M E 4 14

M A C H I N E
D E S I G N 1
B Y E N G R . D E N N IS E . G A N A S
 Lecture 6
COUPLINGS

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 COUPLINGS
A coupling is a device used to connect two shafts together at their ends for the
purpose of transmitting power.
PURPOSES OF COUPLINGS
a. Power Transmission: They transmit power and rotational motion from
one shaft to another.
b. Misalignment Compensation: Couplings can accommodate small
misalignments between shafts (such as angular, parallel, or axial
misalignment).
c. Shock Absorption: Some couplings absorb
shock loads and vibrations, protecting
components from damage.
d. Protection: They can protect the connected
components from damage due to overloads or
mechanical failures by slipping or disengaging
when necessary.

B Y E N G R . D E N N IS E . G A N A S M E4 14
 COUPLINGS
A coupling is a device used to connect two shafts together at their ends for the
purpose of transmitting power.
TYPES OF COUPLINGS
1. RIGID COUPLINGS
 do not allow axial or radial motion between the driving and driven shaft
 provide a solid connection between two shafts, high precision, and torque, but
without misalignment absorption capabilities;
 both shafts should be perfectly aligned to ensure a good performance and avoid
damping transmission and possible breaks in the installation;
 they require lubrication in many times since it allows no movement between the
two shafts .

A. FLANGE COUPLINGS
 two flanged parts are installed on the shafts to be
coupled and are drawn together by a series of bolts;
 the torque places the bolts in shear.

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 RIGID COUPLINGS
B. TAPER LOCK RIGID COUPLINGS
 is a locking mechanism commonly used in
power transmission drives for locating pulleys,
sprockets, and couplings to shafts;
 the outside of the bush is tapered to match the
component bore that is to be located on the
shaft.

C. RIBBED RIGID COUPLINGS

 made of 2 pieces and used to connect shafts of
the same diameter;
 the coupling is keyed to both shafts with bolts for
quick and easy installation and removal;
 ribbed couplings may be used for heavy-duty
services;

B Y E N G R . D E N N IS E . G A N A S M E4 14
 FLEXIBLE COUPLINGS
2. FLEXIBLE COUPLINGS
 designed to transmit torque smoothly while permitting some axial, radial and
angular misalignment;
 when misalignment occurs, parts of the coupling move with little or no resistance;

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 FLEXIBLE COUPLINGS
2. FLEXIBLE COUPLINGS
 designed to transmit torque smoothly while permitting some axial, radial and
angular misalignment;
 when misalignment occurs, parts of the coupling move with little or no resistance;
a. GEAR COUPLINGS
 The most power-dense type of coupling.
 Consists of two hubs with external teeth that engage
internal teeth on a two- or one-piece flanged sleeve.
 Can transmit high torque at high speeds and can
accommodate misalignment with axial clearance
between gear teeth, called backlash.
b. CHAIN COUPLINGS
 These have two hubs with sprockets rather than
teeth. The sprockets are connected by double-
strand roller chain.
 Chain couplings do not transmit as much power as
gear couplings but are usually less costly.
 They are used for low speed, moderate torque
applications.

B Y E N G R . D E N N IS E . G A N A S M E4 14
 METALLIC ELEMENT FC
C. GRID COUPLINGS
 Consist of two hubs with multiple slots, through which a steel grid weaves
back and forth.
 Misalignment and float are accommodated by the sliding motion of the grid
within lubricated slots.
 They are usually composed of metal, require lubrication, and have covers to
retain lubricant and prevent contamination.
 They are used for medium and small equipment applications like pumps and
high-capacity bulk conveyors.

B Y E N G R . D E N N IS E . G A N A S M E4 14
 ELASTOMERIC ELEMENT FC
A. JAW COUPLINGS
 Jaw couplings have two hubs with projecting
jaws that mate with a one-piece elastomeric
spider insert.

B. URETHANE TIRE COUPLINGS

 The elastomeric element is made of urethane,
polyurethane, or polyether material.
 They are split for easy assembly and offer a
high degree of flexibility.
 When the coupling fails, usually only the
elastomeric part is replaced.

C. PIN AND BUSHING COUPLING

 Relative movement is accommodated by
compression of steel-bonded bushings.
 They are used to absorb energy and dampen
loads, especially when equipment is engine-
driven.

