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MECH 344 Problem Set 5-Chapter 10_Power Screws and Fasteners-Selected Problems Concordia University CA$18.80   Add to cart
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MECH 344 Problem Set 5-Chapter 10_Power Screws and Fasteners-Selected Problems Concordia University
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MECH 344 Problem Set 5-Chapter 10_Power Screws and Fasteners-Selected Problems Concordia University

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  • November 24, 2023
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  • 2023/2024
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MECH 344 Problem Set 5-Chapter 10_Power Screws and
Fasteners-Selected Problems Concordia University




SOLUTION (10.1)

Known: A special C-clamp uses a 0.5-inch diameter Acme thread and a collar of
0.625-inch effective mean diameter.

Find: Estimate the force required at the end of a 5-in. handle to develop a 200 lb
clamping force.

Schematic and Given Data:




1/2 in. Acme thread
dc = 5/8 in.




6 in.


Assumptions:
1. Coefficients of running friction are estimated as 0.15 for both the collar and
thescrew.
2. The screw has a single thread.
Analysis:
1. From section 10.3.1, and considering that service conditions may be conducive
torelatively high friction, estimate f = f c 0.15 (for running friction).
2. From Table 10.3, p = 0.1 in., and with a single thread, L = 0.1 in.
3. From Fig. 10.4(a),
p
= 14.5o and dm = d = 0.5 0.05 = 0.45 in.
2
10-1

,4. From Eq. (10.1),
= tan-1 L = tan-1 0. 1 = 4.05o
dm (0. 45)
5. From Eq. (10.6),
-1 -1 o o
n = tan (tan cos ) = tan (tan 14.5 cos 4.05 )
= 14.47o

(Note: with 4o, it is obvious that n≈ and well within the accuracy of
assumed friction coefficients)




10-2

,6. From Eq. (10.4),
f dm Lcos n
T = Wdm + Wf c dc
2 dmcos n f L 2
(150)(0.45) (0 15) (0. 45) 0. 1(cos 14. 47˚) (150)(0.15)(0.625)
T=
2 (0. 45)(cos 14. 47˚) (0. 15)(0. 1) 2

T = 7.70 7.03 = 14.73 lb in. Use T 15 lb in.

At the end of a 6-in. handle, the clamping force required 15/6 = 2.5 lb ■

SOLUTION (10.2)
Known: A double-threaded Acme screw of known major diameter is used in a jack
having a plain thrust collar of known mean diameter. Coefficients of running friction
are estimated as 0.09 for the collar and 0.12 for the screw.

Find:
(a) Determine the pitch, lead, thread depth, mean pitch diameter, and helix angle
ofthe screw.
(b) Estimate the starting torque for raising and for lowering a 4000 N load.
(c) If the screw is lifting a 4000 N load, determine the efficiency of the jack.

Schematic and Given Data:

Double-threaded Acme
Load screw
4,000 N d = 1 in.
dc = 50 mm
fc = 0.09
f
f = 0.12

fc



dm
dc



Assumptions:
1. The starting friction is about 1/3 higher than running friction.
2. The screw is not exposed to vibration.
Analysis:
1. From Table 10.3, there are 5 threads per inch.
p = 1/5 = 0.2 in. = 0.0051 m ■
Because of the double-threaded screw,
L = 2p = 0.4 in. = 0.0102 m ■
10-3

, 2. From Fig. 10.4a,
Threaded depth = 0.5p = 0.10 in. = 0.00254 m ■
dm = d 0.5p = 0.90 in. = 0.02286 m ■
3. From Eq. (10.1),

= tan 1 L 0. 4
= tan 1 = 8. 05o ■
dm 0. 90
4. For starting, increase the coefficient of friction by 1/3:
f c = 0.12, f = 0.16
From Eq. (10.6),
-1 -1 o o
n = tan (tan cos ) = tan (tan 14.5 cos 8.05 )
= 14.36o

5. From Eq. (10.4),

f dm Lcos n
T = Wdm + Wf c dc
2 dmcos n f L 2


=
4000(0.02286)
2 (0.02286)cos
0 16 (0.02286)14.36˚  0.16(0.0102)
0.0102cos 14.36˚

4000(0. 12)(0. 05)
2
T = 7.175 12 = 19.175 N•m. to raise the load ■

6. From Eq. (10.5),

T = Wdm f dm Lcos n + Wf c dc
2 dmcos n f L 2

4000(0.02286) 0 16 (0.02286) 0.0102cos 14.36˚
T=
2 (0.02286)cos 14.36˚ 0.16(0.0102)

4000(0. 12)(0. 05)
2
T = -5.819 12 = 6.18 N•m. to lower the load ■

7. From Eq. (10.4) with fc = 0.09, f = 0.12

4000(0.02286) 0.12 (0.02286) + (0.0102)(cos 14.36˚) 4000(0. 09)(0. 05)
T= +
2 (0.02286)cos 14.36˚ - 0.12(0.0102) 2
10-4

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