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Nonlinear Bending of Strait Beams
CONTENTS
The Euler-Bernoulli beam theory The Timoshenko beam theory
Governing Equations Weak Forms Finite element models Computer Implementation:
calculation of element matrices
Numerical examples
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Von Kármán NONLINEAR STRAINS
12
2 2 2
31 21 1 12 2 2
1
1
1 1 1
12
m m
i
jiij
j i
x
j
x
uuEx x
uE
u
x
ux x
uu ux x x
Green-Lagrange Strain Tensor Components
2
312
31
1 1
1 1
1
( ), ( )
xx xx
uu O Ox
ux x
x
E u
Order-of-magnitude assumption
Nonlinear Problems (1-D) : 3
NONLINEAR ANALYSIS OF EULER-BERNOULLI BEAMS
2 231
1 2 3
21 1
2 22
0( , ) , , ( )
,xx
dwu x z u z u u w xdx
u du d wzx dx dx
u dwx dx
Displacements and strain-displacement relations
•
z
y
Beam cross section
x
q(x) F0
L
z, w
M0
• •fc w
1 3ˆ ˆ) ,( x xu z w dwdx
u e e
M
V
q(x)
V
M•N N
f(x)
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PRINCIPLE OF VIRTUAL DISPLACEMENTSfor the Euler-Bernoulli beams
2
2
2
2
6
1
[( ) ( )]
[( ) ]
e
b
ea
b
a
b b
a a
eI xx xxV
x
xxx A
x
xx xxx
x xe e eE i ix x i
W dV
d u dw d w d wz dAdxdx dx dx dx
d u dw d w d wN M dxdx dx dx dx
W q wdx f udx Q
2
20 0, d dwNdN ddx
Mf qdx dx dx
Equilibrium equations
0 0 0, , d dwNdN dM dVf V qdx dx dx dx dx
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NONLINEAR ANALYSIS OF EULER-BERNOULLI BEAMS
2
20 0, d dwNdN ddx
Mf qdx dx dx
Equilibrium equations
Stress resultants in terms of deflection2 2
1 12 2
212
2
2
2 2
2 2
2
2
xxA A
xxA A
du d w duN dA E Ez dA EAdx dxdx
du d w d wM z dA E Ez z dA EIdx dx dx
dM d d wV EIdx
dw dwdx dx
dwdx
dx dx
σ
σ
= = − =
+ +
= × = + − = −
= = −
∫ ∫
∫ ∫
0 0 0, , d dwNdN dM dVf V qdx dx dx dx dx
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NONLINEAR ANALYSIS OF EULER-BERNOULLI BEAMS
Equilibrium equations in terms of displacements(u,w) 2
12
22 2122 2
0
0
d du dwEA fdx dx dx
d d w d dw du dwEI EA qdx dx dx dxdx dx
FF
( )u L
( )w L,x u
,z w
Clearly, transverse load induces both axial displacement u and transverse displacement w.
EULER-BERNOULLI BEAM THEORY(continued)
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11 1 1 1 4
2 22 2
22 2
2 22 2 3 2 5 6
0
0
( ) ( )
( ) ( )
b
a
b
a
a b
x
a bx
x
x
a bx x
dv N v f dx v x Q v x Qdx
d v dvd w dwEI N v q dxdx dx dx dx
dv dvv x Q Q v x Q Qdx dx
Weak forms2
12
du dwN EAdx dx
5 ( )ebQ V x2 ( )e
aQ V x
1 2
eh
3 ( )eaQ M x
6 ( )ebQ M x
1 ( )eaQ N x 4 ( )e
bQ N x
Beams 7
1 2,v u v w
8
2 ( )aQ V x
1 2eh
5 ( )bQ V x
6 ( )bQ M x3 ( )aQ M x
1 ( )aQ N x 4 ( )bQ N x
2 ( )aw x∆
1 2eh
5 ( )bw x∆
6 ( )bx∆ 3 ( )ax∆
1 ( )au x∆ 4 ( )bu x∆
Generalized displacements
Generalized forces
BEAM ELEMENT DEGREES OF FREEDOM
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9
FINITE ELEMENT APPROXIMATION
4
1 1
( ) ( ), ( ) ( ),n
j j j jj j
w x x u x u x ∆
Primary variables (serve as the nodal variables that must becontinuous across elements) , , dwu w
dxθ = −
Áe1 = 1 ¡ 3
µx ¡ xa
he
¶2
+ 2
µx ¡ xa
he
¶3
Áe2 = ¡(x ¡ xa)
µ1 ¡ x ¡ xa
he
¶2
Áe3 = 3
µx ¡ xa
he
¶2
¡ 2
µx ¡ xa
he
¶3
Áe4 = ¡(x ¡ xa)
"µx ¡ xa
he
¶2
¡ x ¡ xa
he
#
Hermite cubic polynomials
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HERMITE CUBIC INTERPOLATION FUNCTIONS
he
he
he
1
1
xhe
x
xx
x x
x x
slope = 1
slope = 0
slope = 0
slope = 0
slope = 1
slope = 0
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( )i x
1( )x
2( )x
3( )x
4( )x
11
FINITE ELEMENT MODEL
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2 4
1 1
11 12 1
21 22 2
11 12 12
21 1
