Subido por Jorge Armando Anyosa Doza

3. CI587-Beams

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Introduction to The Finite Element
Method (FEM) - CI587
Unit 3
Topic: Euler-Bernoulli and
Timoshenko Beams
The Finite Element Method
Euler-Bernoulli and Timoshenko
Beams
Read: Chapter 5
CONTENTS
Euler-Bernoulli b e a m theory
➢ Governing Equations
➢ Finite element model
➢ Numerical examples
Timoshenko b e a m theory
➢ Governing Equations
➢ Finite element model
➢ Shear locking
➢ Numerical example
UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS
EBT AND TBT
Kinematics of the Linearized
Euler-Bernoulli Beam Theory
q(x )
Strains, displacements, and
rotations are small
z, w
z
f (x )
x
x
Undeformed Beam
𝑑𝑤
−
𝑑𝑥
Euler-Bernoulli
Beam Theory (EBT)
is based on the
assumptions of
(1) straightness,
(2) inextensibility, and
(3) normalitya
u
90∘
Deformed Beam
−
𝑑𝑤
𝑑𝑥
3
UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS
EBT AND TBT
Kinematics of Deformation in the
Euler-Bernoulli Beam Theory (EBT)
𝑑𝑤
−
𝑑𝑥
z
Displacement field (constructed
using the hypothesis)
𝑑𝑤
𝑢1 𝑥, 𝑧 = 𝑢 − 𝑧
,
𝑢2 = 0,
𝑢3 = 𝑤 𝑥
𝑑𝑥
u
Linear strains
w
x
z
 zz
z
y
x
−𝑧
zy
zx
 yx
xz
 xy
 xx
𝑑𝑤
𝑑𝑥
𝜕𝑢1 𝑑𝑢
𝑑2𝑤
𝜀11 = 𝜀𝑥𝑥 =
=
−𝑧 2 ,
𝜕𝑥1 𝑑𝑥
𝑑𝑥
𝜕𝑢1 𝜕𝑢3
𝑑𝑤 𝑑𝑤
𝛾𝑥𝑧 =
+
=−
+
= 0.
𝜕𝑥3 𝜕𝑥1
𝑑𝑥 𝑑𝑥
Constitutive relations
 yy  yz
Notation for stress components
𝑑𝑢
𝑑2𝑤
𝜎𝑥𝑥 = 𝐸𝜀𝑥𝑥 = 𝐸
− 𝐸𝑧 2
𝑑𝑥
𝑑𝑥
𝜎𝑥𝑧 = 𝐺𝛾𝑥𝑧 = 0
4
UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS
EBT AND TBT
Euler-Bernoulli Beam Theory
z, w
Beam
cross section
z
F0
q(x)
M0
•
x
y
•
𝑞 𝑥
M
cf
L
q(x )
V
𝜎𝑥𝑥 + ∆𝜎𝑥𝑥
𝜎𝑥𝑧
V
V
M
cf w
q(x )
𝜎𝑥𝑧 + ∆𝜎𝑥𝑧
𝜎𝑥𝑥
+•
N
M
𝑓 𝑥
𝑐𝑓 𝑤
𝑉 + ∆𝑉
𝑁 + ∆𝑁
𝑀 + ∆𝑀
f (x )
x
x
cf w
Definition of stress resultants
𝑁 = න 𝜎𝑥𝑥 𝑑𝐴,
𝐴
UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS
𝑀 = න 𝜎𝑥𝑥 ∙ 𝑧𝑑𝐴,
𝐴
𝑉 = න 𝜎𝑥𝑧 𝑑𝐴.
𝐴
EBT AND TBT
Euler-Bernoulli Beam Theory
(Continued)
Equilibrium equations
𝑑𝑁
+ 𝑓 = 0,
𝑑𝑥
𝑑𝑀
− 𝑉 = 0,
𝑑𝑥
𝑑𝑉
+ 𝑞 − 𝑐𝑓 𝑤 = 0
𝑑𝑥
න 1 ∙ 𝑑𝐴 = 𝐴,
න 𝑧 ∙ 𝑑𝐴 = 0,
න 𝑧 2 ∙ 𝑑𝐴 = 𝐼
𝐴
𝐴
𝐴
Stress resultants in terms of deflection
𝑑𝑢
𝑑2𝑤
𝑑𝑢
𝑁 = න 𝜎𝑥𝑥 𝑑𝐴 = න 𝐸
− 𝐸𝑧 2 𝑑𝐴 = 𝐸𝐴
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝐴
𝐴
𝑑𝑢
𝑑2𝑤
𝑑2𝑤
𝑀 = න 𝜎𝑥𝑥 × 𝑧 𝑑𝐴 = න 𝐸
− 𝐸𝑧 2 𝑧𝑑𝐴 = −𝐸𝐼 2
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝐴
𝐴
𝑑𝑀
𝑑
𝑑2𝑤
𝑉=
=
−𝐸𝐼 2
𝑑𝑥 𝑑𝑥
𝑑𝑥
UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS
EBT AND TBT
Euler-Bernoulli Beam Theory
(Continued)
Governing equations in terms of the displacements
𝑑
𝑑𝑢
−
𝐸𝐴
− 𝑓 = 0,
𝑑𝑥
𝑑𝑥
Bars
u
0≤𝑥≤𝐿
𝑑2
𝑑2𝑤
𝐸𝐼 2 + 𝑐𝑓 𝑤 − 𝑞 = 0,
𝑑𝑥 2
𝑑𝑥
0≤𝑥≤𝐿
Beams
w
z, w
F0
q(x)
M0
x
cf
L
Axial deformation of a bar
Bending of a beam
Axial displacement is uncoupled from
transverse displacement
UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS
EBT AND TBT
Weak Form of the EB Beam
Theory
Governing
equation
𝑑2
𝑑2 𝑤
𝐸𝐼 2 + 𝑐𝑓 𝑤 − 𝑞 = 0,
𝑑𝑥 2
𝑑𝑥
Weak form
𝑣𝑖
0≤𝑥≤𝐿
set of weight functions
𝑥𝑏
𝑑2
𝑑2𝑤
0 = න 𝑣𝑖
2 𝐸𝐼 𝑑𝑥 2 + 𝑐𝑓 𝑤 − 𝑞 𝑑𝑥
𝑑𝑥
𝑥𝑎
𝑥𝑏
=න
𝑥𝑎
𝑑𝑣𝑖 𝑑
𝑑 2 𝑤ℎ
𝑑
𝑑 2 𝑤ℎ
−
𝐸𝐼
+ 𝑐𝑓 𝑣𝑖 𝑤ℎ − 𝑣𝑖 𝑞 𝑑𝑥 + 𝑣𝑖
𝐸𝐼
𝑑𝑥 𝑑𝑥
𝑑𝑥 2
𝑑𝑥
𝑑𝑥 2
𝑥𝑏
𝑥𝑎
Implies that the primary variable is w
Secondary variable
(displacement)
(shear force)
𝑥𝑏
=න
𝑥𝑎
𝑑𝑣 𝑑
𝑑2𝑤
−
𝐸𝐼 2 + 𝑐𝑓 𝑣𝑤 − 𝑣𝑞 𝑑𝑥 − 𝑣 𝑥𝑎 𝑄1 − 𝑣 𝑥𝑏 𝑄3
𝑑𝑥 𝑑𝑥
𝑑𝑥
UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS
EBT AND TBT
Weak Form
𝑥𝑏
0=න
𝑥𝑎
𝑑 2 𝑣𝑖 𝑑 2 𝑤ℎ
𝐸𝐼
+ 𝑐𝑓 𝑣𝑖 𝑤ℎ − 𝑣𝑖 𝑞 𝑑𝑥 − 𝑣𝑖 𝑥𝑎 𝑄1 − 𝑣𝑖 𝑥𝑏 𝑄3
𝑑𝑥 2 𝑑𝑥 2
𝑥𝑏
0=න
𝑥𝑎
Secondary
variable
(Bending
Moment)
𝑥
Primary
Variable, 
Slope/rotation
(Continued)
𝑑𝑣𝑖
𝑑 2 𝑤ℎ 𝑏
+ −
∙ 𝐸𝐼
𝑑𝑥
𝑑𝑥 2 𝑥
𝑎
𝑑 2 𝑣𝑖 𝑑 2 𝑤ℎ
𝐸𝐼
+ 𝑐𝑓 𝑣𝑖 𝑤ℎ − 𝑣𝑖 𝑞 𝑑𝑥 − 𝑣𝑖 𝑥𝑎 𝑄1 − 𝑣𝑖 𝑥𝑏 𝑄3
𝑑𝑥 2 𝑑𝑥 2
− −
𝑑𝑣𝑖
𝑑𝑣𝑖
∙ 𝑄2 − −
∙ 𝑄4
𝑑𝑥 𝑥
𝑑𝑥 𝑥
𝑎
𝑑
𝑑 2 𝑤ℎ
𝑄1 =
𝐸𝐼
𝑑𝑥
𝑑𝑥 2
= −𝑉ℎ 𝑥𝑎 ,
𝑥𝑎
𝑑 2 𝑤ℎ
𝑄2 = 𝐸𝐼
= −𝑀ℎ 𝑥𝑎 ,
𝑑𝑥 2 𝑥
𝑎
UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS
𝑏
𝑑
𝑑 2 𝑤ℎ
𝑄3 = −
𝐸𝐼
𝑑𝑥
𝑑𝑥 2
= 𝑉ℎ 𝑥𝑏
𝑥𝑏
𝑑 2 𝑤ℎ
𝑄4 = −𝐸𝐼
= 𝑀ℎ 𝑥𝑎
𝑑𝑥 2 𝑥
𝑏
EBT AND TBT
Beam Element Degrees of
Freedom
Generalized displacements
∆3 = 𝑤 𝑥𝑏
∆1 = 𝑤 𝑥𝑎
2 =(xa) 1
2
he
4 =(xb)
Generalized forces
Q1 =−V(xa )
Q2 =−M(xa ) 1
he
UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS
Q3 =V(xb )
2 Q4 =M(xb )
EBT AND TBT
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) requires
the third derivative of w to exist.
UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS
EBT AND TBT
FINITE ELEMENT APPROXIMATION
Primary variables (serve as the nodal variables that must be
𝑑𝑤
continuous across elements)
𝑤, 𝜃 = −
𝑑𝑥
𝑤 𝑥 ≈ 𝑐0 + 𝑐1 𝑥 + 𝑐2 𝑥 2 + 𝑐3 𝑥 3
w
w, 𝜃
●
●
●
× ●
●
Hermite cubic polynomials
●
𝑤 𝑥𝑎 ≈ 𝑐0 + 𝑐1 𝑥𝑎 + 𝑐2 𝑥𝑎2 + 𝑐3 𝑥𝑎3 ≡ ∆1
𝑤 𝑥𝑏 ≈ 𝑐0 + 𝑐1 𝑥𝑏 + 𝑐2 𝑥𝑏2 + 𝑐3 𝑥𝑏3 ≡ ∆3
𝜃 𝑥𝑎
≈ −𝑐1 − 2𝑐2 𝑥𝑎 − 3𝑐3 𝑥𝑎2 ≡ ∆2
𝜃 𝑥𝑏 ≈ −𝑐1 − 2𝑐2 𝑥𝑏 − 3𝑐3 𝑥𝑏2 ≡ ∆4
𝑥 − 𝑥𝑎
𝜑1𝑒 = 1 − 3
ℎ𝑒
2
𝑥 − 𝑥𝑎
+2
ℎ𝑒
𝑥 − 𝑥𝑎
𝜑2𝑒 = −(𝑥 − 𝑥𝑎 ) 1 −
ℎ𝑒
𝑥 − 𝑥𝑎
𝑒
𝜑3 = 3
ℎ𝑒
𝜑4𝑒 = − 𝑥 − 𝑥𝑎
2
2
𝑥 − 𝑥𝑎
−2
ℎ𝑒
𝑥 − 𝑥𝑎
ℎ𝑒
2
3
3
𝑥 − 𝑥𝑎
−
ℎ𝑒
4
𝑤 𝑥 ≈ 𝑐0 + 𝑐1 𝑥 + 𝑐2 𝑥 2 + 𝑐3 𝑥 3 = ෍ ∆𝑗 𝜑𝑗 𝑥
𝑗=1
12
UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS
EBT AND TBT
HERMITE CUBIC
INTERPOLATION FUNCTIONS
𝜙1 𝑥
1
𝜙2 𝑥
slope = 0
slope = 1
x
he
slope = 0
he
x
𝜙3 𝑥
slope = 0
𝜙4 𝑥
1
he
slope = 1
x
slope = 0
he
x
13
UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS
EBT AND TBT
Finite Element Model
4
𝑤 𝑥 = ෍ ∆𝑗 𝜑𝑗 𝑥
𝑗=1
4
෍ 𝐾𝑖𝑗𝑒 ∆𝑗𝑒 − 𝐹𝑖𝑒 = 0
𝐾 𝑒 ∆𝑒 = 𝐹 𝑒
𝑗=1
𝑒
𝐾11
𝑒
𝐾21
𝑒
𝐾31
𝑒
𝐾41
𝑥𝑏
𝐾𝑖𝑗𝑒 = න
𝑥𝑎
𝑒
𝐾12
𝑒
𝐾22
𝑒
𝐾32
𝑒
𝐾42
𝑒
𝐾13
𝑒
𝐾23
𝑒
𝐾33
𝑒
𝐾43
𝑒
𝐾14
𝑒
𝐾24
𝑒
𝐾34
𝑒
𝐾44
∆1𝑒
∆𝑒2
∆𝑒3
∆𝑒4
𝑞1𝑒
𝑄1𝑒
𝑞2𝑒
𝑄2𝑒
=
+
𝑞3𝑒
𝑄3𝑒
𝑞4𝑒
𝑄4𝑒
𝑑 2 𝜙𝑖𝑒 𝑑 2 𝜙𝑗𝑒
𝐸𝐼
+ 𝑐𝑓 𝜙𝑖𝑒 𝜙𝑗𝑒 𝑑𝑥
2
2
𝑑𝑥 𝑑𝑥
w1e  e1
1e e2
w2e  e3
1
he
Q3e , q3e
Q1e , q1e
2  
e
2
𝑥𝑏
𝑒
𝐹𝑖 = න 𝜙𝑖𝑒 𝑞𝑑𝑥 + 𝑄𝑖𝑒
𝑥𝑎
e
4
1e e2
Q2e , q2e
1
2
Q4e , q4e
he
14
UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS
EBT AND TBT
Finite Element Model
For element-wise constant values of 𝐸𝑒 𝐼𝑒 and 𝑞𝑒 (and 𝑐𝑓 = 0)
6
2𝐸𝑒 𝐼𝑒 −3ℎ𝑒
𝐾𝑒 =
−6
ℎ𝑒3
−3ℎ𝑒
−3ℎ𝑒
2ℎ𝑒2
3ℎ𝑒
ℎ𝑒2
−6 −3ℎ𝑒
3ℎ𝑒
ℎ𝑒2
3ℎ𝑒
6
3ℎ𝑒 2ℎ𝑒2
𝐹𝑒 =
𝑞𝑒 ℎ𝑒
12
(Continued)
𝑄1
6
−ℎ𝑒
𝑄2
+
6
𝑄3
ℎ𝑒
𝑄4
Postprocessing
4
2𝜑𝑒
𝑑
𝑑2𝑤
𝑗
𝑀 𝑥 = −𝐸𝐼 2 = −𝐸𝐼 ෍ ∆𝑗𝑒
𝑑𝑥
𝑑𝑥 2
𝑗=1
4
3𝜑𝑒
𝑑
𝑑𝑀
𝑑
𝑑2𝑤
𝑗
𝑉 𝑥 =
=−
𝐸𝐼 2 = −𝐸𝐼 ෍ ∆𝑗𝑒
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑑𝑥 3
𝑗=1
4
2 𝑒
𝑑
𝜑𝑗 𝑥
𝑀 𝑥 𝑧
𝑑2𝑤
𝑒
𝜎𝑥 𝑥, 𝑧 = −
= 𝐸𝑧
=
𝐸𝑧
෍
∆
𝑗
𝐼
𝑑𝑥 2
𝑑𝑥 2
𝑗=1
UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS
EBT AND TBT
ASSEMBLY OF TWO BEAM ELEMENTS
connected end-to-end
6
−3ℎ
2𝐸𝐼 −6
ℎ3 −3ℎ
0
0
−3ℎ
2ℎ2
3ℎ
ℎ2
0
0
−3ℎ
−6
0 0
ℎ2
3ℎ
0 0
3ℎ − 3ℎ −6 −3ℎ
6+6
3ℎ − 3ℎ 2ℎ2 + 2ℎ2 3ℎ ℎ2
6 3ℎ
−6
3ℎ
3ℎ 1
−3ℎ
ℎ2
𝑄11
12
𝑄21
−𝐿
𝑞0 𝐿 24
𝑄31 + 𝑄12
+
48 0
𝑄41 + 𝑄22
12
𝑄32
𝐿
𝑄42
UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS
𝑈1
𝑈2
𝑈3
𝑈4
𝑈5
𝑈6
ℎ = 𝐿 Τ2
𝑄41 + 𝑄22 = 0
𝑄31 + 𝑄12 = 0
𝑄31 , 𝑞31
𝑄11 , 𝑞11
𝑄21 , 𝑞21 1
•
1
𝑄41 , 𝑞41
1
𝑄12 , 𝑞12
𝑄23 , 𝑞23
𝑄22 , 𝑞22
• 2 2•1
2
2
•3
2
EBT AND TBT
𝑄42 , 𝑞42
A SIMPLE EXAMPLE - 1
P
EA, EI
Exact solution (according to the
Euler-Bernoulli beam theory)
L
𝑃𝐿3
𝑤 𝐿 =
3𝐸𝐼
Given problem
One element discretization
P
EA, EI
•
•
U 1, U 2
L
U 3, U 4
Boundary conditions:
𝑈1 = 𝑈2 = 0,
𝑄3 = 𝑃,
𝑄4 = 0
𝐾 𝑒 ∆𝑒 = 𝑞 𝑒 + 𝑄𝑒
6
𝑞𝑒 ℎ𝑒 −ℎ𝑒
𝑒
𝐹 =
+
6
12
ℎ𝑒
6
2𝐸𝑒 𝐼𝑒 −3ℎ𝑒
𝑒
𝐾 =
−6
ℎ𝑒3
−3ℎ𝑒
0
−3ℎ𝑒
2ℎ𝑒2
3ℎ𝑒
ℎ𝑒2
ℎ𝑒 = 𝐿
𝑄1
𝑄2
𝑄3
𝑄4
−6 −3ℎ𝑒
3ℎ𝑒
ℎ𝑒2
3ℎ𝑒
6
3ℎ𝑒 2ℎ𝑒2
17
UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS
EBT AND TBT
A SIMPLE EXAMPLE - 1
Solution using Cramer ’s rule
(continued)
6𝐸𝐼
𝐿2
4𝑃𝐸𝐼
4𝐸𝐼
0
𝑃𝐿3
𝐿
𝐿
𝑃
=
⇒ 𝑈3 =
=
=
0
12 𝐸𝐼 2
3𝐸𝐼
12𝐸𝐼 6𝐸𝐼
𝐿4
𝐿3
𝐿2
6𝐸𝐼 4𝐸𝐼
𝐿
𝐿2
𝑃
12𝐸𝐼
𝐿3
6𝐸𝐼
𝐿2
6𝐸𝐼
𝐿2 𝑈3
4𝐸𝐼 𝑈4
𝐿
12𝐸𝐼
𝑃
𝐿3
6𝐸𝐼
0
6𝑃𝐸 𝐼 Τ𝐿2
𝑃𝐿3
𝐿2
𝑈4 =
=
=−
2
4
12 𝐸𝐼 Τ𝐿
2𝐸𝐼
12𝐸𝐼 6𝐸𝐼
𝐿3
𝐿2
6𝐸𝐼 4𝐸𝐼
EA, EI
𝐿
𝐿2
L
𝑈3
