DEPARTMENT OF
MECHANICAL ENGINEERING
Visca, Baybay City, Leyte, PHILIPPINES
Telefax: (053) 565-0600 local 1029
Email: coe@vsu.edu.ph
Website: www.vsu.edu.ph
Name: NOEL ROGIE M. COLLANO BSME- 4TH
Machine Design II
1. A 3-in. rotating shaft somewhat as shown in figure below carries a bevel gear whose
mean diameter is D in m =10 and which is keyed (profile) to the left end. Acting on the
gear are a radial force G =1570.8 lb, a driving force Q = 3141.6 lb. The thrust force is
taken by the right-hand bearing. Let a = 5 in and L =15 in; material, AISI C1040,
annealed. Base calculations on the Octahedral Shear Stress theory with variable stress.
Compute the indicated design factor N.
Given: For AISI C1040 @ annealed
From Graph of C1040 @ page 572
Assume temperature @ 400 degree Fahrenheit
𝑆𝑦 = 49ksi
𝑆𝑢 = 80ksi
Solving for endurance strength
𝑆𝑛 = 0.6𝑆𝑢
𝑆𝑛 = 0.6(80ksi)
𝑆𝑛 = 48ksi
𝑆𝑛 𝑆
𝑆
= 𝑆 𝑛𝑠
𝑦 𝑦𝑠
𝑆𝑛 48
𝑆𝑦
= 49
Solving for Torque
𝐹𝐷
T = 2𝑚
2(3141.6)(10)
T=
2
T = 31,416 in.lb
Solving for Shear stress
16𝑇
𝑆𝑠 = 𝜋𝐷3
16(31,416)
𝑆𝑠 = 𝜋(3)3
𝑆𝑠 = 5,925.93 psi
𝑆𝑠 = 𝑆𝑚𝑠
𝑆𝑎𝑠 = 0
Solving for equivalent shear stress
𝑆
𝑆𝑒𝑠 = 𝑆 𝑛 𝑆𝑚𝑠 +𝐾𝑓𝑠 𝑆𝑎𝑠
𝑦
48
𝑆𝑒𝑠 = 49 (5,925.93) + 0
48
𝑆𝑒𝑠 = (5,925.93)
49
DEPARTMENT OF
MECHANICAL ENGINEERING
Visca, Baybay City, Leyte, PHILIPPINES
Telefax: (053) 565-0600 local 1029
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𝑆𝑒𝑠 = 5,804.99 psi
QDm 3141.6(10)
=
2 2
QDm
= 15,708 in.lb
2
∑ 𝑀𝐸 = 0
QDm
- 2 - G (5) + 15Av = 0
15,708 = 5 (1,570.8) + 15Av
15708−5(1570.8)
Av = 15
Av = 523.6 lb
Av + Bv = 1,570.8
523.6 + Bv = 1,570.8
Bv = 1047.2 lb
(523.6 lb)(15) = 7,854 in.lb
(1,570.8)(10) = 15,708 in.lb
𝑀𝐶𝑣 = 1,5708 in.lb
𝑀𝐵𝑣 = 7,854 in.lb
∑ 𝑀𝐵 = 0
0 = -15(Ah) + 5(6,283.2)
15(Ah) = 5(6,283.2)
Ah = 2,094.4 lb
∑ 𝐹ℎ = 0
Bh = Ah + F
Bh = 2,094.4 + 6,283.2
Bh = 8,377.6 lb
𝑀𝐶𝑣 = 0
𝑀𝐵ℎ = (15) (2094.4)
𝑀𝐵ℎ = 31,416 in.lb
Maximum Moment
M = √(𝑀𝐵ℎ) 2 + (𝑀𝐵𝑣 )2
M = √(31,416)2 + (7,854)2
M = 32382.87 in.lb
𝑆𝑚𝑠 = 0
32𝑀
𝑆𝑎𝑠 = 𝜋𝐷3
32(32,382.87)
𝑆𝑎𝑠 =
𝜋(3)3
𝑆𝑎𝑠 = 12,216.63 psi
𝑆 𝐾𝑓 𝑆𝑎𝑠
𝑆𝑒 = 𝑆 𝑛 𝑆𝑚 + 𝑆𝐹
𝑦
DEPARTMENT OF
MECHANICAL ENGINEERING
