Chapter 5 Torsion Buckling

8/10/2019 Chapter 5 Torsion Buckling

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Flexual buckling deforms by bending.

Twisting buckling---tensional stiffness of the member

is very small. Thin-walled open sections

Combination of bending and twisting bending and

twisting are coupled so that one necessarily producesthe other. (1)Axially loaded members whose shear

center axis and centroidal axis do not coincide shcu as

angles and channels. (2)Also transversedly loaded

beams.Firstly review some fundamental relationships of

torsional behavior in general.

Chapter 5: Torsion Buckling


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5.1 Elastic torsional and torsional-flexural buckling

of axially loaded columns.

Elastic torsional and torsional-flexural buckling of axially

loaded columns mostly take place in thin-walled columns.Thin-walled columns are divided into two parts:Open

section,Close section

5.1.1 Fundamental relationships of torsional behavior

of thin-walled open section columns.


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Torsional characteristics of thin-walled open section

columns:section warping

As we all know,noncircular sections of the torsional column are

no longer plane during twisting,but warping knaggy sections).Inanother word,the section has a displacement along the axial

direction.

1 The sorts of torsion

Uniform torsion

Nonuniform torsion

Uniform torsion--The member is allowed to warp freely,then

the applied torque is resisted solely by St.Venant shearingstresses.This type of behavior is referred to as uniform

torsion,pure torsion,mean torsion or St.Venant torsion,as

shown in Fig.1-42;


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If a couple of torques with opposite direction are applied to

the both ends of the thin-walled open section (I section)

column,uniform torsion will occur,as shown in Fig.1-42.Uniform torsion has two characteristics:

a.the same magnitude of twisting in every section.Thus,the

longitudinal fibers do not have axial strain,and there is no

normal stress but only shear stress caused by torsion in thesection.The distribution of the shear stresses relates to the shape

of the section and it is the same in each section.

b. the longitudinal fibers do not bend,viz.the longitudinal

fibers of the flanges and the web are still in line,there is just anangle (torsional angle )caused by torsion between the upper

and lower flanges.


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Known as the mechanics of material,to a noncircular sectionbar,the relationships between angle of twist and torque

is similar to a circular section bar, and the torsional rate is

constant,

1.89

t

sv

GI

M

dz

d

==

svM

2 The relationships between andtM

In which torsional rigidity of the section;

G shearing modulus of elasticity;

torsional constant

tGI

tI


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For a thin-walled open section made of rectangular

elements,such as I [ T and Z section the torsional constant

can be approximated by

1.90( )= 33

1iit tbI

In which i the number of the rectangular elements in the

section;

the length and thickness of the i element.ib it

From Eq. 1.89 ,the relationship between the torsionalangle on the left end and the torque in an arbitrary

section z position)is

svM

zGI

M

t

sv= 1.91


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For certain support or loading conditions will prevent

torsional sections from twisting freely,nonuniform torsion occurs.

see Fig.1-44

Fig.1-44

1 Characteristics

As shown in Fig.1-44,the characteristics of nonuniform torsion

are

a. Each section of the column is knaggily ( ) viz. thelongitudinal fibers change in length.b. Torsional rate changes along the z axes.

As shown in Fig.1-44a,the fixed end of the cantilever beam will

restrain other sections from warping freely,and the closer to the

fixed end the stronger the restraint is.For the different warpingconditions in each section, the longitudinal fibers of the column are

no longer in line but bending.


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c. Besides shear stress caused by St.Venant torsion, there

are normal stress and shear stress caused by warping

torsion in the section which are called warping normal stressand warping shear stress, respectively. Sector normal stress

and sector shear stress, respectively because of the sector

coordination

s

2 The relationships between andtM Restrained torsion St.Venant Torsion warping torsion

Restrained torsional tresses

Free torsion shear stress Fig.1-43

Warping normal stress and warping shear stress

s


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St.Venant Torsionacts to resist St.Venant Torsions svM

