ALCORCON ENGINEERING REVIEW CENTER
Cebu Main: 4th floor Coast Pacific Downtown Center, Sanciangko St, Cebu City Tel #(032) 254-33-84
Manila: 3rd floor JPD Bldg 1955, C M Recto corner N. Reyes St, Sampaloc, Manila Tel # (02) 736-4438
MACHINE DESIGN – DAY 2
I.
SHAFTS
1. Shaft - is a rotating member that is used to transmit power.
2. Axle - a stationary member carrying rotating wheels, pulleys.
3. Line shaft - transmission shaft driven by prime mover.
4. Machine shaft - shaft which is an integral part of the machine.
5. Counter shaft - transmission shaft intermediate between the line
shaft and the driven machine.
6. Spindles - are short axles and shafts.
7. Transmission shaft - is a shaft used to transmit power between
the source and the machines absorbing the power, and include
countershafts, line shafts, head shafts, and all factory shafting.
Machine Shaft
MOTOR
Main Shaft
Counter Shaft
Driven Machine 1
Driven Machine 2
FORMULAS:
1. Power Formula in SI unit:
P = 2 p T N where: P = power, KW
P =
2.
Torque, T
2pTN where:
33,000
P = power, Hp
N = speed, rps
T = torque, ft-lb
N = speed, rpm
T = FxR
where: F = applied force
3.
T = torque, KN-m
R = radius = D/2
F
R
Stresses in shaft when subjected to pure torsion (Ss)
A. For solid shaft: Ss = Tc = 16T
J
pD3
B. For hollow shaft: Ss =
16TDo
p(Do 4 - Di 4 )
C. For designed shearing stress and compressive stress with given Sy and Su.
SHEARING:
Ss = 0.18 Su,
Ss = 0.3 Sy
COMPRESSIVE: Sc = 0.36 Su
Sc = 0.6 Sy
Note: Choose whichever is smaller
where:
Do = outside diameter
Di = inside diameter
J = polar moment of inertia
c = is the distance of the farthest fiber from neutral axis
T = torque
d = diameter of shaft
4.
Torsional deflection (q), rad
q = TL , rad
where: L = length of shaft
JG
4
p(Do 4 - Di 4 ) (for hollow)
J = polar moment of inertia = pd (for solid shaft)
J =
32
32
G = modulus of rigidity in shear = 11.5 x 106 psi for steel
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FROM MACHINERIES HANDBOOK:
A. For solid shaft:
Θ =
584 T L
D4 G
, deg
B. Using ALCORCON’S FORMULA for hollow shaft: Θ =
Θ
T
L
D
Do, Di
G
Degrees
in-lb
in
in
in
Psi
Degrees
N-mm
mm
mm
mm
MPa
584 T L
(Do 4 - Di4 ) G
, deg
degrees
KN-m
m
m
m
KPa
5. Stress in shaft when subjected to Torsion and Bending loads:
a.
For solid shaft:
Ss = 16 (K mM)2 + (K t T )2
pd3
St =
16 é
2
2ù
êKmM + (KmM) + (K t T ) úû
pd3 ë
Note: Km and Kt not known, assume 1.0
where:
b.
T = torque
M = moment
Ss = maximum shear stress
St = maximum tensile or compressive stress
For hollow shaft:
16Do
Ss =
p(Do 4 - Di4 )
(K mM)2 + (K t T )2
St =
16Do
4
4
p(Do - Di )
é
2
2ù
êëKmM + (KmM) + (K t T ) úû
6. Strength of shaft with assumed allowable stresses (PSME Code p. 18)
a. For Main Shafts: (S = 4000 psi)
3
P = D N
80
b. For Line Shafts: (S = 6000 psi)
P =
c. For short shafts: (S = 8500 psi)
3
P = D N
38
where: P = power, Hp
D3N
53.5
N = speed, rpm
D = diameter, inch
7. From Machineries Handbook Formula
A. Shaft diameter for 0.08 degrees per foot of length of shaft deflection.
For English units:
Where: D = diameter, in
P = horsepower
D = 0.29 4 T
or D = 4.6
4
P
N
T = torque, in-lb
N = speed, rpm
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D = 2.26 4 T
For SI units:
Where: D = diameter, mm
N = speed, rpm
or D = 125.70
4
P
N
P = power, watts
T = torque, N-mm
B. Shaft deflection of 1 degree for a length of 20 times its diameter.
D = 0.10 3 T
or D = 4.0
Where: P = power, hp
N = speed, rpm
3
P
N
D = diameter, in
T = torque, in-lb
C. Linear deflection of shafting
a. Shafting subjected to no bending action of pulleys except its own weight
b. Shafting subjected to bending action of pulleys, etc
Where: L = shaft length, ft
II.
L = 8.95 3 D2
L = 5.2 3 D2
D = shaft diameter, in
KEYS
A key is a machine member employed at the interface of a pair of
mating male and female circular cross-sectional members to prevent
relative angular motion between these mating members.
Types of keys:
1. Square key
2. Flat key
3. Round key
4. Barth key
5. Woodruff key
6. Gib-head key
7. Saddle key
8. Kennedy key
9. Feather key
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FORMULAS:
1. Power of key:
P = 2p T N, KW
2. Force transmitted, F
where:
F = T = T
r d/ 2
d = shaft diameter
Fc
(h / 2)L
3. Compressive Stress (Sc) of key Sc =
4. Shearing Stress (Ss) of key Ss =
Fs
wL
where: L = length of key
w = width of key
h = height of key
5. Relation of key and shaft for the same material:
w =
D
4
L = 1.2 D
6. Force tangent to pulley rim
T = T’
F . r = F’ . R
7. Ff = force tends to remove key from the hub and shaft = 2 f F
F = force tangent to the key
F’ = force tangent to pulley rim
R = radius of pulley
III.
