Dynamic Analysis of Slider-crank

Question 1
The Whitworth quick return mechanism is shown in the figure with link lengths as follows: OP = 300 mm, OA = 150 mm, AR = 160 mm, RS = 450 mm.

The quick return ratio for the mechanism is ________(round off to one decimal place).
A
1
B
2
C
2.5
D
3.5
GATE ME 2021 SET-1   Theory of Machine
Question 1 Explanation: 




\begin{aligned} \cos \frac{\alpha}{2} &=\frac{150}{300}=\frac{1}{2} \\ \Rightarrow \qquad \frac{\alpha}{2} &=60^{\circ} \\ \alpha &=120^{\circ} \\ \beta &=\left(360^{\circ}-120^{\circ}\right)=240^{\circ} \\ \mathrm{QRR} &=\left(\frac{\beta}{\alpha}\right)=\frac{240}{120}=2 \end{aligned}
Question 2
The crank of a slider-crank mechanism rotates counter-clockwise (CCW) with a constant angular velocity \omega, as shown. Assume the length of the crank to be r.

Using exact analysis, the acceleration of the slider in the y direction, at the instant shown, where the crank is parallel to x-axis, is given by
A
-\omega^2 r
B
2 \omega^2 r
C
\omega^2 r
D
-2 \omega^2 r
GATE ME 2019 SET-2   Theory of Machine
Question 2 Explanation: 


since the velocity of the point A and B are parallel \omega_{A B}=0
\begin{array}{l} \overrightarrow{\mathrm{a}}_{\mathrm{B}}=\overrightarrow{\mathrm{a}}_{\mathrm{A}}+\overrightarrow{\mathrm{a}}_{\mathrm{AB}} \\ \overrightarrow{\mathrm{a}}_{\mathrm{B}}=\mathrm{a}_{\mathrm{B}} \hat{\mathrm{j}} \\ \overrightarrow{\mathrm{a}}_{\mathrm{A}}=-\omega^{2} \mathrm{r} \hat{\mathrm{i}} \\ \overrightarrow{\mathrm{a}}_{\mathrm{AB}}=-\alpha \mathrm{r}_{\mathrm{AB}} \sin 45 \hat{\mathrm{i}}-\alpha \mathrm{r}_{\mathrm{AB}} \cos 45 \hat{\mathrm{j}} \\ \quad\left(\because \omega^{2} \mathrm{r}_{\mathrm{AB}} \text { along link } \mathrm{AB}=0\right) \\ \mathrm{a}_{\mathrm{B}} \hat{\mathrm{j}}=-\omega^{2} \mathrm{r} \hat{\mathrm{r}}-(\alpha \hat{\mathrm{i}}+\alpha \hat{\mathrm{j}}) \quad=-\left(\omega^{2} \mathrm{r}+\alpha\right) \hat{\mathrm{i}}-\alpha \hat{\mathrm{j}} \\ \omega^{2} \mathrm{r}+\alpha=0 \\ \alpha=-\omega^{2} \mathrm{r} \\ \mathrm{a}_{\mathrm{B}}=-\alpha=-\left(-\omega^{2} \mathrm{r}\right)=\omega^{2} \mathrm{r} \end{array}
Question 3
A slider crank mechanism is shown in the figure. At some instant, the crank angle is 45^{\circ} and a force of 40 N is acting towards the left on the slider. The length of the crank is 30 mm and the connecting rod is 70 mm. Ignoring the effect of gravity, friction and inertial forces, the magnitude of the crankshaft torque (in Nm) needed to keep the mechanism in equilibrium is _________ (correct to two decimal places).
A
0.12
B
0.56
C
1.12
D
2.52
GATE ME 2018 SET-1   Theory of Machine
Question 3 Explanation: 


\begin{aligned} 30 \sin 45^{\circ} &=70 \sin \beta \\ \sin \beta &=\frac{30 \sin 45}{70}=0.303045 \\ \beta &=17.6406^{\circ} \\ F_{C} \cos \beta &=40 \\ F_{C} &=\frac{40}{\cos \beta}=41.9737 \mathrm{N} \\ F_{t} &=F_{C} \sin (\theta+\beta) \\ &=41.9737 \times \sin \left(17.6406^{\circ}+45^{\circ}\right) \\ F_{t} &=37.2785 \mathrm{N} \\ T &=F_{t} \times r=47.2785 \times 0.030 \\ &=1.1183 \mathrm{N}-\mathrm{m}=1.118 \mathrm{N}-\mathrm{m} \\ &=1.12 \mathrm{N}-\mathrm{m} \end{aligned}
Question 4
In a certain slider-crank mechanism, lengths of crank and connecting rod are equal. If the crank rotates with a uniform angular speed of 14 rad/s and the crank length is 300 mm, the maximum acceleration of the slider (in m/s^{2}) is ___________
A
117.6m/s
B
215.4m/s
C
125.2m/s
D
458.6m/s
GATE ME 2015 SET-2   Theory of Machine
Question 4 Explanation: 
a=r \omega^{2}\left[\cos \theta+\frac{\cos 2 \theta}{n}\right]
\text{at} \quad \theta=0
\begin{aligned} a &\Rightarrow a_{\max } \\ a_{\max } &=r \omega^{2}\left[1+\frac{1}{n}\right] \\ n &=\frac{l}{r}=1 \\ a_{\max } &=2 r \omega^{2}=2 \times 0.3 \times 14^{2} \\ &=117.6 \mathrm{m} / \mathrm{s}^{2} \end{aligned}
Question 5
A slider-crank mechanism with crank radius 60 mm and connecting rod length 240 mm is shown in figure. The crank is rotating with a uniform angular speed of 10 rad/s, counter clockwise. For the given configuration, the speed (in m/s) of the slider is _______
A
0.2m/s
B
0.9m/s
C
0.6m/s
D
9.5m/s
GATE ME 2014 SET-3   Theory of Machine
Question 5 Explanation: 


