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Solution to Tutorial 1, 2, 3, & 4
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Tutorial 1 Shock Absorber for a Motorcycle
An underdamped shock absorber is to be designed for a motorcycle
of mass 200kg (shown in Fig.(a)). When the shock absorber is
subjected to an initial vertical velocity due to a road bump, the
resulting displacement-time curve is to be as indicated in Fig.(b).
Find the necessary stiffness and dampeing constant of the shock
absorber if the damped period of vibration is to be 2 s and the
amplitude x1 is to be reduced to one-fourth in one half cycle
(i.e., x1.5 = x1/4).
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3
From which can be found as 0.4037. The damped period of
vibration given by 2 s. Hence,
Hence the logarithmic decrement becomes
Solution
16/4/,4/ 15.1215.1 xxxxx ===
( )2
2
1
127726.216lnln
===
=
xx
Since ,
rad/s 4338.3)4037.0(12
21222
2
2
=
=
===
n
ndd
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4
Thus the damping constant is given by:
and the stiffness by:
s/m-N 54.373.1)4338.3)(200(22 === nc mc
s/m-N 4981.554)54.1373)(4037.0( === ccc
N/m 2652.2358)4338.3)(200( 22 === nmk
The critical damping constant can be obtained:
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Tutorial 2 Q1. Equivalent k of a Crane
The boom AB of crane is a uniform steel bar of length 10 m and
x-section area of 2,500 mm2. A weight W is suspended while the
crane is stationary. Steel cable CDEBF has x-sectional area of 100
mm2. Neglect effect of cable CDEB, find equivalent spring constant
of system in the vertical direction.
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m3055.12426.151135cos)10)(3(2103
1
2221
=
=+=
ll
Solution A vertical displacement x of pt B will cause the spring
k2 (boom) to deform by x2 = x cos 45 and the spring k1 (cable) to
deform by an amount x1 = x cos (90 ). Length of cable FB, l1 is as
shown.
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The angle satisfies the relation:
The total potential energy (U):
==
=+
0736.35,8184.0cos10cos)3)((23 21
221
ll
21
22 )]90cos([2
1)45cos(21
+= xkxkU
N/m106822.10355.12
)10207)(10100( 696
1
111 =
==
lEAk
N/m101750.510
)10207)(102500( 796
2
222 =
==
lEAk
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8
Potential Energy of the equivalent spring is:
By setting U = Ueq, hence:
2
21 xkU eqeq =
N/m104304.26 6eq =k
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Tutorial 2
Q2. Find the equivalent mass and stiffness of the system if it
is to be modeled as an SODF system as shown.
??k??m eqeq ==
A
B
C
keq meq
x
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pp r/x=
p11pB r/lxlv ==
cBc r/v=
cp1 rr/lx=
(a) Equivalent Mass:
2xm21 2p
211p lm3
1J21
++ 2B2vm2
1+ 2Bc
2cc vm2
1J21
++ =KE
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pp r/x= p11pB r/lxlv ==
cBc r/v= cp1 rr/lx=
2xm21 2p
211p lm3
1J21
++ 2B2vm2
1+ 2Bc
2cc vm2
1J21
++ =KE
xkeq
meq
2xm21 2
p
2211p r
xlm31J
21
++ 22
p
21
2 xrl
m21 +
22p
21
c2
2c
2p
21
c xrl
m21x
rrl
J21 ++
=KE
2eq xm2
1 =
2p
21c
2c
2p
21c
2p
212
2p
211
2p
peq r
lmrrlJ
rlm
r3lm
rJ
mm +++++=
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(b) Equivalent Stiffness: Potential Energy Should Be the
SAME
Spring k1 compressed by
Spring keq compressed by
Spring k2 compressed by 1p l = p1 r/l=
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Let Potential Energies be the SAME Equivalent Stiffness
Spring k1 compressed by
Spring keq is assumed to be compressed by
Spring k2 compressed by 1p l = p1 r/l=
2eq
22
21 k2
1k21k
21 =+
2eq
22p
21
22
1 k21
rl
k21k
21 =+
22p
21
1eq krl
kk +=
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Tutorial 3 Engine-induced aircraft wing vibration can be modeled
as an SDOF system under harmonic excitation force.
m
k
y(t) F(t)
x
y
t =0F
where m=3,000kg, k=300,000 N/m, c=600Ns/m, F0=1000N. Determine
(a) the static deflection when F(t)=F0=1000N; (b) vibration
amplitudes when =1rad/s, =10rad/s and =50rad/s respectively; (c)
the ratio between the vibration amplitude @ =10Hz and the static
deflection.
c
Solution (a) static deflection when F(t)=F0=1000N:
mmk
FY 333000300
100000 .,
===
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(b) Equation of motion:
( ) ( ) = tYty sin
tFkyycym sin0=++ So, the solution becomes,
= 22
1 2
n
ntan( ) 2222220
4
1
nnmFY
+=
(WHY ???)
mk
n = srad /,, 10
0003000300
==
kmc
2= 010
30000030002600 .=
=
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( ) 2222220
4
1
nnmFY
+=
For the case of =1 rad/s (n):
( ) 222222 1050010450101
30001000
+=
.Y mm140.=
101001021
30001000
=
.
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(c) resonant vibration amplitude / static deflection:
0YYM = 50
333167
==.
= 22
1 2
n
ntan
The phase for =10 rad/s (=n):
The phase for =1 rad/s (n):
= 22
1 2
n
ntan
= 22
1
11010101.02tan o161.=
( )= 1tan o90=
= 22
1
501050100102 .tan o240.=
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Tutorial 4 The landing gear of an airplane can be idealized as a
spring damper system as shown. The runway surface is described
y(t)=0.2sin157t. Assume m=2000kg, k=3MN/m and c=7500Ns/m, determine
the vertical steady-state vibration of the airplane during
landing.
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sradmk
n /71.7020005000000
=== 22.271.70
157===
nr
0375.0500000020002
75002
=
==km
c
( )
( ) ( )YX
222
2
r2r1
r21
+
+=
=
kc 1tan
=
50000001577500tan 1 o25.13=
YX 258.0= m0516.0=
= 2
1
r1r2tan o6.177= ( ) += tXx sin
( )otx 4.164157sin051.0 =
Solution:
Solution to Tutorial 1, 2, 3, & 4Tutorial 1 Shock Absorber
for a Motorcycle SolutionSlide Number 4Tutorial 2 Q1. Equivalent k
of a CraneSolutionSlide Number 7Slide Number 8Slide Number 9Slide
Number 10Slide Number 11Slide Number 12Slide Number 13Slide Number
14Slide Number 15Slide Number 16Slide Number 17Slide Number 18Slide
Number 19
