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Lecture Notes: Introduction to Finite Element Method

Chapter 2. Bar and Beam Elements

Example 2.7

Y

 

P

 

 

 

 

 

 

1

E,I

2

 

 

1

2

3

k

X

 

L

 

L

4

 

Given: P = 50 kN, k = 200 kN/m, L = 3 m,

E = 210 GPa, I = 2×10-4 m4.

Find: Deflections, rotations and reaction forces.

Solution:

The beam has a roller (or hinge) support at node 2 and a spring support at node 3. We use two beam elements and one spring element to solve this problem.

The spring stiffness matrix is given by,

 

v3

v4

ks

k

k

=

 

 

k

k

Adding this stiffness matrix to the global FE equation (see

Example 2.5), we have

© 1997-2002 Yijun Liu, University of Cincinnati

65

Lecture Notes: Introduction to Finite Element Method

Chapter 2. Bar and Beam Elements

v1

θ1

v2

θ2

v3

θ3

12

6L

12

6L

0

0

 

2

6L

2

0

0

 

4L

2L

 

 

24

0

12

6L

 

 

 

EI

 

 

 

2

6L

2

L3

 

 

 

8L

2L

 

 

 

 

 

12 + k'

6L

 

 

 

 

 

 

4L2

 

Symmetry

 

 

 

 

 

 

in which

 

 

 

 

 

 

 

k'=

L3

k

 

 

 

 

EI

 

 

 

 

 

 

 

 

 

is used to simply the notation.

 

 

We now apply the boundary conditions,

 

v1 =θ1 = v2 = v4 = 0,

 

 

M2 = M3 = 0,

 

F3Y = −P

v4

 

 

 

 

 

 

 

 

0

v1

 

F1Y

 

0

 

 

 

 

 

 

 

θ1

 

M1

 

0

v

 

 

F

 

 

0

 

 

2

 

 

2Y

 

θ

 

= M

2

 

 

 

 

2

 

 

 

 

k' v3

 

F3Y

 

0

 

 

 

 

 

 

 

θ3

 

M3

 

k' v4

F4Y

‘Deleting’ the first three and seventh equations (rows and columns), we have the following reduced equation,

8L2

6L 2L2

θ2

 

0

 

EI

6L

12 + k' 6L

 

 

 

 

L3

 

v3

 

= − P

2

 

2

 

 

 

 

 

6L

 

 

2L

4L

θ3

 

0

 

Solving this equation, we obtain the deflection and rotations at node 2 and node 3,

© 1997-2002 Yijun Liu, University of Cincinnati

66

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

θ2

 

 

PL2

 

3

 

 

 

 

 

 

 

 

v3

 

= −

 

 

7L

EI (12 +

 

 

 

 

7k')

9

 

θ3

 

 

 

 

 

 

The influence of the spring k is easily seen from this result. Plugging in the given numbers, we can calculate

θ2 0.002492 rad

v3 = −0.01744 mθ3 0.007475 rad

From the global FE equation, we obtain the nodal reaction forces as,

F1Y

 

 

69.78 kN

 

 

 

 

 

 

 

M1

 

69.78 kN m

 

F

 

=

116.2 kN

 

 

 

 

 

2Y

 

F

 

 

3.488 kN

 

 

4Y

 

 

 

 

Checking the results: Draw free body diagram of the beam

69.78 kN

50 kN

1

 

 

 

 

3

 

2

 

 

 

 

 

 

69.78 kN m

 

 

 

 

 

 

 

 

116.2 kN

3.488 kN

 

 

© 1997-2002 Yijun Liu, University of Cincinnati

67

Lecture Notes: Introduction to Finite Element Method

Chapter 2. Bar and Beam Elements

FE Analysis of Frame Structures

Members in a frame are considered to be rigidly connected. Both forces and moments can be transmitted through their joints. We need the general beam element (combinations of bar and simple beam elements) to model frames.

