Dynamics of structures, Third Edition Jagmohan L. Humar SOLUTION MANUAL

SOLUTION MANUAL DYNAMICS OF STRUCTURES SOLUTIONS MANUAL THIRD EDITION JAGMOHAN L. HUMAR

SOLUTIONS MANUAL FOR Dynamics of Structures Third Edition

by J. Humar

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2

Chapter 2

In a similar manner we get Iy = M üy

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 θ̈ about the center of mass the inertia force on the infinitesimal element is directed along the tangent and is γr2 θ̈dθdr.

The x component of this force is γr2 θ̈dθdr 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 γ θ̈r3 dθdr = γπR2

Mθ = 0

0

R2 R2 θ̈ = M θ̈ 2 2

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 üx is γ üx dxdy. The resultant inertia force in the negative x direction is given by a/2 b/2√1−4x2 /a2 Ix =

Problem 2.2

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 üx the inertia force on the infinitesimal are is γrdθdrüx , 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 = γ üx rdθdr = γπR2 üx = M üx 0

γ üy dydx 1−4x2 /a2

πγab = M üx = 4

b

(a)

√

−a/2

dr R

−b/2

a/2 = γ üx b 1 − 4x2 /a2 dx

dxdy dθ

−a/2

The moment of the x direction inertia force on an element is γ üx 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 üy directed in the negative y direction.

An angular acceleration θ̈ produces a clockwise

moment equal to γr2 θ̈dxdy = γ x2 + y 2 θ̈dxdy.

Integration over the area yields the resultant moment, which is clockwise a/2 b/2√1−4x2 /a2 Iθ =

−a/2

−b/2

γ θ̈ x2 + y 2 dydx

1−4x2 /a2

πab a + b a2 + b2 θ̈ = M θ̈ 4 16 16 2

=γ

√ 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 = γ üx r2 sin θdθdr = 0 0

0

The x and y direction inertia forces produced on the infinitesimal element are −γ θ̈ sin θdxdy and γ θ̈ 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

1 lb.

Equivalent spring stiffness = 0.0297 = 33.67 in.

u

2

mü

2uk

Mü

T

10 Mass = 386.4 = 0.02588 lb.s in.

The equation of motion is given by

0.02588ü + 33.67u = 0 I0θ̈

or ü + 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 + müR + I0 θ̈ − T R = 0

M(L1 R)θ̈

B I0θ̈

T + M ü = 0 Substituting I0 = mR2 /2 and eliminating T gives

3m M+ 2

ü + 4ku = 0

θ cd θ̇

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

2

M (L1 + R) θ̈ +

Wheel

M R2 θ̈ + cd2 θ̇ + kL22 θ = 0 2

u

or

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

M

3 L21 + R2 + 2L1 R 2

θ̈ + cd2 θ̇ + kL22 θ = 0

4

Problem 2.6

Problem 2.8

d

ug(t)

kL θ

Mbθ̈

u

dx

x

mL2 θ̈ 3

θ

A1

cdθ̇ Mb θ̈

A2 Output

Input

MR 2 θ̈ 2

L

Figure S2.8 Figure S2.6 Taking moments about A shown in Figure S2.6

mL2 2 + M b θ̈ + cd2 θ̇ + kL2 θ = 0 3 When the flywheel is braked, an additional rotational inertia term will exist. Moment equilibrium gives

mL2 R2 2 +M b + θ̈ + cd2 θ̇ + 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

= 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

L

∗

k = =

0

EI 0

EI L3

2

{ψ(ξ)} dξ = 0.03016mL

= mL

1 0

d2 ψ(x) dx2

2

d2 ψ(ξ) dξ 2

2

dx dξ = 7.2

ρL {A1 + (A2 − A1 ) ξ} 2

4ξ + ξ 4 − 4ξ 3 dξ

ρL (5A1 + 11A2 ) 30 L 2 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)üg 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

1

ρLüg {A1 + (A2 − A1 ) ξ} 2ξ − ξ 2 dξ 0

1 5 = −ρLüg A1 + A 2 4 12

=−

Problem 2.9 Refer to Figure S2.9 where the inertia force 2¨ut on an infinitesimal element is shown as m̄ ∂∂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 m̄ (z̈(t)ψ(x) + üg ) 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 ∗ m̄L m = m̄[ψ(x)]2 dx = 140 0 L 3 p∗ = −m̄üg ψ(x)dx = − m̄Lüg 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)m̄g is obtained from L 2 3 S(x) ψ (x) dx = m̄g kG = 8 0 The equation of motion including the gravity effect is

3 3EI 33 3 m̄Lz̈ + − m̄g z = − m̄Lg L3 8 8 140

Equation of motion: mL πvt π 4 EI z = F sin z̈ + 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 z̈(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  üg = −1170 sin(6πt) kips 1 The equation of motion is given by 19z̈ + 1250z = −1170 sin(6πt)

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

ml2 θ̈ + 2ka2 + mgl θ = 0

6 A

θ 2 ka θ S

ml θ̈

mg

Figure S2.12

7

Chapter 3

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

Problem 3.1

M∗ ü + C∗ u̇ + K∗ u = 0

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 q̈ g 3

kT q4

w q̈ g 4

mg I0θ̈

mhθ̈

W 2 r q̈ 2 g W q̈ 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 q̈4 + cS (q̇4 − q̇1 − q̇2 a) g + kS (q4 − q1 − q2 a) + kT q4 = 0 w q̈3 + cS (q̇3 − q̇1 + q̇2 a) g + kS (q3 − q1 + q2 a) + kT q3 = 0 For equilibrium of vertical forces on the vehicle body W q̈1 + cS (q̇3 + q̇4 − 2q̇1 ) g + kS (q3 + q4 − 2q1 ) = 0 Taking moments about the mass center of the vehicle body W 2 r q̈2 + cS a (q̇3 − q̇4 + 2aq̇2 ) 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 müt + 2cu̇ + 2ku = 0

(1)

or mü + 2cu̇ + 2ku = −müg Taking moments about A and assuming that θ is small

I0 + mh2 θ̈ + 2cb2 θ̇ + 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

12 0

0 0

0 12

 θ1  θ2    θ3 θ4 

Hence

k21 k31

u1 u2

k2 1

k12

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 0

k2 a 2

Hence 

2.236  0 M= 0 0

k2 a2 k1 b2 2 + 4

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

Mθ̈ + 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. m33 = m11 m11 =

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 Mü + 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 m2 Lü2 + L

m1 Lü1 + cLu̇1 +

= 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

Mü + Cu̇ + Ku = 0 where

c 0 M= C= 0 0 m2

2 ka 1 −1 K= 2 uT = [ u 1 −1 1 L m1

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

1 L (a)

a

L/2 (b)

du1冣 L 冢365 EI 2 L3

L

(c)

1

36 EI du 1 5 L3 ( )

m2ü2

L/4 ( )

42 EI du 1 5 L3

Figure S3.6 ( )

Problem 3.7

( ) ( )

θ1

a

θ2 42 EI du 5 L2 1

ka(θ2 θ1) ka(θ2 θ1)

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|θ̈ 2

m1|θ̈ 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 θ̈1 − ka2 (θ2 − θ1 ) + m1 gl sin θ1 = 0 m2 l2 θ̈2 + 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 θ̈1 + m1 l l k a 2 g θ̈2 + (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 a22 = ds = + × × EI 3EI 2 EI 2 3 7 L = 24 EI

m1 m2 1 L L L3 a12 = ds = − ×L× =− EI 2 EI 2 4EI a11 =

The flexibility matrix is thus given by  1 L3  3 A= EI − 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 ü1 + ü2 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 = m̄ (ψ1 (x))2 dx = 0.2268m̄L ψ1 (x) = 1 − cos

0

L m̄ (ψ2 (x))2 dx = 1.9244m̄L

m22 = 0

L m12 = m21 =

m̄(ψ1 (x)ψ2 (x))dx = 0.5755m̄L 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 ψ1 (x)dx = −0.3634m̄Lüg p1 = −m̄üg 0

L p2 = −m̄üg

ψ2 (x)dx = −1.2122m̄Lüg 0

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

12 The equations of motion are

0.2268 0.5755 z̈1 m̄L 0.5755 1.9244 z̈2

π 4 EI 1 0 0.370 + − m̄g 0.75 32L3 0 81

0.3634 = −m̄Lüg 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=

c + 4EI Lc

2EIb Lb 6EIc L2c

6EIc  L2c 6EIc  L2c  24EIc L3c

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

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

4EIb /Lb

6EIc /Lc2

= [1

4EIc /Lc

= TA θ

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

 θ1 0 a ]  θ2  θ3 

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 2

100×20×20 Slab mass m = 1000×386.4 = 0.1035 kip.s 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(u̇v − η̇)−mt η̈ − mt g. Thus, we have

The equations of motion are Mθ̈ + Kθ = 0

p∗ = k {uv − η} α + c {u̇v − η̇} α − mt η̈α − 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∗ z̈ + k ∗ z = p∗ On combining with the equation for sprung mass and substituting for η, η̇ and η̈, 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 üv 0 mL 2 + m α z̈ 0 t 2

u̇v c −cα + ż −cα cα2 + 2mt θαβ

k −kα − cθβ uv 4 + z −kα kα2 + cθαβ − mt θ2 α2 + π2LEI 3

mv = −g mt α

c{u· v η· (vt)}

k{u η(vt)} mv g

k{u η(vt)} c{u· v η· (vt)} .. mt η

mtg

Problem 3.13

10 × (3.5)2 = 0.3170 lb.in.s2 386.4 lb.s2 10 = 0.02588 m= 386.4 in.

1 1 × EIf = × 30 × 106 = 39, 063 12 64 π × 12 × 106 = 1, 178, 100 GJr = 32 π × 30 × 106 = 1, 472, 620 EIr = 64 I0 =

Figure S3.11 ψ(x) = sin

πx , L

ub = z sin

πx L

Referring to the free body diagram for the sprung mass shown in Figure S3.11 mv üv + c(u̇v − η̇) + k(uv − η) = −mv g where η = zα η̇ = zθβ + żα η̈ = −zθ2 α + 2żθβ + z̈α and θ = (πv/L), α = sin(θt), β = cos(θt).

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

L

1

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 =

1 2 12 × × 12 × 12 × 2 EIf 3

362 12 2 1 × × 36 + × 36 × 12 + × 2 EIr 3 GJr = 0.02970 12 1 × 36 a22 = 1 × ×1+ ×1 EIf GJr = 3.3775 × 10−4 1 12 12 a12 = × × 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 ü 0 0.3170 θ̈

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 L m22 =

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

12EI L3

L2 (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

2EI L

6EI L2 (c)

6EI L2 2EI L

6EI L2

4EI L

6EI L2

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 Mü + 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

0

24EI l3

0

− 6EI l2

0

6EI  l2  

0

4EI l

2EI l

0

− 6EI l2

2EI l

8EI l

2EI l

0

0

2EI l

8EI l

6EI l2

0

0

2EI l

where l = L/3

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 =

EI l3



  0     0    2EI  l   4EI l

24 −12



4 EI  2 KBB =  0 l 0

EI KAB = 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 ü1 u1 + −8.4 9.6 ü2 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

v = l(1 − cos θ) 1 1 1 T = mu̇2 + mv̇ 2 + mx2 2 2 2 1 1 2 = m ẋ + l2 θ̇2 + 2l cos θẋθ̇ + M ẋ2 2 2 1 2 V = 2 × kx + l(1 − cos θ)mg 2

d d ∂T = (mẋ + ml cos θ θ̇) + M ẍ dt dt ∂ ẋ = mẍ − ml sin θ θ̇2 + ml cos θ θ̈ + M ẍ

θ

A ϕ

∂V = 2kx ∂x

d ∂T = ml2 θ̇ + ml cos θẋ dt ∂ θ̇ = ml2 θ̈ + ml cos θẍ − ml sin θẋθ̇ ∂V = mgl sin θ ∂θ

A x

Figure S4.1

M k

k

Kinetic energy

1 1M 2 2 M (R − r2 ) θ̇2 + r2 + r12 φ̇2 2 2 2 1 r22 + r12 1 2 2 2 (R − r2 ) θ̇2 = M (R − r2 ) θ̇ + M 4 r22 2

T =

d dt

∂T ∂ θ̇

M r22 + r12 = M (R − r2 ) θ̈ + (R − r2 ) θ̈ 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 ) θ̈ + 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)ẍ + ml cos θθ̈ − ml sin θθ̇2 + 2kx = 0 ml cos θẍ + ml2 θ̈ − ml sin θẋθ̇ + mgl sin θ = 0 For small x and θ, cos θ ≈ 1, sin θ ≈ θ and θ̇2 , ẋθ̇ are negligible. Hence the equations reduce to (M + m)ẍ + mlθ̈ + 2kx = 0 ẍ + lθ̈ + gθ = 0

18

Problem 4.3

1 πx 1 L dx + M u̇22 mu̇21 sin2 2 0 L 2 1 1 = mLu̇21 + M u̇22 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 = ü1 dt ∂ u̇1 2

d ∂T = M ü2 dt ∂ u̇2 ∂V π4 = EI 3 u1 − k (u2 − u1 ) ∂u1 2L ∂V = k (u2 − u1 ) ∂u2 The equations of motion are m

0 ü1 2 0 M ü2 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 (q̇1 − q̇2 a − q̇3 ) + cS (q̇1 + q̇2 a − q̇4 ) ∂ q̇1 ∂R = −cS a (q̇1 − q̇2 a − q̇3 ) + cS a (q̇1 + q̇2 a − q̇4 ) ∂ q̇2 ∂R = −cS (q̇1 − q̇2 a − q̇3 ) ∂ q̇3 ∂R = −cS (q̇1 + q̇2 a − q̇4 ) ∂ q̇4 Substitution in Lagrange’s equations gives    W q̈1 g W r2   q̈2       g w  q̈3  g w q̈4 g     2 0 −1 −1 q̇1 a −a   q̇2   0 2a + cS     −1 a 1 0 q̇3 −1 −a 0 1 q̇4   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 (q̇2 ) + (q̇3 ) + (q̇4 ) 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 (q̇1 − q̇2 a − q̇3 ) + cS (q̇1 + q̇2 a − q̇4 ) 2 2 Differentiation gives

d ∂T W = q̈1 dt ∂ q̇1 g

W r2 d ∂T = q̈2 dt ∂ q̇2 g

w d ∂T = q̈3 dt ∂ q̇3 g

w d ∂T = q̈4 dt ∂ q̇4 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 cos θ2 θ̇2 2 l2 ẏ = −l1 sin θ1 θ̇1 − sin θ2 θ̇2 2 The kinetic energy is obtained from ẋ = l1 cos θ1 θ̇1 +

1 1 1 mẋ2 + mẏ 2 + I0 θ̇22 2 2 2 Substitution for ẋ and ẏ gives

1 l22 2 2 2 T = m l1 θ̇1 + θ̇2 + l1 l2 θ̇1 θ̇2 cos (θ1 − θ2 ) 2 4 1 + I0 θ̇22 2 Potential energy is obtained from

l2 V = mg l1 + (1 − cos θ2 ) 2 T =

Differentiation gives

∂T l1 l2 2 = m l1 θ̇1 + θ̇2 cos (θ1 − θ2 ) 2 ∂ θ̇1

l1 l2 d ∂T = m l12 θ̈1 + θ̈2 cos (θ1 − θ2 ) dt ∂ θ̇1 2 !

l1 l2 θ̇2 sin (θ1 − θ2 ) θ̇1 − θ̇2 − 2 2

∂T l l1 l2 = m 2 θ̇2 + θ̇1 cos (θ1 − θ2 ) 4 2 ∂ θ̇2 d dt

∂T ∂ θ̇2

+ I0 θ̇2 2 l l1 l2 = m 2 θ̈2 + θ̈1 cos (θ1 − θ2 ) 4 2

l1 l2 − θ̇1 sin (θ1 − θ2 ) θ̇1 − θ̇2 2 + I0 θ̈2

∂V = mgl1 sin θ1 ∂θ1 ∂V mgl2 = sin θ2 ∂θ2 2 Substitution in Lagrange’s equation gives:

l1 l2 m l12 θ̈1 + cos (θ1 − θ2 ) θ̈2 2

l1 l2 sin (θ1 − θ2 ) θ̇1 θ̇2 − θ̇22 − 2 + mgl1 sin θ1 = 0 2 l l1 l2 (θ1 − θ2 ) θ̈1 − m 2 θ̈2 + 4 2

l1 l2 sin (θ1 − θ2 ) θ̇12 − θ̇1 θ̇2 2 mgl2 sin θ2 = 0 + I0 θ̈2 + 2

For small vibrations sin (θ1 − θ2 ) ≈ θ1 −θ2 , sin θ1 ≈ θ1 , sin θ2 ≈ θ2 , cos (θ1 − θ2 ) ≈ 1 and terms containing θ̇1 θ̇2 , θ̇12 , and θ̇22 are negligible. The equations reduce to ml1 l2 ml12 θ̈1 + θ̈2 + mgl1 θ1 = 0 2 2

ml2 ml22 mgl2 ml1 l2 θ̈2 + + θ̈1 + θ2 = 0 12 2 2 4 Canceling out m and substituting l1 = l2 = l

g

1 12 θ̈1 θ1 0 0 l + = g 1 1 0 θ 0 θ̈ 2 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 φθ̇ ẏ1 = − sin φ 2 dt 2 ẏ2 = −l sin φθ̇ l ẋ1 = cos φθ̇ 2 ẋ2 = l cos φθ̇

20 Kinetic energy

Hence

1 1 1 m2 ẏ22 + m1 ẏ12 + m1 ẋ21 2 2 2 2 1 M r2 ẋ2 1 m1 l 2 2 1 2 + θ̇ + M ẋ2 + 2 2 r 2 2 12 2 1 m1 l 1 = m2 l2 sin2 φθ̇2 + sin2 φθ̇2 2 4 2 1 m1 l 2 1 m1 l 2 2 cos2 φθ̇2 + + θ̇ 2 12 2 4 1M 2 1 + M l2 cos2 φθ̇2 + l cos2 φθ̇2 2 2 2

1 m1 l 2 1 1 3M 2 2 2 2 = m2 l sin φ + + l cos φ θ̇2 2 2 3 2 2

ẋ = u̇ + l cos θθ̇

T =

Differentiation gives

1 d ∂T = m2 l2 sin2 φ + m1 l2 3 dt ∂ θ̇

3M 2 l cos2 φ θ̈ + 2

d ∂T = M ü + mü + mθ̈l cos θ − mlθ̇2 sin θ dt ∂ u̇

d ∂T = ml2 θ̈ + mül cos θ − mu̇θ̇l sin θ dt ∂ θ̇ V = ku2 + mgl(1 − cos θ) ∂V = 2ku ∂u ∂V = mgl sin θ ∂θ The Lagrange’s equations are given by

(M + m)ü + ml cos θθ̈ − ml sin θθ̇2 + 2ku = 0 ml cos θü + ml2 θ̈ − ml sin θu̇θ̇ + mgl sin θ = 0

Potential energy V =

ẏ = l sin θθ̇ 1 1 1 T = M u̇2 + mẋ2 + mẏ 2 2 2 2

1 1 = M u̇2 + m u̇2 + l2 θ̇2 + 2u̇θ̇l cos θ 2 2

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 θ̈ 3 2

For small vibrations, θ̇2 and u̇θ̇ are negligible and cos θ ≈ 1, sin θ ≈ θ. The linearized equations become (M + m)ü + mlθ̈ + 2ku = 0 mü + mlθ̈ + 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 θ̇2 + I φ̇2 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

θ̈ +

10, 000 θ=0 2

67 θ̈ + 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 (R − r)2 θ̇δ θ̇ + Iφδ φ̇ 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 θ̇ (δθ) + I φ̇ (δφ)dt g dt dt t1 t2 W t2 2 (R − r) θ̇δθ |t1 − θ̈δθdt = g t1 t2 φ̈δφdt + I φ̇δφ |tt21 − t2 = t1

t1

−

W (R − r)2 θ̈δθ − I φ̈δφ dt g

because variations δθ and δφ vanish at t1 and t2 .

Substitution in δF gives

W δF = − (R − r)2 θ̈ − W (R − r) sin θ g

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

where ug = G sin Ωt. We have ∂T = ml2 θ̈ + mlu̇g sin θ ∂ θ̇

d ∂T = ml2 θ̈ + mlüg sin θ + mlüg cos θ θ̇ dt ∂ θ̇ ∂V = mgl sin θ dθ The equation of motion is

∂V d ∂T + =0 dt ∂ θ̇ dθ or θ̈ +

cos θ g u̇g θ̇ + l l

1+

üg 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)θ̈ + W sin θ = λ g

y = l(1 − cos θ) ẋ = u̇ + l cos θ θ̇

I φ̈ = −λr

ẏ = l sin θ θ̇ Solving these along with the constraint equation yields

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

W r2 + I (R − r)θ̈ + W r2 θ = 0 g Also, λ = − Irφ̈ 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 ẋ = l cos θ θ̇ ẏ = l sin θ θ̇ + u̇g The kinetic energy T and potential energy V are obtained from 2 1 2 1 T = m l cos θ θ̇ + m l sin θ θ̇ + üg 2 2 V = mg(l − l cos θ) + mgug

The kinetic energy is given by 2 1 M R2 u̇ 1 1 1 2 M (u̇) + + m(ẋ)2 + (ẏ)2 2 2 2 R 2 2 3 1 1 2 = M (u̇) + m(u̇)2 + ml2 (θ̇)2 + mu̇θ̇l cos θ 4 2 2

T =

We therefore have ∂T = ml2 θ̇ + mu̇l cos θ ∂ θ̇

d ∂T = ml2 θ̈ − mu̇l sin θ θ̇ + mlü cos θ dt ∂ θ̇ 3 ∂T = M u̇ + mu̇ + m cos θ θ̇ ∂ u̇ 2

3 d ∂T = M + m ü + ml cos θ θ̈ 2 dt ∂ u̇ − ml sin θ (θ̇)2 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 M + m ü + ml cos θ θ̈ − ml sin θ (θ̇)2 2 + ku + cu̇ = 0 ml θ̈ − mu̇l sin θ θ̇ + mlü cos θ + mgl sin θ = 0 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 u̇ + m u̇ + cos u̇ + 2u̇ R R 4 2 R " !

2 3 1 l u ∂T l = M u̇ + m 2u̇ + 2 u̇ + 4 u̇ cos ∂ u̇ 2 2 R R R ! " 2

d ∂T u 3 l 2l ü = M +m+ m + m cos R R R 2 dt ∂ u̇ −

2l u mu̇2 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 ü m + m cos R R

l u 2l u mu̇2 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 ü = −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 =

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 = 25 rad/s 3.2

Maximum displacement = ρ = 1.25 25 = 50 mm (b)

where t = t − 0.1. At end of the era, t = 0.1, and ut = 1.646 in.

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

1 − ξ 2 = 0.995,

u̇t = 10.4 sin ωt + 8.415 cos ωt = 13.3 in/s

ωd t = 1.4708

The damped frequency is given by ωd = ω

u̇t = ω sin ωt =

In the third era ut = 1.646 cos ωt1 + 1.33 sin ωt1 + 0.5(1 − cos ωt1 )

1 − ξ 2 = 24.875 rad/s

Hence t = 0.0591 s and

= 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 umax = 25 = 0.0431 m = 43.1 mm

ut = 2.238 in. u̇t = −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 müt + 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

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), u = 1 − cos 10t, 0 ≤ t ≤ 0.1 t

δ2 = ln(20/2.75),

ξ1 = 0.3441 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,

The velocity of mass m as it strikes the larger mass is given by

√ v̄ = 2gh = 2 × 9.81 × 1 = 4.43 m/s The velocity after impact is obtained from conservation of momentum (M + m)v0 = mm̄ 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

ω1 = 20.07,

k = m1 ω12 = 395, 125 N/m √ 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 ü, the damping force c(a/L)u̇ and the force due to buoyancy Aρgu. Taking moment about the hinge we get a2 M üL + c u̇ + AρguL = 0 or L a 2 M ü + c u̇ + 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 mü + ku = −µN With initial velocity of v0 the solution yields µN v0 sin ωt + (cos ωt − 1) ω k µN ω sin ωt u̇ = 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 u̇ = 0, hence tan ωt =

JM θ̈ + 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 MHθ̈

j0θ̈

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 θ̈ + 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

20.94 Substituting Ω ω = 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 αθ̈

1 mL2θ̈ 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 θ̈ + kL2 θ = 0 3 (

2 ) 1 1 × 40 + 132 θ̈ + 4330θ = 0 3 1.2

Exciting frequency Ω=

2000 × 2π = 209.44 rad/s 60

Frequency ratio β=

Ω 209.44 = = 1.783 ω 117.47

Damping ratio

105θ̈ + 4330θ = 0

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

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 θ̈ + 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

105θ̈ + 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 × 0.12ü + c × 0.08u̇+20 × 0.12 0.12 = −0.6 × 0.12üg

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

0.6ü + c

0.08 0.12

2 u̇ +

20 u = −0.6üg (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

or * + 2 2β (1 − β 2 )2 + η 2 − β2 (4β 3 − 4β) 3/2

{(1 − β 2 )2 + η 2 }

4.094 3.0985 sin(4πt − 0.396) + sin(5πt − 0.568) ω2 ω2

or

Problem 6.10

Substitution of this value in ρ gives

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

=0

β 2 = 1 + η2

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

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 p0 1

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

Table 6.10b

ρ=

Ω

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

β

φ, rad

4.5 1.5 6.0 2.0 9.0 3.0

2.108 2.390 2.658

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

(d)

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

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 12 = 10 kips/in, and ρ = 1.2 Substituting k = 1.2

in. η= u

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 u̇ = k

u=

0<t≤

π ω

At t = ωπ , u = 2P0 /k, u̇ = 0 For the next era of response u = A cos ωt + B sin ωt −

P0 k

Substituting P0 = 10 kips, k = 4π 2 kips/in, ω = √ 4π 2 = 2π rad/s, and t1 = 0.3 s

10 t sin 2πt u= − 4π 2 0.3 2π × 0.3 0 < t ≤ 0.3 = 0.8443t − 0.1344 sin 2πt For t > 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

P0 3 cos ωt − 1 k

10 k F (kips)

u=

t1

π At t = 2π ω , that is at t = ω

u=−

4P0 , k

u̇ = 0 0.3

0.4

u/(P0/k)

8

t(s)

4P0t t1

(t t1)

Figure S7.2 4

The response is therefore given by

0

u = 0.8443t − 0.1344 sin 2πt

40 (t − 0.3) sin 2π(t − 0.3) − − 2 4π 0.3 2π × 0.3

π

2π

3π

4π ωt

4

8

= 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. u̇ = 0.8822 in./s

Figure S7.1 In a similar manner it can be shown that at t = 3π , u = 6Pk 0 and u̇ = 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

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

u=

I = 2P0 t1

sin ωt P0 t − 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

v0 = Hence

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

After some algebraic manipulation this reduces to ω ωt1 4P0 sin (2t − 3t1 ) sin ωt1 sin u= 2 2 kωt1

2P0 t1 m

(a)

For t1 /T = 1/32, ωt1 = 2π/32 = 0.1963 and the exact value of the maximum response becomes 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 = 0.3926−0.3897 × 100 = 0.74%.

0.3897 t1

2t1

P0 t t1 t1

3t1

Time, t

Problem 7.5

P0 t 2t1 t1

Velocity u̇ is given by t u̇ =

Figure S7.3

0

0

For maximum value of u, du dt = 0 which yields 3 cos(ωtp − ωt1 ) = 0 2 3 π ωtp − ωt1 = 2 2 3 T tp = + t 1 4 2

  v δ(τ )dτ 

= v0

 

t≥0

Equation of motion is mü + cu̇ + ku = −müg = −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

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

u=

P0 (1 − cos ωt) k

P0 P0 (1 − cos ωt) − k k

1 ρ − ω sin(ωt + φ) ωt1 t1

=0

(f )

Equation f yields

Figure S7.6

u=

−

0 ≤ t ≤ t1

(a)

t − t1 sin ω(t − t1 ) − t1 ωt1

cos(ωt + φ) = ±

p0 umax = k

(b) 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

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 Equation b can be expressed as

(

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 t

t1 = 0.203 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 − + v0 = u̇(2t1 ) = 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

2t1

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 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

müg sin ω(t − τ )dτ mω

where damping is assumed to be negligible

P0 τ sin ω(t − τ )dτ mω t1 0

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 mü + cu̇ + ku = 0 or mü + cu̇ + ku = −müg = hmΩ2 sin Ωt At time t = 0, both ut and u̇t are zero; however, while ug = h sin Ωt is zero the velocity u̇g = hΩ cos Ω = hΩ. Thus, in terms of the relative displacement and velocity the initial conditions are u0 = u(0) = 0 and v0 = u̇(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

üt =

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 ütmax = 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 30

Relative,

0

0.2

0.25

40 50

0

0.05

0.1

0.15 Time, s

Figure S7.13

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 u̇ 1 T = M u̇2 + mu̇2 + mR2 × 2 2 R 2 2

3 1 M + m u̇2 = 2 2

u

V =

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

4k M + 3m 2

Problem 8.2 Referring to Problem 2.2, 1 = 33.67 lbs/in 0.0297 1 1 V = ku2 = × 33.67u2 2 2 1 10 2 T = u̇ 2 386.4 k=

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

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

# ω=

2k a2 g − 2 m l l

Problem 8.4

11 Wu 2 1 = (20 × 1.2 + 2 × 12 × 0.85) 2 = 22.2 N.m 1 1W 2 Tmax = ω 2 u 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 =

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 θ̇ 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

L Also, ω will be a minimum for sin πξ L = 1 or ξ = 2 .

