2 Chapter Shaft Design
2 Chapter Shaft Design
2 Chapter Shaft Design
Objectives: at the end of this chapter, the students/ readers should be able to:
1. Define shaft;
2. Derive formulas and apply it in solving problems; and
3. Practice exam in the multiple choice test to master the usual question and problem in the licensure examination.
Shaft is a rotating or stationary member having a circular cross section much smaller in diameter than the shaft length and having mounted on it power-transmitting elements as
pulleys, gears, belts, chains cams, flywheels, cranks, sprocket, and rolling element bearing. Shaft could be main shaft, line shaft, transmission shaft, machine shaft and other
similar form.
Main shaft is the receiving its power from the engine or motor and transmitting power to other parts.
Line shaft is a shaft used to transmit power from a central source to individual machines.
Transmission shaft is the shaft transmitting power between the source and the machines absorbing the power.
Machine shaft is the shaft used as an integral part of the machine itself.
Axle is a stationary member of primarily loaded in bending with gears, pulleys and wheels rotating on it.
Counter shaft is a secondary shaft that is driven by a main shaft from which power is supplied to a machine part. It is also known as a spindle.
Jackshaft is a countershaft, especially when used as an auxiliary shaft between two other shafts.
Head shaft is the shaft driven by a chain and mounted at the delivery end of a chain conveyor. It serves as the mount for a sprocket which driven the drag chain.
Formula:
𝐷3 𝑁
𝑃=
80
Where:
P = power transmitted
D = shaft diameter
N = rotative speed
80 = constant value
𝐷3 𝑁
𝑃=
53.5
Where:
P = power transmitted
D = shaft diameter
N = rotative speed
4. Power transmitted
𝑃 = 2𝜋𝑇𝑁
Where:
P = power transmitted
N = rotative speed
T = torque
Where:
T = torque transmitted
𝑆𝑠 = torsional shear stress
𝑍𝑝 = polar section modulus
Where:
T = torque transmitted
D = shaft diameter
𝑆𝑠 = torsional shear stress
5.1 = constant value
𝑤ℎ𝑒𝑟𝑒:
𝑀 = bending moment
D = shaft diameter
𝑆𝑡𝑚𝑎𝑥 = maximum tensile stress
𝑆𝑠𝑚𝑎𝑥 = maximum shearing stress
T = twisting moment
12. Combined torsion and bending considering shock and fatigue factor
16
𝑆𝑡𝑚𝑎𝑥 = [𝐾 𝑀 + √(𝐾𝑚 𝑀)2 + (𝐾𝑠 𝑇)2]
𝜋𝐷3 𝑚
16
𝑆𝑠𝑚𝑎𝑥 = √(𝐾𝑚 𝑀)2 + (𝐾𝑠 𝑇)2
𝜋𝐷3
𝑤ℎ𝑒𝑟𝑒:
M = bending moment
D = shaft diameter
𝑆𝑡𝑚𝑎𝑥 = maximum tensile stress
𝑆𝑠𝑚𝑎𝑥 = maximum shearing stress
T = twisting moment
Km = numerical combined shock and fatigue factor to be applied in every case to the computed bending moment
Ks = numerical combined shock and fatigue factor to be applied in every case to the computed torsional moment
Where:
V = vertical load
A = cross-sectional area
D = diameter
14. Angular deformation (solid)
𝑇𝐿 𝑇𝐿
𝜃= = 𝜋 4
𝐽𝐺 𝐷 𝐺
32
Where:
𝜃 = angular deformation
T = torque
L = length
D = diameter
G = modulus of rigidity
15. Angular deformation (hollow)
𝑇𝐿
𝜃= 𝜋
(𝐷𝑜 4 − 𝐷𝑖 4 )𝐺
32
𝑤ℎ𝑒𝑟𝑒:
𝜃 = angular deformation
T = torque
Do = outside diameter
Di = inside diameter
G = modulus of elasticity
15 3 7 11 15 2
𝑖𝑛 1 𝑖𝑛 1 𝑖𝑛 1 𝑖𝑛 1 𝑖𝑛 2 𝑖𝑛
16 16 16 16 16 16
2 15 7 15 7 15
2 𝑖𝑛 2 𝑖𝑛 3 𝑖𝑛 3 𝑖𝑛 4 𝑖𝑛 4 𝑖𝑛
16
16 16 16 16 16
7 15 1 7 in 1 8in
5 𝑖𝑛 5 𝑖𝑛 6 𝑖𝑛 7 𝑖𝑛
16 16 2 2
Instruction: Choose the correct answer in the problems below and check your answers whether you PASS the TEST or NOT and then answer it again until you master it.
