# Introduction to Finite Element Methods - Constant Strain Triangle(Intro to FEM - CST)

October 5, 2017 | Author: Sathyan Krishnan | Category: Finite Element Method, Deformation (Mechanics), Mathematical Concepts, Mathematical Analysis, Mathematics

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Introduction to Finite Element Methods - Constant Strain Triangle...

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INTRODUCTION TO FINITE ELEMENT METHODS CONSTANT STRAIN TRIANGLE

Introduction to Finite Elements

Constant Strain Triangle (CST)

Summary: • Computation of shape functions for constant strain triangle • Properties of the shape functions • Computation of strain-displacement matrix • Computation of element stiffness matrix • Computation of nodal loads due to body forces • Computation of nodal loads due to traction • Recommendations for use • Example problems

Finite element formulation for 2D: Step 1: Divide the body into finite elements connected to each other through special points (“nodes”) py v3 3 px 4 3 u3 u 1  v4 2 v  v2 Element ‘e’ v 1   1 4 u 2  u u4 ST   v1 2 u2 v 2  y d=  x y u3   Su u1 v 3  1   v x x u 4  u v   4

u (x,y)≈ N1(x,y)u1 + N2(x,y)u2 + N3(x,y)u3 + N4(x,y)u4 v (x,y)≈ N1(x,y)v1 + N2(x,y)v2 + N3(x,y)v3 + N4(x,y)v4

u (x, y)  N 1 u= = v (x, y)  0

u=Nd

0

N2

0

N3

0

N4

N1

0

N2

0

N3

0

u 1  v   1 u 2    0  v 2    N 4  u 3  v 3    u 4  v   4

TASK 2: APPROXIMATE THE STRAIN and STRESS WITHIN EACH ELEMENT

Approximation of the strain in element ‘e’ ∂N3(x, y) ∂N 2 (x, y) ∂N 4 (x, y) ∂u (x, y) ∂N1(x, y) εx = ≈ u1 + u2 + u3 + u4 ∂x ∂x ∂x ∂x ∂x ∂N (x, y) ∂N 2 (x, y) ∂N (x, y) ∂v (x, y) ∂N1(x, y) εy = ≈ v1 + v2 + 3 v3 + 4 v4 ∂y ∂y ∂y ∂y ∂y ∂N (x, y) ∂u (x, y) ∂v (x, y) ∂N1(x, y) γ xy = + ≈ u1 + 1 v1 + ...... ∂y ∂x ∂y ∂x

ε x    ε = ε y  γ xy    u1  v   ∂N1(x, y)  1  ∂N3(x, y) ∂N 2 (x, y) ∂N 4 (x, y) 0 0 0 0  u 2  ∂ x ∂ x ∂ x ∂ x    ∂ N (x, y) ∂ ∂ ∂ N (x, y) N (x, y) N (x, y) 3 1 2 4 v 2  = 0 0 0 0    ∂y ∂y ∂y ∂y u 3   ∂N (x, y) ∂N (x, y) ∂N (x, y) ∂N (x, y) ∂N (x, y) ∂N (x, y) ∂N (x, y) ∂N (x, y) 3 3 1 2 2 4 4  1 v3  ∂x ∂y ∂x ∂y ∂x ∂y ∂x u   ∂y 1 4444444444444444444244444444444444444443 4  B v   4

ε=Bd

Summary: For each element Displacement approximation in terms of shape functions

u=Nd Strain approximation in terms of strain-displacement matrix ε=Bd

Stress approximation σ = DB d Element stiffness matrix

k = ∫ e B D B dV T

V

Element nodal load vector f = ∫ e N X dV + ∫ e N T S dS V ST 1 4243 1 4 4244 3 T

f

T

b

f

S

Constant Strain Triangle (CST) : Simplest 2D finite element v1 1 (x1,y1)

v2 y

u1

v3 (x3,y3)

v

(x,y) 2 (x2,y2)

u

3

u3

u2

x

• 3 nodes per element • 2 dofs per node (each node can move in x- and y- directions) • Hence 6 dofs per element

The displacement approximation in terms of shape functions is u (x,y) ≈ N1u1 + N2 u2 + N3u3 v(x,y) ≈ N1 v1 + N 2 v 2 + N 3 v 3 u (x, y)  N 1 u= = v (x, y)  0

