MESF593 Finite Element Methods

HW #2 Solutions

Prob. #1 (25%)

The element equations of a general tapered beam with a rectangular cross-section are given above. Derive the explicit expression of each term of Kij in terms of the dimensions (a, b, c, L) and Young’s modulus (E) of the beam.

4

3

2

1

44434241

34333231

24232221

14131211

4

3

2

1

4

3

2

1

v

v

v

v

KKKK

KKKK

KKKK

KKKK

q

q

q

q

Q

Q

Q

Q

x = 0

q(x)

x = L

Q1 , v1

Beam Width: b, Beam Thickness: a+(c-a)x/L

v

xQ2 , v2 Q4 , v4

Q3 , v3

L

02

j2

2i

2

ij dxdx

d

dx

dEIK where

L

x

L

xx

L

x2

L

x3

L

x1x

L

x2

L

x31

2

32

2

32

4

3

2

1

Prob. #1 Solution

x = 0

q(x)

x = L

Q1 , v1

Beam Width: b, Beam Thickness: a+(c-a)x/L

v

xQ2 , v2 Q4 , v4

Q3 , v3

L

02

j2

2i

2

3

L

02

j2

2i

2

ij dxdx

d

dx

d

12Lx

acabEdx

dx

d

dx

dEIK

L

1

L

x

L

2x

L

x

L

x

L

x

L

6

L

x

L

6

L

x1

L

x2

L

x1

L

x

L

6

L

x

L

6

dx

ddx

ddx

ddx

d

2

2

2

2

4

3

2

1

L

2

L

x6L

6

L

x12L

4

L

x6L

6

L

x12

dx

ddx

ddx

ddx

d

2

23

2

23

24

2

23

2

22

2

21

2

L

x

L

xx

L

x2

L

x3

L

x1x

L

x2

L

x31

2

32

2

32

4

3

2

1

Prob. #1 Solution

133311 KKK

33

66

23

55

3

44L

0

2

21

23

11acL25

acEb6

acL5

acEb3

acL4

acEb3dx

dx

d

L

xaca

12

EbK

32

66

22

55

2

44L

022

2

21

23

12acL10

acEb

acL20

a7c19Eb

acL4

a2c11Ebdx

dx

d

dx

d

L

xaca

12

EbK

32

66

22

55

2

44L

024

2

21

23

14acL10

acEb

acL40

a19c14Eb

acL12

a3c5Ebdx

dx

d

dx

d

L

xaca

12

EbK

3

66

2

5544L

0

2

22

23

22acL20

acEb

acL10

a2cEb

acL12

a4cEbdx

dx

d

L

xaca

12

EbK

1223 KK

32

66

22

55

2

44L

024

2

22

23

24acL20

acEb

acL40

a19c14Eb

acL12

a3c5Ebdx

dx

d

dx

d

L

xaca

12

EbK

1434 KK

3

66

2

5544L

0

2

24

23

44acL20

acEb

acL10

a2cEb

acL12

ac4Ebdx

dx

d

L

xaca

12

EbK

jiij KK

Prob. #2 (15%)

For the truss structure shown above at the left hand side, is it possible to use the finite elements shown at the right hand side for modeling? Why?

3 Linear Bar Elements:

3 Quadratic Bar Elements:

6 Nodes:

FR1 , u1

S1 , v1

R3 , u3

S2 , v2

R2 , u2

S3 , v3

Prob. #2 SolutionConnection to the mid node of a higher order bar element is not allowed. This is because the degree of freedom (i.e., displacement) of the mid node must be along the axial direction of the bar element as shown in Fig. 1. However, if the mid node is tied to the end node of another linear element, that means this common node can move in any direction. As a result, the higher order bar element may not be a “straight” element.

u1 u3

U2? P3P1 P2?

u1

u3

u2 P3

P1

P2

Fig. 1

Fig. 2

Prob. #3 (30%)

Derive the explicit form of the stiffness matrix Kij of a combined bar-beam element arbitrarily oriented on a 2-D plane.

