Static Dynamic Loading and Failure Theories

Question 1
The von Mises stress at a point in a body subjected to forces is proportional to the square root of the
A
total strain energy per unit volume
B
plastic strain energy per unit volume
C
dilatational strain energy per unit volume
D
distortional strain energy per unit volume
GATE ME 2021 SET-2   Machine Design
Question 1 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.
Question 2
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).
A
1.2
B
2
C
2.5
D
2.9
GATE ME 2021 SET-1   Machine Design
Question 2 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 3
Endurance limit of a beam subjected to pure bending decreases with
A
decrease in the surface roughness and decrease in the size of the beam
B
increase in the surface roughness and decrease in the size of the beam
C
increase in the surface roughness and increase in the size of the beam
D
decrease in the surface roughness and increase in the size of the beam
GATE ME 2019 SET-2   Machine Design
Question 3 Explanation: 
Endurance limit decreases with increase in surface roughness and with increase in size of the beam.
Question 4
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).
A
20.96
B
45.36
C
31.62
D
52.45
GATE ME 2018 SET-2   Machine Design
Question 4 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 5
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 _______
A
1.35
B
1.75
C
2
D
2.25
GATE ME 2017 SET-2   Machine Design
Question 5 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}
Question 6
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.
A
\frac{F}{S_{f}(W-2d)}
B
\frac{F}{S_{f}W}
C
\frac{F}{S_{f}(W-d)}
D
\frac{2F}{S_{f}W}
GATE ME 2017 SET-1   Machine Design
Question 6 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 7
For the given fluctuating fatigue load, the values of stress amplitude and stress ratio are respectively
A
100 MPa and 5
B
250 MPa and 5
C
100 MPa and 0.20
D
250 MPa and 0.20
GATE ME 2015 SET-3   Machine Design
Question 7 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
Question 8
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 _______
A
173.1MPa
B
256.3MPa
C
145.2MPa
D
125.3MPa
GATE ME 2015 SET-2   Machine Design
Question 8 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}
Question 9
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
A
1
B
2
C
2.5
D
3.3
GATE ME 2015 SET-1   Machine Design
Question 9 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}
Question 10
A shaft is subjected to pure torsional moment. The maximum shear stress developed in the shaft is 100 MPa. The yield and ultimate strengths of the shaft material in tension are 300 MPa and 450 MPa, respectively. The factor of safety using maximum distortion energy (von-Mises) theory is _______
A
6.654
B
1.265
C
1.732
D
8.559
GATE ME 2014 SET-4   Machine Design
Question 10 Explanation: 
\begin{aligned} \sigma_{1}&=100 \mathrm{MPa} \\ \sigma_{2}&=-100 \mathrm{MPa} \end{aligned}
As per Von-Mises yield theory:


N^{2}\left[\left(\sigma_{1}-\sigma_{2}\right)^{2}+\left(\sigma_{2}-\sigma_{3}\right)^{2}+\left(\sigma_{1}-\sigma_{3}\right)^{2}\right] \leq 2 \sigma_{y}^{2}
\text{In this case, } \sigma_{3}=0
2 N^{2}\left[\sigma_{1}^{2}+\sigma_{2}^{2}-\sigma_{1} \sigma_{2}\right] \leq 2 \sigma_{y}^{2}
\Rightarrow N^{2}\left[(100)^{2}+(100)^{2}+(100)^{2}\right] \leq(300)^{2}
\Rightarrow N^{2}=\frac{9}{3} \Rightarrow N \leq \sqrt{3}
\therefore \quad N=1.732
There are 10 questions to complete.

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