Test 1 Solution

SULIT

[TPT474]

UNIVERSITI MALAYSIA PERLIS PEPERIKSAAN PERTENGAHAN SEMESTER 20 MAC 2012

TPT474 – ANALISA KEJURUTERAAN IV [ENGINEERING ANALYSIS IV] Masa : 1 Jam 30 Minit _____________________________________________________________________________ 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.]

SULIT

[TPT474]

-2Question 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 + m2 Lθ + 2kx = 0

x g θ + θ + = 0 L L

and

x k

O m1 1

k

1

θ Figure 1

m2

1

(c)

[4 marks]

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]

SULIT

[TPT474]

-3Question 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 (ω n)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] x(t) x(t)

Car body

Electronic control module

Module Engine

m

y(t)

Isolator Fender

Isolation mounting

c

k y(t)

(a) (b) (c) 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. 1.2

ζ

1.0

T.R.

0.8 0.6 0.4 0.2

ζ= 1 ζ = 0.5 ζ = 0.2 ζ = 0.1

r

Figure 3 Plot of0the transmissibility ratio2indicating the 1 3 value of T.R. 4 for a variety of choices of the damping ratio, ζ and the frequency ratio, r. -4-

SULIT

[TPT474]

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 6 L − 12 6 L   12  6 L 4 L2 − 6 L 2 L2   [ k e ] = EI3z  L − 12 − 6 L 12 − 6 L    2 − 6 L 4 L2   6L 2L

(b)

[17 marks]

If support 3 is removed, suggest a method to prove that the beam will not topple. [3 marks] 5000 N/m

1

3

2 5m

4m

Figure 4

ooOoo

-5-

SULIT

[TPT474]

Formulae Q1: For a characteristic equation of s2 +

c k s+ =0 m m

The roots are s1, 2

2 1 c 1  c   k  =    − 4  2 m 2  m   m 

1/ 2

Q2: If x(t ) = X sin(ω nt )

and y (t ) = Y sin(ωt )

the displacement and force transmissibility ratio T.R. are  X  1 + ( 2ζr ) 2 T .R. = =  2 2 2 Y  (1 − r ) + (2ζr ) 

1/ 2

and

where

ωn =

k , m

ζ =

c ω , r= 2mωn ωn

The magnitude of the transmitted force, FT, is given by  1 + (2ζr ) 2  FT = kYr  2 2 2  (1 − r ) + (2ζr ) 

1/ 2

2

Q3: For a uniform load q(x) = -q =constant, the work-equivalent nodal forces and moment is given by  qL  − 2   F1q   qL2   M  −   1q   12    =  qL   F2 q   −  M 2 q   2   qL2   12 

Solution

SULIT

[TPT474]

Q1(a): The free body diagram, x

kx

O

m1

θ

kx

m2 Lθ

L m2

m2θL cos θ

Lsinθ m2g Q1(b): For equilibrium condition, the translational motion in the x direction is given by m1 x + m2 x + m2θL cos θ = − kx − kx (m1 + m2 ) x + m2θL cos θ + 2kx = 0 For small angle, sinθ = θ and cos θ = 1. Therefore, (m1 + m2 ) x + m2θL + 2kx = 0 The rotational motion, taking moment about hinge O, m2θL2 = −m2 gL sin θ − m2 xL cos θ m2θL2 + m 2 gL sin θ + m2 xL cos θ = 0 . Thus, x g θ + θ + = 0 L L Q1(c): Assuming the harmonic motions for the system, we let x = A sin ωt and θ = B sin ωt . Thus, x = −ω 2 A sin ωt and θ = −ω 2 B sin ωt . Substitute into the equations of motion, (m1 + m2 )(−ω 2 A sin ωt ) + (m2 L)(−ω 2 B sin ωt ) + 2kA sin ωt = 0 [−( m1 + m2 )ω 2 A + 2kA − ω 2 m2 LB ] sin ωt = 0 [2k − (m1 + m2 )ω 2 ] A − ω 2 m2 LB = 0 and, − ω 2 B sin ωt +

 − ω2  g  A sin ωt = 0 B sin ωt +  L L  

 − ω 2  g    A +  − ω 2  B  sin ωt = 0  L    L 

SULIT

[TPT474]

