Lecture 19, 20, 21 - Design of Machine Elements

(ME314)

Instructor - Dr. B.S. Reddy

Mechanical Engineering, IIT Guwahati

09 Sep 2019

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Instructor - Dr. B.S. Reddy (IIT Guwahati) Lecture 19, 20 ,21 09 Sep 2019 1 / 19
 Shaft

A shaft is a rotating member, usually of circular cross section, used to

transmit power or motion.
The power is delivered to the shaft by some tangential force and the
resultant torque (or twisting moment) set up within the shaft permits
the power to be transferred to various machines linked up to the shaft.
In order to transfer the power from one shaft to another, the various
members such as pulleys, gears etc., are mounted on it. These
members along with the forces exerted upon them causes the shaft to
bending.
Types of shaft
Transmission shafts - Transmit power between the source and machines
absorbing power. The counter shafts, line shafts, over head shafts and
all factory shafts are transmission shafts. They are subjected to
bending in addition to twisting.
These shafts form an integral part of the machine itself. Eg. The crank
shaft . . . . . . . . . . . . . . . . . . . .
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Instructor - Dr. B.S. Reddy (IIT Guwahati) Lecture 19, 20 ,21 09 Sep 2019 2 / 19
 Design of Shafts
1 Strength
Shafts subjected to twisting moment or torque only,
Shafts subjected to bending moment only,
Shafts subjected to combined twisting and bending moments
Shafts subjected to axial loads, torsional and bending loads.
2 Rigidity -
Torsional - does not twist too much under action of an external torque
Lateral - does not deflect too much under action of external forces and
bending moment

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Instructor - Dr. B.S. Reddy (IIT Guwahati) Lecture 19, 20 ,21 09 Sep 2019 3 / 19
 Design of Shafts
1 Strength
Shafts subjected to twisting moment or torque only,
Shafts subjected to bending moment only,
Shafts subjected to combined twisting and bending moments
Shafts subjected to axial loads, torsional and bending loads.
2 Rigidity -
Torsional - does not twist too much under action of an external torque
Lateral - does not deflect too much under action of external forces and
bending moment

Design based on Torsional Rigidity (Eg. M/c Tool spindles)

Design - permissible angle of twist per metre length of shaft.
Mt l
Angle of twist (rad) - θr =
JG
584Mt l
For solid circular shaft (degrees) - θr =
Gd4
Permissible angle of twist - (i) M/c Tool - 0.250 /meter length, (ii) Line
shaft - 30 /meter length .
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Instructor - Dr. B.S. Reddy (IIT Guwahati) Lecture 19, 20 ,21 09 Sep 2019 3 / 19
 Shaft Design on Strength Basis
Tensile Stress - σaxial = 4P/(πd2 ), Bending - σb = 32Mb /(πd3 ),
Torsion - τ = 16Mt /(πd3 )
Maximum Principal Stress Theory
Maximum Principal
√( Stress -
)2 ( ) √
16Mb 16Mb 16Mt 2 16
σ1 = + + = [M b + M2b + M2t ]
πd3 πd3 πd3 πd3
Syt
Permissible value of max principal stress - σ1 =
fs

Maximum Shear Stress Theory

Maximum√(Shear Stress -
)2 ( )
16Mb 16Mt 2 16 √ 2
τmax = + = [ Mb + M2t ]
πd3 πd3 πd3
Ssy 0.5Syt
Permissible value of max shear stress - τmax = =
fs fs .
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Instructor - Dr. B.S. Reddy (IIT Guwahati) Lecture 19, 20 ,21 09 Sep 2019 4 / 19
 Problem - The layout of a shaft carrying two pulleys 1 and 2, and
supported on two bearings A and B is shown in Fig. The shaft transmits
7.5 kW power at 360 rpm from the pulley 1 to the pulley 2. The diameters
of pulleys 1 and 2 are 250 mm and 500 mm respectively. The masses of
pulleys 1 and 2 are 10 kg and 30 kg respectively. The belt tensions act
vertically downward and the ratio of belt tensions on the tight side to slack
side for each pulley is 2.5:1. The shaft is made of plain carbon steel 40C8
(Syt = 380 N/mm2 ) and the factor of safety is 3. Estimate suitable
diameter of shaft. If the permissible angle of twist is 0.50 per metre
length, calculate the shaft diameter on the basis of torsional rigidity.
Compare with calculation on strength basis. Assume G = 79300 N/mm2 .

