Question

In a strained material, the principal stresses in the X and Y directions respectively are 100 $N/m{m}^2$ (Tensile) and 60 $N/m{m}^2$ (Compressive). On an inclined plane, the normal to which makes an angle of 30° to the X-axis, the major prinipal stress, in $N/m{m}^2$, would be

• A. 60
• B. 80
• C. 20
• D. 40

Answer 1

The correct option is A 60

Major principal stress on the inclined plane

${\sigma }_1=\frac{{\sigma }_x+{\sigma }_y}{2}+\frac{({\sigma }_x-{\sigma }_y)}{2}\cos 2\theta$

$\theta =30{}^{\circ },{\sigma }_x=100N/m{m}^2$

$and{\sigma }_y=-60N/m{m}^2$

$\sigma =\frac{100-60}{2}+\frac{(100+60)}{2}\cos 60{}^{\circ }$

$=60N/m{m}^2$