Brakes and Clutches

Question 1
A short shoe drum (radius 260 mm) brake is shown in the figure. A force of 1 kN is applied to the lever. The coefficient of friction is 0.4.

The magnitude of the torque applied by the brake is ________N.m (round off to one decimal place).
A
150
B
175
C
200
D
250
GATE ME 2021 SET-1   Machine Design
Question 1 Explanation: 



Taking moment about 'O'
\begin{aligned} R_{N}(500)+F_{\gamma}[310-260] &-1000 \times 1000=0 \\ R_{N}(500)+0.4\left(R_{N}\right)(50) &-1000 \times 1000=0 \\ R_{N} &=1923.076 \mathrm{~N} \\ F_{r} &=\mu R_{N}=769.23 \mathrm{~N} \\ T_{f} &=F_{r} \times R=200 \mathrm{~N}-\mathrm{m} \end{aligned}
Question 2
A helical spring has spring constant k. If the wire diameter, spring diameter and the number of coils are all doubled then the spring constant of the new spring becomes
A
k/2
B
k
C
8k
D
16k
GATE ME 2020 SET-2   Machine Design
Question 2 Explanation: 
\begin{aligned} k_{(\text {sping })} &=\frac{G d^{4}}{8 D^{3} n} \\ k_{\text {new }} &=\frac{G(2 d)^{4}}{8(2 D)^{3}(2 n)}=\frac{G d^{4}}{8 D^{3} n} \\ \text{Hence}\qquad k_{\text {new }} &=k \end{aligned}
Question 3
In a disc-type axial clutch, the frictional contact takes place within an annular region with outer and inner diameters 250 mm and 50 mm, respectively. An axial force F_1 is needed to transmit a torque by a new clutch. However, to transmit the same torque, one needs an axial force F_2 when the clutch wears out. If contact pressure remains uniform during operation of a new clutch while the wear is assumed to be uniform for an old clutch and the coefficient of friction does not change, then the ratio F_1/F_2 is_________ (round off to 2 decimal places).
A
0.22
B
0.46
C
0.87
D
0.93
GATE ME 2020 SET-1   Machine Design
Question 3 Explanation: 
\begin{aligned} T &=\mu W \times R_{m} \\ T_{\text {new }} &=\mu \times W \times R_{\text {new }} \\ T_{\text {new }} &=\mu \times W_{\text {new }} \times \frac{2}{3} \frac{\left(R_{0}^{3}-R_{i}^{3}\right)}{R_{0}^{2}-R_{i}^{2}} \\ T_{\text {odd }} &=\mu \times W \times R_{\text {old }} \\ T_{\text {new }} &=T_{\text {dd }} \\ W_{\text {new }} \times \frac{2}{3} \frac{\left(R_{0}^{3}-R_{i}^{3}\right)}{R_{0}^{2}-R_{i}^{2}} &=W_{\text {old }} \times \frac{\left(R_{0}+R_{i}\right)}{2} \\ \frac{W_{\text {new }}}{W_{\text {old }}} &=0.871 \end{aligned}
Question 4
A single block brake with a short shoe and torque capacity of 250 N\cdot m is shown. The cylindrical brake drum rotates anticlockwise at 100 rpm and the coefficient of friction is 0.25. The value of a, in mm (round off to one decimal place), such that the maximum actuating force P is 2000 N, is ________
A
212.5
B
256.4
C
159.8
D
753.5
GATE ME 2019 SET-1   Machine Design
Question 4 Explanation: 


\mu=0.25
76 Torque =250 \mathrm{Nm}
\mu \mathrm{F} \times \mathrm{a}=250
\Rightarrow 0.25 \times \mathrm{F} \times \mathrm{a}=250
\Rightarrow \mathrm{F} \times \mathrm{a}=1000
In FBD of block and lever, \Sigma \mathrm{M}_{\text {fullcrum }}=0
\mathrm{F} \times \mathrm{a}+\mu \mathrm{F} \times \frac{\mathrm{a}}{4}=\mathrm{P} \times 2.5 \mathrm{a}
F \times a\left(1+\frac{\mu}{4}\right)=P \times 2.5
1000\left(1+\frac{0.25}{4}\right)=2000 \times 2.5 \mathrm{a}
\Rightarrow \mathrm{a}=0.2125 \mathrm{m}=212.5 \mathrm{mm}
Question 5
The schematic of an external drum rotating clockwise engaging with a short shoe is shown in the figure. The shoe is mounted at point Y on a rigid lever XYZ hinged at point X. A force F= 100N is applied at the free end of the lever as shown. Given that the coefficient of friction between the shoe and the drum is 0.3, the braking torque (in Nm ) applied on the drum is _______ (correct to two decimal places).
A
3.25
B
5.55
C
7.82
D
8.18
GATE ME 2018 SET-1   Machine Design
Question 5 Explanation: 


\begin{aligned} \Sigma M_{0}&=0 \\ F \times 300\curvearrowleft+f \times 300 \curvearrowleft&=R N \times 200 \curvearrowright\\ 100 \times 300+\mu R_{N} \times 300 &=R_{N} \times 200 \\ 100 \times 300+0.3 \times R_{N} \times & 300=R_{N} \times 200 \\ 300 &=1.1 R_{N} \\ R_{N} &=272.72 \mathrm{N} \\ \text { Braking Torque } &=\mu R_{N} \times R \\&=0.3 \times 272.72 \times 0.100 \\ &=8.18 \mathrm{Nm} \end{aligned}
Question 6
A single - plate clutch has a friction disc with inner and outer radii of 20 mm and 40 mm, respectively. The friction lining in the disc is made in such a way that the coefficient of friction \mu varies radially as \mu=0.01r, where r is in mm. The clutch needs to transmit a friction torque of 18.8 kN\! \! \cdot\! mm. As per uniform pressure theory, the pressure (in MPa) on the disc is ________
A
0.35
B
0.5
C
0.67
D
0.8
GATE ME 2017 SET-2   Machine Design
Question 6 Explanation: 


