Matrix Analysis Stuff

8/10/2019 Matrix Analysis Stuff

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Influence line for statically indeterminate structures

No.1 Prepare an influence line for the vertical reaction at b of given beam.

EI Constant

20

No.2. Prepare an influence line for the vertical reaction at a of given beam.

EI Constant

15

No.3 Compute the ordinates at interval of 2.5 ft of the influence line for the reaction at Afor the beam shown. (EI constant)

No.4. Compute the ordinates at interval of 3 ft of the influence line for the reaction at B forthe beam shown. (EI constant)

No.5. Prepare an influence line for the moment at support b of given continuous beam.

No.6. Prepare an influence line for the moment at support c of given continuous beam.

No.7. Prepare an influence line for the moment at support b of given continuous beam.

No.8. Assuming x sectional area 'A' for all members to be constant. Prepare an influencelines for the forces in the redundant bars of given truss.

ab

a b


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No.9. Assuming x sectional area 'A' of all members to be constant. Prepare an influencelines for the forces in the redundant bars of given truss.

No.10. Prepare an influence line for the moment at support d of given continuous beam.

Analysis of shear - wall structures

No.11. Find the approximate values of the end moments in a column and a shear wall in a

structure which has the same plan as in fig shown and has four stories of equal heighth = b. The frame is subjected to a horizontal force in the x direction of magnitude p/2at top floor and p at each of the other floor levels. The properties of members are asfollows: for any column I = 17 x 10 -6b4, for any beam I = 34 x 10 -6b4, and for anywall I = 87 x 10-3b4. Take E = 2.3 G. The area of wall cross section = 222 x 10-3b2.Consider shear deformation in the wall only.


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Introduction to Structural Dynamics

No.12. A weightless cantilever of length L and constant flexural rigidity EI carries a weightW at its end. Neglecting the moment of inertia I of the mass about its center. (a) Findthe natural angular frequency and the natural period of vibration of the system. (b) Ifthe motion is initiated by displacing the mass in a direction perpendicular to thecantilever by a distance of WL3 and then leaving the system to vibrate freely,

3EIwhat is the max: displacement? What is the displacement at any time t?

No.13. Compute the natural angular frequency of vibration in sidesway for the frame in thefig: and calculate the natural period of vibration. Idealize the frame as a one-degree-of- freedom system. Neglect the axial and shear deformations and the weight of thecolumns. If initially the displacement is 1 in and the velocity is 10 in /sec, what is theamplitude and what is the displacement at t = 1sec?


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No.14.(a) A one-story building is idealized as a rigid girder supported by weightlesscolumns as shown in fig: In order to evaluate the dynamic properties of thisstructure, a free vibration test is made, in which the roof system ( rigid girder ) isdisplaced laterally by a hydraulic jack and then released. During the jackingoperation, it is observed that a force of 20 kip is required to displaced the girder0.2 in. After the instantaneous release of this initial displacement, the max:

displacement on the return swing is only 0.16 in and the period of thisdisplacement cycle is T =1.4 sec.

No.14 (b) The weight W of the building of fig: is 200 kip and the building is set into freevibration by releasing it (at time t = 0) from a displacement of 1.2 in. If the max:displacement on the return swing is 0.86 in at time t = 0.64 s, determine: (a) thelateral spring stiffness. (b) the damping ratio and ( c ) the damping coefficient.

No.15. Assume that the mass and stiffness of the structure of fig: are as follows: m = 2ks2/in,

s = 40 k /in. If the system is set into free vibration with initial conditions D (o) = 0.7in and D (o) = 5.6 in/s, determine the displacement and velocity at t = 1.0 sec,assuming (a) c = 0 ( undamped system), (b) c = 2.8 ks/in.

No.16. Determine the max: steadystate sides way in the frame shown in fig: When it issubjected to a harmonic horizontal force at the level of BC of magnitude of 4 sin 14t(kip) and (a) no damping is present (b) the damping coefficient = 0.1.


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No.17. Determine the natural circular frequencies and characteristic shapes for the two-

degree-of-freedom systems shown in fig.

No.18. Determine the natural circular frequencies and characteristic shapes for the two-degree-of-freedom systems shown in the fig.

No.19. Find the angular frequencies and the normal mode characteristic shapes for the

cantilever in fig shown vibrating freely in the plane of the fig. Neglect the axialdeformations and consider only the lumped masses.


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No.20. Find the response of the system shown in fig to a set of harmonic forces at the threecoordinates shown. P1= 2 P0sin t, P2= P0sin t, and P3= P0sin t.

No.21. Find the response of the system shown in fig to a support motion described by

s= g/5 sin t, where = 4 ( 1/sec ) and g is the acceleration due to gravity.

No.22. Determine the natural frequencies and corresponding mode shapes for the shearbuilding shown in fig.


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No.27. (a) For the spring assemblage shown in fig, obtain the global stiffness matrix by directsuperposition.

(b) If nodes 1 is fixed while node 5 is given a fixed, known displacement of asshown in fig, determine the nodal displacements.

(c) Determine the reactions at the fixed nodes 1 and 5.

No.28. For the spring assemblages shown in fig, determine the nodal displacements, theforces in each element and the reactions. Use the direct stiffness method.

No.29 Describe the flow chart for determining the member forces in given structure by finiteelement analysis.

No.30 (a) If the system is shown has a damping coefficient = 0.1, what are the dampednatural circular frequency and natural period of damped vibration? What is thedisplacement at t = 1 sec, if D0= 1 in and D0 = 10 in/sec?

(EI)AB= (EI)CD= 107k-in2

No.30 (b) If the amplitude of free vibration of a system with one degree of freedom decreaseby 50% in 3 cycle, what is the damping coefficient?


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Answer

1. Prepare an influence line for the vertical reaction at b of this beam.

1 2 EInb = x x(20 (20 )) + (20 - ) x

2 3 2

1 2 20 2

3

= x x + -2 3 2 2

x3

3

= + 10 2 -

3 2

3

= 10 2 -6

1 2 8000EIbb = x 20 x 20( x 20 ) =

2 3 3

EInb x3 3

Xb= = (10x2- ) x

EIbb 6 8000

Influence ordinates for 4ft interval.

43 3At x = 4 Xb= 10 x 4

2 - x = 0.0566 8000

At x = 8'

Xb= 0.208At x = 12' Xb= 0.432At x = 16' Xb= 0. 704At x = 20' Xb= 1


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2.Prepare an influence line for the vertical reaction at a of this beam.

1 2 EIna = x xx + (15 - ) x x

2 3 2

3 3 3

= + 7.5 2

- = 7.5 2

-3 2 6

153

EIaa = 7.5 x 152- = 1125

6

EIna x3

1Xa = = (7.5 x2- ) x

EIaa 6 1125

Influence ordinates for 3ft intervals.

33

1At x = 3 - Xa= 7.5 x 3

2x x = 0.0566 1125

At x = 6' Xa= 0.208At x = 9' Xa= 0.432At x = 12' Xa= 0.704

At x = 15' Xa= 1


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3. Complete the ordinates at internal of 2.5 ft of the influence line for reaction at A for the

continuous beam shown. EI constant.

10 10

XA

1 unit 2 1

1 + SFD

110

+BMD

10M/EI dia:

10 1050 1/2x 10x 10=50

2000/3

250/3 100/3 100/3 50/3

2000 EInAEIAA = ; XA =

3 EIAA

For AB member,

2000 250 1 2000 250 3EInA = + x x = + +

3 3 2 3 3 3 6

For BC member,

-50 1 3 50

EInA = + x x = - 3 2 3 6 3

CB

Primary structure

MB= O ( + )

1 x 10 Cyx 10 = 0

C =1 +


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Influence line ordinate for 2.5 ft interval.

For AB member,

2000 250 0

3At = 0XA= - x 0 + x = 1

3 3 6 2000

At x = 2.5' XA= 0.691At x = 5' XA= 0.406At x = 7.5' XA= 0.168At x = 10' XA= 0

For BC member,2.53 50 3

At x = 2.5' XA= - x 2.5 x = 0.0596 3 2000

At x = 5' XA= - 0.094

At x = 7.5' XA= - 0.082

At x = 10' XA= 0

10.406

0.082 0.059

0.691 0.1680.094

I.L for reaction at A

4. Compute the influence line ordinates at 3 ft intervals for the reaction at support B.

A B C

15 15Primary structure

XB

1/2 1 unit 1/2

BMD

7.5


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x xM/EI dia:

56.25 0.5x 56.25

For AB member,

1 0.5 3

EInB = 56.25 - x x 0.5 = 56.25 -2 3 6

0.5 x 153

EIBB = 56.25 x 15 = 562.56

EInB 0.5 3

1

XB = = ( 56.25 - ) xEIBB 6 562.5

For BC member,

0.5 3

EInB = 56.25 -6

1 0.5 3

XB = ( 56.25 - )562.5 6

Influence line ordinate for 3 ft intervals.

