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Engineering Dynamics Quiz 3 Solutions

This document contains solutions to 4 problems in engineering dynamics: 1) Modal analysis of a 2 degree-of-freedom system to find natural frequencies and mode shapes. 2) Vibration isolation problem analyzing spring-mass system frequency response. 3) Forced vibration of a rotating shaft subjected to periodic torque. 4) Lagrangian analysis of the dynamics of two rigid disks connected by a spring with both translational and rotational degrees of freedom.

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Janatan Choi
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228 views5 pages

Engineering Dynamics Quiz 3 Solutions

This document contains solutions to 4 problems in engineering dynamics: 1) Modal analysis of a 2 degree-of-freedom system to find natural frequencies and mode shapes. 2) Vibration isolation problem analyzing spring-mass system frequency response. 3) Forced vibration of a rotating shaft subjected to periodic torque. 4) Lagrangian analysis of the dynamics of two rigid disks connected by a spring with both translational and rotational degrees of freedom.

Uploaded by

Janatan Choi
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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2.

003SC Engineering Dynamics


Quiz 3 Solutions

Problem 1 Solution:

Modal Analysis Solution:


a) The natural frequencies of the system may be computed directly from the elements of the modal mass
and stifness matrices.
K1 190.4875
ω12 = ⇒ω1 = = 1.972 r/s
M1 48.981
210.5125
ω2 = = 4.269 r/s
11.5579

b) Again, directly from modal values:


C1 9.5244
ζ1 = = = 0.0493
2ω1 M1 2(1.972)(48.981)

q10 q̇10 0.525 −0.0499 1 0.525


c) = 0, = u−1 x
˙ (0) =
�
=
q20 q̇20 0.475 0.0499 0 0.475

d)
(1)
1.0 q̇10 −ζ1 ω1d t
x
�
= uq = e sin(ω1d t)
−9.5125 ω1d
where q̇10 = 0.525, ω1 = 1.972 r/s ≈ ω1d
q̇10
= 0.266

ω1d

1
Problem 2 Solution:

Vibration isolation:

fM = 3 Hz
ζ = 0.13

a) The total equivalent spring constant for springs in parallel is the sum of the individual ks:
Keq = 4k

b) For a linear system, steady state response:


frequency in=frequency out
fin = fout = 10Hz
c)

x (1 + (2ζ ωωn )2 )1/2


= 1/2
y ω2 2
(1 − 2 )
ωn + (2ζ ωωn )2
(1 + .8672 )1/2 f ω 10
= = =
[(1 − ( 10 2 2 2 1/2
3 ) ) + (0.867) ]
fn ωn 3
1.32
= = 0.130
10.148

x
= 0.130
y

2
Problem 3 Solution:

r = 45 RPM
Izz,G = 100kg-m2

τ (t) = τ0 cos ωt, ω = 6π r/s

kt = 1600π 2 N-m/r

ct = 8π (N-m-s)/r

a).

τ/o = τo cos ωtk̂ − cT θ̇k̂ − kt θk̂ = Izz,G θ̈k̂


⇒ Izz,G θ¨ + cT θ̇ + kt θ = τo cos(ωt)

b).

kt 1600π 2 N-m
ωn = = = 4π r/s
Izz,G 100 kg-m2
ωd = ωn 1 − ζ 2

= .99995ωn

� ωn

ct 8π
ζ= = = 0.01
2Izz ωn 2 × 4π × 100

c). ωosc = ω = 6π r/s


d). τ0
kT = 0.2 radians

1/KT
|θ| = |τ0 | · |Hθ/τ | = τ0 1/2
ω2 2
(1 − 2 )
ωn + (2ζ ωωn )2
−1/2
τ0 ω2 ω 6π
θ= (1 − 2 )2 + (2ζ )2 , which when evaluated at ω/ωn = = 1.5 yields:
KT ωn ωn 4π
0.2 rad
= 1/2
[(1 − 1.52 )2 + (2(.01)(1.5))2 ]
.2
=
[1.56 + 0.0009]1/2
= 0.16 = θ

3
Problem 4 Solution:

a) The system has two degrees of freedom for the no-slip condition.
If has four degrees of freedom if slip is allowed.
b) Two generalized conditions are required.
I choose coordinates x1 and x2 , where
x1 = rθ1 , x2 = rθ2 .

c)
1 1 1 1
T = mẋ12 + mẋ22 + Ic1 θ̇12 + Ic2 θ̇22
2 2 2 2

for uniform disk, Ic = mr2 /2.

1 1 1 m r2 ẋ21 1 m r2 ẋ22
T = mẋ21 + mẋ22 + +
2 2 2 2 r2 2 2 r2

3 3

= mẋ21 + mẋ22
4 4

1 2 1 2 1

V = kx1 + kx2 + km (x2 − x1 )2


2 2 2

d) Find EOMs:
From Lagrange d
dt
∂T
∂q̇i − ∂T
∂qi + ∂V
∂qi = Qi
� �
d ∂T 3 ∂T
= mẍ1 , =0
dt ∂ẋ1 2 ∂x1

∂V

= kx1 − km (x2 − x1 )
∂x1
3
⇒ mẍ1 + (k + km )x1 − km x2 = 0
2
� �
d ∂T 3 ∂T
= mẍ2 , =0
dt ∂ẋ2 2 ∂x2

∂V

= kx2 + km (x2 − x1 )
∂x2
3
⇒ mẍ2 − km x1 + (k + km )x2 = 0
2

4
MIT OpenCourseWare
http://ocw.mit.edu

2.003SC / 1.053J Engineering Dynamics


Fall 2011

For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.

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