Question 1 |
A structural member under loading has a uniform
state of plane stress which in usual notations is given
by \sigma _x=3P,\sigma _y=-2P,\tau _{xy}=\sqrt{2}P , where P \gt 0.
The yield strength of the material is 350 MPa. If the
member is designed using the maximum distortion
energy theory, then the value of P at which yielding
starts (according to the maximum distortion energy
theory) is
70 Mpa | |
90 Mpa | |
120 Mpa | |
75 Mpa |
Question 1 Explanation:
Given,
\sigma _x=3P
\sigma _y=-2P
\tau =\sqrt{2}P
According to maximum distortion energy theory,
\begin{aligned} \sqrt{\sigma _x^2-\sigma _x\sigma _y+\sigma _y^2+3\tau _{xy}^2} &=\frac{S_{yt}}{FOS} \\ P\sqrt{3^2-3(-2)+(-2)^2+3(\sqrt{2})^2}&= \frac{350}{1}\\ P \times 5&=350 \\ P&=70\; MPa \end{aligned}
\sigma _x=3P
\sigma _y=-2P
\tau =\sqrt{2}P
According to maximum distortion energy theory,
\begin{aligned} \sqrt{\sigma _x^2-\sigma _x\sigma _y+\sigma _y^2+3\tau _{xy}^2} &=\frac{S_{yt}}{FOS} \\ P\sqrt{3^2-3(-2)+(-2)^2+3(\sqrt{2})^2}&= \frac{350}{1}\\ P \times 5&=350 \\ P&=70\; MPa \end{aligned}
Question 2 |
The von Mises stress at a point in a body subjected to forces is proportional to the square root of the
total strain energy per unit volume | |
plastic strain energy per unit volume | |
dilatational strain energy per unit volume | |
distortional strain energy per unit volume |
Question 2 Explanation:
Condition for failure as per M.D.E.T.
Distortion energy per unit volume under tri-axial state of stress > Distortion energy per unit volume under uni-axial state of stress.
\begin{aligned} &\text { Hence, }\left(\frac{1+\mu}{6 E}\right)\left[\left(\sigma_{1}-\sigma_{2}\right)^{2}+\left(\sigma_{2}-\sigma_{3}\right)^{2}+\left(\sigma_{1}-\sigma_{3}\right)^{2}\right]>\left(\frac{1+\mu}{3 E}\right)\left(S_{y t}\right)^{2}\\ &\left(\sigma_{1}-\sigma_{2}\right)^{2}+\left(\sigma_{2}-\sigma_{3}\right)^{2}+\left(\sigma_{1}-\sigma_{3}\right)^{2}>2\left(S_{y t}\right)^{2}\\ &\sqrt{\sigma_{1}^{2}+\sigma_{2}^{2}+\sigma_{3}^{2}+\sigma_{1} \sigma_{2}-\sigma_{2} \sigma_{3}-\sigma_{1} \sigma_{3}}>\left(S_{y t}\right) \text { (Von Mises effective stress) } \end{aligned}
S_{y t} = Von Mises effective stress is defined as the uni-axial yield stress that would create same distortion energy created by the tri-axial state of stress.
Distortion energy per unit volume under tri-axial state of stress > Distortion energy per unit volume under uni-axial state of stress.
\begin{aligned} &\text { Hence, }\left(\frac{1+\mu}{6 E}\right)\left[\left(\sigma_{1}-\sigma_{2}\right)^{2}+\left(\sigma_{2}-\sigma_{3}\right)^{2}+\left(\sigma_{1}-\sigma_{3}\right)^{2}\right]>\left(\frac{1+\mu}{3 E}\right)\left(S_{y t}\right)^{2}\\ &\left(\sigma_{1}-\sigma_{2}\right)^{2}+\left(\sigma_{2}-\sigma_{3}\right)^{2}+\left(\sigma_{1}-\sigma_{3}\right)^{2}>2\left(S_{y t}\right)^{2}\\ &\sqrt{\sigma_{1}^{2}+\sigma_{2}^{2}+\sigma_{3}^{2}+\sigma_{1} \sigma_{2}-\sigma_{2} \sigma_{3}-\sigma_{1} \sigma_{3}}>\left(S_{y t}\right) \text { (Von Mises effective stress) } \end{aligned}
S_{y t} = Von Mises effective stress is defined as the uni-axial yield stress that would create same distortion energy created by the tri-axial state of stress.
Question 3 |
A machine part in the form of cantilever beam is subjected to fluctuating load as shown in the figure. The load varies from 800 N to 1600 N. The modified endurance, yield and ultimate strengths of the material are 200 MPa, 500 MPa and 600 MPa, respectively.
The factor of safety of the beam using modified Goodman criterion is _______ (round off to one decimal place).
The factor of safety of the beam using modified Goodman criterion is _______ (round off to one decimal place).
