Question

Problem 7 Determine the normal stress and shear stress acting on the inclined plane AB given: Ox = 45 psi Oy = 85 psi Txy = 15 psi O is 40 degrees (Use the stress orientation shown in the figure to determine the sign on the stress values.) a) normal stress: -56 psi, shear stress: 82 psi b) normal stress: -28 psi, shear stress: 75 psi c) normal stress: 17 psi, shear stress: 67 psi d) normal stress: 18 psi, shear stress: -31 psi e) normal stress: 30 psi, shear stress: -31 psi f) normal stress: 56 psi, shear stress: -67 psi

Answer 1

Solution. - Given : = 45 bsi Ty = 85 psi Tay = 15 psi 0 = 20 Take an element of thickesst. A ty eta 16 Distributing forces on axis 4 Ju (ABxt) CACxt) AL In (Acxt Exy (BCxt) Now Take Components along on aris. (Acxt) + (Box) Sino + Czy (ABYB) Sino + xy CBCxo caso - Oy CABxH) Coso so

TA' AC = Ty (AB) COJO – VCBC) Sino – Try (AB) Sin 9 – GeyCBC) Cose 5) on = my (AMES) Cson Tal (BC) sine - Txy ( 206 sino - Tag ( e ) Cose from tigere, cos6 = mesing = On = Jy costo - Jy sin 20 - 2 By Sino. Caso In = Jy Cos? 0-0x Sin 20-tay Sinne On = 05 Cos? 40-45 sine 40 - 15 Sin 80 In = 49.80-18.593-14.772 In = 16.515 psi Now Take components along to axis - YOCACYD)+7x(BCX®) Cos 0+Gry (ABXH) Cos 0 + Oy (ABXH) Sino Try (BCY!) Sino = 0 TO (AO) = - 0x (BO) Ceso – Oy (AB) Sine) – Gary (16) Ces0 +Lwy (BC) Sino 4 = -93 (Beceso - ory (ne sino-Wyl AccesO 4 Txy(Be)sino 7 = -02 Sino-Coso-ag caso sino - Cry Cas20 + Tag Sin? Toro Sim20-gory Singe - Gry CoIO Yo---(45) Sin 80 - $ (85)Sin(80) – 15 cos ao 17 = -99.158 - 41.054 - 9.6097--66.617 ppi

ove sign denotes, direction assumed wrong, so of to is a In So I n = 16.515 bsi s 17bsi To = 66.617 bsi & 67psi so option © is correct.