Dynamics of Structures 5th Edition Chopra Solutions Manual

Dynamics of Structures 5th Edition Chopra Solutions Manual Full clear download (no formatting errors) at: http://testbanklive.com/download/dynamics-of-structures-5th-edition-choprasolutions-manual/ CHAPTER 2 Problem 2.1 A heavy table is supported by flat steel legs (Fig. P2.1). Its natural period in lateral vibration is 0.5 sec. When a 50-lb plate is clamped to its surface, the natural period in lateral vibration is lengthened to 0.75 sec. What are the weight and the lateral stiffness of the table?

Tn = 0.5 sec

Tn = 0.75 sec

Figure P2.1

Solution: Given: Tn = 2π

m = 0. 5 sec k

(a)

Tn′ = 2π

m + 50 g = 0. 75 sec k

(b)

1. Determine the weight of the table. Taking the ratio of Eq. (b) to Eq. (a) and squaring the result gives 2

⎛ ′⎞ T ⎜ n ⎟ = m+50 g ⎜T ⎟ m ⎝ n⎠

50 ⎛ 0.75 ⎞ ⇒ 1+

=⎜

2

⎟ =2.25

mg ⎝ 0.5 ⎠

or mg =

50 = 40 lbs 1. 25

2. Determine the lateral stiffness of the table. Substitute for m in Eq. (a) and solve for k: ⎛ 40 ⎞ k =16π2 m =16π2 ⎜ ⎟ =16.4lbs in. ⎝ 386 ⎠

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Problem 2.2 An electromagnet weighing 400 lb and suspended by a spring having a stiffness of 100 lb/in. (Fig. P2.2a) lifts 200 lb of iron scrap (Fig. P2.2b). Determine the equation describing the motion when the electric current is turned off and the scrap is dropped (Fig. P2.2c).

Figure P2.2

Solution: 1. Determine the natural frequency. k = 100 lb in.

ωn =

k = m

m =

400 lb − sec 2 in. 386

100 = 9. 82 rads sec 400 386

2. Determine initial deflection. Static deflection due to weight of the iron scrap u(0) =

200 = 2 in. 100

3. Determine free vibration. u(t ) = u(0 ) cos ωnt = 2 cos (9. 82t )

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Problem 2.3 A mass m is at rest, partially supported by a spring and partially by stops (Fig. P2.3). In the position shown, the spring force is mg/ 2. At time t = 0 the stops are rotated, suddenly releasing the mass. Determine the motion of the mass.

k

m u Figure P2.3

Solution: 1. Set up equation of motion. ku+m g/2

mü u mg

mu&& + ku =

mg 2

2. Solve equation of motion. u(t ) = A cos ωnt + B sin ωnt +

mg 2k

At t = 0 , u(0) = 0 and u& (0) = 0 mg , B = 0 2k mg (1 − cos ωnt ) u(t ) = 2k

∴ A = −

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Problem 2.4 The weight of the wooden block shown in Fig. P2.4 is 10 lb and the spring stiffness is 100 lb/in. A bullet weighing 0.5 lb is fired at a speed of 60 ft/sec into the block and becomes embedded in the block. Determine the resulting motion u(t) of the block.

vo m k Figure P2.4

Solution: u k

m

v0 m0

10 = 0. 0259 lb − sec2 in. 386 0. 5 m0 = = 1.3 × 10− 3 lb − sec2 in. 386 k = 100 lb in. m =

Conservation of momentum implies m0 v0 = (m + m0 ) u& (0) u& (0) =

m0 v0 = 2.857 ft sec = 34. 29 in. sec m + m0

After the impact the system properties and initial conditions are Mass = m + m0 = 0. 0272 lb − sec2 in. Stiffness = k = 100 lb in. Natural frequency:

ωn =

k = 60. 63 rads sec m + m0

Initial conditions: u(0) = 0, u&( 0) = 34. 29 in. sec The resulting motion is u(t ) =

u& ( 0 )

ωn

sin ωnt = 0.565 sin (60. 63t ) in.

