Aircraft Structures Notes 1

5/26/2009

1

ME4212

Aircraft Structures(formerly Mechanics of Thin-Walled

Structures)

T. E. Tay, Department of Mechanical Engineering, National University of Singapore

Why study thin-walled structures?


5/26/2009

2





5/26/2009

3

Linear Elastic Fracture

Elasto-Plastic Fracture

Damage Tolerance

Fatigue

Delamination

Fracture MechanicsAircraft Structural Analysis

Mechanics of Composite Materials

Stress-Strain Relations of Fiber-Reinforced Composites

Laminate Design and Architecture

Beams

Plates

Shells

Stringers and Panels

Multi-cell Torque Boxes

Mechanics of Thin-Walled Structures


Laminate Design and Architecture

Hygrothermal Effects

Durability and Failure of Composites

Tapered Structures

Buckling and Instability

Vibration

Torsion

Bending

Course Overview:

T.E. TayProfessorRoom EA-7-17

6516 2887Idealized Beams

Multi-cell Sections & Tapered Beams

Circular & Rectangular Plates

Instability of Columns & Plates

[email protected]

S L Toh


Instability of Columns & Plates

Energy Methods in Instability

S.L. TohAssociate Professor

5/26/2009

4

Torsion Non-Circular Shafts

Thin-Walled Open Sections

Thin-Walled Closed Sections

Warping of Unrestrained Sections

B diBending

Unsymmetric Bars

Thin-Walled Open Sections

Thin-Walled Closed Sections

The Shear Center

Idealized Beams Bending & Torsion


dea ed ea s e d g & o s o

Multi-cell Sections Bending & Torsion

Tapered Beams & Beams with Varying Moments of Area

Mechanics of Materials, A.C. Ugural, McGraw-Hill, 1991Ch 6 8 6 9: Torsion Torsion of Thin Walled Shafts

Books:

Ch 6.8, 6.9: Torsion, Torsion of Thin-Walled Shafts

Ch 7: Bending of Beams

Ch 8.6, 8.8: Unsymmetric Bending, Shear Center

Mechanics of Materials, Roy R. Craig Jr., Wiley & Sons, 1996Ch 4 4 7 4 8: Torsion Torsion of Noncircular Sections


Ch 4, 4.7, 4.8: Torsion, Torsion of Noncircular Sections

Ch 6, 6.6, 6.8-6.10, 6.12: Bending, Unsymmetric Bending, Shear Flow, Shear Center

5/26/2009

5

Mechanics of Elastic Structures, J.T. Oden and E.A. Ripperger, McGraw-Hill, 1981

Books (advanced):

Aircraft Structures for Engineering Students, T.H.G. Megson, Edward Arnold, 1990 (2nd Edition)

Analysis of Aircraft Structures, B.K. Donaldson, McGraw-Hill, 1993


Aircraft Structures & Systems, R. Wilkinson, Addison-Wesley Longman, 1996

1


Torsion


Torsion of Circular Sections

Bars of Circular Sections subjected to torsion

Radial lines on cross-section remains straight. But square elements are distorted.

2


r x

The shear strain is defined:

xr

=

And the shear stress is

GrdxdGr

xGr

G

x

=

=

=

=

0lim

where G is the shear modulus.

dxd = is the rate of twist, and is a constant.


rA

The twisting moment (i.e. torque) is

( )

GJTdxdGJ

dArdxdG

dArT

A

A

=

=

=

=

2

(1)

where A is the area of the cross-section, and

JTrGr

=

=

And the shear stress

J is the polar second moment of area.

(2)

dF

3


Torsion of Non-Circular Sections

Bars of Non-Circular Sections subjected to torsion




4




Note: There is no change in shape. Hence

0

0

=

=

yz

yz


Consider a cross-section of arbitrary shape:

A is an arbitrary point.

A

It is displaced to A under torque.

A

There is a point O, that has no displacement when the torque is applied, called the centre of twist.

OLets place the origin of our coordinate system at O.

y

z

5


A

A

O y

z

yw =

zv =

Horizontal displacement of A.

Vertical displacement of A.

xywxzv

==

In general, we write the displacements:

where is the rate of twist.

(3a)

(3b)

),( zyu =

We further assume that the axial displacements are given in this form:

),( zywhere is the warping function, yet to be determined.

(3c)


Axial displacements depend on the warping function ),( zy

y

z

x

Our objective is to determine u, v and w, the complete displacement field.

6


Start with the strain-displacement (compatibility) relations:

xw

zu

yw

zv

xv

yu

zw

yv

xu

xzyzxy

zyx

+

=

+

=

+

=

=

=

=

,,

,,,

Substituting Eqns (3) into above,

+

=

=

====

yz

zy

xz

xy

yzzyx

0 (4a)

(4c)

(4b)

implies cross-section does not distort in its own plane (i.e. all points are simply rotated as in rigid-body rotation).

0=yz


The relevant force equilibrium equation

0,,

0

===

=

+

+

xxzxzxyxy

xzxyx

GG

zyx

This is the Laplace Equation. Clearly, the warping function satisfies both compatibility and equilibrium.

In order to obtain a solution, we need to specify boundary conditions.

with

and Eqns (4) yields

0

0

2

2

2

2

2

=

=

+

zy

(5)

or

7


y

z

x

Consider a small element on the boundary, which is usually traction free.

dz

xz

xy

dy

dx

Force equilibrium in the axial direction provides the equation of the boundary condition.

dydz

dydxdzdx

xzxy

xzxy

=

= 0(6)

In general, it may be necessary to solve the Laplace Equation numerically.


Stress Functions for Torsion

Prandtl assumed a stress function which is twice differentiable, and has the property:

y

z

xz

xy

=

=

(7a)

(7b)

Substituting (7) into (4b) and (4c), differentiating (4b) with respect to z, and (4c) with respect to y,

G zz z z y

G yy y y z

= = +

(8a)

(8b)

8

Ludwig PrandtlFather of Modern Fluid Mechanics

(1875-1953)


Subtracting (8b) from (8a) results in

The solution to this equation satisfies both compatibility and equilibrium.

In order to obtain a solution, we need to specify boundary conditions.

G

Gzy

2

2

2

2

2

2

2

=

=

+ (9)

or

0

0

=

=

d

dyy

dzz

or

For the boundary condition, substituting (7) into (6), we obtain

This means that at the boundary, is a constant. Since the value of can be arbitrary, we shall select = 0 for convenience.

(10)

9


dAxz

dAxyArea dA

z

O y

z

y

( ) =A

xyxz dAzyT

The torque about the x-axis due to the shear stresses is given by

(11)


( )( ) ( )( )dzdyyydydzzzdzdyy

ydydzz

z

dydzzz

dydzyy

dydzzz

yy

T

CCDDAABB

D

C

B

A

=

=

=

+

=

Introducing Eqn (7) , and noting that dA = dydz,

(12)

(13)

(Integration by parts)

10


B

D

O y

z

C

A

B

= dydzT 2

are values at boundary points, which we have chosen to set to zero. Hence the equation reduces to

A C D

(14)


If we retain the expression

GJT = (15)

Only for the very special caseof the circular cross-section is J equal to the second polar moment of area.

= dydzGJ 2

where now we define the symbol J as the torsional constant, then the formula for J is

The product GJ is known as the torsional stiffness.

(16)

11


The Membrane (Soap Bubble) Analogy

yz (into plane)

w

MTTensionPressure p

Good for visualizing .

Equilibrium equation for small deflection w of a flat membrane subjected to internal pressure pclosely resembles the torsion equation.