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Instructor - Dr. B.S. Reddy (IIT Guwahati) Lecture 19, 20 ,21 09 Sep 2019 5 / 19
 Problem - The layout of a transmission shaft carrying two pulleys B and
C and supported on bearings A and D is shown in Fig. Power is supplied
to the shaft by means of a vertical belt on the pulley B, which is then
transmitted to the pulley C carrying a horizontal belt. The maximum
tension in the belt on the pulley B is 2.5 kN. The angle of wrap for both
the pulleys is 1800 and the coefficient of friction is 0.24. The shaft is made
of plain carbon steel 30C8 (Syt = 400 N/mm2 ) and the factor of safety is
3. Determine the shaft diameter on strength basis.

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Instructor - Dr. B.S. Reddy (IIT Guwahati) Lecture 19, 20 ,21 09 Sep 2019 6 / 19
 Keys
Key - machine element which is used to connect the transmission shaft to
rotating machine elements like pulleys, gears, sprockets or flywheels

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Instructor - Dr. B.S. Reddy (IIT Guwahati) Lecture 19, 20 ,21 09 Sep 2019 7 / 19
 Keys...Contd

The primary function - transmit the torque from the shaft to the hub
of the mating element and vice versa.
The second function - prevent relative rotational motion between the
shaft and the joined machine element like gear or pulley. In most of
the cases, the key also prevents axial motion between two elements,
except in case of feather key or splined connection.
A recess or slot machined either on the shaft or in the hub to
accommodate the key is called keyway
Failure - shear and crushing failure
Keys are made of plain carbon steels like 45C8 or 50C8
Indian Standards - steel of tensile strength not less than 600 N/mm2
shall be used as the material for the key.
Selection - Power to be transmitted, tightness of fit, stability of
connection, cost.
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Instructor - Dr. B.S. Reddy (IIT Guwahati) Lecture 19, 20 ,21 09 Sep 2019 8 / 19
 Sunk Keys
Sunk key - key in which half the thickness of the key fits into the keyway
on the shaft and the remaining half in the keyway on the hub. Hence,
keyways are required both on the shaft as well as the hub of the mating
element.
Industrial Practice - to use a square key with sides equal to one-quarter
of the shaft diameter and length at least 1.5 times the shaft diameter

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Instructor - Dr. B.S. Reddy (IIT Guwahati) Lecture 19, 20 ,21 09 Sep 2019 9 / 19
 Design of Flat Keys

Mt 2Mt
P= =
d/2 d . . . . . . . . . . . . . . . . . . . .
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Instructor - Dr. B.S. Reddy (IIT Guwahati) Lecture 19, 20 ,21 09 Sep 2019 10 / 19
 Failure due to Shear

P
τ= =
Area of plane AB
P 2Mt
=
bl dbl

Crushing Failure

P
σc = =
Area of surface AC
P 4Mt
=
(h/2)l dhl

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Instructor - Dr. B.S. Reddy (IIT Guwahati) Lecture 19, 20 ,21 09 Sep 2019 11 / 19
 Problem - The standard cross-section for a flat key, which is fitted on a
50 mm diameter shaft, is 16 × 10 mm. The key is transmitting 475 N-m
torque from the shaft to the hub. The key is made of commercial steel
(Syt = Syc = 230 N/mm2). Determine the length of the key, if factor of
safety is 3.

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Instructor - Dr. B.S. Reddy (IIT Guwahati) Lecture 19, 20 ,21 09 Sep 2019 12 / 19
 Problem - The standard cross-section for a flat key, which is fitted on a
50 mm diameter shaft, is 16 × 10 mm. The key is transmitting 475 N-m
torque from the shaft to the hub. The key is made of commercial steel
(Syt = Syc = 230 N/mm2). Determine the length of the key, if factor of
safety is 3.
Answer
Given - Mt = 475 N − m, Syt = Syc = 230 N/mm2 , fs=3, d=50 mm,
b=16 mm, h=10 mm

STEP I - Permissible Compressive and Shear Stresses

Syc 230 0.5Syt 115
σc = = = 76.67 N/mm2 , τ = = = 38.33 N/mm2
fs 3 fs 3
2Mt 2(475 × 103 )
STEP II (Key Length) - (l)1 = = = 30.98 mm,
τ db 38.33 × 50 × 16
4Mt 4(475 × 103 )
(l)2 = = 49.56 mm.
τ dh 76.77 × 50 × 10
Length of key ≈ 50 mm . . . . . . . . . . . . . . . . . . . .
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Instructor - Dr. B.S. Reddy (IIT Guwahati) Lecture 19, 20 ,21 09 Sep 2019 12 / 19
 ASME Code for Shaft Design
1 For shaft without Keyways
τmax = (0.3Syt , 0.18Sut )min
2 For shaft with keyways, reduce above values by 25 percent

