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

Distribution of Shear Stress
The distribution of shear stress over the entire channel shape is plotted
in the accompanying figure.
623 psi
A
B
623 psi
C
1,038 psi
E
D
623 psi
623 psi
ExAmpLE 9.13
Consider the channel shape of Examples 9.11 and 9.12, shown again
here. Neglecting stress concentrations, determine the maximum shear
stress created in the shape if the load P = 900 lb is applied at the centroid
of the section, which is located 0.643 in. to the left of the web centerline.
Plan the Solution
This example illustrates the considerable additional shear stress created
in the channel when the external load does not act through the
shear center. The distance from the channel centroid to the shear center O
will be calculated and used to determine the magnitude of the torque
that acts on the section. The shear stress created by this torque will be
calculated from Equation (6.25). The total shear stress will be the sum
of the shear stress due to bending, as determined in Equation 9.12, and
the shear stress due to twisting.
SolutioN
Shear Center
From Equation (9.19), the location of the shear center O for the channel is calculated as
d
2
d
2
A
z
E
t f
b
y
P
0.0643 in.
e
B
t w
C
O
x
D
2 2
bdt
2 2
f (3.00 in.) (8.00 in.) (0.125 in.)
e = = = 1.038 in.
4
4I
4(17.333 in. )
z
Equivalent loading
We know that the channel will bend without twisting if load P is applied at the shear
center O, and furthermore, we know how to determine the shear stresses in the channel
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