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Problem 2.5 A mass m 1 hangs from a spring k and is in static equilibrium. A second mass m 2 drops through a height h and sticks to m1 without rebound (Fig. P2.5). Determine the subsequent motion u( t) measured from the static equilibrium position of m1 and k.

k •

m2

•

h m1

Figure P2.5

Solution:

k m2

f S = ku

h

m2 m1

m1 u m 2g

With u measured from the static equilibrium position of m1 and k, the equation of motion after impact is ( m1 + m2 ) u&& + ku = m2 g m2 g k

k m1 + m2

(b)

u(0) = 0 ⇒ A = −

(c)

u&(0) = ωn B ⇒ B =

m2 g k

(e)

2 gh m2 m1 + m2 ωn

(f)

Substituting Eqs. (e) and (f) in Eq. (b) gives

The initial conditions are u& (0) =

2 gh

Impose initial conditions to determine A and B:

u(t ) = A cos ωnt + B sin ωnt +

u(0) = 0

u&2 =

(a)

The general solution is

ωn =

where

m2 m1 + m 2

2gh

(d)

The initial velocity in Eq. (d) was determined by conservation of momentum during impact:

u(t ) =

m2 g (1 − cos ωnt ) + k

2 gh

m2

ωn m1 + m2

sin ωnt

m2u&2 = ( m1 + m2 ) u&( 0)

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Problem 2.6 The packaging for an instrument can be modeled as shown in Fig. P2.6, in which the instrument of mass m is restrained by springs of total stiffness k inside a container; m = 10 lb/g and k = 50 lb/in. The container is accidentally dropped from a height of 3 ft above the ground. Assuming that it does not bounce on contact, determine the maximum deformation of the packaging within the box and the maximum acceleration of the instrument.

u

k/2 m k/2

3’ Figure P2.6

Solution:

4. Compute the maximum acceleration.

1. Determine deformation and velocity at impact. mg 10 u(0) = = = 0.2 in. k 50

2 u&&o = ωn u o = (43.93) 2 (3.8)

= 7334 in./sec2 = 18.98g

u&( 0 ) = − 2 gh = − 2(386 )(36 ) = −166.7 in./sec

2. Determine the natural frequency. ω n=

kg (50)(386) = = 43.93 rad/sec 10 w

3. Compute the maximum deformation. u(t ) = u(0) cos ωn t +

u& (0)

ωn

sin ωnt

⎛ 166.7 ⎞ ⎟ sin 316.8t = (0.2) cos 316.8t − ⎜ ⎝ 43.93⎠ ⎡ u&(0) ⎤ uo = [u(0)]2 + ⎢ ⎥ ⎣ ωn ⎦

2

= 0.2 2 + ( −3.795) 2 = 3.8 in.

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Problem 2.7 Consider a diver weighing 200 lbs at the end of a diving board that cantilevers out 3 ft. The diver oscillates at a frequency of 2 Hz. What is the flexural rigidity EI of the diving board?

Solution: Given: 200 = 6. 211 lb − sec2 ft 32. 2 fn = 2 Hz m =

Determine EI: k = fn =

3EI 3 EI EI = 3 = lb ft 3 9 L3 1

k

2π m

⇒ 2 =

1 EI ⇒ 2 π 55. 90

EI = ( 4 π)2 55. 90 = 8827 lb − ft 2

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Problem 2.8 Show that the motion of a critically damped system due to initial displacement u(0) and initial velocity u˙ (0) is u(t) = {u(0) + [u˙ (0) + ωn u(0)] t} e−ωn t

Solution:

Evaluate Eq. (f) at t = 0 :

Equation of motion: mu&& + cu& + ku = 0

(a)

u&(0) = − ωn A1 + A2 (1 − 0 ) ∴ A2 = u& (0 ) + ωn A1 = u& (0) + ωn u(0)

Dividing Eq. (a) through by m gives u&& + 2 ζωn u& + ωn2u = 0

(b)

Substituting Eqs. (e) and (g) for A1 and A2 in Eq. (d) gives u(t ) = { u (0) + [u& (0) + ωn u(0) ] t} e −ωnt

where ζ = 1.

