MACHINE DESIGN 2
TOPICS:
•
•
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BELTS
POWER CHAIN
FLYWHEEL
BOLTS AND NUTS
POWER SCREW
SPRINGS
SAGER, SHERNELYN P.
SAGER, SHERNELYN P.
BELTS
PROBLEMS
SAGER, SHERNELYN P.
BELTS PROBLEMS
1. A 5 hp motor running at 350 rpm drives an irrigation pump which has a
pulley diameter same as the pulley used in the motor. Determine the angle of
wrap.
a. 170⁰
b. 180⁰
c. 175⁰
Given:
d. 185⁰
Required:
P= 5 hp
𝜃 in degrees
n= 350 rpm
Solution:
𝜃=𝜋±
𝜃=𝜋−
𝐷−𝑑
𝐶
𝐷−𝑑
𝐶
=𝜋 ×
180°
𝜋
𝜽 = 𝟏𝟖𝟎°
2. A power transmitting system uses a flat belt wherein during the operation a
1.5 % slip on each pulley was observed. The driving pulley runs at 750 rpm
with a diameter of 600 mm. Determine the rotative speed of a 2,000 mm
driven pulley.
a. 238.46 rpm
b. 228.28 rpm
c. 208.61 rpm
d. 218.35 rpm
Given:
Required:
N
% 𝑠𝑙𝑖𝑝 = 1.5%
n = 750 rpm
d = 600mm = 0.6 m
D = 2000mm = 2 m
Solution:
𝑉𝑑 = 𝑉𝐷
𝜋𝑑𝑛(1−% 𝑠𝑙𝑖𝑝)
60 𝑠𝑒𝑐
𝑚𝑖𝑛
𝑁=
=
𝑑𝑛(1−% 𝑠𝑙𝑖𝑝)
𝐷(1+% 𝑠𝑙𝑖𝑝)
𝜋𝐷𝑁(1+% 𝑠𝑙𝑖𝑝)
60 𝑠𝑒𝑐
𝑚𝑖𝑛
=
(0.6 𝑚)(750
𝑟𝑒𝑣
)(1−0.015)
min
(2 𝑚)(1+0.015)
𝑵 = 𝟐𝟏𝟖. 𝟑𝟓 𝒓𝒑𝒎
SAGER, SHERNELYN P.
3. An open belting power transmitting system has a center to center distance
between pulley of 3 m. The pulley diameter of the driver and driven are 450
mm and 1000 mm, respectively. Determine the difference in belt length when
using a crossed belting.
a. 130 mm
b. 160 mm
c. 140 mm
d. 150 mm
Given:
Required:
C = 3 m = 3000 mm
belt
belt length difference of open and crossed
d = 450 mm
D = 1000 mm
Solution:
•
𝐵𝑒𝑙𝑡 𝐿𝑒𝑛𝑔𝑡ℎ 𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = 𝐿𝑐𝑟𝑜𝑠𝑠𝑒𝑑 − 𝐿𝑜𝑝𝑒𝑛
𝐿𝑜𝑝𝑒𝑛 =
𝐿𝐷 =
𝜋
2
𝜋
2
𝜋
2
(𝐷−𝑑)2
4𝐶
(1000 + 450)𝑚𝑚 + 2(3000𝑚𝑚) +
𝐿𝑐𝑟𝑜𝑠𝑠𝑒𝑑 =
𝐿𝐷 =
(𝐷 + 𝑑) + 2𝐶 +
𝜋
2
(𝐷 + 𝑑) + 2𝐶 +
(1000−450)2 𝑚𝑚2
4(3000𝑚𝑚)
= 8,302.863 𝑚𝑚
(𝐷+𝑑)2
4𝐶
(1000 + 450)𝑚𝑚 + 2(3000𝑚𝑚) +
(1000+450)2 𝑚𝑚2
4(3000𝑚𝑚)
= 8,452.863 𝑚𝑚
𝐵𝑒𝑙𝑡 𝐿𝑒𝑛𝑔𝑡ℎ 𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = 8,452.863 𝑚𝑚 − 8,302.863 𝑚𝑚
𝑩𝒆𝒍𝒕 𝑳𝒆𝒏𝒈𝒕𝒉 𝑫𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒄𝒆 = 𝟏𝟓𝟎 𝒎𝒎
4. Determine the distance between shaft centers of a 500 mm driving and 1500
mm driven belt pulleys. The arc of contact is 160⁰.
a. 2.865 m
b. 4.432 m
c. 6.211 m
d. 8.422 m
Given:
Solution:
d = 500 mm = 0.5 m
𝜃=𝜋±
D = 1500 mm = 1.5 m
𝜃 = 160⁰
Required:
𝐷−𝑑
𝐷−𝑑
𝐶
𝐶 = 𝜋−𝜃 =
(1.5−0.5)𝑚
𝜋−(160°×
𝜋
)
180°
𝑪 = 𝟐. 𝟖𝟔𝟓 𝒎
C in m
SAGER, SHERNELYN P.
6. The leather material used in an open belt drive has an ultimate strength of
5,600 kPa, consider a factor of safety of 2. The 300 mm wide and 10 mm thick
belt is wire laced by a machine with an efficiency of 88 %. Determine the belt
pull on the tight side.
a. 7.4 KN
b. 9.8 KN
c 5.6 KN
d. 3.2 KN
Given:
Required:
𝑆𝑢 = 5600 𝐾𝑝𝑎
𝐹1 𝑖𝑛 𝐾𝑁
F.S = 2
b = 300 mm = 3 m
t = 10 mm = 0.01 m
J.F = 88%
Solution:
𝑆𝑑 =
𝐹1
𝐹
= 𝑏𝑡1
𝐴
Therefore:
•
𝐹1 = 𝑆𝑑 (𝑏𝑡)
Solving for 𝑆𝑑 :
𝑆𝑑 =
𝑆𝑑 =
•
𝑆𝑢
𝐹.𝑆
× 𝐽. 𝐹
5600 𝐾𝑃𝑎
𝐹1 = 2464
2
𝐾𝑁
𝑚2
× 0.88 = 2,464 𝐾𝑃𝑎
(0.3 𝑚)(0.01𝑚)
𝑭𝟏 = 𝟕. 𝟑𝟗𝟐 𝑲𝑵 ≈ 𝟕. 𝟒 𝑲𝑵
7. Determine the belt tension ratio of an open belt drive when the peripheral belt
speed is zero. The angle of contact on the 200 mm motor is 160⁰. The
coefficient of friction between the belt and pulley is 0.32.
a. 5.4
b. 4.2
c. 2.3
d. 3.6
Given:
v=0
d = 200 mm
𝜃 = 160⁰
𝑓 = 0.32
Required:
Belt tension ratio
SAGER, SHERNELYN P.
Solution:
𝐹1 −𝐹𝑐
= 𝑒 𝑓𝜃
𝐹2 −𝐹𝑐
𝐹1
𝐹2
= 𝑒
𝑭𝟏
𝑭𝟐
𝑓𝜃
=𝑒
where:
(0.32)(160°×
𝐹𝑐 =
𝜌𝑏𝑡𝑣 2
𝑔𝑐
𝑣=0
𝐹𝑐 = 0
𝜋
)
180°
= 𝟐. 𝟒𝟒
8. A 450 mm driving pulley has a rotative speed of 900 rpm. The mass of the
belt is 3 kilograms per meter of belt length. Determine the load on the belt
due to the centrifugal force.
a. 7.24 KN
b. 5.12 KN
c. 3.60 KN
d. 1.35 KN
Given:
d = 450 mm = 0.45 m
n= 900 rpm
M= 3 kg/m
Required:
𝐹𝑐 𝑖𝑛 𝐾𝑁
Solution:
𝐹𝑐 =
•
𝐹𝑐 =
𝜌𝑏𝑡𝑣 2
𝑀
where: 𝜌 = 𝑏𝑡
𝑔𝑐
𝜌𝑏𝑡𝑣 2
𝑔𝑐
=
𝑀𝑣 2
𝑔𝑐
Solving for v:
𝜋𝑑𝑛
𝑣 = 60𝑠/𝑚𝑖𝑛 =
•
𝐹𝑐 =
𝜋(0.45𝑚)(900
𝑟𝑒𝑣
)
𝑚𝑖𝑛
60 𝑠/𝑚𝑖𝑛
= 21.206 𝑚/𝑠
𝑘𝑔
𝑚2
)(21.206)2 2
𝑠
𝑠
𝑘𝑔−𝑚 1000𝑁
(1
)(
)
1 𝐾𝑁
𝑁−𝑠2
(3
𝑭𝒄 = 𝟏. 𝟑𝟒𝟗 𝑲𝑵 ≈ 𝟏. 𝟑𝟓 𝑲𝑵
9. Determine the net belt pull of an 85 kW, 275 rpm motor with 600 mm pulley.
a. 3.15 KN
b. 5.34 KN
c. 9.84 KN
d. 7.68 KN
Given:
P= 85 KW
n = 275 rpm
d = 600 mm = 0.06 m
Required:
F in KN
SAGER, SHERNELYN P.
Solution:
𝑃 = 𝐹 × 𝑉𝑑
•
𝑃
𝐹=𝑉
𝑑
Solving for 𝑉𝑑
𝑉𝑑 =
• 𝐹=
𝜋𝑑𝑛
=
60𝑠/𝑚𝑖𝑛
𝑃
𝑉𝑑
𝜋(0.06𝑚)(275
𝑟𝑒𝑣
)
𝑚𝑖𝑛
60 𝑠/𝑚𝑖𝑛
= 8.639 𝑚/𝑠
𝐾𝑁−𝑚
𝑠
𝑚
8.639
𝑠
85
=
𝑭 = 𝟗. 𝟖𝟑𝟗 𝑲𝑵 ≈ 𝟗. 𝟖𝟒 𝑲𝑵
10. A 300 mm circular saw blade with a peripheral speed of 20 m/s is fastened
to a 600 mm pulley. The power is transmitted thru an open belting with a belt
slip of 3 % on each pulley. Find the relative speed of the driving shaft carrying
a 300 mm pulley.
a. 2,704 rpm
b. 2,528 rpm
c. 2,946 rpm
d. 2,371 rpm
Given:
𝑑𝑠𝑎𝑤 = 300mm = 0.3 m
𝑉𝑠𝑎𝑤 = 20 m/s
D = 600mm = 0.6 m
% 𝑠𝑙𝑖𝑝 = 3%
d = 300mm = 0.3 m
Required:
n
Solution:
𝑉𝑑 = 𝑉𝐷
𝜋𝑑𝑛(1−% 𝑠𝑙𝑖𝑝)
60 𝑠𝑒𝑐
𝑚𝑖𝑛
• 𝑛=
=
𝜋𝐷𝑁(1+% 𝑠𝑙𝑖𝑝)
60 𝑠𝑒𝑐
𝑚𝑖𝑛
𝐷𝑁(1+% 𝑠𝑙𝑖𝑝)
𝑑(1−% 𝑠𝑙𝑖𝑝)
Solving for N:
𝑉𝑠 =
𝑁=
𝜋𝑑𝑠 𝑁
60𝑠/𝑚𝑖𝑛
𝑠
)
𝑚𝑖𝑛
𝑉𝑠 (60
𝜋(𝑑𝑠 )
=
𝑚
𝑠
𝑠
)
𝑚𝑖𝑛
(20 )(60
𝜋(0.3𝑚)
= 1273.24 𝑟𝑝𝑚
SAGER, SHERNELYN P.
• 𝑛=
(0.6𝑚)(1273.24 𝑟𝑝𝑚)(1+0.03)
(0.3𝑚)(1−0.03)
𝒏 = 𝟐𝟕𝟎𝟑. 𝟗𝟗 𝒓𝒑𝒎 ≈ 𝟐𝟕𝟎𝟒 𝒓𝒑𝒎
11. A 40 hp Diesel engine drives an irrigation pump by means of a V-belt. The
adjusted rated horsepower per belt is 5. Determine the number of belts if the
service factor is 1.2.
a. 7
b. 8
c. 9
d.10
Given:
P = 40 Hp
Adjusted rated hp/belt = 5hp/belt
S.F. = 1.2
Required:
-Number of belts, NB
Solution:
NB =
NB =
Design Hp
hp
Adjusted rated
belt
=
(P)(S.F)
djusted rated
hp
belt
(40hp)(1.2)
5
hp
belt
𝐍𝐁 = 𝟗. 𝟔 ≈ 𝟏𝟎 𝐛𝐞𝐥𝐭𝐬
12. An electric motor drives a punch press thru an open belting. When the
clutch was engaged, the belt slips. To correct this condition, an idler pulley
was installed to increase the angle of contact, but same belt and pulley were
used. The original angle of contact and belt tension ratio on the driving motor
pulley are 160⁰ and 2.4, respectively. The presence of idler pulley increases
its transmission capacity by 20 % and the slippage was corrected. What is
the new angle of contact wrap?
a. 215⁰
b. 220⁰
c. 200⁰
d. 210⁰
Given:
𝜃 = 160⁰
Belt Tension Ratio = 2.4
% 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒 = 20%
Required:
- 𝜃 ′ 𝑜𝑓 𝑐𝑜𝑛𝑡𝑎𝑐𝑡 𝑤𝑟𝑎𝑝
Solution:
(𝐹1 −𝐹2 )𝑁
(𝐹1 −𝐹2 )𝑂
(
=
′
𝑒𝑓𝜃 −1
′ )
𝑒𝑓𝜃
𝑒𝑓𝜃 −1
= 1 + % 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒
( 𝑓𝜃 )
𝑒
SAGER, SHERNELYN P.
