Shaft,keys And Coupling 1h5k3c

CHAPTER 04 – Shaft, Keys & coupling

Introduction, Application , Numerical etc.

SHAFT ¢ 

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In machinery, the general term “shaft” refers to a member, usually of circular cross-section, which s gears, sprockets, wheels, rotors, etc. Transmission shaft refers to rotating machine element, which gear, pulley, and sprocket, and transmit power. Shaft are usually cylindrical, may be square or cross-shaped in section. The shaft is always stepped with maximum diameter in the middle & minimum at end to mount the bearing Fillet provided to avoid stress concentration due to abrupt change in cross section

SHAFT q 

An “axle” is a non-rotating member that used to s rotating member wheels, pulleys which is fitted o housing by bearing and carries no torque.

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A “spindle” is a short rotating shaft. Used in al machine tools, such as line shaft, head shaft, transmission shaft, countershaft, are names associated with special usage. q 

Counter shaft – secondary shaft, driven by the main shaft

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Jackshaft – Intermediate shaft between two shaft, function same as countershaft

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Line shaft – Consist no. of shaft, connected in axial direction by coupling, used in group drive

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A good practice for Material selection —  Start with an inexpensive, low or medium carbon steel for the first time through the design calculations. —  Applications requiring greater strength often specify alloy steels, Nickel, chromium corrosion applications call for brass, stainless, Ti, or others. —  The cost of the material and its processing must be weighed against the need for smaller shaft diameters. Manufacturing of Shafts —  For low production, turning is the usual primary shaping process. An economic viewpoint may require removing the least material. —  High production may permit a volume conservative shaping method (hot or cold forming, Forging) —  Shoulder ¢  It allows precise positioning ¢  to minimize deflection. ¢  In cases where the loads are small, positioning is not very important, shoulders can be eliminated.

ELEMENTS ATTACHED TO A SHAFT Shoulders provide axial positioning location, & allow for larger center shaft diameter – where bending stress is highest.

SHAFT DESIGN FOR STRESS It is not necessary to evaluate the stresses in a shaft at every point; a few potentially critical locations will suffice. Critical locations will usually be on the outer surface. ¢  Possible Critical Locations, axial locations where: —  The bending moment is large and/or —  The torque is present, and/or —  Stress concentrations exist. ¢  Stresses due to —  Shear stress (Transmission torque) —  Bending stress (machine element) —  Combined Torsional & Bending loads ¢ 

STRESS DUE TO TORSION Stress should not exceed yield point due to twisting moment ¢  Diameter of shaft obtained by Torsion equation ¢ 

T J

=

Where, ¢  T = Twisting moment ¢  J = Polar moment of inertia = ( / 32) x d4 ¢  τ = Torsional shear stress ¢  r = Distance from NA to outer most fiber = d / 2

τ r T

( / 32) x d4

=

τ d/2

T = ( / 16) x τ x d3 Torque can be obtained by Power formula

SUM 1 ¢  A

line shaft at 200rpm is to transmit 20kw. Shaft is made of up mild steel of shear stress 42MPa. Determine the diameter of shaft by torsion formula

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Solution —  Calculate T = torque ¢  Power = (2 NT / 60 ) —  Calculate d = Shaft Dia

STRESS DUE TO BENDING MOMENT Stress should not exceed yield point due to twisting moment ¢  Diameter of shaft obtained by pure bending equation ¢ 

M

=

σb y

I

Where, ¢  M = Bending moment ¢  I = Moment of inertia = ( / 64) x d4 ¢  σb = Bending stress ¢ 

M

( / 64) x d4

y = Distance from NA to outer most fiber = d / 2

= dσ/ 2b

M = ( / 32) x σb x d3

STRESS DUE TO COMBINED TORSION & BENDING MOMENT ¢ 

When shaft subject to two moment simultaneously various theories are suggested for elastic failure of the material —  — 

Maximum shear stress theory or Guest’s Theory – Ductile Material Maximum normal stress theory or Ranking's Theory – Brittle material

q Maximum shear stress theory or Guest’s Theory - Max shear stress at shaft

!max = (1 / 2) x [(σb)2 + 4(!)2]1/2

!max = (1 / 2) x [(32M / d3 )2 + 4 (16T/ !max = (16 /

d3)2]1/2

d3) x [ M2 +

T2 ]1/2

[ M2 + T2 ]1/2 = Equivalent twisting moment = Denoted by Te

q Maximum normal stress theory or Ranking's Theory

σb(max) = (1 / 2) σb + {(1 / 2) x [(σb)2 + 4(!)2]1/2} Putting σb & ! In above equation

