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Solution Manual for Vibrations 3rd Edition
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Solution Manual for Vibrations 3rd Edition Solutions to Exercises Chapter 2 Section 2.1 2.1 Examine Eqs. (2.1) and (2.5) and verify that the units (dimensions) of the different terms in the respective equations are consistent. Solution 2.1 In Eqs. (2.1), the units of JG are kgm2 and th...

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  • July 20, 2022
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  • 2021/2022
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  • solution manual for vibrations 3rd editionsolution manual for vibrations 3rd editionsolution manual for vibrations 3rd editionsolution manual for vibrations 3rd editionsolution manual for vibrations 3

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Solution Manual for Vibrations 3rd Edition by
Balachandran

,Solutions Manual

Vibrations
3rd Edition


Balakumar Balachandran and Edward B. Magrab

, Solutions to Exercises
Chapter 2
Section 2.1
2.1 Examine Eqs. (2.1) and (2.5) and verify that the units (dimensions) of the different terms in
the respective equations are consistent.
Solution 2.1
In Eqs. (2.1), the units of JG are kgm2 and the units of md2 are kgm2. In Eq. (2.5), the units
of energy are Nm, and the units of mv2 are kg(m/s)2 = kg(m/s2)m = Nm.

Section 2.2

2.2 Consider the slider mechanism of Example 2.2 and show that the rotary inertia JO about the
pivot point O is also a function of the angular displacement .
Solution 2.2
The rotary inertia is
JO  J m ( )  J m  J m  J m
l s b e


where
1 1
J  m e2 J  m b2 J  m b2
me
3 e ms s mb
3 b

 l2   l 
2
l2
J ml ( )  m l  mld 2  m l      a2  al cos( )
12 12 2  
To determine how  depends on , we use geometry. First
r2 ( )  a2  b2  2ab cos
a2  r2 ( )  b2  2r( )b cos(     )
Combining the above equations, we obtain
r2( )  r2( )  b2  2r( )bcos(     )  b2  2abcos
which leads to
b  a cos  r( ) cos(     )
and, hence,
b  a cos
cos(     )  
r( )


1

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