0.5 x 33 1

At x = 3' XB = (56.25 x 3 - ) x = 0.2966 562.5

At x = 6' XB= 0.568At x = 9' XB= 0.792At x = 12' XB= 0.944At x = 15' XB= 1

0.944 1 0.9440.792 0.792

0.568 0.5680.296 0.296

I.L for reaction at B

7.5


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5. Prepare on influence line for the moment at support support b of .given continuousbeam.

a I=1000 b I=900 c I=1200 d I=1600 e

15 12 15 18

MbI 0.9 I 1.2 I 1.6 I primary

structure

1 unit

-0.5 -1 +1 0.5

Member I/L Kadj

ab I/15 x 90/I x 4.5bc 0.9I/12 x 90/I x 5.0625cd 1.2I/15 x 90/I 7.2de 1.6I/18 x 90/I x 6

Joint c d

member end cb cd dc de

Kadj: 5.0625 7.2 7.2 6cycle D.F 0.413 0.587 0.545 0.456

1 FEMBal:M

0.5- 0.207

-- 0.293

--

--

2 COMBal:M

--

--

- 0.1470.08

-0.067

3 COM

Bal:M

-

- 0.017

0.04

- 0.023

-

-

-

-4 COM

Bal:M--

--

- 0.0120.007

-0.005

5 COM

Bal:M

-

- 0.001

0.003

- 0.002

-

-

-

-Final end M 0.275 - 0.275 - 0.072 0.072

1


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1 0.072

BMD0.5 0.275

1.11 0.061 0.045

M/EI dia:0.5 0.229

0.305

.a 1 1 2 1 1

EI 1= = ( x 15 x 1 x x 15 - x 15 x 0. 5 x x 15 ) = 3.7515 15 2 3 2 3

C 1 1 2 1 1EI 2= = ( x 12 x 1.11 x x 12 - x 12 x 0. 305 x x 12 ) = 3.83

12 12 2 3 2 3

EI bb= EI 1+ EI 2= 7.58

For span ab

00.5

1 1 1 2EInb = x x x - x0.5 ( 1 - ) x - xx0.5 x

2 15 3 15 2 2 15 3

= 0.0111 x3- 0.25 x2 + 0.01667 x3 - 0.0111 x3

= - 0.25 x2 + 0.01667 x3

1

1 2

By using moment area theorem

x/15

x

0.5x/15x x

a b

a ba

bx/15

0.5 x(1-x/15)


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EInbMb = = - 0.033 x

2 + 0.0022 x

3

EIbb

For span bc

b c

x

3.83

1 2 1 EInb = - 3.83 + x1.11 ( 1- _) x + xx 1.11 x - x0.305

12 2 2 12 3 2 12

= - 3.83 + 0.555 2 0.0463 3+ 0.0308 3 4.236 x 10-33 x /3

= - 3.83 + 0.555 x2 0.0197 x3

EInbMb = = - 0.5053 x + 0.0732 x

2- 2.6 x10-3x3EIbb

For span cd

d 1 1 1 1 2EI c= = ( x 15 x 0.06 x x 15 - x 15 x 0. 229 x x 15 ) = - 0. 995

15 15 2 3 2 3

1 EInb = 0.995 + x x0.06 x - x0.229 (1 - _) x

2 15 3 15 2

1 2 - x 0.229 x

2 15 3

x

(1.11x/12)

0.305x/12b c

bc

0.995x

x

0.229(1-x/15)

0.229x/15

0.06x/15

x

0.229

0.06

c d

c d c d

1.11(1-x/12)x


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= 0.995 x + 6.667 x 10-4x3- 0.1145 x2+ 7.633 x 10-3x3 - 5.089 x 10-3x3

= 0.995 x - 0.1145 x2+ 3.2107 x 10-3x3

EInbMb = = 0.1313 x - 0.0151 x

2+ 4.236 x10

-4x

3

EIbb

For span de

d

e 1 1 2EI d= = ( x 0.045 x 18 x x 18 ) = 0.27

18 18 2 3

1 2EInb = - 0.27 + x0.045 ( 1- _) x + xx 0.045 x

18 2 2 18 3

= - 0.27 x + 0.0225 x2 1.25 x 10-3x3 + 0.833 x 10-4x3

= -0.27x + 0.0225 x2 4.1667 10

-4x

3

EInb

Mb = = - 0.0356 + 2.968 x 10

-3

2

- 5.497 x10

-5

3

EIbb

I.L for moment at support b

6. Prepare on influence line for the moment at support c of given continuous beam.

I=1000 I=800 I=1000

a b c d14 10 14

0.045

0.045(1-x/18)

x

0.045x/18

0.27

x

de

de

0.

2376

0.7

128

1.0

692

0.

9504

0.

2694

0.3

357

0.2

674

0.1

332

0.

0816

0.

1186

0.1

201

0.

0948

0.

9273

0.

9582

0.

5137

0.

0517


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I 0.8I Mc I

14 10 14

1 unit

-0.5 -1 +1

0.1360.272 1

1

0.136

BMD

1.250.272 1

0.136

M/EI dia

0.272 0.34

b 1 1 2 1 1EI 1= = ( x 10 x 1.25 x x 10 - x 10 x 0.34 x x 10) = 3.602

10 10 2 3 2 3

member I/L kadj:

ab I/14 x 70/I x1 5bc 0.8 I/10 x 70/I x 4.2cd I/14 x 70/I x 3.75

Joint a b

member end ab ba bc

kadj: 5 5 4.2

cycle D.F - 0.543 0.457

1 FEMBal:

--

-- 0.272

- 0.50.228

2 C.O.MBal:

0.136-

--

--

Final end M 0.136 0.272 - 0.272

primary structure

1 2


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d 1 1 2

EI 2= = ( x 14 x 1 x x 14 )= 4.66714 14 2 3

EIcc= EI 1+ EI 2= 8.269

For span ab

1 2 1 EInc= x0.136 ( 1- ) x + x x 0.136 x x - . xx0.272 x14 2 2 14 3 2 14 3

= 0.068 x2- 4.857 x 10

-3x

3+ 3.238 x 10

-3x

3- 3.238 x 10

-3x

3

= 0.068 x2- 4.857 x 10-3x3

EIncMc = = 8.225 x 10

-3x

2 5.87 x 10

-4x

3

EIcc

For span bc

d 1 1 1 1 2

EI c= = ( x 1.25 x 10 x x 10 - x10 x0.34 x x 10 )10 10 2 3 2 3

= 0.95

1 1 0.34 2 EInc = - 0.95 +( x 1.25 x x ) ( x x x ) {0.34 (1 - ) x x }

2 10 3 2 10 3 10 2

0.136

0.136x/14

x

x

0.272x/14x

0.136(1-x/14)

0.272x/14

0.272

a ba b

a b

x

xx

1.25x/10

1.25

0.34(1-x/10

0.34x/10

0.9530.34


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= - 0.95 x + 0. 0208 x3 0.0113 x3 0.17 x2+ 0.017 x3

= - 0.95 x - 0. 17 x2+ 0.0265 x3

EIncMc = = - 0.119 x - 0.0206 x

2+ 0.0032 x3

EIcc

For span cd

1 2EInb = - 4.667 + x( 1- _) x + x x x x

14 2 2 14 3

= - 4.667 x + 0. 5 x2 0.0357x3 + 0.0238x3

= - 4.667 x + 0. 5 x2 0.0119x3

EInbMb = = - 0.5645 x +0.0605 x

2 0.00144 x3

EIcc

I.L for moment at support C

7. Prepare on influence line for the moment at support b of given continuous beam.

I=900 I=1000 I=1000

a b c d12 15 15

1

x/14

1-x/14

x x4.667

0.

2866

0.

5844

0.

7398

0.

5992

0.

0282

0.

094

0.

1692

0.

2256

0.

235

0.

1692

0.

8985

1.

382

1.

52

1.

381

1.

035

0.

5503


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0.9I Mb I I

1 unit

-1 +1 0.5

1 0.286

0.143

1

0.143

BMD

1.11

1 0.2860.143

M/EI diag0.286

a 1 1 2EI 1= = ( x 12 x 1.11 x x 12 ) = 4.44

12 12 2 3

c 1 1 2 1 1EI 2= = ( x 15 x 1 x x 15 - x 15 x 0.286 x x 15 ) = 4.285

15 15 2 3 2 3

21

member I/L kadj:ab 0.9I/12 x 60/I x 3.375bc I/15 x 60/I x 3cd I/15 x 60/I x 1 4

Joint c d

member end cd ed dc

kadj: 3 4 4

cycle D.F 0.429 0.571 -

1 FEMBal:

0.5-0.214

-- 0.286

--

2 C.O.MBal:

--

--

- 0.143-

Final end M 0.286 - 0.286 - 0.143


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EIbb= EI 1+ EI 2= 8.725

For span ab

1.11 a ba

x

x 2.22

b 1 1 1EI a= = ( x 12 x 1.11 x x 12 ) = 2.22

12 12 2 3

1 1

EInb = - 2.22 + x x1.11 x = 2.22 + 0.154 3

2 12 3

EInbMb = = - 0.254 x + 1.765 x 10

-3x3

EIbb

For span bc.