\begin{array}{l} L=240\mathrm{mm}\\ r=60 \mathrm{mm} \\ n=\frac{L}{r}=\frac{240}{60}=4 \\ \omega=10 \mathrm{rad} / \mathrm{s} \end{array}
From the given configuration,
\begin{array}{l} \cos \theta=\frac{\sin \theta}{n} \text { or } \tan \theta=n\\ \text{or }\theta=\tan ^{-1} \\ \quad=75.96^{\circ} \\ \therefore \quad V_{s}=\omega r\left[\sin \theta+\frac{\sin 2 \theta}{2 n}\right] \\ =10 \times 0.06\left[\sin 75.96^{\circ}+\frac{\sin 151.92^{\circ}}{2 \times 4}\right] \\ =0.6[0.97+0.0588] \\ =0.6 \times 1.0288\\ \text{or }V_{0}=\omega \end{array}
since slider is \perp ar to the crank
\begin{aligned} \text{so,} &=\omega r=10 \times 0.06 \\ &=0.6 \mathrm{m} / \mathrm{s} \end{aligned}
Question 6
An offset slider-crank mechanism is shown in the figure at an instant. Conventionally, the Quick Return Ratio (QRR) is considered to be greater than one. The value of QRR is _______
A
1.25
B
2.15
C
3.25
D
4.51
GATE ME 2014 SET-1   Theory of Machine
Question 6 Explanation: 
Length of connecting rod: l_{r}=40 \mathrm{mm}
Crank radius: r=20 \mathrm{mm}
Eccentricity: e=10 \mathrm{mm}
\begin{aligned} &\phi=\cos ^{-1}\left(\frac{e}{l+r}\right)-\cos ^{-1}\left(\frac{e}{l-r}\right) \\ &=\cos ^{-1}\left(\frac{1}{6}\right)-\cos ^{-1}\left(\frac{1}{2}\right)=20.4^{\circ} \\ &\mathrm{QRR}=\frac{\text { Time of advance stroke }}{\text { Time of return stroke }} \\ &=\left(\frac{\theta_{C}}{2 \pi N}\right)\left(\frac{2 \pi N}{\theta_{R}}\right) \\ &\mathrm{QRR} =\frac{\theta_{C}}{\theta_{R}}=\frac{180^{\circ}+\phi}{180^{\circ}-\phi} \\ &=\frac{180^{\circ}+20.4^{\circ}}{180^{\circ}-20.4^{\circ}}=1.25 \end{aligned}
Question 7
A slider crank mechanism has slider mass of 10 kg, stroke of 0.2 m and rotates with a uniform angular velocity of 10 rad/s. The primary inertia forces of the slider are partially balanced by a revolving mass of 6 kg at the crank, placed at a distance equal to crank radius. Neglect the mass of connecting rod and crank. When the crank angle (with respect to slider axis) is 30^{\circ} , the unbalanced force (in newton) normal to the slider axis is _______
A
30N
B
20N
C
10N
D
40N
GATE ME 2014 SET-1   Theory of Machine
Question 7 Explanation: 


\begin{aligned} \omega &=10 \mathrm{rad} / \mathrm{s} \\ \theta &=30^{\circ} \\ r &=\frac{s}{2}=0.1 \mathrm{m} \\ \text { Unbalanced force }&=m \omega^{2} r \sin 30^{\circ} & \\ &=6 \times 100 \times 0.1 \times \frac{1}{2} \\ &=30 \mathrm{N} \end{aligned}
Question 8
For a four-bar linkage in toggle position, the value of mechanical advantage is:
A
0
B
0.5
C
1
D
\infty
GATE ME 2006   Theory of Machine
Question 8 Explanation: 


\omega_{4} of the output link DC becomes zero at the extreme positions. The extreme positions of the linkage are known as "Toggle position".
\therefore Mechanical advantage =\frac{\omega_{\text {input }}}{\omega_{\text {output }}}
\because \quad \omega_{\text {output }}=0
\therefore Mechanical advantage =\infty
Question 9
For a mechanism shown below, the mechanical advantage for the given configuration is
A
0
B
0.5
C
1
D
\infty
GATE ME 2004   Theory of Machine
Question 9 Explanation: 
Mechanical advantage
=\frac{T_{\text {output }}}{T_{\text {input }}}=\frac{\omega_{\text {input }}}{\omega_{\text {output }}}
since output link is a slider
Hence,\omega_{\text {output }}=0
\therefore Mechanical advantage =\infty
There are 9 questions to complete.

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