Dynamics of structures, Third Edition (Solutions, Instructor Solution Manual) [3 ed.] 9780415622967, 9780415620864

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DYNAMICS OF STRUCTURES SOLUTIONS MANUAL THIRD EDITION JAGMOHAN L. HUMAR

SOLUTIONS MANUAL FOR Dynamics of Structures Third Edition

by J. Humar

CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2012 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Printed in the United States of America on acid-free paper Version Date: 20120203 International Standard Book Number: 978-0-415-62296-7 (Paperback) This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com

2

Chapter 2

In a similar manner we get ¨y Iy = M u

Problem 2.1 90 N/mm

60 N/mm

u

40 N/mm

Figure S2.1 Referring to Figure S2.1 the springs with stiffness 60 N/mm and 90 N/mm are placed in series and have an effective stiffness given by k1 =

1 = 36 N/mm 1/60 + 1/90

This combination is now placed in parallel with the spring of stiffness 40 N/mm giving a final effective stiffness of keff = k1 + 40 = 76 N/mm

For an angular acceleration θ¨ about the center of mass the inertia force on the infinitesimal ele¨ ment is directed along the tangent and is γr2 θdθdr. 2¨ The x component of this force is γr θdθdr sin θ. It is easily seen that the resultant of all x direction forces is zero. In a similar manner the resultant y direction force is zero. However, a clockwise moment about the center of the disc exists and is given by


R




2π

¨ 3 dθdr = γπR2 γ θr

Mθ = 0

0

The elliptical plate shown in Figure S2.2(c) is divided into the infinitesimal elements as shown. The mass of an element is γdxdy and the inertia force acting on it when the disc undergoes translation in the x direction with acceleration u ¨x is γu ¨x dxdy. The resultant inertia force in the negative x direction is given by
Ix =

Problem 2.2




a/2

−a/2




= γu ¨x

dxdy dθ

dr R

a (b)

Figure S2.2 The infinitesimal area shown in Figure S2.2(a) is equal to rdθdr. When the circular disc moves in the x direction with acceleration u ¨x the inertia force on the infinitesimal are is γrdθdr¨ ux , where γ ids the mass per unit area. The resultant inertia force on the disc acting in the negative x direction is given by
R
2π Ix = γu ¨x rdθdr = γπR2 u ¨x = M u ¨x 0

b/2

√

−b/2 a/2

b −a/2

1−4x2 /a2

√

γu ¨y dydx 1−4x2 /a2

 1 − 4x2 /a2 dx

πγab = Mu ¨x = 4

b

(a)

R2 ¨ R2 ¨ θ=M θ 2 2

The moment of the x direction inertia force on an element is γ u ¨x ydxdy. The resultant moment obtained over the area is zero. The inertia force produced by an acceleration in the y direction is obtained in a similar manner and is M u ¨y directed in the negative y direction. An angular acceleration θ¨ produces a clockwise   ¨ ¨ moment equal to γr2 θdxdy = γ x2 + y 2 θdxdy. Integration over the area yields the resultant moment, which is clockwise
Iθ =

a/2

−a/2




√

−b/2

1−4x2 /a2

√

  γ θ¨ x2 + y 2 dydx

1−4x2 /a2

πab a + b ¨ a2 + b2 ¨ θ=M θ 4 16 16 2

=γ

b/2

2

0

where M is the total mass of the disc. The resultant moment of the inertia forces about the centre of the disc, which is also the centre of mass is given by
R
2π Mx = γu ¨x r2 sin θdθdr = 0 0

0

The x and y direction inertia forces produced on the infinitesimal element are −γ θ¨ sin θdxdy and γ θ¨ cos θdxdy, respectively. When summed over the area the net forces produced by these are easily shown to be zero.

3

Problem 2.3

The displacement at location A, shown in Figure S2.4, due to a unit load is given by L3b L3s L2 Ls + + b 3EIb 3EIs GJs 3 12 × 768 363 × 64 = + 6 3 × 30 × 10 3 × 30 × 106 × π 2 12 × 36 × 32 + 12 × 106 × π = 0.01475 + 0.01056 + 0.0044 = 0.0297 in.

∆A =

θ

u

Equivalent spring stiffness =

1 0.0297

= 33.67

lb. in.

2

mü

2uk

Mü

T

10 Mass = 386.4 = 0.02588 lb.s in. The equation of motion is given by

0.02588¨ u + 33.67u = 0 I0θ¨

or u ¨ + 1300u = 0 T

Problem 2.5

Figure S2.3 As the center of the pulley moves a distance u, the spring is stretched by 2u and the pulley rotates through u/R. Referring to the free body diagrams shown in Figure S2.3, the conditions of equilibrium are 2uk × 2R + m¨ uR + I0 θ¨ − T R = 0

M(L1  R)θ¨

B I0θ¨

T + Mu ¨=0 Substituting I0 = mR2 /2 and eliminating T gives 

3m M+ 2

 u ¨ + 4ku = 0

θ cd θ˙

Problem 2.4

kL2θ

Figure S2.5 Taking moments about B shown in Figure S2.5, and noting that I0 = M R2 /2

3
0 Light, flexible Light, flexible

1
0

M R2 ¨ 2 M (L1 + R) θ¨ + θ + cd2 θ˙ + kL22 θ = 0 2

Wheel u

or 

Figure S2.4  3 1 1 1 ×1× Ib = = 12 4 768 π π , Js = Is = 64 32

M

L21

 3 2 + R + 2L1 R θ¨ + cd2 θ˙ + kL22 θ = 0 2

4

Problem 2.6

Problem 2.8

d

ug(t)

kL θ

Mbθ¨

u

dx

x

mL2 ¨ θ 3

θ

A1

cdθ˙ Mb θ¨

A2 Output

Input

MR 2 ¨ θ 2

L

Figure S2.8 Figure S2.6 Taking moments about A shown in Figure S2.6   mL2 2 ¨ + M b θ + cd2 θ˙ + kL2 θ = 0 3 When the flywheel is braked, an additional rotational inertia term will exist. Moment equilibrium gives    mL2 R2 2 +M b + θ¨ + cd2 θ˙ + kL2 θ = 0 3 2

1 dξ = , A(ξ) = A1 + (A2 − A1 ) ξ dx L 1 dψ = (2 − 2ξ) , m(ξ) = ρA(ξ) dx L where ρ is the mass density. m∗ =




L




1

ρL {A1 + (A2 − A1 ) ξ}  2  4ξ + ξ 4 − 4ξ 3 dξ

= 0

Problem 2.7 Boundary conditions

where  dψ dψ 1 1 3 = × = 8ξ − 15ξ 2 + 6ξ dx dξ L L 2   d ψ 1 = 2 24ξ 2 − 30ξ + 6 2 dx L All four boundary conditions are satisfied
L 2 m∗ = m {ψ(x)} dx


1




∗

0

EI 0

=



L

k =

2

{ψ(ξ)} dξ = 0.03016mL

= mL

EI L3


0

1



d2 ψ(x) dx2 d2 ψ(ξ) dξ 2

2 dx

2 dξ = 7.2

ρL (5A1 + 11A2 ) 30

2
L dψ(x) ∗ EA(x) dx k = dx 0
E 1 = {A1 + (A2 − A1 ) ξ} L 0   4 + 4ξ 2 − 8ξ dξ   E 1 A1 + A2 = L 3
L m(x)¨ ug ψ(x)dx p∗ = − =

 ψ = 0 x=0 dψ = 0 dx  ψ = 0 x=L d2 ψ EI 2 = 0  dx

0

2

m(x) {ψ(x)} dx 0

EI L3

0




  ρL¨ ug {A1 + (A2 − A1 ) ξ} 2ξ − ξ 2 dξ 0   1 5 = −ρL¨ ug A1 + A 2 4 12 1

=−

Problem 2.9 Refer to Figure S2.9 where the inertia force 2¨ut on an infinitesimal element is shown as m ¯ ∂∂x 2 dx. The total displacement ut = u + ug , where u is the displacement relative to the ground and ug is the displacement of the ground. The inertia force

5 thus becomes m ¯ (¨ z (t)ψ(x) + u ¨g ) dx. The first term within the parentheses produces the generalized mass m∗ while the second term produces the generalized force p∗ as follows
L 33 ∗ 2 mL ¯ m = m[ψ(x)] ¯ dx = 140 0
L 3 ¯ ug p∗ = −m¨ ¯ ug ψ(x)dx = − mL¨ 8 0 The generalized stiffness is given by
L 

2 3EI ∗ k = EI ψ (x) dx = 3 L 0 The geometric stiffness produced by the axial force S(x) = (L − x)mg ¯ is obtained from
L 
2 3 ¯ S(x) ψ (x) dx = mg kG = 8 0 The equation of motion including the gravity effect is   3 3EI 33 3 ¯ z = − mLg ¯ mL¨ ¯ z+ − mg L3 8 8 140

Equation of motion: mL πvt π 4 EI z = F sin z¨ + 2 2L3 L

F v x

Figure S2.10

Problem 2.11 The mass matrix M and the stiffness matrix K are given by 

 2 0 0 M =  0 2 0  kip s2 /in. 0 0 1   1000 −500 0 K =  −500 750 −250  kips/in. 0 −250 250

x

s(x) — m

2ut

d dx dt2

dx

M

ψ(x) x

d2ut  z¨(t) ψ(x) üg d2 ∂2u M EI 2  EI z(t) ψ″(x) ∂t — S(x) (L x) m g

u L

With ψ T = [ 1 2 3 ], the generalized mass m and generalized stiffness k ∗ are obtained from ∗

m∗ = ψ T Mψ = 19 kip s2 /in. k ∗ = ψ T Kψ = 1250 kips/in. The generalized force is obtained from

ug

Figure S2.9

Problem 2.10 u = zψ(x) d2 ψ πx π2 πx , ψ = sin = − 2 sin 2 L dx L L
L πx mL m∗ = dx = m sin2 L 2 0
L 4 πx π π 4 EI k∗ = EI 4 sin2 dx = L L 2L3 0
L πx F δ (x − vt) sin p∗ = dx L 0 πvt = F sin L

  1 p∗ = −ψ T M  1  u ¨g = −1170 sin(6πt) kips 1 The equation of motion is given by 19¨ z + 1250z = −1170 sin(6πt)

Problem 2.12 Taking moments about A shown in Figure S2.12 ml2 θ¨ + 2ka2 θ + mgl sin θ = 0 For small vibrations sin θ ≈ θ, hence   ml2 θ¨ + 2ka2 + mgl θ = 0

6 A

θ 2 ka θ S

ml θ¨

mg

Figure S2.12

7

Chapter 3

Collectively, the four equations can be expressed in the matrix form

Problem 3.1

¨ + C∗ u˙ + K∗ u = 0 M∗ u

q2

a

ks

q3

cs

w

uT = [ q1 W

q1 w, r

 M∗ =  

a

ks

q4

cs(q· 3 q· 1 q· 2a)

· · · cs(q 4  q 1  q 2a) ks(q3  q1  q2a)

q4 ] 

g

  

2

W r g a2

w g

w g

 2 0 −1 −1 2 1 −1   0 C∗ = cS   −1 1 1 0 −1 −1 0 1   2 0 −1 −1  0 2 1 −1   K∗ = kS   −1 1 1 + kkTS 0 −1 −1 0 1 + kkTS

kT

kT

q3



cs

w

aq2

ks(q4  q1  q2a)

Problem 3.2 kT q3

w q¨ g 3

kT q4

w q¨ g 4

mg

mhθ¨

I0θ¨

W 2 r q¨ 2 g W q¨ g 1 cs(q· 3 q· 1 q· 2a) ks(q3 q1 q2a)

h cs(q· 4  q· 1  q· 2a)

θ

 ks(q4  q1  q2a) A

θ

Figure S3.1 The degrees of freedom are identified in Figure S3.1. The free-body diagrams are also shown. For force balance on each of the front and rear axles w q¨4 + cS (q˙4 − q˙1 − q˙2 a) g + kS (q4 − q1 − q2 a) + kT q4 = 0 w q¨3 + cS (q˙3 − q˙1 + q˙2 a) g + kS (q3 − q1 + q2 a) + kT q3 = 0 For equilibrium of vertical forces on the vehicle body W q¨1 + cS (q˙3 + q˙4 − 2q˙1 ) g + kS (q3 + q4 − 2q1 ) = 0 Taking moments about the mass center of the vehicle body W 2 r q¨2 + cS a (q˙3 − q˙4 + 2aq˙2 ) g + kS a (q3 − q4 + 2aq2 ) = 0

· c(u· bθ)

k(u bθ) k(u bθ)

b

· c(u·bθ)

u ug

b

Figure S3.2 Coordinates ut and θ measure the vertical motion and rotation of the base. The forces on the structure are identified in Figure S3.2. Assuming that ut is measured from the position of equilibrium under gravity load, the condition of vertical force balance gives m¨ ut + 2cu˙ + 2ku = 0

(1)

or m¨ u + 2cu˙ + 2ku = −m¨ ug Taking moments about A and assuming that θ is small 

 I0 + mh2 θ¨ + 2cb2 θ˙ + 2kb2 θ − mghθ = 0

(2)

8

Problem 3.3

Problem 3.4 For frame A

a2 + b2 I0 = m 12 Hence the mass matrix is given by 

m M=0 0

u1 = 12θ2 u2 = 12θ4

 0 0  m 0 2 +b2 0 m a 12

or 

k1  1

k1

k11

=



b 2

0 TA = 0 k23

k1  1

k2 k13

k33 k22 k1 k32



0 0

12 0

0 0

0 12



 θ1  θ2    θ3 θ4 

Hence

k21 k31

u1 u2



k12

k2  1

12 0

0 0

0 12



KA = TTA kA TA  0 0 0 540 0 0 = 144  0 0 0 0 −240 0

a 2

b 2

 0 −240   0 240

For frame B K B = KA

Figure S3.3 To obtain the first column of stiffness matrix, give a unit displacement u, with v and w each zero. The resisting forces provided by frames A and B and the external forces required to maintain the displacement are shown in Figure S3.3. For equilibrium k11 = 2k1 , k21 = 0, k31 = 0. For u = 0, v = 1 and θ = 0 equilibrium gives k12 = 0, k22 = k2 , k32 =

k2 a 2

In a similar manner, for u = 0, v = 0, and θ = 1 k13 = 0 k2 a k23 = 2 b2 a2 k33 = 2k1 + k2 4 4

For frame C   1 −15 0 0 Tc = 0 0 1 −15  860 −12900 −360 5400  −12900 193500 KC =  −360 5400 360 5400 −81000 −5400

 5400 −81000   −5400 193500

Assembly gives 

860 −12900  K= −360 5400

−12900 349020 5400 −151020

−360 5400 360 −5400

 5400 −150120   −5400 262620 2

Mass of slab m = 100×30×24 = 2.236 kip.s 32.2 ft . Mass of moment of inertia a2 + b2 302 + 242 = 2.236 × 12 12 = 275.03 kip.ft.s2

I0 = m On assembly, the following stiffness matrix is obtained   2k1 0 0 k2 a  k2 K= 0 2 2 k2 a k2 a2 k1 b 0 2 2 + 4

Hence 

2.236  0 M= 0 0 p1 =

0 275.03 0 0

0 0 2.236 0

 0 0   0 275.03

30 × 12 × p0 sin Ωt = 0.360p0 sin Ωt, p2 = 0 1000

9 30 × 6 × p0 sin Ωt = 0.180p0 sin Ωt, p4 = 0 1000 The equations of motion are p3 =

By symmetry k33 = 440.7 kips/in k23 = −29.3 kips/in k13 = 14.65 kips/in

Mθ¨ + Kθ = p where θ T = [ θ1

θ2

θ3

θ4 ]

pT = [ p1

p2

p3

p4 ]

Assembly gives the stiffness matrix 

 440.7 −29.3 14.7 K =  −29.3 58.6 −29.3  kips/in 14.7 −29.3 440.7

Problem 3.5 Also u2

u1

u3

13750 kip.s2 = 0.0356 386.4 in. = m11

m11 = m33

kip.s2 (27500 + 7500) = 0.0906 386.4  in. 0.0356 0 0 M= 0 0.0906 0  0 0 0.0356

m22 = 1

2

1

∆1

The equations of motion are M¨ u + ku = 0

480

where k12

k22

uT = [ u1

k32

u2

u3 ]

Problem 3.6 Figure S3.5 For the crane, apply a unit force at u1 with restraining forces at u2 and u3 as shown in Figure S3.5.   480 1 2 ∆1 = 2 × × × 480 × × 480 2 EI2 3

Referring to the free body diagram Figure S3.6 and taking moments about A and B respectively ka2 (u1 − u2 ) = 0 L ka2 (u2 − u1 ) = 0 u2 + m2 L¨ L

u1 + cLu˙ 1 + m1 L¨

= 0.0683 in. The equations of motion are Force required for unit displacement = 14.65 kips. Force for unit displacement of crane girder at u1 is 48EI1 = 426.04 kips (240)3 Thus k11 = 426.04 + 14.65 = 440.7 kips/in k21 = −2 × 14.65 = −29.3 kips/in k31 = 14.65 kips/in 48EI2 k22 = = 58.59 kips/in (960)3 k12 = k32 = −29.3 kips/in

M¨ u + Cu˙ + Ku = 0 where  M=

2

K=



m1

ka L2



m2 1 −1



 c 0 C= 0 0  −1 uT = [ u 1 1

u2 ]

10

Problem 3.8 u1

Referring to Figure S3.8(a), the mass matrix is   given by 3m 0 M= 0 m

u2 1

u2 2m

1

u1 m

() L/2

cu· 1 ()

k(u1  u2)

k(u1  u2)

a

()

m1ü1

L

1L (a)

a

L/2 (b)

du1冣 L 冢365 EI 2 L3

L

(c)

1

36 EI du 1 5 L3

L/4

()

()

42 EI du 1 5 L3

m2ü2

Figure S3.6 ()

Problem 3.7

() ()

θ1

a

θ2 42 EI du 5 L2 1

ka(θ2θ1) ka(θ2θ1)

I

du1 L 冢365EI L3 冣 2

L/4

(d)

(e)

Figure S3.8 The moments produced by unit forces acting along d.o.f. 1 and 2 are shown in Figures S3.8 (b) and (c), respectively. The flexibility influence coefficients are given by m2|θ¨2

m1|θ¨1 m1 g




m2g

Figure S3.7 The free-body diagram for the double pendulum is shown in Figure S3.7 when the two pendulums are displaced from the vertical by angles θ1 and θ2 . The forces include reactions at the support hinges A and B, the inertia forces acting on masses m1 and m2 , the gravity forces m1 g and m2 g, and the equal and opposite spring forces ka(θ2 − θ1 ). Taking moments about A and B we get the following two equations of motion m1 l2 θ¨1 − ka2 (θ2 − θ1 ) + m1 gl sin θ1 = 0 m2 l2 θ¨2 + ka2 (θ2 − θ1 ) + m2 gl sin θ2 = 0 Noting that for small θ, sin θ = θ, the two equations reduce to k  a 2 g (θ1 − θ2 ) + θ1 = 0 θ¨1 + m1 l l  a 2 k g θ¨2 + (θ2 − θ1 ) + θ2 = 0 m2 l l

m1 m1 1 L 2 L3 ds = ×L× L= EI 2 EI 3 3EI
m2 m2 (L/2)3 L L L = ds = + × × EI 3EI 2 EI 2 3 7 L = 24 EI
m1 m2 1 L L L3 = ds = − ×L× =− EI 2 EI 2 4EI

a11 = a22

a12

The flexibility matrix is thus given by  L3  A= EI

1 3

− 14

− 14

7 24

 

The stiffness matrix is obtained by inverting A  7 144 EI  24 K= 5 L3 1 4

1 4 1 3

 

11 To determine the force along say d.o.f 1 that is equivalent to the applied force P, we impose displacements δu1 and 0, respectively along the 2 d.o.f. The forces required to impose these displacements are given by     δu1 42/5 EI δu = f1 = K 0 36/5 L3 1

The equations of motion are thus  7    144 EI  24 3m 0 u ¨1 + u ¨2 0 m 5 L3 1 4  9 

The moments produced by these forces are shown in the M diagram Figure S3.8(d). To determine the deflection under the applied load P , we apply a unit load at the point of application of P and in the same direction as P . The m diagram produced by the unit load is shown in Figure S3.8(e). The deflection is then given by
Mm ∆U1 = ds EI  36 L L 2L 1 × × × = − × 2 5EI 4 4 34 L L L 36 × × × − 4 4 8 5EI 1 L 36 − × × L2 × 2 5EI 4 1 42 L EI 2 ×L × + × δu1 2 5 4 L3   15 9 9 21 δu1 = = − − + δu1 160 10 20 160

Problem 3.9

=

160



   u1 u2 1

1 4

3

P

− 17 40 πx 2L 3πx ψ2 (x) = 1 − cos 2L πx π  sin ψ1 (x) = 2L 2L 3π 3πx  ψ2 (x) = sin 2L 2L  π 2 πx  ψ1 (x) = cos 2L 2L  2 3πx 3π ψ2 (x) = cos 2L 2L The mass influence coefficients are given by
L m11 = ¯ m ¯ (ψ1 (x))2 dx = 0.2268mL ψ1 (x) = 1 − cos

0




L

¯ m ¯ (ψ2 (x))2 dx = 1.9244mL

m22 = 0




L

m12 = m21 =

m(ψ ¯ 1 (x)ψ2 (x))dx = 0.5755mL ¯ 0

Applying the principle of virtual displacement we get Q1 δu1 = P ∆U1 which gives 9 P 160 In a similar manner, to determine the equivalent force along d.o.f. 2 we apply the displacements 0 and δu2 along the two d.o.f.. The forces required to impose these displacements are given by     36/5 EI 0 = δu f2 = K 48/5 L3 2 δu2 Q1 =

Using the previous results the corresponding displacement under load P is obtained from     15 48 36 21 9 − δu2 + δu2 ∆U2 = − 36 160 10 42 20 17 = − δu2 40 Using the virtual displacement principle we get Q2 = −

17 P 40

The stiffness influence coefficients are given by
L π 4 EI EI (ψ1  (x))2 dx = k11 = 32L3 0
L π 4 EI k22 = EI (ψ2  (x))2 dx = 81 32L3 0
L EI (ψ1  (x) ψ2  (x))dx = 0 k12 = k21 = 0

The components of the effective load vector are given by
L ¯ ug ψ1 (x)dx = −0.3634mL¨ ¯ ug p1 = −m¨ 0


p2 = −m¨ ¯ ug

L

ψ2 (x)dx = −1.2122mL¨ ¯ ug 0

The influence coefficients in the geometric stiffness matrix are obtained from
L mg ¯ 2 kG11 = mg(L ¯ − x) (ψ1 (x)) dx = 0.3699 L 0
L mg ¯ 2 kG22 = mg(L ¯ − x) (ψ2 (x)) dx = 5.3017 L 0
L mg ¯ kG12 = mg(L ¯ − x) (ψ1 (x)ψ2 (x)) dx = 0.75 L 0 kG21 = kG12

12 The equations of motion are    0.2268 0.5755 z¨1 mL ¯ 0.5755 1.9244 z¨2 4    π EI 1 0 0.370 + − mg ¯ 0.75 32L3 0 81   0.3634 = −mL¨ ¯ ug 1.2122

The other columns of the stiffness matrix are calculated by a similar procedure. The assembled stiffness matrix is given by 0.75 5.302

 

z1 z2



 4EIb Lb

 K=

+

4EIc Lc

2EIb Lb 6EIc L2c

2EIb Lb 4EIb 4EIc Lb + Lc 6EIc L2c

6EIc L2c 6EIc L2c 24EIc L3c

  

On substituting the values of E, Ib , Ic , Lb and Lc

Problem 3.10  6EIb /Lb2

6EIb /Lb2

2EIb /Lb

4EIb /Lb

K = 104

6EIc /Lc2 4EIc /Lc

A 120.0

A   30.0   B 0.9

30.0 120.0 0.9

B  0.900   0.900    0.036

Degrees of freedom 1 and 2 are condensed out from K to give   K∗ = KBB − KBA K−1 AA KAB = 252 kips/in.

