Question 4.1: At a certain point in a strained material, there exists a st...
At a certain point in a strained material, there exists a stress condition as shown in Figure 4.8. Find
(a) the normal and shear stresses on the inclined plane AB;
(b) principal stresses and principal planes;
(c) maximum shear stresses and their planes.
We know that if θ be the angle made by the plane AB with the positive direction of the x-axis, then normal stress on plane AB, \sigma_{n n} using Eq. (4.10) is
\sigma_n=\frac{\sigma_{x x}+\sigma_{y y}}{2}+\frac{1}{2}\left(\sigma_{x x}-\sigma_{y y}\right) \cos 2 \phi-\tau_{x y} \sin 2 \phi (4.10)
\sigma_{n n}=\frac{1}{2}\left(\sigma_{x x}+\sigma_{y y}\right)+\frac{1}{2}\left(\sigma_{x x}-\sigma_{y y}\right) \cos 2 \theta-\tau_{x y} \sin 2 \theta (1)
And the shear stress on plane AB, \tau_{n t} from Eq. (4.11) is
\tau_n=\frac{\sigma_{x x}-\sigma_{y y}}{2} \sin 2 \phi+\tau_{x y} \cos 2 \phi (4.11)
\tau_{n t}=\frac{1}{2}\left(\sigma_{x x}-\sigma_{y y}\right) \sin 2 \theta+\tau_{x y} \cos 2 \theta (2)
(a) Now for the plane AB, clearly θ = 120°. Therefore,
\left.\sigma_{n n}\right|_{ AB }=\frac{1}{2}(-30+120)+\frac{1}{2}(-30-120) \cos 240^{\circ}-40 \sin 240^{\circ}=117.14 MPa
and \left.\tau_{n t}\right|_{ AB }=\frac{1}{2}(-30-120) \sin 240^{\circ}+40 \cos 240^{\circ}=45.0 MPa
(b) Equation (1) can be written as
\sigma_{n n}=\frac{1}{2}\left(\sigma_{x x}+\sigma_{y y}\right)+\sqrt{\frac{1}{4}\left(\sigma_{x x}+\sigma_{y y}\right)^2+\tau_{x y}^2} \cos (2 \theta+2 \phi)
where \tan 2 \phi=2 \tau_{x y} /\left(\sigma_{x x}-\sigma_{y y}\right) . Therefore,
\left(\sigma_{n n}\right)_{\max , \min }=\frac{1}{2}\left(\sigma_{x x}+\sigma_{y y}\right) \pm \sqrt{\frac{1}{4}\left(\sigma_{x x}+\sigma_{y y}\right)^2+\tau_{x y}^2}
And these stresses will occur when
2 \theta+2 \phi=0 \quad \text { or } \quad \theta=-\phi=-\frac{1}{2} \tan ^{-1}\left\lgroup \frac{2 \tau_{x y}}{\sigma_{x x}-\sigma_{y y}} \right\rgroup
and 2 \theta+2 \phi=\pi \quad \text { or } \quad \theta=\frac{\pi}{2}-\frac{1}{2} \tan ^{-1}\left\lgroup \frac{2 \tau_{x y}}{\sigma_{x x}-\sigma_{y y}} \right\rgroup
Therefore,
\left(\sigma_{n n}\right)_{\max }=\frac{1}{2}(-30+120)+\sqrt{\frac{1}{4}(-30-120)^2+40^2}=130 MPa
and \theta_{\max }=-\frac{1}{2} \tan ^{-1}\left(\frac{2 \times 40}{-30-120}\right) \Rightarrow \theta_{\max }=-75.96^{\circ} ⦪
Again, \left(\sigma_{n n}\right)_{\min }=\frac{1}{2}(-30+120)-\sqrt{\frac{1}{4}(-30-120)^2+40^2}=-40 MPa
with \theta_{\min }=90^{\circ}-75.96^{\circ} \Rightarrow \theta_{\min }=14.04^{\circ} ⦨
(c) From Eq. (2), it can be written as
\tau_{ nt }=\sqrt{\frac{1}{4}\left(\sigma_{x x}-\sigma_{y y}\right)^2+\tau_{x y}} \sin (2 \theta+2 \phi)
Substituting values, we get
\left.\tau_{ nt }\right|_{\max }=\sqrt{\frac{1}{4}\left(\sigma_{x x}-\sigma_{y y}\right)^2+\tau_{x y}^2}=85 MPa
with \left(\theta_s\right)_{\max }=\frac{\pi}{4}-\phi=45^{\circ}-75.96^{\circ} \Rightarrow\left(\theta_s\right)_{\max }=-30.96^{\circ} ⦪
The above results can be comprehensively represented in Mohr’s circle as shown in Figure 4.9.
