MCEN 4173/5173

Chapter 3

1D Spring/Truss Elements

Fall, 2006

1
What is Finite Element Method?

P P
undeformed deformed

X2 X2

X1 X1

In a very simple way, a solid mechanics problem is about

Given P, what is u?
For linear elasticity, we want to find an approximate solution by:
[K]{u}={P} Linear equations

[K] is a matrix that shows the relationship between {u} and {P}.

How to formulate [K]?

2
1D Spring/Truss Elements

3
Model the interface as
2 springs

Spring elements: commonly used to model connectors, interface.

In this lecture, we will use spring element to illustrate some basic

features of finite element

Reading Materials: 1.1-1.7 (especially 1.2), 2.1-2.5.

3
Spring/Truss Assembly

F 1 2
F

3 4 5

Spring constant k
Displacements at each node?
Spring element = Truss element 4
Coordinate System

y
ŷ x̂

ẑ
Global Coordinate: (xyz)

z Local Coordinate: (xˆyˆzˆ )

5
1D Spring/Truss Element

x̂
1 2

In a spring element,

dˆ1x fˆ1x dˆ2 x fˆ2 x

x̂
1 2

2 nodes, and each node has 1 DOF.

The element has 2 DOFs.

So, we have two nodal displacements, and two nodal forces.

Sign Rule: the force or the displacement is positive, if it is along the

positive direction of the coordinate.

6
Single Spring Element

dˆ1x fˆ1x dˆ2 x fˆ2 x

x̂
1 2

Constitutive equation (force-displacement equation) fˆ = kΔd

For the spring to be in equilibrium,

(
fˆ2 x = − fˆ1x = k dˆ2 x − dˆ1x )
fˆ1x = kdˆ1x − kdˆ2 x

fˆ2 x = − kdˆ1x + kdˆ2 x

7
Two Spring Elements
dˆ1x dˆ3 x dˆ2 x
k1 k2
3 2 x̂
1
① ②
F1x F3 x F2 x
Fix is the external force on the i-th node.

For element ① For element ②

dˆ1x fˆ1(x1) dˆ3 x fˆ3(x1) dˆ3 x fˆ3(x2) dˆ2 x fˆ2(x2)

k1 k2

1 3 3 2
① ②

8
Two Spring Elements
Force balance at each node

fˆ1(x1) fˆ3(x1) fˆ3(x2) fˆ2(x2)

1 3 3 2
① ②

fˆ3(x1) fˆ3(x2) fˆ2(x2)

fˆ1(x1)
1 3 2
F1x F3 x F2 x

F1x = fˆ1(x1) F3 x = fˆ3(x1) + fˆ3(x2) F2 x = fˆ2(x2)

F: external force, or total force acting on the node

f: internal force, or force contribution from one element

9
Two Spring Elements
Global Stiffness Matrix
For element ① For element ②
⎧⎪ fˆ1(x1) ⎫⎪ ⎡ k1 − k1 ⎤ ⎧⎪ dˆ1x ⎫⎪ ⎧⎪ fˆ3(x2) ⎫⎪ ⎡ k 2 − k 2 ⎤ ⎧⎪dˆ3 x ⎫⎪
⎨ ˆ (1) ⎬ = ⎢ ⎨ ⎬ ⎨ ˆ ( 2) ⎬ = ⎢ ⎥ ⎨ˆ ⎬
⎪⎩ f 3 x ⎪⎭ ⎣− k1 k1 ⎥⎦ ⎪⎩dˆ3 x ⎪⎭ ⎪⎩ f 2 x ⎪⎭ ⎣− k 2 k 2 ⎦ ⎪⎩d 2 x ⎪⎭

