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,

5

[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

7

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

8

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,

10

{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

12

[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

14

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

15

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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