Dynamics of Structures 4th Edition Chopra Solutions Manual

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Dynamics of Structures 4th Edition Chopra Solutions Manual

CHAPTER 2 Problem 2.1 Given: Tn = 2 π

m = 0. 5 sec k

(a)

Tn′ = 2 π

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

⎛ Tn′ ⎞ m + 50 g ⎜⎜ ⎟⎟ = m ⎝ Tn ⎠

2

⇒ 1+

50 ⎛ 0.75 ⎞ =⎜ ⎟ = 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 =16π 2 m =16π 2 ⎜ ⎟ =16.4lbs in. ⎝ 386 ⎠

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Problem 2.2 1. Determine the natural frequency. m =

k = 100 lb in.

ωn =

k m

=

100 400 386

400 386

lb − sec2 in.

= 9. 82 rads sec

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 ω nt = 2 cos ( 9. 82 t )

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Problem 2.3 1. Set up equation of motion. ku+mg/2

mü u mg

mu&& + ku =

mg 2

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

mg 2k

At t = 0 , u( 0 ) = 0 and u& ( 0 ) = 0 ∴ A = − u(t ) =

mg , B = 0 2k

mg (1 − cos ω nt ) 2k

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Problem 2.4 k

m =

u m

v0 m0

10 = 0. 0259 lb − sec2 in. 386

m0 =

0. 5 = 1. 3 × 10 −3 lb − sec2 in. 386

k = 100 lb in.

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 ω nt = 0. 565 sin ( 60. 63t ) in.

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Problem 2.5

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 = m2g

(a)

The general solution is u ( t ) = A cos ω nt + B sin ω nt +

ωn =

m2 g k

k m1 + m2

(b) (c)

The initial conditions are u( 0 ) = 0

u& (0) =

m2 m1 + m 2

2gh

(d)

The initial velocity in Eq. (d) was determined by conservation of momentum during impact: m2u&2 = ( m1 + m2 ) u& ( 0 )

where u&2 =

2 gh

Impose initial conditions to determine A and B: u( 0 ) = 0 ⇒ A = − u& ( 0 ) = ω n B ⇒ B =

m2 g k

(e)

m2 m1 + m2

2 gh

ωn

(f)

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

u(t ) =

m2 g (1 − cos ω nt ) + k

2 gh

ωn

m2 sin ω nt m1 + m2

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Problem 2.6 1. Determine deformation and velocity at impact. u( 0) =

mg 10 = = 0.2 in. k 50

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

2. Determine the natural frequency.

ωn =

kg (50)(386) = = 4393 . rad/sec w 10

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

u& (0)

ωn

sin ω n t

⎛ 166.7 ⎞ = ( 0.2) cos 316.8t − ⎜ ⎟ sin 316.8t ⎝ 4393 . ⎠

⎡ u&( 0) ⎤ uo = [u( 0)]2 + ⎢ ⎥ ⎣ ωn ⎦

2

= 0.2 2 + ( −3.795) 2 = 38 . in.

4. Compute the maximum acceleration.

u&&o = ω n 2 uo = ( 4393 . )2 (38 . ) = 7334 in./sec2 = 18.98g

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Problem 2.7 Given: m =

200 = 6. 211 lb − sec2 ft 32. 2

fn = 2 Hz

Determine EI: k = fn =

3 EI 3 EI EI = lb ft 3 = 3 3 9 L 1

k

2π

m

⇒ 2 =

1

EI

2π

55. 90

⇒

EI = ( 4 π )2 55. 90 = 8827 lb − ft 2

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Problem 2.8 Equation of motion: mu&& + cu& + ku = 0

(a)

Dividing Eq. (a) through by m gives u&& + 2 ζω n u& + ω 2n u = 0

(b)

where ζ = 1.

Equation (b) thus reads u&& + 2 ω n u& + ω 2n u = 0

(c)

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

Because e st is never zero, the quantity within parentheses must be zero: s 2 + 2 ω n s + ω 2n = 0

or s =

− 2ω n ±

( 2 ω n ) 2 − 4 ω 2n 2

= − ωn

(double root) The general solution has the following form: u ( t ) = A1 e − ω n t + A2 t e − ω n 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 ) = − ω n A1 e − ω n t + A2 (1 − ω n t ) e − ω n t

(f)

Evaluate Eq. (f) at t = 0 : u& ( 0 ) = − ω n A1 + A2 (1 − 0 ) ∴ A2 = u& ( 0 ) + ω n A1 = u& ( 0 ) + ω n u ( 0 )

(g)

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

(h)

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Problem 2.9

A2 ω n ⎡−ζ + ζ 2 −1 + ζ + ζ 2 −1 ⎤ = ⎢⎣ ⎦⎥

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

Dividing Eq. (a) through by m gives u&& + 2 ζω n u& +

ω n2 u

= 0

or (b)

where ζ > 1.