(g)

(h)

Equation (b) thus reads u&& + 2 ωn u& + ω2nu = 0

(c)

Assume a solution of the form u(t ) = e st . Substituting this solution into Eq. (c) yields st ( s 2 + 2 ωn s + ωn ) e = 0 2

Because e st is never zero, the quantity within parentheses must be zero: s 2 + 2 ωn s + ω2n = 0 or s =

−2 ωn ± ( 2 ωn ) 2 − 4 ω2 2

n

= − ωn (double root)

The general solution has the following form: u(t ) = A1 e − ωn t + A2 t e − ωn t

(d)

where the constants A1 and A2 are to be determined from the initial conditions: u( 0) and u&( 0) . Evaluate Eq. (d) at t = 0 : u(0) = A1 ⇒ A1 = u(0)

(e)

Differentiating Eq. (d) with respect to t gives u&(t ) = − ωn A1 e

− ωn t

+ A2 (1 − ωn t ) e − ωn t

(f)

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Problem 2.9 Show that the motion of an overcritically damped system due to initial displacement u(0) and initial velocity u˙ (0) is u(t) = e−ζ ωn t A1 e−ωD t + A2 eωD t √ where ωD = ωn ζ 2 − 1 and

√ − u˙ (0) + −ζ + ζ 2 − 1 ωn u(0) 2ωD √ 2 u˙ (0) + ζ + ζ − 1 ωn u(0) A2 = 2ωD A1 =

Solution:

Differentiating Eq. (c) with respect to t gives ⎡⎛ 2 ⎞ t⎤ u& (t ) = A1⎛⎜−ζ − ζ 2 −1⎟⎞ ω n exp ⎢⎜ −ζ − ζ −1 ⎟ ω ⎠ n ⎝ ⎠ ⎣⎝

Equation of motion: mu&& + cu& + ku = 0 Dividing Eq. (a) through by m gives u&& + 2 ζωn u& + ω2nu = 0

(b)

Evaluate Eq. (e) at t = 0 : u& (0) = A 1⎛⎜ −ζ− ζ 2 −1 ⎞⎟ ωn + A 2 ⎛⎜ −ζ + ⎝ ⎠ ⎝

where ζ > 1.

(e)

⎡ ⎤ 2 exp ⎢⎛⎜ −ζ + ζ 2 −1⎟⎞ ωnt + A2 ⎛⎜ −ζ + ζ −1⎟⎞ ω ⎝ ⎠ n ⎣⎝ ⎠

(a)

Assume a solution of the form u(t ) = e . Substituting st

u

A

⎜ ζ

ζ2 −1 ⎞⎟ ωn ⎠

⎟ω A ⎜ ζ

ζ

ζ

2

⎟ω

2

= [ (0) −

this solution into Eq. (b) yields

2]

⎛− − ⎝

−1 ⎞ ⎠

n

+

2

⎛− + ⎝

−1 ⎞ ⎠

n

or

(s 2 + 2ζ ωn s + ω2n ) e st = 0

A2 ωn ⎡ −ζ + ζ 2 −1 + ζ ζ 2 −1⎥⎤ = ⎦ ⎣ + u& (0) + ⎛⎜ζ + ζ 2 −1⎟⎞ ωn u(0) ⎝ ⎠

Because e st is never zero, the quantity within parentheses must be zero: s 2 + 2 ζωn s + ωn2 = 0 or

or s =

−2ζωn ±

(2ζωn ) 2 − 4ωn2 A2 =

2

= ⎛⎜ − ζ ± ζ ⎝

2

−1⎟⎞ ωn ⎠

⎡ + A2 exp ⎢⎛⎜ −ζ + ζ ⎣⎝

⎠

2 ζ

2

(f)

−1 ωn

Substituting Eq. (f) in Eq. (d) gives

The general solution has the following form: ⎡ u(t ) = A1 exp ⎢⎛⎜ −ζ − ζ ⎣⎝

ζ 2 − 1⎞⎟ ωn u(0)

u& (0) + ⎜⎛ ζ + ⎝

2

⎤ −1⎞⎟ωn t ⎠ 2

−1⎞⎟ωn t ⎠

A1 = u(0) −

⎤

2 ζ

(c) 2 ζ =

where the constants A1 and A2 are to be determined from the initial conditions: u( 0) and u& ( 0) .