B Y E N G R . D E N N IS E . G A N A S M E4 14
 SHAFT MISALIGNMENT
Shaft misalignment refers to a condition where the rotational axis of one shaft is
not aligned with the rotational axis of the mating shaft.
1. Parallel Misalignment: This occurs when the rotational axes of the two shafts
are parallel but not perfectly aligned.
2. Angular Misalignment: This occurs when the two shafts are not at the same
angle, causing the shafts to intersect at an angle.
3. Combined Misalignment: This occurs when both parallel and angular
misalignment are present.
4. Axial Misalignment: This occurs when the two shafts are not perfectly aligned
in the axial direction.

B Y E N G R . D E N N IS E . G A N A S M E4 14
 CAUSES & EFFECTS OF MISALIGNMENT
CAUSES OF MISALIGNMENT
1. Improper Installation
2. Poor Maintenance
3. Wear And Tear
4. Foundation Issues
5. Thermal Expansion
6. Pipe Stresses
7. Overloading

EFFECTS OF MISALIGNMENT
1. High/Increased Vibration
2. Noise
3. Premature Equipment Failure
4. Reduced Efficiency
5. Increased Power Consumption
6. Reduced Production

B Y E N G R . D E N N IS E . G A N A S M E4 14
 SHAFT ALIGNMENT

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 STRESS ANALYSIS (Flange Coupling)
BOLT DESIGN Where: 𝐹 – total shear force
𝑇 – torque
𝑇 2𝑇 𝐷𝑏𝑐 – bolt pitch circle diameter
𝐹= = 𝐴𝑠 - shear area of bolt
𝐷𝑏𝑐 𝐷𝑏𝑐 𝑛 – number of bolts
2 𝑑 – bolt diameter
𝐹 𝐹 2𝑇 8𝑇 𝑆𝑠𝑏 - shear stress of bolt
𝑆𝑠𝑏 = = = = 𝑆𝑐𝑏 - comp/crushing stress of bolt
𝐴𝑠 𝜋𝑑 2 𝜋𝑑 2 𝐷𝑏𝑐 𝑛𝜋𝑑 2 𝑡 – thickness of flange
𝑛 4 𝐷𝑏𝑐 𝑛 4
𝐷𝑠 - shaft diameter

𝐹 𝐹 2𝑇
𝑆𝑐𝑏 = = =
𝐴𝑐 𝑡𝑑 𝐷𝑏𝑐 𝑛𝜋𝑡𝑑

Notes: (Conventions)
𝐷𝑏𝑐 = 3 𝐷𝑠
𝑁𝑜. 𝑜𝑓 𝑏𝑜𝑙𝑡𝑠 = 3 for𝐷𝑠 up to 40 mm
= 4 for𝐷𝑠 up to 100 mm
= 6 for𝐷𝑠 up to 180 mm

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 STRESS ANALYSIS (Flange Coupling)
HUB DESIGN
- The hub is designed by considering it as a hollow shaft, transmitting the same torque
as that of the solid shaft.
Where: 𝑆𝑠ℎ - shear stress on hub
16𝑇𝐷ℎ 𝑇 - torque
𝑆𝑠ℎ = 𝐷ℎ – hub diameter
𝜋 𝐷ℎ4 − 𝐷𝑠 4
𝐷𝑠 – shaft diameter
𝐿ℎ – hub length

Notes: (Conventions)
𝐷ℎ = 2𝐷𝑠
𝐿ℎ = 1.5 𝐷𝑠
𝐿ℎ = 𝐿𝑘

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 STRESS ANALYSIS (Flange Coupling)
FLANGE DESIGN
𝑇 = hub circumference x flange thickness x flange shear stress x hub radius
𝐷ℎ
𝑇 = 𝜋 (𝐷ℎ ) (𝑡𝑓 )(𝑆𝑠𝑓 ) Where: 𝑆𝑠ℎ - shear stress on hub
2 𝑡𝑓 – flange thickness
2𝑇 𝐷ℎ – hub diameter
𝑆𝑠𝑓 = 𝑇- torque
𝜋𝐷ℎ2 𝑡𝑓
𝐷𝑓 – flange outside diameter

