Preview Extract

CHAPTER 2
Problem 2.1
Given:
Tn = 2 ฯ
m
= 0. 5 sec
k
(a)
Tnโฒ = 2 ฯ
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
โ Tnโฒ โ m + 50 g
โโ โโ =
m
โ Tn โ
2
โ 1+
50 โ 0.75 โ
=โ
โ = 2.25
mg โ 0.5 โ
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:
โ 40 โ
k =16ฯ 2 m =16ฯ 2 โ
โ =16.4lbs in.
โ 386 โ
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Problem 2.2
1. Determine the natural frequency.
m =
k = 100 lb in.
ฯn =
k
m
=
100
400 386
400
386
lb โ 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 ฯ nt = 2 cos ( 9. 82 t )
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Problem 2.3
1. Set up equation of motion.
ku+mg/2
mรผ
u
mg
mu&& + ku =
mg
2
2. Solve equation of motion.
u ( t ) = A cos ฯ nt + B sin ฯ nt +
mg
2k
At t = 0 , u( 0 ) = 0 and u& ( 0 ) = 0
โด A = โ
u(t ) =
mg
, B = 0
2k
mg
(1 โ cos ฯ nt )
2k
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Problem 2.4
k
u
m
v0
m0
m =
10
= 0. 0259 lb โ sec2 in.
386
m0 =
0. 5
= 1. 3 ร 10 โ3 lb โ 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 โ sec2 in.
Stiffness = k = 100 lb in.
Natural frequency:
ฯn =
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 )
ฯn
sin ฯ 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 ฯ nt + B sin ฯ nt +
ฯn =
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 โ A = โ
m2 g
k
u& ( 0 ) = ฯ n B โ B =
m2
m1 + m2
(e)
2 gh
ฯn
(f)
Substituting Eqs. (e) and (f) in Eq. (b) gives
u(t ) =
m2 g
(1 โ cos ฯ nt ) +
k
2 gh
ฯn
m2
sin ฯ 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 ) = โ 2 gh = โ 2( 386 )( 36 ) = โ 166.7 in./sec
2. Determine the natural frequency.
ฯn =
kg
(50)(386)
=
= 4393
. rad/sec
w
10
3. Compute the maximum deformation.
u(t ) = u(0) cos ฯ n t +
u& (0)
ฯn
sin ฯ n t
โ 166.7 โ
= ( 0.2) cos 316.8t โ โ
โ sin 316.8t
โ 4393
. โ
โก u&( 0) โค
uo = [u( 0)]2 + โข
โฅ
โฃ ฯn โฆ
2
= 0.2 2 + ( โ3.795) 2 = 38
. in.
4. Compute the maximum acceleration.
u&&o = ฯ n 2 uo = ( 4393
. )2 (38
. )
= 7334 in./sec2 = 18.98g
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Problem 2.7
Given:
m =
200
= 6. 211 lb โ 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
2ฯ
m
โ 2 =
1
EI
2ฯ
55. 90
โ
EI = ( 4 ฯ )2 55. 90 = 8827 lb โ 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 ฮถฯ n u& + ฯ 2n u = 0
(b)
where ฮถ = 1.
