Question

For the state of stress shown in figure (a=55 degrees), determine (a) The normal and shear stress acting along the oblique face of the shaded triangle (b) The state of stress, if the element were rotated by an angle 55 degrees clockwise (c) The state of stress acting on the principal normal planes (d) The state of stress acting on the principle shear planes 60 MPa - 80 MPa — x 90 MPa

Answer 1

compo Bompa Jeord gompa Q = - 80 MPa oomla S exy = gompa at oblique klone 2=550 nomal stress no = Story = -80-60 Bury copad + Txy lined - 80+60 cos110° + 90 sin too = -70 - 10000 110° + 90 sio 1100 =-70-3.4202 + 84.572 I no = 11.159 MB shear spend at oblique klane d=550 e = - ( q ry) sinh t try cosza

Date -(-80+60 sinloo t go woll0° - 10 sm 110.+ 90 cablloo Io = -21,384 mpg] ( state of stress, at oblique blone moa. =?!.384mp In 24 some gompa Compa Principle stoess tanzo=1:1-Pyth & klones I a exy 60+80 9190 tanzo = 2xy = - 8x90 +80+ Go I tanzo = 180 PO O= 48017 8 = goto = 138.170

ба с алу , 5-(до с4 - онлаја за 1 -70 + 90.55 6 - 70 — 90. Ss 19, -50 55 тva) са - - По +90 ss L = rooss med 0 0 = -17° "ng = Fo-60 , -о- оо) од КҮЧ-17) - 907 «» = -10 -,- 99,чу | 60 = 0. 55 ml hеnсе d 0 = 98-17 , , occu.

Page No. Date. 48.17 Ky=80 mimor Philocibaſplane hence a major principal plane principal planes are 0 = 48.17°, 138-17° (minar) (major plane) Principal stepped are of = 160.55mpa compra 2 = 20.55 mpa tensile. I state of steps of shear tanao = yra 2 Oy strena a tanzo = -60+80 2xgo tango = Qygo 10. = 93.17° Tocation of maym shear strent plane 0 = 3.41° 493.14°

Page No. Date. Tmax = -(Gry canro + Cry Coxzo - --do-Go) đi (34 ) + No co (Rx37) 1:104282 + 89.49 Temax = 90.55 mpg / 90.55 Soossa 1.90 90 hence - at state of max 0=31170 & shear stress 931170 Tmax = 9o.sompa