SOLUCIONARIO CHOPRA CAP. 1 COMPLETO

CHAPTER 1 Problem 1.1 Starting from the basic definition of stiffness, determine the effective stiffness of the combined spring and write the equation of motion for the spring–mass systems shown in Figs. P1.1 to P1.3. u k 1 m

p(t )

k 2

Figure P1.1

Solution:

If k e is the effective stiffness, fS = keu k

u

1

f S

k u k u 1

f S

2

k

2

Equilibrium of forces: Effective stiffness: Equation of motion:

fS = ( k1 + k2 ) u ke = fS u = k1 + k2 mu&& + keu = p( t )

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Problem 1.2 Starting from the basic definition of stiffness, determine the effective stiffness of the combined spring and write the equation of motion for the spring–mass systems shown in Figs. P1.1 to P1.3. u

•

k 1

m

k 2

p(t )

Figure P1.2

Solution:

If k e is the effective stiffness, fS = keu

(a) u

k 2

k 1

f S

If the elongations of the two springs are u1 and u2 , u = u1 + u2

(b)

Because the force in each spring is f S , fS = k1u1

fS = k2u2

(c)

Solving for u1 and u2 and substituting in Eq. (b) gives f S k e

=

k e =

f S k 1

+

f S k 2

⇒

1 ke

=

1 k1

+

1 k 2

⇒

k1 k 2 k1 + k 2

Equation of motion: mu&& + keu = p( t )


Problem 1.3 Starting from the basic definition of stiffness, determine the effective stiffness of the combined spring and write the equation of motion for the spring–mass systems shown in Figs. P1.1 to P1.3. u k 1 m

p(t )

k 3

k 2

Figure P1.3

Solution:

k 1

k 3

k 2

m

Fig. 1.3(a)

m

Fig. 1.3(b)

⇓ k 3

k 1+ k 2

⇓ u

k e

Fig. 1.3(c)

m

This problem can be solved either by starting from the definition of stiffness or by using the results of Problems P1.1 and P1.2. We adopt the latter approach to illustrate the procedure of reducing a system with several springs to a single equivalent spring. First, using Problem 1.1, the parallel arrangement of k 1 and k 2 is replaced by a single spring, as shown in Fig. 1.3(b). Second, using the result of Problem 1.2, the series arrangement of springs in Fig. 1.3(b) is replaced by a single spring, as shown in Fig. 1.3(c): 1 ke

=

1 k1 + k2

+

1 k3

Therefore the effective stiffness is k e =

( k1 + k2 ) k 3 k1 + k2 + k 3

The equation of motion is mu&& + keu = p( t ) .


Problem 1.4 Derive the equation governing the free motion of a simple pendulum that consists of a rigid massless rod pivoted at point

O with

a mass

m

attached at the tip (Fig. P1.4). Linearize the

equation, for small oscillations, and determine the natural frequency of oscillation. O

L θ

m

Figure P1.4

Solution:

or

1. Draw a free body diagram of the mass. O

&& mLθ + mg sin θ = 0

T

L

3. Linearize for small θ . For small θ, sin θ ≈ θ , and Eq. (a) becomes

θ m

mgsinθ

&& mLθ + mg θ = 0

mgcosθ

⎛ g ⎞ ⎟θ = 0 ⎝ L ⎠

&& θ +⎜

(b)

4. Determine natural frequency.

2. Write equation of motion in tangential direction. Method 1: By Newton’s law.

ω n =

−mg sin θ = ma − mg sin θ = mLθ &&

g L

(a)

&& mLθ + mg sin θ = 0

This nonlinear differential equation governs the motion for any rotation θ . Method 2: Equilibrium of moments about O yields && mL2θ = −mgL sin θ


Problem 1.5 Consider the free motion in the x y plane of a compound pendulum that consists of a rigid rod suspended from a point (Fig. P1.5). The length of the rod is L and its mass m is uniformly distributed. The width of the uniform rod is b and the thickness is t . The angular displacement of the centerline of the pendulum measured from the y -axis is denoted by θ (t ). (a) Derive the equation governing θ (t ). (b) Linearize the equation for small θ . (c) Determine the natural frequency of small oscillations. O

•

x L θ •

y

•

•

b

Figure P1.5

Solution:

1. Find the moment of inertia about O. From Appendix 8, 2

1 ⎛ L ⎞ I 0 = mL + m⎜ ⎟ = mL2 ⎝ 2 ⎠ 12 3 1

2

2. Draw a free body diagram of the body in an arbitrary displaced position.

3. Write the equation of motion using Newton’s second law of motion.

∑ M

x

L

1

2

3

− mg sin θ = 2

mL 3

L/2

&& = I 0θ

0

θ&& +

mgL 2

&& mL2θ

sin θ = 0

(a)

4. Specialize for small θ . For small θ , sinθ ≅ θ and Eq. (a) becomes

θ

mL2

θ&& +

mgL

3 g

=0

3 mg

y

&& θ +

2 L

2

θ = 0

(b)

5. Determine natural frequency. ω n =

3 g 2 L


Problem 1.6 Repeat Problem 1.5 for the system shown in Fig. P1.6, which differs in only one sense: its width varies from zero at O

O to b at

the free end.

•

x L θ •

y

•

•

b

Figure P1.6

Solution:

1. Find the moment of inertia about about O.

3. Write the equation of motion using Newton’s second law of motion.

∑ M

L

∫

I 0 = ρ r 2 d A x

− mg

0

2 L

L

∫

r

α

0

= =

ρ 4 1 2

L 4 α

3

sin θ =

1 2

2 && mL θ

2 mL && 2mgL θ+ sin θ = 0 2 3

= ρ r 2 ( r α dr ) L

&& = I 0θ

0

(a)

4. Specialize for small θ . For small θ , sinθ ≅ θ , and Eq. (a) becomes

mL 2

mL2 && 2mgL θ+ θ = 0 2 3

or 2. Draw a free body diagram of the body in an arbitrary displaced position.

&& + θ

4 g 3 L

=0

(b)

2L/3 x

ω n =

θ

y

5. Determine natural frequency.

mg

4 g 3 L

In each case the system is equivalent to the springmass system shown for which the equation of motion is

⎛ w ⎞ ⎜⎜ ⎟⎟ u&& + ku = 0 ⎝ g ⎠


k

w u

The spring stiffness is determined from the deflection u under a vertical force f S applied at the location of the lumped weight: 3

Simply-supported beam: u

=

fS L

48 EI 3

Cantilever beam: u

=

fS L

⇒

3 EI

k

=

=

fS L

192 EI

⇒

k

=

48 EI 3

L

3 EI 3

L

3

Clamped beam: u

k

⇒

=

192 EI 3

L


Problem 1.7 Develop the equation governing the longitudinal motion of the system of Fig. P1.7. The rod is made of an elastic material with elastic modulus is L. Ignore the mass of the rod and measure

u

E ; its cross-sectional area

is

A and

its length

from the static equilibrium position.

•

L

A, E

•

m u p(t ) Figure P1.7

Solution:

Draw a free body diagram of the mass: f S

mü u p(t )

Write equation of dynamic equilibrium: && + f mu S = p ( t )

(a)

Write the force-displacement relation: f S =

F AE I u GH J L K

(b)

Substitute Eq. (b) into Eq. (a) to obtain the equation of motion: && + mu

F AE I u = GH J L K

p( t )


Problem 1.8

A rigid disk of mass

m is mounted at

the end of a flexible shaft (Fig. P1.8). Neglecting the

weight of the shaft and neglecting damping, derive the equation of free torsional vibration of the disk. The shear modulus (of rigidity) of the shaft is

G.

•

L

d

O

•

θ

•

•

R

Figure P1.8

Solution:

Show forces on the disk: f S O θ

R

Write the equation of motion using Newton's second law of motion:

− fS = I Oθ &&

where I O =

mR 2

2

(a)

(b)

Write the torque-twist relation: f S =

F GJ I θ GH J L K

where

J =

π d 4

32

Substitute Eq. (b) into Eq. (a): I Oθ&& +

F GJ I θ = GH J L K

0

or,

F mR 2 I && F π d 4 G I θ + G θ = GH 2 J K H 32 L J K

0


Problems 1.9 through 1.11 Write the equation governing the free vibration of the systems shown in Figs. P1.9 to P1.11. Assuming the beam to be massless, each system has a single DOF defined as the vertical deflection under the weight w. The flexural rigidity of the beam is E I and the length is

L.

w

EI

L

u

•

Figure P1.9

Figure P1.10

w

EI u

L /2

L /2

Figure P1.11

Solution:

In each case the system is equivalent to the springmass system shown for which the equation of motion is

F w I && + u GH g J K

ku = 0

k

w u

The spring stiffness is determined from the deflection u under a vertical force f S applied at the location of the lumped weight: 3 fS L 48 EI Simply-supported beam: u = ⇒ k = 3 48 EI L f L3 3 EI Cantilever beam: u = S ⇒ k = 3 3 EI L 3 fS L 192 EI Clamped beam: u = ⇒ k = 3 192 EI L


10

Problem 1.12 Determine the natural frequency of a weight w suspended from a spring at the midpoint of a simply supported beam (Fig. P1.12). The length of the beam is L, and its flexural rigidity is E I .

