Bending and Torsion of Thin, Open Sections
Vy
dz
Vz
dy
If the shear forces and their locations are known, then we can calculate
the shear flows assuming there is only bending produced, and from
those shear flows (or through use of the sectorial area function)
determine the location of the shear center.
Then the torsion induced by the shear forces can be calculated:
Vz
Vz
Vy
Vz
Vy
dz
ez
dy
Vy
T
ey
bending + torsion
=
bending only
+
torsion only
T = Vz ( ey − d y ) + Vy ( ez − d z )
and we can solve for the shear stresses generated by T and superimpose
them on the bending stresses to get the total shear stresses
As an example consider our previous problem but now let the 1000 lb load
act through the vertical web:
1000 lb
1000 lb
10
O
3
T = 828.2 in-lb
=
0.828
O
+
O
5
To determine the maximum shearing stress and its location (neglecting
stress concentrations), for the torsional part we have
1
1
3
3
( 8)( 0.1) + (10 )( 0.1)
3
3
= 0.008666 in 4
J eff = 2 ×
τ max =
Ttmax ( 828.2 )( 0.1)
=
= 9556 psi
J eff
0.008666
Now, the maximum shear stress will occur in the vertical web at the center
where the shear flow is a maximum
1000 lb
qmax = 109 lb / in
0.828
O
τ max = qmax / t
T = 828.2 in-lb
O τ = 9556 psi
max
= 1090 psi
bending
torsion
so at point O on the right side we will have
τ = 8466 psi
τ max = 1090 + 9556 = 10, 646 psi
O
Bending and Torsion of a Thin Closed Section
(single cell)
Vz
Vz
q
P
ez
qB
dz
P
=
dy
+
ey
P
T
qT
Vy
Vy
torsion
bending
Note that the torsion that the shear forces generate will induce a constant
shear flow qT as shown. However in the bending induced, the shear flow is
only known up to a constant since we have
qB ( s ) = qB ( 0 )
VI
(
+
z yz
− Vy I yy ) Qz + (Vy I yz − Vz I zz ) Qy
D
and, unlike open sections, we cannot find an end where qB = 0
Thus, the total shear flow will be determined only to within a constant (which
we must determine) that is partly due to bending and partly due to torsion.
Bending and Torsion of a Thin Closed Section
(single cell)
Ω = area contained within the
centerline of the cross section
∑M
Vz
q
P
= Vy d y + Vz d z = ∫ q r⊥ ds
C
dz
r⊥
q ( s ) = qc + Vy f ( s ) + Vz g ( s )
P dy
Vy
this is the unknown constant shear flow due
to both bending and torsion
f (s) =
g (s) =
I yz Qy − I yy Qz
I yy I zz − I yz2
I yz Qz − I zz Qy
I yy I zz − I yz2
Placing the q expression into the moment equation gives:
∑M
P
= 2Ω qc + ∫ ⎡⎣Vy f ( s ) + Vz g ( s )⎤⎦ r⊥ ds
(1)
C
Also, recall we have a relationship between the shear flow and the twist/unit
length induced by that shear flow
φ′ =
1
q
∫C t ds
2GΩ (2)
1. If the shear forces and their positions are known, then qc can be
found directly from Eq.(1) since the left hand side of that equation is
known explicitly and f and g can be found for the given geometry. Then
Eq. (2) can be used (since q is now given completely) to find φ'
2. If the shear forces are known but assumed to act through the shear
center (whose position is unknown), we can set φ' =0, Vy = 0 and solve
Eq.(2) for the unknown qc . Then Eq.(1) gives the location of the shear
center, dz , since
Vz d z = ∫ q r⊥ ds
C
q ( s ) = qc + Vz g ( s )
We can repeat this process by setting φ' =0, Vz = 0 and solving
Eq.(2) again for a new qc . The Eq.(1) gives the location of the shear
center, dy :
Vy d y = ∫ q r⊥ ds
C
q ( s ) = qc + Vy f ( s )
Bending and Torsion of a Thin Closed Section
(multiple cells)
Vz
dz
P
dy
Vy
As in the case of the torsion of a multiple cell closed section, we
need to account for unknown constant shear flows in each cell.