1

x

1 2 1 0.286 EInb = - 4.285 + x ( 1- _) x + x x x - x x

15 2 2 15 3 2 15 3

EInb = - 4.285 x + 0. 5 x

2

0.0333 x

3

+ 0.0222 x

3

3.178 x 10

-3

x

3

= - 4.285 x + 0. 5 x2 0.01428 x

3

EInbMb = = - 0.491 x + 0.0573 x

2 1.637 x 10-3x3

EIbb

1.11x/12

x/15

1-x/15

x0.286x/15

4.285

x

0.286


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For span cd 0.143x

c `dx1.0725

d 1 1 1 1 2EI c= = ( x 15 x 0.143 x x 15 - x 15 x 0.286 x x 15) = - 1.0725

15 15 2 3 2 3

EInb = 1.0725 x + 1. 589 x 10-3x3 0.143 x2+ 9.533 x 10-3x3 6.356 x 10-3x3

= 1.0725 x - 0.143 x2+ 4.766 x 10-3x3

EInbMb = = 0.1229 x - 0.0164 x

2+ 5.462 x 10-4x3

EIbb

I.L for moment at support b

8. Assuming x sectional area A of all members to be constant. Prepare an influencelines for the forces in the redundant bars of given truss.

0 .286(1-x/15

0 .286x/15

0 .286

x

0.143x/15

1.

0015

1.

2367

0.

9711

0.

4695

0.

7143

1.

1428

0.

9993

0.

2358

0.

2650

0.

1759

0.

0570


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Select the redundant members at CD and FGX1 X2

Primary structure

1 1 0.5 0.5

0.5 0.5 0.75 0.75 0.25 0.25

2/3 1 1/3 0

0.5 0.5 1 1

0.25 0.25 0.75 0.75 0.5 0.5

0 1/3 1 2/3

Bar L F1 F2 F12L F1F2 L

ab 15 - 0.5 0 3.75 0

bc 15 - 0.5 0 3.75 0cd 15 - 0.75 - 0.25 8.4375 2.8125

de 15 - 0.75 - 0.25 8.4375 2.8125ef 15 - 0.25 - 0.75 0.9375 2.8125

fg 15 - 0.25 - 0.75 0.9375 2.8125

BC 15 1 0 15 0

CD 15 1 0 15 0

DE 15 0.5 0.5 3.75 3.75

EF 15 0.5 0.5 3.75 3.75

aB 25 0.833 0 17.347 0

0.

417 0

.833

0.

833

0.

417

0.

417

0.

417

0.

833

X1= 1 unit condition


F10.833

0.

417

0.

417

0.

417

0.

417

F2


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25

Bc 25 - 0.833 0 17.347 0

cD 25 - 0.417 0.417 4.347 - 4.347De 25 0.417 - 0.417 4.347 - 4.347

eF 25 - 0.417 0.417 4.347 - 4.347

Fg 25 0.417 - 0.417 4.347 - 4.347

115.832 1.362

F12 L 115.832

1 unit 11= =AE AE

115.83211=

AE

F1F2L 1.362

1 unit 12= =

AE AE

1.36212=

AE

For the symmetrical truss, 11= 22

115.83222=

AE

1.362since 12 21

AE

For unit load applied at b or h

0.625 1 0.625

0.375 0.375

1/2 1/2 0 0

Fb


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Bar L F1 Fb F1Fb Lab 15 - 0.5 0.375 - 2.8125

bc 15 - 0.5 0.375 - 2.8125

aB 25 0.833 - 0.625 - 13.0156

Bc 25 - 0.833 - 0.625 + 13.0156

- 5.625

F1FbL - 5.625

1 unit 1b= =AE AE

F2FbL

1 unit 2b= = 0AE

For unit load applied at d or f

0 1 0

F1FdL - 19.688

1 unit 1d = =AE AE

F2FdL - 14.06

1 unit 2d = =AE AE

19.0881d= -

AE

14.062d= -

AE

Fd

0.375 0.375

0.5625 0.5625 0.1875 0.1875

0.

9375

0.

3125

0.

3125

0.

3125

1


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For unit load applied at e

Bar L F1 F2 Fd Fe F1Fd L F2Fd L F1Fe L F2Fe L

cd 15 - 0.75 - 0.25 0.5625 0.375 - 6.328 - 2.109 - 4.219 - 1.406

de 15 - 0.75 - 0.25 0.5625 0.375 - 6.328 - 2.109 - 4.219 - 1.406

ef 15 - 0.25 - 0.75 0.1875 0.375 - 0.703 - 2.109 - 1.406 - 4.219

fg 15 - 0.25 - 0.75 0.1875 0.375 - 0.703 - 2.109 - 1.406 - 4.219

DE 15 0.5 0.5 - 0.375 - 0.75 - 2.8125 - 2.8125 - 5.625 - 5.625

EF 15 0.5 0.5 - 0.375 - 0.75 - 2.8125 - 2.8125 - 5.625 - 5.625

cD 25 - 0.417 0.417 - 0.9375 - 0.625 -9.773 - 9.773 6.516 - 6.516

De 25 0.417 -0.417 - 0.3125 0.625 - 3.258 3.258 6.516 - 6.516

eF 25 - 0.417 0.417 0.3125 0.625 - 3.258 3.258 -6.516 6.516

Fg 25 0.417 - 0.417 - 0.3125 - 0.625 - 3.258 3.258 - 6.516 6.516

- 19.688 - 14.06 - 22.5 - 22.5

F1FeL - 22.5

1 unit 1e = =AE AE

- 22.51e=

AE

F2FeL - 22.5

1 unit 2e = =AE AE

- 22.51e=

AE

1n+ X111+ X212 = 0

2n+ X121+ X222 = 0

For unit load at b or h

-5.625 + 115.832 X1+ 1.362 X2 = 0

0 + 1.362 X1 + 115.832 X2 = 0

1 unit0

0

0.375 0.375

0.375 0.375 0.375 0.375

0.

625

0.

625

0.

625

0.

62

5

Fe


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28

X1b = 0.0468 = X2h

X2b = -0.00061 = X1h

For unit load at d or f

- 19.688 + 115.832 X1+ 1.362 X2 = 0

- 14.06 + 1.362 X1 + 115.832 X2 = 0

X1d = 0.1686 = X2f

X2d = 0.1194 = X1f

For unit load at e

- 22.5 + 115.832 X1+ 1.362 X2 = 0

- 22.5 + 1.362 X1 +115.832 X2 = 0

X1c = 0.192

X2c = 0.192

0.1920.1686 0.1195

0.0486

0.0006

0.1920.1686

0.1195 0.0486

0.0006

9. Assuming x sectional area A of all members to be constant. Prepare an influencelines for the forces in the redundant bars of given truss.

I.L for barforce in CD member

I.L for barforce in FG member


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29

B C D E F G H

4m

a b c d e f g h i

1 1 0.5 0.5

0.833 0.833 0.417 0.417 0.4170.417

F1

0.5 0.5 0.75 0.75 0.25 0.252/3 1 1/3 0

0.5 0.5 1 1

0.417

0.417 0.417 0.417 0.8330.833

F20.25 0.25 0.75 0.75 0.5 0.5

2/30 1/3 1

Bar L F1 F2 F12 L F1F2 L

ab 3 - 0.5 0 0.75 0bc 3 - 0.5 0 0.75 0

cd 3 - 0.75 - 0.25 1.6875 0.5625

de 3 - 0.75 - 0.25 1.6875 0.5625

ef 3 - 0.25 - 0.75 0.1875 0.5625

fg 3 - 0.25 - 0.75 0.1875 0.5625

BC 3 1 0 3 0

CD 3 1 0 3 0

DE 3 0.5 0.5 0.75 0.75

8 space e 3 m c/c = 24 m

Select forces in bars CD and FG as redundant


primary structure

X1 X2


30/76

30

0.

625

EF 3 0.5 0.5 0.75 0.75

aB 5 0.833 0 3.469 0Bc 5 -0.833 0 3.469 0

cD 5 - 0.417 0.417 0.869 - 0.869

De 5 0.417 -0.417 0.869 - 0.869

eF 5 - 0.417 0.417 0.869 - 0.869

Fg 5 0.417 - 0.417 0.869 - 0.869 23.164 0.274

F12L 23.164

1 unit 11 = =AE AE

F1F2L 0.274

1 unit 12 = =

AE AE

23.16411 =

AE

0.27412 =

AE

For symmetrical truss, 11 = 12

23.164

22=

AE

0.274since 12 = 21 =

AE

For unit load applied at b or h

0.

625

1

0.375 0.375

1 unit 0 0

Fb


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31

0.

