mane 4240 & civl 4240 introduction to finite elements shape functions in 1d prof. suvranu de
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Reading assignment:
Lecture notes, Logan 2.2, 3.1
Summary:
• Linear shape functions in 1D • Quadratic and higher order shape functions• Approximation of strains and stresses in an element
Axially loaded elastic bar
x
y
x=0 x=L
A(x) = cross section at xb(x) = body force distribution (force per unit length)E(x) = Young’s modulus
x
F
Potential energy of the axially loaded bar corresponding to the exact solution u(x)
L)Fu(xdxbudxdx
duEA
2
1(u)
00
2
LL
Finite element formulation, takes as its starting point, not the strong formulation, but the Principle of Minimum Potential Energy.
Task is to find the function ‘w’ that minimizes the potential energy of the system
From the Principle of Minimum Potential Energy, that function ‘w’ is the exact solution.
L)Fw(xdxbwdxdx
dwEA
2
1(w)
00
2
LL
Step 1. Assume a solution
...)()()()( 22110 xaxaxaxw o
Where o(x), 1(x),… are “admissible” functions and ao, a1, etc are constants to be determined.
Step 2. Plug the approximate solution into the potential energy
L)Fw(xdxbwdxdx
dwEA
2
1(w)
00
2
LL
Step 3. Obtain the coefficients ao, a1, etc by setting
,...2,1,0,0(w)
iai
Rayleigh-Ritz Principle
The approximate solution is
...)()()()( 22110 xaxaxaxu o
Where the coefficients have been obtained from step 3
Need to find a systematic way of choosing the approximation functions.
One idea: Choose polynomials!
0)( axw Is this good? (Is ‘1’ an “admissible” function?)
Is this good? (Is ‘x’ an “admissible” function?)xaxw 1)(
Finite element idea:
Step 1: Divide the truss into finite elements connected to each other through special points (“nodes”)
El #1 El #2 El #3
1 2 3 4
L)Fw(xdxbwdxdx
dwEA
2
1(w)
00
2
LL
Total potential energy=sum of potential energies of the elements
El #1 El #2 El #3
x1=0 x2 x3x4=L
dxbwdxdx
dwEA
2
1(w)
2
1
2
1
2
1
x
x
x
x
L)Fw(xdxbwdxdx
dwEA
2
1(w)
00
2
LL
Total potential energy
Potential energy of element 1:
dxbwdxdx
dwEA
2
1(w)
3
2
3
2
2
2
x
x
x
x
Potential energy of element 2:
El #1 El #2 El #3
x1=0 x2 x3x4
Total potential energy=sum of potential energies of the elements
Potential energy of element 3:
(w)(w)(w)(w) 321
L)Fw(xdxbwdxdx
dwEA
2
1(w)
4
3
4
3
2
3
x
x
x
x
Step 2: Describe the behavior of each element
Recall that in the “direct stiffness” approach for a bar element, we derived the stiffness matrix of each element directly (See lecture on Trusses) using the following steps:
TASK 1: Approximate the displacement within each bar as a straight line TASK 2: Approximate the strains and stresses and realize that a bar (with the approximation stated in Task 1) is exactly like a spring with k=EA/L TASK 3: Use the principle of force equilibrium to generate the stiffness matrix
Now, we will show you a systematic way of deriving the stiffness matrix (sections 2.2 and 3.1 of Logan).
TASK 1: APPROXIMATE THE DISPLACEMENT WITHIN EACH ELEMENT TASK 2: APPROXIMATE THE STRAIN and STRESS WITHIN EACH ELEMENT TASK 3: DERIVE THE STIFFNESS MATRIX OF EACH ELEMENT (next class) USING THE PRINCIPLE OF MIN. POT ENERGY
Notice that the first two tasks are similar in the two methods. The only difference is that now we are going to use the principle of minimum potential energy, rather than force equilibrium, to derive the stiffness matrix.
TASK 1: APPROXIMATE THE DISPLACEMENT WITHIN EACH ELEMENT
Simplest assumption: displacement varying linearly inside each bar
xaaw(x) 10 2xd
1xd x
x1 x2
El #1
How to obtain a0 and a1?
