Question 1 |
A steel cubic block of side 200 mm is subjected to hydrostatic pressure of 250 N/mm^2. The elastic modulus is 2 \times 10^5 \; N/mm^2 and Poisson ratio is 0.3 for steel. The side of the block is reduced by _____mm (round off to two decimal places).
0.18 | |
0.22 | |
0.1 | |
0.05 |
Question 1 Explanation:
\begin{aligned} E &=200 \mathrm{GPa} \\ \sigma &=250 \mathrm{MPa} \\ \mu &=0.3 \\ \epsilon_{x} &=\epsilon_{y}=\epsilon_{z}=\frac{\delta a}{a} \\ \frac{1}{E}\left[\sigma_{x}-\mu\left(\sigma_{y}+\sigma_{z}\right)\right] &=\frac{(\delta a)}{a} \\ \left(\frac{-\sigma}{E}\right)(1-2 \mu)&=\frac{\delta a}{a}\\ \delta a &=-\frac{(250)(1-0.6)(200)}{200 \times 10^{3}} \\ \delta a &=(-) 0.10 \mathrm{~mm} \end{aligned}
Reduction in side of cube is 0.10mm.
Question 2 |
A wire of circular cross-section ofdiameter 1.0 mm is bent into a circular arc of radius 1.0 m by application of pure bending moments at its ends. The Young's modulus of the material of the wire is 100 GPa. The maximum tensile stress developed in the wire is ______MPa.
25 | |
50 | |
65 | |
85 |
Question 2 Explanation:
\begin{aligned} &\mathrm{d}=1.0 \mathrm{mm}\\ &\mathrm{y}_{\max }=\frac{\mathrm{d}}{2}=\frac{1.0}{2}=0.5 \mathrm{mm}\\ &\mathrm{R}=1.0 \mathrm{m}=1000 \mathrm{mm}\\ &\mathrm{E}=100 \mathrm{GPa}=100 \times 10^{3} \mathrm{MPa}\\ &\text { From bending equation, } \frac{\mathrm{f}_{\max }}{\mathrm{y}_{\max }}=\frac{\mathrm{E}}{\mathrm{R}}\\ &\mathrm{f}_{\max }=\mathrm{y}_{\max } \cdot \frac{\mathrm{E}}{\mathrm{R}}\\ &=0.5 \times \frac{100 \times 10^{3}}{1000}=50 \mathrm{MPa}\\ &\therefore \mathrm{f}_{\max }=50 \mathrm{MPa} \end{aligned}
Question 3 |
At a critical point in a component, the state of stress is given as \sigma _{xx}=100 MPa, \sigma _{yy}=220 MPa, \sigma _{xy}=\sigma _{yx}=80 MPa and all other stress components are zero. The yield strength of the material is 468 MPa. The factor of safety on the basis of maximum shear stress theory is ______ (round off to one decimal place).
1.7 | |
0.8 | |
1.2 | |
2.6 |
Question 3 Explanation:
\begin{array}{l} \sigma_{x}=100 \mathrm{MPa}, \sigma_{y y}=220 \mathrm{MPa}, \tau_{x y}=80 \mathrm{MPa} \\ S_{y t}=468 \mathrm{MPa}, \mathrm{F.S.}=?(\mathrm{MSST}) \\ \sigma_{1, 2}=\frac{\sigma_{x}+\sigma_{y}}{2} \pm \sqrt{\left(\frac{\sigma_{x}-\sigma_{y}}{2}\right)^{2}+\left(\tau_{x y}\right)^{2}} \\ =\frac{100+220}{2} \pm \sqrt{\left(\frac{100-220}{2}\right)^{2}+80^{2}} \\ \sigma_{1}=257.08 \mathrm{MPa}, \sigma_{2}=62.9 \mathrm{MPa}, \sigma_{3}=0 \end{array}
According to maximum shear stress theory,
\begin{array}{l} \tau_{\max }=\frac{S_{y t}}{2 \times F S}=\operatorname{Max}\left[\left|\frac{\sigma_{1}-\sigma_{2}}{2}\right|,\left|\frac{\sigma_{1}}{2}\right|,\left|\frac{\sigma_{2}}{2}\right|\right]\\ \tau_{\max }=\frac{257.08}{2}=128.54 \mathrm{MPa} \\ 128.54=\frac{468}{2 \times \mathrm{FS}} \\ \mathrm{FS}=1.78 \end{array}
According to maximum shear stress theory,
