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

ExAmpLE 9.11
Derive an expression for the location of the shear center O for the channel
shape shown.
d
2
d
2
e
t f
y
z
b
ds s
e
t w
x
O
P
Plan the Solution
From the concept of shear flow, the horizontal shear force produced in each
channel flange will be determined. The twisting moment produced by these
forces will be counteracted by the moment produced by the vertical external
load P acting at a distance e from the centerline of the channel web.
SolutioN
Since the applied load P is assumed to act at the shear center O, the channel
shape will bend about the z axis (i.e., the neutral axis), but it will not twist
about the x axis. To better understand both the forces that cause twisting in
the thin-walled channel and the forces that counteract this twisting tendency,
consider the rear face of the channel cross section.
The internal shear force V creates a shear flow q in the web and in the
flanges. This shear flow is expressed by
q =
VQ
I z
O
P
t w
x
y
V
t f
z
q
d
2
d
2
The shear force V and the shear flow q act in the directions shown in the
figure to the left. The shear flow at any location in the thin-walled shape
depends only on the first moment of area, Q.
Now, consider the shear flow in the upper flange. This shear flow acts in
the shaded area, which is located a horizontal distance s from the tip of the
flange. The shear flow acting at s can be calculated as
b
VQ
q = =
I
z
V ⎛
⎜st
I ⎝
z
f
d ⎞ Vdt f
⎟ = s
(9.18)
2⎠
2I
z
O
P
e
B
F w V
x
y
Ff
=
b
t f
z
Ff
d
2
d
2
Notice that the magnitude of the shear flow varies linearly from the free
surface at the flange tip, where s = 0, to a maximum value at the web, where
s = b. The total horizontal force acting on the upper flange is determined by
integrating the shear flow over the width of the top flange:
F
f
2
b Vt f d Vb dt f
= ∫ q ds = ∫ s ds = (a)
0
2I
4I
z
The force F f in the lower flange will be the same magnitude as that in the
upper flange; however, it will act in the opposite direction, thus maintaining
equilibrium in the z direction. The couple created by the flange forces F f
tends to twist the channel shape in a clockwise direction, as shown in the
figure in the accompanying figure.
The thickness t f of each flange is thin compared with the overall depth d
of the channel shape; therefore, the vertical shearing force transmitted by
each flange is small and can be neglected. (See Figure 9.16.) Consequently,
z
378