Example 2.8

 

 

 

 

 

Y

 

500 lb/ft

 

 

 

3000 lb

1

1

2

 

 

 

 

 

E, I, A

2

 

3

8 ft

 

 

 

 

 

3

 

4

X

 

 

 

12 ft

 

 

Given:

E = 30 ×106 psi,

I = 65 in.4 ,

A = 68. in.2

 

Find: Displacements and rotations of the two joints 1 and 2.

Solution:

For this example, we first convert the distributed load to its equivalent nodal loads.

© 1997-2002 Yijun Liu, University of Cincinnati

68

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

3000 lb

3000 lb

72000 lb-in.

 

3000 lb

1

1

2

72000 lb-in.

 

 

2

 

3

 

 

3

 

4

 

In local coordinate system, the stiffness matrix for a general 2-D beam element is

 

ui

 

vi

θi

 

uj

 

vj

θj

 

EA

 

0

 

0

EA

 

 

0

 

0

 

 

L

 

 

L

 

 

 

 

 

0

12EI

6EI

 

0

12EI

6EI

 

 

 

L3

 

L2

 

 

L3

 

L2

 

 

 

 

 

 

 

 

 

 

 

 

0

 

6EI

4EI

 

0

6EI

2EI

 

k =

EA

 

L2

 

L

EA

 

 

L2

 

L

 

 

 

0

 

0

 

 

0

 

0

 

L

 

 

 

L

 

 

 

 

 

 

12EI

 

6EI

 

12EI

 

6EI

0

 

0

 

L3

L2

 

 

 

L3

L2

 

 

 

 

 

 

 

 

 

 

 

 

0

 

6EI

2EI

 

0

6EI

4EI

 

 

 

2

 

L

 

2

 

L

 

 

 

 

L

 

 

 

 

 

L

 

 

© 1997-2002 Yijun Liu, University of Cincinnati

69

Lecture Notes: Introduction to Finite Element Method

Chapter 2. Bar and Beam Elements

Element Connectivity Table

Element

Node i (1)

Node j (2)

 

 

 

1

1

2

2

3

1

3

4

2

 

 

 

For element 1, we have

 

 

 

u1

v1

θ1

u2

v2

 

 

1417.

0

0

1417.

0

 

 

 

0

0.784

56.4

0

0.784

 

 

 

0

56.4

5417

0

56.4

k1

= k1

 

' = 104 ×

 

0

0

1417.

0

 

 

1417.

 

 

 

0

0.784

56.4

0

0.784

 

 

 

 

 

 

0

56.4

2708

0

56.4

For elements 2 and 3, the stiffness matrix in local system is,

θ2

0

56.4

2708

0 56.4 5417

 

 

 

ui '

vi '

θi '

uj '

v j '

 

 

212.5

0

0

212.5

0

 

 

 

0

2.65

127

0

2.65

 

 

 

0

127

8125

0

127

k2

' = k3

 

' = 104 ×

 

0

0

212.5

0

 

 

212.5

 

 

 

0

2.65

127

0

2.65

 

 

 

 

 

 

0

127

4063

0

127

where i=3, j=1 for element 2 and i=4, j=2 for element 3.

θj '

0

127

4063

0 127 8125

© 1997-2002 Yijun Liu, University of Cincinnati

70

Lecture Notes: Introduction to Finite Element Method

Chapter 2. Bar and Beam Elements

In general, the transformation matrix T is,

 

l

m

0

0

0

0

m

l

0

0

0

0

 

0

0

1

0

0

 

 

0

T =

0

0

0

l

m

 

 

0

 

0

0

0

m

l

 

 

0

 

0

0

0

0

0

1

We have

l = 0, m = 1

for both elements 2 and 3. Thus,

0

1

0

0

0

0

1

0

0

0

0

0

 

 

0

1

0

0

 

0

0

T =

0

0

0

0

1

 

 

0

 

