Mechanical and Machine Dynamics

This text covers various topics related to Mechanical and Machine Dynamics, including equations of motion, balancing, and natural frequency of systems. It also includes examples and solutions to problems related to these topics.

Running head: MECHANICAL AND MACHINE DYNAMICS 1
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MECHANICAL AND MACHINE DYNAMICS 2
Mechanical and Machine Dynamics
Task 1. CCW-Counter clockwise
A־ω
Au+ω
35°
15°
Au¿ 0.165
Au =(0.165mm, 15 ° ccw ) where ccw is counter clockwise
Auw = ( 0.225, 35° ccw)
Aw = ((Au2 +A2u+w -2AuAu+wcos (∅ −θ)¿ ¿
¿ ¿=0.1735mm
w0=originalimbalance= ( Au
Aw )w= ( 0.165
0.1735 )×50=47.5504 g
∝=cos−1 ¿ ¿)= cos−1 ( 0.1652+ 0.17352−0,2252
2 ×0.165 × 0.1735 )=83.2775°
Which is 83.2775θ counter clockwise.
Flywheel will be stable if a weight of 42.5504g is added at 83.2775 counter clockwise
from the place of the experimental weight or ¿ ¿)=38.2775 clockwise
Task 2
Let Wc=2 Kg , WD =3 Kg ,W E =2 Kg

MECHANICAL AND MACHINE DYNAMICS 3
rc=50 mm , r D=75 mm ,r E =25 mm
θc=90θ ,θD=220θ , θE=−30θ Let W A , r A , θ A∧W G , rG , θG denote the weights added in
planes A and G respectively static balancing;
W c rc cos θ+W D rD cos θD + W E r E cos θE +W G rG cos θG +W A r A cos θA =0
W G rG cos θG +W A r A cos θA =6.2136 (1)
WG rG sin θG + W A r A sin θA +W C rC sin θC+W E r E sin θE +W D rD sin θD =0
W G rG sin θG +W A r A sin θA =5.2136 (2)
Dynamic balancing, we take moments about the left bearing about plane B:
W C rC ∝C cos θC+W D r D ∝D cos θD+W E rE ∝E cos θE +W A r A ∝A cos θ A +W G rG cos θG
¿−16 W A r A cos θ A +88 W G rG cos θG=201.408 (3)
W C rC ∝C sin θC +W D r D ∝D sin θD +W E rE ∝E sin θE +W A +W G rG ∝G sinθG=0
=−16 W A r A sinθ A +88 W G rG sin θG =372.544 (4)
Equation (1) and (2) results
( −16 × 6.594 ) +104 × ( W G rG cos θG ) =201.48
W G rG cos θG=2.9511 (5)
Equation (2) and (4) gives
( −16 ×5.2136 ) + 104 W G rG sin θG =372.544
W G rG sin θG=4.3842 (6)
Equation (5) and ( 6) give
( 2.95112 +4.38422 )
1
2 =5.2849 mm And θG =tan−1 ( 4.3842
2.9511 ¿)=56.054 ¿ ( 7)
Equation (1), (2) and (7) yields ;
W A r A cos θA =6.594−5.2849 cos 56.0549=3.649∧¿ W A r A sin θA =5.2136−5.2849sin 56.0549=0.9007 ¿
(8)

MECHANICAL AND MACHINE DYNAMICS 4
Equation (8) provide
W A r A= ( 3.64292+ 0.90072 )
2
=3.7526 Kg∧θA =tan−1 (¿ 0.9007
3.6429)=13.8877θ ¿
If balancing weight is placed at a radial distance of 50mm in place A and G.
We have rG =r A =50 mm∧thus W A =1.8763 Kg ,θA ¿13.8877θ W G =2.645 Kg ,θG=56.0549θ
Task 3
balanced Primary forces are given by the following equation
Fxp=∑
i=1
6
( Fx ) pi=∑ ( M p +M c )i r ω2 cos (ωt + αi) (1)
F yp=∑
i=1
6
( F y ) pi=∑ (−M c ) i r ω2 sin( ωt+ αi) (2)
unbalanced Secondary force is given by
F xs =∑
i=1
6
( Fx ) si=∑ ( M p )i
r 2 ω2
l cos( 2ωt +2 αi) (3)
Unbalanced primary and secondary moments are given by
( Ms ) p=∑
i=1
6
( F x ) pi li (4)
( Ms )s=∑
i=1
6
( Fx ) si li (5)
( Mx ) p=∑
i=1
6
( F y ) pi li (6)
( M p + M c ) r ω2
∑
i =1
6
cos ( αi ) = ( M p+ Mc ) r ω2 ¿
Equation (2) gives

