Question 1 |
In a two-dimensional stress analysis, the state of stress at a point is shown in the figure. The values of length of PQ, QR, and RP are 4, 3, and 5 units, respectively. The principal stresses are (round off to one decimal place)
\sigma_{x}=26.7 \mathrm{MPa}, \alpha_{y}=172.5 \mathrm{MPa} | |
\sigma_{x}=54.0 \mathrm{MPa}, \sigma_{y}=128.5 \mathrm{MPa} | |
\sigma_{x}=67.5 \mathrm{MPa}, \sigma_{y}=213.3 \mathrm{MPa} | |
\sigma_{x}=16.0 \mathrm{MPa}, \sigma_{y}=138.5 \mathrm{MPa} |
Question 1 Explanation:
\Rightarrow \quad \tan \theta=\frac{\mathrm{QR}}{\mathrm{PQ}}=\frac{3}{4}
\Rightarrow \quad \theta=\tan ^{-1}\left(\frac{3}{4}\right)=36.87^{\circ}
Using transformation equations:
\sigma_{x^{\prime}}=\frac{\sigma_{x}+\sigma_{y}}{2}+\left(\frac{\sigma_{x}-\sigma_{y}}{2}\right) \cos 2 \theta
120=\frac{\sigma_{x}+\sigma_{y}}{2}+\left(\frac{\sigma_{x}-\sigma_{y}}{2}\right) \cos \left(2 * 36.87^{\circ}\right)
\tau_{x^{\prime} y^{\prime}}=-\left(\frac{\sigma_{x}-\sigma_{y}}{2}\right) \sin \left(2 * 36.87^{\circ}\right)
From above equations, we get
\begin{aligned} & \sigma_{x}=67.5 \mathrm{MPa} \\ & \sigma_{y}=213.3 \mathrm{MPa} \end{aligned}
Question 2 |
The infinitesimal element shown in the figure (not to scale) represents the state of stress at a point in a body. What is the magnitude of the maximum principal stress (in \mathrm{N} / \mathrm{mm}^{2}, in integer) at the point?
4 | |
5 | |
7 | |
9 |
Question 2 Explanation:
On plane x^{\prime} \rightarrow \quad \sigma_{x^{\prime}}=5 \mathrm{~N} / \mathrm{mm}^{2}
\tau_{x^{\prime} y^{\prime}}=4 \mathrm{~N} / \mathrm{mm}^{2}
Using transformation equations:
\begin{aligned} & \sigma_{x^{\prime}}=\left(\frac{\sigma_{x}+\sigma_{y}}{2}\right)+\left(\frac{\sigma_{x}-\sigma_{y}}{2}\right) \cos 2 \theta+\tau_{x y} \sin 2 \theta \\ & 5=\left(\frac{\sigma_{x}+6}{2}\right)+\left(\frac{\sigma_{x}-6}{2}\right) \cos \left(2 \times 45^{\circ}\right)+3 \sin \left(2 \times 45^{\circ}\right) \\ & \Rightarrow \sigma_{x}=-2 \mathrm{~N} / \mathrm{mm}^{2} \\ & \sigma_{\text {major } / \text { minor }}=\frac{\sigma_{x}+\sigma_{y}}{2} \pm \sqrt{\left(\frac{\sigma_{x}-\sigma_{y}}{2}\right)^{2}+\tau_{x y}^{2}} \end{aligned}
\sigma_{\text {major } / \text { minor }}=\frac{-2+6}{2} \pm \sqrt{\left(\frac{-2-6}{2}\right)^{2}+3^{2}}
\sigma_{\text {major } / \text { minor }}=2 \pm 5
\Rightarrow \quad \sigma_{\text {major }}=7 \mathrm{~N} / \mathrm{mm}^{2}, \sigma_{\text {minor }}=-3 \mathrm{MPa}
Hence, magnitude of maximum principal stress is 7 \mathrm{~N} / \mathrm{mm}^{2}.