Notes: (Conventions)
𝑡𝑓 = 0.5 𝐷𝑠
𝐷𝑓 = 4 𝐷𝑠

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 STRESS ANALYSIS (Flange Coupling)
TORQUE CAPACITY OF COUPLING (Two Concentric Rows)
𝐷𝑏𝑐1 𝐷𝑏𝑐2
𝑇 = 𝐹1 𝑅1 𝑛1 + 𝐹2 𝑅2 𝑛2 = 𝐹1 𝑛1 + 𝐹2 𝑛2
2 2
SHEAR STRAIN vs RADIAL DISTANCE FROM SHAFT AXIS
𝛾1 𝛾2 (This means that as you move further away
Where: 𝛾 =
𝜏
= from the axis, the shear strain increases 𝐺
𝑅1 𝑅2 proportionally.) 𝐺 = modulus of elasticity in shear

SHEAR STRESS vs RADIAL DISTANCE FROM SHAFT AXIS

𝐹1 𝐹2
𝑆𝑠1 𝑆𝑠2 𝐴1 𝐴2 𝐹1 𝐹2
= or = or =
𝐺1 𝑅1 𝐺2 𝑅2 𝐺1 𝑅1 𝐺2 𝑅2 𝑑12 𝐺1 𝑅1 𝑑22 𝐺2 𝑅2

i) If bolts are of same material:

𝐹1 𝐹2
=
𝑑12 𝑅1 𝑑22 𝑅2
ii) If bolts are of same material and same diameter:
𝐹1 𝐹2
=
𝑅1 𝑅2

B Y E N G R . D E N N IS E . G A N A S M E4 14
 EXAMPLES
EXAMPLE 1: A cast-iron flanged coupling is used to join two steel shafts 2 ½ in., in
diameter, delivering a maximum torque of 60 ft.-lb. The two halves of the couplings are
joined together by 4 bolts, placed equidistant on a bolt circle of 3 in. diameter. If the
design stress is 10,000 psi in shear for the bolt material, evaluate the diameter of the
bolts.
Solution:
12 𝑖𝑛
8𝑇 8(60 𝑓𝑡. 𝑙𝑏) 𝟏
𝑓𝑡
𝑑= = = 𝟎. 𝟏𝟐𝟑𝟔 𝒊𝒏 or "
𝐷𝑏𝑐 𝑛𝜋𝑆𝑠𝑏 𝑙𝑏𝑠 𝟖
(3 𝑖𝑛)(4)(𝜋)(10000 2 )
𝑖𝑛

EXAMPLE 2: A flanged coupling is keyed to a shaft. There are six 25mm-diameter

bolts in a 150mm bolt circle. Calculate the torque that can be transmitted if the
allowable shear stress is 210 MPa in the bolts.
Solution:
8𝑇 𝑆𝑠𝑏 𝐷𝑏𝑐 𝑛𝜋𝑑2 (210)(150)(6)(𝜋)(25)2
𝑆𝑠𝑏 = →𝑇= = = 𝟒𝟔. 𝟒 𝒌𝑵. 𝒎
𝐷𝑏𝑐 𝑛𝜋𝑑 2 8 8

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 EXAMPLES
EXAMPLE 3: A flanged bolt coupling consists of ten 20 mm-diameter bolts spaced
evenly around a bolt circle 400 mm in diameter. Determine the torque capacity of the
coupling if the allowable shearing stress in the bolts is 40 MPa.
Solution:
8𝑇 𝑆𝑠𝑏 𝐷𝑏𝑐 𝑛𝜋𝑑2 (40 𝑁/𝑚𝑚2 )(400𝑚𝑚)(10)(𝜋)(20𝑚𝑚)2
𝑆𝑠𝑏 = →𝑇= =
𝐷𝑏𝑐 𝑛𝜋𝑑 2 8 8
= 𝟐𝟓, 𝟏𝟑𝟐, 𝟕𝟒𝟏. 𝟐𝟑 𝑵. 𝒎𝒎 𝒐𝒓 𝟐𝟓. 𝟏𝟑 𝒌𝑵. 𝒎

EXAMPLE 4: A flange coupling has an outside diameter of 200 mm and connects two
40 mm shafts. There are four 16 mm bolts on a 140 mm bolt circle. The radial flange
thickness is 20 mm. If the torsional stress in the shaft is not to exceed 26 MPa,
determine the shearing stress in the bolts if uniformly distributed.
Solution:
𝑆𝑠 𝜋𝐷3 26 𝜋 403
𝑇= = = 326,725.64 𝑁. 𝑚𝑚
16 16
8𝑇 8(326,725.64 𝑁. 𝑚𝑚)
∴ 𝑆𝑠𝑏 = = = 𝟓. 𝟖𝟎 𝑴𝑷𝒂
𝐷𝑏𝑐 𝑛𝜋𝑑 2 (140𝑚𝑚)(4)(𝜋)(16 𝑚𝑚)2

