103 =126.4MN/m2 WE。HW And since278 Mechanics of Materials the mass falls) it can, for all practical purposes, be ignored in the above calculation: aL 97.18 x 2.6 x lo6 deflection produced (6) = - = E 200 x 109 i.e. elongation of bar = 1.26mm (b) Consider the loading system shown in Fig. 11.8. Let WE be the equivalent force that produces the same deflection and stress when gradually applied as that produced by the falling mass. Then ws 6max 6s -=- where W, is a known load, gradually applied to the beam at mid-span, producing deflection 6, and stress a,. Then w~6, WEX 5 x a,=- - K 1 x 103 .. SOOWE 2.5 W’, 10 10. 1.2+-3-=- By factors, w~=800N or -6oN .. WE = 800N By proportion 6s amax -=- WE and the maximum stress is given by And since WE -=- 6 6