MECHANICAL AND MACHINE DYNAMICS 5
−M c r ω2
∑
i=1
6
sin ( αi )=−M c r ω2 ( 2sin ( 0 )+ 2sin 120+2 sin 240 ) =0
Equation (3) results into
M p r2 ω2
l ∑
i=1
6
cos 2 αi= M p r 2 ω2
l ( 2 cos ( 0 ) +2 cos 240+2 cos 480 )=0
Equation (4) results into;
( M p + M c ) r ω2
∑
i =1
6
li cos ( αi )= ( M p +M c ) r ω2 α ¿
Equation (5) results into
M p r2 ω2
l ∑
i=1
6
li cos 2 αi= M p r2 ω2 α
l ( cos ( 240 ) +2 cos 480+ 3 cos 480+4 cos 240+5 cos 0 )=0
Equation (6) yields:
−M c r ω2
∑
i=1
6
li sin ( αi )=−M c r ω2 α ( sin (120 )+ 2sin 240+3 sin 240+4 sin 120+5 sin 0 )=0
Thus, engine is completely force and moment balanced.
Task 4
a) Equation of motion:
j0 ¨θ1 +2 k1 θ1−k 1 θ2=0
2 j0 ¨θ2−k1 θ1+ k1 θ1….. (1)
Rearranging and substituting in the harmonic solution

MECHANICAL AND MACHINE DYNAMICS 6
θi ( t ) =φi cos (ωt +∅ ¿) ; i=1,2,3 ¿(2)
It brings frequency equation
2 ω4 j0
2+5 ω2 j0 kt +kt
2=0 (3)
The solution of equation (3) gives the natural frequency
b) Mass of the car =m, radius of gyration is r = Jo mr2
Equation of the motion;
yf ( yr )= ground displacements of front wheels, downwards for motion along x
m ¨x +x ( k f +kr ) +θ ( kr l2−kf l1 ) =kf
For motion alongθ;
J0 ¨θ+ x ( l2 kr −l1 kf ) +θ ( kr l2
2 +k f l2
1 ) =(kr l2 yr −k f l1 yf )
Where ground motions can be expressed; yf (t)= ysinωt.
c) using newton’s second law of motion:
For mass m2
m2 ¨x2 + c2 ¿ (1)
For mass m1
m1 ¨x1 + c2 ( ˙x1+ ˙y ) +k1 ( x1+ y )−c1 ( ˙x2 ˙−x1 )−k 2 ( x2−x1 ) =0 (2)
Equation (1) and (2) can be rewritten as;
m1 ¨x1 + ˙x1 ( l1 +l2 ) − ˙x2 c2 + x1 ( k1 +k2 ) −x2 k2=c1 ˙y +k1 y (3)
m2 ¨x2 + ˙x2 c2− ˙x1 c2 + x2 k2−x1 k2=0 (4)

MECHANICAL AND MACHINE DYNAMICS 7
Equations (3) and (4) can be expressed in matrix form as:
[m1 0
0 m2 ] ( ¨x1
¨x2 )+ [c1 +c2 −c2
−c2 c2 ] ( ˙x1
˙x2 )+ [k1+ k2 −k2
−k2 k2 ] (x1
x2 )=[c1 ˙y+ k1 y
0 ] (5)
d) Using x1∧x2, equation of motion of piston
m1 ¨x1 + ( k1 +k2 ) x1− psinθ=0 (1)
Equation of the motion of the pendulum bob
m2 ¨x2 + psinθ=0 (2)
In equation (1) and equation (2) sin θ can be expressed as sin θ ≈ θ= x2−x1
l (3)
This for a small angle θ.
For the vertical equilibrium of the mass m2
p=m2 g cos θ ≈ m2 g (4)
Substituting equation (3) and (4) in (1) and (2) yields:
m1 ¨x1 + ( k1 +k2 ) x1−m2 g ( x2−x1
l )=0 (5)
m2 ¨x2 +m2 g ( x2−x1
l )=0 (6)
Assuming harmonic solutions as
x1 ( t ) =x1 cos(ωt +Φ) (7a)
x2 ( t )=x2 cos( ωt+Φ ) (7b)
Equations (5) and (6) becomes
[−m1 ω2+ ( k1 + k2 ) +( m2 g
l ) y ]x1− m2 g
l x2 =0 (8)
−m2 g
l x1 + [−m1 ω2 + m2 g
l ] x2=0 (9)
By setting the determinants of the coefficient matrix of x1 and x2, we obtain the frequency
equation as