$

x

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

Figure S8.6 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 =

"

L/2

b. The displacement at ξ is given by

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 ξ

L

EI 0.292 0.05725mL + 0.1475M L3

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

L

Problem 8.7

x2 5L x1 − 1 4 2

− (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

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 =

1 L 2 m1 m1 L3 ds = ×L× L= EI 2 EI 3 3EI

(L/2)3 L m2 m2 L L ds = a22 = × + × EI 2 EI 3EI 2 7L3 = 24EI

1 L m1 m2 L L3 ds = a12 = ×L× = 2 EI 2 4EI EI a11 =

B

1

u1

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 4

1 4

 

7 24



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  7 144 EI  24 K= 5 L3 − 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

Exact

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

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, u̇(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

Exact

0.0000 0.0004 0.1673 0.5089 0.9264 1.2757 1.4332 1.3355 1.0246 0.5979 0.1793

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

u̇n

ün

pn+1

un+1

u̇n+1

ün+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 un + u̇n + ün h h2

2 (a) un + u̇n +c h 2 u̇n+1 = −u̇n + (un+1 − un ) h 1 ün+1 = (pn+1 − cu̇n+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 + 8.237u̇n + 0.1ün u̇n+1 = −u̇n + 40 (un+1 − un ) ün+1 = 10pn+1 − 2.366u̇n+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

u̇n

ün

pn+1

un+1

u̇n+1

ün+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 + u̇n + 2ün h2 h

3 h (a) +c un + 2u̇n + ün h 2 h 3 u̇n+1 = (un+1 − un ) − 2u̇n − ün h 2 1 (pn+1 − cu̇n+1 − kun+1 ) ün+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 + 12.473u̇n + 0.2059ün u̇n+1 = 60 (un+1 − un ) − 2u̇n − 0.025ün ün+1 = 10pn+1 − 2.366u̇n+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, u̇0 = 0, and ü0 = 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 ü0 − hu̇0 = 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

u̇n mm/s

pn kN

un+1 mm

u̇n+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

u̇n mm/s

pn kN

un+1 mm

u̇n+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 <t< Ω Ω 0≤t<

The function is symmetric and periodic with period T0 = 2π Ω . The sine terms in the Fourier term expansion vanish because of symmetry, the coefficients of cosine terms are T /2 2 a0 = g(t)dt T0 0 T /2 h 2 tp0 Ω dt = = T0 0 π 2 T0 /2 tp0 Ω 2Ω cos nΩtdt an = 2 π 2π 0 2p0 = (cos nπ − 1) (nπ)2 The series expansion is

p0 4p0 cos 3Ωt cos 5Ωt + + ··· g(t) = − 2 cos Ωt + 2 π 9 25

The series expansion becomes

sin 2Ωt 2p0 sin Ωt − g(t) = π 2

sin 3Ωt sin 4Ωt + − + ··· 3 4 2p 0 sin nΩt = Σ(−1)n−1 nπ n = 1, 2, 3 . . .

The response is given by u(t) =

where βn = nΩ ω . Substitution of Ω = 1.5ω and ξ = 0.1 gives u(t) =

Problem 9.3 From Problem 9.1 1 4 1 u(t) = − 2 cos Ωt p0 /k 2 π 1 − β12 1 + cos 3Ωt 9(1 − β32 ) 1 cos 5Ωt + 25(1 − β52 )

where βn = nΩ ω .

Resonance takes place at βn = 1. Corresponding values of Ω are Ω = ω, 13 ω, 15 ω, . . ..

Problem 9.2 π tp0 Ω 0≤t≤ π Ω The forcing function is unsymmetric and periodic with period T0 = 2π/Ω. The cosine terms in the Fourier series expansion vanish; the coefficients of the sine term are given by

where p0 ≡ h and only the first three harmonic terms have been included in the response.

1.0 u(t)/(p0/k)

1 p0 4p0 cos Ωt − 2 2k π k 1 − β12 1 cos 3Ωt + 9(1 − β32 ) 1 + cos 5Ωt + · · · 25(1 − β52 )

∞ 2p0 1 (−1)n−1 π n=1 1 + 5.063n4 − 4.41n2 * (1 − 2.25n2 ) sin 1.5nωt

−0.3nω cos 1.5nωt}

The response is given by u(t) =

∞ 2p0 1 (−1)n−1 π n=1 (1 − βn2 )2 + (2ξβn )2 * + (1 − βn2 ) sin nΩt − 2ξβn cos nΩt

0.5

g(t) =

2Ω T0 /2 tp0 Ω sin nΩtdt bn = π 0 π 2p0 2p0 cos nπ = (−1)n−1 =− nπ nπ

π/2

π

3π/2

2π

5π/2

ωt

Figure S9.3 Substituting βn = nΩ/ω = 1.5n u(t) 1 = + 0.324 cos 1.5ωt p0 /k 2 + 0.00234 cos 4.5ωt + 0.00029 cos 7.5ωt

51 The response ratio is plotted as a function of ωt in Figure S9.3.

Problem 9.4

Hence I=

a a2 + Ω2

2a a2 + Ω2 With a sampling interval of 0.125 Hz the frequency function is constructed as shown in Table S9.5a.

G(Ω) =

The sampled time function is as shown in Table S9.4a. The Fourier transform of the sampled time function is presented in Table S9.4b and is real. The time function, obtained by taking the inverse transform of the frequency function of Table P9.4b, is identical to the original time function at the sample points.

Table S9.4a Time

g(t) A0

Time

g(t) A0

0.0 0.1 0.2 0.3 0.4

0.0 0.2 0.4 0.6 0.8

0.5 0.6 0.7 0.8 0.9

1.0 0.8 0.6 0.4 0.2

Table S9.5a n

f Hz

Ω = 2πf rad/s

2 F (Ω) = 1+Ω 2

0 1 2 3 4 5 6 7 8 9

0.000 0.125 0.250 0.375 0.500 0.625 0.750 0.875 1.000 1.125

0.000 0.785 1.571 2.356 3.142 3.927 4.712 5.498 6.283 7.068

2.0000 1.2370 0.5768 0.3053 0.1840 0.1218 0.1840 0.3053 0.5768 1.2370

∆t = 0.1, N = 10, T = 1.0, f = 1 Hz

Table S9.4b Frequency f Hz

G(f ) A0

Frequency f Hz

G(f ) A0

0 1 2 3 4

0.50000 -0.20944 0.00000 -0.03056 0.00000

5 6 7 8 9

-0.02000 0.00000 -0.03056 0.00000 -0.20944

Problem 9.5 Fourier transform is given by ∞ G(Ω) = 2 e−at cos Ωtdt 0

The integral is evaluated as follows

e−at sin Ωt I ≡ e−at cos Ωtdt = Ω −at e sin Ωt dt +a Ω e−at sin Ωt ae−at cos Ωt − = Ω2 Ω −at e cos Ωt − a2 dt Ω2 0 0∞ 0

∞ ∞ a2 00 e−at sin Ωt 00 ae−at 00 I 1+ 2 0 = 0 − Ω2 0 Ω Ω 0 0 0 a = 2 Ω

The folding frequency is 0.625 Hz and the samples beyond that frequency are selected to be symmetrical with respect to the folding frequency.

The inverse transform of the frequency function shown in Table S9.5a is compared with the original function in Table S9.5b.

Table S9.5b n

t s

Inverse Fourier Transform

Time Function

0 1 2 3 4 5 6 7 8 9

0.0 0.8 1.6 2.4 3.2 4.0 4.8 5.6 6.4 7.2

0.8410 0.4688 0.1966 0.0985 0.0460 0.0394 0.0460 0.0985 0.1966 0.4688

1.0000 0.4493 0.2019 0.0907 0.0408 0.0183 0.0082 0.0037 0.0017 0.0007

f = 0.125, T =

1 = 8 s, N = 10, ∆t = T /N = 0.8 f

The time functions obtained by taking the inverse transform is periodic with a period 8s. The function values for t > 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<0 0<t<3 3<t

h(t) = 0

A 1.54

Continuous convolution is given by

A 1.8885

t h(t − τ )g(τ )dτ

u(t) = g(t) ∗ h(t) = 0

For t = 0 to 1.5 t τ e−t+τ dτ = t − 1 + e−t u(t) =

A 1.6985

0

A 1.185

For t = 1.5 to 3 1.5 u(t) =

τ e−t+τ dτ +

0

t

(3 − τ )e−t+τ dτ

1.5

= −2e1.5−t + e−t + 4 − t For t = 3 to 4.5 1.5 u(t) =

−t+τ

τe

3 dτ +

t−3

A 0.704

(3 − τ )e−t+τ dτ

A 0.3945

1.5

= −2e1.5−t + e3−t + (4 − t)e−3

3 2

g(T )

1

For t = 4.5 to 6 3 u(t) =

(3 − τ )e−t+τ dτ

t−3

= e3−t + (t − 7)e−3

A 0.187

The discrete convolution values obtained with a sampling interval of 0.5s are shown in Figure S9.6.

A comparison of the continuous and discrete convolution is presented in Table S9.6.

A 0.0125

Figure S9.6

A 0.066

53

Table S9.6

t, s

u(t) Continuous Discrete Convolution Convolution

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0

0.0000 0.1065 0.3680 0.7230 0.9225 0.8465 0.6035 0.3610 0.2035 0.0985 0.0360 0.0075 0.0000

0.0000 0.1250 0.4015 0.7700 0.9445 0.8495 0.5925 0.3520 0.1975 0.0935 0.0330 0.0065 0.0000

The discrete convolution values are in error because of the use of trapezoidal method to approximate integration. The accuracy can be improved by selecting a smaller sampling interval.

Problem 9.7 Table S9.7 t, s

Frequency convolution N = 25 N = 21

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0

0.0000 0.1737 0.5014 0.9225 1.0529 0.9096 0.6004 0.3617 0.2084 0.1067 0.0413 0.0102 0.0000

0.0225 0.1768 0.5014 0.9225 1.0529 0.9096 0.6004 0.3617 0.2084 0.1067 0.0413 – –

Use a sampling interval of ∆t = 0.25 s.

No. of nonzero samples in g(t), p = 12 No. of nonzero samples in h(t), q = 12 Minimum value of T0 = N ∆t, where N = p + q + 1 = 25 The time functions are augmented with sufficient number of zeros and convolved through frequency

domain. The results are presented in Table S9.7.

The frequency analysis is repeated with N = 21 and the new results are also shown in Table S9.7. They are in error over the first second due to overlapping.

Problem 9.8

p0 = 200 kN k = 5000 kN/m # 5000 × 1000 ω= 24000 = 14.43 rad/s 2π Ω= = 6.283 rad/s T Ω = 0.4353 β= ω p0 1 u= (sin Ωt − β sin ωt) k 1 − β2 = 0.0494(sin 6.283t − 0.4353 sin 14.43t) m (a) 0 < t < 1/2 s At t = 0.5s u = 0.0494 × −0.4353 sin(14.43 × 0.5) = −0.0173 m u̇ = 0.0494(6.283 cos 6.283t − 0.4353 × 14.43 cos 14.43t) = −0.4945 m/s Response in the free vibration area u = −0.0173 cos 14.43(t − 0.5) 0.4945 sin 14.43(t − 0.5) − 14.43 = −0.0173 cos 14.43(t − 0.5) − 0.0343 sin 14.43(t − 0.5) m t > 1/2 s The forcing function in kN is given by ( g(t) =

200 sin 2πt 0

0 < t < 0.5 s t > 0.5 s

The unit impulse function is 1 sin ωt mω 1 = sin 14.43t 24 × 14.43

h(t) =

where m is in metric tonnes.

0 < t ≤ 1.0 s

(b)

54 With ∆t = 0.05, the nonzero portion of g(t) lasts for p∆t = 0.5 s so that p = 10. Since the response is required for 1 s, the unit impulse function is truncated at q∆t = 1.0 giving q = 20. To avoid overlapping, the period of time functions should be selected as N ≥ p + q + 1 = 31. A value of N = 32 is selected and the time functions g(t) and h(t) are augmented with sufficient number of zeros. The augmented functions are convolved through the frequency domain using a fast Fourier subroutine. The resulting response is shown in Table S9.8 at intervals of 0.15. For comparison, the exact response obtained from Equations a and b is also shown. Frequency response beyond t = 1 s is not correct and is not shown.

Table S9.8

other parameters are as in Problem 9.8. e−ξωt p0 k (1 − β 2 )2 + (2ξβ)2

Ω 2 2 2ξ − 1 + β sin ωd t + 2ξβ cos ωd t ωd

u̇ = −ξω

+

e−ξωt p0 k (1 − β 2 )2 + (2ξβ)2

2 3 Ω(2ξ 2 − 1 + β 2 ) cos ωd t − 2ξβωd sin ωd t 3 p0 e−ξωt Ω 2 + (1 − β 2 ) cos Ωt + 2ξβ sin Ωt k (1 − β 2 )2 0 < t ≤ 0.5 (b) The displacement and velocity at t = 0.5 are obtained from Equations a and b respectively. u0 = u(0.5) = −0.00108 m v0 = u̇(0.5) = −0.41662 m/s

n

0 2 4 6 8 10 12 14 16 18 20

t, s

Displacement, mm Frequency Exact value Analysis Equations a and b 0.00 5.89 39.65 65.92 38.84 -17.89 -37.49 8.36 39.62 1.70 -39.18

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

0.00 7.70 41.52 66.87 39.49 -17.27 -36.18 8.07 38.23 1.64 -37.81

The response in the free vibration era is given by

u = e−ξω(t−0.5) u0 cos ωd (t − 0.5) t > 0.5 v0 + ωξu0 + sin ωd (t − 0.5) ωd The unit impulse function is given by h(t) =

1 −ξωt e sin ωd t mωd

The details of frequency analysis are similar to those in Problem 9.8. The results are shown in Table S9.9. The exact response obtained from Equations a and c is also shown for comparison.

Table S9.9 Problem 9.9 The closed form solution is given by

u=

p0 e−ξωt 2 k (1 − β )2 + (2ξβ)2

Ω 2 2ξ − 1 + β 2 sin ωd t + 2ξβ cos ωd t ωd 1 p0 + k (1 − β 2 )2 + (2ξβ)2 2 3 (1 − β 2 ) sin Ωt − 2ξβ cos Ωt 0 < t ≤ 0.5

where ξ = 0.1, ωd = ω

(a) 1 − ξ 2 = 14.36 and the

(c)

n,

ts

0 2 4 6 8 10 12 14 16 18 20

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

Displacement, mm Frequency Exact value Analysis Equations a and b 0.00 5.49 35.02 57.89 40.62 -1.38 -26.01 -5.01 18.32 8.01 -11.87

0.00 7.19 36.85 59.10 41.44 -1.08 -25.10 -5.03 17.64 7.86 -11.39

55

Problem 9.10 #

Table S9.10

T = 2π

ωd = ω 1 − ξ 2 = 6.2517 20 × 1000 = 506.61 kg m= 4π 2 = 5.0661 t Unit impulse function as given by h(t) =

1 −ξωt e sin ωd t mωd

In the first round h(t) is truncated at 0.9s. Hence p = 6, q = 9 and N = p + q + 1 = 16. The results of fast convolution with N = 16 are shown in Table S9.10. These results are correct for k = 0 to 9. g(t)

t

k m = 1.5, ω = 2π, ξ = 0.1 k

p 6

500

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Convol. with first section of h(t)

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6

Convol. with second section of h(t)

0.0000 0.0000 4.3383 15.2835 28.1268 38.0340 37.4575 23.5039 2.7614 -16.5227 -27.5989 -27.3281 -19.4290 -10.0520 -2.7375 0.0000 –

Resultant response mm 0.0000 0.0000 4.3383 15.2835 28.1268 38.0340 37.4575 23.5039 2.7614 -16.5227 -27.5989 -27.4607 -17.4819 -2.4065 11.7525 20.0210 20.1267

0.0000 -0.1326 1.9471 7.6455 14.4900 20.0210 20.1267

250 2 h(t)

4

6

8

10

12

14

16

q 9

Problem 9.11 g(t)

0.2

500

0

250

0.2 h(t) 0.2

2 q 6

4

6

8

10

12

14

16

N 16 h(t) 0.2

0 0 0.2 N 13

Figure S9.10 In the second round h(t) values for 1.0 to 1.6s are evaluated and moved to the origin. Now p = 6, q = 6 and N = 13. Fast convolution values for k = 0 to 6 are superposed on those obtained from the previous round as shown in Table S9.10 which also shows the resultant values.

0.2 N 17

Figure S9.11 To obtain response values for 1.6 s, select N = 17 with ∆t = 0.1, T0 = 1.7 s.

The periodic unit impulse function is obtained from Equation 9.79 and convolved with g(t) through the frequency domain. The resulting response is shown in Table S9.11 which indicates u(0) = 14.052 mm. From Equation 9.85 the velocity works out to u̇(0) = −65.947 mm/s.

The corrective responses are obtained from Equations 9.89a and b with ∆u(0) = −14.052 mm and ∆u̇(0) = 65.947. The corrected response obtained by the superposition of periodic response and the corrective response is also shown in Table S9.11.

56

Table S9.11

k

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

t, s

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6

Table S9.12

Periodic Response mm

Corrected Response mm

14.052 5.914 0.953 4.912 15.316 27.671 32.973 25.813 10.233 -7.180 -19.960 -24.066 -19.049 -7.787 4.940 14.391 17.560

0.000 0.235 4.696 15.621 28.326 38.039 37.290 23.244 2.513 -16.671 -27.606 -27.341 -17.293 -2.225 11.863 20.629 20.041

k

t, s

r(t), mm

s(t), mm

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6

0.7994 0.7833 0.4879 0.0523 -0.3507 -0.5802 -0.5743 -0.3630 -0.0464 0.2495 0.4209 0.4210 0.2699 0.0398 -0.1774 -0.3053 -0.3085

-0.03760 0.04370 0.09975 0.11334 0.08465 0.02895 -0.03055 -0.07207 -0.08281 -0.06256 -0.02225 0.02129 0.05205 0.06049 0.04622 0.01705 -0.01480

Problem 9.13 Problem 9.12

g(t)

As in Problem 9.11, T0 = 1.7s. The convolution of the sampled forcing function and h(t) gives the periodic response, which is of course identical to that obtained in Problem 9.11.

The periodic corrective response corresponding to a unit displacement change, r(t) is obtained from Equation 9.107. The periodic corrective response corresponding to a unit velocity change, s(t) is obtained from Equation 9.108. Both r(t) and s(t) are shown in Table S9.12. Now

500

r(0) = 0.79940 s(0) = −0.03757 ṙ(0) = −ω 2 s(0) = 1.48336 ṡ(0) = r(0) − 2ξωs(0) = 0.84661 Also, as in Problem 9.11, ∆u(0) = −14.052 mm ∆u̇(0) = 65.947 mm/s. The weighting factors a and b are obtained from Equation 9.113, which gives a = 12.8584,

b = −100.4237

R2 R1

250 2

4

6

8

10

12

14

16

N 17

Figure S9.13 As in Problem 9.11, N = 17, ∆t = 0.1, T0 = 1.7.

The periodic response to g(t) is obtained by convolving g(t) and h(t) as in Problem 9.11. The initial displacement and velocity in the periodic response are u(0) = 14.052 mm and u̇(0) = −65.947 mm/s. As shown in Figure S9.13, periodic force pulses R1 and R2 are applied at t = 1.5 and t = 1.6 s, respectively. The corrective displacement and velocity produced by R1 are u1 = h(2∆t) = 0.19690 ˙ u̇ (0) = h(2∆t) = 0.71569 1

u2 = h(∆t) = 0.8634 ˙ = 1.43765 u̇ (0) = h(∆t) 2

The corrected response is now obtained from Equation 9.115 and is identical to that calculated in Problem 9.11.

in which h˙ is given by Equation 9.132. The unknown pulses R1 and R2 are obtained from

−14.052 u1 (0) u2 (0) R1 = 65.947 u̇1 (0) u̇2 (0) R2

57 which gives R1 = 117.027, R2 = −104.129. The corrected response is obtained from Equation 9.130 for k = 0 to 14 and is identical to that calculated in Problem 9.11.

Problem 9.14

Problem 9.15 The solution is similar to Problem 9.11 except that in Equation 9.89, ∆u(0) = (20.0 − 14.052) = 5.948 mm and ∆u̇(0) = 100.0 + 65.947 = 165.947 mm/s. The final response is shown in Table S9.15.

Table S9.15

From Problem 9.11, u(0) = 14.052 u(∆t) = 5.914 u(2∆t) = 0.953 The equation to a parabola through these points is u(t) = a0 + a1 t + a2 t2

(b)

Substituting Equations a into b a0 = 14.052 a0 + 0.1a1 + 0.01a2 = 5.914 a0 + 0.2a1 + 0.04a2 = 0.953 Solution of Equations c gives, a0 = 14.052, a1 = −97.261 and a2 = −158.85. From Equation b u̇(0) = a1 = −97.261 mm/s as compared to the more precise value of -65.947 mm/s.

The corrected response is obtained from Equations 9.89 and 9.90 with ∆u(0) = −14.052 mm and ∆u̇(0) = 97.261 mm/s. The results are shown in Table S9.14. The initial velocity estimate and hence the resultant response are in significant error in the present case.

Table S9.14 k

t, s

Corrected response

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6

0.000 2.988 8.889 19.579 30.655 38.097 35.323 20.198 -0.391 -18.408 -27.690 -25.937 -15.081 -0.094 13.158 20.121 19.040

k

t, s

Corrected response

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6

20.000 25.361 25.323 24.877 24.231 23.640 18.971 8.044 -4.479 -13.917 -17.244 -13.989 -6.097 3.052 10.024 12.576 10.312

Problem 9.16 We sample the functions h(t) and g(t) at intervals of 0.05 s and determine the response over 1 s, which therefore is also the period T0 . Parameter a for the exponential window method can be selected from the following expression. a=

p ln(10) = 2.303m T0

The value of a depends on the selection of p. For p lying between 2 and 3, a lies between 4.6 and 6.9.

We select a = 6. The unit impulse function is h(t) =

1 sin ωt mω

The scaled functions ĝ(t) and ĥ(t) are obtained from ĝ(t) = e−at g(t) ĥ(t) = e−at h(t) The scaled functions are convolved through frequency domain using a Fast Fourier Transform algorithm to obtain the scaled response û(t). The true response is obtained from. u(t) = eat û(t)

58 The results are presented in Table S9.16. For comparison, the exact value of the response obtained from the expressions given in the solution to Problem 9.8 are also shown.

Table S9.16 n

t, s

0 2 4 6 8 10 12 14 16 18 20

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

Displacement, mm Frequency Exact value Analysis -0.06 5.91 39.74 65.99 38.85 -17.89 -37.49 8.37 39.62 1.70 -39.18

0.00 7.70 41.52 66.87 39.49 -17.27 -36.18 8.07 38.23 1.64 -37.81

Problem 9.17 A procedure similar to that used in the solution to Problem 9.17 is used except that the damping ratio ξ = 0.1 and the unit impulse function is given by 1 −ξωt h(t) = e sin ωd t mωd

where ωd = ω 1 − ξ 2 is the damped frequency.

The response results are presented in Table S9.17. For comparison, the exact response obtained from the expressions given in the solution to Problem 9.9 are also shown.

Table S9.17 n,

ts

0 2 4 6 8 10 12 14 16 18 20

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

Displacement, mm Frequency Exact value Analysis Equations a and b =-0.02 5.48 35.04 57.91 40.63 -1.38 -26.01 -5.01 18.32 8.01 -11.87

0.00 7.19 36.85 59.10 41.44 -1.08 -25.10 -5.03 17.64 7.86 -11.39

59

Chapter 10

Equations in the normal coordinates are ÿ1 + (24.7)2 y1 = 0

Problem 10.1 Referring to Problem 8.8 the equations of motion are

144 EI 7/24 −1/4 u1 m ü1 + 3 −1/4 1/3 ü2 u m 5 L 2

0 = 0 (a) The characteristic is given by 0 7 0 0 − λ −1 0 0 24 1 1 4 0 = 1 (b) 0 − 0 4 3 −λ 3

Initial conditions in the normal coordinates are y10 =

Selecting q11 = 1 yields q21 = 0.92. Hence

1.000 q1 = 0.920 In a similar manner with λ2 = 0.5634

1.000 q2 = −1.087 EI 5 mL3 Substituting for E, I, L and m 144 ω12 = 0.06163 ×

ω1 = 24.7 rad/s Similarly 144 EI = 5580 5 mL3 ω2 = 74.7 rad/s

ω22 = 0.5634 ×

The modal matrix is given by

1.000 1.000 Q= 0.920 −1.087 A normal coordinate transformation gives

1.846 0 M̃ = QT MQ = m 0 2.181

144 EI 0.1139 0 K̃ = QT KQ = 0 1.2330 5 L3

qT1 Mu0 qT1 Mq1

where uT0 = [ [1 0 ].

14.6 = 0.542 14.6 × 1.846 qT Mu0 y20 = 2T q2 Mq2 14.6 = 0.458 = 14.6 × 2.181

y10 =

2

5 mL ω where λ = 144 EI . Expansion of the determinant gives 5 15 λ+ =0 (c) λ2 = 24 144 On solving Equation c λ1 = 0.0616, λ2 = 0.5634.

For λ1 = 0.0616.

1 7 − 0.0616 q11 − q21 = 0 24 4

Also

ÿ2 + (74.7)2 y2 = 0

Solution of the equations of motion in normal coordinates gives y1 = y10 cos ω1 t = 0.542 cos 24.7t y2 = y20 cos ω2 t = 0.458 cos 74.7t Transforming to physical coordinates

u=

1.00 1.000 0.92 −1.087

0.542 cos 24.7t 0.458 cos 74.8t

or u1 = 0.542 cos 24.7t + 0.458 cos 74.8t in u2 = 0.500 cos 24.7t − 0.500 cos 74.8t in

Problem 10.2 I0 = 1600 × 1.22 = 2304 kg.m2

1600 0 M= 0 2304

2k k(b − a) K= k(b − a) k(a2 + b2 )

70.00 12.25 = 1000 12.25 164.94 The characteristic equation is 0 0 0 70 − 16λ 0 12.25 0 0 0 12.25 164.94 − 23.04λ 0 = 0 2

where λ = ω10 .

60 u1

Expansion of the determinant gives the characteristic equation

u2

A

λ − 11.53λ + 30.91 = 0 2

k

On solving for λ λ2 = 7.297

Figure S10.3

Substituting λ = 7.297

Expansion of the determinant gives

(70 − 16 × 7.297)q12 + 12.25q22 = 0

λ2 − 3.06λ + 1.2656 = 0

Selecting q12 = 1 yields q22 = 3.82 √ ω2 = 7.297 × 10 = 8.542 rad/s T2 = 2π/ω2 = 0.73 s

On solving for λ λ1 = 0.4924, ω1 = 13.79 λ2 = 2.570, ω2 = 31.51

For the second mode to be excited the vehicle should travel 16 m in 0.7355 s. Hence the vehicle speed is

Substitution of λ1 and λ2 , in turn in the frequency equation leads to

16 × 3600 = 78.3 km/hr 0.7355

In order not to experience any acceleration, the passenger should be seated at the node of the second mode which is 1/3.82 = 0.262 m to the rear of the center of gravity.

1.0000 q1 = −0.0151

1.00 q2 = 4.14

M̃ = QT MQ =

M, r

k

K̃ = QT KQ =

k b

a

3.82 0.262

The characteristic equation is 0 0 0 50 − 100λ 0 50 0 0=0 0 50 4100 − 1600λ 0 where λ = ω 2 /386.4.

