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

normal stresses are identical at all points on the cross section, torsional shear stresses are
greatest on the periphery of the shaft. For this reason, critical stresses are normally investigated
on the outer surface of the shaft.
Example 15.1 illustrates the analysis of combined torsional and axial loads in a shaft.
617
COMbINEd AxIAL ANd
TORSIONAL LOAdS
ExAmpLE 15.1
A hollow circular shaft having an outside diameter
of 4 in. and a wall thickness of 0.25 in.
y
is loaded as shown. Determine the principal
96 kip·in.
stresses and the maximum shear stress at points
H and K.
K
z
H
Plan the Solution
A
After computing the required section properties
for the pipe shaft, we will determine the
(1) B
equivalent forces acting at point H. The normal
and shear stresses created by the internal axial force and torque will be computed and
shown in their proper directions on a stress element. Stress transformation calculations
will be used to determine the principal stresses and maximum shear stress for the stress
element at H. The process will be repeated for the stresses acting at K.
SolutioN
Section Properties
The outside diameter D of the pipe is 4 in., and the wall thickness of the pipe is 0.25 in.;
thus, the inside diameter is d = 3.5 in. The cross-sectional area of the pipe will be needed
in order to calculate the normal stress caused by the axial force:
π
A D d
4 [ ] π
= − = [(4 in.) − (3.5 in.) ] = 2.9452 in.
4
2 2 2 2 2
The polar moment of inertia will be required in order to calculate the shear stress caused
by the internal torques in the pipe:
(2)
36 kip·in.
C
25 kips
x
π
J D d
32 [ ] π
= − = [(4 in.) − (3.5 in.) ] = 10.4004 in.
32
4 4 4 4 4
Equivalent Forces at H
The pipe will be sectioned just to the right of a stress element at
H, and the equivalent forces and moments acting at the section
of interest will be determined. This procedure is straightforward
at H, where the equivalent force is simply the 25 kip axial force
and the equivalent torque is equal to the 36 kip · in. torque applied
at C.
Normal and Shear Stresses at H
The normal and shear stresses at H can be calculated from the equivalent forces just
shown. The 25 kip axial force creates a compressive normal stress
F 25 kips
σ axial = = = 8.49 ksi(C)
2
A 2.9452 in.
B
96 kip·in.
(2)
H
36 kip·in.
25 kips