54831754-shear-lag

A consistent beam element formulation considering shear lag effect

This article has been downloaded from IOPscience. Please scroll down to see the full text article.

2010 IOP Conf. Ser.: Mater. Sci. Eng. 10 012211

(http://iopscience.iop.org/1757-899X/10/1/012211)

Download details:

IP Address: 89.122.75.79

The article was downloaded on 06/09/2010 at 17:01

Please note that terms and conditions apply.

View the table of contents for this issue, or go to the journal homepage for more

Home Search Collections Journals About Contact us My IOPscience

http://iopscience.iop.org/page/terms

http://iopscience.iop.org/1757-899X/10/1

http://iopscience.iop.org/1757-899X

http://iopscience.iop.org/

http://iopscience.iop.org/search

http://iopscience.iop.org/collections

http://iopscience.iop.org/journals

http://iopscience.iop.org/page/aboutioppublishing

http://iopscience.iop.org/contact

http://iopscience.iop.org/myiopscience

A Consistent Beam Element Formulation Considering Shear Lag Effect

E Nouchi1, M Kurata1, Buntara S Gan1,3and K Sugiyama2 1 Department of Architecture, College of Engineering, Nihon University, 1-Nakagawara, Koriyama, Fukushima, 963-8642, Japan. 2 Department of Architecture, Graduate School of Engineering, Nihon University, 1-Nakagawara, Koriyama, Fukushima, 963-8642, Japan.

E-mail: [email protected]

Abstract. This paper presents general solutions for beam elements which are derived analytically by considering the effect of shear lag phenomenon. The mass and stiffness matrices which can be obtained from the general solution of displacements can be assembled for general frame analysis by using finite element procedure.

1. Introduction In studying and evaluating the ultimate strength of steel frame structures, it is apparent that the behavior of steel frame structures at failure is fully controlled by consecutive local buckling failures of its members. Hence, to evaluate the local buckling of each member element, especially thin-walled beam, the stress and deformation at section should be considered properly by taking into account the effect of shear lag phenomenon.

The classical theory of beam and thin-walled members was unable to reflect the shear lag phenomenon since it was based on the assumption of cross section remains plane after deformation. The effect of shear lag results in a distribution of direct stresses in the cross section which is different from that predicted by the classical theory of beam. Therefore, the formulations of most analytical and/or semi-empirical methods involved many simplified assumptions. As a result, they can only handle shear lag problems for a particular simple geometry and cannot be easily extended to structures with complex geometry.

The energy approach has been proven to be a simple and practical method in shear lag analysis [1]. Later, the shear lag phenomenon was investigated in box-girders using finite elements [2]. A combination of finite element and transfer matrix techniques are proposed by [3]. Introduction of basis warping function in the finite element was used to explain the shear lag phenomenon [4].

However, all of these approaches are mainly suitable for a particular type of cross section and static analysis, thus it is not applicable for general applications. When the shear lag phenomenon is to be considered for solving dynamic analysis, the mass property has to be introduced and/or approximated. This is due to the short of fundamental theories which can include the effect of shear lag directly in the formulation [5]. Present paper addresses a consistent beam element formulation considering shear lag phenomenon to deal with the difficulties. 3 To whom any correspondence should be addressed.

WCCM/APCOM 2010 IOP PublishingIOP Conf. Series: Materials Science and Engineering 10 (2010) 012211 doi:10.1088/1757-899X/10/1/012211

c© 2010 Published under licence by IOP Publishing Ltd 1

2. General solution for beam with solid cross section considering shear lag In this section, some fundamental equations of an elastic beam with solid cross section lying on a right-handed axes coordinates convention considering shear lag will be derived.

Displacement at any arbitrary point on the cross section beam with member axis in the x direction is given by U. The transversal displacements of the arbitrary point in the y and z direction are expressed by V and W, respectively. The following assumptions are made for the arbitrary point at the cross section of a plane beam in deriving equations.

fyuU ++= β (2.1a) vxvyxVV === )(),( , 0),( == yxWW (2.1b)

where, ),( yxUU = , )(xuu = and )(xββ = in which, ),( yxff = which is defined as a 2nd order or higher function in y direction.