6EIc /Lc2

For frame A 2EIc /Lc

u3 = θ1 + aθ3 6EIb /Lb2

6EIb /Lb2

2EIb /Lb

= [1

6EIc /Lc2

4EIc /Lc

4EIb /Lb

2EIc /Lc 6EIc /Lc2 12EIc /Lb3

 θ1 0 a ]  θ2  θ3 

= TA θ KA = TT K∗ T   1 0 120 = 252  0 0 0 120 0 14400 For frame B

12EIc /Lc3

TB = [ 0

6EIc /Lc2

6EIc /Lc2

6EIc /Lc2

6EIc /Lc2 12EIc /Lc3

12EIc /Lc3

Figure S3.10 The first column of the stiffness matrix of an individual frame is obtained by imposing displacements u1 = 1, u2 = 0, u3 = 0. The external forces required to maintain the displacement pattern are identified in Figure S3.10 and are given by k11 = 4EIb /Lb + 4EIc /Lc k21 = 2EIb /Lb k31 = 6EIc /L2c

1 a]   0 0 0 KB = 252  0 1 120  0 120 14400 For frame C 0 −a ]   1 0 −120 Kc = 252  0 0 0 −120 0 14400 Tc = [ 1

Assembly gives K = KA + K B + K c   2 0 0 = 252  0 1 120  0 120 43200

13 Slab mass m =

100×20×20 1000×386.4

= 0.1035

kip.s2 in. .

For the deck

Mass moment of inertia is given by

∗




L

0

a2 + b2 2402 + 2402 = 0.1035 × 12 12 = 993.79

mL = 2
L 

2 π 4 EI k∗ = EI ψ (x) dx = 2L3 0 ∗ p = pψ(xi )

I0 = m



0.1035 0 M= 0 0.1035 0 0

 0 0  993.8

where xi = vt and p = k(uv −η)+c(u˙ v − η)−m ˙ ¨− tη mt g. Thus, we have

The equations of motion are Mθ¨ + Kθ = 0

p∗ = k {uv − η} α + c {u˙ v − η} ˙ α − mt η¨α − mt gα

where θT = [ θ1 θ2 θ3 ]

Equation of motion of the deck is given by

Problem 3.11 and Problem 3.12 u

mv

m∗ z¨ + k ∗ z = p∗ On combining with the equation for sprung mass and substituting for η, η˙ and η¨, the complete set of equations can be written as

v k

2

m {ψ(x)} dx

m =



c ub  z sin(px/L) mt

L mv üv

  mv u ¨v 0 mL 2 + m α z¨ 0 t 2    u˙ v c −cα + z˙ −cα cα2 + 2mt θαβ  k −kα − cθβ + −kα kα2 + cθαβ − mt θ2 α2 +   mv = −g mt α

 π 4 EI 2L3

uv z



c{u· v  η· (vt)}

k{u  η(vt)} mv g

k{u  η(vt)}  c{u· v  η· (vt)} .. mt η

Problem 3.13

mtg

I0 = Figure S3.11 m= πx , ψ(x) = sin L

πx ub = z sin L

Referring to the free body diagram for the sprung mass shown in Figure S3.11 mv u ¨v + c(u˙ v − η) ˙ + k(uv − η) = −mv g where η = zα η˙ = zθβ + zα ˙ ˙ + z¨α η¨ = −zθ2 α + 2zθβ and θ = (πv/L), α = sin(θt), β = cos(θt).

EIf = GJr = EIr =

10 × (3.5)2 = 0.3170 lb.in.s2 386.4 lb.s2 10 = 0.02588 386.4 in. 1 1 × × 30 × 106 = 39, 063 12 64 π × 12 × 106 = 1, 178, 100 32 π × 30 × 106 = 1, 472, 620 64

The degrees of freedom are identified in Figure S3.13. To derive the flexibility matrix, unit loads are applied as shown and the corresponding moment and torque diagrams are drawn. Flexibility influence coefficients are obtained by integrating the areas

14 k22 is given by

1




1

L

k22 =









EIψ1 (x)ψ1 (x)dx 0

36

12

1

4 × 20000 × 20 4EI = = 2.5 L = 64.00 × 104 N.m.

12

6 kN/m

1

B

u2

C

Figure S3.13

a11 =

a22

a12

1 2 12 × × 12 × 12 × 2 EIf 3

362 12 2 1 × × 36 + × 36 × 12 + × 2 EIr 3 GJr = 0.02970 12 1 × 36 =1× ×1+ ×1 EIf GJr = 3.3775 × 10−4 1 12 12 = × × 12 × 1 + × 36 × 1 2 EIf GJr = 2.2099 × 10−3 

 −4 297.00 22.099 A = 10 22.099 3.378   65.60 −429.2 −1 K=A = −429.2 5768.1 Equations of motion are given by     0.02588 0 u ¨ 0 0.3170 θ¨       65.60 −429.2 u 0 + = −429.2 5768.1 θ 0

3.5 m

u1 A

1.5 m

2.5 m

Figure S3.14 Mass m Due to self weight 138.0 kg/m Due to superimposed load 6000 = 611.6 kg/m 9.81 Total 749.6 kg/m Using this value of total distributed mass
m22 =

L

mψ1 (x)ψ1 (x)dx 0

749.6 × (2.5)3 105 = 111.5 kg.m2

=

For segment AB mass and stiffness matrices are derived from shape functions

Problem 3.14 Considering that the members are axially rigid, end C is fixed and end A can only rotate but not translate in either X or Y direction, the structure has only two degrees of freedom as identified in Figure S3.14 . For segment BC the displaced shape corresponding to unit rotation at B can be taken as  x 2 ψ1 = x 1 − L where x is measured from B along BC and L = 2.5 m is the length of the segment. The stiffness

 x 2 ψ1 = x 1 − L  x2  x −1 ψ2 = L L where x is measured from A along AB, and L = 3.808 m.   x Mass variation is given by m(x) = mA 1 + L where mA = 69 kg. The mass matrix is obtained from
L

mij =

m(x)ψi (x)ψj (x)dx 0

15 which gives



 −9L3 13L3

3

mA 11L 840 −9L3 Substituting mA = 69 kg/m and L = 3.808 m   49.89 −40.82 MAB = −40.82 58.96 MAB =

12EI 12EI L3 6EI 6EI L3 L2 L2

6EI L2

6EI L2

12EI L3 (a)

12EI 12EI 3 L3 6EI 6EI L L2 L2

12EI L3

(b)

The variation of cross sectional flexural rigidity is given by  x EI(x) = EI A 1 + L . The stiffness influence coefficients are obtained

L from kij = 0 EI(x)ψi  (x)ψj (x)dx which gives 5EI A 3EI A 7EI A , k12 = k21 = , k22 = k11 = L L L Substituting 20000 × 10 EIA = = 5.25 × 104 N.m. L 3.808   4 26.26 15.76 N.m. KAB = 10 15.76 36.76 Adding the contributions from AB and BC   49.89 −40.82 M= kg.m2 −40.82 170.51   26.26 15.76 N.m K = 104 15.76 100.76

6EI L2

4EI L

6EI L2

2EI L

(c)

6EI L2 2EI L

6EI L2

6EI L2

4EI L

4EI L

(d)

6EI L2 6EI L2

6EI L2

4EI L

4EI L

2EI L

(e)

6EI L2

2EI L

6EI L2

4EI L

(f)

2EI L

Figure S3.15

The equations of motion are M¨ u + Ku = 0 T

where u = [ u1

u2 ]

Problem 3.15 Figures S3.15 (a) through (f) show the displaced shapes produced by unit displacement applied at one of the d.o.f at a time. The figures also identify the moments and forces required to maintain such displacements. The stiffness matrix is assembled one column at a time. Thus, the first column notes the forces required along d.o.f. u1 , u2 , θ1 , θ2 , θ3 , θ4 , θ5 , and θ6 when a unit displacement is imposed against u1 . 

24EI l3

  12EI  − l3    6EI  − l2     0    6EI  l2  0

− 12EI l3

− 6EI l2

0

6EI l2

24EI l3

0

− 6EI l2

0

0

4EI l

2EI l

0

− 6EI l2

2EI l

8EI l

2EI l

0

0

2EI l

8EI l

0

2EI l

6EI l2

where l = L/3

0

0

We partition the matrix, such that the first 2 by 2 partition, comprising the first two rows and the first two columns, is related to d.o.f. u1 and u2 , while the last 4 by 4 partition, comprising the last four rows and last four columns, relates to d.o.f. θ1 through θ4 . We then have

KAA =



      0     0    2EI  l  



24 −12



6EI l2

4EI l

EI l3

KBB

KAB

4 EI  2 =  0 l 0

EI = 2 l



−6 0

2 8 2 0

0 −6

KBA = KTAB

−12 24



 0 0  2 4

0 2 8 2

6 0

0 6



16 Static condensation of the θ d.o.f. gives the following 2 by 2 d.o.f. K∗ = bf K AA − KAB KBB −1 KBA   9.6 −8.4 = 27EI L 3 −8.4 9.6 The equation of motion is given by 

m 0

0 m



    27EI 9.6 −8.4 u ¨1 u1 + −8.4 9.6 u ¨2 u2 L3   0 = 0

17

Chapter 4

and vertical movements of mass m by u and v respectively.

Problem 4.1

u=x+l

The absolute rotation of the hollow cylinder, φ, is given φ=

Rθ R − r2 −θ =θ r2 r2

θ

A ϕ

v = l(1 − cos θ) 1 1 1 T = mu˙ 2 + mv˙ 2 + mx2 2 2 2  1 1  2 = m x˙ + l2 θ˙2 + 2l cos θx˙ θ˙ + M x˙ 2 2 2 1 2 V = 2 × kx + l(1 − cos θ)mg 2   d d ∂T ˙ + Mx = (mx˙ + ml cos θ θ) ¨ dt dt ∂ x˙ ¨ = m¨ x − ml sin θ θ˙2 + ml cos θ θ¨ + M x ∂V = 2kx  ∂x   d ∂T = ml2 θ˙ + ml cos θx˙ dt ∂ θ˙ x − ml sin θx˙ θ˙ = ml2 θ¨ + ml cos θ¨ ∂V = mgl sin θ ∂θ

A x

Figure S4.1

M k

k

Kinetic energy 1 2 M (R − r2 ) θ˙2 + 2 1 2 = M (R − r2 ) θ˙2 + 2

T =

d dt



∂T ∂ θ˙



 1M  2 r2 + r12 φ˙ 2 2 2 1 r22 + r12 2 (R − r2 ) θ˙2 M 4 r22

M r22 + r12 = M (R − r2 ) θ¨ + (R − r2 ) θ¨ 2 r22

I

Light rigid bar v m

u

2

Figure S4.2

Potential energy The equations of motion are V = M g (R − r2 ) (1 − cos θ) ∂V = M g (R − r2 ) sin θ ∂θ ≈ M g (R − r2 ) θ for small θ Hence the equation of motion is 3r22 + r12 (R − r2 ) θ¨ + gθ = 0 2r22

Problem 4.2 The two degrees of freedom; x and θ, are identified in Figure S4.2. Representing the horizontal

(M + m)¨ x + ml cos θθ¨ − ml sin θθ˙2 + 2kx = 0 ml cos θ¨ x + ml2 θ¨ − ml sin θx˙ θ˙ + mgl sin θ = 0 For small x and θ, cos θ ≈ 1, sin θ ≈ θ and θ˙2 , x˙ θ˙ are negligible. Hence the equations reduce to (M + m)¨ x + mlθ¨ + 2kx = 0 x ¨ + lθ¨ + gθ = 0

18

Problem 4.3
1 πx 1 L dx + M u˙ 22 mu˙ 21 sin2 2 0 L 2 1 1 = mLu˙ 21 + M u˙ 22 4 2 2 
L π2 πx 1 1 2 dx + k (u2 − u1 ) V = EI u1 2 sin L L 2 2 0 π4 1 1 2 = EI 3 u21 + k (u2 − u1 ) L 2 4 Differentiation gives   d ∂T mL = u ¨1 dt ∂ u˙ 1 2   d ∂T = Mu ¨2 dt ∂ u˙ 2 ∂V π4 = EI 3 u1 − k (u2 − u1 ) ∂u1 2L ∂V = k (u2 − u1 ) ∂u2 The equations of motion are m    0 u ¨1 2 0 M u ¨2  4      π EI u1 0 + k −k + 2L3 = 0 u2 −k k T =

where m = mL.

Problem 4.4

∂V = kS (q1 − q2 a − q3 ) + kS (q1 + q2 a − q4 ) ∂q1 ∂V = −kS a (q1 − q2 a − q3 ) + kS a (q1 + q2 a − q4 ) ∂q2 ∂V = kT q3 − kS (q1 − q2 a − q3 ) ∂q3 ∂V = kT q4 − kS (q1 + q2 a − q4 ) ∂q4 ∂R = cS (q˙1 − q˙2 a − q˙3 ) + cS (q˙1 + q˙2 a − q˙4 ) ∂ q˙1 ∂R = −cS a (q˙1 − q˙2 a − q˙3 ) + cS a (q˙1 + q˙2 a − q˙4 ) ∂ q˙2 ∂R = −cS (q˙1 − q˙2 a − q˙3 ) ∂ q˙3 ∂R = −cS (q˙1 + q˙2 a − q˙4 ) ∂ q˙4 Substitution in Lagrange’s equations gives    W q¨1 g W r2   q¨2       g w  q¨3  g w q¨4 g     2 0 −1 −1 q˙1 a −a   q˙2   0 2a + cS     −1 a 1 0 q˙3 −1 −a 0 1 q˙4   2 0 −1 −1  q1   0 2a a −a    q2  + kS   −1 0   q3  = 0 a 1 + kkTS q4 −1 −a 0 1 + kkT S

The degrees-of-freedom have been identified in Problem 3.1. Kinetic energy: 1W 2 1W 2 2 1w 2 1w 2 T = (q1 ) + r (q˙2 ) + (q˙3 ) + (q˙4 ) 2 g 2 g 2g 2g Potential energy: 1 1 1 2 V = kT q32 + kT q42 + kS (q1 − q2 a − q3 ) 2 2 2 1 2 + kS (q1 + a2 a − q4 ) 2 Rayleigh dissipation function: 1 1 2 2 R = cS (q˙1 − q˙2 a − q˙3 ) + cS (q˙1 + q˙2 a − q˙4 ) 2 2 Differentiation gives   d ∂T W = q¨1 dt ∂ q˙1 g   W r2 d ∂T = q¨2 dt ∂ q˙2 g   w d ∂T = q¨3 dt ∂ q˙3 g   w d ∂T = q¨4 dt ∂ q˙4 g

The equations are identical to those obtained in Problem 3.1.

Problem 4.5

θ1

Light in extensible string y

I1 Uniform rigid bar mass m

θ2 x

I2 P(t )

Figure S4.5 The Cartesian coordinates of the centre of the rigid bar are given by: l2 sin θ2 2 l2 y = l1 cos θ1 + cos θ2 2

x = l1 sin θ1 +

19 Differentiating with respect to time l2 x˙ = l1 cos θ1 θ˙1 + cos θ2 θ˙2 2 l2 ˙ y˙ = −l1 sin θ1 θ1 − sin θ2 θ˙2 2 The kinetic energy is obtained from 1 1 1 mx˙ 2 + my˙ 2 + I0 θ˙22 2 2 2 Substitution for x˙ and y˙ gives   1 l22 ˙2 2 ˙2 ˙ ˙ T = m l1 θ1 + θ2 + l1 l2 θ1 θ2 cos (θ1 − θ2 ) 2 4 1 + I0 θ˙22 2 Potential energy is obtained from   l2 V = mg l1 + (1 − cos θ2 ) 2 T =

Differentiation gives   ∂T l1 l2 ˙ 2˙ = m l1 θ 1 + θ2 cos (θ1 − θ2 ) 2 ∂ θ˙1    l1 l2 ¨ d ∂T = m l12 θ¨1 + θ2 cos (θ1 − θ2 ) ˙ dt ∂ θ1 2 !   l1 l2 θ˙2 sin (θ1 − θ2 ) θ˙1 − θ˙2 − 2  2  ∂T l l1 l2 ˙ = m 2 θ˙2 + θ1 cos (θ1 − θ2 ) 4 2 ∂ θ˙2 ˙ + I0 θ2    2 l d ∂T l1 l2 ¨ = m 2 θ¨2 + θ1 cos (θ1 − θ2 ) 4 2 dt ∂ θ˙2   l1 l2 ˙ ˙ ˙ − θ1 sin (θ1 − θ2 ) θ1 − θ2 2 + I0 θ¨2 ∂V = mgl1 sin θ1 ∂θ1 ∂V mgl2 = sin θ2 ∂θ2 2 Substitution in Lagrange’s equation gives:  l1 l2 m l12 θ¨1 + cos (θ1 − θ2 ) θ¨2 2   l1 l2 sin (θ1 − θ2 ) θ˙1 θ˙2 − θ˙22 − 2  m

+ mgl1 sin θ1 = 0 l22

l1 l2 (θ1 − θ2 ) θ¨1 − θ¨2 + 4 2   l1 l2 sin (θ1 − θ2 ) θ˙12 − θ˙1 θ˙2 2 mgl2 sin θ2 = 0 + I0 θ¨2 + 2

For small vibrations sin (θ1 − θ2 ) ≈ θ1 −θ2 , sin θ1 ≈ θ1 , sin θ2 ≈ θ2 , cos (θ1 − θ2 ) ≈ 1 and terms containing θ˙1 θ˙2 , θ˙12 , and θ˙22 are negligible. The equations reduce to ml1 l2 ¨ ml12 θ¨1 + θ2 + mgl1 θ1 = 0 2  2  ml2 ml22 ¨ mgl2 ml1 l2 ¨ θ2 + + θ1 + θ2 = 0 12 2 2 4 Canceling out m and substituting l1 = l2 = l     g     1 12 θ¨1 θ1 0 0 l + = g 1 1 ¨2 0 θ 0 θ 2 2l 2 3

Problem 4.6 Let c be the extension of the spring when in the position of equilibrium.  m1  g kc = m2 + 2

m2

c

y2

m1

φ

l y1

r

M

x1 x2

Figure S4.6 Referring to Figure S4.6 l cos φ 2 y2 = l cos φ l x1 = sin φ 2 x2 = l sin φ y1 =

For small vibration, if angle φ changes to φ + θ, θ = dφ. Hence dφ l l = − sin φθ˙ y˙ 1 = − sin φ 2 dt 2 y˙ 2 = −l sin φθ˙ l x˙ 1 = cos φθ˙ 2 x˙ 2 = l cos φθ˙

20 Kinetic energy

Hence

1 1 1 m2 y˙ 22 + m1 y˙ 12 + m1 x˙ 21 2 2 2  2 1 M r2 x˙ 2 1 m1 l 2 ˙ 2 1 2 + θ + M x˙ 2 + 2 2 r 2 2 12 2 1 m1 l 1 = m2 l2 sin2 φθ˙2 + sin2 φθ˙2 2 4 2 1 m1 l 2 1 m1 l 2 ˙ 2 cos2 φθ˙2 + + θ 2 12 2 4 1M 2 1 + M l2 cos2 φθ˙2 + l cos2 φθ˙2 2 2 2   1 m1 l 2 1 1 3M 2 2 2 2 = m2 l sin φ + + l cos φ θ˙2 2 2 3 2 2

x˙ = u˙ + l cos θθ˙ y˙ = l sin θθ˙

T =

Differentiation gives    1 d ∂T = m2 l2 sin2 φ + m1 l2 ˙ 3 dt ∂ θ  3M 2 l cos2 φ θ¨ + 2 Potential energy V =

1 1 2 k (c + l sin φθ) − kc2 2 2 l − m1 g sin φθ − m2 gl sin φθ 2

∂V = kcl sin φ θ + kl2 sin2 φ θ ∂θ  m 1 + m2 gl sin φ θ − 2 = kl2 sin2 φ θ The equation of motion becomes   1 3M 2 2 2 2 2 m2 l sin φ + m1 l + l cos φ θ¨ 3 2

1 1 1 M u˙ 2 + mx˙ 2 + my˙ 2 2 2 2   1 1 ˙ cos θ = M u˙ 2 + m u˙ 2 + l2 θ˙2 + 2u˙ θl 2   2 d ∂T ¨ cos θ − mlθ˙2 sin θ = Mu ¨ + m¨ u + mθl dt ∂ u˙   d ∂T ˙ sin θ = ml2 θ¨ + m¨ ul cos θ − mu˙ θl dt ∂ θ˙ V = ku2 + mgl(1 − cos θ) ∂V = 2ku ∂u ∂V = mgl sin θ ∂θ T =

The Lagrange’s equations are given by (M + m)¨ u + ml cos θθ¨ − ml sin θθ˙2 + 2ku = 0 ml cos θ¨ u + ml2 θ¨ − ml sin θu˙ θ˙ + mgl sin θ = 0 For small vibrations, θ˙2 and u˙ θ˙ are negligible and cos θ ≈ 1, sin θ ≈ θ. The linearized equations become (M + m)¨ u + mlθ¨ + 2ku = 0 m¨ u + mlθ¨ + mgθ = 0

Problem 4.8 The constraint equation is r(θ + φ) = Rθ The kinetic and potential energies are given by

+ kl2 sin2 φθ = 0

1W 1 (R − r)2 θ˙2 + I φ˙ 2 2 g 2 V = (R − r)(1 − cos θ)W T =

Substituting m1 = 1kg, m2 = 3kg, m = 5kg, k = 10, 000 N/m and φ = 45o 

3 1 15 + + 2 3 4



10, 000 θ¨ + θ=0 2 67 ¨ θ + 5000θ = 0 12

The equations of motion are obtained by minimizing the functional


t2

F =

T − V + λ {(R − r)θ − rφ} dt

t1

so that

Problem 4.7 The Cartesian displacements of mass m are given by x = u + l sin θ y = l(1 − cos θ)




t2

δF = t1

w ˙ θ˙ + Iφδ φ˙ (R − r)2 θδ g

− W (R − r) sin θδθ + λ(R − r)δθ − λrδφ dt =0

21 Integration of the first two terms is carried out as follows
t2 W d d (R − r)2 θ˙ (δθ) + I φ˙ (δφ)dt g dt dt t1


t2 W t2 2 ˙ ¨ (R − r) θδθ |t1 − θδθdt = g t1


t2 ¨ ˙ |t2 − φδφdt + I φδφ t1


t2

= t1

t1

−

W ¨ − I φδφ ¨ (R − r)2 θδθ dt g

because variations δθ and δφ vanish at t1 and t2 . Substitution in δF gives
 W δF = − (R − r)2 θ¨ − W (R − r) sin θ g

 + λ(R − r) δθ − (I φ¨ + λr)δφ dt = 0 from which the following two equations are obtained

where ug = G sin Ωt. We have ∂T = ml2 θ¨ + mlu˙ g sin θ ˙ ∂ θ   d ∂T = ml2 θ¨ + ml¨ ug cos θ θ˙ ug sin θ + ml¨ dt ∂ θ˙ ∂V = mgl sin θ dθ The equation of motion is   ∂V d ∂T + =0 ˙ dt ∂ θ dθ or

cos θ ˙ g θ¨ + u˙ g θ + l l

For small vibrations this reduces to   W r2 + I (R − r)θ¨ + W r2 θ = 0 g ¨

Also, λ = − Irφ which can be interpreted as the tangential friction force at the contact between the wheel and the rail.

