M E 414

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 5
KEYS

B Y E N G R . D E N N IS E . G A N A S ME 414
 KEYS
- a machine element which is used to connect the transmission shaft to rotating
machine elements like pulley, gear, sprocket or flywheel;
- keys provide a positive means of transmitting torque between shaft and hub of the
mating element;
- a slot is machined in the shaft or in the hub or both to accommodate the key and is
called keyway;
- keyway reduces the strength of the shaft as it results in stress concentration.

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 TYPES OF KEYS
1. RECTANGULAR (FLAT) KEY – has a rectangular
cross-section with the smaller dimension placed in the
radial direction with half sunk in the shaft and half in
the hub and is used where the weakening of the shaft
by the keyway is serious.
2. SQUARE KEY – has a square cross-section with
half of its depth sunk in the shaft and half in the hub.
3. ROUND KEY – has a circular cross-section.

4. BARTH KEY – is modified from the square key and

has a bevel shaped smooth edges at the upper part.
This bevel shaped smooth edges help the key to fit
tightly.
5. WOODRUFF KEY – consists of one-half of a circular disk fitting into a rectangular
keyway in the female member and a semi-circular keyway in the male member

6. TAPERED KEY - A key with a tapered shape that provides a stronger connection
between the shaft and the component, offering a better fit and higher torque capacity
than a standard square or rectangular key.

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 TYPES OF KEYS
7. GIB-HEAD TAPER KEY – is a flat key with a special gib-head
to facilitate easy driving and removal of the key.
8. SADDLE KEY – is a flat key without a keyway in the shaft.

9. KENNEDY KEY – are tapered square keys with the diagonal

dimension in a circumferential direction.
10. FEATHER KEY – is one which has a tight fit into one member
and a loose sliding fi in the other mating member thus allowing the
hub to move along the shaft but prevents rotation on the shaft

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 MATERIALS USED FOR KEYS

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 MATERIALS USED FOR KEYS

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 KEY SIZE VS SHAFT DIAMETER

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 KEY SIZE VS SHAFT DIAMETER

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 STRESS ANALYSIS
2 BASIC MODES OF FAILURES IN KEYS
 Shear across the shaft/hub interface
 Compression failure due to the bearing action between the sides of the key
and the shaft or hub material

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 SHEAR STRESS ANALYSIS
SHEAR FORCE
𝑇 Where 𝑇 = torque of shaft
𝐹= 𝐷 = bore or shaft diameter
𝐷
𝑆𝑠𝑘 = design shear stress (key)
2
𝐿𝑠 = key length (shear)
SHEAR STRESS 𝑊 = width of key
𝑆𝑦 = yield stress of shaft/key
𝐹 𝐹 𝑇 2𝑇
𝑆𝑠𝑘 = = = = 𝐹𝑆 = factor of safety
𝐴𝑠 𝑊𝐿 𝐷 𝐷𝑊𝐿
(𝑊𝐿)
2

MINIMUM REQUIRED KEY LENGTH FOR SHEAR

2𝑇
𝐿𝑠 =
𝑆𝑠𝑘 𝐷𝑊

MAXIMUM SHEAR STRESS THEORY OF FAILURE

0.5 𝑆𝑦
𝑆𝑠 = (for both shaft and key)
𝐹𝑆

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 COMPRESSIVE STRESS ANALYSIS
DESIGN STRESS FOR COMPRESSION
𝑆𝑦𝑘 Where 𝑇 = torque of shaft
𝑆𝑐𝑘 = Note: 𝑆𝑐 = 2 𝑆𝑠𝑘 𝐷 = bore or shaft diameter
𝐹𝑆 𝑆𝑐𝑘 = design comp stress (key)
𝐿𝑐 = key length (compression)
COMPRESSIVE (CRUSHING) STRESS 𝐻 = height or thickness of key
𝐹 𝐹 𝑇 4𝑇 𝑆𝑦𝑘 = yield stress of key
𝑆𝑐𝑘 = = = = 𝐹𝑆 = factor of safety
𝐴𝑐 𝐿 𝐻 𝐷 𝐻
(𝐿) 𝐷𝐻𝐿
2 2 2
MINIMUM REQUIRED KEY LENGTH FOR COMPRESSION
4𝑇
𝐿𝑐 =
𝑆𝑐 𝐷𝐻

NOTES:
𝟐
1. If width is too small, for rectangular key use 𝑾 = 𝑫/𝟒 and 𝑯 = 𝑾 = 𝑫/𝟔.
𝟑
For square key, use 𝑾 = 𝑯 = 𝑫/𝟒. (see Examples 10 & 11)
2. For square key, 𝑳𝒔 = 𝑳𝒄 .
3. If key type is not indicated, use Table 1 as reference (based on diameter size). (see Examples 8,9 & 11)
4. If key material is not given, use SAE 1018 (low cost, readily available and useful for most applications).
(see Example 9)