A2 =

Assume a solution of the form u ( t ) = e st . Substituting this solution into Eq. (b) yields ( s 2 + 2ζω n s + ω n2 ) e st = 0

Because e st is never zero, the quantity within parentheses must be zero: s

2

+ 2 ζω n s +

ω 2n

= 0

or s =

−2ζω n ±

u& (0) + ⎛⎜ ζ + ζ 2 −1 ⎞⎟ ω n u (0) ⎝ ⎠

(a)

u& (0) + ⎛⎜ ζ + ⎝

2 ζ −1 ω n

Substituting Eq. (f) in Eq. (d) gives u& (0) + ⎛⎜ ζ + ζ 2 − 1 ⎞⎟ω nu (0) ⎝ ⎠ A1 = u (0) − 2 2 ζ −1ω n 2 ζ 2 − 1 ω nu (0) − u& (0) − ⎛⎜ ζ + ζ 2 − 1 ⎞⎟ ω nu (0) ⎝ ⎠ = 2 2 ζ −1ω n

2 ζ 2 − 1ω n

(g)

ζ 2 −1 ⎞⎟ ω n ⎠

The solution, Eq. (c), now reads:

The general solution has the following form:

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

u (t ) = e −ζω nt

ω ′D =

(d)

Differentiating Eq. (c) with respect to t gives

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

1

−ω ′D t

′

+ A2 e ω D t

)

ζ2 − 1 ω n

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

Evaluate Eq. (c) at t = 0 :

⎡ ⎤ u& (t ) = A1 ⎛⎜ −ζ − ζ 2 −1 ⎞⎟ ω n exp⎢⎛⎜ −ζ − ζ 2 −1 ⎞⎟ ω nt ⎥ ⎝ ⎠ ⎝ ⎠ ⎣ ⎦

(A e

where (c)

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

(f)

−u& (0) + ⎛⎜ −ζ + ζ 2 − 1 ⎞⎟ ω nu (0) ⎝ ⎠ =

(2ζω n ) 2 − 4ω n2

⎡ ⎤ u (t ) = A1 exp ⎢⎛⎜ −ζ − ζ 2 −1 ⎞⎟ω n t ⎥ ⎠ ⎣⎝ ⎦

⎠

2

2

= ⎛⎜ − ζ ± ⎝

ζ 2 −1 ⎞⎟ ω n u (0)

A2 =

u& (0) + ⎛⎜ ζ + ⎝

ζ 2 −1 ⎞⎟ ω n u (0) ⎠

2ω ′D

(e)

Evaluate Eq. (e) at t = 0 : u& (0) = A 1 ⎛⎜ −ζ − ζ 2 −1 ⎞⎟ ω n + A 2 ⎛⎜ −ζ + ζ 2 −1 ⎞⎟ ω n ⎠ ⎠ ⎝ ⎝ = [u (0) − A2 ] ⎛⎜ −ζ − ζ 2 −1 ⎞⎟ ω n + A2 ⎛⎜ −ζ + ζ 2 −1 ⎞⎟ ω n ⎝ ⎠ ⎝ ⎠

or

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Problem 2.10

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

Equation of motion: u&& + 2ζω n u& + ω n2 u = 0

(a)

u(t) = e

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

Assume a solution of the form

(i)

A2 = u& (0)

(j)

st

Substituting in Eq. (i) gives

Substituting this solution into Eq. (a) yields:

u(t ) = u& (0) t e −ω n t

FH s2 + 2ζω n s + ω 2n IK e st = 0

(c) Overdamped Systems, ζ>1 The roots of the characteristic equation [Eq. (b)] are:

st

Because e is never zero s 2 + 2ζω n s + ω 2n = 0

(b)

The roots of this characteristic equation depend on ζ.

s1,2 = ω n

1−ζ

IK

(l)

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

The two roots of Eq. (b) are 2

FH

s1,2 = ω n −ζ ± ζ 2 − 1

The general solution is:

(a) Underdamped Systems, ζ