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Mechanical Key Design Guide

A flat key mounted on a 4 in diameter motor shaft is subjected to a maximum twisting moment of 45,000 lbs-in. To withstand these stresses, the key should be 1 in wide, 3/4 in thick, and 2.5 in long. A flat key is used to prevent slipping of a cast-iron gear mounted on a 2 in diameter shaft. The gear delivers 125 hp at 250 rpm and subjects the key to a torque of 31,500 lbs-in. To withstand the stresses from this torque, the key should be 1/2 in wide, 3/8 in thick, and 4.6 in long.

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1K views2 pages

Mechanical Key Design Guide

A flat key mounted on a 4 in diameter motor shaft is subjected to a maximum twisting moment of 45,000 lbs-in. To withstand these stresses, the key should be 1 in wide, 3/4 in thick, and 2.5 in long. A flat key is used to prevent slipping of a cast-iron gear mounted on a 2 in diameter shaft. The gear delivers 125 hp at 250 rpm and subjects the key to a torque of 31,500 lbs-in. To withstand the stresses from this torque, the key should be 1/2 in wide, 3/8 in thick, and 4.6 in long.

Uploaded by

Christian de Asis
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as DOCX, PDF, TXT or read online on Scribd
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A flat key mounted on a motor shaft 4 in.

diameter is
subjected to a maximum twisting moment of 45,000 lbs-
in.
Design a suitable key assuming the design stress in
bearing
to be 24,000 psi and in shear at 9000 psi.

Solution: From table 7 – 1 (see prep), it is obvious that for a 4 in. diameter
shaft, a flat key of dimensions, width = 1 in. and thickness = 3/4 in. is a
suitable choice. The force acting on the key is given by,
F=
T/r (1)
where, T = torque
r = radius of the shaft.
Substituting values in (1) gives
F = {(45,000) / (2.0)}
F = 22,500 lbs.
The length of the key considering shearing is given by,
lks = (F / bτd) (2)
where, b = width of the key
τd = design stress in shear.
Substituting values of known quantities in (2) gives,
lks = {(22,500) / (1 × 9000)} = 2.5 in.
The length of the key considering bearing is given by,
lkb = (2F / tSb) (3)
where, t = thickness of key
Sb = design stress in bearing.
Substituting values of known quantities in (3) gives,
lkb = [{2 × 22,500} / {(3/4) × 24,000}] = 2.5 in.
Thus the key should have the following dimensions,
thickness = 3/4 in.
width = 1 in.
length = 2.5 in.
A flat key is used to prevent slipping of a cast-iron gear
mounted on a shaft, 2 in. in diameter. Determine the
appropriate dimensions of the key if the gear delivers 125
hp
at 250 rpm. Assume design stresses for shear and
bearing
as 19,000 psi and 37,000 psi.

Solution: The torque acting on the gear T is given by,


T = [{63,000(hp)} /
N] (1)
where, hp = horsepower transmitted by the gear
N = rpm of the gear.
Substituting values of known quantities in (1) gives,
T = [{63,000 × 125} / {250}] = 31,500 lbs.-in.
From table 7 – 1 (see prep), for a shaft diameter of 2 in., a flat key
of width = (1/2) in. and thickness = 3/8 in. is suggested.
The length of the key, considering shearing, is given by,
lks = {2T / (τdbd)} (2)
where, T = torque
τd = design stress in shear
b = key width
d = shaft diameter.
Substituting values of known quantities in (1) gives,
lks = [{2 × 31,500} / {(19,000) × (0.5) × (2.0)}] ≈ 3.32 in.
Similarly, the length of the key considering bearing is given by,
lkb = {4T / (Sdtd)} (3)
where, Sb = design stress in bearing
t = key thickness.
Substituting values of known quantities in (2) gives,
lkb = [{4 × 31,500} / {(37,000) × (3/8) × (2.0)}] = 4.54 in.
Thus the dimensions of the key should be:
key thickness = 3/8 in.
key width = 1/2 in.
key length = 4.6 in.

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