( ) ( ), ( ) ( )
{ }[ ] [ ] { }{ }[ ] [ ] { }
, ,
,
b b
a a
b
a
j j j jj j
x xj ji iij ijx x
x jiij ix
u x u x w x x
uK K FK K F
d dd dK EA dx K EA dxdx dx dx dx
ddK EA
dwdx
d dx F fdx dx
wdx
∆
∆
1 4
2222
2 2
22 5 3 6
2
( ) ( )
,
( ) ( )
b
a
b b
a a
b
aa b
x
i i a i bx
x xj ji iij x x
xi i
i i i a i bxx x
dx x Q x Q
d dd dK EI dx EA dxdx dx dx dx
d dF q dx x Q x Q Q Qdx d
dwdx
x
Finite Element Equations5eQ2
eQ
1 23eQ 6
eQ1eQ 4
eQ
5e∆2
e∆
1 23e∆ 6
e∆1e∆ 4
e∆
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MEMBRANE LOCKING
20 1
2xxdu dwdx dx
( )q x
20 1
2
212
0xxdu dwdx dx
du dwdx dx
Membrane strain Beam on roller supports
2
Remedy
make to behave like a constantdwdx
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SOLUTION OF NONLINEAR EQUATIONSDirect Iteration
Direct Iteration Method
Non-Linear Finite Element Model [ ( )] assembled [ ( )]e e e eK F K U U F
th 1
1
Solution { } at iteration is known and solve for{ }
[ ({ } )]{ } { }
r r
r r
U r U
K U U F
K(U)U ≡ F(U)F
U
FC
UCU0
K(U0)
U1
K(U1)
U2
K(U2)
•
•• •
U3
°UC - Converged
solution
U0 - Initial guesssolution
Nonlinear Problems: (1-D) - 13
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SOLUTION OF NONLINEAR EQUATIONS(continued)
Direct Iteration Method
Convergence Criterion
Possible convergence
21
1
21
1
specified tolerance
NEQr rI I
INEQ
rI
I
U U
U
th 1
1
Solution { } at iteration is known and solve for{ }
[ ({ } )]{ } { }
r r
r r
U r U
K U U F
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SOLUTION OF NONLINEAR EQUATIONSNewton’s Iteration Method
Taylor’s series
21 1 1 2
2
1 2 1
1{ ( )} { ( )} ( ) ( )
2 !
{ ( )} ( ) ( ) ,
rrr r r r r r
rr r r r r
R RR U R U U U U U
U U
RR U U U O U U U U
U
1 st
tan
2tan
1 1
Requiring the residual { } to be zero at the 1 iteration, we have
[ ({ } )]{ } { } { } [ ( )] { }
The tangent matrix at the element level is
r
r r r r r r
ni
ij ip p ij j p
R r
K U U R F K U U
RK K F
1Residual, { } [ ({ } )]{ } { }r r rR K U U F
SOLUTION OF NONLINEAR EQUATIONSNewton’s Iteration (continued)
1[ ({ } )]{ } { } [ ( )] { } , { } { } { }r r r r r r rT F K
2 2
1 1 1 1
n nipi
ij ip p i ij p ijp pj j j
KRT K F K T
K(∆) ∆ − F ≡ R(∆)F
∆
FC
∆0
T(∆0)
T(∆1)
T(∆2)
•
••
∆C = ∆3
°∆C - Converged
solution
∆0 - Initial guesssolution
δ ∆1 δ ∆2
∆1 = δ ∆1 + ∆0 ∆2 = δ ∆2 + ∆0Nonlinear Problems: (1-D) - 16
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R®i =
2X
°=1
X
p=1
K®°ip ¢°
p ¡ F®i =
nX
p=1
K®1ip up +
4X
P=1
K®2iP ¢P ¡ F®
i
T®¯ij =
Ã@R®
i
@¢¯j
!= K®¯
ij +nX
p=1
@
@¢¯j
¡K®1
ip
¢up+
4X
P=1
@
@¢¯j
¡K®2
iP
¢¢P
T 11ij = K11
ij +nX
p=1
@K11ip
@ujup +
4X
P=1
@K12iP
@uj¢P
= K11ij +
nX
p=1
0 ¢ up +4X
P=1
0 ¢ ¢P
Summary of the N-R Method
Computation of tangent stiffness matrix
[T (f¢g(r¡1)]f¢gr = ¡fR(f¢g(r¡1))g
f¢gr = f¢g(r¡1) + f±¢g
18
Undeformed Beam
Euler-Bernoulli Beam Theory (EBT)Straightness, inextensibility, and normality
Timoshenko Beam Theory (TBT)Straightness and inextensibility
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z, w
x, u
x
z
dwdx−
dwdx−
dwdx−
φx
u
Deformed Beams
( )q x
( )f x
THE TIMOSHENKO BEAM THEORY
19
KINEMATICS OF THE TIMOSHENKO BEAM THEORY
z
xw
dwdx−
z
φ
zφ
u
Constitutive Equations
,xx xx xz xzE G
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Displacement field
1
2 3
( , ) ( ) ( ),0, ( , ) ( )
u x z u x z xu u x z w x
1 3ˆ ˆ)( xu z w u e e
231 1
21
212
31
3 12 2
xx xx
x
xz xz xz
x
uuEx x
ddu dw zdx dx dx
uuEx x
dwdx
20
Equilibrium Equations
Beam Constitutive Equations
212
21
212
2
xxx
A A
x xxx
A A
s xz s x s xA A