This can be accomplished by conceptually decomposing our
problem into two simpler problems as shown on the next page.
q= q1
Vz
dz
P
dy
q=q2
Vy
q= 0
=
q1
q2
+
q=0
constant shear flows from constant
parts of q due to bending and constant
q's due to torsion.
bending shear flows q(s)
due to this "open" section
q= 0
q1
+
q2
q=0
2
Vy d y + Vz d z = 2Ω1q1 + 2Ω 2 q2 + ∑ ∫ ⎡⎣Vy f ( s ) + Vz g ( s )⎤⎦ r⊥ ds (1)
m =1 Cm
φ′ =
1
q
∫C t ds
2GΩ1 1
φ′ =
1
2GΩ 2
q
∫C t ds
2
(2)
(3)
2
Vy d y + Vz d z = 2Ω1q1 + 2Ω 2 q2 + ∑ ∫ ⎡⎣Vy f ( s ) + Vz g ( s )⎤⎦ r⊥ ds (1)
m =1 Cm
φ′ =
1
q
ds
∫
2GΩ1 C1 t
φ′ =
1
2GΩ 2
q
∫C t ds
2
(2)
(3)
1. If the shear forces and their locations are known, then q1 and q2
are first found in terms of the unknown φ' from Eqs. (2) and (3).
These qm 's are then placed into Eq.(1) which is solved for the
unknown φ' . Once φ' is known in this manner, the qm 's are
completely determined.
2. If the shear forces are known but assumed to act through the shear
center (whose position is unknown), we can set φ' =0, Vy = 0 and solve
Eqs. (2) and (3) for the unknowns q1 and q2 . Then Eq.(1) gives the
location of the shear center, dz , since
2
Vz d z = 2Ω1q1 + 2Ω 2 q2 + ∑ ∫ ⎡⎣Vz g ( s )⎤⎦ r⊥ ds
m =1 Cm
We can repeat this process by setting φ' =0, Vz = 0 and solving
Eqs.(2) and (3) again for new values q1 and q2 . Then Eq.(1) gives the
location of the shear center, dy :
2
Vy d y = 2Ω1q1 + 2Ω 2 q2 + ∑ ∫ ⎡⎣Vy f ( s )⎤⎦ r⊥ ds
m =1 Cm
Example; Calculate the shear flows in the walls of the box beam shown.
The beam has a uniform wall thickness of t = 0.1 in. all dimensions are in
inches
z
1000 lb
2
9
3
5.21
C
y
3.5
4
12
1
4
I yy( ) = 1.433 I zz( ) = 18.44 I yz( ) = −4.074
1
1
1
I yy( 2) = ? I zz( 2) = ? I yz( 2) = ?
I yy( ) = 6.975 I zz( ) = 24.43 I yz( ) = −4.689
3
3
3
I yy( ) = 14.70 I zz( ) = 15.15 I yz( ) = −3.318
4
4
1
To find the I’s for 2
5
12
13
ds
s
y
5.21
5
z = 0.5 + s
13
12
y = 6.79 − s
13
z
6.79
0.5
I
( 2)
yy
13
= ∫ z 2 dA
0
13
= ∫ ( 0.5 + 5s /13) ( 0.1) ds
2
0
= 14.41 in 4
similarly
( 2)
13
I zz = ∫ y 2 dA = 16.41 in 4
0
( 2)
13
I yz = ∫ yzdA = −3.419 in 4
0
4
so total area
moments are
I yy = ∑ I yy( ) = 37.52 in 4
m
m =1
4
I zz = ∑ I zz( ) = 74.43 in 4
m
m =1
4
I yz = ∑ I yz( ) = −15.50 in 4
m =1
m
Vz = 1000 lb, Vy = 0
∆q =
− ( I zz Qy − I yz Qz )Vz
I yy I zz − I yz2
−74.43Q
(
=
y
− 15.5Qz ) (1000 )
( 37.52 )( 74.43) − (15.5)
2
∆q = −29.16 Qy − 6.073 Qz
Now, use this relation for each section
z
6.79
y
3.5
z
s
q1
Qy(1) = ∫ zdA = − ( 3.5 − s / 2 )( 0.1)( s )
= −0.35s + 0.05s 2
z
A
Qz(1) = ∫ ydA = ( 6.79 )( 0.1)( s )
= 0.679 s
y
A
1
q ( ) ( s ) = q1 − 29.16 ⎡⎣ −0.35s + 0.05s 2 ⎤⎦ − 6.073 [ 0.679 s ]
= q1 + 6.082 s − 1.458s 2
q ( ) ( 4 ) = q1 + 1 lb / in
1
lb / in
z
5
13
s/2
s/2
12
z
6.79 − y 12
=
s/2
13
z − 0.5 5
=
s/2
13
q1 + 1
y
6.79
0.5
y
6
s
13
5
z = 0.5 + s
26
y = 6.79 −