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 SHAFT STRENGTH FACTOR
According to HF Moore, the shaft strength factor is the ratio of the strength of the shaft
with keyway to the strength of the same shaft without keyway.
𝑊 𝐻 Where 𝑒 = shaft strength factor
𝑒 = 1 − 0.2 − 1.1
𝐷 2𝐷 𝑊 = width of key
𝐷 = shaft diameter
𝐻 = height or thickness of key

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 EXAMPLES
EXAMPLE 1: A keyed sprocket deliver a torque of 778.8 N.m thru the shaft of 54 mm
diameter. The key thickness is 1.5875 cm and the width is 1.10 cm. Compute the
length of the same key. The permissible stress for shear is 60MPa.
Solution:
2𝑇 2𝑇 2(778.8𝐸3)
𝑆𝑠𝑘 = → 𝐿𝑠 = = = 𝟒𝟑. 𝟕 𝒎𝒎
𝐷𝑊𝐿 𝐷𝑊𝑆𝑠𝑘 54(11)(60)

EXAMPLE 2: A 20 mm square key with a length of 150 mm was used in a 100 mm

shaft to fasten a 250 mm spur gear. Determine the force that will cause the key to
shear if the shearing stress is 25 MPa.
Solution:

𝐹
𝑆𝑠 = → 𝐹 = 𝑆𝑠 𝐴𝑠 = 𝑆𝑠 𝑊𝐿 = 25 20 150 = 𝟕𝟓 𝒌𝑵
𝐴𝑠

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 EXAMPLES
EXAMPLE 3: Design the rectangular key for a shaft of 50 mm diameter. The shearing
and crushing stresses for the key material are 42 MPa and 70 MPa, respectively.
Solution:
From Table 13, 𝑾 = 𝟏𝟔 𝒎𝒎 and 𝒕 = 𝟏𝟎 𝒎𝒎

16𝑇 𝑆𝑠 𝜋𝐷3 42𝜋(50)3

𝑆𝑠𝑠 = →𝑇= = = 1,030,835 𝑁. 𝑚𝑚
𝜋𝐷3 16 16
2𝑇 2𝑇 2(1,030,835)
𝑆𝑠𝑘 = → 𝐿𝑠 = = = 𝟔𝟏. 𝟑 𝒎𝒎
𝐷𝑊𝐿 𝑆𝑠 𝐷𝑊𝐿 42(50) 16

4𝑇 4𝑇 4(1,030,835)
𝑆𝑐𝑘 = → 𝐿𝑐 = = = 𝟏𝟏𝟕. 𝟖 𝒎𝒎
𝐷𝐻𝐿 𝑆𝑐 𝐷𝐻𝐿 70(50)(10)

Use longer length, 𝑳 = 𝟏𝟏𝟕. 𝟖 𝒎𝒎

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 EXAMPLES
EXAMPLE 4: A 45 mm diameter shaft is made of steel with a yield strength of 400
MPa. A parallel key of size 14 mm wide and 9 mm thick made of steel with a yield
strength of 340 MPa is to be used. Find the required length of the key, if the shaft is
loaded to transmit the maximum permissible torque. Use maximum theory and
assume a factor of safety of 2.
Solution:

𝑆𝑠𝑠 𝜋𝐷3 100𝜋(45)3 0.5 (400)

𝑇= = = 1,789,235 𝑁. 𝑚𝑚 𝑆𝑠𝑠 = = 100 𝑀𝑃𝑎
16 16 2
2𝑇 2(1,789,235)
𝐿𝑠 = = = 𝟔𝟕 𝒎𝒎 0.5 (340)
𝑆𝑠 𝐷𝑊 85(45)(14) 𝑆𝑠𝑘 = = 85 𝑀𝑃𝑎
2

4𝑇 4(1,789,235) 𝑆𝑐𝑘 = 2𝑆𝑠𝑘 = 170 𝑀𝑃𝑎

𝐿𝑐 = = = 𝟏𝟎𝟒 𝒎𝒎
𝑆𝑐𝑘 𝐷𝐻 170(45)(9)

Use longer length, 𝑳 = 𝟏𝟎𝟒 𝒎𝒎

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 EXAMPLES
EXAMPLE 5: A cast iron pulley is to be keyed to a 2 ½ in. shaft, made of 1040, and it is
to transmit 100hp at 200 rpm. A flat key of cold-finished C1020 is to be used. The drive
is expected to quiet minor vibrations, so that a design factor of 1.75 appears reasonable.
Specify a suitable key.
Solution:
63000 ℎ𝑝 63000 (100) 𝑆𝑐𝑘 = 2 𝑆𝑠𝑘 = 2 14,571.43
𝑇= = = 31,500 𝑖𝑛. 𝑙𝑏
𝑁 200 = 29,142.86 𝑝𝑠𝑖
From Appendix 3, 𝑆𝑦𝑘 of C1020 (key) = 51 ksi or 51,000 psi
0.5𝑆𝑦𝑘 0.5 (51000)
𝑆𝑠𝑘 = = = 14,571.43 𝑝𝑠𝑖
1.75 1.75

