The Finite Element Method - TAMU...
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The Finite Element Method
Read: Chapter 5
Euler-Bernoulli beam theory Governing Equations Finite element model Numerical examples
Timoshenko beam theory Governing Equations Finite element model Shear locking Numerical example
Euler-Bernoulli and Timoshenko Beams
CONTENTS
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KINEMATICS OF THE LINEARIZED EULER-BERNOULLI BEAM THEORY
Undeformed Beam
Euler-Bernoulli Beam Theory (EBT)is based on the assumptions of(1)straightness, (2)inextensibility, and(3)normality
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z,
x
x
z
dwdx−
dwdx−
w
u
Deformed Beam
( )q x
( )f x
Strains, displacements, and rotations are small
90
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z
xw
dwdx−z
dwdx−
z
u
Kinematics of Deformation in the Euler-Bernoulli Beam Theory (EBT)
1 2 30= − = =( , ) , , ( )dwu x z u z u u w xdx
Displacement field (constructed using the hypothesis)
21
11 21
31
3 1
0
∂= = = −∂∂∂= + = − + =
∂ ∂
ε ε
γ
,
.
xx
xz
u du d wzx dx dxuu dw dw
x x dx dx
Linear strains
2
2
0
= = −
= =
σ ε
σ γ
,xx xx
xz xz
du d wE E Ezdx dx
G
Constitutive relationsx
z yxzσ
zzσyzσ
yyσ zyσ
xyσxxσ
zxσyxσ
Notation for stress components
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Euler-Bernoulli Beam Theory
, , .σ σ σ= = ⋅ = xx xx xzA A A
N dA M z dA V dA
Definition of stress resultants
x
q(x) F0
L
z, w
M0
• •
z
y
Beamcross section
cf
+ MV
q(x)
VM •
fc w
( )q x
( )f xfc w
xxσ xx xxσ σΔ+xzσ
xz xzσ σΔ+
xΔ
( )q x
( )f xfc w
N N NΔ+
V V VΔ+
xΔ
M MΔ+M
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Equilibrium equations
0, 0, 0fdN dM dVf V q c wdx dx dx
+ = − = + − =
Euler-Bernoulli Beam Theory (Continued)
Stress resultants in terms of deflection2
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 dx dx
= = − =
= × = − = −
= = −
σ
σ
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20,1 ,A AA
z ddA A zA dA I⋅ = ⋅= =⋅
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x
q(x) F0
L
z, w
M0
Bending of a beam
Beams
cf
w
Axial deformation of a bar
Barsu
6
Governing equations in terms of the displacements
2 2
2 2
0 0
0 0
,
,f
d duEA f x Ldx dx
d d wEI c w q x Ldx dx
− − = < <
+ − = < <
Euler-Bernoulli Beam Theory (Continued)
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Axial displacement is uncoupled from transverse displacement
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Weak Form of the EB Beam Theory2 2
2 2 0 0,fd d wEI c w q x Ldx dx
+ − = < <
Weak form
22
2 2
2 2
2 2
0 b
a
bb
aa
xh
i f hx
xx
i h hf i h i ix
x
d wdv EI c w q dxdx dx
dv d w d wd dEI c v w v q dx v EIdx dx dxdx dx
é ùæ ö÷çê ú÷ç= + -÷ê úç ÷÷çè øê úë ûé ù é ùæ ö æ ö÷ ÷ç çê ú ê ú÷ ÷ç ç= - + - + ⋅÷ ÷ê ú ê úç ç÷ ÷÷ ÷ç çè ø è øê ú ê úë û ë û
ò
ò
Secondary variable (shear force)
Implies that the primary variable is w(displacement)
2
1 320 ( ) ( )b
a
x
f a bx
dv d d wEI c vw vq dx v x Q v x Qdx dx dx
é ùæ ö÷çê ú÷= - + - - -ç ÷ê úç ÷çè øê úë ûò
{ } set of weight functionsiv −
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Governing equation
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2 2
1 32 2
2
2
0 ( ) ( )b
a
b
a
xi h
f i h i i a i bx
x
i h
x
d v d wEI c v w v q dx v x Q v x Q
dx dx
dv d wEI
dx dx
é ùê ú= + - - -ê úê úë û
é ùæ ö÷çê ú÷+ - ⋅ç ÷ê úç ÷çè øê úë û
ò
Primary Variable, θSlope/rotation
Secondary variable(Bending Moment)
2 2
1 32 2
2 4
0 ( ) ( )b
a
a b
xi h
f i h i i a i bx
i i
x x
d v d wEI c v w v q dx v x Q v x Q
dx dxdv dv
Q Qdx dx
é ùê ú= + - - -ê úê úë û
æ ö æ ö÷ ÷ç ç÷ ÷- - ⋅ - - ⋅ç ç÷ ÷ç ç÷ ÷ç çè ø è ø
ò
2 2
1 32 2
2 2
2 42 2
( ), ( )
( ) , ( )
a b
a b
h hh a h b
x x
h hh a h b
x x
d w d wd dQ EI V x Q EI V xdx dxdx dx
d w d wQ EI M x Q EI M x
dx dx
é ù é ùæ ö æ ö÷ ÷ç çê ú ê ú÷ ÷ç ç= =- = - =÷ ÷ê ú ê úç ç÷ ÷÷ ÷ç çè ø è øê ú ê úë û ë ûæ ö æ ö÷ ÷ç ç÷ ÷ç ç= =- = - =÷ ÷ç ç÷ ÷÷ ÷ç çè ø è ø
Weak Form (Continued)
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1 ( )=− aQ V x
1 2eh
3 ( )= bQ V x
4 ( )= bQ M x2 ( )=− aQ M x
1 ( )aw xΔ =
1 2eh
3 ( )bw xΔ =
4 ( )bxθΔ =2 ( )axθΔ =
Generalized displacements
Generalized forces
Beam Element Degrees of Freedom
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FINITE ELEMENT APPROXIMATION:Some Remarks
Continuity requirement based on the weak form, which requires that the second derivative of w exists and square-integrable.
