Bearings, Shafts and keys

Question 1
A shaft AC rotating at a constant speed carries a thin pulley of radius r = 0.4 m at the end C which drives a belt. A motor is coupled at the end A of the shaft such that it applies a torque M_z about the shaft axis without causing any bending moment. The shaft is mounted on narrow frictionless bearings at A and B where AB = BC = L = 0.5 m. The taut and slack side tensions of the belt are T_1= 300 N and T2 = 100 N, respectively. The allowable shear stress for the shaft material is 80 MPa. The selfweights of the pulley and the shaft are negligible. Use the value of \pi available in the on-screen virtual calculator. Neglecting shock and fatigue loading and assuming maximum shear stress theory, the minimum required shaft diameter is _______ mm (round off to 2 decimal places).

A
45.32
B
32.78
C
23.94
D
18.25
GATE ME 2022 SET-2   Machine Design
Question 1 Explanation: 


\begin{aligned} M_{max}&=400 \times L =400 \times 0.5 \times 10^3\\ &=200 \times 10^3 \; N.mm\\ T_{max}&=M_z\\ &=(T_1-T_1) \times r\\ &=(200 \times 0.4) \times 10^3\\ T_{max}&=80 \times 10^3 \; N.mm \end{aligned}
Here section-B is critical due to loading,

Critical particle :
According to maximum shear stress theory,
\begin{aligned} \frac{16}{\pi d^3}\sqrt{M_{max}^2+T_{max}^2}&=\frac{S_{{ys}}}{FOS}\\ \frac{16 \times 10^3}{\pi d^3}\sqrt{200^2+80^2}&=\frac{80}{1}\\ \Rightarrow d&=23.94mm \end{aligned}
Question 2
A shaft of length L is made of two materials, one in the inner core and the other in the outer rim, and the two are perfectly joined together (no slip at the interface) along the entire length of the shaft. The diameter of the inner core is d_i and the external diameter of the rim is d_o, as shown in the figure. The modulus of rigidity of the core and rim materials are G_i and G_o, respectively. It is given that do d_o=2d_i and G_i=3G_o. When the shaft is twisted by application of a torque along the shaft axis, the maximum shear stress developed in the outer rim and the inner core turn out to be \tau _o and \tau _i, respectively. All the deformations are in the elastic range and stress strain relations are linear. Then the ratio \tau _i /\tau_o is ______ (round off to 2 decimal places).

A
1.15
B
2.65
C
1.85
D
1.5
GATE ME 2022 SET-2   Machine Design
Question 2 Explanation: 
Given G_i=3G_o, d_o=2d_i, l_i=l_o,\theta _i=\theta _o \text{ (It is a rigid joint)}
Find \frac{T_i}{T_o}=?
\begin{aligned} \frac{T}{J}&=\frac{\tau }{r}=\frac{G\theta }{L}\\ \tau _{max}(\text{in core})&=\frac{G\theta \times r_{max}}{L}\\ &=\frac{G_i \times \theta _i \times \frac{d_i}{2}}{L_i}=\tau _i\\ \tau _{max}(\text{rim})&=\frac{G_o \times \theta _o\times \frac{d_o}{2} }{L_o}=\tau _o\\ \frac{T_i}{\tau _o}&=\frac{G_i \times \theta _i \times \frac{d_i}{2}}{L_i} \times \frac{L_o}{G_o \times \theta _o\times \frac{d_o}{2} }\\ \frac{T_i}{T_o}&=\frac{G_i}{G_o} \times \frac{d_i}{d_o}=3 \times \frac{1}{2}=1.5 \end{aligned}
Question 3
A short shoe external drum brake is shown in the figure. The diameter of the brake drum is 500 mm. The dimensions a= 1000 mm, b = 500 mm and c = 200 mm. The coefficient of friction between the drum and the shoe is 0.35. The force applied on the lever F= 100 N as shown in the figure. The drum is rotating anti-clockwise. The braking torque on the drum is______ N.m (round off to two decimal places).
A
12.75
B
8.55
C
20.35
D
26.85
GATE ME 2019 SET-2   Machine Design
Question 3 Explanation: 
W.r.t FBD of link
\mathrm{F}_{\mathrm{s}}=\mu_{\mathrm{s}} \mathrm{N}=0.35 \times \mathrm{N}

