Shape Functions in 1D
Summary:
• Linear shape functions in 1D
• Quadratic and higher order shape functions
• Approximation of strains and stresses in an element
Prepared by : Sachin Chaturvedi
Axially loaded elastic bar
x
y
x=0 x=L
A(x) = cross section at x
b(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
Prepared by : Sachin Chaturvedi
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
Prepared by : Sachin Chaturvedi
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)
i
ai
Rayleigh-Ritz Principle
Prepared by : Sachin Chaturvedi
The approximate solution is
...)()()()( 22110 xaxaxaxu o
Where the coefficients have been obtained from step 3
Prepared by : Sachin Chaturvedi
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)(
Prepared by : Sachin Chaturvedi
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
Prepared by : Sachin Chaturvedi
El #1 El #2 El #3
x1=0 x2
x3
x4=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:
Prepared by : Sachin Chaturvedi
El #1 El #2 El #3
x1=0 x2
x3
x4
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
Prepared by : Sachin Chaturvedi
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
Prepared by : Sachin Chaturvedi
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.
Prepared by : Sachin Chaturvedi
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
Prepared by : Sachin Chaturvedi
2x2102
1x1101
dxaa)w(x
dxaa)w(x
Solve simultaneously
2x
12
1x
12
1
2x
12
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) Prepared by : Sachin Chaturvedi
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)
Prepared by : Sachin Chaturvedi
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 x2 El #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
Prepared by : Sachin Chaturvedi
2. Compatibility: The displacement approximation is continuous
across element boundaries
2x3x
23
222x
23
232
(2)
2x2x
12
121x
12
222
(1)
ddxx
x-xd
xx
x-x)x(xw
ddxx
x-xd
xx
x-x)x(xw
xx1 x2 El #1
2x
12
11x
12
2(1) dxx
x-xd
xx
x-x(x)w
3x
23
22x
23
3(2) dxx
x-xd
xx
x-x(x)w
x3 El #2
At x=x2
Hence the displacement approximation is continuous across elements
Prepared by : Sachin Chaturvedi
3. Completeness
xallforx(x)xN(x)xN
xallfor1(x)N(x)N
2211
21
Use the expressions
And check
12
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
2
12
11
12
22211
12
1
12
221
Prepared by : Sachin Chaturvedi
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
Prepared by : Sachin Chaturvedi
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 x
x
Let us see whether the displacement approximation allows this.
Prepared by : Sachin Chaturvedi
Completeness = Rigid body modes + Constant Strain states
Compatibility + Completeness Convergence
Ensure that the solution gets better as more elements are introduced
and, in the limit, approaches the exact answer.
Prepared by : Sachin Chaturvedi
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 x2 El #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
Prepared by : Sachin Chaturvedi
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
Prepared by : Sachin Chaturvedi
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)! Prepared by : Sachin Chaturvedi
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
Prepared by : Sachin Chaturvedi
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.
Prepared by : Sachin Chaturvedi
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
Prepared by : Sachin Chaturvedi
For a linear element
Displacement approximation in terms of shape functions
2x
1x
12
1
12
2
d
d
xx
x-x
xx
x-xu(x)
Strain approximation
Stress approximation
Summary
2x
1x
12d
d11
xx
1ε
2x
1x
12d
d11
xx
E
Prepared by : Sachin Chaturvedi