937

0.

3125

0.

3125

0.

3125

1 Fd

Fe

0

.625

0.

625

0

.625

0

.625

For unit load applied at d or f

For unit load applied at e

Bar L F1 F2 Fd Fe F1Fd L F2Fd L F1Fe L F2Fe L

cd 3 - 0.75 - 0.25 0.5625 + 0.375 - 1.266 - 0.422 - 0.844 - 0.281

de 3 - 0.75 - 0.25 0.5625 + 0.375 - 1.266 - 0.422 - 0.844 - 0.281

ef 3 - 0.25 - 0.75 0.1875 0.375 - 0.1406 - 0.422 - 0.281 - 0.844fg 3 - 0.25 - 0.75 0.1875 0.375 - 0.1406 - 0.422 - 0.281 - 0.844

DE 3 0.5 0.5 - 0.375 - 0.75 - 0.5625 - 0.5625 - 1.125 - 1.125

EF 3 0.5 0.5 - 0.375 - 0.75 - 0.5625 - 0.5625 - 1.125 - 1.125

cD 5 - 0.417 0.417 - 0.9375 - 0.625 1.9547 1.9547 1.303 - 1.303

De 5 0.417 -0.417 - 0.3125 0.625 - 0.6516 0.6516 1.303 - 1.303

eF 5 - 0.417 0.417 0.3125 0.625 - 0.6516 0.6516 - 1.303 1.303

Fg 5 0.417 - 0.417 - 0.3125 - 0.625 - 0.6516 0.6516 - 1.303 1.303 - 3.938 - 2.819 - 4.5 - 4.5

Bar L F1 Fb F1FbL

ab 3 - 0.5 0.375 - 0.5625

bc 3 - 0.5 0.375 - 0.5625

aB 5 0.833 - 0.625 - 2.603

Bc 5 - 0.833 - 0.625 2.603

- 1.125

F1FbL 1.125Unit 1b= =

AE AE

1.1251b =

AE

F2FbL 0Unit 2b= =

AE AE

2b = 0

0.375 0.375

0.5625 0.5625 0.1875 0.1875

0 3/4 1/4 01 unit

0.375 0.375

0.375 0.375 0.375 0.375

0 1/2 1/2 01 unit


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32

F1FdL - 3.398

1 unit 1d = =AE AE

- 3.3981d=AE

F2FdL - 2.183

1 unit 2d = =AE AE

- 2.1832d=

AE

F1FeL - 4.5

1 unit 1e = =AE AE

- 4.51e= = 2e

AE

1n+ X111+ X212 = 0

2n+ X121+ X222 = 0

For point b or h

-1.125 + 23.164 X1+ 0.274 X2 = 0

0 + 0.274 X1 + 23.164 X2 = 0

X1b = 0.0486

X2b =- 0.0006

For point d or f

- 3.398 + 23.164 X1+ 0.274 X2 = 0

- 2.813 + 0.274 X1 + 23.164 X2 = 0

X1d = 0.1687

X2d = 0.1195


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33

For point e

- 4.5 + 23.164 X1+ 0.274 X2 = 0

- 4.5 + 0.274 X1+ 23.164 X2 = 0

X1e = 0.192

X2e = 0.192

0.1920.1687

0.04860.1195

0.0006

0.1920.1195

0.16870.0486

0.0006

10. Prepare an influence line for the moment at support d of given continuous beam.

MdI 0.9I 1.2I 1.6I

1 unit

-0.5 -1 +1

I.L for barforce in CD member

I.L for barforce in FG member

member I/L kadj:

ab I/15 x 90/I x1 6bc 0.9 I/12 x 90/I x 1 6.75cd 1.2 I/15 x 90/I x 5.4de 1.6 I/18 x 90/I x 6

primary structure


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34

0.036 0.07 0.261 1

a b c d e

10.07

BMD

0.036 0.078 0.2610.07 0.833

0.625

M/EI dia

0.2180.036 0.29

1 2

c 1 1 2 1 1EI 1= = ( x 15 x 0.833 x x 15 - x 15 x 0.218 x x 15) = 3.62

15 15 2 3 2 3

e 1 1 2EI 2= = ( x 18 x 0.625 x x 18 ) = 3.75

18 18 2 3

EI dd= EI 1+ EI 2= 7.37

Joint a b c

member end ab ba bc cb cd

kadj: 6 6 6.75 6.75 5.4

cycle D.F - 0.471 0.529 0.556 0.444

1 FEMBal:M

--

--

--

-0.278

- 0.50.222

2 C.O.MBal:M

--

-- 0.065

0.139- 0.074

--

--

3 C.O.MBal:M

- 0.033-

-

-

-

-

- 0.0370.021

-0.016

4 C.O.MBal:M

-

-

-- 0.005

0.011- 0.006

-

-

-

-

5 C.O.MBal:M

- 0.003-

-

-

-

-

- 0.0030.002

-0.001

Final end M - 0.036 - 0.07 0.07 0.261 - 0.261


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35

For span ab

0.036 x

1 1 EInd = x x 0.07 x - x 0.36 (1 x ) x

2 15 3 15 2

1 2- x x 0.036 x

2 15 3

= 7.778 x 10-4

x3 0.018 x

2+ 1.2 x 10

-3x

3 0.8 x 10

-3x

3

= - 0.018 x2 + 1.178 x 10-3 x3

EIndMd= = - 2.442 x 10

-3x

2+

1.598 x 10

-4x

3

EIdd

For span bc

b c

0.29

c 1 1 2 1 1

EI b= = ( x 12 x 0.078 x x 12 - x 12 x 0.29 x x 12) = 0.26812 12 2 3 2 3

1 2EInd = 0.268 + x 0.078 ( 1- _) x + x 0.078 x x x

2 2 2 12 31 1

- x x 0.29 x 2 12 3

x

0.036(1-x/15

0.036x/15

0.07x/15

0.07

x

x

x

x

0.078x/12

0.078

0.29x/12

0.268

0.078(1-x/12)

0.29x/12


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36

= 0.268 x + 0.039 x2 - 3.25 x 10-3 x3+ 2.167 x 10-3 x3 - 4.028 x 10-3 x3

= 0.268 x + 0.039 x2 - 5.111 x 10-3 x3

EInd

Md = = 0.0364 x + 5.292 x 10-3

x2

- 6.935 x 10-4

x3

EIdd

For span cd

d 1 1 1 1 2EI c= = ( x 15 x 0.833 x x 15 - x 15 x 0.218 x x 15) = 0.9925

15 15 2 3 2 3

1 1 xEInd = - 0.9925 + x x 0.833 _ x - x 0.218 (1 - ) x

2 15 3 15 2

1 2- x x 0.218 x

2 15 3

= - 0.9925 x + 9.256 x 10-3

x3 - 0.109 x

2+ 7.267 x 10

-3x

3 - 4.844 x 10-

3x

3

= - 0.9925 x - 0.109 x2

+. 11.679 x 10-3

x3

EIndMd = = - 0.1347 x - 0.0148 x

2 + 1.585 x 10-3x3EIdd

xx

x

0.833

0.9925

cd

c d0.833x/15

x

0.218x/15

0.218

dc

0.218(1-x/15)


37/76

37

For span de

d e

1 2EInd = - 3.75 + x 0.625 ( 1- _) x + x x 0.625 x

18 2 2 18 3

= - 3.75 x + 0.3125 x2- 0.01736 x3 + 0.01157 x3

= - 3.75 x + 0.3125 x2- 5.786 x 10-3x3

EIndMd = = - 0.5088 x + 0.0424 x2 - 7.851 x 10-4x3

EIdd

I.L for moment at support d

11. Find the approximate values of the end moments in a column and a shear wall in astructure which has the same plan as in fig shown and has four stories of equal heighth = b. The frame is subjected to a horizontal force in the x direction of magnitude p/2 attop floor and p at each of the other floor levels. The properties of members are asfollows : for any column I = 17 x 10-6b4, for any beam I = 34 x 10 -6b4, and for any

wall I = 87 x 10-3b4. Take E = 2.3 G. The area of wall cross section = 222 x 10 -3b2.Consider shear deformation in the wall only.

0.625x/18

x

0.625(1-x/18)

3.75x

0.625

0.0

1766

0.0

534

0.

0813

0.0

755

0.4

95

0.

999

1.

256

1.

009

1.1

66

1.

696

1

.717

1

.357

0.1

381

0.

2591

0.

2507

0.

742


38/76

38

soln:In actual structure three are 16 columns, 4 walls 12 beams of length 1.6b and 4 beams

of length 2 b.