F1x = fˆ1(x1) = k1dˆ1x − k1dˆ3 x = + k1dˆ1x + 0dˆ2 x − k1dˆ3 x

F2 x = fˆ2(x2) = k 2 dˆ2 x − k 2 dˆ3 x = 0dˆ1x + k 2 dˆ2 x − k 2 dˆ3 x

F3 x = fˆ3(x1) + fˆ3(x2) = − k1dˆ1x + k1dˆ3 x + k 2 dˆ3 x − k 2 dˆ2 x

= − k dˆ − k dˆ + (k + k )dˆ
1 1x 2 2x 1 2 3x

10
Two Spring Elements
Global Stiffness Matrix

F1x = + k1dˆ1x + 0dˆ2 x − k1dˆ3 x ⎧ F1x ⎫ ⎡ k1 0 − k1 ⎤ ⎧ dˆ1x ⎫

⎪ ⎪ ⎢ ⎥ ⎪ˆ ⎪
F2 x = 0dˆ1x + k 2 dˆ2 x − k 2 dˆ3 x ⎨ F2 x ⎬ = ⎢ 0 k2 − k 2 ⎥ ⎨d 2 x ⎬
⎪ F ⎪ ⎢− k − k2 k1 + k 2 ⎥⎦ ⎪ dˆ3 x ⎪
⎩ 3x ⎭ ⎣ 1 ⎩ ⎭
F3 x = − k1dˆ1x − k 2 dˆ2 x + (k1 + k 2 )dˆ3 x

⎧ F1x ⎫
⎪ ⎪
⎨ F2 x ⎬ Global (external) nodal force matrix
⎪F ⎪
⎩ 3x ⎭ ⎡ k1 0 − k1 ⎤
⎢ 0 k2 − k 2 ⎥⎥ Global stiffness
⎢ matrix
⎢⎣− k1 − k2 k1 + k 2 ⎥⎦
⎧ dˆ1x ⎫
⎪ˆ ⎪
⎨d 2 x ⎬ Global nodal displacement matrix
⎪ dˆ ⎪
⎩ 3x ⎭
11
Global Stiffness Matrix
Direct Stiffness Method

Element stiffness matrix

For element ① For element ②
⎧⎪ fˆ1(x1) ⎫⎪ ⎡ k1 − k1 ⎤ ⎧⎪ dˆ1x ⎫⎪ ⎧⎪ fˆ3(x2) ⎫⎪ ⎡ k 2 − k 2 ⎤ ⎧⎪dˆ3 x ⎫⎪
⎨ ˆ (1) ⎬ = ⎢ ⎨ ⎬ ⎨ ˆ ( 2) ⎬ = ⎢ ⎥ ⎨ˆ ⎬
⎪⎩ f 3 x ⎪⎭ ⎣− k1 k1 ⎥⎦ ⎪⎩dˆ3 x ⎪⎭ ⎪⎩ f 2 x ⎪⎭ ⎣− k 2 k 2 ⎦ ⎪⎩d 2 x ⎪⎭

Expanded Stiffness matrix

⎧ fˆ1(x1) ⎫ ⎡ k1 0 − k1 ⎤ ⎧ dˆ1x ⎫ ⎧ fˆ1(x2) ⎫ ⎡0 0 0 ⎤ ⎧ dˆ1x ⎫

⎪ ˆ (1) ⎪ ⎢ ⎥ ⎪ˆ ⎪ ⎪ ˆ ( 2) ⎪ ⎢ ⎥ ⎪ˆ ⎪
⎨ f 2 x ⎬ = ⎢ 0 0 0 ⎥ ⎨d 2 x ⎬ ⎨ f 2 x ⎬ = ⎢0 k 2 − k 2 ⎥ ⎨d 2 x ⎬
⎪ fˆ (1) ⎪ ⎢− k 0 k ⎥ ⎪dˆ ⎪ ⎪ fˆ ( 2) ⎪ ⎢0 − k k 2 ⎥⎦ ⎪dˆ3 x ⎪
⎩ 3x ⎭ ⎣ 1 1 ⎦⎩ 3x ⎭
⎩ 3x ⎭ ⎣ 2 ⎩ ⎭

