ME3112_part2_Solution_2_3[1]

ME3112-PART 2 TUTORIAL 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…

Review (tutorial 2) • Forced Vibration • Force frequency • Natural frequency • No damping • Damped vibration • Under damped • Critical damping • Over damped

= 1−

=

⁄

1−

+

2

Tutorial 2-Q1 Assumptions: • Uniform mass distribution • Natural vibration Given: • AB = AC = 0.5 and BC = 0.6

Required: Natural frequency ???

=

Tutorial 2-Q1 y x

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 m Mass per unit length : m/L=1.6/1.6 = 1 kg/m Mass of AB = 0.5 kg Mass of AC = 0.5 kg Mass of BC = 0.6 kg

h1=h3=0.2 G1

Center of Mass: dCG Let d be the distance from CG (the center of mass from A), =∑ : = ℎ + ℎ + ( ) 1.6 (d) = 0.5 (0.2) + 0.5 (0.2) + 0.6 (0.4) d = 0.44/1.6 = 0.275 m

d

G3 CG

G2

Tutorial 2-Q1 Moment of Inertia: = + + =(1/3)(0.5)(0.5)2

h1=h3=0.2 (1/3)(0.5)(0.5)2

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

Natural Frequency: =

=

G1

d

G3 CG

G2

Tutorial 2-Q2 At max potential energy T1 = 0 V1 = (1/2) ((kB/4) + kC)L2 qm2 At max kinetic energy, V2 = 0 T2 = (1/2) (1/3)mL2 ( qm )2

T1 = 0 V1 ≠ 0

T1 + V1 = T2 + V2 2= (3/m) ((k /4) + k ) B C =3

=

= 31.72

/

4

+

V2 = 0 T2 ≠ 0

Tutorial 2-Q3 =

A 25 kg motor is supported by four springs, each having a constant of 200 kN/m. The unbalance of the rotor is equivalent to a mass of 30g located 125 mm from the axis of rotation. Knowing that the motor is constrained to move vertically, determine the amplitude of the steady state vibration of the motor at a speed of 1800 rpm, assuming that (a) no damping is present. (b) the damping factor is 0.125. Answers: (a) 1.509 mm (b) 0.583 mm.

sin

=

=

= 1800

= 1800 ×

2 60



=

= 1−

⁄

1−

+

2

Tutorial 2-Q4 A machine element having a mass of 500 kg is supported by two springs, each having a constant of 48 kN/m. A periodic force of maximum magnitude equal to 150 N is applied to the element with a frequency of 2.4 Hz. Knowing that the coefficient of damping is 1600 Ns/m, determine the amplitude of the steady state vibration of the element. Answer: 5.02 mm.

=2

= =2 =

=

⁄

1−

+

2

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

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)

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

Tutorial 3-Q1 DOF =3(n-1) -2L -h

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

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

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

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

Tutorial 3 - Q2 and Q3 Summary: = =

× +

= ×

=

=

×

× +

=

× ×

×

+ ×

× +

×

+

× ×

Final answers: Q2

Q3

=4 = 6.67

.

.

. .

= 124 .

rad

= 28.9

C. C. W.

rad

C. W.

= 29.33

.

= 11.29

. .

.

= 356.15 .

= 860.08

rad rad

C. W. C. C. W.