SHAFT

INTRODUCTION

 A shaft is a rotating machine element which is used to transmit power from one place to another. The power is delivered to the shaft by some tangential force and the resultant torque (or twisting moment) set up within the shaft permits the power to be transferred to various machines linked up to the shaft. In order to transfer the power from one shaft to another, the various members such as pulleys, gears etc.. are mounted on it. These members along with the forces exerted upon them causes the shaft to bending. In other words, we may say that a 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 crass-shaped in section. They are solid in cross-section but sometimes hollow shafts an also used.

Material :

The material used for shafts should have the following properties 

1. It should have high strength.

2. It should have good machinability.

3. It should have low notch sensitivity factor.

 4. It should have good heat treatment properties.

5. It should have high wear resistant properties. 

The material used for ordinary shafts is carbon steel of grades 40C 8.45 C 8.50 C4 and 50C 12. The mechanical properties of these grades of carbon steel are given in the following table.

When a shaft of high strength is required then an alloy steel such a. Nickel, nickel-chromium or chrome-vanadium steel is used. 

Manufacturing of Shafts :

Shafts are generally manufactured by hot rolling and finished to size by cold drawing or turning and grinding. The cold rolled shafts are stronger than hot rolled shafts but with higher residual stresses. The residual stresses may cause distortion of the shaft when it is machined especially when slots or keyways are cut. Shafts of larger diameter are usually forged and turned to size in a lathe.

Stresses in shafts

The following stresses are induced in the shafts: 

1. Shear stresses due to the transmission of torque (i.e. due to torsional load).

2. Bending stresses (tensile or compressive) due to the forces acting upon machine elements like gears. pulleys etc. as well as due to the weight of the shaft itself. 

3. Stresses due to combined torsional and bending stresses

Design of Shaft 

The shafts may be designed on the basis of 

 1. Strength 

 2. Rigidity and stiffness

In designing of shafts on the basis of strength, the following cases may be considered

(a) Shafts subjected to twisting moment or torque only

(b) Shafts subjected to bending moment only

(c) Shafts subjected to combined twisting and bending moments

(d) Shafts subjected to axial loads in addition to combined torsional and bending loads.

DESIGN OF SHAFTS SUBJECTED TO TWISTING MOMENT / TORQUE ONLY:

We have the general Torsion equation as   T / J = Ƭ / r

Where T = Torsional moment / Twisting Moment / Torque - N-mm

            J = Polar Moment Inertia of cross sectional area about the axis of rotation - mm⁴

            Ƭ = Torsional Shear stress of the shaft – MN / mm²

             r = Radius of the outer most fabric from the axis of the rotation

              = d/2, where d = dia. of the shaft.

Also J = πd⁴ / 32

T = πd⁴ x Ƭ       =       πd³ x Ƭ  

      32     d/2               16

If d₀ = Outer diameter of a hollow shaft

    dᵢ = Inside dia. of the hollow shaft

Then r = d₀ / 2   &       J =      π x [d₀⁴ - dᵢ⁴]

                                 32

Then T = π  x [d₀⁴ -  dᵢ⁴] x 2 Ƭ     =     π  x   Ƭ  [d₀⁴ -  dᵢ⁴]    =   π  x Ƭ  x   d₀⁴   1 –   dᵢ  ⁴   

   32                      d₀          16                 d₀             16             d₀          d₀              

            =  π/16 x Ƭ x d₀³ [ 1 – k⁴]                                   Where k/d₀  =     dᵢ

              

Ie.   π d³ Ƭ  =  π  Ƭ    d₀ [ 1 – k⁴]   =  d³ =  d₀³ [1 – k⁴] 

                    16           16                   

               

We also know that Torque T is given by    T = P x 60/ 2 π n

Where P = Power generated / transmitted in kW   &     n = Revolutions per Minute (r. p. m).