12
Global Stiffness Matrix
Direct Stiffness Method

⎧ F1x ⎫ ⎧ f1x ⎫ ⎧ f1x ⎫

(1) ( 2)

⎪ ⎪ ⎪ (1) ⎪ ⎪ ( 2) ⎪
⎨ F2 x ⎬ = ⎨ f 2 x ⎬ + ⎨ f 2 x ⎬
⎪ F ⎪ ⎪ f (1) ⎪ ⎪ f ( 2) ⎪
⎩ 3x ⎭ ⎩ 3x ⎭ ⎩ 3x ⎭

⎡ k1 0 − k1 ⎤ ⎧ dˆ1x ⎫ ⎡0 0 0 ⎤ ⎧ dˆ1x ⎫
⎢ ⎥ ⎪ˆ ⎪ ⎢ ⎥ ⎪ˆ ⎪
= ⎢ 0 0 0 ⎥ ⎨d 2 x ⎬ + ⎢ 0 k 2 − k 2 ⎥ ⎨d 2 x ⎬
⎢⎣− k1 0 k1 ⎥⎦ ⎪dˆ3 x ⎪ ⎢⎣0 − k 2 k 2 ⎥⎦ ⎪dˆ3 x ⎪
⎩ ⎭ ⎩ ⎭

⎡ k1 0 − k1 ⎤ ⎧ dˆ1x ⎫
⎪ˆ ⎪
= ⎢⎢ 0 k2 ⎥
− k 2 ⎥ ⎨d 2 x ⎬
⎢⎣− k1 − k2 k1 + k 2 ⎥⎦ ⎪dˆ3 x ⎪
⎩ ⎭

13
Global Stiffness Matrix
Properties of Global Stiffness Matrix

⎧ F1x ⎫ ⎡ k1 0 − k1 ⎤ ⎧ dˆ1x ⎫
⎪ ⎪ ⎢ ⎥ ⎪ˆ ⎪
⎨ F2 x ⎬ = ⎢ 0 k2 − k 2 ⎥ ⎨d 2 x ⎬
⎪ F ⎪ ⎢− k − k2 k1 + k 2 ⎥⎦ ⎪ dˆ3 x ⎪
⎩ 3x ⎭ ⎣ 1 ⎩ ⎭

1. The product of the i-th row of the global stiffness matrix and the
global displacement matrix gives the external force on the i-th DOF
of the system.

14
Global Stiffness Matrix
Properties of Global Stiffness Matrix

⎧ F1x ⎫ ⎡ k1 0 − k1 ⎤ ⎧ dˆ1x ⎫
⎪ ⎪ ⎢ ⎥ ⎪ˆ ⎪
⎨ F2 x ⎬ = ⎢ 0 k2 − k 2 ⎥ ⎨d 2 x ⎬
⎪ F ⎪ ⎢− k − k2 k1 + k 2 ⎥⎦ ⎪ dˆ3 x ⎪
⎩ 3x ⎭ ⎣ 1 ⎩ ⎭
Cannot find a solution because

det[K ] = k1k 2 (k1 + k 2 ) − k1k 22 − k 2 k12 = 0

15
Boundary Conditions

dˆ1x = 0 dˆ3 x dˆ2 x

k1 k2
3 2 x̂
1
① ②
F1x F3 x F2 x

⎧ F1x ⎫ ⎡ k1 0 − k1 ⎤ ⎧ 0 ⎫
⎪ ⎪ ⎢ ⎥ ⎪ˆ ⎪
⎨ F2 x ⎬ = ⎢ 0 k2 − k 2 ⎥ ⎨d 2 x ⎬
⎪ F ⎪ ⎢− k − k2 k1 + k 2 ⎥⎦ ⎪⎩ dˆ3 x ⎪⎭
⎩ 3x ⎭ ⎣ 1