Msv

s

Nonuniform torsion

Warping normal stress flange bending moment Mf

Mf flange shear force VfV

f torsional moment M=Vfh


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Mf Vf

Mf

Warping normal stressflange bending moment M

f

Mf flange shear force V

f

Vfh

M

Vf torsional moment M

M=Vfh 1.93

Mf constitute

double moment B

B=Mfh


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Relationships between Mfand the displacement u of thef lange along the axis x

VfMf

x

y

z

dz

u

2

hu=

2

2

2

2

2 dz

dh

dz

ud = a moment of inertia of theflange about axis yf

I

2

2

2

2

2 dz

dEI

h

dz

udEIM fff

== 1.94

xh/2


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VfMf

x

y

z

dz

From the relationships between the bending moment in the upper

flange Mfand the shear force Vfthat

dz

dMV

f

f =

Substitution of the Eq. 1.94)gives

3

3

2 dz

dEI

hV ff

= 1.95


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Eq 1.93 becomes

3

32

2 dz

dhEIM

f

= 1.96

or

1.97

in which

3

3

dz

dEIM

=

2442

2322 htbhIhII

yf === (1.98)

is a significant property of the cross section in the

calculation of the restrained torsion called the warpingconstant or sector moment of inertia. Eq. 1.98 can beused for a doubly symmetric I section.

I


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Introducing the notation as the applied torque of the restrained

torsion,and for the equilibrium conditiontM

MMM svt += 1.99a

Substitution of Eq. 1.89 and Eq. 1.97 into theequation above gives

1.99b = EIGIM tt

Eq. 1.99b) is the general expression for any restrained

torsioned open thin-walled section.In which and are called

the free torsional rigidity and the warping rigidity of the section.

tGI EI


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In Eq. 1.99b ,the first term represent the resistance of thesection to twist and the second term represent the resistance to

warping.But it is important to point out that the second term iscaused not by the warping of the member but by its resistance to

warping.

4 Strain energy of torsion

The strain energy stored in a twisted member can be

broken down into two parts

St.Venant strain energy of torsion

warping strain energy of torsion

The first part is caused by St.Venant torsion,and the secondpart is caused by warping torsion.

1)St.Venant strain energy of torsion Usv


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The increment of strain energy stored in an element dz of a

twisted member due to St.Venant torsion is equal to one half

the product of the torque and the change in angle of

twist.Hence

dMdU svsv2

1=

Substitution of Eq. 1.89 gives

dzGI

MdU

t

svsv

2

2

1=

dzdGIM tsv =

or

dzdz

dGIdU tsv

2

2

1

=


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Integrating over the length gives the strain energy in the entire

member due to St.Venant torsion.Thus

( )

=

t

tsv dzGIU0

2

2

1 1.100

2)Warping TorsionFor an I beam,the strain energy stored in the member due

to its resistance to warping is assumed to be equal to the bending

energy of the flanges,and the shear energy is assumed to be

negligible.The bending energy stored in an element dz of one ofthe flanges is equal to the product of one half the moment

EIfu) and the rotation(u).Thus

( ) dzuEIdU f2

2

1 =

Notice that and so the expression becomes2

hu = ,

2

2hII

f=


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( ) dzEIdU 24

1 =

Integrating over the length and multiplying by 2 to account

for both flanges.That is

( ) = l

dzEIU0

2

2

1

1.101

The total strain energy is

UUU sv+=

( ) ( ) += l

o

l

ot dzEIdzGI

22

2

1

2

1

1.102

The shear center It is a special point in the cross section,whichis also the center of rotation when a pure torque is applied.when

there are symmetrical axes the shear center must be on the

symmetrical axes.


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5.1.2 Elastic torsional and torsional-flexural buckling of

axially loaded columns.

1 Torsional bucklingThe mode of buckling is torsional deformation.

It usually takes place in the axially loaded columns with

lower torsional stiffness and doubly symmetric section (Eg.+

section).This kind of compressive columns may occur

torsional buckling before the load reaches to Euler load.

A axially compressive column with doubly symmetric I

section,shown as Fig.1-48,is fixed by simply supports at the

ends of the member.The coordinates x y z are shown asgiven.


