Shafting(Lec15)

Shafting ME 147P Machine Design I

• Shaft: is a rotating machine element which is used to transmit power from one place to another and subjected to torsion or a combination of torsion, bending and axial loading. • In order to transfer the power from one shaft to another, the various members such as pulleys, gears etc., are mounted on it. • Shaft is used for the transmission of torque and bending moment. • The various members are mounted on the shaft by means of keys or splines. • The shafts are usually cylindrical, but may be square or cross-shaped in section. They are solid in cross-section but sometimes hollow shafts are also used.

• An axle, though similar in shape to the shaft, is a stationary machine element and is used for the transmission of bending moment only. • It simply acts as a support for some rotating body such as hoisting drum, a car wheel or a rope sheave. • A spindle is a short shaft that imparts motion either to a cutting tool (e.g. drill press spindles) or to a work piece (e.g. lathe spindles).

Types of Shafts • 1. Transmission shafts. These shafts transmit power between the source and the machines absorbing power. The counter shafts, line shafts, over head shafts and all factory shafts are transmission shafts. Since these shafts carry machine parts such as pulleys, gears etc., therefore they are subjected to bending in addition to twisting. • 2. Machine shafts. These shafts form an integral part of the machine itself. The crank shaft is an example of machine shaft. Standard Sizes of Transmission Shafts • For SI Units: 25 mm to 60 mm with 5 mm steps; 60 mm to 110 mm with 10 mm steps ; 110 mm to 140 mm with 15 mm steps ; and 140 mm to 500 mm with 20 mm steps. • For English Units: Use the preferred sizes in our previous lecture.

• Pure Torsion

Tr J 16T Ss   solid shaft 3 D 16TDo Ss   hollow shaft 4 4  Do  Di Ss 



• Pure Bending



Mc Sf  I 32 M Sf   solid shaft 3 D 32 MDo Sf   hollow shaft 4 4  Do  Di





Stresses in Shafts

• Angular Deflection   

TL JG

 

TL

 solid shaft

4

32

DG TL

32

D

4 o



D G 4 i

 hollow shaft

Deflections of Shafts

• Lateral Deflection • For Simply Supported Shaft: FL3 y 48 EI

• For Cantilever: FL3 y 3EI

• Axial and Flexural Loads:

S max S max

Sy S Mc F    S allow  or u I A N N 32 M 4 F 32  FD     M 3 2 3  D D D  8 

Combined Loads on Shafts

• Axial and Torsional Loads:

F 4F Tr 16T S  ; Ss   2 A D J D 3 Based on Max. Shear Stress Theory : S smax 

S s 2   S 

2



16  FD  2 T    3 D  8 

2

2 Based on Max. Normal Stress Theory : S max

S   2

2

16 S 2 S s      3 D 2

2  FD FD     T2     8  8   

• Flexural and Torsional Loads:

Mc 32 M Tr 16T  ; S   s 3 I D J D 3 Based on Max. Shear Stress Theory : S

2

S smax

16 S 2 2 2  S s      T  M D 3 2

Based on Max. Normal Stress Theory : S max

S   2

2



16 S 2 S s      3 M  T 2  M 2 D 2



• Axial, Flexural and Torsional Loads

Mc F 32  FD  Tr 16T S   M  ; S s  3  I A D  8  J D 3 Based on Max. Shear Stress Theory : 16 FD   2 T  M    3 2  D 8     Based on Max. Normal Stress Theory : S smax 

S max

S s 2   S 

2

S   2

2



2

S s 2   S   163 D 2

2  FD FD   M   T2 M    8 8      

• Maximum Shear Stress Theory (Guest’s Theory) • This theory is applicable for ductile materials. • The equation shows the maximum shear stress that would result in a combination of shear stress and normal stress applied to a machine member. • This is also based on the assumption that a ductile material is most probably going to fail by shearing.

S smax 

S s 

2

S   2

2

For design considerations, Ssmax is equated to the allowable stress.

Maximum Stress Theories

• Maximum Normal Stress Theory (Rankine’s Theory) • This theory is applicable for brittle materials. • The equation shows the maximum normal stress that would result in a combination of shear stress and normal stress applied to a machine member. • This is also based on the assumption that a brittle material is most probably going to fail by normal stress (e.g. bending)

S max

S   2

S s 

2

S   2

2

For design considerations, Smax is equated to the allowable stress.