Problem 4.9 The x an y displacements and velocities of the point mass m are given by x = l sin θ y = l(1 − cos θ) + ug x˙ = l cos θ θ˙ y˙ = l sin θ θ˙ + u˙ g The kinetic energy T and potential energy V are obtained from 2 1  2 1  ¨g T = m l cos θ θ˙ + m l sin θ θ˙ + u 2 2 V = mg(l − l cos θ) + mgug

1+

u ¨g g

 sin θ = 0

Problem 4.10 The x an y displacements and velocities of the point mass m are given by x = u + l sin θ

W (R − r)θ¨ + W sin θ = λ g I φ¨ = −λr Solving these along with the constraint equation yields   W r2 + I (R − r)θ¨ + W r2 sin θ = 0 g



y = l(1 − cos θ) x˙ = u˙ + l cos θ θ˙ y˙ = l sin θ θ˙ The kinetic energy is given by 1 2 M (u) ˙ + 2 3 2 ˙ + = M (u) 4

T =

 2 1 M R2 u˙ 1 1 2 ˙ 2 + (y) ˙ + m(x) 2 2 R 2 2 1 1 ˙ 2 + mu˙ θl ˙ cos θ m(u) ˙ 2 + ml2 (θ) 2 2

We therefore have ∂T ˙ cos θ = ml2 θ˙ + mul ˙ ∂ θ   d ∂T = ml2 θ¨ − mul ˙ sin θ θ˙ + ml¨ u cos θ dt ∂ θ˙ 3 ∂T = M u˙ + mu˙ + m cos θ θ˙ ∂ u ˙ 2     3 d ∂T = M +m u ¨ + ml cos θ θ¨ 2 dt ∂ u˙ ˙ 2 − ml sin θ (θ) The potential energy V is given by 1 V = mg(l − l cos θ) + ku2 2 Differentiation with respect to u and θ gives ∂V = ku ∂u ∂V = mgl sin θ ∂θ

22 The virtual work equation is δW = (−cu)δu from which the generalized forces are obtained as Q1 = −cu and Q2 = 0. The equations of motion are   3 ˙ 2 M +m u ¨ + ml cos θ θ¨ − ml sin θ (θ) 2 + ku + cu˙ = 0 ˙ sin θ θ˙ + ml¨ u cos θ + mgl sin θ = 0 ml θ − mul 2¨

When the pendulum is locked at the hinge θ = u/R, so that ! "  2 3 u 1 l 2 l 2 2 2 T = M u˙ + m u˙ + cos u˙ + 2u˙ R R 4 2 R " !  2 3 1 l u ∂T l = M u˙ + m 2u˙ + 2 u˙ + 4 u˙ cos ∂ u˙ 2 2 R R R !  "  2  d ∂T u 3 l 2l u ¨ = M +m+ m + m cos R R R 2 dt ∂ u˙ −

2l u mu˙ 2 sin R2 R

The potential energy and its derivative with respect to u are given by  1 2 u ku + mgl 1 − cos 2 R ∂V l u = ku + mg sin ∂u R R V =

The equation of motion is "

3 M +m+ 2

−



l R

2

! u 2l u ¨ m + m cos R R

l u 2l u mu˙ 2 sin + ku + mg sin + cu = 0 R2 R R R

23

Chapter 5

Velocity at touch down, v0 , is given by  √ v0 = 2gh = 2 × 1.6 × 10 = 5.66 m/s

Problem 5.1 During recoil the damper is not engaged, hence equating the maximum kinetic energy to maximum potential energy

The displacement from the position of touch down to that of rest under gravity ∆=

1 1500 1 × 642 k × 42 = × 32.2 2 2 k = 11, 925 lb./ft.

# √ 11, 925 × 1500 c = ccr = 2 km = 2 32.2 = 1490 lb.s/ft.

Thus u0 = −0.016 m/s and the dynamic displacement from the position of rest under gravity is given by u(t) = ρe−ξωt sin(ωd t + φ) where $ ρ=

Problem 5.2 # ω=

k = m

#

e−ξωt e−ξω(2Td +t) 4πξ



1 − ξ2

386.4 = 22.70 rad/s 3/4

=4 = ln 4

ξ = 0.1096  √ ωd = ω 1 − ξ 2 = 22.70 1 − 0.10962 = 22.56 rad/s



u20 +

tan φ =

$ g = ∆

4500 × 1.6 = 0.016 m 450 × 1000

v0 + ωξu0 ωd

2 = 0.5746

u0 ω d = −0.02786 v0 + u0 ωξ

which gives φ = −0.02785 The maximum displacement occurs when  sin(ωd t + φ) = 1 − ξ 2 = 0.9798 ωd t + φ = 1.3694 so that t = 0.1426. The maximum displacement is given by umax = 0.5746×e(−0.2×10×0.1426) ×0.9798 = 0.4233 m The total displacement from the position at touch down is 0.4233 + 0.016 = 0.4393 m.

Problem 5.3 Problem 5.4 u0 mg/k

V0

Initial velocity of mass m on contact  v0 = 2gh

m u c

k

Also

v0 sin ωt ω u ¨ = −v0 ω sin ωt u=

Touch down

Deflection under gravity

Figure S5.3 The position of the capsule at touch down as well as position of rest under gravity is shown in Figure S5.3. #

450 × 1000 = 10 rad/s 4500   ωd = ω 1 − ξ 2 = 10 1 − 0.22 ω=

= 9.798 rad/s

Hence the maximum acceleration = # 2ghk = m

√

2gh ω

Problem 5.5 Because momentum is conserved on impact (M + m)v0 = mvb 0.2 × 20 = 1.25 m/s v0 = 3.2

24 At t = 0.1 ut = 0.4597 in. 8.415 in/s. In the second era

(a) The displacement is given by u = ρ sin (ωt + φ) where % Frequency ω =

= 25 rad/s

Maximum displacement = ρ = (b)

8.415 sin ωt + 1.5(1 − cos ωt) 10 = 1.5 − 1.04 cos ωt + 0.8415 sin ωt 0.1 < t ≤ 0.2 s

ut = 0.4597 cos ωt +

v0 ρ= ω 2×1000 3.2

1.25 25

= 50 mm

where t = t − 0.1. At end of the era, t = 0.1, and ut = 1.646 in. u˙ t = 10.4 sin ωt + 8.415 cos ωt

u = ωv0d e−ωξt sin ωd t The maximum displacement occurs when sin ωd t =

 1 − ξ 2 = 0.995,

= 13.3 in/s ωd t = 1.4708

The damped frequency is given by ωd = ω



1 − ξ 2 = 24.875 rad/s

Hence t = 0.0591 s and umax

u˙ t = ω sin ωt =

In the third era ut = 1.646 cos ωt1 + 1.33 sin ωt1 + 0.5(1 − cos ωt1 ) = 0.5 + 1.146 cos ωt1 + 1.33 sin ωt1 0.2 < t ≤ 0.3 where t1 = t − 0.2. At the end of era t1 = 0.1, and

1.25 −25×0.1×0.0591 e = 25 = 0.0431 m = 43.1 mm

ut = 2.238 in. u˙ t = −11.46 sin ωt1 + 13.3 cos ωt1 = −2.457 in/s

u

k  2 N/mm

3 kg

20 m/s

0.2 kg u, in

C O3

Figure S5.5

B

57.3° O1

Problem 5.6 The phase plane diagram is shown in Figure S5.6. To obtain an analytical solution, solve the equation of motion

O2

A t, s

Figure S5.6 m¨ ut + kut = kug

Problem 5.7

which yields ut = u0 cos ωt +

v0 sin ωt + ug (1 − cos ωt) ω

In the first era u0 = v0 = 0, ug = 1, ω = 10 u = 1 − cos 10t, 0 ≤ t ≤ 0.1 t

Let δ = ln(u1 /u2 ) represent the logarithmic decrement, u1 and u2 being the amplitudes during two consecutive cycles, then δ

ξ=

(4π 2 + δ 2 )

δ1 = ln(20/2), ξ1 = 0.3441 δ2 = ln(20/2.75), ξ2 = 0.3011

25 Let m1 be the mass of the vehicle without the passenger and m2 = m1 + 300 the mass with the passenger. If c represents the damping constant we have c = 0.3441 ξ1 = √ 2 km1 c = 0.3011 ξ2 = √ 2 km2 # ξ1 m1 + 300 0.3441 = = = 1.1428 ξ2 m1 0.3011

%

1 − ξ12 = 18.85,

ω1 = 20.07,

m1 ω12

= 395, 125 N/m

k=

The velocity of mass m as it strikes the larger mass is given by  √ v¯ = 2gh = 2 × 9.81 × 1 = 4.43 m/s The velocity after impact is obtained from conservation of momentum (M + m)v0 = mm ¯ 30v0 = 5 × 4.43 or v0 = 0.738 m/s

which gives m1 = 980.4 kg. Also, T1 = 1/3 s, ωd1 = 2π/T1 = 18.85 rad/s. We thus have ω1

Problem 5.9

√ and c = 2ξ1 km1 = 13, 545 N.s/m.

The displacement of the combined masses due to static application of mass m will produce a downward displacement of m/k = 5/1920 = 0.0026 m. The problem can thus be considered as that of free vibration of mass M + m about its position of rest under gravity with an initial displacement of −0.0026 and an initial velocity of 0.738 m/s. The characteristics of the system # k 1920 ω= = = 8 rad/s M +m 30 c 48 ξ=  = 0.1 = √ 2 1920 × 30 2 k(M + m) #

Problem 5.8 ca/Lu· a u

The free vibration response is given by

L

Aρgu

mü

u = ρe−ωξt sin(ωd t + φ) Figure S5.8 where We use tip deflection as the single degree of freedom. The free body diagram is shown in Figure S5.8. The forces include the inertia force M u ¨, the damping force c(a/L)u˙ and the force due to buoyancy Aρgu. Taking moment about the hinge we get a2 Mu ¨L + c u˙ + AρguL = 0 or L  a 2 Mu ¨+c u˙ + Aρgu = 0 L

 ωd = ω 1 − ξ 2 = 7.96 $ 2  v0 + ωξu0 = 0.0925 ρ = u20 + ωd u0 ωd φ = tan−1 = −1.613◦ v0 + ωξu0 The time to maximum displacement is obtained by equating the time derivative of u to zero. This gives sin(ωd t + φ) =

k = Aρg,

1 − ξ2

ωd t + φ = 84.26◦

Thus we have ∗



∗

 a 2

c =c L  √ ∗ ∗ c = ccrit = 2 k M = 2 AρgM  2  L c= 2 AρgM a

ωd t = 85.874◦ = 1.499 rad 1.499 = 0.1883 s. t= 7.96 Substituting this value of t in the expression for u we get umax = 0.0796 m. The displacement from the initial position of mass M is 0.0026 + 0.0796 = 0.0822 m.

26

Problem 5.10

Problem 5.12

If u1 and un represent the displacements at the interval of 10 cycles, we have 2π × 10ξ u1 150 =  ln = ln 80 un 1 − ξ2

For critical damping u = (u0 + ωu0 t)e−ωt Substituting u = u0 /10 and ω = 20 we get 0.1 = (1 + 20t)e−20t

which gives ξ = 0.01. Also  ωd = 3 × 2π = ω 1 − ξ 2 so that ω = 18.85 rad/s. The generalized stiffness corresponding to the tip deflection is 3EI/l3 from which we get the flexural rigidity EI. Thus k = mω 2 or 3EI/l3 = 75 × 18.852 and EI = 8883 N/m.

Problem 5.11 Let J0 represent the mass moment of inertia about the center of gravity and JM that about the meta-center, then we have

Solution by trial and error gives t = 0.195 s

Problem 5.13 # ω=

For movement to the right, the equation of motion is m¨ u + ku = −µN With initial velocity of v0 the solution yields µN v0 sin ωt + (cos ωt − 1) ω k µN ω sin ωt u˙ = v0 cos ωt − k

JM = J0 + M H 2 where H is the distance between the meta-center and the center of gravity. Referring to Figure S5.11 which identifies the gravity force M g and the inertial moment for vibration about the meta-center, the equation of motion is given by

2 × 1000 = 25.8 rad/s. 3

u=

At extreme right u˙ = 0, hence tan ωt =

JM θ¨ + M gH sin θ = 0

v0 k ω µN

Substituting ω = 25.8, v0 = 1.0 m/s, µ = 0.2, k = 2000 N/m and N = 3 × 9.81 N. tan ωt = 13.17 sin ωt = 0.9970 cos ωt = 0.0757

M θ MHθ¨

j0θ¨

Hence 0.2 × 3 × 9.81 1 × 0.997 + (0.0757 − 1) 25.8 2000 = 0.0359 m = 35.9 mm

u= Figure S5.11 For small small vibrations we have JM θ¨ + M gHθ = 0 M gH ω2 = JM   2 2 ω J0 + M H − M gH = 0 Solving the quadratic equation in H we get ' & # 4 ω 4J 1 0 g + g2 − H= M 2ω 2

In a half cycle of motion amplitude decreases by 2

µN 2 × 0.2 × 3 × 9.81 = k 2000 = 5.89 × 10−3 m = 5.89 mm

Maximum displacement on return swing = 35.9 − 5.89 = 30.0 mm

27

Problem 5.14

Problem 5.15 (a)

40

e−ξωt

= 3.52 e−ξω(4Td +t) ω = ln 3.52 = 1.258 8πξ ωd 8πξ  = 1.258, ξ = 0.05 1 − ξ2

30

Displacement, mm

20 10 2.943 0 2.943

10

With change in m, ω changes to ω1 and ξ changes to ξ1 while k and c remain constant

20 30 40

c ξ= √ , 2 km 0

0.1

0.2

0.3

0.4 Time, s

0.5

0.6

0.7

0.8

Hence

ξ = 0.0408 ξ1 = √ 1.5

Figure S5.14 The displacement response is shown in Figure S5.14. As shown in Problem 5.13 the first peak is reached when tan ωt = 13.17, or ωt = 1.495 s and since ω = 25.8, t = 0.05794 s. Each subsequent cycle takes T = 2π/ω = 0.2435 s. It is of interest to note that while the time required for each half cycle of motion consisting, for example, the motion from zero position to the extreme right and back to zero takes a time T /2 = π/ω = 0.1217 s, the time to move from zero to extreme right is not T /4 = 0.0609 s, but just 0.05794 s. The time it takes to return from extreme right back to zero is thus 0.0638 s. The system comes to rest in a cycle in which the peak displacement u is such that ku < µmg = 0.2 × 3 × 9.81 × 1000 = 2.943 mm From the solution to Problem 5.13 we found that the displacement at the first peak was 35.9 mm and that the amplitude decreased by 4µN/k = 11.78 mm in each cycle. As shown in Figure S5.14 the amplitude at the second positive peak it is 35 − 9 − 11.78 = 24.12 mm, at the third peak it is 12.34 mm and at the fourth peak 0.56 mm. The last displacement is less than 2.943 mm so that the mass block will come to rest at this displacement, the spring force being insufficient to overcome friction. The total time before the moving mass comes to rest is given by t = 3T + 0.05794 = 0.7885 s

c ξ1 = √ 2 k × 1.5m

Decrease in amplitude over four cycles =

e−ξ1 ω1 t e−ξ1 ω1 (8π/ωd1 +t) ω

8πξ1 ω 1

=e

d1

=e

√8πξ1

1−ξ2

= 2.79

(b) With hysteretic damping, equivalent damping ξh = η2 remains constant; hence there is no change in the decrease in amplitude.

28

Chapter 6

The amplitude of resulting harmonic response is given by 1 120.56 ρ= 4330 1 − (Ω/ω)2

Problem 6.1

Substituting

Ω ω

20.94 6.42

=

= 3.26

ρ = 2.892 × 10−3 rad k M

Displacement of end B is given by

m, L

∆ = 2.892 × 10−3 × 1200 = 3.4704 mm α

Problem 6.2 Deflection under self weight ∆=

θ A

2240 = 0.028 in 4 × 20000

Natural frequency M αθ¨

1 mL2θ¨ 3

#

kLθ

ω=

g = ∆

#

386.4 = 117.47 rad/s 0.028

Figure S6.1 The forces acting on the bar, including inertia forces, are shown in Figure S6.1. Taking moments about A, the free vibration equation becomes   1 2 2 ¨ mL + M a θ + kL2 θ = 0 3 (  2 ) 1 1 × 40 + 132 θ¨ + 4330θ = 0 3 1.2 105θ¨ + 4330θ = 0

Exciting frequency Ω=

Frequency ratio β=

= 20.94 Substituting e = 3 mm, and Ω = 200×2π 60 rad/s 3 × (20.94)2 p0 = 132 × 1000 = 173.6 N

Ω 209.44 = = 1.783 ω 117.47

Damping ratio ξ = 0.2

The frequency is given by # 4330 ω= = 6.42 rad/s 105 The unbalanced force due to rotating flywheel is given by p0 = M eΩ2 sin Ωt

2000 × 2π = 209.44 rad/s 60

Transmission ratio is given by % 2 1 + (2ξβ) TR = % = 0.536 2 2 2 (1 − β ) + (2ξβ) Also ρ=

1

p0 % k

2

2

(1 − β 2 ) + (2ξβ 2 )

or 0.0025 =

p0 1 80000 2.293

The equation of forced motion is   1 mL2 + M a2 θ¨ + kL2 θ = p0 a sin Ωt 3

Hence p0 = 458.56 lbs. Transmitted force = p0 × T R = 458.56 × 0.536 = 245.6 lbs.

or

Problem 6.3

105θ¨ + 4330θ =

173.6 × 1 sin 20.94t (1.2)2

= 120.56 sin 20.94t

Natural frequency ω = rad/s.

%

9.81×1000 0.76

= 113.60

29 Exciting frequency at operating speed Ω=

1600 × 2π = 167.55 rad/s 60

Frequency ratio β = Ω/ω = 1.475 The maximum amplitude is given by ρmax =

e m0 m 2ξ 1 − ξ 2

Amplitude of payload, χ, is given by $ 1 + (2ξβ)2 χ=G (1 − β 2 )2 + 4ξ 2 β 2 where G = 0.1 in. and β = 1. $ 1 + (2 × 0.0123)2 χ = 0.1 4 × (0.0123)2 = 4.07 in

Hence   em0 = ρmax 2ξ 1 − ξ 2 = 2.5 × 2ξ 1 − ξ 2 m The amplitude of response at the operating speed is obtained from ρ= % or

2 0 em m β

2.5 × 2ξ

0.5 = %

2

2

(1 − β 2 ) + (2ξβ) 

1 − ξ 2 × 1.4752 2

(1 − 1.4752 ) + 4ξ 2 × 1.4752

or 473.03ξ 4 − 464.33ξ 2 + 1.380 = 0 Solving for ξ 2 yields ξ = 0.0546 Damping coefficient is given by c = 2ξωm 1160 = 2 × 0.0546 × 113.6 × 9.81 N.s = 1467 m Eccentricity e is obtained from e

 m0 = 2.5 × 2ξ 1 − ξ 2 = 0.2726 m 1160 = 0.805 mm e = 0.2726 × 40 × 9.81

Problem 6.4 # g 386.4 = = 31.64 rad/s ω= ∆ 0.386 e−ξωt 0.2 = =4 −ξω(t+3.57) 0.05 e 3.57ξω = ln 4 = 1.386 1.386 ξ= = 0.0123 3.57 × 31.64 #

Problem 6.5

900 = 1.286 mm 700 # # g 9.81 × 1000 = = 87.35 rad/s ω= ∆ 1.286 3000 × 2π Ω= = 314.16 rad/s 60 Ω β= = 3.597 ω 1 p0  ρ= k (1 − β 2 )2 + (2ξβ)2 1 360  = 700 (1 − 3.5972 )2 + (2 × 0.2 × 3.597)2 1 360 × = 0.0428 mm = 700 12.025

∆=

Transmissibility is given by  1 + (2ξβ)2 TR =  (1 − β 2 )2 + (2ξβ)2  1 + (2 × 0.2 × 3.597)2 = 0.1457 = 12.025 Transmitted force = 0.1457 × 360 = 52.46 N.

Problem 6.6 Neglect damping and, as a conservative estimate, take the total amplitude as the direct sum of the amplitudes due to component excitations. Thus 0.018 0.030 0.045  2 +  2 +  2 20 35 |1− f | |1− f | | 1 − 50 | f = 0.004 Trial and error gives f = 6.5 Hz. Hence # # k k ω = 2π × 6.5 = = m 45 k = 75, 058 N/m

30

Problem 6.7 With u as the vertical displacement of mass at B relative to the frame, the forces acting on the arm ACB are as shown in Figure S6.7. Taking moments about A   0.08 u 0.6 × 0.12¨ u+c × 0.08u+20 ˙ × 0.12 0.12 = −0.6 × 0.12¨ ug

Since the scale reading will be 2ρ, the calibration factor to convert the scale reading to the amplitude of ground acceleration is 12 × 2314.35 = 1157.2. At a frequency of 5 Hz 2π × 5 = 0.653 48.11 3/(48.11)2 ρ= % 2 {1 − (0.653)2 } + (2 × 0.7 × 0.653)2

β=

= 1.2 × 10−3 m

or  0.6¨ u+c

0.08 0.12

2 u˙ +

20 u = −0.6¨ ug (0.12)2

The scale reading will show an acceleration of a = 2 × 1.2 × 10−3 × 1157.2 = 2.777 m/s2 Percentage error = 100 × (3.0 − 2.777)/3.0 = 7.4.

Problem 6.8 From Problem 6.7 we have the natural frequency ω = 48.11 rad/s. For Ω = 1 Hz, the frequency ratio is given by 20 

ü 0.12

0.08 cu· 0.12

0.6(ü  üg)

Figure S6.7

β=

1 × 2π Ω = = 0.1306 ω 48.11

Using the damping ratio ξ = 0.7 we get

Hence ∗

m = 0.6 2  0.08 ∗ c = c = 0.444c 0.12 20 = 1388.9 k∗ = (0.12)2 # # k∗ 1388.9 = 48.11 rad/s = ω= m∗ 0.6

tan φ =

2ξβ = 0.186 1 − β2

Hence φ = 10.54◦ For Ω = 5 Hz we get 5 × 2π = 0.653 48.11 2 × 0.7 × 0.653 tan φ = 1 − 0.6532 β=

which gives φ = 57.89◦

The damping coefficient is now obtained as follows c∗ = 2m∗ ωξ = 2 × 0.6 × 48.11 × 0.7 = 40.415 40.415 N.s c= = 90.94 0.444 m At input frequency of 1 Hz 2π Ω = = 0.1306 β= ω 48.11 If the amplitude of ground acceleration is a, the response amplitude is given by ρ=

m∗ a 1 m∗ a  = × 1.0002 k∗ k∗ (1 − β 2 )2 + (2ξβ)2

Hence a=

ω2 ρ = 2314.35ρ 1.0002

Problem 6.9 The measured amplitude is given by ρ=

a 1  2 2 ω (1 − β )2 + (2ξβ)2

which for β approaching zero becomes ρ = a/ω 2 The measure value ρ should thus be multiplied by ω 2 to obtain the ground acceleration a. The calibration factor is then 1600 When the input signal is 4 sin(4πt) and the frequency ratio β1 = 4π/40 = 0.3146, the amplitude of output signal is given by 1 ρ1 = 4ω 2  2 2 (1 − 0.3141 ) + (2 × 0.6 × 0.3141)2 4 = 2 × 1.0236 ω

31 The instrument reading is ω 2 ρ1 = 4.094 m/s2 . The phase angle is given by 2ξβ1 = 0.4182 tan φ1 = 1 − β12 This gives φ1 = 0.3961 rad. For the input signal 3 sin 5πt we have β2 = 5π/40 = 0.3927 and the amplitude of output signal becomes 1

ρ2 = 3ω 2  (1 − 0.39272 )2 + (2 × 0.6 × 0.3927)2 3 = 2 × 1.0328 ω The instrument reading is ω 2 ρ2 = 3.0985 m/s2 . The phase angle is obtained from tan φ2 =

Problem 6.11 The amplitude of response of a system excited by rotating unbalanced mass and having viscous damping is given by Equation 6.68. ρ= 

(em0 /m)β 2 (1 − β 2 )2 + (2ξβ)2

In the case of hysteretic damping ξ should be ren . Substitution gives placed by ξh = 2β (em0 /m)β 2

ρ= 

(1 − β 2 )2 + η 2

For maximum amplitude dρ =0 dβ

2ξβ2 = 0.5572 1 − β22

which gives φ2 = 0.508 rad. The combined output signal is 4.094 3.0985 sin(4πt − 0.396) + sin(5πt − 0.568) ω2 ω2

or * + 2β (1 − β 2 )2 + η 2 − {(1 −

β 2 )2

β2 3 2 (4β 2 3/2

− 4β)

+η }

=0

or β 2 = 1 + η2 Substitution of this value in ρ gives

Problem 6.10 If the input to the displacement measuring instrument is G sin Ωt, the output amplitude is given by β2 ρ = G (1 − β 2 )2 + (2ξβ)2

em0 1  em0 1 + η 2  = 1 + η2 4 2 m m η η +η

ρ=

Problem 6.12

and the phase shift is obtained from tan φ =

2ξβ 1 − β2

e2π×5ξω/ωd =

Using ω = 3 × 2π ra/s. the computations for different input frequencies are tabulated below

β

ρ/G

%error

4.5 1.5 0.9207 6.0 2.0 0.9747 9.0 3.0 0.4960 Table 6.10b Ω

β

4.5 1.5 6.0 2.0 9.0 3.0

φ, rad 2.108 2.390 2.658

1 − ξ2

= ln 5.333 = 1.674

ξ2 = 1 − ξ2

Table P6.10a Ω

ξ

10π 

8.0 2.5 0.4

32 = 5.333 6 

1.674 10π

2

= 2.839 × 10−3

ξ = 0.0532 From Problem 6.1, k = 4330 N/m, m = 105 kg, p0 = 120.56 N/m, β = 3.26. Hence ρ=

p0 1  = 2.8901 × 10−3 rad k (1 − β 2 )2 + (2ξβ)2

Maximum displacement at B = 2.8901 × 10−3 × 1200 = 3.4681 mm.