MTp

zw

yw

=

+

2

2

2

2

(17)


Comparing Eqns (9) and (17),

MTp

zw

yw

=

+

2

2

2

2

(17)Gzy22

2

2

2

=

+ (9)

= dydzT 2

Furthermore, according to Eqn (14), the volume under the membrane is proportional to the torsion T.

(14)

w is analogous to MTp

G2is analogous to

12


The slope of the membrane is proportional to the shear stresses.

yz (into plane)

w

yw

y

zw

z

xz

xy

=

=

yw

Local slope

zw

Similarly for


If the section is hollow,

yz (into plane)

0=

yw

yz

0=d at both inner and outer boundaries.

= constant

= 0

13


Torsion of Thin-Walled Open Sections

y

z

Profile of w or in x-y plane

0=y

over much of y

y

x

Consider a narrow rectangular section:

Profile of w or in x-z plane

0=z

only at the centerline of the section (at z = 0).

x

z


Gzy

222

2

2

=

+ (9)

0=y

Since , Eqn (9) reduces to

Gz

222

=

(18)

212 CzCzG ++=

Integrating twice with respect to z yields

(19)

Applying the boundary condition that = 0 at z = + t/2 and t/2, we obtain and .4/0 221 tGCC ==

=

4

22 tzG (20)

14


0

2

=

=

=

=

y

zGz

xz

xy

The stresses are therefore

(21b)

Eqn (21a) shows that xy varies linearly with z and is zero at the centerline of the section (at z =0).

(21a)


3

22

2

3

2/

2/

2/

2/

22

btJ

dzdytz

dydzG

J

b

b

t

t

=

=

=

From Eqn (16),

(22)The torsional constant for a thin rectangular section.

zJTzGxy

22 ==

The shear stress

(23)

JTt

MAXxy=

has maximum values at z = + t/2 and t/2, i.e.

(24)

15


y

z

Linear distribution of shear stress across thickness:xy


Now, consider the warping displacements:

= zy

Gxy

From Eqns (4),

(25a)

+

= yz

Gxz (25b)

From Eqn (21b), xz = 0, so Eqn (25b) becomes

yz

=

yz= (26)

Clearly, C1(y) = C2(z) = 0 and

Upon integration,

)(1 yCyz +=

From Eqn (21a), xy = - 2G, so Eqn (25a) becomes

zy

zy

z

=

=

2

Upon integration,

)(2 zCyz +=

16


y

z

x

A hyperbolic-paraboloid surface

yzu = (27)

The warping displacement is therefore


General Open Sections

yz

Linear distribution of shear stress across thickness

xy

In general,

3

3BtJ =

s = 0

s = B

17


b1

b2

b3

b4

b5

b6

t1

t2

t3t4

t5

t6

333333

366

355

344

333

322

311 tbtbtbtbtbtbJ +++++=

For a section consisting of n elements,

=

=n

iiitbJ

1

3

31 (28)

Maximum shear stress is

i

iMAXiiMAX J

tT=

where the proportion of torque T acting on element i is

TJJT ii

=

(29)

(30)


Consider a thin-walled closed section:

yz

yz (into plane)

yw

Local slope almost constant

Torsion of Thin-Walled Closed Sections

18


yz

Constant distribution of shear stress across thickness

xy

Closed Sections


y

z

x

x

t1

t2

1

2

xt 11

xt 22

Taking force equilibrium in the axial direction,

2211 tt = (31)

Convenient to define shear flow

tq = (32) constant21 == qq

19


Choose an arbitrary point on the wall, at s = 0.

s = 0qds

Force due to shear flow on a small element of wall is qds.

O

r

Moment about an arbitrary point O is therefore qrds, where r is the perpendicular distance.

== rdsqrqdsT

The total moment then equals the torque,

(33)


s = 0

O

qdsr

=

=

=1

01 2

1)(ss

s

rdss

As we move from s = 0 to s = s1, the area traced is known as the sectorial swept area, and is defined as

(34)

20


O

s = 0

=

=

=1

01 2

1)(ss

s

rdss

As we move from s = 0 to s = s1, the area traced is known as the sectorial swept area, and is defined as

= rds21

The total area enclosed by the centerline of the wall is

(35)qT = 2

and the torque, from Eqn (33) is

(36)


x =The section does not distort under torsion. Points on the wall experience a rotation about the center of twist.

x

is the tangential displacement component in the s direction.

xr =

OCenter of twist

(37)

21


y

z

x

xConsider a small element of the wall

s

y

z

x

x


s

Displacement in the sdirection is

22


y

z

x

x

su

Displacement in the xdirection is u


y

z

x

x

sTherefore, the shear strain in the wall is

rsu

xsu

xs

+

=

+

=

(38)

23


Since it follows that Gxsxs / =

rGs

u xs = (39)

Multiplying both sides by ds and integrating around the total periphery S of the tube,

= rdstds

Gqdu (40)

The integral on the left-hand-side is zero, since u at s = 0 and u at s = S are the same (i.e. same point), so that the difference is zero.

Substituting for from Eqn (35),

=

=

tds

Gqt

dsGq

2

20

(41)


Substituting for q from Eqn (36),

= tds

GT

24 (42)

Since T = GJ , the torsional constant J for a closed section is

=

tdsJ

24(43)

If the thickness t is constant,

StJ

24= (44)

24


=

+

=n

iiitb

tdsJ 1

32

314

Hybrid Sections

(45)


Example 1

The round tube shown has an average radius R and a wall thickness t. Compare the torsional strength and rigidity of this tube with that of a similar tube which is slit along its entire length. Assume R/t = 20.

R

t

R

t

Closed tube Open section

25


Torsional constant

tRtRt

dsJ Closed

332

2

22

4

4

==

=


323

3

3

Rt

btJ Open

=

=

For a given angle of twist, the closed section resists 1200 times the torsion of the open section.

1200

32

=

=

tR

JJ

Open

Closed

For a given torque, the open section twists through an angle 1200 times as great as the closed section.


Maximum shear stress

tRT

tT

Closed

22

2

=

=


223RtT

JTt

MAXOpen

=

=

For a given torque, the shear stress in the open section is 60 times as high as in the closed tube.

3

60

Open

Closed

Rt

=

=

If the allowable shear stresses are equal, the closed tube is 60 times as strong as the open section.

26


Example 2

A torque of T = 6000 Nmm is applied to the section shown. Determine the rate of twist and maximum shear stress. Assume shear modulus G = 80 GPa.

Dimensions in mm

1

2

1

2 215

20

19


5.582

192

201

19120

=+++= tds

The torsional constant is

( ) ( )( ) 4332

32

mm10913.921531

5.583804

314

=+=

+

=

bt

tdsJ

Total area enclosed: 2mm3801920 ==

Since T = GJ , the rate of twist is

( )( )rad/mm1057.7

108010913.96000

6

33

=

=

27


The contributions to the torsional constant are

( )

( )( ) 43

432

mm4021531

mm10873.95.58

3804

==

==

Fin

PartClosed

J

J } Compare !The proportions of torque carried are

Nmm2.2460009913

40

Nmm59766000913.9873.9

==

==

Fin

PartClosed

T

T

The maximum stresses are

( )( )( )( ) MPa21.1

4022.24

MPa86.713802

59762

===

==

=

Fin

FinMAXFin

MIN

PartClosedMAXPartClosed

JtTt

T


If the section is slit longitudinally at A as shown, what are the rate of twist and maximum shear stress?

