Shear stresses in the bending of thin, unsymmetrical sections
ds
z
s
Mz
dx
y
t(s)
My
Vy
Vz
x
consider an element of a thin beam, as shown
σ xx
q =τt
q + dq
ds
dx
dA
∑F
x
dσ xx
σ xx +
dx
dx
=0
dσ xx ⎞
⎛
dx ⎟ dA − σ xx dA = 0
( q + dq ) dx − qdx + ⎜ σ xx +
dx
⎝
⎠
dσ
dq = − xx dA
dx
If we integrate from s1 to s2
s1
s2
q ( s1 )
q ( s2 )
A
s2
dσ xx
dA
dx
A
∫ dq = −∫
s1
dσ xx
dA
dx
A
q ( s2 ) − q ( s1 ) = Δq = − ∫
Since
σ xx
M
(
=
I + M z I yz ) z − ( M z I yy + M y I yz ) y
y zz
Vz
(I
2
−
I
I
yy zz
yz )
−Vy
−Vy
Vz
dM y
⎛ dM y
⎞ ⎛ dM z
⎞
dM z
I zz +
I yz ⎟ z − ⎜
I yy +
I yz ⎟ y
⎜
dx
dx
dx
dσ xx ⎝ dx
⎠
⎝
⎠
=
dx
( I yy I zz − I yz2 )
dσ xx (Vz I zz − Vy I yz ) z − (Vz I yz − Vy I yy ) y
=
2
dx
I
I
I
−
( yy zz yz )
placing this in our shear flow expression gives
VI
(
Δq =
z yz
where
− Vy I yy ) Qz + (Vy I yz − Vz I zz ) Qy
D
D = I yy I zz − I yz2
Qz = ∫ ydA
A
Qy = ∫ zdA
A
If we have a symmetrical expression and if Vy = 0
Δq =
−Vz Qy
I yy
If q(s1) =0 then
Δq = τ ( s1 ) t ( s1 )
τ=
and we obtain
−Vz Qy
I yy t
the minus sign is here since we have taken
+
Vz
Note: in our expression for Δq the q flows "out" from the end of the
section under consideration, whether the section is cut from the left or
right of the cross section. At the beginning of the section the q flows "in".
Examples:
q ( s1 )
s1
q ( s1 )
s1
A
A
q ( 0)
in both these cases
Δq = q ( s1 ) − q ( 0 )
q ( 0)
VI
(
=
z yz
− Vy I yy ) Qz + (Vy I yz − Vz I zz ) Qy
D
Just as in the torsion of closed sections, the shear flows
generated by bending must be conserved at a junction. This
follows from equilibrium:
dx
q3
σ xx
q1
q2
∑F
x
dσ xx
dx
σ xx +
dx
dA
=0
dσ xx ⎞
⎛
σ
q
+
q
+
q
dx
+
+
dx ⎟ dA − σ xx dA = 0
( 1 2 3 ) ⎜ xx
dx
⎝
⎠
dA
( q1 + q2 + q3 ) + dσ xx = 0
dx
so as
dA → 0
q1 + q2 + q3 = 0
i.e. the net shear flow out of (or into) a junction must be zero
A 1000 lb load produces bending but no twisting in the thin open section
shown below. Determine the shear flow distributions in this section and the
location of the shear center. Assume the thickness of all sections is 0.1 in. All
dimensions are measured to the mid planes of the walls.