Hence, strain and displacement relationships can be given as,

xfy

xU

x ∂∂

++=∂∂

= κεε , yf

yU

xV

xy ∂∂

+=∂∂

+∂∂

= 0γγ (2.2)

0=∂∂

=yV

yε , 0=∂∂

=z

Wzε , 0=

∂∂

+∂∂

=zV

yW

yzγ , 0=∂∂

+∂∂

=zU

xW

xzγ

which are obtained from the assumption that the cross section remain straight after deformation

where, dxdu

=ε , dxdβκ = and [ ] βγγ +==

= dxdv

yxy 00. Here, zyx εεε ,, are normal strains in x, y and

z directions and xzyzxy γγγ ,, are engineering shear strains on x-y, y-z and x-z planes. It can be noted from equations (2.1) that the displacement U and V are independent of z axis and y-z plane, respectively.

Assuming elastic material following the Hooke’s law, the shear strain xyγ can be related to the

shear stress xyτ from the relationship xyxy Gγτ = , hence the function f from equation (2.2) can be obtained as,

( )dyG

fy

xy∫ −=0 0

1 ττ (2.3)

where, G is the shear modulus of elasticity and 0γ is given by G

00

τγ = .

Substituting the function f into equation (2.1a) and (2.2), the normal strain and stress equations are then given as,

( )dyxG

yxU y

xyx ∫ −∂∂

++=∂∂

=0 0

1 ττκεε (2.4a)

( )dyxG

EEyEy

xyx ∫ −∂∂

++=0 0ττκεσ (2.4b)

where, E is the normal modulus of elasticity By integrating the normal stress xσ for the entire cross sectional area, the following axial force and

bending moment can be obtained.

( )∫ ∫ ∫ −∂∂

+==A A

y

xyx dydAxG

EEAdAN0 0ττεσ (2.5a)

( )∫ ∫∫ −∂∂

+==A

y

xyA x dydAx

yGEEIdAyM

0 0ττκσ (2.5b)

where, dAyIydASdAAA AA ∫ ∫∫ ==== 20, .


2

By substitution of ε andκ from equations (2.5a) and (2.5b) into (2.4b), results in the following normal stress equation.

τσσ ++= yI

MAN

x (2.6)

Here, τσ is the effect of shear lag in the normal stress which is given as follow.

( ) ( ) ( )dyxG

EdydAx

yGIEydydA

xGAE y

xyA A

y

xy

y

xy ∫∫ ∫ ∫∫ −∂∂

+−∂∂

−−∂∂

−=0 00 00 0 ττττττστ (2.7)

The dynamic equilibrium equation at an arbitrary point can be given as follow.

0=−+∂∂

+∂∂

+∂∂ UX

zyxxzxyx &&ρττσ

with 0== xzxz Gγτ (2.8)

By integration, the shear stress xyτ in the equation (2.8) can be given as,

∫ ⎟⎠⎞

⎜⎝⎛ −+∂∂

=e

yx

xy bdyUXxb

&&ρστ 1 (2.9)

where, X is the body force, ρ is the material density, 2

2

tUU

∂∂

=&& is the acceleration with t in time,

)(ybb = is width of cross section as a function of y, e is defined as a distance to the outermost fiber of cross section where [ ] 0=

=eyxyτ .

Dynamic governing differential equations for the analysis of a beam by considering first order solution of normal stress xσ which correspond to displacement u,v and deflection β are given by,

0=−+′ uAqN x &&ρ , 0=−+−′ βρ &&ImQM , 0=−+′ vAqQ y &&ρ (2.10)

where, ( ) ( )dx

d=′ , ∫= A xydAQ τ , ∫ ==

Ax XAXdAq , ∫= Ay YdAq and ∫ ==A

yXdAm 0 .

The solution of member forces can be obtained by integrating the equations (2.10) to reach the following equations.