Continuity based on the primary variables, which requires carrying w and its first derivative as the nodal variables, requires cubic approximation w.
Post-computation of secondary variables (bending moment and shear force) requiresthe third derivative of w to exist.
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FINITE ELEMENT APPROXIMATION
2 30 1 2 3( )w x c c x c x c x» + + +
2 30 1 2 3 1
2 30 1 2 3 3
21 2 3 2
21 2 3 4
2 3
2 3
( )( )( )( )
a a a a
b b b b
a a a
b b b
w x c c x c x c xw x c c x c x c x
x c c x c xx c c x c x
q
q
Δ
Δ
Δ
Δ
» + + + º
» + + + º
»- - - º
»- - - º
42 3
0 1 2 31
( ) ( )j jj
w x c c x c x c x x=
» + + + = Då f
Primary variables (serve as the nodal variables that must becontinuous across elements) , dww
dxθ = −
φe1 = 1− 3x− xahe
2
+ 2x− xahe
3
φe2 = −(x− xa) 1− x− xahe
2
φe3 = 3x− xahe
2
− 2 x− xahe
3
φe4 = −(x− xa)x− xahe
2
− x− xahe
Hermite cubic polynomials
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● ● ● ●● ●,w q w ×
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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 xf1( )xf
2( )xf
3( )xf
4( )xf
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13
1 2ee21 Δ≡θ ee
42 Δ≡θ
eew 11 Δ≡ eew 32 Δ≡
eh
ee q,Q 22 ee q,Q 44
ee q,Q 11ee q,Q 33
1 2ee21 Δ≡θ
eh
4
1
( ) ( )j jj
w x xfΔ=
Ȍ
FINITE ELEMENT MODEL
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4
j=1
Keij∆
ej − F ei = 0 or [Ke]{∆e} = {F e}
⎡⎢⎣Ke11 Ke
12 Ke13 Ke
14
Ke21 Ke
22 Ke23 Ke
24
Ke31 Ke
32 Ke33 Ke
34
Ke41 Ke
42 Ke43 Ke
44
⎤⎥⎦⎧⎪⎨⎪⎩∆e1∆e2∆e3∆e4
⎫⎪⎬⎪⎭ =
⎧⎪⎨⎪⎩qe1qe2qe3qe4
⎫⎪⎬⎪⎭+⎧⎪⎨⎪⎩Qe1Qe2Qe3Qe4
⎫⎪⎬⎪⎭Keij =
xb
xa
EId2φeidx2
d2φejdx2
+ cf φeiφej dx F ei =
xb
xa
φei q dx+Qei
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14
For element-wise constant values of EeIe and qe:
[Ke] =2EeIeh3e
⎡⎢⎣6 −3he −6 −3he−3he 2h2e 3he h2e−6 3he 6 3he−3he h2e 3he 2h2e
⎤⎥⎦ {F e} = qehe12
⎧⎪⎨⎪⎩6−he6he
⎫⎪⎬⎪⎭+⎧⎪⎨⎪⎩Q1Q2Q3Q4
⎫⎪⎬⎪⎭
Finite Element Model (Continued)
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M(x) = −EI d2w
dx2= −EI
4
j=1
∆ejd2φejdx2
V (x) =dM
dx= − d
dxEId2w
dx2= −EI
4
j=1
∆ejd3φejdx3
σx(x, z) = −M(x)zI
= Ezd2w
dx2= Ez
4
j=1
∆ejd2φej(x)
dx2
Postprocessing
(and cf = 0):
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15
2EI
h3
⎡⎢⎢⎢⎢⎢⎣6 −3h −6 −3h 0 0−3h 2h2 3h h2 0 0−6 3h 6 + 6 3h− 3h −6 −3h−3h h2 3h− 3h 2h2 + 2h2 3h h2
0 0 −6 3h 6 3h0 0 −3h h2 3h 2h2
⎤⎥⎥⎥⎥⎥⎦
⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩
U1U2U3U4U5U6
⎫⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎭
+
⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩
Q11Q12
Q13 +Q21
Q14 +Q22
Q23Q24
⎫⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎭=q0L
48
⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩
12−L24012L
⎫⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎭+
•1 21 2
11
11, qQ
12
12, qQ
14
14, qQ
13
13, qQ
2 3
21
21 , qQ
22
22 , qQ
23
23 , qQ
24
24 , qQ
21• • •
Q13 +Q21 = 0, Q14 +Q
22 = 0
1
ASSEMBLY OF TWO BEAM ELEMENTSconnected end-to-end
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2/h L=