F1x = 0 +0 − k1dˆ3 x
F2 x = 0 + k 2 dˆ2 x − k dˆ
2 3x
F3 x = 0 − k dˆ 2 2x + (k1 + k 2 )dˆ3 x

16
Boundary Conditions

F1x = 0 +0 − k1dˆ3 x
F2 x = 0 + k 2 dˆ2 x − k dˆ2 3x
F3 x = 0 − k dˆ 2 2x + (k1 + k 2 )dˆ3 x

⎧ F1x ⎫ ⎡ k1 0 − k1 ⎤ ⎧ 0 ⎫
⎪ ⎪ ⎢ ⎪ ⎪
⎨ F2 x ⎬ = ⎢ 0 k2 − k 2 ⎥⎥ ⎨dˆ2 x ⎬
⎪ F ⎪ ⎢− k − k2 k1 + k 2 ⎥⎦ ⎪⎩ dˆ3 x ⎪⎭
⎩ 3x ⎭ ⎣ 1

⎡ k2 − k2 ⎤
det ⎢ ⎥ ≠0
⎣− k 2 k1 + k 2 ⎦

How to find F1x ?

17
Boundary Conditions

dˆ1x = 0 dˆ3 x dˆ2 x

k1 k2
3 2 x̂
1
① ②
F1x F3 x F2 x

⎧ F1x ⎫ ⎡ k1 0 − k1 ⎤ ⎧ 0 ⎫
⎪ ⎪ ⎢ ⎪ ⎪
⎨ F2 x ⎬ = ⎢ 0 k2 − k 2 ⎥⎥ ⎨dˆ2 x ⎬
⎪ F ⎪ ⎢− k − k2 k1 + k 2 ⎥⎦ ⎪⎩ dˆ3 x ⎪⎭
⎩ 3x ⎭ ⎣ 1

F1x = 0 + 0 − k1dˆ3 x
F2 x = 0 + k 2 dˆ2 x − k 2 dˆ3 x (dˆ ˆ
2 x , d3x )
F3 x = 0 − k 2 dˆ2 x + (k1 + k 2 )dˆ3 x
18
Boundary Conditions

(dˆ ˆ
2 x , d3x ) F1x = 0 + 0 − k1dˆ3 x

⎧ F1x ⎫ ⎡ k1 0 − k1 ⎤ ⎧ 0 ⎫
⎪ ⎪ ⎢ ⎪ ⎪
⎨ F2 x ⎬ = ⎢ 0 k2 − k 2 ⎥⎥ ⎨dˆ2 x ⎬
⎪ F ⎪ ⎢− k − k2 k1 + k 2 ⎥⎦ ⎪⎩ dˆ3 x ⎪⎭
⎩ 3x ⎭ ⎣ 1

19
Global Stiffness Matrix & Direct Stiffness Method

For the system with two springs: ⎧ F1x ⎫ ⎡ k1 0 − k1 ⎤ ⎧ dˆ1x ⎫

⎪ ⎪ ⎢ ⎥ ⎪ˆ ⎪
⎨ F2 x ⎬ = ⎢ 0 k2 − k 2 ⎥ ⎨d 2 x ⎬
⎪ F ⎪ ⎢− k − k2 k1 + k 2 ⎥⎦ ⎪dˆ3 x ⎪
⎩ 3x ⎭ ⎣ 1 ⎩ ⎭

