Question 4.5: Two mutually perpendicular planes of an element are subjecte...
Two mutually perpendicular planes of an element are subjected to normal stresses of 10.5 MPa (tensile) and 3.5 MPa (compressive) and shear stress of 7 MPa (Figure 4.15). Find out the magnitudes and direction of principal stresses.
Here \sigma_x=10.5 MPa , \sigma_y=-3.5 MPa \text { and } \tau_{x y}=7.0 MPa . The principal stresses are given by
\sigma_{1,2}=\frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left\lgroup \frac{\sigma_x-\sigma_y}{2} \right\rgroup^2+\tau_{x y}^2}
Substituting the given values, we get
\begin{aligned} \sigma_{1,2} & =\frac{10.5-3.5}{2} \pm \sqrt{\left\lgroup \frac{10.5+3.5}{2}\right\rgroup^2+7.0^2} \\ & =3.5 \pm \sqrt{7^2+7^2}=3.5 \pm 9.9 MPa \end{aligned}
Therefore, \sigma_1=13.4 MPa \text { and } \sigma_2=-6.4 MPa . Hence, the maximum principal stress is 13.4 MPa and minimum is 6.4 MPa (compressive).
The directions of the principal planes are given by
\tan 2 \theta=\frac{2 \tau_{x y}}{\sigma_x-\sigma_y}
or 2 \theta=45^{\circ} \text { or } 225^{\circ}
Therefore, \theta_1=22^{\circ} 30^{\prime} \text { and } \theta_2=112^{\circ} 30^{\prime}