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Given problem
,EA EI
L
P
3
3( ) PLw L
EI=
Exact solution (according to the Euler-Bernoullibeam theory)
,EA EI
L
P
• •
1 2,U U 3 4,U U
{F e} = qehe12
⎧⎪⎨⎪⎩6−he6he
⎫⎪⎬⎪⎭+⎧⎪⎨⎪⎩Q1Q2Q3Q4
⎫⎪⎬⎪⎭
[Ke] =2EeIeh3e
⎡⎢⎣6 −3he −6 −3he−3he 2h2e 3he h2e−6 3he 6 3he−3he h2e 3he 2h2e
⎤⎥⎦
eh L=
0
Boundary conditions:
1 2 3 40 0, ,U U Q P Q= = = =
A SIMPLE EXAMPLE - 1
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[ ]{ } { } { }e e e eK q QD = +One element discretization
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A SIMPLE EXAMPLE – 1 (continued)
2
3 33 2
3 2 4
3 22 4
2
6
12 6 40 412 6 36 4 12
0
6 4
( / )( ) /
EIPL
EI EI EIU PPEI L PLL L LU
EI EI EIEI EI EI LU L LL LEI EIL L
é ù ì ü ì üï ï ï ïê ú ï ï ï ïï ïê ú ï ïï ï ï ï= = = =í ý í ýê ú é ùï ï ï ïê ú ï ï ï ï ê úë ûï ï ï ïê ú ï ïî þï ïî þë û
3
2 22
4 2 4
3 2
2
12
6 0 612 6 212
6 4
( / )( ) /
EI PLEI
PEI L PLLUEI EI EIEI L
L LEI EIL L
-= = =-
é ùê úë û
,EA EI
L 4 x LdwUdx ==-
3U
2
2PLEI
3
3( ) PLw L
EI=
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Solution using Cramer’s rule
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EXAMPLE – 2: A determinate frame structure
Given structure
a
b
F
A B
C P
ab
F
A B
C
2
1
12
3
1 2
2
1
P
Finite element discretization
• •
•
PP bx
1A B
F
••
F
PP b
x
P
2
C
B•
•
F
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EXAMPLE – 2 (continued)
Bar element, AB1 1
1 1
1 11 1
0
A A
B B
A B B,
u QE Au Qa
Pau Q P uE A
ì ü ì üé ù- ï ï ï ïï ï ï ïê ú =í ý í ýê ú ï ï ï ï-ë ûï ï ï ïî þ î þ
= =- =-
Beam element, AB1
2 221 1
31
2 22
1 2
6 3 6 33 2 326 3 6 3
3 3 2
0 0
AA
AA
BB
BB
B BA A, , ,
wa a Qa a a a QE I
wa a a Qa a a a Q
w Q F Q Pb
q
q
q
ì üì üé ù ï ï- - - ï ï ï ïï ïê ú ï ïï ï ï ïê ú ï ï- ï ïï ïï ï ï ïê ú =í ý í ýê ú ï ï ï ï- ï ï ï ïê ú ï ï ï ïê ú ï ï ï ï- ï ï ï ïë û ï ïî þ ï ïî þ= = =- =
PP b
x
1A B
F
••a
F
PP b
x
P
2
C
B•
•
F
Bar element, BC2 2
CFbu
E A=
12 2
22 23
12 2
2
1 2
6 3 6 33 2 326 3 6 33 3 2
0 0 0
BB
BB
CC
CC
C CB B, , ,
wb b Qb b b b QE I
wb b b Qb b b b Q
w Q P Q
q
q
q
ì üì üé ù ï ï- - - ï ï ï ïï ïê ú ï ïï ï ï ïê ú ï ï- ï ïï ïï ï ï ïê ú =í ý í ýê ú ï ï ï ï- ï ï ï ïê ú ï ï ï ïê ú ï ï ï ï- ï ï ï ïë û ï ïî þ ï ïî þ= = =- =
Beam element, BC
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Displacements at C relative to point B
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xk
L
q0
z, w
k w (L )
k
k w (L )
1 2• •1
U4 0¹
14 0Q =
11Q 1
2Q
1 2 0U U= =
3U
=- =-13 3
( )Q kw L kU
=3
( )kw L kU
13Q
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ì ü ì üé ù ï ï ï ï- ï ï ï ïï ï ï ïê ú = = = =í ý í ýê ú ï ï ï ï-ê ú ï ï ï ïë û ï ï ï ïî þ î þ
1 11 2 3 2 3
2 2
1 1, 0,
1 1
s ss s s
s s
u Qk u u U Q kU
u Q
Alternatively,
EXAMPLE – 3: Handling of a vertical spring
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JN Reddy Beams 21
SOLUTION TO THE SPRING-SUPPORTED BEAM
1 1 12 2
2 1 203
3 2 32 2
4 2 4
6 3 6 3 63 2 32
126 3 6 3 63 3 2
Boundary conditio