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x

y

z

N

o s

xs

N

s

N

Fig.1-48


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2 The property of the problem

restrained torsion symmetric section).

3)The buckling differential equation

Introducing the applied torque Mt1, the

equation stand by the condition of

equilibrium is

01 = MMM svt a

Equilibrium equation

1 simply supported

zero warping, zero rotation

x

y

z

o s

N

N


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The applied torsion Mt1The applied torsion Mt1is caused by axial force N.So lets

analyze how does N cause Mt1

Pick out an element dz from Fig.1-48 c) analyze the fiber

DE on dz.

01 = MMM svt a

1tM

svM

M

the applied torsion

St.Venant torsion

warping torsion

Substitution of Eq.(1.89)and Eq. 1.97)gives

01 =+ EIGIM tt b


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D

x

y

z

N

o s x

dAD

s

N

z

dzD

E

dA

s

D

N

E

D

Fi g. 1-48

a) b) c)


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dz

s

s

D

E

D

E

d

E

dAF1

F2

Torsion of fiber DE DE

D D torsional angle

E E torsional angle +d

The angle between DE and the

plumb line DE) .

dz

d=

Force on line DE decomposed by F1 and F2 along

DE and EE F2 along s creates torsion torque dMt1

dA

dAdzddAFdMt 221 ===

Integrating,we have


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In which i0 is the polar radius of gyration.

AIIAdAi yxA

/)(/220 +== Buckling differential equation

Substitution of Eq. 1.103 into a ,gives the buckling

differential equation

==== 2021 NidzdI

AIdz

ddA

dz

dM

At (1.103)

0)( 20 = NiGIEI t 1.105The boundary conditions

) z=0, =0, no torsion

) z=0, =0, free warping ) z=l, =0, no torsion

) z=l, =0 free warping


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The explanation about free warping

normal stress without torsion

the flange bending moment

0=

0=fM

02

== ff EIh

M

thus 0=

The critical load NSolving Eq. 1.105 with the boundary condition gives

the critical load of torsional buckling:

2

02

2

/)( iGIl

EIN t+=

242

232 htbhII

f ==


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Exampl e 14

I section compressive column with simple supports at the

ends, l=8m the section is shown in Fig.1-49

x

y

b=250

h=218 tw=6

t=10

h0=200

t=10

E=206 103N/mm2,

G=79 103N/mm2

Calculate the bending critical load and

torsional critical load

Solving2

2

l

EINcr

=

2

02

2

/)( iGIl

EIN t+=

A=2 25 1+20 0.6=62cm2

Ix=0.6 203/12+2 25 1 10.52=5913 cm4

Iy=2 1 253/12=2604cm4


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22

0 37.13762/8517/)( cmAIIi yx ==+=cmAIi yy 48.662/2604/ ===

433

11.183/)6.0201252( cmIt =+= 4232 4.287109)212/(212512/ cmhII f ===

Bending critical load

222

432

2

2

, /4.13310628000

10260410206mmNl

EIyycr = ==

Torsional critical load

22

0

2

2

, /6.2758517/)1430690912083( mmNAi

GI

l

EIt

cr =+=+

=

,, crycr < Bending buckling first


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2 Torsional-flexural bucklingIt mostly takes place in singly symmetric thin-walled

compressive columns.There is bending deformation besidetorsional deformation when the element is buckling. seeFig.1.51 .

1)Torsional buckling differential equation

Torque by St.Venant shear stressThe warping torsion

The applied torque

Torsional differential equation

= tsv GIM = EIM

= 201 NiMt

0)( 20 = NiGIEI t 1.105 2 Because of the bending axial load N

produces torque Mt2


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y

x os a

u

au0

y

xo

s

z

N N

z

=du/dz

z dz

x

N N

u

Bending moment My by

axial force in plane xz

NuMy

dz

dMQ

y=Shear force Q acting on the centroid andcreating torsion torque Mt2

Q


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adz

duNM t =2 In the same direction with

torsional angle

Thus the equilibrium equation becomes

021 =+ MMMM svtt 3 Differential equation of torsional-flexural buckling

On the basis of Eq. 1.105),we introduce the notation Mt2,thus

0)( 20 =+ uNaNiGIEI t 1.108

1.108 two unknown u so we need one more equationwhich is the differential equation of bending deformation:

