fea unit wise imp.formulae.docx
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ALPHA COLLEGE OF ENGINEERINGTHIRUMAZHISAI, CHENNAI – 600124
DEPARTMENT OF MECHANICAL ENGINEERING
ME 6603 - FINITE ELEMENT ANALYSIS
UNIT WISE IMPORTANT FORMULAE
UNIT – I (INTRODUCTION)
1. Initial and Boundary value Problems:
(i) When two roots (m1, m2) are real and un-equal
The Complementary function y(x) = C1em1x+ C2em2x
(ii) When two roots (m1, m2) are real and equal (m1=m2=m)
The Complementary function y(x) = (C1+ C2x) emx
(iii) When two roots are having real and imaginary part (α±iβ)
The Complementary function y(x) = eαx (C1Cosβx+ C2Sinβx)
2. Weighted Residual Methods
(i) Point Collocation method
Residual(R) = 0
(ii) Sub-domain Collocation method
ʃ R dx = 0
(iii) Least Square Method
ʃ R2 dx = 0 (or) ʃ R (dR/da)dx = 0
(iv) Galerkin Method
ʃ wi R dx = 0
1
3. Rayleigh-Ritz Method
Total potential energy = Strain Energy – Work done by external forces
π = U – H
(i) For beam problem,
U = (EI/2) ʃ (d2y/dx2)2dx
H = ʃ w y dx (udl load) H = W ymax (point load) y = a1 Sin (πx/l) + a2 Sin (3πx/l)
B.M (M)= EI (d2y/dx2)
(ii) For bar Problem,
U = (EA/2) ʃ (du/dx)2dx
H = Fu (or) Pu y = a0+a1x+a2x2
(iii) For Spring Problem
U = ½ K 𝛿2 (Where, 𝛿 = u2-u1) H = Fu
UNIT – II2
ONE DIMENSIONAL FINITE ELEMENT ANALYSIS
1-D BAR ELEMENT,
1. Linear Polynomial Equation,u = ao+a1x
2. Shape functions
N1 = l−xl
N2 = xl
3. Stiffness matrix,
[K ]=∫V
❑
[B ]T [D ] [B ]dv
Where, [B] → Strain displacement relationship matrix.[D] → Elasticity matrix or Stress-strain relationship matrix.[D] = [E] = E = Young’s modulus. dv = A dx
∴Stiffness matrix [K ]= AEl [ 1 −1
−1 1 ]4. General FEA equation is,
{F} = [K] {u}Where,
{F} is an element force vector [Column matrix].[K] is a stiffness matrix [Row matrix].{u} is a nodal displacement [Column matrix].
⇒{F1F2}= AE
l [ 1 −1−1 1 ]{u1
u2}5. 1D Displacement equation,
u=N1u1+N2u2
6. Force vector due to self weight,
[Fe ]= ρAl2 {11}
7. Reaction force, [R ]=[K ] {u }− {F }
3
8. Stress,
σ=E dudx
Where,
E = Young’s modulus
dudx =
u2−u1l
9. Temperature effect,Force,
{F }=EAα ∆T {−11 }
Stress,
σ=E( dudx )−Eα ∆T
Where,
A = Area of cross section of bar element.∆T = Temperature difference.α = Coefficient of thermal expansion.
