Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

162
TORSION
Another common measure of rotational speed is revolutions per minute (rpm). Rotational
speed ω can be expressed in terms of revolutions per minute, n, as
π
ω = ⎛ n rev⎞
⎛ 2 rad⎞
⎝ ⎠ ⎝
⎜
⎠
⎟ ⎛ min rev ⎝ ⎜
1min
60 s
Equation (6.17) can be written in terms of rpm n as
⎞
⎠
⎟ =
2πn
60 rad/s
2π
nT
P = Tω
= (6.20)
60
ExAmpLE 6.6
A solid 0.75 in. diameter steel shaft transmits 7 hp at 3,200 rpm. Determine the maximum
shear stress magnitude produced in the shaft.
Plan the Solution
The power transmission equation [Equation (6.17)] will be used to calculate the torque in
the shaft. The maximum shear stress in the shaft can then be calculated from the elastic
torsion formula [Equation (6.5)].
SOLUTION
Power P is related to torque T and rotation speed ω by the relationship P = Tω.
Since information about the power and rotation speed is given, this relationship can be
solved for the unknown torque T. The conversion factors required in the process, however,
can be confusing at first. We have
T
P
= =
ω
⎛
⎝
⎛ 550 lb⋅ft/s⎞
(7 hp)
⎝
⎜
1hp ⎠
⎟
⎞ ⎛ 2π
rad⎞
⎛
⎠ ⎝ 1rev ⎠ ⎝
3,200 rev
min
1 min
60 s
⎞
⎠
=
lb⋅ft
3,850
s
335.1032 rad
s
= 11.4890 lb⋅ft
The polar moment of inertia for a solid 0.75 in. diameter shaft is
π
4 4
J = (0.75 in.) = 0.0310631 in.
32
Therefore, the maximum shear stress produced in the shaft is
Tc (11.4890 lb⋅ft)(0.75 in./2)(12in./ft)
τ = =
= 1,664 psi Ans.
4
J
0.0310631 in.
mecmovies
ExAmpLES
m6.16 A 2 m long hollow steel [G = 75 GPa] shaft has an
outside diameter of 75 mm and an inside diameter of 65 mm.
If the maximum shear stress in the shaft must be limited to
50 MPa and the angle of twist must be limited to 1°, determine
the maximum power that can be transmitted by this shaft
when it is rotating at 600 rpm.
=
=
=
=
τ =
φ =