Solve for f:
Belt Tension Ratio = 𝑒 𝑓𝜃
Ln BTR = Ln 𝑒 𝑓𝜃
𝜋
Ln 2.4 = 𝑓(160° × 180°)
𝑓 = 0.3135
′
𝑒 𝑓𝜃 −1
(
𝑒
𝑓𝜃′
𝑒 𝑓𝜃 −1
)=(
) (1 + % 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒)
′
𝑒 𝑓𝜃
𝑒 𝑓𝜃 −1
2.4−1
′
𝑒 𝑓𝜃
2.4
(
)=(
) (1 + 0.20)
′
𝑒 𝑓𝜃 −1
𝑒 𝑓𝜃
′
= 0.70
′
′
𝑒 𝑓𝜃 − 1 = 0.70(𝑒 𝑓𝜃 )
′
𝑒 𝑓𝜃 (1 − 0.70) = 1
1
′
𝑒 𝑓𝜃 = 1−0.70
′
𝑒 𝑓𝜃 = 3.333
′
𝐿𝑛 𝑒 𝑓𝜃 = 𝐿𝑛 3.333
𝑓𝜃 ′ = 𝐿𝑛 3.333
𝜃′ =
𝐿𝑛 3.333
𝑓
=
𝐿𝑛 3.333
0.3135
= 3.84 𝑟𝑎𝑑 ×
180°
𝜋
𝜽′ = 𝟐𝟐𝟎°
13. The sheave diameter of the driver and driven are 5.4 inches and 15.4
inches, respectively. Determine the center to center distance between shafts.
a. 10.7 in
b. 13.6 in
c. 18.5 in
d. 15.8 in
Given:
d = 5.4 in
D = 15.4 in
Required:
- center to center distance, C
Solution:
𝐶=
𝐶=
𝑑−𝐷
2
+𝑑
5.4 𝑖𝑛−15.4 𝑖𝑛
2
+ 5.4 𝑖𝑛
𝑪 = 𝟏𝟓. 𝟖 𝒊𝒏
SAGER, SHERNELYN P.
14. A 55 kW, 750 rpm motor with 600 mm pulley is connected by a flat belt to
a 2,000 mm flywheel of a machine. The allowable design stress of the belt is
2 MPa. The distance between shaft centers is 3 m. The coefficient of friction
between the belt and pulley is 0.3. Angle of contact on the pulley is 140⁰. Find
the width of an 8 mm medium leather belt to be used.
a. 333 mm
b. 353 mm
c. 383 mm
d. 403 mm
Given:
P = 55 KW
n = 750 rpm
d = 600 mm = 0.6 m
D = 2000 mm = 2 m
𝑆𝑑 = 2 Mpa
C=3m
f = 0.3
7
𝜃 = 160⁰ = 9 𝜋
Required:
-width, b
t = 8 mm
Solution:
𝐹 = 𝑏𝑡 [𝑆𝑑 −
•
𝑏=
𝜌𝑣 2
][
𝑔𝑐
𝑒 𝑓𝜃 −1
𝑒 𝑓𝜃
]
𝐹
𝜌𝑣2 𝑒𝑓𝜃 −1
𝑡[𝑆𝑑 −
][ 𝑓𝜃 ]
𝑔𝑐
𝑒
Solve for v & F:
𝜋𝑑𝑛
𝑣 = 𝑉𝑑 = 60𝑠/𝑚𝑖𝑛 =
𝜋(0.6𝑚)(750
𝑟𝑒𝑣
)
𝑚𝑖𝑛
60 𝑠/𝑚𝑖𝑛
= 23.562 𝑚/𝑠
𝑃 = 𝐹 × 𝑉𝑑
𝑃
(55
𝐾𝑁−𝑚
)
1000𝑁
𝐹 = 𝑉 = 23.562𝑠𝑚/𝑠 ×
1 𝐾𝑁
𝑑
•
𝑏=
𝑏=
= 2,334.267 𝑁
2,334.267 𝑁
7
𝑘𝑔
𝑚2
(965 3 )(23.562)2 2 𝑒(0.3)(9𝜋) −1
𝑁
𝑚
𝑠
][
]
8 𝑚𝑚[2
−
7
𝑚𝑚2
𝑘𝑔−𝑚 1000 𝑚𝑚 2
(0.3)( 𝜋)
(1
)(
)
9
𝑒
1𝑚
𝑁−𝑠2
2,334.267 𝑁
𝑁
2.081−1
][
]
2.081
𝑚𝑚2
8 𝑚𝑚[1.464
𝒃 = 𝟑𝟖𝟑. 𝟔𝟕𝟕 𝒎𝒎
SAGER, SHERNELYN P.
POWER CHAIN
PROBLEMS
SAGER, SHERNELYN P.
POWER CHAIN PROBLEMS
1. The driving sprocket has 18 teeth rotates at 700 rpm and pitch of chain is 1
inch. The velocity is 5/3. Determine the rotative speed of the driven
sprocket.
a. 400 rpm
b. 420 rpm
c. 440 rpm
d. 460 rpm
Given:
𝑁𝑑 = 18 𝑡𝑒𝑒𝑡ℎ
𝑛𝑑 = 700 𝑟𝑝𝑚
P = 1 𝑖𝑛𝑐ℎ
5
𝑉𝑅 = 3
Required:
- 𝑛𝐷
Solution:
𝑉𝑑 = 𝑉𝐷
𝑃𝑁𝑑 𝑛𝑑
12
=
𝑃𝑁𝐷 𝑛𝐷
12
𝑁𝑑 𝑛𝑑 = 𝑁𝐷 𝑛𝐷
𝑉𝑅 =
𝑛𝑑
𝑛𝐷
=
𝑁𝐷
𝑁𝑑
5
𝑛𝑑
3
𝑛𝐷
𝑉𝑅 = =
• 𝑛𝐷 =
3𝑛𝑑
5
=
3(700 𝑟𝑝𝑚)
5
𝒏𝑫 = 𝟒𝟐𝟎 𝒓𝒑𝒎
2. The driving sprocket has 18 teeth rotates at 700 rpm and pitch of chain is 1
inch. The velocity ratio is 5/3. Determine the number of teeth of the driven
sprocket.
a. 32 teeth
b. 34 teeth
c. 30 teeth
d. 28 teeth
Given:
𝑁𝑑 = 18 𝑡𝑒𝑒𝑡ℎ
𝑛𝑑 = 700 𝑟𝑝𝑚
P = 1 𝑖𝑛𝑐ℎ
5
𝑉𝑅 = 3
Required:
- 𝑁𝐷
SAGER, SHERNELYN P.
Solution:
𝑉𝑑 = 𝑉𝐷
𝑃𝑁𝑑 𝑛𝑑
12
=
𝑃𝑁𝐷 𝑛𝐷
12
𝑁𝑑 𝑛𝑑 = 𝑁𝐷 𝑛𝐷
𝑉𝑅 =
𝑛𝑑
𝑛𝐷
=
𝑁𝑑
5
𝑁𝐷
3
𝑁𝑑
𝑉𝑅 = =
•
𝑁𝐷
𝑁𝐷 =
5𝑁𝑑
3
=
5(18 𝑡𝑒𝑒𝑡ℎ)
3
𝑵𝑫 = 𝟑𝟎 𝒕𝒆𝒆𝒕𝒉
3. The driving sprocket has 18 teeth rotates at 700 rpm and pitch of chain is 1
inch. The velocity ratio is 5/3. Determine the pitch diameters of driving and
driven sprockets in inches.
a. 6.43; 10.04
b. 7.82; 11.95
c. 8.64; 12.85
d.5.73; 9.55
Given:
𝑁𝑑 = 18 𝑡𝑒𝑒𝑡ℎ
𝑛𝑑 = 700 𝑟𝑝𝑚
P = 1 𝑖𝑛𝑐ℎ
5
𝑉𝑅 = 3
Required:
- 𝐷𝑑 𝑎𝑛𝑑 𝐷𝐷
Solution:
Solving for pitch diameter of
Solving for pitch diameter of driven
driving sprocket
sprocket
𝐷𝑑 =
𝑃
180°
)
𝑠𝑖𝑛(
𝑁𝑑
=
𝑃
𝑠𝑖𝑛(
180°
)
18
𝐷𝐷 =
𝑃
180°
)
𝑁𝐷
𝑠𝑖𝑛(
5
𝑁𝐷
3
𝑁𝑑
𝑉𝑅 = =
𝑫𝒅 = 𝟓. 𝟕𝟔 𝒊𝒏𝒄𝒉𝒆𝒔
𝑁𝐷 =
5𝑁𝑑
3
=
5(18 𝑡𝑒𝑒𝑡ℎ)
3
=
30 𝑡𝑒𝑒𝑡ℎ
𝐷𝐷 =
𝑃
𝑠𝑖𝑛(
180°
)
30
𝑫𝑫 = 𝟗. 𝟓𝟕 𝒊𝒏𝒄𝒉𝒆𝒔
SAGER, SHERNELYN P.
4. The driving sprocket has 18 teeth rotates at 700 rpm and pitch of chain is 1
inch. The velocity ratio is 5/3. Determine the outside diameters of driving
and driven sprockets in inches.
a. 6.27; 10.11
b. 8.27; 14.11
Given:
𝑁𝑑 = 18 𝑡𝑒𝑒𝑡ℎ
𝑛𝑑 = 700 𝑟𝑝𝑚
P = 1 𝑖𝑛𝑐ℎ
5
𝑉𝑅 = 3
c.4.27; 8.11 d.10.27; 14.11
Required:
- 𝐷𝑑 𝑜 𝑎𝑛𝑑 𝐷𝐷 𝑜
Solution:
Solving for outside diameter of
Solving for outside diameter of
driving sprocket
driven sprocket
180°
180°
𝐷𝑑 𝑜 = 𝑃 [(0.6 + 𝑐𝑜𝑡 ( 𝑁 ))]
𝐷𝐷 𝑜 = 𝑃 [(0.6 + 𝑐𝑜𝑡 ( 𝑁 ))]
𝐷
𝑑
𝐷𝑑 𝑜 = 𝑃 [(0.6 + 𝑐𝑜𝑡 (
180°
18
5
𝑁𝐷
3
𝑁𝑑
𝑉𝑅 = =
))]
𝑁𝐷 =
𝑫𝒅 𝒐 = 𝟔. 𝟐𝟕 𝒊𝒏𝒄𝒉𝒆𝒔
5𝑁𝑑
3
=
5(18 𝑡𝑒𝑒𝑡ℎ)
3
= 30 𝑡𝑒𝑒𝑡ℎ
180°
𝐷𝐷 𝑜 = 𝑃 [(0.6 + 𝑐𝑜𝑡 (
30
))]
𝑫𝑫 𝒐 = 𝟏𝟎. 𝟏𝟏 𝒊𝒏𝒄𝒉𝒆𝒔
5. The driving sprocket has 18 teeth rotates at 700 rpm and pitch of chain is 1
inch. The velocity ratio is 5/3. Determine the peripheral speed velocity or
chain speed.
a. 1,050 fpm
b. 1,250 fpm
c. 1,450 fpm
d. 850 fpm
Given:
𝑁𝑑 = 18 𝑡𝑒𝑒𝑡ℎ
𝑛𝑑 = 700 𝑟𝑝𝑚
P = 1 𝑖𝑛𝑐ℎ
5
𝑉𝑅 = 3
Required:
-V
Solution:
𝑃𝑁𝑛
𝑃𝑁𝑑 𝑛𝑑
𝑃𝑁𝐷 𝑛𝐷
𝑉 = 12 = 12
= 12
𝑉𝑑 =
𝑟𝑒𝑒𝑣
)
𝑚𝑖𝑛
(1 𝑖𝑛𝑐ℎ)(18 𝑡𝑒𝑒𝑡ℎ)(700
12 𝑖𝑛𝑐ℎ
1 𝑓𝑜𝑜𝑡
𝑽𝒅 = 𝟏𝟎𝟓𝟎 𝒇𝒑𝒎 = 𝑽
SAGER, SHERNELYN P.
6. The driving sprocket has 18 teeth rotates at 700 rpm and pitch of chain is 1
inch. The velocity ratio is 5/3. Determine the center to center distance
between sprockets in inches.
a. 10.45
b. 12.415
c. 16.45
d. 14.45
Given:
𝑁𝑑 = 18 𝑡𝑒𝑒𝑡ℎ
𝑛𝑑 = 700 𝑟𝑝𝑚
P = 1 𝑖𝑛𝑐ℎ
5
𝑉𝑅 = 3
Required:
-C
Solution:
• 𝐶 = 𝐷𝐷 +
𝐷𝑑
2
Solving for pitch diameter of
Solving for pitch diameter of driven
driving sprocket
sprocket
𝐷𝑑 =
𝑃
180°
𝑠𝑖𝑛(
)
𝑁𝑑
=
𝑃
𝐷𝐷 =
180°
𝑠𝑖𝑛(
)
18
𝑃
180°
)
𝑁𝐷
𝑠𝑖𝑛(
5
𝐷𝑑 = 5.76 𝑖𝑛𝑐ℎ𝑒𝑠
𝑉𝑅 = 3 =
𝑁𝐷 =
5𝑁𝑑
𝐷𝐷 =
3
𝑁𝐷
𝑁𝑑
=
5(18 𝑡𝑒𝑒𝑡ℎ)
3
= 30 𝑡𝑒𝑒𝑡ℎ
𝑃
𝑠𝑖𝑛(
180°
)
30
𝐷𝐷 = 9.57 𝑖𝑛𝑐ℎ𝑒𝑠
• 𝐶 = 9.57 𝑖𝑛 +
5.76 𝑖𝑛
2
𝑪 = 𝟏𝟐. 𝟒𝟓 𝒊𝒏𝒄𝒉𝒆𝒔
SAGER, SHERNELYN P.
