Solution Manual – Engineering Vibration, 5th Edition (Daniel J. Inman) – Complete Problem Solutions (Chapters 1–8)

ENGINEERING VIBRATION
Solution Manual – Engineering Vibration, 5th Edition (Daniel J.

Inman) – Complete Problem Solutions (Chapters 1–8)

Bonie314 Stuvia

,Problems and Solutions Section 1.1 (1.1 through 1.19)
h h h h h h h




1.1 The spring of Figure 1.2 is successively loaded with mass and the corresponding (static)
h h h h h h h h h h h h h


displacement is recorded below. Plot the data and calculate the spring's stiffness. Note
h h h h h h h h h h h h h


that the data contain some error. Also calculate the standard deviation.
h h h h h h h h h h h




m(kg) 10 11 12 13 14 15 16
x(m) 1.14 1.25 1.37 1.48 1.59 1.71 1.82

Solution:

Free-body diagram: h
From the free-body diagram and static
h h h h h


equilibrium:
h


kx
kx mg h h (g 9.81m/ s2)
h h h h




k k mg/ x
h h h




k
m  i 86.164
n
mg

20
The sample standard deviation in
h h h h


computed stiffness is:
h h h
n

m 15 i1(k  )2
i h
h h


 h h h 0.164 h

n 1 h h




10
0 1 2
x
Plot of mass in kg versus displacement in m
h h h h h h h h


Computation of slope from mg/x h h h h


m(kg) x(m) k(N/m)
10 1.14 86.05
11 1.25 86.33
12 1.37 85.93
13 1.48 86.17
14 1.59 86.38
15 1.71 86.05
16 1.82 86.24

, 1.2 Derive the solution of m˙x˙ kx 0 and plot the result for at least two periods for the case
h h h h h h h h h h h h h h h h h h



with n = 2 rad/s, x0 = 1 mm, and v0 = 5 mm/s.
h h h h h h h h h h h




Solution:

Given:
m!x!kx 0 (1) h h


Assume: x(t) ae . Then: x! are and !x! ar e . Substitute into equation (1) to
rt h
2 rt h
rt
h h h h h h h h h h h


get:
h



mar2ert kaert 0 h h



mr2 k 0 h h



k
r  hi h


m
Thus there are two solutions:
h h h h

 k h  h kh h
 i ht   h i t
x1 c1e h
m
h h , and h
2 c2 e h
m
h 
h x


k
2 rad/s where  n  h h h h

m
The sum of x1 and x2 is also a solution so that the total solution is:
h h h h h h h h h h h h h h h




x x x c e2it c e2it
h h
h
h
h
1 2 1 2



Substitute initial conditions: x0 = 1 mm, v0 =
h h h h h h h h 5 mm/s

x0c1c2 x0 1 c2 1 c1, and v0x!02ic1  2ic2 v0  5 mm/s
h h h h h h h h h h h h h h h h h h




 2c1 2c2  5 i. Combining the two underlined expressions (2 eqs in 2 unkowns):
h h h h h h h h h h h h h h h



1 5 1 5
2c 2  2c  5 i  c   i, and  
h h h h h

h h h i h h h h h h h h



hc
1 1 1 2
2 4 2 4

Therefore the solution is: h h h




xe  1  5 i 2it  1  5 i e2it
h
h
h
h
h
h h h h


2 4 
h
h
2 4 h


   
h h h




Using the Euler formula to evaluate the exponential terms yields:
h h h h h h h h h


1 5  1 5 
i cos2t isin2t  i cos2t  isin2t
h h h h

x   h
h
h
h
h h h h h
h
h
h
h h h h h h

2 4 2 4
   
h h h h




5 3
 x(t) cos2t  sin2t  sin2t 0.7297
h h

h h h h h h h h h h


2 2

, Using Mathcad the plot is:
h h h h




5. h


x t
h h cos 2. t h sin 2. t
h
h


2


2




x t
h


0 5 10



2

t