Solution to Test 16
1. C 3000 psi
From machinery’s handbook
D = 0.29 4√𝑇
𝑖𝑛.
D = 0.29√3142𝑓𝑡. 𝑙𝑏 𝑥 12
𝑓𝑡
D = 4.00 in.
2. C 192.5 lbs
W = wiegth
2.205𝑙𝑏𝑠
W = 87.3 kg x
1𝑘𝑔
W= 192.5 lbs
3. S 45 hp
P = power
𝐷3𝑁
P=
80
7 3
(28) (150)
P=
80
P = 45 hp
4. D 2.38 in.
D = shaft diameter
3 80𝑃
D=√
𝑁
3 (80 (30))
D=√
180
D = 2.38 in
5. A 67 mm
D = shaft diameter
3 80𝑃
D=√
𝑁
3 100
80( )
D=√ 0.746
600
D = 67 mm
6. C 76 mm
D = shaft diameter
3 80𝑃
D=√
𝑁
3 100
80( )
D=√ 0.746
400
D = 2.9929 in
D = 76 mm
7. A 23,132 watts
P = power transmitted
𝐷3𝑁
P=
38
1𝑖𝑛 3
(40𝑚𝑚 𝑥 25.4𝑚𝑚) (300)
P=
38
746 𝑤𝑎𝑡𝑡𝑠
P =30.833 hp x
1ℎ𝑝
P = 23,132 watts
8. D 77 mm
3 80𝑃
D=√
𝑁
3 100
80( )
D = √ 0.746
800
25.4𝑚𝑚
D = 2.9929 in x
1𝑖𝑛
D = 77 mm
7
9. A 1 𝑖𝑛.
8
T = torque transmitted
𝑃
T=
2𝜋𝑁
25 𝑥 33000 𝑥 12
T=
2𝜋(225)
T = 7002.82 in.lb
D = diameter
3 16𝑇
D=√
𝜋𝑆𝑠
3 10(7002.82)
D=√
𝜋(6000)
D = 1.81 in.
7
D = 1 𝑖𝑛
8
10. B 1.75
D = diameter
3 53.5𝑃
D=√
𝑁
3 53.5(20)
D=√
200
D = 1.75 in
11. B 2.75 in
From machinery’s handbook
D = shaft diameter
4 𝑃
D = 4.6 √
𝑁
4 15
D = 4.6√
120
D = 2.74 in.
12. B 2.75 in
From machinery’s handbook
D = shaft diameter
4 𝑃
D = 4.6 √
𝑁
4 15
D = 4.6√
120
D = 2.74 in.
13. C 72 kw
P = power transmitted
𝐷3𝑁
P=
38
55 3
(25.4) (360)
P=
38
0.746𝑘𝑤
P = 96.18 hp x
1ℎ𝑝
P = 72 kw
14. C 61 mm
D = shaft diameter
3 (80𝑃)
D=√
𝑁
3 100
80( )
D=√ 0.746
800
25.4 𝑚𝑚
D = 2.375 in x
1𝑖𝑛
D = 60.34 mm
Solving for 𝜃
0.08°0 𝜋
𝜃= 𝑥
𝑓𝑡 1800
𝜃 = 0.001396 𝑟𝑎𝑑
Solving for polar moment of inertia
𝜋𝐷 4
J=
32
5 4
𝜋( )
12
J=
32
J = 0.00295906 𝑖𝑛4
16. C 296 kg
W = weight of a high chrome abrasion resistant steel
1𝑘𝑔
W = 652 lbs x
2.205 𝑙𝑏𝑠
W = 296 kg
17. C 84 mm
D = shaft diameter
3 80𝑃
D=√
𝑁
3 200
80( )
D=√ 0.746
600
25.4𝑚𝑚
D = 3.29 in. x
1𝑖𝑛.
D = 84mm
18. A 0.29
𝜃 = 𝑡𝑜𝑟𝑠𝑖𝑜𝑛𝑎𝑙 𝑑𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛
𝑇𝐿
𝜃=
𝐽𝐺
(28,000)(48) 180
𝜃= 7 4
𝑥
𝜋(3 ) 𝜋
8
(12 𝑥 106 )
32
𝜃 = 0.29°
Type equation here.