7. The driving sprocket has 18 teeth rotates at 700 rpm and pitch of chain is 1
inch. The velocity ratio is 5/3. Determine the length of the chain in pitches.
a. 52
b. 54
c. 56
d. 50
Given:
𝑁𝑑 = 18 𝑡𝑒𝑒𝑡ℎ
𝑛𝑑 = 700 𝑟𝑝𝑚
P = 1 𝑖𝑛𝑐ℎ
5
𝑉𝑅 = 3
Required:
-L
Solution:
2𝐶
• 𝐿=
𝑃
𝑁𝐷 +𝑁𝑑
+
𝐶 = 𝐷𝐷 +
2
+
𝑃(𝑁𝐷 −𝑁𝑑 )2
40𝐶
𝐷𝑑
2
Solving for pitch diameter of
Solving for pitch diameter of driven
driving sprocket
sprocket
𝐷𝑑 =
𝑃
180°
)
𝑁𝑑
𝑠𝑖𝑛(
=
𝑃
𝑠𝑖𝑛(
𝑃
𝐷𝐷 =
180°
)
18
𝐷𝑑 = 5.76 𝑖𝑛𝑐ℎ𝑒𝑠
180°
)
𝑁𝐷
𝑠𝑖𝑛(
5
𝑁𝐷
3
𝑁𝑑
𝑉𝑅 = =
𝑁𝐷 =
𝐷𝐷 =
5𝑁𝑑
3
=
5(18 𝑡𝑒𝑒𝑡ℎ)
3
= 30 𝑡𝑒𝑒𝑡ℎ
𝑃
𝑠𝑖𝑛(
180°
)
30
𝐷𝐷 = 9.57 𝑖𝑛𝑐ℎ𝑒𝑠
𝐶 = 9.57 𝑖𝑛 +
• 𝐿=
5.76 𝑖𝑛
2(12.45 𝑖𝑛)
1
2
+
= 12.45 𝑖𝑛𝑐ℎ𝑒𝑠
(30+18)𝑖𝑛
2
+
𝑃(30−18)2 𝑖𝑛2
40(12.45 𝑖𝑛)
𝑳 = 𝟓𝟎 𝒑𝒊𝒕𝒄𝒉𝒆𝒔
SAGER, SHERNELYN P.
8. The three strand No. 80 roller chain has a rating of 25 kW for a single
strand. The service factor may be taken equal to 1.2 and multiple-strand
factor is 2.7. Determine the power that can be transmitted in kW.
a. 56.25
b. 61.5
c. 66.5
d. 51.25
Given:
Capacity/strand = 25 KW
S.F.= 1.2
Multiple-strand factor = 2.7
Required:
- Power transmitted, KW
Solution:
𝐻𝑝 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑝𝑒𝑟 𝑠𝑡𝑟𝑎𝑛𝑑 =
𝐻𝑝 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑝𝑒𝑟 𝑠𝑡𝑟𝑎𝑛𝑑 =
𝑃𝑜𝑤𝑒𝑟 𝑇𝑟𝑎𝑛𝑠𝑚𝑖𝑡𝑡𝑒𝑑 =
𝐷𝑒𝑠𝑖𝑔𝑛 𝐻𝑝
𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑒 𝑠𝑡𝑟𝑎𝑛𝑑 𝑓𝑎𝑐𝑡𝑜𝑟
𝑃𝑜𝑤𝑒𝑟 𝑇𝑟𝑎𝑛𝑠𝑚𝑖𝑡𝑡𝑒𝑑 (𝑆.𝐹)
𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑒 𝑠𝑡𝑟𝑎𝑛𝑑 𝑓𝑎𝑐𝑡𝑜𝑟
(𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑝𝑒𝑟 𝑠𝑡𝑟𝑎𝑛𝑑)(𝑚.𝑠,𝑓)
𝑆.𝐹
=
(25 𝐾𝑊)(2.7)
1.2
𝑷𝒐𝒘𝒆𝒓 𝑻𝒓𝒂𝒏𝒔𝒎𝒊𝒕𝒕𝒆𝒅 = 𝟓𝟔. 𝟐𝟓 𝑲𝑾
9. A 4 inches diameter shaft is driven at 3,600 rpm by a 400 hp motor. The
shaft drives a 48 inches diameter chain sprocket having an output efficiency
of 85%. The output force of the driving sprocket and the output of the driven
sprocket are?
a. 200 lb & 250 hp
c. 291.66 lb & 340 hp
b. 261.6 lb & 300 hp
d. 180 lb & 200 hp
Given:
𝐷𝑠 = 4 𝑖𝑛𝑐ℎ𝑒𝑠
𝑛𝐷 = 3,600 𝑟𝑝𝑚
𝑃 = 400 ℎ𝑝
𝐷𝑑 = 48 𝑖𝑛𝑐ℎ𝑒𝑠
𝑒 = 85%
Required:
- output force of the driving sprocket, F
- output power of the driven sprocket, P
Solution:
For output force of the driving sprocket
•
𝐹=
2𝑇
𝐷
Solve for T:
𝑃=
𝑇𝑛
𝑙𝑏−𝑖𝑛
ℎ𝑝−𝑚𝑖𝑛
63,025
SAGER, SHERNELYN P.
𝑇=
•
𝐹=
𝑃(63,025
𝑙𝑏−𝑖𝑛
)
ℎ𝑝−𝑚𝑖𝑛
𝑛
=
(400 ℎ𝑝)(63,025
𝑙𝑏−𝑖𝑛
)
ℎ𝑝−𝑚𝑖𝑛
𝑟𝑒𝑣
(3600𝑚𝑖𝑛)
= 7002.778 𝑙𝑏 − 𝑖𝑛
2(7002.778 𝑙𝑏−𝑖𝑛)
48 𝑖𝑛𝑐ℎ𝑒𝑠
𝐹 = 291. 782 𝑙𝑏
For output power of the driven sprocket
𝑒=
•
𝑃𝑜𝑢𝑡𝑝𝑢𝑡
𝑃𝑖𝑛𝑝𝑢𝑡
𝑃𝑜𝑢𝑡𝑝𝑢𝑡 = 𝑒(𝑃𝑖𝑛𝑝𝑢𝑡 ) = 0.85 (400 ℎ𝑝)
𝑷𝒐𝒖𝒕𝒑𝒖𝒕 = 𝟑𝟒𝟎 𝒉𝒑
Situational Problem (10 - 12)
A roller chain and sprocket is to drive vertical centrifugal discharge bucket
elevator; the pitch of the chain connecting sprockets is 1.75 inches. The
driving sprocket is rotating at 120 rpm and as 11 teeth while the driven
sprocket is rotating at 38 rpm. Determine:
10.
The number of teeth of the driven sprocket.
a. 40
b. 35
c. 50
d. 4
11. The length of the chain in pitches if the minimum center distance is
equal to the diameter of the bigger sprocket.
a. 60
b. 54
c. 48
d. 42
12.
The roller chain speed in fpm.
a. 172.5
b. 162.5
Given:
𝑃 = 1.75 𝑖𝑛𝑐ℎ𝑒𝑠
𝑛𝑑 = 120 𝑟𝑝𝑚
𝑁𝑑 = 18 𝑡𝑒𝑒𝑡ℎ
𝑛𝐷 = 38 𝑟𝑝𝑚
c. 182.5
d. 192.5
Required:
10. 𝑁𝐷
11. L
12. V
SAGER, SHERNELYN P.
Solution:
Solving for number of teeth of the driven sprocket
𝑉𝑑 = 𝑉𝐷
𝑃𝑁𝑑 𝑛𝑑
𝑃𝑁𝐷 𝑛𝐷
=
12
12
𝑁𝑑 𝑛𝑑 = 𝑁𝐷 𝑛𝐷
•
𝑁𝐷 =
𝑁𝑑 𝑛𝐷
=
𝑛𝐷
(11 𝑡𝑒𝑒𝑡ℎ)(120 𝑟𝑝𝑚)
38 𝑟𝑝𝑚
= 34.74 𝑡𝑒𝑒𝑡ℎ
𝑵𝑫 = 𝟑𝟓 𝒕𝒆𝒆𝒕𝒉
Solving for length of the chain
2𝐶
• 𝐿=
𝑃
+
𝑁𝐷 +𝑁𝑑
2
+
𝑃(𝑁𝐷 −𝑁𝑑 )2
40𝐶
IF: 𝐶 = 𝐷𝐷
𝐷𝐷 =
𝑃
180°
)
𝑠𝑖𝑛(
𝑁𝐷
=
1.75 𝑖𝑛𝑐ℎ𝑒𝑠
𝑠𝑖𝑛(
180°
)
35
𝐷𝐷 = 19.523 𝑖𝑛𝑐ℎ𝑒𝑠 = 𝐶
2(19.523)
• 𝐿=
(1.75)
+
35+11
2
+
(1.75)(35−11)2
40(19.523)
𝐿 = 46.6 𝑝𝑖𝑡𝑐ℎ𝑒𝑠
𝑳 = 𝟒𝟖 𝒑𝒊𝒕𝒄𝒉𝒆𝒔
Solving for roller chain speed
•
𝑉=
𝑃𝑁𝑛
𝑉𝑑 =
12
=
𝑃𝑁𝑑 𝑛𝑑
12
=
𝑃𝑁𝐷 𝑛𝐷
12
(1.75 𝑖𝑛𝑐ℎ)(11 𝑡𝑒𝑒𝑡ℎ)(120
𝑟𝑒𝑒𝑣
)
𝑚𝑖𝑛
12 𝑖𝑛𝑐ℎ
1 𝑓𝑜𝑜𝑡
𝑽𝒅 = 𝟏𝟗𝟐. 𝟓 𝒇𝒑𝒎
SAGER, SHERNELYN P.
FLYWHEEL
PROBLEMS
SAGER, SHERNELYN P.
FLYWHEEL PROBLEMS
1. A machine is to be equipped with a cast iron flywheel 1,520 mm in diameter
and 305 mm wide. The arms and hub of the flywheel are to be considered
equivalent to 5 % of the rim weight concentrated at the mean diameter. The
change in energy to be handles is 3,400 N - m and the coefficient of fluctuation
is to be 0.03. The running speed of the wheel is 250 rpm. Find the rim
thickness in mm.
a. 16
b. 36
c. 21
d. 26
Given:
D = 1,520 mm = 1.52 m
b = 305 mm = .305 m
mH + mA = 5%m
∆𝐸 = 3400 𝑁 − 𝑚
𝐶𝑓 = 0.03
n= 250 𝑟𝑝𝑚
𝜌𝑖𝑟𝑜𝑛 = 7200 𝑘𝑔/𝑚3
Required:
- thickness, t in mm
Solution:
𝑉 = 𝑏𝑡(𝐿) = 𝑏𝑡(𝜋𝐷)
• 𝑡=
∆𝐸 =
𝑚=
𝑚=
𝑉=
𝑉
𝑏(𝜋𝐷)
𝑚𝑤
𝑔𝑐
(𝐶𝑓 )(𝑉𝑎 2 ) =
∆𝐸(𝑔𝑐)
2
(1.05)(𝐶𝑓 )(𝑉𝑎 )
(𝑚+0.05𝑚)
𝑔𝑐
(1.05)(𝐶𝑓 )(
𝜌
• 𝑡=
272.649 𝑘𝑔
7200 𝑘𝑔/𝑚3
𝑔𝑐
(𝐶𝑓 )(𝑉𝑎 2 )
2
𝜋𝐷𝑛
)
60 𝑠/𝑚𝑖𝑛
𝑘𝑔−𝑚
)
𝑁−𝑠2
𝑟𝑒𝑣 2
𝜋(1.52 𝑚)(250
)
𝑚𝑖𝑛 )
(1.05)(0.03)(
60 𝑠/𝑚𝑖𝑛
=
1.05𝑚
∆𝐸(𝑔𝑐)
=
(3400 𝑁−𝑚)(1
𝑚
(𝐶𝑓 )(𝑉𝑎 2 ) =
= 272.649 𝑘𝑔
= 0.0379 𝑚3
0.0379 𝑚3
(0.305 𝑚)(𝜋)(1.52 𝑚)
= 0.026 𝑚 ×
1000 𝑚𝑚
1𝑚
𝒕 = 𝟐𝟔 𝒎𝒎
SAGER, SHERNELYN P.
2. It is found that a shearing machine requires 2.05 kJ of energy to shear a
specific gauge of sheet steel. The mean diameter of the flywheel is to be 760
mm. The normal operating speed is 200 rpm, it slows down to 180 rpm during
the shearing process. The rim width is to be 305 mm. The mass density of
cast iron is 7,197 kg/m3. Find the thickness of the rim assuming that the
hub and arm accounts for 10% of the rim weight concentrated at the mean
diameter.
a. 58.77 mm
b. 47.88 mm
c. 38.57 mm
d. 65.78 mm
Given:
∆𝐸 = 2.05 𝐾𝐽
𝐷𝑚 = 760 𝑚𝑚 = 0.76 𝑚
𝑛1 = 200 𝑟𝑝𝑚
𝑛2 = 180 𝑟𝑝𝑚
𝑏 = 305 𝑚𝑚 = 0.305 𝑚
𝜌 = 7,197 𝑘𝑔/𝑚3
mH + mA = 10%m
Required:
- thickness, t in mm
Solution:
𝑉 = 𝑏𝑡(𝐿) = 𝑏𝑡(𝜋𝐷)
• 𝑡=
∆𝐸 =
𝑉
𝑏(𝜋𝐷)
𝑚𝑤
2𝑔𝑐
(𝑣2 2 − 𝑣1 2 ) =
(𝑚+0.1𝑚)
2𝑔𝑐
(𝑣2 2 − 𝑣1 2 ) =
1.1 𝑚
2𝑔𝑐
(𝑣2 2 − 𝑣1 2 )
∆𝐸(2)(𝑔𝑐)
𝑚 = 1.1(𝑣 2−𝑣 2)
2
1
Solve for 𝑣2 𝑎𝑛𝑑 𝑣1 :
𝑣2 =
𝑣1 =
𝑚=
𝑉=
𝜋𝐷𝑛2
60
𝜋𝐷𝑛1
60
=
=
𝑟𝑒𝑣
)
𝑚𝑖𝑛
𝜋(0.76𝑚)(200
60 𝑠/𝑚𝑖𝑛
𝑟𝑒𝑣
)
𝑚𝑖𝑛
𝜋(0.76𝑚)(180
60 𝑠/𝑚𝑖𝑛
𝑘𝑔−𝑚 1000 𝑁
)(
)
1 𝐾𝑁
𝑁−𝑠2
2
2 𝑚2
1.1(7.96 −7.16 ) 2
𝑠
(2.05 𝐾𝑁−𝑚)(2)(1
𝑚
𝜌
= 7.96 𝑚/𝑠
= 7.16 𝑚/𝑠
= 308.141 𝑘𝑔
308.141 𝑘𝑔
= 7197 𝑘𝑔/𝑚3 = 0.0428 𝑚3
• 𝑡=
0.0428 𝑚3
(0.305 𝑚)(𝜋)(0.76 𝑚)
= 0.05877 𝑚 ×
1000 𝑚𝑚
1𝑚
𝒕 = 𝟓𝟖. 𝟕𝟕 𝒎𝒎
SAGER, SHERNELYN P.