M = ( / 32) x σb x d3 T = ( / 16) x τ x d3 We get,

σb(max) = (32 /

d3) x { (1/2) M + [M2 + T2 ]1/2

{(1/2) M +[ (M2 + T2) ½] } = Equivalent Bending moment = Denoted by Me

STRESS DUE TO AXIAL, COMBINED TO TORSION & BENDING LOAD ¢  Shaft

is subjected to axial load (tensile or compressive) P = F / A = F / [( / 4) x d2] ¢  Resultant Stress

32 M 4 F σx = + 3 πd πd2

CH-18

Static or Quasi-Static Loading on Shaft

LEC 29

Normal stress

Shear stress

Slide 13

The stress at an element located on the surface of a solid round shaft of diameter d subjected to bending, axial loading, and twisting is

32 M 4 F σx = + 3 πd πd2 16T τ xy = 3 πd 2

Non-zero principal stresses

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⎤ ⎛ σ x + σ y ⎞ ⎡⎛ σ x − σ y ⎞ 2 σ A ,σ B = ⎜ ⎟ ± ⎢⎜ ⎟ + τ xy ⎥ ⎥⎦ ⎝ 2 ⎠ ⎢⎣⎝ 2 ⎠

keys ¢ 

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A key is the piece inserted in an axial direction between a shaft and hub of the mounted machine element such as pulley or gear etc.  A key is used for temporary fastening, Key t consist of shaft, hub, and key The shaft and rotating machine element must have a keyway, also known as a key seat, which is a slot or pocket the key fits in. The whole system is called a keyed t 2 basic function of key —  — 

Primary function is to transmit the torque from shaft to the hub of mating element and vice versa Secondary function is to prevent rotational motion between the shaft and the hub like gear or pulley.

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Slot machined on the shaft & hub called keyway, usually cut by milling cutter Drawback – keyway result in stress concentration in the shaft & part becomes weak Material – usually plain carbon steel to withstand shear and compressive stresses Types of keys are —  Saddle key & sunk key —  Square key & flat key —  Key with & without Gib head —  Special types

Square & rectangular key ¢ 

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The sunk keys are provided half in the keyway of the shaft and half in the keyway of the hub or boss of the pulley. 1.  Rectangular sunk key -  Also called flat key -  More stability than square key -  Used in machine tool 2.  Square sunk key - Used in general industrial machinery The only difference between a rectangular sunk key and a square sunk key is that its width and thickness are equal, i.e. b = h Selecting key without stress analysis – Thumb Rule —  Square, b = h = (d / 4), l = 1.5 d —  Rectangular b = d / 4, h = (2b / 3) = (d / 6), l = 1.5 d Where, d = Diameter of shaft (mm) b = Width of key (mm) h = Height of key (mm) l = Length of key (mm)

Design of Keys – stress analysis to determine required length 1. Forces ( P ) due to the torque transmitted by the shaft. 2. Forces ( P’ ) due to fit of the key in its keyway, as in a tight fitting straight key driven in place.

Force (P) = Mt / (d/2) Where Mt = Transmitted torque d = diameter of shaft P = force on key

Failure of key in shear: occurs at Plane AB

Strength of a Sunk Key Considering shearing of the key, the tangential shearing force acting at the circumference of the shaft:

P = Area resisting shearing × Shear stress

τ

=

P

=

Plane AB

2 Mt = dxbxl bxl P

Where, d = Diameter of shaft (mm) b = Width of key (mm) h = Height of key(mm) l = Length of key (mm)

Failure of key in Compressive

Crushing between shaft and key : Considering crushing of the key, the tangential crushing force acting at the circumference of the shaft: P = Area resisting crushing × Crushing stress

σc

=

P Surface AC

=

P (h/2) x l

Square Key h = b

σc =

4 Mt bxlxd

=

2 Mt (h/2) x l x d

=

4 Mt hxlxd

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permissible crushing stress for the usual key material is at least twice the permissible shearing stress

