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Chapter 2
Torsion Stresses in Thin-Walled
Multi-Cell Box-Girders
2.1 Torsion of Uniform Thin-Walled Two-Cell Box-Girders
The thin-walled box section with uniform thickness t as shown in Fig. 2.1, is
subjected to a torsion moment T.
The shear flow and angle of twist for the thin-walled two cell structure shown in
Fig.2.1could be determined as follows.
The flexural warping coefficients are given by
d111=GIs1
ds=t
1=Gt ACCDDBBA
d22 1=Gt DCCFFHHD
d12 1=GCD=t
Since the angle of twist is the same for the two cells, then the basic equations
are given by
d11q1 d12q22A1 h 0 2:1
d12q1 d22q22A2 h 0 2:2
From equations (2.2.1) and (2.2.2) we get
d11 d22q1 d12 d22q22d22A1 h
d212q1 d12 d22q2 2d12A2 h
Hence
q1d11 d22 d212 2d22A1 h2d12A2 h
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The solution of equations (2.1) and (2.2) gives
q1 h 2d22A12d12A2 d11 d22 d
212
D1h
q2 h 2A1 D1d11 =d12
D2 h
where
D1 2d22 A12d12A2 d11 d22d
212
D2 2A1 D1d11 =d12
The equilibrium condition gives
T 2A1q12A2q2 D3 h
where
D3 2A1q12A2q2
Hence
h 1=D3T
q1 D1=D3T
q2 D2=D3T
q12q1 q2 D1 D2 =D3T:
Example 2.1 Determine the torsion shear stress and angle of twist for the two
uniform thickness thin-walled box-girder shown in Fig. 2.2.
Solution The shear flow and angle of twist for the thin-walled two cell structure
shown in Fig.2.2could be determined as follows.
Fig. 2.1 Shear flow due to
torsion of a thin-walled box
girder with two unequal cells
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The torsional moment is given by
T 2 q1A1q2A2 2:3
The angle of twist for cells 1 and 2 are given by
h1 1
2GA1 q1
I1
ds=tq2
Z12
ds=t
0@
1A 2:4
h2 1
2GA2q2
I2
ds=t
I q1
Z21
ds=t
0
@
1
A 2:5
Since the angle of twist is the same for the two cells, then we have
h1 h2 h3
Reformulating equations (2.4) and (2.5), we get
1
G q1
I1
ds=tq2
Z12
ds=t
0@
1A 2A1h 2:6
1
Gq2
I2
ds=tq1
I21
ds=t
0@
1A 2A2h 2:7
Let
d warping flexibility
d11 1
GI1
ds=t
d22 1
G
I2
ds=t
Fig. 2.2 A uniform thin-
walled box-girder with two
cells
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d12 d21 1
G
Z12
ds=t
Substituting in equations (2.2.4) and (2.2.5), we get
d11q1 d12q22A1h 0 2:8
d12q1 d22q22A2h 0 2:9
Solving equations (2.2.3), (2.2.8) and (2.2.9), q1 and q2 could be determined.
Example2.2 Determine the torsion shear stresses and the rate of twist for the thin-
walled 2-cell box-girder shown in Fig.2.3. The girder is subjected to a constant
torque T.