• This is based on the two aforementioned stress theories described. • The ff. expressions are used as design stresses based on ASME Code for Design of Transmission Shafting:

Shear Design Stress : S sd  0.3S y or 0.18Su  use whichever is smaller. If shaft is with a keyway, S sd '  0.75S sd Normal Design Stress : S d  0.6 S y or 0.36 Su  use whichever is smaller. If shaft is with a keyway, S d '  0.75S d

Shaft Design by ASME Code

For Solid Shaft

Tr 16T Mc 32 M Ss   ; Sf   3 J D I D 3 If material is ductile, max. shear stress theory applies : S smax  S smax

S s 

2

2

2

S  16T   32 M      3   3 2  D   2D 

16 2 2  T  M ; 3 D

2

T 2  M 2  equivalent twisting moment

Combined Torsion and Bending on Shaft

Considering factors due to combined shock and fatigue from Table 9.1 K s  for shear; K m  for bending S smax

K sT 2  K m M 2  S s

16  D 3

d

If material is brittle, max. normal stress theory applies :

S s 

2

32 M S  16T   32 M       3  3 3 2 2  D  D 2  D      

S max

S   2

S max

16 2 2  M  T  M D 3

M 

2



2





T 2  M 2  equivalent bending moment

2

Considering factors due to combined shock and fatigue from Table 9.1 K s  for shear; K m  for bending S max

16  D 3

K M   m

Table 9.1, Recommended Values of Km and Ks, from the textbook by Faires is similar to the table below:

K sT 2  K m M 2   S d 

For Hollow Shaft

16TDo Di Tr Ss   ; let j   Di  jDo 4 4 J  Do  Di Do





16TDo 16T Ss   4 4 4  Do  j Do Do3





 1  1  j 4   

32 MDo Mc S  Sf   I  Do4  Di4



32 MDo 32 M S  4 4 4  Do  j Do Do3

 1  1  j 4   







If material is ductile, max. shear stress theory applies, use : S smax

16  Do3

 K sT    K m M  2

2

 1  1  j 4   S sd  

If material is brittle, max. normal stress theory applies, use : S max

16  Do3

K M   m

 K sT 

2

 1    K m M    Sd  1  j 4  2

If shaft is solid, the equation above can also be referred to by using Do=D and j=0 since Di=0.

FD   K M   S d for solid shaft   m  8  32  1   FD 1  j 2   K M   S d for hollow shaft  3  4  m Do 1  j   8 

S max  S max

32 D 3





  factor for column action   1, if axial load is tensile

Combined Axial and Bending Loads on Shaft

Column Factor In case of long shafts (slender shafts) subjected to compressive loads, a factor known as column factor (α) must be introduced to take the column effect into account. 2

L S y  e  k Le  fr. Euler' s formula  A. For  120,   2 k  E Le 1 fr. Johnson' s formula  B. For 30   120,   2 k  Le     Sy k   1   2   4 E    

From Max. Shear Stress theory : S smax 

16  1    3  4  Do  1  j 

K sT 2   K m M  FDo 1  j 

 From Max. Normal Stress theory : S max



2

  S sd 

8



 K T 

 FDo 1  j 2 16  1     KmM    3  4  Do  1  j   8 

s

2





2 2    FD 1  j 2 o  Km M     S d 8   

If shaft is solid, the equation above can also be referred to by using Do=D and j=0 since Di=0.

Combined Axial, Torsional and Bending Loads on Shaft

• A solid circular shaft is to transmit 20 hp at 600 rpm. It also supports a bending load of 2000 in-lbs. The shaft is made of AISI C1020, as rolled steel. A gear is fastened at the midpoint by means of a key. If the load is gradually applied with a reversed bending, determine the shaft diameter. • Solve the above if shaft is hollow where Do=2Di. P63025 2063025 T   2100.83in  lbs. N rpm 600 M  2000in  lbs Properties of AISI C1020, as rolled steel S y  48000 psi Su  65000 psi

Sample Problem 1

If material is ductile, max. shear stress theory : 16 2 2  K sT    K m M   S s d ' S smax  3 D S sd  0.3S y  0.348000  14400 psi S sd  0.18Su  0.1865000  11700 psi

Use whichever is smaller, that is, 11700psi. But shaft is with keyway, S sd '  0.75S sd  0.7511700  8775 psi For combined shock and fatigue factors : From Table 9.1, p.279, gradually applied load with reversed bending : K s  1.0; K m  1.5

Solving values : 16 2 2 1.02100.83  1.52000  S sd '  8775 S smax  3 D D

3

16

1.02100.83  1.52000  8775 2

2

 1.286in  1 3 in 8

For brittle material : 16  2 2     S max  K M  K T  K M  Sd ' s m 3  m   D  16  2 2             S max  1 . 5 2000  1 . 0 2100 . 83  1 . 5 2000  Sd ' 3    D  S d  0.6 S y  0.648000  28800 psi S d  0.36 Su  0.3665000  23400 psi

Use whichever is smaller, that is, 23400psi. But shaft is with keyway, S d '  0.75S d  0.7523400  17550 psi 16 1.52000   D3  D  1.24in  1 1 in 4

1.02100.832  1.520002   17550

Given : Transmission System Shaft material properties : S y  60000 psi Su  78000 psi

Sample Problem 2

Pressure angle of gears  20 Load is gradually applied. Both shafts have keyways. Required : diameters of shafts x and y

shaft x : Assume material is ductile, 16 2 2  K sT    K m M   S s d ' S smax  3 D P 3063025 T   3151.25in  lbs. Nx 600 For M : 2T 23151.25 Ft    1050.42lbs. Dx 6