32

Problem 6.13

90°

In free vibration η = 2ξh . Since ξh = 0.0532 as in Problem 6.12,

2.0 Hz

20

η = 0.1064 Also

43.5°, 1.8 Hz

133.2°, 2.2 10

1 p0  k (1 − β 2 )2 + η 2 1 120.56  = 2 4330 (1 − 3.26 )2 + 0.10642

ρ=

= 2.8918 × 10−3 rad

10

Displacement at B = 2.8418×10−3 ×1200 = 3.4702 mm.

Figure S6.14b

Problem 6.15 Problem 6.14

For hysteretic damping 1 ρ = (p0 /k) (1 − β 2 )2 + η 2

ρ, mm 25

and tan φ =

20

η 1 − β2

or (1 − β 2 ) =

15

(a)

(b)

η tan φ

Substituting for (1 − β 2 ) in Equation a 10

ρ=

1 ρ = sin φ (p0 /k) η

(c)

5

Taking ρ and φ as the polar coordinates, the Cartesian coordinates can be expressed as 0.5

1.0

1.5

2.0

β

1 1 sin φ cos φ = sin 2φ η 2η 1 1 (1 − cos 2φ) y = ρ sin φ = sin2 φ = η 2η

x = ρ cos φ = Figure S6.14a The relationship between the amplitude and frequency ratio is as shown in Figure S6.14a. The amplitude at resonance (β = 1) is 25 mm. The half power points are at an amplitude of √12 ×25 = 17.68 mm. The corresponding frequency ratios can be measured as βA = 0.88, βB = 1.09. Hence 1 ξ = (βB − βA ) = 0.105 2 The Nyquist plot is shown in Figure S6.14b. Relevant frequencies measured from the plot are as follows: corresponding to φ = 45o , Ω1 = 1.807; and for φ = 135o , Ω2 = 2.209. The damping ratio is given by 1 1 Ω2 − Ω1 (β2 − β1 ) = 2 2 ω 1 2.209 − 1.807 = 2 2 = 0.1005

ξ=

or y− Hence

1 1 = − cos 2φ 2η 2η

2  2  1 1 = x2 + y − 2η 2η

This is the equation of a circle with radius 1 center 0, 2η . Also

(d) 1 2η

and

ρmax 1 = (p0 /k) η The half power points are obtained by substituting ρ = √12 η1 in Equation c, which yields 1 sin φ = √ , 2

φ = 45o or 135o

33

Problem 6.17

90°, 1.0

The hysteresis loop obtained by plotting force versus displacement is shown in Figure S6.17. The area enclosed by the hysteresis loop is measured as 4.46 in.kip and represents the energy loss per cycle. Thus W = ηkπρ2 = 4.46 W η= kπρ2

4 3 46.5°, 0.9

136.4°, 1.1 2 1

1

2

3

f, kips

Figure S6.15

12

The Nyquist plot obtained from the given date is shown in Figure S6.15.

Problem 6.16 1.2

Representing the system by an equivalent single degree-of-freedom system and denoting the friction force by F , the amplitude decrease per cycle is 4F k . Hence

1.2

u, in.

12

3 1 4F = − = 0.7344 6× k 4 64 0.7344 × 20 4F = = 2.448 6

Figure S6.17 Substituting k = in. η= u

12 1.2

= 10 kips/in, and ρ = 1.2

4.46 = 0.0986 10 × π × (1.2)2

The maximum amplitude at resonance is given by k

ρmax = M

F

Figure S6.16 6 Natural frequency f = 1.5 = 4Hz Exciting frequency = 6 Hz Frequency ratio β = 64 = 1.5 The response amplitude is given by

ρ=

, 2  4F p0 . 1 − p0 π

(1 − β 2 )2 , 2  2.448 1 − 3×π 3. = = 0.1159 in. 20 {1 − (1.5)2 }2 k

1.2 p0 1 = = 12.2 in. k η 0.0986

34

Chapter 7

Problem 7.2 Response to ramp function load is given by   sin ωt P0 t − u= ωt1 k t1

Problem 7.1 We have  P0  (1 − cos ωt)  k P0 ω   sin ωt u˙ = k

u=

0 0.3 s, the forcing function is a superposition of two ramp functions given by

where t = t − ωπ . Substituting the displacement and velocity at t = 0

p=

P0 t 4P0 − (t − t1 ) t1 t1

3P0 P0 2P0 or A = =A− k k k P0t

and B = 0. Hence

At t =

2π ω ,

 P0  3 cos ωt − 1 k

that is at t = u=−

10 k F (kips)

u=

t1

π ω

4P0 , k

u˙ = 0 0.3

0.4 

u/(P0/k)

8

t(s)

4P0t t1

(t  t1)

Figure S7.2 4

0

The response is therefore given by

π

2π

3π

4π ωt

4

8

u = 0.8443t − 0.1344 sin 2πt   40 (t − 0.3) sin 2π(t − 0.3) − − 2 4π 0.3 2π × 0.3 = 1.013 − 2.533t − 0.1344 sin 2πt + 0.5375 sin 2π(t − 0.3) 0.3 < t ≤ 0.4 At t = 0.4 u = 0.2369 in. u˙ = 0.8822 in./s

Figure S7.1 In a similar manner it can be shown that at t = 3π , u = 6Pk 0 and u˙ = 0. The relationship between ω u and t is shown in the accompanying figure. The displacement is given by u = (−1)n−1 2nP0 /k where n is the number of half cycles. The response grows indefinitely with n.

For t > 0.4, the system undergoes free vibrations and the response is given by 0.8822 sin 2π(t − 0.4) 2π = 0.2369 cos 2π(t − 0.4) + 0.1404 sin 2π(t − 0.4)

u = 0.2369 cos 2π(t − 0.4) + t > 0.4

35

Problem 7.3

Problem 7.4

The forcing function can be represented by the superposition of four ramp function loads as shown in the figure. The response for t > 3t1 is given by

sin ωt P0 t − u= k t1 ωt1

P0 t − t1 sin ω(t − t1 ) − − k t1 ωt1

P0 t − 2t1 sin ω(t − 2t1 ) − − k t1 ωt1

sin ω(t − 3t1 ) P0 t − 3t1 − + t1 ωt1 k

The impulse of the force is given by area under the force displacement curve

After some algebraic manipulation this reduces to

For t1 /T = 1/32, ωt1 = 2π/32 = 0.1963 and the exact value of the maximum response becomes

ω ωt1 4P0 sin (2t − 3t1 ) sin ωt1 sin u= 2 2 kωt1

(a)

I = 2P0 t1 v0 =

2P0 t1 m

Hence v0 2P0 t1 sin ωt = sin ωt ω mω 2p0 t1 2P0 ωt1 = = mω k

u= umax

4P0 sin 0.1963 sin 0.0982 1.963k P0 = 0.3897 k

umax = P0 t t1 Force

The approximate value of maximum response P0 t  3t1 t1

P0

umax =

P0 2P0 × 0.1963 = 0.3926 k k

Percentage error = t1

2t1

P0 t  t1 t1

3t1

Problem 7.5 Velocity u˙ is given by
u˙ =

Figure S7.3 For maximum value of u,

0

= v0

= 0 which yields

3 cos(ωtp − ωt1 ) = 0 2 3 π ωtp − ωt1 = 2 2 3 T tp = + t 1 4 2

× 100 = 0.74%.

Time, t

P0 t  2t1 t1

du dt

0.3926−0.3897 0.3897

t

  v0 δ(τ )dτ  t≥0  

Equation of motion is m¨ u + cu˙ + ku = −m¨ ug = −mv0 δ(t)

(b)

The Duhamel integral solution is

where T = 2π/ω and tp is the time at maximum. If maximum is to occur at tp > 3t1


t 1 −mv0 δ(τ )e−ξω(t−τ ) sin ωd (t − τ )dτ mωd 0 v0 = − e−ξωt sin ωd t ωd

u=

3 T + t1 > 3t1 4 2

For maximum

or

T t1 < 6 The maximum value is obtained by substituting t = tp from Equation b into Equation a umax =

4P0 ωt1 sin ωt1 sin kωt1 2

(c)

du v0 ξω −ξωt v0 −ξωt = e sin ωd t − e ωd cos ωd t = 0 dt ωd ωd $

or ωd = tan ωd t = ξω

1 − ξ2 ξ

36 Hence, sin ωd t =

 1 − ξ2

where √ −1

 v0 − ωξω sin 1−ξ 2 e d 1 − ξ2 ωd √ v0 √ −ξ sin−1 1−ξ2 = e 1−ξ2 ω

umax =

ρ=

If the maximum occurs in the era 0 < t < t1

v0 = 0.1475v0 T ω

P0 du = ω sin ωt = 0 k dt ωt = π

and that for ξ = 0.1 by umax = 0.8626

(d)

cos ωt1 ωt1 + sin ωt1

tan φ =

The response spectrum for ξ = 0.05 is given by umax = 0.9267

% ω 2 t21 + 2ωt1 sin ωt1 + 1

v0 = 0.1373v0 T ω

and the maximum value is umax =

Problem 7.6

2P0 k

(e)

For this to be valid

Force

tp = P0

or

π < t1 ω

1 t1 > T 2

If the maximum occurs in the era t1 < t < 2t1 , then by differentiating Equation c t1

P0

2t1

Time, t

P0 du = k dt

t  t1 t1

−

1 ρ − ω sin(ωt + φ) = 0 ωt1 t1

(f )

Equation f yields

Figure S7.6

ρ sin(ωt + φ) = 1

The forcing function can be represented by the superposition of a rectangular pulse and a ramp function. Hence the response is P0 (1 − cos ωt) 0 ≤ t ≤ t1 (a) k

sin ω(t − t1 ) P0 P0 t − t1 − u= (1 − cos ωt) − k k t1 ωt1 u=



cos(ωt + φ) = ±

ρ2 − 1 ρ

The maximum is obtained by substituting for cos(ωt + φ) in Equation c and taking the negative sign in the expression for cos(ωt + φ).

t1 < t ≤ 2t1

umax (b)

Equation b can be expressed as t sin ωt cos ωt1 P0 2 − − cos ωt + u= k t1 ωt1

cos ωt sin ωt1 − ωt1 P0 t cos ωt1 = 2− + sin ωt t1 ωt1 k

ωt1 + sin ωt1 − cos ωt ωt1

t P0 ρ 2− − = cos(ωt + φ) k t1 ωt1



p0 = k

(

t 2− + t1



ρ2 − 1 ωt1

) (g)

The limiting condition under which the maximum will occur in the free vibration era is obtained by requiring that du dt = 0 at t = 2t1 . This yields − (c)

cos ωt1 ωt1 + sin ωt1 1 + cos 2ωt1 + sin 2ωt1 = 0 t1 t1 t1

or ωt1 sin 2ωt1 + cos ωt1 = 1 A trial and error solution of Equation(h) gives ωt1 = 1.275

(h)

37 or


t1 = 0.203 T

t

−v0 δ(τ ) sin ω(t − τ )dτ

u= 0

Maximum will occur in the free vibration era provided t1 /T ≤ 0.203. The initial conditions at the start of free vibrations era are obtained by substituting t = 2t1 in Equation c and its derivative

P0 sin ωt1 − cos 2ωt1 + ωt1 k

1 cos ωt1 P0 ω sin 2ωt1 − + ˙ 1) = v0 = u(2t k ωt1 ωt1

u0 = u(2t1 ) =

Maximum response in the free vibration era is # umax =

u20 +

 v 2 0

ω

(i)

v0 sin ωt ω 0 < t ≤ t1
t v0 v0 u = − sin ωt + sin ω(t − τ )dτ ω t1 ωt1 v0 v0 1 − cos ω(t − t1 ) = − sin ωt + ω ω ωt1 t1 < t ≤ 2t1
2t1 v0 v0 sin ω(t − τ )dτ u = − sin ωt + ωt1 ω t1 v0 = − sin ωt ω

v0 cos ω(t − 2t1 ) − cos ω(t − t1 ) + ω ωt1 =−

2t1 < t

Equations e, g and i give the response spectrum for the complete range of t1 /T . For P0 = 10 kN, k=250 kN/m, t1 =0.375, and T = 1 the maximum response is in the era t1 < t < 2t1 and is obtained as follows. ωt1 = 2π × 0.375 = 2.3562 % ρ = ω 2 t21 + 2ωt1 sin ωt1 + 1 = 3.1439 tan φ = −0.2308, φ = −0.2268 1 sin(ωt + φ) = = 0.3180 ρ ωt + φ = 2.8179 ωt = 3.0448 ( ) √ 3.0448 3.14392 − 1 10 2− + umax = 250 2.3562 2.3562 = 0.079 m

. ug

0

t1 .. ug

0( )

0 t1

Figure S7.7

Problem 7.8 For t ≤ t1 the Duhamel’s integral gives the following value for response u(t)


t

u(t) =

Problem 7.7

0

Corresponding to the given ground velocity, the ground acceleration is as shown in Figure S7.7. The response u relative to the ground is obtained by using Duhamel’s integral


t

−

u= 0

m¨ ug sin ω(t − τ )dτ mω

where damping is assumed to be negligible

2t1

P0 τ sin ω(t − τ )dτ mω t1

Integrating by parts "0 ! 0t
t cos ω(t − τ ) P0 00 τ cos ω(t − τ ) 00 dτ u(t) = 0 − mω 0 ωt1 ωt1 0 0 " 0 ! 0 0 sin ω(t − τ ) 0t P0 t 0 = −0 0 k t1 0 ωt1 0   P0 t sin ωt = − ωt1 k t1

38 For t > t1 the response as obtained from After completing the integration and simple manipDuhamel’s integral is ulation we get
t1 "   P0 τ 2 sin ω t − t21 t 2P0 u(t) = sin ω(t − τ )dτ 1− + u(t) = mω t1 0 t1 ωt1 k
t P0  sin ω(t − τ )dτ + sin ωt t1 mω − "0 ! ωt1 0t
t1 P0 00 τ cos ω(t − τ ) 00 1 cos ω(t − τ ) = dτ 0 − mω 0 ωt1 ωt1 0 0 Fot t > t1 the response is obtained from 0 0t 0 P0 00
t1 /2 cos ω(t − τ )00 + 2P0 τ mω 2 0 sin ω(t − τ )dτ u(t) = t1 " mω t1 0t1 ! 0 0   0 sin ω(t − τ ) 0
t1 P0 0 τ 2P0 cos ω(t − t1 ) + 00 = 0 1 − sin ω(t − τ )dτ + k ωt1 0 t1 t1 /2 mω P0 [1 − cos ω(t − t1 )] + k Integration by parts gives   P0 sin ω(t − t1 ) sin ωt "   +1 − = ωt1 ωt1 k  t1  sin ω t − t21 2P0 1 cos ω t − + u(t) = k 2 2 ωt1 "0 0t1  Problem 7.9 0 2P0 00 sin ωt cos ω(t − τ )00 + − 0 ωt1 k t1 /2 For t ≤ t1 /2 the Duhamel’s integral gives the ! 0 0
t t1 following value for response u(t) 0 τ cos ω(t − τ ) 0 1 cos ω(t − τ ) 0 dτ − 00 +
t 0 t1 t1 t1 /2 2P0 τ t1 /2 sin ω(t − τ )dτ u(t) = 0 mω t1 which leads to Integrating by parts "   "0 ! 0t
t 2 sin ω t − t21 2P sin ω(t − t1 ) 0 cos ω(t − τ ) 2P0 00 τ cos ω(t − τ ) 00 u(t) = − − dτ u(t) = k ωt ωt1 1 0 mω 0 ωt1 ωt1 0 0  " sin ωt 0 ! 0 0 sin ω(t − τ ) 0t − 2P0 t 0 ωt1 = −0 0 k t1 0 ωt1 0   2P0 t sin ωt = − Problem 7.10 k t1 ωt1 For t1 /2 < t ≤ t1 the response as obtained from Duhamel’s integral is
t1 /2 2P0 τ sin ω(t − τ )dτ u(t) = mω t1 0  
t τ 2P0 1− sin ω(t − τ )dτ + t1 t1 /2 mω Integration by parts gives "    t1  sin ω t − t21 2P0 1 cos ω t − + u(t) = k 2 2 ωt1 " 0 0t  0 2P0 00 sin ωt 0 cos ω(t − τ ) + − 0 0 k ωt1 t1 /2 ! 0t 0
t 0 τ cos ω(t − τ ) 0 cos ω(t − τ ) 0 dτ − 00 + 0 t1 t1 t1 /2 t1 /2

The displacement response is obtained from the Duhamel’s integral as follows


t1

P0 sin ω(t − τ )dτ mω 0
2t1 +t0 P0 sin ω(t − τ )dτ mω t1 +t0 "0 0t P0 00 cos ω(t − τ ) 00 1 = 0 mω 0 ω 0 ! 0 0 0 cos ω(t − τ ) 02t1 +t0 0 0 0 0 ω t1 +t0  P0 = cos ω(t − t1 ) − cos ωt mω 2

u(t) =

 + cos ω(t − 2t1 − t0 ) − cos ω(t − t1 − t0 )

39 On substituting t0 = π/ω − t1 we get  P0 cos ω(t − t1 ) − cos ωt u(t) = k

 + cos {−π + ω(t − t1 )} − cos(−π + ωt) = 0

The absolute displacement is obtained by noting that ut = u + ug so that   1 hβ 2 sin Ωt − sin ωt + h sin Ωt u = 1 − β2 β h = (sin Ωt − β sin ωt) 1 − β2 t

Problem 7.11 The vertical motion of the point of contact of the tire with the ground is ug = h sin(πvt/L). Assuming that ut (t) represents the absolute vertical displacement of the vehicle and u(t) its displacement relative to that of the point of contact and setting Ω = (πv/L) the equation of motion becomes m¨ u + cu˙ + ku = 0 or m¨ u + cu˙ + ku = −m¨ ug = hmΩ2 sin Ωt At time t = 0, both ut and u˙ t are zero; however, while ug = h sin Ωt is zero the velocity u˙ g = hΩ cos Ω = hΩ. Thus, in terms of the relative displacement and velocity the initial conditions are u0 = u(0) = 0 and v0 = u(0) ˙ = −hΩ. The solution to the equation of motion is now given by the superposition of the following steady state component, up and transient part uc (Refer to Equations 6.26 and 6.31) 1 hmΩ2 2 2 k (1 − β ) + (2ξβ)2 + * (1 − β 2 ) sin Ωt − 2ξβ cos Ωt   hΩ −ξωt uc = −e sin ωd t ωd

up =

e−ωξt hmΩ2 k (1 − β 2 )2 + (2ξβ)2   + ω * 2 2 2ξβ cos ωd t + 2βξ − β(1 − β ) sin ωd t ωd +

where β = Ω/ω. The corresponding solutions for an undamped system are hβ 2 sin Ωt 1 − β2

uc = −hβ sin ωt − =

−hβ sin ωt 1 − β2

hβ 2 1 − β2

#

# k 40000 = = 5.345 rad/s m 1400 πv π × 5.555 Ω= = = 11.636 L 1.5 Ω = 2.177 β= ω The time at which the relative displacement is a maximum is obtained by equating the differential of u(t) to zero, which gives Ω cos Ωt −

hβ 3 sin ωt 1 − β2

 sin Ωt −

ω cos ωt = 0 β

Solution by trial and error gives t = 0.1204. The corresponding value of u = −45 mm The maximum spring force is 40000 × 0.045 = 1800 N. It is easily verified that ut increases monotonically during the 1.5/5.555 = 0.27 s it takes for the vehicle to travel over the hump. At 0.27 s the absolute displacement is ut (0.27) = 28.873 mm. The velocity at this time is obtained by substituting t = 0.27 s in the following   h v (t) = Ω cos Ωt − βω cos ωt 1 − β2 t

= 175.42 mm/s After this the vehicle undergoes simple harmonic vibrations with an amplitude given by "

 2

ρ = 28.873 +

175.42 5.345

2 !1/2 = 43.71 mm

The absolute acceleration experienced by the vehicle while it is on the hump is obtained by taking the derivative of the velocity expression

Superposition gives u=

The speed of vehicle is v = 20 km/hr or 5.56 m/s. The natural and imposed frequencies are given by ω=

t

up =

Problem 7.12

1 sin ωt β

 u ¨t =

h 1 − β2



 − Ω2 sin Ωt + βω 2 sin ωt

40 The maximum value is seen to occur at 0.1204 s and it is 1285.6 mm/s2 The maximum acceleration experienced by the passengers during the free vibration era is given by u ¨tmax = 43.71ω 2 = 1248.8 mm/s2

Problem 7.13 The relative displacement of the vehicle is given by the expressions derived in Problem 7.11. It is plotted in Figure S7.13. The maximum numerical value is at 0.1029 s and is −34.276 mm. However, the displacement at 0.27 s, when the vehicle is exiting the hump, is 36.045 mm. The maximum total displacement is at 0.253 s and its value is 36.513 mm. The total displacement at 0.27 s is 36.045 mm. As would be expected, the total and relative displacements at 0.27 s are identical because the displacement of the contact point is zero at this time. The total velocity at exit from the hump, obtained by taking the derivative of the total displacement expression and substituting t = 0.27 s, is -55.257 mm/s. In the free vibration era the amplitude of vibration is given by substituting u0 = 36.045 mm and v0 = −55.257 mm/s in the following expression. $ ρ =  u20 +



v0 + ωξu0 ωd

2

1/2 

= 36.32 mm

40 30

Displcement, mm

20

Total,  0.4

10 Total,  0

0 10

Relative,  0.4

20 Relative,  0

30 40 50

0

0.05

0.1

0.15 Time, s

Figure S7.13

0.2

0.25

0.3

41

Chapter 8

y

Problem 8.1 As the center of the pulley moves a distance u, the spring is stretched by 2u and the pulley rotates through u/R. The potential and kinetic energies are given by 1 k(2u)2 = 2ku2 2  2 1 1 1 u˙ 1 T = M u˙ 2 + mu˙ 2 + mR2 × 2 2 R 2 2   3 1 M + m u˙ 2 = 2 2

u

V =

With u = A sinωt, Vmax = 2kA2 , and Tmax = 1 2 2 m + 32 m . Since energy is conserved 2A ω

Figure S8.3 Assuming that θ = A sin ωt and that Vmax = Tmax ω=

ω=

1 = 33.67 lbs/in 0.0297 1 1 V = ku2 = × 33.67u2 2 2 1 10 2 T = u˙ 2 386.4 k=

Using the principle of conservation of energy 33.67 × 386.4 10 ω = 36.07 rad/s

11 Wu 2 1 = (20 × 1.2 + 2 × 12 × 0.85) 2 = 22.2 N.m 1 1W 2 u = ω2 2 g 1 10−3 (20 × 1.22 + 2 × 12 × 0.852 ) = ω2 2 9.81 = ω 2 2.352 × 10−3 N.m

Vmax =

Problem 8.2 Referring to Problem 2.2,

2k a2 g − 2 m l l

Problem 8.4

Tmax = Vmax $

4k M + 3m 2

#

Tmax

On equating Vmax to Tmax # ω=

22.2 = 97.15 rad/s 2.353 × 10−3

ω2 =

Problem 8.5 Cross-sectional area =100 × 12 + 113 × 12

Problem 8.3

=2556 mm2

Mass m moves down a distance y = l(1 − cos θ) For small θ, y ≈ lθ2 /2 and the potential and kinetic energies are given by V =2 T =

1 k(aθ)2 2

1 2 ˙2 ml θ 2

− mgl

θ2 1 = (2ka2 − mgl)θ2 2 2

Mass per unit length m=

2556 × 2770 = 7.08 kg/m 106

Cross-sectional properties y = 85.84 mm I = 3.944 × 106 mm4

42 j

due to a unit vertical load at C is obtained by integrating diagram c on diagram b

W



 L x x 1 x 2 L −x + x x+ Lx 2 EI 2 2 EI 3 2EI x3 xL2 Lx2 − + = 4EI 6EI 2EI

u1 = 100

12

In a similar manner, the horizontal displacement of a point on leg AB due to a unit vertical load at C is obtained by integrating diagram d on diagram b

125 y

Figure S8.5 Assume displaced shape ψ = sin πx L . π2 πx ψ (x) = − L2 sin L .

u2 =

Lx2 L x x = 4EI 2EI 2

Also






L  W πx πx 2 δ(x − ξ) sin2 dx m sin dx + L g L 0 0 πξ mL W + sin2 = 2 g L
L 

2 π 4 EI EI ψ (x) dx = k∗ = 2L3 0 π 4 EI/2L3 ω2 = (a) mL/2 + (W/g) sin2 πξ L

m∗ =




L

Substituting ξ = 1.2 m, m = 7.08 kg/m, L = 3.6 m, E = 69000 × 106 N/m2 , I = 3.944 × 10−6 m4 , 2 W = 450 N and g = 9.81 m/s . ω 2 = 6025,

L/2EI 1 B C

A x/EI

1

x

ω = 77.62 rad/s 1

It is apparent Equation a that ω will be a maximum when sin πξ L = 0 or ξ = 0. $ ωmax =

ωmin =

x/EI

π 2 EI/2L3 = 149.3 rad/s mL/2

Also, ω will be a minimum for sin πξ L = 1 or ξ = $

x

Figure S8.6 L 2.