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 EXAMPLES
EXAMPLE 5: A flange coupling is to connect two 57 mm shafts. The hubs of the
coupling are each 111 mm in diameter and 92 mm thick and the flange webs are 19
mm thick. Six – 16 mm in a 165 mm diameter circle connects the flanges. The keyway
is 6 mm shorter than the hub’s thickness and the key is 14 mm x 14 mm. The coupling
is to transmit 45 kW at 165 rpm. The yield point in tension or compression is 448 MPa.
Find the shear stress in bolts.
Solution:
𝑃 45,000
𝑇= = = 2604.4 𝑁. 𝑚
2𝜋𝑁 2𝜋 165
60
2𝑇 2(2604.4)
𝑆𝑠𝑏 = = = 𝟐𝟔. 𝟏𝟕 𝑴𝑷𝒂
𝜋𝑑 2 𝜋(0.016)2
𝐷𝑏𝑐 𝑛 0.165(6)
4 4

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 EXAMPLES
EXAMPLE 6: A flange coupling connects two 2” diameter shafts. The flanges are fitted
with 6 bolts of SAE 1040 steel on a 7” bolt circle. The shaft runs at 300 rpm and
transmits 45 hp. Assume a factor of safety of 5, ultimate tension of 70,000 psi and
ultimate shear of 55,000 psi. How thick should the flange be?
Solution:
63000 45
8𝑇 8(9,450 𝑖𝑛. 𝑙𝑏) 𝑇= = 9,450 𝑖𝑛. 𝑙𝑏
𝑑= = = 𝟎. 𝟐𝟐𝟖 " 300
𝐷𝑏𝑐 𝑛𝜋𝑆𝑠𝑏 55,000
(7 𝑖𝑛)(6)(𝜋)
5
𝐹 2𝑇
𝑆𝑐𝑏 = =
𝑡𝑑 𝐷𝑏𝑐 𝑛 𝑡𝑑

2𝑇 2(9,450)
𝑡= = = 𝟎. 𝟏𝟒𝟏 "
𝐷𝑏𝑐 𝑛𝑆𝑐 𝑑 (7)(6) 70000 (0.228)
5

Thickness is too small (thin), use 𝒕 = 𝟎. 𝟓𝑫𝒔 = 𝟎. 𝟓 𝟐 = 𝟏”

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 EXAMPLES
EXAMPLE 7: A torque of 700 lb-ft is to be carried by a flanged bolt coupling that
consists of eight ½-in.-diameter steel bolts on a circle of diameter 12 in. and six ½-in.-
diameter steel bolts on a circle of diameter 9 in. Determine the shearing stress in the
bolts.
Solution:
For same material and diameter,
𝐷𝑏𝑐1 𝐷𝑏𝑐2
𝑇 = 𝐹1 𝑛1 + 𝐹2 𝑛2 𝐹1 𝐹1 𝐷𝑏𝑐2
2 2 𝐹2 = 𝑅2 =
𝑅1 𝐷𝑏𝑐1 2
𝐷𝑏𝑐1 𝐹1 𝐷𝑏𝑐2 𝐷𝑏𝑐2 2
= 𝐹1 𝑛1 + 𝑛2
2 𝐷𝑏𝑐1 2 2
2
12 𝐹1 9 9
700 12 = 𝐹1 8+ 6 → 𝑭𝟏 = 𝟏𝟐𝟑. 𝟏𝟐 𝐥𝐛𝐬 ∴ 𝑭𝟐 = 𝟗𝟐. 𝟑𝟒 𝐥𝐛𝐬
2 12 2 2
2
i) For outer bolts: ii) For inner bolts:
𝐹1 𝐹1 123.12 𝐹2 𝐹2 92.34
𝑆𝑠𝑏1 = = = 𝑆𝑠𝑏2 = = =
𝐴1 𝜋𝑑12 𝜋𝑥0.52 𝐴2 𝜋𝑑22 𝜋𝑥0.52
4 4 4 4
= 𝟔𝟐𝟕. 𝟎𝟒 𝒑𝒔𝒊 = 𝟒𝟕𝟎. 𝟐𝟖 𝒑𝒔𝒊

