An element in a stressed body is subjected to normal stresses on mutually perpendicular directions as shown in fig. Determine the normal tangential and resultant stresses on plane P-P inclined as shown. Either use analytical method or graphical method.

From above figure, we get,

$P_1 = 100 \ Mpa$ (tensile)

$P_2 = 60 \ Mpa$ (tensile)

$\theta = 30°$

A] Normal stress on P – P Plane.

$\sigma_n = \frac{P_1 + P_2}{2} + (\frac{p_1 –p_2}{2}) \ cos \ 2 \ \theta $

$= \frac{100 + 60}{2} + (\frac{100 – 60}{2}) \ cos (2 \times 30)$

$\sigma_n = 91.75 \ Mpa$ (tensile).

B] Tangential stress on P – P Plane.

$\sigma_t = (\frac{p_1 – p_2}{2}) sin \ 2 \ \theta$

$= (\frac{100 – 60}{2}) \ sin (2 \times 30) $

$\sigma_t = 16.18 \ Mpa$

C] Resultant stress and its direction.

$\sigma_R = \sqrt{\sigma n^2 + \sigma t^2} = 93.27 \ Mpa$

$\phi = tan^{-1} (\frac{\sigma t}{\sigma n}) = 11.11°$