49.43 0 0 70736.0

ÿ1 + (13.79)2 y1 = 0

1

ÿ2 + (31.51)2 y2 = 0 (a) In this case, u1 = −1 and u2 = 0. Hence y10 =

qT1 Mu0 0.2588 =− = −0.9965 0.2597 qT1 Mq1

y20 =

qT2 Mu0 0.2588 =− = −0.00363 T 71.23 q2 Mq2

Problem 10.3

1 100 0 386.4 0 1600

2k k(b − a) K= k(b − a) k(a2 + b2 )

50 50 = 50 4100

0.2597 0 0 71.23

Equations in the normal coordinates are

Figure S10.2

M=

Transforming to normal coordinates

u1 u2

B k

b

a

λ1 = 4.236,

V =

M, r

The response in the normal coordinates is given by y1 = −0.9965 cos 13.79t y2 = −0.00363 cos 31.51t Transformation to physical coordinates gives

1.0000 1.00 u= −0.0151 4.14

y1 y2

u1 = −0.9965 cos 13.79t − 0.00363 cos 31.51t u2 = 0.01505 cos 13.79t − 0.01503 cos 31.51t

61 (b) In this case, u10 = −1/9 and u20 = 1/9.

0.0357 = −0.1375 0.2597 1.876 = 0.0263 y20 = 71.23

where λ = ω 2 /1000.

Expansion of the determinant gives

y10 = −

λ2 − 1.825λ + 0.50 = 0 Solving for λ λ1 = 0.3357, λ2 = 1.4893,

Response in normal coordinates is given by y1 = −0.1375 cos 13.79t y2 = 0.0263 cos 31.51t Response in physical coordinates is obtained from u = Qy which yields u1 = −0.1375 cos 13.79t + 0.0263 cos 31.51t u2 = 0.00208 cos 13.79t + 0.1089 cos 31.51t

0.2231 = −0.8590 0.2597 2.1348 = −0.0300 y20 = − 71.23 y1 = −08590 cos 13.79t y2 = −0.0300 cos 31.51t

The response in physical coordinates is given by

The modal masses are qT1 Mq1 = 19.93 qT2 Mq2 = 6.674 The mass orthonormal mode shapes are

1 0.2240 q1 = φ1 = √ 0.4838 19.93

1 0.3871 q2 = φ2 = √ −0.2799 6.674 Initial conditions in the normal coordinates are given by y0 = 0

u = Qy

ẏ0 = ΦT Mv0

0.2240 0.4838 5 0.0 = 0.3877 −0.2799 0 3.2

0.00 × −1.25

−1.9352 = 1.1196

or u1 = −0.8590 cos 13.79t − 0.0300 cos 31.51t u2 = 0.0130 cos 13.79t − 0.1242 cos 31.51t

Problem 10.4 If v is the velocity of 3 kg mass on impact, then because momentum is conserved

ω2 = 38.58

Substitution of λ1 and λ2 , in turn, in the frequency equation gives the mode shapes

1.00 q1 = 2.16

1.0000 q2 = −0.7232

1 8 (c) In this case, u10 = − 2×8 18 = − 9 and u20 = − 9

y10 = −

ω1 = 18.32

Response in the normal coordinates is given by

ẏ0i sin ωi t ωi 1.9352 y1 = − sin 18.32t = −0.1056 sin 18.32t 18.32 1.1196 sin 38.58t = 0.02902 sin 38.58t y2 = 38.58

The characteristic equation is

6 − 5λ −2 =0 −2 2 − 3.2λ

Transforming to physical coordinates

0.2240 0.3871 u= 0.4838 −0.2799

−0.1056 sin 18.32t × 0.02902 sin 38.58t

−0.02366 sin 18.32t + 0.01123 sin 38.58t = m −0.0511 sin 18.32t − 0.00812 sin 38.58t

3.2v = 0.2 × 20 v = 1.25 m/s After impact

5 0 M= kg.

0 3.2

6 −2 K = 1000 N/m −2 2

yi =

62

Problem 10.5

Problem 10.6 By substituting for q1 and M, the normalized mass in the first mode works out to

Damping matrix

C=

48 −16 −16 16

qT1 Mq1 = 1.214

Since the damping matrix is proportional to the stiffness matrix, normal coordinate transformation will diagonalize it and the transformed equations will be uncoupled. The transformed damping matrix is given by

Mass orthonormal mode shape is given by   0.9077 1 φ1 = √ q1 =  2.0695  1.214 2.8682 In a similar manner

C̃ = ΦT CΦ

 2.3819 φ2 =  1.8293  −2.0710   2.6153 φ3 =  −2.3851  0.8970

Substitution gives

C̃ =

2.686 0 0 11.91

2ξ1 ω1 = 2.686 2.686 ξ1 = = 0.0733 2 × 18.32

Initial condition in the normal coordinates are y0 = ΦT Mu0

In a similar manner ξ2 =

11.91 = 0.1543 2 × 38.59

Response in the first normal coordinate is given by y1 =

y10 −ξ1 ω1 t e sin ωd1 t ωd1

Since uT0 = [ 0.00 0.75 0.00 ]   0.1164 y0 =  0.1029  −0.1342 Response in the normal coordinates is

where ωd1 = ω1 1 − ξ12 = 18.273.

Substitution gives y1 = −



y1 = 0.1164 cos ω1 t y2 = 0.1029 cos ω2 t

1.9352 −1.343t e sin 18.273t 18.273

In a similar manner % ωd2 = ω2 1 − ξ22 = 38.129 1.1196 −5.955t y2 = sin 38.129t e 38.124 Transformation to physical coordinates yields

0.2240 0.3871 u= 0.4838 −0.2799

−0.1059e−1.343t sin 18.273t × 0.0294e−5.955t sin 38.129t

y3 = −0.1342 cos ω3 t Transformation to physical coordinates gives u = Φy or u1 = 0.1056 cos ω1 t + 0.2451 cos ω2 t − 0.3510 cos ω3 t m u2 = 0.2409 cos ω1 t + 0.1882 cos ω2 t + 0.3201 cos ω3 t m u3 = 0.3339 cos ω2 t + 0.2137 cos ω2 t − 0.1204 cos ω3 t m where ω1 = 10.6, ω2 = 33.7, ω3 = 59.2.

which yields u1 = −0.02372e−1.343t sin 18.273t −5.955t

+ 0.01138e −1.343t

u2 = −0.05123e

sin 18.273t

−5.955t

− 0.0082e

sin 38.129t

sin 38.129t

Problem 10.7

ωd1 = 10.6 1 − 0.12 = 10.55 rad/s

ωd2 = 33.7 1 − 0.12 = 33.53 rad/s

ωd3 = 59.2 1 − 0.12 = 58.90 rad/s

63 Response in the normal coordinates is obtained as follows

ω1 ξ1 y0 −ξ1 ω1 t y1 = e y10 cos ωd1 t + sin ωd1 t ωd1 = e−1.06t (0.1164 cos 10.55t + 0.0117 sin 10.55t)

which gives 

 0.9077 2.3819 2.6153 Φ =  2.0695 1.8293 −2.3851  2.8682 −2.0770 0.8970 The vector of initial velocities is

In a similar manner

v0T = [ 5 10 5 ] in./s

y2 = e−3.37t (0.1029 cos 33.53t + 0.0103 sin 33.53t) y3 = e−5.92t (−0.1342 cos 58.90t −0.0135 sin 58.90t) Transformation to physical coordinates gives u1 = e−1.06t (0.1056 cos 10.55t + 0.0106 sin 10.55t)

The initial velocities in modal coordinates are given by   0.3384 ẏ0 = ΦT Mv0 =  0.6792  in./s 0.7044 The response in modal coordinates is obtained from

+ e−3.37t (0.2451 cos 33.53t + 0.0245 sin 33.53t) −5.92t

+e

−1.06t

u2 = e

−3.37t

+e

−5.92t

+e

−1.06t

u3 = e

−3.37t

+e

(−0.3510 cos 58.90t −0.0353 sin 58.90t) (0.2409 cos 10.55t + 0.0242 sin 10.55t) (0.1882 cos 33.53t + 0.0188 sin 33.53t) (0.3201 cos 58.90t + 0.0322 sin 58.90t) (0.3339 cos 10.55t + 0.0336 sin 10.55t) (−0.2137 cos 33.53t −0.0214 sin 33.53t)

+ e−5.92t (−0.1204 cos 58.90t −0.0121 sin 58.90t)

yi =

ẏ0i sin ωi t ωi

where ωi is the ith frequency. Thus we have   0.2800 sin 10.6t y =  0.0441 sin 33.7t  −0.0080 sin 59.2t The response in physical coordinates is obtained from u = Φy =   0.254 sin 10.6t + 0.105 sin 33.7t − 0.021 sin 59.2t  0.580 sin 10.6t + 0.081 sin 33.7t + 0.019 sin 59.2t  0.803 sin 10.6t − 0.092 sin 33.7t − 0.007 sin 59.2t

Problem 10.8 The mass matrix is   0.075 0 0 M =  0 0.075 0  kip.s2 /in.

0 0 0.075

Problem 10.9 The Rayleigh damping matrix is given by C = (a0 M + a1 K) where a0 and a1 are obtained from

The modal matrix is   1.000 1.000 1.000 Q =  2.280 0.768 −0.912  3.160 −0.872 0.343 To mass orthonormalize the mode shapes we obtain the scale factors to be applied to Q s1 = qT1 Mq1 = 1.2138 s2 = qT2 Mq2 = 0.1763 s3 = qT3 Mq3 = 0.1462 The normalized mode shapes are obtained from qi φi = √ si

a0 + a1 ω12 = 2ξ1 ω1 a0 + a1 ω32 = 2ξ3 ω3 or a0 + 112.4a1 = 1.06 a0 + 3504.6a1 = 5.92 Solving for a0 and a1 , a0 = 0.8990 a1 = 0.001433 Substitution of these values in Equation a gives   0.31310 −0.1347 0.02335 C =  −0.13470 0.2627 −0.08980  0.02335 −0.0898 0.13700

(a)

64 Also

The required damping matrix is obtained from

a0 + a1 ω22 = 2ξ2 ω2 a0 a1 ω2 ξ2 = + 2ω2 2 = 0.0375

C = MΦĈΦT M where

Problem 10.10 Equation 10.151 gives



 0.777 0 0 0 0  0 1.898 0 0 0    Ĉ =  0 0 0 0 0   0 0 0 0 0 0 0 0 0 0

Using the given modal matrix we get T

C = MΦĈΦ M 

where

M = 0.075 



1



1 1

and from Problem 10.6   0.9077 2.3819 2.6153 Φ =  2.0695 1.8293 −2.3851  2.8682 −2.0770 0.8970 Also

 Ĉ = 



2ξ1 ω1



2ξ2 ω2





2ξ3 ω3

1.06 0 0 = 0 0 0  0 0 5.92 Substitution leads to   0.2327 −0.1965 0.0937 C =  −0.1965 0.2150 −0.0359  0.0937 −0.0359 0.0758 To verify that the damping in the second modes is zero, we note that c2 = φT2 Cφ2 Substitution gives c2 = 0.

Problem 10.11 The mass matrix is 

 2 0 0 0 0 0 2 0 0 0   M = 0 0 2 0 0   0 0 0 2 0 0 0 0 0 1 For the first and second mode shapes 2ξ1 ω1 = 2 × 0.05 × 7.77 = 0.777 2ξ2 ω2 = 2 × 0.05 × 18.98 = 1.898

C=   0.367 0.652 0.583 0.033 −0.458  0.652 1.167 1.062 0.127 −0.752     0.583 1.062 1.023 0.301 −0.502    0.033 0.127 0.301 0.632 0.526 −0.458 −0.752 −0.502 0.526 1.080 The damping constants for the 3rd, 4th and 5th modes are given, respectively, by φT3 Cφ3 , φT4 Cφ4 and φT5 Cφ5 . Substitution shows that each of them is zero.

Problem 10.12 The stiffness matrix is given by K=   2800 −1200 0 0 0 0   −1200 2400 −1200 0   −1200 2000 −800 0   0   0 0 −800 1200 −400 0 0 0 −400 400 Stiffness proportional damping C = αbf K will produce 1% damping in the 5th mode when α=

2 × 0.01 2ξ5 = = 0.00043751 ω5 45.71

The damping produced in the other modes is given by ξj = ξi (ωj /ωi ), so that 7.77 = 0.0017 45.71 18.98 ξ2 = 0.01 × = 0.0042 45.71 27.66 ξ3 = 0.01 × = 0.00605 45.71 36.390 ξ4 = 0.01 × = 0.00796 45.71

ξ1 = 0.01 ×

To build up the damping to 5% in the first and second modes, we need to increase the damping ratios in these modes by ξ¯1 = 0.05 − 0.0017 = 0.0483 and ξ¯2 = 0.05 − 0.0042 = 0.0458. Thus we have 2ω1 ξ¯1 = 0.7506 2ω2 ξ¯2 = 1.7386

65 which gives  0.7506 0 0 0 0 1.7386 0 0 0   0   Ĉ =  0 0 0 0 0   0 0 0 0 0 0 0 0 0 0 

The damping matrix is now given by C = αK + MΦĈΦT M =   1.563 0.077 0.539 0.037 −0.416  0.077 2.128 0.461 0.132 −0.680     0.539 0.461 1.830 −0.052 −0.447    0.037 0.132 −0.052 1.134 0.324 −0.416 −0.680 −0.447 0.324 1.176 The damping in the third mode is 0.605 % and that in the fourth mode 0.796 %.

66

Chapter 11

rocal theorem

Problem 11.1

a23 = a21 = a12 =

The flexibility matrix is derived first. The application of a unit load along coordinate 1 produces the m1 diagram; a unit load along coordinate 2 produces the m2 diagram; and that along 3 produces the m3 diagram.

The flexibility influence coefficients are given by L

L a22 =

0

m1 m1 dx EI

a13 =

0

m1 m2 dx EI

a11 = L a12 =

0

m2 m2 dx EI

0

m1 m3 dx EI

L

11L3 768EI

3L3 256EI

a33 = a11 =

Hence the flexibility matrix is given by 

 9 11 7 L  11 16 11  A= 768EI 7 11 9 3

1

2

L/4

3

L/4

L/4

L/4

Using area integration 1 2 3L 1 3L L × × × × a11 = × 2 16 4 EI 3 16 1 2 3L 1 3L 3L × × × × + × 2 16 4 EI 3 16 3L3 = 256EI

1 2 L 1 L L × × × × × a22 = 2 2 4 2 EI 3 4 L3 = 48EI

1 2 L 1 3L L × × × × × a12 = 2 16 4 EI 3 8

1 L L 1 2 L + × × × × × 2 8 2 EI 3 4

L 1 3L L L + × + 4 6EI 16 4 4

L L L + + 8 2 8 3 11L = 768EI

2 1 3L L 1 L × a13 = × × × × 2 16 4 EI 3 16

1 L L L 2 3L + × × × × × 2 16 4 EI 3 16

L 3L L 3L 1 + × + 2 6EI 16 8 16

L 3L L + + 16 8 16 7L3 = 768EI On account of symmetry and Maxwell’s recip-

L/16

3L/16

L/8

L/8

L/4

L/8

L/8

3L/16

L/16

m1

m2

m3

Figure S11.1

The mass matrix is  M = m



1



1 1

Hence D = K−1 M = AM = mA, which is symmetric. The first 3 steps in the Jacobi iteration are shown in Table S11.1 The trigonometric functions used in the iteration steps are obtained as follows aii − ajj 2R aii − ajj 2 sin α = 0.5 − 2R 1/2

2 R = (aii − ajj ) + 4a2ij

cos2 α = 0.5 +

Function sin α is chosen as positive and cos α takes the sign of aij .

At the end of three iterations, the off-diagonal elements have reduced in magnitude but are still significant. However, for a rough estimate of the

67 #

frequencies, assume that the off-diagonal elements are negligible. Then mL3 γ1 = 31.55 , #768EI EI ω1 = 4.93 mL3 mL3 γ2 = 2.00 , 768EI

λ1 =

λ2 =

EI mL3 mL3 , γ3 = 0.447 768EI # EI ω3 = 41.45 mL3 ω2 = 19.60

768 EI 31.55 mL3

768 EI 2 mL3

λ3 =

768 EI 0.447 mL3

At the end of six iterations the off diagonal elements would be found to be zero to the fourth place of decimal.

Table S11.1 Iteration Row Column No. i j 1

1

2

2

1

3

3

2

3

cos α

sin α 

Transformed Matrix NT AN

 24.04 0.0000 13.0100  0.00 0.9566 0.8429  0.5903 0.8072 13.01 0.8429 9.0000   31.5500 0.4213 0.0000 0.8662 0.4998  0.4213 0.9566 0.7301  0.0000 0.7301 1.4931   31.5500 0.2411 −0.3455 0.5723 0.8200  0.2411 2.0020 0.0000  −0.3455 0.0000 0.4470

Problem 11.2 The stiffness matrix is obtained by inverting the flexibility matrix and can be shown to be   23 −22 9 768 EI  K= −22 32 −22  28 L3 9 −22 23

The first 4 iteration steps in the generalized Jacobi method are shown in Table S11.2. At this stage the elements in K and M matrix have grown considerably in magnitude. If iterations are continued without scaling the matrices in some way, a numerical overflow is likely to occur. In one way of scaling the third rows and columns of M and M may be divided by m33 . This will change the eigenvectors

Table S11.2 Iteration i j

α

γ

1

2 3

1.2253

1.2253

2

1 3

1.1099

-2.7761

3

2 3

0.7414

-3.0257



 4

1 3 -7.6339

61.844

NT KN

 23.000 −10.973 35.956  −10.973 12.618 0.000  35.956 0.000 124.951   256.74 −10.973 0.000  −10.973 12.618 30.462  0.000 30.462 102.575   256.739 −10.973 33.200  −10.973 114.163 0.000  33.200 0.000 33.750



NT MN

 1.000 0.000 0.000  0.000 2.501 0.000  0.000 0.000 2.501   4.081 0.000 0.000  0.000 2.501 0.000  0.000 0.000 10.208   4.081 0.000 0.000  0.000 8.112 0.000  0.000 0.000 33.107

   1712.172 −10.973 0.007 1928.367 0.000 0.000  −10.973 114.163 −678.605   0.000 8.112 0.000  0.007 −678.605 986103.3 0.000 0.000 15642.83

68 but will leave the eigenvalues unaltered. The scaled matrices are   1712.172 −10.973 0.000 K =  −10.973 114.163 −0.043  0.000 −0.043 63.039   1928.367 0.000 0.000 M= 0.000 8.112 0.000  0.000 0.000 1.000 Assuming that iterations have converged 768 1712.172 EI EI × = 24.35 ω12 = 3 mL3 28 # 1928.367 mL EI ω1 = 4.93 mL3 768 114.163 EI EI × ω22 = = 386.0 28 # 8.112 mL3 mL3 EI ω2 = 19.65 mL3 EI 768 63.039 EI × = 1729 ω32 = 28 # 1 mL3 mL3 EI ω3 = 41.58 mL3

The trial vector for the second mode should satisfy the relationship u1 + 1.414u2 + u3 = 0 

    ũ1 0 −1.414 −1 u1  ũ2  =  0 1 0   u2  0 0 1 ũ3 u3

or

The modified dynamic matrix is given by D2 = DS

   9 11 7 0 −1.414 −1 mL3  11 16 11   0 1 0 = 768EI 7 11 9 0 0 1   0 −1.7278 −2 mL3  = 0 0.4438 0  768EI 0 1.1006 2

It is obvious from the form of D2 that a trial vector given by uT = [ 1 0 −1 ] will give convergence in one step as shown in Table S11.3b.

Table S11.3b k

Problem 11.3

uk−1   1  0 −1

From Problem 11.1

  9 11 7 3 mL  11 16 11  D = K−1 M = 768EI 7 11 9

The iterations for the first mode are commenced with a unit trial vector and are shown in Table S11.3a.

Table S11.3a k 1  2

3

uk−1   1 1 1

 1.000  1.407  1.000   1.000  1.414  1.000

ūk = Duk−1   27  38  27   31.48  44.52  31.48   31.556  44.628  31.556



uk

 1.000  1.407  1.000   1.000  1.414  1.000   1.000  1.414  1.000

Convergence has been achieved at this stage mL3 γ1 = × 31.556 768EI 1 EI = 24.338 λ1 = γ1 mL2 # EI ω1 = 4.933 mL3

1

ūk = Duk−1   2  0 −2



uk

 1  0 −1

Hence mL3 EI × 2 λ2 = 384 768EI# mL3 EI ω2 = 19.396 mL3 The third mode should satisfy the orthogonality relationship γ2 =

u1 + 1.414u2 + u3 = 0 u1 − u3 = 0 Selecting u1 = 1, the other two elements work out to u2 = −1.414, u3 = 1. Iterations are not essential for the last mode. The frequency is obtained from the relationship Dq3 = γ3 q3   9 11 7 1.000 mL3  11 16 11   −1.414  768EI 7 11 9 1.000   0.4436 =  −0.6274  0.4436 

  1.000 0.4436 mL3  = −1.414  768 EI 1.000

69 Hence mL3 EI , λ3 = 1731.3 3 1731.3EI mL # EI ω3 = 41.61 mL3

The iterations are started with a trial vector uT0 = [ 1 2 3 ] and are shown in Table S11.4.

γ3 =

Table S11.4

k

Problem 11.4

1 

3

2 2

1

Figure S11.4

3

ūk = Duk−1   0.6049 −  1.0546  1.4149   0.5121 −  0.8934  1.1997   0.5126 −  0.8942  1.2004

uk−1   1 2 3

 1.0000  1.7435  2.3392   1.0000  1.7447  2.3430



uk

 1.0000  1.7435  2.3392   1.0000  1.7447  2.3430   1.0000  1.7447  2.3429

Assume that convergence has been achieved at this stage



 1000 −500 0 K =  −500 750 −250  0 −250 250     2 1 M =  2  u = 2 1 3

γ̂1 = −0.5126 δ̂ = −1.9510 λ = 65.789 − 1.951 = 63.838 ω1 = 7.99

uT Ku = 1250

Problem 11.5

uT Mu = 19

Operating speed Ω = 250×2π = 26.18 rad/s 60 To obtain the frequency closest to the operating speed use inverse iteration with a shift given by µ = Ω2 = 685.39

uT Ku = 65.789 uT Mu ω1 = 8.111 µ=

Forming the eigenvalue problem with a shifted origin (K − µM) q = δMq

This yields 

 119.44 −93.40 16.20 [K − µM] =  −93.40 84.34 −63.30  16.20 −63.30 −2.76

where 

 868.42 −500.00 0 [K − µM] =  −500.00 618.42 −250  0 −250.00 184.21

Also −1

Inversion gives 

 0.03675 0.06583 0.08934 −1 [K − µM] = −  0.06583 0.11434 0.15518  0.08934 0.15518 0.20517 Hence −1

D = [K − µM] M   0.07350 0.13166 0.08934 = −  0.13166 0.22868 0.15518  0.17869 0.31035 0.20517

D = [K − µM] M   13.571 4.123 −14.550 0.0745  4.123 1.896 −19.356  = 1000 −14.550 −19.356 −4.320 Iterations are carried out with an initial vector uT0 = [ 1 1 −1 ]. The computations are shown in Table S11.5.

70

Table S11.5 k 1

2

3

4

5

6

7

uk−1   1  1 −1   1.00  0.79  −0.92   1.000  0.776  −0.856   1.000  0.758  −0.885   1.000  0.767  −0.859   1.000  0.760  −0.879   1.000  0.765  −0.863

ūk = Duk−1   32.24  25.38  −29.59   30.21  23.43  −25.87   29.23  22.16  −25.87   29.57  22.69  −25.40   29.23  22.20  −25.69   29.49  22.57  −25.46   29.28  22.28  −25.63



uk

 1.00  0.79  −0.92   1.000  0.776  −0.856   1.000  0.758  −0.885   1.000  0.767  −0.859   1.000  0.760  −0.879   1.000  0.765  −0.863   1.000  0.761  −0.875

To obtain a better precision, the iterations must be continued. However, for the purpose of illustration it will be assumed that convergence has taken place 29.28 × 0.0745 1000 1 γ̂ = = 458.12 8 λ = 685.39 + 458.12 = 1143.5 ω = 33.8 rad/s δ̂ =

Problem 11.6 Table S11.6 j

x̃j   1 1 1

βj 

xj = x̃j /βj

xj+1 = K−1 Mxj

αj = xTj Mxj+1

   2.1153 0.009214  2.1153   0.019623  1 0.47276 0.008624 2.1153 0.025890       −0.009028 −2.7768 0.001429  0.001381   0.4247   0.006948  2 0.003257 0.002072 0.007647 2.3521 0.012255       3.0418 × 10−4 1.1129 −0.00028148 3  −8.0935 × 10−4  2.7533 × 10−4  −2.9611   −0.00115275  0.00042839 1.8482 0.00143392 5.0517 × 10−4

The Lanczos vectors are obtained by using Equations 11.137 with a starting vector xT = [ 1 1 1 ]. The computations are shown in Table S11.6 The Lanczos vectors are given by   2.1153 −2.7768 1.1129 X =  2.1153 0.4247 −2.9611  2.1153 2.3521 1.8482

x̃j+1 = xj+1 −βj xj+1 − αj xj   −0.009028  0.001381  0.007647   3.0418 × 10−4  −8.0935 × 10−4  5.0517 × 10−4

The tridiagonal matrix is T = XT MK−1 MX   8.624 3.2510 0.0000 1  3.251 2.0720 0.2753  = 1000 0.000 0.2753 0.4284

71

Problem 11.7 Reduction of the tridiagonal matrix to an upper triangular matrix proceeds as shown in Table S11.7a. The factor 1/1000 has been taken out from the computations shown in the table.

Table S11.7b

i

Table S11.7a

i

cos α

sin α 

cos α

sin α 

Transformed matrix Ai+1 = NTi A(i)

 9.9590 0.3214 0.0031 0.8581 0.1050  1 0.99956 0.02970  0 0 0.1051 0.3023   9.959 0.3214 0.0031 2 0.99258 0.12157  0 0.8645 0.1410  0 0 0.2873

Transformed matrix Ai+1 = NTi A(i)

 9.216 3.7730 0.0971 1 0.93572 0.35274  0 0.7921 0.2576  0 0.2753 0.4284   9.216 3.7730 0.0971 2 0.94457 0.32829  0 0.8386 0.3840  0 0 0.3201

Q1 = N1 N2   0.9996 −0.0295 0.0036 =  0.0297 0.9922 −0.1215  0 0.1216 0.9926

The transformation matrix Q1 is obtained from

A2 = QT1 AQ1   9.9640 0.0257 0 =  0.0257 0.8748 0.0349  0 0.0349 0.2852

Q1 = N1 N2   0.9357 −0.3332 0.1158 =  0.3527 0.8838 −0.3072  0 0.3283 0.9446

Assume that the off-diagonal elements are negligible at this stage and the iterations have converged

The first transformation is completed by

9.964 1 , = λ1 1000 1 0.8748 , = λ2 1000 1 0.2852 , = λ3 1000

A2 = QT1 AQ1 = R1 Q1   9.9550 0.2958 0 =  0.2958 0.8672 0.1051  0 0.1051 0.3023 where R1 = A(3) was obtained earlier. The second iteration proceeds in a similar manner and is shown in Table S11.7b.

λ1 = 100.36 λ2 = 1143.1 λ3 = 3506.3

For the purpose of comparison, after 10 iterations the off-diagonal elements are zero to five places of decimal and λ1 = 100.356, λ2 = 1140.5, λ3 = 3532.3.

Problem 11.8

Table S11.8a k

µk−2

pk−2

µk−1

1 -10.00 0.1853×106 0.00 2 0.00 0.1665×106 88.95 3 88.95 0.1701×105 99.07 4 99.07 0.1893×104 100.33

pk−1

µk = −µk−2 µk−1 µpk−1 pk−1 k−1 −pk−2

pk

0.1665×106 0.1701×105 0.1893×104 0.2661×102

88.95 99.07 100.33 100.35

0.1701×105 0.1893×104 0.2661×102 0.03906

72 The search for the first eigenvalue is commenced with µk−2 = −10 and µk−1 = 0. A new estimate is obtained by using Equation 11.139 with η = 1. The iterations are repeated until a zero value of the polynomial is found. The computations are shown in Table S11.8a.

Iterations may be assumed to have converged at this stage. Thus λ1 = 100.35

Determinant search for the second eigenvalue is carried out on a deflated polynomial given by p(2) =

p µ − 100.35

Iterations are commenced with µk−2 = 150, µk−1 = 200 and are shown in Table S11.8b.

