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Lecture Notes: Introduction to Finite Element Method

Chapter 2. Bar and Beam Elements

Example 2.1

1

2A,E

 

2

A,E

 

1

 

2

P

3

x

 

L

 

 

L

 

Problem: Find the stresses in the two bar assembly which is loaded with force P, and constrained at the two ends, as shown in the figure.

Solution: Use two 1-D bar elements.

Element 1,

 

 

 

 

u1

u2

 

k1

=

2EA

1

1

L

 

 

1

 

 

 

1

 

Element 2,

 

 

 

 

u2

u3

k 2

=

EA

1

1

 

 

 

 

L

 

 

 

1

1

Imagine a frictionless pin at node 2, which connects the two elements. We can assemble the global FE equation as follows,

© 1997-2002 Yijun Liu, University of Cincinnati

32

Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements

2 2

0 u

 

F

 

EA

1

 

1

 

2 3

1 u2

 

= F2

 

L 0 1

1 u

 

F

 

 

3

 

3

 

Load and boundary conditions (BC) are,

u1 = u3 = 0,

F2 = P

 

FE equation becomes,

 

 

 

 

2 2

0 0

 

F

 

EA

 

 

1

 

2 3

1 u2

 

= P

 

L 0 1

1 0

 

F

 

 

 

 

3

 

Deleting the 1st row and column, and the 3rd row and column, we obtain,

EAL [3]{u2 } = {P}

Thus,

u2 = 3PLEA

and

u

 

 

 

0

1

 

 

PL

 

u2

 

=

 

1

 

 

 

 

3EA

 

u3

 

 

 

0

Stress in element 1 is

© 1997-2002 Yijun Liu, University of Cincinnati

33

Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements

σ1 = Eε1 = EB1u1 = E[1/ L

1

u

 

/ L] 1

 

 

 

u2

 

= E u2 L u1 = EL 3PLEA 0 = 3PA

Similarly, stress in element 2 is

σ2 = Eε2 = EB2u2 = E[1/ L

1

u

 

/ L] 2

 

 

 

u3

 

= E

u u

2

=

E

PL

= −

P

3

0

 

 

 

L

 

 

3A

 

 

 

L

 

3EA

 

which indicates that bar 2 is in compression.

Check the results!

Notes:

In this case, the calculated stresses in elements 1 and 2 are exact within the linear theory for 1-D bar structures. It will not help if we further divide element 1 or 2 into smaller finite elements.

For tapered bars, averaged values of the cross-sectional areas should be used for the elements.

We need to find the displacements first in order to find the stresses, since we are using the displacement based FEM.

© 1997-2002 Yijun Liu, University of Cincinnati

34

Lecture Notes: Introduction to Finite Element Method

Chapter 2. Bar and Beam Elements

Example 2.2

 

 

 

 

1

A,E

 

2

 

1

2

P

3

x

L

 

 

L

 

Problem: Determine the support reaction forces at the two ends of the bar shown above, given the following,

P = 6.0 ×104 N,

E = 2.0 ×104 N / mm2 ,

A = 250 mm2 ,

L = 150 mm, =1.2 mm

Solution:

We first check to see if or not the contact of the bar with the wall on the right will occur. To do this, we imagine the wall on the right is removed and calculate the displacement at the right end,

∆ = PL = (6.0 ×104 )(150) = > ∆ =

0 18.mm 12. mm EA (2.0 ×104 )(250)

Thus, contact occurs.

The global FE equation is found to be,

© 1997-2002 Yijun Liu, University of Cincinnati

35

Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements

 

1 1

0 u

 

F

 

EA

1

 

1

 

L

1 2

1 u2

 

= F2

 

0 1

1 u

 

F

 

 

 

3

 

3

 

The load and boundary conditions are,

 

F = P = 6.0 ×104 N

 

 

 

2

 

 

 

 

 

u1

= 0,

u3 = ∆ = 1.2 mm

FE equation becomes,

 

 

 

 

 

1 1

0 0

 

F

 

EA

 

 

1

 

L

1 2

1 u2

 

= P

 

0 1

1

 

F

 

 

 

 

 

3

 

The 2nd equation gives,

EAL [2 1] u2 = {P}

that is,

EA

[2]{u2

 

EA

 

L

}= P +

L

 

 

 

Solving this, we obtain

u2 = 21 EAPL + ∆ = 15.mm

and

© 1997-2002 Yijun Liu, University of Cincinnati

36

Lecture Notes: Introduction to Finite Element Method Chapter 2. Bar and Beam Elements

u

 

 

0

 

1

 

 

 

(mm)

u2

 

= 15.

 

 

 

 

 

u3

 

12.

 

To calculate the support reaction forces, we apply the 1st and 3rd equations in the global FE equation.

The 1st equation gives,

 

EA

 

 

u1

 

 

EA

 

 

 

 

 

 

 

 

 

 

 

 

4

 

F1 =

 

 

[1

1

0] u2

 

=

 

 

(u2 ) = −5.0

× 10

 

N

L

L

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

u3

 

 

 

 

 

 

 

 

and the 3rd equation gives,

 

 

 

 

 

u

 

 

 

 

 

 

 

F =

EA

[

0 1

1

u1

 

=

EA

(u

 

+ u

)

 

 

 

 

3

L

 

] 2

 

L

2

3

 

 

 

 

 

 

u3

 

 

 

 

 

 

 

 

= −1.0 ×10

4 N

 

 

 

 

 

 

 

 

Check the results.!

© 1997-2002 Yijun Liu, University of Cincinnati

37

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