From Table 1, for flat key of D = 2.5 in. use 𝑾 = 𝟓/𝟖” and 𝑯 = 𝟕/𝟏𝟔”
2𝑇 2(31500)
𝐿𝑠 = = = 𝟐. 𝟕𝟕"
𝑆𝑠𝑘 𝐷𝑊 (14571.43)(2.5) 5
8 Use longer length, L = 4in
4𝑇 4(31500) with 5/8 x 7/16 in. cross-
𝐿𝑐 = = = 𝟑. 𝟗𝟓" section
𝑆𝑐𝑘 𝐷𝐻 (29142.86)(2.5) 7
16

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 EXAMPLES
EXAMPLE 6: A ½ “ shaft transmits 5 hp at 1750 rpm. The shaft is made from cold-
drawn steel; the hub is made from hot-rolled 1213 steel and SAE 1045 steel key is to
be used. Design a key for a factor of safety of 2.5.
Solution:
From Table 1, for D = ½ ” use square key, thus: 𝑊 = 𝐻 = 1/8” = 0.125”
63000ℎ𝑝 63000(5)
𝑇= = = 180 𝑖𝑛. 𝑙𝑏
𝑁 1750
From Appendix 3: CD 1045: 𝑆𝑦𝑘 = 77 𝑘𝑠𝑖 From Appendix 3,
2𝑇 2(180) CD 1045: 𝑆𝑦𝑘 = 77 𝑘𝑠𝑖
𝐿𝑠 = = = 𝟎. 𝟑𝟕𝟒"
𝑆𝑠𝑘 𝐷𝑊 15,400 (0.5)(0.125) 0.5𝑆𝑦𝑘 0.5 (77,000)
𝑆𝑠𝑘 = =
𝐹𝑆 2.5
= 15,400 𝑝𝑠𝑖
4𝑇 4(180)
𝐿𝑐 = = = 𝟎. 𝟑𝟕𝟒"
𝑆𝑐𝑘 𝐷𝐻 30,800 (0.5) 0.125 𝑆𝑐𝑘 = 2 𝑆𝑠𝑘 = 30,8000 𝑝𝑠𝑖

𝟏
Use 𝑾 = 𝑯 = 𝟖 ”, 𝑳 = 𝟑/𝟖”.

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 EXAMPLES
EXAMPLE 7: A 1 ¼ -in. shaft, of cold-drawn SAE 1117, has a pulley keyed to it. Compute
the standard length of a flat key if its is made of SAE 1018. Assume a safety factor of 1.5
for the key.
Solution:
0.5 𝑆𝑦𝑠 0.5 65,000 From Table 1: for D = 1 ¼ in
𝑆𝑠𝑠 = = = 32,500 𝑝𝑠𝑖
𝐹𝑆 1 𝑊 = 1/4 “, 𝐻 = 3/16”
𝜋𝐷3 𝑆𝑠𝑠 𝜋 1.25 3 (32,500) From Appendix 3,
𝑇= = = 12,463.59 𝑙𝑏. 𝑖𝑛 CD 1117: 𝑆𝑦𝑠 = 65 𝑘𝑠𝑖
16 16
1018: 𝑆𝑦𝑘 = 54 𝑘𝑠𝑖
2𝑇 2(12,463.59)
𝐿𝑠 = = = 𝟒. 𝟒𝟑"
𝑆𝑠𝑘 𝐷𝑊 18,000 1.25 1/4 0.5𝑆𝑦𝑘 0.5 (54,000)
𝑆𝑠𝑘 = =
𝐹𝑆 1.5
4𝑇 4(12,463.59) = 18,000 𝑝𝑠𝑖
𝐿𝑐 = = = 𝟓. 𝟗𝟏"
𝑆𝑐𝑘 𝐷𝐻 (36,000)(1.25) 3/16 𝑆𝑐𝑘 = 2 𝑆𝑠𝑘 = 36,000 𝑝𝑠𝑖

Use longer length, L = 𝟓. 𝟗𝟏“ or 𝟔”.

