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Chapter 2

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.

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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