Dimensions in mm

1

2

1

2 215

20

19

A

28


( )( )

( )( ) rad/mm1078.415710806000

MPa4.76157

26000

43

=

==

===

GJT

JTt MAX

MAX

The torsional constant is

( )( ) ( )( ) ( )( ) ( )( )[ ] 43333 mm15723521912011931

=+++=J

Note the increase in stress and rate of twist.

and


Multiplying both sides by ds and integrating from s = 0 to some other point on the wall,

where u0 is the displacement of the point s = 0 in the x-direction.

Warping of Unrestrained Thin-Walled Sections

Recall from Eqn (39) that

= rdsdsGuu xs 1

0

rGs

u xs =

(46)

29


s = 0

O

r

dsrs = 21)(Recall the sectorial swept area (Eqn (34)), defined as

The equation, valid for both open and closed sections, can be rewritten as

0

)(21

uuu

sdsG

u xs

=

=

where r is measured from an arbitrary point O (called the pole).

)(210 sdsGuu xs = (47)


O

s = 00=u

Here, is a function of s, and is zero when the integration is completed around the closed tube.

u

For closed sections, since , we havetqxs /=

2= tds

Gqu (48)

30


For open sections, the contribution of the shearing strain xs to the warping displacement u is negligible because the walls are very thin, so that

0 dsxs

2=u

Hence the equation reduces to

(49)

(50)


Example 3

The closed box section shown is subjected to a torque T. Determine the warping displacements.

th

b

h

th

tb

tb

T

12

43

+=

+=

+++=

=

bh

bh

hbhb

tb

th

GhbT

tb

th

GhbT

tds

tds

tds

tds

GhbT

tds

GT

22

22

2

1

1

4

4

3

3

222

2

2

224

4

4

Rate of twist:

31


Select the pole at O the center of section, and starting point at I.

12

43

OI

bhTTq

22=

= and

hI h th

tds

2

1

=We find that

For closed sections, the relative warping displacement is

2= tds

Gqu

1

1

1 2 II ht

dsGqu =

4222 1

bhhbI =

=Also,

=

bh tb

th

bhGTu

81Therefore,

Interestingly, there will be no warping if

bh tb

th=


12

43

OI

12

222

8

43

222

uuth

tb

bhGT

bhtb

th

GhbT

tb

th

bhGTu

hb

bhbh

=

+=

+

+=

43

242 2

bh

bhbhI

=

+=Now,

1413 , uuuu ==Similarly,

bh

bI hI

tb

th

tds

tds

tds

+=

+=

2

2

1

12

and

32

1

2

4

3

I


Due to symmetry, we know that the axial displacement at I is zero. Since we have .11 uu =

,11 Iuuu =


Example 4

What would the warping displacements be if the section is slit longitudinally along its entire length at point I ?

12

43

OI ( )332

3

bh bthtGT

GJT

+==Rate of twist:

( )3332

bh bthtJ +=Torsional constant:

2=uFor an open section,

33


12

43

OI

( ) ( )33331 83

23

4 bhbh bthtGbhT

bthtGTbhu

+=

+=From I to 1,

( ) 1332 389 u

bthtGbhTu

bh

=+

=From I to 2,

==

4525 313bhuu IFrom I to 3,

==

4727 414bhuu IFrom I to 4,

==

4828 31'bhuu II From I to I,

1

2

4

3

I


Longitudinal slit

34


Example 5

Let us return to the closed section. What will happen if we select another point O as pole?

12

43

O I

22 1

bhI =From I to 1,

=

+

=

b

bhh

tb

bhGT

bhtb

th

GhbT

th

bhGTu

4

2222 221

hI th

tds

2

1

=and


12

43

O I

bhI =22From I to 2,

bhI tb

th

tds

+= 22

and

=

+

+=

h

bhbh

th

bhGT

bhtb

th

GhbT

tb

th

bhGTu

4

222 222

1423 , uuuu ==

Similarly, we can show that

Try it yourself !

35


10,30 ==bh tb

thFor illustration purposes, let 0.1

2=

bhGT

and

5.2,5.7,5.7,5.2 4321 ==== uuuuThen

1

2

4

3

I

1

2

4

3

I

From Example 3: Present Example 5:


Hence changing the pole does not alter the problem (i.e. the stresses are also unaffected), but merely changes the reference plane from which the axial displacements are measured.

In this example, the new reference plane is rotated about the OI axis.

1


Bending


Bending of Bars

Bars of Non-Circular Sections subjected to pure bending

Bending Moment M

2


Pure bending occurs only when the force F acts through the Shear Center E.

Built-in end

When the force F acts through any other point, there is combined bending and twisting.


t

Let the loads Vz and Vy act through the shear centre E. Hence pure bending.

Vy

Vz

Shear Centre

E

y

z

xCentroid O

x-y-z are the centroidal axes

3


t

y

z

xCentroid O

x-y-z are the centroidal axes

The shear loads give rise to the applied moments (positive right hand screw rule)

My

Mz


y (into page)

z

x

Contribution of Vzto bending moment

dx

Vz

E

Hence

xM

V

dxVdM

yz

zy

=

=

(51a)

+ dMy

4


z (out of page)

y

x

dx

Vy

E

Hence

xMV

dxVdM

zy

yz

=

=+ 0

(51b)

Contribution of Vyto bending moment

+ dMz


t

y

z

x

dx

ds

Let x denote the normal stress due to bending. Since there is no nett axial load,

0= dAA

x (52a)

xArea dA = t ds

5


t

y

z

x

Consider a small element of the wall (in positive z)

Consider applied moment about the y-axis

dx

ds My

x

Area dA = t ds

z

The moment about the y-axis contributes to the normal stress xdue to bending.


y (into page)

z

x

+ dMy

dx

dAzMA

xy = (52b)

x

Area dA = t ds

z

Contribution of axial force dF due to x to the bending moment

(Tensile on positive z)

6


t

y

z

x

Consider a small element of the wall (in positive y)

Consider applied moment about the z-axis

dx ds

Mz

x

y

The moment about the z-axis also contributes to the normal stress xdue to bending.


z (out of page)

y

x

+ dMz

dx

x

y

Area dA = t dsdAyM

Axz = (52c)

Contribution of axial force dF due to x to the bending moment

(Compressive on positive y)

7


Plane sections remain plane assumption implies that the axial strain x (and axial stress x ) is a linear function of y and z. The cross-section rotates about a neutral axis whose orientation is dependent upon the loading and geometry of the cross-section.

y

z

Cross-sectionx

Side view along the neutral axis


Let zCyCCx 321 ++=

and substituting this into Eqns (52) and solving for the constants C1, C2, C3, we have C1 = 0 and

zIII

IMIMy

IIIIMIM

yzzzyy

yzzzzy

yzzzyy

yzyyyzx

++

+= 22 (53)

8


Eqn (53) can be rewritten in the form

+

=

+

=

+=

zzyy

yz

zz

yzzy

y

zzyy

yz

yy

yzyz

z

yy

y

zz

zx

III

IIM

MM

III

IIM

MM

IzM

IyM

2

2

1

1


The neutral axis can be located by setting x = 0 in Eqn (53), and find, for angle ,

yz

yzzzzy

yzyyyz

IMIMIMIM

yz

++

==tan (54)

9


y

z

x

Neutral axis Compression

Tension

xGeneral distribution of normal bending stress


If the section is symmetric about either the y- and z-axes, Iyz = 0 and Eqn (53) reduces to

yy

y

zz

zx I

zMI

yM+= (55)

10

A word about sign convention for moments:

Applied Moment MyPositive into page

yy

y

zz

zx I

zMI

yM+=

y (into page)

z

x

z (out of page)

y

xPositive out of page

Applied Moment Mz

This causes tensile (+) stresses for positive z

This causes compressive (-) stresses for positive y


Note that Equation (53) for normal bending stresses is valid for all general sections(whether solid or thin-walled, open or closed sections).

