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An element in a stressed body is subjected to normal stresses on mutually perpendicular directions as shown in fig.

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.

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From above figure, we get,

P1=100 Mpa (tensile)

P2=60 Mpa (tensile)

θ=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°

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