16 √
σ1 = 3
[kb Mb + (kb Mb )2 + (kt Mt )2 ]
πd
16 √
τmax = [ (kb Mb )2 + (kt Mt )2 ]
πd3 .
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Instructor - Dr. B.S. Reddy (IIT Guwahati) Lecture 19, 20 ,21 09 Sep 2019 13 / 19
 Problem - The layout of an intermediate shaft of a gear box supporting
two spur gears B and C is shown in Fig. The shaft is mounted on two
bearings A and D. The pitch circle diameters of gears B and C are 900 and
600 mm respectively. The material of the shaft is steel FeE 580 (Sut =
770 and Syt = 580 N/mm2). The factors kb and kt of ASME code are 1.5
and 2.0 respectively. Determine the shaft diameter using the ASME code.
Assume that the gears are connected to the shaft by means of keys.

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Instructor - Dr. B.S. Reddy (IIT Guwahati) Lecture 19, 20 ,21 09 Sep 2019 14 / 19
 Design for lateral rigidity

1 A component is considered as rigid when it does not deflect or bend

too much due to external forces or moments
2 Permissible deflection is the criterion for design
3 The maximum permissible deflection for the transmission shaft -
0.001 L ≤ δ ≤ 0.003 L
4 Max permissible deflection at gear - δ = 0.01 m, m is module of gear
teeth
5 Lateral Rigidty can be improved by
reduce the span length
increase the number of supports
reduce the number of joints
assemble the components with pre-load
lubricate the contact surfaces with high viscosity oil
select a cross-section in which the crosssectional area is away from the
neutral axis, such as an I or tubular section
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Instructor - Dr. B.S. Reddy (IIT Guwahati) Lecture 19, 20 ,21 09 Sep 2019 15 / 19
 Shaft Stresses
Solid Shaft with round cross-section
Assumption - Axial loads are usually comparatively very small at critical
locations where bending and torsion dominate, hence they will be left out

32Ma 32Mm
σ a = Kf 3
, σ m = Kf
πd πd3
32Ta 32Tm
τa = Kfs , τm = Kfs
πd3 πd3
′ ′
Fluctuating von-mises stresses - σa = (σa2 + 3τa2 )1/2 , σm = (σm
2 + 3τ 2 )1/2
m
′ ′
1 σ σ
Fatigue failure (Modified Goodman) - = a+ m
n Se Sut
′ Syt
Check against yielding - σmax = [(σm + σa )2 + 3(τm + τa )2 ]1/2 , ny = ′
σmax
′ ′ ′
Conservative check - σmax = σa + σm . . . . . . . . . . . . . . . . . . . .
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Instructor - Dr. B.S. Reddy (IIT Guwahati) Lecture 19, 20 ,21 09 Sep 2019 16 / 19
 Problem - At a machined shaft shoulder the small diameter d is 1.100 in,
the large diameter D is 1.65 in, and the fillet radius is 0.11 in. The
bending moment is 1260 lbf-in and the steady torsion moment is 1100
lbf-in. The heat-treated steel shaft has an ultimate strength of Sut = 105
kpsi and a yield strength of Syt = 82 kpsi. The reliability goal for the
endurance limit is 0.99.
(a) Determine the fatigue factor of safety of the design using Modified
Goodman Criteria
(b) Determine the yielding factor of safety.

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Instructor - Dr. B.S. Reddy (IIT Guwahati) Lecture 19, 20 ,21 09 Sep 2019 17 / 19
 First Iteration Estimates for Kt and Kts

Figure: Warning: These factors are only estimates for use when actual dimensions
are not yet determined. Do not use these once actual dimensions are available.

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Instructor - Dr. B.S. Reddy (IIT Guwahati) Lecture 19, 20 ,21 09 Sep 2019 18 / 19
 Problem - Rotating solid steel shaft is simply supported by bearings at B
and C and is driven by a gear (not shown) which meshes with spur gear at
D, which has 150-mm pitch dia. Force F from drive gear acts at pressure
angle of 200 . The shaft transmits a torque to point A of TA = 340 N-m.
Shaft is machined from steel with Syt = 420 MPa and Sut = 560 MPa.
Using a factor of safety of 2.5, determine minimum allowable dia. of
250-mm section of shaft based on (a) static yield analysis using DE theory
and (b) a fatigue-failure analysis. Assume sharp fillet radii at bearing
shoulders for estimating stress-concentration factors.

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Instructor - Dr. B.S. Reddy (IIT Guwahati) Lecture 19, 20 ,21 09 Sep 2019 19 / 19