( ) ( )( )∫ ∫ ∫

∫∫+−−+=

−−=−−=x x x

y

x

y

x

x

dxIdxdxvAqxQMM

dxvAqQQdxuAqNN

0 0 0

00

)0()0(

)0(,)0(

βρρ

ρρ

&&&&

&&&& (2.11)

Defining difference of shear stresses in the following form,

( )bIQSSS

bIQ

bISQ

bISQ

xy −=−=−=− 00

0ττ (2.12)

where, bI

SQxy =τ and

bISQ 0

0 =τ . Here, first moment of areas from 0 and or y to e of the cross

section are defined by ∫=e

yybdyS , ∫=

eybdyS

00 and ∫=e

yybdyS .

Substituting equation (2.12) into equations (2.5a) and (2.5b), resulted in,

( )

( )∫ ∫

∫ ∫

+−−=

+−−=

A

y

y

A

y

y

dASdyyvAqGbI

EEIM

dASdyvAqGbI

EEAN

0

0

&&

&&

ρκ

ρε (2.13)


3

from where, ε and κ can be obtained as follow.

( )

( )∫ ∫

∫ ∫

+−+==

+−+==

A

y

y

A

y

y

SdydAyvAqGbIEI

Mdxd

dASdyvAqGAbIEA

Ndxdu

02

0

1

1

&&

&&

ρβκ

ρε (2.14)

Finally, the displacements u, β and v of a beam considering shear lag phenomenon can be obtained as,

( ) )0(00

0 udxvAqGAk

dxEANu x

yx ++−+= ∫∫ &&ρ (2.15a)

( ) )0(01

0 βρβ ++−+= ∫∫x

yx dxvAq

GAk

dxEIM

&& (2.15b)

( )∫ ∫∫ ∫ +−+−++−=x x

y

x xvxdxdxvAq

GAkxQ

GAkdxdx

EIMv

0 010

0 0)0()0()0( βρ && (2.15c)

where, ∫ ∫∫ ∫ == Ay

Ay SdydAy

bIAkdydAS

bIk 02100 ,1

, bI

ASk 00 = and 1

01 k

bIASk −= .

It can be shown that for a solid rectangular beam section with dimension of hb× , the values of normal and bending shear lag coefficients 0.00 =k , 3.01 =k , 5.10 =k and 2.11 =k computed are agree with the common shear lag coefficients reported elsewhere.

By using the equations in (2.11) which are based on the linear first order approximation, the dynamic governing equations, i.e. mass matrix, stiffness matrix and loading vector, of a beam element considering shear lag effect which connecting beam’s nodal displacements with displacements along the member axis can be derived explicitly by using the displacements given in equations (2.15).

2.1. Cantilever beam example Cantilever beams, shown in figure 1, are subjected to concentrated load P at the free end and subjected to distributed load q are considered. It is noted that the transverse vertical displacement v result for both cases are different with the classical beam theory. It is also shown that the difference between stresses computed by the present method and the classical beam theory is quite significant, i.e. the effect of shear lag must be considered in engineering design.

500 mm 500 mm50 mm

100

mm

E = 205000 N/mm2

P = 1000 N q = 10 N/mm

Rectangular beam deflection ( concentrated load )

0.00

0.01

0.02

0.03

0.04

0.05

0.06

0 100 200 300 400 500x ( mm )

v ( m

m )

Considering shear lag effect

Classical beam theory

Rectangular beam deflection ( distributed load )

0.00

0.02

0.04

0.06

0.08

0.10

0.12

0 100 200 300 400 500

x ( mm )

v ( m

m )



Normal stress distribution

-50

-25

0

25

50

-20 0 20

σx ( MPa )

y ( m

m )

Function f(y)

-50

-25

0

25

50

-0.1 -0.05 0 0.05 0.1

f(×10-5)

y ( m

m )

Figure 1. Rectangular section cantilever beam example.


4

3. General solution for thin-walled beam considering shear lag In this section, the fundamental equations of an elastic plane thin-walled beam either open or close section which lying on a right-handed zyx ,, Cartesian and an auxiliary s,ς coordinate systems by considering shear lag will be derived.