Visca, Baybay City, Leyte, PHILIPPINES
Telefax: (053) 565-0600 local 1029
Email: coe@vsu.edu.ph
Website: www.vsu.edu.ph
𝑆 𝐾𝑓 𝑆𝑎𝑠
𝑆𝑒 = 𝑆 𝑛 𝑆𝑚 +
𝑦 𝑆𝐹
𝐾𝑓 = 1.0 @ assume value
1.0
𝑆𝑒 = 0 + (0.85 ) (12,216.63)
𝑆𝑒 = 14,372.50 psi
Maximum shear theory 𝑆𝑛𝑠 = 0.5𝑆𝑛
1
1 14,372.50 2 5,804.99
= [( ) +( ) 2 ]2
𝑁 48,000 0.5(48,000)
1
1
= [(0.0896) + (0.0585)]2
𝑁
1
( = 0.3848) N
𝑁
N = 2.598
2. A cast-steel (SAE 080, N & T) pulley is keyed to a 2 1/2-in. shaft by means of a
standard square key, 3 ½ -in. long, made of cold-drawn SAE 1015. The shaft is made of
cold-drawn AISI 1045. If the shaft is in virtually pure torsion, and turns at 420 rpm, what
horsepower could the assembly safely transmit (steady loading)?
Given:
For Pulley:
Cast-Steel @ SAE 080, N & T @ Table 6 page 570 – 571
𝑆𝑦 = 40ksi
For Key:
SAE 1015, cold drawn @Table 7 page 576 – 577
𝑆𝑦 = 63ksi
For Shaft:
D = 2 ½ in
L = 3 ½ in = 3.5 in
N = 420 rpm
Table 19 @ D = 2.5 in page 594
5
b = 8 𝑖𝑛
7
t = 16 𝑖𝑛
5
For Square Key @ b = 8 𝑖𝑛
N = 1.5 for steady loading (smooth)
For Shaft: Solving forTorque
𝑆𝑠 π𝐷3
T= 16
(0.6)𝑆𝑦 π𝐷3
T= 𝑁16
(85)0.6π(2.5)3
T= (1.5)16
DEPARTMENT OF
MECHANICAL ENGINEERING
Visca, Baybay City, Leyte, PHILIPPINES
Telefax: (053) 565-0600 local 1029
Email: coe@vsu.edu.ph
Website: www.vsu.edu.ph
T = 104.31 in.lb
For Key: torque, shear, compression
𝑆 𝑙𝑏𝐷
T = 𝑠2
𝑆𝑦𝑠
𝑆𝑠 = 𝑁
𝑆𝑦 (0.6)
𝑆𝑠 = 1.5
(63)(0.6)
𝑆𝑠 =
1.5
𝑆𝑠 = 25.2ksi
5
(25.3)(3.5)( )(2.5)
8
T= 2
T = 69 in.kips < 104.31 in.kips from shaft Torque
By compression:
𝑆 𝐿𝑡𝐷
T = 𝑐4
𝑆𝑦(𝐹𝑜𝑟 𝑝𝑢𝑙𝑙𝑒𝑦)
𝑆𝑐 = 𝑁
5
(26.7)(3.5)( )(2.5)
8
T= 4
T = 36.50 in.kips < 104.31 in.kips from shaft Torque
Therefore used T = 36,503.90 in.kips
𝑇𝑛
Hp = 63,000
(36,503.90)(420)
Hp =
63,000
Hp = 243.359 hp
3. A bolt, 1 1/8 in.-7-UNC-2, is subjected to a tensile load of 10,000 lb. The head has a
thickness of ¾ in. and the nut a thickness of 1 in. If the material is SAE grade 2 (Table
5.2), find the design factor as based on ultimate stresses (a) of the threaded shank, (b)
of the head against being sheared off, and (c) of the bearing surface under the head.