Assuming that convergence has taken place

ω1 = 10.02 rad/s

λ2 = 1143.04

ω2 = 33.81 rad/s

Table S11.8b (2)

k

µk−2

pk−2

(2)

µk−1

pk−1

(2)

µk =

pk

1 2 3 4

150.00 200.00 925.53 1089.98

-0.1380×104 -0.1291×104 -0.2326×103 -0.5835×102

200.00 925.53 1084.98 1138.36

-0.1291×104 -0.2326×103 -0.5835×102 -0.4698×10

925.53 1084.98 1138.36 1143.04

-0.2326×103 -0.5835×102 -0.4698×10 -0.1122

where T is a transformation matrix. Transformation of stiffness and mass matrices gives

Problem 11.9 

 1 −1 0 K = 15000  −1 2 −1  0 −1 1   1 60000  M= 1  386.4 1

1

2

5 1 1 2

60000 2 1 M̃ = TT MT = 3864 1 2 K̃ = TT KT = 15000

The eigenvalue problem in the reduced space is K̃q̃ = λM̃

3

The characteristic equation is given by

Figure S11.9 The rigid body motion is a lateral translation of the three cars given by   1 q = 1 1 The orthogonality relationship to be satisfied by other modes is

0 0 0 5 − 2λ 1 − λ 0 0 0 0 1 − λ 2 − 2λ 0 = 0 where λ = λ × 60000/(15000 × 386.4).

Solution of the characteristic equation gives λ2 = 1

λ3 = 3

Substitution of λ in the frequency equation yields the two mode shapes in the reduced space

0 q̃2 = 1

q̃3 =

1

− 12

q̃1 + q̃2 + q̃3 = 0 Using the transformation equation q = Tq̃ or



   q̃1 −1 −1  q̃2  =  1 0  q̃2 = Tq̃ q̃3 0 1 q̃3



   −1 −1 −1 0 q2 =  1 0  =  0 1 0 1 1

73  1 

−2 −1 −1 1 =  1 q3 =  1 0  1 −2 − 12 0 1 

The sweeping matrix for the first mode shape is given by q1 qT M S1 = I − T 1 q1 Mq1   0.6704 −0.3318 −0.3330 =  −0.3318 0.6659 −0.3352  −0.3330 −0.3352 0.6637

The frequency corresponding to the rigid body mode is ω1 = 0. The other two frequencies are given by 386.4 λ2 = 96.6, 4 386.4 λ3 = λ3 = 289.8, 4

λ2 =

ω2 = 9.83 ω3 = 17.02

The dynamic matrix for convergence on the second mode shape is obtained from D2 = DS1

 5.755 −1.127 −4.574 = 10−3  −1.127 2.297 −1.173  −4.574 −1.173 5.698

Problem 11.10 We add a fictitious stiffness of 0.01 ∗ 15000 to d.o.f. 1. The modified stiffness matrix is given by 

 1.01 −1 0 K = 15000  −1 2 −1  0 −1 1

Iterations are started with a trial vector uT0 = [ 1 2 3 ] and are shown in Table S11.10b

Table S11.10b

The mass matrix remains unchanged. We use inverse iteration to determine the frequencies and mode shapes. The dynamic matrix is given by D = K−1 M   1.0352 1.0352 1.0352 =  1.0352 1.0455 1.0455  1.0352 1.0455 1.0559

k 1  2

3 Iterations are started with a trial vector uT0 = [ 1 2 3 ] and are shown in Table S11.10a

Table S11.10a

k 1  2

3

uk−1   1 2 3

 1.000  1.008  1.013   1.000  1.007  1.010

ūk = Duk−1   6.211  6.263  6.294   3.128  3.149  3.159   3.123  3.149  3.154





uk

 1.000  1.008  1.013   1.000  1.007  1.010   1.000  1.007  1.010

uk−1   1 2 3

 1.000  0.005  −0.995   1.000  0.005  −0.995

ūk = D2 uk−1   −0.0102  −0.0001  0.0102   0.0103  0.0001  −0.0102   0.0103  0.0001  −0.0102



uk

 1.000  0.005  −0.995   1.000  0.005  −0.995   1.000  0.005  −0.995

We assume that the iterations have converged at this stage so that the second eigenvalue is given by λ2 = 1/0.0103 = 97.08. The first mode shape is qT1 = [ 1.000 0.005 −0.995 ] The third mode shape should satisfy the following two conditions of orthogonality, qT i Mq3 for i = 1, 2 q31 + 1.007 q32 + 1.010 q33 = 0 q31 + 0.005 q32 − 0.995 q33 = 0 Solution of the above two equations with q31 = 1 gives the following mode shape qT3 = [ 1 −1.982 0.995 ] The third eigenvalue is obtained from

We assume that the iterations have converged at this stage so that the first eigenvalue is given by λ1 = 1/3.123 = 0.320. The first mode shape is qT1 = [ 1.000 1.007 1.010 ]

λ3 =

qT3 Kq3 qT3 Mq3

= 289.96

74 The results obtained are approximate; the accuracy can be improved by using a smaller value for the fictitious stiffness.

The dynamic matrix for convergence on the second mode shape is obtained from D̂2 = D̂S1

 6.049 −1.171 −4.879 = 10−3  −1.165 2.341 −1.176  −4.868 −1.170 6.039

Problem 11.11 Since the system is unrestrained and the stiffness matrix is singular we introduce a shift µ = 5.0 so that the eigenvalue problem takes the form 1 q δ qD̂q = γ̂

−1

(K − µM)

Iterations are started with a trial vector uT0 = [ 1 2 3 ] and are shown in Table S11.11b

Mq =

Table S11.11b

in which γ̂ = 1/delta and λ = µ + δ = 5 + δ The dynamic matrix is now given by

k

 0.0606 0.0678 0.0715 D̂ = −  0.0678 0.0643 0.0678  0.0715 0.0678 0.0606

1



Iterations are started with a trial vector uT0 = [ 1 2 3 ] and are shown in Table S11.11a

Table S11.11a

k 1  2

3

4

uk−1   1 2 3

 1.000  0.973  0.947   1.000  1.002  1.003   1.000  1.000  1.000

ūk = D̂uk−1   0.4109 −  0.4000  0.3891   0.1944 −  0.1947  0.1950   0.2003 −  0.2003  0.2003   0.2000  0.2000  0.2000





uk

 1.000  0.973  0.947   1.000  1.002  1.003   1.000  1.000  1.000   1.000  1.000  1.000

 2

uk−1   1 2 3

 1.000  0.001  −0.998

ūk = D̂2 uk−1   −0.0109  −0.0000  0.0109   0.0109  0.0000  −0.0109

q1 qT M S1 = I − T 1 q1 Mq1   0.6666 −0.3334 −0.3333 =  −0.3334 0.6667 −0.3333  −0.3333 −0.3333 0.6667

uk

 1.000  0.001  −0.998   1.000  0.001  −0.998

We assume that the iterations have converged at this stage so that the shifted second eigenvalue is given by δ2 = 1/0.0109 = 91.60. The true second eigenvalue is λ2 = 91.60 + 5.0 = 96.60. The second mode shape is qT1 = [ 1.000 0.000 −0.998 ] The third mode shape should satisfy the following two conditions of orthogonality, qT i Mq3 for i = 1, 2 q31 + 1.000 q32 + 1.000 q33 = 0 q31 + 0.000 q32 − 0.998 q33 = 0 Solution of the above two equations with q31 = 1 gives the following mode shape qT3 = [ 1 −2.002 1.002 ] The third eigenvalue is obtained from λ3 =

We assume that the iterations have converged at this stage so that the first shifted eigenvalue is given by δ1 = −1/0.2 = −5.0. so that λ1 = µ + δ1 = 0.0 The first mode shape is qT1 = [ 1.000 1.000 1.000 ] The sweeping matrix for the first mode shape is given by



qT3 Kq3 qT3 Mq3

= 289.80

Problem 11.12 The three translational degrees of freedom are as shown in Figure S11.12. The stiffness matrix corresponding to the three d.o.f. is derived by standard methods of structural analysis. Referring to the displaced configurations in Figure S11.12, the stiffness matrix becomes   2 −1 −1 3EI  K= 3 −1 12 12  L −1 12 12

75 2

1

Hence

3

m

m

3EI

3EI

L3

L3 6EI

Problem 11.13

L3 1

2m 3EI k∗ 1+ ω = ∗ = mL3 M m # 3EI M + 2m ω= L3 M m 2

M

3EI

3EI 2L3

2L3

Iteration 1 3EI L3

3EI 2L3

3EI

3EI

2L3

L3

1

Figure S11.12 The mass matrix is  M M= m

  m

The rigid body modes corresponding to translation and rotation are respectively

V1 = K−1 MV0   0.3125 × 10−2 −0.1250 × 10−2 −2 −2 −0.2083 × 10   0.6958 × 10   =  0.1013 × 10−1 −0.1250 × 10−2    0.1338 × 10−1 0.1250 × 10−2 −1 −2 0.1588 × 10 0.3750 × 10 T

K∗1 = V1 KV1

0.5624 × 10−1 0.2083 × 10−2 = 0.2083 × 10−2 0.1167 × 10−1 T

M∗1 = V1 MV1

0.9312 × 10−3 0.3085 × 10−4 = 0.3085 × 10−4 0.3212 × 10−4

qT1 = [ 1 1 1 ] qT2 = [ 0 1 −1 ]

Solution of second order eigenvalue problem gives

If the third mode is q3T = [u1 u2 u3 ], orthogonality relationships give

λ1 = 60.39 λ2 = 372.7

32.79 5.806 Z1 = −0.741 −179.330

qT1 Mq3 = 0

V1 = V1 Z1   0.1034 0.2423  0.2297 0.4140    =  0.3330 0.2829    0.4377 −0.1465 0.5178 −0.5802

qT2 Mq3 = 0 or M u1 + mu2 + mu3 = 0 mu2 − mu3 = 0 The two equations give the transformation relationship    2m  −M u1  u2  =  1  u3 u3 1 or u = Tu3 The transformed matrices are

2m ∗ T m = T MT = 2m 1 + M 2

3EI 2m k ∗ = TT KT = 3 2 1 + L M

Iteration 2

V2 = K−1 MV1   0.1703 × 10−2 0.6282 × 10−3  0.3802 × 10−2 0.1062 × 10−2    =  0.5518 × 10−2 0.8058 × 10−3    0.7260 × 10−2 −0.2859 × 10−3 0.8554 × 10−2 −0.1737 × 10−3 T

K∗2 = V2 KV2

0.1656 × 10−1 0.4880 × 10−5 = 0.4880 × 10−5 0.2731 × 10−2

76 T

M∗2 = V2 MV2

0.2742 × 10−3 0.1016 × 10−6 = 0.1016 × 10−6 0.2731 × 10−2

Problem 11.14 Iteration 1 V1 = K−1 MV0   0.5717 × 10−2 0.1067 × 10−2  0.1060 × 10−1 0.1300 × 10−2  =  0.1382 × 10−1 −0.1333 × 10−3 0.1567 × 10−1 −0.2158 × 10−2

λ1 = 60.39 λ2 = 363.10

60.39 0.14 Z2 = −0.09 −364.59 V2 = V2 Z2   0.1028 −0.2288  0.2295 −0.3867    =  0.3332 −0.2930    0.4384 0.1053 0.5168 0.6344

T

K∗1 = V1 KV1

0.8299 × 10−1 0.1550 × 10−3 = 0.1550 × 10−3 0.7177 × 10−2 T

M∗1 = V1 MV1

0.9173 × 10−3 0.2257 × 10−5 = 0.2257 × 10−5 0.1035 × 10−4

Iteration 3

Solve the second order eigenproblem K∗1 Z1 = M∗1 Z1 Γ

V3 = K−1 MV2   0.1703 × 10−2 −0.6075 × 10−3 −2 −2 −0.1036 × 10   0.3802 × 10   =  0.5519 × 10−2 −0.8204 × 10−3    0.7261 × 10−2 0.2357 × 10−3 0.8553 × 10−2 0.1822 × 10−2

This gives

Z1 =

V1 = V1 Z1   0.1890 −0.3271  0.3503 −0.3958  =  0.4561 0.0524 0.5167 0.6835

T

K∗3 = V3 KV3

0.1656 × 10−1 0.7241 × 10−6 = 0.7241 × 10−6 0.2765 × 10−2 Iteration 2

T

M∗3 = V3 MV3

0.2742 × 10−3 −0.1737 × 10−8 = −0.1737 × 10−3 0.7661 × 10−5 The required eigenvalues and frequencies λ1 = 60.39, λ2 = 361.3 ω1 = 7.77, ω2 = 19.01

Z3 =

33.02 0.79 0.26 −310.93

60.39 0.006 0 361.300

V2 = K−1 MV1   0.2090 × 10−2 −0.5479 × 10−3 −2 −3 −0.5507 × 10   0.3864 × 10 =  0.5055 × 10−2 0.1063 × 10−3 −2 −3 0.9606 × 10 0.5701 × 10 T

K∗2 = V2 KV2

0.1106 × 10−1 0.3380 × 10−6 = 0.3380 × 10−6 0.1462 × 10−2 λ1 = 90.46, ∗

M

0.12220 × 10−3 0.4903 × 10−8 0.04903 × 10−8 0.2152 × 10−5

90.46 0.03 Z2 = 0.79 −681.62   0.1886 0.3736  0.3491 0.3755  V2 = V2 Z2 =   0.4574 −0.0723 0.5165 −0.6546 =

V3 = V3 Z3   0.1028 −0.2195  0.2296 −0.3743    =  0.3333 −0.2964    0.4385 0.0852 0.5165 0.6582 These are the eigenvectors or mode shapes.

λ2 = 679.32

= V2T MV2

77 Iteration 3  0.2089 × 10−2 0.5823 × 10−3 −2 −3 0.5421 × 10   0.3863 × 10 V3 =   0.5056 × 10−2 −0.1239 × 10−3 −2 −3 0.5702 × 10 −0.9422 × 10

−1 −0.1755 × 10−5 0.1106 × 10 K∗3 = −0.1755 × 10−5 0.1477 × 10−2

0.1222 × 10−3 −0.3128 × 10−8 M∗3 = −0.3128 × 10−8 0.2184 × 10−5 

λ1 = 90.46

λ2 = 676.1

ω1 = 9.51,

ω2 = 26.00

Assuming that convergence has taken place, these are the required eigenvalues and frequencies.

90.46 0.00 Z3 = 0.00 676.6   0.1890 0.3940  0.3495 0.3668  V3 = V3 Z3 =   0.4574 −0.0838 0.5158 −0.6375

To tridiagonalize the E matrix we reduce elements e13 and e31 to zero by using the Householder method. The steps involved in the computation are as follows

−1253.7 c1 = 217.4 r12 = cT1 c1 = 1.619 × 106 2h21 = r12 − e12 r1 = 1.619 × 106 − (−1253.7) × 1272.4 = 3.2142 × 106

1272.4 g1 = 0 v1 = c1 − g1

−2526.1 = 217.4 1 R1 = I − 2 v1 v1T 2h1

−0.9853 0.1709 0.1709 0.9853

These are the required mode shapes.

The transformation matrix N1 is given by

Problem 11.15



The stiffness and mass matrices given in Problem 11.5 are   170.5 −93.4 16.2 K =  −93.4 135.4 −63.3  16.2 −63.3 48.3   0.0745 0 0 M= 0 0.0745 0  0 0 0.0745

 1 0 0 N1 =  0 −0.9853 0.1709  0 0.1709 0.9853 On carrying out the transformation we have E2 = NT1 EN1   2288.6 1272.4 0.000 =  1272.4 2069.4 603.2  0.0000 603.2 396.3

For diagonal mass matrix we have M = LM LTM where LM = LTM = M1/2 On transforming the eigenvalue to the standard form we get T −1 T L−1 LM q = λLTM q M K LM Eq̃ = λq̃ where LTM q = q̃. On carrying out the indicated operations we get   2288.6 −1253.7 217.4 E =  −1253.7 1817.4 −849.7  217.4 −849.7 648.3

Problem 11.16 The first transformation matrix Q1 in the series of orthogonal transformation to be performed on the tridiagonal matrix obtained in Problem 11.15 is itself the product of two other matrices and is given by Q1 = N̄1 N̄2 in which N̄1 and N̄2 are obtained from Equation 11.48. Computations for N̄1 and N̄2 are shown in Table S11.16.

78 Table S11.16

i

cos α

sin α

1

0.874

0.486

2

0.892

0.452

Transformed matrix, E(i+1) = N̄Ti E(i)   2618.5 2117.7 293.1  0 1190.4 527.2  0 603.2 396.3   2618.5 2117.7 293.1  0 1334.5 649.4  0 0 115.3

The transformation matrix Q1 is given by   0.8740 −0.4335 0.2196 Q1 = N̄1 N̄2 =  0.4859 0.7796 −0.3950  0 0.4520 0.8920 The last matrix shown in Table S11.16 is QT1 E and is upper triangular. The first transformation is now completed as follows: E2 = QT1 EQ1   3317.6 648.5 0 =  648.5 1333.9 52.1  0 5.21 102.8 At the end of eight transformations we get   3510.9 0.3 0 E9 =  0.3 1143.1 0  0 0 100.4 At the stage the transformed matrix is almost diagonal and the entries on the diagonal represent the eigenvalues of the original matrix. The eigenvectors of the tridiagonal matrix are given by T = Q1 Q2 Q8 I   0.7148 −0.6538 0.2481 =  0.6865 0.5887 −0.4267  0.1329 0.4754 0.8697 The eigenvectors of the standardized eigenvalue problem are obtained from Z = N1 T where N1 is the transformation matrix from Problem 11.15. Finally, the eigenvectors of the original linearized eigenvalue problem are given by −1

Z V = LTM   2.6190 −2.3952 0.9091 =  −2.3950 −1.8274 2.0850  0.9097 2.0847 2.8721

79

Chapter 12

Problem 12.2 The modal steady state response is given by

Problem 12.1 Given ω1 = 10.6, ω2 = 33.7, ω3 = 59.2.

The normalized mode shapes were obtained in Problem 10.6 and are given by   0.9077 2.3819 2.6153 Φ =  2.0695 1.8293 −2.3551  2.8682 −2.0770 0.8970 Exciting frequency Ω = 400×2π = 41.88 rad/s.

60 Applied force = p0 sin Ωt where 12 × (41.88)2 800 × 1000 386.4 = 43.59 kips     0 125.0 p = Φ  0  =  −90.5  kips 43.59 39.1

p0 = meΩ2 =

The modal equations are 2

ÿ1 + (10.6) y1 = 125.0 sin 41.88t ÿ2 + (33.7)2 y2 = −90.5 sin 41.88t ÿ3 + (59.2)2 y3 = 39.1 sin 41.88t The modal steady state responses are obtained from pon 1 sin Ωt k 1 − βn2 41.88 β1 = = 3.950 10.6 41.88 β2 = = 1.243 33.7 41.88 = 0.707 β3 = 59.2 1 125.0 y1 = sin 41.88t (10.6)2 1 − (3.95)2

yn =

= −0.0762 sin 41.88t 1 90.5 sin 41.88t y2 = − (33.7)2 1 − (1.243)2 = 0.1462 sin 41.88t 1 39.1 sin 41.88t y3 = (59.2)2 1 − (0.707)2 = 0.0223 sin 41.88t Transforming to physical coordinates u = Φy u1 = 0.3374 sin 41.88t in u2 = 0.05656 sin 41.88t in u3 = −0.5022 sin 41.88t in

yi =

1 p0 2 ki (1 − βi2 ) + (2ξi βi )2 2

3 1 − βi2 sin Ωt − 2ξi βi cos Ωt

First mode 41.88 = 3.95, ξ1 = 0.05, p01 = 125.0 10.6 k1 = ω12 = (10.6)2 y1 = −0.07607 sin Ωt − 0.002058 cos Ωt

β1 =

Second mode 41.88 = 1.248, ξ2 = 0.05, 33.7 k2 = ω22 = (33.7)2 y2 = 0.1390 sin Ωt + 0.03175 cos Ωt

β2 =

p02 = −90.5

Third mode 41.88 = 0.707, ξ3 = 0.05, p02 = 39.1 59.7 k3 = ω32 = (59.2)2 y3 = 0.02188 sin Ωt − 0.00309 cos Ωt

β3 =

Transformation to physical coordinates gives u = φ1 y1 + φ2 y2 + φ3 y3 Using the values of normalized modes from Problem 12.1 u1 = 0.3193 sin Ωt + 0.0656 cos Ωt in u2 = 0.04466 sin Ωt + 0.06119 cos Ωt in u3 = −0.4872 sin Ωt − 0.07462 cos Ωt in

Problem 12.3 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 An eigenvalue solution using Matlab routine provides eigenvalues λ1 = 63.91, λ2 = 339.24, and λ3 = 723.13. The frequencies in rad/s are ω1 =

80



 0.0649 0.0966 0.1226 Φ =  0.1132 0.0623 −0.1090  0.1520 −0.1752 0.0577 The exciting frequency is Ω = 5π/3 = 5.236 rad/s. The frequency ratios, Ω/ωi , are β1 = 0.6549, β2 = 0.2843, and β3 = 0.1947 The load vector is given by 

 0  kN p= 0 45 sin(5πt/3) The load vector in the modal coordinates is given by   6.840 p∗ = ΦT p =  −7.886  sin(5πt/3) 2.596 The modal equations are ÿ1 + ω12 y1 = 6.840 sin(5πt/3) ÿ2 + ω22 y2 = −7.886 sin(5πt/3) ÿ3 + ω32 y3 = 2.596 sin(5πt/3) The solution of the modal equations is of the form 1 p∗ (sin(5πt/3) − βn sin ωn t) yn = n2 ωn 1 − βi2 Substitution of the values of various parameters gives y1 = 0.1874 {sin(5πt/3) − 0.6549 sin 7.995t} y2 = −0.0253 {sin(5πt/3) − 0.2843 sin 18.419t} y3 = 0.0037 {sin(5πt/3) − 0.1947 sin 26.891t} The response in the physical coordinates is obtained from 1 φi yi u(t) = i=1:3

The contribution from the first mode is     12.159 7.963  21.213  sin(5πt/3) −  13.893  sin(7.995t) mm 28.487 18.657 The second mode contribution is     −2.444 −0.695  −1.575  sin(5πt/3)−  −0.448  sin(18.419t) mm 4.432 1.260

The contribution from the third mode is     0.458 0.089  −0.407  sin(5πt/3)−  −0.079  sin(26.819t) mm 0.215 0.042 The total displacement at the roof level is ur (t) = 33.1 sin(5πt/3) − 18.657 sin(7.995t) − 1.260 sin(18.419t) − 0.042 sin(26.819t) mm The modal contributions and the total displacement at the roof are plotted in Figure S12.3.

The maximum value of the displacement takes place at 0.949 s and its value is -48.83 mm.

40 Total

30 Mode 1

20 Top floor displacement, mm

7.995, ω2 = 18.419, and ω3 = 26.891. The mass orthonormal mode shapes are

10

Mode 2

0

Mode 3

10 20 30 40 50

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Time, s

Figure S12.3

Problem 12.4 From the plots of modal contribution shown in Figure S12.3, in fact, from the computations leading to them we find the following results.

The maximum in the first mode is at 0.949 s and its value is -45.55 mm. The maximum in the second mode is at 0.266 s and its value is 5.6 mm.

The maximum in the third mode is at 0.981 s and its value is -0.235 mm.

The sum of the absolute value of the three modal maxima gives 51.387 mm for the maximum displacement. It is an overestimate because the three maxima occur at different instances of time and they are not in the same direction. In the present case the first mode dominates the response so that the time at which the maximum displacement occurs is close the time at the maximum in the first mode.

Problem 12.5 The story level forces in mode n are given by fn = Kun = ωn2 Mun

81 where un is the displacement in the nth mode.

The base shear in a mode is obtained by taking the sum of the story level forces.

The base shear histories for each mode as well as the total base shear are plotted in Figure S12.5.

The maximum base shear in the first mode occurs at 0.95 s and its value is -170.32 kN. The maximum in the second mode is at 0.266 s and its value is -27.05 kN. The maximum base shear in the third mode is at 0.98 s and its value is -4.37 kN. The maximum total base shear is at 0.952 s and its value is -159.73 kN. It is of interest to note that at the time of maximum the base shears in the first, second and third mode are, respectively -170.31, 14.73, and -4.15 kN and the base shear in the second mode counteracts that in the other two modes.

The total steady statel displacement at the roof level is us r(t) = 33.134 sin(5πt/3) mm The maximum values of the modal contributions are all at 0.3 s and their values are, respectively, 28.487, 4.432, and 0.215 mm, for the first, second, and third modes. The maximum total displacement is also at 0.3 s and its value is 33.134 mm The modal contributions and the total roof displacement are plotted in Figure S12.6 0.04 Total

0.03 Top floor displacement, m

150 Mode 1

100 Total

Base shear, kN

50 Mode 3 0 Mode 2

50

0.02

Mode 1

0.01

Mode 2

0

Mode 3

0.01 0.02 0.03

100

0.04

0

0.4

0.2

150 200 0

0.6

0.8

1

Time, t

0.2

0.4

0.6

0.8

Figure S12.6

1

Time, s

Problems 12.7 and 12.8 Figure S12.5

Problem 12.6 Referring to the solution for Problem 12.3 the steady state part of the modal contributions to the displacements are as follows.

Contribution from the first mode   12.159  21.213  sin(5πt/3) mm 28.487 Second mode contribution   −2.444  −1.575  sin(5πt/3) mm 4.432 Contribution from the third mode   0.458  −0.407  sin(5πt/3) mm 0.215

The modal frequencies and orthonormal mode shapes were obtained in the solution to Problem 12.3. The modal force vector is given by 

   0 5.6592 p∗ = ΦT  50  =  3.1142  kN 0 −5.4501 The duration of the rectangular pulse is t1 = 0.5 s. From Equations 7.29 and 7.32 the modal displacement responses are given by p∗i (1 − cos ωi t) t ≤ t1 ωi2 p∗ yi = i2 {cos ωi (t − t1 ) − cos ωi 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 qT1 Mu0 m2 g = q1 Mq1 M +mk T q Mu0 m2 g =− y20 = 2 q2 Mq2 M +mk

M

y10 =

u1

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)ÿ1 = (M + m)g

M

1+

M m

ÿ2 + c

M m

M ẏ2 m

M M y2 = 0 1+ +k m m

1+

% where ω =

k √c m , ξ = 2 km , and ωd = ω

1 − ξ2.

Transforming to physical coordinates u = Qy

qT1 = [ 1 1 ]

Modal matrix

m2 g gt2 + M +mk 2

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

y1 =

mg

gt2 m2 g + 2 M +mk

m2 g −ξωt ωξ − e cos ωd t + sin ωd t M +mk ωd 2 2 m g gt + u2 = 2 M +mk

M m g −ξωt ωξ + e cos ωd t + sin ωd t M +mk ωd u1 =

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 ü1 + 3 −41.1 150.9 ü2 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 q̈ + 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

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

u2 = 4.185 sin 30t mm Total displacement

or (402.4 − λ) − (109.6) = 0 2

2

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

which yields λ1 = 292.8,

λ2 = 512

(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 sin Ωt = 2700 sin Ωt 0

1 p2 = qT2 2700 sin Ωt = 2700 sin Ωt 0 p1 = qT1 2700

The modal equations are 2700 sin 30t 600 2700 sin 30t ÿ2 + 512.0y2 = 600

ÿ1 + 292.8y1 =

üb = −

900 × 10 1 2 sin 30t m/s 0 1000

Also

300 0 −Maa Rüb = 9 0 300

0.59340 0.40570 1 × sin 30t 0.09335 −0.09426 0 The equations of motion are

300 0 0 300

ü1 150.9 −41.1 u1 + 800 −41.1 150.9 ü2 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 ÿ2 + 512.0y2 = 600 = 2.250 sin 30t

ÿ1 + 292.4y1 =

85 The steady state modal responses are y1 =

The modal equations are p0 4m p0 ÿ2 + ω22 y2 = 2m p0 ÿ3 + ω32 y3 = 4m

ÿ1 + 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

fs = ΣKqi yi = Σωi2 Mqi yi

M1 = 4m , M2 = 2m , M3 = 4m

Since yi = Mpi 0ω2 (1 − cos ωi t)

The modal forces are given by

i

pn = qTn p

p2 = p0 ,

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

fs = Σ

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

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

p3 = p0

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 0 0 0 2 0 0 M=  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 Ψ̃ produced by them.   2.079 0.4090  3.862 0.6001  Ψ̃ = K−1 MΨ = 10−3   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  Ψ̄ =   0.8843 0.0567 1.0000 1.0000 We now use the vectors Ψ̄ in the Rayleigh Ritz procedure similar to that given in the solution to Problem 13.1.

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

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

3.7429 −0.1809 M = Ψ̄ MΨ̄ = −0.1809 2.4510

338.6 −13.90 ∗ T K = Ψ̄ KΨ̄ = −13.90 1689.4 ∗

T

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 = Ψ̄Z =   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

u = z1 + z2 sin

πx

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

ψ1 =

Hence

K=

0 0 4 0 π2LEI 3

+

2k 0 2k 0 = 0 0 0 k

1 T (ż) 2

L m 0

1 [ 1 sin πx L ] dx ż 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 2 π where λ = ω 2 (mL/k).