− ω2 g  A +  − ω 2 B = 0 L L  We can transform the two equations into matrix form, such as 2k − (m1 + m2 )ω 2 − ω 2 m2 L  A     = 0 − ω2 g 2    − ω   B  0 L L   For non-trivial solution of A and B, the determinant for the coefficient of A and B must be zero. 2k − (m1 + m2 )ω 2 − ω 2 m2 L det − ω2 g 2 =0  −ω  L L  Using the values given, the determinant is (1000 − 5ω 2 )(19.6 − ω 2 ) − (1.5ω 2 )( 2ω 2 ) = 0 19600 − 1000ω 2 − 98ω 2 + 5ω 4 − 3ω 4 = 0 ω 4 − 549ω 2 + 9800 = 0

[

]

1/ 2 549 1 2  ( 549 ) − 4( 9800 ) 2 2 1 1/ 2 = 274 .5  [ 301401 − 39200 ] = 274.5 256 .03 2

ω12 , ω22 =

ω12 = 18.47 and ω22 = 530.53 ω1 = 4.30 rad/s and ω2 = 23.03 rad/s Q1(d):

Q2(a):

SULIT

[TPT474]

The response amplitude X is 0.005 m. The amplitude Y is 0.01 m. Thus, the displacement transmissibility ratio is X 0.005 T .R. = = = 0.5 Y 0.01 From the plot of the transmissibility ratio, there are several possible solutions for ζ and ω n. 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 /ω n = 1.65 and ω = 35 rad/s. Therefore, the isolation system’s natural frequency is ω n = 35/1.65 = 21.212 rad/s. The mass of the module is m = 3 kg. Thus, k = mω n2 = (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 = 2ζmω n = 2(0.02)(3) (21.212) = 25.45 kg/s. Q2(b):  1 + (2ζr ) 2  FT = kYr  2 2 2  (1 − r ) + (2ζr )  2

1/ 2

= 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.

Q3(a):

SULIT

[TPT474]

Element stiffness matrix for element 1 is

[k ] = 210 × 10125× 2 × 10 (1)

9

−4

30 − 12 30  0.24 − 0.096 0.24   12  0.096  30 100 − 30 50   0.24 0.8 − 0.24 0.4    = 420 × 10 5  − 12 − 30 12 − 30 − 0.096 − 0.24 0.096 − 0.24      50 − 30 100  0.4 − 0.24 0.8   30  0.24

Element stiffness matrix for element 2 is

[k ] = 210 ×1064× 2 ×10 ( 2)

9

−4

24 − 12 24  0.375 − 0.1875 0.375   12  0.1875  24   64 − 24 32  1 − 0.375 0.5  5  0.375  = 420 × 10 − 12 − 24 12 − 24 − 0.1875 − 0.375 0.1875 − 0.375      32 − 24 64  0.5 − 0.375 1   24  0.375

The assembled global stiffness matrix is 0.24 − 0.096 0.24 0 0   0.096  0.24 0.8 − 0.24 0.4 0 0   − 0.096 − 0.24 0.2835 0.135 − 0.1875 0.375  [ K ] = 420 × 105   0.4 0.135 1.8 − 0.375 0.5   0.24  0 0 − 0.1875 − 0.375 0.1875 − 0.375   0 0.375 0.5 − 0.375 1   0 The global equilibrium equations become  v1   − 12500 + F1  θ   − 10416 .666   1   v2  − 22500 + F2  [ K ]  =   3750 θ 2     v3   − 10000 + F3      θ3   6666.6666  Applying the boundary conditions, we get 0.8 0.4 0  θ1  − 10416 .666      5 420 × 10 0.4 1.8 0.5 θ 2  =  3750      0 0.5 1  θ3   6666 .667  Using the Gaussian Elimination approach, the equations become 0.8 0.4 0 − 10416 .666   5 420 × 10  0 1.6 0.5 8958.333   0 0 0.861 4178.241 

Solving the equations, we get θ3 = 1.155 × 10 −4 rad

θ 2 = 0.972 × 10 −4 rad

SULIT

θ1 = −3.586 × 10 −4 rad The reaction forces are F1 = 9865 N F2 = 28485 N F3 = 6650 N Q3(b): To check the reaction force at node 1.

[TPT474]