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

Note that, since the shape is thin walled, the centerline dimensions can be used in this
calculation. Furthermore, the term containing t 3 f can be neglected because it is very small;
thus, the moment of inertia is calculated as
I
z
3 2
twd
tbd f
= +
12 2
(0.125 in.)(8.00 in.)
= +
12
4
= 17.333 in.
(0.125 in.)(3.00 in.)(8.00 in.)
2
3 2
d
2
d
2
A
E
t f
z
b
y
t w
B
C
D
Shear Stress in the Flanges
The shear stress in the flanges will be distributed linearly, from zero at the flange tips
(i.e., A and E) to a maximum value at the junction of the flange and the web (i.e., B
and D). The first moment of area for point B can be calculated as
Q
and the shear stress at point B is thus
B
d
= ( bt f) 2
= (3.00 in.)(0.125 in.)(4.00 in.) = 1.50 in.
VQ
τ B =
It
z f
B
3
(900 lb)(1.50 in. )
= = 623 psi
4
(17.333 in. )(0.125 in.)
3
A
t f
b
B
Shear Stress in the Web
The shear stress in the web will be distributed parabolically, from minimum values
at points B and D to its maximum value at point C. The shear stress at point B in the
web is
d
2
z
y
C
d
4
VQ
τ B =
It
B
z w
3
(900 lb)(1.50 in. )
= = 623 psi
4
(17.333 in. )(0.125 in.)
d
2
t w
The first moment of area for point C can be calculated as
E
D
Q
d d d
= ( bt ) + ⎛ t 2 ⎝ ⎜ ⎞
2 ⎠
⎟ 4
c f w
= 1.50 in. + (0.125 in.)(4.00 in.)(2.00 in.) = 2.50 in.
3 3
and the shear stress at point C is
VQ
τ c =
It
c
z w
3
(900 lb)(2.50 in. )
= = 1, 038 psi
4
(17.333 in. )(0.125 in.)
380