𝑃𝐿3
2𝐸𝐼
𝑃𝐿3
𝑤 𝐿 =
3𝐸𝐼
𝑑𝑤
𝑈4 = −
ቤ
𝑑𝑥 𝑥=𝐿
18
UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS
EBT AND TBT
EXAMPLE – 2: A determinate frame
structure
B
A
1
A
1•
a
1
b
P
1
x
2
•
2
C 3
F
F
Given structure
A
•
2 1
a
b
C
B
•2
P
Finite element discretization
B
Pb
•
P
F
P
F
Pb
•B
x
2
P
•C
F
UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS
EBT AND TBT
EXAMPLE – 2
A
•
x
1
B
a
Pb
Bar element, A B
•
(continued)
P
•B
x
2
F
𝑄𝐴
−1 𝑢𝐴
=
𝑄𝐵
1 𝑢𝐵
𝑃𝑎
𝑢𝐴 = 0, 𝑄𝐵 = −𝑃 ⇒ 𝑢𝐵 = −
𝐸1 𝐴1
Beam element, A B
P
P
•C
F
Bar element, B C
𝐹𝑏
𝑢𝑐 =
𝐸2 𝐴2
Beam element, B C
−6 −3𝑎
3𝑎 𝑎2
6
3𝑎
3𝑎 2𝑎2
𝑄1𝐴
𝑤𝐴
𝑄2𝐴
𝜃𝐴
𝑤𝐵 = 𝑄𝐵
1
𝜃𝐵
𝑄2𝐵
𝑤𝐴 = 0, 𝜃𝐴 = 0, 𝑄1𝐵 = −𝐹,
𝑄1𝐵 = 𝑃𝑏
−3𝑎
2𝑎2
3𝑎
𝑎2
Pb
Displacements at C
relative to point B
𝐸1 𝐴1 1
𝑎3 −1
6
2𝐸1 𝐼1 −3𝑎
−6
𝑎3
−3𝑎
F
𝑄1𝐵
𝑤𝐵
𝑄2𝐵
𝜃𝐵
𝑤𝐶 = 𝑄𝐶
1
𝜃𝐶
𝑄2𝐶
𝑤𝐵 = 0, 𝜃𝐵 = 0, 𝑄1𝐶 = −𝑃, 𝑄2𝐶 = 0
6
2𝐸2 𝐼2 −3𝑏
−6
𝑏3
−3𝑏
−3𝑏
2𝑏2
3𝑏
𝑏2
−6 −3𝑏
3𝑏 𝑏2
6
3𝑏
3𝑏 2𝑏2
20
UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS
EBT AND TBT
EXAMPLE – 3: Handling of a
vertical spring
z, w
𝑞0
𝓀𝑤(𝐿)
𝓀
1•
𝑄21
1
𝓀𝑤(𝐿)
𝓀
L
𝑄11
x
𝑈3
𝑄31
1
•2 𝑄4 = 0
𝑄31 = −𝑘𝑤 𝐿 = −𝑘𝑈3
𝑈4 ≠ 0
𝑈1 = 𝑈2 = 0
𝑘𝑤 𝐿 = 𝑘𝑈3
Alternatively,
1 −1
𝑘
−1 1
𝑢1𝑠
𝑄1𝑠
=
,
𝑢2𝑠
𝑄2𝑠
𝑢1𝑠 = 0, 𝑢2𝑠 = 𝑈3 ⇒ 𝑄2𝑠 = 𝑘𝑈3
21
UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS
EBT AND TBT
SOLUTION TO THE SPRINGSUPPORTED BEAM
6
2𝐸 𝐼 −3𝐿
𝐿3 −6
−3𝐿
−3𝐿
2𝐿2
3𝐿
𝐿2
−6 −3𝐿
3𝐿 𝐿2
6
3𝐿
3𝐿 2𝐿2
0
𝑈1 = 𝑤1
𝑄1
0 𝑞 𝐿 6
𝑈2 = 𝜃1
0
−𝐿 + 𝑄2 − 𝓀𝑈
=
3
𝑈3 = 𝑤2
𝑄3
12 6
𝐿
𝑄4 0
𝑈4 = 𝜃2
Boundary conditions
𝑤1 = 0, 𝜃1 = 0, 𝑄3 = −𝓀𝑈3 , 𝑄4 = 0
Condensed equations for the unknown generalized
nodal displacements
12𝐸𝐼
+𝓀
𝐿3
6𝐸𝐼
𝐿2
UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS
6𝐸𝐼
𝐿2 𝑈3 = 𝑞0 𝐿 6
4𝐸𝐼 𝑈4
12 −𝐿
𝐿
EBT AND TBT
HANDLING OF A POINT SOURCES
INSIDE AN ELEMENT
ℎ
s0
𝑞𝑖 = න 𝑞 𝑠 𝜑𝑖 𝑠 𝑑𝑠
1
0
𝑞 𝑠 = 𝐹0 𝛿(𝑠 − 𝑠0 )
0
ℎ
𝑑𝜑𝑖
𝑞𝑖 = න 𝑞 𝑠 𝜑𝑖 𝑠 𝑑𝑠 = −𝑀0
ቤ
, 𝑖 = 1,2,3,4
𝑑𝑠
0
𝑠=𝑠
s0 M 0
0
𝜑1 𝑠 = 1 − 3
𝑠
ℎ
+2
𝑠
ℎ
𝑠 2
𝑠 3
𝜑3 𝑠 = 3
−2
,
ℎ
ℎ
,
𝜑2
for F0 placed
at s 0 = 0.5h
UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS
𝜑2 𝑠 = −𝑠 1 −
q =
1
q2 = -
𝑠
ℎ
F0
2
F0
1
●
s
2
h
𝑞 𝑠 = 𝑀0 𝛿 ´ (𝑠 − 𝑠0 )
q3 =
F0h
8
1
2
𝑠 2 𝑠
−
ℎ
ℎ
𝑠 = −𝑠
2
h
𝑞𝑖 = න 𝑞 𝑠 𝜑𝑖 𝑠 𝑑𝑠 = 𝐹0 𝜑𝑖 𝑠0 , 𝑖 = 1,2,3,4
3
●
s
ℎ
2
F0
2
F0
2
q4 =
F0h
8
EBT AND TBT
EXAMPLE – 4: A simply-supported
beam
(a) Find the center deflection using one
Euler-Bernoulli element in full beam
6
2𝐸 𝐼 −3𝐿
𝐿3 −6
−3𝐿
−3𝐿
2𝐿2
3𝐿
𝐿2
−6 −3𝐿
3𝐿 𝐿2
6
3𝐿
3𝐿 2𝐿2
0
𝑈1 = 𝑤1
𝑈2 = 𝜃1
𝑈3 = 𝑤2
𝑈4 = 𝜃2
𝑄1
𝑞1
𝑞
𝑄2
= 𝑞2 +
𝑄3
3
𝑞4
𝑄4
𝑈2 =
