The stress system at the point in the material may be represented as shown in Fig. 1.13 by considering the stresses to act uniformly over the sides of a triangular element ABC of unit thickness. Suppose that the direct stress on the principal plane AB is \sigma . For horizontal equilibrium of the element,
\sigma AB \cos \theta=\sigma_{x} BC +\tau_{x y} AC
which simplifies to
\tau_{x y} \tan \theta=\sigma-\sigma_{x} (i)
Considering vertical equilibrium gives
\sigma AB \sin \theta=\sigma_{y} AC +\tau_{x y} BC
or
\tau_{x y} \cot \theta=\sigma-\sigma_{y} (ii)
Hence, from the product of Eqs. (i) and (ii),
\tau_{x y}^{2}=\left(\sigma-\sigma_{x}\right)\left(\sigma-\sigma_{y}\right)
Now, substituting the values \sigma_{x}=160 N / mm ^{2}, \sigma_{y}=-120 N / mm ^{2}, \text { and } \sigma=\sigma_{1}=200 N / mm ^{2}, we have
\tau_{x y}=\pm 113 N / mm ^{2}
Replacing cot\theta in Eq. (ii) with 1//tan\theta from Eq. (i) yields a quadratic equation in \sigma:
\sigma^{2}-\sigma\left(\sigma_{x}-\sigma_{y}\right)+\sigma_{x} \sigma_{y}-\tau_{x y}^{2}=0 (iii)
The numerical solutions of Eq. (iii) corresponding to the given values of \sigma_{x}, \sigma_{y} and \tau_{x y} are the principal stresses at the point, namely,
\sigma_{1}=200 N / mm ^{2}
given
\sigma_{ II }=-160 N / mm ^{2}
Having obtained the principal stresses, we now use Eq. (1.15) to find the maximum shear stress, thus
\tau_{\max }=\frac{\sigma_{ I }-\sigma_{ II }}{2} (1.15)
\tau_{\max }=\frac{200+160}{2}=180 N / mm ^{2}
The solution is rapidly verified from Mohr’s circle of stress (Fig. 1.14). From the arbitrary origin O , OP _{1}, and OP _{2} are drawn to represent \sigma_{x}=160 N / mm ^{2} and \sigma_{y}=-120 N / mm ^{2}. The mid-point C of P _{1} P _{2} is then located. Next, OB =\sigma_{1}=200 N / mm ^{2} is marked out and the radius of the circle is then CB. OA is the required principal stress. Perpendiculars P _{1} Q _{1} and P _{2} Q _{2} to the circumference of the circle are equal to \pm \tau_{x y} (to scale), and the radius of the circle is the maximum shear stress.
Using the element shown in Fig. 1.13, derivations of the principal stresses and maximum shear stress are obtained through the following MATLAB file:
