TORSION OF SHAFTS CLASS NOTES FOR MECHANICAL ENGINEERING

TORSION OF SHAFTS CLASS

NOTES FOR MECHANICAL ENGINEERING

Shafts transmit power. These are

subjected to torsion and hence to shear

stress. But due to self weight, weight of

flywheel/pulley, shafts are subjected to

bending. The shaft is in rotation, there

comes reversal of stresses. It causes

fatigue. Fatigue reduces the strength of

shaft. Maximum failure in machinery

having shaft occur due to fatigue.

Therefore, design of shaft requires

special attention.

 VARIOUS FORCES/MOMENTS ACTING ON A SHAFT 

There are three types of forces acting on a shaft. These are torque, bending moment and axial thrust. All the three acts on a propeller shaft. Torque and bending acts on all other shafts except the propeller shafts. Propeller shafts are those in which a vehicle or machine is mobile i.e. going from one place to another place. For example, a car, bus, truck, ship etc. Shafts under torque only is just theoretical.

1. Turning moment OR Torque to rotate the shaft and it causes variable shear stress.

Fig. (a) Torsion of a shaft (b) Shaft cross section

It is a must for every shaft.

Use torsion equation  Tmax / J = τ/rmax = Gθ/L

Where Tmax = maximum torque in N mm

J= Polar moment of inertia

= (π/32) d4 in mm4 for a solid shaft

= ( π/32)(do 4 — di4) for a hollow shaft

τ = shear stress in the shaft, N/mm2

r =radius of the solid shaft, Or  outer radius of the hollow shaft in mm

G = modulus of rigidity in N /mm= 80 x 1000 N/mm2 for steel shafts

θ = Angle of twist in radians

L = length of the shaft in mm

2. Bending moment due to the weight of the shaft, weight of the pulley and due to the belt tensions etc.

Uses bending equation

M / I = σ/y = E/R

M = Maximum bending moment in N mm found from the bending moment diagram

I = Moment of inertia of the entire beam about the centroid axis in mm4

=  (π/64 ) ( d 4 ) for a solid shaft

= (π/64 )  (do 4 — di4) for a hollow shaft

σ = Bending stress in the outer radius

y =outer radius = do/2

E = Young’s modulus  in N /mm= 200 x 1000 N/mm2 for steel shafts

R = radius of curvature in mm

3. Axial thrust due to the forward motion of a machine like all vehicles, boat, ship and an aero-plane

Uses σP = P/A

σP = Axial stress due to axial thrust ‘P’ in N /mm2

P   = Axial thrust in N

A  = area of cross section of the shaft in mm2

=  (π/4)  d2 for a solid shaft

= (π/4)   (do2 — di2) for a hollow shaft

STRESSES AND STRAINS IN SHAFTS

Shaft is a power transmitting device. There are normally two shafts for power transmission.

1. Electric motor shaft or Diesel engine shaft

2. Machine shaft say shaft of a lathe machine, drilling machine, shaper machine, any other machine.

High speed electric motors are most efficient. Therefore RPM of the motor shaft is different from the machine shaft RPM. Hence a belt/chain/ gear drive is required to transfer power from the motor shaft to the machine shaft.

VARIOUS SHAPES OF SHAFTS

1. Solid circular like a rod

2. Hollow circular like a pipe

POWER TRANSMISSION EQUATION

π N Tav /60 = W =power in watts

Where N = RPM= revolutions per minute

Tav = Average torque acting on the shaft in Nm

= Actual torque acting on the shaft

1 HP = 735.5 watts in SI Units

1 kW = 1000 W

MAXIMUM TORQUE ‘T’max

It is the torque for which shaft is designed or fabricated.

Tmax > Tav

Normally

Tmax   = 1.0 to 2.0  Tav

In the absence of any data, take Tmax =  Tav

NOTE:  Tav MUST BE CONVERTED TO TMAX  BEFORE PROCEEDING FURTHER.

DESIGN OF SHAFTS

(i) Only T is acting

Torsion equation is used i.e.  Tmax / J = τ/r = Gθ/L

(ii) When T and M are acting.

(a)  Find  Equivalent torque

Teq =( T2 + M2)0.5

Equivalent torque is one which acting alone produces the combined shear effect of T and bending effect of M together

Equation uses Equivalent torque in the design of shafts when allowable shear stress is given.

