ME3112-PART 2TUTORIAL 2 & 3

Shahrokh

Review (tutorial 2)• Natural Vibration

• Finding natural frequency and mode shapes• Equation of motion

• Newton’s second law• Energy conservation equation• Other methods…

• Forced Vibration• Force frequency• Natural frequency• No damping• Damped vibration

• Under damped• Critical damping• Over damped

Review (tutorial 2)

푋 =푃푘

1 − 휔휔 + 2휉휔

휔

⁄

푋 =푃푘

1 − 휔휔

Tutorial 2-Q1Assumptions:• Uniform mass distribution• Natural vibration

Given:• AB = AC = 0.5 and BC = 0.6

Required:Natural frequency ???

휔 =푚푔푦퐼

CGG1 G3

G2

y

x

Tutorial 2-Q1

h1=h3=0.2

From the geometry, AC = AB = 0.5 m, BC = 0.6 m, therefore AG2 = 0.4 m

Total length of wireframe : 0.5 + 0.5 + 0.6= 1.6 mMass per unit length : m/L=1.6/1.6 = 1 kg/mMass of AB = 0.5 kgMass of AC = 0.5 kgMass of BC = 0.6 kg

Center of Mass: dCG

Let d be the distance from CG (the center of mass from A),푚 푦 = ∑ 푚 푦 : 푚 푑 = 푚 ℎ + 푚 ℎ + 푚 (퐴퐺 )

1.6 (d) = 0.5 (0.2) + 0.5 (0.2) + 0.6 (0.4)d = 0.44/1.6 = 0.275 m

d

Tutorial 2-Q1

CGG1 G3

d

h1=h3=0.2

G2Natural Frequency: 휔

휔 =퐾푚 =

푚푔푑퐼

Moment of Inertia: 퐼퐼 = 퐼 + 퐼 + 퐼

퐼 =(1/3)(0.5)(0.5)2 + (1/3)(0.5)(0.5)2 +(1/12)(0.6) (0.6)2 + 0.6 (0.4)2 = 0.1973 kg m2

Tutorial 2-Q2

At max potential energy T1 = 0 V1 = (1/2) ((kB/4) + kC)L2 qm

2

At max kinetic energy, V2 = 0T2 = (1/2) (1/3)mL2 (휔 qm )2

T1 + V1 = T2 + V2휔 2= (3/m) ((kB/4) + kC)

휔 =퐾푚

퐾 = 3푘4 + 푘

T1 = 0 V1 ≠ 0

V2 = 0 T2 ≠ 0

휔 = 31.72푟푎푑/푠

Tutorial 2-Q3

푋 =푃푘

1 − 휔휔 + 2휉휔

휔

⁄푋 =푃푘

1 − 휔휔

푃 = 푚푟휔

푃 = 푃 sin 휔푡

휔 =푘푚

휔 = 1800퐶푟푚푖푛 = 1800 ×

2휋60

푟푎푑푠

A 25 kg motor is supported by four springs,each having a constant of 200 kN/m. Theunbalance of the rotor is equivalent to amass of 30g located 125 mm from the axisof rotation. Knowing that the motor isconstrained to move vertically, determinethe amplitude of the steady state vibration ofthe motor at a speed of 1800 rpm, assumingthat (a) no damping is present. (b) thedamping factor is 0.125.Answers: (a) 1.509 mm (b) 0.583 mm.

Tutorial 2-Q4

푋 =푃푘

1 − 휔휔 + 2휉휔

휔

⁄

휔 =푘푚

푐 = 2푚휔

휉 =푐푐

A machine element having a mass of 500kg is supported by two springs, eachhaving a constant of 48 kN/m. A periodicforce of maximum magnitude equal to 150N is applied to the element with afrequency of 2.4 Hz. Knowing that thecoefficient of damping is 1600 Ns/m,determine the amplitude of the steady statevibration of the element. Answer: 5.02 mm.

휔 = 2휋푓

Review (tutorial 3)• Degrees of freedom of a Mechanism

• The 4-bar mechanism analysis• 2-point formula (or I.C.R.)• Starting with the link with known angular velocity

DOF=3(n-1) -2L -h

Where DOF = total degrees of freedom in the mechanism n = number of links(including the frame) L = number of lower pairs(one degree of freedom) h = number of higher pairs(two degrees of freedom)

Tutorial 3-Q1

n= 4L =4h=0DOF = 1

DOF =3(n-1) -2L -h

n= 4L =4h=0DOF = 1

n= 4L =4h=0DOF = 1

n= 9L =11h=0DOF = 2

Tutorial 3 - Q2 and Q3푣 = 휔 × 퐴퐵

푣 = 푣 + 휔 × 퐵퐷푣 = 휔 × 퐸퐷

푎 = 휔 × 휔 × 퐴퐵 + 훼 × 퐴퐵푎 = 푎 + 휔 × 휔 × 퐵퐷 + 훼 × 퐵퐷푎 = 휔 × 휔 × 퐸퐷 + 훼 × 퐸퐷

휔 = 4푟푎푑푠퐶.푊.

휔 = 6.67푟푎푑푠퐶.퐶.푊.

Summary:

Final answers:

훼 = 28.9rad푠

C. W.

훼 = 124rad푠

C. C. W. 휔 = 29.33푟푎푑푠퐶.푊.

휔 = 11.29푟푎푑푠퐶.퐶.푊. 훼 = 860.08

rad푠

C. C. W.

훼 = 356.15rad푠

C. W.

Q2 Q3