Dynamics of-machines-formulae-sheet
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The document provides formulae for various mechanical systems including friction, screw threads, clutches, disk brakes, belt drives, bearings, gears, flywheels, rotors, and reciprocating machinery. Key formulae include equations for frictional force, torque required to tighten a nut, axial force in a clutch or brake given a pressure distribution, transmitted power in a belt drive, bearing characteristic number, gear module and ratios, flywheel energy fluctuations given input and output torques, out-of-balance forces and moments in rotors, and primary and secondary out-of-balance forces and moments in reciprocating systems.
![1
2
1
2
2
1
N
N
D
D
p
p
==−
ω
ω
where ω1,2 are their angular velocities (measured in the
same sense on both gears), D1,2 are their pitch circle
diameters and N1,2 are their numbers of teeth.
If a gearbox with multiple input and output shafts
denoted by the subscript i, is assumed lossless then:
0=∑i
iiT ω
where the torque, Ti, and angular velocity, ωi, on a
particular shaft are measured in the same direction.
An epicyclic gearbox operating at constant speed must
be in moment equilibrium, hence:
0=++ csa TTT
where Ta, Ts and Tc are the torques on the annulus, sun
and carrier respectively, measured in the same sense.
FLYWHEELS
For a flywheel rotating from angle θ1 to θ2, subjected to
input and output torques, Tin and Tout respectively:
( ) ( ) ( )[ ] θθθωω
θ
θ
dTTI outin∫ −=−
2
1
2
1
2
22
1
where I is the moment of inertia of the flywheel and ω1
and ω2 are its angular velocities at angles θ1 and θ1
respectively.
The maximum fluctuation in kinetic energy (MFKE)
over one revolution is defined as:
( ) ωωωω ∆≈−= meanIIMFKE 2
min
2
max2
1
where ωmean, ωmin and ωmax are the mean, minimum and
maximum angular velocities respectively and ∆ω =
ωmax - ωmin. The maximum percentage fluctuation in
speed (MPFS) is defined as:
%100minmax
×
−
=
ω
ωω
MPFS
ROTORS
The amplitude of the out-of-balance rotating force, F,
generated by a rotor represented as a number of point
masses, mi, is given by:
( ) ( )222
sincos iiiiii rmrmF θθω ∑∑ +=
where ω is the angular velocity and ri and θi are the
radial and angular positions of the ith
mass.
The amplitude of the out-of-balance rotating moment,
M is given by:
( ) ( )222
sincos iiiiiiii armarmM θθω ∑∑ +=
where ai is the axial position of the ith
mass.
RECIPROCATING MACHINERY
The oscillating force, F, required to accelerate and
decelerate a slider of mass m in a slider-crank
mechanism is usually approximated by:
+−= t
q
tmrF ωωω 2cos
1
cos2
where r is the length of the crank, ω is the angular
velocity of the crank, t is time and q is the ratio of
connecting rod length, d, to crank length (i.e. q = d / r).
The two terms are usually considered separately as
primary and secondary out-of-balance forces. The
amplitude of the primary out-of-balance force, F1, is
therefore:
2
1 ωmrF =
and the amplitude of the secondary out-of-balance
force, F2, is given by:
q
mr
F
2
2
ω
=
In a machine with multiple parallel sliders driven or
driving a common crankshaft, the amplitude of the total
out-of-balance primary force, F1
Total
, is given by:
( ) ( )222
1 sincos ∑∑ += iiiiii
Total
rmrmF θθω
where mi, ri and θi are the slider masses, crank lengths
and relative angular positions of the cranks for each
slider. Similarly, the amplitude of the total out-of-
balance secondary force, F2
Total
, is given by:
22
2
2
2sin2cos
+
= ∑∑
i
iii
i
iiiTotal
q
rm
q
rm
F
θθ
ω
where qi are the connecting rod to crank length ratios
for each slider. The amplitudes of the total out-of-
balance primary moment, M1
Total
, and secondary
moment, M2
Total
, are respectively:
( ) ( )222
1 sincos ∑∑ += iiiiiiii
Total
armarmM θθω
and
22
2
2
2sin2cos
+
= ∑∑
i
iiii
i
iiiiTotal
q
arm
q
arm
M
θθ
ω
where ai is the axial position of the slider along the
crankshaft.](https://image.slidesharecdn.com/dynamics-of-machines-formulae-sheet-151126183423-lva1-app6892/85/Dynamics-of-machines-formulae-sheet-2-320.jpg)
Dynamics of-machines-formulae-sheet
- 1. Dynamics of Machines Formulae FRICTION For solid friction the frictional force, F, at the point of slip is given by: NF µ= where µ is the coefficient of friction and N is the normal contact force SCREW THREADS The torque, T, required to tighten a nut against an axial force, W, is given by: 2sincos cossin md WT − + = φµφ φµφ where φ is the helix angle, µ is the coefficient of friction and dm is the mean thread diameter. The torque to loosen the nut is given by: 2sincos cossin md WT + +− = φµφ φµφ The efficiency of tightening, η, is given by: T Wt π η 2 = where t is the pitch of the thread. CLUTCHES AND DISK BRAKES The axial force, W, associated with a radial pressure distribution, P(r), between one pair of contacting surfaces in a clutch or disk brake (assuming that the brake pad is a sector of an annulus) is: ( )∫= 2 1 R R drrrPW θ where θ is the angle subtended by the brake pad (θ = 