Design of Machine Elements Test 3

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Design of Machine Elements Test 3

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Weekly Quiz Competition

Question 1
1 / -0

A roller bearing with bore diameter of 40 mm is subjected to radial force of 25 kN. The coefficient of friction is 0.0012 and the speed of rotation is 1440 rpm. The power lost in friction in W is ________.

A
88-93

B
884-931

C
887-936

D
889-932

Solution

Concept:
The power lost in friction for a bearing is given by:
\({P_{loss}} = \mu WV = \mu W \times \frac{{{\rm{\pi DN}}}}{{60}}\)
\({P_{loss}} = \mu WV = \mu W \times \frac{{{\rm{\pi DN}}}}{{60}}\)
Calculation:
Given: W = 25 kN, μ = 0.0012, N = 1440 rpm, D = 40 mm = 0.04 m
\({P_{loss}} = 0.0012 \times 25000 \times \frac{{{\rm{\pi }} \times 0.04 \times 1440}}{{60}} = 90.477\;W = 0.09\;kW\)
Question 2
1 / -0

Calculate the torque transmitted by the clutch having four shoes, with spring force of 700 N and centrifugal force of 4000 N. Rim diameter is 560 mm and coefficient of friction is 0.3

A
1108.8 N-m

B
1100 N-m

C
1000 N-m

D
1579 N-m

Solution

Concept:
P_c = Centrifugal force acting on each shoe
P_s = inward force on each shoe exerted by the spring
Net outward radial force (i.e. centrifugal force) with which the shoe presses against the rim at the running speed: P_c - P_s
Frictional force acting tangentially on each shoe: F = μ(Pc - Ps)
Frictional torque acting on each shoe = F × R = μR(Pc - Ps)
Total frictional torque transmitted = nμR(Pc - Ps)
where n is number of shoes.
Calculation:
Total frictional torque transmitted:
\(T = n\mu R\left( {{P_c} - {P_s}} \right) = 4 \times 0.3 \times \left( {\frac{{0.56}}{2}} \right) \times \left( {4000 - 700} \right) = 1108.8\:Nm\)
Question 3
1 / -0

A 6203 single row deep groove ball bearing has a basic dynamic load rating is 7.34 kN and life of bearing expectancy is 37 million revolution cycles, then equivalent static load will be

A
3.67 kN

B
2.2 kN

C
1.2 kN

D
0.98 kN

Solution

Concept:
For ball bearing, n = 3
C = 7.35 kN = Basic dynamic load capacity
L₁₀ = 37 × 10⁶ rev = life of bearing
\({L_{10}} = {\left( {\frac{C}{P}} \right)^3}\)
Calculation:
\(\therefore 37 \times {10^6} = {\left( {\frac{{7.35}}{P}} \right)^3}\)
∴ P = 2.205 kN
Question 4
1 / -0

A shoe-brake makes an angle of 120° at drum-center and the coefficient of friction between them is 0.25. The equivalent coefficient of friction is ______

A
0.2-0.3

B
0.24-0.31

C
0.27-0.36

D
0.29-0.32

Solution

Explanation:
Given:
2θ = 120°, μ = 0.25
Equivalent coefficient of friction,
\(\mu ' = \mu \left[ {\frac{{4\sin \theta }}{{2\theta + \sin \theta }}} \right]\)
\(\mu ' = 0.25\left[ {\frac{{4\sin 120^\circ }}{{\frac{{2\pi }}{3} + \sin 120^\circ }}} \right]\)
∴ μ' = 0.292
Question 5
1 / -0

A hydrodynamic journal bearing is subjected to a radial load of 50 kN at a rotational speed of 1440 rpm. It has a journal diameter and bearing length of 100 mm both. The radial clearance is 0.12 mm. The Sommerfeld number is 0.013. The viscosity of the lubricating oil will be _____.

A
15.6 mPa.s

B
3.9 mPa.s

C
15.6 Pa.s

D
3.9 Pa.s

Solution

Concept:
The Sommerfeld number is given by:
\(S = {\left( {\frac{r}{c}} \right)^2}\frac{{\mu {n_s}}}{P}\)
Where,
c = radial clearance (mm)
R = radius of bearing (mm)
r = radius of journal (mm)
μ = viscosity of the lubricant (N.S/mm²) or (MPa.s)
n_s = journal speed (rev/s)
P = unit bearing pressure (N/mm²)
Given:
r = 50 mm, L = 50 mm, W = 50 kN = 50000 N, N = 1440 rpm = 24 rps
Calculation
\(P = \frac{W}{{Ld}} = \frac{{50000}}{{100 \times 100}} = 5\;N/m{m^2}\)
\(S = {\left( {\frac{r}{c}} \right)^2}\frac{{\mu \;{n_s}}}{p}\)
\(0.013 = {\left( {\frac{{50}}{{0.12}}} \right)^2} \times \frac{{\mu \times 24}}{5}\)
μ = 1.56 × 10^-8 N.S/mm² = 1.56 × 10^-8 MPa.s = 15.6 × 10^-9 MPa.s
μ = 15.6 mPa.s = 15.6 centipoise
Question 6
1 / -0

A 1500 kg automobile moving on a level road at 108 km/hr is to be stopped in a distance of 100 meter, the wheel of diameter being 80 cm. All frictional energy except for the brake is to be neglected.
What is total average breaking torque must be applied (kNm)?