U w QL LU QL L L L Lq LEIU w QL L LU QL L L L L
q
q
ì ü ì üé ù ì ü=- - - ï ï ï ïï ïï ï ï ïï ïê ú ï ï ï ïï ïê ú ï ï ï ïï ï=- -ï ï ï ïï ïï ï ï ïê ú = +í ý í ý í ýê ú ï ï ï ï ï ï=- ï ï ï ï ï ïê ú ï ï ï ï ï ïê ú ï ï ï ï ï ï=- ï ï ï ï ï ïë û î þï ï ï ïî þ î þ
1 1 3 3 40 0 0ns, , ,w Q kU Qq= = =- =
3kU-
0
00
33 20
2 4
12 6 6
126 4
EI EI Uk q LL LEI EI
LUL L
é ù ì ü ì üï ï ï ïê ú+ ï ï ï ïï ïê ú ï ïï ï ï ï=ê ú í ý í ýï ï ï ïê ú ï ï ï ïê ú ï ï ï ï-ï ïî þï ïî þê úë û
Condensed equations for the unknown generalized nodal displacements
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HANDLING OF A POINT SOURCES INSIDE AN ELEMENT
2 3 2
1 2
2 3 2
3 4
( ) 1 3 2 , ( ) 1
( ) 3 2 , ( )
s s ss s s
h h h
s s s ss s s
h h h h
f f
f f
æ ö æ ö æ ö÷ ÷ ÷ç ç ç= - + = - -÷ ÷ ÷ç ç ç÷ ÷ ÷è ø è ø è øé ùæ ö æ ö æ ö÷ ÷ ÷ê úç ç ç= - = - -÷ ÷ ÷ç ç ç÷ ÷ ÷ê úè ø è ø è øë û
f= ò0( ) ( )
h
i iq q s s ds
f f= = =ò 0 00( ) ( ) ( ), 1,2, 3, 4
h
i i iq q s s ds F s i
1 2
0F= 01 2
Fq = 0
3 2
Fq
=- 02 8
F hq = 0
4 8
F hq
=0
0
for placedat 0.5Fs h
JN Reddy 22
ff
=
== =-ò 00
0
1, 2, 3, 4( ) ( ) ,h
ii i
s s
id
q q s s ds Mds
1 2
0F0s
hs
●
d= -0 0
( ) ( )q s F s s
1 20M
0s
hs
●
d= -0 0
( ) '( )q s M s s
![Page 23: The Finite Element Method - TAMU Mechanicsmechanics.tamu.edu/wp-content/uploads/2016/09/03_Ch5Beams.pdfJN Reddy The Finite Element Method Read: Chapter 5 Euler-Bernoulli beam theory](https://reader030.fdocuments.us/reader030/viewer/2022040113/5ed97a8a1b54311e7967a351/html5/thumbnails/23.jpg)
JN Reddy
xL
0F
EXAMPLE – 4: A simply-supported beam
(a) Find the center deflection using oneEuler-Bernoulli element in full beam
1 1 1 12 2
2 1 2 23
3 2 3 32 2
4 2 4 4
6 3 6 33 2 326 3 6 3
3 3 2
U w q QL LU q QL L L LEIU w q QL L LU q QL L L L
q
q
ì ü ì ü ì üé ù =- - - ï ï ï ï ï ïï ï ï ï ï ïê ú ï ï ï ï ï ïê ú ï ï ï ï ï ï=- ï ï ï ï ï ïê ú = +í ý í ý í ýê ú ï ï ï ï ï ï=- ï ï ï ï ï ïê ú ï ï ï ï ï ïê ú ï ï ï ï ï ï=- ï ï ï ï ï ïë û î þ î þ î þ
0
0= 0
1 2
Fq = 0
3 2
Fq
= 04 8
F Lq
1 2= 0
2 8
F Lq
0
8
F L
- 0
8
F L
2 22 0
3 2 24
1228 12
UL L F LEIUL L L
é ù ì ü ì üï ï ï ïï ï ï ïê ú =í ý í ýê ú ï ï ï ï-ï ïî þï ïî þë û
Condensed equations
2 20 0
2 416 16,F L F LU U
EI EI= =-
1 1 2 2 3 3 4 4
2 2 4 4
2 220
2 30 0
116
0 516 8 8 64
( ) ( ) ( ) ( ) ( )( ) ( )
( . )
w x U x U x U x U xU x U x
F L x x xx xEI L L L
F L F LL Lw LEI EI
f f f f
f f
= + + +
= +
ì üé ù é ùï ïæ ö æ ö æ öï ïï ïê ú ê ú÷ ÷ ÷ç ç ç= - - + -÷ ÷ ÷í ýç ç çê ú ê ú÷ ÷ ÷ç ç çï ïè ø è ø è øê ú ê úï ïë û ë ûï ïî þæ ö÷ç= - - =-÷ç ÷çè ø
0
0
![Page 24: The Finite Element Method - TAMU Mechanicsmechanics.tamu.edu/wp-content/uploads/2016/09/03_Ch5Beams.pdfJN Reddy The Finite Element Method Read: Chapter 5 Euler-Bernoulli beam theory](https://reader030.fdocuments.us/reader030/viewer/2022040113/5ed97a8a1b54311e7967a351/html5/thumbnails/24.jpg)
JN Reddy
xL
0F
EXAMPLE – 4: A simply-supported beam
(b) Find the center deflection using oneEuler-Bernoulli element in half beam
1 1 12 2
2 1 23
3 2 32 2
4 2 4
6 1 5 6 1 51 5 0 5 1 5 0 2516
6 1 5 6 1 53 0 25 1 5 0 5
. .. . . .