Qy( ) = ( 0.5 + 5s / 26 )( 0.1)( s ) = 0.05s + 0.1923s 2
2
Qz( ) = ( 6.79 − 6 s /13)( 0.1)( s ) = 0.679 s − 0.04615s 2
2
q ( 2) ( s ) = q1 + 1 − 29.16 ⎡⎣0.05s + 0.01923s 2 ⎤⎦ − 6.073 ⎡⎣ 0.679s − 0.04615s 2 ⎤⎦
( s ) = q1 + 1 − 5.58s − 0.2802s 2
2
q ( ) (13) = q1 − 118.9 lb / in
q(
2)
lb / in
q1 − 118.9
z
s
5.5
5.21
y
Qy( ) = ( 5.5 − s / 2 )( 0.1)( s ) = 0.55s − 0.05s 2
3
Qz( ) = ( −5.21)( 0.1)( s ) = −0.521s
3
( s ) = q1 − 118.9 − 29.16 ⎡⎣0.55s − 0.05s 2 ⎤⎦ − 6.073[ −0.521s ]
3
q ( ) ( s ) = q1 − 118.9 − 12.87 s + 1.458s 2 lb / in
3
q ( ) ( 9 ) = q1 − 116.6 lb / in
q(
3)
z
y
5.21
3.5
q1 − 116.6
q1
s
Qy( 4) = ( −3.5 )( 0.1)( s ) = −0.35s
Qz( 4) = − ( 5.21 − s / 2 )( 0.1)( s ) = −0.521s + 0.05s 2
( s ) = q1 − 116.6 − 29.16 [ −0.35s ] − 6.073 ⎡⎣−0.521s + 0.05s 2 ⎤⎦
4
q ( ) ( s ) = q1 − 116.6 + 13.37 s − 0.3036 s 2 lb / in
q ( 4) (12 ) = q1 − 116.6 + 160.44 − 43.72 = q1 (approximately)
q(
4)
total force in each wall
F
(1)
4
= ∫ ( q1 + 6.082 s − 1.458s 2 ) ds = 4q1 + 17.55 lb
0
F
( 2)
4
= ∫ ( q1 + 1 − 5.58s − 0.2802s 2 ) ds = 13q1 − 663.7 lb
0
F
( 3)
4
= ∫ ( q1 − 118.9 − 12.87 s + 1.458s 2 ) ds = 9q1 − 1237 lb
0
F
( 4)
4
= ∫ ( q1 − 116.6 + 13.37 s − 0.3036 s 2 ) ds = 12q1 + 611.4 lb
0
A
9q1 − 1237
13q1 − 663.7
9
12
12q1 − 611.4
4q1 + 17.55
A
A
13q1 − 663.7
9q1 − 1237
9
1000 lb
=
12
4q1 + 17.55
12q1 − 611.4
to find q1 we must equate moment about any point generated by the
shear flows to the moment due to the applied load
∑M
A
=0
( 2q1 − 116.4 )( 9 ) + ( 4q1 + 17.55)(12 ) = 0
q1 = 33.92 lb / in
then the shear flows are explicitly
q (1) ( s ) = 33.9 + 6.08s − 1.46 s 2 lb / in
q ( 2) ( s ) = 34.9 − 5.58s − 0.208s 2 lb / in
q ( 3) ( s ) = −85.0 − 12.9s + 1.46 s 2 lb / in
q(
4)
-85
( s ) = −82.7 + 13.4s − 0.304s 2
lb / in
-13.2 (midpoint)
-85
34.9
-113
(midpoint)
34.9
40.2 (midpoint)
33.9 33.9
-82.7
-82.7
-13.5 (midpoint)
Now, find the location of the shear center for the vertical load
We need to set
φ′ =
1
q
ds = 0
∫
2GΩ t
and find the q1 that meets this requirement
Since t is a constant here we have equivalently
∫ qds = 0
which is just the sum of the forces we already computed
( 4q1 + 17.55) + (13q1 − 663.7 ) + ( 9q1 − 1237 ) + ( 2q1 − 611.4 ) = 0
which gives
q1 = 65.65 lb / in
q1 = 65.65 lb / in
A
A
1000 lb
13q1 − 663.7
9q1 − 1237
e
9
=
12
4q1 + 17.55
12q1 − 611.4
∑M
A
= 1000e
( 4q1 + 17.55)(12 ) + (12q1 − 611.4 )( 9 ) = 1000e
e = 4.95 in
If we want to find the vertical location of the shear center we would have to
solve this problem for a horizontal applied shear force
1000 lb
1000 lb
T
+
=
4.95
bending
torsion
shear flow due to torsion (constant around the whole cross section)
q=
T
4950
=
= 31.7 lb / in
2Ω 2 ⎡⎣( 4 )(12 ) + ( 5 )(12 ) / 2 ⎤⎦
Thus, the q1 in our original problem is due partially to the bending part of the
solution, partially due to torsion
1000 lb
1000 lb
=
T
+
4.95
q1 = 33.9
q1 = 65.5
q1 = −31.7