1.2 | |
2 | |
2.5 | |
2.9 |
Question 3 Explanation:
A : Critical Point
\begin{aligned} \sigma_{\max , A} &=\sigma_{b, \max } \text { at } A \text { due to } 1600 \mathrm{~N}=\frac{6 \mathrm{M}}{b d^{2}}=\frac{6 \times 1600 \times 200}{12 \times(2 \sigma)^{2}} \\ \sigma_{\max } &=200 \mathrm{MPa} \\ \sigma_{\min , A} &=\sigma_{b, \max } \text { at } A \text { due to } 800 \mathrm{~N} \\ &=\frac{6 \times 800 \times 200}{12 \times(20)^{2}}=100 \mathrm{MPa} \\ \text { Modified Goodman } &=\frac{\sigma_{m}}{S_{y t}}+\frac{\sigma_{a}}{\sigma_{e}} \leq \frac{1}{N} \\ \sigma_{m} &=\left|\frac{\sigma_{\max }+\sigma_{\min }}{2}\right|=150 \mathrm{MPa} \\ \sigma_{a} &=\left|\frac{\sigma_{\max }-\sigma_{\min }}{2}\right|=50 \mathrm{MPa} \\ \frac{150}{600}+\frac{50}{200} & \leq \frac{1}{N} \\ N & \leq 2 \\ N & \approx 2\\ \text { Langer, } \qquad \frac{\sigma_{m}}{S_{y t}}+\frac{\sigma_{a}}{S_{y t}} & \leq \frac{1}{N} \\ \frac{150}{500}+\frac{50}{500} & \leq \frac{1}{N} \qquad \qquad \qquad \qquad (N\leq2.5)\\ N &=2.5 \end{aligned}
Modified Goodman = Safe result of [Goodman or Langer]
Question 4 |
Endurance limit of a beam subjected to pure bending decreases with
decrease in the surface roughness and decrease in the size of the beam | |
increase in the surface roughness and decrease in the size of the beam | |
increase in the surface roughness and increase in the size of the beam | |
decrease in the surface roughness and increase in the size of the beam |
Question 4 Explanation:
Endurance limit decreases with increase in surface roughness and with increase in size of the
beam.
Question 5 |
A bar is subjected to a combination of a steady load of 60 kN and a load fluctuating between -10 kN and 90 kN. The corrected endurance limit of the bar is 150 MPa, the yield strength of the material is 480 MPa and the ultimate strength of the material is 600 MPa. The bar cross-section is square with side a. If the factor of safety is 2, the value of a (in mm), according to the modified Goodman's criterion, is ________ (correct to two decimal places).
20.96 | |
45.36 | |
31.62 | |
52.45 |
Question 5 Explanation:
\begin{array}{l} P_{m}=\frac{P_{\max }+P_{\min }}{2} \\ P_{a}=\frac{P_{\max }-P_{\min }}{2} \\ P_{m}=100 \mathrm{kN} \\ \mathrm{Pa}=50 \mathrm{kN} \\ \sigma_{\mathrm{m}}=\frac{100 \times 10^{3}}{a^{2}} \mathrm{MPa} \\ \sigma_{a}=\frac{50 \times 10^{3}}{a^{2}} \mathrm{MPa} \end{array}
Solution by Goodman Equation,
\begin{aligned} \frac{\sigma_{m}}{S_{u t}}+\frac{\sigma_{a}}{\sigma_{e}} &=\frac{1}{N} \\ 100\left[\frac{100}{a^{2} \times 600}+\frac{50}{150 a^{2}}\right] &=\frac{1}{2} \\ a^{2} &=1000 \\ a &=31.62 \mathrm{mm} \end{aligned}
Solution by Langar equation,
\begin{aligned} \frac{\sigma_{m}}{S_{y t}}+\frac{\sigma_{a}}{S_{y t}}&=\frac{1}{N}\\ 100\left[\frac{100}{480 a^{2}}+\frac{50}{480 a^{2}}\right] &=\frac{1}{2} \\ a^{2} &=625 \\ a &=25 \mathrm{mm} \end{aligned}
Hence final answer by modified Goodman's Griterion is 31.62 mm.
Question 6 |
A machine component made of a ductile material is subjected to a variable loading with \sigma _{min}=-50 MPa and \sigma _{max}=50 MPa .If the corrected endurance limit and the yield strength for the material are \sigma\, '_{e}=100 MPa and \sigma_{y}=300 MPa , the factor of safety is _______
1.35 | |
1.75 | |
2 | |
2.25 |
Question 6 Explanation:
Given variable loading is a completely reversed fatigue or variable loading because
\begin{array}{c} \sigma_{\max }=-\sigma_{\min } \\ \text { Hence, } \sigma_{\text {mean }}=\sigma_{m}=0 ; \sigma_{\text {variable }}=\sigma_{v}=\sigma_{\max } \end{array}
For completely reversed fatigue loading Soderberg, Goodman, Gerber and strength criterion will give same results.