Fr  Ft tan   1050.42tan 20  382.32lbs 2 2     FR  Ft  F  1050.42  382.32  1117.83lbs. FR Lx 1117.83lbs 8in  M   2235.66in  lbs 2

4

2 r

4

Design Stress : S sd  0.3S y  0.360000  18000 psi

S sd  0.18Su  0.1878000  14040 psi

Use whichever is smaller, that is, 14040psi. But shaft is with keyway, S sd '  0.75S sd  0.7514040  10530 psi For combined shock and fatigue factors : From Table 9.1, p.279, gradually applied load with reversed bending : K s  1.0; K m  1.5

Solving values : 16 2 2           S smax  1 . 0 3151 . 25  1 . 5 2235 . 66  S sd '  10530 3 D D3

16

1.03151.252  1.52235.662  10530

 1.3in  1 3 in 8

If material is brittle : 16  2 2     S max  K M  K T  K M  Sd ' s m 3  m   D  16  2 2             S max  1 . 5 2235 . 66  1 . 0 3151 . 25  1 . 5 2235 . 66  Sd ' 3    D  S d  0.6 S y  0.660000  36000 psi S d  0.36 Su  0.3678000  28080 psi

Use whichever is smaller, that is, 28080psi. But shaft is with keyway, S d '  0.75S d  0.7528080  21060 psi 16 1.52235.66    D3  D  1.24in  1 1 in 4

1.03151.252  1.52235.662   21060

shaft y : Assume material is ductile, 16 2 2     S smax  K T  K M  S sd ' s m 3 D Dx  D y N x Dx P T ; Ny  ;C  Ny Dy 2 D y  29"  6"  12"; N y 

6006   300rpm 12

3063025 T  6302.5in  lbs. 300 For M : Ft  1050.42lbs; Fr  382.32lbs; FR  1117.83lbs. M

FR Ly 4

 1117.83lbs 12in    3353.49in  lbs 4

Design Stress : S sd  0.3S y  0.360000  18000 psi

S sd  0.18Su  0.1878000  14040 psi

Use whichever is smaller, that is, 14040psi. But shaft is with keyway, S sd '  0.75S sd  0.7514040  10530 psi For combined shock and fatigue factors : From Table 9.1, p.279, gradually applied load with reversed bending : K s  1.0; K m  1.5

Solving values : 16 2 2           S smax  1 . 0 6302 . 5  1 . 5 3353 . 49  S sd '  10530 3 D D3

16

1.06302.52  1.53353.492  10530

 1.57in  1 5 in 8

If material is brittle : 16  2 2     S max  K M  K T  K M  Sd ' s m 3  m   D  16  2 2             S max  1 . 5 3353 . 49  1 . 0 6302 . 5  1 . 5 3353 . 49  Sd ' 3    D  S d  0.6 S y  0.660000  36000 psi S d  0.36 Su  0.3678000  28080 psi

Use whichever is smaller, that is, 28080psi. But shaft is with keyway, S d '  0.75S d  0.7528080  21060 psi 16 1.53353.49    D3  D  1.47in  1 1 in 2

1.06302.52  1.53353.492   21060

• A solid circular shaft is subjected to a bending moment of 3000 N-m and a torque of 10 000 N-m. The shaft is made of a material having ultimate tensile stress of 700 MPa and an ultimate shear stress of 500 MPa. Assuming a factor of safety as 6, determine the diameter of the shaft.

More Problems

• A shaft supported at the ends by ball bearings carries a straight tooth spur gear at its mid span and is to transmit 7.5 kW at 300 rpm The pitch circle diameter of the gear is 150 mm. The distances between the centre line of bearings and gear are 100 mm each. If the shaft is made of steel and the allowable shear stress is 45 MPa, determine the diameter of the shaft. The pressure angle of the gear may be taken as 20° and the load may be assumed as gradually applied.

• A line shaft is driven by means of a motor placed vertically below it. The pulley on the line shaft is 1.5 metre in diameter and has belt tensions 5.4 kN and 1.8 kN on the tight side and slack side of the belt respectively. Both these tensions may be assumed to be vertical. If the pulley be overhung from the shaft, and the distance of the centre line of the pulley from the centre line of the bearing being 400 mm, find the diameter of the shaft. Assuming that the yield strength of the shaft material is 187 MPa and the load is gradually applied with reversed bending. The pulley is also mounted on the shaft with a key.

• A mild steel shaft transmits 20 kW at 200 r.p.m. It carries a central load of 900 N and is simply supported between the bearings 2.5 metres apart. Determine the size of the shaft, if the allowable shear stress is 42 MPa and the maximum tensile or compressive stress is not to exceed 56 MPa. What size of the shaft will be required, if it is subjected to gradually applied loads?