Structural Dynamics

CHAPTER 18 Problem 18.1

~ = m 12 I(x) = I ( 1 –

x 2L

0

)

LM N . ~ = mL L 013376 m MN0.363296

3π x

ω% 1 = 4. 3178

2L

EI 1 −

z

F x IJ LMFG 3π IJ EI G 1 − H 2 L K MNH 2 L K

0

~ k 22 =

L

0

~ k12 =

IJ LMFG π IJ K MNH 2 L K

z

FG H

z

FG H

L

EI 1 −

0

= 2.7758

x 2L

x 2L

IJ LMFG π IJ K MNH 2 L K

OP Q

0.363296 . 12978

%2 m % z k% z = ω

2L

For the selected ψ j ( x ) , the stiffness coefficients are computed from Eq. (18.1.4a): ~ k11 =

OP Q

2. Solve eigenvalue problem.

πx

% 1. Set up k% and m.

L

IJ FG1 − cos π x IJ FG1 − cos 3π x IJ dx KH 2LK H 2L K

EI 2.59145 2.7758 ~ k = 3 2.7758 187.7 L

The shape functions selected are

ψ 2 ( x ) = 1 − cos

x 2L

Thus

x

ψ1 ( x ) = 1 − cos

FG H

m 1 −

= 0.363296 mL

x ) 2L

m(x) = m ( 1 –

L

z L

2

cos

2

OP 2 L PQ

πx

2

dx = 2.59145

3π x cos 2L

2

cos

OP PQ

2

dx = 187.7

OP LMF 3π I G J 2 L PQ MNH 2 L K

πx

2

cos

EI L3

z1 =

EI

ω% 2 = 24. 8415

m L4

LM0.9997OP N0.0244Q

z2 =

EI m L4

LM− 0.9406OP N 0.3396Q

3. Determine natural modes from Eq. (18.1.8). ~

FG H

πx

IJ K

FG H

IJ K

FG H

IJ K

3π x 2L πx 3π x . = 10241 − 0.9997 cos − 0.0244 cos 2L 2L

φ 1 ( x ) = 0.9997 1 − cos

2L

+ 0.0244 1 − cos

EI L3

OP PQ

3π x dx 2L

~

FG H

πx

IJ K

3π x 2L πx 3π x = − 0.6010 + 0.9406 cos − 0.3396 cos 2L 2L

φ 2 ( x ) = − 0.9406 1 − cos

2L

+ 0.3396 1 − cos

EI L3

Similarly the mass coefficients are determined from Eq. (18.1.4b): ~ = m 11

z

m 1 −

FG H

x 2L

z

m 1 −

FG H

x 2L

L

0

~ = m 22

L

0

IJ FG1 − cos π x IJ KH 2LK

2

IJ FG1 − cos 3π x IJ KH 2L K

dx = 013376 . mL

2

dx = 12978 mL .

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Problem 18.2

Thus

x L

m(x)=2mo

m(x)=2mo (1 –

LM N

0 EI 48.705 ~ k = 3 0 3945.07 L

x ) L

%. 3. Set up m

mass distribution EI

x L/2

~ = m 11

ψ1′ ( x ) =

π

ψ 1′′( x ) = −

πx

cos

L

ψ 2 ( x ) = sin

L

ψ ′2 ( x ) =

L

FG π IJ H LK

2

sin

πx L

3π L

ψ 2′′ ( x ) = −

3π x

cos

FG 3π IJ H LK

3π x

=

l 2

sin

3π x L

z

FG π IJ EI sin π x dx H LK L F π I L π x L − sin bπ x Lg cos bπ x Lg OP = EI G J M H L K MN 2 2 PQ F π I π = π EI = 48.705 EI = EI G J H LK 2 2 L L 4 L

2mo L

3

L

= 0

k%21

= 0 = k%

12

z

FG 3π IJ EI sin 3π x dx H LK L F 3π I L 3π x L − sin b6 π x Lg OP = EI G J M H L K MN 2 4 PQ F 3π I 3π = (3π ) EI = EI G J H LK 2 2 L