3
5
z
1000 lb
y
e
S
3
5
all dimensions are in inches
10
3
5
y1
A
yc =
B
A1
C
z
=
yc
A3
10
2 × ( y1 A1 ) + ( 0 ) A3
2 A1 + A3
( 2 )(8)( 0.1)(1)
( 2 )( 8)( 0.1) + (10 )(.1)
= 0.6154 in
y
A2
E
F
D
so we have
y2
3
Vz = 1000 lb, Vy = 0, I yz = 0
Δq =
5
−Vz Qy
I yy
negligible
1
3
2⎤
3
⎡1
I yy = 2 × ⎢ ( 8 )( 0.1) + ( 8 )( 0.1)( 5 ) ⎥ + ( 0.1)(10 )
⎣12
⎦ 12
= 48.33 in 4
Δq =
for AB
s1
q=0
−1000
Qy = −20.69Qy
48.33
q ( s1 )
z
5
y
Δq = q ( s1 ) − 0 = −20.69 ⎡⎣( 0.1)( s1 )( 5 ) ⎤⎦
q ( s1 ) = −10.35 s1 lb / in
q1B = −31.03 lb / in
A
3
B
for BC
z
q ( s2 )
s2
q=0
5
y
Δq = q ( s2 ) − 0 = −20.69 ⎡⎣( 0.1)( s2 )( 5 ) ⎤⎦
q ( s2 ) = −10.35 s2
qB2 = −51.75 lb / in
B
5
C
qB3
for BD
s3
z
5
q ( s3 )
y
Δq = q ( s3 ) − qB3 = −20.69 ⎡⎣( 0.1)( s3 )( 5 − s3 / 2 ) ⎤⎦
q ( s3 ) = qB3 − 10.35 s3 + 1.035 s32
31.03
51.75
qB3
qB3 + 51.75 − 31.03 = 0
qB3 = −82.78 lb / in
q ( s3 ) = −82.78 − 10.35 s3 + 1.035 s32
At the middle
q ( 5 ) = −109 lb / in
-82.78
q ( s3 ) = −82.78 − 10.35 s3 + 1.035 s32
-109
-82.78
For the lower flange Qy is the same as for the upper flange but
with opposite sign so we have
+31.03
E
3
D
+51.75
D
5
F
the total forces carried by each
section:
R2
R1
the entire shear stress
distribution:
10
3
5
3
R1 = ∫ 10.35s1ds1 = 46.58 lb
0
5
R2 = ∫ 10.35s2 ds2 = 129.38 lb
0
10
10
R3
R1
R2
3
5
Note: we have used the
negative of the shear flow
expressions to agree with
the directions assumed for
the forces shown
R3 = ∫ ( 82.78 + 10.35s3 − 1.035s32 ) ds3 = 1000 lb
0
to find the shear center
R2
R1
1000 lb
e
3
R3
10
O
10
O
5
R1
R2
3
5
We obtained the shear flow distributions under the assumption that there
was only bending. These shear flows will generate the shear forces and
their moments about a particular point (i.e. the force locations). Thus, we
can determine the shear center here by requiring that the moment of the
1000 lb load about any point produce the same moments as the shear flows,
∑M
O
= 1000 e
∑M
1000e = 828
e = 0.828 in
O
= 10 R2 − 10 R1 = (129.38 − 46.58 )(10 )
= 828 in − lb
location of the shear center by sectorial area
s2
C
B
A
s1
S
e
O
5
5
D
E
F
s3
BD
ω = ω0 + es1
AC
ω = ω0 + 5e + 5s2
EF
ω = ω0 − 5e + 5s3
take the starting point for the
integration at O
∫ ω dA = 0
3
5
⎡5
⎤
t ⎢ ∫ (ω0 + es1 ) ds1 + ∫ (ω0 + 5e + 5s2 ) ds2 + ∫ (ω0 − 5e + 5s3 ) ds3 ⎥ = 0
−5
−3
⎣ −5
⎦
odd
10ω0 + 8ω0 + 8ω0 = 0
ω0 = 0
so the sectorial area distribution looks like:
+
Thus,
+
z
s1
∫ yω dA = 0
automatically and we need only consider
∫ zω dA = 0
-
∫ zω dA = 0
3
5
⎡ +5
⎤
t ⎢ ∫ ( es1 ) s1ds1 + 5 ∫ ( 5e + 5s2 ) ds2 + ( −5 ) ∫ ( −5e + 5s3 ) ds3 ⎥ = 0
−5
−3
⎣ −5
⎦
z = s1 z = 5
which gives
z = −5
250
e + 400e − 400 = 0
3
e=
400
= 0.828 in
400 + 250 / 3