Given:
For SAE grade 2 @ Table 5.2 page 161
𝑆𝑢 = 55ksi
𝑆𝑢𝑠 = 0.75𝑆𝑢
𝑆𝑢𝑠 = 0.75(55)
𝑆𝑢𝑠 = 41.25ksi
𝟏
For 1𝟖 in -7- UNC – 2. @ Table 14 page 587
𝐴𝑠 = 0.763 sq.in.
𝟏𝟏
Head @ A = 1𝟏𝟔
F = 10,000lb
𝑭
s=𝐴
𝑠
DEPARTMENT OF
MECHANICAL ENGINEERING
Visca, Baybay City, Leyte, PHILIPPINES
Telefax: (053) 565-0600 local 1029
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a.)
𝟏𝟎,𝟎𝟎𝟎
s = 0.763
s = 13,106.2psi
𝑠
N = 𝑠𝑢
𝑑
55,000
N = 13,106.2
N = 4.19
b.)
𝐹
𝑠𝑠 = 𝜋𝐷𝑡
3
t = 4 in
10,000
𝑠𝑠 = 𝟏 𝟑
𝜋(𝟏 )( )
𝟖 𝟒
𝑠𝑠 = 3,773psi
𝑠𝑢𝑠
N= 𝑠𝑠
(0.75)(55,000)
N= 3,773
N = 11
c.)
360𝑂
Ɵ = 12
Ɵ = 30𝑂
1 𝐴 𝐴
A = 6(2) (2) ( 2 )( 2 tan Ɵ)
𝟏𝟏2
1 1
A = 6(2) ( ) ( tan 30𝑂 ) 𝟏𝟔
2 2
A = 2.5 sq.inches
𝐹
𝑠𝑏 = 𝐴−𝐴
𝑏
10,000
𝑠𝑏 = 𝜋 𝟏
2.5 − (𝟏 )2
4 𝟖
𝑠𝑏 = 6,640psi
𝑆
N= 𝑢
𝑠𝑏
55,000
N = 6,640
N = 8.28
DEPARTMENT OF
MECHANICAL ENGINEERING
Visca, Baybay City, Leyte, PHILIPPINES
Telefax: (053) 565-0600 local 1029
Email: coe@vsu.edu.ph
Website: www.vsu.edu.ph
4. pair of steel gears is defined by Pd = 8, b = 1.5 in, Np = 25, Ng = 75, e = 0.001 in,
200 F.D. If these gears may transmit continuously and without failure 75 hp at 1140 rpm
of the pinion, what horsepower would be satisfactory for np =1750 rpm?
Given:
𝑃𝑑 = 8
B = 1.5 in
𝑁𝑝 = 25
𝑁𝑔 = 75
E = 0.001 in
Solving:
𝑁𝑝
𝐷𝑝 = 𝑃
𝑑
25
𝐷𝑝 = 8
𝐷𝑝 = 3.125 in
𝜋𝐷𝑝 𝑛𝑝
𝑉𝑚 = 12
𝜋(3.125)(1,140)
𝑉𝑚 = 12
𝑉𝑚 = 932.66 fpm
Solving:
33,000ℎ𝑝
𝐹𝑡 = 𝑉
𝑚
33,000(75)
𝐹𝑡 = (932.66)
𝐹𝑡 = 2,653.7lb
Solving @ Table 25, page 601, e = 0.001 in, 20𝑜 F.D
k = 0.111e
C = 1,660
0.05𝑉𝑚 (𝐶𝑏+𝐹𝑡 )
𝐹𝑑 = 𝐹𝑡 + 1
0.05𝑉𝑚 +(𝐶𝑏+ 𝐹𝑡 ) 2
0.05(932.66)(1660)(1.5)+2,653.7 )
𝐹𝑑 = 2,653.7 + 1
0.05(932.66)+((1660)(1.5)+ 2,653.7) 2
118,769.87
𝐹𝑑 = 2,653.7 + 118.35
𝐹𝑑 = 3,657.24lb
Solving for 𝑛𝑝 = 1,750rpm
𝜋𝐷𝑝 𝑛𝑝
𝑉𝑚 =
12
𝜋(3.125)(1,750)
𝑉𝑚 = 12
𝑉𝑚 = 1,431.71 fpm
DEPARTMENT OF
MECHANICAL ENGINEERING
Visca, Baybay City, Leyte, PHILIPPINES
Telefax: (053) 565-0600 local 1029
Email: coe@vsu.edu.ph
Website: www.vsu.edu.ph
0.05𝑉𝑚 (𝐶𝑏+𝐹𝑡 )
𝐹𝑑 = 𝐹𝑡 + 1
0.05𝑉𝑚 +(𝐶𝑏+ 𝐹𝑡 ) 2
71.5855(2,490+𝐹𝑡 )
3,657.24= 𝐹𝑡 + 1
71.5855+(2,490+ 𝐹𝑡 ) 2
𝐹𝑡 = 2,260lb
𝐹𝑡 𝑉𝑚
hp = 33,000
(2,260)(1,431.71)