1 2 3

dˆ1x dˆ2 x dˆ3 x

⎡ k1 0 − k1 ⎤ F1x 1

[K ] = ⎢⎢ 0 k2 − k 2 ⎥⎥ F2 x 2
⎢⎣− k1 − k2 k1 + k 2 ⎥⎦ F3 x 3

⎡ K11 K12 K13 ⎤

[K ] = ⎢⎢ K 21 K 22 K 23 ⎥⎥ K ij = K ji
⎢⎣ K 31 K 32 K 33 ⎥⎦

20
Global Stiffness Matrix & Direct Stiffness Method

General Form of Global Stiffness Matrix

For a system with N DOFs

1 … i … j … N

⎡ ⎤ 1
⎢ ⎥ …
⎢ ⎥
⎢ K ij ⎥ i
⎢ ⎥
[K ] = ⎢ ⎥ …
⎢ ⎥ j
⎢ ⎥
⎢ ⎥ …
⎢ ⎥
⎣ ⎦ N

N by N matrix
K ii K jj K ij

21
Global Stiffness Matrix & Direct Stiffness Method

Direct Stiffness Method

Element stiffness matrix

For element ① For element ②

⎡ k1 − k1 ⎤ ⎡ k2 − k2 ⎤
⎢− k ⎢− k k 2 ⎥⎦
⎣ 1 k1 ⎥⎦ ⎣ 2

Expanding Stiffness matrix

⎡ k1 0 − k1 ⎤ ⎡0 0 0 ⎤
⎢ 0 0 0 ⎥⎥ ⎢0 k
⎢ ⎢ 2 − k 2 ⎥⎥
⎢⎣ − k1 0 k1 ⎥⎦ ⎢⎣0 − k 2 k 2 ⎥⎦

Adding element matrixes together

⎡ k1 0 − k1 ⎤ ⎡0 0 0 ⎤ ⎡ k1 0 − k1 ⎤
⎢ 0 0 0 ⎥ + ⎢0 k − k 2 ⎥⎥ [K ] = ⎢⎢ 0 k2 − k 2 ⎥⎥
⎢ ⎥ ⎢ 2
⎢⎣− k1 0 k1 ⎥⎦ ⎢⎣0 − k 2 k 2 ⎥⎦ ⎢⎣− k1 − k2 k1 + k 2 ⎥⎦
22
Global Stiffness Matrix & Direct Stiffness Method

Direct Stiffness Method

k2
k1

1 ① 3 3 ② 2

Element number
1 3 2 3
⎡k11(1) ⎡ k2 − k2 ⎤ 2 ⎡k 22 ⎤
⎡ k1 − k1 ⎤ 1 k13(1) ⎤
( 2) ( 2)
k 23
⎢− k ⎢ ( 2) ( 2) ⎥
⎢− k
⎣ 1 k1 ⎥⎦ ⎢ (1) (1) ⎥ ⎣ 2 k 2 ⎥⎦ ⎣k 23 k33 ⎦
3
3
⎣k13 k33 ⎦

Nodal number Nodal number

23
Global Stiffness Matrix & Direct Stiffness Method

Direct Stiffness Method

⎡k11(1) k13(1) ⎤
⎢ (1) ⎡k33
( 2) ( 2)
k32 ⎤
(1) ⎥
⎣k13 k33 ⎦ ⎢ ( 2) ( 2) ⎥
⎣k32 k 22 ⎦
Expanding Stiffness matrix

1 2 3 1 2 3
⎡ k11(1) 0 k13(1) ⎤ 1 ⎡0 0 0 ⎤ 1
⎢ ⎥ ⎢0 k ( 2 ) ( 2) ⎥
⎢ 0 0 0 ⎥ 2
⎢ 22 k 23 ⎥ 2
⎢ k13(1) (1) ⎥
0 k 33 ⎢⎣0 k 23
( 2) ( 2)
⎥⎦ 3
⎣ ⎦ 3 k 33

⎡k11(1) 0 k13(1) ⎤
⎢ ( 2) ( 2) ⎥
⎢ 0 k 22 k 23 ⎥
⎢k13(1) ( 2)
k 23 (1)
k33 + k33 ( 2) ⎥
⎣ ⎦
24
Global Stiffness Matrix & Direct Stiffness Method

Direct Stiffness Method Element I

i j
N by N matrix ⎡kii( I ) kij( I ) ⎤
⎢ (I ) ⎥
1 … i … j … N ⎣⎢kij k (jjI ) ⎦⎥
⎡ ⎤ 1
⎢ ⎥ …
⎢ ⎥
⎢ ⎥ i
[K ] = ⎢⎢ ⎥
⎥ …
⎢ ⎥ j
⎢ ⎥
⎢ ⎥ …
⎢ ⎥
⎣ ⎦ N