2π for a clutch), r is the radial position, R1 is the inner radius of the brake pad or clutch and R2 is its outer radius. Similarly, the maximum torque, T, at the point of slip due to one pair of contacting surfaces in a clutch or disk brake is given by: ( )∫= 2 1 2 R R drrPrT θµ where µ is the coefficient of friction. In a new clutch or disk brake the pressure distribution can be assumed to be equal to a constant pressure, P0, so: ( ) 0PrP = For a worn clutch or disk brake a common assumption is that the pressure distribution is inversely proportional to radial position: ( ) r k rP = where k is a constant. BELT DRIVES For a belt or rope wrapped around a drum or pulley, the maximum possible ratio of tensions between the tight, T1, and slack, T2, sides at slip is given by: L e T T µφ = 2 1 where µ is the coefficient of friction and φL is the angle of lap. If the rope and pulley are rotating then dynamic effects should be included and the previous equation becomes: L e mvT mvT µφ = − − 2 2 2 1 where m is the mass per unit length of the belt and v is the belt velocity. The power, P, transmitted by a belt drive is given by: ( )21 TTvP −= and the initial belt tension, T0, is related to the running belt tensions by: 2 21 0 TT T + = BEARINGS The bearing characteristic number, n, for a journal bearing may be defined (other definitions possible) as: P n γω = where γ is the viscosity of the lubricant, ω is the angular velocity of the shaft and P is the nominal pressure (the radial load divided by projected area). GEARS The module, m, of a gear wheel is defined as: π P N D m p == where Dp is the pitch circle diameter, N is the number if teeth and P is the circular pitch. In order for a pair of gears to mesh correctly, they must have the same module. For a pair of meshing spur gears:
- 2. 1 2 1 2 2 1 N N D D p p ==− ω ω where ω1,2 are their angular velocities (measured in the same sense on both gears), D1,2 are their pitch circle diameters and N1,2 are their numbers of teeth. If a gearbox with multiple input and output shafts denoted by the subscript i, is assumed lossless then: 0=∑i iiT ω where the torque, Ti, and angular velocity, ωi, on a particular shaft are measured in the same direction. An epicyclic gearbox operating at constant speed must be in moment equilibrium, hence: 0=++ csa TTT where Ta, Ts and Tc are the torques on the annulus, sun and carrier respectively, measured in the same sense. FLYWHEELS For a flywheel rotating from angle θ1 to θ2, subjected to input and output torques, Tin and Tout respectively: ( ) ( ) ( )[ ] θθθωω θ θ dTTI outin∫ −=− 2 1 2 1 2 22 1 where I is the moment of inertia of the flywheel and ω1 and ω2 are its angular velocities at angles θ1 and θ1 respectively. The maximum fluctuation in kinetic energy (MFKE) over one revolution is defined as: ( ) ωωωω ∆≈−= meanIIMFKE 2 min 2 max2 1 where ωmean, ωmin and ωmax are the mean, minimum and maximum angular velocities respectively and ∆ω = ωmax - ωmin. The maximum percentage fluctuation in speed (MPFS) is defined as: %100minmax × − = ω ωω MPFS ROTORS The amplitude of the out-of-balance rotating force, F, generated by a rotor represented as a number of point masses, mi, is given by: ( ) ( )222 sincos iiiiii rmrmF θθω ∑∑ += where ω is the angular velocity and ri and θi are the radial and angular positions of the ith mass. The amplitude of the out-of-balance rotating moment, M is given by: ( ) ( )222 sincos iiiiiiii armarmM θθω ∑∑ += where ai is the axial position of the ith mass. RECIPROCATING MACHINERY The oscillating force, F, required to accelerate and decelerate a slider of mass m in a slider-crank mechanism is usually approximated by: +−= t q tmrF ωωω 2cos 1 cos2 where r is the length of the crank, ω is the angular velocity of the crank, t is time and q is the ratio of connecting rod length, d, to crank length (i.e. q = d / r). The two terms are usually considered separately as primary and secondary out-of-balance forces. The amplitude of the primary out-of-balance force, F1, is therefore: 2 1 ωmrF = and the amplitude of the secondary out-of-balance force, F2, is given by: q mr F 2 2 ω = In a machine with multiple parallel sliders driven or driving a common crankshaft, the amplitude of the total out-of-balance primary force, F1 Total , is given by: ( ) ( )222 1 sincos ∑∑ += iiiiii Total rmrmF θθω where mi, ri and θi are the slider masses, crank lengths and relative angular positions of the cranks for each slider. Similarly, the amplitude of the total out-of- balance secondary force, F2 Total , is given by: 22 2 2 2sin2cos + = ∑∑ i iii i iiiTotal q rm q rm F θθ ω where qi are the connecting rod to crank length ratios for each slider. The amplitudes of the total out-of- balance primary moment, M1 Total , and secondary moment, M2 Total , are respectively: ( ) ( )222 1 sincos ∑∑ += iiiiiiii Total armarmM θθω and 22 2 2 2sin2cos + = ∑∑ i iiii i iiiiTotal q arm q arm M θθ ω where ai is the axial position of the slider along the crankshaft.