UNIVERSITI MALAYSIA PERLIS
PEPERIKSAAN PERTENGAHAN SEMESTER 20 MAC 2012
TPT474 – ANALISA KEJURUTERAAN IV [ENGINEERING ANALYSIS IV]
Masa : 1 Jam 30 Minit
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Sila pastikan kertas peperiksaan ini mengandungi LIMA (5) muka surat yang bercetak termasuk muka hadapan sebelum anda memulakan peperiksaan ini.
[Please make sure that this examination paper has FIVE (5) printed pages including this front page before you start the examination.]
Kertas peperiksaan ini mengandungi TIGA (3) soalan. Jawab semua soalan. Markah bagi setiap soalan adalah 20 markah.
[This examination paper has THREE (3) questions. Answer all questions. Each question contributes 20 marks.]
ζ = 1 ζ = 0.5 ζ = 0.2 ζ = 0.1
-2-Question 1
(a) A simple pendulum of length L, and mass m2 is pivoted to the mass m1 which slides without any friction on a horizontal plane as shown in Figure 1. Draw the free-body diagram of the mass and pendulum system.
[2 marks] (b) If the system is vibrating freely without any damping effect, show that the equations of
motion of the two-degrees-of-freedom system are
(m1+m2)x+m2Lθ+2kx=0 and + + =0 L x L g θ θ Figure 1 [4 marks] (c) Determine the natural frequencies of the system, given that m1 = 2 kg, m2 = 3 kg, k = 500
N/m, and L = 0.5 m.
[10 marks] (d) Suggest two (2) design changes to the system in order to reduce the natural frequency of the
pendulum. [4 marks] x θ m 2 1 m1 1 k k O 1 ζ = 1 ζ = 0.5 ζ = 0.2 ζ = 0.1
-3-Question 2
An electronic control system for an automobile engine is to be mounted on top of the fender inside the engine compartment of the automobile as shown in Figure 2. The control module electronically computes and controls the engine timing, fuel-air mixture, and so on, and completely controls the engine. To protect it from fatigue and breakage, it is desirable to isolate the module from the vibration induced in the car body by road and engine vibration. Hence the module is mounted on an isolator.
(a) Based on the transmissibility curve given as in Figure 3, select suitable values c and k if the mass of the module is 3 kg and the dominant vibration of the fender is approximated by y(t) = (0.01) sin (35)t m. Here, it is desired to keep the displacement of the module less than 0.005 m at all times, given that x(t) = 0.005 sin (ωn)t m.
[14 marks] (b) Once the design values for isolators are chosen, calculate the magnitude of the force
transmitted to the module through the isolator and comment on the amount of the calculated force with respect to the design of the electronic module.
[6 marks]
Figure 2 (a) Cutaway sketch of the engine compartment of an automobile illustrating the location of the car’s electronic control module. (b) Close-up of the control module mounted on the inside fender on an isolator. (c) Vibration model of the module isolator system.
Figure 3 Plot of the transmissibility ratio indicating the value of T.R. for a variety of choices of the damping ratio, ζ and the frequency ratio, r.
-4-Engine Isolation mounting Electronic control module Car body (a) m (c) c k x(t) y(t) (b) Module Isolator Fender x(t) y(t) 1.2 1.0 0.8 0.6 0.4 0.2 1 2 3 4 ζ = 1 ζ = 0.5 ζ = 0.2 ζ = 0.1 0 ζ r T.R.
Question 3
(a) For the beam shown in Figure 4, determine the nodal slopes and the reactions given that the modulus of elasticity, E and the area moment of inertia about the z axis, Iz are 210 GPa and 2 × 10-4 m4 respectively. Given that the complete stiffness matrix for the flexure element is
[ ]
− − − − − − = 2 2 2 2 3 4 6 2 6 6 12 6 12 2 6 4 6 6 12 6 12 L L L L L L L L L L L L L EI k z e [17 marks] (b) If support 3 is removed, suggest a method to prove that the beam will not topple.[3 marks] Figure 4 ooOoo -5-1 2 3 5000 N/m 5 m 4 m
Formulae Q1:
For a characteristic equation of
The roots are
Q2:
If x(t)= Xsin(ωnt) and y(t)=Ysin(ωt)
the displacement and force transmissibility ratio T.R. are 2 / 1 2 2 2 2 ) 2 ( ) 1 ( ) 2 ( 1 . . + − + = = r r r Y X R T ζ ζ and where m k n = ω , n m c ω ζ 2 = , n r ω ω =
The magnitude of the transmitted force, FT, is given by 2 / 1 2 2 2 2 2 ) 2 ( ) 1 ( ) 2 ( 1 + − + = r r r kYr FT ζ ζ Q3:
For a uniform load q(x) = -q =constant, the work-equivalent nodal forces and moment is given by