Solution Area of cell (1) is given by
A12a2
Area of cell (2) is given by
A2a2
Let
d warping flexibility
d111=G
I ds=t 6a=Gt
d224a=Gt
d12 a=Gt
The basic equations are
d11q1 d12q22A1h 0 2:10
d12q1 d22q22A2h 0 2:11
Fig. 2.3 A thin-walled box-
girder with two unequal cells
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The equilibrium equation gives
T 2 q1A1q2A2 2a2 2q1q2 2:12
From equations (2.10) and (2.11), we get
d11q1A2 d12q2A2 d12q1A1 d22q2A10
q1 d11A2 d12A1 q2 d12A2 d22A1 0
Hence
q1q2d22A1 d12A2d11A2 d12A1
q22d22 d12d112d12
From equation (2.12), we get
T 2a2q2 22d22 d12d11 2d12
1
T2a2q24d224d12 d11
d11 2d12
From which q2 is given by
q2
T
2a2
d112d12
4d224d12 d11
q1 T
2a22d22 d12
4d224d12 d11
Substituting in equation (2.10), we get
h 1
2A1 d11q1 d12q2
1
4a2 T
2a2 2d11d22 d11d12 d12d112d
2
124d22 4d12 d11
T
4a4
d11d22 d212
4d224d12 d11
Substituting for d12, d11 and d22, we get
q1 T
2a2G8a=ta=t = 16a=t4a=t6a=t
T
2a2G
9
26
q2 T2a2G 6a=t2a=t = 26a=t T
2a2G 8
26
h T
4a4G24a2
t2 a2
t2
26a=t
T
4a4G23=26a=t
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But
h T=GJ
Hence
J T=Gh 104=23a3 t:
2.2 The General Case of a Uniform Two-Cell Box Girder
This is an indeterminate structural problem and its solution is based on the
assumption that the rate of twist for each cell is the same as for the whole section,see Fig.2.4.
i.e.,
h1h20 and h du=dz
h T=GJ
The torque T is given by
T 2q1A12q2A2
h1 1
2GA1
I1
q=tds
and
h2 1
2GA2
I2
q=tds
Fig. 2.4 Idealized section and torsion shear flow of a thin-walled two cell structure
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i.e.,
1
G q1 I
1
q=tdsq2 ds=t 12
24
35 2A1h1 2:13
1
Gq82
I2
ds=tq1 ds=t 21
24
35 2A2h2 2:14
Equations (2.13) and (2.14) are simplified to
d11q1 d12q22A1h
d21q1 d22q22A2h
d11 d12
d21 d22
q1
q2
A1A2
2h
or
d qf g 2h Af g
The shear flow in each cell is given by
qf g d 1 2h Af g
i.e.,
qf g 2h d 1 Af g
d11 1
G
I1
ds=t
d22 1
GI2
ds=t
d12 d21
1
Gds=t 12
2:15
The torque is given by
T 2q1A12q2A2 2:16
Solving equations (2.15) and (2.16), we get q1, q2 and h
The torque T is given by
T GJh
2.2 The General Case of a Uniform Two-Cell Box Girder 27
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Hence J is given by
J T=Gh:
Example 2.3 Determine the torsion shear stresses and angle of twist for the thin-walled box section having uniform thickness t as shown in Fig. 2.5. The section is
subjected to a torsion moment T.
Solution Condition for Compatibility (Consistent Deformation). The warping
flexibilities are given by
d111=Gt ABBCCDDA
d221=Gt CHHFFNNC d12 1=GCD=td12 1=GCD=t
The basic equations of consistent deformation are given by
d11 q1 d12q2 2A1h 2:17
d12 q1 d22q2 2A2h 2:18
Solving equations (2.17) and (2.18) we get
q1 2d22A12d12 A2
d11 d22 d212
h D1h
q2 2A1
d12h D1
d11
d12h D2h
Equation for equilibrium condition is given by
T 2A1q12A2q2 h D3h 2:19
The solution of equations (2.17), (2.18) and (2.19) gives
h T=D3
q1T D1=D3
Fig. 2.5 Shear flow due to
torsion of a box-girder with
two cells
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2.3.3 Rate of Twist h
The rate of twist is given by
h T=GJ
where
J 4A2I
ds=t
I ds=t 2B=tB2D=tS
2.4 Torsion of Three-Cell Box-Girder
Following the same principle that the angle of twist is the same for the three cell
box-girder shown in Fig. 2.7.