11


The simply-supported beam has an unsymmetrical cross-section. Locate the neutral axis and determine the largest tensile and compressive stresses in terms of load Q and where they occur. (Assume Q acts through the shear center.)

Example 6

1000 1500

Q

y

z

20

80

20 60

Q

Cross-section


The maximum bending moment occurs at the point of application of load Q and is given by

My = - (0.6Q)(1000) = -600 Q Nmm.

Mz = 0.

1000 1500

Q

0.6 Q 0.4 Q

1000

0.6 Q

MyFree-body diagram:

Positive into page

12


First, locate the centroid C, as the equations are related to centroidal axes.

20

80

20 60

QThe total cross-sectional area:

= (20)(100) + (60)(20) = 3200 mm2

Cy

z

AD

B

Define the points A, B and D.

mm510020110

20

60

0

=

+==

dyydyyA

ydAA

y

mm151 == zdAAz

Therefore,

measured from point B,and

measured from point D.

15

5


20

80

20 60

Q

Cy

z

15

5

Section properties (second moments of area):

46

23

23

mm10907.2

)25)(1200(12

)20)(60(

)15)(2000(12

)100)(20(

=

++

+=yyI

46 mm10627.1 =zzI

46 mm102.1

)25)(25)(1200()15)(15)(2000(

=

+=yzI

13


Cy

z

Neutral axis:

=

==

++

=

4.36627.1

2.1

tan

zz

yz

yzzzzy

yyzyzy

II

IMIMIMIM

since Mz = 0.Compression

Tension

36.4

A (-25,35)

B (-5,-65)

Points A and B are furthest from the neutral axis, and therefore experience the greatest stresses.

From Eqn (53), we find

( )( ) MPa0182.0

MPa0158.0Q

Q

Bx

Ax

=

=


There is also a shear stress distribution due to bending.

We will now discuss shear stresses due to bending in thin-walled members only.

14


y

z

x

Shear stress distribution due to bending

qdx

dsds

sqq

+

x

dxx

xx

+

xt

sq

dxdssqqqdxtdsdx

xtds

x

xxx

=

=

++

++

0

Force equilibrium in x-direction:

(56)


Recall that Eqn (53) holds for normal bending stresses. Differentiating with respect to x,

where the relations in Eqns (51) have been used.

zIIIIVIV

yIIIIVIV

x yzzzyyyzyzzz

yzzzyy

yzzyyyx

+

+

=

22

(57)

( ) ( )( )tIII

QIVIVQIVIVtq

yzzzyy

yyzyzzzzyzzyyyxs 2

+== (58)

Substituting Eqn (57) into Eqn (56) and integrating with respect to dA (= t ds), we have

zAdAzQ

yAdAyQ

Ay

Az

==

==

where (59a)

(59b)

( and are coordinates of the centroid)

zy

15



=

=

+=

zzyy

yz

zz

yzyz

z

zzyy

yz

yy

yzzy

y

yy

yz

zz

zy

IIIIIV

VV

III

IIV

VV

IQV

IQV

q

2

2

1

1


If the section is symmetric about either the y- and z-axes, Iyz = 0 and Eqn (58) reduces to

+=

yy

yz

zz

zy

IQV

IQV

q (60)

16


y

z

x

q is a function of s and may even change direction, depending upon loading

General distribution of bending shear flow q for open sections

s

Bending of open sections

q = 0

q = 0


Vy

Vz

E

y

z

xCentroid O

Bending of closed sections

Select an arbitrary point m on the wall for s = 0

smx

dxx

xx

+

ds

sqqm

+

qm

We already know that . This gives the same solution for the shear flow distribution as in the open section case. But qm is not zero because the section is closed. However, qm can be found by equating to zero the sum of moments about x-axis, since no torque is applied.

xt

sq x

=

17


Example 7

Determine the shear flow distribution in the z-section of uniform thickness t, due to a shear load Vz applied through the shear center of the section.

First thing to note is that due to anti-symmetry, the centroid and the shear center are the same point O. This is a special case.h/2

t

h/2

y

z

O

h/2

h/2

8422422

12232

312222

3

33

332

thhhhthhhtI

thhtI

ththhhtI

yz

zz

yy

=

+

=

=

=

=+

=

Vz


Hence

z

zzyy

yz

zzz

zzyy

yz

zz

yzz

y V

III

VVV

IIIIIV

V 28.21

86.01

22=

==

=

The shear flow due to bending:

=

=

+=

ssz

ss

z

s

yy

zs

zz

y

zdsydshV

ztdsth

ytdsth

V

ztdsIVytds

IV

q

003

03

03

00

85.63.10

3/28.2

12/86.0

18


h/2

t

h/2

y

z

O

h/2

h/2

For the bottom flange, define a coordinate s1 such that z = - h/2 and y = - h/2 + s1, where 2/0 1 hs

( )chsshV

dshdshshVq

z

ssz

+=

=

1213

01

011312

72.115.5

285.6

23.10

11

1 2

s1

Boundary condition: At s1 = 0, q12 = 0; hence c = 0

( )121312 72.115.5 hsshVq z =


y

z

O

1 2

s1

( )121312 72.115.5 hsshVq z =

For values of s1 < 0.334h, q12 is negative and therefore in opposite direction to s1.

At 2, s1 = h/2, q12 = 0.43 Vz/h.

Shear flow distribution is parabolic, with a change in sign at s1 = 0.334h.

0.334h

The shear flow distribution on the top flange is similar.

19


( ) 220

2323

2

85.643.3 qdsshhVq

sz +=

where q2 is the shear flow at point 2, i.e. q2 = 0.43 Vz/h.

y

z

O

1 2

s2

3Along the vertical web 2-3, z = s2 h/2, y = 0, and hs 20

This shear flow distribution is also parabolic, with a maximum at s2 = 0.5h.

( )2222323 43.343.343.0 shshhVq z +=


The Shear Center

There is a point where the resultant force must pass through in order that no twisting moments (torque) are developed. This point is called the shear center.

Why do we wish to find the shear center?

So that we will know, for a given load, how to proportion the stresses (and deformations) due to bending and torsion.

How do we find the shear center ?

We make use of its definition. A general procedure:

1. Find the centroid.

2. Find section properties.

3. Apply Vz or Vy, in turn.

4. The moment due to shear flows is due to application of the shear force through the shear center. The position of the shear center with respect to a convenient point can thus be obtained.

20


Example 8

Determine the position of the shear center of the section.

First thing to note is that since the section is symmetric, the positions of the centroid and the shear center must lie somewhere on the horizontal axis of symmetry.

Vz

Assume the shear center E exists and is located distance ye from vertical web, as shown. h/2

tw

tf

y

z

O

b

h/2

tfApply a shear load Vz through the shear center.

yeE

02122

212

2323

=

+=

+=

yz

fwf

wyy

I

bhththbthtI

Section properties:

because section is symmetric.

Izz is not relevant.


Vzh/2

tw

tf

y

z

O

b

h/2

tf

yeE

At the top flange, z = h/2, y = b s1, where s1 defined.

Hence 0, == yzz VVV

dstzIVq

s

yy

z =0

s1

.0,0

2

2

1

1

1

1

1

10

==

=

=

s

yy

fz

s

fyy

zs

qs

sI

htV

dshtIVq

At (free edge).

