Introduction to FEM
13
Variational Formulation of Plane Beam Element
IFEM Ch 13 – Slide 1
Department of Engineering Mechanics
- PhD. TRUONG Tich Thien
13 Variational Formulation of Plane Beam Element IFEM Ch 13 - - PDF document
Department of Engineering Mechanics PhD. TRUONG Tich Thien Introduction to FEM 13 Variational Formulation of Plane Beam Element IFEM Ch 13 Slide 1 Department of Engineering Mechanics PhD. TRUONG Tich Thien Introduction to FEM Beams
Introduction to FEM
IFEM Ch 13 – Slide 1
Department of Engineering Mechanics
Introduction to FEM
IFEM Ch 13 – Slide 2
Department of Engineering Mechanics
Introduction to FEM
Neutral surface Compressive stress Tensile stress
IFEM Ch 13 – Slide 3
Department of Engineering Mechanics
Introduction to FEM
IFEM Ch 13 – Slide 4
Department of Engineering Mechanics
Introduction to FEM
z
Beam cross section Symmetry plane Neutral surface Neutral axis
q(x) L x, u y, v y, v
IFEM Ch 13 – Slide 5
Department of Engineering Mechanics
Introduction to FEM
Department of Engineering Mechanics
Introduction to FEM
u(x, y y y y y y y y y y ) v(x, )
− ∂v(x) ∂x v(x) v(x)
− v′ = − θ v(x)
∂x = − ∂2v ∂x2 = − d2v dx2 = − κ σ = Ee = −E d2v dx2 = −E κ
Plus equilibrium equation M'' = q (not used specifically in FEM)
IFEM Ch 13 – Slide 7
Department of Engineering Mechanics
Introduction to FEM
Transverse displacements Distributed transverse load Prescribed end displacements Curvature Bending moment Prescribed end loads
v(x) q(x) κ(x) M(x) κ = v'' M = EI κ M''=q
Kinematic Constitutive Displacement BCs Force BCs Equilibrium
IFEM Ch 13 – Slide 8
Department of Engineering Mechanics
Introduction to FEM
2
2
2
2
IFEM Ch 13 – Slide 9
Department of Engineering Mechanics
Introduction to FEM
i
θi θj
j
vi vj
IFEM Ch 13 – Slide 10
Department of Engineering Mechanics
Introduction to FEM
(e)
IFEM Ch 13 – Slide 11
Department of Engineering Mechanics
Introduction to FEM
N (e)
θi (ξ) = 1 8ℓ(1 − ξ)2(1 + ξ)
θ(e)
i
= 1 θ(e)
j
= 1 ξ = −1 ξ = 1 N (e)
θ j (ξ) = − 1 8ℓ(1 + ξ)2(1 − ξ)
v(e)
i
= 1 v(e)
j
= 1 N (e)
vi (ξ) = 1 4(1 − ξ)2(2 + ξ)
N (e)
vj (ξ) = 1 4(1 + ξ)2(2 − ξ)
IFEM Ch 13 – Slide 12
Department of Engineering Mechanics
Introduction to FEM
v(e) = [ N (e)
vi
N (e)
θi
N (e)
vj
N (e)
θj ]
v(e)
i
θ(e)
i
v(e)
j
θ(e)
j
= Nu(e) ξ = 2x ℓ − 1 N (e)
vi = 1 4(1 − ξ)2(2 + ξ),
N (e)
θi
= 1
8ℓ(1 − ξ)2(1 + ξ),
N (e)
vj = 1 4(1 + ξ)2(2 − ξ),
N (e)
θj
= − 1
8ℓ(1 + ξ)2(1 − ξ).
IFEM Ch 13 – Slide 13
Department of Engineering Mechanics
Introduction to FEM
2u(e)T K(e)u(e) − u(e)T f(e)
−1
2ℓ dξ
−1
2ℓ dξ
IFEM Ch 13 – Slide 14
Department of Engineering Mechanics
Introduction to FEM
K(e) = E I 2ℓ3 1
−1
36ξ 2 6ξ(3ξ − 1)ℓ −36ξ 2 6ξ(3ξ + 1)ℓ (3ξ − 1)2ℓ2 −6ξ(3ξ − 1)ℓ (9ξ 2 − 1)ℓ2 36ξ 2 −6ξ(3ξ + 1)ℓ symm (3ξ + 1)2ℓ2 dξ = E I ℓ3 12 6ℓ −12 6ℓ 4ℓ2 −6ℓ 2ℓ2 12 −6ℓ symm 4ℓ2
IFEM Ch 13 – Slide 15
Department of Engineering Mechanics
Introduction to FEM
Corroborates the result from hand integration.
Ke for prismatic beam:
EI
EI
EI
EI
EI
EI
EI
EI
EI
EI
EI
EI
EI
EI
EI
B={{6*Ξ,(3*Ξ-1)*l,-6*Ξ,(3*Ξ+1)*l}}/l^2; Ke=(EI*l/2)*Integrate[Transpose[B].B,{Ξ,-1,1}]; Ke=Simplify[Ke]; Print["Ke for prismatic beam:"]; Print[Ke//MatrixForm];
IFEM Ch 13 – Slide 16
Department of Engineering Mechanics
Introduction to FEM
2qℓ
−1
2qℓ
−1
1 4(1 − ξ)2(2 + ξ) 1 8ℓ(1 − ξ)2(1 + ξ) 1 4(1 + ξ)2(2 − ξ)
8ℓ(1 + ξ)2(1 − ξ)
1 2 1 12ℓ 1 2
12ℓ
"fixed end moments"
IFEM Ch 13 – Slide 17
Department of Engineering Mechanics
Introduction to FEM
ClearAll[q,l,Ξ] Ne={{2*(1-Ξ)^2*(2+Ξ), (1-Ξ)^2*(1+Ξ)*l, 2*(1+Ξ)^2*(2-Ξ),-(1+Ξ)^2*(1-Ξ)*l}}/8; fe=(q*l/2)*Integrate[Ne,{Ξ,-1,1}]; fe=Simplify[fe]; Print["fe^T for uniform load q:"]; Print[fe//MatrixForm];
fe^T for uniform load q:
l q
l q
l q
Force vector printed as row vector to save space.
IFEM Ch 13 – Slide 18
Department of Engineering Mechanics