The spring stiffness is

k .

Assume the beam to be massless. L

•

•

EI

k w

Figure P1.12

Solution:

1. Write the equation of motion.

L

Equilibrium of forces in Fig. 1.12c gives EI

simply supported

k

m u&& + f s = w + p (t )

(a)

where

w

f s = k e u Fig. 1.12a

(b)

The equation of motion is: mu&& + k e u = w + p (t )

(c)

2. Determine the effective stiffness. Undeformed position

δst

f s = k eu

(d)

where Static Equlibrium

u u

Deformed position

Fig. 1.12b

u = δ spring + δ beam

(e)

f s = k δ spring = k beam δ beam

(f)

Substitute for the δ’s from Eq. (f) and for u from Eq. (d): f s k e

f s

=

k e = k e =

f s ..

mu p(t) w

f s k

+

f s k beam

kk beam k + k beam

(

k 48 EI / L3 48 EI k + L3

)

3. Determine the natural frequency. ω n =

k e m

Fig. 1.12c


Problem 1.13 Derive the equation of motion for the frame shown in Fig. P1.13. The flexural rigidity of the beam and columns is as noted. The mass lumped at the beam is

; otherwise, assume the

m

frame to be massless and neglect damping. By comparing the result with Eq. (1.3.2), comment on the effect of base fixity. EI b =

•

h

m

∞

EI c

p(t ) EI c

•

L

•

•

Figure P1.13

Solution:

Compute lateral stiffness: 1 3 EI c /h

3

h

k = 2 × k column = 2 ×

3 EI c h

3

=

6 EI c h3

Equation of motion: && + ku = p( t ) mu

Base fixity increases k by a factor of 4.


Problem 1.14 Write the equation of motion for the one-story, one-bay frame shown in Fig. P1.14. The flexural rigidity of the beam and columns is as noted. The mass lumped at the beam is m; otherwise, assume the frame to be massless and neglect damping. By comparing this equation of motion with the one for Example 1.1, comment on the effect of base fixity. EI c

•

h

m

p(t )

EI c

EI c

•

L = 2h

•

•

Figure P1.14

Solution:

1. Define degrees of freedom (DOF).

3. Form structural stiffness matrix. u1 = 1, u2 = u3 = 0 3

2

1

k 31

k 21

k 11 = 2 2. Reduced stiffness coefficients.

3 EI c h

3

k 21 = k 31 =

Since there are no external moments applied at the pinned supports, the following reduced stiffness coefficients are used for the columns.

=

6 EI c h3

3 EI c h2

u2 = 1, u1 = u3 = 0

Joint rotation: 3 EI

EI

L

1

3 EI

k 23

k 22

3 EI

L2

2

L L

Joint translation: EI

3 EI 1

3 EI

2

L

3 EI

3

L

L

L3

k 22 =

3 EI c

k 32 =

2 EI c

k 12 =

3 EI c

h ( 2h )

+

4 EI c

=

EI c

( 2h )

=

5 EI c h

h

h2


u3 = 1, u1 = u2 = 0

where

k 23

k tt =

k 33 k 13

EI c h3 EI c

k t 0 =

3 EI c

k 23 =

2 EI c

k 13 =

3 EI c

h ( 2h )

+

4 EI c

=

EI c

( 2h )

=

5 EI c

−1

k 00 =

5h h

2

5h 2 ⎥⎦

⎡ 5 − 1⎤ ⎢ ⎥ 24 EI c ⎣− 1 5 ⎦ h

⎥

h ⎥ 2 5h ⎥⎦

The lateral stiffness k of the frame can be obtained by static condensation since there is no force acting on DOF 2 and 3: 2

⎥

2

we get

3h ⎤

4. Determine lateral stiffness.

3h

⎢

h 3 ⎢⎣ h

h2 ⎤

Since

Hence

⎡6 EI c ⎢ 3h 3 h ⎢ ⎢⎣3h

EI c ⎡5h 2

−1 T k = k tt − k t 0 k 00 k t 0

h

2

3h]