21


Vzh/2

tw

tf

y

z

O

b

h/2

tf

yeE

Therefore, the shear center is to the left of the section.

Take moments about the point A,

zyy

f

b

zyy

f

b

sez

VI

thb

sVIth

dsqhyV

4

2222

0

21

2

10

1

=

=

= s1

( )fwf

yy

fe

bthttb

Ithb

y

63

42

22

+=

=

A

b

s dsq0

11


Example 9

Determine the position of the shear center of the semi-circular section.

The section is symmetric, the positions of the centroid and the shear center must lie somewhere on the horizontal axis of symmetry.

Vz

Assume the shear center E exists and is located distance ye from vertical web, as shown.

y

z

OApply a shear load Vz through the shear center.

yeE

02

3

=

=

yz

yy

I

trI

Section properties:


Izz is not relevant.

t

r

22


Vz

y

z

Oye

E

t

r

Define coordinate s.

where.,cos rddsrz ==

sin

cos

2

0

2

0

yy

z

yy

z

s

yy

z

ItrV

drI

tV

dstzIVq

=

=

=

s


Vz

yeEy

z

O

t

r

s

Take moments about the point O,

[ ]

z

yy

z

S

ez

Vr

ItrV

dqr

dsqryV

4

cos 04

0

2

0

=

=

=

=

The shear center is to the leftof the section.

r

rye

27.1

4

=

=

23


Example 10

Determine the position of the shear center of the closed section and the distribution of shear flow.

Vz

Assume the shear center E exists and is located distance ye from vertical web, as shown.

t1

150

t1Apply a shear load Vz = 1 106 N through the shear center.

.mm6,mm3

0

mm1078.25

21

46

==

=

=

tt

I

I

yz

yy

Section properties:


Izz is not relevant.t1

250

t2

yeE


Define coordinates and points as follows.

Let the shear flow at A be qm, an unknown quantity to be determined.

t1 t1

t1

t2

y

z

O

s4

s2

s1

s3

A

B C

D1251 += zsFrom A to B,

1253 += zsFrom C to D,

From B to C, z = 125

From D to A, z = 125

24


( )

( )211116

6

11

5.0125116.0

1251078.25

)3)(10(

ssq

dssq

dsztIVqq

m

m

yy

zmAB

=

=

=

At B, qAB = qm

( )2

2

21

5.14

125116.0

sq

dsq

dsztIVqq

m

m

yy

zmBC

+=

=

=

At C, qBC = 2175 + qm

t1 t1

t1

t2

y

z

O

s4

s2

s1

s3

A

B C

Dqm


qm ( )

( ) 21751255.0233.0125

1078.25)6)(10(

323

336

6

32

+=

=

=

ssq

dssq

dsztIVqq

m

m

yy

zmCD

At D, qCD = qm + 2175

( )21755.14

125116.0

4

4

41

+=

=

=

sq

dsq

dsztIVqq

m

m

yy

zmDA

t1 t1

t1

t2

y

z

O

s4

s2

s1

s3

A

B C

D

25


Rate of twist = 0, since load acts through shear center.

0

02

1

41

32

21

11

=+++

=

=

dst

qdst

qdst

qdst

q

dstq

Gdxd

AB

DA

AB

CD

BC

BC

AB

AB

which simplifies to

1.8860598125675

==+

m

m

qq


4

323

2

211

5.14128929116.01289

5.141.886058.05.141.886

sqssq

sqssq

DA

CD

BC

AB

=+=

+=+=

Therefore,A

B C

D

qm = - 886.1

qm = - 886.1 q = 1288.9

q = 1288.9

26


Take moments about point O to obtain shear center.

( )

( )

mm7.939364625075525

29116.09.1288150

5.141.886250

150250

3

250

03

23

2

150

02

32

=+=

++

+=

=+

e

ez

ez

D

CCD

C

BBC

y

dsss

dssyV

yVdsqdsq

Vz

150

250

yeE

O

A

B C

D


Combined Bending and

Torsion

27


General Case: Superposition of Bending and Torsion

P

E

P

=Pure Bending through shear center

+Pure Torsion T = Pd

E

d

T


Pure Torsion of Open Sections

T

Linear distribution of shear stress across thickness

xy

28


Pure Torsion of Closed Sections

T

Shear flow q is constant anywhere in the section, but shear stress is NOT constant because of varying wall thickness

xs

Constant distribution of shear stress across the thickness

xs


Pure Bending of Open Sections

Bending stress exists in walls

x

E

P

Distribution of shear flow q in walls due to bending such that nett contribution of q must be due to P, applied force.

29


Pure Bending of Closed Sections

E

P

Bending stress exists in walls

x

Distribution of shear flow q in walls is similar to that of an open section, except for a constant q0superimposed.

1


Idealized Beams


2



3



y

z

xCentroid O tn

An

The section is made up of concentrated stringers of area An, joined by thin sheets of thickness tn.

zn

yn

The nth stringer is located at positions yn and zn away from the origin at the centroid.

Let the loads Vz and Vy act through the shear centre E. Hence pure bending.

VyVz

Shear Centre

E

4


Taking moments as before (see Eqns (51)), we have

xVMx

MV

zy

yz

=

=

(61a)

and

xVMx

MV

yz

zy

=

=

(61b)


y

z

xCentroid O

An

Let n denote the normal stress due to bending in the nthstringer. Since there is no nett axial load,

01

==

N

nnn A (62a)

n where N is the total number of stringers.

5


y

z

xCentroid O

An

zn

yn

=

=

=

=

N

nnnn

N

nyny

zA

MM

1

1

(62b)

Myn

n

The moment about the y-axis contributes to the normal stress ndue to bending.


y (into page)

z

x

+ Myn

dx

n

Contribution of axial force dF due to n to the bending moment

(Tensile on positive z)

nnnyn zAM =Area An

zn

6


y

z

xCentroid O

An

zn

yn

=

=

=

=

N

nnnn

N

nznz

yA

MM

1

1

(62c)

The moment about the z-axis contributes to the normal stress ndue to bending.

Mzn

n


z (out of page)

y

x

dMzn

dx

Contribution of axial force dF due to n to the bending moment

(Compressive on positive y)

nnnzn yAM =Area An

n

yn

7


Let nnn zCyCC 321 ++=

and substituting this into Eqns (62) and solving for the constants C1, C2, C3, we have C1 = 0 and

nyzzzyy

yzzzzyn

yzzzyy

yzyyyzn zIII

IMIMy

IIIIMIM

++

+= 22 (63)

Invoke again the assumption that plane sections remain plane, which implies that the axial strain (and axial stress ) is a linear function of y and z.

Note the similarity with Eqn (53).

The expression for the neutral axis remains the same as Eqn (54).


However, the section properties are now

=

=

=

=

=

=

N

nnnnyz

N

nnnzz

N

nnnyy

zyAI

yAI

zAI

1

1

2

1

2 (64a)

(64b)

(64c)

They are easier to evaluate than integrals!