Expansion of the determinant gives the characteristic equation λ2 − 21.116λ + 21.116 = 0

1−

so that u = z1 ψ 1 + z 2 ψ 2

z = [ ψ1 ψ2 ] 1 z2

Kinetic energy T is given by T =

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 Hence

2k 0 0 0 4 + 0 0 0 8πL3EI

2k 0 = 0 16k

K=

89 Kinetic energy T is given by T =

1 T ż 2

L m 0

L"

1 − 2x L 2πx [ 1 − 2x L sin L ] dx ż sin 2πx L 2

4x 1 + 4x L2 − L

2πx 1 − 2x 0 L sin L

2πx 1 − 2x L sin L dx ż sin2 2πx L 1 1 1 T = ż mL 31 π1 ż 2 π 2

1 = żT m 2

Hence

1 M = mL

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

x=0

EI ∂∂xu2 = 0 3 EI ∂∂xu3 = −ku

x=L

EI ∂∂xu2 = 0 3 EI ∂∂xu3 = ku

2

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(ż)2 2 0 1 = (4.234mL)(ż)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

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. p1 = 5.019

The modal equations are ÿ1 + (10.6)2 y1 = 5.019 sin Ωt ÿ2 + (33.7)2 y2 = −3.634 sin Ωt ÿ3 + (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 K̃ = T KT = −480 480

4 0 M̃ = M = 0 2

T



 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 K̃q̃ = λM̃q̃ This yields λ1 = 84.63, λ2 = 425.4.

q̃1 =

1.000 1.545

q̃2 =

1.0 −1.3

The expanded eigenvector are obtained from

Renumbered d.o.f.

4

2

3

4

2

1

1

3

q = Tq̃     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 M̃ = TT MT −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 0 Mtt = 0 1

2 0 Mθθ = 0 2

I

−K−1 θθ Kθt

1 2400

0

0

1 2000

−1200 0 −1200 −800

On substituting the appropriate values

M̃ =

3.22 0.48 0.48 1.32

92 The revised eigenproblem becomes

1080 −480 3.22 0.48 q̃ = λ q̃ −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 q̃2 = q̃1 = 1.478 −1.617

Modal equations are 1.1912 p0 = 0.1584p0 7.522 0.0468 ÿ2 + 759.4y2 = − p0 = −0.00914p0 5.119

ÿ1 + 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 = Tq̃ 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

1 −K−1 θθ Kθt 3.22 0.48 0.48 1.32

 1 0 0 1 =1  2 0 

1.000 1.000 1.478 −1.617

y1 y2

3 5

2 5

ũ2 = 0.5000 − 0.4962 cos 9.71t − 0.0039 cos 27.56t Finally, transformation to original physical coordinates gives

kip.s2 /in

1080 −480 K̃ = kips/in −480 480

M̃ =

=

+ 0.0024 cos 27.56t

From Problem 13.8 T=

ũ1 = 0.3333 − 0.3357 cos 9.71t

Problem 13.9

ũ1 ũ2

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  p̃ = 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 M̃1 = q̃T1 M̃q̃1 = 7.522 M̃2 = q̃T2 M̃q̃2 = 5.119

0.6 p = 1.1912p0 0.4 0

0.6 p̃2 = [ 1 −1.617 ] p = −0.0468p0 0.4 0 p̃1 = [ 1 1.478 ]

u = Tũ  1 0 u2

0 1  u4    ũ1 1  =  2 0  ũ2 u1 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 x̃2 = x2 − c1 x1 =  0.4765  1.7559 Normalize with respect to M α22 = x̃T2 Mx̃2 = 7.1   −0.7306 x̃2 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 M̃Z = m 0 122.529

768 EI 0.8946 0 ∗ T K = Z K̃Z = 0 1874.7 28 L3

0.5021 p∗ = ZT p̃ = p0 8.6553 ∗

The modal equations are 768 EI p0 × 0.8946y1 = 0.5021 1.0082ÿ1 + 28 mL3 m 768 EI p0 122.529ÿ2 + × 1874.7y2 = 8.6553 28 mL3 m or EI p0 ÿ1 + 24.338 y1 = 0.4980 3 mL m EI p0 ÿ2 + 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 = Qũ

1 0 0 1

EI 768 1.004 −1.292 K̃ = QT KQ = 28 L3 −1.292 15.185

0.5681 p p̃ = QT p = −0.7306 0

M̃ = QT MQ = m

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

ũ = Zy ũ1 =

ũ2 =

¨ + K̃ũ = p̃ M̃ũ 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 = Qũ 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 , u̇0 and ü0 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 + u̇n + ün 0 8 0 0.1 ün+1 = −ün + 1600(un+1 − un − 0.05u̇n ) (b) u̇n+1 = −u̇n + 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 u̇n + 2 ün+1 0 0.1 0 0.1 ün+1 = −2ün + 2400 (un+1 − un − 0.05u̇n ) (b) u̇n+1 = −2u̇n − 0.025ün + 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 + u̇ −0.116 12.237 n

0.4072 −0.0014 + ü −0.0014 0.2029 n ün+1 = −2ün + 2400(un+1 − un − 0.05u̇n ) (d) u̇n+1 = −2u̇n − 0.025ün + 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 + u̇n −0.116 8.802

0.4101 −0.0020 + ün −0.0020 0.2040 The velocity and acceleration at time n + θh are obtained from Equations 13.117 and 13.116. u̇n+θ = 42.857(un+θ − un ) − 2u̇n − 0.035ün (b) ün+θ = −2ün + 1224.5(un+θ − un − 0.07u̇n ) (c) 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

(a)

+ 0.2ün2 + 15un+1 1

Table S13.17

Table S13.16 u1 in

255un+1 = pn+1 + 240un2 + 12u̇n2 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

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

ÿ1 + ω12 y1 = 5.164 g(t) ÿ2 + ω22 y2 = 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)

Mü + Ku = f g(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 ĝ(t) = e−at g(t) and determine its discrete transform Ĝ(Ω).

2. From the discrete version of the frequency function Hj sampled at the frequency interval ∆Ω obtain Ĥj for samples up to N/2 by noting that Ĥ(Ω) = H(Ω − ia). Samples of Ĥj

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 ûj (t) by taking the inverse discrete Fourier transform of the product of Ĝ(Ω) and Ĥ(Ω). Calculate the true response from uj (t) = ûj (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

x=0

∂ u = ku ∂x3

x=L

Problem 14.1 3

EI

Kinetic energy is given by L T = 0

1 m 2

∂u ∂t

2 dx

Problem 14.2

Potential energy is given by L V = 0

1 EI 2

∂2u ∂x2

2

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

(d) (e)

(d)

For section AB x1 = 0

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

x1 = L

(e)

f1 (x1 ) = 0

For section BC x2 = 0

f2 (x2 ) = 0 f2 (x2 ) = 0 f2 (x2 ) = 0

x2 = 2L

(f )

For section DB 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

L V = 0

t2 δ

2 dx

(b)

( )

)

(

)dx

( )

Acs

d2u d2u d

Acs dtdx dtdx dx

d2ut dx dt 2 d 2u

(

EA c du dx dt

du dx

d

dx

t2 " L

du EA dx dx

dx

Boundary conditions are u = 0 at x = 0, ∂u ∂x = 0 at x = L.

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

L 2 ∂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

2 ∂θ 1 dx I0 2 ∂t 0

1 M1 R 2 ∂θ 2 + M2 R + ∂t L 2 2 2 !

L ∂θ 1 GJ − dx = 0 2 ∂x 0

2

(c)

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

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

(T − V )dt = 0

or

t1 du EA dx

dtdx

∂θ ∂x

t1

δ

Acs

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

1 GJ 2

t2 L δ 0

t1

1 I0 2

2 ∂θ dx = ∂t ) t2 ( L ∂2θ I0 2 δθdx dt − ∂t t1 0

(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 L δ

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

is the mass moment of inertia about a diameter.

t2 " L 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

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 00 ∂VS = −S δu0 + ∂x 0 ∂x 0

Problem 14.5

S

∂u ∂x

δu dx

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 x=0 x=L

∂u ∂x = 0

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

(b) (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)

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

(c)

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

Equations a, b and c yield C1 = 0, C3 = 0 and C4 = C2

sinh βL sin βL

3 The boundary condition 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

gives

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

B2 = −0.9852 × 2.898A1 = −2.8552A1 x2

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

L yn (0)

fm (x)I0 fn (x)dx 0

106 Using the orthogonality relationship

ym (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 2L3 L3

x 2 x 1− θ =− 2 L

u(x, 0) =

L fm (x)θ(x, 0)dx 0 L 2 {fm (x)} dx 0 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 1 − sin 2βm L 2βm L

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 L yn (0) =

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.

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(L x)/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 πv πvt × sin ω1 t − sin L Lω1

u(x, t) =

Also

ÿn + ωn2 yn =

2A0 mΩ2 L nπ

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

Hence 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, · · ·

πx L

T =

The modal equation becomes #

sin

2P 1 1

2 mL ω1 πv 2 Lω1

−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

2

L

m1 {f1 (x1 )} dx +

M3 = 0

fn (x)ÿn dx

n=1

fi (x)

=

0

∞ 1

EAfn (x)yn dx

n=1

2

fn (x)ẏn 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 L

∞ 1

fi (x)m

fi (x)p dx

= 3m1 L Using orthogonality condition

The modal equations are EA1 0.4082 ÿ1 + (0.6155)2 y1 = P0 m1 L2 m1 L EA1 0.4082 ÿ2 + (2.5261)2 y2 = − P0 m1 L2 m1 L ÿ3 + (3.7571)2

Mi ÿi + a0 Mi ẏi + Mi ωi2 yi = pi

(c)

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

EA1 0.4082 y3 = − P0 m1 L2 m1 L

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 EA1

y1 = 1.0776

−ξ1 ω1 t

× 1−e

P0 L EA1

y2 = −0.06397

× 1 − e−ξ2 ω2 t

ξ1 ω1 cos ωd1 t + sin ωd1 t ωd1

ξ2 ω2 cos ωd2 t + sin ωd2 t ωd2

P0 L EA1

y3 = −0.02892

The modal equations are given by

ÿi = ω12 yi = or

× 1 − e−ξ3 ω3 t cos ωd3 t +

ξ3 ω3 sin ωd3 t ωd3

1.662 × 400 × 1000 cos 5.01 148, 960 × 52.5 = 0.08504 cos 5.01t 1.4678 × 400 × 1000 ÿ2 + 27.92y2 = − cos 5.01t 131, 93 × 52.5 = −0.08524 cos 5.01t

ÿ1 + 6.275y1 =

where %

#

EA m1 L2 # % EA 2 ωd2 = ω2 1 − ξ2 = 2.5620 m1 L2 # % EA1 2 ωd3 = ω3 1 − ξ3 = 3.7570 m1 L2 ωd1 = ω1

1 − ξ12 = 0.6147

The modal steady state responses are

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

(x) + y2 (t)f

(2)

(x) + y3 (t)f

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)

pi Mi

(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 52.5 0.0297 × 2.0464 N + 52.5

V (0)max = 2.52 × 109

= 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<x<c σ=

2500 2500 kN = = 8.842 MPa A 0.283 m2

113 To maximize Pg differentiate with respect to ξ and equate to 0. This gives

6.156 Incident wave

8.845

Resultant

30.0 Reflected wave

cos

10.0 3.845

2π 2π (ξ − 5) = − cos (ξ + 5) 35.38 35.38 2π 2π (ξ + 5) = π − (ξ − 5) 35.38 35.38 2ξ + 10 = 35.38 − 2ξ + 10 ξ = 35.38/4

5.0

Figure S17.2a (b) For rigid soil the force wave is reflected with the same sign as the incident wave arriving at the pile tip. On arrival at the pile cap, the wave is reflected with an opposite sign.

The maximum force at mid-height is still ± 2500 kN. To find the maximum force at the lower gage, assume that the rear end of the wave is at a distance ξ from the tip. The incident and reflected waves in this position are shown in Figure S17.2b. The equation of the wave in terms of coordinate x shown is given by 2π x P = 2500 sin 35.38

Substituting this value in Equation a

π 10π (Pg )max = 2500 sin − 2 35.38

π 10π + sin + 2 35.38 = 31.55 kN The stress measured by the gage is 31.55/0.283 = 11.15 MPa. A tensile stress of similar magnitude will be induced when the force wave arrives back at the tip after being reflected from the free end.

Problem 17.3 The displacement is given by x+ct v0 (η)dη u(x, t) = x−ct

30.0

Incident wave Reflected wave

5.0

ξ

x

For x+ct ≤ − π2 , v0 (η) = 0 and hence u(x, t) = 0. In other words u(x, t) = 0 for x ≤ − π2 −ct or x ≤ − 76 π.

Similarly, u(x, t) = 0 for x − ct ≥ π2 or x ≥ 76 π.

For intermediate values of x, the displacement is obtained as shown in Table P17.3.

Table P17.3 x Figure S17.2b The force at gage location due to incident wave is given by 2π (ξ − 5) P1 = 2500 sin 35.38 while that due to reflected wave is 2π P2 = 2500 sin (ξ + 5) 35.38 The total force is obtained from Pg = P1 + P2 = 2500 sin

2π 2π (ξ − 5) + sin (ξ + 5) 35.38 35.38 (a)

Lower Upper Adjus- Adjus- u(x, t) Limit Limit ted ted Lower Upper Limit Limit b 1 x − ct x + ct a b 2 sin x|a

−π − 53 π − 56 π − 96 π − 23 π − 43 π − 12 π − 76 π − 13 π −π − 16 π − 56 π 0 − 23 π

− 13 π − 16 π 0 1 6π 1 3π 1 2π 2 3π

− 12 π − 12 π − 12 π − 12 π − 12 π − 12 π − 12 π

− 13 π − 16 π 0 1 6π 1 3π 1 2π 1 2π

0.067 0.250 0.500 0.750 0.933 1.000 1.000

The displaced shape is symmetrical above the y axis.

114

Problem 17.4

Hence

The initial displacement shape and the position of forward and backward moving waves at ct = 0.75 are shown in Figure S17.4. The net displacement at ct = 0.75 is obtained by superposing the two latter wave shapes.

and the refracted wave is obtained from c1 2 (x − c2 t) f3 (x − c2 t) = f1 c2 1+α

u 1

2

1

= 1.4776f1 {1.414(x − c2 t)} 1

x

2

1 Forward wave

1 u 3

1 Backward wave

2

3

x

3

x

1 1

Resultant

u 3

2

1

2

1

Figure S17.4 The displacement at x = −0.25 is measured at u = −0.25.

Problem 17.5 Idealizing the building as a shear beam, for the base portion 12EI 12 × 24 × 104 = 2 h 12 × 12 = 2 × 104 kips 24 m= = 2 kip.s2 /ft2 12 # k = 100 ft/s c= m k=

For the tower portion 12 × 6 × 104 = 0.5 × 104 kips 12 × 12 12 m= = 1 kip.s2 /ft2 12

c = 0.5 × 104 = 70.71 ft/s k=

# # 1 k2 m2 0.5 × 104 × 1 = = α= k1 m1 2 × 104 × 2 8 The reflected wave is given by 1−α f2 (x + c1 t) = f1 (−x − c1 t) 1+α = 0.4776f1 (−x − c1 t) #

The position of incident, refracted and reflected waves for the first six seconds are shown in Figure S17.5.

The displacement at the level of set back at t = 2 s is measured as 0.8866 A0 ft.

At 3 s. the refracted wave is entirely in the tower portion while the reflected wave is in the base portion. Call them Waves 3 and 2 respectively.

The refracted wave moving in the tower portion reaches the free end and is reflected there with the same sign. At 4 s this reflection is entirely within the tower portion and is moving towards the junction of the tower and the base. Call this Wave No. 4. In the meantime the first reflected wave moving in the base portion arrived at the foundation and was reflected there with an opposite sign.

Call this reflected wave, Wave No. 5.

At 5 s. Wave 4 is still in the tower portion while Wave 5 is entirely in the base portion and moving upwards as shown in Figure S17.5.

It will take another 49.08/70.71 = 0.694 s. for the tail of Wave 4 to reach the set-back level.

For wave 4 $ √ 2 × 104 × 2 α1 = = 8 1 4 2 × 10 × 1 Reflected wave f6 =

1 − α1 f4 = −0.4776f4 1 + α1

Refracted wave 2 f4 = 0.5224f4 1 + α1 The positions of Waves 6 and 7 at 6 s. are shown in Figure S17.6.

In the meantime Wave 5 will reach the setback level and will be refracted and reflected there giving rise to Waves 8 and 9. f8 = 0.4776f5 f7 =

f9 = 1.4776f5 At 6 s. the position of Waves 8 and 9 will be as shown.

115 f1

f2 and f3

20

120

Resultant

14.14 f3

80 1.0

1 1.478 2 0.887 3 0.287 4 0.478

20

180

150

50

1

2 43

1

1 0.6

50 t 1.0 s.

t 2.0 s.

35.35 f5

f4 and f2 f3

35.15 120

1

1 1.478 2 0.478 264.44

119.50 180

f4

35.35

60.0

1

1 1.478

1 −0.478

2 0.098

50.0 f2

Resultant

f5

2

1 40.0 2 10.0

t 3.0 s.

t 4.0 s.

f4 and f5

f6 and f7

70.92

63.02

35.35

35.35 f6

120 53.73 180

50.0

f4 2 f5

1

52.23

1 1.478 50.0 2 0.478

1

Resultant

77.58 1 35.35 f9 1 −0.70517.07 2 0.772 50.0 f 8

99.4

90.0

f8 and f9

t 5.0 s.

Figure S17.5

f7

1 −0.706 2 −0.228 2

2 120.0

t 6.0 s.

116

Problem 17.6 The force wave propagating upwards from the base is given by p1 = g1 (x − c1 t) = k1 f1 (x − ct1 ) The refracted and reflected waves are obtained as follows d f2 (x + c1 t) dx 1−α k1 f1 (−x − c1 t) =− 1+α 1−α =− g1 (−x − c1 t) 1+α d p3 = k2 f3 (x − c2 t) dx 2 c1 c1 f (x − c2 t) = 1 + α c2 1 c2 2 c1 k2 c1 = (x − c2 t) g1 c2 1 + α c2 k1 c1 2α g1 (x − c2 t) = 1+α c2

p2 = k1

The remaining solution now proceeds as in Problem 17.5 except that when a force wave arrives at the top of the building it is reflected with the same amplitude but with opposite sign so that the net force at the free end is zero. When a force wave arrives at the base it is reflected with the same sign and the same magnitude so that the net force at the fixed base is double that caused by the incident wave alone.

117

20

120

Resultant

g2 and g3

g1

14.14 1 g3

80

2

1 400

180

20 150

50

1 209.0 2 −191.0

1 400.0

50 t 1.0 s.

t 2.0 s.

35.35 g4 and g2 35.15 120

g4

35.35

1 g3

119.50

1 209.0 2 −191.0

1

50.0

1 −191.0

1 −209.0 2 191.0

264.44 180

Resultant

g5

g2 1

2

60.0

2

g5

40.0

10.0 t 3.0 s.

t 4.0 s.

g4 and g5 63.02

70.92 1

120 53.73 50.0

2 g6

1

35.35 g6

35.35 g4

180

g6 and g7

1 −209.0 52.23 2 −191.0 50.0

2

Resultant

g8 and g9 77.58

35.35 17.07 1== 99.8 2 = −306.8 50.0

1 g9 2 g8

1 −99.8 2 91.2

g7 99.4

90.0

t 5.0 s.

120.0

t 6.0 s.

Figure S17.6

118

Chapter 18

global matrices

2EA 2 −1 −1 32 L "2 1! mL 3 6 M= 1 1 2

Problem 18.1

K=

q1

q1

(a) A

B

6

C R

L u2

(b) A

u1

B

u2

B

2

C

3

Figure S18.1, (a) Global dof, (b) Local dof Figure S18.1 shows the uniform bar, its division into 2 elements, and the local as well as the global degrees of freedom. The stiffness and mass matrices for Element 1 are given by Equation 18.25.

K1 =

2EA (−) L (1)

mL M1 = 2

"1 3

1 6

1 6

1 3

&

(−) (1) 1 −1 −1 1

!

1 6

mL  1 2 6 3 3 0 16

0



(e)

1  6  1 3

Eigenvalue solutions using Matlab gives the following frequencies: Table S18.1 Axial vibration frequencies of a bar % EA mL2

multiples of

(a)

Mode 2-Element 3-element Exact No.

When writing the virtual work equation for Element 2 we must include the term δq2 ku2 . As a result, the stiffness matrix becomes

2EA 1 −1 0 0 + (b) K2 = −1 1 0 k L On substituting k = EA/L we get

M=

'

The numbers along the borders of the stiffness matrix indicate the correspondence between the local and the global degrees of freedom. Since there are no global degrees of freedom at the fixed end, the first row and the first column of the stiffness as well as the mass matrix are omitted in the process of assembly. Term (2,2) of the element stiffness matrix is entered in location (1,1) of the global stiffness matrix, and similarly for the mass matrix.

2EA 1 K2 = L 2

3

The stiffness and mass matrices for division into three elements, each of length L/3, are obtained in a manner similar to that for the twoelement case and are   2 −1 0 3EA  K= −1 2 −1  L 0 −1 43

L

u1

(d)

1 2

1 −1 −1 32

(c)

The mass matrix for Element 2 is identical to that for Element 1. On assembly we get the following

1 2 3

2.119 6.054 -

2.068 5.474 9.723

2.029 4.913 7.979

Problem 18.2 The bar is divided into two finite elements, as shown in Figure S18.2, which also shows the global and local degrees of freedom. The stiffness and mass matrices for the two elements are given by

EA 1 −1 K1 = L −1 1

2EA 1 −1 K2 = −1 1 L "1 1! (a) M1 = mL

3

6

1 6

1 3

"1 M2 = 2mL

3

1 6

1 6

1 3

!

119 where L = l1 = l2 , m1 = m, m2 = 2m, A1 = A, and A2 = 2A. Assembly of the element matrices gives the following system mass and stiffness matrices. 1

1 6

0

0 13

2 3

3

Problem 18.3 B

C



x

  M = mL  16 1 13  

Element 1

Figure S18.3 We have for each of the two sections x N = [1 − L

B= q1

q1 B

C

L u2

A

B

(a)

1 [ −1 1 ] L

EA L −1 x 2 K= 2 [ −1 1 ] 2 − dx 1 L 0 L

7EA 1 −1 = −1 1 3L

u2

u1

B

x L ]

Referring to Figure S18.3 Let the cross-sectional 2 area at C be denoted as A = πd4 . Thus the crosssectional area at distance x from the left-hand end is given by

x 2 A(x) = 2 − (b) L The stiffness matrix for Section BC is given by

L

u1

Element 2

(b)



1 −1 0 EA  K= −1 3 −2  L 0 −2 2

A

D

C

(c)

Figure S18.2

To satisfy the boundary condition at the clamped end where u = 0, the first row and first column of each of the mass and stiffness matrices are omitted; so that the free vibration equation becomes. " mL

1

1 3

1 3

2 3

! ü1 ü2

+

EA L

"

3 −2 −2

ω1 = 0.625

0

=

2

u1 u2

(c)

EA and ω2 = 3.035 mL2

#

EA mL2

The exact frequencies as obtained in Example 15.2 are # # EA EA and ω2 = 2.526 ω1 = 0.616 mL2 mL2

L

AρL 62 23 60 23 32

x L ]

2−

x 2 dx L

(d) The mass and stiffness matrices for Section CD

!

Standard eigenvalue solution gives the following values for the frequencies #

and the mass matrix is L x 1− L x M = Aρ [1 − L x

are K=

EA 1 −1 L −1 1 "1 1!

M = ρAL

3

6

1 6

1 3

(e)

On assembling the element matrices and applying the clamped-end boundary condition we get the following system equation " 10 ! " 13 1 !

EA 3 −1 ü1 u1 15 6 + ρAL 1 1 ü2 u2 L −1 1 6 3 (f ) =

0 P

120 (a) When P is applied statically the accelerations ü1 and ü2 are zero. Solving the resulting Equation f for u1 and u2 we get u1 =

3 PL 10 P L and u2 = 7 EA 7 EA

u(y) = u1 1 −

y L

+ u2

y L

(h)

where x is measured from B and y from C. The exact values of the displacement are determined from x u(x) = 0

1 P

dx AE 2 − x 2 L

x PL = 2AE 2L − x PL Py Py = + u(y) = u1 + AE 2AE AE

(g)

σ1 = E

(h)

The strains and stresses are constant throughout a section because we have used linear interpolation functions as given in Equation (a). In comparison, the exact values are given by σ1 =

P

x 2 2− L

σ2 =

P A

φ2 = ω12 = 1.162

0.5996 1.1683 −0.9577 1.3979

E E ω22 = 7.689 2 2 ρL ρL

(l)

The modal equations are given by P ρAL P ÿ2 + ω22 y2 = 1.3979 ρAL

Transformation to the physical coordinates gives u1 u2

PL EA PL + EA =

0.6027 1.1744

(1 − cos ω1 t)

−0.1741 0.2542

(1 − cos ω2 t)

ÿ1 + ω12 y1 = 1.1683

(m)

(o)

Problem 18.4 The stiffness and mass matrices for a single linear element are

EA 1 −1 K1 = L −1 1 "1 1! (a) 3

6

1 6

1 3

An eigenvalue solution gives two frequencies; the first one is zero and corresponds to the rigid body mode, the second frequency is given by # EA ω2 = 3.464 mL2 The stiffness and mass matrices for two linear elements are   1 −1 0 2EA  −1 2 −1  K2 = L 0 −1 1 1 3

M2 =

(k)

(n)

PL (1 − cos ω2 t) y2 = 0.1818 EA

(j)

(b) The eigenvalues obtained by standard method from the stiffness and mass matrices are φ1 =

PL (1 − cos ω1 t) EA

M1 = mL

The approximate values of stresses in the two sections are obtained from d x 3 P u1 = dx L 7A

y y P d u1 1 − + u2 = σ2 = E dy L L A

y1 = 1.0052

(g)

The variation of displacements along the length of sections BC and CD are given by x u(x) = u1 L

Solution of the modal equations gives

1 6

mL  1 2 6 3 2 0 16

0



(b)

1  6  1 3

The two non-zero frequencies are # # EA EA ω2 = 3.464 ω3 = 6.928 mL2 mL2

(c)

The stiffness amd mass matrices for a quadratic element are given by Equations 18.87 and 18.89, respectively.   7 −8 1 EA  −8 16 −8  K3 = 3L 1 −8 7 (d)   4 2 −1 mL  M3 = 2 16 2  30 −1 2 4

121 A standard eigenvalue solution using a Matlab routine gives the following non-zero frequencies #

#

EA mL2

ω2 = 3.464

EA mL2

ω3 = 7.746

Table S18.5b Correspondence between the local and global degrees of freedom

(e)

Local dof

ω2 = 3.142

#

EA mL2

ω3 = 6.283

EA mL2

2

3

4

Global dof

The corresponding exact frequencies are #

1

Element 1 Element 2 Element 3 Element 4 Element 5

(f )

5 1 1

2 2

3 3 3 1 5

4 4 4 2 -

Problem 18.5 The stiffness matrix for Element No. 1 is obtained from Equation d of Example 3.6 with angle α = 30o

4 3 1

K1 = 

2 3

2 5

1

30 4 L

5

 0.6495 0.3750 −0.6495 −0.3750 EA  0.3750 0.2165 −0.3750 −0.2165    −0.6495 −0.3750 0.6495 0.3750 L −0.3750 −0.2165 0.3750 0.2165

L

The stiffness matrix for Element No. 2 is obtained with α = 150o .

Figure S18.5

K2 =  The global degrees of freedom are shown in Figure S18.5, which also shows the numbers assigned to the five elements. The properties of the various elements are listed in Table S18.5a.

Table S18.5a Properties of the Elements of Truss Element Length Area α No. Deg. 1 2 3 4 5

1.155L A 1.155L A 0.577L 0.5A L 0.7A L 0.7A

30 150 90 0 0

The correspondence between the local and global degrees of freedom is given in Table S18.5b.

