MACHINE DESIGN FORMULAS:
CLUTCHES:
Uniform Wear = old clutch
Uniform Pressure = new clutch
T = torque (N-m) (lbs-in)
Pa = load (N/m2) (pascals)
N−m
s
P = power (
or W) (
ft−lb
or
s
(N)
Hp) = (2π)(60 seconds)(T)
F = Axial force (N) (lbf)
Dm = mean diameter (m) (in.)(ft)
Do = outside diameter (m) (in.)(ft)
ro = outside radius (m) (in.) (ft)
Di = inside diameter (m) (in.) (ft)
ri = inside radius (m) (in.) (ft)
μ = coefficient of friction
z = contact pairs = z1 + z2 -1
z1 = # of disk on drive shaft
z2 = # of disk on driven shaft
L = Face width or length of conical element in contact
α = cone angle
ω = ? (rad/s)
rg = radius of the center of gravity (m)
rd = radius of the inside drum (m)
m = mass (kg)
FCF = centrifugal force acting on each shoe (N)
n = number of shoes
l = Contact length of shoes (m)
b = Width of the shoes (m)
ft−lb
s
note:1Hp=550
NEW PLATE CLUTCHES
OLD PLATE CLUTCHES
Use Uniform Pressure if not stated in problem
which to use
Use Uniform Wear if problem stated a mean
diameter
T=
F=
L=
(z)(π)(μ)(Pa )
12
(π)(Pa )
4
(Do 3 − Di 3 )
T=
(Do 2 − Di 2 )
F=
Do −Di
2
L=
1(D 3 −D 3 )
Frictional Radius = 3(Do 2 −Di2 )
o
(z)(μ)(π)(Pa )(Di )
8
(π)(Pa )(Di )
2
(Do 2 − Di 2 )
(Do − Di )
Do −Di
2
Frictional Radius =
i
Do +Di
4
Do +Di
2
Dm=
NEW CONE CLUTCHES
OLD CONE CLUTCHES
Uniform Pressure
Uniform Wear
*assign X = other diameter in calculator*
*assign X = other diameter in calculator*
F=
T=
(π)(Pa )
4
(F)(μ) 1(Do 3 −Di 3 )
[
]×
sin(α) 3(Do 2 −Di 2 )
D −D
o
i
L = 2sin
(α)
Use Uniform Wear if problem stated a mean
diameter
(Do 2 − Di 2 )
z
D
2
F = (π)(Pa )( i)(Do − Di )
(F)(μ) (Do +Di )
]×
4
T = sin(α) [
D −D
o
i
L = 2sin
(α)
(F)(μ) Dm
]×
2
z = sin(α) [
z
CENTRIFUGAL CLUTCHES
*Use
(θ)(π)
180
to convert degrees to radians*
ω1 = 2πN1
ω2 = 2πN2
T = (n)(μ)(m)(rg )(rd )(ω2 2 − ω1 2 )
FCF1 = (m)( ω1 2 )(rg )
FCF2 = (m)( ω2 2 )(rg )
F = (μ)(FCF2 − FCF1 )
l=[
(θ)(π)
180
] (rd )
(Pa )(l)(b) = (m)(rg )(ω2 2 − ω1 2 )
BRAKES:
m = mass (kg)
v1 = initial velocity (m/s)
v2 = final velocity (m/s)
KE = kinetic energy (J)
rad
)
s
ω = angular velocity (
I = mass moment of inertia (kg-m2)
k = radius of gyration
g = gravity (9.807 m/s2)
E = energy that has dissipated through angular displacement (J)
Mt = braking torque (N-m)
θ = angle brake drum rotated or anglular displacement (rad)
t = thickness
d = diameter
ρ = density(kg/m3)
MECHANICAL BRAKES
ENERGY DISSIPATION
KE =
1
× m × (v1 2 − v2 2 )
2
∗ v2 = 0 when full stop ∗
KE =
1
× I × (ω1 2 − ω2 2 )
2
KE =
1
× m × k 2 × (ω1 2 − ω2 2 )
2
Radius of gyration =
d
D2
=
8
√8
= k2
PE = m × g × h
E = Mt × θ
m = A× t × ρ
ρ=
m
volume
volume = A× t
ω=
rpm
×
60s
2π
average velocity =
ω1 + ω2
2
ω1 + ω2
θ=(
)(time in seconds)
2