19. B 0.148
𝜃 = 𝑡𝑜𝑟𝑠𝑖𝑜𝑛𝑎𝑙 𝑑𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛
𝑇𝐿
𝜃=
𝐽𝐺
(3000)(1.4) 180°
𝜃= 120 4
𝑥
𝜋( ) 𝑁 1000𝑚𝑚 2 𝜋
1000
(80,000 )( )
32 𝑚𝑚2 1𝑚
𝜃 = 0.148°
20. A 27
P = power transmitted
𝐷3𝑁
P=
53.5
(2) 3 (180)
P=
53.5
P = 27 hpType equation here.
5
21. B2
8
𝑇 = 𝑡𝑜𝑟𝑞𝑢𝑒 𝑡𝑟𝑎𝑛𝑠𝑚𝑖𝑡𝑡𝑒𝑑
𝑃
𝑇=
2𝜋𝑁
𝑓𝑡−𝑙𝑏
(12ℎ𝑝) (3300 )
ℎ𝑝.𝑚𝑖𝑛
𝑇= 𝑟𝑒𝑣
2𝜋𝑟𝑒𝑣 (120 )
𝑚𝑖𝑛
𝑇 = 525.21 𝑓𝑡. 𝑙𝑏
𝐷 = 𝑠ℎ𝑎𝑓𝑡 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟
4 𝑇𝐿
𝐷= √ 𝜋
𝜃( )𝐺
32
4 (525.21)(12)(12)
𝐷= √ 𝜋 𝜋
(0.08) ( ) (12,000,000)
180 32
𝐷 = 2.604 𝑖𝑛
5
𝐷 = 2 𝑖𝑛
8
1
22. D3
2
𝐷 = 𝑠ℎ𝑎𝑓𝑡 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟
3 80𝑃
𝐷= √
𝑁
3 80(200)
𝐷= √
400
1
𝐷 = 3.42 𝑖𝑛. 𝑜𝑟 3 𝑖𝑛.
2
23. A 68 mm
𝑇 = 𝑡𝑜𝑟𝑞𝑢𝑒 𝑡𝑟𝑎𝑛𝑠𝑚𝑖𝑡𝑡𝑒𝑑
10,000
𝑇= 120
2𝜋 ( )
60
𝑇 = 795.77 𝑁. 𝑚
𝐷 = 𝑠ℎ𝑎𝑓𝑡 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟
4 𝑇𝐿
𝐷= √ 𝜋
𝜃( )𝐺
32
4 795.77(1)
𝐷= √ 𝜋 𝜋 101.325
0.26 ( ) ( ) (12 𝑥 106 ) ( ) (1000)
180 32 14.7
1000𝑚
𝐷 = 0.068𝑚 𝑥
1𝑚
𝐷 = 68 𝑚𝑚
24. A 2.34in
𝑇 = 𝑡𝑜𝑟𝑞𝑢𝑒 𝑡𝑟𝑎𝑛𝑠𝑚𝑖𝑡𝑡𝑒𝑑
𝑓𝑡.𝑙𝑏
(20ℎ𝑝) (33,000 )
ℎ𝑝.𝑚𝑖𝑛
𝑇= 𝑟𝑒𝑣
2𝜋 (300 )
𝑚𝑖𝑛
𝑇 = 350.141 𝑓𝑡. 𝑙𝑏
4 350.14(1)
𝐷 = √𝜋 𝜋
(0.80) ( ) (12 𝑥 106 )(144)
32 180
𝐷 = 2.353𝑖𝑛
25. B 2.8 hp
𝑇 = 𝑡𝑜𝑟𝑞𝑢𝑒
𝑇 = 𝐹. 𝑟. 𝑓
𝑙𝑏
𝑇 = (200 ) (3𝑖𝑛)(6𝑖𝑛)(0.20)
𝑖𝑛
𝑇 = 720 𝑖𝑛. 𝑙𝑏
𝑃 = ℎ𝑜𝑟𝑠𝑒𝑝𝑜𝑤𝑒𝑟 𝑡𝑟𝑎𝑛𝑠𝑚𝑖𝑡𝑡𝑒𝑑
𝑃 = 2𝜋𝑇𝑁
1 1 1
𝑃 = 2𝜋(720) (240𝑥 ) ( ) ( )
60 12 550
𝑃 = 2.74 ℎ𝑝
Test 37
Instruction: Choose the correct answer in the problems below and check your answers whether you PASS the TEST or NOT and then answer it again until you master it.