Taking summation of moment about O = zero
\begin{array}{l} \Sigma M_{0}=0 \\ N \times 500-F_{s} \times 200-F \times 1000=0 \\ N \times 500-0.35 \times N \times 200-100 \times 1000=D \\ N=\frac{100 \times 1000}{(500-70)}=232.558 \mathrm{N} \end{array}
With respect to FBD of drum
\text { Braking torque } \mathrm{M}=\mathrm{F}_{\mathrm{s}} \times \mathrm{r}=\frac{0.35 \times 232.558 \times 500}{2 \times 1000}
\mathrm{M}=20.35 \mathrm{N}-\mathrm{m}
Question 4
A self-aligning ball bearing has a basic dynamic load rating ( C_{10} for 10^{6} revolutions) of 35 kN. If the equivalent radial load on the bearing is 45 kN, the excepted life (in revolutions) is
A
below 0.5
B
0.5 to 0.8
C
0.8 to 1.0
D
above 1.0
GATE ME 2018 SET-1   Machine Design
Question 4 Explanation: 
\begin{aligned} C &=35 \mathrm{kN} \\ \mathrm{P}_{C} &=45 \mathrm{kN} \\ L_{90} &=\left(\frac{C}{P_{C}}\right)^{3}=\left(\frac{35}{45}\right)^{3}=0.4705 \mathrm{MR} \end{aligned}
Question 5
Which of the bearings given below SHOULD NOT be subjected to a thrust load?
A
Deep groove ball bearing
B
Angular contact ball bearing
C
Cylindrical (straight) roller bearing
D
Single row tapered roller bearing
GATE ME 2016 SET-3   Machine Design
Question 5 Explanation: 
Correct option is (C)
Question 6
For ball bearings, the fatigue life L measured in number of revolutions and the radial load F are related by FL^{1/3}= K , where K is a constant. It withstands a radial load of 2 kN for a life of 540 million revolutions. The load (in kN) for a life of one million revolutions is ________
A
16.2865kN
B
98.2658kN
C
15.2547kN
D
98.2658kN
GATE ME 2015 SET-3   Machine Design
Question 6 Explanation: 
\begin{aligned} F L^{1 / 3} &=k \\ 2 k N \times(540)^{1 / 3} &=F \times(1)^{1 / 3} \\ F &=16.2865 \mathrm{kN} \end{aligned}
Question 7
Ball bearings are rated by a manufacturer for a life of 10^{6} revolutions. The catalogue rating of a particular bearing is 16 kN. If the design load is 2 kN, the life of the bearing will be p\times 10^{6} revolutions, where p is equal to _______
A
512
B
651
C
245
D
124
GATE ME 2014 SET-4   Machine Design
Question 7 Explanation: 
L_{10}=\left(\frac{C}{W}\right)^{3}
where
L_{10}= Basic life rating in millions of revolutions
C= Basic dynamic load rating
P= Equivalent dynamic radial load factor
\therefore \;L_{10}=\left(\frac{16}{2}\right)^{3}=512
Question 8
A hydrodynamic journal bearing is subject to 2000 N load at a rotational speed of 2000 rpm. Both bearing bore diameter and length are 40 mm. If radial clearance is 20 \mum and bearing is lubricated with an oil having viscosity 0.03 Pa.s, the Sommerfeld number of the bearing is _______
A
1.2
B
0.8
C
0.5
D
3.5
GATE ME 2014 SET-1   Machine Design
Question 8 Explanation: 
\begin{aligned} r &=20 \mathrm{mm}, \quad N=2000 \mathrm{rpm} \\ c &=20 \times 10^{-3}, \quad \mu=0.03 \mathrm{Ns} / \mathrm{m}^{2} \\ p &=\frac{2000}{1600} \mathrm{N} / \mathrm{mm}^{2} \qquad \left(p=\frac{W}{A}\right)\\ &=1.25 \mathrm{MPa}, S=? \\ S &=\frac{Z N}{60 p}\left(\frac{r}{c}\right)^{2} \\ &=\frac{0.03 \times 2000}{60 \times 1.25 \times 10^{6}}\left(\frac{20}{20 \times 10^{-3}}\right)^{2} \\ &=0.8 \end{aligned}
Question 9
A solid circular shaft needs to be designed to transmit a torque of 50N.m. If the allowable shear stress of the material is 140MPa, assuming a factor of safety of 2, the minimum allowable design diameter in mm is
A
8
B
16
C
24
D
32
GATE ME 2012   Machine Design
Question 9 Explanation: 
\begin{aligned} T &=50 \mathrm{Nm}=140 \mathrm{MPa} \quad N=2 \\ \tau_{\text {permissible }} &=\frac{140}{N}=70 \mathrm{MPa} \\ \therefore \quad \tau_{\text {per }} &=\frac{16 T}{\pi d^{3}} \\ d & \geq\left(\frac{16 T}{\pi \tau_{\text {per }}}\right)^{1 / 3} \geq\left(\frac{16 \times 50}{\pi \times 70 \times 10^{6}}\right)^{13} \\ d & \geq 0.0153 \mathrm{m} \\ d & \geq 15.37 \mathrm{mm} \approx 16 \mathrm{mm} \end{aligned}
Question 10
Two identical ball bearings P and Q are operating at loads 30kN and 45kN respectively. The ratio of the life of bearing P to the life of bearing Q is
A
81/16
B
27/8
C
9/4
D
3/2
GATE ME 2011   Machine Design
Question 10 Explanation: 
L = \left(\frac{C}{P}\right)^{3}
Where L = Life of bearing
P = Load
\frac{(L i f e)_{P}}{(L i f e)_{Q}}=\frac{(L o a d)_{\Omega}^{3}}{(Load)_{P}^{3}}=\left[\frac{45}{30}\right]^{3}=3.375 \; or \; \frac{27}{8}
There are 10 questions to complete.

1 thought on “Bearings, Shafts and keys”

  1. great work guys.
    I want to some suggestions
    1. add a bookmark or favourite questions option after login.
    2. keep it up you guys are awesome.

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