ENME 361 Homework 1 2019 Solution PDF

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ENME 361 - Spring 2016: Vibrations, Controls, & Optimization I

ENME 361

Homework#1 Due Date: 2/7/2019

1. Consider the inverted pendulum system shown in Figure (1). Determine the equation of motion of the system using: a. x - y Fixed coordinate system fixed at point O. b. e1 -e2 Body-fixed moving coordinate system fixed at point G. c. energy method Compare & comment on the results. e2 y

e1

x

Figure (1)

2.

Determine the equation of motion that govern the deflection x of the masses m of the governor system shown in Figure (2). The masses are supported by massless rods which are connected from one end to a spring k and from the other end to a massless sliding sleeve. The assembly rotates around the y axis at a constant angular speed Ω.

y Ω Sliding Sleeve

x

Assume that the length of each rod is l, the mass of each ball is m, and the free length of the spring is h. x Figure (2)

ENME 361 - Spring 2016: Vibrations, Controls, & Optimization I

3. A cylinder of mass m and mass moment of inertia Jo is free to roll without slipping but is restrained by two springs of stiffnesses and as shown in Figure (3). Find the equation of motion of the cylinder.

Figure (3)

4. Determine the equivalent mass and stiffness for the system shown in Figure (5) as measured at the mass m. Assume that the pulley system has a mass moment of inertia Jp.

Jp

Figure (4)

5.

Determine the degrees of freedom (DOF) of the systems shown in Figure (5).

Identify ALL the links (N), the ground link, ALL the joints whether revolute (R) or prismatic (P), and use Grubler’s criterion to determine the DOF.

ENME 361 - Spring 2016: Vibrations, Controls, & Optimization I

(a) - Transom mechanism

Sliding Sleeve

(b) - Whitworth Quick Return Mechanism

Fixed bearing

(c) - Scotch Yoke Mechanism Figure (5)

ENME 361 - Spring 2016: Vibrations, Controls, & Optimization I

SOLUTIONS A. Fixed Frame

y

y

may

=

rG x

Rx

max

rG x

Rx

Ry

Ry

Free Body Diagram As Then,

and

Kinetic Diagram

1 1 rG = l sin θ i + l cos θ j , 2 2 1 1 v G = l cos θ θ i − l sin θ θ j 2 2 1 1 1 1 aG =  − l sin θ θ 2 + l cos θ θ i −  l cos θ θ 2 + l sin θ θ j 2 2  2  2 

Equations of Motion:

∑ Fx = ma x → Rx − 2 kl sin θ = m  − 2 l sin θ θ 2 + 2 l cos θ θ 

(1)

∑ Fx = ma x → R y − mg = − m  2 l cosθ θ 2 + 2 l sin θ θ 

(2)

1

1





1

1





∑ M O = I o θ → −2 kl 2 sin θ cos θ + 2 mgl sin θ =  3 ml 2  θ 1

1



Equation (3) gives the equation of motion



(3)

ENME 361 - Spring 2016: Vibrations, Controls, & Optimization I

Divide by

θ + 3

1 2 ml gives: 3

k 3 g sin 2θ − sin θ = 0 m 2l

For small oscillations: θ +  6

k 3 g − θ = 0  m 2 l

Equations (1) and (2) can yield Rx and Ry.

B. Body-Fixed Frame Taking the moment around the pivot G gives:

a. ∑ M G = I Gθ 1 2 1 ml θ = − R1 l / 2 − 2k (l sin θ ) l cos θ 12 2

(1)

and equilibrium in the e1 direction gives:

b. ∑ F1 = ma1 = m ( lθ / 2) l  m  θ = mg sin θ + R1 − 2kl sin θ cos θ 2 

e2 (2)

Eliminate R1 between (1) & (2) gives:

1 2  1 ml θ = −2 k ( l sin θ ) l cos θ + mgl sin θ 3 2 e1 Divide by

1 2 ml gives: 3

R1

R2

ENME 361 - Spring 2016: Vibrations, Controls, & Optimization I

θ + 3

k 3g sin 2θ − sin θ = 0 m 2l

For small oscillations: θ +  6

k 3 g − θ = 0  m 2 l

C. Energy Method: 1 1 K .E. = T =  ml 2 θ 2 2 3  1 2 1 P.E. =U = 2k (l sin θ ) − mgl ( 1 −cos θ ) 2 2

and Total Energy:

K.E. + P.E. = T +U =

1  1 2  2 2 1 ml θ + (kl sin θ ) − mgl (1− cos θ ) = cons tan t  23 2 

Differentiate wrt time gives:

(

)

 1 ml 2    + 2 k l 2 sin cos  − 1 mgl sin  = 0 θ θ θ θθ 3 θ θ 2   1 2 Divide by  ml  θ gives: 3 

θ+ 3

k 3g sin 2θ  − sin θ = 0 m  2 l

For small angles θ:

θ+  6

k 3 g − θ = 0  m 2l 

ENME 361 - Spring 2016: Vibrations, Controls, & Optimization I

y

2.

Ω Sleeve

x

x

N=3, R=3, P=1 DOF=3x3-2x3-2=1 R

R 1 1 R

2lcosθ

2l

R 2

P R

2 P

Δ 3

R

hf h

3

Δ=2l(1-cosθ) Deflected

A. Energy Method Deflection of spring Δ: ∆ = 2l (1− cos θ )

Undeflected

ENME 361 - Spring 2016: Vibrations, Controls, & Optimization I

1  1  K.E. = T = 2  ml 2θ2  + 2  ml 2 sin 2 θ Ω 2  2 2     1 2 P.E. =U = k ×4l 2 (1 −cos θ ) + 2mgl (1 − cos θ ) 2

and Total Energy:

K .E. + P.E. = T + U = ml 2θ 2 + ml 2 sin 2 θ Ω 2

(

(

)

)

+ 2 kl 2 ( 1 − cos θ ) + 2 mgl (1 − cos θ ) = cons tan t 2

Differentiate wrt time gives:

(

)

2ml 2 θ+ sin θ cos θ Ω 2 θ + 4kl 2 (1 − cos θ ) sin θ θ + 2mgl sin θ θ = 0

(

)

Divide by 2ml 2 θ gives:

( θ + sinθ cos θ Ω2 ) + 2 mk ( 1− cos θ) sin θ + gl sin θ= 0 Or

k g   + 1 2  θ 2 sin 2θ Ω  + 2 m (1 − cos θ ) sin θ + l sin θ = 0  

For small angle approximation:

θ + 

g k + Ω 2  θ + θ 3 = 0 m l 

As x = l sin θ → x ≅ lθ or θ = x / l :

g k x +  + Ω2  x + 2 x3 = 0 ml l 

ENME 361 - Spring 2016: Vibrations, Controls, & Optimization I

B. Fixed Frame

y Ω Sliding Sleeve

A B

x Rod

rm

x

O

z Position Vector

rm = OA + AB = l sinθ i + ( [ h + 2l ] − l cosθ ) j

Differentiate wrt time to determine the velocity vm: d  l sinθ i + ([ h + 2 l] − l cos θ ) j  + ω × rm dt  = [l cos θ i + l sin θ j]θ + Ω j × l sin θ i + ([h + 2l ] − l cos θ ) j

vm =

= [l cos θ i + l sin θ j]θ − Ωl sin θ k

v 2m = l 2θ 2 + ( Ωl sinθ )

Note that:

2

same as in Energy Method.