0

0

0

1

0

 

 

0

 

0

0

0

0

0

1

Using the transformation relation,

k = TT k'T

we obtain the stiffness matrices in the global coordinate system for elements 2 and 3,

© 1997-2002 Yijun Liu, University of Cincinnati

71

Lecture Notes: Introduction to Finite Element Method

Chapter 2. Bar and Beam Elements

k2 = 104

and

k3 = 104

 

u3

2.65

 

0

 

127

 

×

 

2.65

 

0

 

 

127

 

u4

2.65

 

0

 

127

 

×

 

2.65

 

0

 

 

127

v3

θ3

u1

v1

θ1

 

0

127

2.65

0

127

212.5

0

0

212.5

0

 

0

8125

127

0

 

 

4063

0

127

2.65

0

127

 

 

212.5

0

0

212.5

0

 

 

0

4063

127

0

8125

 

v4

θ4

u2

v2

θ2

 

0

127

2.65

0

127

212.5

0

0

212.5

0

 

0

8125

127

0

 

 

4063

0

127

2.65

0

127

 

 

212.5

0

0

212.5

0

 

 

0

4063

127

0

8125

 

Assembling the global FE equation and noticing the following boundary conditions,

u3 = v3 = θ3 = u4 = v4 =θ4 = 0

F1X

= 3000lb, F2 X = 0,

F1Y = F2Y = −3000lb,

M1

= −72000lb in., M2

= 72000lb in.

we obtain the condensed FE equation,

© 1997-2002 Yijun Liu, University of Cincinnati

72

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

144.3

 

 

0

 

127

1417.

0

 

0

u

 

 

0

 

213.3

56.4

0

0.784

56.4

v1

 

 

127

 

56.4

 

13542

0

56.4

 

 

 

1

 

 

 

 

2708 θ

 

104 ×

 

 

 

0

 

0

144.3

0

 

127

 

1

 

1417.

 

 

 

u2

 

0

 

0.784

56.4

0

213.3

56.4

v2

 

 

0

 

56.4

 

2708

127

56.4

 

 

 

 

 

 

 

 

13542 θ2

 

 

 

 

 

 

 

 

 

 

 

3000

 

 

 

 

 

 

 

 

 

 

 

 

3000

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

72000

 

 

 

 

 

 

 

 

 

 

=

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3000

 

 

 

 

 

 

 

 

 

 

 

 

72000

 

 

Solving this, we get

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

u1

 

 

 

0.092 in.

 

 

 

 

 

 

 

 

v

 

 

0.00104in.

 

 

 

 

 

 

 

 

1

 

 

0.00139 rad

 

 

 

 

 

 

 

 

θ

 

 

 

 

 

 

 

 

 

 

1

 

=

 

0.0901in.

 

 

 

 

 

 

 

 

u2

 

 

 

 

 

 

 

 

 

 

v2

 

 

0.0018in.

 

 

 

 

 

 

 

 

θ

 

388. ×105 rad

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

To calculate the reaction forces and moments at the two ends, we employ the element FE equations for element 2 and element 3. We obtain,

© 1997-2002 Yijun Liu, University of Cincinnati

73

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

F

 

 

672.7lb

 

3 X

 

 

2210 lb

 

F3Y

 

=

 

 

 

 

 

 

M3

 

60364 lb in.

and

 

 

 

 

F

 

 

2338lb

 

4 X

 

 

3825 lb

 

F4Y

 

=

 

 

 

 

 

 

M4

 

112641 lb in.

Check the results:

Draw the free-body diagram of the frame. Equilibrium is maintained with the calculated forces and moments.

3000 lb

3000 lb

72000 lb-in.

 

3000 lb

 

 

60364 lb-in.

 

672.7 lb

2210 lb

3825 lb

72000 lb-in.

112641 lb-in.

2338 lb

© 1997-2002 Yijun Liu, University of Cincinnati

74

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