25
Global Stiffness Matrix & Direct Stiffness Method

Direct Stiffness Method

Example
k1 k2 k3 k4 k5 k6

1 2 3 4 5 6 7

26
27
Global Stiffness Matrix & Direct Stiffness Method

Direct Stiffness Method

Example

⎡ k1 − k1 0 0 0 0 0 ⎤
⎢ −k k1 + k 2 − k2 0 0 0 0 ⎥
⎢ 1 ⎥
⎢ 0 − k2 k 2 + k3 − k3 0 0 0 ⎥
⎢ ⎥
⎢ 0 0 − k3 k3 + k 4 − k4 0 0 ⎥
⎢ 0 0 0 − k4 k 4 + k5 − k5 0 ⎥
⎢ ⎥
⎢ 0 0 0 0 − k5 k5 + k 6 − k6 ⎥
⎢ 0 − k6 ⎥
⎣ 0 0 0 0 k6 ⎦

28
Global Stiffness Matrix & Direct Stiffness Method

Physical Implication of K ij
The product of the i-th row of the global stiffness matrix and the global
displacement matrix gives the external force on the i-th DOF of the system.

⎧ ⎫ ⎡ ⎤ ⎧ d1 ⎫
⎪ ⎪ ⎢ ⎥ ⎪ ... ⎪
⎪⎪ ⎪⎪ ⎢ ⎥ ⎪⎪ ⎪⎪
⎨ Fi ⎬ = ⎢ K i1 ... K ij ... K iN ⎥ ⎨ d j ⎬ Fi = K i1d1 + ... + K ij d j + ... + K iN d N
⎪ ⎪ ⎢ ⎥⎪ ⎪
⎪ ⎪ ⎢ ⎥ ⎪ ... ⎪
⎪⎩ ⎪⎭ ⎢⎣ ⎥⎦ ⎪⎩d N ⎪⎭

Now, we consider a special case, where

29
Global Stiffness Matrix & Direct Stiffness Method

Physical Implication of K ij

Now, we consider a special case, where

d1 = d 2 = d j −1 = d j +1 = d N = 0 d j =1

Fi = K ij

So, Kij is equal to the reaction force on the i-th DOF due to
a unit displacement on the j-th DOF whereas all the other
DOFs are fixed.

30
Global Stiffness Matrix & Direct Stiffness Method

Physical Implication of K ij
km kn

i j

km kn

i j

31
Global Stiffness Matrix & Direct Stiffness Method

The next step in FEA is to solve the following linear equations

⎡ ⎤ ⎧ d1 ⎫ ⎧ F1 ⎫
⎢ ⎥⎪ d ⎪ ⎪ F ⎪
⎢ 0 ⎥⎪ 2 ⎪ ⎪ 2 ⎪
⎢ ⎥ ⎪ ... ⎪ ⎪ ... ⎪
⎢ ⎥⎪ ⎪ ⎪ ⎪
⎢ ⎥ ⎪ ... ⎪ = ⎪ ... ⎪
⎢ ⎥ ⎨ ... ⎬ ⎨ ... ⎬
⎢ ⎥⎪ ⎪ ⎪ ⎪
⎢ Symm. ⎥ ⎪ ... ⎪ ⎪ ... ⎪
⎢ ⎥ ⎪ ... ⎪ ⎪ ... ⎪
⎢ ⎥⎪ ⎪ ⎪ ⎪
⎢⎣ ⎥⎦ ⎪⎩d N ⎪⎭ ⎪⎩ FN ⎪⎭

In FEA, the global stiffness is symmetric and sparse. This offers

FEA a big advantage because the linear equations with such
properties can be solved with a very high efficiency.
32