Ic = Ici = 16 x 17 x 10-6b4= 272 x 10-6b4

12 4(I/L)b = 4 (I/L)bi = 4 x 34 x 10

-6b4( + ) = 1292 x 10-6b31.6b 2b

Iw = Iwi = 4 x 87 x 10-3b4= 348 x 10-3b4

arw = arwi = 4 x 5/6 x 222 x 10-3b2= 740 x 10-3b2

wall substitute frame

12.9s 0.423 Eh

3

t - 0.273 Eh3

c - 0.65

-

01.09 x 10-3Eh3

0.54 x 10-3

Eh3

0.53.56

3E= ( I/L )b

S

( 4 + ) E Is =

(1 + ) h

( 2 - ) E It =

(1 + ) h

c = t/s

p

h

h

h

h

p2

p

p

2

3

1

8 4

5

6

7

2

3

1

8 4

5

6

7


39/76


40/76

40

[ S22]w = 0.423

Eh3 1 -0.65 0 0-0.65 2 -0.65 00 -0.65 2 -0.650 0 -0.65 2

Condensed the stiffness matrix

[S*]w= [ S11]w - [ S12]w [ S22]w-1

[ S21]w

= 10-2 Eh 13.98 -22.76 5.7 2.33

-22.76 51.22 -31.84 2.555.7 -31.84 53.77 -30.19

2.33 2.55 -30.19 56.33

[S*] = [S*]w+ [S*]r= 10-4

Eh 1421.4 symmetry-2303.05 5177.15

574.22 -3215.76 5433.34232.28 259.64 -3051.41 5692.92

{F*} = P 0.5111

{D*} = [S*]-1 {F*}

= 10 P 11.47

Eh 8.325.03

2.03

Forces resisted by wall {F*}w= [S*]w{D*}

= P 0.43681.0051.00411.0083

Forces resisted by substitute frame

{F*}r= [S*]r{D*}

= P 0.0632-0.005-0.0041-0.0083


41/76

41

Introduction to Structural Dynamics

12. A weightless cantilever of length L and constant flexural rigidity EI carries a weight W

at its end. Neglecting the moment of inertia I of the mass about its center. (a) Find the naturalangular frequency and the natural period of vibration of the system. (b) If the motion is

initiated by displacing the mass in a direction perpendicular to the cantilever by a distance ofand WL

3/ 3EI and then leaving the system to vibrate freely, what is the max: displacement?

What is the max: displacement at any time t?

Soln:W

W = mg ; m =g

1 unit stiffness, S=3EI/L3

3 EI/L3

(a) = s/m = 3 EIg ; T = 2 = 2x WL3L3xW 3EIg

(b) D0= WL3 : D0

= 0

3 EI

Dmax:= ( D0/)2+ D0

2 = D0= WL

33EI

D = D0sint + D0cost

D = W L3 cost3EI

13. Compute the natural angular frequency of vibration in sidesway for the frame in the fig:and calculate the natural period of vibration. Idealize the frame as a one-degree-of-freedom system. Neglect the axial and shear deformations and the weight of thecolumns. If initially the displacement is 1 in and the velocity is 10 in /sec, what is theamplitude and what is the displacement at t = 1sec?


42/76

42

Soln:

SAB= 12 EI = 12 x 107 = 2.572 k/in

L3 (30 x 12)3

SCD= 12 EI = 12 x 107 = 5.023 k/in

L3

(24 x 12)3

total stiffness S = SAB+ SCD= 7.595 k/in

W = 60 k , m = W = 60 = 0.1554 k s2/in

g 386

(a) = s/m = 7.595 = 7 rad/s; T = 2 = 2 = 0.898 sec0.1554 7

D0= l in, D0= 10 in/s

= ( D0/)2+ D0

2 = (10/7)2 + 12 = 1.744 in

D = D0sin t + D0cos t

= (10/7) sin 7 x 1 + 1 cos 7 x 1 = 1.692 in #

14.( a )A one-story building is idealized as a rigid girder supported by weightless columns asshown in fig: In order to evaluate the dynamic properties of this structure, a free vibration test is made, in which the roof system ( rigid girder ) is displaced laterally by ahydraulic jack and then released. During the jacking operation, it is observed that a forceof 20 kip is required to displaced the girder 0.2 in. After the instantaneous release of thisinitial displacement, the max: displacement on the return swing is only 0.16 in and the

period of this displacement cycle is T =1.4 sec.


43/76

43

Soln:

T = 1.4 sec , S = 20 = 100 k/in0.2

= 2 = 2 = 4.48 rad /sec

T 1.4

f = 1 = 1 = 0.714 HzT 1.4

2 = ( S/m) ; m = (S/ 2) = ( 100/4.482) = 4.982 k s2/in

W = mg = 4.982 x 386 = 1923.23 kip

= Ln (Un/ Un+1) = Ln (0.2/0.16) = 0.223

0.223

= = = 3.55 %

2 2

C = Ccr = 2 m = 3.55 x 2 x 4.982 x 4.48 = 1.584 k s/in100

D= 1 - 2 = ( 1- 3.55/100 )

2 = 4.48 rad/s

No.14 (b)The weight W of the building of fig: is 200 kip and the building is set into freevibration by releasing it (at time t = 0) from a displacement of 1.2 in. If the max:

displacement on the return swing is 0.86 in at time t = 0.64 s, determine: (a) the lateralspring stiffness. (b) the damping ratio and ( c ) the damping coefficient.

soln:

W = 200 k , m = W = 200 = 0.518 ks2/ing 386

D0= 1.2 in , T = 0.64 sec

= 2/T = 2/ 0.64 = 9.817 rad/sec


44/76

44

(a) 2 = s/m s =

2m = 9.817

2x 0.518

= 49.93 k/in

(b) = Ln Un = Ln (1.2/0.86) = 0.3331%Un+1

= = 0.3331 = 5.302 %

2 2

(c) C = Ccr = 2 m= 5.302x 2 x 0.518 x 9.817 = 0.5393 ks/in100

15. Assume that the mass and stiffness of the structure of fig: are as follows: m = 2k s2/in,

s = 40 k /in. If the system is set into free vibration with initial conditions D(o) = 0.7 inand D

(o) = 5.6 in /s, determine the displacement and velocity at t = 1.0 sec, assuming

(a) c = 0 (undamped system), (b) c = 2.8 ks/in.

soln:

= s/m = 40/2 = 4.472 rad/sec

(a)undamped system

D0= 0.7 in , D0= 5.6 in/s

D = D0sin t + D0 cos t = 5.6 sin (4.472 x 1) + 0.7 cos (4.472 x 1)

4.472= 1.383 in

D= D0

cos t - D0sin t = 5.6 cos 4.472 x 1 0.7 x 4.472x sin 4.472 x 1

= 1.0771 in/sec

(b) Damped system , C = 2.8 ks/in

= C = 2.8 = 0.1565 = 15.65 %

2 m 2 x 2 x 4.472


45/76

45

D= 1 - 2 = 4.472 1 0.15652 = 4.417 rad/s

D0+ D0

D = e-t x sin Dt + D0 cos Dt

D

= = 0.1565 x 4.472 = 0.6999 0.7

e-t = e-0.7 x 1 = 0.4966

D0+ D0 5.6 + 0.7 x 0.7

= = 1.3788

D 4.417

D = 0.4966 [ (1.3788 sin 4.417 x 1 ) + (0.7 cos 4.417 x 1) ]

= 0.7563 in

D0+ D0 D0

+ D0

D= - e

-t sin Dt + D0 cos Dt + e

-t xDcos Dt

D D

- D0Dsin Dt

= - 0.7 x 0.7563 + 0.4966 [ (1.3788 x 4.417 cos 4.417 x 1) (0.7 x 4.417 x sin 4.417 x

1) ]

= 0.0591 in/s

16. Determine the max: steady state sides-way in the frame shown in fig: When it issubjected to a harmonic horizontal force at the level of BC of magnitude of 4 sin 14t

(kip) and (a) no damping is present (b) the damping coefficient = 0.1.

soln:

12 EI 12 x 107SAB= = = 2.572 k/in

L3 (30 x 12)3


46/76

46

12 EI 12 x 107SCD= = = 5.023 k/in

L3 (24 x 12)3

Total S = SAB+ SCD= 7.595 k/in

W = 60 k , m = (W/g) = (60/386) = 0.1554 k s2/in

= s/m = 7.595 = 7 rad/s0.1554

(a) no damping is present

P(t) = P0sin t = 4 sin 14 t

P0= 4 k , = 14

P0 1 sin tDmax=

S 1 (/)2

P0 1 4 1Dmax= =

S 1 (/)2 7.595 1 (14/7)2

= 0.176 in

(b) damping coeff: = 0.1

= 0.1 = ( / )

= 0.1 = 0.1 x 7 = 0.7

P0 x 1Dmax = 2 2

S 1 (/)2 + (2/2)

4 x 1Dmax = 2 2

7.595 1 14 2 + (2 x 0.7 x 14 )/497

= 0.174 in


47/76

47

17. Determine the natural circular frequencies and characteristic shapes for the two-degree-

of- freedom systems shown in fig.