π 2 EI/2L3 = 69.62 rad/s mL/2 + W/g

Problem 8.6 Figure S8.6 b shows the M/EI diagram produced by a unit vertical load applied at C. Figure S8.6 c shows the same diagram for a unit vertical load applied at a distance x from B. Figure S8.6 d shows the moment diagram produced by a unit horizontal load applied at a distance x from point A. The vertical displacement of a point on leg BC

To obtain the strain energy, we need the vertical displacement at C

u1

 2  3   L L L L 1 = − 2 4EI 2 6EI 2   2 L L + 2EI 2 14L3 = 48EI

Strain energy is given by V =

1 L3 1 14L3 = × 0.292 ×1× 2 48EI 2 EI

43 Kinetic energy is obtained from 1 2 ω 2

T =

"


b. The displacement at ξ is given by




L/2

2

m (u1 ) dx + 0

 2

+ M {u1 (L/2)} + 2

2

m (u2 ) dx 0

=

6

Equating the kinetic energy to the strain energy ω2 =

EIu(ξ) =

  mL + m {u2 (L)}2 2

1ω L (0.05725mL + 0.1475M ) 2 (EI)2

=

  ξ 5L x2 1− x1 dx1 x1 − 1 2 2L 4 0  
2L  ξx1 5L x21 x1 − ξ− dx1 + 4 2 2L ξ


L

EI 0.292 0.05725mL + 0.1475M L3



ξ

5Lξ 3 ξ4 L3 ξ − + 2 24 24 0 ≤ ξ ≤ 2L

The moment diagram produced by a unit load applied at a distance η from C is shown in Figure S8.7 c. The displacement at η is given by




2L

−

EIu(η) = 0




Problem 8.7

x2 5L x1 − 1 4 2

L

− (x2 − η)

+ η

A

B

x1

x2

C

(a)

C

B

j

(L − η)

x1 dx1 2L

x22 dx2 2

19 5 η4 = − L4 + L3 η − 24 6 24 0≤η≤L Strain energy

1

A



j(2L  )/2L

(b) 1

 5Lξ 4 ξ4 L3 ξ − + dξ 2 24 24 0 
L η4 19 4 5L3 η 1 + dη L − + 6 24 24 2EI 0 49 L5 L5 = = 0.4083 120 EI EI

1 V = 2EI




2L



Kinetic energy A (L  h)

B

h

(c)

"
2 2L  3 ω2 m L ξ 5Lξ 4 ξ4 T = dξ − + 2(EI)2 0 2 24 24 2 !
L 19 4 5L3 η η4 + dη L − + 24 6 24 0 = 0.1575

Figure 8.7 Obtain the deflected shape produced by a unit load per unit length applied as shown in Figure 8.7 a. With coordinates x1 and x2 measured as indicated x2 5Lx1 − 1 4 2 x22 M (x2 ) = 2

M (x1 ) =

ω 2 mL9 (EI)2

Equating the kinetic energy to potential energy 0.4083 EI EI = 2.593 4 4 0.1575 mL mL # EI ω = 1.61 mL4

ω2 =

0 ≤ x1 ≤ 2L 0 ≤ x2 ≤ L

The moment diagram produced by a unit load applied at a distance ξ from A is shown in Figure 8.7

Problem 8.8 Mass matrix is given by   1 0 M=m 0 1

44 The moment diagrams produced by the application of unit loads along coordinates 1 and 2 are shown in Figures S8. b and c, respectively. The flexibility influence coefficients are given by


m1 m1 ds = EI
m2 m2 ds = = EI 7L3 = 24EI
m1 m2 ds = = EI

a11 = a22

a12

B

1 L ×L× 2 EI (L/2)3 L + 3EI 2

2 L3 L= 3 3EI L L × × EI 2

3

1 L L L ×L× = 2 EI 2 4EI

1

u1

Using the defection shape uT = [1 1] 9EI 1 T u ku = 2 5L3 1 2 T T = ω u Mu = ω 2 m 2

V =

Equating V to T # ω=

9EI = 1.342 5mL3

#

EI mL3

An improved estimate of vibration shape is given by u1 = K−1 Mu = AMu   L3 7 = 12EI 13 2 Using

 u1 =

C u2

7



13 2

the potential and kinetic energies and hence the frequency are obtained as follows 1 T 81EI u Ku1 = L3 2 1 1 2 T 363 2 ω m T = ω u1 Mu1 = 2 8 # # 8 × 81 EI EI ω= = 1.332 365 mL3 mL3

V =

A L (b)

(a) L/ 2

1

Problem 8.9 # ω= (c)

k = m

#

14 = 11.832 rad/s 0.1

Duhamel’s integral for undamped system gives
t 1 p(τ ) sin ω(t − τ )dτ u= mω 0 = A sin wt − B cos ωt

Figure S8.8

The flexibility matrix is obtained by assembling the influence coefficients.

where


t 1 p(τ ) cos ωτ dτ mω 0
t 1 p(τ ) sin ωτ dτ B= mω 0 A=

1 L3  3 A= EI 1

1 4

 

7 24

4



The calculations for numerical evaluation of the Duhamel’s integral are shown in Table S8.9.   For a forcing function of p = 50 e−4t − e−8t , the particular integral works out to



up (t) = 3.205e−4t − 2.451e−8t

By inversion  144 EI  K= 5 L3

7 24

− 14

− 14

1 3

45

Table S8.9 t or τ

p(τ )

p(τ ) cos ωτ

p(τ ) sin ωτ

Trapezoidal Sum A

Trapezoidal Sum B

(A sin ωt −B cos ωt) ∆τ 1 2 × mω

0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50

0.0 7.4 11.0 12.4 12.4 11.6 10.1 9.3 8.1 6.9 5.9

0.0000 6.1423 4.1574 -2.5124 -8.8574 -11.4051 -9.2707 -5.2707 0.1658 3.9646 5.5069

0.0000 4.1270 10.1841 12.1428 8.6779 2.1173 -4.0079 -7.8240 -8.0983 -5.6473 -2.1176

0.0000 6.1423 16.4421 18.0871 6.7173 -13.5453 -34.2212 -48.5193 -53.3810 -49.2506 -39.7791

0.0000 4.1270 18.4380 40.7649 61.5856 72.3808 70.4901 58.6582 42.7359 28.9903 21.2254

0.0000 0.0000 0.1744 0.5488 1.0288 1.4514 1.6541 1.5324 1.1092 0.4997 -0.1169

Exact

0.0000 0.0353 0.2310 0.6117 1.0823 1.4840 1.6627 1.5315 1.1053 0.4991 -0.1090

0.05 1 ∆τ 1 = × = 0.02113 × 2 mω 2 0.1 × 11.832 The complete solution is

where −4t

u = C cos ωt+D sin ωt+3.205e

−8t

−2.451e


1 eξωτ p(τ ) cos ωd τ dτ A= mωd 0
1 B= eξωτ p(τ ) sin ωd τ dτ mωd 0 ξ = 0.1  √ ωd = ω 1 − ζ 2 = 11.832 1 − 0.01 = 11.773

(a)

where the constants C and D, worked out by applying the initial conditions u(0) = 0, u(0) ˙ = 0, are C = −0.754 and D = −0.5737. For the purpose of comparison, the exact solution is given by Equation a is also tabulated.

The calculations for numerical integration of the Duhamel’s integral are shown in Table S8.10. The closed form solution can be shown to be

Problem 8.10 The Duhamel’s integral for damped system is u(t) =

1 mωd




t

e−ξω(t−τ ) p(τ ) sin ωd (t − τ )dτ

0 −ξωt

= Ae

sin ωd t − Be−ξωt cos ωd t

u = e−ξωt {−0.7105 cos ωd t − 0.7480 sin ωd t} + 3.4121e−4t − 2.7017e−8t For the purpose of comparison, the exact values are also shown in the table.

Table S8.10 t or τ

p(τ )

eξωτ p(τ ) cos ωd τ

eζωτ p(τ ) sin ωd τ

Trapezoidal A

Trapezoidal B

0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50

0.0 7.4 11.0 12.4 12.4 11.6 10.1 9.3 8.1 6.9 5.9

0.0000 6.5296 4.7475 -2.8712 -11.0911 -15.2870 -13.3209 -7.8508 -0.0422 6.4930 9.8325

0.0000 4.3591 11.4353 14.5271 11.1271 3.0731 -5.4797 -11.6776 -13.0025 -9.7947 -4.1196

0.0000 6.5296 17.8068 19.6831 5.7208 -20.6572 -49.2650 -70.4368 -78.3299 -71.8792 -55.5537

0.0000 4.3591 20.1536 46.1160 71.7703 85.9705 83.5639 66.4065 41.7264 18.9291 5.0148

∆τ 0.05 1 1 = × = 0.02124 × 2 0.1 × 11.773 2 mωd



Ae−ξωt sin ωd t  −Be−ξωt cos ωd t 1 × ∆t 2 × mωd 0.0000 0.0004 0.1673 0.5089 0.9264 1.2757 1.4332 1.3355 1.0246 0.5979 0.1793

Exact

0.0000 0.0342 0.2175 0.5598 0.9654 1.2981 1.4429 1.3494 0.0462 0.6258 0.2096

46

Problem 8.11

Table S8.11 n

Time

un

u˙ n

u ¨n

pn+1

un+1

u˙ n+1

u ¨n+1

0 1 2 3 4 5 6 7 8 9 10

0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50

0.0000 0.0403 0.2048 0.5176 0.9029 1.2388 1.4109 1.3587 1.0976 0.7051 0.2893

0.0000 1.6134 4.9667 7.5470 7.8631 5.5720 1.3149 -3.4058 -7.0358 -8.6668 -7.9636

0.000 66.536 69.575 33.673 -21.010 -70.612 -99.644 -89.156 -56.022 -9.204 37.339

7.4 11.0 12.4 12.4 11.6 10.1 9.3 8.1 6.9 5.9

0.0403 0.2048 0.5176 0.9029 1.2388 1.4109 1.3587 1.0976 0.7051 0.2893

1.6134 4.9661 7.5470 7.8631 5.5720 1.3149 -3.4058 -7.0358 -8.6668 -7.9636

66.536 69.575 33.673 -21.010 -70.612 -99.644 -89.156 -56.022 -9.204 37.339

The integration equations for average acceleration method are   4m 2c + k un+1 = + h2 h   4 4 pn+1 + m ¨n un + u˙ n + u h h2   2 (a) un + u˙ n +c h 2 u˙ n+1 = −u˙ n + (un+1 − un ) h 1 u ¨n+1 = (pn+1 − cu˙ n+1 − kun+1 ) m

Also c = 2ξωm = 2 × 0.1 × 11.832 × 0.1 = 0.2366 Substituting m = 0.1, c = 0.2366, k = 14 and h = 0.05 in Equation a. 183.47un+1 = pn+1 + 169.46un + 8.237u˙ n + 0.1¨ un u˙ n+1 = −u˙ n + 40 (un+1 − un ) u ¨n+1 = 10pn+1 − 2.366u˙ n+1 − 140un+1 Response calculations for the first 0.5 seconds are shown in Table S8.11.

Problem 8.12 Table S8.12 n

Time

un

u˙ n

u ¨n

pn+1

un+1

u˙ n+1

u ¨n+1

0 1 2 3 4 5 6 7 8 9 10

0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50

0.0000 0.0276 0.1950 0.5215 0.9223 1.2662 1.4340 1.3635 1.0800 0.6695 0.2478

0.0000 1.6555 5.0779 7.6661 7.8926 5.4443 1.0357 -3.7451 -7.2942 -8.7201 -7.7562

0.000 66.220 70.686 32.854 -23.790 -74.145 -102.213 -89.033 -52.946 -4.097 42.653

7.4 11.0 12.4 12.4 11.6 10.1 9.3 8.1 6.9 5.9

0.0276 0.1950 0.5215 0.9223 1.2662 1.4340 1.3635 1.0800 0.6695 0.2478

1.6555 5.0779 7.6661 7.8926 5.4443 1.0357 -3.7451 -7.2942 -8.7201 -7.7562

66.220 70.686 32.854 -23.790 -74.145 -102.213 -89.033 -52.946 -4.097 42.653

47 The integration equations for linear acceleration method are   6m 3c + + k un+1 = h2 h   6 6 pn+1 + m un + u˙ n + 2¨ un h2 h   3 h (a) +c un + 2u˙ n + u ¨n h 2 h 3 u˙ n+1 = (un+1 − un ) − 2u˙ n − u ¨n h 2 1 (pn+1 − cu˙ n+1 − kun+1 ) u ¨n+1 = m Substituting m = 0.1, c = 0.2366 and k = 14 in Equation a

Table S8.13

Time

pn

un−1

un

un+1

0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50

0.0 7.4 11.0 12.4 12.4 11.6 10.1 9.3 8.1 6.9 5.9

0.0000 0.0000 0.0000 0.1747 0.5318 0.9660 1.3252 1.4802 1.3671 1.0344 0.5882

0.0000 0.0000 0.1747 0.5318 0.9660 1.3252 1.4802 1.3671 1.0344 0.5882 0.1603

0.0000 0.1747 0.5318 0.9660 1.3252 1.4802 1.3671 1.0344 0.5882 0.1603

268.2un+1 = pn+1 + 254.2un + 12.473u˙ n + 0.2059¨ un u˙ n+1 = 60 (un+1 − un ) − 2u˙ n − 0.025¨ un u ¨n+1 = 10pn+1 − 2.366u˙ n+1 − 140un+1

Substituting m = 0.1, c = 0.2366, k = 14 and h = 0.05.

The response calculations for the first 0.5 s are shown in Table S8.12.

Problem 8.13

42.366un+1 = pn + 66un − 37.634un−1 Also, since u0 = 0, u˙ 0 = 0, and u ¨0 = 0

The central difference expression is   c  2m un+1 = pn + −k + 2 un + h2 2h h  c m − un−1 + 2h h2

u−1 = u0 +

m

h2 u ¨0 − hu˙ 0 = 0 2

The response calculations for the first 0.5 s are shown in Table S8.13.

Problem 8.14

Table S8.14

n

Time

pn+1

un−2

un−1

un

un+1

2 3 4 5 6 7 8 9 10

0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50

12.4 12.4 11.6 10.1 9.3 8.1 6.9 5.9

0.0000 0.0403 0.2048 0.4824 0.8101 1.1008 1.2718 1.2803 1.1258

0.0403 0.2048 0.4824 0.8101 1.1008 1.2718 1.2803 1.1258 0.8472

0.2048 0.4824 0.8101 1.1008 1.2718 1.2803 1.1258 0.8422 0.5111

0.4824 0.8101 1.1008 1.2718 1.2803 1.1258 0.8472 0.5111

48 The Houbolt’s expression is   2m 11c + k un+1 = + h2 6h   5un 4un−1 4n−2 pn+1 + +m − + 2 h2 h2 h   3un 3un−1 un−2 +c − + h 2h 3h

When the displacement is represented by u = zψ the generalized mass m∗ and generalized stiffness are given by m∗ = ψ T Mψ = 251.56 tonne k ∗ = ψ T Kψ = 16151.0 kN/m

Substituting m = 0.1, c = .2366, k = 14 and h = 0.05 102.675un+1 = pn+1 + 214.196un − 167.098un−1 + 41.577un−2

The generalized force is obtained from 

 0  = 112.5 sin(5πt/3) kN 0 p∗ = ψ T  45 sin(5πt/3)

(a)

To start the iterations, u0 , u1 , u2 are needed so that u3 can be calculated. From the results of Problem 8.11 and the given initial conditions, u0 = 0, u1 = 0.0403, u2 = 0.2048. Using these values and Equation a, the displacement response is calculated for the first 0.5S. The results are shown in Table S8.14.

Problem 8.15 The mass and stiffness matrices are given by   35.0 0 0 M =  0 35.0 0  tonne 0 0 17.5   17520 −8760 0 K =  −8760 13140 −4380  kN/m 0 −4380 4380 The assumed displaced shape is   1.00 ψ =  1.75  2.50

The force amplitude for the first 1.0 s at interval of 0.1 s is p∗ (nh) = 0, 56.25, 97.43, 112.50, 97.43, 56.25, 0.00, −56.25, −97.43, −112.5, −97.43 Recurrence formulas given by Equations 8.78 and 8.81 are used to calculate the response. Coefficients A, B, C, D, and A1, B1, C1, D1 are obtained from Equation 8.83 with ξ = 0 and h = 0.1. A = 0.6958, B = 0.0896 C = 1.242 × 10−5 , D = 6.416 × 10−6 A1 = −5.755, B1 = 0.6958 C1 = 1.680 × 10−4 , D1 = 1.884 × 10−4 The response in physical coordinates is obtained from the computed generalized coordinate z by using the transformation u = zψ. Thus the roof displacement is 2.5 times z. Roof displacements and velocities are tabulated for the first 1 s at intervals of 0.1 s in Table S8.15.

Table S8.15 n

Time s

un mm

u˙ n mm/s

pn kN

un+1 mm

u˙ n+1 mm/s

uexact mm

0 1 2 3 4 5 6 7 8 9 10

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

0 0.902 6.311 16.637 26.951 29.625 19.330 -2.726 -28.179 -45.244 -44.893

0 26.489 82.734 115.130 77.482 -33.796 -170.390 -256 .290 -232.130 -93 .234 102.390

0 0.5625 0.9743 1.1250 0.9743 0.5625 0 -0.5625 -0.9743 -1.1250 -0.9743

0.902 6.311 16.637 26.951 29.625 19.330 -2.726 -28.179 -45.244 -44.893

26.489 82.734 115.130 77.482 -33.796 -170.390 -256.290 -232.130 -93.234 102.390

0 0.932 6.470 17.033 27.579 30.307 19.769 -2.797 -28.834 -46.289 -45.928

49 The displacement at the first floor level is equal to 1 × z. The maximum value of this displacement is 11.85 mm at 0.5 s. The maximum base shear is thus V = 8760 × 0.1185 = 103.8 kN. For the given forcing function an exact solution exists and is given by Equations6.26 and 6.31 with u0 = 0, v0 = 0 and ξ = 0. The exact response is also tabulated in Table S8.15.

Problem 8.16 Computations similar to those in the solution to Problem 8.15 are carried out with ξ = 0.1. The coefficients in the recurrence formulas are given by

A = 0.7112, B = 0.0828 C = 1.171 × 10−5 , D = 6.170 × 10−6 A1 = −5.318, B1 = 0.5785 C1 = 1.505 × 10−4 , D1 = 1.788 × 10−4 The results are tabulated in Table S8.16. The maximum first story displacement is 10.32 mm at 0.5 s. The corresponding base shear is 10.32 × 8760 = 90.39 kN.

Table S8.16 n

Time s

un mm

u˙ n mm/s

pn kN

un+1 mm

u˙ n+1 mm/s

uexact mm

0 1 2 3 4 5 6 7 8 9 10

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

0 0.868 5.849 14.929 23.615 25.796 17.765 0.869 -18.744 -33.040 -35.987

0 25.142 74.637 99.008 63.753 -26.907 -131.584 -195.735 -182.560 -92.868 36.109

0 0.5625 0.9743 1.1250 0.9743 0.5625 0 -0.5625 -0.9743 -1.1250 -0.9743

0.868 5.849 14.929 23.615 25.796 17.765 0.869 -18.744 -33.040 -35.987

25.142 74.637 99.008 63.753 -26.907 -131.584 -195.735 -182.560 -92.868 36.109

0 0.896 5.996 15.284 24.165 26.391 18.171 0.883 -19.181 -33.806 -36.821

50

Chapter 9 Problem 9.1

g(t) =

tp0 Ω π

g(t) = p0 2 −

tΩ π



π Ω π 2π 4.0 or n > 5 should be interpreted as pertaining to the corresponding negative time. The difference between the inverse transform and time function arises because of the truncation

52 of the frequency function so that frequencies higher than 0.625 Hz are not included. The accuracy can be improved by increasing the value of N .

3

g( )

2 1

h(t  )

Problem 9.6 The two time functions are given by

g( ) .h(t  ) A  0.25

g(t) = 0 g(t) = t g(t) = 3 − t

A  0.803

t < 0.0 0 < t < 1.5 1.5 < t < 3.0 3.0 < t

g(t) = 0 h(t) = 0 h(t) = e−t

t t1 ωi yi =

The response in the physical coordinates is obtained from 1 u(t) = φi yi i=1:3

82 The contribution from the first mode is   5.745 u1 =  10.022  (1 − cos ω1 t) t ≤ t1 13.459   5.745 u1 =  10.022  {cos ω1 (t − t1 ) − cos ω1 t} t > t1 13.459

is determined by the value of t1 /Ti . If this value is greater than 0.5, the maximum occurs at time tpi = π/ωi and its value is 2p∗i /ki∗ , where for orthonormal modes ki∗ = ωi2 . For the present case T1 = 2π/7.995 = 0.785 so that t1 /T1 = 0.636 which is greater than 0.5. Thus tpi = π/7.995 = 0.393 s and hence y1max = 2 × 5.659/63.9 = 0.1771. The contribution to the roof level displacement is therefore 0.1771φ13 = 0.1771 × 0.1520 = 0.0269 m. Similar calculations can be performed for the other two modes; in each case the maximum would be found to occur in the forced vibration era.

The contribution from the second mode is   0.887 u2 =  0.572  (1 − cos ω2 t) t ≤ t1 −1.609   0.887 u2 =  0.572  {cos ω2 (t − t1 ) − cos ω2 t} t > t1 −1.609

Problem 12.9 The modal story displacements were obtained in the solution to Problem 12.8. For mode No. n the inter-story drift in the second story is given by dn2 = un (2)−un (1), where un (2) is the second story displacement in the nth mode and un (1) is the first story displacement in that mode. The time histories of the modal contributions and the total inter-story drift in the second story are plotted in Figure S12.9. The maximum total drift occurs at 0.357 s and its value is 11.83 mm. The shear in the second story is 8760×0.0118 = 103.6 kN.

The modal contributions to the roof displacement and the total roof displacement are plotted in Figure S12.7. The maximum in the first mode is at 0.393 s and its value is 26.92 mm. The maximum in the second mode is at 0.171 s and its value is −3.22 mm. The maximum in the third mode is at 0.117 s and its value is −0.87 mm. The maximum total displacement takes place at 0.839 s and its value is -27.61 mm 30

15 Total 10 Interstory drift, 2nd story, mm

The contribution from the third mode is   −0.924 u3 =  0.822  (1 − cos ω3 t) t ≤ t1 −0.432   −0.924 u3 =  0.822  {cos ω3 (t − t1 ) − cos ω3 t} t > t1 −0.432

Mode 1 5

Mode 3

0 Mode 2

5

Mode 1

Top floor displacement, mm

20 10 0

Total

10

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Figure S12.9

Mode 2

Problem 12.10

10

The rigid box and the instrument can be modeled as shown in Figure S12.10.

20

30 0

0.2

Time, s

Mode 3

0

0.1

 0.1

0.2

0.6

0.4

0.8

1

Time, s

Figure S12.7 From discussion in Section 7.7 we know that the maximum in a mode occurs at a time which

M= 

M 0 0 m



 k −k −k k   c −c C= −c c

K=

83 Mg

The initial displacements in the modal coordinates are given by

M u1

y20

The modal responses are obtained as

m mg/k

u2

Figure S12.10 The applied forces in free fall are M g and mg as shown. The initial displacements are u10 = 0, u20 = mg k ; and the initial velocities are zero. Thus uT0 = [ 0 mg/k ]. The rigid body mode of the system is given by

The second mode is derived from the orthogonality relationship qT1 Mq2 = 0 or M q12 + mq22 = 0 Hence qT2 = [ 1 − M m ] 1 1 Q= 1 −M m



The transformed property matrices are given by   M +m  0 M M∗ = QT MQ = M 1+ m   0 0   K∗ = QT KQ = k M 0 M m 1+ m   0 0   C∗ = QT CQ = c M 0 M m 1+ m     Mg 1 1 p∗ = QT P = mg 1 −M   m (M + m)g = 0 The modal equations reduce to (M + m)¨ y1 = (M + m)g  M

1+

M m

 y¨2 + c

M m



 M y˙ 2 m   M M y2 = 0 1+ +k m m

1+

% where ω =

k m,

ξ=

√c , 2 km

and ωd = ω

 1 − ξ2.