Question 3 |
A hanger is made of two bars of different sizes. Each bar has a square cross-section. The hanger is loaded by three-point loads in the mid vertical plane as shown in the figure. Ignore the self-weight of the hanger. What is the maximum tensile stress in \mathrm{N} / \mathrm{mm}^{2} anywhere in the hanger without considering stress concentration effects?
15 | |
25 | |
35 | |
45 |
Question 3 Explanation:
\sigma_{A B}=\frac{P_{A B}}{A_{A B}}=\frac{250 \times 10^{3}}{100 \times 100}=25 \mathrm{~N} / \mathrm{mm}^{2}
\sigma_{B C}=\frac{P_{B C}}{A_{B C}}=\frac{50 \times 10^3}{50 \times 50}=20 \mathrm{N} / \mathrm{mm}^{2}
\sigma_{\max }=\sigma_{\mathrm{AB}}=25 \mathrm{~N} / \mathrm{mm}^{2}
Question 4 |
Stresses acting on an infinitesimal soil element are shown in the figure (with \sigma _z \gt \sigma _x). The major and minor principal stresses are \sigma _1
and \sigma _3, respectively. Considering the compressive stresses as positive, which one of the
following expressions correctly represents the angle between the major
principal stress plane and the horizontal plane?
\tan ^{-1}\left ( \frac{\tau _{zx}}{\sigma _1-\sigma _x} \right ) | |
\tan ^{-1}\left ( \frac{\tau _{zx}}{\sigma _3-\sigma _x} \right ) | |
\tan ^{-1}\left ( \frac{\tau _{zx}}{\sigma _1+\sigma _x} \right ) | |
\tan ^{-1}\left ( \frac{\tau _{zx}}{\sigma _1+\sigma _3} \right ) |
Question 4 Explanation:
\begin{aligned} \Sigma F_x &=0 \\ \sigma _x(BC)-\tau _Z \times (AB)\sigma _1 \sin \theta &= 0\\ \sigma _x\left ( \frac{AC \sin \theta }{\cos \theta } \right )+\tau _{zx}\left ( \frac{AC \cos \alpha }{\cos \theta } \right ) &=\sigma _1 \frac{AC \sin \theta }{\cos \theta }\\ \sigma _x \tan \theta +\tau _{zx} &=\sigma _1 \tan \theta \\ \tan \theta(\sigma _1-\sigma _2) &= \tau _{zx}\\ \tan \theta &= \left ( \frac{\tau _{zx}}{\sigma _1-\sigma _x} \right ) \end{aligned}
Question 5 |
The state of stress in a deformable body is shown in the figure. Consider transformation of the stress from the x-y coordinate system to the X-Y coordinate system. The angle \theta, locating the X-axis, is assumed to be positive when measured from the x-axis in counter-clockwise direction.
The absolute magnitude of the shear stress component \sigma_{\mathrm{xy}} (in MPa,round off to one decimal place) in x-y coordinate system is ________________
The absolute magnitude of the shear stress component \sigma_{\mathrm{xy}} (in MPa,round off to one decimal place) in x-y coordinate system is ________________
96.2 | |
54.6 | |
48.2 | |
28.7 |
Question 5 Explanation:
\begin{aligned} \sigma_{x}^{\prime}&=\frac{\sigma_{x}+\sigma_{y}}{2}+\left(\frac{\sigma_{x}-\sigma_{y}}{2}\right) \cos 2 \theta+\tau_{x y} \sin 2 \theta \\ \text { Here } \theta=60^{\circ} \\ \sigma_{x}&=40 \mathrm{MPa}, \sigma_{y}=35.6, \sigma_{x}^{\prime}=120, \tau_{x^{\prime} y^{\prime}}=-50 \end{aligned}
Substituting the values in above equation, we get
\tau_{x y}=96.186 \mathrm{MPa}
There are 5 questions to complete.
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Didn’t understand question 4