\begin{array}{l} \tau_{\max }=\frac{S_{y t}}{2 \times F S}=\operatorname{Max}\left[\left|\frac{\sigma_{1}-\sigma_{2}}{2}\right|,\left|\frac{\sigma_{1}}{2}\right|,\left|\frac{\sigma_{2}}{2}\right|\right]\\ \tau_{\max }=\frac{257.08}{2}=128.54 \mathrm{MPa} \\ 128.54=\frac{468}{2 \times \mathrm{FS}} \\ \mathrm{FS}=1.78 \end{array}
Question 4 |
In a linearly hardening plastic material, the true stress beyond initial yielding
increases linearly with the true strain | |
decreases linearly with the true strain | |
first increases linearly and then decreases linearly with the true strain
| |
remains constant |
Question 4 Explanation:
Question 5 |
A rectangular region in a solid is in a state of plane strain. The (x,y) coordinates of the corners of the under deformed rectangle are given by P(0,0), Q (4,0), S (0,3). The rectangle is subjected to uniform strains,\varepsilon _{xx}=0.001 , \varepsilon _{yy}=0.002,\gamma _{xy}=0.003.The deformed length of the elongated diagonal, up to three decimal places, is _________ units.
4.547 | |
4.845 | |
5.015 | |
6.348 |
Question 5 Explanation:
\tan \theta=\frac{4}{3}
\begin{aligned} \theta &=53.13^{\circ} \\ 2 \theta &=106.26^{\circ} \\ \cos 2 \theta &=-0.27999 \\ \sin 2 \theta &=0.96 \end{aligned}
Normal strain on a oblique plane (i.e. RP) inclined
at an angle (\theta) is given by
\begin{aligned} \left(\varepsilon_{n}\right)_{\theta}=& \frac{1}{2}\left[\varepsilon_{x}+\varepsilon_{y}\right]+\frac{1}{2}\left[\varepsilon_{x}-\varepsilon_{y}\right] \cos 2 \theta \\ & +\left(\frac{\gamma x y}{2}\right) \sin 2 \theta \\ =& \frac{1}{2}[0.001+0.002]+\frac{1}{2}[0.001-0.002] \\ &[-0.27999]+\left(\frac{0.003}{2}\right)(0.96) \\ &\left(\varepsilon_{n}\right)_{\theta=53.13^{\circ}}=0.00308 & \\ =& \frac{\text { Change in length of diagonal }}{\text { Original length of diagonal }}=0.00308 \end{aligned}
Hence, change in length of diagonal
=(0.00308) 5=0.0154 \mathrm{mm}
deformed length of diagonal
=5+0.0154=5.0154 \mathrm{mm}
\begin{aligned} \theta &=53.13^{\circ} \\ 2 \theta &=106.26^{\circ} \\ \cos 2 \theta &=-0.27999 \\ \sin 2 \theta &=0.96 \end{aligned}
Normal strain on a oblique plane (i.e. RP) inclined
at an angle (\theta) is given by
\begin{aligned} \left(\varepsilon_{n}\right)_{\theta}=& \frac{1}{2}\left[\varepsilon_{x}+\varepsilon_{y}\right]+\frac{1}{2}\left[\varepsilon_{x}-\varepsilon_{y}\right] \cos 2 \theta \\ & +\left(\frac{\gamma x y}{2}\right) \sin 2 \theta \\ =& \frac{1}{2}[0.001+0.002]+\frac{1}{2}[0.001-0.002] \\ &[-0.27999]+\left(\frac{0.003}{2}\right)(0.96) \\ &\left(\varepsilon_{n}\right)_{\theta=53.13^{\circ}}=0.00308 & \\ =& \frac{\text { Change in length of diagonal }}{\text { Original length of diagonal }}=0.00308 \end{aligned}
Hence, change in length of diagonal
=(0.00308) 5=0.0154 \mathrm{mm}
deformed length of diagonal
=5+0.0154=5.0154 \mathrm{mm}
Question 6 |
In the engineering stress-strain curve for mild steel, the Ultimate Tensile Strength (UTS) refers to
Yield stress | |
Proportional limit | |
Maximum stress | |
Fracture stress. |
Question 6 Explanation:
Ultimate tensile strength represents the max. stress that a material can withstand without fracture.
