torsion stresses in thin-walled tubes

8/11/2019 Torsion Stresses in Thin-Walled Tubes

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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

M. Shama, Torsion and Shear Stresses in Ships, 21


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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

22 2 Torsion Stresses in Thin-Walled Multi-Cell Box-Girders


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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

2.1 Torsion of Uniform Thin-Walled Two-Cell Box-Girders 23


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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

2.1 Torsion of Uniform Thin-Walled Two-Cell Box-Girders 25


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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


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


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.


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


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

2.5 Torsion of Uniform Thin-Walled Multi-Cell Box-Girder 33


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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

2.6 Combined Open and Closed Thin-Walled Sections 35


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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.



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torsion stresses in thin-walled tubes

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