~ k 22 =

L

0

3

3

L

sin 2

2

−

cos (2 π x L) 8

0

=

2

L

L2

o 2

= 0 4 L

πx

LM N 2m L L (π x L) M 4 π MN

0

2

z L

+

−

FG π x IJ d FG π x IJ H LK H LK

OP Q

L2

+ 0

sin (2 π x L) 2mo π x L − π L 2 4

z

2

L2


LM MN

3

FG π IJ FG 3π IJ EI sin π x sin 3π x dx H LK H L K L L F π I F 3π I L sin b− 2π x Lg − sin b4 π x Lg OP = EI G J G J M H L K H L K MN 2 b− 2π Lg 2 b4π Lg PQ

~ k12 =

sin 2


cos (2 π x L) 8

2 L

2

z

sin 2

2mo L (π x L) 2 (π x L) sin (2 π x L) − − 2 4 4 π

4

3

L

L

π π

0

3

2

πx

0

1 L

2mo

2

z

L2

π π

2mo π L

2. Set up k% . ~ k11 =

FG H

2mo 1 −

L2

L

FG IJ H K xI J sin FGH πLx IJK dx LK

πx x dx + sin 2 L L

2mo

0 L

1. Select shape functions.

ψ1 ( x ) = sin

z z

L2

L/2

πx

OP Q

LM MN 2m L L π M π MN 4

2

−

L

− L2

(π x L) sin (2 π x L) − 4

OP Q

L L2

2mo L (π 2) 2 1 1 + + 2 4 8 8 π o 2

OP Q

OP + PQ

LM N

2mo L π π − π 2 4

(π 2) 2 1 1 − − 8 4 8

OP PQ

OP − Q

0

4

3

= 3945. 07

EI L3


=

LM MN LM MN

(π 2) 2 1 + 4 4

2mo L π 2 1 = + 2 8 2 π

OP PQ

~ = m 22

OP PQ

z z

~ = m 12

2mo L = 3π 3π

2mo

L2

=

2mo L

z

L2

0

R| S| T

LM sin (2π x L) N 2 (2π L)

z L

L2

−

z

L2

sin (2π x L) dx + 2 (2π L)

2mo

2mo − L

R| L sin (2π x L) S| x MN 2 (2π L) T

0

sin (4π x L) 2 (4π L)

sin (2π x L) dx + 2 (2π L)

z L

OP Q

= −

0

sin (4π x L) dx 2 (4π L)

sin (4π x L) 2 (4π L)

−

OP Q

3π x 3π x 3π x d sin 2 L L L

0

FG IJ FG IJ H K H K

sin 2

L2

z

−

3π x 3π x 3π x d sin 2 L L L

L2

LM MN

(3π x L) sin (6 π x L) 2mo L (3π x L) 2 − − 2 4 4 (3π )

U| V| W

L2

2

0

0

o

L

L

o

2

2

= − 0.1013 mo L

% 21 = m % 12 = − 0.1013 mo L m

2

L2

+ 0

L

L2

2

2

o

o

OP Q

LM 3π x L − sin (6π x L) OP − 4 Q N 2 (3π x L) sin(6 π x L) 2m L L (3π x L) − M 4 4 (3π ) MN cos(6 π x L) O PQ 8 2m L L (3π 2) 1 1O = + + P + M 8 8 PQ (3π ) MN 4 2m L L 3π 3π O − − M 3π N 2 4 PQ (3π 2) 2m L L (3π ) 1 1O − − − P M 8 4 8 PQ (3π ) MN 4 2m L L (3π 2) (3π ) (3π ) 1 = + + − M 4 4 4 (3π ) MN 4 (3π 2) O 1 + P 4 4 PQ 2m L L 9π 1O = + P M 2 PQ (3π ) MN 8 o

L2

L2

L2

−

2mo 3π L

L

L cos (4 π x L) OP 2m R|L cos (2 π x L) O = − M P M S L |MN 2 (2π L) PQ MN 2 (4π L) PQ T 2m l0q − R L cos (4 π x L) OP 2m |L cos (2 π x L) O − M P M S L |NM 2 (2π L) PQ MN 2 (4π L) PQ T O 2m L − 2 1 2 gP = − b M L MN 2(2π L) 2(2π L) PQ 2

FG 3π x IJ d FG 3π x IJ H LK H LK

+

FG IJ FG IJ H K H K

L

L2

−

L

L2

2

o

2

2

cos (6 π x L) 8

U| V| + W

sin (4π x L) dx 2 (4π L)

