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The principal stresses at a point across two perpendicular planes are 100 MPa (tensile and horizontal) and 60 MPa (tensile and vertical)

The principal stresses at a point across two perpendicular planes are 100 MPa (tensile and horizontal) and 60 MPa (tensile and vertical). Determine the normal shear and resultant stresses in magnitude and direction in a plane, the normal of which makes an angle of 30 degree with the direction of maximum principal stress. Use either analytical method or graphical method.

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Analytical Method:-

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Data:

Principal stresses:

$P_1 = 100 \ Mpa$ (tensile)

$P_2 = 60 \ Mpa$ (tensile)

The above fig. shows the principal stresses at the point. On a plane at 30° with the maximum principal plane.

A] Normal stress is $\sigma_n/Pn$

$\therefore \ Pn/ \sigma_n = \frac{P_1 + P_2}{2} + [\frac{P_1 – P_2}{2}] \ cos \ 2 \ \theta$

$= \frac{100 + 60}{2} + [\frac{100 – 60}{2}] \ cos \ 2 \times 30$

$\sigma_n = 90 \ Mpa$ (tensile).

B] Tangential stress is $\sigma_t / pt$

$pt / \sigma_t = [ \frac{P_1 – P_2}{2}] \ sin \ 2 \ \theta = (\frac{100 – 60}{2}) \ sin \ 2 \times 30$

$\sigma_t = 17.32 \ Mpa$

C] Resultant stress is $\sigma R/P_R$

$\therefore \ \sigma_R = \sqrt{\sigma_{n^2} + \sigma_{t^2}}$

$\sigma_R = \sqrt{90^2 + 17.32^2} = 91.65 \ Mpa$

And its inclination $\phi = tan^{-1} (\frac{\sigma_t}{\sigma_n})$

$tan^{-1} (\frac{17.32}{90})$

$\phi = 10.89°$ or 10°53’