. .. . .
U w QL LU QL L L LEIU w QL L LU QL L L L
q
q
ì ü ì üé ù =- - - ï ï ï ïï ï ï ïê ú ï ï ï ïê ú ï ï ï ï=- ï ï ï ïê ú =í ý í ýê ú ï ï ï ï=- ï ï ï ïê ú ï ï ï ïê ú ï ï ï ï=- ï ï ï ïë û î þ î þ
0
0
0
2
F
==-
4
3 0
0,
0.5
U
Q F
1==
1
2
0,
0
U
Q
0
-0
0.5F
22
033
00 5 1 516 0 511 5 6
. . ..
UL LEI FUL L
é ù ì ü ì üï ï ï ïï ï ï ïê ú =í ý í ýê ú ï ï ï ï-ï ïî þï ïî þë û
Condensed equations
3 20 0
2 2
3 320 0
3 2
4 1 532 3 1 16
4 0 532 3 1 48
. ,
.
F L F LLUEI L EI
F L F LLUEI L EI
= =
= =
2
![Page 25: The Finite Element Method - TAMU Mechanicsmechanics.tamu.edu/wp-content/uploads/2016/09/03_Ch5Beams.pdfJN Reddy The Finite Element Method Read: Chapter 5 Euler-Bernoulli beam theory](https://reader030.fdocuments.us/reader030/viewer/2022040113/5ed97a8a1b54311e7967a351/html5/thumbnails/25.jpg)
JN Reddy Beams 25
EXERCISE PROBLEM
2 2 2
2 2 2 0 0,d d w d wEI P x Ldx dx dx
+ = < <
Problem: Develop weak form and the finite element model of the following equation, where w and P are unknowns:
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JN Reddy Beams 26
EXERCISE PROBLEM
dF0
hh k
Linear elasticspring,
Rigid loading frame
q0
2EI EI
Problem: Use the minimum number of EBT elements to find the compression in the spring, reactions at the fixed support, and spring force.
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JN Reddy Beams 27
Governing Equations Finite element model Shear locking Numerical example
TIMOSHENKO BEAM THEORYand its Finite Element Model
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28
Kinematics of Timoshenko Beam Theory
Undeformed Beam
Euler-Bernoulli Beam Theory (EBT)Straightness, inextensibility, and normality
Timoshenko Beam Theory (TBT)Straightness and inextensibility
JN Reddy
z, w
x, u
x
z
dwdx−
φx
u
dwdx−
dwdx−
Deformed Beams
( )q x
( )f x
90
![Page 29: The Finite Element Method - TAMU Mechanicsmechanics.tamu.edu/wp-content/uploads/2016/09/03_Ch5Beams.pdfJN Reddy The Finite Element Method Read: Chapter 5 Euler-Bernoulli beam theory](https://reader030.fdocuments.us/reader030/viewer/2022040113/5ed97a8a1b54311e7967a351/html5/thumbnails/29.jpg)
29
Timoshenko Beam Theory
Kinematic Relations
1
2 3
1
1
31
3 1
( , ) ( ) ( ),0, ( , ) ( )
,
x
xxx
xz x
u x z u x z xu u x z w x
du du zx dx dx
uu dwx x dx
f
fe
g f
= +
= =
¶= = +
¶
¶¶= + = +
¶ ¶
z
x
w
dwdx−
z
u
Constitutive Equations
xxx xx
xz xz x
dduE E zdx dx
dwG Gdx
fs e
s g f
æ ö÷ç= = + ÷ç ÷÷çè øæ ö÷ç= = + ÷ç ÷÷çè ø
JN Reddy
xzf
xf
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30
Equilibrium Equations
Beam Constitutive Equations
0 0 0fdN dV dMf , q c w , V .dx dx dx
+ = − − + = − + =
xxx
A A
xxA A
s xz s sA A
ddu duN dA E z dA EAdx dx dx
du d dM z dA E z z dA EIdx dx dx
dw dwV K dA GK dA GAKdx dx
fs
f fs