8


Eqn (63) can, of course, be rewritten in terms of equivalent moments:

+

=

+

=

+=

zzyy

yz

zz

yzzy

y

zzyy

yz

yy

yzyz

z

yy

ny

zz

nzn

III

IIM

MM

III

IIM

MM

IzM

IyM

2

2

1

1


y

z

x

nGeneral distribution of normal bending stress

Neutral axis

Compression

Tension

9


y

z

x

Now, consider the shear flows in the sheets or webs.

qn

nnA

+ dxx

A nnn

nnn

n

nnnn

nnn

qAx

q

dxqdxqAdxx

A

+

=

=+

++

+

+

1

1 0

Force equilibrium in x-direction:

(65)

qn+1

Select a small length dx of a typical nth stringer with two adjacent nthand (n + 1)th sheets.

dx


Substituting Eqn (63) into Eqn (65) and using Eqns (61),

( ) ( )( ) nnyzzzyy

nyzyzzznyzzyyyn qAIII

zIVIVyIVIVq +

+=+ 21 (66)

This equation relates the shear flow of the (n + 1)th sheet to the shear flow of the adjacent nth sheet.

If the section is open, starting from one free edge, the shear flows in successive sheets may be easily found.

Note that the shear flow does not vary with s within each sheet or web.

10



1

2

2

1

1

y n z nn n n

zz yy

z yzy

yyy

yz

yy zz

y yzz

zzz

yz

yy zz

V y V zq A qI I

V IV

IV

II I

V IV

IV

II I

+

= + +

=

=


If the section is closed, we first analyze the problem as if one of the sheets (pick one arbitrarily) is missing.

The unknown shear flow is then superposed onto the problem and solved by taking moments of the combined shear flows about a convenient point.

q1

q2

q3

q4

q5

Replace the missing sheet with the unknown shear flow.

q0

+ q0

+ q0

+ q0

+ q0

+ q0

11


Determine the shear flow distribution in the idealized channel section produced by a vertical load of 48 kN acting through its shear center. Assume the skin is effective only in resisting shear stresses, while the stringers, each of area 300 mm2, resist all the direct stresses.

Example 11

y

z2000

2000

2000

48 kN

2

43

1

4924

1

2 mm108.4)2000)(300(4 ====i

iiyy zAI

Dimensions in mm


N/mm612)2000)(300(108.4

48000

N/mm126)2000)(300(108.4

48000

N/mm6)2000)(300(108.4

48000

0

9

233334

9

122223

9

1112

=

=

+=

=

=

+=

=

=

+=

qzAIVq

qzAIVq

zAIVq

yy

z

yy

z

yy

z

y

z2000

2000

2000

2

43

1

Dimensions in mm

12 N/mm

6 N/mm

6 N/mm

Note that we can check vertical and horizontal force equilibrium and show that the resultants are 48 kN and zero, respectively.

12


Find the shear flows in the webs of the unsymmetrical closed beam section.Example 12

Vy = 4 kN

Vz = 10 kN

Vz

Vy

A

200 mm

100 mm

CD

B300 mm2 100 mm2

300 mm2 100 mm2

y

z


46

4

1

46224

1

2

46224

1

2

mm100.2)100)(50)(100()100)(50)(100(

)50)(100)(300()50)(100)(300(

mm100.8)100)(100(2)100)(300(2

mm100.2)50)(100(2)50)(300(2

=++

+==

=+==

=+==

=

=

=

iiiiyz

iiizz

iiiyy

zyAI

yAI

zAI

42

42

2 .04000 100002.0 1.8667 10

( 2.0)1(2.0)(8 .0)

2 .010000 40008.0 1.4667 10

( 2.0)1(2.0)(8 .0)

y

z

V

V

= =

= =

13


( )

1

2 2

3

1.8667 1.46678.0 10 2.0 10

2.33 7.33 10

y n z nn n n

zz yy

n nn n

n n n n

V y V zq A qI I

y z A q

y z A q

+

= + +

= + +

= +


Let qAD be the unknown shear flow, i.e. qAD = q0.

( )0

03

60)100(10)50(33.7)100(33.2

qqqDC

+=+=

A

CD(-100,-50)

B

q0

Consider stringer D:

100 mm2 60 + q0

14


( )0

3

100)300(10)50(33.7)100(33.2

qqq DCCB

+=+=

A

C(100,-50)D

B

q0

Consider stringer C:

300 mm2 60 + q0

100 + q0


( )0

3

40)100(10)50(33.7)100(33.2

qqq CBBA

+=+=

A

CD

B(100,50)

q0

Consider stringer B:

100 mm2

60 + q0

100 + q0

40 + q0

15


N/mm502004

0)50)(200()100)(100()50)(200()100)(100(

0

0

==

=+++

qq

qqqq DCCBBAAD

A

CD

B

q0

Let us now take moments about, say O:

60 + q0

100 + q0

40 + q0

O

qAD = - 50 N/mm qDC = 10 N/mm

qCB = 50 N/mm qBA = - 10 N/mm


Find the shear center.Example 13

Area of each stringer = A mm2

y

zr Vz

yeE

rV

IAzVq

ArzAI

z

yy

nnz

iiiyy

2

2 22

1

2

==

== =

Taking moments about O,

2

2

20

2

0

ry

rVqr

dqr

dsqryV

e

z

r

ez

=

=

=

=

=

O

s

16


Find the shear center.Example 14

A

B

45

F

120

60

C

D

60 mm2

60 mm2

120 mm2

120 mm2 All dimensions in mm

40360

)90)(120()60)(60(1

40360

)120)(60(21

4

1

4

1

=+

==

===

=

=

iii

iii

zAA

z

yAA

y

First, remember to locate the centroid!

measured from C

measured from C

y

z

O40

40

(-40,-50)

(80,-20)

(80,40)(-40,40)


46

4

1

46224

1

2

462224

1

2

mm10144.0)50)(40)(120(

)40)(40)(120()40)(80)(60()20)(80)(60(

mm10152.1)40)(120(2)80)(60(2

mm10612.0)50)(120()20)(60()40)(12060(

=+

++==

=+==

=+++==

=

=

=

iiiiyz

iiizz

iiiyy

zyAI

yAI

zAI

Section properties:

17


Let us first find the z-coordinate position of shear center E.

A

B

F

C

D

Assume an arbitrary shear force Vy applied to the shear center.

y

z

O

Vy

6

6

0.894 10

0.210 10

yy

zz

zy

yy

VV

IV VI

=

=

Hence

ze

E


( )

( )

6

3

6 3

3

3

0

0.894(80) 0.210( 20) 60 10

4.543 10

0.894(80) 0.210(40) 60 10 4.543 10

8.33 10

5.68 10

y A z AAB A

zz yy

y

y

y B z BBC B AB

zz yy

y y

y

CD y

V y V zq AI I

V

V

V y V zq A qI I

V V

V

q V

= + +

=

=

= + +

=

=

=

A

B

F

C

D

y

z

O

qAB

qBC

qCD

18


A

B

F

C

D

y

z

O

qBC

qABqCD

Taking moments about a convenient point F,

2 3( 5) [ (45) ( 5.68) ( 4.543)(60)(120) ( 8.330)(120)(45)] 10

113.83

108.8 mm

y e y

y

e

V z V

V

z

+ = + +

=

= measured from O.

5

Vy

ze

E


Let us now find the y-coordinate position of shear center E.

A

B

F

C

D

Assume an arbitrary shear force Vz applied to the shear center.

y

z

O

yyy

z

yzz

y

VIV

VIV

6

6

1068.1

1021.0

=

=

Now,

Vz

ye

E

19


zCD

BC

zAB

Vq

qVq

3

3

10056.7

010024.3

=

==

A

B

F

C

D

y

z

O

qABqCD

Taking moments about F,

mm7.267.66

10)]120)(60)(024.3()056.7()45([)40( 32

==

+=+

e

z

zez

yV

VyV

measured from O.

40

Vz

ye

E


The idealized tube is loaded by a vertical shear force of 10 kN acting in the plane of the section. Assuming the direct stresses are carried by the stringers while the webs are effective only in carrying the shear stresses, calculate the distribution of shear flow around the section.