Solution to Test 17
1. C 0.20
𝜃 = 𝑡𝑜𝑟𝑠𝑖𝑜𝑛𝑎𝑙 𝑑𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛
𝑇𝐿
𝜃=
𝐽𝐺
(3𝑥106 )(1400)
𝜃= 𝜋
( ) (100) 4(83,000)
32
180°
𝜃 = 0.00352 𝑟𝑎𝑑
𝜋𝑟𝑎𝑑
𝜃 = 0.20°
2. B 1.0𝑥106
𝑇 = 𝑡𝑜𝑟𝑞𝑢𝑒 𝑡𝑟𝑎𝑛𝑠𝑚𝑖𝑡𝑡𝑒𝑑
𝑃
𝑇=
2 𝜋𝑁
20(1𝑥106 )
𝑇= 200
2𝜋 ( )
60
𝑇 = 0.9549929𝑥106
3. C 1050 in.lb
𝑇 = 𝑡𝑜𝑟𝑞𝑢𝑒 𝑡𝑟𝑎𝑛𝑠𝑚𝑖𝑡𝑡𝑒𝑑
𝑃
𝑇=
2𝜋𝑁
(20)(33,000)(12)
𝑇=
2𝜋(1200)
𝑇 = 1050 𝑖𝑛. 𝑙𝑏
4. A 0.22°
𝜃 = 𝑡𝑜𝑟𝑠𝑖𝑜𝑛𝑎𝑙 𝑑𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛
𝑇𝐿
𝜃=
𝐽𝐺
(3.1𝑥106 )(1400) 180
𝜃= 𝜋 𝑥
(110) 4(80,000) 𝜋
32
𝜃 = 0.22°
5. D 3 in.
𝐷 = 𝑠ℎ𝑎𝑓𝑡 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟
3 (38)(200)
𝐷= √
300
𝐷 = 3.00 𝑖𝑛,
6. D 78
𝑃 = 𝑝𝑜𝑤𝑒𝑟 𝑡𝑟𝑎𝑛𝑠𝑚𝑖𝑡𝑡𝑒𝑑
𝐷3 𝑁
𝑃=
38
(2)3 (500)
𝑃=
38
0.746𝑘𝑤
𝑃 = 105.26 ℎ𝑝 𝑥
1ℎ𝑝
𝑃 = 78 𝑘𝑤
7. D 3.25
𝑇 = 𝑡𝑜𝑟𝑞𝑢𝑒
2𝜋𝑇𝑁
𝑃=
33,000
2𝜋𝑇(120)
20 =
33,000
𝑇 = 875.35 𝑓𝑡 − 𝑙𝑏𝑠
(875.35)(12)(12)
𝐷= √ 𝜋 𝜋
( ) (0.06) ( ) (12𝑥106 )
32 180
𝐷 = 3.18 𝑖𝑛. 𝑜𝑟 3.25 𝑖𝑛.
5
8. C2
8
𝑇 = 𝑡𝑜𝑟𝑞𝑢𝑒
𝑃 = 2𝜋𝑇𝑁
2𝜋𝑇(150)
10 =
33,000(12)
𝑇 = 4201.69 𝑖𝑛. 𝑙𝑏
𝐷 = 𝑠ℎ𝑎𝑓𝑡 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟
𝑇𝐿
𝐷= √ 𝜋
( ) (𝜃)(𝐺)
32
4201.69(12)
𝐷= √ 𝜋 𝜋
( )( ) (0.06)(12𝑥106 )
32 180
5
𝐷 = 2.53 𝑖𝑛 𝑜𝑟 2 𝑖𝑛
8
9. D 2.27 in
𝑇 = 𝑡𝑜𝑟𝑞𝑢𝑒
𝑃 = 2𝜋𝑇𝑁
2𝜋𝑇(180)
10 =
33,000(12)
𝑇 = 3501.41 𝑖𝑛. 𝑙𝑏
𝑆𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑠ℎ𝑎𝑓𝑡 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟
𝐷 = 𝑠ℎ𝑎𝑓𝑡 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟
4 𝑇𝐿
𝐷= √ 𝜋
( ) (𝜃)(𝐺)
32
4 3501.41(12)
𝐷 = √𝜋 𝜋
(0.08) ( ) (11.5𝑥106 )
32 180
1
𝐷 = 2.27 𝑖𝑛 𝑜𝑟 2 𝑖𝑛
4
7
10. D3
16
𝐷 = 𝑠ℎ𝑎𝑓𝑡 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟
3 80𝑃
𝐷= √
𝑁
3 80(200)
𝐷= √
400
7
𝐷 = 3.42 𝑖𝑛 𝑜𝑟 3 𝑖𝑛
16
11. C 0.22
𝜃 = 𝑡𝑜𝑟𝑠𝑖𝑜𝑛𝑎𝑙 𝑑𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛
𝑇𝐿
𝜃=
𝐽𝐺
3.1𝑥106 (1400)
𝜃 = 𝜋(110)
(80,000)
32
180
𝜃 = 0.037742 ( )
𝜋
𝜃 = 0.22 deg
12. B 1.50
𝑇 = 𝑡𝑜𝑟𝑞𝑢𝑒
𝑃
𝑇=
𝜔
13.31(550)(12)
𝑇=
15.7
𝑇 = 5591.08 𝑖𝑛. 𝑙𝑏
𝐷 = 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟
3 16𝑇
𝐷= √
𝜋𝑆
3 16(5591.08)
𝐷= √
𝜋(8500)
𝐷 = 1.50 𝑖𝑛
13. C 2265 rpm
𝑁 = 𝑟𝑜𝑡𝑎𝑡𝑖𝑣𝑒 𝑠𝑝𝑒𝑒𝑑
80𝑃
𝑁= 3
𝐷
80(250)
𝑁=
52.5 3
( )
25.4
𝑁 = 2265 𝑟𝑝𝑚
5
14. C1 𝑖𝑛
16
𝑇 = 𝑡𝑜𝑟𝑞𝑢𝑒
𝑃 = 2𝜋𝑇𝑁
2𝜋𝑇(1800)
75 =
33,000(12)
𝑇 = 2626.056𝑖𝑛. 𝑙𝑏
𝑆𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑠ℎ𝑎𝑓𝑡 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟
𝐷 = 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟
3 16𝑇
𝐷=√
𝜋𝑆
3 16(2626.056)
𝐷= √
𝜋(6000)
5
𝐷= 1 𝑖𝑛
16
15. C 13.31 hp
𝑁 = 𝑟𝑜𝑡𝑎𝑡𝑖𝑣𝑒 𝑠𝑝𝑒𝑒𝑑
15.7𝑥60
𝑁=
2𝜋
𝑁 = 150 𝑟𝑝𝑚
𝑃 = 𝑝𝑜𝑤𝑒𝑟 𝑡𝑟𝑎𝑛𝑠𝑚𝑖𝑡𝑡𝑒𝑑
𝐷3 𝑁
𝑃=
38
(1.5)3 (150)
𝑃=
38
𝑃 = 13.32 ℎ𝑝
3
16. D 2 in
4
𝐷 = 𝑠ℎ𝑎𝑓𝑡 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟
3 16𝑇
𝐷= √
𝜋𝑆𝑠
3 16(36,000)
𝐷= √
𝜋(10,000)
𝐷 = 2.636 𝑖𝑛
17. D 14.6
𝐹𝑟𝑜𝑚 𝑚𝑎𝑐ℎ𝑖𝑛𝑒𝑟𝑦 ′ 𝑠𝐻𝑎𝑛𝑑𝑏𝑜𝑜𝑘
2 𝐿
𝐷3 =
8.95
𝐷 = 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟
3
16 2
𝐷= ( )
8.95
𝐷 = 2.39 𝑖𝑛.
