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

The equivalent moments at the section of interest are found from the sum of M A and M C :
M = M + M = 30,500 lb⋅ft i − 24,000 lb⋅ ft j + 28,800 lb⋅ft
k
A
C
Section Properties
The outside diameter of the pipe column is D = 9.0 in., and the inside diameter is d = 8.0 in.
The area, the moment of inertia, and the polar moment of inertia for the cross section are,
respectively, as follows:
z
y
12,900 lb
H
K
966 psi
x
π
A D d
4 [ 2 2 ] π
2 2 2
= − = [(9.0 in.) − (8.0 in.) ] = 13.352 in.
4
π
I D d
64 [ ] π
= − = [(9.0 in.)
64
− (8.0 in.) ] = 121.00 in.
π
J D d
32 [ ] π
= − = [(9.0 in.)
32
− (8.0 in.) ] = 242.00 in.
4 4 4 4 4
4 4 4 4 4
Stresses at H
The equivalent forces and moments acting at the section of interest will be evaluated sequentially
to determine the type, magnitude, and direction of any stresses created at H.
The 12,900 lb axial force creates compressive normal stress, which acts in the
y direction:
Fy
12,900 lb
σ y = = = 966 psi(C)
2
A 13.352 in.
y
3,000 lb
448 psi
Although shear stresses are associated with the 3,000 lb shear force, the shear stress
at point H is zero.
H
K
z
x
13,612 psi
The 30,500 lb · ft bending moment about the x axis creates compressive normal
stress at H:
H
K
30,500
lb·ft
Mc x
(30,500 lb⋅ft)(4.5 in.)(12in./ft)
σ y = =
= 13,612 psi ( C)
4
I
121.0 in.
x
z
x
y
24,000 lb·ft
The 24,000 lb · ft torque acting about the y axis creates shear stress at H. The magnitude
of this shear stress can be calculated from the elastic torsion formula:
5,355 psi
H
K
Tc
τ = =
J
(24,000 lb⋅ft)(4.5 in.)(12in./ft)
242.0 in.
4
= 5,355 psi
z
x
644