Normal load on elemental ring (d M)=p 2 \pi r d r
Frictional force on elemental ring \left(d F_{f}\right)=\mu d W
Frictional torque transmitted by elemental ring,
d T_{f}=\left(d F_{f}\right) r
Hence, \quad d T_{f}=\mu p 2 \pi r^{2} d r
Total frictional torque transmitted by clutch plate
\begin{aligned} \left(T_{f}\right)&=\int_{R_{i}}^{R_{0}} d T_{f} \\ T_{f}&=\int_{R_{i}}^{P_{0}} \mu p 2 \pi r^{2} d r\\ \text{As per U.P.T.}\\ T_{f} &=2 \pi p \int_{R_{i}}^{R_{0}} \mu r^{2} d r \\ T_{f} &=2 \pi p \int_{R_{i}}^{R_{0}}(0.01 r) r^{2} d r \\ \left(T_{f}\right)_{U . P . T .} &=(2 \pi p)(0.01)\left[\frac{R_{0}^{4}-R_{i}^{4}}{4}\right] \\ 18.85 \times 10^{3} &=2 \pi p(0.01)\left[\frac{40^{4}-20^{4}}{4}\right] \\ p &=0.5 \mathrm{MPa} \end{aligned}
Question 7
For the brake shown in the figure, which one of the following is TRUE?
A
Self energizing for clockwise rotation of the drum
B
Self energizing for anti-clockwise rotation of the drum
C
Self energizing for rotation in either direction of the drum
D
Not of the self energizing type
GATE ME 2016 SET-2   Machine Design
Question 7 Explanation: 


For clockwise direction rotation of drum direction of moment of forces F_{\mathrm{r}_{2}} and F will be same. Hence, self energizing for clockwise rotation of the drum.
Question 8
The forces F1 and F2 in a brake band and the direction of rotation of the drum are as shown in the figure. The coefficient of friction is 0.25. The angle of wrap is 3\pi /2 radians. It is given that R = 1 m and F2 = 1 N. The torque (in N-m) exerted on the drum is _________
A
2.24Nm
B
6.25Nm
C
8.9Nm
D
4.5Nm
GATE ME 2016 SET-2   Machine Design
Question 8 Explanation: 


\begin{aligned} F_{1} &>F_{2} \\ \frac{F_{1}}{F_{2}} &=e^{\mu \theta} \\ \frac{F_{1}}{F_{2}} &=e^{0.25 \times \frac{3 \pi}{2}} \\ F_{1} &=F_{2} \cdot e^{0.25 \times \frac{3 \pi}{2}} \\ &=1 \times e^{0.25 \times \frac{3 \times \pi}{2}} \\ &=3.2482 \mathrm{N} \\ \text { Torque } &=\left(F_{1}-F_{2}\right) \times r \\ &=(3.2482-1) \times 11 \\ &=2.2482 \mathrm{Nm} \end{aligned}
Question 9
A four-wheel vehicle of mass 1000 kg moves uniformly in a straight line with the wheels revolving at 10 rad/s. The wheels are identical, each with a radius of 0.2 m. Then a constant braking torque is applied to all the wheels and the vehicle experiences a uniform deceleration. For the vehicle to stop in 10 s, the braking torque (in N.m) on each wheel is _______
A
10Nm
B
20Nm
C
30Nm
D
50Nm
GATE ME 2014 SET-3   Machine Design
Question 9 Explanation: 
Initial velocity,
u=\omega r=10 \times 0.2=2 \mathrm{m} / \mathrm{s}
For vehicle to stop,
\begin{aligned} v &=u-\mathrm{at} \\ \Rightarrow \quad 0 &=u-\mathrm{at}\\ \therefore \quad a&=\frac{u}{t}=\frac{2}{10}=0.2 \mathrm{m} / \mathrm{s}^{2} \end{aligned}
Braking force = Total mass x deceleration
\begin{aligned} F &=1000 \times 0.2 \\ &=200 \mathrm{N} \\ F_{\text {each wheel }} &=\frac{200}{4}=50 \mathrm{N} \end{aligned}
Braking torque on each wheel
=50 \times 0.2=10 \mathrm{Nm}
Question 10
A drum brake is shown in the figure. The drum is rotating in anticlockwise direction. The coefficient of friction between drum and shoe is 0.2. The dimensions shown in the figure are in mm. The braking torque (in N.m) for the brake shoe is _______
A
25Nm
B
98Nm
C
64Nm
D
48Nm
GATE ME 2014 SET-3   Machine Design
Question 10 Explanation: 


Taking \Sigma M_{0}=0
\begin{aligned} \Rightarrow \quad 1000 \times 800&=F_{t} \times 100+R_{N} \times 480 \\ \Rightarrow \quad 8000 &=F_{t}+4.8 R_{N} \\ F_{t} &=\mu R_{N}=0.2 R_{N} \\ \therefore 0.2 R_{N}+4.8 R_{N} &=8000\\ \text{or }\quad R_{N}&=\frac{8000}{5}=1600 \mathrm{N}\\ \therefore \qquad F_{t} &=\mu R_{N}=0.2 \times 1600 \\ &=320 \mathrm{N}\\ \text{Braking torque,}\\ T &=F_{t} \times r \\ &=320 \times 0.2=64 \mathrm{Nm} \end{aligned}
There are 10 questions to complete.

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