A
2.5-2.9

B
2.54-2.91

C
2.57-2.96

D
2.59-2.92

Solution

Concept:
\({\rm{Kinetic\;energy\;}} = \frac{1}{2}m{V^2}\)
\({\rm{Tangential\;Breaking\;force\;}} = \frac{{Kinetic\;Energy}}{{distance}}\)
Breaking torque = Tangential force × radius
Calculation:
V = 108 km/hr = 30 m/s, m = 1500 kg, d = 80 cm = 800 mm
\(Kinetic\;Energy\;\left( {K.E.} \right) = \frac{1}{2}m{V^2} = \frac{1}{2} \times 1500 \times {30^2}\)
∴ KE = 675 kN.m
Now,
\({\rm{Tangential\;breaking\;force\;}} = \frac{{KE}}{{100}}\)
\({\rm{Tangential\;breaking\;force\;}} = \frac{{675\; \times \;{{10}^3}N.m}}{{100\;m}}\)
∴ F_r = 6750 N
Now,
\(\therefore {\rm{\;Breaking\;torque\;}} = {F_r} \times \frac{d}{2}\)
\(\therefore {\rm{\;Breaking\;torque\;}} = 6750 \times \frac{{800}}{{1000\; \times \;2}}\)
∴ T = 2.7 kN.m
Question 7
1 / -0

A roller bearing with a radial load of 9 kN and axial load of 3 kN rotates at 900 rpm. The bearing has an average life of 6000 hours. The service factor is 1.5 and the inner race rotates. For radial and axial load factors equal to 0.6 and 0.9 respectively, the dynamic capacity of bearing (in kN) is ______

A
42-43

B
424-431

C
427-436

D
429-432

Solution

Explanation:
Given:
F_r = 9 kN, F_a = 3 kN, N = 900 rpm, C_s = 1.5, X = 0.6, Y = 0.9
For inner race rotation, V = 1
Now,
Average life, L_avg = 5 × rated life (L₉₀)
\({L_{90}} = \frac{{{L_{avg}}}}{5} = \frac{{6000}}{5} = 1200\;hours\)
\({L_{90}} = 60\;N\;{\left( {{L_{90}}} \right)_{hours}} = 60 \times 900 \times 1200\)
L₉₀ = 64.8 × 10⁶ rev
∴ L₉₀ = 64.8 million revolutions
Now,
Equivalent load (P) = (XVf_r + Y.F_a) C_s
P = 1.5 (0.6 × 1 × 9 + 0.9 × 3)
∴ P = 12.15 kN
For roller bearing,
\({L_{90}} = {\left( {\frac{C}{P}} \right)^{\frac{{10}}{3}}}\)
\(64.8 = {\left( {\frac{C}{{12.15}}} \right)^{\frac{{10}}{3}}}\)
∴ C = 42.46 kN
Question 8
1 / -0

Find the maximum force which can be applied on a helical compression spring of mean coil diameter 150 mm and of 20 mm wire diameter. The permissible shear stress of spring wire is 400 N/mm²

A
6.99 kN

B
7.49 kN

C
5.32 kN

D
5.90 kN

Solution

Explanation:
Given:
D = 150 mm, d = 20 mm, τ_per = 400 N / mm²
Now,
Spring index (C) is:
\(C = \frac{D}{d} = \frac{{150}}{{20}}\)
C = 7.5
\({\rm{Wahl\;factor\;}}\left( {{k_w}} \right) = \frac{{4c - 1}}{{4c - 4}} + \frac{{0.615}}{c}\)
\({k_w} = \frac{{4\left( {7.5} \right) - 1}}{{4\left( {7.5} \right) - 4}} + \frac{{0.615}}{{7.5}}\)
∴ k_w = 1.197
Now,
Permissible shear stress,
\(\tau = {k_w}\left( {\frac{{8PC}}{{\pi {d^2}}}} \right)\)
\(400 = 1.197\left( {\frac{{8P \times 7.5}}{{\pi \times {{20}^2}}}} \right)\)
P = 6998.81 N
∴ P = 6.998 kN

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Re-Attempt Weekly Quiz Competition

Design of Machine Elements Test 3

Result

Design of Machine Elements Test 3

Score

-

Rank

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SHARING IS CARING

Question 1

88-93

884-931

887-936

889-932

Solution

Question 2

1108.8 N-m

1100 N-m

1000 N-m

1579 N-m

Solution

Question 3

3.67 kN

2.2 kN

1.2 kN

0.98 kN

Solution

Question 4

0.2-0.3

0.24-0.31

0.27-0.36

0.29-0.32

Solution

Question 5

15.6 mPa.s

3.9 mPa.s

15.6 Pa.s

3.9 Pa.s

Solution

Question 6

2.5-2.9

2.54-2.91

2.57-2.96

2.59-2.92

Solution

Question 7

42-43

424-431

427-436

429-432

Solution

Question 8

6.99 kN

7.49 kN

5.32 kN

5.90 kN

Solution

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