E, A, I, L

Y1 , v1

Y2 , v2

M1 , 1

X1 , u1

M2 , 2

X2 , u2

x-

2

2

2

1

1

1

ij

2

2

2

1

1

1

v

u

v

u

K

M

Y

X

M

Y

X

Note: The plane coordinate transformation matrix for a rotation about the z-axis is

100

0cossin

0sincos

Prob. #3 Solution

2

'2

'2

1

'1

'1

22

2323

22

2323

2

'2

'2

1

'1

'1

'ij

2

'

'1

'

'

v

u

v

u

L

EI4

L

EI60

L

EI2

L

EI60

L

EI6

L

EI120

L

EI6

L

EI120

00L

AE00

L

AEL

EI2

L

EI60

L

EI4

L

EI60

L

EI6

L

EI120

L

EI6

L

EI120

00L

AE00

L

AE

v

u

v

u

K

M

Y

X

M

Y

X

2

2

1

1

E, A, I, LY’1 , v’1 Y’2 , v’2

M1 , 1

X’1 , u’1

M2 , 2

X’2 , u’2

From the lecture notes

Prob. #3 Solution

2

1

2

'

'1

'

'

M

Y

X

M

Y

X

100000

0cossin000

0sincos000

000100

0000cossin

0000sincos

M

Y

X

M

Y

X

2

2

1

1

2

2

1

1

2

'

'1

'

'

2

1

M

Y

X

M

Y

X

100000

0cossin000

0sincos000

000100

0000cossin

0000sincos

M

Y

X

M

Y

X

2

2

1

1

2

2

1

1

2

1

2

'

'

1

'

'

2

2

1

1

2

2

1

1

v

u

v

u

100000

0cossin000

0sincos000

000100

0000cossin

0000sincos

v

u

v

u

2

2

2

1

1

1

2

2

2

1

1

1

''

2

'

'1

'

'

''

2

'

'1

'

'

'

2

2

2

1

1

1

v

u

v

u

K

v

u

v

u

T K T

v

u

v

u

K T

M

Y

X

M

Y

X

T

M

Y

X

M

Y

X

ijij

2

2

1

1

ij

2

2

1

1

Prob. #3 Solution

L

EI4

L

EI6C

L

EI6S

L

EI2

L

EI6C

L

EI6S

L

EI6C

L

EI12C

L

AES

L

EI12CS

L

AECS

L

EI6C

L

EI12C

L

AES

L

EI12CS

L

AECS

L

EI6S

L

EI12CS

L

AECS

L

EI12S

L

AEC

L

EI6S

L

EI12CS

L

AECS

L

EI12S

L

AEC

L

EI2

L

EI6C

L

EI6S

L

EI4

L

EI6C

L

EI6S

L

EI6C

L

EI12C

L

AES

L

EI12CS

L

AECS

L

EI6C

L

EI12C

L

AES

L

EI12CS

L

AECS

L

EI6S

L

EI12CS

L

AECS

L

EI12S

L

AEC

L

EI6S

L

EI12CS

L

AECS

L

EI12S

L

AEC

sinS ,cosC TKTK

2222

2322

32322

3

23322

23322

2222

2322

32322

3

23322

23322

''ij ij

L

EI40

L

EI6

L

EI20

L

EI6

0L

AE00

L

AE0

L

EI60

L

EI12

L

EI60

L

EI12L

EI20

L

EI6

L

EI40

L

EI6

0L

AE00

L

AE0

L

EI60

L

EI12

L

EI60

L

EI12

K

1 S &0 C ,2/ if

22

2323

22

2323

ij

L

EI40

L

EI6

L

EI20

L

EI6

0L

AE00

L

AE0

L

EI60

L

EI12

L

EI60

L

EI12L

EI20

L

EI6

L

EI40

L

EI6

0L

AE00

L

AE0

L

EI60

L

EI12

L

EI60

L

EI12

K

1 S &0 C ,2/ if

22

2323

22

2323

ij

Prob. #4 (30%)

Use the finite element method to solve for all reaction forces and moments at the boundaries of the anti-symmetric Z-frame structure given above.

(All three segments have the same E, A, I, L)

F

F

Prob. #4 Solution

2

2

2

1

1

1

22

2323

22

2323

I

I

I

I

I

I

v

u

v

u

L

EI4

L

EI60

L

EI2

L

EI60

L

EI6

L

EI120

L

EI6

L

EI120

00L

AE00

L

AEL

EI2

L

EI60

L

EI4

L

EI60

L

EI6

L

EI120

L

EI6

L

EI120

00L

AE00

L

AE

M

V

P

M

V

P

2

2

2

1

1

1

For Element I

(from the lecture notes)