𝐹𝑟𝑜𝑚 𝑚𝑎𝑐ℎ𝑖𝑛𝑒𝑟𝑦 ′ 𝑠ℎ𝑎𝑛𝑑𝑏𝑜𝑜𝑘
𝐷 4
𝐻𝑝 = ( ) 𝑁
4.6
𝐻𝑝 = ℎ𝑜𝑟𝑠𝑒𝑝𝑜𝑤𝑒𝑟
2.39 4
𝐻𝑝 = ( ) (200)
4.6
𝐻𝑝 = 14.6
18. C 62
𝑇 = 𝑡𝑜𝑟𝑞𝑢𝑒
𝑆𝑆 𝜋𝐷3
𝑇=
16
𝑇 = 13.548𝐷3
𝑆𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑠ℎ𝑎𝑓𝑡 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟
𝐷 = 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟
𝑇𝐿
𝜃=
𝐽𝐺
1
𝜋 13.548𝐷3 ( )
2
0.08 ( )= 𝜋 4
180 𝐷 (79,300)
32
𝐷 = 0.0623𝑚
𝐷 = 62.32𝑚𝑚
19. B 384.81 lbs
𝑊 = 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓𝑐ℎ𝑟𝑜𝑚𝑢𝑖𝑚
𝑙𝑏𝑠
𝑊 = 174.5𝑘𝑔 𝑥 2.205
𝑘𝑔
𝑊 = 384.0 𝑙𝑏𝑠
20. A 163 in.lbs
𝑇 = 𝑡𝑜𝑟𝑞𝑢𝑒 𝑡𝑟𝑛𝑎𝑠𝑚𝑖𝑡𝑡𝑒𝑑
𝑃
𝑇=
2𝜋𝑁
𝑓𝑡−𝑙𝑏 𝑖𝑛
11ℎ𝑝𝑥33,000 𝑥 12
ℎ𝑝−𝑚𝑖𝑛 𝑓𝑡
𝑇= 𝑟𝑎𝑑 𝑟𝑒𝑣
2𝜋 (4250)
𝑟𝑒𝑣 𝑚𝑖𝑛
𝑇 = 163.12 𝑖𝑛. 𝑙𝑏
3
21. D 1080𝑖𝑛
𝑉 = 𝑣𝑜𝑙𝑢𝑚𝑒
𝑉 =𝑡𝑥𝑤𝑥𝐿
𝑖𝑛
𝑉 = 0.75𝑖𝑛 𝑥 6𝑖𝑛 𝑥 20𝑓𝑡 𝑥 12
𝑓𝑡
𝑉 = 1080 𝑖𝑛3
22. A 95 kw
𝑃 = 𝑝𝑜𝑤𝑒𝑟 𝑡𝑟𝑎𝑛𝑠𝑚𝑖𝑡𝑡𝑒𝑑
𝐷3 𝑁
𝑃=
38
1𝑖𝑛 3
(55𝑚 𝑥 ) (480)
25.4𝑚𝑚
𝑃=
38
𝑘𝑤
𝑃 = 128.24 ℎ𝑝 𝑥 0.746
ℎ𝑝
𝑃 = 95.67 𝑘𝑤
23. A 202
𝑃 = 𝑝𝑜𝑤𝑒𝑟 𝑡𝑟𝑎𝑛𝑠𝑚𝑖𝑡𝑡𝑒𝑑
𝑃 = 2𝜋𝑇𝑁
𝑟𝑎𝑑 𝑟𝑒𝑣 1𝑚𝑖𝑛
𝑃 = 2𝜋 (1200𝑁. 𝑚) (1200 )( )
𝑟𝑒𝑣 𝑚𝑖𝑛 60𝑠𝑒𝑐
𝑃 = 95.67 𝑘𝑤
24. C 1.53 in.
𝐷 = 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟
3 53.5𝑃
𝐷= √
180
3 53.5(12)
𝐷= √
180
𝐷 = 1.53
25. D 380
𝑇 = 𝑡𝑜𝑟𝑞𝑢𝑒
𝑃
𝑇=
2𝜋𝑁
750
𝑇= 1500
2𝜋 ( )
60
𝑇 = 4.7746 𝐾𝑁. 𝑚
𝑇 = 4.7746 𝑥 106 𝑁. 𝑚𝑚
𝑆𝑜𝑙𝑣𝑖𝑛𝑔 𝑓𝑜𝑟 𝑡𝑜𝑟𝑠𝑖𝑜𝑛𝑎𝑙 𝑠𝑡𝑟𝑒𝑠𝑠
𝑆𝑠 = 𝑡𝑜𝑟𝑠𝑖𝑜𝑛𝑎𝑙 𝑠𝑡𝑟𝑒𝑠𝑠
16𝑇
𝑆𝑠 =
𝜋𝐷3
16(4.7746 𝑥 106 )
𝑆𝑠 =
𝜋(40)3
𝑆𝑠 = 380 𝑀𝑃𝑎
Test 38
Instruction: Choose the correct answer in the problems below and check your answers whether you PASS the TEST or NOT and then answer it again until you master it.
Answer to test 38
1. B upper deviation is zero
2. D all of the above
3. C eight times
(𝑑𝐴 )3
4. C
(𝑑𝐵 )3
5. D twisting moment of both the shaft is the same
6. A maximum normal stress
7. Improves the fatigue life
8. C √𝑀 2 − 𝑇 2
9. B Ductile materials
10. A brittle materials
11. B rankine’d theory
12. B elastic limit
13. A modulus of elasticity
14. D 210 KN/𝑚𝑚2
15. B directly proportional
16. B 𝜋𝑠𝑇𝑆𝑢
17. C factor of safety
18. B4
19. A yield stress
20. B temperature gradient
21. C bearing stress
22. B 0.25 to 0.33
23. B double
24. C strain energy
25. B proof of resilience
Test 39
Instruction: Choose the correct answer in the problems below and check your answers whether you PASS the TEST or NOT and then answer it again until you master it.
