40 Mechanics of Materials 2 $2.6 Therefore dividing through,top and bottom,by A, y=[(co）/(设-】 But P/A =o and (2E/)/(L2A)=oe(the Euler stress for pin-ended struts) y= (o。-g) Therefore total deflection at any point is given by πr y+0= (oe-o)」 Cocos +Cocos L L(oe-o) Co cosL (2.13) Maximum deflection (when x =0)= Co (2.14) L(Oe-o) maximum B.M.=P (2.15) (e-0) My P maximum stress owing to bending= 1(oe-0) where h is the distance of the outside fibre from the N.A.of the strut. Therefore the maximum stress owing to combined bending and thrust is given by P[Oe P Omax Coh+ (2.16) Γ1(oe-o)] A P「oe P Ak2(oe-)] CohA noe Coh =0 +1 (e-a) where n= k2 If omax=o.the compressive yield stress for the material of the strut,the above equation when solved for o gives 。=l,++1oJ (2.17) 2 This is the Perry-Robertson formula required.If the material is brittle,however,and failure is likely to occur in tension,then the sign between the two square-bracketed terms becomes positive and ov is the tensile yield strength.40 Mechanics of Materials 2 $2.6 Therefore dividing through, top and bottom, by A, y = [ ( fCo cos 7) / (g - 31 But PIA = a and (r2EZ)/(L2A) = a, (the Euler stress for pin-ended struts) Therefore total deflection at any point is given by .. Maximum deflection (when x = 0) = maximum B.M. = P ~ [(De: a)] c0 maximum stress owing to bending = 9 = [A] Coh I I (ae - a) where h is the distance of the outside fibre from the N.A. of the strut. Therefore the maximum stress owing to combined bending and thrust is given by (2.13) (2.14) (2.15) (2.16) If amax = a,,, the compressive yield stress for the material of the strut, the above equation when solved for a gives (2.17) This is the Perry-Robertson formula required. If the material is brittle, however, and failure is likely to occur in tension, then the sign between the two square-bracketed terms becomes positive and av is the tensile yield strength