F

F

1 2

3 4

I

II

III

X4, u4

Y4, v4

X3, u3

Y3, v3

X1, u1

Y1, v1

X2, u2

Y2, v2

M4, 4M3, 3

M2, 2M1, 1

Prob. #4 Solution

4

4

4

3

3

3

22

2323

22

2323

III

III

III

III

III

III

v

u

v

u

L

EI4

L

EI60

L

EI2

L

EI60

L

EI6

L

EI120

L

EI6

L

EI120

00L

AE00

L

AEL

EI2

L

EI60

L

EI4

L

EI60

L

EI6

L

EI120

L

EI6

L

EI120

00L

AE00

L

AE

M

V

P

M

V

P

2

2

2

1

1

1

For Element III

For Element II

3

3

3

2

2

2

22

2323

22

2323

II

II

II

II

II

II

v

u

v

u

L

EI40

L

EI6

L

EI20

L

EI6

0L

AE00

L

AE0

L

EI60

L

EI12

L

EI60

L

EI12L

EI20

L

EI6

L

EI40

L

EI6

0L

AE00

L

AE0

L

EI60

L

EI12

L

EI60

L

EI12

M

P

V

M

P

V

2

2

2

1

1

1

(use the result of Prob. #4, let = -/2)

(from the lecture notes)

Prob. #4 Solution

0

0

0

v

u

v

u

0

0

0

L

EI4

L

EI60

L

EI2

L

EI60000000

L

EI6

L

EI120

L

EI6

L

EI120000000

00L

AE00

L

AE000000

L

EI2

L

EI60

L

EI4

L

EI4

L

EI6

L

EI6

L

EI20

L

EI6000

L

EI6

L

EI120

L

EI6

L

EI12

L

AE00

L

AE0000

00L

AE

L

EI60

L

AE

L

EI12

L

EI60

L

EI12000

000L

EI20

L

EI6

L

EI4

L

EI4

L

EI6

L

EI6

L

EI2

L

EI60

0000L

AE0

L

EI6

L

AE

L

EI120

L

EI6

L

EI120

000L

EI60

L

EI12

L

EI60

L

EI12

L

AE00

L

AE

000000L

EI2

L

EI60

L

EI4

L

EI60

000000L

EI6

L

EI120

L

EI6

L

EI120

00000000L

AE00

L

AE

M

Y

X

0

F

0

0

F

0

M

Y

X

3

3

3

2

2

2

22

2323

2222

2323

2323

2222

2323

2323

22

2323

4

4

4

1

1

1

2

2

2

2

23 v

u

L

EI2

L

EI60

L

EI6

L

EI120

00L

AE

M

Y

X

1

1

1

3

3

3

2

23 v

u

L

EI2

L

EI60

L

EI6

L

EI120

00L

AE

M

Y

X

4

4

4

Prob. #4 Solution

3

3

3

2

2

2

222

23

2323

222

23

2323

v

u

v

u

L

EI8

L

EI6

L

EI6

L

EI20

L

EI6L

EI6

L

EI12

L

AE00

L

AE0

L

EI60

L

AE

L

EI12

L

EI60

L

EI12L

EI20

L

EI6

L

EI8

L

EI6

L

EI6

0L

AE0

L

EI6

L

AE

L

EI120

L

EI60

L

EI12

L

EI60

L

EI12

L

AE

0

F

0

0

F

0

(due to anti-symmetry, u2 = -u3, v2 = v3, 2 = -3)

2

2

2

2

2

2

222

23

2323

222

23

2323

v

u

v

u

L

EI8

L

EI6

L

EI6

L

EI20

L

EI6L

EI6

L

EI12

L

AE00

L

AE0

L

EI60

L

AE

L

EI12

L

EI60

L

EI12L

EI20

L

EI6

L

EI8

L

EI6

L

EI6

0L

AE0

L

EI6

L

AE

L

EI120

L

EI60

L

EI12

L

EI60

L

EI12

L

AE

0

F

0

0

F

0

Prob. #4 Solution

L

11

0L

11

0

EI6

FL

v

u

v

u

v

L

EI6

L

EI6L

EI6

L

EI12

0

F & 0uv

u

L

EI6

L

EI6

L

EI6L

EI6

L

EI120

00L

AE

L

EI24L

EI6

L

EI6

L

EI12L

EI6

L

EI120

00L

EI24

L

AE

0

F

0

0

F

0

3

3

3

3

2

2

2

2

2

2

23

2

2

2

2

22

23

3

22

23

3

F3

2F

0

v

u

L

EI2

L

EI60

L

EI6

L

EI120

00L

AE

M

Y

X

2

2

2

2

23

1

1

1

F3

2F

0

L

EI2

L

EI60

L

EI6

L

EI120

00L

AE

M

Y

X

2

23

4

4

4