 0.6495 −0.3750 −0.6495 0.3750 EA  −0.3750 0.2165 0.3750 −0.2165    −0.6495 0.3750 0.6495 −0.3750 L 0.3750 −0.2165 −0.3750 0.2165 For Element 3, with α = 90o   0 0 0 0 EA  0 0.8660 0 −0.8660  K3 =   0 0 0 0 L 0 −0.8660 0 0.8660

Finally for Elements 4 and 5 we have α = 0o so that   0.7000 0 −0.7000 0 EA  0 0 0 0 K4 = K5 =   −0.7000 0 0.7000 0 L 0 0 0 0 On assembling the element matrices using the correspondence between the local and global degrees of freedom we get the following global stiffness matrix EA K= × L   1.400 0 0 0 −0.700 0.866 0 −0.866 0   0   0 1.299 0 −0.650   0   0 −0.866 0 1.299 0.375 −0.700 0 −0.650 0.375 1.350

122 Part (a) The mass matrices remain unaltered under transformation, and are thus given by

unaltered under coordinate transformation. As an example the mass term at Node D, is given by

M1 = M2 =   0.3849 0 0.1925 0 0.3849 0 0.1925   0 ρAL   0.1925 0 0.3849 0 0 0.1925 0 0.3849

M (3, 3) = M (4, 4) = 0.5 (ρA1 L1 + ρA2 L2 + ρA3 L3 ) = 0.5 ρAL (1.156 + 1.156 + 0.577 × 0.5) = 1.299 ρAL The lumped mass matrix is given by

M3 = 

 0.0962 0 0.0481 0 0.0962 0 0.0481   0 ρAL   0.0481 0 0.0962 0 0 0.0481 0 0.0962 M4 = M5   0.2333 0 0.1167 0 0.2333 0 0.1167   0 ρAL   0.1167 0 0.2333 0 0 0.1167 0 0.2333

M = ρAL×   0.844 0 0 0 0 0 0   0 0.844 0   0 1.299 0 0   0   0 0 0 1.299 0 0 0 0 0 0.927 An eigenvalue solution using the stiffness matrix from Part (a) and the lumped mass matrix gives the following frequencies and mode shapes

Assemby of the local mass matrices yields the following global matrix

$ ω1 = 0.2842

M = ρAL×   0.5629 0 0.0481 0 0.1167 0.5629 0 0.0481 0   0   0 0.8660 0 0.1925   0.0481   0 0.0481 0 0.8660 0 0.1167 0 0.1925 0 0.6182 Standard eigenvalue solution using the global stiffness and mass matrices and a Matlab routine gives the following frequencies and massorthonormal mode shapes $ ω1 = 0.3263 $ ω3 = 1.4436 $ ω5 = 2.2696

$ E ω2 = 0.7367 ρL2

E ρL2

E ω4 = 1.7156 ρL2

E ρL2

$

E ρL2

Φ=   0.247 0.400 0.906 0.261 −0.860  0.707 0.561 0.142 −0.971 −0.087     −0.258 0.519 −0.811 0.038 −0.496    0.654 0.353 −0.045 0.762 0.160 −0.463 0.556 0.304 −0.069 1.086 Part (b) The lumped mass matrix is obtained by assigning half of the total mass of an element to each of its end node, in each x and y direction. It may be noted that the lumped mass matrix remains

$ ω3 = 1.1193 $ ω5 = 1.5877

$ E ω2 = 0.6780 ρL2 E ω4 = 1.3509 ρL2

$

E ρL2 E ρL2

E ρL2

Φ=   −0.206 0.386 0.656 −0.308 −0.684  0.626 0.438 0.152 0.748 −0.133     −0.213 0.517 −0.633 −0.038 −0.234    0.577 0.242 −0.034 −0.583 0.193 −0.391 0.559 0.320 0.062 0.712

Problem 18.6 The revised stiffness and mass matrices are obtained from similar matrices obtained in Problem 18.5 by deleting row and column numbers 5, giving 

 1.400 0 0 0 EA  0 0.866 0 −0.866  K=   0 0 1.299 0 L 0 −0.866 0 1.299 

 0.5629 0 0.0481 0 0.5629 0 0.0481   0 M = ρAL   0.0481 0 0.8660 0 0 0.0481 0 0.8660 An eigenvalue solution gives the following frequencies and mode shapes

123 $ ω1 = 0.5107 $ ω3 = 1.5864

On carrying out the integration according to Equation 18.103 we get a stiffness matrix that is identical to that given by Equation 18.35. and a mass matrix that is identical to the one given by Equation 18.39.

$ E ω2 = 1.2204 ρL2

E ρL2

E ω4 = 1.7216 ρL2

E ρL2

$

Problem 18.8



 0.0000 0.1354 −1.3292 0.0000  0.9046 0.0000 0.0000 −0.9832  Φ=  0.0000 1.0615 0.1828 0.0000 0.7405 0.0000 0.0000 0.7822

5

3 4

2 1

Modes 1 and 4, which involve vertical motions of nodes B and D, but no horizontal motion, are symmetric modes. Modes 2 and 4 that involve only horizontal motions of nodes B and D are antisymmetric modes.

6

4

2 3

1

2 3

1

2

1

4

2

4

1

3 3

Figure S18.8

Problem 18.7 Coordinate x is given by x = L1 x1 + L2 x2

(a)

Noting that L1 + L2 = 1 so that L2 = 1 − L1 and L1 = 1 − L2 , we have ∂x = x1 − x2 = −L ∂L1 ∂x = −x1 + x2 = L ∂L2

(b)

The inverse relationship is thus given by ∂ ∂L2 d ∂ ∂L1 + = dx ∂L1 ∂x ∂L2 ∂x

∂ 1 ∂ = − L ∂L2 ∂L1

where l is the length of the segment.

For segment 1

Bq =

(d)

in which we have used Equation c to carry out the differentiations.

The stiffness and mass matrices are now obtained from

BT B dL K = EI L

(e) T M = m N N dL L

x 2 ψ1 = x 1 − l x 3 x 2 +2 ψ2 = 1 − 3 ll l

x2 x −1 ψ3 = l l x 2 x 3 ψ4 = 3 −2 l l

(c)

The curvature d2 w/dx2 plays the role of strain and the strain displacement relationship is therefore given by d2 w d2 N = q dx2 dx2 1 = 2 [ (6 − 12L1 ) L(2L2 − 4L1 ) L (6 − 12L2 ) L(4L2 − 2L1 ) ] q

The three finite elements and the local degrees of freedom are shown in Figure S18.8. The shape functions are

2x EI(x) = EI 3 − l

x m(x) = m 2 − l l

kij = EIψi (x)ψj (x)dx 0

10EI 14EI , k12 = 2 k11 = l l 4EI 14EI k13 = , k14 = − 2 l l 24EI 10EI k23 = 2 k22 = 3 , l l 24EI k24 = − 3 l 6EI 10EI , k34 = − 2 k33 = l l

124 Assembling and noting that l = L4 2



−

1

2

3

 224

− L2     224 1536 160 1536  − − L3 L2 L3   L2  K1 = EI    16 24 160  160 2  − L2 L L2   L   1536 160 1536 3 − 224 − − L2 L3 L2 L3 1

40 L

224 L2

16 L

The numbers along the borders of the stiffness matrix indicate the correspondence between the local and the global degrees of freedom.

Noting that mij =

l m(x)ψi (x)ψj (x)dx 0

2 3 2L

 1

3 11 2 2 L

4 − 38 L3

5



13 2 4 L

   11 2  13 2  3 78L − 4 L 27L  2 L  m    M2 =   420  3 3 13 2 1 3 11 2  4  −8L − 4 L −2L  2L   13 2 11 2 5 L 27L − L 78L 4 2 For segment 3

x I(x) = I 1 + 2 l

x m(x) = m 1 + l l

kij = EI(x)ψi (x)ψj (x)dx 0

l

13 37 2 l ml3 , m12 = 420 840 9 19 2 ml3 , l m13 = − m14 = 840 420 23 1 ml, m23 = − ml2 m22 = 35 21 27 m24 = ml 140 29 11 ml3 , ml2 m34 = − m33 = 840 420 16 ml m44 = 35

mij =

m11 =

Calculations similar to those shown earlier give

4

37 2 8 L

9 3 − 64 L

19 2 8 L

81 2 L

2 − 29 8 L

19 2 8 L



   37 2  81 5 2 L 138L − L L   2 2 m  8  M1 =   840  9 3 5 2 11 3 2 L  − 64 L − 2 L 64 L − 29  8   96L

The correspondence numbers have not been shown against matrix M1 but they are identical to those on the stiffness matrix K1 .

For the uniform segment 2, the stiffness and consistent mass matrices are obtained from standard expressions with l = L/2 

2

3

4

8 L

24 L2

4 L

  24 96 24 3 L3 L2  L2  K2 = EI  4 8 24 4 L2 L  L  5 − L242 − L963 − L242

4

5

6

24 L

160 L2

16 L

 11

3 64 L

2



  160 1536 224 5  L3 L2  L2  K3 = EI  16 40 224 6  L2 L  L  1536 224 − − 160 L2 − L3 − L2

Assembling and substituting l = L4  13

EI(x)ψi (x)ψj (x)dx 0

5  − L242   96  − L3    24  − L2   96 L3

−  − 160 L2   1536  − L3    224  − L2   1536 L3

3 64 L

29 2 8 L

9 3 − 64 L

5 2 2L

81 2 L

2 − 37 138L 8 L

5 2 2L



   29 2  L2 81 L  L 96L − 19  8 2 8 m   M3 =    840  9 3 19 2 13 3 L2   − 64 L − 8 L 64 L − 37 8  

The correspondence numbers for M3 are identical to those for K3 .

The global matrices are now assembled using the correspondence numbers. 

13L3 −9L3 152L2 0  −9L3 75L3 472L3 −48L3 m   152L2 472L2 16128L −416L2 M=  −48L3 −416L2 75L3 53760  0  0 416L2 3456L −472L2 0 0 0 −9L3  0 0 2 0 416L   3456L 0  2 3  −472L −9L   16128L −152L2 2 3 −152L 13L

125 

16 L

40 L

− 224 L2

0

  16 32 − 136  L L L2    224 1632  − L2 − 136 L2 L3  K = EI   4 24  0 L L2     0 − L242 − L963  0

0

0

4 L

− L242

24 L2

− L963

32 L

136 L2

136 L2

1632 L3

16 L

224 L2

0

0



Element AB has both axial and flexural deformations. The stiffness matrix is therefore expanded to include the axial degrees of freedom as shown in Figure S18.9b. The stiffness matrix is given by

  0     0    16  L    224  L2  

 EA L

 EA − L    0    0    0 

40 L

Problem 18.9 (a)

0

30°

2

A

6

4 5

3

C

B 2000

0

EA L

0

0

0

0

4EI L

6EI L2

2EI L

0

6EI L2

12EI L3

6EI L2

0

2EI L

6EI L2

4EI L

0

0



    6EI  − L2    − 12EI L3    − 6EI L2  0

12EI − 6EI − 6EI L2 − L3 L2

12EI L3



 15020 11.27 7510 −11.27  11.27 0.01127 11.27 −0.01127  E  7510 11.27 15020 −11.27 −11.27 −0.01127 −11.27 0.01127

3

2 1 3

0

The flexural component is given by

1000

4

(b)

0

The axial and flexural deformations are uncoupled. The axial component of the stiffness matrix, obtained by substituting the various parameter values in units of Newton and mm is

0.62 −0.62 E −0.62 0.62

D

1

− EA L

4

5

1

4

2

6

3

1

2

(c)

5 kN 10 kN.m

The stiffness matrix of strut BD is   0.0650 −0.0375 −0.0650 0.0375  −0.0375 0.0217 0.0375 −0.0217  E  −0.0650 0.0375 0.0650 −0.0375 0.0375 −0.0217 −0.0375 0.0217

5 kN 10 kN (d)

10 kN.m

10 kN (e)

The stiffness matrix of element BC is   30040 45.06 15020 −45.06  45.06 0.09012 45.06 −0.09012  E  15020 45.06 30040 −45.06 −45.06 −0.09012 −45.06 0.09012

15 kN

25.98 kN

The correspondence between the local and global degrees of freedom are given in Table S18.9.

Table S18.9 Correspondence between the local and global degrees of freedom

25.98 kN 15 kN

Local dof

1

3

4

5

6

2 -

4 -

Global dof

Figure S18.9

The crane is shown in Figure S18.9a, which also identifies the global degrees of freedom.

2

Element AB Element BC Element BD

2 3

3 4 4

1 5 -

6 -

126 The local matrices are assembled using the correspondence between the local and global degrees of freedom to obtain the global stiffness matrix 

15020 7510 0 −11.27 0 −33.80  7510 45060  0 0.6850 −0.0375  0 K=E 0.1230  −11.27 33.80 −0.0375  0 15020 0 45.06 0 −45.06 0 −0.09012  0 0 15020 −45.06   0 0   45.06 −0.09012   30040 −45.06 −45.06 0.09012

and the corresponding forces obtained from pBD = KBD uBD are shown in Figure S18.9e.

The largest stress in arm ABC occurs in element AB at B. It is compressive stress at the bottom of the section and is given by 25981 10 × 106 × 100 P Md + = 154.1 MPa = + 7.51 × 106 A 2Ib 1240

Problem 18.10 The structure can be modeled as a singledegree-of-freedom system, with the vertical displacement at the tip C being treated as the selected degree of freedom. The stiffness along ths d.o.f is obtained from the ratio of applied load at C to the deflection at C. Thus, we have

The global load vector is given by PT = [ 0 0 0 0 0 −10000 ] N The equilibrium equation is given by Ku = P

The displacement vector for element BD is   −0.2095     −3.8270 mm uBD = 0     0

(a) k=

On solving Equation a we get the vector of global displacements   0.0003058      −0.006352        −0.2095 u= mm   −3.8270         −0.009680   −12.3983 The tip deflection is -12.40 mm. The corresponding displacements in element AB are       

 0   −0.2095     0.0003058 uAB = mm 0           −0.006352   −3.827 The element forces are obtained from pAB = KAB uAB and are shown in Figure S18.9c.

For element BC the displacement vector is   −0.006352     −3.8270 uBC = mm   −0.009681   −12.398 The element forces obtained from pBC = KBC uBC are shown in Figure S18.9d.

10000 = 806.56 N/mm 12.3983

The concentrated mass at C is m = 5000/9.81 = 509.7 kg, and the vibration frequency is given by ω = k/m = 1.258 rad/s.

The crane will undergo free vibrations about the position of rest under the gravity load of 5000 N, that is with a tip deflection of 6.199 mm. The displacement relative to this position is given by ur = 6.199 cos ωt = 6.199 cos (1.258t) The stiffness along the vertical d.o.f. at C can also be obtained by condensing out all other d.of. from the system stiffness matrix. Thus k = K22 − K21 K−1 11 K12

(a)

where K11 is a sub-matrix comprising the first 5 rows and the first 5 columns of the global stiffness matrix K obtained in Problem 18.9, K12 is formed from the first five rows and Column No. 6 of matrix K, K21 = KT12 , and K22 is the single element K(6, 6). On carrying out the operations in Equation a we have k = 806.56, which is the same as obtained directly.

Problem 18.11 The applied load can be expressed as p = p1 +

p2 − p1 x L

127 where L is the length of the beam. The consistent load vector is given by

L NT

p= 0

p2 − p1 p1 + x dx L

3

2

1

(a)

C

B

(a) A

where D

N = [ N1 N2 N3 N4 ] and

x 2 N1 (x) = x 1 − L x 2 x 3 N2 (x) = 1 − 3 +2 L L

x2 x N3 (x) = −1 L L x 2 x 3 −2 N4 (x) = 3 L L

2

(b) 1

4

1 B

2

3

1

1 1 2 2 20 p1 L + 30 p2 L

3

A 4



  7 3 p2 L   20 p1 L + 20   f =  1 1 p2 L2  p1 L2 − 20  − 30

2

(b)

3

On carrying out the integration in Equation a, we get the following equivalent loads 

C

C

B

D 4

Figure S18.12

(c)

3 7 20 p1 L + 20 p2 L

The applied load is now expressed as

the local and global degrees of freedom is shown in Table S18.12.

Table S18.12 Correspondence between the local and global degrees of freedom in the frame

p = p1 L1 + p2 L2 Local dof and the consistent load vector is obtained from  2  L1 (3 − 2L1 )    L L21 L2 L   f=  (p1 L1 + p2 L2 ) dL (d) 2  L (3 − 2L ) 2 0   2   2 −L1 L2 L When integrations in Equation d are carried out using Equation 18.103 we get the same load vector as in Equation c.

Problem 18.12 As shown in Figure S18.12a, the frame has three degrees of freedom, rotations at Nodes B and C, and lateral translation along BC. The stiffness and mass matrices for individual elements corresponding to their local degrees of freedom, shown in Figure 18.12b, are obtained from Equations 18.35 and 18.39. In addition the mass of beam BC contributes inertia force along the lateral translation degree of freedom u3 . The correspondence between

1

2

3

4

Global dof Element BA Element BC Element CD

1 1 2

3 3

2 -

-

The element stiffness matrices are assembled in accordance with the correspondence between the local and global degrees of freedom to obtain the following global stiffness matrix  8 L

 K = EI  L2 6 L2

2 L

6  L2

8 L

3  L2 

3 L2

15 L3

(a)

The global mass matrix is obtained in a similar manner. However, the term 10mL must be added in the position (3,3) to account for the inertia contributed by the vibration of beam BC as a rigid body along degree of freedom 3. The resulting mass

128 matrix is given by   0.3905L2 −0.2857L2 0.0524L   M = mL  −0.2857L2 0.4952L2 0.3143L  (b) 0.0524L

where

8 2 X= 2 8

−1

The transformed matrices are obtained from Equations 3.80 and 3.81   8 2 6 EI 7 = 2 8 3  K L3 6 3 15 (d)   0.3905 −0.2857 0.0524 8 = mL  −0.2857 0.4952 0.3143  M 0.0524 0.3143 11.4857

0.7 0.2

Transformation of the global stiffness and mass matrices gives EI 7 k = TT KT = 10.2 3 L m 7 = TT MT = 11.42 mL The corresponding frequency and mode shape are $ ω=

7 k = 0.9452 m 7

Standard eigenvalue solution using the transformed matrices gives the following frequencies % and

EI mode shapes. Frequencies as multiples of mL4 are: ω1 = 0.9439, ω2 = 2.8986, ω3 = 8.1392. Mode shapes in coordinates u1 , u2 , and u3 are

φT1 = [ −0.2174 −0.0413 0.2951 ] φT2 = [ −0.7641 0.8803 −0.0482 ] −1.6897 0.0337 ]

If a lumped mass matrix is used, the only non-zero term in the mass matrix will be along the translational degree of freedom; its value being 12 mL + 5m × 2L + 12 (1.5m × 2L) = 12mL. An eigenvalue solution with the lumped mass matrix gives # EI ω = 0.9220 φT = [ −0.7 −0.2 1.0 ] mL4 The transformation matrix in Guyan’s reduction is given by −X 1

#

EI mL4

φT = [ −XT 1 ] = [ −0.7 − 0.2 1.0 ]

Problem 18.13 p2y

6 C

p2x

5

A

p1y

4

2

T=

=

0.3143L 11.4857

Since the stiffness and mass matrices carry terms with differing powers of the variable L, a numerical solution of the eigenvalue problem is difficult. This can be resolved if all displacements along all of the coordinates are measured in the same units. We carry out a coordinate transformation in which the rotations θ1 and θ2 are expressed, respectively, as u1 /L and u2 /L, where coordinates u1 and u2 are measured in units of length. The transformation is expressed as   1    θ1  L 0 0  u1  =  0 L1 0  u2 = Tu (c) θ  2   u3 u3 0 0 1

φT3 = [ −1.9743

6 3

1

B

3

p1x

Figure S18.13 The degrees of freedom are shown in Figure S18.13. From Equation 18.67 we have a1 = 508, a2 = −8, a3 = −100 b1 = −16, b2 = 16, b3 = 0 , c1 = −15, c2 = −10, c3 = 25 G = a1 + a2 + a3 = 400, and A = 200 cm2 The strain-displacement matrix is given by   −16 0 16 0 0 0 1  0 −15 0 −10 0 25  B= 400 −15 −16 −10 16 25 0 The elasticity matrix is given by 

 1.0 0.3 0 D = 2.198 × 107  0.3 1.0 0  0 0 0.35

129 The stiffness matrix is now obtained from Equation 18.73

BT DBt dA K= 

Problem 18.14 (a)

4

6 D

C

A

5

1.8393 0.8571 −1.1181 −0.1978  0.8571 1.7286 −0.0879 0.3319   −1.1181 −0.0879 1.5989 −0.5714 107   −0.1978 0.3319 −0.5714 1.0418  −0.7212 −0.7692 −0.4808 0.7692 −0.6593 −2.0604 0.6593 −1.3736  −0.7212 −0.6593 −0.7692 −2.0604   −0.4808 0.6593   N/cm 0.7692 −1.3736   1.2019 0 0 3.4341 The mass matrix is given by

M= NT Nρt dA A

 2 0 1 0 1 0 0 2 0 1 0 1   1 0 2 0 1 0 = 0.26   0 1 0 2 0 1   1 0 1 0 2 0 0 1 0 1 0 2

The nodal loads are given by

100 200 + = 2.924 × 105 N p1 = St 3 6

100 200 + = 3.655 × 105 N p2 = St 6 3

These loads act in a direction perpendicular to side BC. They can be resolved in the x and y directions.

Let θ be the angle between the y axis and the perpendicular to side BC, then sin θ =

2

1

16 15 cos θ = 21.93 21.93

p1x = p1 sin θ = 2.133 × 105 p1y = p1 cos θ = 2.0 × 105 p2x = p2 sin θ = 2.667 × 105 p2y = p2 cos θ = 2.5 × 105

2

8

A

B

7

1

Figure S18.14a Global d.o.f. (b) (0,8)

4

6 C

D

6 (10,8)

C

3

5



The length of side BC is obtained from

S = (13 − 28)2 + (20 − 4)2 = 21.93 cm

3

(10,8) 5

2

1 2

2

4

A

(0,0) A

1

(0,0)

B (10,0) 3

1

Figure S18.14b Local d.o.f.

The the local degrees of freedom for each of the two elements as well as the global degrees of freedom are shown in Figure S18.14. For Element 1 a1 = 80, a2 = 0, a3 = 0 b1 = −8, b2 = 8, b3 = 0 , c1 = 0, c2 = −10, c3 = 10 G = a1 + a2 + a3 = 80, and A = 40 in2 The strain-displacement matrix for Element 1 is given by   −8 0 8 0 0 0 1  0 0 0 −10 0 10  B1 = 80 0 −8 −10 8 10 0 The elasticity matrix is given by 

 1.0 0.3 0 D = 3.297 × 107  0.3 1.0 0  0 0 0.35

130 The stiffness matrix for Element 1 as obtained from Equation 18.73 is

K= BT DBt dA 

A

1.3187 0 −1.3187 0.4945 0 0.4615 0.5769 −0.4615   −1.3187 0.5769 2.0398 −1.0714  107  0.4945 −0.4615 −1.0714 2.5220   0 −0.5769 −0.7212 0.5769 −0.4945 0 0.4945 −2.0604  0 −0.4945 −0.5769 0   −0.7212 0.4945   lb/in 0.5769 −2.0604   0.7212 0 0 2.0604 For Element 2 a1 = 80, a2 = 0, a3 = 0 b1 = 0, b2 = 8, −8

The x direction loads acting at nodes B and C are given by

900 1200 + = 4000 lbs 3 6

900 1200 p2 = 8 × 1 = 4400 lbs + 3 6

p1 = 8 × 1

The nodal load vector for the unrestrained nodes is therefore given by pT = [ 4000 0 4400 0 ] lbs The global equilibrium equation Ku = p is now solved to get uT = 10−4 [ 3.432 0.5130 3.225 −0.2540 ] in Using the correspondence between the local and global degrees of freedom we have

, c1 = −10, c2 = 0, c3 = 10

uT1 = 10−4 [ 0 0 3.432 0.513 3.225 −0.254 ]

G = a1 + a2 + a3 = 80, and A = 40 in2

uT2 = 10−4 [ 0 0 3.225 −0.2540 0 0 ] in

The strain-displacement matrix for Element 1 is given by   0 0 8 0 −8 0 1  B2 = 0 −10 0 0 0 10  80 −10 0 0 8 10 −8 The stiffness matrix for Element 2 is K2 = 

0.7212 0 0 −0.5769 0 2.0604 −0.4945 0   0 −0.4945 1.3187 0 7 10  0 0 0.4615  −0.5769  −0.7212 0.4945 −1.3187 0.5769 0.5769 −2.0604 0.4945 −0.4615  −0.7212 0.5769 0.4945 −2.0604   −1.3187 0.4945   lb/in 0.5769 −0.4615   2.0398 −1.0714 −1.0714 2.5220 Using the correspondence between the local and global degrees of freedom shown in Figure S18.14 the element stiffness matrices are assembled to get the following global stiffness matrix corresponding to the unrestrained degrees of freedom K= 

 2.0398 −1.0714 −0.7212 0.4945  −1.0714 2.5220 0.5769 −2.0604  lb 107   −0.7212 0.5769 2.0398 0 in 0.4945 −2.0604 0 2.5220

The element strains are obtained as follows For Element 1    0.3432  1 = B1 u1 = 10−4 −0.0958   0.0254 For Element 2    0.3225  2 = B2 u2 = 10−4 0   −0.0254 The stresses for the two elements are given by    1.0367  psi σ1 = D 1 = 103 0.0235   0.0293    1.0633  σ2 = D 2 = 103 0.3190 psi   −0.0293 The strains and stresses are constant throughout an element.

131

Problem 18.15 (a)

11

17

8 C

D

18

10

12

15

4

6

7

9

16

A

2

5

14

13

3

B

1

Figure S18.15a Global d.o.f.

(b)

9

11 (10,8)

(5,8)

(0,8) D

2

(0,4)

9

C

C (10,8) 2

5

3 12

8

3

12

10

11

(5,4)

(5,4) 6

(10,4)

4

5

6 1

7

7

10

8 B

A

(0,0) A

1

(0,0)

1

(5,0)

4

(10,0) 2

Figure S18.15b Local d.o.f.

The the local degrees of freedom for each of the two elements as well as the global degrees of freedom are shown in Figure S18.15. The element stiffness matrices are obtained from Equations 18.123, 18.124,

132 and 18.125, and are given by 

1.3187 0.4396 0 −1.7582 0 0 0 −0.1648 0 −0.1923 −1.0714  0.4396 2.0398 0.2404 −1.7582 −0.9615  0 0.2404 0.7212 0 −0.9615 0 0.1923 −0.1923   0 5.4396 0 −1.9231 0.7692 0.6593  −1.7582 −1.7582  0 −0.9615 −0.9615 0 5.4396 −3.5165 0 0.7692   0 0 0 −1.9231 −3.5165 5.4396 −0.7692 0 7 K1 = 10  0 −0.1923 0.1923 0.7692 0 −0.7692 0.4615 0.1538   0 0.1538 2.5220  −0.1648 −1.0714 −0.1923 0.6593 0.7692  0 0 0.6593 −0.6593 0 0.6868  0.1648 −0.1648  0.6593 0.7692 0 −1.4286 −1.4286 1.4286 −0.6154 −0.6154   0 0.6593 0.7692 −1.4286 −1.4286 1.4286 0 −2.7473 −0.6593 0 −0.7692 1.4286 1.4286 −1.4286 0 0  0.1648 0.6593 0 −0.6593 −0.1648 0.7692 0.6593 0   0 0 0.7692 −0.7692   0 −1.4286 −1.4286 1.4286   0.6593 −1.4286 −1.4286 1.4286   −0.6593 1.4286 1.4286 −1.4286   lb/in 0 −0.6154 0 0   0.6868 −0.6154 −2.7473 0   2.0604 0 −2.7473 0   0 6.7253 0 −5.4945   −2.7473 0 6.7253 −1.2308 0 −5.4945 −1.2308 6.7253 

0.7212 0 0.2404 0 0 −0.9615 0 0.1923 0 1.3187 0.4396 0 −1.7582 0 0.1648 0   0 −1.7582 −0.9615 −0.1648 −0.1923  0.2404 0.4396 2.0398  0 0 0 5.4396 −1.9231 −3.5165 −0.6593 −0.7692   0 −1.7582 −1.7582 −1.9231 5.4396 0 0 0.7692   0 −0.9615 −3.5165 0 5.4396 0.6593 0 7  −0.9615 K2 = 10  0 0.1648 −0.1648 −0.6593 0 0.6593 2.0604 0   0 −0.1923 −0.7692 0.7692 0 0 0.4615  0.1923  0 0.6593 0.7692 0.6868 0.1538  −0.1923 −0.1648 −1.0714  0 −1.4286 1.4286 1.4286 0 0  −0.7692 −0.6593  0 0.6593 0.7692 1.4286 −1.4286 −1.4286 0 −0.6154 0.7692 0 0.6593 1.4286 −1.4286 −1.4286 −2.7473 0  −0.1923 −0.7692 0 0.7692 −0.1648 −0.6593 0.6593 0   −1.0714 0 0.7692 0.6593   0 −1.4286 1.4286 1.4286   0.6593 1.4286 −1.4286 −1.4286   0.7692 1.4286 −1.4286 −1.4286   lb/in 0.6868 0 0 −2.7473   0.1538 0 −0.6154 0   2.5220 0 −0.6154 −2.7473   0 6.7253 −5.4945 −1.2308   −0.6154 −5.4945 6.7253 0 −2.7473 −1.2308 0 6.7253 The coorespondence between the local and global degrees of freedom are shown in Table S18.15.