s f f
æ ö÷ç= = + =÷ç ÷ç ÷è ø
æ ö÷ç= = + =÷ç ÷÷çè ø
æ ö æ ö÷ ÷ç ç= = + = +÷ ÷ç ç÷ ÷÷ ÷ç çè ø è ø
ò ò
ò ò
ò ò
(1)
0 (2)
s f
s
d dwGAK c w qdx dx
d d dwEI GAKdx dx dx
f
ff
é ùæ ö÷çê ú- + + =÷ç ÷÷ê úçè øë ûæ ö æ ö÷ ÷ç ç- + + =÷ ÷ç ç÷ ÷÷ ÷ç çè ø è ø
Governing Equations in terms of the displacements
Timoshenko Beam Theory (Continued)
JN Reddy
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31
Weak Form of Eq. (1)
1
11 1 1
1
0b
a
bb
aa
b
a
x
s fx
xx
s f sxx
x
sx
d dwv GAK c w q dxdx dx
dv dw dwGAK c v w v q dx v GAKdx dx dx
dv dwGAKdx dx
f
f f
f
ì üé ùï ïæ öï ï÷çê ú= - + + -÷í ýç ÷÷ê úçï ïè øï ïë ûî þì üé ù é ùï ïæ ö æ öï ï÷ ÷ç çê ú ê ú= + + - - ⋅ +÷ ÷í ýç ç÷ ÷÷ ÷ê ú ê úç çï ïè ø è øï ïë û ë ûî þ
é æ ö÷ç= + ÷ç ÷÷çè ø
ò
ò
ò 1 1
1 1
11 1 1 1 1 3
( ) ( )
( ) ( )
a b
b
a
f
a s b sx x
x
s f a bx
c v w v q dx
dw dwv x GAK v x GAKdx dx
dv dwGAK c v w v q dx v x Q v x Qdx dx
f f
f
ì üùï ïï ïê ú + -í ýê úï ïï ïë ûî þé ù é ùæ ö æ ö÷ ÷ç çê ú ê ú- ⋅ - + - ⋅ +÷ ÷ç ç÷ ÷÷ ÷ê ú ê úç çè ø è øë û ë û
ì üæ öï ïï ï÷ç= + + - - ⋅ - ⋅÷í ýç ÷÷çï ïè øï ïî þò
WEAK FORMS OF TBT
JN Reddy
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32
Weak Form of Eq. (2)
2
22 2
22
0b
a
bb
aa
b
a
x
sx
xx
sx x
x
sx
d d dwv EI GAK dxdx dx dx
dv d dw dEI GAK v dx v EIdx dx dx dx
dv d dwEI GAK vdx dx dx
ff
f ff
ff
é ùæ ö æ ö÷ ÷ç çê ú= - + +÷ ÷ç ç÷ ÷÷ ÷ê úç çè ø è øë ûé ùæ ö æ ö é ù÷ ÷ç çê ú ê ú= + + - ⋅÷ ÷ç ç÷ ÷÷ ÷ê úç ç ê úè ø è ø ë ûë ûé æ ö æ ö÷ ÷ç ç= + +÷ ÷ç ç÷ ÷÷ ÷ç çè ø è øë
ò
ò
ò 2 2
22 2 2 2 4
( ) ( )
0 ( ) ( )
a b
b
a
a bx x
x
s a bx
d ddx v x EI v x EIdx dx
dv d dwEI GAK v dx v x Q v x Qdx dx dx
f f
ff
ù æ ö æ ö÷ ÷ç çê ú - ⋅ - - ⋅÷ ÷ç ç÷ ÷÷ ÷ê ú ç çè ø è øûé ùæ ö æ ö÷ ÷ç çê ú= + + - ⋅ - ⋅÷ ÷ç ç÷ ÷÷ ÷ê úç çè ø è øë û
ò
Total Potential Energy2 2
2
1 3 2 4
( )2 2 2
( ) ( ) ( ) ( )
b
a
b
a
x fsx x
x
a b a bx
cG A KE I d dww , w dxdx dx
w q dx w x Q w x Q x Q x Q
φφ φ
φ φ
Π = + + +
− + + + +
Weak Forms of TBT (continued)
JN Reddy
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33
Finite Element Approximation
{ }{ }
{ }{ }
11 12 1
21 22 2
K K FwSK K F
é ù ì üé ù é ù ï ïì ü ï ïï ïê úê ú ê úë û ë û ï ïï ïê ú =í ý í ýï ï ï ïê úé ù é ù ï ï ï ïî þê úê ú ê ú ï ïë û ë û î þë û
11 12 21
22 21
1 21 3 2( ) ( ) ( ) ( )
b b
a a
b b
a a
b
a
x xji iij s f i j ij s j jix x
x xj jiij s i j ij s ix x
x
i i i a i b i i a i bx
dd dK GAK c dx, K GAK dx Kdx dx dx
d ddK EI GAK dx, K GAK dxdx dx dx
F q dx x Q x Q , F x Q x
yy yy y j
j yjj j j
y y y j j
æ ö÷ç ÷ç= + = =÷ç ÷ç ÷çè øé ùê ú= + =ê úê úë û
= + + = +
ò ò
ò ò
ò 4Q
1 1( ), ( )
m n
j j j jj j
w w x S xy f j= =
» »å å
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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34
Shear Locking in Timoshenko Beams
Linear interpolation of both , :xw f