Teq / J = τ/r = Gθ/L

(b) Find  Equivalent bending moment

Meq =  (M + (M2 + T2)0.5)

Equivalent bending moment is one which acting alone produces the combined bending effect of T and M together

It is important to notice that equivalent bending moment is used when allowable normal stress (σ) is given.  Meq / I = σ/y=E/R

(iii) When T, M and P are acting

Stress due to P     σP = P/A

Stress due to M      σM =  I σ /y

Resultant stress due to P and M,  σR = σP + σM

Find MR from σR using

MR = (I σR ) / y

Now it becomes Case 2 with MR and T

Teq =( T2 + MR2)0.5 = Equivalent torque

Use Equivalent torque equation is used when allowable shear stress is given. Teq /J = τ/r = Gθ/L

Meq =  (MR + (MR2 + T2)0.5) = Equivalent bending moment

Use Equivalent bending moment when allowable normal stress (σ) is given.   Meq /I = σ/y=E/R

NOTE: Uses both equations separately when allowable shear stress as well as allowable normal stresses are given.

Two values of shaft diameter will be obtained. Use the bigger diameter as final selection as it will satisfy both stress conditions.

 Torsion equation 

T / J =  τ / r = G θ / L

T = torque acting on the shaft, Nm

J polar moment of inertia, m4

 τ is the shear stress, N/mm2

G = modulus of rigidity, GN/m2

θ = Angle of twist, radiations

L = length of the shaft, m

 Power equation for a shaft

The power equation is

2 πNT/60 =Power in Watts

Where N is RPM

T is torque in Nm

 Assumptions used in torsion of shafts

(i)  Stresses and strains are within elastic limit.

(ii)  Shaft is circular in section.

(iii)   A transverse plane remains a transverse plane before and after the torque is applied.

(iv) Material is homogeneous, isotropic and continuous.

(v)  Shaft is initially straight.

(vi)  There is no twisting of the radial lines of the shaft.

(vii) A constant turning moment (torque) is applied over the entire shaft.

 Hollow shaft superior to a solid shaft in torsion

Hollow shaft: Stress variation is less in a hollow shaft. There is material saving in a hollow shaft.

Solid shaft: Shear stress at the center is zero and it is quiet less for some radius. It is therefore not the economic  use of  the material. Thus hollow shafts are better.

VARIATION OF SHEAR STRESS AND SHEAR STRAIN IN TORSION OF A SOLID SHAFT

These will be maximum at the outer radius and minimum (Zero) at the zero radius i.e.at the center as shown in the figure.

AB is the stress at the outer radius and zero stress at the center O.

CD is the strain at the outer radius and zero shear strain at the center E.

Fig. Shear stress              Fig. Shear strain variation

variation in a solid shaft        in a solid shaft

 VARIATION OF SHEAR STRESS AND SHEAR STRAIN IN TORSION OF A HOLLOW SHAFT

These will be maximum at the outer radius and minimum (Zero) at the inner radius as shown in the figure.

GH is the shear stress at the outer radius and  IJ is the shear stress at the inner radius.

KL is the shear strain at the outer radius and  MN is the shear strain at the inner radius.

Fig. Shear Stress Variation   Fig. Shear Strain Variation

in a Hollow Shaft                       in a Hollow Shaft

Problem

Find the ratio of the torsional moments of resistance of a solid circular shaft of diameter ‘D’ and a hollow circular shaft having external diameter ‘D’ and internal diameter ‘d’.

Torsional moment is torque’ T ’.

T solid shaft = (πD3/16) z

T hollow = π(D4—d4)/16 D) z

Ratio T solid / T hollow = [(πD3/16) z]/[ π(D4—d4)/16 D) z]= D4/(D4—d4)

T solid / T hollow = D4/(D4—d4)

Problem

A solid circular shaft is subjected to a bending moment of M and a torque T. What is ratio of maximum bending stress to the maximum shear stress?

A.

Equivalent bending moment is

Meq = (1/2) [M+(M2 +T2)0.5]

Maximum bending stress σ = (Meq/I) (D/2)

Equivalent torque is

Teq = (M2 +T2)0.5

Maximum shear stress z = (Teq/ 2I) (D/2)

Therefore ratio σ/z = 2 Meq/Teq

 Difference between torsional rigidity and lateral rigidity.

Torsional rigidity= GJ

Torsional rigidity is resistance against rotational deflection i.e. angle of twist is minimum i.e.

Angle of twist = θ =TL/GJ

Therefore for θ to be minimum, GJ should be maximum. Hence torsional rigidity is GJ.

Lateral rigidity = EI

Lateral rigidity is least lateral deflection, ’y’

Deflection to be minimum, radius of curvature should be large.

WE know M/I = E/R

R = EI/M

For R to be maximum, EI should be maximum for a given value of M. Therefore EI is lateral rigidity.

https://mesubjects.net/wp-admin/post.php?post=7685&action=edit            Polar & moment of inertia

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