3. Find the capacity of a 600 mm cast iron flywheel which has a mass of 300 kg.
The peripheral velocity of the flywheel decreases 10% from its rated speed of
20 m/s while performing a shearing operation.
a. 13.5 KN-m
b. 11.4 KN-m
c. 9.3 KN-m
d. 7.2 KN-m
Given:
D= 600 𝑚𝑚 = 0.6 𝑚
m = 300 𝑘𝑔
𝑣2 = 20 𝑚/𝑠
𝑣1 = 20 𝑚⁄𝑠 − (0.10)(20 𝑚⁄𝑠) = 18 𝑚/𝑠
Required:
- ∆𝐸
Solution:
∆𝐸 =
∆𝐸 =
𝑚
2𝑔𝑐
(𝑣2 2 − 𝑣1 2 )
300 𝑘𝑔(202 −182 )𝑚2 ⁄𝑠 2
𝑘𝑔−𝑚 1000 𝑁
2(1
)(
)
1 𝐾𝑁
𝑁−𝑠2
∆𝑬 = 𝟏𝟏. 𝟒 𝑲𝑵 − 𝒎
4. Find the coefficient of stability of a 600 mm cast iron flywheel which has a
mass of 300 kg. The peripheral velocity of the flywheel decreases 10% from
its rated speed of 20 m/s while performing a shearing operation.
a. 6.5
b. 7.5
c. 8.5
d. 9.5
Given:
D= 600 𝑚𝑚 = 0.6 𝑚
m = 300 𝑘𝑔
𝑣2 = 20 𝑚/𝑠
𝑣1 = 20 𝑚⁄𝑠 − (0.10)(20 𝑚⁄𝑠) = 18 𝑚/𝑠
Required:
- coefficient stability, 𝐶𝑠
Solution:
•
1
𝐶𝑠 = 𝐶
𝑓
𝐶𝑓 =
𝑣2 −𝑣1
𝑤ℎ𝑒𝑟𝑒:
𝑣𝑎
𝑣𝑎 =
𝐶𝑓 =
•
(20−18)𝑚/𝑠
19𝑚/𝑠
𝑣2 −𝑣1
2
=
20−18
2
= 19 𝑚/𝑠
= 0.1053
1
𝐶𝑠 = 0.1053
𝑪𝒔 = 𝟗. 𝟓
SAGER, SHERNELYN P.
5. A cast iron flywheel has inside diameter of 510 mm and an outside diameter
of 610 mm. The mass of rim is 100 kg. Find the flywheel effect.
a. 0.03123 KN-m2
b. 0.01369 KN-m2
Given:
𝐷𝑖 = 510 𝑚𝑚 = 0.51 𝑚
𝐷𝑜 = 610 𝑚𝑚 = 0.61 𝑚
𝑚 = 100 𝑘𝑔 = 𝐹
c. 0.07688 KN-m2
d. 0.05344 KN-m2
Required:
- Flywheel Effect
Solution:
•
𝐷 2
𝐹𝑙𝑦𝑤ℎ𝑒𝑒𝑙 𝐸𝑓𝑓𝑒𝑐𝑡 = 𝐹𝑤 𝑅𝑤 2 = 𝐹𝑤 ( 2 )
Solve for D:
𝐷=
𝐷𝑜 +𝐷𝑖
2
=
(0.61+0.51)𝑚
2
= 0.56 𝑚
9.807 𝑁
𝐹𝑙𝑦𝑤ℎ𝑒𝑒𝑙 𝐸𝑓𝑓𝑒𝑐𝑡 = (100 𝑘𝑔) (
1 𝑘𝑔
1 𝐾𝑁
0.56𝑚 2
) (1000 𝑁) (
2
)
𝑭𝒍𝒚𝒘𝒉𝒆𝒆𝒍 𝑬𝒇𝒇𝒆𝒄𝒕 = 𝟎. 𝟎𝟕𝟔𝟖𝟗 𝑲𝑵 − 𝒎𝟐
6. A flywheel with a coefficient of fluctuation of 0.1 and a mean velocity of 1,200
ft/min produces an energy of 450 ft-lb. Find the weight in lb.
a. 322.50 lb
b. 382.25 lb
c. 342.50 lb
d. 362.25 lb
Given:
𝐶𝑓 = 0.1
𝑉𝑎 = 1200 𝑓𝑡/𝑚𝑖𝑛
∆𝐸 = 3400 𝑓𝑡 − 𝑙𝑏
Required:
- m in lb
Solution:
∆𝐸 =
𝑚
𝑔𝑐
• 𝑚=
(𝐶𝑓 )(𝑉𝑎 2 )
∆𝐸(𝑔𝑐)
(𝐶𝑓 )(𝑉𝑎 2 )
=
𝑓𝑡−𝑙𝑏
)
𝑙𝑏𝑓−𝑠2
1 𝑚𝑖𝑛 2
(450𝑓𝑡⁄𝑙𝑏)(32.2
𝑓𝑡
(0.1)(1200𝑚𝑖𝑛× 60 𝑠 )
𝒎 = 𝟑𝟔𝟐. 𝟐𝟓 𝒍𝒃
SAGER, SHERNELYN P.
7. A 600 mm cast iron flywheel has a capacity of 2.15 KN-m. When used in a
shearing process its normal operating speed of 20 m/s slows down to 18 m/s.
Determine the thickness of the rim if the width is 300 mm.
a. 14.265 mm
b. 13.896 mm
c. 16.124 mm
d. 18.357 mm
Given:
D= 600 𝑚𝑚 = 0.6 𝑚
Required:
-thickness, t
∆𝐸 = 2.15 𝐾𝑁 − 𝑚
𝑣2 = 20 𝑚/𝑠
𝑣1 = 18 𝑚/𝑠
𝑏 = 300 𝑚𝑚 = 0.3 𝑚
𝜌𝑖𝑟𝑜𝑛 = 7200 𝑘𝑔/𝑚3
Solution:
𝑉 = 𝑏𝑡(𝐿) = 𝑏𝑡(𝜋𝐷)
• 𝑡=
∆𝐸 =
𝑚=
𝑉
𝑏(𝜋𝐷)
𝑚
2𝑔𝑐
(𝑣2 2 − 𝑣1 2 )
∆𝐸(2)(𝑔𝑐)
(𝑣2 2 −𝑣1 2 )
=
1000 𝑁
𝑘𝑔−𝑚
)(2)(1
)
1 𝐾𝑁
𝑁−𝑠2
2
2
2
2
(20 −18 )𝑚 ⁄𝑠
(2015 𝐾𝑛−𝑚)(
𝑚 = 56.579 𝑘𝑔
𝑉=
𝑚
𝜌
• 𝑡=
56.579 𝑘𝑔
= 7200 𝑘𝑔/𝑚3 = 0.007858 𝑚3
0.007858 𝑚3
(0.3 𝑚)(𝜋)(0.6 𝑚)
= 0.013896 𝑚 ×
1000 𝑚𝑚
1𝑚
𝒕 = 𝟏𝟑. 𝟖𝟗𝟔 𝒎𝒎
SAGER, SHERNELYN P.
8. A 600 mm cast iron flywheel has a capacity of 2.15 KN-m. When used in a
shearing process its normal operating speed of 20 m/s slows down to 18 m/s.
Determine the thickness of the rim if the width is 300 mm.
a. 14.265 mm
b. 13.896 mm
c. 16.124 mm
d. 18.357 mm
Given:
∆𝐸 = 1.5 𝐾𝐽
𝑛1 = 3 𝑟𝑒𝑣/𝑠
𝑛2 = 2.75 𝑟𝑒𝑣/𝑠
FA + FH = 15%F
𝐷𝑚 = 1 𝑚
Required:
- Force of the rim, F
Solution:
𝐹
𝑤
(𝑣2 2 − 𝑣1 2 ) =
∆𝐸 = 2𝑔𝑜
(𝐹+0.15𝐹)
2𝑔𝑜
(𝑣2 2 − 𝑣1 2 ) =
1.15 𝐹
2𝑔𝑜
(𝑣2 2 − 𝑣1 2 )
• 𝐹 = 1.15(𝑣
∆𝐸(2)(𝑔𝑜)
2
2
2 −𝑣1 )
Solve for 𝑣2 𝑎𝑛𝑑 𝑣1 :
𝑣2 = 𝜋𝐷𝑛2 = 𝜋(1𝑚)(3 𝑟𝑒𝑣⁄𝑠) = 9.4248 𝑚/𝑠
𝑣1 = 𝜋𝐷𝑛1 = 𝜋(1𝑚)(2.75 𝑟𝑒𝑣⁄𝑠) = 8.6394 𝑚/𝑠
• 𝐹=
(1.5 𝐾𝑁−𝑚)(2)(9.807 𝑚/𝑠 2 )
2
2
1.15(9.4248 −8.6394 )𝑚2 ⁄𝑠2
𝑭 = 𝟏. 𝟖𝟎𝟑 𝑲𝑵
9. The flywheel of a punching machine can punch a 75 mm hole at 10 holes per
minute from a 40 mm thick steel plate. Determine the power required to drive
the punching machine if the ultimate strength in shear is 300 MPa.
a. 9.425 kW
b. 7.425 kW
c. 5.425 kW
d. 11.425 Kw
Given:
d= 75 𝑚𝑚
no. of holes = 10 ℎ𝑜𝑙𝑒𝑠/𝑚𝑖𝑛
Required:
- Power
𝑡𝑝 = 40 𝑚𝑚
𝑆𝑢 = 300 𝑀𝑃𝑎
SAGER, SHERNELYN P.
Solution:
•
𝑃𝑜𝑤𝑒𝑟 = 𝐸 × 𝑛𝑜. 𝑜𝑓 ℎ𝑜𝑙𝑒𝑠
𝑁
1 𝐾𝑁
1𝑚
2
𝑆𝑢 × 𝜋𝑑𝑡𝑝 (300 𝑚𝑚2 ) (𝜋)(75 𝑚𝑚)(40 𝑚𝑚) (1000 𝑁) (1000 𝑚𝑚)
𝐸=
=
2
2
𝐸 = 56.549 𝐾𝑁 − 𝑚 𝑜𝑟 𝐾𝐽
•
𝑃 = 56.549 𝐾𝐽 × 10
ℎ𝑜𝑙𝑒𝑠
𝑚𝑖𝑛
1
× 60 𝑠
𝑷 = 𝟗. 𝟒𝟐𝟓 𝑲𝑱⁄𝒔 𝒐𝒓 𝑲𝑾
10. A shearing machine requires 150 kg – m of energy to shear a steel sheet,
and has a normal speed of 3.0 rev/sec, slowing down to 2.8 rev/sec during
the shearing process. The flywheel has a mean diameter of 75 cm and weighs
0.018 kg/cm3. The width of the rim is 30 cm. If the hub and arms of the
flywheel account for 15 % of its total weight, find the thickness of the rim.
a. 6.12 cm
b. 5.12 cm
c. 4.12 cm
d. 3 cm
Given:
∆𝐸 = 150 𝑘𝑔 − 𝑚
𝑛1 = 3.0 𝑟𝑒𝑣/𝑠
𝑛2 = 2.8 𝑟𝑒𝑣/𝑠
𝐷𝑚 = 75 𝑐𝑚 = 0.75 𝑚
𝜌 = 0.018 𝑘𝑔/𝑐𝑚3
𝑏 = 30 𝑐𝑚
mH + mA = 15%m
Required:
- thickness, t
Solution:
𝑉 = 𝑏𝑡(𝐿) = 𝑏𝑡(𝜋𝐷)
• 𝑡=
∆𝐸 =
𝑉
𝑏(𝜋𝐷)
𝑚𝑤
2𝑔𝑐
(𝑣2 2 − 𝑣1 2 ) =
(𝑚+0.15𝑚)
2𝑔𝑐
(𝑣2 2 − 𝑣1 2 ) =
1.15 𝑚
2𝑔𝑐
(𝑣2 2 − 𝑣1 2 )
∆𝐸(2)(𝑔𝑐)
𝑚 = 1.15(𝑣 2 −𝑣 2)
2
1
Solve for 𝑣2 𝑎𝑛𝑑 𝑣1 :
𝑣2 = 𝜋𝐷𝑛2 = 𝜋(0.75𝑚)(3.0 𝑟𝑒𝑣⁄𝑠) = 7.069 𝑚/𝑠
𝑣1 = 𝜋𝐷𝑛1 = 𝜋(0.75𝑚)(2.8 𝑟𝑒𝑣⁄𝑠) = 6.597 𝑚/𝑠
SAGER, SHERNELYN P.
𝑚=
𝑉=
9.807𝑁
𝑘𝑔−𝑚
)(2)(1
)
1 𝑘𝑔
𝑁−𝑠2
2
2 𝑚
2
1.15(7.069 −6.597 ) 2
𝑠
(150 𝑘𝑔−𝑚)(
𝑚
𝜌
= 396.743 𝑘𝑔
396.743 𝑘𝑔
= 0.018 𝑘𝑔/𝑐𝑚3 = 22,041.2778 𝑐𝑚3
• 𝑡=
22,041.2778 𝑚3
(30 𝑐𝑚)(𝜋)(75 𝑐𝑚)
𝒕 = 𝟑. 𝟏𝟐 𝒄𝒎
11. Find the working pressure required to punch 4’’ diameter hole on ¼’’ thick
steel plate.
a. 60 tons
b. 70 tons
c. 75 tons
d. 80 tons
Given:
D= 4 𝑖𝑛𝑐ℎ𝑒𝑠
t = ¼ 𝑖𝑛𝑐ℎ𝑒𝑠
Required:
- Pressure, P
Solution:
Empirical Formula
•
𝑃 = 𝐷𝑡(80)
NOTE: 𝐷 & 𝑡 𝑚𝑢𝑠𝑡 𝑏𝑒 𝑖𝑛 𝑖𝑛𝑐ℎ𝑒𝑠
1
𝑃 = (4 𝑖𝑛𝑐ℎ𝑒𝑠) (4 𝑖𝑛𝑐ℎ𝑒𝑠) (80)
𝑷 = 𝟖𝟎 𝒕𝒐𝒏𝒔
SAGER, SHERNELYN P.