Transforming to physical coordinates u = Qy

qT1 = [ 1 1 ]



m2 g gt2 + M +mk 2   2 g m ωξ −ξωt cos ωd t + y2 = −e sin ωd t ωd M +mk

y1 =

mg

Modal matrix

qT1 Mu0 m2 g = q1 Mq1 M +mk T q Mu0 m2 g = 2 =− q2 Mq2 M +mk

y10 =

gt2 m2 g + 2 M +mk  m2 g −ξωt − e cos ωd t + M +mk m2 g gt2 + u2 = 2 M +mk  M m g −ξωt + e cos ωd t + M +mk u1 =

ωξ sin ωd t ωd

ωξ sin ωd t ωd





Problem 12.11 Letting subscript a refer to structure degrees of freedom 1 and 2 and b to support degrees of freedom 3 and 4   EI 150.9 −41.1 Kaa = 3 −41.1 150.9 L   EI −85.7 −65.1 Kab = 3 10.3 30.9 L   EI 57.4 28.3 Kbb = 3 28.3 36.9 L   3 1 L 150.9 41.1 −1 Kaa = EI 21081.6 41.1 150.9   0.59340 0.40570 −1 R = −Kaa Kab = 0.09335 −0.09426 (a) The equations of motion in terms of displacements relative to the supports are given by 

300 0 0 300



    EI 150.9 −41.1 u1 u ¨1 + 3 −41.1 150.9 u ¨2 u2 L   300 × 10 × 900 1 = sin 30t 0 1000

84 where the units used are N and m

Second mode response

EI 200000 × 0.5 × 106 = = 0.8 N/mm 3 L (5000)3

30 = 1.3257 512.0 1 4.5 y2 = sin 30t 512.0 1 − (1.3257)2

β2 = √

= 800 N/m

= −0.01160 sin 30t

The free vibration equations are 

   8 150.9 −41.1 1 0 q ¨+ q=0 0 1 3 −41.1 150.9

Transforming to physical coordinates 

The characteristic equation is

u1 u2





1 1 = 1 −1



y1 y2



u1 = −19.02 sin 30t mm u2 = 4.185 sin 30t mm

0 0 0 402.4 − λ −109.6 0 0 0 0 −109.6 402.4 − λ 0 = 0

Total displacement

or (402.4 − λ) − (109.6) = 0 2

2

which yields λ1 = 292.8,

λ2 = 512

ut1 = (−19.02 + 10) sin 30t = −9.02 sin 30t mm ut2 = 4.185 sin 30t mm (b) Support acceleration

The mode shapes obtained by substituting λ1 and λ2 in the frequency equation are q1 =

  1 1

 q2 =

1 −1



Modal masses are given by M1 = qT1 Mq1 = 600 M2 = qT2 Mq2 = 600 Modal forces are   1 p1 = sin Ωt = 2700 sin Ωt 0   1 p2 = qT2 2700 sin Ωt = 2700 sin Ωt 0 qT1 2700

The modal equations are 2700 sin 30t 600 2700 sin 30t y¨2 + 512.0y2 = 600

y¨1 + 292.8y1 =

u ¨b = −

  900 × 10 1 2 sin 30t m/s 0 1000

Also 

 300 0 ub = 9 −Maa R¨ 0 300    0.59340 0.40570 1 × sin 30t 0.09335 −0.09426 0 The equations of motion are 

300 0 0 300



    u ¨1 150.9 −41.1 u1 + 800 −41.1 150.9 u ¨2 u2   0.59340 = 2700 sin 30t 0.09335

Modal forces are given by 

  1 1 0.59340 p = 2700 sin 30t 1 −1 0.09335   0.6868 = 2700 sin 30t 0.5000 The modal equations become

First mode response 30 β1 = √ = 1.753 292.8 1 4.5 y1 = sin 30t 292.8 1 − (1.753)2 = −0.007414 sin 30t

2700 × 0.6868 sin 30t 600 = 3.091 sin 30t 2700 × 0.5 sin 30t y¨2 + 512.0y2 = 600 = 2.250 sin 30t

y¨1 + 292.4y1 =

85 The steady state modal responses are y1 =

The modal equations are p0 4m p0 y¨2 + ω22 y2 = 2m p0 y¨3 + ω32 y3 = 4m

y¨1 + ω12 y1 =

3.091 1000 sin 30t 292.4 1 − (1.753)2

= − 5.10 sin 30t mm 1000 2.250 sin 30t y2 = 512.0 1 − (3.257)3 = − 5.80 sin 30t mm

The modal responses are given by pn (1 − cos ωn t) Mn ωn2 p0 L3 (1 − cos ω1 t) y1 = 97.34EI p0 L3 y2 = (1 − cos ω2 t) 768EI p0 L3 (1 − cos ω3 t) y3 = 6924EI

yn =

Transforming to physical coordinates u1 = −10.9 sin 30t mm u2 = 0.7 sin 30t mm To obtain the total displacement note that the pseudo-static displacements are    0.5934 0.4057 1 usa = Rub = sin 30t 0.09335 −0.09426 0   5.934 = sin 30t 0.934 Hence

uta = ua + usa   −4.966 = sin 30t 1.634

Transformation to physical coordinates gives u = Qy so that p0 L3 (0.01172 − 0.01027 cos ω1 t EI − 0.0013 cos ω2 t − 0.00014 cos ω3 t)

u1 =

p0 L3 (0.01433 − 0.01453 cos ω1 t EI + 0.00020 cos ω3 t)

u2 =

Problem 12.12 From Problems 11.1 through 11.3

p0 L3 (0.00911 − 0.01027 cos ω1 t EI + 0.0013 cos ω2 t − 0.00014 cos ω3 t)

u3 =



 √1 1 √1 Q =  2 0 − 2 1 −1 1 # EI ω1 = 4.933 mL3 # EI ω2 = 19.596 mL3 # EI ω3 = 41.606 mL3 Modal masses are obtained from

where #

$ EI = mL3

= 6.152 ω1 = 4.933 × 6.152 = 30.35 rad/s ω2 = 19.596 × 6.152 = 120.55 rad/s ω3 = 41.606 × 6.152 = 255.96 rad/s The elastic forces are given by

Mn = qTn Mqn M1 = 4m , M2 = 2m , M3 = 4m

fs = ΣKqi yi = Σωi2 Mqi yi

The modal forces are given by

Since yi =

pn = qTn p

fs = Σ

Since pT = [ p0 0 0 ], p1 = p0 ,

p2 = p0 ,

200000 × 10.90 × 106 × 1000 900 × (4000)3

p3 = p0

p0 (1 Mi ωi2

− cos ωi t)

p0 Mqi (1 − cos ωi t) Mi p0 p0 = q1 (1 − cos ω1 t) + q2 (1 − cos ω2 t) 4 2 p0 + q3 (1 − cos ω3 t) 4

86 Substituting the values of ω1 , ω2 , ω3 , t and the mode shapes q1 , q2 , q3   1 p0  √  2 (1 − cos 15.175) fs = 4 1   1 p0  + 0  (1 − cos 60.275) 2 −1   1 p0  √  + − 2 (1 − cos 127.98) 4 1   1.8018 =  0.0622  p0 kN −0.0284 The bending moment under the load works out to 1.375 p0 kN.m.

87

Chapter 13

Problem 13.2

Problem 13.1 The mass an stiffness matrices are 

2 0 M= 0 0 

0 2 0 0

0 0 2 0

 0 0  0 1

 2400 −1200 0 0 0  −1200 2400 −1200 K=  0 −1200 2000 −800 0 0 −800 800

Using the transformation q = Ψz the transformed matrices are given by 

11.00 −0.80 M = Ψ MΨ = −0.80 4.77   1100 −620 ∗ T K = Ψ KΨ = −620 3620 ∗



T

We apply inertia forces proportional to the orthonormal mode shapes determined in the solution to Problem 13.1 and determine the displacements ˜ produced by them. Ψ   2.079 0.4090 3.862 0.6001  ˜ = K−1 MΨ = 10−3  Ψ   5.052 −0.0517 5.712 0.9119 Normalizing the last set of vectors such that the largest element in each vector is 1 we get   0.3639 −0.5378 0.6760 −0.6581  ¯ = Ψ   0.8843 0.0567 1.0000 1.0000 ¯ in the Rayleigh Ritz We now use the vectors Ψ procedure similar to that given in the solution to Problem 13.1.

 z1 =

   1.0000 −0.00178 , z2 = 0.1717 1.0000

Converting the mode shapes back to the original coordinates we get 

 0.5859 0.4991  1.1718 0.9982  Q=  1.4842 −0.1027 1.7424 −1.5036 To mass orthonormalize the mode shapes we obtain the product 

10.866 0 Q MQ = 0 2.185



T

The mass orthonormal mode shapes are given by



 0.1777 q1  0.3555  φ1 = √ =  0.4498 10.866 0.5286   0.2285 q2  0.4569  φ2 = √ =  −0.0470 2.185 −0.6882



¯ = ¯ T MΨ M =Ψ

Solution of the eigenvalue problem K∗ z = λM∗ z gives the following frequencies and mode shapes λ1 = 91.46, ω1 = 9.563, λ2 = 758.92, ω2 = 27.55



3.7429 −0.1809 −0.1809 2.4510   338.6 −13.90 ∗ T ¯ ¯ K = Ψ KΨ = −13.90 1689.4 ∗

Solution of the eigenvalue problem K∗ z = λM z gives the following frequencies and mode shapes λ1 = 90.459, ω1 = 9.511, λ2 = 691.50, ω2 = 26.30     1.0000 0.0494 , z2 = z1 = −0.00167 1.0000 ∗

The mode shapes in the original coordinates are obtained from   0.3648 −0.5198 0.6771 −0.6247  ¯ = Q = ΨZ   0.8842 0.1004 0.9983 1.0494 Mass orthonormalization gives   0.1886 −0.3326  0.3499 −0.3997  Φ=  0.4570 0.0642 0.5160 0.6715 The exact values of the four frequencies are 9.51, 25.98, 37.04, and 45.41. The normalized mode shapes are   0.1890 0.4117 0.4265 −0.3359  0.3495 0.3602 −0.1223 0.4829    0.4573 −0.0966 −0.3915 −0.3581 0.5156 0.6187 0.5475 0.2296

88

Problem 13.3

Solution of the characteristic equation yields #

The displacement can be expressed as  πx 



u = z1 + z2 sin

πx = 1 sin L L

z1 z2



k ω1 = 1.025 mL # k ω2 = 4.48 mL

λ1 = 1.052 λ2 = 20.06

Strain energy V is given by


1 1 2 2 EI {u (x)} dx + k {u(0)} 2 2 1 2 + k {u(L)} 2 
L 1  π 20 = EI zT − L sin πx 2 0 L 3 2  π 2 dx z × 0 − L sin πx L   1 1 [1 0]z + 2k zT 0 2   1 0 T 0 4 = EI z z 2 πx π 0 L 2 4 sin L   1 2k 0 + zT z 0 0 2     1 T 0 0 1 T 2k 0 = z z+ z z π4 0 2L 0 0 2 2 3

V =

The mode shapes are obtained by substituting the values of λ in the frequency equation 

1.000 z1 = 1.414

πx f1 (x) = 1 + 1.414 sin   L 1.000 z2 = −1.414 πx f2 (x) = 1 − 1.414 sin L

Problem 13.4 Select



2x L 2πx ψ2 = sin L



0 0



0

 +

4

π EI 2L3

   2k 0 2k 0 = 0 0 0 k

so that u = z1 ψ 1 + z 2 ψ 2   z = [ ψ1 ψ2 ] 1 z2

Kinetic energy T is given by T =

1 T (˙z) 2






L

m 0

 1 [ 1 sin πx ˙ L ] dx z sin πx L

Hence


L

M=m  =m

0

L 2L π



 1 sin πx L 2 πx dx sin πx L sin L  2L π L 2

The characteristic equation is det(K − ω 2 M) = 0 0 0 02 − λ −2λ 0 π 0 0 0 −2λ 1 − λ 0 = 0 π



1−

ψ1 =

Hence K=



Strain energy is given by 
L 1 0 T 2 V = EIz 2πx − 4π 2 0 L2 sin L 2 3 2 2πx dx z × 0 − 4π L2 sin L   1 1 + k zT [1 0] z 0 2   1 −1 + k zT [ −1 0 ] z 0 2     1 T 1 T 2 0 0 0 = z EI z+ kz z π4 0 0 0 8L 2 2 3

2

where λ = ω 2 (mL/k). Expansion of the determinant gives the characteristic equation λ2 − 21.116λ + 21.116 = 0

Hence

 K=  =

0 0



0

4

8π EI L3

2k 0 0 16k



 +

2k 0 0 0



89 Kinetic energy T is given by T =

1 T z˙ 2






L

m 0




"

 1 − 2x L [1 − sin 2πx L

2x L

˙ sin 2πx L ] dx z

2

4x 1 + 4x L2 − L  2πx 1 − 2x 0 L sin L   2πx  1 − 2x L sin L dx z˙ sin2 2πx L 1 1 1 T = z˙ mL 31 π1 z˙ 2 π 2

1 = z˙ T m 2

L

Hence

 M = mL

1 3 1 π

1 π 1 2



The characteristic equation is 0 0 0 2 − 1λ − 1 λ 0 3 π 0 0 0 − 1 λ 16 − 1 λ 0 = 0 π 2 where λ = ω 2 (mL/k). Expansion of the determinant and solution of the resulting equation gives λ1 = 5.348

λ2 = 91.57

ω1 = 2.312

k ω2 = 9.569 mL

#

#

k mL

The mode shapes are obtained by substituting λ1 and λ2 in the frequency equation  q1 =

1.0000 0.1277



 q2 =

1.0000 −0.9786



The corresponding vibration shapes are  2πx 2x f1 (x) = 1 − + 0.1277 sin L L   2πx 2x − 0.9786 sin f2 (x) = 1 − L L 

Problem 13.5 The four boundary conditions are

2

EI ∂∂xu2 = 0 3 EI ∂ u3 = −ku ∂x∂ 2 u EI ∂x2 = 0 x=L 3 EI ∂∂xu3 = ku

x=0

The two moment conditions are automatically satisfied by the selected shape function. The condition

on shear at x = 0 gives  π 3 d = −k −EI L kL3 π 4 EI L3 π d= 3 = = π EI 2L3 π 3 EI 2 π πx ψ(x) = 1 + sin L 2 3 π πx ψ  (x) = − 2 sin 2L L u = zψ(x) Strain energy "
1 L 1 2 2 V = EI {ψ  (x)} dx + k {ψ(0)} 2 2 0  1 2 + k {ψ(L)} z 2 2   6 1 π EI 1 = + 2k z 2 2 8L3 2 6 π EI π2 k∗ = + 2k = k + 2k = 4.467k 8L3 4 Kinetic energy
1 L 2 T = m {ψ(x)} dx(z) ˙ 2 2 0 1 = (4.234mL)(z) ˙ 2 2 m∗ = 4.234mL # # k∗ k ω= = 1.027 ∗ m mL

Problem 13.6 Given



 1.00 1.000 1.000 Q =  2.28 0.768 −0.912  3.16 −0.872 0.343

The modal masses are given by Mn = qTn Mqn M1 = 1.206, M2 = 0.1751, M3 = 0.1452 The mode shapes are now mass orthonormalized so that qn φn = √ M  n  0.9107 φ1 =  2.0760  2.8778   2.3898 φ2 =  1.8354  −2.0839   2.624 φ3 =  −2.393  0.900

90 Exciting frequency

where Ω = 8.378.

80 × 2π = 8.378 rad/s 60 p = p0 sin Ωt

Ω=

where p0 = emΩ2 =

12 × 800 × (8.375)2 = 1.744 kips 386.4 × 1000

From Problem 13.6, story forces in the first mode are   0.907 fs =  2.067  sin Ωt kips 2.865 Also

p1 = 5.019 p2 = −3.634

 0.9107 φ1 =  2.0760  2.8778   0 f = 0  p0

p3 = 1.570 kips

γ1 = φT1 f = 2.8778p0

Modal forces are given by pn = φTn p where pT = [ 0 0 1.744 ] sin Ωt.

The modal equations are y¨1 + (10.6)2 y1 = 5.019 sin Ωt y¨2 + (33.7)2 y2 = −3.634 sin Ωt y¨3 + (59.2)2 y3 = 1.570 sin Ωt The modal responses are obtained next. y1 =

Problem 13.7

1 5.019  sin Ωt  (10.6)2 1 − 8.378 2 10.6

= 0.1189 sin Ωt 1 3.634 y2 = −  8.378 2 sin Ωt 2 (33.7) 1 − 33.7

= −0.00341 sin Ωt 1 1.570 y3 =  sin Ωt  (59.2)2 1 − 8.378 2 59.2

= 0.000457 sin Ωt The elastic forces are given by f3 =

3 1

ωi2 Mφi yi

i=1

On substituting the values     0.907 −0.689 fs =  2.067  sin Ωt +  −0.530  sin Ωt 2.865 0.601   0.312 +  −0.285  sin Ωt 0.107   0.530 =  1.251  sin Ωt kips 3.573





 0.9107 γ1 Mφ1 = 2.8778p0 × 0.0745  2.0760  2.8778   0.1953 =  0.4451  p0 0.6170     0 0.1953 eM = p0  0  − p0  0.4451  1 0.6170   −0.1953 = p0  −0.4451  0.3830 f T eM = 0.383 fT f On including the static correction for the second and the third modes the spring force vector is given by Equation 13.58 fS = p + ω 2 Mφ1 y1 − Mφ1 φT1 p where, referring to Problem 13.6   0 p =  0  sin Ωt 1.744   1 0 0 M = 0.075  0 1 0  0 0 1 y1 = 0.1189 sin Ωt Substitution in Equation a gives   0.570 fS =  1.300  sin Ωt kips 3.544

(a)

91 This may be compared with the more precise result obtained in Problem 13.6.

Substitution gives  1 0 0 1  T=  12 0  

Problem 13.8

3 5

With the degrees of freedom renumbered as indicated in Figure S13.8, the stiffness and mass matrices are as shown below.

  1080 −480 T ˜ K = T KT = −480 480   ˜ =M= 4 0 M 0 2



 2400 0 −1200 −1200 0 800 0 −800   K=  kips/in −1200 0 2400 0 −1200 −800 0 2000   4 0 0 0 0 2 0 0 2 M=  kip.s /in 0 0 0 0 0 0 0 0

2 5

Solve the eigenproblem ˜ q = λM˜ ˜q K˜ This yields λ1 = 84.63, λ2 = 425.4.  q ˜1 =

1.000 1.545



 q ˜2 =

1.0 −1.3



The expanded eigenvector are obtained from

Renumbered d.o.f. 4

2

3

4

2

1

1

3

q = T˜ q     1.000 1.00  1.545   −1.30  q1 =   q2 =   0.500 0.50 1.218 0.08 With reference to the original degrees of freedom these vectors become 

θ

A

B

   0.500 0.50  1.000   1.00  u1 =   u2 =   1.218 0.08 1.415 −1.30

C

Figure S13.8 Denoting degrees of freedom 1 and 2 as t and 3 and 4 as θ, a static condensation is performed as follows. Derive a transformation matrix T such that   ut = Tut uθ

Guyan’s Reduction Using the transformation matrix derived above reduce the original mass matrix as follows ˜ = TT MT M −1 −1 = Mtt − Mtθ K−1 θθ Kθt + Ktθ Kθθ Mθθ Kθθ Kθt 3 2 T − Mtθ K−1 θθ Kθt

where

where

 T=

Also K−1 θθ = Kθt =

Mtθ = 0  2 Mtt = 0  2 Mθθ = 0



I

−K−1 θθ Kθt  

1 2400

0

0



1 2000

−1200 0 −1200 −800

0 1 0 2

 

On substituting the appropriate values 

  ˜ = 3.22 0.48 M 0.48 1.32

92 The revised eigenproblem becomes     1080 −480 3.22 0.48 ˜=λ q q ˜ −480 480 0.48 1.32 On solving the eigenproblem by the characteristic equation method, λ1 = 94.34, λ2 = 759.4 and     1.000 1.000 ˜2 = q q ˜1 = 1.478 −1.617

Modal equations are 1.1912 p0 = 0.1584p0 7.522 0.0468 y¨2 + 759.4y2 = − p0 = −0.00914p0 5.119

y¨1 + 94.34y1 =

The modal responses are obtained from 0.1584 p0 (1 − cos 9.71t) 94.34 = 0.3357(1 − cos 9.71t) 0.00914 p0 (1 − cos 27.56t) y2 = − 739.4 = −0.0024(1 − cos 27.56t)

y1 =

The expanded eigenvectors obtained from q = T˜ q are     1.000 1.000  1.478   −1.617  q1 =   q2 =   0.500 0.500 1.191 −0.047 With reference to the original degrees of freedom     0.500 0.500  1.000   1.000  u1 =   u2 =   1.191 −0.047 1.478 −1.617

Transforming to physical coordinates in the reduced space 

From Problem 13.8 T= 



 =

1.000 1.000 1.478 −1.617



y1 y2



u ˜1 = 0.3333 − 0.3357 cos 9.71t + 0.0024 cos 27.56t

Problem 13.9



u ˜1 u ˜2

1 −K−1 θθ Kθt





 1 0 0 1 =1  2 0 

3 5

2 5

˜ = 3.22 0.48 kip.s2 /in M 0.48 1.32   ˜ = 1080 −480 kips/in K −480 480 Eigenvectors in the reduced space are   1.000 1.000 1.478 −1.617 while the eigenvalues are: λ1 = 94.34, λ2 = 759.4. Also   0  1 3  1 0 T 2 5  0  p ˜=T p=   0 0 1 0 25 p0   0.6 = p 0.4 0 where p0 = 200 kips. Obtain a solution of the problem in the reduced space using mode superposition method ˜1 = q ˜ q1 = 7.522 ˜T1 M˜ M ˜2 = q ˜ q2 = 5.119 M ˜T2 M˜   0.6 p˜1 = [ 1 1.478 ] p = 1.1912p0 0.4 0   0.6 p˜2 = [ 1 −1.617 ] p = −0.0468p0 0.4 0

u ˜2 = 0.5000 − 0.4962 cos 9.71t − 0.0039 cos 27.56t Finally, transformation to original physical coordinates gives u = T˜ u  1 0 u2   0 1 ˜1  u4    u 1  = 2 0 u u1 ˜2 3 2 u3 5 5 

or in inch units u1 = 0.1667 − 0.1679 cos 9.71t + 0.0012 cos 27.56t u2 = 0.3333 − 0.3357 cos 9.71t + 0.0024 cos 27.56t u3 = 0.4000 − 0.3999 cos 9.71t − 0.0001 cos 27.56t u4 = 0.5000 − 0.4962 cos 9.71t − 0.0039 cos 27.56t

Problem 13.10 The first load dependent Ritz vector is given by

x1 = K−1 p

   9 11 6 1 L3  11 16 11   0  = 768EI 7 11 9 0   9 L3 p0   = 11 768EI 7

93 Omit the multiplier and normalize with respect to M α12 = xT1 Mx1 = 251   0.5681 x1 x1 = =  0.6943  α1 0.4418      9 11 7 0.5681 15.843 x2 =  11 16 11   0.6943  =  22.218  7 11 9 0.4418 15.591 Mass orthogonalize with respect to the first vector c1 = xT1 Mx2 = 31.314   −1.9466 ˜ 2 = x2 − c1 x1 =  0.4765  x 1.7559 Normalize with respect to M α22 = x x2 = 7.1 ˜T2 M˜   −0.7306 ˜2 x x2 = =  0.1788  α2 0.6590 Thus



The corresponding mode shapes are     1.00000 1.000 z1 = z2 = 0.09033 −11.069 Transformation gives

  1.0082 0 ˜ M = Z MZ = m 0 122.529   768 EI 0.8946 0 ∗ T ˜ K = Z KZ = 0 1874.7 28 L3   0.5021 p∗ = ZT p ˜ = p0 8.6553 ∗