𝐿2
2𝐿2
𝐹0 𝐿2
16𝐸𝐼
,
𝑈2
𝑈4
𝐹0 𝐿 1
=
8 −1
𝑈4 = −
𝐹0 𝐿2
x
L
q2 =
q3 =
F0L
F0L
=
q4
8
8
2
𝑤 𝑥 = 𝑈1 𝜑1 𝑥 + 𝑈2 𝜑2 𝑥 + 𝑈3 𝜑3 𝑥 + 𝑈4 𝜑4 (𝑥)
= 𝑈2 𝜑2 𝑥 + 𝑈4 𝜑4 𝑥
16𝐸𝐼
𝐹0 𝐿2
16𝐸𝐼
𝑥 2
−𝑥 1 −
+𝑥
𝐿
𝑥 2
𝑥
−
𝐿
𝐿
𝐹0 𝐿2
𝐿 𝐿
𝐹0 𝐿3
𝑤 0.5𝐿 =
− −
=−
16𝐸𝐼
8 8
64𝐸𝐼
UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS
F0
2
F0
2
q1 =
1
Condensed equations
2𝐸 𝐼 2𝐿2
𝐿3 𝐿2
F0
EBT AND TBT
EXAMPLE – 4: A simply-supported
beam
(b) Find the center deflection using one
Euler-Bernoulli element in half beam
0
6
16𝐸 𝐼 −1.5𝐿
−6
𝐿3
−3𝐿
−1.5𝐿
0.5𝐿2
1.5𝐿
0.25𝐿2
−6 −1.5𝐿
1.5𝐿 0.25𝐿2
6
1.5𝐿
1.5𝐿 0.5𝐿2
𝑈1 = 𝑤1
𝑈2 = 𝜃1
𝑈3 = 𝑤2
0
𝑈4 = 𝜃2
𝑄1
0
𝑄2
=
− 0.5𝐹0
𝑄3
𝑄4
F0
Condensed equations
16𝐸 𝐼 0.5𝐿2
𝐿3 1.5𝐿
x
L
1.5𝐿 𝑈2 = 0.5𝐹 0
0
𝑈3
−1
6
𝐹0 𝐿3 4 1.5𝐿 𝐹0 𝐿2
𝑈2 =
=
32𝐸𝐼 3𝐿3 1
16𝐸𝐼
𝐹0 𝐿3 4 0.5𝐿 𝐹0 𝐿3
𝑈2 =
=
32𝐸𝐼 3𝐿3 1
48𝐸𝐼
UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS
𝐹0
2
1
2
𝑈1 = 0,
𝑄2 = 0,
𝑈4 = 0,
𝑄3 = −0.5𝐹0
EBT AND TBT
EXERCISE PROBLEM
Problem: Develop weak form and the finite element model of the
following equation, where w and P are unknowns:
𝑑2
𝑑2 𝑤
𝑑2 𝑤
𝐸𝐼 2 + 𝑃 2 = 0,
𝑑𝑥 2
𝑑𝑥
𝑑𝑥
UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS
0<𝑥<𝐿
EBT AND TBT
EXERCISE PROBLEM
d
F0
Problem: Use the minimum number of E B T
elements to find the compression in the spring,
reactions at the fixed support, and spring force.
q0
Rigid loading frame
2E I
h
EI
h
UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS
Linear elastic
spring,
k
EBT AND TBT
TIMOSHENKO BEAM THEORY
and its Finite Element Model
➢ Governing Equations
➢ Finite element model
➢ Shear locking
➢ Numerical example
UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS
EBT AND TBT
Kinematics of Timoshenko Beam
Theory
𝑞 𝑥
z, w
z
𝑥, 𝑢
Undeformed Beam
𝑓 𝑥
x
90∘
𝑑𝑤
−
𝑑𝑥
Deformed Beams
Euler-Bernoulli
𝑑𝑤 Beam Theory (EBT)
Straightness,
−
𝑑𝑥
inextensibility, an d
normality
x
−
u
UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS
𝑑𝑤
𝑑𝑥
Timoshenko Beam
Theory (TBT)
Straightness an d
inextensibility
EBT AND TBT
Timoshenko Beam Theory
Kinematic Relations
z
𝑢1 𝑥, 𝑧 = 𝑢 𝑥 + 𝑧𝜙𝑥 (𝑥)
𝑢2 = 0,
u
𝑢3 (𝑥, 𝑧) = 𝑤(𝑥)
𝜕𝑢1 𝑑𝑢
𝑑𝜙𝑥
𝜀𝑥𝑥 =
=
+𝑧
,
𝜕𝑥1 𝑑𝑥
𝑑𝑥
w
z
x
𝜕𝑢1 𝜕𝑢3
𝑑𝑤
𝛾𝑥𝑧 =
+
= 𝜙𝑥 +
,
𝜕𝑥3 𝜕𝑥1
𝑑𝑥
Constitutive Equations
𝑑𝑢
𝑑𝜙𝑥
𝜎𝑥𝑥 = 𝐸𝜀𝑥𝑥 = 𝐸
+𝑧
𝑑𝑥
𝑑𝑥
𝑑𝑤
𝜎𝑥𝑧 = 𝐺𝛾𝑥𝑧 = 𝐺 𝜙𝑥 +
𝑑𝑥
UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS
EBT AND TBT
Timoshenko Beam Theory
𝑑𝑁
𝑑𝑉
+ 𝑓 = 0, −
− 𝑞 + 𝑐𝑓 𝑤 = 0,
𝑑𝑥
𝑑𝑥
Equilibrium Equations
(continued)
−
𝑑𝑀
+ 𝑉 = 0.