Then
h1h2h3h
The equations of consistent deformation are given by
d11q1 d12q22A1h10 2:20
d12q1 d22q22A2h20 2:21
d32q1 d33q22A3h30 2:22
The torque is given by
T 2A1q12A2q22A3q3h 2:23
Solving equations (2.20)(2.23), we get q1, q2, q3, q4 and h.
Fig. 2.7 Torsion of a three-
cell box-girder
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Hence
d11 d12 0d21 d22 d23
0 d
32 d
33" #
q1q2
q3( ) 2h
A1A2
A3( ) 2:24
d qf g 2h Af g
Hence the shear flow in each cell is given by
qf g d1
Af g2h
and
T 2h XAiqiwhere
d11
I1
ds=t
d22
I2
ds=t
d33 I3
ds=t
d12 d21 ds=t 12d23d32 ds=t 23:
Example2.4 Determine the shear flow, shear stress and rate of twist for the three-
cell box girder shown in Fig.2.8.
Fig. 2.8 Shear flow due to
torsion of a 3-cell thin-walledbox girder
2.4 Torsion of Three-Cell Box-Girder 31
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Solution Following the same principle that the angle of twist is the same for all
cells, see Fig. 2.8, we get
d11q1 d12q22A1h
d21q1d22q2d23q32A2h2d32q2d33q32Ah3
2:25
but
h1h2h3h
Then
d11 d12 0d21 d22 d23
0 d32 d33" #
q1q2
q3( ) 2h
A1A2
A3( )
d qf g 2h Af g 2:26
qf g d1
Af g2h 2:27
The torque is given by
T 2q1A12q2A2 2q3A3 2:28
Solving equations (2.26) and (2.28) we get
q12hh1A1
q22hh1A2
q32hh1A3
Substituting in equation (2.25), we get
h 1=2A1 d11q1d12q2
where
d111=G
I1
ds=t
d221=G
I2
ds=t
d331=G I3
ds=t
d12 d21 1=G ds=t 12d23 d32 1=G ds=t 23:
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2.5 Torsion of Uniform Thin-Walled Multi-Cell Box-Girder
The multi-cell thin-walled structure when subjected to pure torsion is a statically
indeterminate problem; see Fig. 2.9.The torque T is given by
TXni1
2Aiqi GJhJ
where T = applied uniform torque; Ai = enclosed area of the ith cell; J = torsion
constant
J 4Xni1
Aid1
Ai
The angle of twist per unit length
h du=dz
hi hj hijhjn
where
hi1=2GAi qiI
ds=t
The angle of twist for cell i is given by
hi1=2GAi qi
I ds=tqi1
Z ds=tqi1
Z ds=t
2:29
Equation (2.29) represents a series of simultaneous equations which gives
q1; q2; q3;. . .; qn.The set of equations of consistent deformation is given by
d11q1 d12q22A1h 0 2:30
Fig. 2.9 Torsion of a multi-cell thin-walled box-girder
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d12q1 d22q22A2h 0 2:31
d32q1 d33q22A3h 0 2:32
This set of equations could be put in the following formd qf g 2h Af g
Hence, for a multi cell box girder, the shear flow in each cell is given by
qf g d 12h Af g i 1; 2;. . .; n
where
d
d11 d12 0 0 0 0d21 d22 d23 0 0 0
0 d32
d33
d34
0 00 0 d43 d44 d45 00 0 0 d54 d55 d560 0 0 0 d65 d66
0BBBB@ 1CCCCAThe torsion shear stresses are given by
s1 q1=t1; s2q2=t2; s3q3=t3:
2.6 Combined Open and Closed Thin-Walled Sections
For the combined open and closed section, see Fig. 3.1, the angle of twist is the
same for the whole section whether it is an open or closed section.
2.6.1 Combined Open Section with One Closed Cell
The total torque T for the thin-walled section shown in Fig. 2.10is given by
T =X2I1
Ti G Jh
Fig. 2.10 Combined open
and closed one-cell thin-
walled section
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Hence
h T=GJ
whereT1 GJ1 h
J1 = torsion constant of the open section; J2 = Torsion constant of the closed
section.