hp = 33,000
hp = 98.050hp
5. A 10-in. medium double leather belt, cemented joints, transmits 60 hp from a 9-in.
paper pulley to a 15-in. pulley on a mine fab; dusty conditions. The compensator-started
motor turns 1750 rpm; C = 42 in. This is an actual installation. (a) Using the general
equation, determine the horsepower for this belt. (b) Estimate the service factor from
Table 17.7 and apply it to the answer in (a).
Given:
From Table 17.1 & 17.2 page 450 - 451
𝐶𝑚 = 0.67 @ squirrel cage, compressor starting
𝐶𝑝 = 0.7 @ pulley size from 9 to 12
𝐶𝑓 = 0.74 @ Oily, wet or dusty atmosphere
b = 10 in
Solving:
𝜋𝐷𝑝 𝑛𝑝
a.) 𝑉𝑚 =
12
𝜋(9)(1750)
𝑉𝑚 = 12
𝑽𝒎 = 4,123 fpm
ℎ𝑝 = (11.15) (10) (0.67) (0.7) (0.74)
hp = 38.69hp
12𝑝𝑉𝑠2 𝑒 𝑓Ɵ −1
b.) 𝐹1 - 𝐹2 = bt (s - )( )
32.2 𝑒 𝑓Ɵ
b = 10 in
p = 0.035 lb/cu.in
s = 400ƞ
ƞ = 1.0 for cement joint
s = 400 psi
𝐷 −𝐷
Ɵ=π- 2 1
𝐶
DEPARTMENT OF
MECHANICAL ENGINEERING
Visca, Baybay City, Leyte, PHILIPPINES
Telefax: (053) 565-0600 local 1029
Email: coe@vsu.edu.ph
Website: www.vsu.edu.ph
𝐷2 − 𝐷1
Ɵ=π- 𝐶
15− 9
Ɵ = π - 42
Ɵ = 2.99Rad
At leather on paper pulley, f = 0.5
f Ɵ = (0.5) (2.9987)
f Ɵ = 1.499
𝑒 𝑓Ɵ −1
( ) = 0.77689
𝑒 𝑓Ɵ
4,123
𝑣𝑠 = 60
𝑣𝑠 = 68.71 fps
12𝑝𝑉𝑠2 𝑒 𝑓Ɵ −1
c.) 𝐹1 - 𝐹2 = bt (s - )( )
32.2 𝑒 𝑓Ɵ
20 12(0.035)(68.71)2
d.) 𝐹1 - 𝐹2 = (10)(64) (400 - 32.2
) (0.77689)
𝐹1 - 𝐹2 = 821.176
(𝐹 − 𝐹2 )𝑉𝑚
ℎ𝑝 = 133,000
(821.176)(4,123)
ℎ𝑝 = 33,000
𝒉𝒑 = 102.59hp
Test II.
1. b
2. a
3. a
4. c
5. a
6. a
7. c
8. a
9. b
10. d
11. c
12. b
13. c
14. b
15. a
16. c
17. b
18. a
19. d
20. a
DEPARTMENT OF
MECHANICAL ENGINEERING
Visca, Baybay City, Leyte, PHILIPPINES
Telefax: (053) 565-0600 local 1029
Email: coe@vsu.edu.ph
Website: www.vsu.edu.ph
