Machine Design I(EME-501)
UNIT 3
Dr. Suvandan Saraswat
1
Shafts
• Connect power transmission components.
• Inherently subjected to transverse loads and
torsion.
Introduction
In machinery, the general term “shaft” refers to a member, usually of
circular cross-section, which supports gears, sprockets, wheels, rotors,
etc., and which is subjected to torsion and to transverse or axial loads
acting singly or in combination.
An “axle” is a non-rotating member that supports wheels, pulleys,… and
carries no torque.
A “spindle” is a short shaft. Terms such as lineshaft, headshaft, stub
shaft, transmission shaft, countershaft, and flexible shaft are names
associated with special usage.
Causes of Failure
• Due to Axial load
• Due to torsion load
• Due to bending load
• Due to combination of above
• Due to fatigue load
Considerations for Shaft Design
1. Deflection and Rigidity
(a) Bending deflection
(b) Torsional deflection
(c) Slope at bearings and shaft supported elements
(d) Shear deflection due to transverse loading of shorter shafts
2. Stress and Strength
(a) Static Strength
(b) Fatigue Strength
(c) Reliability
Stresses
Combined Axial, Bending and Torsion
ASME Code Design
Fatigue Strength
Ma= (Mmax-Mmin)/2, Mm=(Mmax+Mmin)/2
Ta= (Tmax-Tmin)/2, Tm=(Tmax+Tmin)/2
The fluctuating stresses due to bending and torsion are given by
Assuming a solid shaft with round cross section, appropriate geometry terms can be
introduced for c, I, and J resulting in
For Soderbergs Criteria (0.6y = ye)
(a y / e + m y/ y ) = y /n
n( a / 0.6+ m ) = eq
similarly( 0.4τy = τ e)
n(τa / 0.4 + τ m ) = τeq
For Goodman’s Criteria
(a u / e + m u/ u ) = u /n
(0.5u = e)
(2 a + m ) = eq
similarly ( 0.3τy = τ e)
Design of Shaft for Torsional
Rigidity
It is based on the permissible angle of twist. The amount of twist permissible depends
on the particular application, and varies about 0.3 degree/m for machine tool shafts to
about 3.0 degree/m for line shafting.
MtL
584 4
For solid shaft
Gd
MtL
584 For hollow shaft
4
G do di
4
Where:
angleof twist, deg
L length of the shaft, m
G Torsional modulus of elasticity, N m2
21
Design of Key
if
Q. A shaft 30mm diameter is transmitting power at a
maximum shear stress of 80N/mm2. If a pulley is
connected to the shaft by means of key, find the
dimension of the key so that the shear stress in the key
is not to exceed 50N/mm2 and the length of key is 4
times of width the key.
Q. Design a square key for fixing a gear on the shaft
having 25 mm diameter. The rotates at 550 rpm and
transmits 12 kW power to the meshing gear. The key is
made of steel having yield stress in tension as 400 Mpa.
The yield stress in compression and tension may be
taken equal to each other. Assume factor of safety 2.5.
Couplings
Rigid couplings
Sleeve coupling
Sleeve / muff coupling Design
Torque T = π τsleeve (do4-d4)/16d
do=2d+13
L=3d
Length of key l= L/2
T=w l τkey d/2 = t l σckey d/4
Shaft
T= π τs d3/16
Force on each bolt F = π db2 σt/4
Force exerted on each shaft = F n/2 = π σt db2 /4* n/2
(n is no. of bolt)
Let p be the pressure on shaft sleeve by
p =force/projected area=F/(Ld/2)
= (π db2 σt/4* n/2) /(Ld/2)
Frictional force between each shaft and muff = μ p (πdL/2)
= (μ π2 σt db2 n /8)
Torque
T= frictional force *d/2
=(μ π2 σt db2 nd /16)
(The design of sleeve, key and shaft will be same as for sleeve coupling)
Length of hub L=3d
Diameter of hub d1=2d+13
Pitch circle diameter of bolt d2= 3d
Outside diameter of flange d3 = 4d
Thickness of flange t2=0.5d
No. of bolt n = 2 for d up to 40mm
=4 for d up to 100 mm
Design of Shaft
T= π τs d3/16
Design of Key
Length of key l= L/2
T=w l τkey d/2 = t l σckey d/4
Design of Hub
T = π τsleeve (d14-d4)/16d
Thickness of the flange
T= π d1t2 τflnge d1/2
Diameter of bolt
Shear force on each bolt F = π db2 τbolt/4
T= F *n*d2/2 =π db2 τbolt*n*d2/8
Crushing force on bolt
F=n σt dbt2
T = n σt dbt2 *d2/2
Flexible rubber – bushed pin couplings
dbc= diameter of pin, dc = diameter of bush
l = length of bush in each flange
Bearing load on each pin W = Pb dr l
Total bearing WT = Pb dr l n
T = WT dc/2= Pb dr l n dc/2
τ = 4WT /π dbc2
M = WT l/2, σ = M/Z= 32M/π dbc3
Q. Design a muff coupling to connect two shafts transmitting 40kW at 150rpm.
The allowable shear and crushing stresses for the shaft and key are 37N/mm2
and 96.25N/mm2 respectively. The permissible shear stress for the muff is
17.5N/mm2 . Assume that the maximum torque transmitted is 20% more than
the mean torque. Take the width and depth of the parallel key is 22mm and
14mm respectively.
Q. Design clamp coupling to transmit 25 kW at 100 rpm. The allowable shear
stress for shaft and key is 40 MPa and no of bolts connecting the two halves is
six. The coefficient of friction between the muff and the shaft is 0.25. the
permissible tensile stress for the bolts is 70 Mpa.
Q. Design a cast iron protected type flange coupling to connect two shafts of 36
mm diameter, transmitting 15 kW at 720 rpm. The overload capacity is 1.25
times the average torque. The bolts and key are made of C20 steel and flanges
are made of cast iron.