Soln:

[ m ] = m1 0 = m 00 m2 0 m

= m 1 00 1

From appendix E-1

[ s ] = EI 9.6 - 8.49L3 - 8.4 9.6

[ m ]-1 = 1 1 0m 0 1

[ B ] = [ m ]-1

[ s ]

= EI 1 0 9.6 - 8.49 = EI 9.6 - 8.49mL3 0 1 - 8.4 9.6 mL3 - 8.4 9.6

[ B ] - 2[ I ] = 0

[ B ] - 2 1 0 = EI 9.6-2 -8.4

0 1 mL3 -8.4 9.6-

2

[ B ] - 2 = ( 9.6 - 2 )2 - 8.42 = 0

92.16 19.2 2+ 4 70.56 = 0

4 19.2

2+ 21.6 = 0

(2)

2- 19.2 (

2) + 21.6 = 0

12

= 1.2 EI

mL3


48/76

48

22

= - 18 EImL3

For 12 = 1.2 EI

mL3

[ B ] - 12[I] D = { 0 }

EI 9.6-1.2 -8.4 D1 = 0mL3 -8.4 9.6-1.2 D2

Let D1 = 1 ,

D2 = 1

D(1)

= 11

For 22 = 18 EI /mL3

EI 9.6-18 - 8.4 D1 = 0mL

3- 8.4 9.6-18 D2

Let D1 = 1

9.6 D1 + 9.6 D2 = 0

D2 = -1

D(2) = 1-1

18. Determine the natural circular frequencies and characteristic shapes for the two-degree-of-freedom systems shown in the fig.

soln:

2 1


49/76

49

[ m ] = m1 0 = m 0 = m 1 0

0 m2 0 m 0 1

From appendix E-3

[ s ] = EI 1.71428 -4.28571L3 -4.28571 13.71428

[ m ]-1

= 1 1 0m 0 1

[ B ] = [ m ]-1

[ s ]

= EI 1 0 1.71428 -4.28571mL3 0 1 -4.28571 13.71428

= EI 1.71428 -4.28571mL3 -4.28571 13.71428

[ B ] - 2 [ I ] = 0

[ B ] - 2 [ I ] = EI 1.71428- 2 -4.28571

mL3 -4.28571 13.71428 -

2

[ B ] - 2 [ I ] = (1.71428 - 2) (13.71428 - 2) 18.36731 = 0

23.51012 15.42856 2+ 4- 18.36731 = 0

4 15.42856

2+ 5.14281 = 0

12= 0.341 EI

mL3

22= 15.088 EI

mL3

For 12 = 0.341 EI

mL3

( [ B ] - 12 [ I ] ) { D} = {0}


50/76

50

1.37328 -4.28571 D1 = 0-4.28571 13.3733 D2 0

Let D1= 1 , D2= 0.32

D(1) = 1

0.32

For 22 = 15.088 EI

mL3

( [ B ] - 22 [ I ] ) { D} = {0}

-13.7372 -4.28571 D1 = 0

-4.28571 -1.37372 D2 0

Let D1= 1 , D2= -3.12

D(2)

= 1-3.12

19. Find the angular frequencies and the normal mode characteristic shapes for thecantilever in fig shown vibrating freely in the plane of the fig. Neglect the axialdeformations and consider only the lumped masses.

soln:

The system has three degree of freedom indicated by the three coordinates.

3 2 1


51/76

51

From appendix E-3

[ s ] = EI 1.6154 -3.6923 2.7692L3 -3.6923 10.1538 -10.6154

2.7692 -10.6154 18.4615

[ m ] = w 4 0 0g 0 1 0

0 0 1

[ B ] = [ m ]-1 [ s ]

[ m ]-1

= g 0.25 0 0w 0 1 0

0 0 1

[ B ] = EIg 0.25 0 0 1.6154 -3.6923 2.7692wL3 0 1 0 -3.6923 10.1538 -10.6154

0 0 1 2.7692 -10.6154 18.4615

[ B] { D } = 2 { D } ; [B ] EIg 0.4039 -0.9231 0.6923wL3 -3.6923 10.1538 -10.6154

[ B ] - 2 [ I ] = 0 2.7692 -10.6154 18.4615

[ B] - 2 [ I ] = EIg 0.4039 - 2 -0.9231 0.6923

wL3 -3.6923 10.1538 - 2 -10.6154

2.7692 -10.6154 18.4615 - 2

[ B ] - 2 [ I ] = 0.4039 - 2 -0.9231 0.6923

-3.6923 10.1538 - 2 -10.6154 = 0

2.7692 -10.6154 18.4615 - 2

(0.4039 - 2) [(10.1538 - 2) (18.4615 - 2) - 10.61542]

+0.9231 -3.6923 (18.4625 -2) + 2.7692 x 10.6154

+0.6923 3.6923 x 10.6154 - 2.7692 (10.1538 - 2) = 0

6- 29.0192

4+ 81

2 2.0797 = 0

(2

)3- 29.0192 (

2)

2 + 81 (

2) 2.0797 = 0


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52

12

= 0.02588 EIg/wL3

22

= 3.09908 EIg/wL3

32

= 25.89415 EIg/wL3

1 = 0.1609 EIg/wL3 ; 2 = 1.7604 EIg/wL3 ; 3 = 5.0886 EIg/wL3

For 12 = 0.02588 EIg

wL3

{[ B ] - 12 [ I ] }{ D} = {0}

0.37802 -0.9231 0.6923 D1 0-3.6923 10.12792 -10.6154 D2 = 0

2.7692 -10.6154 18.43562 D3 0(1)

Let D1= 1 ,D2= 0.5224D3= 0.1506

1D(1) = 0.5224

0.1506

For 22 = 3.09908 EIg

wL3

{ [ B ] - 22 [ I ] } { D} = {0}

-2.69518 -0.9231 0.6923 D1 = 0-3.6923 7.05472 -10.6154 D2 02.7692 -10.6154 15.36242 D3 0

Let D1= 1 ,D2= -6.3414D3= -4.5622

D(2)

= 1-6.3414 -4.5622

-6.3414

1

1

0.5224

1.506

1= 0.1609 3EIg

wl

1 3

EIg = 1.7604

wl


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For 32

= 25.89415 EIgwL3

{ [ B ] - 32 [ I ] } { D} = {0}

-25.49025 -0.9231 0.6923 D1 = 0-3.6923 -15.74035 -10.6154 D2 02.7692 -10.6154 -7.43265 D3 0

Let D1= 1 ,

D2= -13.1981D3= 19.2222

1D

(3) = -13.1981

19.22221.0

13.1981

193.222

20. Find the response of the system shown in fig to a set of harmonic forces at the three

coordinates shown P1= 2 P0 sin t, P2= P0 sin t and P3= P0 sin t.

soln:

From appendix E-3

[ s ] = EI 1.6154 -3.6923 2.7692L3 -3.6923 10.1538 -10.6154

2.7692 -10.6154 18.4615

3 = 5.0866 EIg/wL3


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[ m ] = w 4 0 0g 0 1 0

0 0 1

[ B ] = [ m ]-1 [ s ]

[ B] = [ m ]-1 [ s ] = EIg 0.4039 -0.9231 0.6923

wL3

-3.6923 10.1538 -10.61542.7692 -10.6154 18.4615

[ B ] - 2

[ I ] = 0

12

= 0.02588 EIgwL

3

22 = 3.09908 EIg

wL3

32

= 25.89415 EIgwL3

[ B ] - 2 [ I ] D = 0

[ ] = 1.0 1.0 1.00.5224 -6.3414 -13.19810.1506 -4.5622 19.2222

1 = {( 1)}T[ m ] {(1)} = [ 1 0.5224 0.1506 ] W 4 0 0 1

g 0 1 0 0.52240 0 1 0.1506

= 4.296 W/g

2 = {(2)}T[ m ] {(2)} = [ 1 -6.3414 -4.5622] W 4 0 0 1

g 0 1 0 -6.34140 0 1 -4.5622

= 65.072 W/g

3= {( 3)

}T[ m ] {

(3)} = [1 -13.1981 19.2222] W 4 0 0 1

g 0 1 0 - 13.1981

0 0 1 19.2222

= 547.68 W/g

2

P = P0 1 sin t1


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2

L1 = {( 1)}T{ P } = [ 1 0.5224 0.1506 ] P0sin t 1

1

= 2.673 P0sin t

2

L2 = {( 2)}T{ P } = [ 1 -6.3414 -4.5622 ] P0sin t 11

= -8.904 P0sin t

2

L3 = {( 3)

}T{ P } = [ 1 -13.1981 19.2222 ] P0sin t 1

1

= 8.024 P0sin t

The uncouple eqn:of motion is

r

+ r2

r= Lrr

1 + 1

21= L1

1

1 + 0.02588 EIg 1= 0.6222 P0sin t

wL3 w/g

2 + 3.09908 EIg2= - 0.1369 P0sin t

wL3

w/g

3 + 25.89415 EIg3= 0.0147 P0sin t

wL3 w/gThe steady state displacement is,

1= 0.6223 P0 g/W sin t 2= - 0.1369 P0 g/W sin t

0.02588 EIg -2 ; 3.09908 EIg -

2 ;