DESIGN OF SHAFTS SUBJECTED TO ONLY BENDING MOMENT:

We have the General bending equation          M   =    σ 

                                                                    I        r

Where M = Bending Moment – Nmm                       I = Second moment of area / Moment of Inertia - mm⁴

            σ = Bending Stress – MN / mm²                  r = radius of the shaft = d/2, where d = dia. of the shaft

Also I of a solid shaft =     πd⁴  

                                       64

Then        M/  πd⁴ / 64         =      σ /d/2         (or)    M =    σ πd³ /32                            ……….. Eqn. (1)

                                                              

For a hollow shaft 

d₀ = Outer diameter                             dᵢ = Inside diameter

M =    σ π [d₀³ - dᵢ³]        from eqn. (1)  

                32

M =    σ π d₀³  1-    dᵢ  ³                       M =     σ   π   d₀³ [1- k³]     where            k = dᵢ

32               d₀                                32                                                      d₀

SHAFTS SUBJECTED TO BOTH TWISTING & BENDING MOMENTS

1.  The shaft must be designed on the basis of two moments simultaneously.

2.  Materials are subjected to elastic failure when subjected to multiple forces

Shafts under multiple forces are analyzed by two types of theories.

1. Maximum shear stress theory – Guests theory – Used for ductile materials
2. Maximum normal stress theory – Rankine’s theory – Used for brittle materials

Maximum shear stress theory:

We have, for a solid shaft, the max. shear stress (τmax) is given by

Ƭ max. =  1  √(σb)² + 4 (Ƭ)²   ……… eqn. II         

              2

σ  = Bending stress – MN / mm²          Ƭ  = Shear stress MN / mm²                   D = Diameter of the shaft.

But we have  σ = σb  = 32 M    &  Ƭ =   16 T

                                         πd³               πd³

Substituting these values in eqn (II) Ƭmax. =        1   √   32M    ²  + 4    16T     ²

                                                                                2       πd³                 πd³

      =        1       16     √ (2m)² + 4 (T)²

    2       πd³

       =      1         16      2 √ M² + T²

                                                                           2         πd³

Ƭmax.  = 16    √ M² + T² …………………………… (2)

               πd³

Ƭmax.  πd³     = √ M² + T² ……………………….. (3)

              16

• The value √ M² + T² is called “Equivalent twisting moment” – Te.
• Te is defined as the Twisting Moment, which when acting alone, will produce the same shear stress (τ)as the actual Twisting Moment

If Max. Shear stress is equal to allowable shear stress then Ƭ max.   = Ƭ  =  Te

√ M² + T²   =    Ƭ π d³   = Te ………………………………………………….. (4)

     16

Maximum normal stress theory – Rankine’s theory

We have, for a solid shaft, the max. Bending stress (τmax) is given by

 σb (max). =  1   σb +  1  √(σb) ² + 4 (Ƭ) ²   ……… ………………………… (IV)      

                     2            2

Substituting  σb  = σ =           32M       &              Ƭ  =   16T      in the above equation

                                             πd³                              πd³ 

σb max. =  1        32M  +  1 √    32M   ² + 4   16T    ²

                  2       πd³       2         πd³               πd³ 

             

                = 16 M  +   1       16      √ (2M)² + 4 T²

                     πd³      2        πd³ 

                 =  16 M  + 1  x    16   x  2 √ M² +  T²

                    πd³      2        πd³    

             =  16     [M + √ M² +  T²]  = σb max (πd³) 

                 πd³     

               =  σb(max.) π d³  =  1 [M + √ M² +  T²]

                                  32         2

v  The term 1  [M + √ M ² +  T²] is called equivalent bending moment (Me)

  2

v  Me is defined as the bending moment, which when acts alone, will produce the same bending stress as the normal bending moment.

If σbmax. = σ,   then, σb πd³  =  1    [M + √ M² +  T²]  = Me …………………….. (V)

                                      32            2

FOR A HOLLOW SHAFT:          If d₀ = Outer diameter & dᵢ = Inside diameter

Then   √ M² + T²   =   Ƭ   π d₀³ [1 - k⁴] = Te  ………………………. (VI)

                                         16

σb  π d₀³ [1 - k⁴]    =   1    [M + √ M² +  T²]  = Me ……………………. (VII),   where k = dᵢ / d₀

        32                         2