Practice Problems

Special Topics: Finite Element Method

Question 1: Consider the below. A piping system is idealized as a number of ‘springs’. The system has 4 elements

and 4 nodes below.

Using the principle of minimum potential energy, find the stiffness matrix, the load vector and displacement vector and state the problem in the form [𝐾]{𝑢} = {𝐹}

a. Compute the reaction at node 1

b. Find an expression for the potential energy: 𝐼 = 𝜋 − 𝑊𝑝 in terms of nodal displacements 𝑢1, 𝑢2, 𝑢3 𝑎𝑛𝑑 𝑢4

c. Find the derivative of the potential energy w.r.t. the nodal displacements and set to zero

d. Write the resulting equations in matrix form

e. Apply the boundary condition 𝑢1 = 0

Solution:

a. Defining the positive x-axis to be towards the right, the reaction at 1 is – 𝑃

b. Potential Energy is given by:

𝐼 =1

2𝑘(𝑢2 − 𝑢1)2 +

1

2𝑘(𝑢3 − 𝑢2)2 +

1

2𝑘(𝑢3 − 𝑢2)2 +

1

2𝑘(𝑢4 − 𝑢3)2 − 𝑃𝑢4 + 𝑃𝑢1

c. The derivate can be computed as:

𝜕𝐼

𝜕𝑢1= −𝑘(𝑢2 − 𝑢1) + 𝑃 = 0

𝜕𝐼

𝜕𝑢2= 𝑘(𝑢2 − 𝑢1) − 𝑘(𝑢3 − 𝑢2) − 𝑘(𝑢3 − 𝑢2) = 0

𝜕𝐼

𝜕𝑢3= 𝑘(𝑢3 − 𝑢2) + 𝑘(𝑢3 − 𝑢2) − 𝑘(𝑢4 − 𝑢3) = 0

𝜕𝐼

𝜕𝑢4= 𝑘(𝑢4 − 𝑢3) − 𝑃 = 0

d. Rewriting in Matrix Form: [𝐾]{𝒖} = {𝑭}

𝑘 [

1−100

−13

−20

0−23

−1

00

−11

] {

𝑢1

𝑢2

𝑢3

𝑢4

} = {

−𝑃00𝑃

}

Notice that the stiffness matrix is symmetric, as it should be

e. Applying Boundary Condition:

𝑘 [3 −2 0

−2 3 −10 −1 1

] {

𝑢2

𝑢3

𝑢4

} = {00𝑃

}

The displacements 𝑢1, 𝑢2 and 𝑢3 may be found by solving the system of equations, (by taking the inverse of

the stiffness matrix and multiplying it with the load vector)

Question 2: Find the matrix of shape functions [𝑁(𝑥)] for a one-dimensional quadratic interpolation function:

𝑢(𝑥) = 𝑎1 + 𝑎2𝑥 + 𝑎3𝑥2

The constants 𝑎1, 𝑎2 and 𝑎3 can be evaluated by ensuring

𝑢(𝑥) = 𝑞𝑖 𝑎𝑡 𝑥 = 0

𝑢(𝑥) = 𝑞𝑗 𝑎𝑡 𝑥 =𝑙

2

𝑢(𝑥) = 𝑞𝑘 𝑎𝑡 𝑥 = 𝑙

Find the shape function [𝑁(𝑥)] can be found by rearrangement such that:

𝑢(𝑥) = [𝑁(𝑥)]{𝒒(𝒆)} = [𝑁1 𝑁2 𝑁3] {

𝑞𝑖

𝑞𝑗

𝑞𝑘

}

Solution:

The systematic method of obtaining the shapes functions is as follows:

The constants 𝑎1, 𝑎2 and 𝑎3 need to be evaluated using:

𝑢(0) = 𝑞𝑖 = 𝑎1

𝑢 (𝑙

2) = 𝑞𝑗 = 𝑎1 + 𝑎2 (

𝑙

2) + 𝑎3 (

𝑙

2)

2

𝑢(𝑙) = 𝑞𝑘 = 𝑎1 + 𝑎2𝑙 + 𝑎3𝑙2

Solving for 𝑎1, 𝑎2 and 𝑎3:

𝑎1 = 𝑞𝑖

𝑎2 + 𝑎3 (𝑙

2) = (𝑞𝑗 − 𝑞𝑖)

2

𝑙

𝑎2 + 𝑎3𝑙 = (𝑞𝑘 − 𝑞𝑖)1

𝑙

Solving simultaneously:

𝑎2 = [(𝑞𝑗 − 𝑞𝑖)2

𝑙− (𝑞𝑘 − 𝑞𝑖)

1

2𝑙] ÷ [1 −

1

2] → 𝑎2 =

(4𝑞𝑗 − 4𝑞𝑖 − 𝑞𝑘 + 𝑞𝑖)

𝑙

→ 𝑎2 =1

𝑙(4𝑞𝑗 − 3𝑞𝑖 − 𝑞𝑘)

𝑎3 = [(𝑞𝑘 − 𝑞𝑖)1

𝑙− (𝑞𝑗 − 𝑞𝑖)

2

𝑙] ÷ [𝑙 −

𝑙

2] → 𝑎3 =

2(𝑞𝑘 − 𝑞𝑖 − 2𝑞𝑗 + 2𝑞𝑖)

𝑙2

→ 𝑎3 =2

𝑙2 (𝑞𝑘 + 𝑞𝑖 − 2𝑞𝑗)