In figure 2, point O is the centroid of a thin-walled beam cross section and lying along the x coordinate axis. This point O will become the origin of the y and z coordinates axes. Axis s is defined along the center line of the thickness t of the curvature thin-walled beam, from where axis n which is drawn normal to this curvature line is defined as )(st which is a function of s.

Displacements at any arbitrary point ),,( zyxP on the cross section of thin-walled beam are given by WVU ,, in the zyx ,, directions and WVU

))),, in the ς,,sx directions. Here, the displacement u in

the x direction is referred to a point where s=0. The transversal displacements cc wv , in the zy, directions are originated at point C which is the shear center of the cross section.

z

y

s

ς o

θ

cz

sr

ςr V

WV)

0=s

t

y

z

cy

c

Ux,

Up)

,

p′cv

cw

W)

−

s

Figure 2. Cross section of a thin-walled beam.

The strain and displacement relationships are given as follow.

xU

x ∂∂

=)

ε and ( ) srU

xV

xs ∂−∂

+∂∂

=ς

γ1

))

(3.1)

where, the radius curvature of the thin-walled beam cross section is given by )(srr = . The displacements of the point ),,( zyxP can be assumed as follow.

fzyuU yz +++= ββ , ( )φcc zzvV −−= and ( )φcc yywW −+= (3.2) From figure 2, the displacements can then be expressed in terms of ς,, sx coordinates system,

fzyuUU yz +++== ββ)

(3.3a)

φθθθθ ςrwvWVV cc ++=+= sincossincos)

(3.3b)

φθθθθ scc rwvWVW ++−=+−= cossincossin)

(3.3c)

where, ( ) ( ) θθς cossin cc yyzzr −+−= and ( ) ( ) θθ sincos ccS yyzzr −−−= . The following assumptions can be assumed for any arbitrary point at the cross section of the thin-

walled beam.

sU

xV

xs ∂∂

+∂∂

=))

γ (3.4)


5

By substituting equation (3.4), the function f in equation (3.3a) can be solved as follow.

( )( )( ) )0(1sincos)(0

fdsrsfs

zyxs ++−+−= ∫ ωχςθγθγγ (3.5)

In which, dxd

dxdw

dxdv

yc

zczc

yφχβγβγ −=+=+= ,, and ( )∫ −= s dsrr0 1 ςω ς are defined.

Accordingly, the average normal displacement of the thin-walled beam which can be expressed by

using ∫=A

dAUA

u)1

is given as,

FzyuUU yz ++++== ωχββ)

(3.6)

where, ( )( )( ) dAdsrA

FA

s

zyxs∫ ∫ −+−−=0

1sincos1 ςθγθγγ .

The normal strain is then given by, Fzyu yzx ′+′+′+′+′= χωββε (3.7)

Hence, the normal stress xσ for an elastic material following Hooke’s law can be obtained by,

FEEEzEyuEE yzxx ′+′+′+′+′== χωββεσ (3.8)

The first order solution of normal stress xσ in terms of axial force and moments of the beam can be expressed by,

ωσω

ω

IMz

IIIMIMI

yIII

MIMIAN

yzzy

zyzyz

yzzy

yyzzyx +

−−

+−−

+= 22 (3.9)

where, the internal forces and section properties are given as follow: ∫= A xdAN σ ;

∫= A xz dAyM σ ; ∫= A xy dAzM σ ; ∫= A xdAM ωσω ; ∫= AdAA ; ∫ ==

Az ydAS 0 ; ∫ ==Ay zdAS 0 ;

∫ ==A

dAS 0ωω ; dAyIAz ∫= 2 ; ∫= Ay dAzI 2 ; ∫= Ayz yzdAI ; ∫ ==

Ay dAyI 0ωω and

∫ ==Az dAzI 0ωω .

The dynamic equilibrium equation of any arbitrary point on the cross section is given as below.