BOLTS &
NUTS
PROBLEMS
SAGER, SHERNELYN P.
BOLTS AND NUTS PROBLEMS
1. A cylinder head of a steam engine is held on by 14 bolts. The effective diameter
of the cylinder is 35 cm and the steam pressure is 8.5 kg/cm2. Assuming that
the bolts are not initially stressed, find the size of bolts if the tensile stress is
not to exceed 200 kg/cm2.
a.19.50 mm
b.19.15 mm
c.19. 65 mm
d. 19.30 mm
Given:
𝑁𝐵 = 14 𝑏𝑜𝑙𝑡𝑠
𝐷𝑐 = 35 𝑐𝑚
𝑃𝑐 = 8.5 𝑘𝑔/𝑐𝑚2
𝑆𝑡 = 200 𝑘𝑔/𝑐𝑚2
Required:
- diameter of bolts, D
Solution:
𝑆𝑡 =
𝐹𝑒
𝐹
= 𝜋 𝑒2
𝐴𝑠
• 𝐷=√
4
𝐷
4𝐹𝑒
𝑆𝑡 𝜋
Solve for 𝐹𝑒 :
𝐹
𝑁𝐵 = 𝐹𝑐
𝑒
𝐹
∴ 𝐹𝑒 = 𝑁𝑐
𝐵
𝐹
where: 𝑃𝑐 = 𝐴𝑐
𝑐
𝜋
∴ 𝐹𝑐 = 𝑃𝑐 𝐴𝑐 = (8.5 𝑘𝑔⁄𝑐𝑚2 ) ( 4 ) (35 𝑐𝑚)2 = 8177.9584 𝑘𝑔
𝐹𝑒 =
8177.9584 𝑘𝑔
14
= 584.1399 kg
4(584.1399 𝑘𝑔)
𝐷=√
𝑘𝑔
(200 2 )(𝜋)
𝑐𝑚
= 1.9284 𝑐𝑚 ×
10 𝑚𝑚
1 𝑐𝑚
𝑫 = 𝟏𝟗. 𝟐𝟖𝟒 𝒎𝒎
SAGER, SHERNELYN P.
2. A steam engine cylinder has an effective diameter of 35 cm and the maximum
steam pressure acting on the cover is 12.5 kg/cm2. Calculate the number of
2 cm diameter bolts required to fix the cylinder. Assume the permissible
stress in the bolts to be 330 kg/cm2.
a.12 bolts
b.10 bolts
c.13 bolts
d.9bolts
Given:
𝐷𝑐 = 35 𝑐𝑚
𝑃𝑐 = 12.5 𝑘𝑔/𝑐𝑚2
𝐷 = 2 𝑐𝑚
𝑆𝑡 = 330 𝑘𝑔/𝑐𝑚2
Required:
-no. of bolts, 𝑁𝐵
Solution:
• 𝑁𝐵 =
𝐹𝑐
𝐹𝑒
Solve for 𝐹𝑐 and 𝐹𝑒
𝐹
𝑃𝑐 = 𝐴𝑐
𝑐
𝜋
∴ 𝐹𝑐 = 𝑃𝑐 𝐴𝑐 = (12.5 𝑘𝑔⁄𝑐𝑚2 ) ( ) (35 𝑐𝑚)2 = 12,026.4094 𝑘𝑔
4
𝑆𝑡 =
𝐹𝑒
𝐴𝑠
𝜋
∴ 𝐹𝑒 = 𝑆𝑡 𝐴𝑠 = (330 𝑘𝑔⁄𝑐𝑚2 ) ( ) (2 𝑐𝑚)2 = 1036.7256 𝑘𝑔
4
𝑁𝐵 =
12,026.4094 𝑘𝑔
1036.7256 𝑘𝑔
𝑵𝑩 = 𝟏𝟏. 𝟔 𝒃𝒐𝒍𝒕𝒔 ≈ 𝟏𝟐 𝒃𝒐𝒍𝒕𝒔
SAGER, SHERNELYN P.
3. Calculate the stress area (in mm2) of each of eighteen bolts used to fasten the
hemispherical joints of a 480 mm pressure vessel; with an internal pressure
of 10 MPa and a bolt tensile strength of 150 MPa.
a.775
b. 667.5
c. 750
d. 670.25
Given:
𝑁𝐵 = 18 𝑏𝑜𝑙𝑡𝑠
𝐷𝑐 = 480 𝑚𝑚
𝑃𝑐 = 10 𝑀𝑃𝑎
𝑆𝑡 = 150 𝑀𝑃𝑎
Required:
- stress area of bolts in mm2, 𝐴𝑠
Solution:
𝑆𝑡 =
𝐹𝑒
𝐴𝑠
• 𝐴𝑠 = 𝑆𝑒
𝐹
𝑡
Solve for 𝐹𝑒 :
𝐹
𝑁𝐵 = 𝐹𝑐
𝑒
∴ 𝐹𝑒 =
𝐹𝑐
𝑁𝐵
𝐹
where: 𝑃𝑐 = 𝐴𝑐
𝑐
𝑁
𝜋
∴ 𝐹𝑐 = 𝑃𝑐 𝐴𝑐 = (10 𝑚𝑚2 ) ( 4 ) (480 𝑚𝑚)2 = 1,809,557.368 𝑁
𝐹𝑒 =
• 𝐴𝑠 =
1,809,557.368 𝑁
18
= 100,530.9649 𝑁
100,530.9649 𝑁
𝑁
(150
)
𝑚𝑚2
𝑨𝒔 = 𝟔𝟕𝟎. 𝟐𝟎𝟔𝟒 𝒎𝒎𝟐
SAGER, SHERNELYN P.
4. A 12 cm x 16 cm air compressor is operated with a maximum pressure of 10
kg/cm2. There are 5 bolts with yield strength of 64 ksi holding the cylinder
head to the compressor. Determine the tensile stress of the bolt.
a. 26.59 MPa
b. 29.69 MPa
c. 28.54 MPa
d. 32.56 MPa
Given:
𝐷 = 12 𝑐𝑚
𝐿 = 16 𝑐𝑚
𝑃𝑐 = 10 𝑘𝑔/𝑐𝑚2
𝑁𝐵 = 5 𝑏𝑜𝑙𝑡𝑠
𝑆𝑦 = 64 𝑘𝑠𝑖
Required:
-𝑆𝑑
Solution:
• 𝑆𝑑 = 𝐴𝑒
𝐹
𝑠
Solve for 𝐹𝑒 & 𝐴𝑠 :
𝐹
𝑁𝐵 = 𝐹𝑐
𝑒
𝐹
∴ 𝐹𝑒 = 𝑁𝑐
𝐵
𝐹
where: 𝑃𝑐 = 𝐴𝑐
𝑐
𝑘𝑔
𝜋
∴ 𝐹𝑐 = 𝑃𝑐 𝐴𝑐 = (10 𝑐𝑚2 ) ( 4 ) (12 𝑐𝑚)2 = 1130.9734 𝑘𝑔
𝐹𝑒 =
1130.9734 𝑘𝑔
5
= 226.1947 𝑘𝑔
Since 𝑆𝑦 and D is given (Use Empirical Formula)
𝑆𝑦
𝐹𝑒 = ( 6 ) (𝐴𝑠 )
∴ 𝐴𝑠 =
3⁄
2
2⁄
3
6𝐹𝑒
(𝑆 )
𝑦
=[
2.205 𝑙𝑏
)
1 𝑘𝑔
𝑙𝑏
64,000 2
𝑖𝑛
6(226.1947 𝑘𝑔)(
2⁄
3
]
= 0.1298 𝑖𝑛2
9.81 𝑁
• 𝑆𝑑 =
226.1947 𝑘𝑔( 1 𝑘𝑔 )
25.4 𝑚𝑚 2
(0.1298 𝑖𝑛2 )(
1 𝑖𝑛
)
= 26.5
𝑁
𝑚𝑚2
𝑺𝒅 = 𝟐𝟔. 𝟓𝟎 𝑴𝒑𝒂
SAGER, SHERNELYN P.
5. The manhole cover of an ammonia storage tank is to be held by 25 stud bolts.
If the pressure inside the storage tank will remain constant at 12.5 kg/cm2
and the manhole diameter is 508 mm, what should be the stress area of the
carbon steel bolt?
a. 0.606 in2
b. 0.566 in2
c. 0.452 in2
d.0.645in2
Given:
𝑁𝐵 = 25 𝑏𝑜𝑙𝑡𝑠
𝑃𝑐 = 12.5 𝑘𝑔/𝑐𝑚2
𝐷𝑐 = 508 𝑚𝑚 = 50.8 𝑐𝑚
Material= Carbon Steel bolt
𝐶 𝑓𝑜𝑟 𝑐𝑎𝑟𝑏𝑜𝑛 𝑠𝑡𝑒𝑒𝑙 = 5000 𝑝𝑠𝑖
Required:
- stress area of carbon steel bolts, 𝐴𝑠
Solution:
Since bolt material and D is given (use empirical formula)
𝐹𝑒 = 𝐶𝐴𝑠 1.418
•
1
𝐴𝑠 =
𝐹 1.418
( 𝐶𝑒 )
Solve for 𝐹𝑒 :
𝑁𝐵 =
𝐹𝑐
𝐹𝑒
𝐹
∴ 𝐹𝑒 = 𝑁𝑐
𝐵
𝐹
where: 𝑃𝑐 = 𝐴𝑐
𝑐
𝑘𝑔
𝜋
∴ 𝐹𝑐 = 𝑃𝑐 𝐴𝑐 = (12.5 𝑐𝑚2 ) ( 4 ) (50.8 𝑐𝑚)2 = 25,335.374 𝑘𝑔
𝐹𝑒 =
• 𝐴𝑠 = (
25,335.374 𝑘𝑔
25
= 1013.415 𝑘𝑔
2.205 𝑙𝑏
)
1 𝑘𝑔
𝑙𝑏
5000 2
𝑖𝑛
1013.415 𝑘𝑔(
1
1.418
)
𝑨𝒔 = 𝟎. 𝟓𝟔𝟔𝟕 𝒊𝒏𝟐
SAGER, SHERNELYN P.
6. What is the working strength of a 1-inch bolt which is screwed up tightly in
a packed joint when the allowable working stress is 10,000 psi?
a. 3,000 lb
b. 4,000 lb
c. 5,000 lb
d. 6,000 lb
Given:
𝐷 = 1 𝑖𝑛𝑐ℎ
𝑆𝑡 = 10,000 𝑝𝑠𝑖
Required:
- working stress, W
Solution:
𝑊 = 𝑆𝑡 (0.55𝐷2 − 0.25𝐷)
𝑊 = (10,000
𝑙𝑏
𝑖𝑛2
) [0.55(1 𝑖𝑛)2 − 0.25(1 𝑖𝑛)]
𝑾 = 𝟑, 𝟎𝟎𝟎 𝒍𝒃
7. What is the working strength of a 2’’ bolt which is screwed up tightly in a
packed joint when the allowable working stress is 12,000 psi?
a. 20.4 kip
b. 22.4 kip
c. 23.4 kip
d. 18 kip
Given:
𝐷 = 2 𝑖𝑛𝑐ℎ
𝑆𝑡 = 12,000 𝑝𝑠𝑖
Required:
- working stress, W
Solution:
𝑊 = 𝑆𝑡 (0.55𝐷2 − 0.25𝐷)
𝑙𝑏
𝑊 = (12,000 𝑖𝑛2 ) [0.55(2 𝑖𝑛)2 − 0.25(2 𝑖𝑛)]
1 𝑘𝑖𝑝
𝑊 = 20,400 𝑙𝑏 × 1000 𝑙𝑏
𝑾 = 𝟐𝟎. 𝟒 𝒌𝒊𝒑
8. Compute the working strength of a 1’’ bolt which is screwed up tightly in a
packed joint when the allowable working stress is 13,000 psi.
a. 3,900 lb
b. 3,700 lb
c. 3,800 lb
d. 3,600 lb
Given:
𝐷 = 1 𝑖𝑛𝑐ℎ
𝑆𝑡 = 13,000 𝑝𝑠𝑖
Required:
- working stress, W
SAGER, SHERNELYN P.
Solution:
𝑊 = 𝑆𝑡 (0.55𝐷2 − 0.25𝐷)
𝑙𝑏
𝑊 = (13,000 𝑖𝑛2 ) [0.55(1 𝑖𝑛)2 − 0.25(1 𝑖𝑛)]
𝑾 = 𝟑, 𝟗𝟎𝟎 𝒍𝒃
9. Compute the working strength of a bolt having a 1 1/2’’ diameter under a
tensile stress of 8,000 psi.
a. 3,060 lb
b. 4,560 lb
c. 6,900 lb
d. 7,500 lb
Given:
𝐷 = 1 1⁄2 𝑖𝑛 = 1.5 𝑖𝑛
𝑆𝑡 = 8,000 𝑝𝑠𝑖
Required:
- working stress, W
Solution:
𝑊 = 𝑆𝑡 (0.55𝐷2 − 0.25𝐷)
𝑙𝑏
𝑊 = (8,000 𝑖𝑛2 ) [0.55(1.5 𝑖𝑛)2 − 0.25(1.5 𝑖𝑛)]
𝑾 = 𝟔, 𝟗𝟎𝟎 𝒍𝒃
10. Determine the working strength of a 1.25’’ bolt screwed up tightly with a
tensile stress of 8,000 psi.
a.4,375 lb
b. 3,475 lb
c. 4,175 lb
d. 7,543 lb
Given:
𝐷 = 1.25 𝑖𝑛
𝑆𝑡 = 8,000 𝑝𝑠𝑖
Required:
- working stress, W
Solution:
𝑊 = 𝑆𝑡 (0.55𝐷2 − 0.25𝐷)
𝑙𝑏
𝑊 = (8,000 𝑖𝑛2 ) [0.55(1.25 𝑖𝑛)2 − 0.25(1.25 𝑖𝑛)]
𝑾 = 𝟒, 𝟑𝟕𝟓 𝒍𝒃
SAGER, SHERNELYN P.
