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
- devices 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;

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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.

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 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.

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 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.

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 STRESS ANALYSIS
SHEAR FORCE (on the bolts) Where: 𝐹 – total shear force
𝑇 2𝑇 𝑇 – torque
𝐹= = 𝐷𝑏𝑐 – bolt circle diameter
𝐷𝑏𝑐 𝐷𝑏𝑐
2 𝐴𝑠 - shear area of bolt
𝑛 – number of bolts
SHEAR STRESS per bolt 𝑑 – bolt diameter
𝐹 𝐹 2𝑇 8𝑇 𝑆𝑠 - shear stress
𝑆𝑠 = = = = 𝑆𝑐 - compressive stress
𝐴𝑠 𝜋𝑑2 𝜋𝑑 2 𝐷𝑏𝑐 𝑛𝜋𝑑 2
𝑛 4 𝐷𝑏𝑐 𝑛 4 𝑡 – thickness of flange

COMPRESSIVE STRESS ON FLANGE

𝐹 𝐹
𝑆𝑐 = =
𝐴𝑐 𝑡𝑑

REQUIRED BOLT DIAMETER & NO. OF BOLTS

8𝑇 8𝑇
𝑑= 𝑛=
𝐷𝑏𝑐 𝑛𝜋𝑆𝑠 𝑆𝑠𝐷𝑏𝑐 𝜋𝑑 2

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 STRESS ANALYSIS
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 𝜏
Where: 𝛾 =
= 𝐺
𝑅1 𝑅2 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

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 EXAMPLES
EXAMPLE 1: 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)
𝑆𝑠 = = = 𝟐𝟔. 𝟏𝟕 𝑴𝑷𝒂
𝜋𝑑 2 𝜋(0.016)2
𝐷𝑏𝑐 𝑛 0.165(6)
4 4

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 EXAMPLES
EXAMPLE 2: 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 𝑛𝜋𝑆𝑠 𝑙𝑏𝑠 𝟖
(3 𝑖𝑛)(4)(𝜋)(10000 2 )
𝑖𝑛

EXAMPLE 3: 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 𝑆𝑠 𝐷𝑏𝑐 𝑛𝜋𝑑 2 (210)(150)(6)(𝜋)(25)2
𝑆𝑠 = →𝑇= = = 𝟒𝟔. 𝟒 𝒌𝑵. 𝒎
𝐷𝑏𝑐 𝑛𝜋𝑑 2 8 8

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 EXAMPLES
EXAMPLE 4: 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 𝑆𝑠𝐷𝑏𝑐 𝑛𝜋𝑑2 (40 𝑁/𝑚𝑚2 )(400𝑚𝑚)(10)(𝜋)(20𝑚𝑚)2
𝑆𝑠 = →𝑇= =
𝐷𝑏𝑐 𝑛𝜋𝑑 2 8 8
= 𝟐𝟓, 𝟏𝟑𝟐, 𝟕𝟒𝟏. 𝟐𝟑 𝑵. 𝒎𝒎 𝒐𝒓 𝟐𝟓. 𝟏𝟑 𝒌𝑵. 𝒎

EXAMPLE 5: 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:
16𝑇 𝑆𝑠 𝜋𝐷3 26 𝜋 403
𝑆𝑠𝑠 = → 𝑇= = = 326,725.64 𝑁. 𝑚𝑚
𝜋𝐷3 16 16
8𝑇 8(326,725.64 𝑁. 𝑚𝑚)
∴ 𝑆𝑠 = = = 𝟓. 𝟖𝟎 𝑴𝑷𝒂
𝐷𝑏𝑐 𝑛𝜋𝑑 2 (140𝑚𝑚)(4)(𝜋)(16 𝑚𝑚)2

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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 70000 psi and
ultimate shear of 55000 psi. How thick should the flange be?
Solution:
𝐹 𝐹
𝑆𝑐 = →𝑡= = 𝟎. 𝟖𝟒 𝒊𝒏
𝑡𝑑 𝑆𝑐 𝑑

EXAMPLE 7: 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 bolts is 60 MPa, determine the torque
capacity of the coupling.
Solution:
i) For outer circle:
𝜋𝑑12 𝜋(102 )
𝑭𝟏 = 𝑆𝑠 𝐴1 = 𝑆𝑠 = 60 → 𝑭𝟏 = 𝟒, 𝟕𝟏𝟐. 𝟑𝟗 𝑵
4 4

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 EXAMPLES
i) For inner circle (same material):
𝐹1 𝐹2 𝐹1 4712.39 100
2 = 2 → 𝐹2 = 2 (𝑑22 𝑅2 ) = 202
𝑑1 𝐺1 𝑅1 𝑑2 𝐺2 𝑅2 𝑑1 𝑅1 150 2
102 2
𝑭𝟐 = 𝟏𝟐, 𝟓𝟔𝟔. 𝟑𝟕 𝐍

∴ The torque capacity of the coupling is:

𝑇 = 𝐹1 𝑅1 𝑛1 + 𝐹2 𝑅2 𝑛2
𝐷𝑏𝑐1 𝐷𝑏𝑐2
= 𝐹1 𝑛1 + 𝐹2 𝑛2
2 2
= 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 8: 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:
𝐷𝑏𝑐1 𝐷𝑏𝑐2 𝐹1 𝐹1 𝐷𝑏𝑐2
𝑇 = 𝐹1 𝑛1 + 𝐹2 𝑛2 𝑏𝑢𝑡 𝐹2 = 𝑅 =
2 2 𝑅1 2 𝐷𝑏𝑐1 2
2
𝐷𝑏𝑐1 𝐹1 𝐷𝑏𝑐2 𝐷𝑏𝑐2 Same material and diameter!
= 𝐹1 𝑛1 + 𝑛2
2 𝐷𝑏𝑐1 2 2
2
12 𝐹1 9 9
700(12) = 𝐹1 8+ 6 ∴ 𝑭𝟏 = 𝟏𝟐𝟑. 𝟏𝟐 𝐥𝐛𝐬 and 𝑭𝟐 = 𝟗𝟐. 𝟑𝟒 𝐥𝐛𝐬
2 12 2 2
2
i) For outer bolts: i) For inner bolts:
𝐹1 𝐹1 123.12 𝐹2 𝐹2 92.34
𝑆𝑠1 = = = 𝑆𝑠2 = = =
𝐴1 𝜋𝑑12 𝜋𝑥0.52 𝐴2 𝜋𝑑22 𝜋𝑥0.52
4 4 4 4
= 𝟔𝟐𝟕. 𝟎𝟒 𝒑𝒔𝒊 = 𝟒𝟕𝟎. 𝟐𝟖 𝒑𝒔𝒊

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 EXAMPLES
EXAMPLE 9: 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
𝑆𝑠𝐷𝑏𝑐 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
𝑆𝑠𝐷𝑏𝑐 𝜋𝑑 2 (70)(240)𝜋(12)2 ∴ Use n = 5 bolts !

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