Now, we want to rewrite 𝑢(𝑥) = 𝑎1 + 𝑎2𝑥 + 𝑎3𝑥2 … as 𝑢(𝑥) = 𝑁1𝑞𝑖 + 𝑁2𝑞𝑖 + 𝑁3𝑞𝑖

Where 𝑁1, 𝑁2 and 𝑁3 are shape functions, which are a function of 𝑥

𝑢(𝑥) = 𝑞𝑖 +1

𝑙(4𝑞𝑗 − 3𝑞𝑖 − 𝑞𝑘)𝑥 +

2

𝑙2 (𝑞𝑘 + 𝑞𝑖 − 2𝑞𝑗)𝑥2

Gathering terms with 𝑞𝑖

𝑁1(𝑥) = 1 −3𝑥

𝑙+

2𝑥2

𝑙2

Gathering terms with 𝑞𝑗

𝑁2(𝑥) =4𝑥

𝑙−

4𝑥2

𝑙2

Gathering terms with 𝑞𝑘

𝑁3(𝑥) = −𝑥

𝑙+

2𝑥2

𝑙2

And finally the matrix of shape functions is:

[𝑁] = [𝑁1(𝑥) 𝑁2(𝑥) 𝑁3(𝑥)]

Question 3:

Identify the displacements and stresses generated due to the load hanged in the truss structure as shown

below (use minimum potential energy theorem).

All Elements have E =200 GPa and A = 5*10-4 m2.

The transformation between local and global coordinates and the stiffness matrix for 2D Truss elements in global

coordinate system is given by:

Solution:

The structure is divided into three elements and four nodes, as marked on the figure. Let’s set the global coordinate

system as shown, with the origin at node 1

The elements (1), (2) and (3) are oriented at an angle of 90°, 180° and 240° w.r.t. the global coordinate system.

For simplicity, let me just change the question so the lengths of all elements is 2 meters. Defining:

𝑘𝑒 =𝐴𝑒𝐸𝑒

𝑙𝑒=

(5 × 10−4)(200 × 109)

2= 50 × 106 𝑁 𝑚⁄

The direction cosines for all elements are shown below:

𝑋

𝑌

Element (1) Element (2) Element (3) 𝑙 = cos 90° = 0 𝑚 = sin 90° = 1

𝑙 = cos 180° = −1 𝑚 = sin 180° = 0

𝑙 = cos 240° = −1/2

𝑚 = sin 240° = −√3/2 The element stiffness matrix in global coordinate system:

[𝐾1] = 𝑘𝑒 [

0000

010

−1

0000

0−101

]

The element stiffness matrix in global coordinate system:

[𝐾2] = 𝑘𝑒 [

10

−10

0000

−1010

0000

]

The element stiffness matrix in global coordinate system:

[𝐾3] =𝑘𝑒

4[

1

√3−1

−√3

√33

−√3−3

−1

−√31

√3

−√3−3

√33

]

Make sure the resulting matrix is symmetric

Next we must assemble the Global Stiffness Matrix. To do this, we must recognize which terms in the global

stiffness matrix refer to which global displacements. Lets stick to the conventional notation for using capital U to

designate deflection along the x-axis, and capital V to designate the deflection along the y-axis in the global

coordinate system, with the subscript donating the node number. For example 𝑈1 designates deflection of node

along x-axis, 𝑉4 designates the deflection along y-axis of node 4, and so on

Element 1, is located between nodes 1 and 2:

[𝐾1] = 𝑘𝑒 [

0000

010

−1

0000

0−101

]

𝑈1

𝑉1

𝑈2

𝑉2

Element 2, is located between nodes 1 and 3:

[𝐾2] = 𝑘𝑒 [

10

−10

0000

−1010

0000

]

𝑈1

𝑉1

𝑈3

𝑉3

Element 3, is located between nodes 1 and 4:

[𝐾3] =𝑘𝑒

4[

1

√3−1

−√3

√33

−√3−3

−1

−√31

√3

−√3−3

√33

]

𝑈1

𝑉1

𝑈4

𝑉4

𝑈1 𝑉1 𝑈2 𝑉2

𝑈1 𝑉1 𝑈3 𝑉3

𝑈1 𝑉1 𝑈4 𝑉4

The global stiffness matrix may be compiled by superimposing the correct terms in the right places:

U1 V1 U2 V2 U3 V3 U4 V4

1.25 0.433 0 0 -1 0 -1 -

0.433 U1

0.433 1.75 0 -1 0 0 -0.43 0 V1

0 0 0 0 0 0 0 0 U2

[𝐾𝐺] = 𝑘𝑒 0 -1 0 1 0 0 0 0 V2

-1 0 0 0 1 0 0 0 U3

0 0 0 0 0 0 0 0 V3

-0.25 -0.43 0 0 0 0 0.25 0.433 U4

-0.43 0 0 0 0 0 0.433 0.75 V4

Applying the boundary conditions, we’ll get rid of all rows and columns, except those corresponding to 𝑈1 𝑎𝑛𝑑 𝑉1

[𝐾]{𝒖} = {𝑭} will reduce to:

𝑘𝑒 [1.25 0.433

0.433 1.75] {

𝑈1

𝑉1} = {

0−40,000

}

Solving:

𝑈1 = 0.1732 𝑚𝑚

𝑉1 = −0.5 𝑚𝑚