0

N2

0

N3

N1

0

N2

0

u 2×1 = N 2×6 d 6×1

 N1 N= 0

0 N1

N2 0

0 N2

N3 0

0 N 3 

 u1  v   1 0  u 2    N 3   v 2  u 3     v 3 

Formula for the shape functions are v1 v3 1 u1 (x3,y3) (x1,y1) u3 v2 v u 3 y (x,y)

where

2 (x2,y2)

u2

x

a1 + b1 x + c1 y 2A a2 + b2 x + c2 y N2 = 2A a3 + b3 x + c3 y N3 = 2A N1 =

1 x 1 1  A = area of triangle = det 1 x 2 2 1 x 3

y1   y2  y 3 

a1 = x 2 y 3 − x3 y 2 a 2 = x3 y1 − x1 y 3

b1 = y 2 − y 3 b2 = y 3 − y1

c1 = x3 − x 2 c 2 = x1 − x3

a 3 = x1 y 2 − x 2 y1

b3 = y1 − y 2

c3 = x 2 − x1

Properties of the shape functions: 1. The shape functions N1, N2 and N3 are linear functions of x and y N2

N1 1

N3 1

1 1

1

3 3

y

1

2

3

2

x

1 at node ' i ' Ni =  0 at other nodes

2

2. At every point in the domain

3

∑N i =1

i

=1

3

∑N x

i

=x

∑N y

=y

i =1

i

3

i =1

i

i

3. Geometric interpretation of the shape functions At any point P(x,y) that the shape functions are evaluated,

P (x,y)

1 A3

y

A2 A1

2 x

3

A1 N1 = A A2 N2 = A A3 N3 = A

Approximation of the strains ∂u   ∂x  εx      ∂v ε = ε y  =  ∂y γ    xy   ∂ u ∂ v  ∂y + ∂x 

     ≈ Bd    

∂N1(x,y) ∂N2(x,y) 0  ∂x  ∂x ∂N1(x,y)  B= 0 0  ∂y ∂N (x,y) ∂N (x,y) ∂N (x,y) 1 2  1 ∂x ∂y  ∂y b1 0 b2 0 b3 0  1 =  0 c1 0 c2 0 c3  2A c1 b1 c2 b2 c3 b3 

 0 0   ∂N3(x,y) ∂N2(x,y) 0 ∂y ∂y  ∂N2(x,y) ∂N3(x,y) ∂N3(x,y)  ∂x ∂y ∂x  ∂N3(x,y) ∂x

Inside each element, all components of strain are constant: hence the name Constant Strain Triangle

Element stresses (constant inside each element)

σ = DB d

IMPORTANT NOTE: 1. The displacement field is continuous across element boundaries 2. The strains and stresses are NOT continuous across element boundaries

Element stiffness matrix t

k = ∫ e B D B dV T

V

Since B is constant

A

k = B D B∫ e dV = B D B At T

T

V

t=thickness of the element A=surface area of the element

f = ∫ e N X dV + ∫ e N T S dS V ST 1 4243 1 4 4244 3 T

f

T

b

f

S

Element nodal load vector due to body forces f b = ∫ e N X dV = t ∫ e N X dA T

T

V

A

fb1y 1

y

fb2y

fb1x Xb (x,y)

Xa

fb2x

2 x

fb3y

3

fb3x

t N X 1 a ∫ Ae   f b1x   f   t ∫ e N1 X b  b1 y   A  f b 2 x  t ∫ e N 2 X a A fb = =  f b 2 y  t ∫Ae N 2 X b  f b3 x     t ∫Ae N 3 X a  f b3 y   t ∫Ae N 3 X b

dA   dA   dA  dA  dA  dA 

EXAMPLE: If Xa=1 and Xb=0 t N X 1 a ∫ Ae   f b1x   f  t ∫ e N1 X b  b1 y   A  f b 2 x  t ∫ e N 2 X a A fb = =  f b 2 y  t ∫Ae N 2 X b  f b3 x     t ∫Ae N 3 X a  f b3 y   t ∫Ae N 3 X b

dA   tA     t ∫ e N1 dA 3  A   dA   0   0  dA t N dA  tA   =  ∫Ae 2  =  3  0 dA   0   t N dA   tA  dA  ∫Ae 3      3  0   0 dA 

Element nodal load vector due to traction f

= ∫ e N T S dS T

S

ST

EXAMPLE:

fS1y

fS3y

1

fS1x 3

y 2 x

f

fS3x

S

= t∫

l1− 3

N e

T along 1−3

T S dS

Element nodal load vector due to traction EXAMPLE:

f S = t∫

fS2y

l2−3

(2,2) 2 y

fS3y 1 (0,0)

fS2x

1 TS =  0

3 f (2,0) S3x

x

N e

f S2 x = t ∫

l 2−3

T along 2 −3

Similarly, compute

f S3 x = t f S3 y = 0

1

N 2 along 2−3 (1) dy e

1 = t   × 2 ×1 = t  2

f S2 y = 0

T S dS

2

Recommendations for use of CST 1. Use in areas where strain gradients are small 2. Use in mesh transition areas (fine mesh to coarse mesh) 3. Avoid CST in critical areas of structures (e.g., stress concentrations, edges of holes, corners) 4. In general CSTs are not recommended for general analysis purposes as a very large number of these elements are required for reasonable accuracy.

