## Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## Tip our Team

Our Website is free to use.
To help us grow, you can support our team with a Small Tip.

## Holooly Tables

All the data tables that you may search for.

## Holooly Help Desk

Need Help? We got you covered.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Products

## Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

## Holooly Help Desk

Need Help? We got you covered.

## Q. 2.9

Tensile stresses of 160 MPa and 40 MPa are acting on two perpendicular planes in a body. Determine the location of a plane on which the resultant stress is most inclined to its normal.

## Verified Solution

Let the stresses be $\sigma_{\acute{x}}$ , and $\tau _{\acute{x}\acute{y}}$ on the plane at which the resultant stress $\sigma _r$ is most inclined to its normal. If the inclination of $\sigma_{\acute{x}}$ , is $\theta$ w.r.t. the stress 160 MPa, then,

$\sigma_{\acute{x}} =\frac{160+40}{2}+\frac{160-40}{2} \cos 2 \theta$

$=100+60 \cos 2 \theta$            (i)

$\tau_{\acute{_x} \acute{_y}} =\frac{160-40}{2} \sin 2 \theta=60 \sin 2 \theta$             (ii)

$\sigma_r=\sqrt{\sigma_ {\acute{x}}^2+\tau_{\acute{_x} \acute{_y}}^2}$

$=\sqrt{(100+60 \cos 2 \theta)^2+(60 \sin 2 \theta)^2}$         (iii)

Inclination of $\sigma_r$ with respect to normal $\sigma_{\acute{x}}$,

$\phi=\tan ^{-1} \frac{\tau_{\acute{x} \acute{y}}}{\sigma _{\acute{x}}}=\tan ^{-1} \frac{60 \sin 2 \theta}{100+60 \cos 2 \theta}$

For maximum inclination $\phi$, differentiate with respect to $\theta$,

$\frac{d \phi}{d \theta}=0=2 \times 3600+12000 \cos 2 \theta$

or, $\theta=63.435^{\circ}$, and corresponding $\phi=36.87^{\circ}$

Substituting $\theta$ in Eq. (iii), $\sigma_r$ = 80 MPa