POWER SCREW
PROBLEMS
SAGER, SHERNELYN P.
POWER SCREW PROBLEMS
1. Find the horsepower lost when a collar is loaded with 1,000 lb rotates at 25
rpm and has a coefficient of friction of 0.15. The outside diameter of the collar
is 4 inches and the inside diameter is 2 inches.
a. 0.045
b. 0.89
c. 0.093
d. 0.56
Given:
𝑊 = 1,000 𝑙𝑏
𝑛 = 25 𝑟𝑝𝑚
𝑓𝑐 = 0.15
𝐷𝑜𝑐 = 4 𝑖𝑛𝑐ℎ𝑒𝑠
𝐷𝑖𝑐 = 2 𝑖𝑛𝑐ℎ𝑒𝑠
Required:
- Power in collar in hp, P
Solution:
• 𝑃𝑐𝑓 =
𝑇𝑐𝑓 𝑛
𝑙𝑏−𝑖𝑛
ℎ𝑝−𝑚𝑖𝑛
63,025
Solve for 𝑇𝑐𝑓
𝑇𝑐𝑓 = 𝑓𝑐 𝑊𝑟𝑓
1 𝐷 3
𝐷 3
1 (4)3
(2)3 𝑖𝑛3
𝑟𝑓 = ( 𝑜𝑐2 − 𝑖𝑐 2) = [ 2 − 2] 2 = 1.5556 𝑖𝑛
3 𝐷
𝐷
3 (4)
(2) 𝑖𝑛
𝑜𝑐
𝑖𝑐
𝑇𝑐𝑓 = (0.15)(1,000 𝑙𝑏)(1.5556 𝑖𝑛) = 233.34 𝑙𝑏 − 𝑖𝑛
• 𝑃𝑐𝑓 =
(233.34 𝑙𝑏−𝑖𝑛)(25
𝑟𝑒𝑣
)
𝑚𝑖𝑛
𝑙𝑏−𝑖𝑛
63,025ℎ𝑝−𝑚𝑖𝑛
𝑷𝒄𝒇 = 𝟎. 𝟎𝟗𝟑 𝒉𝒑
SAGER, SHERNELYN P.
2. What is the frictional hp acting on a collar loaded with 100 kg weight? The
collar has an outside diameter of 100 mm and an internal diameter of 40 mm.
The collar rotates at 1,000 rpm and the coefficient of friction between the
collar and the pivot surface is 0.15.
a. 0.8
b. 0.3
c. 0.5
d. 1.2
Given:
𝑊 = 100 𝑘𝑔
𝐷𝑜𝑐 = 100 𝑚𝑚
𝐷𝑖𝑐 = 40 𝑚𝑚
𝑛 = 1,000 𝑟𝑝𝑚
𝑓𝑐 = 0.15
Required:
- Power in collar in hp, P
Solution:
• 𝑃𝑐𝑓 =
𝑇𝑐𝑓 𝑛
𝑙𝑏−𝑖𝑛
ℎ𝑝−𝑚𝑖𝑛
63,025
Solve for 𝑇𝑐𝑓
𝑇𝑐𝑓 = 𝑓𝑐 𝑊𝑟𝑓
1 𝐷 3
𝐷 3
1 (100)3
(40)3 𝑚𝑚3
1 𝑖𝑛
𝑟𝑓 = ( 𝑜𝑐2 − 𝑖𝑐 2) = [
−
]
×
= 1.4623 𝑖𝑛
3 𝐷
𝐷
3 (100)2
(40)2 𝑚𝑚2
25.4 𝑚𝑚
𝑜𝑐
𝑖𝑐
2.205 𝑙𝑏
𝑇𝑐𝑓 = (0.15)(100 𝑘𝑔)(1.4623 𝑖𝑛) (
1 𝑘𝑔
) = 48.3656 𝑙𝑏 − 𝑖𝑛
𝑟𝑒𝑣
• 𝑃𝑐𝑓 =
(48.3656 𝑙𝑏−𝑖𝑛)(1000𝑚𝑖𝑛)
𝑙𝑏−𝑖𝑛
63,025ℎ𝑝−𝑚𝑖𝑛
𝑷𝒄𝒇 = 𝟎. 𝟕𝟔𝟕𝟒 𝒉𝒑 ≈ 𝟎. 𝟖 𝒉𝒑
SAGER, SHERNELYN P.
3. An ACME single-threaded power screw is lifting a load of 70 KN and has a
pitch of 6mm. The combined efficiency of the screw and collar is 15%. Find
the power input in kW if the screw turns at 60 rpm.
a. 2.8
b. 4.6
c. 6.7
d. 8.5
Given:
𝑊 = 70 𝐾𝑁
𝑃 = 6𝑚𝑚 = 0.006𝑚 = 𝐿(𝑠𝑖𝑛𝑔𝑙𝑒)
𝑒𝑂𝑅 = 15%
𝑛 = 60 𝑟𝑝𝑚
Required:
-power input
Solution:
𝑒𝑂𝑅 =
𝑃𝑜𝑢𝑡𝑝𝑢𝑡
𝑃𝑖𝑛𝑝𝑢𝑡
• 𝑃𝑖𝑛𝑝𝑢𝑡 =
𝑃𝑜𝑢𝑡𝑝𝑢𝑡
𝑒𝑂𝑅
Solve for 𝑃𝑜𝑢𝑡𝑝𝑢𝑡
60𝑟𝑒𝑣⁄𝑚𝑖𝑛
)]
⁄𝑚𝑖𝑛
𝑛
𝑃𝑜𝑢𝑡𝑝𝑢𝑡 = 𝑊𝑣 = 𝑊 [𝐿 ( )] = 70𝐾𝑁 [0.006𝑚 ( 𝑠𝑒𝑐
60
60
• 𝑃𝑖𝑛𝑝𝑢𝑡 =
= 0.42 𝐾𝐽/𝑠
0.42 𝐾𝑊
0.15
𝑷𝒊𝒏𝒑𝒖𝒕 = 𝟐. 𝟖 𝑲𝑾
4. A power screw with collar has an efficiency of 15% lifts a load of 70 KN. The
lead is 18 mm. Find the operating force when it is applied at a radius of 915
mm.
a. 8.08 KN
b. 6.54 KN
c. 2.82 KN
d. 1.46 KN
Given:
𝑒𝑂𝑅 = 15%
𝑊 = 70𝐾𝑁
𝐿 = 18 𝑚𝑚 = 0.018 𝑚
𝑟 = 915 𝑚𝑚 = 0.915 𝑚
Required:
-operating force, 𝐹𝑅
Solution:
𝑒𝑂𝑅 =
𝑊𝐿
(2𝜋)𝐹𝑅 𝑟
• 𝐹𝑅 = (2𝜋)𝑒
𝑊𝐿
𝑂𝑅 𝑟
(70 𝐾𝑁)(0.018 𝑚)
= (2𝜋)(0.15)(0.915
𝑚)
𝑭𝑹 = 𝟏. 𝟒𝟔 𝑲𝑵
SAGER, SHERNELYN P.
5. A square thread power screw has an efficiency of 12.5 %. The coefficient of
thread friction is 0.13. When lifting a load, determine the lead angle required.
a. 5.1⁰
b. 3.1⁰
c. 1.1⁰
d. 7.1⁰
Given:
𝑒𝑇𝑅 = 12.5%
𝑓 = 0.13
Required:
- lead angle, 𝛼
Solution:
Considering frictional torque of thread only
𝑒𝑇𝑅 =
(𝑡𝑎𝑛𝛼)(𝑐𝑜𝑠∅−𝑓𝑡𝑎𝑛𝛼)
(𝑐𝑜𝑠∅𝑡𝑎𝑛𝛼+𝑓)
SQUARE THREAD: cos ∅ = cos 0° = 1
𝑒𝑇𝑅 =
(𝑡𝑎𝑛𝛼)(1−𝑓𝑡𝑎𝑛𝛼)
(𝑡𝑎𝑛𝛼+𝑓)
(𝑡𝑎𝑛𝛼)(1−0.13𝑡𝑎𝑛𝛼)
0.125 =
(𝑡𝑎𝑛𝛼+0.13)
0.125(𝑡𝑎𝑛𝛼 + 0.13) = (𝑡𝑎𝑛𝛼)(1 − 0.13𝑡𝑎𝑛𝛼)
0.125 𝑡𝑎𝑛𝛼 + 0.01625 = 𝑡𝑎𝑛𝛼 − 0.13𝑡𝑎𝑛2 𝛼
0.13𝑡𝑎𝑛2 𝛼 − 0.875 𝑡𝑎𝑛𝛼 + 0.01625 = 0
Using quadratic formula:
𝑡𝑎𝑛 𝛼 =
𝑡𝑎𝑛 𝛼 =
−𝑏±√𝑏2 −4𝑎𝑐
2𝑎
=
−(−0.875)±√(−0.875)2 −4(0.13)(0.01625)
2(0.13)
−(−0.875)±√(−0.875)2 −4(0.13)(0.01625)
2(0.13)
tan 𝛼(+) = 6.7121
tan 𝛼(−) = 0.0186
use small number:
𝛼 = 𝑡𝑎𝑛−1 (0.0186)
𝜶 = 𝟏. 𝟏°
SAGER, SHERNELYN P.
6. A single-threaded square power screw has a mean diameter of 33 mm and of
the collar is 90 mm. The lead is 6 mm. The coefficient of friction on threads
is 0.13 and for the collar is 0.1. Determine the combined efficiency of screw
and collar.
a. 12.5%
b. 14.5%
c. 10.5%
d. 16.5%
Given:
𝐷𝑚 = 33 𝑚𝑚
𝐷𝑚𝑐 = 90 𝑚𝑚
𝐿 = 6 𝑚𝑚
𝑓 = 0.13
𝑓𝑐 = 0.1
Required:
-𝑒𝑂𝑅
Solution:
𝑒𝑂𝑅 =
(𝑡𝑎𝑛𝛼)(𝑐𝑜𝑠𝜃−𝑓𝑡𝑎𝑛𝛼)
𝑓 𝐷
(𝑐𝑜𝑠∅ 𝑡𝑎𝑛𝛼+𝑓)+( 𝑐 𝑚𝑐 )(𝑐𝑜𝑠∅−𝑓𝑡𝑎𝑛𝛼)
𝐷𝑚
SQUARE THREAD: cos ∅ = cos 0° = 1
• 𝑒𝑂𝑅 =
(𝑡𝑎𝑛𝛼)(1−𝑓𝑡𝑎𝑛𝛼)
𝑓 𝐷
(𝑡𝑎𝑛𝛼+𝑓)+( 𝑐 𝑚𝑐 )(1−𝑓𝑡𝑎𝑛𝛼)
𝐷𝑚
Solve for lead angle
𝐿
tan 𝛼 = 𝜋𝐷
𝑚
𝛼 = 𝑡𝑎𝑛−1 (
𝐿
𝜋𝐷𝑚
𝑒𝑂𝑅 =
6 𝑚𝑚
) = 𝑡𝑎𝑛−1 [𝜋(33 𝑚𝑚)] = 3.3123°
(tan 3.3123°)(1−0.13 tan 3.3123°)
(0.1)(90 𝑚𝑚)
(tan 3.3123°+0.13)+[ (33
](1−0.13 tan 3.3123°)
𝑚𝑚)
× 100
𝒆𝑶𝑹 = 𝟏𝟐. 𝟓 %
SAGER, SHERNELYN P.
7. The power input required to raise a load of 70 KN is 3.36 kW. The linear speed
of the screw is 0.006 m/s. Find overall efficiency of the screw.
a. 12.5%
b. 14.5%
c. 10.5%
d. 16.5%
Given:
𝑊 = 70 𝐾𝑁
𝑃𝑖𝑛𝑝𝑢𝑡 = 3.36 𝐾𝑊
𝑣 = 0.006 𝑚/𝑠
Required:
-𝑒𝑂𝑅
Solution:
• 𝑒𝑂𝑅 =
𝑃𝑜𝑢𝑡𝑝𝑢𝑡
𝑃𝑖𝑛𝑝𝑢𝑡
Solve for 𝑃𝑜𝑢𝑡𝑝𝑢𝑡
𝑃𝑜𝑢𝑡𝑝𝑢𝑡 = 𝑊𝑣 = (70 𝐾𝑁)(0.006 𝑚⁄𝑠)
• 𝑒𝑂𝑅 = 3.36 𝐾𝑊 × 100
0.42 𝐾𝑊
𝒆𝑶𝑹 = 𝟏𝟐. 𝟓 %
8. A ¾’’ diameter square thread power screw has six threads per inch. It is to be
used to raise a load of 4,000 lbf. For a coefficient of friction of 0.15, calculate
the torque required to rotate the screw in ft- lbf.
a. 25.8
b. 26.8
c. 28.8
d. 27.8
Given:
𝐷𝑜 = ¾ inches
𝑁𝑇 = 6 𝑡ℎ𝑟𝑒𝑎𝑑𝑠/𝑖𝑛𝑐ℎ
𝑊 = 4,000 𝑙𝑏𝑓
𝑓 = 0.15
Required:
-𝑇𝑇𝑓 in ft-lbf
Solution:
𝐷𝑚
𝑇𝑇𝑓 = 𝑊 (
2
𝑐𝑜𝑠∅𝑡𝑎𝑛𝛼+𝑓
) (𝑐𝑜𝑠∅−𝑓𝑡𝑎𝑛𝛼 )
SQUARE THREAD: cos ∅ = cos 0° = 1
• 𝑇𝑇𝑓 = 𝑊 ( 2𝑚) (1−𝑓𝑡𝑎𝑛𝛼)
𝐷
𝑡𝑎𝑛𝛼+𝑓
SAGER, SHERNELYN P.