00=+ NuuEIy Deformation of the shapecenter


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auu +=00=++ aNNuuEIy

Conclusion:

0)( 20 =+ uNaNiGIEI t 0=++ NauNuEIy 1.111

1.110

4 Critical loadAssume:

l

zA

sin=

l

zBu

sin=

Thus

1)())((22

0

= crcrcry Ni

aNNNN


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2

2

l

EIN

y

y

= Critical bending load

along axis y

)(12

2

2

0

tGIlEI

iN += Critical load of the

torsional buckling

IIai

yx++= 220

So we can obtain the critical load Ncr

5 Discussion Double symmetric section

a=0 use the smaller one between Ncr=Ny and Ncr=N

Simple symmetric sectionNcr is smaller than both Ny and N ,the larger a/i0 the

smaller Ncr is.

5 1 3 Design of steel columns in Chinese code


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5.1.3 Design of steel columns in Chinese code

Base on three column strength-slenderness ratio curve, modify it

(curve C) to design torsion buckling steel column.

y yf


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5.2 Lateral buckling of beams

A lateral buckling of a beam is a combination of twisting

and lateral bending brought about by the instability of thecompression flange.

5.2.1 Lateral buckling of rectangular beams in pure

bending

Assume:

the material obeys Hookes law

the deformations remain small

the geometry of the cross section does not changeduring buckling

Consider the rectangular beam in pure bending.


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The ends of the member are assumed to be simply

supported as far as bending about the x and y axes is

concerned.

Hence

at2 2

2 20

d u d vu v

dz dz = = = =

In addition, the ends of the member are prevented fromrotating about the z axis but are free to warp.

Thus

at2

20

d

dz

= = 0,z l=

0,z l=


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The differential equations of bending and twisting are2

2

2

2

x x

y

z

uEI M

GJ M

d vEI

dz

d

dz

d

dz

=

=

=

z

The third equation is an analogous expression for twisting

about the axis.

The moment about the and axes, denoted by vectors

in the figure, are given byx xy

cos

cos( 90) sinx x x

y x x x

M

M M M M

= == + = =

(5.59)

(5.60)

(5.61)

(5.62)

(5.63)


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cos(90 ) sinz x x xM M Mdu du du

Mdz dz dz

= = = (5.64)

Substitution of the expression in (5.62), (5.64) into Eqs.(5.59),

(5.60) leads to the following differential equations:

(5.65)

(5.66)

(5.67)

2

2

2

2

0

0

0

x x

y x

x

M

EI M

GJ M

d v

EI dz

d u

dz

d du

dz dz

+ =

+ =

=

H th fi t ti d ib b di i th ti l


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Hence the first equation describes bending in the vertical

plane. The second and third equations describe lateral bending

and twisting.

The variable can be eliminated between Eqs.(5.66) and (5.67)and one obtains

u

22

20x

y

M

EI

dGJ

dz

+ =

or

22

20k

d

dz

+ =

Where The solution of Eq.(5.69) is2 2 / .x yGJEIk =

sin cosA kz B kz= +

(5.68)

(5.69)

(5.70)

S b tit ti f th b d diti t i t


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Substitution of the boundary condition at into

Eq.(5.70) gives0= 0z=

0=

And from the condition at one obtains0= 0z=sin 0kl= (5.71)

Equilibrium in a deformed configuration is possible only when

sin 0kl=This gives kl =

For whichcr yGJEI

lM

= (5.72)

Obviously, the critical moment is proportional to the

torsional stiffness GJ and the bending stiffnessyI

The extreme fiber stress:


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The extreme fiber stress:

crcr

xS =

cr y

x

GJEIlS =or (5.73)