L2

z

2mo L 3π 3π

L2

z

L2

L

2mo 3π L

FG IJ FG IJ H K H K FG1 − x IJ sin FG π x IJ sin FG 3π x IJ dx H LK H L K H L K

FG H

FG IJ H K xI J sin FGH 3πLx IJK dx LK

3π x x sin 2 dx + L L

2mo 1 −

L2

πx 3π x x sin dx + 2mo sin L L L

0 L

2mo

0 L

= 0. 3513mo L

L2

z z

L2

2mo L (π 2) 2 π2 π2 1 + + − + 4 4 4 4 π2

U| V| + W U| V| W

2

o

2

2

o

2

2

o

2

2

2

+

2

2

o

2

= 0. 2613 mo L


Thus ~ = m L m o

LM 0.3513 . N− 01013

OP Q

. − 01013 0.2613

4. Solve reduced eigenvalue problem. %2 m % z k% z = ω

ω% 1 = z1 =

11. 765 2

L

EI mo

RS 10. UV T−0.0036W

ω% 2 = z2 =

130. 467 2

L

EI mo

RS0.2790UV T0.9603W

5. Determine natural vibration modes from Eq. (18.1.8).

FG π x IJ − 0.0036 sin FG 3π x IJ H LK H LK ~ F π xI F 3π x IJ φ ( x ) = 0.2790 sin G J + 0.9603 sin G H LK H LK ~

φ 1 ( x ) = 10 . sin

2

Note that the first mode of this beam with nonuniform mass is only slightly different from that for a uniform beam; the second mode differs more between the two beams.


LM 0.2196 − 0.6898 L Φ = M MM 0 L N 0.6898 L

Problem 18.3 1. Stiffness and mass matrices. 5

2

6

1

4

3

L/2

LM0.5 m = mL M MM N

Fig. P18.3a 1 3

k =

m =

3

LM 312 m L M− 6.5 L 840 M 0 MN 6.5L

OP PP LM PQ N

0 − 3L L2 2 0 k tt = 2 2 2L L 2 k 0t 2 2 L 2 L

− 6.5 L

0

L2 − 0.75 L2

− 0.75 L2 2 L2

0

− 0.75 L2

OP 0 P − 0.75 L P PQ L 6.5 L

2

2

2. Solve eigenvalue problem. 2

( k − ω m )φ = 0

ω1 = 9. 9086

ω 3 = 110.14

EI m L4

EI 4

mL

0 0

OP PP P 0Q

48 EI −1 k$lat = ktt − kt 0 k00 k0 t = L3

The 6 × 6 stiffness and mass matrices with reference to the DOFs in Fig. P18.3a are given in Example 18.2. Impose boundary conditions u1 = u5 = 0 by eliminating the corresponding rows and columns to obtain k and m with reference to the DOFs in Fig. P18.3b: 3L L2 L2 2 0

0 L 0.7066 L

Eliminate the three rotational DOFs by static condensation to obtain

2

Fig. P18.3b

LM 24 8 EI M 3 L L M 0 MN− 3L

0.5774 L − 0.5774 L

3. Solution using lumped mass matrix

L/2

4

OP P − 0.5774 L P P − 0.5774 L Q

0 0 − 0.0386 − 0.5774 L − 0.7066 L − 0.5774 L

ω 2 = 43. 818

ω 4 = 200. 80

EI m L4

kt0 k 00

Natural frequency:

ω1 =

k$lat 0. 5 m L

Natural mode equations:

OP Q

EI

= 9. 798

after

m L4

using

static

condensation

φ1T = 0. 2196 − 0. 6588 L 0 L 0. 6588 L 4. Compare with exact frequencies. For a simply-supported beam, the exact values of the first four natural frequencies are

ω1 = 9. 8696 ω 3 = 88. 826

EI 4

mL EI

4

mL

ω 2 = 39. 478 ω 4 = 157. 914

EI m L4 EI m L4

The finite element method using consistent mass provides an excellent result for the fundamental frequency, but the accuracy deteriorates for higher modes. The lumped mass approximation provides only the fundamental frequency and the result is less accurate. Note that the finite element method overestimates the fundamental frequency whereas the lumped mass approximation provides an under estimate.