SPLINE SHAFT
Is recommended when the power transmitted is too high that three keys are not enough.
1. Types of Fits
Types of
Fits
Permanent
Fit
To slide
when not
under load
To slide
when
under load
6 Splines
4 Splines
10 Splines
d = 0.9D
w = 0.25D
h = 0.05D
d = 0.85D
w = 0.25D
h = 0.075D
d = 0.80D
w = 0.25D
h = 0.10D
d = 0.85D
w = 0.241D
h = 0.075D
d = 0.75D
w = 0.24D
h = 0.125D
d = 0.91D
w = 0.156D
h = 0.045D
d = 0.86D
w = 0.156D
h = 0.07D
d = 0.81D
w = 0.156D
h = 0.095D
Where: d = minor diameter, in or mm
D = major diameter, in or mm
r = radius, in or mm = d/2; R = D/2
w = width, in or mm
L = length of spline
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2. Rm = mean radius = r + R
2
Note:
If L is not given use the recommended, L = 1.5 D,
from Vallance, Machine Design
3. Calculations for safe dimensions of splines:
3.1 Based on shearing of splines
Ss =
F
F
=
A s Ns w L
Since Ts = Fæç d ö÷ or F = 2 T
d
è2ø
3.2 Based on Compression between splines and hub.
Sc =
Flange
F
F
=
Ac h L Ns
T
F
NS = number of splines
Dc
Ac = compressive area
r +R )
Tc = F Rm = F(
2
or F = 2Tc / (
r +R)
2
Shaft
Bolt
F = compressive force
3.3. Based on Torsion of splined Shaft
Ss =
Note: For torque capacity, T for one spline is:
16 Tt
p d3
T=
Tt
(1.1)
Ns
This recommendation is based by shearing one spline only.
IV.
T
FLANGE COUPLING
Fb
1. Coupling - is a mechanical device which is used to connect length of shafts permanently.
Dc
FORMULAS:
1. Power transmitted:
P = 2 p T N , KW
2. Total force transmitted, (F)
F =
T
T
=
r Dc / 2
Shear Area
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3. Force transmitted per bolt, (Fb) Fb =
F
n
Flange
where: n = no. of bolts Dc = bolt circle diameter
4. Shearing of bolts(Ss)
Ss =
Fb
p 2
d
4
Dc
5. Compressive stress on bolts and flange (Sc)
F
Sc = b
td
where: t = thickness of flange
Compressed Area
d = bolt diameter
PROBLEMS:
SHAFTING
1.
A line shaft is to transmit 200 HP at 900 rpm. Find the diameter of the shaft.
A. 2.18 in
B. 2.28 in
C. 3.18 in
D. 3.28 in
2.
A 1/4 in diameter solid shaft has an allowable stress of 6000 psi and is subjected to pure torsion. If its rotational
speed is 1800 rpm, how much horsepower can it transmitted safely?
A. 0.425 hp
B. 0.225 hp
C. 0.316 hp
D. 0.525 hp
3.
Determine the thickness of hollow shaft having an outside diameter of 100 mm if it is subjected to a maximum
torque of 5,403.58 N-m without exceeding a shearing stress of 60 MPa or a twist of 0.5 degree per meter length
of shaft G=83,000 MPa
A. 15 mm
B. 86 mm
C. 76.8 mm
D. 69.96 mm
4.
An 80 mm solid shaft is to be replaced with a hallow shaft of equal torsional strength. Find percentage of weight
saved, if the outside of the hallow shaft is 100 mm.
A. 56.53%
B. 67.31%
C. 48.49%
D. 52.90%
5.
A round steel shaft to a torque 200 rpm and is subjected to a torque of 226 N-m, the allowable shearing stress
41.4 MPa. It is also subjected to a bending moment of 339 N-m. The allowable tensile stress is 55 MPa. Find the
diameter.
A. 37 mm
B. 41 mm
C. 45 mm
D. 51 mm
KEYS & COUPLING
1.
A rectangular key was used in a pulley connected to a line shaft with a power of 10 KW at a speed of 1200 rpm.
If the shearing stress of the shaft and key are 30 N/mm2 and 240 N/mm2, respectively. What is the key length if
the width is 10 mm?
A. 18.7 mm
B. 21.7 mm
C. 25.8 mm
D. 30.2 mm
2.
A flange bolt coupling consists of eight steel 20 mm diameter steel bolts spaced evenly around a bolt circle 300
mm in diameter. if the coupling is subjected to a torque of 15.1 KN-m, determine the maximum shearing stress in
the bolts?
A. 40450 kPa
B. 63320 kPa
C. 40054 kPa
D. 31298 kPa
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3.
A flange coupling having 180 mm bolt circle and 19 mm thick uses 8 bolts, 16 mm diameter to connect two shafts.
It is used to transmit 60 kw at 180 rpm. Determine the factor of safety in compression if yield point in compression
is 448 MPa.
A. 15.6
B. 30.8
C. 18.5
D. 25.4
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