written 3.9 years ago by
teamques10
★ 57k

•
modified 3.9 years ago

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’