TRUSS ELEMENTS,
1. Stiffness matrix,
[K ]= AE¿ [ l ² lm −l ² −lm
lm m ² −lm −m²−l ²−lm
−lm−m ²
l ² lmlm m²]
Where,A = Area of the truss elementE = Young’s modulus of elementle = Equivalent length
l=cosθ= x 2−xı¿
m=sin θ= y2− yı¿
¿=√ (x 2−x 1 )2+( y 2− y 1) ²2. Strain energy,
U=12
{u }T {u }[K ]
4
3. Finite element general equation, {F }=[K ]{u }
Where,[K] = stiffness matrix{U} = nodal displacement matrix
4. Stress,
σ= E¿ [−l −m l m ]{u1
u2
u3
u4}
SPRINGS
1. Stiffness matrix,
[K ]=k [ 1 −1−1 1 ]
2. Tensile force,T=k .∆u
Where,
k = spring constant
∆u = change in deformation
∆u=u2−u1
BEAMS
1. Shape functions,
N 1=1L3 (2 x3−3 x2 L+x L3 )
N 2=1L3 (x3L−2x2 L2+x L3 )
N 3=1L3 (−2 x3+3x2 L )
N 4=1L3 (x3 L−x2L2 )
2. Stiffness matrix,
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[K ]= EIL3 [ 12 6 L −12 6 L
6 L 4 L ² −6 L 2L ²−126 L
−6 L2L ²
12−6 L
−6 L4 L² ]
3. Finite element equation, {F }=[K ]{u }
⇒{F1 y
m1
F2y
m2}= EI
L3 [ 12 6 L −12 6 L6 L 4 L ² −6L 2L ²−126 L
−6 L2 L²
12−6 L
−6 L4 L ² ]{
d1 y
∅ 1
d2 y
∅ 2}
Where,L = length of the beam elementE = Young’s modulusI = Moment of inertia
LONGITUDINAL & TRANSVERSE VIBRATION PROBLEMS
6
ONE DIMENSIONAL HEAT TRANSFER PROBLEMS
1. Finite Element Equation For 1D Heat Conduction with free end Convection
2. Finite Element Equation For 1D Heat Conduction, Convection and Internal Heat Generation
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UNIT III (2D SCALAR VARIABLE PROBLEMS
CONSTANT TRIANGULAR ELEMENT (CST),
1. Shape functions,
N1=p1+¿ q1x+ r1 y
2 A¿
N2=p2+¿ q2x+ r2 y
2 A¿
N 3=p3+¿q3 x+r3 y
2 A¿
Where,
p1=x2 y3− y2 x3
p2=x3 y1− y3 x1
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p3=x1 y2− y1 x2
q1= y2− y3
q2= y3− y1
q3= y1− y2
r1=x3−x2
r2=x1−x3
r3=x2− x1
A=12 |1 x1 y1
1 x2 y2
1 x3 y3|
2. Displacement functions,
u=[N1 0 N 2 0 N 3 00 N1 0 N2 0 N3]{
u1
v1
u2
v2
u3
v3
}3. Stiffness matrix,
[K] = [B]T [D] [B] A t
4. Strain displacement matrix,
[B ]= 12 A [q1 0 q2 0 q3 0
0 r 1 0 r2 0 r3
r1 q1 r2 q2 r3 q3]
5. Stress - strain matrix in general 2D form,
D= E(1+v )(1−2v ) [
1−v v v 0 0 0v 1−v v 0 0 0
v000
v000
1−v000
01−2 v
200
0 00 0
1−2 v20
01−v
2]
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Plane stress condition,
D= E1−v2 [1 v 0
v 1 0
0 0 1−v2 ]
Plane strain condition,
D= E(1+v )(1−2v ) [1−v v 0
v 1−v 0
0 0 1−2v2 ]
6. Element stress,{σ }=[D ] [B ]{u }
⇒{σ x
σ y
τ xy}=[D ] [B]{u1
v1
u2
v2
u3
v3
}Maximum stress,
σ max¿σ1=σ x+σ y
2+√( σ x−σ y
2 )2
+ τ xy2
Minimum stress,
σ min¿σ2=σ x+σ y
2−√( σx−σ y
2 )2
+ τxy2
7. Principle angle,
tan2θp=2 τxyσx−σ y
8. Element strain, {e }=[B ] {u }
TEMPETATURE EFFECT OF CST ELEMENT,
1. Initial strain,Plane stress,
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{e0 }={α ΔTα ΔT0 }
Plane strain,
{e0 }=(1+v){α ΔTα ΔT0 }
2. Element temperature force,{F }=[B ]T [D ] {e0 }AtWhere,
t = thicknessA= area of the element.