− − − = 12 2 12 2 2 2 2 2 1 1 qL qL qL qL M F M F q q q q Solution 0 2+ + = m k s m c s 2 / 1 2 2 , 1 4 2 1 2 1 − = m k m c m c s
Q1(a):
The free body diagram,
Q1(b):
For equilibrium condition, the translational motion in the x direction is given by
kx kx L m x m x m1 + 2 + 2θ cosθ =− − 0 2 cos ) (m1+m2 x+m2θL θ + kx=
For small angle, sinθ = θ and cos θ = 1. Therefore, 0
2 )
(m1+m2 x+m2θL+ kx=
The rotational motion, taking moment about hinge O, θ θ θ 2 2 sin 2 cos 2 L m gL m xL m =− − 0 cos sin 2 2 2 2θL +m gL θ +m xL θ = m . Thus, 0 = + + L x L g θ θ Q1(c):
Assuming the harmonic motions for the system, we let
t A
x= sinω and θ =Bsinωt. Thus, x=−ω2Asinωt and θ=−ω2Bsinωt. Substitute into the equations of motion,
0 sin 2 ) sin )( ( ) sin )( ( 2 2 2 2 1+m − A t + m L − B t + kA t = m ω ω ω ω ω 0 sin ] 2 ) ( [ 2 2 2 2 1+ + − = − m m ω A kA ω m LB ωt 0 ] ) ( 2 [ 2 2 2 2 1+ − = − m m A m LB k ω ω and, 0 sin sin sin 2 2 = − + + − A t L t B L g t B ω ω ω ω ω 0 sin 2 2 = − + − B t L g A L ω ω ω x kx O m1 kx m2 m2g θ L Lsinθ θ L m2 θ θ cos 2 L m
[
301401 39200]
274.5 256.03 2 1 5 . 274 − 1/2 = = 0 2 2 = − + − B L g A L ω ωWe can transform the two equations into matrix form, such as = − − − + − 0 0 ) ( 2 2 2 2 2 2 2 1 B A L g L L m m m k ω ω ω ω
For non-trivial solution of A and B, the determinant for the coefficient of A and B must be zero. 0 ) ( 2 det 2 22 2 2 2 1 = − − − + − ω ω ω ω L g L L m m m k
Using the values given, the determinant is 0 ) 2 )( 5 . 1 ( ) 6 . 19 )( 5 1000 ( − ω2 −ω2 − ω2 ω2 = 0 3 5 98 1000 19600− ω2 − ω2 + ω4 − ω4 = 0 9800 549 2 4− ω + = ω
( )
(
)
[
2]
1/2 2 2 2 1 549 49800 2 1 2 549 ,ω = − ω 47 . 18 2 1 = ω and 2 530.53 2 = ω 30 . 4 1 =ω rad/s and ω2 =23.03rad/s Q1(d):
The response amplitude X is 0.005 m. The amplitude Y is 0.01 m. Thus, the displacement transmissibility ratio is 5 . 0 01 . 0 005 . 0 . . = = = Y X R T
From the plot of the transmissibility ratio, there are several possible solutions for ζ and ωn. A straight horizontal line through T.R. crosses at several values ζ and r.
If ζ = 0.2, r = 1.65 for T.R. = 0.5. These values provide one possible solution. We know that r = ω/ωn = 1.65 and ω = 35 rad/s.
Therefore, the isolation system’s natural frequency is ωn = 35/1.65 = 21.212 rad/s.
The mass of the module is m = 3 kg. Thus, k = mωn2 = (3)(21.212)2 = 1350 N/m. The isolation mount must be made of a material with this stiffness.
The damping ratio, ζ is related to the damping constant c by this equation, c = 2ζmωn = 2(0.02)(3) (21.212) = 25.45 kg/s. Q2(b): 2 / 1 2 2 2 2 2 ) 2 ( ) 1 ( ) 2 ( 1 + − + = r r r kYr FT ζ ζ = kYr2(T.R.) = (1350)(0.01)(1.65)2(0.5) = 18.38 N. If this force happens to be too large, the design must be redone.
Other factors that must be considered for a good design are cost, ease of assembly, temperature range, reliability of vendor, availability of product, and quality required.
Element stiffness matrix for element 1 is
[ ]
− − − − − − × = − − − − − − × × × = − 8 . 0 24 . 0 4 . 0 24 . 0 24 . 0 096 . 0 24 . 0 096 . 0 4 . 0 24 . 0 8 . 0 24 . 0 24 . 0 096 . 0 24 . 0 096 . 0 10 420 100 30 50 30 30 12 30 12 50 30 100 30 30 12 30 12 125 10 2 10 210 9 4 5 ) 1 ( kElement stiffness matrix for element 2 is
[ ]
− − − − − − × = − − − − − − × × × = − 1 375 . 0 5 . 0 375 . 0 375 . 0 1875 . 0 375 . 0 1875 . 0 5 . 0 375 . 0 1 375 . 0 375 . 0 1875 . 0 375 . 0 1875 . 0 10 420 64 24 32 24 24 12 24 12 32 24 64 24 24 12 24 12 64 10 2 10 210 9 4 5 ) 2 ( kThe assembled global stiffness matrix is
[ ]
− − − − − − − − − − × = 1 375 . 0 5 . 0 375 . 0 0 0 375 . 0 1875 . 0 375 . 0 1875 . 0 0 0 5 . 0 375 . 0 8 . 1 135 . 0 4 . 0 24 . 0 375 . 0 1875 . 0 135 . 0 2835 . 0 24 . 0 096 . 0 0 0 4 . 0 24 . 0 8 . 0 24 . 0 0 0 24 . 0 096 . 0 24 . 0 096 . 0 10 420 5 KThe global equilibrium equations become
[ ]
+ − + − − + − = 6666 . 6666 10000 3750 22500 666 . 10416 12500 3 2 1 3 3 2 2 1 1 F F F v v v K θ θ θApplying the boundary conditions, we get
− = × 667 . 6666 3750 666 . 10416 1 5 . 0 0 5 . 0 8 . 1 4 . 0 0 4 . 0 8 . 0 10 420 3 2 1 5 θ θ θ
Using the Gaussian Elimination approach, the equations become
− × 241 . 4178 333 . 8958 666 . 10416 861 . 0 0 0 5 . 0 6 . 1 0 0 4 . 0 8 . 0 10 420 5
Solving the equations, we get 4 3 =1.155×10− θ rad 4 2 =0.972×10− θ rad
4 1=−3.586×10−
θ rad
The reaction forces are 9865 1 = F N 28485 2 = F N 6650 3 = F N Q3(b):