MECHANICAL AND MACHINE DYNAMICS 8
|−m1 ω2
( k1 + k2 ) +( m2 g
l )
−m2 g
l −m2 ω2 + m2 g
l |=0 (10)
i.e m1 m2 ω4− m1 m2 g ω2
l + m2 g
l ( k1 +k2 + m2 g
l ) =0 (11)
The roots of the equation (11) give the natural frequency of the system.
e) Taking moments about 0 and mass m1
m1 l1
2 ¨θ1=−ω1 ( l1 sin θ1 ) +α sin θ2 ( l1 cos θ1 )−α cos θ2 ( l1 sin θ1 ) =−ω1 l1 θ1+ ω2 l1 (θ2−θ¿¿ 1)❑¿
(1)
Assuming that α ≈ ω2
m2 l2
2 ¨θ2+m2 l2 l1 ¨θ1=−ω2 ( l2 sin θ2 )=−ω2 l2 θ2 (2)
Using the relations θ1= x1
l1
and θ2= x2−x1
l2
equations (1) and (2) becomes
m1 l1
2 ¨x1 +
( ω1 +ω2 ( l1 +l2
l2 ) ) x1− ω2 l1 x2
l2
=0 (3)
m2 l2 ¨x2 +ω2 x1 +ω2 x2=0 (4)
When m1=m2=m, l1=l2=l and ω1=ω2=mg equations (3) and (4) gives
ml ¨x1 +3 mg x1−mg x2=0 (5a)
ml ¨x2−mg x1+ mg x2=0 (5b)
For harmonic motion xi ( t ) =xi cosωt; where i=1 , 2 ,3 , … equation (5) becomes
−ω2 ml x1 +3 mg x1−mg x2=0
−ω2 ml x2−mg x1+ mg x2=0 (6)
From which the frequency equation can be obtained as
ω4 m2 l2−(4 m2 lg)ω2 +2 m2 g2=0

MECHANICAL AND MACHINE DYNAMICS 9
Let R1 , R2∧R3 be the restoring forces in the spring. Equations of motion ot mass m in x
and y directions are:
m ¨x =∑
i=1
3
Ri cos α i (1)
m ¨y=∑
i=1
3
Ri sin αi (2)
Where Ri=−ki ( x cos αi+ y sin αi) (3)
Equations (1) to (3) gives
m ¨x +∑
i=1
3
ki ( x cos2 α i + y sin αi cos αi )=0 (4)
m ¨y+∑
i=1
3
ki ( x sin αi cos α i+ y sin2 α i )=0 (5)
For α 1=45 °, α 2=135 °, α 3=270 ° and k1 =k2=k3=k, equations (4) and (5)
m ¨x + kx=0 (6)
m ¨y+ 2ky =0 (7)
These two equations are uncoupled for harmonic motion.
x (t)=x cos (ωt +Φ)
y (t )= y cos(ωt+Φ )
And hence ω1= √ k
m for motion in x-direction.
ω2= √ 2 k
m for motion in the y-direction.
Natural modes are given by
x (t)=x cos ( √ k
m t +Φ1 )
y (t )= y cos( √ 2 k
m t+ Φ2 )
Where x , y , Φ1∧Φ2 can be determined by initial conditions.

MECHANICAL AND MACHINE DYNAMICS
10
Task 5.
(a) Equation of motion;
Assume that θ1, θ2 are small.
Moment equilibrium eqns of the masses about P and Q:
ml2 ¨θ+ mgl θ1 +kd2 ¿) =0 ……..(1)
ml2 ¨θ−mgl2 −kd2 ( θ1+θ2 )=0 … … … …(2)
(b)Natural frequencies and mode shapes.
Assume: Harmonic motion with;
θ j (t)=φj cos (ωt−∅ ¿); j=1 , 2 ,3 … … ..(3) ¿
Whereφ1∧φ2 are the amplitudes of θ1∧θ2 respectively,
ω is the natural frequency ,∧∅ is the phase angle:
Using equation (3), equation (1) and equation (2), they can be expressed in matrix form as;
ω2 ml2
[1 0
0 1 ]φ1
φ2
+[mgl +kd2 −k d2
−k d2 mgl +kd2 ]φ1
φ2
=0
0 ………….(4)
Frequency equation:
[ ω2 ml2 +mgl +kd2 −k d2
−kd2 −ωml2 +mlg+ kd2 ]=0 ………..(5)
or
ω4−ω2 +
( 2 g
l
2 kd2
ml2 )+ g2
l2 + 2 gkd2
ml2 =0 ………(6)
By substituting for ω1
2∧ω2
2 in to equation (4), we obtain
φ2
φ1
(1)
=1∨φ1
φ2
(1)
=1
1 φ1
(1) and φ2
φ1
(2)
=−1∨φ1
φ2
(2 )
= 1
−1 φ1
(2)
Thus, the motion of the masses in the two modes is given by;⃗
θt
(1 ) ( t )=¿ 41
(1) = φ1
(1) 1
1 cos( ω1 t−∅1 ¿) ¿…… (7)⃗
θt
(2)=φ1
(2) 1
−1(cos ω2 +∅ 2)…………..(8)
c) Free vibration response.
Using linear superposition of natural modes, the free vibration response is given by;

