Kinematics, Dynamics and Vibrations

Dr. Mustafa Arafa Mechanical Engineering Department American University in Cairo [email protected] Kinematics, dynamics and vibration

• Kinematics: study of motion (displacement, velocity, acceleration, time) without reference to the cause of motion (i.e. regardless of forces). • Dynamics: study of forces acting on a body, and resulting motion. • Vibration: Oscillatory motion of bodies & associated forces. Outline

A. Kinematics of mechanisms B. Dynamics C. Vibration: natural frequency and resonance D. Balancing Kinematics of mechanisms Four-bar mechanism

Single Degree of Freedom Slider-crank mechanism

Single Degree of Freedom Position analysis

Given: a,b,c,d, the ground position, q2.

Find: q3 and q4 b B c

b A q3 c

q4 q2 d

O2 O4 Graphical solution

• Draw an arc of radius b, c centered at A • Draw an arc of radius c, B1 centered at O4 b • The intersections are the A q3 two possible positions for c

the linkage, open and a

d q4 crossed q2

O2 O4

B2 Analytical solution Obtain coordinates of point A:

Ax  acosq 2

Ay  asinq 2 Obtain coordinates of point B:

2 2 2 b  Bx  Ax   By  Ay  2 2 2 c  Bx  d  By

These are 2 equations in 2 unknowns: Bx and By

See “position analysis” on page 242 Dynamics Types of motion

Rectilinear Curvilinear Overview

Kinematics: equations for constant velocity and acceleration

 d  Kinetics: Newton’s second law of motion: F  mv   dt   For constant mass: F ma

Kinetic energy: 1 2 T 2 mv Potential energy: U mgh Gravity

1 2 U 2 kx Elastic Friction

F P N Basic equations Projectile Projectile y v

a g

v x a a v Plane Motion of a Rigid Body

 Fx ma x  F y  ma y  M G  I g

For rotation about a fixed axis:   MIOO 

17 Example

At a forward speed of 30 ft/s, the truck brakes were applied, causing the wheels to stop rotating. It was observed that the truck skidded to a stop in 20 ft. Determine the magnitude of the normal reaction and the friction force at each wheel as the truck skidded to a stop.

18 Solution

22 v v00 2 a x  x  0 302 2a 20 a 22.5 ft s

Free-body diagram:

F ma  x Gx FAB F  m 22.5 F ma  y Gy W mg  NAB  N MI   GG 7NFFNBBAA 4  4  5

But FNFNAABB,

Unknowns: NNAB,, N N  m 22.5 AB   0.699

NAB N mg

19 NAB0.35 mg , N 0.65 mg Vibrations Natural frequency and resonance Overview f Equation of motion: m mx cx   kx  f x k c Natural frequency:n  km  2 f rad/s Hz

 cm/2 n  cccritical  1 overdamping

Damping Damping cccritical  1 critical damping ratio coeff. cccritical  1 underdamping Free vibrations with no damping Equation of motion: m mx  kx 0 x k Solution:

x C12cosnn t C sin t

C1 and C2 are constants to be determined from initial conditions x0 and v0

x x00cosn t v  n sin  n t Model of a vibrating system Spring and mass Spring, mass and damper Forced vibration Balancing of machinery Static unbalance Couple unbalance Dynamic unbalance