Equation (b) thus reads
u&& + 2 ฯ n u& + ฯ 2n u = 0
(c)
Assume a solution of the form u ( t ) = e st . Substituting
this solution into Eq. (c) yields
( s 2 + 2 ฯ n s + ฯ 2n ) e st = 0
Because e st is never zero, the quantity within parentheses
must be zero:
s 2 + 2 ฯ n s + ฯ 2n = 0
or
s =
โ 2ฯ n ยฑ
( 2 ฯ n ) 2 โ 4 ฯ 2n
2
= โ ฯn
(double root)
The general solution has the following form:
u ( t ) = A1 e โ ฯ n t + A2 t e โ ฯ 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 โ A1 = u (0)
(e)
Differentiating Eq. (d) with respect to t gives
u& ( t ) = โ ฯ n A1 e โ ฯ n t + A2 (1 โ ฯ n t ) e โ ฯ n t
(f)
Evaluate Eq. (f) at t = 0 :
u& ( 0 ) = โ ฯ n A1 + A2 (1 โ 0 )
โด A2 = u& ( 0 ) + ฯ n A1 = u& ( 0 ) + ฯ n u ( 0 )
(g)
Substituting Eqs. (e) and (g) for A1 and A2 in Eq. (d) gives
u (t ) = { u (0) + [u& (0) + ฯ n u (0) ] t} e โฯ nt
(h)
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Problem 2.9
A2 ฯ n โกโฮถ + ฮถ 2 โ1 + ฮถ + ฮถ 2 โ1 โค =
โขโฃ
โฅโฆ
Equation of motion:
mu&& + cu& + ku = 0
Dividing Eq. (a) through by m gives
u&& + 2 ฮถฯ n u& +
ฯ n2 u
= 0
or
(b)
where ฮถ > 1.
A2 =
Assume a solution of the form u ( t ) = e st . Substituting
this solution into Eq. (b) yields
( s 2 + 2ฮถฯ n s + ฯ n2 ) e st = 0
Because e st is never zero, the quantity within parentheses
must be zero:
s
2
+ 2 ฮถฯ n s +
ฯ 2n
โ2ฮถฯ n ยฑ
(2ฮถฯ n ) 2 โ 4ฯ n2
= 0
u& (0) + โโ ฮถ +
โ
โ
2
2 ฮถ โ1 ฯ n
(f)
Substituting Eq. (f) in Eq. (d) gives
u& (0) + โโ ฮถ + ฮถ 2 โ 1 โโฯ nu (0)
โ
โ
A1 = u (0) โ
2
2 ฮถ โ1ฯ n
2 ฮถ 2 โ 1 ฯ nu (0) โ u& (0) โ โโ ฮถ + ฮถ 2 โ 1 โโ ฯ nu (0)
โ
โ
=
โu& (0) + โโ โฮถ + ฮถ 2 โ 1 โโ ฯ nu (0)
โ
โ
=
2 ฮถ 2 โ 1ฯ n
2
= โโ โ ฮถ ยฑ
โ
ฮถ 2 โ1 โโ ฯ n u (0)
2 ฮถ 2 โ1ฯ n
or
s =
u& (0) + โโ ฮถ + ฮถ 2 โ1 โโ ฯ n u (0)
โ
โ
(a)
(g)
ฮถ 2 โ1 โโ ฯ n
โ
The solution, Eq. (c), now reads:
The general solution has the following form:
โก
โค
u (t ) = A1 exp โขโโ โฮถ โ ฮถ 2 โ1 โโฯ n t โฅ
โ
โฃโ
โฆ
โก
โค
+ A2 exp โขโโ โฮถ + ฮถ 2 โ1 โโฯ n t โฅ
โ
โฃโ
โฆ
u (t ) = e โฮถฯ nt
ฯ โฒD =
(d)
Differentiating Eq. (c) with respect to t gives
โก
โค
+ A2 โโ โฮถ + ฮถ 2 โ1 โโ ฯ n expโขโโ โฮถ + ฮถ 2 โ1 โโ ฯ nt โฅ
โ
โ
โ
โฃโ
โฆ
โฯ โฒD t
โฒ
+ A2 e ฯ D t
)
ฮถ2 โ 1 ฯ n