Then compute the lateral stiffness k from

h

h2

⎡ 6 3h EI c ⎢ 2 k = 3 ⎢3h 5h h ⎢⎣3h h 2

[3h

h3

k 00 =

k 33 =

[6]

3h ⎤ ⎧ u1 ⎫

⎧ f S ⎫ ⎥⎪ ⎪ ⎪ ⎪ h ⎥ ⎨u2 ⎬ = ⎨ 0 ⎬ 2 ⎪ ⎪ ⎪ 5h ⎥⎦ ⎪ ⎩u3 ⎭ ⎩ 0 ⎭

k =

6 EI c

k =

EI c

k =

h3

EI c h3

[3h

3h] ⋅

⎡ 5 − 1⎤ EI c ⎡3h⎤ ⎢ ⎥⋅ ⎢ ⎥ 24 EI c ⎣− 1 5 ⎦ h 3 ⎣3h⎦ h

[6 − 3]

h3 3 EI c h3

5. Equation of motion. && + mu

2

−

3 E I c h3

u = p ( t )

First partition k as

⎡6 EI c ⎢ k = 3 ⎢3h h ⎢⎣3h

3h 5h h

2

2

3h ⎤

2 ⎥ ⎡ k tt h ⎥=⎢ ⎣k t 0 2 5h ⎥⎦

k t 0 ⎤

⎥

k 00 ⎦


Problem 1.15 Write the equation of motion of the one-story, one-bay frame shown in Figs. P1.15 and P1.16. The flexural rigidity of the beam and columns is as noted. The mass lumped at the beam is ; otherwise, assume the frame to be massless and neglect damping. Check your result from

m

Problem 1.15 against Eq. (1.3.5). Comment on the effect of base fixity by comparing the two equations of motion. m

EI c /2

•

h

EI c

p(t ) EI c

•

•

L = 2h

•

Figure P1.15

Solution:

1

Ib = I c / 2 h

I c

k 22

k 32

k

k 12

I c 2h

k 22 =

4 EI c

k 32 =

2 EI b

k 12 =

6 EI c

Define degrees of freedom (DOF): 2

3 1

h (2 h )

+

4 EI b

=

EI c

(2h )

=

4 EI c h

+

k 21

k21 = k 31 =

6 EI c

u2 = 1, u1 = u3 = 0

5 EI c

k 33 k 13

k 11

=

=

h

2h

k 23

k 31

h3

5 EI c

h2

u1 = 1, u2 = u3 = 0

k 11 = 2

h

=

u3 = 1, u1 = u2 = 0

Form structural stiffness matrix:

12 EI c

EI c

24 EI c

k 33 =

4 EI c

k 23 =

2 EI b

k 13 =

6 EI c

h3

h2

h (2 h )

+

4 EI b

=

EI c

(2 h )

=

4 EI c h

+

EI c h

h

2h

h2


Hence

Since

⎡24 EI c ⎢ k = 6h 3 ⎢ h ⎢ ⎣6h

6h ⎤

6h 5h

2

1 2

2

h

1 2

2

−1

⎥

k 00

h ⎥ 2 5h ⎥⎦

we get

The lateral stiffness k of the frame can be obtained by static condensation since there is no force acting on DOF 2 and 3:

⎡ 24 EI c ⎢ 6h 3 ⎢ h ⎢ ⎣6 h

6 h ⎤ ⎧ u1 ⎫

⎧ f S ⎫ ⎥ ⎪ ⎪ ⎪ ⎪ 1 2 h ⎥ ⎨u 2 ⎬ = ⎨ 0 ⎬ 2 2 ⎪ ⎪ ⎪ 5h ⎥⎦ ⎪ ⎩u 3 ⎭ ⎩ 0 ⎭

6h 5h

2

1 2

2

h

First partition k as

⎡24

k =

EI c ⎢ h

⎢6 h ⎢6 h ⎣

3

⎡ 5 − 12 ⎤ = ⎢ ⎥ 5⎥⎦ 99 EI c ⎣⎢− 12 4h

k =

24 EI c EI c h3

−

=

EI c

=

h 120 EI c

3

11

h3

( 24 −

h

[6h

144 11

6h ] ⋅

4h ⎡ 5

⎢

99 EI c ⎣⎢− 12

− 12 ⎤ EI c ⎡6h⎤ ⎥⋅ ⎢ ⎥ 5⎦⎥ h3 ⎣6h⎦

)