1 1 2 2 1 1 2 2, xw( x ) w ( x ) w ( x ) ( x ) S ( x ) S ( x )y y f y y» + » +
2he
1he
1 21w 2w
1S 2S
(1) Thick beam experiences shear deformation,
(2) Shear deformation is negligible in thin beams,
xdwdx
f ¹-
xdwdx
f =-
Thus, in the thin beam limit it is not possible for the element to realize the requirement
xdwdx
f =-JN Reddy
2 2
1 1( ), ( )e e e e
j j x j jj j
w w x S xy f y= =
» »å å
![Page 35: The Finite Element Method - TAMU Mechanicsmechanics.tamu.edu/wp-content/uploads/2016/09/03_Ch5Beams.pdfJN Reddy The Finite Element Method Read: Chapter 5 Euler-Bernoulli beam theory](https://reader030.fdocuments.us/reader030/viewer/2022040113/5ed97a8a1b54311e7967a351/html5/thumbnails/35.jpg)
35
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”.
(1)(1)11 (1) (1)
(1)12 (2) 21
(2)(2)22 (2) (2)
b
a
b
a
b
a
x jiij s f i jx
xi
ij s j jix
x jiij s i jx
ddK GAK c dxdx dx
dK GAK dx Kdx
ddK EI GAK dxdx dx
yyy y
yy
yyy y
æ ö÷ç ÷ç= + ÷ç ÷ç ÷çè ø
= =
é ùê ú= +ê úê úë û
ò
ò
ò
JN Reddy
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JN Reddy Beams 36
STIFFNESS MATRICES OF TIMOSHENKO BEAM ELEMENT
(for constant EI and GA)
1 1 12 2
1 2 23
20 3 32 2
2 4 4
6 3 6 33 326 3 6 3
3 3
e ee e
e ee e e e e ee e
e e ee ee
e ee e e e e e
h h w q Qh h h h q QE I
h h wh q Qh h h h q Q
x z fm
z x f
ì ü ì üé ù ì ü- - - ï ï ï ïï ï ï ï ï ïï ïê ú ï ï ï ïï ï ï ï ï ïê ú ï ï- ï ï ï ïï ïï ï ï ï ï ïê ú = +í ý í ý í ýê ú ï ï ï ï ï- ï ï ï ï ïê ú ï ï ï ï ïê ú ï ï ï ï ï- ï ï ï ï ïê ú ï ïî þë û ï ï ïî þ î
021 5 6 1 5 6 12. , . , , ee ee e e e e e
e e s e
E IG A K h
x z m
ïïïïïïþ
= + L = - L L = = L
Reduced integration linear element (RIE)
1 1 12 2
1 2 23
2 3 32 2
2 4 4
6 3 6 33 32
2
26 3 6 3
3 3
e ee e
e ee e e e e ee e
e ee ee e
e ee e e e e e
h h w q Qh h h h q QE I
h h wh q Qh h h h q Q
fm
f
ì ü ì üé ù ì ü- - - ï ï ï ïï ï ï ï ï ïï ïê ú ï ï ï ïï ï ï ï ï ïê ú ï ï- S Q ï ï ï ïï ïï ï ï ï ïê ú = +í ý í ý í ýê ú ï ï ï ï ï- ï ï ï ï ïê ú ï ï ï ï ïê ú ï ï ï ï ï- Q S ï ï ï ï ïê ú ï ïî þë û ï ï ïî þ î
21 0 1 0 6 1 13 2. , . , ,e ee e e e e e e
e e s e
E IG A K h
m
ïïïïïïïþ
S = + L Q = - L L = = + L
Consistent interelement element (CIE)
Linear approximation of both w and f
of w and dependent quadratic approximation of Hermite cubic approximation
f
![Page 37: The Finite Element Method - TAMU Mechanicsmechanics.tamu.edu/wp-content/uploads/2016/09/03_Ch5Beams.pdfJN Reddy The Finite Element Method Read: Chapter 5 Euler-Bernoulli beam theory](https://reader030.fdocuments.us/reader030/viewer/2022040113/5ed97a8a1b54311e7967a351/html5/thumbnails/37.jpg)
3
3( ) FLw L
EI=
Exact solution (according to the E-B beam theory)
One element discretization using the RIE element
,EA EI
L
F
• •
1 2,U U 3 4,U U
Boundary conditions:
