Lecture-2- mechanical vibrations, Mohamed salem.pdf

Vibration course for mechanical engineering by Dr. mohamed Sameh

1. Dr. Mohamed Sameh Salem
Contact Info:-
Email: [email protected]
Mobile: [phone]
1Lecture (1)
Mechanical Vibrations
ME 242
Dr. Mohamed S. Salem
ةعمجملا ةعماج
ةسدنهلا ةيلك
ةيعانصلا و ةيكيناكيملا ةسدنهلا مسق

2. -A mass, m, of a system is subjected to a repeated motion (or
vibration ) due to external excitation. This excitation could affect only
once or continuously.
-If not prevented, this motion may damage the system to which the
mass is attached.
-Therefore, a spring and a damper are added to minimize (or remove
) the effect of that motion (or vibrations) on the mass system.
-The vibration system is then composed of:
MASS + SPRING + DAMPER
The purpose of studying Mechanical Vibrations

3. Mass – Spring – Dashpot system
Kinetic energy storing elements, inertia
element: mass m or mass moment of inertia J
Potential energy storing elements: spring
element and/or gravity
Energy dissipating elements: damper
Force Element (F)

4. Inertia Elements

5. Inertia elements
•Inertia element in vibrating systems stores the Kinetic Energy (T)
and it could be:
–Mass, m for rectilinear or curvilinear vibrations of particles or
rigid bodies.
•??????=
1
2
�ሶ�
2
is the kinetic energy (in Joule) for the mass m and
ሶ� is the velocity of the mass (m/s).
–Mass moment of inertia, J, for rotation about a fixed axis for
rigid bodies.
•[Remember that particles do not rotate about fixed axis
passing through it since particles have negligible size]
•[We used to refer to the mass moment of inertia as I, in
dynamics. Here, in vibrations, we use the letter J instead]

6. Mass Element (M or m) (kg)
•MassorInertiaElements:
•Themassorinertiaelementisassumedtobearigidbody.Itcangainorlose
potentialandkineticenergies.

7. •Motion
•Force �=�.??????
°°
•PotentialEnergy(PE) ??????�=±�??????ℎ
•KineticEnergy(KE)
Translational
Rotational
1
��=
2
�??????
°
2
1
2
��=J�??????
°
2
Mass Element (M or m) (kg)

8. 12
�
2 3
�.�
12
1 �
J=��
2
�=
J�=J�.�+��
2
=
1
2
��
2
+�
2
1
=��
2
•MassmomentofInertia(J
o)(kg.m
2) J�=J�.�+��
2
J
C.G:massmomentofInertiaatthecenterofgravityoftheelement
d:thedistancebetweenthecenterofrotationandthecenterofgravityoftheelement
•LumpedMass:
misconcentratedata pointanditsradiusiszero
J�=J�.�+��
2
=0+��
2
=��
2
•RodMass:
When(M)cannotbe neglected
Mass Element (M or m) (kg)

9. •Rodwithlumpedmass:
J�=J??????��+J�.�=
1
3
��
2
+��
2
•CylinderorDiskmass:
�.�
1
2
J=��
2�=�
�J=J�.�+ ��
2 2 2
31
=��+��=��
2 2
2
�.�
•Spheremass:
2
5
J=��
2�=�
�J=J
2 2 2
2 7
�.�+��=
5
��+��=
5
��
2
Mass Element (M or m) (kg)

10. Combinationofmasses(m
eq)
Case1:TranslationalMassesConnectedbyaRigidBar.
Mass Element (M or m) (kg)

11. Combinationofmasses(m
eq)
Case1:TranslationalandRotationalMassesCoupledTogether
Mass Element (M or m) (kg)

12. Spring Elements

13. •Aspringisatypeofmechanicallink,whichinmostapplicationsisassumedtohave
negligiblemassanddamping.
•Themostcommontypeofspringisthehelical-coilspringusedinretractablepens
andpencils,staplers,andsuspensionsoffreighttrucksandothervehicles.
•Severalothertypesofspringscanbeidentifiedinengineeringapplications.Infact,
anyelasticordeformablebodyormember,suchasacable,bar,beam,shaftorplate
canbeconsideredasaspring.

14. Types of Mechanical Springs

15. Types of Mechanical Springs

16. •Force-Deformationdiagramofthehelicalspring
LinearSpring
�=????????????
SoftorHardSpring
�=�??????+�??????
3

17. ??????.??????
a)Translational Spring(N/m)
•Force �=????????????
•PotentialEnergy(PE)
1
??????�=
2
????????????
2
????????????.??????
b)TorsionalSpring(N.m/deg)
•Torque ??????=??????????????????
•PotentialEnergy(PE)
1
??????�=
2
??????????????????
2

18. •Thestiffnessofthehelicalspring
��
4
??????=
8�
3�
G:ModulusofRigidity or Elasticity (Shear modulus)(N/m2)
d:Diameterofthespringwire(m)
D:Centraldiameterofthespring(m)
n:No.ofactivecoils

19. Helical Spring(cont.)
Spring force, F = �� ; where k is the spring stiffness (N/m)and x is the spring deflection
Spring potential energy, U
s =
1
2
��
2
; where k is the spring stiffness and x is the spring deflection
measured from unstretched length.