3

This result can be checked against Eq. 1.3.5: 6h ⎤

6h 5h

2

1 2

2

h

⎡k tt h ⎥ = ⎢ T ⎣k t 0 2 5h ⎥⎦

1 2

2⎥

k t 0 ⎤

⎥ k 00 ⎦

k =

24 EI c ⎛ 12 ρ +1 ⎞ h3

⎜⎜ ⎟⎟ ρ 12 4 + ⎝ ⎠

Substituting ρ = Ib 4 I c = 1 8 gives

where

12 +1 ⎞ 24 EI c ⎛ ⎜ 8 ⎟ 24 EI c ⎛ 5 ⎞ 120 EI c 1

k tt =

EI c

k t 0 =

EI c

h

h3

EI k 00 = c h3

k =

24

3

6h

⎡ 5h 2 ⎢1 2 ⎢⎣ 2 h

⎜ 12 1 + 4 ⎟ ⎝ 8 ⎠

=

h3

⎜ ⎟= ⎝ 11 ⎠ 11 h 3

Equation of motion:

6h 1 2

h3

h2 ⎤

⎛ 120 EI c ⎞ m u&& + ⎜⎜ ⎟ u = p(t ) 3 ⎟ ⎝ 11 h ⎠

⎥ 5h ⎥⎦ 2

Then compute the lateral stiffness k from −1

T

k = ktt − kt 0 k 00 kt 0


Problem 1.16 Write the equation of motion of the one-story, one-bay frame shown in Figs. P1.15 and P1.16. The flexural rigidity of the beam and columns is as noted. The mass lumped at the beam is ; otherwise, assume the frame to be massless and neglect damping. Check your result from

m

Problem 1.15 against Eq. (1.3.5). Comment on the effect of base fixity by comparing the two equations of motion. m

EI c /2

•

h

p(t )

EI c

EI c

•

L = 2h

•

•

Figure P1.16

Solution:

1. Define degrees of freedom (DOF). u3

u4

m

EI c

EI c

u2

u5

k 12 =

h 2 EI c

k 32 =

u1

EIc /2 h

4 EI c

k 22 =

6 EI c h

2

k 42 = k 52 = 0

h

u3 = 1 , u1 = u2 = u4 = u5 = 0 u 3 =1

2h

2. Form the structural stiffness matrix. u1 = 1 , u2 = u3 = u4 = u5 = 0 u 1 =1

k 33 = k 13 = k 11 = 2

12 EI c h

3

=

24 EI c h

k 43 =

3

k 21 = k 31 = k 41 = k 51 =

4 EI c h

+

4 EI c 2(2h)

6 EI c

k 23 =

2

h 2 EI c 2(2h)

=

=

EI c 2h

5 EI c h 2 EI c h k 53 = 0

6 EI c 2

h

u2 = 1 , u1 = u3 = u4 = u5 = 0

u 2 =1


u4 = 1 , u1 = u2 = u3 = u5 = 0

3. Determine the lateral stiffness of the frame. First partition k . u 4 =1

k 44 = k 14 = k 34 =

4 EI c h

+

4 EI c 2(2h)

=

5 EI c

6 EI c h2 2 EI c 2(2h)

24 6h 6h 4h 2 EI  2 k = 3c 6 h 2 h h  6h 0 6h 0

h k 24 = 0

=

EI c

k 54 =

2h

6h 2h

6h

2

0

5h 2 1 2

h

1 2

2

0

h2

5h

2

2h

2

6h  0 

  k tt

0  =  T k t 0 2 2h 

k t 0 



k 00 

2 4h 

Compute the lateral stiffness.

2 EI c h

k = k tt − k t 0k −001k T t 0 k =

u5 = 1 , u1 = u2 = u3 = u4 = 0

24 EI c h

3

−

22 EI c h

3

=

2 EI c h

3

4. Write the equation of motion. u5 =1

mu&& + ku = p(t )

⎛ 2 EI c ⎞ ⎟ u = p (t ) 3 ⎟ ⎝ h ⎠

mu&& + ⎜⎜

k 55 = k 45 =

4 EI c

k 15 =

h 2 EI c

6 EI c h

2

k 25 = k 35 = 0

h

Assemble the stiffness coefficients:

⎡24 6h ⎢6 h 4h 2 EI ⎢ 2 k = 3c ⎢6 h 2h h ⎢ 6h 0 ⎢ ⎢⎣6 h 0

6h

6h

2

2h

5h 2 1 2

1 2

6h ⎤

0

0 ⎥

h2

0 ⎥

⎥

h2

5h 2

2h2 ⎥

0

2h 2

4h2 ⎥⎦

⎥


Problem 1.17 A heavy rigid platform of weight

w

is supported by four columns, hinged at the top and

the bottom, and braced laterally in each side panel by two diagonal steel wires as shown in Fig. P1.17. Each diagonal wire is pretensioned to a high stress; its cross-sectional area is A and elastic modulus is E . Neglecting the mass of the columns and wires, derive the equation of motion governing free vibration in (a) the x -direction, and (b) the y-direction. ( Hint: Because of high pretension, all wires contribute to the structural stiffness, unlike Example 1.2, where the braces in compression do not provide stiffness.) •

h

y x

• •

h

•

h •

•

Figure P1.17

Solution:

(a) Equation of motion in the x-direction. The lateral stiffness of each wire is the same as the lateral stiffness of a brace derived in Eq. (c) of Example 1.2:

⎛ AE ⎞ 2 k w = ⎜ ⎟cos θ ⎝ L ⎠ ⎛ AE ⎞ 2 1 AE ⎟cos 45 = = ⎜⎜ ⎟ 2 2 h ⎝ h 2 ⎠ o

Each of the four sides of the structure includes two wires. If they were not pretensioned, under lateral displacement, only the wire in tension will provide lateral resistance and the one in compression will go slack and will not contribute to the lateral stiffness. However, the wires are pretensioned to a high stress; therefore, under lateral displacement the tension will increase in one wire, but

decrease in the other; and both wires will contribute to the lateral direction. Consequently, four wires contribute to the stiffness in the x-direction: k x = 4 k w =

2

AE h

Then the equation of motion in the x-direction is mu&& x + kxux = 0

(b) Equation of motion in the y-direction. The lateral stiffness in the y direction, k y = k x , and the same equation applies for motion in the y-direction: mu&& y + kyuy = 0


Problem 1.18 Derive the equation of motion governing the torsional vibration of the system of Fig. P1.17 about the vertical axis passing through the center of the platform. Solution:

z h h

k w h 2

f O

I

O

uθ = 1

f

S

Fig. 1.18(a)

Fig. 1.18(b)

1. Set up equation of motion. The elastic resisting torque f S and inertia force f I are shown in Fig. 1.18(a). The equation of dynamic equilibrium is f I + fS = 0 or IOu&&θ + f S = 0

(a)

where IO = m

h 2 + h2 12

=

mh2 6

(b)

2. Determine torsional stiffness, k θ . fS = kθuθ

(c)

Introduce uθ = 1 in Fig. 1.18(b) and identify the resisting forces due to each wire. All the eight forces are the same; each is kw h 2 , where, from Problem 1.17, k w =

1

AE

2 2

h

The torque required to equilibrate these resisting forces is kθ = 8k w

h h

2 2 AEh = 2

= 2 kw h2 =

2

AE 2 ( )h 2 2 h (d)

3. Set up equation of motion. Substituting Eq. (d) in (c) and then Eqs. (c) and (b) in (a) gives the equation of motion: mh2 6

u&&θ +

AEh 2

uθ = 0


Problem 1.19 An automobile is crudely idealized as a lumped mass

m

supported on a spring–damper sys-

tem as shown in Fig. P1.19. The automobile travels at constant speed v over a road whose roughness is known as a function of position along the road. Derive the equation of motion.

m

v c

k

t

u (t )

ug( x ) x

Figure P1.19

Solution:

u t m k

f

v

I

u t - u

c

g

f f S

D

u

g

Fig. 1.19(a)

Fig. 1.19(b)

Displacement ut is measured from the static equilibrium position under the weight mg . From the free-body diagram in Fig. 1.19(b) f I + fD + f S = 0

(a)

where f I = mu&&t f D = c ( u& t − u& g )

(b)

fS = k ( ut − u g ) Substituting Eqs. (b) in Eq. (a) gives t && mu + c (u& t − u& g ) + k (u t − u g ) = 0

Noting that x = vt and transferring the excitation terms to the right side gives the equation of motion: t && mu + cu& t + ku t = cu& g ( vt ) + kug ( vt )


SOLUCIONARIO CHOPRA CAP. 1 COMPLETO

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