W L3

W L3

3= 0.0147 P0 g/W sin t

25.89415 EIg - 2

W L3


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21. Find the response of the system shown in fig to a support, motion described by

s = g/5 sin t, where = 4 (1/sec) and g is the acceleration due to gravity.

soln:

From appendix E-3

[ s ] = EI 1.6154 -3.6923 2.7692

L3 -3.6923 10.1538 -10.61542.7692 -10.6154 18.4615

[ m ] = W 4 0 0g 0 1 0

0 0 1

[ B ] = [ m ]-1 [ s ]

[ B ] - [ m ]-1 [ s ] = EIg 0.4039 -0.9231 0.6923

WL3 -3.6923 10.1538 -10.61542.7692 -10.6154 18.4615

[ B ] - 2 [ I ] = 0

12

= 0.02588 EIgWL

3

22 = 3.09908 EIg

WL3

32

= 25.89415 EIgWL3

[ B ] - 2 [ I ] D = 0


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57

[ ] = 1.0 1.0 1.00.5224 -6.3414 -13.19810.1506 -4.5622 19.2222

m* = W/g 4

11

4

L1 = - s {( 1)}T{ m*} = - s [ 1 0.5226 0.15063 ] 1 W/g

1

= - g/5 x W/g sin t x 4.67323

= - 0.9346 W sin t

L2 = - s {( 2)

}T{ m

*}

4

= - g/5 sin t x W/g [ 1 -6.3414 -4.5622 ] 11

= 1.381 W sin t

L3 = - s {( 3)

}T{ m

*}

4

= - g/5 sin t x W/g [ 1 -13.1981 19.2222 ] 11

= - 2.005 W sin t

1 = {( 1)}T{ m }{( 1)}

4 0 0 1= [ 1 0.5224 0.1506 ] W/g 0 1 0 0.5224

0 0 1 0.1506

= 4.296 W/g

2 = {( 2)}T{ m }{( 2)}

4 0 0 1= [ 1 -6.3414 -4.5622 ] W/g 0 1 0 -6.3414

0 0 1 -4.5622

= 65.027 W/g


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58

3 = {( 3)

}T{ m }{

( 3)}

4 0 0 1= [ 1 -13.1981 19.2222 ] W/g 0 1 0 -13.1981

0 0 1 19.2222

= 547.68 W/g

The uncouple eqn:

of motion is

r + r

2r= Lr

r

1 + 121= L1

1

1 + 0.02588 EIg1= - 0.2174 g sin t

WL3

2 + 3.09908 EIg 2= 0.0212 g sin t

WL3

3 + 25.89415 EIg 3= - 0.0037 g sin t

W L3

The steady state uncouple displacement is

1= - 0.2174 g sin t

0.02588 EIg - 2

WL3

2= 0.0212 g sin t

3.09908 EIg - 2WL

3

3= - 0.0037 g sin t

25.89415 EIg - 2

wL

3


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59

22. Determine the natural frequencies and corresponding mode shapes for the shearbuilding shown in fig.

Soln:

m1 0 0 0.025 0 0[ m ] = 0 m2 0 = 0 0.05 0 ks

2/in0 0 m3 0 0 0.1

k1+ k2 - k2 0 40 -10 0[ k ] = - k2 k2+ k3 - k2 = -10 20 -10 k/in

0 - k3 k3 0 -10 10

[ m ]-1 = 40 0 00 20 00 0 10

40 0 0 40 - 10 0[ B ] = [ m ]-1 [ k ] = 0 20 0 x -10 20 -10

0 0 10 0 -10 10

1600 -400 0

= -200 400 -2000 -100 100

1600 -2 -400 0

[ B ] - 2[ I ] = - 200 400 -2 -200

0 -100 100 -2

[ B ] - 2[ I ] = (1600 -2 ) (400 -2 ) (100 -2 ) 2 x 104 + 400 - 200 (100 - 2 ) = 0

(1600 -2 ) 4 x 104 500 2 + 4 2 x 104 - 8 x 106 + 8 x 1042= 0

32 x 106 80 x 10

4

2+ 1600

4 2 x 10

4

2+ 500

4

6- 8 x 10

6+ 8 x 10

4

2= 0

- 6

+ 2100 4 74 x10

4

2+ 24 x10

6= 0


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60

- (2}

3+ 2100 (

2}

2- 74 x10

4(

2} + 24 x10

6= 0

12= 36.06 ; 1= 6.01 rad/s

22= 400 ; 2= 20 rad/s

32= 1663.94 ; 3= 40.79 rad/s

Natural frequencies

1= = 0.96 Hz ; 2= 3.183 Hz ; 3= 6.492 Hz2

For 12= 36.06

[ B ] - 12 [ I ] D = 0

1563.94 -400 0 D1 0-200 363.94 -200 D2 = 0

0 -100 63.94 D3 0

Let D1= 1 , 6.12D2= 3.91D3= 6.12 3.91

1 1D

(1) = 3.91

6.12 1= 6.01rad/sec

For 22 = 400

( [ B ] - 22 [ I ] ) { D} = {0}

1200 - 400 0 D1 = 0- 200 0 - 200 D2 0

0 - 100 - 300 D3 0

Let D1= 1 ,D2= 3D3= - 1

1D(2) = 3

- 1

2 = 20rad/sec

3.0

1.0

-1.0


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61

For 32

= 1663.94

([ B ] - 32 [ I ] ) { D} = {0}

- 63.94 - 400 0 D1 = 0

- 200 - 1263.94 -200 D2 00 - 100 - 1563.94 D3 0

Let D1= 1 ,D2= - 0.16

D3= 0.01

1D

(3) = - 0.16

0.01

3 = 40.79 rad.s

FINITE ELEMENT METHOD

23. For the spring assemblage with arbitrarily numbered nodes shown in fig. obtain (a) theglobal stiffness matrix (b) the displacements of nodes 3 and 4. (c) the reaction forces atnode 1 and 2 (d) the forces in each spring . A force of 5000 lb is applied at node 4 in the

x direction. The spring constants are given in the fig. Nodes 1 and 2 are fixed.

Soln:

For element 1,

k(1)

= k1 - k1 = 1000 - 1000 lb/in

- k1 k1 -1000 1000

For element 2,

k(2) = k2 - k2 = 2000 - 2000 lb/in- k2 k2 -2000 2000

1.00

0.01

-0.16


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63

1x= -10,000 lb

11

3x= 10,000 lb

11

110000/11 10000/11

For element 2,

3x = 2000 -2000 d3x 10/11

4x -2000 2000 d4x 15/11

3x

= -10,000 lb

11

4x= 10,000 lb11

210000/11 10000/11

For element 3,

4x = 3000 -3000 d4x 15/112x -3000 3000 d2x 0

4x= 45,000 lb

11

2x= -45,000 lb11

345000/11 lb

45000/11 lb


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24. For the spring assemblages shown in fig, obtain (a) the global stiffness matrix, (b) thedisplacement of nodes 2, 3 and 4 (c) the global nodal forces, and (d) the local elementforces. Node 1 is fixed while node 5 is given a fixed, known displacement = 20 mm .The spring constants are all equal to k = 200 kN/m.

Soln:For element 1,

k(1)

= k1 - k1 = 200 - 200 kN/m- k1 k1 -200 200

k(1) = k(2)= k(3)= k(4)

(a) k = 200 -200 0 0 0-200 400 -200 0 00 -200 400 -200 00 0 - 200 400 -2000 0 0 -200 200

(b) k d = F

200 -200 0 0 0 d1x F1x-200 400 -200 0 0 d2x F2x0 -200 400 -200 0 = d3x F3x0 0 -200 400 -200 d4x F4x0 0 0 200 200 d5x F5x

Boundary condition d1x= 0 , d5x= 20 mm = 0.02 mF2x= 0 , F3x= 0 , F4x= 0

400 -200 0 d2x = F2x-200 400 -200 d3x F3x

0 -200 400 d4x F3x + 200 x

400 -200 0 d2x = 0

-200 400 -200 d3x 00 -200 400 d4x 4

D2x= 0.005 m, D3x= 0.01 m, D4x= 0.015 m


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(c)

200 -200 0 0 0 0 F1x-200 400 -200 0 0 0.005 F2x0 -200 400 -200 0 0.01 = F3x0 0 -200 400 -200 0.015 F4x

0 0 0 200 200 0.02 F5x

F1x= -1 kN, F2x= 0, F3x= 0, F4x= 0, F5x= 1 kN

( d ) For element 1,

= k d

1x = 200 -200 d1x 0

2x -200 200 d2x 0.005

1x= - 1 kN

2x= 1 kN

1

1 kN 1 kN

For element 2,

2x = 200 -200 d2x 0.005

3x -200 200 d3x 0.01

2x= - 1 kN

3x= 1 kN

2

1 kN 1 kN

For element 3,

3x = 200 -200 d3x 0.01

4x -200 200 d4x 0.015


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66

3x= - 1 kN

4x= 1 kN

3

1 kN 1 kN

For element 4,

4x = 200 -200 d4x 0.015

5x -200 200 d5x 0.01

4x= - 1 kN

5x= 1 kN

41 kN 1 kN

25. (a)Obtain the global stiffness matrix K of the assemblage shown in fig by superimposingthe stiffness matrices of the individual springs. Here k1, k2, and k3are the stiffness ofthe springs as shown.(b) If nodes 1 and 2 are fixed and a force P acts on node 4 in the positive x direction,

find an expression for the displacements of nodes 3 and 4.(c) Determine the reaction forces at nodes 1 and 2.