As per strength criterion,
\begin{aligned} \sigma_{\max } & \leq \sigma_{\text {per }} \text { or } \frac{\text { failure stress }}{\text { F.O.S. }} \\ \sigma_{\max } &=\frac{\text { Endurance limit }}{N} \\ N &=\frac{\text { Endurance limit }}{\text { Max. stress }}=\frac{100}{50}=2 \end{aligned}
\begin{array}{c} \sigma_{\max }=-\sigma_{\min } \\ \text { Hence, } \sigma_{\text {mean }}=\sigma_{m}=0 ; \sigma_{\text {variable }}=\sigma_{v}=\sigma_{\max } \end{array}
For completely reversed fatigue loading Soderberg, Goodman, Gerber and strength criterion will give same results.
As per strength criterion,
\begin{aligned} \sigma_{\max } & \leq \sigma_{\text {per }} \text { or } \frac{\text { failure stress }}{\text { F.O.S. }} \\ \sigma_{\max } &=\frac{\text { Endurance limit }}{N} \\ N &=\frac{\text { Endurance limit }}{\text { Max. stress }}=\frac{100}{50}=2 \end{aligned}
Question 7 |
Consider the schematic of a riveted lap joint subjected to tensile load F, as shown below. Let d be the diameter of the rivets, and S_{f} be the maximum permissible tensile stress in the plates. What should be the minimum value for the thickness of the plates to guard against tensile failure of the plates? Assume the plates to be identical.
\frac{F}{S_{f}(W-2d)} | |
\frac{F}{S_{f}W} | |
\frac{F}{S_{f}(W-d)} | |
\frac{2F}{S_{f}W} |
Question 7 Explanation:
Safe condition to avoid tearing failure of the joint
along the row of rivets,
\begin{aligned} \left(\sigma_{\text {max }}\right)_{\text {tensile }} & \leq\left(\sigma_{t}\right)_{\text {per }} \\ \frac{F}{(W-2 d) t} &=S_{f} \\ t & \geq \frac{F}{S_{f}(W-2 d)} \end{aligned}
Question 8 |
For the given fluctuating fatigue load, the values of stress amplitude and stress ratio are respectively
100 MPa and 5 | |
250 MPa and 5 | |
100 MPa and 0.20 | |
250 MPa and 0.20 |
Question 8 Explanation:
\sigma_{a}=\frac{\sigma_{\max }-\sigma_{\min }}{2}=\frac{250-50}{2}=100 \mathrm{MPa}
\text{Stress ratio }=\frac{\sigma _{max}}{\sigma _{min}}=\frac{50}{250}=0.2
\text{Stress ratio }=\frac{\sigma _{max}}{\sigma _{min}}=\frac{50}{250}=0.2
Question 9 |
The uniaxial yield stress of a material is 300 MPa. According to von Mises criterion, the shear yield stress (in MPa) of the material is _______
173.1MPa | |
256.3MPa | |
145.2MPa | |
125.3MPa |
Question 9 Explanation:
According to Von-Mises criterion,
\begin{aligned} \tau_{y} &=\frac{\sigma_{y}}{\sqrt{3}} \\ &=\frac{300}{\sqrt{3}}=173.1 \mathrm{MPa} \end{aligned}
\begin{aligned} \tau_{y} &=\frac{\sigma_{y}}{\sqrt{3}} \\ &=\frac{300}{\sqrt{3}}=173.1 \mathrm{MPa} \end{aligned}
Question 10 |
A machine element is subjected to the following bi-axial state of stress: \sigma _{x} = 80 MPa; \sigma _{y} = 20 MPa; \tau _{xy} = 40 MPa. If the shear strength of the material is 100 MPa, the factor of safety as per Tresca's maximum shear stress theory is
1 | |
2 | |
2.5 | |
3.3 |
Question 10 Explanation:
\begin{array}{c} \sigma_{1,2}=\frac{1}{2}\left[\left(\sigma_{x}+\sigma_{y}\right) \pm \sqrt{\left(\sigma_{x}+\sigma_{y}\right)^{2}+4 \tau_{x y}^{2}}\right] \\ =\frac{1}{2}[(80+20) \pm \sqrt{(60)^{2}+4 \times 40^{2}}] \\ \sigma_{1}=100 \mathrm{MPa} \\ \sigma_{2}=0 \mathrm{MPa} \\ \left(\tau_{\max }\right)=\frac{\sigma_{1}}{2}=50 \mathrm{MPa} \\ F(\text { Factor of safety })=\frac{\tau_{y s}}{\tau_{\max }}=\frac{100}{50}=2 \end{array}
There are 10 questions to complete.
Please refer to Question 16
The solution to the question is incorrect W/4 should be the correct option.
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