𝑑𝑥
Beam Constitutive Equations
𝑑𝑢
𝑑𝜙𝑥
𝑑𝑢
𝑁 = න 𝜎𝑥𝑥 𝑑𝐴 = න 𝐸
+𝑧
𝑑𝐴 = 𝐸𝐴
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝐴
𝐴
𝑑𝑢
𝑑𝜙
𝑑𝜙
𝑀 = න 𝜎𝑥𝑥 𝑧 𝑑𝐴 = න 𝐸
+𝑧
𝑧 𝑑𝐴 = 𝐸𝐼
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝐴
𝐴
𝑉 = 𝐾𝑠 න 𝜎𝑥𝑧 𝑑𝐴 = 𝐺𝐾𝑠 𝜙 +
𝐴
𝑑𝑤
𝑑𝑤
න 𝑑𝐴 = 𝐺𝐴𝐾𝑠 𝜙 +
𝑑𝑥
𝑑𝑥
𝐴
Governing Equations in terms of the displacements
𝑑
𝑑𝑤
−
𝐺𝐴𝐾𝑠 𝜙 +
𝑑𝑥
𝑑𝑥
𝑑
𝑑𝑤
−
𝐺𝐴𝐾𝑠 𝜙 +
𝑑𝑥
𝑑𝑥
UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS
+ 𝑐𝑓 𝑤 = 𝑞
+ 𝑐𝑓 𝑤 = 0
31
EBT AND TBT
WEAK FORMS OF TBT
Weak Form of Eq. (1)
𝑥𝑏
0=න
𝑥𝑎
𝑥𝑏
=න
𝑥𝑎
𝜈1 −
𝑑
𝑑𝑤
𝐺𝐴𝐾𝑠 𝜙 +
𝑑𝑥
𝑑𝑥
𝑑𝜈1
𝑑𝑤
𝐺𝐴𝐾𝑠 𝜙 +
𝑑𝑥
𝑑𝑥
+ 𝑐𝑓 𝑤 − 𝑞 𝑑𝑥
+ 𝑐𝑓 𝜈1 𝑤 − 𝜈1 𝑞 𝑑𝑥 − 𝜈1 ∙ 𝐺𝐴𝐾𝑠
𝑥𝑏
𝑑𝑤
𝜙+
𝑑𝑥
𝑥𝑎
𝑥𝑏
𝑑𝜈1
𝑑𝑤
=න ቊ
𝐺𝐴𝐾𝑠 𝜙 +
𝑑𝑥
𝑑𝑥
𝑥𝑎
+ 𝑐𝑓 𝜈1 𝑤
− 𝜈1 𝑞 ቋ 𝑑𝑥 − 𝜈1 𝑥𝑎 ∙ 𝐺𝐴𝐾𝑠 𝜙 +
𝑥𝑏
=න
𝑥𝑎
𝐺𝐴𝐾𝑠
𝑑𝑤
𝑑𝑥
− 𝜈1 (𝑥𝑏 ) ∙ 𝐺𝐴𝐾𝑠 𝜙 +
𝑥𝑎
𝑑𝑤
𝑑𝑥
𝑥𝑏
𝑑𝜈1
𝑑𝑤
𝜙+
+ 𝑐𝑓 𝜈1 𝑤 − 𝜈1 𝑞 𝑑𝑥 − 𝜈1 𝑥𝑎 ∙ 𝑄1 − 𝜈1 𝑥𝑏 ∙ 𝑄3
𝑑𝑥
𝑑𝑥
32
UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS
EBT AND TBT
WEAK FORMS OF TBT
(continued)
Weak Form of Eq. (2)
𝑥𝑏
0=න
𝑥𝑎
𝑥𝑏
=න
𝑥𝑎
𝑥𝑏
=න
𝑥𝑎
𝑥𝑏
0=න
𝑥𝑎
𝜈2 −
𝑑
𝑑𝜙
𝑑𝑤
𝐸𝐼
+ 𝐺𝐴𝐾𝑠 𝜙 +
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑥
𝑑𝜈2
𝑑𝜙
𝑑𝑤
𝐸𝐼
+ 𝐺𝐴𝐾𝑠 𝜈2 𝜙 +
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑑𝜙 𝑏
𝑑𝑥 − 𝜈2 ∙ 𝐸𝐼
𝑑𝑥 𝑥
𝑑𝜈2
𝑑𝜙
𝑑𝑤
𝐸𝐼
+ 𝐺𝐴𝐾𝑠 𝜈2 𝜙 +
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑑𝜙
𝑑𝜙
𝑑𝑥 − 𝜈2 𝑥𝑎 ∙ −𝐸𝐼
− 𝜈2 𝑥𝑏 ∙ 𝐸𝐼
𝑑𝑥 𝑥
𝑑𝑥 𝑥
𝑑𝜈2
𝑑𝜙
𝑑𝑤
𝐸𝐼
+ 𝐺𝐴𝐾𝑠 𝜈2 𝜙 +
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑑𝑥 − 𝜈2 𝑥𝑎 ∙ 𝑄2 − 𝜈2 𝑥𝑏 ∙ 𝑄4
𝑎
𝑎
𝑏
Total Potential Energy
𝑥𝑏
Π 𝑊, 𝜙𝑥 = න
𝑥𝑎
𝐸𝐼 𝑑𝜙
2 𝑑𝑥
2
𝐺𝐴𝐾𝑠
𝑑𝑤
+
𝜙+
2
𝑑𝑥
2
+
𝑐𝑓 2
𝑤 𝑑𝑥
2
𝑥𝑏
− න 𝑤𝑞𝑑𝑥 + 𝑤 𝑥𝑎 𝑄1 + 𝑤 𝑥𝑏 𝑄3 + 𝜙 𝑥𝑎 𝑄2 + 𝜙 𝑥𝑏 𝑄4
𝑥𝑎
33
UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS
EBT AND TBT
FINITE ELEMENT MODELS OF
TIMOSHENKO BEAMS
𝑚
Finite Element Approximation
w1
1
w1
1
s1
w2
2
he
w2
2
he
m=n=2
1
s1
w3
3
1
m=n=3
s2
he
s2
𝑛
𝑤 ≈ ෍ 𝑤𝑗 𝜓𝑗 𝑥 ,
𝑗=1
𝑗=1
2
s3
2
3
𝐾 11
𝐾 21
𝐾 12
𝐾 22
𝑤
𝑆
=
𝐹1
𝐹2
he
𝑥𝑏
𝑥𝑏
𝐾𝑖𝑗11 = න
𝑥𝑎
𝑑𝜓𝑖 𝑑𝜓𝑗
𝐺𝐴𝐾𝑠
+ 𝑐𝑓 𝜓𝑖 𝜓𝑗 𝑑𝑥
𝑑𝑥 𝑑𝑥
𝐾𝑖𝑗12 = න
𝑥𝑎
𝑥𝑏
22
𝐾𝑖𝑗 = න
𝑥𝑎
𝑑𝜑𝑖 𝑑𝜑𝑗
𝐸𝐼
+ 𝐺𝐴𝐾𝑠 𝜑𝑖 𝜑𝑗 𝑑𝑥
𝑑𝑥 𝑑𝑥
𝑥𝑏
𝑥𝑏
1