Differentiate wrt time to determine the acceleration am:

d  ( l cosθ i + l sinθ j ) θ − Ω l sinθ k  dt  = ( − l sinθ i + l cosθ j) θ2 + ( l cosθ i + l sinθ j) θ− Ω l cos θ θ k

am =

(

+ Ω j× l cos θ θ i − Ω l sin θ k

(

)

) (

)

 j − 2 Ωl cos θ θ k = −l sin θ θ2 + l cos θ θ − Ω2 l sin θ i + l cos θ θ2 + l sin θ θ =

ax i

+

ay j

+

az k

ENME 361 - Spring 2016: Vibrations, Controls, & Optimization I

y

y max T1

=

x T2

T2

T2

x

max m

mg

k

x

x

Free Body Diagram

Kinetic Diagram

Equilibrium of Spring: 2T2 cosθ = − 2 lk (1 − cosθ ) In x - Direction: In y - Direction:

→

(

2 2 m −l sin θ θ + l cos θθ − Ω l sin θ

(

)

T2 cosθ = lk ( 1 − cosθ )

(1)

)

(2)

= − (T1 + T 2 ) sin θ

= (T1 − T2 ) cos θ − mg (3)

m l cos θθ2 + l sin θθ

Multiplying eqn. (2) by cos θ and eqn. (3) by sin θ. Add the two equations gives: 2 ml θ + ml Ω sin θ = −2T 2 sin θ cos θ − mg sin θ

From equations (1) and (4):

ml θ + ml Ω 2 sin θ = −2lk (1 − cos θ )sin θ − mg sin θ Divide by ml gives:

g k θ +  Ω2 +  sin θ + 2 (1 − cos θ )sin θ = 0 

l

m

ENME 361 - Spring 2016: Vibrations, Controls, & Optimization I

For small angle approximation:

θ + 

g k + Ω 2  θ + θ 3 = 0 m l 

As x = l sin θ → x ≅ lθ or θ = x / l :

g k x +  + Ω2  x + 2 x3 = 0 l ml  

ENME 361 - Spring 2016: Vibrations, Controls, & Optimization I

3.

x

θO

R

C

Energies 1 1  mR 2 + mR 2  θ2  2 2  1 2 P.E. =U = (k1 + k 2 )(R +a ) θ 2 2

K.E. = T = and

Total Energy:

K.E. + P.E. = T +U =

1  3 2  2 1 2 mR θ + (k 1 + k 2 )(R + a ) θ 2 = cons tan t  2 2 2 

Differentiate wrt time gives:

 3 mR 2    + k + k R + a 2  = ) θθ 0 2  θ θ ( 1 2 )(   3  Divide by  mR 2  θ gives: 2  

θ+

(k 1 + k 2 )(R + a )2 θ = 0 2 1 .5 mR

ENME 361 - Spring 2016: Vibrations, Controls, & Optimization I

4.

x 2= r 2 θ

θ

Jp

x1 = r1 θ

x1

Equivalent Stiffness:

U=

1 2 1 k1x1 + k 2x 22 2 2

But,

x1 = r1θ

Then,

x1 =

and x2 = r2θ

r1 x2 r2

1 r U = k1 + k 2  1  2  r2  

where

2

1  x 12 = k eq x 12 2 

2   r1   k eq = k1 + k 2      r2  

Equivalent Mass:

1 1 T = J pθ 2 + m1 x12 2 2

ENME 361 - Spring 2016: Vibrations, Controls, & Optimization I

x1 r1

But,

x1 = r1θ → θ =

Then,

T =

where

1  m eq =  2 J p + m1   r1 

 2 1 1 1 2  2 J p + m1  x1 = meq x1 2  r1 2 

ENME 361 - Spring 2016: Vibrations, Controls, & Optimization I

5. Degrees of Freedom P N=4, R=3, P=1

R 3

DOF=3(4-1)-2*3-2*1=1 2 4

R

1 R (a) – Transom Mechanism

R

5 6

P 4

R

R 3

2

N=6, R=5, P=2 R

R

DOF=3(6-1)-2*5-2*2=1 1 (b) – Quick return mechanism

P 1

ENME 361 - Spring 2016: Vibrations, Controls, & Optimization I

4

P

P 3 2

R

N=4, R=2, P=2 DOF=3(4-1)-2*2-2*2=1

1

R

(c) - Scotch Yoke Mechanism...