# 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 T_{max }/ J = τ/r_{max} = Gθ/L

#### Where T_{max} = maximum torque in N mm

#### J= Polar moment of inertia

#### = (**π/32)** d^{4} in mm^{4} for a solid shaft

#### = ( **π/32)**(d_{o}^{ 4} — d_{i}^{4}) for a hollow shaft

#### τ = shear stress in the shaft, N/mm^{2}

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

#### G = modulus of rigidity in N /mm^{2 }= 80 x 1000 N/mm^{2} 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 mm^{4}

#### = (**π/64 ) **( d^{ 4 )} for a solid shaft

#### = (**π/64 )** (d_{o}^{ 4} — d_{i}^{4}) for a hollow shaft

#### σ = Bending stress in the outer radius

#### y =outer radius = d_{o}/2

#### E = Young’s modulus in N /mm^{2 }= 200 x 1000 N/mm^{2} 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 σ_{P} = P/A

#### σ_{P} = Axial stress due to axial thrust ‘P’ in N /mm^{2}

#### P = Axial thrust in N

#### A = area of cross section of the shaft in mm^{2}

#### = (**π/4) ** d^{2} for a solid shaft

#### = (**π/4) ** (d_{o}^{2} — d_{i}^{2}) 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**

#### 2 **π** N T_{av} /60 = W =power in watts

#### Where N = RPM= revolutions per minute

#### T_{av} = 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}

_{max}

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

#### T_{max > }Tav

#### Normally

#### T_{max } = 1.0 to 2.0 _{ }Tav

#### In the absence of any data, take T_{max }= _{ }Tav

**NOTE: **_{ }T_{av} MUST BE CONVERTED TO T_{MAX } BEFORE PROCEEDING FURTHER.

_{ }T

_{av}MUST BE CONVERTED TO T

_{MAX }BEFORE PROCEEDING FURTHER.

**DESIGN OF SHAFTS**

#### (i) Only T is acting

#### Torsion equation is used i.e. T_{max }/ J = τ/r = Gθ/L

**(ii) When T and M are acting.**

#### (a) Find Equivalent torque

#### T_{eq} =( T^{2} + M^{2})^{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.**

#### T_{eq }/ J = τ/r = Gθ/L

#### (b) Find Equivalent bending moment

#### M_{eq} = (M + (M^{2} + T^{2})^{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. M_{eq / I} = σ/y=E/R

**(iii) When T, M and P are acting**

#### Stress due to P σ_{P} = P/A

#### Stress due to M σ_{M} = I σ /y

#### Resultant stress due to P and M, σ_{R} = σ_{P} + σ_{M}

#### Find M_{R} from σ_{R using}

#### M_{R} = (I σ_{R ) / y}

#### Now it becomes Case 2 with M_{R} and T

#### T_{eq} =( T^{2} + M_{R}^{2})^{0.5} = Equivalent torque

#### Use Equivalent torque equation is used when **allowable shear stress is given. T**_{eq }/J = τ/r = Gθ/L

_{eq }/J = τ/r = Gθ/L

#### M_{eq} = (M_{R} + (M_{R}^{2} + T^{2})^{0.5}) = Equivalent bending moment

#### Use Equivalent bending moment when allowable normal stress (σ) is given. M_{eq /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, m^{4}

** **τ is the shear stress, N/mm^{2}

#### G = modulus of rigidity, GN/m^{2}

#### 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} = (πD^{3}/16) z

#### T _{hollow} = π(D^{4}—d^{4})/16 D) z

#### Ratio T _{solid} / T _{hollow} = [(πD^{3}/16) z]/[ π(D^{4}—d^{4})/16 D) z]= D^{4}/(D^{4}—d^{4})

#### T _{solid} / T _{hollow} = D^{4}/(D^{4}—d^{4})

**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

#### M_{eq} = (1/2) [M+(M2 +T2)^{0.5}]

#### Maximum bending stress σ = (M_{eq}/I) (D/2)

#### Equivalent torque is

#### T_{eq} = (M2 +T2)^{0.5}

#### Maximum shear stress z = (T_{eq}/ 2I) (D/2)

#### Therefore ratio σ/z = 2 M_{eq}/T_{eq}

** 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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