Solving for 𝐷𝑚 & 𝛼
𝑃
𝐷𝑚 = 𝐷𝑜 + 2
𝑃=
1
𝑁𝑇
=
1
6 𝑡ℎ𝑟𝑒𝑎𝑑𝑠/𝑖𝑛𝑐ℎ
3
0.1667 𝑖𝑛
4
2
𝐷𝑚 = 𝑖𝑛𝑐ℎ𝑒𝑠 +
tan 𝛼 =
= 0.1667 𝑖𝑛𝑐ℎ
= 0.6665 𝑖𝑛𝑐ℎ
𝐿
𝜋𝐷𝑚
where L for single thread
𝐿 = 1(𝑃) = 0.1667 𝑖𝑛𝑐ℎ
𝛼 = 𝑡𝑎𝑛−1 (
𝐿
𝜋𝐷𝑚
0.1667 𝑖𝑛
) = 𝑡𝑎𝑛−1 [𝜋(0.6665 𝑖𝑛)] = 4.5519°
• 𝑇𝑇𝑓 = 4000 𝑙𝑏𝑓 (
0.6665 𝑖𝑛
2
tan 4.5519°+0.15
) (1−0.15 tan 4.5519°)
𝑇𝑇𝑓 = 309.7737 𝑙𝑏𝑓 − 𝑖𝑛 ×
1 𝑓𝑡
12 𝑖𝑛
𝑻𝑻𝒇 = 𝟐𝟓. 𝟖 𝒇𝒕 − 𝒍𝒃𝒇
Situational Problem (9-10)
A single-threaded square power screw is to raise a load of 70 KN. The screw
has a major diameter of 36 mm and a pitch of 6 mm. The coefficient of thread
friction and collar friction are 0.13 and 0.10, respectively. If the collar mean
diameter is 90 mm and the screw turns at 60 rpm.
9. Find the combined efficiency of screw and collar.
a. 12.526%
b. 10.042%
c. 14.348%
10. Find the power input to the screw in kW.
a. 5.604
Given:
𝑊 = 470 𝐾𝑁
𝐷𝑂 = 36 𝑚𝑚
𝑃 = 6 𝑚𝑚 = 𝐿(𝑠𝑖𝑛𝑔𝑙𝑒)
𝑓 = 0.13
b. 3.353
c. 7.258
d. 16.648%
d. 9.404
𝑓𝑐 = 0.10
𝐷𝑚𝑐 = 90 𝑚𝑚
𝑛 = 60 𝑟𝑝𝑚
SAGER, SHERNELYN P.
Required:
9. 𝑒𝑂𝑅
10. Power input
Solution:
Solving for 𝑒𝑂𝑅
𝑒𝑂𝑅 =
(𝑡𝑎𝑛𝛼)(𝑐𝑜𝑠𝜃−𝑓𝑡𝑎𝑛𝛼)
𝑓 𝐷
(𝑐𝑜𝑠∅ 𝑡𝑎𝑛𝛼+𝑓)+( 𝑐 𝑚𝑐 )(𝑐𝑜𝑠∅−𝑓𝑡𝑎𝑛𝛼)
𝐷𝑚
SQUARE THREAD: cos ∅ = cos 0° = 1
• 𝑒𝑂𝑅 =
(𝑡𝑎𝑛𝛼)(1−𝑓𝑡𝑎𝑛𝛼)
𝑓 𝐷
(𝑡𝑎𝑛𝛼+𝑓)+( 𝑐 𝑚𝑐 )(1−𝑓𝑡𝑎𝑛𝛼)
𝐷𝑚
𝑃
𝐷𝑚 = 𝐷𝑜 + 2 = 36 𝑚𝑚 +
𝐿
6 𝑚𝑚
2
= 33 𝑚𝑚
6 𝑚𝑚
𝛼 = 𝑡𝑎𝑛−1 (𝜋𝐷 ) = 𝑡𝑎𝑛−1 [𝜋(33 𝑚𝑚)] = 3.3123°
𝑚
• 𝑒𝑂𝑅 =
(tan 3.3123°)(1−0.13 tan 3.3123°)
(tan 3.3123°+0.13)+[
(0.1)(90 𝑚𝑚)
](1−0.13 tan 3.3123°)
(33 𝑚𝑚)
× 100
𝒆𝑶𝑹 = 𝟏𝟐. 𝟓𝟐𝟔%
Solving for Power input
𝑒𝑂𝑅 =
𝑃𝑜𝑢𝑡𝑝𝑢𝑡
𝑃𝑖𝑛𝑝𝑢𝑡
• 𝑃𝑖𝑛𝑝𝑢𝑡 =
𝑃𝑜𝑢𝑡𝑝𝑢𝑡
𝑒𝑂𝑅
Solve for 𝑃𝑜𝑢𝑡𝑝𝑢𝑡
𝑛
60𝑟𝑒𝑣⁄𝑚𝑖𝑛
)]
⁄𝑚𝑖𝑛
𝑃𝑜𝑢𝑡𝑝𝑢𝑡 = 𝑊𝑣 = 𝑊 [𝐿 (60)] = 70𝐾𝑁 [0.006𝑚 (60𝑠𝑒𝑐
= 0.42 𝐾𝐽/𝑠
• 𝑃𝑖𝑛𝑝𝑢𝑡 = 0.12526
0.42 𝐾𝑊
𝑷𝒊𝒏𝒑𝒖𝒕 = 𝟑. 𝟑𝟓𝟑 𝑲𝑾
SAGER, SHERNELYN P.
SPRING
PROBLEMS
SAGER, SHERNELYN P.
SPRING PROBLEMS
1. A coil spring has 10 coils. The ends are squared and ground. Determine the
number of active coils.
a. 5
b. 6
c. 7
d. 8
Given:
𝑛 = 10
Required:
-𝑛𝑐
Solution:
𝑛 = 𝑛𝑐 + 2
𝑛𝑐 = 𝑛 − 2 = 10 − 2
𝒏𝒄 = 𝟖 𝒂𝒄𝒕𝒊𝒗𝒆 𝒄𝒐𝒊𝒍𝒔
2. A coil spring has 10 coils. The ends are squared. Determine the number of
active coils.
a. 5
b. 6
c. 7
d. 8
Given:
𝑛 = 10
Required:
-𝑛𝑐
Solution:
𝑛 = 𝑛𝑐 + 2
𝑛𝑐 = 𝑛 − 2 = 10 − 2
𝒏𝒄 = 𝟖 𝒂𝒄𝒕𝒊𝒗𝒆 𝒄𝒐𝒊𝒍𝒔
3. The solid and working deflection of a coil spring are 12 mm and 10 mm,
respectively Determine the clash allowance.
a. 15 %
b. 20 %
c. 25 %
d. 10 %
Given:
𝛿𝑠 = 12 𝑚𝑚
𝛿𝑤 = 10 𝑚𝑚
Required:
-clash allowance, CA
Solution:
𝐶𝐴 =
𝐶𝐴 =
𝛿𝑠 −𝛿𝑤
𝛿𝑤
12 𝑚𝑚−10 𝑚𝑚
10 𝑚𝑚
× 100
𝑪𝑨 = 𝟐𝟎%
SAGER, SHERNELYN P.
4. A coil spring has a free length of 350 mm and solid length of 338 mm.
Determine the working deflection if the clash allowance is 20 %.
a. 10 mm
b. 15 mm
c. 20 mm
d. 5 mm
Given:
𝐿𝑓 = 350 𝑚𝑚
𝐿𝑠 = 338 𝑚𝑚
𝐶𝐴 = 20%
Required:
-working deflection,𝛿𝑤
Solution:
𝐶𝐴 =
𝛿𝑠 −𝛿𝑤
𝛿𝑤
𝛿
= 𝛿𝑠 − 1
𝑤
𝛿
𝐶𝐴 + 1 = 𝛿 𝑠
𝑤
𝑠
• 𝛿𝑤 = 𝐶𝐴+1
𝛿
where:
𝛿𝑠 = 𝐿𝑓 − 𝐿𝑠 = (350 − 338)𝑚𝑚
𝛿𝑠 = 12 𝑚𝑚
• 𝛿𝑤 = 0.20+1
12 𝑚𝑚
𝜹𝒘 = 𝟏𝟎 𝒎𝒎
5. A helical coil spring has a wire diameter of 1/8 inch and a spring index of 8.
Determine the outer diameter of the coil.
a. 2 5/8’’
b. 1 1/8’’
c. 1 5/8’’
d. 2 1/8’’
Given:
𝑑𝑤 = 1⁄8 𝑖𝑛𝑐ℎ
𝐶=8
Required:
-outer diameter of coil,𝐷𝑜
Solution:
𝐶=
𝐷𝑚
𝑑𝑤
=
𝐶+1=
𝐷𝑜 −𝑑𝑤
𝐷𝑜
𝑑𝑤
=
𝐷𝑜
𝑑𝑤
−1
𝑑𝑤
1
• 𝐷𝑜 = 𝑑𝑤 (𝐶 + 1) = 𝑖𝑛(8 + 1)
8
𝑫𝒐 =
𝟗
𝟏
𝒊𝒏𝒄𝒉 ≈ 𝟏
𝒊𝒏𝒄𝒉
𝟖
𝟖
SAGER, SHERNELYN P.
6. A helical coil spring has a mean diameter of 1 inch and a wire diameter 0f
1/8 inch. Determine the stress concentration factor.
a. 1.2
b. 1.4
c. 1.6
d. 1.8
Given:
𝐷𝑚 = 1 𝑖𝑛𝑐ℎ
𝑑𝑤 = 1⁄8 𝑖𝑛𝑐ℎ
Required:
-stress concentration factor,𝐾𝑤
Solution:
•
4𝐶−1
𝐾𝑤 = ⌊4𝐶−4 +
𝐶=
•
𝐷𝑚
𝑑𝑤
0.615
𝐶
⌋
1 𝑖𝑛
=1
4(8)−1
𝐾𝑤 = ⌊4(8)−4 +
⁄8𝑖𝑛
= 8 𝑖𝑛
0.615
(8)
⌋
𝑲𝒘 = 𝟏. 𝟏𝟖𝟒 ≈ 𝟏. 𝟐
7. A helical coil spring has a mean coil diameter of 1 inch and a wire diameter
of 1/8 inch. The maximum stress is 60 ksi and the stress concentration factor
is 1.19. Determine the direct shear load it can support.
a. 35.2 lb
b. 38.7 lb
c. 43.3 lb
d. 46.6 lb
Given:
𝐷𝑚 = 1 𝑖𝑛ch
𝑑𝑤 = 1⁄8 𝑖𝑛𝑐ℎ
𝑆𝑠 𝑚𝑎𝑥 = 60 𝑘𝑠𝑖
𝐾𝑠 = 1.19
Required:
-F
Solution:
𝑆𝑠𝑚𝑎𝑥 =
𝜋𝑑𝑤 3
𝑆𝑠 𝑚𝑎𝑥 (𝜋)(𝑑𝑤 3 )
• 𝐹=
𝐹=
8𝐾𝑠 𝐹𝐷𝑚
8𝐾𝑠 𝐷𝑚
𝑘𝑖𝑝𝑠
1 3
1000 𝑙𝑏
)(𝜋)( ) 𝑖𝑛3 (
)
8
1 𝑘𝑖𝑝
𝑖𝑛2
(60
8(1.19)(1 𝑖𝑛)
𝑭 = 𝟑𝟖. 𝟔𝟕 𝒍𝒃 ≈ 𝟑𝟖. 𝟕 𝒍𝒃
SAGER, SHERNELYN P.
8. A helical coil spring has a mean coil diameter of 1 inch and a wire diameter of 1/8
inch. It can support a load of 38.7 lb. Determine the number of an active coils if the
axial direction is 1 inch (G = 12.5 x 106 psi).
a. 7
b. 8
c. 9
Given:
𝐷𝑚 = 1 𝑖𝑛ch
𝑑𝑤 = 1⁄8 𝑖𝑛𝑐ℎ
𝐹 = 38.7 𝑙𝑏
𝛿 = 1 𝑖𝑛𝑐ℎ
𝐺 = 12.5 × 106 𝑝𝑠𝑖
d. 10
Required:
- 𝑛𝑐
Solution:
𝛿 = 𝛿𝑐𝑠 =
• 𝑛𝑐 =
8𝐹𝐷𝑚 3 𝑛𝑐
𝑑𝑤 4 𝐺
𝛿𝑑𝑤 4 𝐺
8𝐹𝐷𝑚 3
4
=
𝑙𝑏
(1 𝑖𝑛)(1⁄8) 𝑖𝑛4 (12.5×106 2 )
𝑖𝑛
8(38.7 𝑙𝑏)(1 𝑖𝑛)3
𝒏𝒄 = 𝟗. 𝟖𝟔 ≈ 𝟏𝟎 𝒂𝒄𝒕𝒊𝒗𝒆 𝒄𝒐𝒊𝒍𝒔
9. A coil spring is to have a spring index of 6 and is to deflect1/2 inch under a load of
50 lb. The shear modulus of elasticity is 12 x 103 ksi and the spring have 12 active
coils. Find the diameter of the coil.
a. 2.566 inches b. 2.038 inches
c. 1.566 inches
d. 1.038 inches
Given:
𝐶=6
𝛿 = ½ 𝑖𝑛𝑐ℎ
𝐹 = 50 𝑙𝑏
𝐺 = 12 × 103 𝑘𝑠𝑖
𝑛𝑐 = 12 𝑎𝑐𝑡𝑖𝑣𝑒 𝑐𝑜𝑖𝑙𝑠
Required:
-𝐷𝑚
Solution:
𝐶=
𝐷𝑚
𝑑𝑤
• 𝐷𝑚 = 𝐶(𝑑𝑤 )
Solve for 𝑑𝑤
𝛿=
8𝐹𝐷𝑚 3 𝑛𝑐
𝑑𝑤 4 𝐺
SAGER, SHERNELYN P.