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 EXAMPLES
EXAMPLE 8: Eight 10 mm diameter steel bolts on a bolt circle 150 mm in radius and
six 20 mm diameter steel bolts on a concentric bolt circle 100 mm in radius were used
in a rigid coupling. If the design stress in the outer bolts is 60 MPa, determine the
torque capacity of the coupling.
Solution:

i) For outer circle: ii) For inner circle (same material):

𝜋𝑑12 𝐹1 𝐹2
𝑭𝟏 = 𝑆𝑠𝑏 𝐴1 = 𝑆𝑠𝑏 =
4 𝑑12 𝑅1 𝑑22 𝑅2
𝜋(102 )
= 60 → 𝑭𝟏 = 𝟒, 𝟕𝟏𝟐. 𝟑𝟗 𝑵 𝐹1
4 𝐹2 = 𝑑22 𝑅2
𝑑12 𝑅1
∴ The torque capacity of the coupling is: 4712.39 100
2
𝑇 = 𝐹1 𝑅1 𝑛1 + 𝐹2 𝑅2 𝑛2 = 20
150 2
102
𝐷𝑏𝑐1 𝐷𝑏𝑐2 2
= 𝐹1 𝑛1 + 𝐹2 𝑛2
2 2 𝑭𝟐 = 𝟏𝟐, 𝟓𝟔𝟔. 𝟑𝟕 𝐍
= 4,712.39 150 8 + 12,566.37 100 6
= 𝟏𝟑, 𝟏𝟗𝟒, 𝟔𝟗𝟎 𝐍. 𝐦𝐦 or 𝟏𝟑. 𝟏𝟗 𝐤𝐍. 𝐦

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 EXAMPLES
EXAMPLE 9: A flanged bolt coupling consists of six ½ -in. steel bolts evenly spaced
around a bolt circle 12 in. in diameter, and four ¾ -in. aluminum bolts on a concentric
bolt circle 8 in. in diameter. What torque can be applied without exceeding 9000 psi in
the steel or 6000 psi in the aluminum? Assume Gst = 12 × 106 psi and Gal = 4 × 106 psi.
Solution:
𝑇 = 𝐹𝑠 𝑅𝑠 𝑛𝑠 + 𝐹𝑎 𝑅𝑎 𝑛𝑎 = (𝐴𝑠 𝑆𝑠 )𝑅𝑠 𝑛𝑠 + (𝐴𝑎 𝑆𝑎 )𝑅𝑎 𝑛𝑎 For different material and diameters,
𝜋(0.52 ) 𝜋(0.752 ) 𝑆𝑠 𝑆𝑎 𝑆𝑠 𝑆𝑎
= (𝑆𝑠 )(6)(6) + (𝑆𝑎 )(4)(4) 𝐺𝑠 𝑅𝑠 = 𝐺𝑎 𝑅𝑎 → 12𝐸6(6) = 4𝐸6(4)
4 4
9
𝑇 = 7.07𝑆𝑠 + 7.07𝑆𝑎 → eq. 1 𝑆𝑠 = 𝑆𝑎 → eq. 2a
2
Combining eq. 1 and eq. 2a, 2
𝑆𝑎 = 𝑆𝑠 → eq. 2a
9
9
𝑇1 = 7.07 6000 + 7.07 6000 = 233,310 𝑙𝑏𝑠. 𝑖𝑛
2
Combining eq. 1 and eq. 2b,
2
𝑇2 = 7.07 9000 + 7.07 9000 = 77,770 𝑙𝑏𝑠. 𝑖𝑛
9
∴ Use smaller torque, 𝑻 = 𝟕𝟕, 𝟕𝟕𝟕𝟎 𝒍𝒃𝒔. 𝒊𝒏 𝐨𝐫 𝟕𝟕. 𝟖 𝒌𝒊𝒑𝒔

B Y E N G R . D E N N IS E . G A N A S M E4 14
 EXAMPLES
EXAMPLE 10: A solid circular shaft 90 mm in diameter is connected by a rigid
coupling to a hollow shaft 100 mm in outside diameter and 90 mm in inside diameter. If
the allowable shear stress in the shafts and bolts is 70 MPa, determine the number of
12-mm diameter steel bolts to be used on a 240 mm diameter bolt circle so that the
coupling will be as strong as the weaker shaft.
Solution:
i) For solid shaft:
16𝑇 𝑆𝑠𝑠 𝜋𝐷3 (70)𝜋(90)3
𝑆𝑠𝑠 = → 𝑇= = = 10,019,717.07 𝑁. 𝑚𝑚
𝜋𝐷3 16 16
8𝑇 8 10,019,717.07
𝑛= = = 10.55 𝑜𝑟 𝟏𝟏 𝒃𝒐𝒍𝒕𝒔
𝑆𝑠𝑠 𝐷𝑏𝑐 𝜋𝑑 2 (70)(240)𝜋(12)2

ii) For hollow shaft: → Weaker shaft!