1. The strain energy stored in a body, when suddenly loaded is ____the strain energy stored when same load is applied gradually.
A. Equal to
B. One half
C. Twice
D. Four times
2. When a machine member is subjected to torsion, the torsional shear stress set up in the member is
A. Zero at both the centroidal axis outer surface of the member
B. Maximum both the centroidal axis and outer surface of the member
C. Zero at the centroidal axis and maximum at the outer surface of the member
D. None of the above
3. The torsional shear stress on any cross-section normal to the axis is _____ the distance from the center of the axis.
A. Directly proportional to
B. Inversely proportional to
C. Equal to
D. Not equal to
4. The neutral axis of a beam is subjected to
A. Zero stress
B. Maximum tensile stress
C. Maximum compressive stress
D. Maximum shear stress
5. At the neutral axis of a beam
A. The layers are subjected to a maximum bending stress
B. The layers are subjected to tension
C. The layers are subjected to compression
D. The layers do not undergo any strain
6. The bending stress in a curved beam is
A. Zero at the centroid axis
B. Zero at the point other than centroid axis
C. None of the above
7. The maximum bending stress, in curved beam having symmetrical section, always, occur, at the
A. Centroid axis
B. Neutral axis
C. Inside fiber
D. Outside fiber
8. If d = diameter of solid shaft and S = permissible shear in shear for the shaft material, then torsional strength of shaft is written as
𝜋
A. 𝑑4 𝑆
32
B. 𝑑𝐿𝑜𝑔𝑒 𝑆
𝜋 3
C. 𝑑 𝑆
16
𝜋
D. 𝑑3 𝑆
32
9. If 𝑑𝑖 𝑎𝑛𝑑 𝑑𝑜 are the inner and outer diameter of a hollow shaft, then its polar moment of inertia is
𝜋
A. [𝑑𝑜 4 − 𝑑𝑖 4 ]
32
𝜋
B. [𝑑𝑜 3 − 𝑑𝑖 3 ]
32
𝜋
C. [𝑑𝑖 2 − 𝑑𝑖 2]
32
𝜋
D. [𝑑𝑖 − 𝑑𝑜 ]
32
10. Two shafts are under pure torsion are of identical length and identical weight and made of the same materials. The shaft A is solid and the shaft B is hollow, we can say
that
A. Shaft B is better than shaft A
B. Shaft A is better than shaft A
C. Both the shaft are equally good
D. None of the above
11. A slid shaft transmits a torque T. the allowable shear stress is 𝑆𝑠 . The siameter of the shaft is
3 16𝑇
A. √ 𝜋𝑆
𝑠
3 64𝑇
B. √ 𝜋𝑆
𝑠
3 32𝑇
C. √ 𝜋𝑆
𝑠
3 16𝑇
D. √ 𝑆𝑠
12. When a machine member is subjected to a tensile stress St due to direct load or bending and a shear stress S s due to torsion, then the maximum shear stress induces in
the member will be
1
A. √𝑆𝑡 2 + 4𝑆𝑠 2
2
1
B. √𝑆𝑡 2 − 4𝑆𝑠 2
2
C. √𝑆𝑡 2 + 4𝑆𝑠 2
D. 𝑆𝑡 2 + 4𝑆𝑠 2
1
12. A √𝑆𝑡 2 + 4𝑆𝑠 2
2
13. A brittle materials
14. B ductile materials
15. B maximum
16. C 1.5 times
17. C fluctuating stress
18. D completely reversed load
19. D below yield point
20. D endurance limit
21. A higher
22. C endurance limit to the working stress
23. C 0.50
24. C 0.55
25. C decrease
Test 40
Instruction: Choose the correct answer in the questions below and check your answers whether you PASS the TEST or NOT and then answer it again until you master it.
1. ME Board October 1999
Torsional deflection is a significant consideration in the design of shaft and the limit should be in the range of _________.
A. 0.004 to 0.006
B. 0.08 to 1
C. 0.006 to 0.008
D. 0.008 to 1
4. Supplementary Question
It is rotating machine member that transmits power.