Example 15

100

All dimensions in mm

y

z

O 60200

240120 240

1

23

4

56 7 8

10 kNA1 = A8 = 200 mm2

A2 = A7 = 250 mm2

A3 = A6 = 400 mm2

A4 = A5 = 100 mm2

20


( )

0mm1086.13

)50)(100()100)(400()100)(250()30)(200(2

46

22228

1

2

==

+++== =

yz

iiiyy

I

zAI

Section properties:

because the section is symmetric about the horizontal axis.

Izz does not enter into calculation because Vy = 0.


Select the shear flow in web 2-3 to be the unknown shear flow, q23.

100 y

z

O 60200

240120 240

1

23

4

56 7 8

10 kNA1 = A8 = 200 mm2

A2 = A7 = 250 mm2

A3 = A6 = 400 mm2

A4 = A5 = 100 mm2

23

236

3

2333

34

9.28

)400(1086.13

)100)(1010(

q

q

qAI

zVqyy

z

+=

+

=

+

=

q23q23 28.9

21


100 y

z

O 60200

240120 240

1

23

4

56 7 8

10 kNA1 = A8 = 200 mm2

A2 = A7 = 250 mm2

A3 = A6 = 400 mm2

A4 = A5 = 100 mm2

23

236

3

3444

45

5.32

9.28)100(1086.13

)50)(1010(

q

q

qAI

zVqyy

z

+=

+

=

+

=

2367

233456 9.28qq

qqq=

+==

Due to symmetry,

q23q23 28.9

q23 32.5

q23 28.9 q23


100 y

z

O 60200

240120 240

1

23

4

56 7 8

10 kN

A1 = A8 = 200 mm2

A2 = A7 = 250 mm2

A3 = A6 = 400 mm2

A4 = A5 = 100 mm2

23

236

3

81

23

236

3

78

4.22

1.18)200(1086.13

)30)(1010(

1.18

)250(1086.13

)100)(1010(

q

qq

q

qq

+=

++

=

+=

+

=

237812 1.18 qqq +==Due to symmetry,

q23q23 28.9

q23 32.5

q23 28.9 q23q23 + 18.1

q23 + 22.4

q23 + 18.1

22


100 y

z

O 60200

240120 240

1

23

4

56 7 8

10 kN

Take moments about the point of intersection of the line of action of the shear load and the horizontal axis of symmetry.

q23q34

q45

q12

q81

A


100 y

z

O 60200

240120 240

1

23

4

56 7 8

10 kN

0)480)(30()240)(70()100)(240240120()120)(50(

8112

12233445

=+++++

qqqqqq


50q45

50q34

120q34 240q23240q12

70q12

30q81

A

23


N/mm7.12N/mm0.17N/mm7.12N/mm4.5N/mm3.34N/mm9.37N/mm3.34

N/mm4.5

12

81

78

67

56

45

34

23

========

qqqqqqqqSubstituting for q23 and solving, we obtain

1

23

4

56 7 8

10 kN

5.434.3

37.9

12.7

17.0

34.35.4

12.7


Question to ponder:

How do we determine the separate proportions of these shear flows due to bending and torsion?

1

23

4

56 7 8

10 kN

5.434.3

37.9

12.7

17.0

34.35.4

12.7

24


Determine the position of the shear center.

(Let the thickness of each web = 1 mm.)

Example 15

100

All dimensions in mm

y

z

O 60200

240120 240

1

23

4

56 7 8

A1 = A8 = 200 mm2

A2 = A7 = 250 mm2

A3 = A6 = 400 mm2

A4 = A5 = 100 mm2


100 y

z

O 60200

240120 240

1

23

4

56 7 8

eyqq

qqqq

)1010()480)(30()240)(70(

)100)(240240120()120)(50(3

8112

12233445

=++

+++


50q45

50q34

120q34 240q23240q12

70q12

30q81

A10 kN

ye

E

25


e

e

yqy

qqqqqq

=+=+

+++++++

24.5192.97)1010(322560

1440043440034680030408016800240002400012000)5.32(6000

23

32323

23232323

Substituting for the unknown shear flow,

The additional equation is supplied by the rate of twist equation.


y

z

O

1

23

4

56 7 8

A


0222

02

1

8118 81

8145

54 45

4534

43 34

3423

32 23

2312

21 12

12 =++++

=

=

dstqds

tqds

tqds

tqds

tq

dstq

Gdxd

26



0)4.22()5.32(

)9.28(22)1.18(2

8118

234554

23

3443

232332

231221

23

=+++

+++

dsqdsq

dsqdsqdsq

264.03701400

0)60)(4.22()100)(5.32()130)(9.28(2)240(2)250)(1.18(2

23

23

2323

232323

==

=++++++

qq

qqqqq

Substituting this into the moment equation, we have

mm9.544=ey

5/26/2009

1

Multicell Thin-walled SectionsThe multicell tube in pure torsion is a statically indeterminate problem. The condition of consistent twist must be used for each cell of the tube in order to obtain the solution.

Consider the n-celled tube of general shape subjected to a torque T. Clearly, the shear flow in the web between cell i and cell j is qj-qi. Similarly, in the web between cell j and cell k, the shear flow is qj-qk. H th t f t i t f ll j bHence, the rate of twist of cell j becomes

where j is the area enclosed by cell j.

=

j ji jks s skij

jj tdsq

tdsq

tdsq

Gdxd

21

n

j

21 , , qi qjqk

(1)

T.E.Tay,DepartmentofMechanicalEngineering, NationalUniversityofSingapore

The first integral is evaluated over the entire cell parameter sj, the second integral over the web between cell i and cell j (i.e. sji), and the third integral over the web between cell j and cell k (i.e. sjk).

1

ki

Although cross-sections warp, they do not distort in their own plane. This means that

the entire cross-section and each cell rotate at the same rate of twist . Thusdxd

Therefore, for a general case where cell j is bounded by m cells,

nj dxd

dxd

dxd

dxd

dxd

==

==

=

=

......21

(2)

= m

jdsqdsqd 1 (3)


For n cells, there are n such equations.

2

=

j jrs r srj

j tq

tq

Gdx 12(3)

5/26/2009

2

However, the number of unknowns involved is (n+1), i.e. q1, q2, q3, , qj, , qnand . The additional equation required is the moment equilibrium equation, i.e.

dxd

(4) n

T 2

where qi is the shear flow in the ith cell.

The foregoing procedure can also be used for multicell thin-walled tubes loaded by

transverse forces, in which case the moments induced by the transverse forces

(4)=

=i

iiqT1

2


t a sve se o ces, w c case t e o e ts duced by t e t a sve se o ces

should be taken into account in the moment equilibrium equation, and the shear

flow expressions for bending should be used.

3

Example 1:The section shown has constant thickness t throughout. It is subjected to a constant torque T. Determine the rate of twist.

a 2a

Consider cell 1: Consider cell 2:

4 61 ddd aa

aq1

1 2

q2

T.E.Tay,DepartmentofMechanicalEngineering, NationalUniversityofSingapore 4

(2)( )21

0

122

4

0

112

42

1

21

qqtGa

tdsq

tdsq

Gadxd aa

=

=

( )12

0

121

6

0

222

64

1

41

qqtGa

tdsq

tdsq

Gadxd aa

=

=

(1)

5/26/2009

3

Moment equilibrium:

2211

2

1

22

2

qq

qTi

ii

+=

= =

From (1) and (2) we have:

d

(3)( )212 22 qqa +=

dxdaGtq

2316

1 =

Substituting these into (3),

dxdGtaT 3

23104

=


and

5

dxdaGtq

2318

2 =

3104

23Gta

Tdxd

=

Example 2. Calculate the shear stress distribution in the walls of the three-cell wing

section shown when it is subjected to an anticlockwise torque of 11.3kNm.