Soln:

For element 1,

k(1) = k1 - k1- k1 k1

For element 2,

k(2)

= k2 - k2- k2 k2


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For element 3,

k(3) = k3 - k3- k3 k3

(a) k1 0 - k1 0k = 0 k3 0 - k3

- k1 0 k1 + k2 - k20 - k3 - k2 k2 + k3

(b) k d = F

k1 0 - k1 0 d1x F1x

0 k3 0 - k3 d2x = F2x- k1 0 k1 + k2 - k2 d3x F3x

0 - k3 - k2 k2 + k3 d4x F4x

Boundary condition d1x= 0 , d2x= 0 , F3x= 0 , F4x= 0 ,

k1 + k2 - k2 d3x = F3x 0- k2 k2+ k3 d4x F4x P

(k1 + k2) d3x - k d4x = 0

(k1 + k2)d4x = d3x

k2

- k2d3x+ (k2 + k3) d4x = P

(k1 + k2)- k2d3x+ (k2 + k3) x d3x = P

k2

- k22+ k1k2+ k1k3+ k2

2+ k2 k3

d3x ( ) = Pk2

k2d3x = P

k1k2+ k2 k3 + k1k3

k1+ k2d4x= P

k1k2+ k2 k3 + k1k3


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68

(c)k1 0 - k1 0 0 F1x0 k3 0 - k3 0 F2x

- k1 0 k1 + k2 - k2 k2 P =k1k2+ k2 k3 + k1k3 F3x

0 - k3 - k2 k2 + k3 (k1+k2) P

k1k2+ k2 k3 + k1k3 F4x

- k1k2F1x = P

k1k2+ k2 k3 + k1k3

- k3( k1k2)F2x = P

k1k2+ k2 k3 + k1k3

26. For the spring assemblage shown in fig, determine the displacement at node 2 and theforces in each spring element. Also determine the force F3. Given node 3 displaces anamount = 1 in the positive x direction because of the force F3 and k1 = k2= 1000 lb/in.

Soln:For element 1,

k(1) = k1 - k1 = 1000 -1000 lb/in- k1 k1 -1000 1000

For element 2,

k(2) = k2 - k2 = 1000 -1000 lb/in

- k2 k2 -1000 1000

Assemblage the spring elements

k = 1000 - 1000 0- 1000 2000 - 1000

0 - 1000 1000

(b) k d = F

1000 - 1000 0 d1x F1x- 1000 2000 - 1000 d2x = F2x

0 - 1000 1000 d3x F3x


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69

Boundary condition d1x= 0 , d3x= 1 in , F2x= 0

2000 d2x - 1000 d3x = F2x

2000 d2x = 1000 x 1

d2x = 0.5 in

1000 - 1000 0 0 F1x- 1000 2000 - 1000 0.5 = F2x

0 - 1000 1000 1 F3x

F1x = - 500 lb

F3x = - 500 + 1000 = 500 lbFor element 1,

= k d

1x = 1000 -1000 d1x 0

2x -1000 1000 d2x 0.5

1x = - 500 lb ;

2x = 500 lb

1

500 500

For element 2,

= k d

2x = 1000 -1000 d2x 0.5

3x -1000 1000 d3x 1

2x = - 500 lb ;

3x = 500 lb

2

500 lb 500 lb


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27. (a) For the spring assemblage shown fig, obtain the global stiffness matrix by directsuperposition.

(b) In nodes 1 in fixed while node 5 is given a fixed, known displacement of as shownin fig, determine the nodal displacements.

(c) Determine the reactions at the fixed nodes 1 and 5.

Soln:For element 1,

k(1) = k - k- k k

k(2)

= k(3)

= k(4)

= k(1)

= k - k

- k k

(a) k = k -k 0 0 0- k 2 k - k 0 00 - k 2 k - k 00 0 - k 2 k - k0 0 0 - k k

(b) k d = F

k -k 0 0 0 d1x F1x-k 2 k -k 0 0 d2x = F2x0 -k 2 k -k 0 d3x F3x0 0 - k 2 k -k d4x F4x0 0 0 -k k d5x F5x

Boundary condition d1x= 0 , d5x=

F2x = 0 , F3x = 0 , F4x = 0

2 k - k 0 d2x F2x 0- k 2 k - k d3x = F3x 00 - k 2 k d4x F4x + k k

2 k d2x- k d3x = 0d3x = 2 d2x

- k d2x+ 2 k d3x - k d4x= 0


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71

- k d2x+ 4 k d2x - k d4x= 0

3 k d2x = k d4x

d4x = 3 d2x

- k d3x+ 2 k d4x= k

- 2 k d2x+ 6 k d2x= k

4 k d2x = k

d2x = /4

d2x = /4 ; d3x = /2 ; d4x = 3/4

( c ) k d = F

k -k 0 0 0 0 F1x-k 2 k -k 0 0 0.25 = F2x0 -k 2 k -k 0 0.5 F3x0 0 - k 2 k -k 0.75 F4x0 0 0 -k k F5x

F1x = - 0.25 k

F5x = - 0.25 k

28. For the spring assemblages shown in fig, determine the nodal displacements, the forces

in each element and the reactions. Use the direct stiffness method.

Soln:For element 1,

k(1) = k1 - k1 = 500 -500 lb/in

- k1 k1 -500 500

For element 2,

k(2) = k2 - k2 = 1000 -1000 lb/in- k2 k2 -1000 1000


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For element 3,

k(3) = k3 - k3 = 1500 -1500 lb/in- k3 k3 -1500 1500

(b) k d = F

500 500 0 0 d1x F1x

-500 1500 -1000 0 d2x = F2x0 -1000 2500 -1500 d3x F3x

0 0 -1500 1500 d4x F4x

Boundary condition d1x= 0

F2x = -1000 lb, F4x = 4000 lb , F3x = 0

1500 - 1000 0 d2x -1000- 1000 2500 - 1500 d3x = 0

0 - 1500 1500 d4x 4000

d2x = 6 in

d3x = 10 in

d4x = 12.667 in (38/3)

F1x = 500 d1x - 500 d2x - = - 3000 lb

For element 1,

= k d

1x = 500 -500 d1x 0

2x -500 5 00 d2x 6

1x = - 3000 lb ;

2x = 3000 lb

1

3000lb 3000lb


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73

For element 2,

2x = 1000 -1000 d2x 6

3x -1000 1000 d3x 10

2x = - 4000 lb ;

3x = 4000 lb

2

4000 4000lb

For element 3,

3x = 1500 -1500 d3x 10

4x -1500 1500 d4x 38/3

3x = - 4000 lb ;

4x = 4000 lb

3

4000 lb 4000lb


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29. Describe the flow chart for determining the member forces in given structure by finiteelement analysis.


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30.(a) If the system is shown has a damping coefficient = 0.1, what are the dampednatural circular frequency and natural period of damped vibration? What is thedisplacement at t = 1 sec, if D0= 1 in and D0 = 10 in/sec?

(EI)AB= (EI)CD= 107

k-in2

12 EI 12 x 107

SAB= = = 2.572 k/in

L3 (30 x 12)3

12 EI 12 x 107SCD= = = 5.025 k/in

L3 (24 x 12)3

Total S = SAB+ SCD=7.595 k/in

W 60m = = = 0.1554 ks

2/in

g 386

s 7.595

= = = 7 rad/sm 0.1554

= = 0.1 x 7 = 0.7

d2 = 2 - 2 = 72 - 0.72

= 6.965 rad/s

2 2

Td= = = 0.902 sec

d 6.965


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If D0= 1 in, D0 = 10 in/sec

D02+ D0

D = e-t

sin d+ D0 cos dt

d

10+ 0.7 x 1

D = e-0.7 x 1

sin 6.969 x 1 + 1 cos 6.965 x 16.956

D = 0.8663 in

30 (b) If the amplitude of free vibration of a system with one degree of freedom decrease by50% in 3 cycle, what is the damping coefficient?

1 Dn = in

P Dn+p

1 1 = ln

3 0.5

= 0.2311

0.2311Damping coefficient = = = 0.0368 = 3.68%2 2

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