For the rectangular cross section being considered,

3

3

hb

J=

2 xx hS =

3

12x

bh

I =In view of these expressions, Eq.(5.73) becomes

23

3

12

3 4 /

y y

cr

x x

I h Ihb GE GE

l I bh l b I

= =

(5.74)


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/ xII

ratio of span to width

critical stress

ratio of the principal

rigidities

5 22 B kli f I b b h d


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5.22 Buckling of I beams by energy method

1 Uniform Bending-Simple SupportsConsider the simply supported I beam, subject to a uniform

bending moment M.

boundary conditions:

2 2

2 2

2

2

0

0

d u d vu v dz dz

d

dz

= =

= =

= = at

at

0,z l=

0,z l=

The condition indicates that the section is free towarp tat the supports.

2 2/ 0dzd =


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The strain energy: due to bending about the y axis and the


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The strain energy: due to bending about the y axis and the

energy due to twisting about the z axis.

22 2

2 2

2 20 0 01 1 1( ) ( ) ( )2 2 2

l l ly

d u d d I dz GJ dz E dzdz dz dz

U + + = (5.76)

And the potential energy is

2V M= Where is the angle of rotation about the axis at each end

of the beam.

x

(5.77)

0

1

2

ldu ddz

dz dz

= (5.85)

0

ldu dV M dz

dz dz

= (5.86)

Thus the potential energy of the external loads given byEq.(5.77) becomes


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Thus the total potential energy of the system

222

20 0

1 1( ) ( )

2 2

l l

y

d u dU V EI dz GJ dz

dz dz

+ = +

22

20 0

1( )

2

l ld du d dz M dz

dz dz dz

+

(5.87)

The boundary conditions will be satisfied if and areu


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The boundary conditions will be satisfied if and are

approximated by

u

sin

sin

zu A

l

zB

l

=

=

(5.88)

(5.89)

Using (5.89) and (5.92), the total potential energy becomes

2 2 2

2 2 220 0

1 1sin cos2 2

l l

y

M z zU V dz GJB dz I l l l

+ = + 4 2 2

2 2 2

4 0 0

1sin cos

2

l l

y

z M B zB dz dz

l l EI l

+

(5.93)


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and since2 2

0 0

1sin cos

2

l lz zdz dz

l l

= =

Eq.(5.93) reduces to

2 2 2 4 2 2

3

1

4 y

GJB E B M B l U V

l l EI

+ = +

The critical moment is reached when neutral equilibrium is

possible. Hence

( ) 2 4 23

02 y

d U V B GJ E M l

dB l l EI

+ = + =

(5.95)

(5.94)

Thus


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2 4 2

30

y

GJ E M l

l l EI

+ =

from which2

2cr yM EI GJ E

l l

= +

(5.96)

2 Uniform Bending-Fixed Ends


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2 Uniform Bending-Fixed Ends

Figue.Boundary condition:

0

0

0

v v

u u

= == == =

at

at

at

0,

0,

0,

z l

z l

z l

===

(5.97)

The conditions in (5.97)will be satisfied if and are

approximated by

u

21 cos

21 cos

zu A

l

zB

l

=

=

(5.98)

S b tit ti f th h i t th i i


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Substitution of these shapes into the energy expression given

by Eq.(5.87) leads to

2 42

4 0

1 16 2cos

2

l

y

A zU V EI dz

l l

+ =

2 22

2 0

1 4 2sin

2

l zGJ dz

l l

+

2 4 2

4 0

1 16 2cos2

lB z dzl l

+ 2

2

2 0

4 2sin

lAB zdz

l l

(5.99)

To determine the critical moment, the equilibrium is


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, q

( )

( )

2 2

2

2 2

2

8 2 0

2 8 2 0

y

U V

EI A MBA l l

U VGJBV E B MA

B l l

+ = =

+ = + =

(5.101)

or

( )

( )

2

2

2

2

4 0

4 0

yEI A M Bl

M A GJ E Bl

=

+ + =

(5.102)

Hence


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2 22

2 24 4 0yEI GJ E M

l l

+ =

from which

2

2

24cr yM EI GJ El l

= + (5.103)

Compare a fixed beam with a hinged beam, it is evident that the

critical moment of the fixed beam can be anywhere from two to

four times as large as the critical moment of the hinged beam.