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 EXAMPLES
EXAMPLE 8: It is required to design a square key for fixing a gear on a shaft of 25 mm
diameter. The shaft is transmitting 15 kW power at 720 rpm to the gear. The key is
made of steel 50C4 (Sy = 460 MPa) and the factor of safety is 3. If the crushing strength
of the key is equal to its shearing strength, determine the dimensions of the key.
Solution:

𝑃 15,000,000 0.5 𝑆𝑦𝑘 0.5(460)

𝑇= = = 198,943.7 𝑁. 𝑚𝑚 𝑆𝑆𝑘 = 𝑆𝑐𝑘 = =
2𝜋𝑁 720 𝐹𝑆 3
2𝜋 60
= 76.67 𝑀𝑃𝑎
4𝑇 4(198,943.7)
𝐿𝑐 = = = 𝟒𝟏. 𝟓 𝒎𝒎
𝐷𝑊𝑆𝑐𝑘 25 10 76.67 From Table 13.1:
for 𝐷 = 25𝑚𝑚 we use square key,
∴ 𝑊 = 𝐻 = 10𝑚𝑚
∴ 𝑾 = 𝑯 = 𝟏𝟎 𝒎𝒎, 𝑳 = 𝟒𝟏. 𝟓 𝒎𝒎

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 EXAMPLES
EXAMPLE 9: A huge pulley is keyed to a 7-in shaft that transmits 1000 hp at 150 rpm
to a jaw crusher. Design a key if the safety factor is 2.5.
Solution:

63000(1000)
𝑇= = 420,000 𝑖𝑛. 𝑙𝑏 Using SAE 1018, from Table 11−4:
150 𝑆𝑦𝑘 54 𝑘𝑠𝑖
2𝑇 2(420,000) 𝑆𝑐𝑘 = = = 21.6 𝑘𝑠𝑖
𝐹𝑆 2.5
𝐿𝑠 = = = 𝟔. 𝟑𝟓"
𝑆𝑠𝑘 𝐷𝑊 7 1.75 21,600
From Table 1:
4𝑇 4(420,000) for 𝐷 = 7" we use rectangular key,
𝐿𝑐 = = = 𝟕. 𝟒 𝒊𝒏 3 1
𝐷𝐻𝑆𝑐𝑘 7 1.5 21,600 ∴ 𝑊 = 1 4 ", 𝐻 = 1 2 "
𝟑 𝟏
∴ 𝑾 = 𝟏 ", 𝑯 = 𝟏 “, 𝑳 = 𝟕. 𝟒"
𝟒 𝟐

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 EXAMPLES
EXAMPLE 10: A rectangular key is used to connect a shaft with a coupling in a machine.
The shaft has a diameter of 50 mm, and the key is made from steel with a shear strength
of 60 MPa. The key's length is 100 mm, and it is positioned in a keyway in the shaft. The
key is designed to transmit a torque of 200 Nm. Determine the required key dimensions
to safely transmit the torque, assuming the key is designed to shear along its length.
Solution:
2𝑇 2𝑇 2(200,000)
𝐿𝑠 = →𝑊= = = 1.33 𝑚𝑚
𝐷𝑊𝑆𝑠𝑘 𝐷𝐿𝑆𝑠𝑘 50 100 60

The width is too short for the shaft diameter of 50 mm, (From Table 1)
𝐷 50
𝑊 = = = 𝟏𝟐. 𝟓 𝒎𝒎
4 4
𝐷 50
𝐻= = = 𝟖. 𝟑 𝐦𝐦
6 6

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 EXAMPLES
EXAMPLE 11: A 15 kW, 960 rpm motor has a mild steel shaft of 40 mm diameter and the
extension (length) being 75 mm. The permissible shear and crushing stresses for the mild
steel key are 56 MPa and 112 MPa. Design the key. Check the shear strength of the key
against the normal strength of the shaft.
Solution:
𝑃 15,000 2𝑇 2(149,207.76)
𝑇= = = 149,207.76 𝑁. 𝑚𝑚 𝑊= = = 1.8 𝑚𝑚
2𝜋𝑁 2𝜋 960 𝑆𝑠𝑘 𝐷𝐿𝑠 56 40 75
60
𝑫
Too short, for a 40 mm shaft we will use a square key so that 𝑾 = 𝑯 = = 𝟏𝟎 𝒎𝒎 .
𝟒
𝑊 𝐻 10 10
𝑒 = 1 − 0.2 − 1.1 = 1 − 0.2 − 1.1 = 0.8125
𝐷 2𝐷 40 2(40)
𝜋 𝜋
Shear strength of shaft w/ keyway = 16 𝑆𝑠𝑠 𝐷3 𝑒 = 16 56 40 3
0.8125 = 571,770 N. mm

𝐷𝑊𝐿𝑆𝑠𝑠 40(10)(75)(56)
Shear strength of key = = = 840,000 𝑁. 𝑚𝑚
2 2

Shear strength of key 840,000

∴ = = 𝟏. 𝟒𝟕
Shear strength of shaft 571,770

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B Y E N G R . D E N N IS E . G A N A S ME 414