( ) 01

=−+∂

∂+

∂−∂

+∂∂

UXsrx

xxsx &&))ρ

ςτ

ςτσ ς (3.10)

Substitution of equation (3.9) into equation (3.10), the shear stress xsτ can be solved as,

( ) ( ) ( )∫ ∫∫ +−+−−−∂∂

−=s

xs

ss xxs xdsrUdsrXdsr

x 0 00),0,(111 ςτςρςςστ &&) (3.11)

where, ),0,( ςτ xxs is the indeterminate shear stress.

The body force )(xXX))

= and applied moment per unit length terms are 0,, =zyx mmm , the first order approximation of shear stress can be given as,

∑=

=4

1

~i

iixs Qgτ (3.12)

where, )(~~),(~~),(~~,~~,,,, 43214321 θθθχ ωω ggggggggTQQQQQQ zyzy ====′=′′=′′=′′−=′ .


6

Herewith, all the functions in equation (3.12) are given explicitly as:

( )∫ −Ω

=θς rdr

Gg1

2~ ; 22 )(yzzy

yyzzy

IIISISI

g−−

=

))

θ ; 23 )(yzzy

zyzyz

IIISISI

g−−

=

))

θ ; ωθ Sg)

=)(4 ;

[ ] zzyydxdf

dxdxsy QgQg )0(~)0(~

0,0,0 +== === θφττ ; [ ] zzyydxdf

dxdxsz QgQg )2(~)2(~

2,0,0 ππττ πθφ +== === ;

( )∫ −= e rdrySz

θ

θθςθ 1)(

); ( )∫ −= e rdrzSy

θ

θθςθ 1)(

); ( )∫ −= e rdr

IS

θ

θω

ω θςωθ 11)()

;

( )∫ −=Ω θςτ rdrr 12 ; ( )yczz wGG βγτ +′== and ( )zcyy vGG βγτ +′== .

Furthermore, F ′ in the equation (3.8) can be solved as,

∑=

′=′4

1

1i

iiQaG

F (3.13)

Herewith, all the functions in equation (3.13) are given explicitly as: )(1 θaa = ; )(2 θyaa = ;

)(3 θzaa = ; )(4 θωaa = ; ( ) ( ) ( ))(ˆ)()2(~)(ˆ)()0(~)(ˆ)( 32 θθπθθθθ SSgCCgGGa iiiii −−−−−= ;

0)2(~)0(~)2(~)0(~43421312 ==== ππ gggg ; ( )∫ −=

θθςθ

01~)( rdrgG ii ; ∫=

A ii dAGA

G )(1)(ˆ θθ ;

( ) θςθθθ

rdrC −= ∫ 1cos)(0

; ∫=A

dACA

C )(1)(ˆ θθ ; ( ) θςθθθ

rdrS −= ∫ 1sin)(0

;

∫=A

dASA

S )(1)(ˆ θθ .

The differential governing equations of x-derivative displacements can also be given as,

( )χω ′−′+′+′−=′ 111413121 kTkQkQk

GAEANu zy (3.14a)

22

ˆˆ,

ˆˆ

yzzy

yzzzyy

yzzy

yzyyzz III

IMIMIII

IMIM−

−=′

−

−=′ ββ (3.14b)

( ) ( )( )ωωωω

ωχ TkQkQkkGIGIkEI

Mzy ′+′+′

−−

−=′ 444342

4141

11

(3.14c)

( )zyzc QkQkGA

v 21111

++−=′ β (3.14d)

( )zyyc QkQkGA

w 22121

++−=′ β (3.14e)

where, yM̂ and zM̂ are defined by ( )χω ′−′+′+′−= 21242322ˆ kTkQkQk

GEMM zyzz and

( )χω ′−′+′+′−= 31343332ˆ kTkQkQk

GEMM zyyy .

The shear lag coefficients are then can be given as follow.