20. Measuring stiffness of helical springs
Free length of the
spring
(unstretched
length)
m
1
m
2
m
3
1.Hang a spring freely
2.Hang a mass m
1 to the end of the spring and measure the deflection x
1
3.Add more mass such that the total mass is m
2 and the spring deflection is x
2
4.Add more mass such that the total mass is m
3 and the spring deflection is x
3.
5.Report your results in the following table:

21. Measuring stiffness of helical springs
Mass, kg*
Spring force, F (N)
[Weight in this case]
Deflection, x (m)
0.0 0.0 0.0
0.2 0.2*9.81=1.962 0.01
0.4 0.4*9.81=3.924 0.02
0.6 0.6*9.81=5.886 0.03
6.Plot the weight and the deflection on a graph as the following:
F
The slope of the linear part of the curve is
the stiffness of the spring, K
* Numbers are for illustration only

22. Torsional springl
EI
k=
Spring Moment, M
t = �?????? ; where k is the spring torsional stiffness (N.m/rad) and θ is the spring
angular deflection in rad.
Spring potential energy, U
s =
1
2
�θ
2

23. Longitudinal stiffness of bars
Where
F = is the applied force (N)
E=Modulus of elasticity ( a material property) (N/m
2
)
A=bar cross sectional area (m
2
)
l = bar length (m)

24. Torsional stiffness of rods (or shafts)
Where G=shear modulus of rigidity of the rod
J
p=Polar moment of inertia of the rod
= πd
4
/32; d=diameter of the rod
l = length of the rod
Spring Moment, M
t = �?????? ; where k is the rod torsional stiffness (N.m/rad) and θ is the spring
angular deflection in rad.

25. This corresponding to leaf spring
Stiffness, k, is given by:- 3
3
l
EI
k=
Transverse stiffness of cantilevered beam

26. Spring equivalent to cantilever beam

27. Theequivalentstiffnessofmechanicalelements(K
eq)
•Rodundertensionorcomparison(Translational)
??????��=
�??????
�
??????
??????
−
��=
•Rodundertorsion(Rotational)
��
�
32
??????
�=�
4
E:ModulusofElasticity(N/m
2)
A:CrossSec.Area(m
2)
L:RodLength(m)
G:ModulusofRigidity(N/m
2)
J:PolarMomentofInertia(m
4)
L:RodLength(m)

28. Theequivalentstiffnessofmechanicalelements(K
eq)
•Fixed-FreeBeam(Cantilever)
•SimplysupportedBeam(Hinged-Roller)
??????��=
3��
�
3
E:ModulusofElasticity(N/m
2)
I:2
ndMomentofArea(m
4)
L:RodLength(m)
�ℎ
3
�=
12
??????
�=�
64
4
??????��=
48��
�
3
E:ModulusofElasticity(N/m
2)
I:2
ndMomentofArea(m
4)
L:RodLength(m)

29. Deflection of Beams and Plates
�=
��??????� ??????
������????????????�� ��??????ℎ� ��??????� ��??????�?????? �(�)
x is the distance form the left hand to the load P.

30. �=
��??????� ??????
������????????????�� ��??????ℎ� ��??????� ��??????�?????? �(�)
x is the distance form the left hand to the load P.
Deflection of Beams and Plates

31. Deflection of Beams and Plates

33. Example 1
In Figure (a), a particle of mass m is attached to the midspan of a simply supported beam of
length L, elastic modulus E, and area moment of inertia I. determine the beam stiffness
At the mid span, x=a=L/2, and either equation can be used to determine the deflection. If we
use the first equation:
y
L
2
=
??????
??????
2
??????
2
6????????????�
??????
2
−
??????
2
2
−
??????
2
2
=
????????????
3
48????????????
�=
??????
�
=
48????????????
??????
3
;stiffness is

34. If we use the second equation:
y
L
2
=
??????
??????
2
??????−
??????
2
6????????????�
2??????
??????
2
−
??????
2
2
−
??????
2
2
=
????????????
3
48????????????
�=
??????
�
=
48????????????
??????
3
;stiffness is
Both equations give the same deflection at load point and hence either of them can be used.
The stiffness k given:
�=
48????????????
??????
3
And the beam can modeled as a mass-spring system

35. Dr. Mohamed Sameh Salem
Associate professor, Mechanical Power Engineering
[email protected]
Tel: [phone]