(d)

2

ζ

2

2

− 1⎟⎞ωn u (0) ⎠

−1ω n

⎛ −1 ω n u (0) − u& (0) − ⎜ ζ + ζ ⎝ 2 2 ζ −1ωn

−u& (0) + ⎜⎛ −ζ + =

Evaluate Eq. (c) at t = 0 : u(0)= A1 + A2 ⇒ A1 + A2 =u(0)

u& (0) + ⎛⎜ζ + ζ ⎝

⎝

2

2

− 1⎞⎟ nωu (0) ⎠

− 1 ⎟⎞ ωnu (0) ⎠

2 ζ

2

− 1ωn

(g)

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The solution, Eq. (c), now reads: u(t ) = e −ζωnt

(A e

−ω′D t

1

′

+ A2 e ωD t

)

where

ω′D =

ζ 2 − 1 ωn

−u& (0) + ⎛⎜ −ζ + ζ 2 − 1⎟⎞ωn u (0) ⎝ ⎠ A1 = 2ω u& (0) + ⎛⎜ζ + ζ 2 −1 ⎞ ω u (0) ⎝ ⎠⎟ A2 = n ′ 2ωD

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Problem 2.10 Derive the equation for the displacement response of a viscously damped SDF system due to initial velocity u˙ (0) for three cases: (a) underdamped systems; (b) critically damped systems; and (c) overdamped systems. Plot u(t) ÷ u˙(0)/ωn against t/Tn for ζ = 0.1, 1, and 2.

Solution:

Substituting A and B into Eq. (f) gives u(t ) =

Equation of motion: u& + 2ζ ωn u& + ω2nu = 0

(a)

2

IK t

(g)

(b) Critically Damped Systems, ζ = 1 s1 = −ωn

s2 = −ωn

(h)

The general solution is u( t) = A1 e − ωn t + A2 t e −ωnt

IK

s 2 + 2ζω s + ω2 e st = 0 n n

A1 = 0

st

s 2 + 2ζωn s + ω2 = 0 n

u(t ) = u& (0) t e −ωn t

The roots of the characteristic equation [Eq. (b)] are: 2

(c)

FH

s1,2 = ωn −ζ ± ζ

Hence the general solution is s t

j

(d)

ωD = ωn 1 − ζ 2

(A cos ωD t + Bsin ω D t)

(f)

Determine A and B from initial conditions u(0) = 0 and u& (0) : B=

(m)

FH −ζ +

IK

ζ 2 −1 ωn t

+ A 2e

IK

ζ 2 −1 ωn t

Determined from the initial conditions u(0) = 0 and u&(0) : − A1 = A2 =

u& (0)

(o)

2ωn ζ 2 − 1

Substituting in Eq. (n) gives u(t ) =

u&(0) e −ζωn t

F GH e

ωt

n

ωD

FH −ζ −

(n) (e)

Rewrite Eq. (d) in terms of trigonometric functions:

u& (0)

(l)

which after substituting Eq. (l) becomes u(t ) = A1e

where

−ζωnt

−1

u(t ) = A1e s1t + A2 e s2 t

which after substituting in Eq. (c) becomes u(t ) = e−ζωn t A1e iωDt + A 2e −iωDt

2

The general solution is:

u( t) = A1e 1 + A2e 2

e

(k)

(c) Overdamped Systems, ζ>1

The two roots of Eq. (b) are

st

(j)

Substituting in Eq. (i) gives

(a) Underdamped Systems, ζ