133 Table S18.15 Correspondence between the local and global degrees of freedom

Local dof

1

2

3

4

5

6

7

8

9

10

11

12

6 16 7 15 16 8

8 17

9 10 12 12 11 18

Global dof Element 1 Element 2

13 13

1 2

2 3 4 14 6 5

Using the correspondence between the local and global degrees of freedom the element stiffness matrices are assembled to get the global stiffness matrix. The following matrix is obtained corresponding to the unrestrained degrees of freedom. 

0.2040 0.0240 −0.1758 −0.0962 0 0 −0.1071 −0.0165 0.0240 0.2040 0 −0.0962 −0.1758 0 −0.0192 0   −0.1758 0 0.5440 0 0 −0.1923 0.0659 0   0 0.5440 0 −0.3516 0.0769 0.0659  −0.0962 −0.0962  0 −0.1758 0 0 0.5440 −0.1923 0 0.0769   0 0 −0.1923 −0.3516 −0.1923 1.0879 0 −0.1429  K = 108  0 0 0.2522 0.0687  −0.1071 −0.0192 0.0659 0.0769  −0.0165 0 0 0.0659 0.0769 −0.1429 0.0687 0.2522   0.0769 0 −0.1429 −0.1429 0 0.1429 −0.0615 0   0 0.1429 −0.2747 −0.2747  0.0659 0.0769 −0.1429 −0.1429  0 0.0659 0 0 −0.1429 0.1429 0 −0.0615 0 −0.1429 0.1429 0.1429 0.1429 −0.2857 0 0  0.0769 0.0659 0 0 0 0.0769 0.0659 −0.1429   −0.1429 −0.1429 0 0.1429   −0.1429 −0.1429 0 0.1429   0 0 −0.1429 0.1429   0.1429 0.1429 0.1429 −0.2857   lb/in −0.0615 −0.2747 0 0   0 −0.2747 −0.0615 0   0.6725 0 0 −0.5495   0 0.6725 0 −0.1231   0 0 0.6725 −0.5495 −0.5495 −0.1231 −0.5495 1.3451 The nodal forces that are equivalent to the traction on surface BC are obtained from Equation 18.132.

On substituting pB = (pA + pC )/2 we get pTS = tS [ p6A

pC pA 3 + 3

pC 6 ] = [ 1200

5600 1600 ]

The nodal load vector for unrestrained nodes becomes pT = [ 1200 1600 0 5600 0 0 0 0 0 0 0 0 ] On solving the equilibrium equation Ku = p we get the nodal displacements uT = 10−3 [ 0.2986 0.3951 0.1524 0.3419 0.1893 0.1651 −0.0078 −0.0984 0.0272 −0.0497 −0.0504 −0.0096 ] in

134 Noting the correspondence between the local and global degrees of freedom we obtain the displacements along element nodes uT1 = 10−3 [ 0 0.2986 0.3951 0.1524 0.3419 0.1651 0 −0.0078 −0.0984 0.0272 −0.0497 −0.0096 ] in uT2 = 10−3 [ 0 0.3951 0 0.1651 0.1893 0 0 −0.0984 0 −0.0096 −0.0504 0 ] in The strain-displacement matrix is not constant but depends on the coordinates of the point at which strains are desired. The centroid of a triangular element, where we want to calculate the strains and stresses, is located at L1 = 1/3, L2 = 1/3, and L3 = 1/3. The strain displacement matrices are now obtained from Equations 18.118, 18.121 and 18.122, and are given by  B(1) =

−8

8 0

0

32 −32 0

1  0 0 0 0  0 240 0 −10 10 −40 0 

B(2) =

0

8 −8

0

1  0 0  0 0 0 240 −10 0 10 −40 40

0

0

0

0 −10 10 −40 0

40 −8

32 0 −32

0

0



0

 40 

8 0

0

32 −32

0 0

0

0

0

−10 0 10 −40 40

0

0

8 −8

0

 0 

32 0 −32

The strains are obtained from i = B(i) ui which gives      0.3354   0.3518  2 = 10−4 −0.0681 1 = 10−4 −0.0991     0.0052 −0.0052 The corresponding stresses are obtained from σi = D i      1007.6   1092.4  σ1 = 4.8 σ2 = 123.5 psi     6.1 −6.1



135

Chapter 19

where Ψaj is the jth attachment mode shape and Φn represents the nth normal mode. To obtain coefficients αnj we premultiply both sides of Equation (e) by ΦTn M. Using the orthogonality property of the mode shapes we get

Problem 19.1 3

2

4

αnj =

1

I L

Figure S19.1 The stiffness matrix corresponding to the 4 d.o.f. shown in Figure S19.1 is given by 

(a)

The consistent mass matrix is given by  312 0 13 −13 ml  0 8 −3 −3  M=   13 −3 4 0 420 −13 −3 0 4



(h)

Since Ψak is not equal to Ψa , the normal modes need to be supplemented by residual flexibility attachment modes. The residual flexibility matrix is given by Equation 19.41 T Gd = G − Gk = G − Φk Λ−1 k Φk

(i)

EI EI G = 3 K−1 3 l l   0.1667 0 0.2500 −0.2500 0 0.1667 −0.0833 −0.0833   =  0.2500 −0.0833 0.6667 −0.3333 −0.2500 −0.0833 −0.3333 0.6667

(j)

(c) We have

The two attachment modes corresponding to d.o.f.

3 and 4 are obtained from 

(d)

where we have omitted the multiplier l3 /(EI) since only the relative values of the elements of a mode shape are of interest.

Each attachment mode can be expressed as a superposition of the four normal modes, which are independent of each other and span a 4-dimensional space. Thus Ψaj = α1j Φ1 + α2j Φ2 + α3j Φ3 + α4j Φ4

 0.2600 −0.2600  −0.1250 −0.1250  =  0.5333 −0.2833 −0.2833 0.5333

(b)



   0 0 0.2500 −0.2500  0 0   −0.0833 −0.0833  Ψa = K−1  =  1 0 0.6667 −0.3333 0 1 −0.3333 0.6667

Evidently, when all normal modes are retained there is no need to supplement them by the attachment modes.

When only the first two modes are retained the attachment modes are only partially represented as would be seen from

0.2580 −0.2580 Ψak = [ Φ1 Φ2 ] 0.0323 0.0323 

In both the stiffness and mass matrices the rotational d.o.f. θ are instead represented by lθ.

The four normal modes obtained from a Matlab routine are  1.008 0 −0.911 0 −3.873 0 10.247   0 Φ=  1.583 3.873 8.336 10.247 −1.583 3.873 −8.336 10.247

(f )

Coefficients α as evaluated from Equation (f) work out to   0.2580 −0.2580  0.0323 0.0323  αnj =  (g)  0.0110 −0.0110 0.0041 0.0041

I

 24 0 −6 6 EI  0 8 2 2  K= 3   −6 2 4 0 l 6 2 0 4

ΦTn MΨaj ΦTn MΦn

(e)



 1.008 0 −3.873   0 Φk =   1.583 3.873 −1.583 3.873

EI 6.136 0 Λ= 0 120.0 ml4

(k)

(l)

Substitution of Equations (j), (k), and (l) in Equation (i) gives

136

ml 312 0 420 0 8

ml −13 54 = MTcf Mf c = 420 −3 13

ml 4 −22 Mcc = 420 −22 156

Mf f = Gd = 

 0.0011 0 −0.0100 0.0100 l  0 0.0417 0.0417 0.0417    EI −0.0100 0.0417 0.1333 −0.0500 0.0100 0.0417 −0.0500 0.1333 3

(m)

The residual flexibility attachment modes are now obtained from 

   0 0 −0.0100 0.0100  0 0   0.0417 0.0417  Ψad = Gd  =  1 0 0.1333 −0.0500 0 1 −0.0500 0.1333

(n)

(b)

where the rotational d.o.f. θ are instead represented by lθ.

The fixed interface normal modes obtained from Kf f and Mf f are

1.160 0 (c) Φf = 0 7.246

where again we omit the multiplier l3 /(EI).

Adding Equations (h), and (n) and comparing the result with Equation (d) we get

The two constrained modes are obtained from

−0.25 0.50 −1 (d) Ψf c = −Kf f Kf c = −0.25 0.75

Ψak + Ψad = Ψa

The transformation matrix obtained from one retained mode and two constrained modes is given by   1.160 −0.25 0.50  0 −0.25 0.75  T= (e)  0 1 0 0 0 1

(o)

Problem 19.2 EI, m

The transformed stiffness and mass matrices are obtained from   32.308 0 0 EI TT KT = 3  0 2.000 −1.500  (f ) l 0 −1.500 1.500

L 3

2

4

1

2 3

I

I



 1.0000 −0.2514 0.5801 TT MT = ml  −0.2514 0.0762 −0.2095  (g) 0.5801 −0.2095 0.7429

1

4

I

Component 1

I

Component 2

Component 2

Figure S19.2 For each component, the stiffness and mass matrices are subdivided along the free d.o.f f , (1 and 2), and constrained d.o.f. c, (3 and 4). The computations then proceed as follows.

Component 1 EI Kf f = 3 l

24 0 0 8

6 −12 = KTcf 2 −6

EI 4 −6 Kcc = 3 −6 12 l

EI Kf c = 3 l

(a)

Kf f =

EI l3

24 0 0 8

−6 −12 = KTcf 2 6

EI 4 6 Kcc = 3 6 12 l

ml 312 0 Mf f = 420 0 8

ml 13 54 = MTcf Mf c = 420 −3 −13

ml 4 22 Mcc = 420 22 156 EI Kf c = 3 l

(h)

(i)

137 The fixed interface normal modes obtained from Kf f and Mf f are

1.160 0 Φf = 0 7.246

(j)

The two constrained modes are obtained from

0.25 0.50 −1 Ψf c = −Kf f Kf c = (k) −0.25 −0.75 The transformation matrix obtained from one retained mode and two constrained modes is given by   1.160 0.25 0.50  0 −0.25 −0.75  T= (l)  0 1 0 0 0 1 The transformed stiffness and mass matrices are obtained from   32.308 0 0 EI 2.000 1.500  (m) TT KT = 3  0 l 0 1.500 1.500 

 1.0000 0.2514 0.5801 TT MT = ml  0.2514 0.0762 0.2095  0.5801 0.2095 0.7429

(n)



 1.0000 −0.2514 0.5801 0 0 0.2514   −0.2514 0.1524 M̃ = ml   (p) 0.5801 0 1.4857 0.5801 0 0.2514 0.5801 1.0000 An eigenvalue solution using matrices K̃ and M̃ gives # ω = [ 22.41 62.28 124.76 289.84 ]

# ω = [ 22.38 61.69 120.91 ]

The free-interface normal modes and the residual flexibility attachment modes are first derived for each component. The transformation matrix for a component consists of the one retained normal mode and two residual flexibility modes. The transformed property matrices for the two components are obtained next; they are then coupled so as to ensure the compatibility of displacements on the interface.

The stiffness matrix is given by  24 0 6 −12 EI  0 8 2 −6  K= 3   6 2 4 −6 l −12 −6 −6 12

1 4

2 2

3

The consistent mass matrix is 

 312 0 −13 54 ml  0 8 −3 13  M=   4 −22 420 −13 −3 54 13 −22 156

(b)



 −0.4806 −1.0308 0.1616 0.6753  −0.8232 0.3102 −6.0734 6.9406  Φ=  (c) −0.9743 3.4379 7.6593 25.7789 −1.4156 1.4281 1.5883 2.6672

3 3

On assembling the matrices we get  32.308 0 0 0 EI  0 4.000 0 0  K̃ = 3   0 0 3.000 0 l 0 0 0 32.308

(a)

where the rotational d.o.f. θ are instead represented by lθ.

The free-interface normal modes and the corresponding eigenvalues are obtained from K and M

Global coordinate Component 1 Component 2

(r)

Problem 19.3

Table S19.2 Correspondence between the local and global generalized coordinates

2

EI mL4



The system matrices are obtained from direct assembly noting the correspondence between the local and global d.o.f. The d.o.f are identified in Figure S19.2b and the correspondence is shown in Table S19.2

1

(q)

where L = 4l. The exact values of the frequencies for the first 3 modes of a fixed-fixed uniform beam as given in Figure 15.3 are

Coupling of component matrices

Local coordinate

EI mL4





(o)

 0.7734 0 0 0 EI  0 30.86 0 0  Λ=   0 0 353.0 0 ml4 0 0 0 2974.0

(d)

138 Of the four normal modes only one is retained.

The residual flexibility matrix is therefore given by

EI EI T Gd = 3 K−1 − Φk Λ−1 k Φk 3 l l   0.0347 −0.0116 −0.1055 −0.0464  −0.0116 0.1238 −0.0370 −0.0067  =  −0.1055 −0.0370 0.7726 0.2167 −0.0464 −0.0067 0.2167 0.0756

(d)

where the rotational d.o.f. θ are instead represented by lθ.

The free-interface normal modes and the corresponding eigenvalues are obtained from K and M 

 −0.4806 −1.0308 0.1616 0.6753  0.8232 −0.3102 6.0734 −6.9406  Φ=  (i) 0.9743 −3.4379 −7.6593 −25.7789 −1.4156 1.4281 1.5883 2.6672 

 0.7734 0 0 0 EI  0 30.86 0 0  Λ=   4 0 0 353.0 0 ml 0 0 0 2974.0

The residual flexibility attachment modes are obtained from     0 0 −0.1055 −0.0464  0 0   −0.0370 −0.0067  Ψa = Gd  =  (e) 1 0 0.7726 0.2167 0 1 0.2167 0.0756

Of the four normal modes only one is retained.

The residual flexibility matrix is therefore given by

where we have dropped the multiplier l3 /(EI). It will be noted that the two attachment modes are just the last two columns of Gd .

The coordinate transformation matrix is comprised of the retained normal mode and the two residual flexibility attachment modes

EI EI T Gd = 3 K−1 − Φk Λ−1 k Φk 3 l l   0.0347 0.0116 0.1055 −0.0464  0.0116 0.1238 −0.0370 0.0067  =  0.1055 −0.0370 0.7726 −0.2167 −0.0464 0.0067 −0.2167 0.0756



 −0.4806 −0.1055 −0.0464  −0.8232 −0.0370 −0.0067  T=  −0.9743 0.7726 0.2167 −1.4156 0.2167 0.0756

(e)

The transformed property matrices are given by K̃(1) = TT KT   0.7734 0 0 EI = 3  0 0.7726 0.2167  l 0 0.2167 0.0756

(f )

M̃(1) = TT MT   1.0000 0 0 = ml  0 0.0130 0.0053  0 0.0053 0.0022

(g)

(j)

(k)

The residual flexibility attachment modes are obtained from     0 0 0.1055 −0.0464  0 0   −0.0370 0.0067  Ψa = Gd  (l) =  1 0 0.7726 −0.2167 0 1 −0.2167 0.0756 where we have dropped the multiplier l3 /(EI). As for component 1, the attachment modes are just the last two columns of Gd .

The coordinate transformation matrix is comprised of the retained normal mode and the two residual flexibility attachment modes 

 −0.4806 0.1055 −0.0464  0.8232 −0.0370 0.0067  T=  0.9743 0.7726 −0.2167 −1.4156 −0.2167 0.0756

(m)

Component 2 The transformed property matrices are given

The stiffness matrix is given by  24 0 −6 −12 EI  0 8 2 6 K= 3   −6 2 4 6 l −12 6 6 12

by



The consistent mass matrix is   312 0 13 54 ml  0 8 −3 −13  M=   13 −3 4 22 420 54 −13 22 156

(h)

(h)

K̃(2) = TT KT   0.7734 0 0 EI  = 3 0 0.7726 −0.2167  l 0 −0.2167 0.0756

(n)

M̃(2) = TT MT   1.0000 0 0 = ml  0 0.0130 −0.0053  0 −0.0053 0.0022

(o)

139 Coupling of component matrices The generalized coordinates for component 1 (1) (1) are pk and pa , in which the former has only one element while the latter has two elements. In a similar manner the generalized coordinate for compo(2) (2) nent 2 are pk and pa . Not all of these coordinates (2) are independent. We select pa as the dependent coordinates in order to satisfy compatibility of displacements along d.o.f. 3 and 4. We denote the (1) (1) independent coordinates as q1 = pk , q2 = pa , (2) and q3 = pk . The transformation between the p set and q set is given by p = Sq

(p)

Compatibility of displacements requires that (2) u(1) a = ua

(q)

or (2) (2) (1) (1) (2) (2) (1) Φak pk + Ψ(1) aa pa = Φak pk + Ψaa pa

On setting Γ =

or

(2) paa

−1

(r)

The transformed system matrices are obtained from

(1) 0 T K̃ S K̃ = S 0 K̃(2)   193.91 −50.22 −15.77 −128.58 EI  −50.22 14.18 4.41 32.33  = 3   −15.77 4.41 1.39 10.21 l −128.58 32.33 10.21 89.81 (aa)

M̃(1) 0 M̃ = S S 0 M̃(2)   3.536 −0.537 −0.175 −2.072 (bb)  −0.537 0.127 0.042 0.436  = ml   −0.175 0.042 0.014 0.142 −2.072 0.436 0.142 2.700 T

Solution of an eigenvalue problem with matrices K̃ and M̃ gives the following frequencies # ω = [ 22.40 62.45 129.56 318.97 ]

we get

(1) (1) (2) (2) (1) (1) p(2) a = Γ Φak pk + Ψaa pa − Φak pk

(s)

(1) (2) (1) p(2) a = Γ Φak q1 + Ψaa q2 − Φak q3

(t)

The transformation matrix S is thus given by   I 0 0 I 0   0 (u) S=  0 0 I (1) (1) (2) ΓΦak ΓΨaa −ΓΦak From Equations (c), (e), (i) and (l) we get

−0.9743 (1) Φak = (v) −1.4156

0.7726 0.2167 (1) (w) Ψaa = 0.2167 0.0756

0.9743 (2) (x) Φak = −1.4156

0.7726 −0.2167 (2) (y) Ψaa = −0.2167 0.0756 On substituting Equations (v) through (y) in Equation (u) we get   1.0000 0 0 0 0 1.0000 0 0     0 0 1.0000 0   S= (z)  0 0 0 1.0000     −33.1290 9.1769 2.8540 20.295 −113.634 29.157 9.1769 76.864

EI mL4

(cc)

where L = 4l. These values of the frequencies may be compared with the exact frequencies for a fixedfixed beam given in the solution to Problem 19.2.

Problem 19.4 The normal modes, the residual flexibility attachment mode, and the constrained mode are first derived for each component. The transformation matrix for a component consists of the one retained normal mode, the residual flexibility modes, and the constrained mode. The transformed property matrices for the two components are obtained next; they are then coupled so as to ensure the compatibility of displacements on the interface.

The stiffness and mass matrices corresponding to the free d.o.f. are given by 

 24 0 6 EI Kf f = 3  0 8 2  l 6 2 4   312 0 −13 ml  Mf f = 0 8 −3  420 −13 −3 4

(a)

(b)

where the rotational d.o.f. θ are instead represented by lθ. The mode shapes and eigenvalues obtained from the above stiffness and mass matrices are   1.0410 −0.4793 0.5875 Φ =  0.5613 4.5722 7.5228  (c) −2.0532 −5.1123 12.2331

140   15.13 0 0 EI   Λ= 0 213.20 0 ml4 0 0 1513.97

(d)

  312 0 13 ml  0 8 −3  Mf f = 420 13 −3 4

(l)

Of the three normal modes only one is retained.

The residual flexibility matrix is therefore given by

EI EI −1 −1 T K = − Φ Λ Φ G k d k ff k l3  l3  (e) 0.0013 −0.0074 0.0162 =  −0.0074 0.1354 −0.0488  0.0162 −0.0488 0.2214

where the rotational d.o.f. θ are instead represented by lθ. The mode shapes and eigenvalues obtained from the above stiffness and mass matrices are   1.0410 0.4793 −0.5875 Φ =  −0.5613 4.5722 7.5228  (m) 2.0532 −5.1123 12.2331

The residual flexibility attachment mode is obtained from      0   0.0162  Ψf a = Gd 0 = −0.0488 (f )     1 0.2214

(n)

where we have dropped the multiplier l3 /(EI) The constrained mode is obtained from      −12   0.3125  Ψf s = K−1 (g) f f  −6  =  0.5625  −6 0.7500 The coordinate transformation matrix comprising the retained normal mode, the residual flexibility attachment mode, and the constrained mode is given by   1.0410 0.0162 0.3125  0.5613 −0.0488 0.5625  T= (h)  −2.0532 0.2214 0.7500 0 0 1.0000 The transformed stiffness and mass matrix are now obtained from K̃(1) = TT KT   15.1337 0 0 EI = 3  0 0.2214 0  l 0 0 0.3750 M̃(1) = TT MT   1.0000 0 0.4927 = ml  0 0.0006 −0.0093  0.4927 −0.0093 0.4714

(i)

Of the three normal modes only one is retained.

The residual flexibility matrix is therefore given by

EI EI −1 T Gd = 3 K−1 f f − Φk Λk Φk 3 l  l  0.0013 0.0074 −0.0162 =  0.0074 0.1354 −0.0488  −0.0162 −0.0488 0.2214

(o)

The residual flexibility attachment mode is obtained from      0   −0.0162  Ψf a = Gd 0 = −0.0488 (p)     1 0.2214 where we have dropped the multiplier l3 /(EI) The constrained mode is obtained from      −12   0.3125  6 = −0.5625 (q) Ψf s = K−1 ff     6 −0.7500 The coordinate transformation matrix comprising the retained normal mode, the residual flexibility attachment mode, and the constrained mode is given by 

(j)

 1.0410 −0.0162 0.3125  −0.5613 −0.0488 −0.5625  T=  2.0532 0.2214 −0.7500 0 0 1.0000

(r)

The transformed stiffness and mass matrix are now obtained from

Component 2 The stiffness and mass matrices corresponding to the free d.o.f. are given by 

Kf f =

  15.13 0 0 EI   Λ= 0 213.20 0 ml4 0 0 1513.97



24 0 −6 EI  0 8 2 l3 −6 2 4

(k)

K̃(2) = TT KT   15.1337 0 0 EI  = 3 0 0.2214 0  l 0 0 0.3750

(s)

141 M̃(2) = TT MT   1.0000 0 0.4927 = ml  0 0.0006 0.0093  0.4927 0.0093 0.4714

(t)

Coupling of component matrices (1)

(1)

The six constrained coordinates are pk , pa , Compatibility of dsipalcements at the interface requires that the following relationships be satisfied

(1) (2) (2) (2) ps , pk , pa , and ps .

(2) u(1) a = ua

(u)

(2) u(1) s = us

(v)

Equation (u) gives (1) (1)

(1) (1) (1) Φak pk + Ψ(1) aa pa + Ψas ps (2) (2)

(2) (2) (2) = Φak pk + Ψ(2) aa pa + Ψas ps

(w)

Equation (v) gives (2) p(1) s = ps

On setting

Γ=

Ψ(2) aa

−1

(x)

(y)

we get from Equations (w) and (x) 2 (1) (1) (1) (1) p(2) a = Γ Φak pk + Ψaa pa

(1) (2) (1) (2) (2) + Ψas − Ψas ps − Φak pk

(z)

The transformed system matrices are obtained from (1)

0 T K̃ S K̃ = S 0 K̃(2)   34.17 −2.053 −13.91 19.04 EI  −2.053 0.443 1.500 −2.053  = 3   −13.91 1.500 10.91 −13.91 l 19.04 −2.053 −13.91 34.17 (cc)

M̃(1) 0 S 0 M̃(2)   1.055 −0.0059 0.366 0.055 (dd)  −0.006 0.0013 0.004 −0.006  = ml   0.366 0.0043 1.099 0.366 0.055 −0.0059 0.366 1.055

M̃ = ST

Solution of an eigenvalue problem with matrices K̃ and M̃ gives the following frequencies # ω = [ 22.41 62.24 125.79 297.55 ]

EI mL4

(ee)

where L = 4l. These values of the frequencies may be compared with the exact frequencies for a fixedfixed beam given in the solution to Problem 19.2.

Problem 19.5

We select the following as independent coordinates: (1) (1) (1) (2) q1 = pk , q2 = pa , q3 = ps , and q4 = pk The transformation matrix S relating coordinates p and q is given by

EA, m

L

S= 

 I 0 0 0  0  I 0 0    0  0 I 0    0  0 0 I

  (1) (1) (2) (2) (1)  ΓΦak ΓΨaa Γ Ψas − Ψas −ΓΨak  0 0 I 0 (aa)

1

2 1 2

3 1 2

(1)

From Equations (h) and (r) we have Φak = (1) (1) (2) −2.0352, Ψaa = 0.2214, Ψas = 0.75, Φak = (2) (2) 2.0532, Ψaa = 0.2214, and Ψas = −0.75. Substituting these values in Equations (y) and (aa) we get   1.0000 0 0 0 0 1.0000 0 0     0 0 1.0000 0   S=  (bb) 0 0 0 1.0000     −9.2727 1.0000 6.7742 −9.2727 0 0 1.0000 0

I Component 1

I

I

I

Component 2

I

I

Component 3

Figure S19.5 The transformed stiffness and mass matrices are obtained for all three components. The properties for components 1 and 3 are identical. Component 2 is a free-free component and must be treated differently.

142 

 2/3 1/6 1/6 M(2) = ml  1/6 1/3 0  1/6 0 1/3

Components 1 and 3 The stiffness and consistent mass matrices are given by

EA 2 −1 K= −1 2 l (a)

2/3 1/6 M = ml 1/6 1/3 An eigenvalue solution gives the following normal mode shapes

−0.7444 −1.0771 Φ= (b) −1.0527 1, 5233 The matrix of corresponding eigenvalues is

EA 0.6492 0 Λ= 0 7.9223 ml2

(c)

We retain the first of the two modes. The residual flexibility matrix is then given by T Gd = K−1 − Φk Λ−1 k Φk

l 0.1464 −0.2071 = EA −0.2071 0.2929

(d)

The residual flexibility attachment is obtained from Ψ a = Gd

0 I

=

−0.2071 0.2929

(e)

where we have dropped the multiplier l/(EA).

The coordinate transformation is thus obtained from

−0.7444 −0.2071 T = [ Φk Ψa ] = (f ) −1.0527 0.2929 where we have dropped the scale factors on the modes.

The transformed stiffness and mass matrices are now given by

EA 0.6492 0 K̃(i) = TT K(i) T = (g) 0 0.2929 l

1.0000 0 T (i) (i) M̃ = T M T = ml (h) 0 0.0370 Components 2 The stiffness and consistent mass matrices are given by   2 −1 −1 EA  −1 1 0  K(2) = (i) l −1 0 1

(j)

An eigenvalue solution gives the following normal mode shapes and the corresponding eigenvalues.   −0.7071 0 −1.2247 (2) Φ =  −0.7071 −1.2247 1.2247  (k) −0.7071 1.2247 1.2247   0 0 0 EA  0 3.0 0  (l) Λ(2) = ml2 0 0 12.0 The first mode shape in Equation (k) is the rigid body mode. The inertia relief projection matrix is obtained from

−1 PI = I − MΦr M̃rr ΦTr   (m) 0.25 −0.50 −0.50 =  −0.25 0.75 −0.25  −0.25 −0.25 0.75 To obtain the inertia relief attachment mode corresponding to interface d.o.f. 2, we apply a unit force along that d.o.f. while restraining d.o.f. 3.

The stiffness matrix corresponding to the free d.o.f.

1 and 2 is

EA 2 −1 Kf f = (m) −1 1 l so that K−1 ff =

l 1 1 EA 1 2

(n)

and the augmented flexibility matrix is given by   1 1 0 l  Gr = 1 2 0 (o) EA 0 0 0 The inertia-relief attachment mode is obtained from Equation 19.35 with faT = [ 0 1 0 ] ΨTa = [ −0.125 0.625 −0.375 ]

(p)

where subscript a on Ψ denotes the attachment mode corresponding to the interface d.o.f 2 and we have dropped the multiplier l/(EA).

The inertia relief attachment mode corresponding to d.o.f. 3 is obtained in a similar manner.

Thus we have

EA 2 −1 Kf f = −1 1 l

l 1 1 K−1 = ff (q) EA 1 2   1 0 1 l  Gr = 0 0 0 EA 1 0 2

143 and with faT = [ 0 0 1 ] ΨTr = [ −0.125 −0.375 0.625 ]

(r)

where subscript r on Ψ denotes the attachment mode corresponding to the interface d.o.f 3 and we have dropped the multiplier l/(EA).