Example y

3

1000 lb 300 psi

2

El 2 2 in El 1 1 4

Thickness (t) = 0.5 in E= 30×106 psi ν=0.25

x

3 in

(a) Compute the unknown nodal displacements. (b) Compute the stresses in the two elements.

Realize that this is a plane stress problem and therefore we need to use

  ν 1 0 3.2 0.8 0    E 0.8 3.2 0  ×107 psi   D= 1 0 = ν  1 −ν 2  1 −ν   0 1.2 0 0   0 2   Step 1: Node-element connectivity chart ELEMENT

Node 1

Node 2

Node 3

Area (sqin)

1

1

2

4

3

2

3

4

2

3

Node

x

y

1

3

0

2

3

2

3

0

2

4

0

0

Nodal coordinates

Step 2: Compute strain-displacement matrices for the elements Recall

b1 0 b2 0 b3 0  1 B =  0 c1 0 c2 0 c3  2A c1 b1 c2 b2 c3 b3 

For Element #1:

2(2)

with

b1 = y 2 − y3

b2 = y3 − y1

b3 = y1 − y 2

c1 = x3 − x2

c2 = x1 − x3

c3 = x2 − x1

y1 = 0; y 2 = 2; y3 = 0 x1 = 3; x2 = 3; x3 = 0 Hence

b1 = 2

b2 = 0 b3 = −2

c1 = −3 c2 = 3 4(3) 1(1) Therefore (local numbers within brackets)

For Element #2:

−2 0 0 0 2 0 1 (2) B =  0 3 0 −3 0 0 6  3 − 2 −3 0 0 2

c3 = 0

 2 0 0 0 −2 0  1 (1) B =  0 −3 0 3 0 0  6 −3 2 3 0 0 − 2

Step 3: Compute element stiffness matrices (1) T

(1) T

k = AtB D B = (3)(0.5)B D B (1)

(1)

(1)

0.9833 − 0.5 − 0.45 0.2 − 0.5333  1.4 0.3 −1.2 0.2   0.45 0 0 = 1.2 − 0.2   0.5333   u1

v1

u2

v2

u4

0.3  − 0.2 − 0.3 7  ×10 0  0   0.2  v4

k

( 2)

( 2) T

= AtB

( 2) T

D B = (3)(0.5)B ( 2)

( 2)

DB

0.9833 − 0.5 − 0.45 0.2 − 0.5333  1.4 0.3 −1.2 0.2   0.45 0 0 = 1.2 − 0.2   0.5333   u3

v3

u4

v4

u2

0.3  − 0.2 − 0.3 7  ×10 0  0   0.2  v2

Step 4: Assemble the global stiffness matrix corresponding to the nonzero degrees of freedom Notice that

u3 = v3 = u 4 = v4 = v1 = 0

Hence we need to calculate only a small (3x3) stiffness matrix

u1  0.983 − 0.45 0.2 K =  − 0.45 0.983 0  × 10 7 u2  0.2 0 1.4  v2 u1

u2

v2

Step 5: Compute consistent nodal loads

 f1 x   0      f =  f2x  =  0  f  f   2y   2y  f 2 y = −1000 + f S 2 y The consistent nodal load due to traction on the edge 3-2

f S2 y = ∫

3

x =0

N 3 3− 2 ( −300 )tdx

= ( −300 )(0.5) ∫

3

x =0

x 3

N 3 3− 2 dx

x = −150 ∫ dx x =0 3 3

3

N 2 3− 2 =

 x2  9 = −50   = −50   = −225 lb  2  2 0

3

2

Hence

f 2 y = −1000 + f S 2 y = −1225 lb Step 6: Solve the system equations to obtain the unknown nodal loads

Kd = f  0.983 − 0.45 0.2  u1   0      10 7 ×  − 0.45 0.983 0  u 2  =  0   0.2 0 1.4   v2  − 1225  Solve to get −4  u1   0.2337 × 10 in      −4 u 0 . 1069 10 in = ×  2    v  − 0.9084 × 10 − 4 in   2  

Step 7: Compute the stresses in the elements In Element #1

σ (1) = D B(1) d(1) With

d

(1) T

= [u1 v1 u2 v2 u4 v4 ]

[

]

= 0.2337×10−4 0 0.1069×10−4 − 0.9084×10−4 0 0 Calculate

σ (1)

 − 114.1  = − 1391.1 psi  − 76.1 

In Element #2

σ (2) = D B(2) d(2) With

d

( 2)T

= [u3 v3 u4 v4 u2 v2 ]

[

= 0 0 0 0 0.1069×10−4 − 0.9084×10−4 Calculate

σ ( 2)

 114.1  =  28.52  psi − 363.35

Notice that the stresses are constant in each element

]