โu& (0) + โโ โฮถ + ฮถ 2 โ 1 โโฯ n u (0)
โ
โ
A1 =
2ฯ โฒD
Evaluate Eq. (c) at t = 0 :
โก
โค
u& (t ) = A1 โโ โฮถ โ ฮถ 2 โ1 โโ ฯ n expโขโโ โฮถ โ ฮถ 2 โ1 โโ ฯ nt โฅ
โ
โ
โ
โ
โฃ
โฆ
1
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 โ A1 + A2 =u (0)
(A e
A2 =
u& (0) + โโ ฮถ +
โ
ฮถ 2 โ1 โโ ฯ n u (0)
โ
2ฯ โฒD
(e)
Evaluate Eq. (e) at t = 0 :
u& (0) = A 1 โโ โฮถ โ ฮถ 2 โ1 โโ ฯ n + A 2 โโ โฮถ + ฮถ 2 โ1 โโ ฯ n
โ
โ
โ
โ
= [u (0) โ A2 ] โโ โฮถ โ ฮถ 2 โ1 โโ ฯ n + A2 โโ โฮถ + ฮถ 2 โ1 โโ ฯ n
โ
โ
โ
โ
or
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Problem 2.10
The general solution is
u(t) = A1 e โ ฯn t + A2 t e โฯ nt
Equation of motion:
u&& + 2ฮถฯ n u& + ฯ n2 u = 0
(a)
u(t) = e
A2 = u& (0)
(j)
st
Substituting in Eq. (i) gives
Substituting this solution into Eq. (a) yields:
u(t ) = u& (0) t e โฯ n t
FH s2 + 2ฮถฯ n s + ฯ 2n IK e st = 0
The roots of the characteristic equation [Eq. (b)] are:
st
s 2 + 2ฮถฯ n s + ฯ 2n = 0
(b)
The roots of this characteristic equation depend on ฮถ.
FH
s1,2 = ฯ n โฮถ ยฑ ฮถ 2 โ 1
IK
(l)
The general solution is:
(a) Underdamped Systems, ฮถ1
Because e is never zero
1,2
Determined from the initial conditions u(0) = 0 and u& (0) :
A1 = 0
Assume a solution of the form
(i)
(c)
F โฮถ + ฮถ โ1IK ฯ t
u(t ) = A1e H
2
n
F โฮถ โ ฮถ โ1IK ฯ t
+ A 2 eH
2
n
(n)
Hence the general solution is
st
Determined from the initial conditions u(0) = 0 and u& (0) :
s t
u(t) = A1e 1 + A2e 2
which after substituting in Eq. (c) becomes
e
u(t ) = e โฮถฯ n t A1e iฯ D t + A2 e โiฯ D t
โ A1 = A2 =
j
(d)
(o)
ฮถ 2 โ1
2ฯ n
Substituting in Eq. (n) gives
where
1โ ฮถ 2
(e)
u(t) = e โฮถฯ nt (A cos ฯ D t + Bsin ฯ D t)
B=
ฯ n 1โฮถ
2
0.8
FH
IK
(g)
(b) Critically Damped Systems, ฮถ = 1
The roots of the characteristic equation [Eq. (b)] are:
s2 = โฯ n
โ ฯ n t ฮถ 2 โ1
IJ
K
(p)
ฮถ = 0.1
ฯD
e โฮถฯ n t sin ฯ n 1 โ ฮถ 2 t
โe
Plot Eq. (g) with ฮถ = 0.1; Eq. (k), which is for ฮถ = 1;
and Eq. (p) with ฮถ = 2.
u& (0)
Substituting A and B into Eq. (f) gives
u& (0)
ฯ n t ฮถ 2 โ1
2
(d) Response Plots
(f)
Determine A and B from initial conditions u(0) = 0 and
u& (0) :
s1 = โฯ n
FG e
2ฯ ฮถ โ 1 H
u& (0) e โฮถฯ n t
n
Rewrite Eq. (d) in terms of trigonometric functions:
A=0
u( t ) =
.
