The Timoshenko Beam · Timoshenko theory Equilibrium equations in terms of deflections Combining...

Institute of Structural Engineering

Method of Finite Elements I

Chapter 6

The Timoshenko Beam

Book Chapters[O] V2/Ch2[F] Ch13



Today’s Lecture

• Timoshenko beam theory

• Weak form

• Discretization

• Stiffness matrix – load vector

• Shear locking

Timoshenko beam elements



Reminder: Euler-Bernoulli theory

Assumptions:

Uniaxial Element

The longitudinal direction is sufficiently larger than the other two

Prismatic Element

The cross-section of the element does not change along the element’s length


Method of Finite Elements I30-Apr-10


Euler/ Bernoulli assumption

Upon deformation, plane sections remain plane AND perpendicular to the beam

axis

z

z

dwu y y

dx

dw

dx

2

2

0

xx

xy

du wy

dx x

dw du dw dw

dx dy dx dx




Valid for:

Slender beams: / 1 /100h L



Timoshenko theory

Assumptions:

Uniaxial Element

The longitudinal direction is sufficiently larger than the other two

Prismatic Element

The cross-section of the element does not change along the element’s length



Timoshenko theory

Timoshenko assumption

Upon deformation, plane sections remain plane but not necessarily perpendicular

to the beam axis

z

z

dw

dx

z z

dw

dx



Timoshenko theory

Valid for:

Slender beams:

Thick beams:

/ 1 /100h L

/ 1 /10h L



Timoshenko theoryDisplacements

Strains

z z

dwu y y

dx

w w x

, 0

0

zxx yy

xy z

u wy

x x y

w u w w w

x y x x x

Stresses

,

2 1

xx xx

xy xy

E

EG G



Timoshenko theoryCross sectional forces – Bending Moment

2z zxx xx z z

A A A A

d dM y dA E y dA E y y dA E y dA EI

dx dx

M

xx

x

y

Normal stress distribution

where:

2

z

A

zz

I y dA

d

dx

Or:

b z

b z

M D

D EI

→ moment of inertia

→ bending strain/curvature



Timoshenko theoryCross sectional forces – Shear Force

xy xy xy xy

A A A

V dA G dA G dA GA

x

y

xy

Vx

y

xy

V

Actual shear stress distribution Assumed shear stress distribution

,

2 1A

EA dA G

where:

Or:

,s xy sV D D kGA where is introduced to account for the actual stress distribution 1k



Timoshenko theoryShear Correction parameter for standard cross sections

5 / 6k 6 / 7k 0.32k

0.416k 0.5k 0.358k



Timoshenko theoryEquilibrium equations (identical to Bernoulli beams)

Force equilibrium

Moment equilibrium

y

dVf x

dx

dM

V xdx

Combined

2

2 y

d Mf x

dx

yf x



2

22 3

const

2 3

y

EIz zz z z y z y

zz

d Mf x

dxd d d

M EI k EI f x EI f xdx dx dx

dk

dx

Timoshenko theoryEquilibrium equations in terms of deflections

Equilibrium + Moment-curvature relation + Curvature definition

Force equilibrium + Shear force-Shear strain relation + Shear strain definition

2

const

2

y

ykGA zxy z y

xy z

dVf x

dxf xd dw d d w

V kGA kGA f xdx dx dx dx kGA

dw

dx



3

243

4 22

2

zz y

yzz y

yz

dEI f x

d f xd w EIdxEI f x

dx kGA dxf xd d w

dx dx kGA


Combining the two expressions



3

243

4 22

2

zz y

yzz y

yz

dEI f x

d f xd w EIdxEI f x

dx kGA dxf xd d w

dx dx kGA


Combining the two expressions

Comparison to Bernoulli-Euler theory

24

4 2

yzz

d f xd w EIEI p x

dx kGA dx

4

4z y

d wEI f x

dx

Timoshenko Bernoulli-Euler



Weak formPrinciple of Virtual Work

0

InternalVirtualWork ExternalVirtualWork

L

xx xx xy xy y z i yi zi i

i iV

dV w f m dx w F M

where:

yf

m

yiF

iM

→ distributed force

→ distributed moment

→ nodal force

→ nodal moment



Weak form

Bending Shear

0 0

0 0

xx xx xy xy xx xx xy xy

V V V

zxx xy xy

V V

L L

zxx xy xy

A A

M V

L L

zxy

dV dV dV

dy dV dV

dx

dy dA dx dA dx

dx

dM dx V

dx

0 0

L L

z zb xy s xy

dx

d dD dx D dx

dx dx

Internal Virtual Work



2 2

2 2

0 0b

z zzzizz

L L

z y i yi i

i i iDdd

dxdx

d w d w dw dwEI dx w f m dx w F M

dx dx dx dx

0 0 0

z z

L L L

z zb xy s xy y z i yi zi i

i i

d dD dx D dx w f m dx w F M

dx dx


Principle of Virtual Work – Euler Bernoulli theory




0 0 0

L L L

z b z xy s xy y z i yi zi i

i i

D dx D dx w f m dx w F M

Principle of Virtual Work – Euler Bernoulli theory

0 0

L L

z b z y z i yi zi i

i i

D dx w f m dx w F M



Discretization

Two noded beam element

Isoparametric coordinates are used

Independent assumptions need to be made for the deflections and rotations

4 dofs are available in total (1 deflection and 1 rotation per node)