EI m L4


Problem 18.4

4. Compare with exact frequencies.

1. Stiffness and mass matrices.

For a beam clamped at both ends, the exact values of the first two natural frequencies are

1

ω1 = 22. 37

2

Fig. P18.4 The 6 × 6 stiffness and mass matrices are given in Example 18.2. Impose boundary conditions u1 = u2 = u5 = u6 = 0 by eliminating the corresponding rows and columns to obtain k and m with reference to the two DOFs in Fig. P18.4: k =

LM N

8 EI 24 0 2 L3 0 2 L

OP Q

m =

LM N

mL 312 0 840 0 2 L2

EI m L4

ω 2 = 61. 67

EI m L4

The finite element method using consistent mass provides an excellent result for the fundamental frequency, but the accuracy deteriorates for the second mode. The lumped mass approximation provides only the fundamental frequency, but the result is less accurate and lower than the exact value.

OP Q


( k − ω m )φ = 0

ω1 = 22. 74 φ1 =

EI 4

mL

RS1UV T0W

ω 2 = 81. 98 φ2 =

EI m L4

RS 0 UV T1 LW

3. Solution using lumped mass matrix. m = mL

LM0.5 OP N 0Q

Eliminate the DOF 2 by static condensation to obtain 8 EI 192 EI k$lat = 3 ( 24 ) = L L3

Natural frequency:

ω1 =

k$lat 0. 5 m L

= 19. 60

EI m L4

Natural mode after calculating rotations using static condensation equations:

φ1T = 1 0


Problem 18.5

Natural mode after calculating rotation using static condensation equations:

1. Stiffness and mass matrices.

φ1T = 0. 2375 − 0. 8143 L 0. 2036 L 4. Compare with exact frequencies.

1 3

For a beam clamped at one end and simply supported at the other, the exact values of the first three natural frequencies are

2

Fig. P18.5

ω1 = 15. 42 The 6 × 6 stiffness and mass matrices are given in Example 18.2. Impose boundary conditions u1 = u5 = u6 = 0 by eliminating the corresponding rows and columns to obtain k and m with reference to the three DOFs in Fig. 18.5: k =

m =

LM 24 3L 0 OP L k k O L 2P = M 3L L P L M MN 0 L 2 2 L PQ Nk k Q 0O L 312 − 6.5L mL M − 6.5 L L − 0.75 L PP M 840 MN 0 − 0.75L 2 L PQ

8 EI

2

3

2

2

2

tt

t0

0t

00

2

2

2

2

ω 3 = 104. 2

EI m L4

ω 2 = 49. 97

EI m L4

EI m L4

The finite element method using consistent mass provides an excellent result for the fundamental frequency, but the accuracy deteriorates increasingly for the second and third modes. The lumped mass approximation provides only the fundamental frequency; the result is less accurate and lower than the exact value.


( k − ω m )φ = 0

ω1 = 15. 56

EI mL

ω 3 = 155. 64

Φ =

ω 2 = 58. 41

4

EI m L4

EI m L4

LM 0.2375 MM− 0.9370 L N 0.2561 L

OP PP Q

− 0.0349 − 0.0205 − 0.7449 L − 0.8516 L 0.6662 L − 0.5237 L

3. Solution using lumped mass matrix.

LM0.5 m = mL M NM

0

OP P 0QP

Eliminate the two rotational DOFs by static condensation to obtain 109. 71EI −1 k$lat = ktt − kt 0 k00 k0 t = L3 Natural frequency:

ω1 =

k$lat 0. 5 m L

= 14. 81

EI m L4


Problem 18.6 2 h

2. Mass matrix. 2

22 mh /420

3

2m , I/2 m, I

1

2m (2 h) + 2(156 mh /420)

m, I

&u&1 = 1

2h

4mh 3/420 + 4(2 m )(2 h)3/420

1. Stiffness matrix.

3

– 3(2 m )(2 h ) /420 2

22 mh /420

6EI h2

6EI h2

24 EI h3

&u&2 = 1 3

4mh /420 +

u 1= 1

4(2 m )(2 h)3/420

4EI 4E(I/2) + 2h h

2E(I/2) 2h

3

– 3(2 m )(2 h) /420 2

22 mh /420

6EI h2

&u&3 = 1

u 2= 1

Thus

4EI

2E(I/2)

h

2h

+

4E(I/2)

mh m = 420

2h 6EI h2

LM 1992 MM N(sym)