𝑑𝑤 =
8𝐹𝐶 3 𝑛𝑐
𝑑𝑤 𝐺
=
1 𝑘𝑖𝑝
)
1000 𝑙𝑏
1
𝑘𝑖𝑝𝑠
( 𝑖𝑛𝑐ℎ)(12×103 2 )
2
𝑖𝑛
8(50 𝑙𝑏)(6)3 (12)(
= 0.1728 𝑖𝑛
• 𝐷𝑚 = 6(0.1728 𝑖𝑛)
𝑫𝒎 = 𝟏. 𝟎𝟑𝟔𝟖 𝒊𝒏𝒄𝒉
10. A helical steel spring has a maximum load of 800 lb and a corresponding
deflection of 2 inches, if it has 8 active coils and an index of 6, what minimum
shear strength of the spring material is required?
a. 57 ksi
b. 47 ksi
c. 67 ksi
d. 37 ksi
Given:
𝐹 = 800 𝑙𝑏
𝛿 = 2 𝑖𝑛𝑐ℎ𝑒𝑠
𝑛𝑐 = 8 𝑎𝑐𝑡𝑖𝑣𝑒 𝑐𝑜𝑖𝑙𝑠
𝐶=6
𝐺𝑠𝑡𝑒𝑒𝑙 = 11.5 × 106 𝑝𝑠𝑖
Required:
- 𝑆𝑠
Solution:
• 𝑆𝑠 =
8𝐾𝑤 𝐹𝐶
𝜋𝑑𝑤 2
Solve for 𝑑𝑤
𝛿=
8𝐹𝐶 3 𝑛𝑐
𝑑𝑤 =
𝑑𝑤 𝐺
8𝐹𝐶 3 𝑛𝑐
𝛿𝐺
=
8(800 𝑙𝑏)(6)3 (8)
𝑙𝑏
(2 𝑖𝑛)(11.5×106 2 )
𝑖𝑛
= 0.4808 𝑖𝑛
Solve for 𝐾𝑤
𝐾𝑤 = ⌊
4𝐶−1
4𝐶−4
• 𝑆𝑠 =
+
0.615
𝐶
⌋=⌊
4(6)−1
4(6)−4
+
0.615
6
⌋ = 1.2525
8(1.2525)(800 𝑙𝑏)(6)
𝜋(0.4808 𝑖𝑛)2
𝑺𝒔 = 𝟔𝟔. 𝟐𝟑 𝒌𝒔𝒊
SAGER, SHERNELYN P.
11. A helical spring is compressed by 30 mm. The spring scale is 18 KN/m
while its allowable shear stress is 345 MPa and the spring index is 8. What is
the diameter of the spring wire?
a. 3.12 mm
b. 4.23 mm
c. 9.24 mm
d. 6.14 mm
Given:
𝛿 = 30 𝑚𝑚
𝑘 = 18 𝐾𝑁/𝑚
𝑆𝑠 = 345 𝑀𝑃𝑎
𝐶=8
Required:
-wired diameter, 𝑑𝑤
Solution:
8𝐾𝑤 𝐹𝐶
𝑆𝑠 = 𝑆𝑠𝑚𝑎𝑥 =
• 𝑑𝑤 = √
𝜋𝑑𝑤 2
8𝐾𝑤 𝐹𝐶
𝜋𝑆𝑠
Solve for 𝐾𝑤 & 𝐹
𝐾𝑤 = ⌊
𝑘=
4𝐶−1
+
4𝐶−4
0.615
𝐶
4(8)−1
4(8)−4
+
0.615
8
⌋ = 1.1840
𝐹
𝛿
𝐹 = 𝑘(𝛿) = 18
•
⌋=⌊
𝐾𝑁
𝑚
(0.03 𝑚) = 0.54 𝐾𝑁 ≈ 540 𝑁
8(1.840)(540 𝑁)(8)
𝑑𝑤 = √
𝜋(345
𝑁
)
𝑚𝑚2
𝒅𝒘 = 𝟔. 𝟏𝟒 𝒎𝒎
12. A coil spring is subjected to a direct shear load of 40 KN. The deflection
caused was 254 mm. Determine the spring gradient.
a. 157.5 KN/m b. 150.1 KN/m c. 164.3 KN/m
d. 171.7 KN/m
Given:
𝐹 = 40 𝐾𝑁
𝛿 = 254 𝑚𝑚 = 0.254 𝑚
Required:
-spring gradient, k
Solution:
𝑘=
𝐹
𝛿
=
40 𝐾𝑁
0.254 𝑚
𝒌 = 𝟏𝟓𝟕. 𝟒𝟖 𝑲𝑵/𝒎 ≈ 𝟏𝟓𝟕. 𝟓 𝑲𝑵/𝒎
SAGER, SHERNELYN P.
13. Two extension springs are hooked in series and supports a load of 0.45
KN. One spring has a constant of 0.009 KN/mm and the other have a
constant of 0.018 KN/mm. Determine the deflection of the load.
a. 60 mm
b. 70 mm
c. 75 mm
d. 80 mm
Given:
𝑘1 = 0.009 𝐾𝑁/𝑚𝑚
𝑘2 = 0.018 𝐾𝑁/𝑚𝑚
𝐹 = 0.45 𝐾𝑁
Required:
-𝛿
Solution:
𝐹
𝑘𝑒 = 𝛿
•
𝐹
𝛿=𝑘
𝑒
where: 𝐹 = 𝐹1 = 𝐹2
• 𝛿=
𝐹1
𝑘1
+
𝐹2
𝑘2
=
0.45 𝐾𝑁
𝐾𝑁
0.009
𝑚𝑚
+
0.45 𝐾𝑁
𝐾𝑁
𝑚𝑚
0.018
𝜹 = 𝟕𝟓 𝒎𝒎
14. Three extension springs are hooked in parallel and supports a load of 0.45
KN. One spring has a constant of 0.009 KN/mm and the other have a
constant of 0.018 KN/mm. Determine the deflection of the load.
a. 5 mm
b. 15 mm
c. 10 mm
d. 20 mm
Given:
𝑘1 = 0.009 𝐾𝑁/𝑚𝑚
𝑘2 = 0.018 𝐾𝑁/𝑚𝑚
𝑘3 = 𝑘2
𝐹 = 0.45 𝐾𝑁
Required:
-𝛿
Solution:
𝐹
𝑘𝑒 = 𝛿
•
𝛿=
𝐹
𝑘𝑒
where: 𝑘𝑒 = 𝑘1 = 𝑘2 = 𝑘3
• 𝛿=
𝐹
𝑘1 +𝑘2 +𝑘3
=
0.45 𝐾𝑁
(0.009+0.018+0.018)
𝐾𝑁
𝑚𝑚
𝜹 = 𝟏𝟎 𝒎𝒎
SAGER, SHERNELYN P.
15. A three extension coil springs are hooked in series that support a single
weight of 100 kg. The first spring is rated at 0.40 kg/mm and the other two
springs are rated at 0.64 kg/mm. Compute the total deflection.
a. 563 mm
b. 268 mm
c. 156 mm
d. 250 mm
Given:
𝑘1 = 0.40 𝑘𝑔/𝑚𝑚
𝑘2 = 0.64 𝑘𝑔/𝑚𝑚
𝑘3 = 𝑘2
𝐹 = 100 𝑘𝑔
Required:
-𝛿
Solution:
𝐹
𝑘𝑒 = 𝛿
•
𝐹
𝛿=𝑘
𝑒
where: 𝐹 = 𝐹1 = 𝐹2 = 𝐹3
• 𝛿=
𝐹1
𝑘1
+
𝐹2
𝑘2
+
𝐹3
𝑘3
=
100 𝑘𝑔
𝑘𝑔
0.40
𝑚𝑚
+
100 𝑘𝑔
𝑘𝑔
0.64
𝑚𝑚
+
100 𝑘𝑔
0.64
𝑘𝑔
𝑚𝑚
𝜹 = 𝟓𝟔𝟐. 𝟓 𝒎𝒎 ≈ 𝟓𝟔𝟑 𝒎𝒎
16. All four-compression coil spring support one load of 400 kg. All four
springs are arranged in parallel and rated same at 0.709 kg/mm. Compute
the deflection in mm.
a. 564
b. 1457
c. 171
d. 141
Given:
𝐹𝑇 = 400 𝑘𝑔
𝐹1 = 100 𝑘𝑔
𝑘 = 0.709 𝑘𝑑/𝑚𝑚 = 𝑘1 = 𝑘2 = 𝑘3 = 𝑘4
Required:
-𝛿
Solution:
𝛿=
𝐹1
𝑘1
𝐹𝑒𝑎𝑐ℎ =
𝛿=
400 𝑘𝑔
4
= 100 𝑘𝑔
100 𝑘𝑔
0.709 𝑘𝑔/𝑚𝑚
𝜹 = 𝟏𝟒𝟏. 𝟎𝟒 𝒎𝒎
SAGER, SHERNELYN P.
17. A weight of 800 lb falls freely from a distance of 40 inches, then strikes
and deflects a coil spring at 10 inches. Determine the resisting force of the
spring.
a. 7,500 lb
b. 8,000 lb
c. 8,500 lb
d. 9,000 lb
Given:
𝑊 = 800 𝑙𝑏
ℎ = 40 𝑖𝑛𝑐ℎ𝑒𝑠
𝛿 = 10 𝑖𝑛𝑐ℎ𝑒𝑠
Required:
-F
Solution:
𝐹
=[
𝑊
2(ℎ+𝛿)
𝛿
𝐹 = 𝑊[
]
2(ℎ+𝛿)
2(40+10)
𝛿
10
] = 800 𝑙𝑏 [
]
𝑭 = 𝟖, 𝟎𝟎𝟎 𝒍𝒃
18. A weight of 160 lb falls from a height of 2.5 ft to the center of a horizontal
platform mounted on four helical springs. At impact each spring deflects 3
inches. Calculate the wire diameter if the maximum design stress is 60,000
psi and spring index of 5.
a. 0.53 in
b. 0.8 in
c. 0.57 in
d. 0.88 in
Given:
𝑊 = 160 𝑙𝑏
ℎ = 2.5 𝑓𝑡
no. of helical springs= 4
𝛿 = 3 𝑖𝑛𝑐ℎ𝑒𝑠
𝐶=5
𝑆𝑠𝑚𝑎𝑥 = 60,000 𝑝𝑠𝑖
Required:
- wired diameter, 𝑑𝑤
Solution:
8𝐾𝑤 𝐹𝐶
𝑆𝑠𝑚𝑎𝑥 =
• 𝑑𝑤 = √
𝜋𝑑𝑤 2
8𝐾𝑤 𝐹𝐶
𝜋𝑆𝑠
Solve for 𝐾𝑤 & 𝐹
𝐾𝑤 = ⌊
4𝐶−1
4𝐶−4
+
0.615
𝐶
⌋=⌊
4(5)−1
4(5)−4
+
0.615
5
⌋ = 1.3105
SAGER, SHERNELYN P.
𝐹 = 𝑊[
2((2.5 𝑓𝑡×
2(ℎ+𝛿)
] = 160 𝑙𝑏 [
𝛿
12 𝑖𝑛
)+3 𝑖𝑛)
1 𝑓𝑡
3 𝑖𝑛
] = 3250 𝑙𝑏
3520 𝑙𝑏
𝐹𝑒𝑎𝑐ℎ = 4 ℎ𝑒𝑙𝑖𝑐𝑎𝑙 𝑠𝑝𝑟𝑖𝑛𝑔𝑠 = 880 𝑙𝑏
•
8(1.3105)(880 𝑙𝑏)(5)
𝑑𝑤 = √
𝑙𝑏
𝜋(60,000 2 )
𝑖𝑛
𝒅𝒘 = 𝟎. 𝟒𝟗 𝒊𝒏 ≈ 𝟎. 𝟓 𝒊𝒏
19. Compute the deflection of an 18 coils helical spring having a load of 100
kg. The modulus of elasticity in shear of spring is 96.62 GPa, OD of 9.256 cm
and with diameter of 9.525 mm.
a. 9 cm
b. 100 mm
c. 11 mm
d. 14 mm
Given:
𝑛 = 18
𝐹 = 100 𝑘𝑔
𝐺 = 96.62 𝐺𝑃𝑎 = 96,620 𝑀𝑃𝑎
𝐷𝑜 = 9.256 𝑐𝑚
𝑑𝑤 = 9.525 𝑚𝑚 = 0.9525 𝑐𝑚
Required:
-𝛿
Solution:
• 𝛿=
8𝐹𝐷𝑚 3 𝑛𝑐
𝑑𝑤 4 𝐺
Solve for 𝐷𝑚 and 𝑛𝑐
𝐷𝑚 = 𝐷𝑜 − 𝑑𝑤 = 9.256 𝑐𝑚 − 0.9525 𝑐𝑚 = 8.3035 𝑐𝑚
𝑛𝑐 = 𝑛 − 2 = 18 − 2 = 16 𝑎𝑐𝑡𝑖𝑣𝑒 𝑐𝑜𝑖𝑙𝑠
9.81 𝐾𝑁
• 𝛿=
8(100 𝑘𝑔× 1 𝑘𝑔 )(8.3035)3 𝑐𝑚3 (16)
(0.9525)4 𝑐𝑚4 [96,620
𝑁
(10 𝑚𝑚)2
×
]
2
𝑚𝑚
1 𝑐𝑚2
𝜹 = 𝟗. 𝟎𝟒 𝒄𝒎
SAGER, SHERNELYN P.
20. A spring with 12 active coils and a spring index of 9 supports a static load
of 220 N with a deflection of 12 mm. The shear modulus of the spring material
is 83 GPa. What is the theoretical wire diameter?
a. 18 mm
b. 16 mm
c. 14 mm
d. 20 mm
Given:
𝑛𝑐 = 12
𝐶=9
𝐹 = 220 𝑁
𝛿 = 12 𝑚𝑚
𝐺 = 83 𝐺𝑃𝑎 = 83,000 𝑀𝑃𝑎
Required:
- wired diameter, 𝑑𝑤
Solution:
𝛿=
8𝐹𝐶 3 𝑛𝑐
𝑑𝑤 =
𝑑𝑤 =
𝑑𝑤 𝐺
8𝐹𝐶 3 𝑛𝑐
𝛿𝐺
8(220 𝑁)(9)3 (12)
𝑁
)
𝑚𝑚2
(12)(83,000
𝒅𝒘 = 𝟏𝟓. 𝟒𝟔 𝒎𝒎 ≈ 𝟏𝟔 𝒎𝒎
SAGER, SHERNELYN P.