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## Q. 2.7

Two-dimensional stresses at a point on a plane P are 67.2 MPa (tension) and 84.64 MPa (shear, clockwise) and on another plane Q the stresses are 61.35 MPa (compression) and 68.57 MPa (shear, clockwise). Find the stresses on a plane R inclined at 20° clockwise to Q.

## Verified Solution

Express the stresses on plane P in terms of Eqs (2.1) and (2.2).

$\sigma _{x^{\prime}}=\frac{\sigma _x+\sigma _y}{2}+\frac{\sigma _x-\sigma _y}{2}\cos 2\theta -\tau _{xy}\sin 2 \theta$           (2.1b)

$\tau _{x^{\prime}y^{\prime}}=\frac{\sigma_x- \sigma_y}{2}\sin 2\theta +\tau _{xy}\cos 2\theta$                  (2.2)

$67.2= \frac{-61.35+\sigma_y}{2}+\frac{-61.35-\sigma_y}{2} \cos 2 \theta$

$-68.57 \sin 2 \theta$            (i)

$84.64= \frac{-61.35-\sigma_y}{2} \sin 2 \theta+68.57 \cos 2 \theta$           (ii)

where $\sigma_y$ = normal stress on the plane perpendicular to Q,

$\theta$ = angle between the planes P and Q.

From Eq. (i) and (ii),

$\tan \theta=-\frac{128.55}{153.21}$

or, $\theta \approx-40^{\circ}$,  i.e. plane P is at $40^{\circ}$ clockwise to plane Q.

Substituting in Eq. (iii), $\sigma_y=86.36 MPa.$

Angle of inclination of plane R w.r.t. Q,

$2 \theta=2 \times\left(-20^{\circ}\right)=-40^{\circ}$

Normal stress on plane R,

$\sigma_{\acute{x} }= \frac{-61.35+86.36}{2}-\frac{61.35+86.36}{2} \times \cos \left(-40^{\circ}\right)-68.57 \sin \left(-40^{\circ}\right) \simeq 0$

$\cos \left(-40^{\circ}\right)-68.57 \sin \left(-40^{\circ}\right) \simeq 0$

$\tau_{\acute{x} \acute{y}}= \frac{-61.35-86.36}{2} \sin \left(-40^{\circ}\right)+68.57 \cos \left(-40^{\circ}\right)=100 MPa$