Wall Length (mm) Thickness (mm) G (N/mm2) Cell Area (mm2)

12o 1650 1.22 24200 AI = 258000

12i 508 2 03 27600 A = 355000

Note: The superscript symbols o and i are used to distinguish between outer and inner walls connecting the same points.

12i 508 2.03 27600 AII = 355000

13, 24 775 1.22 24200 AIII = 161000

34 380 1.63 27600

35, 46 508 0.92 20700

56 254 0.92 20700

11300Nm


I II III

6

5

4

3

2

1

5/26/2009

4

Let andGPa 6.27=refG tGGt

ref

= Hence

07.122.16.272.24

12==t o

508

154207.1

1650

12 12

12 ==

ds

tds

o o

o

69.0

07.16.27

564635

2413

===

==

ttt

tt

736

233

725

25003.2

508

4635

34 34

34

24 24

24

13 13

13

12 12

12

=

==

==

dsdst

dst

dst

ds

tds

i i

i


368

736

56 56

56

46 46

46

35 35

35

=

==

tds

tt

For cell I,( )

( ) ( )( )IIIref

IIIIref

qqG

tds

qt

dsq

tds

qGdx

di i

i

i i

i

o o

o

250250154225800021

25800021

12 12

12

12 12

12

12 12

12

+=

+=

For cell II,

For cell III,

( ) ( )( )IIIIIIrefqqq

Gdxd 233725233725250250

35500021

++++=

( ) ( )( )IIIIIrefqq

Gdxd 368233736736233

16100021

++++=

531


I II III

642qI qII qIII

5/26/2009

5

From the moment equilibrium equation

( ) 6103.111610003550002580002 =++ IIIIII qqq

These equations are solved to give

N/mm 2.4N/mm 9.8N/mm 1.7

===

III

II

I

qqq

7.18.9 4.21 3 5


4.2

4.2

4.7

8.9

1.8

2 46

The Method of Successive Approximations for Pure Torque

Consider a two-cells section free to warp and subjected to torque T.

I II

If the cells were separated, we see that and this violates the constant rate of twist condition.

qIqII

III dxd

dxd


qI

I

qII

II

5/26/2009

6

If we arbitrarily let , then 1=

=

III dxdG

dxdG

For cell I, For cell II,

12

12

=

=

=

II

II

I

I

qt

dsqdxdG

12

12

=

=

=

IIII

IIII

II

II

qt

dsqdxdG


2 I 2 II

Considering the separated cell I, we see that at the wall shared with cell II, the shear

flow must be corrected for the shear flow due to cell II.

qI

qIII


5/26/2009

7

The contribution of qII to the rate of twist is represented by an equivalent shear flow qI

over the perimeter of cell I.

IIIII

II t

dsqt

dsq,

=

q


I IIq I Iq

Or,

where for the wall common to cells I and II

I

IIIIII qq

,=

= tds

III ,

The correction carry-over factor is then defined as

t,

IIIIII

I

IIIIII

Cqq

C

,

,,

=

=

q

I Iq


qI

5/26/2009

8

Since the correction carry-over factor is less than unity, if the procedure is repeated,

the correction become successively smaller until they are finally negligible. The

number of iterations depends on the accuracy required.

where

( ) K++++= IIIIFinalI qqqqq

IIIIII

IIIIII

CqqCqq

,

,

=

=


Similarly for cell II,

IIIIIIC

,, =

IIqI

IIIIII

II

Cqq ,=

IIq

qII


qIIIIIIq

5/26/2009

9

Example 3:Let us solve the problem of the previous example using the method of successive approximations. 1212 +=

=

i i

i

o o

o

tds

tdst

dsI

I II III

6

531

11300Nm 17922501542

12 1212 12

=+=

i io o tt

1933725233725250

=+++=II

368736233736 +++=III


642 2073=

250, =III

233, =IIIII

The carry-over factors are:

531

1400250

129.01933250

,

,, ===

III

II

IIIIII

C

C

The initial assumed values of shear flows:

I II III

642

121.01933233

112.02073233

140.01792

,

,

,,

==

==

===

IIIII

IIIII

IIII

C

C

C

( ) 9.287179225800022

==

=I

IIq


( )

( ) 3.1552073

16100022

3.367193335500022

==

=

==

=

III

IIIIII

II

IIII

I

q

q

5/26/2009

10

Cell I Cell II Cell III

CI,II = 0.129 CII,I = 0.140 CII,III = 0.112 CIII,II = 0.121

A d 287 9 367 3 155 3Assumed q 287.9 367.3 155.3

q (0.140)(367.3)=51.4 (0.129)(287.9)=37.14 (0.121)(155.3)=18.8 (0.112)(367.3)=41.1

q (0.140)(37.14)=5.2 (0.129)(51.4)=6.63 (0.121)(41.1)=4.97 (0.112)(18.8)=2.10

q 0.93 0.67 0.25 0.56

q 0.09 0.12 0.07 0.03

Final q 345 5 435 9 199 1


Final q 345.5 435.9 199.1

We still need the torque equilibrium equation to arrive at the correct solution.

From the torque equilibrium equation,

( )( ) ( )( ) ( )( )1610001.19923550009.43522580005.3452222

++=++= IIIIIIIIIIII qqqT

But the actual applied torque is 0.113x108

So the values of q must be scaled down by a factor of 0.113/5.52=0.0205

81052.5 =

Cell I Cell II Cell III


Final q 345.5 435.9 199.1

Scaled q 7.07 8.92 4.1

5/26/2009

11

Bending Shear Stresses in Multi-cell Beams

Consider a three-cells section carrying a shear load through the shear centre

(no twist):

I II III

VZ


If each cell were cut, the open-section shear flow would be given by

( )

( )2yzzzyy

zyzzzyzb

VdsztIdsytI

IIIVQIQI

q

=

Distribution of qb:

( )2yzzzyy

zzzyz

III

VdsztIdsytI

=


I II III

5/26/2009

12

If we now close the cells, the constant shear flows qI, qII and qIII must be superposed

onto the shear flow qb in the respective cells:

qI qII qIII

qb


Considering each cell separately, we now have to correct for the shear flows in the

shared walls. Consider, for example, cell II, where qI acts on the left wall, and qIII on the

right wall:

qI

qII

qIII


qb

5/26/2009

13

Since the twist is zero,

From the figure, taking positive torque anti-clockwise,

0+ dsdsdsds

02

1=

=

IIIIII t

dsqdxdG 0=

IItdsqor

The other two cells have similar equations, i.e.:

and

0,,

=

+

IIIIIIII

IIII

IIII

IIb t

qt

qt

qt

q

0,,

=

+

IIII

IIIIII

III

II

b tdsq

tdsq

tdsq

tdsq

0,,

=

+

IIIII

IIIIIII

IIIIII

IIIb t

dsqt

dsqt

dsqt

dsq


These three equations can be solved to obtain

The above equations are also applicable to idealized multicell beams.

,,

qI , qII and qIII

The Method of Successive Approximations for Bending Shear Flows

Since the twist is zero,If shear force acts through shear centre

02

1=

=

IIIIII tdsq

dxdG

d

Hence, from the figure, taking positive torque anti-clockwise, for cell II,

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