22h 68h 2

22h − 48h 2 68h 2

OP PP Q


( k − ω m )φ = 0 u 3= 1

ω1 = 1. 5354

Thus k =

EI h3

LM 24 MM N(sym)

6h 5h

2

OP h 2P 5h PQ 6h

ω 3 = 10. 7471

2

2

Φ

EI mh

ω 2 = 4. 0365

4

EI mh 4

EI mh 4

LM 0.5440 = M − 0.5933 h MN− 0.5933 h

OP 2 hP 2 hPQ

0 − 0.0001 −1 1

2h 2h

1 1


4. Plot modes.

– 0.5933/ h

0.5440

– 0.5933/ h

– 1/ 2 h

– 0.001

1/ 2 h

1/ 2 h 1/ 2 h


Problem 18.7 1. Determine lateral stiffness. Statically condense joint rotations in the stiffness matrix of Problem 18.6: −1 k 0t k$lat = k tt − k t 0 k 00

=

EI h3

F GG 24 − H

= 10.9091

6 6

LM 5 N1 2

OP LM6OPIJ 5 Q N6QJK

12

−1

EI h3

2. Calculate lumped mass. mlat = 2 m ( 2 h ) + (1 2 ) ( mh + mh ) = 5 mh 1 424 3 144 42444 3 beam columns

3. Calculate natural frequency.

ω1 =

k$lat mlat

= 1. 477

EI mh 4

The exact value is

ω1 = 1. 5354

φ

L = M N− k

EI mh 4

LM MM N

0.544 0.544 = − 0.594 h −1 00 k 0t ( 0.544) − 0.594 h

OP Q

OP PP Q

4. Comment on accuracy. The accuracy of the lumped mass procedure is satisfactory; the fundamental frequency is estimated with approximately 4% error and the first mode is very accurate (less than 0.03% error). Using the lumped mass procedure we have reduced the order of the system to obtain only the fundamental frequency and mode.


Problem 18.8

This is the same as determined in Problem 18.6.

2m, I/2

2

3

h

1

m, I

u2

1

3

m, I 2

4. Element mass matrix

u1 L

u4

u3

2h

me =

1. Identify DOFs. The frame (assemblage) DOFs that correspond to the element DOFs are identified for each element in Table P18.8.

LM 156 mL M 420 M MN(sym)

22 L

54

2

− 6L

4L

156

OP P − 22 L P P 4L Q − 13 L − 3 L2 2

1

2

0

0

2

0

3

1

3

0

0

1 0 1 2 2 3 0 0 0 0 3 0

0

Table P18.8 Element 1

2

3

1 2 0 0

0 2 0 3

1 3 0 0

2. Element stiffness matrix

ke =

LM 12 EI M 6 L L M− 12 MN 6 L 3

For elements L = 2h.

1

6 L − 12 6L 2 4 L − 6 L 2 L2 − 6L 12 − 6 L 2 2 L − 6 L 4 L2

1

2

0

0

0 1

2 3

0 0

3 0

and

2

OP PP PQ

1 0 1 2 2 3 0 0 0 0 3 0

, L = h ; for element

3

,

3. Assemble element stiffness matrices.

LM12 EI + 12 EI M h 6EI h k = M MM 6hEI MN h 3

3

2 2

=

EI h3

LM24 MM6h N6h

6 EI 2

h 4 E ( I 2) 4 EI + h h 2 E ( I 2) 2h

6h 6h 5h 2 h 2 2 h 2 2 5h 2

OP h PP 2 E ( I 2) 2h P 4 E ( I 2) 4 EI P + h h PQ 6 EI 2

OP PP Q


5. Assemble element mass matrices.

LM156mh + 156mh 420 MM 420 m = M MM ( sym) MN L 1992 22h mh M = 68h 420 M MN(sym) 2

+ 2m ( 2h) 4mh 3 420

22mh 2 420 4 ( 2m) ( 2h) 3 + 420

22mh 2 420 3 (2m) ( 2h) 3 − 420 4 ( 2m) ( 2h) 3 4mh 3 + 420 420

OP PP PP PP Q

OP − 48h P 68h PQ 22h

2

2

where 2 m ( 2 h ) in m11 is due to rigid translation of the beam. This mass matrix is the same as determined in Problem 18.6.


Structural Dynamics

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