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).
45.32 | |
32.78 | |
23.94 | |
18.25 |
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).
1.15 | |
2.65 | |
1.85 | |
1.5 |
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}
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).
12.75 | |
8.55 | |
20.35 | |
26.85 |
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}
\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
below 0.5 | |
0.5 to 0.8 | |
0.8 to 1.0 | |
above 1.0 |
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?
Deep groove ball bearing | |
Angular contact ball bearing | |
Cylindrical (straight) roller bearing | |
Single row tapered roller bearing |
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 ________
16.2865kN | |
98.2658kN | |
15.2547kN | |
98.2658kN |
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 _______
512 | |
651 | |
245 | |
124 |
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
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 _______
1.2 | |
0.8 | |
0.5 | |
3.5 |
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
8 | |
16 | |
24 | |
32 |
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
81/16 | |
27/8 | |
9/4 | |
3/2 |
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}
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.
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I want to some suggestions
1. add a bookmark or favourite questions option after login.
2. keep it up you guys are awesome.