A. Cam
B. Plate
C. Shaft
D. Flywheel
5. Supplementary Question
A stationary member carrying pulleys, wheels and etc. that transmit power.
A. Axle
B. Propeller shaft
C. Turbine shaft
D. Machine shaft
6.Supplementary Question
A line shaft is also known as
A. Counter shaft
B. Jackshaft
C. Main shaft
D. Head shaft
7.Supplementary Question
Which of the following shaft intermediate between a line shaft and a driven shaft?
A. Counter shaft
B. Jackshaft
C. Head shaft
D. All of the above
8.Supplementary Question
Short shafts on machine are called
A. core shafts
B. head shafts
C. medium shafts
D. spindles
9.Supplementary Question
For shafts, the shear due to bending is a maximum at the neutral plane where the normal stress is
A. constant
B. maximum
C. minimum
D. zero
10.Supplementary Question
Criteria for the limiting torsional deflection vary from 0.08 per foot of length for machinery shafts to _______ per foot.
A. 1°
B. 2°
C. 3°
D. 4°
11.Supplementary Question
For transmission shafts the allowable deflection is one degree in a length of _________ diameters.
A. 10
B. 15
C. 20
D. 25
12.Supplementary Question
An old rule thumb for transmission shafting is that the deflection should not exceed ________ of length between supports.
A. 0.01 in. per foot
B. 0.02 in. per foot
C.0.03 in. per foot
D. 0.04 in. per foot
13.Supplementary Question
In general for machinery shafts, the permissible deflection may be closer to
A. 0.02 in/ft
B. 0.01 in/ft
C. 0.002 in/ft
D. 0.030 in/ft
14.Supplementary Question
The speed at which the center of mass will equal the deflecting forces on the shaft with its attached bodies will then vibrate violently, since the centrifugal force changes its direction as the
shaft turns.
A. Critical speed
B. Geometrical speed
C. Means speed
D. Unit Speed
15.Supplementary Question
For shaft, the minimum value of numerical combined shock and fatigue factor to be applied in every case to the computed bending moment is
A. 1.0
B. 1.3
C. 1.5
D. 1.8
16.Supplementary Question
It is suggested that the design factor on the yield strength be about 1.5 for the smooth load, about 2 to 2.25 for minor shock loads, and _________ when the loading reverses during
operation.
A. 3.0
B. 4.0
C. 3.5
D. 4.5
17.Supplementary Question
A column is called short column when
A. the length is more than 30 times the diameter
B. slenderness ratio is more than 120
C. length is less than 8 times the diameter
D. the slenderness ratio is more than 32
18.Supplementary Question
For a circular shaft subjected to torque the value of shear stress
A. is uniform throughout
B. has maximum value at the axis
C. has maximum value at the surface
D. is zero at the axis and linearly increases to a maximum value at the
surface of the shaft
19.Supplementary Question
The compression members tend to buckle in the direction of
A. axis of load
B. perpendicular to the axis of load
C. minimum cross section
D. least radius of gyration
20.Supplementary Question
A reinforced concrete beam is considered to be made up of
A. clad materials
B. composite materials
C. homogeneous material
D. heterogeneous material
21.Supplementary Question
The column splice is used for increasing
A. strength of the column
B. cross-sectional area of the column
C. length of the column
D. all of the above
22.Supplementary Question
A simply supported beam has a uniformly distributed load on it, the bending moment is
A. triangle
B. parabola
C. semi-circle
D. rectangle
23.Supplementary Question
The power obtained by piston reaches flywheel through the
A. countershaft
B. crank shaft
C. transmission shaft
D. line shaft
24.Supplementary Question
There are two types of crankshafts
A. single piece and built up
B. forged and burned
C. rotary and stationary
D. none of these
25.Supplementary Question
Engines valve get open by means of
A. cam shaft
B. rocker shaft
C. crank shaft
1. B 0.08 to 1
2. C 15
3. B 0.010
4. C Shaft
5. A Axle
6. C main shaft
7. D All of the above
8. D spindles
9. D zero
10. A 1°
11. C 20
12. A 0.01 in. per foot
13. C 0.002 in/ft
14. A critical speed
15. C 1.5
16. D 4.5
17. C the length is less than 8 times the diameter
18. D is zero at the axis and linearly increase to a maximum value at the surface of the shaft
19. D least radius of gyration
20. D heterogeneous material
21. C length of the column
22. B parabola
23. B crank shaft
24. A single piece and built up
25. A cam shaft