2x2102
1x1101
dxaa)w(x
dxaa)w(x
2x2102
1x1101
dxaa)w(x
dxaa)w(x
Solve simultaneously
2x12
1x12
1
2x12
11x
12
20
dxx
1d
xx
1a
dxx
xd
xx
xa
2x21x12x
(x)N
12
11x
(x)N
12
210 (x)dN(x)dNd
xx
x-xd
xx
x-xxaaw(x)
21
Hence
“Shape functions” N1(x) and N2(x)
In matrix notation, we write
dNw(x)
Vector of nodal shape functions
12
1
12
221 xx
x-x
xx
x-x(x)N(x)NN
Vector of nodal displacements
2x
1x
d
dd
(1)
NOTES: PROPERTIES OF THE SHAPE FUNCTIONS
1. Kronecker delta property: The shape function at any node has a value of 1 at that node and a value of zero at ALL other nodes.
xx1 x2El #1
12
21 xx
x-x(x)N
12
12 xx
x-x(x)N
1 1
0xx
x-x)x(xNand
1xx
x-x)x(xN
xx
x-x(x)N
12
2221
12
1211
12
21
Check
2. Compatibility: The displacement approximation is continuous across element boundaries
2x3x23
222x
23
232
(2)
2x2x12
121x
12
222
(1)
ddxx
x-xd
xx
x-x)x(xw
ddxx
x-xd
xx
x-x)x(xw
xx1 x2El #1
2x12
11x
12
2(1) dxx
x-xd
xx
x-x(x)w
3x23
22x
23
3(2) dxx
x-xd
xx
x-x(x)w
x3El #2At x=x2
Hence the displacement approximation is continuous across elements
3. Completeness
xallforx(x)xN(x)xN
xallfor1(x)N(x)N
2211
21
Use the expressions
And check12
12
12
21
xx
x-x(x)N
;xx
x-x(x)N
xxxx
x-xx
xx
x-xx(x)Nx(x)Nand
1xx
x-x
xx
x-x(x)N(x)N
212
11
12
22211
12
1
12
221
Rigid body mode
What do we mean by “rigid body modes”?
Assume that d1x=d2x=1, this means that the element should translate in the positive x direction by 1. Hence ANY point (x) on the bar should have unit displacement. Let us see whether the displacement approximation allows this.
1(x)N(x)N(x)dN(x)dNw(x) 212x21x1
YES!
1 2N (x) N (x) 1 for all x
Constant strain states
1 1 2 2N (x)x N (x)x x at all x
What do we mean by “constant strain states”?
Assume that d1x=x1 and d2x=x2. The strain at ANY point (x) within the bar is
1dx
dw(x)(x) Hence,
x(x)xN(x)xN(x)dN(x)dNw(x) 22112x21x1
YES!
2x 1 2 1
2 1 2 1
d d x x(x) 1
x x x xx
Let us see whether the displacement approximation allows this.
Completeness = Rigid body modes + Constant Strain states
Compatibility + Completeness ConvergenceEnsure that the solution gets better as more elements are introducedand, in the limit, approaches the exact answer.
4. How to write the expressions for the shape functions easily (without having to derive them each time):
12
1
21
12
12
21
x-x
x-x
x-x
x-x(x)N
x-x
x-x(x)N
xx1 x2El #1
12
21 xx
x-x(x)N
12
12 xx
x-x(x)N
1 1
Start with the Kronecker delta property (the shape function at any node has value of 1 at that node and a value of zero at all other nodes)
Notice that the length of the element = x2-x1
Node at which N1 is 0
The denominator is the numerator evaluated at the node itself
A slightly fancier assumption: displacement varying quadratically inside each bar
3231
213
2321
312
1312
321
x-xx-x
x-xx-x(x)N
x-xx-x
x-xx-x(x)N
x-xx-x
x-xx-x(x)N
xx1 x2
El #1
(x)N1(x)N3
x3
1
(x)N2
3x32x21x1 (x)dN(x)dN(x)dNw(x)
This is a quadratic finite element in 1D and it has three nodes and three associated shape functions per element.
TASK 2: APPROXIMATE THE STRAIN and STRESS WITHIN EACH ELEMENT
dNw(x)
From equation (1), the displacement within each element
dx
dwε Recall that the strain in the bar
Hence
dBddx
Ndε
(2)
The matrix B is known as the “strain-displacement matrix”
dx
NdB
For a linear finite element
11xx
1
xx
1
xx
1-B
121212
12
1
12
221 xx
x-x
xx
x-x(x)N(x)NN
Hence
12
1x2x
2x
1x
1212
xx
d-d
d
d
xx
1
xx
1-dBε
Hence, strain is a constant within each element (only for a linear element)!
2xd
1xd x
x1 x2
El #1
x
x1 x2
El #1
xaaw(x) 10 Displacement is linear
Strain is constant
12
1x2x
xx
d-dε
dx
duEEε Recall that the stress in the bar
Hence, inside the element, the approximate stress is
dBE (3)
For a linear element the stress is also constant inside each element. This has the implication that the stress (and strain) is discontinuous across element boundaries in general.
Inside an element, the three most important approximations in terms of the nodal displacements (d) are:
dBE
(1)
Displacement approximation in terms of shape functions
dNu(x)
dBε(x)
Strain approximation in terms of strain-displacement matrix
(2)
Stress approximation in terms of strain-displacement matrix and Young’s modulus
(3)
Summary
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