A linear assumption is made for both the deflections and the rotations:

2l

1 1 0

0 1 0 1, zw a a b b



DiscretizationShape functions – Linear isoparametric

2l

1 1 0

1N

-1 1

2N

-1 1

Geometry

1

1 1

2

xx N N

x

2

2 2,

1 11 ,

2 2

dNN N

d

1

1 1,

1 11 ,

2 2

dNN N

d

1

1, 1,

2

1 2 1

2

1 1

2 2 2 2

xdxJ N N

xd

x x x L

x

1 2dJ

dx L



DiscretizationDisplacements – Virtual Displacements

1

1 2

2

1

1

1 2

2

2

1

1

1 2

2

2

or

0 0

0 0

z

z

z

z

ww N N

w

w

w N Nw

w

w N Nw

w

w

w

N

u

w

N

u

N u

N u

1

1 2

2

1

1

1 2

2

2

1

1

1 2

2

2

or

0 0

0 0

z

z

z

z

z

z

z

z

z

N N

w

N Nw

w

N Nw

θ

θ

θ

N

u

θ

N

u

N u

N u

Rotations – Virtual Rotations



DiscretizationBending strain – Virtual bending strain

1 1

1 1

1 1

1, 2, 1, 2,

2 2

2 2

1

1

2

2

0 0 0 0

1 2 1 20 0

2 2

1 1, , 0 0

z zzz x x

J Jz z

z

z

z z

w w

d d dN N N N

w wdx dx dx

w

wL L

L L

bB

u

b b bB u B u B



DiscretizationShear strain – Virtual shear strain

1 1

1 1

1, 2, 1 2

2 21 1

2 2

1

1

2

2

0 0 0 0

1 1 1 11 1

2 2

1 1, , 1

2

z z

xy x x

L Lz z

z

z

xy xy

w w

dwN N N N

w wdx

w

wL L

L

sB

u

s s sB u B u B 1 1

12L



Stiffness Matrix – Load VectorPrinciple of Virtual Work (distributed moments and nodal forces ignored)

0 0 0

L L L

z b z xy s xy yD dx D dx w f dx

Substituting the discretized fields

0 0 0

L L L

T T T T T T

b s y

T

D dx D dx f dx

b b s s wu B B u u B B u u N

u0

L

T T

bD dx b bB B u u

0

L

T T

sD dx s sB B u u

0

0 0 0

L

T

y

L L L

T T T

b s y

f dx

D dx D dx f dx

w

b b s s w

fK

N

B B B B u N



Stiffness Matrix – Load VectorStiffness matrix

0 0

, ,

L L

T T

b sD dx D dx b s b b b s s sK K K K B B K B B

Load Vector

0

L

T

yf dx wf N

1 10 0 ,

1 1 1 11 1 ,

2 2

b z

s

D EIL L

D kGAL L

b

s

B

B

1 1

1 0 1 02 2

wN



Stiffness Matrix – Load VectorStiffness matrix – Analytical integration

1

0 1

0 0 0 0

0 1 0 1

0 0 0 0

0 1 0 1

L

T T

b b

b

z

D dx D Jd

D

L

EI

L

b b b b b

b

bb

K B B B B

K

K K




1

0 1

0 0 0 0

0 1 0 1

0 0 0 0

0 1 0 1

L

T T

b b

b

z

D dx D Jd

D

L

EI

L

b b b b b

b

bb

K B B B B

K

K K

→ Similar to a bar element!




1

0 1

2 2

2 2

2 2

2 2

1 1 1 11 1

2 2 2 2

1 1 1 1

2 3 2 6 2 3 2 6

1 1 1 11 1

2 2 2 2

1 1 1 1

2 6 2 32 6 2 3

L

T T

s s

ss

D dx D Jd

L L

L L L L

L L L L

D L LD L kGAL

L LL

L L L L

L L L L

L L

s s s s s

ss

K B B B B

K K



Stiffness Matrix – Load VectorLoad vector– Analytical integration

0 0

2

0

2

0

L L

T T

y y

y

y

f dx f Jd

f

f

w wf N N

f



Stiffness Matrix – Load VectorLoad vector– Analytical integration

0 0

2

0

2

0

L L

T T

y y

y

y

f dx f Jd

f

f

w wf N N

f → Also similar to a bar element!



