Question

XC Question Completion Status: a. Determine the equivalent states of stress on an element at the same point oriented 30 counterclockwise with respect to the element shown. 75MPa 040 MPa + ↑ 35 MPa ox'= Oxtoy 2 Ox-by cos 20+ Key since 2 oyz oxtoy 2 Xxy'= - Ox Oy Oxby cos 20 - Exysinal sin 20+ Teu cos 20 2

Answer 1

Stress is - 75 MPa Solution Given system of Normal stress in x-direction (ox) = + 40 MPa Normal stress in y direction (og) Shear stress (Try-Tyx) = – 35 MPa sign convention - Tensile stress is taken as positive compressive stress is taken as negative бу = = 75 MPa ►ox- ox=40 MPa łny = 35 mpa Stresses at a plane which is inclined at an angle 0 = 30° counteyclockwise direction is ontby + ou by costo + singo Tny Ta' 2. 40+(-75) + 40-(-15) cos (2x30)+(-35) sin(2x30) Ox' 40475) २ ha's -17.5 + $8.75 - 30.31 19.06 MPa oi shows it is compressive Negative sign

Stress in Og' Entry - On-by Losao y direction is given by - Try singo Sy's 40+(-75) og' 60 40-(-75) LS (60) - (-35) sin 2 -17.5 28.75 + 30.31 - 15.94 Mpa oy This stress is also compressive in nature is shear stress given by Los go Tay Tny' 2 - Om-) singo + Try' = -(40-4-75)) sin60° + 1-35) (0560 २ - 49.80 -17.5 Tny' Try' - 67.30 MPa Thus, stress on a plane which is in clined at 30° counterclockwise are - ox' - 19.06 MPa - 15.94 MPa oyl 67,30 MPa Tay'

The correct representation is ox' Exy Ex бу! The drawn stress diagram in the question is wrong due to following following reason U normal stress in direction is shown as tensile whereas it is compressive (ii shear stress representation is shown as positive direction whereas it is negative