u(t) รท (u(0)) / ฯn )
ฯ D = ฯn
u( t ) =
u& (0)
ฮถ = 1.0
ฮถ = 2.0
0.4
0
0.25
0.5
0.75
1
1.25
1.5
t/Tn
-0.4
-0.8
(h)
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Problem 2.11
F
GH
u1
1
ln
j
uj +1
I โ 2ฯฮถ โ 1 ln F 1 I โ 2ฯฮถ
GH 01. JK
JK
j
10%
โด j10% โ ln (10 ) 2 ฯฮถ โ 0. 366 ฮถ
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Problem 2.12
โ
โ
ui
โ 2ฯฮถ โ
= exp โ
u i +1
โ 1โฮถ 2 โโ
โ
โ
(a) ฮถ = 0. 01:
(b) ฮถ = 0. 05 :
(c) ฮถ = 0. 25 :
ui
ui + 1
ui
ui + 1
ui
ui + 1
= 1. 065
= 1. 37
= 5. 06
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Problem 2.13
Given:
w = 20.03 kips (empty); m = 0.0519 kip-sec2/in.
k = 2 (8.2) = 16.4 kips/in.
c = 0.0359 kip-sec/in.
(a) Tn = 2 ฯ
(b) ฮถ =
m
0. 0519
= 2ฯ
= 0. 353 sec
16. 4
k
c
2 km
=
0. 0359
2 (16. 4 ) ( 0. 0519 )
= 0. 0194
= 1. 94%
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Problem 2.14
(a) The stiffness coefficient is
k=
3000
= 1500 lb/in.
2
The damping coefficient is
c = ccr = 2 km
c = 2 1500
3000
= 215.9 lb – sec / in.
386
(b) With passengers the weight is w = 3640 lb. The
damping ratio is
ฮถ=
c
2 km
=
215.9
3640
2 1500
386
= 0.908
(c) The natural vibration frequency for case (b) is
ฯ D = ฯ n 1โ ฮถ 2
1500
1 โ (0.908) 2
3640 / 386
= 12.61 ร 0.419
= 5.28 rads / sec
=
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Problem 2.15
1. Determine ฮถ and ฯ n .
ฮถ โ
โ u โ
1
โ 1 โ
lnโ 1 โ =
lnโ
โ = 0.0128 = 1.28%
โ
โ
2ฯ j โ u j +1 โ
2ฯ (20) โ 0.2 โ
1
Therefore the assumption of small damping implicit in the
above equation is valid.
TD =
3
= 0.15 sec ; Tn โ TD = 0.15 sec ;
20
ฯn =
2ฯ
= 41. 89 rads sec
0.15
2. Determine stiffness coefficient.
k = ฯ 2n m = ( 41. 89 )2 0.1 = 175. 5 lbs in.
3. Determine damping coefficient.
ccr = 2 mฯ n = 2 ( 0.1) ( 41. 89 ) = 8. 377 lb โ sec in.
c = ฮถ ccr = 0. 0128 ( 8. 377 ) = 0.107 lb โ sec in.
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Problem 2.16
250
= 312. 5 lbs in.
0. 8
(a) k =
w
m =
g
=
250
386
= 0. 647 lb โ sec2 in.
k
= 21. 98 rads sec
m
ฯn =
(b) Assuming small damping,
F u I โ 2 jฯฮถ โ
GH u JK
F u IJ = ln (8) โ 2 (2) ฯ ฮถ โ ฮถ = 0.165
ln G
H u 8K
1
ln
j +1
0
0
This value of ฮถ may be too large for small damping
assumption; therefore we use the exact equation:
ln
F u I = 2 jฯ ฮถ
GH u JK 1 โ ฮถ
1
j +1
2
or,
ln ( 8) =
2 (2) ฯ ฮถ
1 โ ฮถ
2
โ
ฮถ
1 โ ฮถ2
= 0.165 โ
ฮถ2 = 0. 027 (1 โ ฮถ2 ) โ
ฮถ =
0. 0267 = 0.163
(c) ฯ D = ฯ n 1 โ ฮถ2 = 21. 69 rads sec
Damping decreases the natural frequency.
