Transverse Shear/Nonuniform bending
(Chapter 5 in Gere)
KML 9/30/09
Shear forces produce nonuniform bending
of beams since they generate shear stresses and
bending stresses that vary over the length of beams. Curvature is not constant over the beam.
Since the shear stress on a differential element has the same magnitude on all four planar faces,
the shear stresses generated in the vertical direction are equivalent to the horizontal shear
stresses, which will be derived here.
The following is a differential beam element with the accompanying moment and bending stress
that vary due to nonuniform bending.
y
Beam element: side view
Differential element with bending stress
y
x
σ
1
σ
2
Differential subelement with horizontal stresses
τ
y
1
h/2
Right end view of subelement
y
z
b
dA
y
F
1
F
2
F
3
I
y
M
1

=
σ
I
y
dM)
(M
2
+

=
σ
dx
x
M
V
M + dM
V + dV
x
M + dM
σ
1
σ
2
M
The magnitudes of the horizontal forces due to
1
σ
and
2
σ
are:
∫
∫
=
σ
=
dA
I
My
dA
F
1
1
∫
∫
+
=
σ
=
dA
I
y
dM)
(M
dA
F
2
2
For horizontal force equilibrium:
∫
∫
=
=

=
dA
y
I
dM
dA
I
y
dM
F
F
F
1
2
3
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View Full DocumentThe force due to the shear load distributed over the width of the beam (assuming the shear stress
is uniformly distributed over the beam width, which holds approximately for b < h):
dx
b
F
3
τ
=
(
A
F
⋅
τ
=
, where dx is the length, and b is the width over which
τ
acts)
Equating the expressions for
3
F
and solving for shear stress,
τ
:
∫
∫
⋅
=
⋅
=
τ
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 Fall '08
 Sharpe
 Force, Shear Stress, Shear

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