𝐹𝑖 = න 𝑞𝜓𝑖 𝑑𝑥 + 𝜓𝑖 (𝑥 𝑎 ൯𝑄1 + 𝜓𝑖 (𝑥 𝑏 ൱ 𝑄3 ,
𝑥𝑎
UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS
𝜙 ≈ ෍ 𝑆𝑗 𝜑𝑗 𝑥 ,
𝐾𝑖𝑗21 = න
𝑥𝑎
𝑑𝜓𝑖
𝐺𝐴𝐾𝑠
𝜑𝑗 𝑑𝑥 = 𝐾𝑖𝑗21
𝑑𝑥
𝐺𝐴𝐾𝑠 𝜑𝑖
𝑑𝜓𝑖
𝑑𝑥
𝑑𝑥
𝐹𝑖2 = 𝜑𝑖 (𝑥 𝑎 ൯𝑄2 + 𝜑𝑖 (𝑥 𝑏 ቁ 𝑄4
34
EBT AND TBT
Shear Locking in Timoshenko
Beams
(1) Thick beam experiences shear deformation,
𝑑𝑤
𝜙𝑥 ≠ −
𝑑𝑥
(2) Shear deformation is negligible in thin beams,
𝜙𝑥 = −
Linear interpolation of both: 𝑤, 𝜙𝑥
2
2
𝑤 ≈ ෍ 𝑤𝑗𝑒 𝜓𝑗𝑒 𝑥 ,
𝜙 ≈ ෍ 𝑆𝑗𝑒 𝜑𝑗𝑒 𝑥 ,
𝑗=1
𝑤 𝑥 = 𝑤1 𝜓1 𝑥 + 𝑤2 𝜓2 𝑥 ,
w2
w1
S2
2
1
he
𝑗=1
𝜙𝑥 𝑥 = 𝑆1 𝜓1 𝑥 + 𝑆2 𝜓2 𝑥
S1
2
1
𝑑𝑤
𝑑𝑥
he
Thus, in the thin beam limit it is not possible for the element to
realize the requirement
𝑑𝑤
𝜙𝑥 = −
𝑑𝑥
UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS
EBT AND TBT
SHEAR LOCKING-REMEDY
In the thin beam limit,  should become constant so that it
matches dw/dx. However, if  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  and d  /dx as constants. The terms highlighted should
be evaluated using “reduced integration”.
𝑥𝑏
𝐾𝑖𝑗11 = න
𝑥𝑎
𝑥𝑏
1
1
𝑑𝜓𝑖 𝑑𝜓𝑗
1
1
𝐺𝐴𝐾𝑠
+ 𝑐𝑓 𝜓𝑖 𝜓𝑗
𝑑𝑥
𝑑𝑥
1
𝑑𝜓
2
𝐾𝑖𝑗12 = න 𝐺𝐴𝐾𝑠 𝑖 𝜓𝑗 𝑑𝑥 = 𝐾𝑖𝑗21
𝑑𝑥
𝑥𝑎
1 𝑑𝜓 2
𝑥𝑏
𝑑𝜓
𝑗
1
2
𝐾𝑖𝑗22 = න
𝐸𝐼 𝑖
+ 𝐺𝐴𝐾𝑠 𝜓𝑖 𝜓𝑗
𝑑𝑥 𝑑𝑥
𝑥𝑎
UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS
𝑑𝑥
𝑑𝑥
EBT AND TBT
STIFFNESS MATRICES OF
TIMOSHENKO BEAM ELEMENT
(for constant EI and GA)
Reduced integration linear element (RIE) Linear approximation of
6
2𝐸𝑒 𝐼𝑒 −3ℎ𝑒
𝜇0𝑒 ℎ𝑒3 −6
−3ℎ𝑒
−3ℎ𝑒
ℎ𝑒2 𝜉𝑒
3ℎ𝑒
ℎ𝑒2 𝜁𝑒
𝜉𝑒 = 1.5 + 6Λ𝑒 ,
both w and f
𝑞1𝑒
𝑄1𝑒
𝑤1
𝑞2𝑒
𝑄2𝑒
𝜙1
𝑤2 = 𝑞 𝑒 + 𝑄𝑒
3
3
𝑒
𝜙2
𝑞4
𝑄4𝑒
𝐸𝑒 𝐼𝑒
𝑒
𝜁𝑒 = 1.5 − 6Λ𝑒 ,
Λ𝑒 =
2 , 𝜇0 = 12Λ 𝑒
𝐺𝑒 𝐴𝑒 𝐾𝑠 ℎ𝑒
−6 −3ℎ𝑒
3ℎ𝑒 ℎ𝑒2 𝜁𝑒
3ℎ𝑒
6
3ℎ𝑒 ℎ𝑒2 𝜉𝑒
Consistent interelement element (CIE) Hermite cubic approximation
of w and dependent quadratic approximation of f
6
2𝐸𝑒 𝐼𝑒 −3ℎ𝑒
𝜇𝑒𝑒 ℎ𝑒3 −6
−3ℎ𝑒
−3ℎ𝑒
2ℎ𝑒2 Σ𝑒
3ℎ𝑒
ℎ𝑒2 Θ𝑒
−6 −3ℎ𝑒
3ℎ𝑒 ℎ𝑒2 Θ𝑒
3ℎ𝑒
6
3ℎ𝑒 2ℎ𝑒2 Σ𝑒
Σ𝑒 = 1.0 + 3Λ𝑒 , Θ𝑒 = 1.0 − 6Λ𝑒 ,
UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS
𝑞1𝑒
𝑄1𝑒
𝑤1
𝑞2𝑒
𝑄2𝑒
𝜙1
𝑤2 = 𝑞 𝑒 + 𝑄𝑒
3
3
𝑒
𝜙2
𝑞4
𝑄4𝑒
𝐸𝑒 𝐼𝑒
Λ𝑒 =
, 𝜇 = 1 + 12Λ𝑒
𝐺𝑒 𝐴𝑒 𝐾𝑠 ℎ𝑒2 𝑒
EBT AND TBT
AN EXAMPLE of TBT
EA, EI