3 Concentrated Load-Simple Supports


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The strain energy stored in the member during buckling

has the same form it had in the preceding computations.Consider an element of the beam located a distance

from the right support, as shown in Fig.5-14c.dz

2 2/ 2

00 2

l

y

z dzw I

= (5.106)

2 2 2/ 2

0 0 2

l

y

P zV Pw dz

EI

= =

(5.107)

the potential energy of P is

the total potential energy of the system


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22

20 0

2 2

l lyEI d u GJ d U V dz dz

dz dz

+ = +

22 2 2 2/ 2

20 02 2

l l

y

E d P zdz dz

dz EI

+

(5.108)

Substitution of as given by (5.105) into Eq.(5.60) leads toy


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g y ( ) q ( )y

2

2 2y

d u Pz

EI dz

=or 2

2 2 y

d u Pz

dz EI

= (5.109)

Making use of this relationship to rewrite the first term in

Eq.(5.108) in terms of , one obtains

222 2 2 2/ 2

20 0 04 2 2

l l l

y

P z GJ d E dU V dz dz dz

EI dz dz

+ = + +

(5.110)

assume that :


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sin z

Bl

= (5.111)

Thus2 2 2 2

/ 22 2 2

20sin cos

4 2

l

y

B z GJB zU V z dz dz

I l l l

+ = +

2 42

4 0sin

2

l

y

B zdz

l l

+

(5.112)

Using the definite integrals

3 2/ 2

2 2

0

2 2

0 0

sin 18 6

sin cos2

l

l l

z lz dz

l

z z ldz dz l l

= +

= =

(5.113)

Hence


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e ce

( ) 2 3 2 2 42 3

1 0

2 8 6y

d U V B P l GJ E

dB EI l l

+ = + + + =

from which

2 2

2 2 2

4 3

6cr yP EI GJ El l

= + +

(5.115)

5.2.4 Design simplifications for lateral buckling


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for a simply supported I beam subject to uniform bending:

2

2cr y

EM EI GJ

l l

= +

for the same beam, bent by a concentrated load at midspan:

2 2

2 2 24 3

6cr y

EP EI GJl l

= + +

(5.139)

(5.140)

A critical internal bending moment for that of a critical

applied loading

2

21.36

4cr y

Pl EM EI GJ

l l

= = +

(5.141)

a suitable design relationship is

2


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2

1 2cr y

EM C EI GJ

l l

= +

Where loading coefficient

1C

(5.142)

1.0 for uniform bending

1.04 for concentrated loads at the third points

1.13 for a uniformly distributed load

1.36 for a concentrated load at midspan

applicable to different boundary conditions.


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( )

2

1 2cr yEM C EI GJ

kl kl = +

where is an effective-length factor.k

(5.144)

k

1.0 for simply supported ends

0.5 for fully fixed ends.

Stress was assumed to be proportional to strain. Inelastic

critical moment becomes a smaller.

Chinese code


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(1)If there are slabs connected with the beam, no

stability need to be calculated.(2)Otherwise for welded simple supported I beam:

1 2

2

,

4320 235[ 1 ( ) ]

4.4

x

b x

y

b b b

y x y

fW

tAh

W h f

= + +

Where stability coefficient, coefficients relatedloading and loading point, coefficient related to

symmetric of section

bb

b

(3)For rolled I beam can be get directly from the table.b


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(4)For rolled channel beam,a simple equation is deducted:

(5)When is greater than 0.6, inelastic parametershould be used:b

'

3/ 2

0.4646 0.12691.1 ( )b

b b= +

For simple supported box beams no stability needed to

calc late if the follo ing condition be satisfied:


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calculate if the following condition be satisfied:

0

1 0

(1) / 6(2) / : 95,65,57

h bl b

Chapter 5 Torsion Buckling

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