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

the resultant force F w of the shear flow in the web must equal V. Moreover, the upward
internal shear force V must equal the downward external load P to satisfy equilibrium in
the y direction; hence, P = V.
The forces P and V, which are separated by a distance e, create a couple that tends to
twist the channel shape in a counterclockwise direction. A moment equilibrium equation
about point B can thus be written as
M =- F d + Pe = 0
B
f
In this equation, substitute P = V and replace F f with the expression derived in Equation
(a) to get
and then solve for e:
2
Vb dt f
Ve = ⎛ ⎞
d
⎝ ⎜ 4I
⎠
⎟
z
2 2
bdt f
e = (9.19) Ans.
4I
The distance e from the centerline of the channel web defines the location of the shear
center O. Notice that the shear center location is dependent only on the dimensions and
geometry of the cross section.
z
ExAmpLE 9.12
For the channel shape of Example 9.11, assume that d = 8.00 in., b = 3.00 in.,
t f = 0.125 in., and t w = 0.125 in. Determine the distribution of shear stress
produced in the channel if a load P = 900 lb is applied at the shear center.
Plan the Solution
The moment of inertia of the thin-walled channel shape will be determined.
The shear stress produced in each channel flange is linearly distributed; thus,
only the maximum value, which occurs at points B and D, will need to be
determined. The distribution of shear stress in the flange is parabolically distributed,
with its minimum value occurring at points B and D, and its maximum
value occurring at point C.
SolutioN
Moment of inertia
The moment of inertia for the channel shape can be expressed by the
following equation:
d
2
d
2
A
z
E
t f
b
y
e
B
t w
C
x
D
O
P
I
z
twd ⎡ bt f d
= + 2 + ⎛ 12 12 ⎝ ⎜ ⎞
⎢ 2 ⎠
⎟
⎣
3 3 2
bt
f
⎤
⎥
⎦
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