MECHANICAL AND MACHINE DYNAMICS
11⃗
θ ( t )=C1⃗ θ ( 1 ) (t)+C2⃗ θ (2 ) (t)………….(9)
By using C1=C2 =1, without loss of originality (7) and (9) lead to ;
θ1 ( t )=φ(1)
1 cos ¿ + φ1
(2) cos ¿…..(10)
θ2 ( t ) =φ(1)
1 cos ¿ - φ1
(2) cos ¿…..(11)
Where φ1
(1), ∅ 1, φ1
2 and ∅ 2 are constants to be determined from initial conditions. Where
θ1 ( 0 ) =a ,θ2 ( 0 ) =0, ˙θ1 ( 0 )=0∧ ˙θ2 ( 0 ) =0, eqn (10) and (11) yield
a=φ1
(1) cos ∅ 1+ φ1
(2) cos ∅2
0=φ1
(1)cos ∅1 −φ1
( 2 ) cos ∅2
σ =−ω1 φ1
(1 ) sin θ1−ω2 φ1
2 sin ∅ 2 …(12)
σ =−ω1 φ1
(1 ) sin θ1 +ω2 φ1
2 sin ∅ 2
Eqn (12) can be solved for φ1
(1) ,∅ 1 , φ1
2∧∅ 2 to obtain
∅ 1 ( t ) =a cos ω2−ω1
2 t . cos ω2 +ω1
2 t
…..(13)
∅ 2 ( t ) =a sin ω2−ω1
2 t . sin ω2+ω1
2 t
Task 6.
(a) The sheer building can be modeled as shown below.
k 2( x2−x1 ) k3 ( x3 −x2 ) k 4 (x4 −x3 )
F1(t) F2( t) F3 (t) F4 (t)Re m1 ¨x1
----
Gs m2 ¨x2
----
Jh m3 ¨x3
----
Bs m4 ¨x4
----

MECHANICAL AND MACHINE DYNAMICS
12
c2 ( ˙x2− ˙x1) c3 ( ˙x3− ˙x2) c4 ( ˙x4− ˙x3)
(a) Equations of motion from the above diagram;
m1 ¨x1 +c1 ˙x1 +k1 x1 −c2 ( ˙x2− ˙x1 ) −k2 ( x2−x1 )=F1 (t )
m2 ¨x +c2 ( ˙x2 − ˙x1 ) + k2 ( x2−x1 ) −c3 ( ˙x3−x2 ) −k3 ( x3 −x2 ) =F2( t)
m3 ¨x3 +c3 ( ˙x3− ˙x2 ) + k3 ( x3 −x2 ) −c4 ( ˙x4 − ˙x3 ) −k 4 ( x4−x3 )=F3 (t)
m4 ¨x4 +c4 ( ˙x4 − ˙x3 ) + k4 ( x4 −x3 ) =F4 (t)…………………(1)
(b) Langrage’s equations are;
d
dt ( ∂ T
∂ Qj )− ∂ K
∂Qj + ∂ P
∂ Qj + ∂ R
∂ Qj =Qj ………………(2)
Where K is the kinetic energy, P is the potential energy and R is the Rayleigh’s dissipation
function, Q j= jth, generalized coordinate.
K= 1
2 ( m1 ˙x1
2 +m2 ˙x2
2 +m3 ˙x3
2+ m4 ˙x4
2)
P= 1
2 ¿
R= 1
2 ¿
Q j=M j ; j=1,2,3,4 … … .
Using Q j=x j ; j=1,2,3,4 … … … .
The application equation (2 yields the equation of motion in the equation (1).

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