ยฉ 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication
is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical, photocopying,
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Problem 2.17
Reading values directly from Fig. 1.1.4b:
Peak
1
31
Time,
t i (sec)
0.80
7.84
Peak, u&&i (g)
0.78
0.50
7.84 โ 0.80
= 0.235 sec
30
โ 0.78g โ
1
ฮถ=
lnโ
โ = 0.00236 = 0.236%
2ฯ (30) โ 0.50g โ
TD =
ยฉ 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication
is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical, photocopying,
recording, or likewise. For information regarding permission(s), write to:
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Problem 2.18
1. Determine buckling load.
wcr
ฮธ
L
k
wcr ( L ฮธ ) = k ฮธ
wcr =
k
L
2. Draw free-body diagram and set up equilibrium
equation.
fI
w
ฮธ
L
fS
O
where
โ MO = 0 โ fI L + fS = w Lฮธ
fI =
w 2 &&
L ฮธ
g
fS = k ฮธ
Substituting Eq. (b) in Eq. (a) gives
w 2 &&
L ฮธ + (k โ w L) ฮธ = 0
g
(a)
(b)
(c)
3. Compute natural frequency.
ฯ nโฒ =
k โ wL
=
( w g) L2
k
wL โ
โ
โ1 โ
โ
2
k โ
( w g) L โ
or
ฯ โฒn = ฯ n 1 โ
w
wcr
(d)
ยฉ 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication
is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical, photocopying,
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Problem 2.19
For motion of the building from left to right, the
governing equation is
mu&& + ku = โ F
(a)
for which the solution is
u ( t ) = A2 cos ฯ nt + B2 sin ฯ nt โ uF
(b)
With initial velocity of u& ( 0 ) and initial displacement
u( 0 ) = 0 , the solution of Eq. (b) is
u(t ) =
u& ( 0 )
ฯn
sin ฯ nt + uF (cos ฯ nt โ 1)
u& ( t ) = u& ( 0 ) cos ฯ nt โ uFฯ n sin ฯ nt
(c)
(d)
At the extreme right, u& ( t ) = 0 ; hence from Eq. (d)
tan ฯ nt =
u& ( 0 ) 1
ฯ n uF
Substituting ฯ n = 4 ฯ , uF = 0.15 in.
20 in. sec in Eq. (e) gives
tan ฯ nt =
(e)
and
u& ( 0 ) =
20 1
= 10. 61
4 ฯ 0.15
or
sin ฯ nt = 0. 9956; cos ฯ nt = 0. 0938
Substituting in Eq. (c) gives the displacement to the right:
u =
20
( 0. 9956 ) + 0.15 ( 0. 0938 โ 1) = 1. 449 in.
4ฯ
After half a cycle of motion the amplitude decreases by
2 uF = 2 ร 0.15 = 0. 3 in.
Maximum displacement on the return swing is
u = 1. 449 โ 0. 3 = 1.149 in.
ยฉ 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication
is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical, photocopying,
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Problem 2.20
Given:
F = 0.1w , Tn = 0. 25 sec
0.1w
0.1mg
0.1g
0.1g
F
=
=
=
=
( 2 ฯ Tn )2
k
k
k
ฯ 2n
0.1g
=
= 0. 061 in .
( 8 ฯ )2
uF =
The reduction in displacement amplitude per cycle is
4uF = 0. 244 in.
The displacement amplitude after 6 cycles is
2.0 โ 6 (0.244) = 2.0 โ 1.464 = 0.536 in.
Motion stops at the end of the half cycle for which the
displacement amplitude is less than uF . Displacement
amplitude at the end of the 7th cycle is 0.536 โ 0.244 =
0.292 in.; at the end of the 8th cycle it is 0.292 โ 0.244 =
0.048 in.; which is less than uF . Therefore, the motion
stops after 8 cycles.
ยฉ 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication
is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical, photocopying,
recording, or likewise. For information regarding permission(s), write to:
20
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