)4,3,2,1,( =∫= kjkA jjk dAak λ and )2,1,(~~

== nmmnmn gAk (3.15)

here, 11 =λ , y=2λ , z=3λ , ωλ =4 , )(1 θaa = , )(2 θyaa = , )(3 θzaa = , )(4 θωaa = ,

)0(~~11 ygg = , )0(~~

21 zgg = , )2(~~12 πygg = , )2(~~

22 πzgg = .


7

Since the internal forces ωMMMQQN yzzy ,,,,, can be determined from the first order

approximation of the dynamic governing equation, the bimoment ωM can be solved from,

( ) ( )

( ) ( ) ( ) ( )

( )φρ

χρ ωωωω

ωωωωωω

ωωωω

ω

ωωωω

&&&&&&

&&

pcczccycx

mmzzmyym

m

mmm

IwSvSm

ImkGI

GJkQkQkkGI

GJ

MGIkEI

GJdxMd

kGIGJk

+++−

′+′⎟⎟⎠

⎞⎜⎜⎝

⎛−

+−′+′−

−=

−−⎟⎟

⎠

⎞⎜⎜⎝

⎛−

+

1

11 2

2

(3.16)

where, ( ) dArrSA syc ∫ −−= θθς cossin , ( )∫ −=

A szc dArrS θθς sincos , ( )∫ +=A spc dArrI φς

&&22 .

From the above equation, the right hand side of the second order differential equation are known

values, thus the general solution of the x-derivative twist χ , which is defined by χφφ −==′dxd

, can

be solved. The general solution of the χ can then be used to derive the element mass and stiffness matrices in the finite element procedure.

4. Pipe beam example considering shear lag To examine a close section example of thin-walled beam, a solution for pipe beam shown in figure 3 is derived by following the same procedures for obtaining the general solution of thin-walled beam,

z

y

ς

sW)

V)

y

x

UU)

=

θ

s

r

),,( ςsxP

W

V

V)

W)

−

UU)

=

P

θ

z

y

sς

θς dr )( −

dy

dzθ

Figure 3. Thin-walled circular beam

The displacements of the centroid cross section of the pipe beam and the yx, coordinate system

can be expressed in terms of rotational angles zyx ββφβ ,,= as follow. fyuU ++= β , vV = and 0=W (4.1)

where, ββ =z , 0=yβ . By assuming 0=w , 0=φ , θς sin)( −= ry and θς cos)( −−= rz , the displacements become UU = , vV = and 0=W . Hence, equation (4.1) yields to the following equations.

fyuUU ++== β)

, θcosvV =)

and θsinvW −=)

(4.2) The strain and displacement relationships for pipe beam can further be expressed as,

xU

x ∂∂

=)

ε , ( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛−

∂∂

−= WV

r)

)

θςεθ

1 and ( ) θς

γ θ ∂∂

−+

∂∂

=U

rxV

x

))1

(4.3)

Hence the function f from equation (4.1) can be solved for,

( )( )∫ −−=θ

θ θςθγγθ0

cos)( drf x (4.4)


8

Substitution of equation (4.2) into equation (4.3) results in the following normal strain and stress equations.

( )( )∫ −′−′+′+′=∂∂

=θ

θ θςθττβε0 0 cos1 dr

Gyu

xU

xx (4.5a)

( )( )∫ −′−′+′+′==θ

θ θςθττβεσ0 0 cos dr

GEEyuEE xxx (4.5b)

By integrating the normal stress xσ for the entire cross sectional area, the following axial force and bending moment can be obtained.

( )( )∫ ∫∫ −′−′+′==A xA x dAdr

GEuEAdAN

θ

θ θςθττσ0 0 cos (4.6a)

( ) ( )( )∫ ∫∫ ⎟⎠⎞⎜

⎝⎛ −′−′−+′==

A xA x dAdrrGEEIdAyM

θ

θ θςθττθςβσ0 0 cossin (4.6b)

For pipe beam section, 0=τσ , the first order solution of normal stress xσ becomes,

yI

MAN

x +=σ (4.7)

The dynamic equilibrium equation at an arbitrary point can be given as follow.

