For shear stress
W
at a section
y
1
from the neutral axis, the incremental horizontal force
dN
over
the slice of thickness
t
is
dN
=
W
tdx
or
W
±µ¶
t
²
dN
/
dx
The total horizontal (normal) forces due to the bending moment are given as
࠵? = ∫
࠵?࠵?࠵? and
࠵? + ࠵?࠵? = ∫
(࠵? + ࠵?࠵?)࠵?࠵?
࠵?
2
࠵?
1
࠵?
2
࠵?
1
Subtracting the two terms, we have:
࠵?࠵? = ∫
࠵?࠵?࠵?࠵? = ∫
࠵?࠵?
࠵?
࠵?࠵?࠵? =
࠵?࠵?
࠵?
∫
࠵?࠵?࠵? =
࠵?࠵?
࠵?
࠵?
࠵?
2
࠵?
1
࠵?
2
࠵?
1
࠵?
2
࠵?
1
where
࠵?࠵? =
࠵?࠵?
࠵?
࠵?
and
࠵? = ∫
࠵?࠵?࠵? (= ∑ ࠵?
࠵?
̅ ࠵?
࠵?
)
࠵?
2
࠵?
1
is termed the first moment of area.
The shear stress is thus given as
࠵? =
1
࠵?
࠵?࠵?
࠵?࠵?
=
1
࠵?
࠵?࠵?
࠵?࠵?
࠵?
࠵?
=
࠵?࠵?
࠵?࠵?
where the shear force (from mechanics of materials, is defined as the change of moment)
࠵? =
࠵?࠵?
࠵?࠵?
.
It should be noted that the shear stress
W
is defined over the slice with thickness
t
.

113
Example
: A W 16
×
100 (
I
x
= 1490
in
4
) is subjected to a vertical shear force of 80 kips. Determine the
maximum shear stress and plot the distribution of shear stress for the cross section.
Solution
:
The maximum shear stress will be at the neutral axis where
Q
, the statical moment about the
neutral axis, is maximum.
4
࠵? = 10.425(0.985) (7.50 +
0.985
2
) + 7.50(0.585) (
7.50
2
) = 98.5 in
3
Shear stress at NA: ࠵?
࠵?
=
࠵?࠵?
࠵?࠵?
=
80(98.5)
1490(0.585)
= 9.04 ksi
The shear stresses over the cross section can be calculated as shown.
_______________________
Note that the flanges resist very low shear stresses. It is the web that predominantly resists the
shear in wide flange beams. AISC allows the use of average web shear for calculating shear
stress:
࠵?
࠵?
=
࠵?
࠵?࠵?
࠵?
=
80
16.97(0.585)
= 8.06 ksi
where
d
is the full depth of beam and
t
w
is the web thickness of the beam. This is less than the
shear stress of 9.04 ksi calculated by the general shear shear stress formula!
4
Shear stress between flange and web:
࠵? = 10.425(0.985) (7.50 +
0.985
2
) = 82.07 in
3
Bottom of flange: ࠵?
࠵?
=
80(82.07)
1490(10.425)
= 0.42 ksi
Top of web: ࠵?
࠵?
=
80(82.07)
1490(0.585)
= 7.53 ksi
b
f
= 10.425 in
t
f
= 0.985 in
7.50 in
t
w
= 0.585 in
d
= 16.97 in
N.A.
0.42 ksi
7.53 ksi
9.04 ksi
Cross section
Shear distribution

114
Design Shear Strength
:
V
u
≤ φ
v
V
n
,
φ
v
=1.0 (for W-shapes, see AISC Pg. 16.1-67).
V
n
=
0.6
F
y
A
w
C
v
(Eq. G2-1)
where
C
v
is termed web shear Coefficient. (Note: Shear yield strength
F
v
=
0.6
F
y
.)
For webs of rolled I-shape members with
ℎ/࠵?
࠵?
≤ 2.24
√
࠵?/࠵?
࠵?
(which applies to all
F
y
= 50
ksi W-shapes and most W-shapes tabulated in the manual, see pg. 16.1-68),
C
v
=1.
Generally, web shear strength may need to consider possible shear buckling limit state. For
example, for webs of doubly symmetric shapes (except rolled W-shapes), etc. (see AISC Section
G2.1(b)),
•
Web yielding (plastic zone):
ℎ
࠵?
࠵?
≤ 1.10
√
࠵?࠵?
࠵?
࠵?
࠵?
(≈
418
√
࠵?
࠵?
)
C
v
=1 (Eq. G2-3)
• Inelastic buckling of web :
1.10
√
࠵?࠵?
࠵?
࠵?
࠵?
<
ℎ
࠵?
࠵?
≤ 1.37
√
࠵?࠵?
࠵?
࠵?
࠵?
(≈
522
√
࠵?
࠵?
)
࠵?
࠵?
=
1.10√࠵?࠵?

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- Spring '14
- GregoryG.Deierlein
- Shear Stress, FY, Shear strength