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## Q. 2.10

Solve Example 2.9 using Mohr’s circle.

## Verified Solution

Plot principal stress points (160, 0) and (40, 0) and draw a circle through its center C as shown in Fig. 2.17. The resultant is given by, tangent OD = 80 MPa. Its inclination with its normal = $\phi$ and with major principal stress 160 MPa = $\theta$ , can be read from Fig. 2.17. Angle of inclination of resultant stress $\sigma _r$ with normal stress $\sigma _x$,

$\sin \phi=\frac{D C}{O C}=\frac{60}{100}=0.6$ $\phi=36.874$

Inclination with major principal plane

$\theta=\frac{1}{2}\left(90^{\circ}+\phi\right)=63.435^{\circ}$