( ) 0=−+∂∂

+∂−

∂+

∂∂ UX

rxxxx &&ρςτ

θςτσ ςθ (4.8)

By integration, the shear stress θτ x in the equation (4.8) yields to,

( ) )0(0 θ

θ

θ τθςτ xx dryttIQ

+−−= ∫ (4.9)

where, due to symmetric geometry of the cross section in y axis, from 0)2( =πτ θx condition, the

equation (4.9) resulted in tISQ

x =θτ and tISQ 0

0 =τ at [ ] 0=θθτ x in which, ( )∫ −=2π

θθς drytS and

( )∫ −=2

00

πθς drytS are denoted.

From which, the following equations can be obtained as,

∫∫

∫′

+′=′

+′==

′+′==

A mA x

nA x

kGA

IQEEIdAygGtQEEIdAyM

QrkGEuEAdAN

βθβσ

σ

)( (4.10)

where, ( )( )∫ −−=θ

θςθθ0 0 cos1)( drSS

Ig , ∫=

An dAgtr

k )(1 θ and ∫=Am dAyg

tIAk )(θ are given.

Finally, the displacements u, β and v of the pipe beam considering shear lag phenomenon can be obtained as,

( ) )0(00

udxvAqGArkdx

EANu

x

yn

x+−+= ∫∫ &&ρ (4.11a)

( )∫∫ +−+=x

ym

xdxvAq

GAkdx

EIM

00)0(βρβ && (4.11b)

( ) )0()0()0(0 00 0

vxdxdxvAqGAkxQ

GAkdxdx

EIMv

x x

ym

x x+−−−+−= ∫ ∫∫ ∫ βρτ && (4.11c)


9

where, tISAk 0=τ and τkkk mm += .

Thus, the shear lag coefficients in the displacement solutions can be obtained as 0=nk , 0=mk ,

0.2≈τk and 0.2≈mk .

4.1. Cantilever beam example Cantilever beams, shown in figure 4, are subjected to concentrated load P at the free end and subjected to distributed load q are considered. It is noted that the transverse vertical displacement v results for both cases are different with the classical beam theory. Thus, the effect of shear lag must be considered in engineering design.

t

r

Pipe beam deflection ( concentrated load )

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0 50 100 150 200 250x ( mm )

v ( m

m )



Pipe beam deflection ( distributed load )

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0 50 100 150 200 250

x ( mm )

v ( m

m )



Figure 4. Pipe beam section cantilever beam example.

5. Conclusions Based on the classical beam theory where the effect of shear lag is included in the formulations of displacements, strains and stresses of solid, thin-walled and pipe beam consistently without using any kind of stress approximation functions. The shear lag phenomenon is now can be considered directly for solving dynamic problems in which the mass and stiffness matrices can be derived consistently.

References [1] Hadji-Argyris J and Cox H L 1944 Diffusion of load into stiffened panels of varying section Br.

Aeron. Res. Council Reports. Mem. 1969 Reissner E 1946 Analysis of shear lag in box beams by the principle of minimum potential

energy Q. Appl. Math. 4 268-78 [2] Malcolm D J and Redwood R G 1970 Sehar Lag in stiffened box-girders J. Struct. Div. ASCE

96 ST7 1403-15 Moffatt K R and Dowling P J 1975 Shear lag in steel box-girder bridges Struct. Engineer 53

439-48 [3] Tesar A 1996 Shear lag in the behavior of thinwalled box bridges Comp. and Struct. 4 607-12 [4] Prokić A 1996 New warping function for thin-walled beams I: Theory J. Struct. Eng. ASCE 122

ST12 1437-42 Prokić A 1996 New warping function for thin-walled beams II: Finite element method and

applications J. Struct. Eng. ASCE 122 ST12 1443-52 Prokić A 2002 New finite element for analysis of shear lag Comp. and Struct. 80 1011-24

[5] Kurata M 2009 General solution of beam considering shear lag phenomenon (in Japanese) Summaries of Tech. Papers of Research Annual Meeting Nihon Univ. College of Eng. 2009 1-2


10

Documents

54831754-shear-lag