2D HEAT TRANSFER PROBLEMS
Stiffness Matrix for both Conduction and Convection
UNIT IV (2D VECTOR VARIABLE PROBLEMS
AXI-SYMMETRIC ELEMENT,
11
1. Shape function,
N 1=α 1+¿ β1r+γ 1 z
2 A¿
N 2=α 2+¿ β2r+γ 2 z
2 A¿
N 3=α 3+¿ β3r+ γ3 z
2 A¿
Where,
α 1=r2 z3−r3 z2
α 2=r3 z1−r1 z3
α 3=r1 z2−r2 z1
β1=z2−z3
β2=z3−z1
β3=z1−z2
γ1=r3−r2
γ2=r1−r3
γ3=r 2−r1
A=12 |1 r1 z1
1 r2 z2
1 r3 z3|
2. Strain displacement matrix,
[B ]= 12A
¿
3. Stress strain relation
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[D ]= E(1+v )(1−2v) [(1−v ) v v 0
v (1−v ) v 0
v0
v0
(1−v) 0
0 1−2 v2
]4. Stiffness matrix,
[K ]=2πrA [B ]T [D ] [B ]
TEMPETATURE EFFECT OF Axisymmetric Element,
3. Initial strain,
{e0 }={α ΔTα ΔT0
αΔT}
4. Element temperature force,{F }=[B ]T [D ] {e0 }2πrAWhere,
t = thicknessA= area of the element.
UNIT V-ISOPARAMETRIC FORMULATION13
Iso Parametric Quadrilateral Element
1. Shape functions,
N 1=14
[1−ε ] [ 1−η ]
N 2=14
[1+ε ] [ 1−η ]
N 3=14
[ 1+ε ] [ 1+η ]
N 4=14
[ 1−ε ] [ 1+η ]
2. Strain displacement matrix,
[B ]= 14|J|[ J22 −J12 0 0
0 0 −J 21 J 11
−J21 J 11 J22 J12]×[− [1−η ] 0 [ 1−η ] 0 [ 1+η ] 0 −[ 1+η ] 0
−[ 1−ε ] 0 −[ 1+ε ] 0 [ 1+ε ] 0 [1−ε ] 000
−[1−η ]−[1−ε ]
0 [1−η ] 0 [ 1+η ] 0 −[ 1+η ]0 −[ 1+ε ] 0 [ 1+ε ] 0 [ 1−ε ] ]
3. Displacement function,Rectangular element,
u=[N1 0 N 2 0 N3 00 N1 0 N2 0 N3
N4 00 N4]{
u1
v1
u2
v2
u3
v3
u4
v4
}u=N1u1+N2u2+N 3u3+N 4u4
v=N1 v1+N 2 v2+N3 v3+N 4 v4
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u=[N1 0 N 2 0 N3 00 N1 0 N2 0 N3
N4 00 N4]{
x1
y1
x2
y2
x3
y3
x4
y4
}x=N1 x1+N2 x2+N3 x3+N 4 x4
y=N 1 y1+N2 y2+N3 y3+N4 y4
4. Jacobian matrix,
[J ]=[J 11 J 12
J21 J 22] Where: J11=
14
{−[1−η ] x1+[ 1−η ] x2+ [1+η ] x3−[ 1+η ] x4 }
J12=14 {− [1−η ] y1+[ 1−η ] y2+ [1+η ] y3− [1+η ] y4 }
J21=14 {− [1−ε ] x1− [1+ε ] x2+ [1+ε ] x3+ [ 1+ε ] x4 }
J22=14
{−[ 1−ε ] y1− [1+ε ] y2+ [ 1+ε ] y3+ [1+ε ] y 4}
5. Force vector,
{F }e=[N ]T {F x
F y}Where,
ε , η = natural co-ordinates
[B] = strain-displacement relationship matrix
[D] = stress strain relationship matrix
N = shape function
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F x = load or force on x direction
F y = force on y direction
6. Element stress,{σ }=[D ] [B } {u }
Gaussian Quadrature (Or) Numerical Integration
(i) For 2 point Quadrature
∫−1
1
f ( x )dx=w1 f ( x 1 )+w 2 f (x 2)
Where, w1 = w2 =1 and x1= √1/3, x2 = -√1/3
(ii) For 3 point Quadrature
∫−1
1
f ( x ) dx=w 1 f ( x 1 )+w 2 f ( x2 )+w3 f (x 3)
Where, w1 = w3 = 5/9, w2 = 8/9 and x1= √3/5, x2 = 0, x3 = -√3/5
(iii) For double Integration,
∬−1
1
f ( x , y ) dxdy=¿¿w12f(x1,y1) + w1w2f(x1,y2) + w2w1f(x2,y1) + w2
2f(x2,y2)
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