The modal equations are 768 EI p0 × 0.8946y1 = 0.5021 1.0082¨ y1 + 28 mL3 m 768 EI p0 122.529¨ y2 + × 1874.7y2 = 8.6553 28 mL3 m or EI p0 y¨1 + 24.338 y1 = 0.4980 3 mL m EI p0 y¨2 + 419.657 y2 = 0.07062 3 mL m Solution of the modal equation leads to p0 L3 (1 − cos ω1 t) EI p0 L3 (1 − cos ω2 t) y2 = 0.1683 × 10−3 EI

y1 = 0.02046



0.5681 −0.7306 Q =  0.6943 0.1788  0.4418 0.6590

Carry out a coordinate transformation given by u = Q˜ u

  ˜ = QT MQ = m 1 0 M 0 1   EI 768 1.004 −1.292 ˜ = QT KQ = K 28 L3 −1.292 15.185   0.5681 ˜ = QT p = p p −0.7306 0 The equations of motion are

where

#

EI , ω2 = 20.486 ω1 = 4.933 mL3 Modal superposition yields

where λ = (28/768)(mL3 /EI)λ, and λ is an eigenvalue. Solution of the characteristic equation gives λ1 = 0.8873 λ2 = 15.300

EI mL3 EI λ2 = 419.657 mL3 λ1 = 24.338

#

EI mL3

u ˜ = Zy u ˜1 =

u ˜2 =

˜u ˜u = p ¨ ˜ + K˜ ˜ M To solve these equations use a mode superposition method with frequencies and mode shapes corresponding to the reduced shape. The characteristic equation is 0 0 0 1.004 − λ −1.292 0 0 0 0 −1.292 15.185 − λ 0 = 0

T

p0 L3 (0.02063 − 0.02046 cos ω1 t EI − 0.1683 × 10−3 cos ω2 t) p0 L3 (−0.1440 × 10−4 EI − 0.1849 × 10−3 cos ω1 t + 0.1863 × 10−3 cos ω2 t)

Finally, transformation to the physical coordinates gives u = Q˜ u u1 =

u2 =

u3 =

p0 L3 (0.01170 − 0.01027 cos ω1 t EI − 0.001457 cos ω2 t) p0 L3 (0.01432 − 0.01454 cos ω1 t EI + 0.00022 cos ω2 t) p0 L3 (0.00911 − 0.01026 cos ω1 t EI + 0.00114 cos ω2 t

94 This may be compared with the response obtained in Problem 12.12.

Problem 13.11 The mass matrix M and the stiffness matrix K are given by   0.2 0 M= kip.s2 /in 0 0.1   45 −15 K= kips/in −15 15 Since the system starts from rest and the applied forces at t = 0 are zero, u0 , u˙ 0 and u ¨ 0 are all null vectors. The displacement vector obtained from Equation 13.108 is also null. Equation 13.107 is used to obtain the response history of displacement. 1 2 1 Mun+1 = pn + (−K + 2 M)un − 2 Mun−1 2 h h h or 





80 0 45 −15 un+1 = pn − un 0 40 −15 15   160 0 + un 0 80   80 0 − un−1 0 40

which leads to the following two uncoupled equations 80u1(n+1) = p1(n) + 115u1(n) + 15u2(n) 40u2(n+1) = p2(n) + 15u2(n) + 65u2(n)

As in Problem 13.11, the initial displacement, velocity and acceleration vectors are zero. Iterations are carried out with Equations 13.112, 13.109 and 13.110. For h = 0.05 these equations reduce, respectively, to     320 0 45 −15 un+1 + un+1 0 160 −15 15   320 0 = pn+1 + un (a) 0 160     16 0 0.2 0 + u˙ n + u ¨n 0 8 0 0.1 u ¨ n+1 = −¨ un + 1600(un+1 − un − 0.05u˙ n ) (b) u˙ n+1 = −u˙ n + 40(un+1 − un ) (c) The response obtained from Equations a, b, and c is shown in Table S13.12 for the first 0.5 s

Table S13.12



− 80u1(n−1)

Problem 13.12

(a)

− 40u2(n−1)

Time s

u1 in

u2 in

0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50

0.042 0.219 0.568 1.025 1.478 1.818 1.982 1.962 1.777 1.450

0.046 0.250 0.697 1.384 2.227 3.065 3.695 3.932 3.665 2.900

Problem 13.13

The response for the first 0.5 s calculated from Equation a is shown in Table S13.11.

Table S13.11 Time s

u1 in

u2 in

0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50

0.185 0.576 1.073 1.538 1.863 2.003 1.983 1.818 1.492 0.989

0.185 0.645 1.389 2.325 3.255 3.916 4.092 3.680 2.742 1.483

The initial conditions are zero as in Problem 13.12. Equations 13.115, 13.113, and 13.114 are used in the iterations. For h = 0.05 they reduce as follows:     480 0 45 −15 u + un+1 0 240 n+1 −15 15   480 0 = pn+1 + un (a) 0 240     0.2 0 0.2 0 + 120 u˙ n + 2 u ¨ n+1 0 0.1 0 0.1 ¨ n+1 = −2¨ u un + 2400 (un+1 − un − 0.05u˙ n ) (b) u˙ n+1 = −2u˙ n − 0.025¨ un + 60 (un+1 − un ) (c) The response results obtained from Equations a, b and c are shown in Table S13.13 for the first 0.5 s.

95

Table S13.13

Table S13.14

Time s

u1 in

u2 in

Time s

u1 in

u2 in

0.05 0.10 0.15 0.20 0.25 0.30 0.40 0.45 0.50

0.029 0.208 0.569 1.040 1.499 1.836 1.967 1.785 1.459

0.031 0.229 0.681 1.386 2.257 3.123 3.987 3.678 2.855

0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50

0.028 0.201 0.546 0.990 1.417 1.730 1.875 1.854 1.687 1.389

0.030 0.224 0.660 1.329 2.137 2.922 3.486 3.670 3.395 2.693

Problem 13.15

Problem 13.14 The damping matrix is given by C = αM + βK

(a)

where α and β are obtained from the following equations α + βω12 = 2ξ1 ω1 (b) α + βω22 = 2ξ2 ω2 Substituting ω1 = 8.66, ω2 = 17.30, and ξ1 = ξ2 = 0.05 and solving Equations b for α and β yields α = 0.5771,

β = 3.852 × 10−3

Substitution in Equation a gives 

0.2888 −0.0578 C= −0.0578 0.1155



Equations 13.115, 13.113 and 13.114 now reduce as follows   542.326 −18.467 u −18.467 261.929 n+1   497.326 −3.467 un = pn+1 + −3.467 246.929   (c) 24.578 −0.116 + u˙ −0.116 12.237 n   0.4072 −0.0014 + u ¨ −0.0014 0.2029 n un + 2400(un+1 − un − 0.05u˙ n ) (d) u ¨ n+1 = −2¨ u˙ n+1 = −2u˙ n − 0.025¨ un + 60(un+1 − un ) (e) The response results obtained from Equations c, d and e are shown in Table S13.14 for the first 0.5 s

The stiffness and mass matrices are given in the solution to Problem 13.11. The damping matrix is as derived in the solution to Problem 13.15. Using these values, h = 0.05 and θ = 1.4 Equation 13. 118 reduces to   302.27 −17.47 un+1 −17.47 14240   257.27 −2.47 = pn+θ + un −2.47 127.40   (a) 17.720 −0.116 + u˙ n −0.116 8.802   0.4101 −0.0020 + u ¨n −0.0020 0.2040 The velocity and acceleration at time n + θh are obtained from Equations 13.117 and 13.116. u˙ n+θ = 42.857(un+θ − un ) − 2u˙ n − 0.035¨ un (b) un + 1224.5(un+θ − un − 0.07u˙ n ) (c) u ¨ n+θ = −2¨ The response results obtained from Equations a, b and c are shown in Table S13.15 for the first 0.5 s.

Table S13.15 Time s

u1 in

u2 in

0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50

0.027 0.190 0.520 0.952 1.383 1.714 1.881 1.875 1.711 1.414

0.030 0.220 0.644 1.289 2.065 2.824 3.393 3.626 3.440 2.840

96

Problem 13.16 The mass, stiffness and damping matrices have been provided in the solutions to Problems 13.11 and 13.15. The displacements it time steps 0, 1 and 2 are taken from the solution to Problem 13.14. Thus we have       0 0.0283 0.2014 u0 = u1 = u2 = 0 0.0302 0.2242

 215.58 −17.12 un+1 −17.12 99.23   417.32 −3.46 = pn+1 + un −3.46 206.93   328.66 −1.73 − un−1 −1.73 163.46   81.925 0.385 + un−2 −0.385 40.770

u2 in

0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50

0.028 0.201 0.508 0.890 1.268 1.566 1.739 1.766 1.650 1.407

0.030 0.224 0.624 1.208 1.896 2.561 3.072 3.313 3.218 2.783

+

0.2¨ un2

(a)

15un+1 1

+

Table S13.17

Table S13.16 u1 in

255un+1 = pn+1 + 240un2 + 12u˙ n2 2 2

(a)

The response results obtained from iterations with Equation a are shown in Table S13.16.

Time s

= pn1 + 115un1 + 15un2 − 80un−1 80un+1 1 1

The velocity and acceleration along d.o.f 2 are obtained from Equations 13.114 and 13.113, respectively. Iterations with Equation a and Equations 13.114 1nd 13.113 give the results presented in Table S13.17.

Using h = 0.05 Equation 13.123 reduces to 

To start the iterations, we need u−1 1 . This is obtained from Equation 13.108 and is readily seen to be zero. Equations 13.126 and 13.127 now reduce to

Problem 13.17

Time s

u1 in

u2 in

0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50

0.000 0.190 0.590 1.091 1.545 1.845 1.954 1.915 1.760 1.479

0.029 0.218 0.660 1.369 2.266 3.170 3.835 4.045 3.687 2.805

Problem 13.18 The mass and stiffness matrices are given by 

 0.2 0 M= 0 0.1   45 −15 K= −15 15 An eigenvalue solution gives the following frequencies and orthonormal mode shapes.

We have ω1 = 8.660 ω2 = 17.321 rad/s M11 = 0.2 M22 = 0.1 K11 = 45 K22 = 15 K12 = K21 = −15 p1 (t) = [ 0 14.8 22.0 24.8 24.8 23.2 20.2 18.6 16.2 13.8 11.8 ] p2 (t) = [ 0 7.4 11.0 12.4 12.4 11.6 10.1 9.3 8.1 6.9 5.9 ] The displacements, velocities, and accelerations at time 0 s are all zero.

 φ1 =

1.291 2.582



 φ2 =

1.826 −1.826



The modal forces are given by   2 g(t) = 5.164 g(t) 1   2 p∗2 = φT2 g(t) = 1.8257 g(t) 1 p∗1 = φT1

97 where the values of g(t) are specified at intervals of 0.05 s as follows g(t) = [ 0 7.4 11.0 12.4 12.4 11.6 10.1 9.3 8.1 6.9 2.95 ] It may be noted that the last value of g(t) is taken as half the specified value because at a discontinuity the Fourier transform converges to the average of the values on either side of the discontinuity. The modal equations are y¨1 + y¨2 +

ω12 y1 ω22 y2

The results are presented in Table S13.18. They provide only approximate estimates of the response because of the errors involved in using rectangular summation to represent integration. The accuracy of the results can be improved by using a smaller sampling interval.

Problem 13.19 The equations of motion are

= 5.164 g(t) M¨ u + Ku = f g(t)

= 1.8257 g(t)

The response in each mode is obtained by Fast Convolution between the effective force sampled at 0.05 s for n = 0 to n = 10 and the following unit impulse function. 1 sin ωn t mn ω n where mn is 1 for each mode because the mode shapes have been mass orthonormalized. To obtain the correct response for 0 ≤ t ≤ 1.0 s, the unit impulse response function is truncated at t = 1.0 s and then sampled at intervals of 0.05 s. The number of samples to be used in Fast Convolution must be greater than p + q + 1 where p = 11 is the number of samples in g(t) and q = 11 is the number of samples in h(t). We select the total number of samples in each g(t) and h(t) equal to 25. Each of the two functions is padded with zeros to make up the samples to 25. The period T0 = N × 0.05 = 1.25 s. Frequency sampling rate is given by ∆Ω = 2π/T0 = 5.027 rad/s or 0.8 Hz. The augmented time functions 5.164 g(t) and h1 (t) are convolved through the frequency domain using a Fast Fourier Transform routine to obtain y1 (t). Similarly, 1.8257 g(t) and h2 (t) are convolved to get y2 (t). The response in the physical coordinates is then obtained from hn (t) =

u(t) = φ1 y1 (t) + φ2 y2 (t)

where the mass matrix M, stiffness matrix K and time variation of the forcing function g(t) have been provided in the solution to Problem 13.18. Vector f is given by fT = [ 2 1 ] If g(t) = eiΩt , we have u = HeiΩt and the equation of motion becomes (K − MΩ2 )H = f or 

45 − 0.2Ω2 −15 −15 15 − 0.1Ω2

u2 in

0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50

0.000 0.174 0.546 1.029 1.494 1.830 1.979 1.957 1.783 1.460

0.000 0.185 0.638 1.362 2.265 3.164 3.818 4.023 3.669 2.792

 =

  2 1

(a)

1 (45 − 0.2Ω2 ) D 1 H2 = (75 − 0.2Ω2 ) D H1 =

where D = 450 − 7.5Ω2 + 0.02Ω4 We use a time sampling interval of 0.05 s and obtain the response over N = 20 samples using the exponential window method. The period is given by T0 = 20 × 0.05 = 1.0 s. The decay parameter a is chosen from a=

u1 in

H1 H2

On solving Equationa for H1 and H2 we get

Table S13.18 Time s



p ln 10 = 2.303p T0

where the value of p may be selected between 2 and 3. We will use a = 6. The frequency sampling interval is given by ∆Ω = 2π/T0 . The analysis is now carried out in the following steps. 1. Obtain the scaled function gˆ(t) = e−at g(t) and ˆ determine its discrete transform G(Ω). 2. From the discrete version of the frequency function Hj sampled at the frequency interˆ j for samples up to N/2 by val ∆Ω obtain H ˆ ˆj noting that H(Ω) = H(Ω − ia). Samples of H

98 beyond N/2 are taken as being complex conjugates of the corresponding samples below N/2 that are placed symmetrically with respect to N/2 so that Hj (n∆Ω) = Hj∗ (N/2 − n)∆Ω for N/2 ≤ n ≤ N . 3. Obtain the scaled response u ˆj (t) by taking the inverse discrete Fourier transform of the prodˆ ˆ uct of G(Ω) and H(Ω). Calculate the true response from uj (t) = u ˆj (t)eat The response results over 1 s are presented in Table S13.19.

Table S13.19 Time s

u1 in

u2 in

0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00

0.019 0.201 0.576 1.059 1.521 1.852 1.998 1.971 1.794 1.460 0.910 0.138 -0.707 -1.444 -1.854 -1.899 -1.482 -0.943 0.040

0.023 0.216 0.674 1.396 2.295 3.187 3.835 4.034 3.676 2.789 1.488 -0.027 -1.478 -2.658 -3.365 -3.548 -3.050 -2.112 -0.431

99

Chapter 14

Equation e is satisfied by u=0

Problem 14.1

x=0

3

EI

Kinetic energy is given by


L

T = 0

1 m 2



∂u ∂t

2 dx


0

L

1 EI 2



∂2u ∂x2

2

x=L

Problem 14.2

Potential energy is given by V =

∂ u = ku ∂x3

The equation of motion for Section AB is given by 1 2 dx + k {u(L)} 2

EI

∂ 4 u1 ∂ 2 u1 +m 2 =0 4 ∂t ∂x1

(a)

Let u1 = f1 (x)g1 (t). Then for free vibrations at frequency ω

u x

∂ 2 u1 = −ω 2 f1 (x1 )g1 (t) ∂t2

EI, m

k

Substitution in Equation a gives EI

L

Figure S14.1

d4 f1 (x1 ) = ω 2 mf1 (x) dx41

(b)

In a similar manner for sections BC and DB respectively.

Hamilton’s variational formulation gives


t2

δ

d4 f2 (x2 ) = ω 2 mf2 (x) dx42 d4 f3 (x3 ) EI = ω 2 mf3 (x) dx43

2EI (T − V )dt = 0

(a)

t1

Proceedings as in Section 14.3, Equation a reduces to  
t2 "
L ∂2u ∂2u ∂2 −m 2 − EI δu dx ∂t ∂x2 ∂x2 0 t1   0L  0L 0 ∂2u ∂2u ∂u 00 ∂ − EI 2 δ EI 2 δu00 + ∂x ∂x 00 ∂x ∂x 0  − ku(L)δu(L) dt = 0 (b) Since δu is arbitrary, Equation b yields for a uniform beam ∂2u ∂4u + EI =0 ∂t2 ∂x2 0  L ∂u 00 ∂2u =0 −EI 2 δ ∂x ∂x 00 0L ∂ 2 u 00 EI 3 δu0 − ku(L)δu(L) = 0 ∂x 0 m

For section AB x1 = 0

(e)

f1 (x1 ) = 0 f1 (x1 ) = 0

x1 = L

(e)

f1 (x1 ) = 0

For section BC x2 = 0

f2 (x2 ) = 0 f2 (x2 ) = 0 x2 = 2L f2 (x2 ) = 0

(f )

For section DB

(d)

(d)

x3 = 0

f3 (x3 ) = 0 f3 (x3 ) = 0

x3 = L

(g)

f3 (x3 ) = 0

For slope continuity at B

Equation d is satisfied by ∂2u EI 2 = 0 ∂x

(c)

(c)

f1 (L) = f2 (0) x = 0 or L

f3 (L) = f1 (L)

(h)

100 For moment equilibrium at B EIf1 (L) − 2EIf2 (L) + EIf2 (L) = M

(i)

where IW is the mass moment of inertia of the flywheel. The potential energy is given by

Equation b, c, and d are the equations of motion while Equations e through i provide the 12 boundary conditions.

Problem 14.3


V = 0




( ) EA

c du dx dt d2ut dx dt 2 Acs

d 2u dtdx



∂θ ∂x

2 dx

(b)

or


"


 2 ∂θ 1 dx I0 2 ∂t 0     2 1 M1 R 2 ∂θ 2 + M2 R + ∂t L 2 2  2 !
L ∂θ 1 GJ − dx = 0 2 ∂x 0

t2

(dudx )  dxd (EA dudx )dx (

d2u d2u d  Acs dtdx dtdx dx

)dx

L

(c)

The first term in Equation c is reduced by partial integration to the form

dx

Figure S14.3




For equilibrium of the forces acting on the elemental section     ∂u ∂u ∂ ∂ ∂ EA + cs A ∂x ∂x ∂t ∂x ∂x ∂ 2 ut ∂u −m 2 −c =0 ∂t ∂t or     ∂u ∂u ∂2 ∂ EA + cs A ∂x ∂x ∂t∂x ∂x ∂2u ∂ 2 ug ∂u −m 2 −c =m 2 ∂t ∂t ∂t Boundary conditions are u = 0 at x = 0, x = L.

(T − V )dt = 0

t1

t1 du EA dx

t2

δ

δ

Acs

1 GJ 2

Hamilton’s variational formulation gives

The diagram shows the forces acting on an infinitesimal element of length x. In it ut is the displacement with reference to a fixed frame of axis and u is the displacement relative to the ground, so that ut (x, t) = u(x, t) + ug (t)

m

L

∂u ∂x

= 0 at

Problem 14.4 Let θ(x, t) represent the angle of rotation of the circular shaft. The kinetic energy is given by  2  

2
L ∂θ ∂θ 1 1 I0 T = dx + IW ∂t 2 ∂t x=L 0 2  

2 ∂θ 1 + M2 R 2 ∂t x=L  2
L ∂θ 1 dx = I0 ∂t 0 2     2 1 ∂θ R2 M1 + M2 R 2 + (a) 2 2 ∂t L

t2




L

δ 0

t1

1 I0 2



∂θ ∂t
−

2

t2

t1

dx = (


L

0

) ∂2θ I0 2 δθdx dt ∂t

(d)

The second term is reduced as follows    2 M1 R 2 ∂θ 2 dt + M2 R δ 2 ∂t L t1 0    
t2 0 ∂θ M1 R 2 ∂θ 2 + M2 R δ dx00 = 2 ∂t ∂t t1 x=L 0t2   2 0 ∂θ M1 R = + M2 R 2 δθ0 2 ∂t 0t1 0  2
t2  0 M1 R 2 ∂ θ 0 − δθ dt + M2 R 2 0 2 2 ∂t t1 x=L 0 
t2  2 0 M1 R 2 ∂ θ 2 =− δθ dt00 + M2 R 2 2 ∂t


t2

1 2



t1

x=L

(e) The third term in Equation c is also reduced by partial integration and yields


t2




δ



L

GJ t1

0

∂θ ∂x

2

0L ∂θ 00 GJ δθ0 dx = ∂x 0 t1 )
L ∂2θ − GJ 2 δθdx dt ∂x 0 (f )


t2



101 Substituting Equations d, e and f in Equation c


t2

t1

"


 ∂2θ ∂2θ −I0 2 + GJ 2 δθ dx ∂t ∂x 0 0L ∂θ 0 − GJ δθ00 ∂x 0  2   ∂ θ M1 R 2 + M2 R 2 − δθ dt = 0 2 ∂t2 L

is the mass moment of inertia about a diameter. π 4 π D = (300)4 = 397.6 × 106 mm4 64 64 EI = 25000 × 397.6 = 9.94 × 106 N.m2



I=

(g)

Since δθ is arbitrary, Equation g gives 2

2

∂ θ ∂ θ + GJ 2 = 0 ∂t2 ∂x   M1 ∂2θ ∂θ (h) + M2 R2 2 = 0 at x = L + GJ 2 ∂t ∂x −I0

θ = 0 at x = 0

π (300)2 × × 2400 = 169.6 kg/m 106 4 π M0 = D2 tρ 4 π = × 16 × 0.075 × 2400 = 2262 kg 4 S = 2262 × 9.81 = 22190 N m=

I0 = M0

R2 4

= 2262 ×

22 = 2262 kg m2 4

Variation of the potential energy due to axial force effect is given by Equation 14.82. 0L
L

∂u ∂ ∂u 00 S δu dx ∂VS = −S δu0 + ∂x ∂x 0 ∂x 0

Problem 14.5

The equation of motion gets modified to 2.5 m

2.5 m 75 mm

∂4u ∂2u ∂2u +S 2 +m 2 =0 4 ∂x ∂x ∂t

The boundary conditions at the end x = L change to (  ∂u  2 ∂2 EI ∂∂xu2 + I0 ∂t 2 ∂x = 0 At x = L 2 ∂u ∂3u S ∂x + EI ∂x2 − M0 ∂∂t2u = 0

u

300 mm E  25,000 MPa

3.5 m

EI

x

Problem 14.6 The parameters of the shear beam are given by k1 = 720, 000 × 1000 × 3.5

Figure S14.5 When the axial load effect is neglected, the solution is identical to that in Example 14.1. Equation of motion m

∂4u ∂2u + EI =0 ∂u4 ∂t2

(a)

= 2.52 × 109 N 4700 × 1000 m1 = = 136, 887 kg/m 9.81 × 3.5 k2 = 260, 000 × 1000 × 3.5 = 0.91 × 109 N 2000 × 1000 = 58, 250 kg/m m2 = 9.81 × 3.5

The boundary conditions are The equations of motion are: for the base

∂u

=0 u=0

x=0 x=L

∂x

(b)

 ∂u  2 ∂2 EI ∂∂xu2 + I0 ∂t 2 ∂x = 0 2 ∂3u EI ∂x3 − M0 ∂∂t2u = 0

(c)

Where M0 is the mass of the umbrella roof and I0

k1

∂2u ∂2u − m =0 1 ∂x21 ∂t2

and for the tower k2

∂2u ∂2u − m2 2 = 0 2 ∂x2 ∂t

102 The boundary conditions are given by x1 = 0

u=0 ∂u x2 = 52.5 k2 =0 ∂x2 At the junction of base and tower, for continuity of displacement and the equilibrium of shear forces respectively. x1 = 52.5, x2 = 0

u(x1 ) = u(x2 ) ∂u ∂u k1 ∂x = k2 ∂x 2 1

103

Chapter 15

The fundamental mode shape is given by

Problem 15.1

f1 (x) = sinh β1 x +

sinh β1 L sin β1 x sin β1 L

The vibration shape of the beam is given by Equation 15.51

Problem 15.2

f (x) = C1 cosh βx + C2 sinh βx

Since the torsional vibrations of a shaft are governed by an equation that is similar to the one for axial vibrations, the vibration shape is given by Equation 15.109.

