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1. 1
MANE 4240 & CIVL 4240
Introduction to Finite Elements
Shape functions in 1D
Prof. Suvranu De
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 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(xdxbudx
dx
du
EA
2
1
(u)
00
2
=−−⎟
⎠
⎞
⎜
⎝
⎛
=Π ∫∫
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(xdxbwdx
dx
dw
EA
2
1
(w)
00
2
=−−⎟
⎠
⎞
⎜
⎝
⎛
=Π ∫∫
LL

2. 2
Step 1. Assume a solution
...)()()()( 22110 +++= xaxaxaxw o ϕϕϕ
Where ϕo(x), ϕ1(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(xdxbwdx
dx
dw
EA
2
1
(w)
00
2
=−−⎟
⎠
⎞
⎜
⎝
⎛
=Π ∫∫
LL
Step 3. Obtain the coefficients ao, a1, etc by setting
,...2,1,0,0
(w)
==
∂
Π∂
i
ai
Rayleigh-Ritz Principle
The approximate solution is
...)()()()( 22110 +++= xaxaxaxu o ϕϕϕ
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(xdxbwdx
dx
dw
EA
2
1
(w)
00
2
=−−⎟
⎠
⎞
⎜
⎝
⎛
=Π ∫∫
LL
Total potential energy=sum of potential energies of the elements

3. 3
El #1 El #2 El #3
x1=0 x2 x3 x4=L
dxbwdx
dx
dw
EA
2
1
(w)
2
1
2
1
2
1 ∫∫ −⎟
⎠
⎞
⎜
⎝
⎛
=Π
x
x
x
x
L)Fw(xdxbwdx
dx
dw
EA
2
1
(w)
00
2
=−−⎟
⎠
⎞
⎜
⎝
⎛
=Π ∫∫
LL
Total potential energy
Potential energy of element 1:
dxbwdx
dx
dw
EA
2
1
(w)
3
2
3
2
2
2 ∫∫ −⎟
⎠
⎞
⎜
⎝
⎛
=Π
x
x
x
x
Potential energy of element 2:
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 Π+Π+Π=Π
L)Fw(xdxbwdx
dx
dw
EA
2
1
(w)
4
3
4
3
2
3 =−−⎟
⎠
⎞
⎜
⎝
⎛
=Π ∫∫
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.

4. 4
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
2x
12
1x
12
1
2x
12
1
1x
12
2
0
d
xx
1
d
xx
1
a
d
xx
x
d
xx
x
a
−
+
−
−=
−
−
−
=
2x21x12x
(x)N
12
1
1x
(x)N
12
2
10 (x)dN(x)dNd
xx
x-x
d
xx
x-x
xaaw(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
2
21
xx
x-x
xx
x-x
(x)N(x)NN
Vector of nodal displacements
⎭
⎬
⎫
⎩
⎨
⎧
=
2x
1x
d
d
d
(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 x2
El #1
12
2
1
xx
x-x
(x)N
−
=
12
1
2
xx
x-x
(x)N
−
=
1 1
0
xx
x-x
)x(xNand
1
xx
x-x
)x(xN
xx
x-x
(x)N
12
22
21
12
12
11
12
2
1
=
−
==
=
−
==⇒
−
=
Check

5. 5
2. Compatibility: The displacement approximation is continuous
across element boundaries
2x3x
23
22
2x
23
23
2
(2)
2x2x
12
12
1x
12
22
2
(1)
dd
xx
x-x
d
xx
x-x
)x(xw
dd
xx
x-x
d
xx
x-x
)x(xw
=
−
+
−
==
=
−
+
−
==
xx1 x2
El #1
2x
12
1
1x
12
2(1)
d
xx
x-x
d
xx
x-x
(x)w
−
+
−
=
3x
23
2
2x
23
3(2)
d
xx
x-x
d
xx
x-x
(x)w
−
+
−
=
x3El #2
At 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 check
12
1
2
12
2
1
xx
x-x
(x)N
;
xx
x-x
(x)N
−
=
−
=
xx
xx
x-x
x
xx
x-x
x(x)Nx(x)Nand
1
xx
x-x
xx
x-x
(x)N(x)N
2
12
1
1
12
2
2211
12
1
12
2
21
=
−
+
−
=+
=
−
+
−
=+
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
1
dx
dw(x)
(x)Hence,
x(x)xN(x)xN(x)dN(x)dNw(x) 22112x21x1
==
=+=+=
ε YES!
2x 1 2 1
2 1 2 1
d d x x
(x) 1
x x x x
x
ε
− −
= = =
− −
Let us see whether the displacement approximation allows this.

6. 6
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.
4. How to write the expressions for the shape functions easily
(without having to derive them each time):
( )
( )
( )
( )
( )
( )12
1
21
1
2
12
2
1
x-x
x-x
x-x
x-x
(x)N
x-x
x-x
(x)N
==
=
xx1 x2
El #1
12
2
1
xx
x-x
(x)N
−
=
12
1
2
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
21
3
2321
31
2
1312
32
1
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
dw
ε =Recall that the strain in the bar
Hence
dBd
dx
Nd
ε =⎥
⎦
⎤
⎢
⎣
⎡
= (2)
The matrix B is known as the “strain-displacement matrix”
⎥
⎦
⎤
⎢
⎣
⎡
=
dx
Nd
B

7. 7
For a linear finite element
[ ]11
xx
1
xx
1
xx
1-
B
121212
−
−
=⎥
⎦
⎤
⎢
⎣
⎡
−−
=
[ ] ⎥
⎦
⎤
⎢
⎣
⎡
−−
==
12
1
12
2
21
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-d
ε
−
=
dx
du
EEε ==σRecall that the stress in the bar
Hence, inside the element, the approximate stress is
dBE=σ (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=σ
(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

8. 8
For a linear element
Displacement approximation in terms of shape functions
⎭
⎬
⎫
⎩
⎨
⎧
⎥
⎦
⎤
⎢
⎣
⎡
−−
=
2x
1x
12
1
12
2
d
d
xx
x-x
xx
x-x
u(x)
Strain approximation
Stress approximation
Summary
[ ]
⎭
⎬
⎫
⎩
⎨
⎧
−
−
=
2x
1x
12 d
d
11
xx
1
ε
[ ]
⎭
⎬
⎫
⎩
⎨
⎧
−
−
=
2x
1x
12 d
d
11
xx
E
σ