The transformation matrix is thus given by T = [ Φr Ψa Ψr ]   −0.7071 −0.1250 −0.1250 =  −0.7071 0.6250 −0.3750  −0.7071 −0.3750 0.6250

where 2 3 (1) (2) (2) (2) (3) (3) (aa) pT = p(1) pa pk pa pr pk pa k 

(s)

The transformed stiffness and mass matrices are given by   0 0 0 EA  0 0.6250 −0.3750  (t) K̃(2) = l 0 −0.3750 0.6250  1 0 0 M̃(2) = ml  0 0.1771 −0.1563  0 −0.1563 0.1771

in q can be treated as the independent coordinates, so that the transformation from p to q is expressed as p = Sq (z)



 1.000 0 0 0 0 1.000 0 0 0   0   0 1.000 0 0   0   S =  −2.632 0.732 2.828 −1.579 0.439  (aa)    −1.579 0.439 2.828 −2.632 0.732    0 0 0 1.000 0 0 0 0 0 1.000 The transformed system stiffness matrix is given by  (1)  K̃ 0 0 K̃ = ST  0 K̃(2) 0  S 0 0 K̃(3) 

(u)

3.4197 −0.7708 −2.9775 −0.7708 0.5074 0.8284  EA  =  −2.9775 0.8284 4.0000 l  1.6623 −0.4625 −2.9775 −0.4625 0.1287 0.8284  1.6623 −0.4625 −0.4625 0.1287   −2.9775 0.8284   3.4197 −0.7708 −0.7708 0.5074

Coupling of component matrices The compatibility of displacements at the interface of components requires that (2) (2) (3) u(1) a = ua , and ur = ua

(v)

which gives (1) (1)

(2) (2)

The transformed system mass matrix is given

(1) (2) (2) Φak pk + Ψ(1) aa pa = Φak pk + Ψaa pa (2) + Ψ(2) ar pr (3) (3) (2) (2) (3) (2) (2) Φak pk + Ψ(3) aa pa = Φrk pk + Ψrr pr

by (w)

(2) + Ψ(2) ra pa



 M̃(1) 0 0 M̃ = ST  0 M̃(2) 0  S 0 0 M̃(3) 

Equation (w) yields

(2) 0.6250 −0.3750 pa (2) −0.3750 0.6250 pr

−1.0527 0.2929 0.7071 = 0 0 0.7071

(cc)

(x)

0 0 q −1.0527 0.2929

1.3694 −0.1028 −0.2481  −0.1028 0.0656 0.0690  = ml  −0.2481 0.0690 1.3333  0 0 −0.2481 0 0 0.0690  0 0 0 0   −0.2481 0.0690   1.3694 −0.1028 −0.1028 0.0656

(dd)

where 3 2 (1) (2) (3) (3) qT = p(1) pa pk pk pa k

(y)

Relationship in Equation (y) shows that of the seven generalized coordinates p the five included

An eigenvalue solution using the system matrices gives the following frequencies # ω = [ 3.18 6.57 10.39 14.70 18.85 ]

EA mL2

(ee)

144 where L = 6l.

It may be note that component 1 has two d.o.f.

The two modes shapes, a normal mode and a residual flexibility mode, that we are using to model it thus span the entire 2-dimensional space. A similar consideration applies to component 3. Component 2 has three d.o.f., but we are also using three modes, a normal mode and two inertia-relief attachment modes. Thus the selected mode shapes span the entire 3-dimensional space. As a consequence, the results obtained from the CMS should be identical to those obtained from a finite element solution for the frequencies of the bar treated as a single component having six finite elements. To verify this, note that the stiffness matrix of the bar is given by  2 −1 0 0 0 −1 2 −1 0 0  EA    K=  0 −1 2 −1 0  l   0 0 −1 2 −1 0 0 0 −1 2

Denoting the four retained d.o.f. by t and the two condensed d.o.f. by θ, the stiffness matrix for the component is given by 

 24 −12 0 0 EI  −12 24 −12 6  Ktt = 3   0 −12 12 −6 l 0 6 −6 4   0 6 EI  −6 0  Ktθ = KTθt = 3   0 −6 l 0 2

EI 8 2 Kθθ = 3 2 8 l

Ktt Ktθ K= Kθt Kθθ



The mass matrix is   2/3 1/6 0 0 0 0   1/6 2/3 1/6 0   M = ml  0 1/6 2/3 1/60    0 0 1/6 2/3 1/6 0 0 0 1/6 2/3

(ff )

where the rotational d.o.f. θ is represented by lθ.

El, m

L

(gg) 5

6

1

Eigenvalue solution using the matrices in Equations (ff) and (gg) yield the same values of frequencies as in Equation (gg).

In practical apllications the number of modes or Ritz vectors used to model a component will be significantly smaller than the number of d.o.f. in the component.

The exact values of the frequencies of a continuous uniform fixed-fixed bar vibrating in the axial direction are given by # ωn = nπ

4

2

3 7

6

8

4 3

I

I

1

I

2

I

Component 1

I

5

I

Component 2

Figure S19.6

The consistent mass matrix is given by

EA mL2

(hh)

Hence # ω = [ 3.14 6.28 9.42 12.57 15.71 ]

(a)

EA mL2

(ii)

Problem 19.6



 312 54 0 0 ml  54 312 54 −13  Mtt =   0 54 156 −22 420 0 −13 −22 4   0 −13 ml  13 0 T Mtθ = Mθt =   0 13 420 0 −3

Component 1 The 6 d.o.f. for component 1 are shown in Figure S19.6. Of these we will condense out the rotational d.o.f. 5 and 6, leaving the two interior d.o.f. 1 and 2 and two interface d.o.f. 3 and 4.

Mθθ =

ml 8 −3 420 −3 8

M=

Mtt Mtθ Mθt Mθθ

(b)

145 The transformation required to condense out rotational d.o.f. 5 and 6 is given by Equation 13.60

T=

I

−K−1 θθ Kθt

 1 0 0 0 1 0 0   0   0 1 0  0  =  0 0 1  0    0.2 0.8 −0.2 0.0667 −0.8 −0.2 0.8 −0.2667 

(c)

The transformed stiffness matrix is given by T K(1) c = T KT 

 19.2 −13.2 4.8 −1.6 EI  −13.2 19.2 −10.8 5.6  = 3   4.8 −10.8 7.2 −4.4 l −1.6 5.6 −4.4 3.47

(d)

T M(1) c = T MT   0.8076 0.1519 −0.0648 0.0190  0.1519 0.8076 0.1052 −0.0238  = ml   −0.0648 0.1052 0.4362 −0.0714 0.0190 −0.0238 −0.0714 0.0150 (e)

An eigenvalue solution using the transformed (1) (1) stiffness and mass matrices, Kc and Mc , gives the following frequencies. $ EI mL41

(f )

where L1 = 3l These frequencies may be compared with those for a uniform cantilever beam, of which the first three are $ Ω = [ 3.516 22.03 61.70 ]

EI mL41



 24 −12 −12 −6 0 0 0 0 −12 6   −12 24  EI  0 12 6 0 0  −12 Ktt = 3   0 6 4 0 0 l  −6   0 −12 0 0 12 −6 0 6 0 0 −6 4   0 6  −6 0    EI  6 0 Ktθ = KTθt = 3   l  2 0   0 −6 0 2

EI 8 2 Kθθ = 3 2 8 l

Ktt Ktθ K= Kθt Kθθ

(h)

where the rotations θ have been represented by lθ.

The consistent mass matrix is given by

The transformed mass matrix is

ω = [ 3.516 22.17 62.67 251.90 ]

given by

(g)



 312 54 54 13 0 0  54 312 0 0 54 −13   ml  0 156 22 0 0  54 Mtt =   0 2 4 0 0 420  13   0 54 0 0 156 −22 0 −13 0 0 −22 4   0 −13 0  13  ml  0  −13 T Mtθ = Mθt =   0 420  −3   0 13 0 −3

ml 8 −3 Mθθ = −3 8 420

Mtt Mtθ M= Mθt Mθθ

The transformation required to condense out rotational d.o.f. 7 and 8 is given by Equation 13.60

T=

Component 2 From the d.o.f. shown in Figure S19.6 we condense out the two rotational d.o.f. 7 and 8, leaving two interior d.o.f. 1 and 2, two interface d.o.f. 3 and 4 and two d.o.f. at the free end 5 and 6. Denoting again the retained d.o.f. by t and the condensed d.o.f by θ, the stiffness matrix for the component is

(i)

I −K−1 θθ Kθt

=

 1 0 0 0 0 0 1 0 0 0 0   0   0 1 0 0 0   0   0 0 1 0 0   0   0 0 0 1 0   0   0 0 0 0 1   0   0.2 0.8 −0.8 −0.267 −0.2 0.067 −0.8 −0.2 0.2 0.067 0.8 −0.267 

(j)

146 The transformed stiffness matrix is given by T K(2) c = T KT  19.2 −13.2 −10.8 −5.60 4.8 1.60  −13.2 19.2 EI  4.8 7.2 4.40  −10.8 = 3  1.6 4.4 3.47 l  −5.6  4.8 −10.8 −1.2 −0.40 (k) −1.6 5.6 0.40 0.13  4.8 −1.60 −10.8 5.60   −1.2 0.40   −0.4 0.13   7.2 −4.40 −4.4 3.47

subsets of the stiffness and mass matrices. From Equation (d) of Problem 19.6 the subset of stiffness matrix corresponding to d.o.f. 1 and 2 shown in Figure S19.7 is given by 19.2 −13.2 −13.20 19.20

(a)

El, m

L 4 1

The transformed mass matrix is obtained from T M(2) c = T MT =  0.8076 0.1519 0.1052 0.0238  0.1519 0.8076 −0.0648 −0.0190   0.1052 −0.0648 0.4362 0.0714 ml   0.0238 −0.0190 0.0714 0.0150  −0.0648 0.1052 0.0233 0.0071 0.0190 −0.0238 −0.0071 −0.0022  −0.0648 0.0190 0.1052 −0.0238   0.0233 −0.0071   0.0071 −0.0022   0.4362 −0.0714 −0.0714 0.0150

EI Kf f = 3 l

2

3 6 4 3

I

I

I

I

5

I

Component 2

Figure S19.7

(l)

where L1 = 3l These frequencies may be compared with those for a uniform cantilever beam, of which the first five are $ EI (g) ω = [ 0 0 22.38 61.68 120.91 ] mL41

2

I

Component 1

An eigenvalue solution using the transformed (2) (2) stiffness and mass matrices, Kc and Mc , gives the following frequencies. $ EI ω = [ 0 0 22.53 62.06 216.8 ] (f ) mL41

1

The corresponding mass matrix is

0.8076 0.1519 Mf f = ml 0.1519 0.8076

(b)

An eigenvalue solution gives the following mode shapes

−0.7219 −0.8732 Φf = (c) −0.7219 0.8732 The constrained mode shapes are obtained from

4.8 −1.6 Kf s = (d) −10.8 5.6

l0.2593 −0.2222 Ψf = −Kf f Kf s = (e) 0.7407 −0.4444 The transformation between the constrained coordinates shown in Figure S19.7 and the generalized coordinates is given by

Problem 19.7

u = Tp

(f )

Component 1 Referring to the solution for Problem 19.6, the retained d.o.f. for component 1 are shown in Figure S19.7. For modelling of the component 1 we select one fixed interface normal mode and two constrained modes. The normal mode is obtained from

where 

 −0.7219 0.2593 −0.2222  −0.7219 0.7407 −0.4444  T = [Φ Ψ] =   (g) 0 1.0000 0 0 0 1.0000

147 The transformed stiffness matrix is given by K̃(1) = TK(1) c T   6.2531 0 0 EI  = 3 0 0.4444 −0.6667  l 0 −0.6667 1.3333

(h)

The transformed mass matrix is given by M̃(1) = TM(1) c T   1.0000 −0.7219 0.4652 = ml  −0.7219 1.1143 −0.4714  0.4652 −0.4714 0.2571

(i)

The transformed mass matrix is given by M̃(2) = TM(2) c T  1.0000 −0.7219 −0.4652  −0.7219 1.1143 0.4714  = ml  −0.4652 0.4714 0.2571  −0.7219 0.3857 0.2786 0.4652 −0.2786 −0.1929  −0.7219 0.4652 0.3857 −0.2786   0.2786 −0.1929   1.1143 −0.4714 −0.4714 0.2571

(n)

Component 2 The retained d.o.f. for component 1 are shown in Figure S19.7. For modelling of the component 2 we select one fixed interface normal mode and four constrained modes. The normal modes are obtained from subsets of the stiffness and mass matrices which are are identical to those for component 1.

The constrained mode shapes are obtained from

−10.8 −5.6 4.8 −1.6 Kf s = (j) 4.8 1.6 −10.8 5.6 Ψf = −Kf f Kf s

0.7407 0.4444 0.2593 −0.2222 = 0.2593 0.2222 0.7407 −0.4444

Coupling of components The total number of generalized coordinates is 8, of which 3 are for component 1 and 5 are for component 2. Of these only 6 are independent because of the requirement of the compatibility of displacements at the interface of the two components. The (1) (1) 8 constrained coordinates are denoted by p1 , p2 , (2) (2) (2) (2) (1) (2) p3 , p1 , p2 , p3 , p4 , p5 . The selected indepen(1) (1) (1) dent coordinates are q1 = p1 , q2 = p2 , q3 = p3 , (2) (2) (2) q4 = p1 , q5 = p4 , q6 = p5 . The compatibility conditions require that

(k) (1)

(2)

u3 = u3 ,

The transformation matrix between the constrained coordinates shown in Figure S19.7 and the generalized coordinates is given by T = [Φ Ψ] =   −0.7219 0.7407 0.4444 0.2593 −0.2222  −0.7219 0.2593 0.2222 0.7407 −0.4444    (l) 0 1.0000 0 0 0     0 0 1.0000 0 0     0 0 0 1.0000 0 0 0 0 0 1.0000

(1)

(2)

p2 = p 2 ,

(o)

(1)

(2)

p3 = p3

(p)

The component matrices can be coupled either by direct stiffness assembly or by noting the following transformation between coordinates p and q

where

(m)

(4)

Equation (o) along with Equations (g) and (l) gives the following relationship between the generalized coordinates

p = Sq

(q)

 1 0 0 0 0 0 0 1 0 0 0 0   0 0 1 0 0 0   0 0 0 1 0 0 S=  0 1 0 0 0 0   0 0 1 0 0 0   0 0 0 0 1 0 0 0 0 0 0 1

(q)

The transformed stiffness matrix is given by K̃(2) = TK(2) c T  6.2531 0 0 0 0.4444 0.6667  EI  0.6667 1.3333 = 3  0 l  0 −0.4444 −0.6667 0 0.6667 0.6667  0 −0.0000 −0.4444 0.6667   −0.6667 0.6667   0.4444 −0.6667 −0.6667 1.3333

(1)

u4 = u 4



The transformed system stiffness matrix is

148 given by

K̃ = ST     =  

K̃(1) 0 S 0 K̃(2)

6.2531 0 0 0 0 0.8889 0 0 0 0 2.6667 0 0 0 0 6.2531 0 −0.4444 −0.6667 0 0 0.6667 0.6667 0  0 0 −0.4444 0.6667   −0.6667 0.6667   0 0   0.4444 −0.6667 −0.6667 1.3333

(r)

The transformed system mass matrix is given

by The transformed system stiffness matrix is given by

M̃ = ST

M̃(1) 0 S 0 M̃(2)



1.0000 −0.7219 0.4652 0 0 −0.7219  −0.7219 2.2286  0 0.5143 −0.4652  0.4652 = 0 −0.7219 −0.4652 1.0000   0 0.3857 0.2786 −0.7219 0 −0.2786 −0.1929 0.4652  0 0 0.3857 −0.2786   0.2786 −0.1929   −0.7219 0.4652   1.1143 −0.4714 −0.4714 0.2571 (s) An eigenvalue solution using K̃ and M̃ gives the following frequencies # ω = [ 3.516 22.07 62.06 124.4 ]

EI mL4

(t)

where L = 6l. The first three exact frequencies for a uniform cantilever beam are # EI (u) ω = [ 3.516 22.03 61.70 ] mL4

149

Chapter 20 Acceleration ün+1 is then obtained by satisfying the equation of motion at time tn+1 .

Problem 20.1

mun+1 + cu̇n+1 + fS = pn+1

The initial elastic stiffness and frequency are determined as follows

The results are tabulated in Table S20.1. The force displacement relation for the first 0.8 s is plotted in Figure S20.1. Also shown for comparison is the specified force displacement relation. The events identified by numbers show the response at each integration step. Evidently, the time step is quite coarse so that the displacement change during each time step is comparatively large. As a consequence, while the resisting spring force reaches the plastic limit at a displacement of 0.024 m, that is somewhere between 0.3 and 0.4 s, the excursion into the plastic state is detected only at 0.4 s.

35.0 k= = 1458.3 kN/m 0.024 # # k 1458.3 = = 9.128 rad/s ω= m 17.5 The corresponding period T = 0.688 s. The time step used for integration h = 0.1 is thus rather coarse.

Using the incremental form of linear acceleration method of direct integration the incremental displacement and velocity are determined from the following expressions.

6m 3c + kT = ∆p + h2 h

6u̇n h +m + 3ün + c 3u̇n + ün h 2 h 3 ∆u̇ = ∆u − 3u̇n − ün h 2

40 67 35

5

4

30

Force, kN

25

where kT is the tangent stiffness. Substituting the parameter values we have

3 20 15 8

(10500 + kT ) ∆u = ∆p + 17.5(60u̇n + 3ün )

Analytical Zero velocity Integration steps Specified

10

= ∆p + 1050u̇n + 52.5ün ∆u̇ = 30∆u − 3u̇n − 0.05ün

0

5 0

On adding the increments we have

2

1 0

0.01

0.02

0.03

un+1 = un + ∆u u̇n+1 = u̇n + ∆u̇

0.04 0.05 0.06 Displacement, m

0.07

0.08

Figure S20.1

Table S20.1 t

un

u̇n

ün 2

s

m

m/s

m/s

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

0 0.0004 0.0039 0.0154 0.0361 0.0590 0.0775 0.0790 0.0627

0 0.0125 0.0661 0.1716 0.2143 0.2286 0.1143 -0.0857 -0.2187

0 0.2509 0.8119 1.2908 0.5714 -0.2657 -2.0000 -2.0000 -0.6413

pn

kT

∆u

∆u̇

un+1

u̇n+1

fs

ün+1

kN

kN/m

m

m/s

m

m/s

kN

m/s2

0 5 20 45 45 30 0 0 0

1458.3 1458.3 1458.3 1458.3 0 0 0 1458.3

0.0004 0.0035 0.0115 0.0207 0.0229 0.0186 0.0014 -0.0163

0.0125 0.0535 0.1055 0.0427 0.0143 -0.1143 -0.2000 -0.1321

0.0004 0.0039 0.0154 0.0361 0.0590 0.0775 0.0790 0.0627

0.0125 0.0661 0.1716 0.2143 0.2286 0.1143 -0.0857 -0.2178

0.6097 5.6513 22.411 35.000 35.000 35.000 35.000 11.229

0.2509 0.8119 1.2908 0.5714 -0.2857 -2.0000 -2.0000 -0.6413

0.09

150

Problem 20.3

The velocity at step 0.6 s is positive so that the tank is still moving to the right. At step 0.7 s the velocity is negative, which implies that tank is now moving to the left. In fact, the velocity would cross the zero mark sometime between 0.6 and 0.7 s, and simultaneously unloading would proceed along a line parallel to the force displacement relation in the elastic stage. As a consequence of the finite time step, the change from zero stiffness to a finite stiffness is detected too late resulting in an overshoot.

Errors are thus introduced at every change in stiffness. A number of methods exist for minimizing such errors, including the use of a sufficiently small time step in the numerical integration.

Using the incremental form of average acceleration method of direct integration the incremental displacement and velocity are determined from the following expressions.

4m 2c 4u̇n + kT = ∆p + m + 2ün + h2 h h 2 ∆u̇ = ∆u − 2u̇n h where kT is the tangent stiffness. Substituting the parameter values we have

Problem 20.2

(7000 + kT ) ∆u = ∆p + 17.5(40u̇n + 2ün ) = ∆p + 700u̇n + 35ün

The results obtained with h = 0.02 s are tabulated in Table S20.2. There is substantial difference between these results and those obtained in Problem 20.1, where we used a rather coarse interval of h = 0.1 s. The event u > 0.024 is now detected at 0.34 s when the displacement value is u = 0.0241.

The velocity is positive at 0.66 s, but negative at 0.68 s. Thus the stiffness should have been set equal to 1458.3 kN/m sometime between 0.66 and 0.68 s.

Because of the finite time step the change is made only at 0.68 s, but this should give better results than in Problem 20.1, where the change in velocity was detected only at 0.7 s.

∆u̇ = 20∆u − 2u̇n The results are tabulated in Table S20.3 and are similar to those obtained for Problem 20.1 where a linear acceleration method was used in the solution. Evidently, the time step is quite coarse so that the displacement change during each time step is comparatively large. As a consequence, while the resisting spring force reaches the plastic limit at a displacement of 0.024 m, that is somewhere between 0.3 and 0.4 s, the excursion into the plastic state is detected only at 0.4 s.

Table S20.2 t

u(t)

u̇(t)

ü(t)

p(t)

kT

∆u

∆u̇

u(t + h)

u̇(t + h)

fs

ü(t + h)

2

kN

kN/m

m

m/s

m

m/s

kN

m/s2

0 5 20 45 45 45 45 30 0 0 0 0 0

1458.3 1458.3 1458.3 1458.3 1458.3 0 0 0 0 0 1458.3 1458.3

0.0000 0.0003 0.0016 0.0038 0.0041 0.0044 0.0051 0.0052 0.0026 0.0002 -0.0006 -0.0014

0.0006 0.0064 0.0175 0.0214 0.0148 0.0114 0.0097 -0.0091 -0.0400 -0.0400 -0.0395 -0.0378

0.0000 0.0008 0.0057 0.0199 0.0241 0.0285 0.0430 0.0694 0.0888 0.0917 0.0911 0.0897

0.0006 0.0197 0.0870 0.1994 0.2142 0.2256 0.2582 0.2536 0.1085 -0.0115 -0.0510 -0.0888

0.0055 1.1381 8.2904 29.056 35.000 35.000 35.000 35.000 35.000 35.000 34.086 32.041

0.0568 0.3921 0.9548 0.9111 0.5714 0.5714 0.4000 -0.6286 -2.0000 -2.0000 -1.9477 -1.8309

s

m

m/s

m/s

0 0.10 0.20 0.30 0.32 0.34 0.40 0.50 0.60 0.66 0.68 0.70 0.80

0 0.0005 0.0041 0.0161 0.0199 0.0241 0.0380 0.0642 0.0862 0.0915 0.0917 0.0911 0.0776

0 0.0133 0.0694 0.1780 0.1994 0.2142 0.2485 0.2628 0.1485 0.0285 -0.0115 -0.0510 -0.1997

0 0.2478 0.7990 1.2265 0.9111 0.5714 0.5714 -0.2857 -2.0000 -2.0000 -2.0000 -1.9477 -0.8244

151

Table S20.3 t s

un m

u̇n m/s

ün m/s2

pn kN

kT kN/m

∆u m

∆u̇ m/s

un+1 m

u̇n+1 m/s

fs kN

ün+1 m/s2

0 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80

0 0.0006 0.0043 0.0157 0.0346 0.0566 0.0735 0.0748 0.0593

0 0.0118 0.0628 0.1650 0.2126 0.2269 0.1126 -0.0874 -0.2228

0 0.2365 0.7827 1.2621 0.5714 -0.2857 -2.0000 -2.0000 -0.7074

0 5 20 45 45 30 0 0 0

14.583 1458.3 1458.3 1458.3 0 0 0 1458.3

0.0006 0.0037 0.0114 0.0189 0.0220 0.0170 0.0013 -0.0155

0.0118 0.0510 0.1022 0.0475 0.0143 -0.1143 -0.2000 -0.1354

0.0006 0.0043 0.0157 0.0346 0.0566 0.0735 0.0748 0.0593

0.0118 0.0628 0.1650 0.2126 0.2269 0.1126 -0.0874 -0.2228

0.8620 6.3019 22.912 35.000 35.000 35.000 35.000 12.380

0.2365 0.7827 1.2621 0.5714 -0.2857 -2.0000 -2.0000 -0.7074

The velocity at step 0.6 s is positive so that the tank is still moving to the right. At step 0.7 s the velocity is negative, which implies that tank is now moving to the left. In fact, the velocity would cross the zero mark sometime between 0.6 and 0.7 s, and simultaneously unloading would proceed along a line parallel to the force displacement relation in the elastic stage. As a consequence of the finite time step, the change from zero stiffness to a finite stiffness is detected too late resulting in an overshoot. Errors are thus introduced at every change in stiffness.

Problem 20.4 The following recursion equations are used to obtain the displacement and velocity response m c 2m c un+1 = pn + 2 un + + un−1 2 h 2h h 2h (a) m − 2 un−1 − fS (tn ) h u̇n =

1 (un − un−1 ) 2h

(b)

On substituting m = 17.5 tonne, c = 0, and h = 0.02 s in Equations (a) and (b) we get 1750un+1 = pn + 3500un − 1750un−1 − fn

(c)

u̇n = 5(un − un−1 )

(d)

The response results obtained from Equations (c) and (d) are tabulated in Table S20.4. Between t = 0.3 s and 0.4 s the displacement changes from 0.0148 m to 0.0401 m. Since the latter is greater than 0.024 m, the spring force is set as 35 kN. Obviously the time step is very coarse resulting in a large overshoot. Between 0.6 s and 0.7 s the velocity changes from 0.1817 m/s to -0.0183 m/s. Since the system is unloading the stiffness should change to 1458.3 kN/m during this interval. However, in the central difference algorithm, unloading is detected only when the displacements at a time step is smaller than that at the previous time step. This occurs between 0.7 and 0.8 s. Hence the spring force at 0.8 s is calculated from fS = 35 − 1458.3(0.1075 − 0.0956) = 17.75 kN. Again, the rather coarse time step introduces large errors.

Table S20.4 Time s 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

pn kN

un−1 m

0 0 5.0 0 20.0 0 45.0 0.0029 45.0 0.0148 30.0 0.0401 0 0.0711 0 0.0993 0 0.1075

un m

fn kN

0 0 0 0 0.0029 4.1667 0.0148 21.5278 0.0401 35.0000 0.0711 35.0000 0.0993 35.0000 0.1075 35.0000 0.0956 17.7546

un+1 m

u̇n m/s

0 0 0.0029 0.0143 0.0148 0.0738 0.0401 0.1862 0.0711 0.2817 0.0993 0.2960 0.1075 0.1817 0.0956 -0.0183 0.0737 -0.1690

152 0.1

Problem 20.5

Central difference Linear acceleration

0.09 0.08 0.07 Displacement, m

The solution follows that for Problem 20.4 except that h = 0.02 s. The results are tabulated in Table S20.5. It will be noted that the errors caused by overshoots have been substantially reduced. The transition to the flat plateau on the force-displacement relationship is detected at u = 0.0241 m rather than at 0.0240 m. Unloading is detected at 0.70 s when the velocity is -0.0504. The results obtained here are quite close to those obtained for Problem 20.2 using linear acceleration method and a time step of 0.02 s as shown in Figure S20.5.

0.06 0.05 0.04 0.03 0.02 0.01 0

0

0.1

0.2

0.3

0.4

0.5

Time, s

Figure S20.5

Table S20.5 Time s

p(t) kN

u(t − h) m

u(t) m

f (t) kN

0.00 0.10 0.20 0.30 0.32 0.34 0.40 0.50 0.60 0.68 0.70 0.80

0 5.0 20.0 45.0 45.0 45.0 45.0 30.0 0 0 0 0

0 0.0002 0.0029 0.0128 0.0161 0.0199 0.0332 0.0591 0.0832 0.0919 0.0921 0.0818

0 0.0004 0.0041 0.0161 0.0199 0.0241 0.0380 0.0644 0.0866 0.0921 0.0915 0.0780

0 0.6436 5.9684 23.5119 29.0798 35.0000 35.0000 35.0000 35.0000 35.0000 34.0986 14.4134

u(t + h) m

u̇(t) m/s

0 0 0.0008 0.0134 0.0056 0.0697 0.0199 0.1786 0.0241 0.2000 0.0285 0.2148 0.0431 0.2491 0.0696 0.2634 0.0892 0.1491 0.0915 -0.0109 0.0901 -0.0504 0.0739 -0.1991

0.6

0.7

0.8

0.9

K15235

9 780415 622967