+ C3 cos βx + C4 sin βx 

f (x) = C1 β sinh βx + C2 β cosh βx − C3 β sin βx + C4 β cos βx f  (x) = C2 β 2 cosh βx + C2 β 2 sinh βx

f (x) = C1 cos βx + C2 sin βx

− C3 β 2 cos βx − C4 β 2 sin βx f  (x) = C1 β 3 sinh βx + C2 β 3 cosh βx

where

+ C3 β 3 sin βx − C4 β 3 cos βx

β2 =

Since f (x) = 0 at x = 0 C1 + C3 = 0

(a)

Also, f  (x) = 0 at x = 0 and L, yielding C1 − C3 = 0

(b)

ω 2 I0 GJ

At x = 0, f (x) = 0 yielding C1 = 0. Since θ(x, t) = g(t)f (x) ∂θ df = g(t) ∂x dx = g(t)C2 β cos βx

and C1 cosh βL + C2 sinh βL − C3 cos βL − C4 sin βL = 0

(c)

Equations a, b and c yield C1 = 0, C3 = 0 and C4 = C2 The boundary condition gives

sinh βL sin βL 3 EI ∂∂xu3

Applying the boundary condition given by Equation h of Problem 14.4 GJg(t)C2 β cos βx   M1 d2 g(t) + C2 sin βx = 0 + M2 R 2 2 dt2 at x = L

= ku at x = L

  sinh βL EI C2 β 3 cosh βL − C2 β 3 cos βL sin βL   sinh βL = k C2 sinh βL + C2 sin βL sin βL = 2kC2 sinh βL

(d) On substituting k = π 4 EI/(2L3 ), Equation d reduces to (cosh βL sin βL − cos βL sinh βL) (f ) π4 sin βL sinh βL = 0 − (βL)3 When the left hand side of Equation f is plotted as a function of βL, the first zero of the function is obtained at β1 L = 2.83. The fundamental frequency is ω 2 mL4 β14 L4 = 1 EI # # EI EI 2 = 8.01 ω1 = (β1 L) 4 mL mL4

(a) For free vibrations g(t) is a harmonic function so 2 that d dtg(t) = −ω 2 g(t). Substituting in Equation a 2 and using the given value of (M1 /2 + M2 )R2 GJβ cos βL = I0 Lω 2 sin βL 1 tan βL = βL

(b)

The solutions to Equation b are obtained as in Example 15.3 and give β1 L = 0.8605 β2 L = 3.4265 The corresponding mode shapes are given by x x = sin 0.8605 L L x x f2 (x) = sin β2 L = sin 0.34265 L L

f1 (x) = sin β1 L

104

Problem 15.3 Considering that the building structure is idealized as a shear beam, the vibration shapes are given by

Hence

f1 (x1 ) = A1 cos β1 x1 + B1 sin β1 x1 f2 (x2 ) = A2 cos β2 x2 + B2 cos β2 x2 where

β1 L ω1 = L

m1 ω 2 k1 2 m 2ω β21 = k2

The coefficients A1 , B1 , A2 , B3 are obtained by applying the following boundary equations f1 (0) = 0 f1 (L) = f2 (0) 0 0 df1 00 df2 00 k1 = k2 dx1 0x1 =L dx2 0x2 =0 0 df2 00 k2 =0 dx2 0x2 =L These conditions lead to

−A2 β2 sin β2 L + B2 β2 cos β2 L = 0

(a) (b) (c)

Equations a, b and c give 

 sin β1 L −1 0  k1 β1 cos β1 L 0 −k2 β2  0 − sin β2 L cos β2 L     B1 0 ×  A2  =  0  0 B2

(d)

For a nontrivial solution the determinant of the coefficient matrix in Equation d should be zero, yielding β2 k2 tan β1 L tan β2 L − 1 = 0 (e) β1 k1 where β2 = β1 =

# $

m2 k1 m1 k2 58, 250 × 2.52 × 109 = 1.0855 136, 887 × 0.91 × 109

β2 k2 = 0.3920 β1 k1

#

k1 m $1 0.9692 2.52 × 109 = 2.505 rad/s = 52.5 136, 890 $ 2.0464 2.52 × 109 ω2 = = 5.289 rad/s 52.5 136, 890

β12 =

A1 = 0 β1 sin β1 L = A2 k1 B1 β1 cos β1 L = k2 B2 β2

When the left hand side of Equation e is plotted as a function of β1 L the first two zeros are obtained as follows β1 L = 0.9692 β2 L = 2.0463

The mode shapes are obtained as follows First mode shape A2 = B1 sin 0.9692 = 0.8244B1 k1 β1 B2 = B1 cos 0.9692 k2 β2 1 = × 0.5660B1 = 1.4438B1 0.392 x1 f1 (x1 ) = B1 sin 0.9692 L  x2 f2 (x2 ) = B1 0.8244 cos 1.0521 L x2  + 1.4438 sin 1.0521 L Second mode shape A2 = B1 sin 2.0464 = 0.8890B1 1 cos 2.0464 = −1.168B1 B2 = B1 × 0.392 x1 f1 (x1 ) = B1 sin 2.0464 L  x1 f2 (x2 ) = B1 0.8890 cos 2.2214 L x2  −1.168 sin 2.2214 L

Problem 15.4 The vibration shapes are given by f1 (x1 ) = A1 cosh βx1 + B1 sinh βx1 + C1 cos βx1 + D1 sin βx1 f2 (x2 ) = A2 cosh βx2 + D2 sinh βx2 + C2 cos βx2 + B2 sin βx2 The boundary conditions are f1 (0) = 0 f1 (0) = 0

f2 (0) = 0 f1 (L) = 0 f2 (0) = 0 f1 (L) = f2 (2L) f2 (2L) = 0 f1 (L) + f2 (2L) = 0

105 x1

The remaining coefficients are now derived as follows

EI, m

x2

B2 = −0.9852 × 2.898A1 = −2.8552A1 C2 = −A1 D1 = −B1 = 1.4201A1

EI m

C2 = −A2 = −2.898A1 D2 = −B2 = 2.8552A1

Figure S15.4 Application of boundary conditions gives A1 (cosh βL − cos βL) + B1 (sinh βL − sin βL) = 0 (a) A2 (cosh 2βL − cos 2βL) (b) + B2 (sinh 2βL − sin 2βL) = 0

The mode shape is given by  x1 x1 f1 (x1 ) = A1 cosh βL − 1.4201 sinh βL L L x1 x1  − cos βL + 1.4201 sin βL L  L x2 x2 f2 (x2 ) = A1 2.898 cosh βL − 2.855 sinh βL L L x2  x2 − 2.898 cos βL + 2.855 sin βL L L where βL = 2.1878.

A1 (sinh βL + sin βL) + B1 (cosh βL − cos βL) = A2 (sinh 2βL + sin 2βL) + B2 (cosh 2βL − cos 2βL)

Problem 15.5 (c)

A1 (cosh βL + cos βL) + B1 (sinh βL + sin βL)

The frequency equation and its solutions are as follows

+ A2 (cosh 2βL + cos 2βL)

1 βL β1 L = 0.8605 β2 L = 3.4256

tan βL =

+ B2 (sinh 2βL + sin 2βL) = 0 (d) Simultaneous solution of Equations a, b, c and d yields ( cosh βL cos βL − 1) × (sinh 2βL cos 2βL − cosh 2βL sin 2βL) + (cosh 2βL cos 2βL − 1)

β3 L = 6.4313 βn L ≈ (n − 1)π for n large (e)

Also

∆y x R L and the mode shapes are given by θ(x, 0) =

× (sinh βL cos βL − cosh βL sin βL) = 0 On solving Equation e by trial and error βL = 2.1878. Hence # # EI EI 2 2 = 4.7863 ω=β L mL4 mL4

fn (x) = sin βn x Using the mode superposition method θ(x, t) = Σyn (t)fn (x)

From Equation a B1 = −A1

cosh βL − cos βL = −1.4201A1 sinh βL − sin βL

(f )

From Equation b cosh 2βL − cos 2βL B2 = −A2 = −0.9852A2 (g) sinh 2βL − sin 2βL Substituting Equations f and g in Equation c sinh 2βL − sin 2βL cosh βL cos βL − 1 A2 = A1 sinh βL − sin βL cosh 2βL cos 2βL − 1 = 2.898A1

where yn (t) is given by yn (t) = yn (0) cos ωn t and yn (0) is obtained as follows 1 θ(x, 0) = yn (0)fn (x)


n L

fm (x)I0 θ(x, 0)dx 0

=

1 n


yn (0)

L

fm (x)I0 fn (x)dx 0

106 Using the orthogonality relationship


L 0

ym (0) =

L 0

Mm dx EI 1 3x 2 x(L − x) =− θ×x× × 2 L2 3 L   x(L − x) 3 6x θ − + × (L − x) × 6L L2 L x3 (L − x) x(L − x)2 (2x + L) =− θ − θ 3 2L3 L   x 2 x 1− θ =− 2 L

u(x, 0) =

fm (x)θ(x, 0)dx 2

{fm (x)} dx L x sin βm x dx 0 L sin2 βm x du 0

=

∆y RL

=

1 2∆y sin βm L − βm L cos βm L   RL (βm L)2 mL 1 − sin2β2β mL

The response is thus given by

θ(x, t) =

1 2∆y

sin βn L − βn cos βn L   RL (βn L) nL 1 − sinβ2β nL 1

2

n

The orthonormal mode shapes for transverse vibrations are # 2 nπx f (x) = sin mL L The response to initial displacement is given by

× cos ωn t sin βn x

u(x, t) = Σyn (t)φn (x) Taking the first three modes where ∆y  x 1.3216 cos ω1 t sin 0.8605 θ(x, t) = R L x + 0.5563 cos ω2 t sin 3.4256 L x − 0.3068 cos ω3 t sin 6.4373 L

yn (t) = yn (0) cos ωn t and


u(x, 0)mφn (x)dx  
L # x3 nπx 2m θ x − 2 sin dx − = L 2 L L 0 # 2m θ 6L2 = cos nπ L 2 (nπ)3 0

Problem 15.6 A moment of 3EI L θ applied at the right-hand end of the beam will produce a rotation θ. The corresponding M diagram is shown in Figure S15.6. To obtain the displacement of the beam at a distance x from the left end apply a unit load as shown. The resulting m diagram is also shown.

L

yn (0) =

Thus u(x, t) =

1

yn (0)

2 nπx cos ωn t sin mL L

(−1)n

6Lθ nπx cos ωn t sin (nπ)3 L

n

=

1 n

EI, m

#

where ωn is given by Equation 15.59 # 2 2

ωn = n π

L (3EI/L)  1 x () x(Lx)/L

Figure S15.6

(3EI/L)

EI mL4

107

Chapter 16 Problem 16.1 (a) The equation of motion for identical support motion is EI

The steady state response is given by # 2 A0 mΩ2 L 1 sin Ωt yn (t) = nπωn2 1 − βn2 mL   nπx 2A0 1 βn2 1 sin sin Ωt u(x, t) = π 1 − βn2 n L n = 1, 2, 3, · · ·

∂4u ∂2u + m = A0 mΩ2 sin Ωt ∂x4 ∂t2

where u is the displacement relative to the support. For the simply supported beam # EI 2 2 ωn = n π mL4 # 2 nπx φn = sin mL L Hence the modal force is given by #
L nπx 2 pn = sin dx A0 mΩ2 sin Ωt mL L 0 # nπx  2 L  2 1 − cos sin Ωt = A0 mΩ mL nπ L # 2A0 mΩ2 L 2 = sin Ωt nπ mL n = 1, 3, 5 · · ·

Problem 16.2 Taking the first term in the series given by Equation 16.31 1 2P mL (πv/L)2 − ωn2

πx πv πvt sin × sin ω1 t − sin L Lω1 L

u(x, t) =

Also

2π ω1 πv vT =α = Lω1 2L π 4 EI ω12 = mL4 T =

Hence

The modal equation becomes y¨n + ωn2 yn =

2A0 mΩ2 L nπ

#

u(x, t) = 2 sin Ωt mL

Hence the steady state response is # 2 2A0 mΩ2 L 1 yn (t) = sin Ωt 2 2 nπωn 1 − βn mL n = 1, 3, 5, · · · 1 yn (t)φn (x) u(x, t) = n

4A0 = π

1

βn2 1 nπx sin 1 − βn2 n L

 sin Ωt

n = 1, 3, 5, · · · (b) The pseudo-static displacement of the beam due to a unit vertical motion of the left hand end is  x us = 1 − L Hence the modal force is derived from #
L nπx x 2 pn = sin 1− A0 mΩ2 sin Ωt dx mL L L 0 # 2 L 2 A0 mΩ sin Ωt = nπ mL n = 1, 2, 3, · · ·

2P 1  mL ω12

πv Lω1

1 2

−1

vT t πx πv 2πt − sin 2π sin sin T 2L T L Lω1 4 1 2P mL = mL π 4 EI α2 − 1

t πx 2πt − sin 2πα sin × α sin T T L 2P L3 = 4

π EI α 1 t t − × sin 2πα sin 2π 1 − α2 T 1 − α2 T πx × sin L

×

The crawl deflection is obtained by setting α = 0 us (x, t) =

2P L3 t πx sin 2πα sin 4 π EI T L

Note that 2πα Tt = π vt L is finite because the product vt is finite. The maximum value of us is at mid-span when the load is also at mid-span (us )max =

P L3 2P L3 = 4 π EI 48.7EI

108 This value may be compared with the exact value of P L3 /48EI. The deflection index is given by us (L/2, t) (us )max α t α2 t sin 2π = sin 2πα − T 1 − α2 T 1 − α2

DI =

Problem 16.3

where T Tv +* + * ∆ = 1 − (φ + α)2 1 − (φ − α)2 + * D1 = 1 − (φ2 + α2 ) /∆ φ=

The alternating force can be represented as F = W Aδ(x − vt) cos 2π

After integration and algebraic manipulation this reduces to #  t 2 WA t D1 sin 2πα cos 2π y1 (t) = mL ω12 T Tv  t t t − αD3 sin 2π + D2 cos 2πα sin 2π T Tv T

t Tv

The corresponding modal force is given by #
L 2 nπx t W Aδ(x − vt) sin pn = dx cos 2π mL 0 L Tv Taking the first mode # πvt t 2 W A sin cos 2π p1 = Tv mL L # t t 2 W A sin 2πα cos 2π = Tv mL T The modal response is obtained by using the Duhamel’s integral #
2 WA t τ τ sin 2πα cos 2π y1 (t) = mL ω1 0 T Tv 2π (t − τ )dτ × sin T The integrand can be reduced by using the trigonometric identities 1 sin A cos B = {sin(A + B) + sin(A − B)} 2 1 sin A sin B = {cos(A − B) − cos(A + B)} 2 so that # 
 τ 2 WA t τ y1 (t) = sin 2πα + 2π mL 2ω1 0 T Tv   τ 2π τ + sin 2πα − 2π sin (t − τ )dτ T Tv T # 
τ τ 2 WA t cos 2πα + 2π = mL 4ω1 0 T Tv  t τ −2π + 2π T T   τ t τ τ − cos 2πα + 2π + 2π − 2π T Tv T T   τ τ τ t + cos 2πα − 2π − 2π + 2π T Tv T T   τ τ τ t − cos 2πα − 2π dτ + 2π − 2π T Tv T T

D2 = 2αφ/∆ * + D3 = 1 + (φ2 − α2 ) /∆ The response in the physical coordinate is given by ua (x, t) = y1 (t)φ1 (x) # πx 2 sin = y1 (t) mL L  t t 2W L3 A D1 sin 2πα cos 2π = T Tv π 4 EI t t + D2 cos 2πα sin 2π T Tv  πx t sin − αD3 sin 2π T L

Problem 16.4 The mode shapes and frequencies were derived in Example 15.2. Mode 1 β1 L = β2 L = 0.6155 # EA1 ω1 = 0.6155 m1 L1 2 x1 f1 (x1 ) = 1.732 sin 0.6155 L x2 x2 f2 (x2 ) = cos 0.6155 + 0.7071 sin 0.6155 L L Modal force = P0 f2 (L) = 1.2247P0 . Modal mass M1 is given by
2 M1 = m {f (x)} dx
L x1 2 1.732 sin 0.6155 m1 dx = L 0
L x2 cos 0.6155 + L 0 x2 2 m2 dx +0.7071 sin 0.6155 L = 3m1 L

109 Mode 2

Response in the physical coordinates is obtained from

β1 L = β2 L = 2.5261 # EA1 ω2 = 2.5261 m1 L2 f1 (x1 ) = 1.7332 sin 2.5261

u = y1 f (1) (x) + y2 f (2) (x) + y3 f (3) (x) x1 L

Problem 16.5

x2 x2 − 0.7071 sin 2.5261 f2 (x2 ) = cos 2.5261 L2 L2

The equation of damped vibration is given by

Modal force = P0 f2 (L) = −1.2247P0 . The modal mass is given by





L

2

L

m2 {f2 (x2 )} dx

m1 {f1 (x1 )} dx +

M2 = 0

2

0

m

∂2u ∂2u ∂u − EA + c =p ∂t2 ∂t ∂x2

(a)

where c = a0 m. Representing u in terms of modal coordinates of the associated undamped system

= 3m1 L

∞ 1

u=

Mode 3

yn (t)fn (x)

(b)

n=1

β1 L = β2 L = 3.7571 # EA1 ω3 = 3.7571 m1 L2 f1 (x1 ) = −1.732 sin 3.7511 f2 (x2 ) = cos 3.7571

Substituting Equation b in Equation a, multiplying both sides by fi (x) and integrating over the length


x1 L

x2 x2 + 0.7071 sin 3.7571 L L




L

2

m1 {f1 (x1 )} dx +

M3 = 0

L

fn (x)¨ yn dx

n=1




fi (x)

=

0

∞ 1

EAfn (x)yn dx

n=1




2

fn (x)y˙ n dx

n=1


−

m2 {f2 (x2 )} dx

∞ 1

fi (x)a0 m

+

Modal force = P0 f2 (L) = −1.2247P0 . The modal mass is given by


∞ 1

fi (x)m

fi (x)p dx

= 3m1 L Using orthogonality condition

The modal equations are EA1 0.4082 y¨1 + (0.6155) y1 = P0 m1 L2 m1 L EA1 0.4082 y¨2 + (2.5261)2 y2 = − P0 m1 L2 m1 L EA1 0.4082 y¨3 + (3.7571)2 y3 = − P0 m1 L2 m1 L

Mi y¨i + a0 Mi y˙ i + Mi ωi2 yi = pi

2

(c)

where Mi is the modal mass for the ith mode. On dividing through by Mi , Equation c reduces to y¨i + a0 y˙ i + ωi2 yi =

pi Mi

(d)

Thus

The modal responses are

2ξi ωi = a0 2

0.4082 P0 m1 L y1 = (1 − cos ω1 t) (0.6155)2 m1 L EA1 P0 L = 1.0776 (1 − cos ω1 t) EA1 0.4082 P0 L (1 − cos ω2 t) y2 = − (2.5261)2 EA1 P0 L (1 − cos ω2 t) = −0.06397 EA1 0.4082 P0 L y3 = − (1 − cos ω3 t) (3.7571)2 EA1 p0 L (1 − cos ω3 t) = −0.02892 EA1

For the first mode ξ1 = 0.05, ω1 = 0.6155 Hence a0 = 0.1ω1

%

EA1 m1 L2 .

For the second mode a0 ω1 = 0.05 2ω2 ω2 % 1 and since ω2 = 2.5261 mEA 2 , substitution for ω1 1L and ω2 gives ξ2 =

ξ2 = 0.05 ×

0.6155 = 0.0122 2.5261

110 Similarly ξ3 = 0.05

Modal force p1 = f2 (L)p = 1.6626p. Modal mass M1 is given by
L
L 2 2 m2 {f1 (x)} dx m1 {f1 (x)} dx + M1 =

ω1 0.6155 = 0.0082 = 0.05 × ω3 3.7571

0

0

The modal responses are given by yi =

pi ωi2

= 148, 960L



  ξi ωi 1 − e−ξi ωi t cos ωdi t + sin ωdi t ωdi

Second mode ω2 = 5.289 rad/s x1 f1 (x1 ) = sin 2.0464 L

Using the modal force values derived in Problem 16.4

x2 L x2 − 1.1680 sin 2.2214 L p2 = −1.4670p M2 = 131, 193L

f2 (x2 ) = 0.8890 cos 2.2214 P0 L y1 = 1.0776 EA1   ξ1 ω1 −ξ1 ω1 t cos ωd1 t + × 1−e sin ωd1 t ωd1 P0 L y2 = −0.06397 EA1   ξ2 ω2 −ξ2 ω2 t cos ωd2 t + × 1−e sin ωd2 t ωd2 P0 L y3 = −0.02892 EA1   ξ3 ω3 × 1 − e−ξ3 ω3 t cos ωd3 t + sin ωd3 t ωd3 where %

EA m1 L2 # % EA 2 = ω2 1 − ξ2 = 2.5620 m1 L2 # % EA1 2 = ω3 1 − ξ3 = 3.7570 m1 L2

ωd1 = ω1 ωd2 ωd3

#

1−

ξ12

= 0.6147

u(x, t) = y1 (t)f

(x) + y2 (t)f

(2)

(x) + y3 (t)f

y¨i = ω12 yi =

pi Mi

or 1.662 × 400 × 1000 cos 5.01 148, 960 × 52.5 = 0.08504 cos 5.01t 1.4678 × 400 × 1000 y¨2 + 27.92y2 = − cos 5.01t 131, 93 × 52.5 = −0.08524 cos 5.01t

y¨1 + 6.275y1 =

The modal steady state responses are 0.08504 1 cos 5.01t 6.275 1 − 4 = −0.004516 cos 5.01t 1 0.08524 cos 5.01t y2 = − 27.97 1 − (0.947)2

y1 =

The response in the physical coordinate is obtained from (1)

The modal equations are given by

(3)

(x)

= −0.02967 cos 5.01t 5.01 in which frequency ratio β2 = 5.239 = 0.947. The displacement response is given by

u(x, t) = y1 (t)f (1) (x) + y2 (t)f (2) (x)

Problem 16.6 The mode shapes and frequencies of the building were derived in Problem 15.2. Using those results First mode ω1 = 2.505 rad/s x1 f1 (x1 ) = sin 0.9692 L

x2 L x2 + 1.4438 sin 1.0521 L

f2 (x2 ) = 0.8244 cos 1.0521

The shear in the base portion is given by V (x, t) = k1

∂u ∂x

0.9692 L x1 × cos 5.01t cos 0.9692 L 2.0464 − 0.02967 × L x1  × cos 5.01t cos 2.0464 L

= k1 −0.004516 ×

111 The maximum base shear is obtained by setting cos 5.01t = −1 and x1 = 0 0.004516 × 0.9692 V (0)max = 2.52 × 109 52.5

0.0297 × 2.0464 N + 52.5 = 3124 kN

112

Chapter 17

The wave shape at t = 2 s is given by 

2πx − 4π u(x, 2) = −A0 sin c 2πx = A0 sin c

Problem 17.1 28.28 1.0 0.5 t  1.0 s. 0.5 1.0

60.00

0 < x < 2c

and that at t = 3 by u(x, 3) = A0 sin

t  2.0 s.



2πx c

0 < x < 3c

At t = 3 s the wave has gone past the right hand end. As the wave arrives at that end it is reflected back so that the net displacement at the fixed end is zero. The reflected shape and the net displacement are shown in Figure S17.1.

Problem 17.2 3.44

The force impulse is given by

3.44 t  2.0 s.

P (0, t) = 2500 sin

Reflected wave 1.5 1.0 0.5

2π t 0.012

The wave velocity is obtained from

0.5 1.0 1.5

#

EA m # 20000 × A × 106 = 2300 × A × 1 = 2949 m/s

Resultant

c=

24.84

Figure S17.1 Given T = 400 N

where A is the cross-sectional area in m2 . The force pulse can now be expressed as

m = 0.5 kg/m % T = 28.28 m/s. Wave velocity c = m

P (0, t) = −2500 sin

u(0, t) = A0 sin 2πt   2π (−ct) = f (−ct) = A0 sin − c   2π u(x, t) = f (x − ct) = A0 sin − (x − ct) c 2π (x − ct) = −A0 sin c Comparing with Equation 17.10c, wave length λ = c = 28.28 m. The time taken by the wave to arrive at the right end = 60/28.28 = 2.12 s. The equation of wave shape at t = 1 s is given by 

2πx − 2π c 2πx = A0 sin c



u(x, 1) = −A0 sin

2π (−2949t) 35.38

The equation of force wave is P (x − ct) = −2500 sin

2π (x − 2949t) 35.38

The wave length = 35.38 m. (a) For soft soil, the pressure wave is reflected with a sign opposite to the incident wave. Figure S17.2a shows the position of incident and reflected waves when a wave crest arrives at a gage location. It is readily seen that the maximum tensile and compressive forces at the gage locations are 2500 kN. Hence the stress recorded by the gages is given by

0