Question 4.8: Direct tensile stresses of 120 MN/m² and 70 MN/m² act on a b...
Direct tensile stresses of 120 MN/m² and 70 MN/m² act on a body on mutually perpendicular planes. What is the magnitude of shearing stress that can be applied so that major principal stress at the point does not exceed 135 MN/m²? Determine the value of major principal stress and the maximum shear stress.
Let us refer to Figure 4.18(a) of the previous problem in which we have
\sigma_{x x}=+120 MN / m ^2 \quad \text { and } \quad \sigma_{y y}=+70 MN / m ^2
Let \tau_{x y} be the shear stress acting on the body. Therefore,
\begin{aligned} \sigma_1 & =\frac{1}{2}\left(\sigma_{x x}+\sigma_{y y}\right)+\sqrt{\left\lgroup \frac{\sigma_{x x}-\sigma_{y y}}{2} \right\rgroup^2+\tau_{x y}^2} \\ 135 & =0.5(120+70)+\sqrt{\{0.5(120-70)\}^2+\tau_{x y}^2} \end{aligned}
This implies
\tau_{x y}=\pm 31.22 MPa \Rightarrow\left|\tau_{x y}\right|=31.22 MN / m ^2
Let \sigma_2 be the other principal stress, then
\sigma_1+\sigma_2=\sigma_{x x}+\sigma_{y y}=120+70=190
or \sigma_2=190-135=55 MPa
If \tau_{\max } be the maximum shear stress, then
\tau_{\max }=\sqrt{\left\lgroup \frac{\sigma_{x x}-\sigma_{y y}}{2} \right\rgroup^2+\tau_{x y}^2}
or \tau_{\max }=\sigma_1-\frac{1}{2}\left(\sigma_{x x}+\sigma_{y y}\right)=[135-0.5(120+70)]=40 MPa