ddu dw duN dA E z dA EAdx dx dx dx
d ddu dwM z dA E z z dA EIdx dx dx dx
dwV K dA GK dA GAKx
wdx
d
d
dwdx
TIMOSHENKO BEAM THEORY (continued)
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0 0 0, ,dN dM dV d dwf V N qdx dx dx dx dx
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11 1 1 1 4
11 1 12 1 4
21
0 ( ) ( )
( ) ( )
b
a
b
a
x
a bx
x
a bx
dv N v f dx v x Q v x Qdx
dv duEA v f dx v x Q v x Qdx d
wdxxd
WEAK FORMS OF TBT
2 22
2 2 2 5
0
( ) ( )
b
a
x
s xx
a b
dv dvdw dwGAK v q dxdx dx dx dx
v x Q v x
N
Q
33
3 3 3 6
0
( ) ( )
b
a
xx
s xx
a b
dv d dwEI GAK v dxdx dx dx
v x Q v x Q
22
Finite Element Approximation
11 12 13 1
21 22 23 2
331 32 33
K K K FuK K K w F
S FK K K
(1) (2) (3)
1 1 1( ), ( ), ( )
pm n
j j j j j jj j j
u u x w w x S x
FINITE ELEMENT MODELS OFTIMOSHENKO BEAMS
2he
1 2he
1
w w s s221 1
3
he
1 3he
1
w1
2 2
w w s ss2 23 312m n= =
3m n= =JN Reddy
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SHEAR LOCKING IN TIMOSHENKO BEAMS
2he
1he
1 2
Linear interpolation of both , xw
1 1 2 2 1 1 2 2, xw( x ) w ( x ) w ( x ) ( x ) S ( x ) S ( x )
1w 2w1S 2S
(1) Thick beam experiences shear deformation,
(2) Shear deformation is negligible in thin beams,
xdwdx
xdwdx
In the thin beam limit it is not possible for the element to realize the requirement
xdwdx
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SHEAR LOCKING - REMEDY
In the thin beam limit, φ should become constant so that it matches dw/dx. However, if φ is a constant then the bending energy becomes zero. If we can mimic the two states (constant and linear) in the formulation, we can overcome the problem. Numerical integration of the coefficients allows us to evaluate both φ and dφ/dx as constants. The terms highlighted should be evaluated using “reduced integration”.
(2)(2)
(2)(
22
23 32
(3)(3
3)
(3))
(333 )
b
a
b
a
b
a
x
ij x
x
ij jix
x jii
jis
is j
s ij jx
K ... dx
K dx K
ddK EI dxd
ddGAKdx dx
dGAKd
x dK
x
x
GA
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0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0Load,
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
1.10
1.20D
efle
ctio
n,
Clamped-clamped
Pinned-pinned
q0
w0
NUMERICAL EXAMPLESPinned-pinned beam (EBT)
Nonlinear Problems: (1-D) - 25
• •
• •
q0
q0
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q
w L = Length, H = Height of the beam
= 100LH
= 50LH
= 10LH
(in.
)
(psi.)
Pinned-pinned beam (TBT)
Nonlinear Problems: (1-D) - 26
• •
q0
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Pinned-pinned beam (EBT, TBT)
0 1 2 3 4 5 6 7 8 9 10Load (lb/in),
0.0
0.2
0.4
0.6
0.8
1.0De
flect
ion
3
4(0.5 ) EHw w LL
=
0q
/ 100(TBT,EBT)L H =
/ 80(TBT,EBT)L H =
/ 50(TBT,EBT)L H =
w/ 10(TBT)L H =
/ 10(EBT)L H =
H = beam heightL = beam length
• • 0q
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Load, q0 (psi)0.0
0.1
0.2
0.3
L/H=100
L/H=10
EBT = Euler−Bernoulli beam theoryTBT = Timoshenko beam theory
EBTTBT
TBT
w =
wEH
3 /qL
4
Non
dim
ensi
onal
def
lect
ion
Nonlinear Problems: (1-D) - 28
Hinged-Hinged beam (EBT and TBT)
• •
q0
29
SUMMARY
In this lecture we have covered the following topics:• Derived the governing equations of the
Euler-Bernoulli beam theory• Derived the governing equations of the
Timoshenko beam theory• Developed Weak forms of EBT and TBT• Developed Finite element models of EBT
and TBT• Discussed membrane locking (due to the
geometric nonlinearity)• Discussed shear locking in Timoshenko beam
finite element• Discussed examples
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