16𝑇𝐷𝑜 𝑆𝑠ℎ 𝜋 𝐷𝑜 4 − 𝐷𝑖 4 (70)𝜋 1004 − 904
𝑆𝑠ℎ = → 𝑇= = = 4,726,722.50 𝑁. 𝑚𝑚
𝜋 𝐷𝑜 4 − 𝐷𝑖 4 16𝐷𝑜 16(100)
8𝑇 8 4,726,722.50
𝑛= = = 𝟓 𝒃𝒐𝒍𝒕𝒔
𝑆𝑠𝑠 𝐷𝑏𝑐 𝜋𝑑 2 (70)(240)𝜋(12)2 ∴ Use n = 5 bolts !

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 EXAMPLES
EXAMPLE 11: Two 35 mm shafts are connected by a flanged coupling. The flanges
are fitted with 6 bolts on 125 mm bolt circle. The shafts transmit a torque of 800 N.m at
a 350 rpm. For the safe stresses mentioned below, calculate 1. bolt diameters, 2.
flange thickness and diameter, 3. flat key dimensions, 4. hub diameter and length, and
5. power transmitted.
Safe shear stress for shaft material = 63 MPa
Safe stress for bolt material = 56 MPa
Safe stress for cast iron coupling = 10 MPa
Safe stress for key material = 46 MPa
Solution:
1. Bolt Design:
8𝑇 8(800𝐸3)
𝑑= = = 𝟔. 𝟗𝟔 𝑠𝑎𝑦 𝟖 𝒎𝒎
𝐷𝑏𝑐 𝑛𝜋𝑆𝑠𝑏 (125)(6)(𝜋)(56)

2. Flange Design: Use 𝐷ℎ = 2𝐷𝑠 = 70 𝑚𝑚

2𝑇 2(800𝐸3)
𝑡𝑓 = 𝜋𝐷 2𝑆 = 𝜋 = 𝟏𝟎. 𝟒 𝒎𝒎 say 𝟏𝟐 𝒎𝒎
ℎ 𝑠𝑓 702 (10)

𝐷𝑓 = 4 𝐷𝑠 = 4 35 = 𝟏𝟒𝟎 𝒎𝒎

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 EXAMPLES
3. Key Design: from Table 13.1 for a 35 mm flat key, 𝑾 = 𝟏𝟐 𝒎𝒎 & 𝑯 = 𝟖 𝒎𝒎
2𝑇 2(800𝐸3)
𝐿𝑠 = 𝑆 = 46(35)(12) = 𝟖𝟐. 𝟖 𝒎𝒎 say 𝟖𝟓 𝒎𝒎
𝑠𝑘 𝐷𝑊

Note: if 𝐿𝑠𝑘 = 1.5𝐷𝑠 = 1.5 (35) = 52.5 𝑚𝑚

2𝑇 2(800𝐸3)
Then 𝑆𝑠𝑘 = = = 𝟕𝟐. 𝟔 𝑴𝑷𝒂 > 𝟒𝟔 𝑴𝑷𝒂 (not acceptable)
𝐿𝑠𝑘 𝐷𝑊 52.5(35)(12)

4. Hub Design:
𝐷ℎ = 2𝐷𝑠 = 2 35 = 𝟕𝟎 𝒎𝒎
𝐿ℎ = 𝐿𝑠𝑘 = 𝟖𝟓 𝒎𝒎
5. Power Transmitted:
350
𝑃 = 2𝜋𝑇𝑁 = 2𝜋 800 = 𝟐𝟗. 𝟑𝟐 𝒌𝑾
60

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 QUIZ 6

ME3A – April 14, 2025 (Monday)

ME3B – April 15, 2025 (Tuesday)

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B Y E N G R . D E N N IS E . G A N A S M E4 14