For the open part of the structure, the shear flow q1 is given by
q1T1t21=J1
T1GJ1h
For the closed section of the structure, the shear flow q2 is given by
q1T2=2A2
T22A2q2 GJ2h:
2.6.2 Combined Open Section with Two Closed Cells
The applied torque T for the thin-walled structure shown in Fig. 2.11is given by
TXnj1
2qjAjXmi1
GJih
In the above particular example
T 2q1A12q2A2GJ3h
where J3 = the torsion constant of the open section part of the structure and is
given by J3
T 1=3Xni1
bit3i
Fig. 2.11 Combined thin-
walled open and closed
two-cell structure
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The shear flow in the two cells is given by
d11q1 d12q22A1h
d21
q1 d
22q
22A
2h
d111=G
I1
ds=t
d221=G
I2
ds=t
d12 d21 1=G ds=t 12
In the general case, for combined open and closed sections, the shear flow in
each cell is given by
qi d12hhi i 1; 2;. . .; n
And in each open member the shear flow is given by
qi T=Jt2i
And the angle of twist is given by
h T=GJ
where T = the torque and is given by
T 4Xni1
A0id1Ai
1
3
Xmj1
bjt3j:
Example 2.5 Determine the shear flow distribution and rate of twist for the ide-
alized ship section shown in Fig. 2.12. The ship section is subjected to a torque T.
Solution The torque T is distributed among the thin-walled structural members of
the ship section as follows
TX4i1
Ti
where
T12A1q1; T22A2q2; T3 2A3q3;
T4GJ4 h q4J4
t24; J41=3St
34
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because of symmetry of the ship section, we have
T1 T2; T4T5
Hence
T1 T22A1q1G J1h
Thus
T 2GJ1h2GJ4hGJ3hX3j1
GJh
where
G E
21t
E
26
J 2J12J4J3
where
J14A21
I ds
t4ab)2
2b=t2a=t
J3A(Bh)2
2B=t32h=t3
J41=3S4t341=3 Dah)t
34
Hence
h T.X
GJ
Substituting, we get the torque carried by each structural element.
Fig. 2.12 Idealized ship
section
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Hence
T1 GJ1h
T3 GJ3h
T4 GJ4h
Substituting, we get the shear flow in each structural element as follows
q1T1=2A1; q3T3=2A3; and q4T4=2A4:
Example 2.6 Determine the shear flow and rate of twist for the ship section of
bulk carrier shown in Fig. 2.13.
Solution The torque T is given by
T 2 2A1q12q2A2GJ6h2A3q32A4q4 2A5q5
The torsion constant J is given by
J 2 X5
i1
4A0id1Ai1=3k6t
36
( )
The rate of twist h is given by
h T=GJ
The set of equations of consistent deformation for cells (1) and (2) is given
by
d11q1 d12q2 2A1h 2T=JA1
d21q1 d22q2 2A2h 2T=JA2
Fig. 2.13 An idealized sec-
tion of a bulk carrier
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This set of equations can be put in the matrix form as follows
d11 d12
d21 d22
q1
q2
2h
A1A2
i.e.,
d qf g 2T=J Af g
Hence, the torsion shear flow in cells (1) and (2) are given by
q1 d1 2T=JA1
q2 d1 2T=JA2
Similarly, the torsion shear flow in cells (3), (4) and (5) are given by
q3 d1 2T=JA3
q4 d1 2T=JA4
q5 d1 2T=JA5
where
dd
33 d
34 0d43 d44 d45
0 d54 d55
24 35
dii 1=GPmJ1
kj
tj
; i 1; 2;. . .; n N of cells
drj 1=G krj
trj
i
where r, and j are cells, having a common boundary; i = cell No. i.
2.6 Combined Open and Closed Thin-Walled Sections 39
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