Q21GATE 2005MCQ

If principal stresses in a two-dimensional case are -10 MPa and 20 MPa respectively, then maximum shear stress at the point is

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📖 Explanation

\begin{aligned} \text { Maximum shear stress } &=\frac{\sigma_{1}-\sigma_{2}}{2} \\ &=\frac{20-(-10)}{2}=15 \mathrm{MPa} \end{aligned}

Q22GATE 2004MCQ

In a two dimensional stress analysis, the state of stress at a point is shown below. If $\sigma$ = 120 MPa and $\tau$ = 70 MPa, then $\sigma _{x} \; and \; \sigma _{y}$ are respectively

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📖 Explanation

\begin{array}{l} \text { Let } \angle \mathrm{CAB}=\theta \\ \therefore \sin \theta=\frac{3}{5} ; \cos \theta=\frac{4}{5} ; \tan \theta=\frac{3}{4} \end{array}

Thus from force equilibrium,

\begin{aligned} \sigma_{x} \times A B&=A C \times (\sigma \cos \theta-\tau \sin \theta)\\ \Rightarrow \quad \sigma_{x}&=\frac{5}{4} \times \left(120 \times \frac{4}{5}-70 \times \frac{3}{5}\right) \\ \Rightarrow \quad \sigma_{x}&=67.5 \mathrm{MPa} \\ \text{And, }\sigma_{y} \times B C&=A C \times (\sigma \sin \theta+\tau \cos \theta) \\ \Rightarrow \quad \sigma_{y}&=\frac{5}{3} \times \left(120 \times \frac{3}{5}+70 \times \frac{4}{5}\right) \\ \Rightarrow \quad \sigma_{y}&=213.3 \mathrm{MPa} \end{aligned}

Q23GATE 2003MCQ

The state of two dimensional stresses acting on a concrete lamina consists of a direct tensile stress, $\sigma _{x}$ =1.5 $N/mm^{2}$ , and shear stress, $\tau$ = 1.20 $N/mm^{2}$ , which cause cracking of concrete. Then the tensile strength of the concrete in $N/mm^{2}$ is

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📖 Explanation

Maximum principal stress

\begin{aligned} &=\frac{\sigma_{\mathrm{x}}}{2}+\sqrt{\left(\frac{\sigma_{\mathrm{x}}}{2}\right)^{2}+\tau^{2}} \\ &=\frac{1.5}{2}+\sqrt{\left(\frac{1.5}{2}\right)^{2}+(1.20)^{2}} \\ &=2.17 \mathrm{N} / \mathrm{mm}^{2} \end{aligned}