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

H
y
3,000 lb
448 psi
K
The 3,000 lb shear force acting horizontally at the section of interest is also associated
with shear stress at point K. From Equation (9.10), the first moment of area for the
hollow circular cross section is
1
Q D d
12 [ ] 1
= − = [(9.0 in.) − (8.0 in.) ] = 18.083 in.
12
3 3 3 3 3
z
x
The shear stress formula [Equation (9.2)] is used to calculate the shear stress:
13,612 psi
H
K
30,500 lb·ft
3
VQ (3,000 lb)(18.083 in. )
τ = =
= 448 psi
4
It (121.0 in. )(9 in. − 8 in.)
x
The 30,500 lb · ft bending moment about the x axis creates bending stresses at the
section of interest. Point K, however, is located on the neutral axis for this bending
moment, and consequently, the bending stress at K is zero.
z
x
y
24,000 lb·ft
The 24,000 lb · ft torque acting about the y axis creates shear stress at K. The magnitude
of this shear stress can be calculated from the elastic torsion formula:
5,355 psi
Tc
τ = =
J
(24,000 lb⋅ft)(4.5 in.)(12in./ft)
242.0 in.
4
= 5,355 psi
H
K
z
x
12,853 psi
The 28,800 lb · ft bending moment about the z axis creates tensile normal stress
at K:
28,800 lb·ft
H
K
Mc z (28,800 lb⋅ft)(4.5 in.)(12in./ft)
s y = =
= 12,853 psi ( T)
4
I
121.0 in.
z
z
x
Combined Stresses at K
The normal and shear stresses acting at point K can be summarized on a stress element.
Torsion shear
5,355.4 psi
y
Bending stress
12,852.8 psi
y
11,887 psi
5,804 psi
z
Axial stress
966.2 psi
K
Beam shear
448.3 psi
z
K
646