Question 7 |
In a metal forming operation when the material has just started yielding, the principal stresses are \sigma _{1}=+180 Mpa,\sigma _{2}=-100 Mpa,\sigma _{3}=0. Following Von Mises criterion, the yield stress is ________ MPa.
245.76 | |
240.12 | |
248.57 | |
251.98 |
Question 7 Explanation:
Safe condition for design as per von-Mises criterion,
\sigma_{1}^{2}+\sigma_{2}^{2}-\sigma_{1} \sigma_{2} \leq\left(\frac{S_{y t}}{N}\right)^{2}
When yielding occurs, N=1
\begin{array}{r} \sigma_{1}^{2}+\sigma_{2}^{2}-\sigma_{1} \sigma_{2}=\left(S_{y t}\right)^{2} \\ (180)^{2}+(-100)^{2}-(180)(-100)=\left(S_{y t}\right)^{2} \\ S_{t y}=245.7641 \mathrm{MPa} \end{array}
\sigma_{1}^{2}+\sigma_{2}^{2}-\sigma_{1} \sigma_{2} \leq\left(\frac{S_{y t}}{N}\right)^{2}
When yielding occurs, N=1
\begin{array}{r} \sigma_{1}^{2}+\sigma_{2}^{2}-\sigma_{1} \sigma_{2}=\left(S_{y t}\right)^{2} \\ (180)^{2}+(-100)^{2}-(180)(-100)=\left(S_{y t}\right)^{2} \\ S_{t y}=245.7641 \mathrm{MPa} \end{array}
Question 8 |
The poisson's ratio for a perfectly incompressible linear material is
1 | |
0.5 | |
0 | |
infinity |
Question 8 Explanation:
For perfectly incompressible material, poisson's ratio =0.5
(if change in vol. should be zero, \mu=0.5)
(if change in vol. should be zero, \mu=0.5)
Question 9 |
The state of stress at a point on an element is shown in figure (a). The same state of stress is shown in another coordinate system in figure (b).
The components (\tau _{xx},\tau _{yy},\tau _{xy}) are given by
The components (\tau _{xx},\tau _{yy},\tau _{xy}) are given by
(p/\sqrt{2},-p/\sqrt{2},0 ) | |
(0,0,p ) | |
(p,-p,p/\sqrt{2} ) | |
(0,0,p/\sqrt{2} ) |
Question 9 Explanation:
The given plane is principal plane
\left(\sigma_{1}=P, \sigma_{2}=-P\right) . At 45^{\circ} from principal plane, plane of max shear occurs. On the plane of max shear.
\begin{aligned} \operatorname{both}, \quad \tau_{x y} &=\tau_{y y}=\frac{\sigma_{1}+\sigma_{2}}{2}=0 \\ \tau_{x y} &=\frac{\sigma_{1}+\sigma_{2}}{2}=P \end{aligned}
\left(\sigma_{1}=P, \sigma_{2}=-P\right) . At 45^{\circ} from principal plane, plane of max shear occurs. On the plane of max shear.
\begin{aligned} \operatorname{both}, \quad \tau_{x y} &=\tau_{y y}=\frac{\sigma_{1}+\sigma_{2}}{2}=0 \\ \tau_{x y} &=\frac{\sigma_{1}+\sigma_{2}}{2}=P \end{aligned}
Question 10 |
A shaft with a circular cross-section is subjected to pure twisting moment. The ratio of the maximum shear stress to the largest principal stress is
2 | |
1 | |
0.5 | |
0 |
Question 10 Explanation:
For the case of pure shear
\begin{aligned} \sigma_{1} &=\tau_{\max } \\ \sigma_{2} &=-\tau_{\max } \\ \text { Required ratio } &=\tau_{\max } / \sigma_{1}=\sigma_{1} / \sigma_{1}=1 \end{aligned}
\begin{aligned} \sigma_{1} &=\tau_{\max } \\ \sigma_{2} &=-\tau_{\max } \\ \text { Required ratio } &=\tau_{\max } / \sigma_{1}=\sigma_{1} / \sigma_{1}=1 \end{aligned}
There are 10 questions to complete.
Calculation mistake in question 18,
Sigma x is negative, and sigma y is positive, and in under root how there is 25+64 …there should be 1+64 so root 65
Thanks
Check out question 24, it should be root((150²)+(100²))