$2.9 Struts 43 where n=√(P/E) This is a second-order differential equation,the solution of which is as follows: y=Asin nx +Bcosnx-e Now when x=0,y=0 B=e L dy and when x= 2'd =0 0=nA cosn2 -ne sinn nL A e tan- 2 nL y+e=etan2 sinnx+ecosnx .'maximum deflection,when x=L/2 and y=8,is sin2 nL ste=e-nt nL +ecos 2 CoS 2 Sin2 nt +cos2 nL 2 2 nL =e (2.23) nL =esec- cos 2 nL maximum B.M.P(8+e)=Pe sec (2.24) 2 nL h maximum stress owing to bending My Pesec"2x 1 where h is the distance from the N.A.to the highest stressed fibre. Therefore the total maximum compressive stress owing to combined bending and thrust, assuming a ductile materialT,is given by a-+(e）)月 P =l+袋e (2.25) +袋() For a brittle material which is relatively weak in tension it is the maximum tensile stress which becomes the criterion of failure and the bending and direct stress components are opposite in sign.$2.9 Struts 43 where n = J(P/EI) This is a second-order differential equation, the solution of which is as follows: y = Asin nx + Bcosnx - e Now when x = 0, y = 0 .. B=e L dy and when x = - - = 0 2’ dx .. .. :. maximum deflection, when L L 0 = nAcosn- - nesinn- 2 2 nL A = e tan - 2 nL . y + e = etan - sinnx + ecosnx 2 x = L/2 and y = 8, is nL maximum B.M. = P(6 + e) = Pe sec - 2 .. (2.24) MY nL h maximum stress owing to bending = - = Pesec - x - I 21 .. where h is the distance from the N.A. to the highest stressed fibre. assuming a ductile material?, is given by Therefore the total maximum compressive stress owing to combined bending and thrust, (2.25) For a brittle material which is relatively weak in tension it is the maximum tensile stress which becomes the criterion of failure and the bending and direct stress components are opposite in sign