τ

2 Mt = dxbxl

¢  From

σc =

4 Mt dxbxl

above equation, We get

σc = 2 τ Sum 1 required to design a square key for fixing gear on a shaft of 25mm diameter. The shaft transmitting 15kw power at 720rpm to the gear. Ultimate tensile stress is 460MPa, factor of safety 3. Determine the dimensions of the key. Solution : 1.  find out permissible stress 2.  Find out torque (Mt) by power formula 3.  Calculate b = h = (d / 4) 4.  Calculate l by shearing 5.  Check the crushing stress by putting above value

Coupling ¢ 

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All manufacturing plants have at least 2 things in common. These are: —  Electric motors and. —  Couplings - There are many different types, sizes and styles. All couplings; 1. Connect 2 shafts together 2. Require safety guards. Most couplings; 1. Do more than 1 function at a time. 2. Connect 2 shafts together. 3. Allow for minor misalignment. 4. Allow for axial movement.

¢  Two —  — 

major classification

Solid couplings (also called rigid or sleeve couplings) Flexible couplings

q  Rigid

coupling : It is used to connect two shafts which are perfectly aligned. Types of rigid coupling are ü  ü  ü 

Sleeve or muff coupling. Clamp or split-muff or compression coupling, Flange coupling

q  Flexible

coupling : It is used to connect two shafts having both lateral and angular misalignment. Types of flexible coupling are ü  ü  ü 

Bushed pin type coupling, Universal coupling, and Oldham coupling

Muff Coupling ¢  ¢  ¢  ¢  ¢ 

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Also called sleeve or box coupling It is the simplest type of rigid coupling, made of cast iron. It consist of hollow cylinder or sleeve, shaft (input & output) & key Sleeve inner diameter is the same as that of the shaft diameter. It is fitted over the ends of the two shafts by means of a sunk key, the power is transmitted from one shaft to the other shaft by means of a key and a sleeve. Simple in design & manufacture.

Design procedure 1. Calculated the diameter of shaft (d) : Shaft subjected to pure Torsional moment,

τ

=

16 Mt d3

2. Usually defined on shop floor. Empirical relationship were developed by engineer on the basis their past experience then check for shearing D = 2d + 13 L = 3.5d Where, D = Outer diameter of the sleeve L = axial length of sleeve

Mt x r τ = J 3. Determine the dimension of key l = L / 2 – each shaft 4. Check shear & compressive stress in the key

τ

2 Mt = dxbxl

σc =

4 Mt dxbxl

J= D =r 2

(D4 – d4)

32

SUM 1 ¢ 

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Design muff coupling to connect two steel shaft transmitting 25kw power at 360rpm. The shaft and key are made of plain carbon steel of ultimate yield stress 400MPa. Factor of safety for shaft & key 4. the sleeve made of grey cast iron, ultimate yield stress 200MPa. Factor of safety for shaft & key 6. Solution : —  Find out permissible stress —  Diameter of shaft —  Dimensions of sleeve —  Dimensions of key —  Key for stresses in key

CLAMP COUPLING ¢ 

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Second type of rigid coupling, Also called split muff coupling or compression coupling In this, sleeve is made of two halves, which are split along a plane ing trough axes of shaft The two halves of the sleeve are clamped together by means of bolts No. of bolts can be four or eight It is easy to assemble & dismantle

Design procedure 1. Calculated the diameter of shaft (d) : Shaft subjected to pure Torsional moment,

Mt x r τ= J 2. Usually defined on shop floor. Empirical relationship were developed by engineer on the basis their past experience D = 2.5 d

L = 3.5 d

Where, D = Outer diameter of the sleeve L = axial length of sleeve

3. Determine the dimension of key l = L / 2 – each shaft 4. Check shear & compressive stress in the key

τ

=

2 Mt dxbxl

σc =

4 Mt dxbxl

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Calculate the diameter of clamping bolts by using, (d1) 1.

By empirical, when d < 55 mm ------- d = diameter of shaft d1 = 0.2d + 10 mm d > 55 mm d1 = 0.15d + 15 mm

4. Clamping force (P1) : Each bolt P1 = ( / 4) d12 x !t 2. Clamping force (N) : Each Shaft (half bolts) N = (P1 x n) / 2 where n = Total no. of bolts 3. Frictional force : between sleeve halves & shaft Mt =

f x d x P1 x n 2

P1 =

2 x Mt fxdxn

FLANGE COUPLING