θ
=
TL
GJ
=
(4.10
*
10
3
N
#
mm)(250
mm)
(80
*
10
3
N
>
mm
2
)(982
mm
4
)
=
0.013
rad
Using
π
rad
=
180
∘
,
θ
=
(0.013 rad)(180
°
>
π
rad)
=
0.75
°
3–10
TORSION IN MEMBERS
HAVING NONCIRCULAR
CROSS SECTIONS
The behavior of members having noncircular cross sections
when subjected to torsion is radically different from that for
members having circular cross sections. However, the fac-
tors of most use in machine design are the maximum stress
and the total angle of twist for such members. The formulas
for these factors can be expressed in similar forms to the for-
mulas used for members of circular cross section (solid and
hollow round shafts).
The following two formulas can be used:
➭
Torsional Shear Stress
τ
max
=
T
>
Q
(3–12)
➭
Deflection for Noncircular Sections
θ
=
TL
>
GK
(3–13)
Note that Equations (3–12) and (3–13) are similar to
Equations (3–10) and (3–11), with the substitution of
Q
for
Z
p
and
K
for
J
. Refer to Figure 3–10 for the methods of deter-
mining the values for
K
and
Q
for several types of cross sec-
tions useful in machine design. These values are appropriate
only if the ends of the member are free to deform. If either
end is fixed, as by welding to a solid structure, the resulting
stress and angular twist are quite different. (See References
1–3, 6, and 7.)
Example Problem
3–8
A 2.50-in-diameter shaft carrying a chain sprocket has one end milled in the form of a square to
permit the use of a hand crank. The square is 1.75 in on a side. Compute the maximum shear stress
on the square part of the shaft when a torque of 15 000 lb · in is applied.
Also, if the length of the square part is 8.00 in, compute the angle of twist over this part. The
shaft material is steel with
G
=
11.5
*
10
6
psi.
Solution
Objective
Compute the maximum shear stress and the angle of twist in the shaft.
96
PART ONE
Principles of Design and Stress Analysis
Colored dot (
) denotes
location of
H9270
max
FIGURE 3–10
Methods for determining values for
K
and
Q
for several types of cross sections
Given
Torque
=
T
=
15000
lb
#
in; length
=
L
=
8.00
in.
The shaft is square; thus,
a
=
1.75
in.
G
=
11.5
*
10
6
psi.
Analysis
Results
Figure 3–10 shows the methods for calculating the values for
Q
and
K
for use in Equations (3–12)
and (3–13).
Q
=
0.208
a
3
=
(0.208)(1.75
in)
3
=
1.115
in
3
K
=
0.141
a
4
=
(0.141)(1.75
in)
4
=
1.322
in
4
Now the stress and the deflection can be computed.
τ
max
=
T
Q
=
15 000
lb
#
in
(1.115
in
3
)
=
13 460 psi
θ
=
TL
GK
=
(15 000
lb
#
in)(8.00
in)
(11.5
*
10
6
lb
>
in
2
)(1.322
in
4
)
=
0.0079
rad
Convert the angle of twist to degrees:
θ
=
(0.0079 rad)(180
°
>
π
rad)
=
0.452
°
CHAPTER THREE
Stress and Deformation Analysis
97
The shear stress computed by this approach is the
aver-
age stress
in the tube wall. However, if the wall thickness
t
is small (a thin wall), the stress is nearly uniform through-
out the wall, and this approach will yield a close approxi-
mation of the maximum stress. For the analysis of tubular
sections having nonuniform wall thickness, see References
1–3, 6, and 7.