Stiffness Matrix – Load VectorStiffness matrix – Numerical integration

1

0 1

L

T T T

b b i i b i

i

D dx D Jd w D J

b b b b b b bK B B B B B B

Stiffness matrix – Required number of Gauss points




1

0 1

L

T T T

b b i i b i

i

D dx D Jd w D J

b b b b b b bK B B B B B B


1 10 0 constant

constant

constant

constant 0

b

T

i b i

L L

J

D

D J p

b

b b

B

B B

→ 1 Gauss point is enough




1

0 1

L

T T T

s s i i s i

i

D dx D Jd w D J

s s s s s s sK B B B B B B





1

0 1

L

T T T

s s i i s i

i

D dx D Jd w D J

s s s s s s sK B B B B B B


1 1 1 11 1 linear

2 2

constant

constant

quadratic 2

s

T

i s i

L L

J

D

D J p

s

s s

B

B B

→ 2 Gauss points are

required



Stiffness Matrix – Load VectorLoad vector – Numerical integration

0 0

L L

T T T

y y i i y

i

f dx f Jd w f J w w wf N N N

Load vector – Required number of Gauss points



Stiffness Matrix – Load VectorLoad vector – Numerical integration

1

0 1

L

T T T

y y i i y

i

f dx f Jd w f J

w w wf N N N

Load vector – Required number of Gauss points

1 1

1 0 1 0 linear2 2

constant

constant

linear

y

T

y

J

f

f J

w

w

N

N

→ 1 Gauss point is enough



Shear lockingExplanation

Consider a 2 noded beam under pure bending:

For a slender beam no shear strain should occur, however since

1constzz z

dM EI b

dx

1 1 0xy z

dwa b

dx

0 1w a a

→ shear strain is not zero for any value of 1a



Shear lockingGeometrical interpretation

Element deformation for a slender beam Actual deformation of a slender beam

1z 2z

1 2

xy0xy



Shear lockingIn terms of the system of equations

2

3

z zEI EI kGAkGAL L

L L L

b s b sb s

K K K K K K K

Ku f

Assuming an orthogonal cross section:

3

3

3

2

2

5

6 2 112

5

12 12 1

h

EbhbhE

LL L

Eb Eb h

LL

L

b s

b s

K K K

K K

Since for slender beams ,

the shear term dominates the bending

one leading to artificially stiff systems

3h h

L L



Shear lockingIn terms of the system of equations

As the beam becomes more slender , the coefficient of the shear term

increases much faster than the one of the bending term

The higher coefficient of the shear term can be seen as penalty parameter rendering

the system stiffer

At the limit, the system becomes extremely stiff leading to a “locking” effect

For general cross sections, the thickness to length ratio affects the results similarly

3

2 5

12 12 1

zEI Eb EbkGAL L

h h

L LL

b s b sb sK K K K K K K

Ku f

0h

L



Shear lockingA first remedy: Reduced integration

Locking is essentially caused by the “spurious” linear term in the shear strain:

Ignoring the energy produced by that term would remove locking effects

A simple way of improving the performance of the element, is to simply reduce the

integration order used for the shear term

Using a single Gauss point can accurately integrate functions of order 2x1-1=1, thus the

quadratic terms produced by the “spurious” linear term cannot be accurately integrated

1 1xy z

dwa

dxb




Shear part of the stiffness matrix using full and reduced integration:

2 2

2 2

1 12 2

2 2

1 12 2

4

2 2

4

4 4

s

L L

L L

D

L LL

L

L L

L L L

r

sK

2 2

2 2

1 12 2

2 3 2 6

1 12 2

2 6 2 3

s

L L

L L L L

D

L LL

L L L L

sK

Reduced integrationFull integration




Resulting forces for a displacement vector corresponding to pure bending for the fully

integrated shear contribution of the stiffness matrix:

2 2 2

2

2 2

1 102 2

0

2 3 2 6 6

0 01 1

2 2

6

2 6 2 3

s s

L L

L L L L L

D D

L LL L

L

L L L L

sK u

1 1 2 2 0 0TT

z zw w u




Resulting forces for a displacement vector corresponding to pure bending for the shear

contribution of the stiffness matrix obtained using reduced integration:

2 2

2 2

1 12 2

0 0

02 4 2 4

0 01 1

2 2 0

2 4 2 4

s

L L

L L L L

D

L LL

L L L L

r

sK u

1 1 2 2 0 0TT

z zw w u

→ The shear term yields no forces!



Shear lockingAn alternative remedy: Assumed shear strain field

An alternative interpolation scheme can be used for the shear strain:

where:

DI

xy γN γ

γN

DIγ

→ a vector of interpolation functions used for the shear stain

→ a vector of shear strain values computed at specific interpolation points



Shear lockingAn alternative remedy: Assumed shear strain field

For the two noded beam element the shear strain should be constant, thus only one

point is required

The midpoint of the element is selected, where the strain assumes the value:

The interpolation function is set constant:

The resulting element is equivalent to the one using reduced integration

1

1 2 1 1 2

2

2

1 1 1 10

2 2 2

zDI z zxy

z

w

w w

wL L L

1N

The Timoshenko Beam · Timoshenko theory Equilibrium equations in terms of deflections Combining...

Documents

Transcript of The Timoshenko Beam · Timoshenko theory Equilibrium equations in terms of deflections Combining...