Question

A rectangular block of material is subjected to a tensile stress of 110 n/mm2 on one plane and a tensile stress of 47 n/mm2 on a plane at right angle to the former. Each of the above stresses is accompanies by a shear stress of 63 n/mm2 . Determine the principal stresses, principal planes and the maximum shear stress.

Answer 1

Principal stresses = $148.93 N/mm^2\textrm {and} \ 8.07 N/mm^2$

Principal plane = 0.2318(anticlock wise)

Principal shear stresses = $70.43 N/mm^2$

Step by step Explanation:

Given:

Let the stress in x direction, $\sigma_x = 110 N/mm^2$

The stress in y direction, $\sigma_y = 47 N/mm^2$

The shear stress,$\tau_{xy} = 63 N/mm^2$

Now we can determine the principal stresses as :

$\sigma_{1,2} = \dfrac{\sigma_x+\sigma_y} {2} \pm \sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}$

By substituting all the values we get:

$\sigma_{1,2} = \dfrac{110+47} {2} \pm \sqrt{\left(\dfrac{110-47}{2}\right)^2+63^2}\\\sigma_{1,2} = 78.5 \pm \sqrt{\left(31.5\right)^2+63^2}\\\sigma_{1,2} = 78.5 \pm \sqrt{4961.25}\\\\\sigma_{1,2} = 78.5 \pm 70.43\\\sigma_{1,2} = 148.93 N/mm^2, 8.07N/mm^2$

So we have principal stresses = $148.93 N/mm^2\textrm {and} \ 8.07 N/mm^2$

Now we can calculate the principal plane as:

$tan 2\theta_p = -\dfrac{\sigma_x-\sigma_y}{2}$

$tan2 \theta_p = -\dfrac{110-47}{2\times 63}\\tan2 \theta_p = -\dfrac{63}{2\times 63}\\tan2 \theta_p = -0.5\\2\theta_p = tan^{-1} (-0.5)\\2\theta_p = -0.4636\\\theta_p = \dfrac{-0.4636}{2}\\\theta_p = -0.2318^o$

We have principal plane = 0.2318(anticlock wise)

Maximum  shear stress can be calculated as:

$\tau_{max} = \pm{1}{2} (\sigma_1 - \sigma_2)\\\tau_{max} = \pm{1}{2} (148.93 - 8.07)\\\\\tau_{max} = \pm 70.43 N/mm^2$

We have principal shear stresses = $70.43 N/mm^2$