•
U1, U2
L
F
Exact solution (according
to the E- B beam theory)
•
𝐹𝐿3
𝑤 𝐿 =
3𝐸𝐼
U 3, U 4
One element discretization using the RIE element
6
2𝐸𝑒 𝐼𝑒 −3ℎ𝑒
𝜇0𝑒 ℎ𝑒3 −6
−3ℎ𝑒
𝜉𝑒 = 1.5 + 6Λ𝑒 ,
−3ℎ𝑒
ℎ𝑒2 𝜉𝑒
3ℎ𝑒
ℎ𝑒2 𝜁𝑒
−6 −3ℎ𝑒
3ℎ𝑒 ℎ𝑒2 𝜁𝑒
3ℎ𝑒
6
3ℎ𝑒 ℎ𝑒2 𝜉𝑒
𝑞1𝑒
𝑄1𝑒
𝑤1
𝑞2𝑒
𝑄2𝑒
𝜙1
𝑤2 = 𝑞 𝑒 + 𝑄𝑒
3
3
𝑒
𝜙2
𝑞4
𝑄4𝑒
𝜁𝑒 = 1.5 − 6Λ𝑒 ,
Λ𝑒 =
𝑈1 = 𝑈2 = 0, 𝑄3 = 𝐹,
𝑄4 = 0
𝐸𝑒 𝐼𝑒
𝑒
2 , 𝜇0 = 12Λ 𝑒
𝐺𝑒 𝐴𝑒 𝐾𝑠 ℎ𝑒
Boundary conditions:
UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS
EBT AND TBT
AN EXAMPLE (TBT)
2𝐸𝐼 6
𝜇0 𝐿3 3𝐿
(continued)
𝜇0 𝐿3
𝐹𝐿2 𝜉
12Λ𝐹𝐿3 1.5 + 6Λ
3𝐿 𝑈3
𝐹
=
⇒ 𝑈3 =
=
𝜉𝐿2 𝑈4
0
2𝐸𝐼 6𝐿2 𝜉 − 9𝐿2
6𝐸𝐼 12Λ
When
𝐸𝐼
1.5𝐹𝐿3 𝐹𝐿3
Λ=
= 0 ⇒ 𝑈3 =
=
𝐺𝐴𝐾𝑠 𝐿2
6𝐸𝐼
4𝐸𝐼
When Λ ≠ 0 then
𝐹𝐿3 1.5 + 6Λ
𝐹𝐿3
𝑈3 =
= 0.75 + 3Λ
6𝐸𝐼
3𝐸𝐼
𝐸𝐼
2 1 + 𝜈 𝐻2
1+𝜈
Λ=
=
=
𝐺𝐴𝐾𝑠 𝐿2
12𝐿2 𝐾𝑠
6𝐾𝑠
𝐻2
𝐿
2
1.3 𝐻
=
5 𝐿
2
𝐻
= 0.26
𝐿
2
39
UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS
EBT AND TBT
AN EXAMPLE (TBT)
(continued)
One element discretization using the CIE element
6
2𝐸𝐼 −3𝐿
𝜇𝐿3 −6
−3𝐿
Σ = 1.0 + 3Λ ,
−3𝐿
2𝐿2 Σ𝑒
3𝐿
𝐿2 Θ𝑒
−6 −3𝐿
3𝐿 𝐿2 Θ𝑒
3𝐿
6
3𝐿 2𝐿2 Σ𝑒
Θ = 1.0 − 6Λ,
Λ𝑒 =
𝑤1
𝑄1
𝜙1
𝑄2
=
𝑤2
𝑄3
𝜙2
𝑄4
𝐸𝐼
,
𝐺𝐴𝐾𝑠 𝐿2
𝜇 = 1 + 12Λ
Condensed equations for the unknown displacements
2𝐸𝐼 6
𝜇𝐿3 3𝐿
𝜇𝐿3
2𝐹𝐿2 Σ
𝜇𝐹𝐿3 Σ
3𝐿 𝑈3
𝐹
=
⇒ 𝑈3 =
=
2Σ𝐿2 𝑈4
0
2𝐸𝐼 12𝐿2 Σ − 9𝐿2
𝐸𝐼 12Σ − 9
40
UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS
EBT AND TBT
AN EXAMPLE (TBT)
When
Λ = 0 ⇒ Σ = 1 and;
𝜇=1
(continued)
then
𝜇𝐹𝐿3 Σ
𝐹𝐿3
𝑈3 =
=
𝐸𝐼 12 σ −9
3𝐸𝐼
𝜇𝐹𝐿3 Σ
𝐹𝐿3 1 + 3Λ 1 + 12Λ
𝐹𝐿3
When Λ ≠ 0, 𝑈3 =
= 𝑈3 =
= 1 + 3Λ
𝐸𝐼 12 σ −9
1 + 12Λ
3𝐸𝐼
𝐸𝐼
2 1 + 𝜈 𝐻2
1+𝜈
Λ=
=
=
𝐺𝐴𝐾𝑠 𝐿2
12𝐿2 𝐾𝑠
6𝐾𝑠
𝐻2
𝐿
2
1.3 𝐻
=
5 𝐿
2
𝐻
= 0.26
𝐿
2
41
UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS
EBT AND TBT
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 E B T and TBT
• Developed Finite element models of E B T
and TBT
• Discussed shear locking in Timoshenko beam
finite element
• Discussed assembly of beam elements
• Discussed examples
42
UNIVERSIDAD PERUANA DE CIENCIAS APLICADAS
EBT AND TBT
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