1 2 3 40 0, ,U U Q F Q= = = =
AN EXAMPLE of TBT
JN Reddy 37
1 1 12 2
1 2 23
20 3 32 2
2 4 4
6 3 6 33 326 3 6 3
3 3
e ee e
e ee e e e e ee e
e e ee ee
e ee e e e e e
h h w q Qh h h h q QE I
h h wh q Qh h h h q Q
x z fm
z x f
ì ü ì üé ù ì ü- - - ï ï ï ïï ï ï ï ï ïï ïê ú ï ï ï ïï ïê ú ï ï ï ïï ï- ï ï ï ïï ï ï ï ï ïê ú = +í ý í ý í ýê ú ï ï ï ï ï ï- ï ï ï ï ï ïê ú ï ï ï ï ï ïê ú ï ï ï ï ï- ï ï ï ï ïê ú î þë û ï ï ïî þ î þ
021 5 6 1 5 6 12. , . , , ee ee e e e e e
e e s e
E IG A K h
x z m
ïïï
= + L = - L L = = L
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AN EXAMPLE (TBT) (continued)
( )
( )
3 2 33 0
33 2 2 240
3 3
3
32
3
3
6 32 12 1 5 62 6 123 0 6 9
1 506 4
1 5 036
75 36
( . )( )
.When too stiff
( . )When 0, then ( . )
s
UL F LEI FL FLUUL EI EIL L L L
EI FL FLUGAK L EI EI
FLUEI
EI
FLE
GA
I
m xm x x
ì üé ù ì üï ï ï ï L + Lï ï ï ïê ú = = =í ý í ýê úï ï ï ï L-ï ïë û î þï ïî þ
L = = = =
+ LL ¹ = =
L =
+ L
2 2 22
2 2
2 1 1 1 3 0 2612 6 5( ) ( ) . .
s s s
H H H HK L L K K L L L
n n æ ö æ ö æ ö+ + ÷ ÷ ÷ç ç ç= = = =÷ ÷ ÷ç ç ç÷ ÷ ÷ç ç çè ø è ø è ø
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One element discretization using the CIE element
AN EXAMPLE of TBT
JN Reddy 39
1 12 2
1 23
2 32 2
2 4
2
6 3 6 33 2 326 3 6 3
3 3 2
1 0 3 1 0 6 1 12. , . , ,s
L L w QL L L L QEI
L L wL QL L L L Q
E IG A K L
fm
f
m
ì üé ù ì ü ï ï- - - ï ï ï ïï ïê ú ï ïï ï ï ïê ú ï ï- S Q ï ïï ïï ï ï ïê ú =í ý í ýê ú ï ï ï ï- ï ï ï ïê ú ï ï ï ïê ú ï ï ï ï- Q S ï ï ï ïê ú ï ïë û î þ ï ïî þ
S = + L Q = - L L = = + L
( )3 2 3
333 2 2 2
4
6 32 22 12 93 2 0 12 9 ( )
UL FEI L FL FLUUL EI EIL L L L
m mm
ì üé ù ì üï ï ï ï S Sï ï ï ïê ú = = =í ý í ýê úï ï ï ï S-S S-ï ïë û î þï ïî þ
Condensed equations for the unknown displacements
![Page 40: The Finite Element Method - TAMU Mechanicsmechanics.tamu.edu/wp-content/uploads/2016/09/03_Ch5Beams.pdfJN Reddy The Finite Element Method Read: Chapter 5 Euler-Bernoulli beam theory](https://reader030.fdocuments.us/reader030/viewer/2022040113/5ed97a8a1b54311e7967a351/html5/thumbnails/40.jpg)
AN EXAMPLE (TBT) (continued)
3 3
3
3 3
3
2
3
2 22
2 2
0 1 112 9 3
1 3 1 1212 9 3 1 12
2 1 1 1 3 0 261
1 3
2 6
3
5
When and ; then( )
( )( )When 0, ( ) ( )
( ) ( ) .
( )
.s s s
FL FLUEI EI
FL FLUEI EI
EI H H H HGAK L L K K L L L
FLEI
mm
m
n n
SL = S= = = =
S-
S + L + LL ¹ = = =
S- + L
æ ö æ ö æ ö+ + ÷ ÷ ÷ç ç çL = = = = =÷ ÷ ÷ç ç ç÷ ÷ ÷ç ç çè è
L
ø ø è ø
+
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41
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 theTimoshenko beam theory
• Developed Weak forms of EBT and TBT• Developed Finite element models of EBT
and TBT• Discussed shear locking in Timoshenko beam
finite element• Discussed assembly of beam elements• Discussed examples
JN Reddy