Question 1 |
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 1 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 2 |
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 2 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 3 |
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 3 Explanation:
Correct option is (C)
Question 4 |
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 4 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 5 |
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 5 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 6 |
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 6 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 7 |
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 7 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 8 |
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 8 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}
Question 9 |
A lightly loaded full journal bearing has a journal of 50mm, bush bore of 50.05mm and bush length of 20mm. if rotational speed of journal is 1200rpm and average viscosity of liquid lubricant is 0.03 Pa s, the power loss (in W) will be
37 | |
74 | |
118 | |
237 |
Question 9 Explanation:
Tangential velocity of shaft,
\begin{aligned} u &=\frac{\pi D N}{60}=\frac{\pi \times 50 \times 10^{-3} \times 1200}{60} \\ &=3.14 \mathrm{m} / \mathrm{s} \end{aligned}
Clearance,
\begin{aligned} y&=\frac{50.05-50}{2}=0.025 \mathrm{mm}\\ By \quad \tau&=\mu \cdot \frac{d u}{d y} \text{ shear stress on shaft}\\ \tau&=0.03 \times \frac{3.14}{0.025 \times 10^{-3}}=3768 \mathrm{N} / \mathrm{m}^{2} \end{aligned}
Shear force on shaft,
\begin{aligned} F &=\tau \times \text { Area }=3768 \times \pi D \times L \\ &=3768 \times \pi \times 50 \times 10^{-3} \times 20 \times 10^{-3} \\ &=11.83 \mathrm{N} \end{aligned}
\begin{aligned} \text { Torque: } T &=F \times \frac{D}{2}=11.83 \times \frac{50 \times 10^{-3}}{2} \\ &=0.2953 \mathrm{Nm} \end{aligned}
\begin{aligned} \text { Power loss } &=\frac{2 \pi N T}{60}=\frac{2 \pi \times 1200 \times 0.2953}{60} \\ &=37.1 \mathrm{W} \end{aligned}
\begin{aligned} u &=\frac{\pi D N}{60}=\frac{\pi \times 50 \times 10^{-3} \times 1200}{60} \\ &=3.14 \mathrm{m} / \mathrm{s} \end{aligned}
Clearance,
\begin{aligned} y&=\frac{50.05-50}{2}=0.025 \mathrm{mm}\\ By \quad \tau&=\mu \cdot \frac{d u}{d y} \text{ shear stress on shaft}\\ \tau&=0.03 \times \frac{3.14}{0.025 \times 10^{-3}}=3768 \mathrm{N} / \mathrm{m}^{2} \end{aligned}
Shear force on shaft,
\begin{aligned} F &=\tau \times \text { Area }=3768 \times \pi D \times L \\ &=3768 \times \pi \times 50 \times 10^{-3} \times 20 \times 10^{-3} \\ &=11.83 \mathrm{N} \end{aligned}
\begin{aligned} \text { Torque: } T &=F \times \frac{D}{2}=11.83 \times \frac{50 \times 10^{-3}}{2} \\ &=0.2953 \mathrm{Nm} \end{aligned}
\begin{aligned} \text { Power loss } &=\frac{2 \pi N T}{60}=\frac{2 \pi \times 1200 \times 0.2953}{60} \\ &=37.1 \mathrm{W} \end{aligned}
Question 10 |
A journal bearing has a shaft diameter of 40mm and a length of 40 mm. the shaft is rotating at 20
rad/s and the viscosity of the lubricant is 20mPa-s. the clearance is 0.020 mm. the loss of torque
due to the viscosity of the lubricant is approximately:
0.040 Nm | |
0.252 Nm | |
0.400 Nm | |
0.6562 Nm |
Question 10 Explanation:
\begin{aligned} \tau &=\frac{\mu v}{c}=\frac{20 \times 10^{-3} \times 20 \times 20 \times 10^{-3}}{0.02 \times 10^{-3}} \\ &=400 \mathrm{N} / \mathrm{m}^{2}\\ \text { Force }&=\tau A=400 \times 3.14 \times 0.04 \times 0.04 \mathrm{N} \\ &\qquad(\because F=\tau \cdot \pi d l) \\ \text { Torque }&=400 \times 3.14 \times 0.04 \times 0.04 \times 0.02 & \\ &\qquad(\because\text { Torque }=F \times r)\\ &=0.040 \mathrm{Nm} \end{aligned}
There are 10 questions to complete.
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.