TORSION OF SHAFTS CLASS NOTES FOR MECHANICAL ENGINEERING - ME Subjects

TORSION OF SHAFTS CLASS

NOTES FOR MECHANICAL ENGINEERING

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

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

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

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 /mm2 = 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 /mm2 = 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

2 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

Use torsion equation T_max/ J = τ/r_max = Gθ/L

Where T_max = maximum torque in N mm

= (π/32) d⁴ in mm⁴ for a solid shaft

= ( π/32)(d_o⁴ — d_i⁴) for a hollow shaft

τ = shear stress in the shaft, N/mm²

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

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

= (π/64 ) ( d^{4 )} for a solid shaft

= (π/64 ) (d_o⁴ — d_i⁴) for a hollow shaft

y =outer radius = d_o/2

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

Uses σ_P = P/A

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

A = area of cross section of the shaft in mm²

= (π/4) d² for a solid shaft

= (π/4) (d_o² — d_i²) for a hollow shaft

2 π N T_av /60 = W =power in watts

T_av = Average torque acting on the shaft in Nm

MAXIMUM TORQUE ‘T’_max

T_{max >}Tav

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.

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

T_eq =( T² + M²)^0.5

T_eq/ J = τ/r = Gθ/L

M_eq = (M + (M² + T²)^0.5)

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

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 =( T² + M_R²)^0.5 = Equivalent torque

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

M_eq = (M_R + (M_R² + T²)^0.5) = Equivalent bending moment

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

J polar moment of inertia, m⁴

τ is the shear stress, N/mm²

G = modulus of rigidity, GN/m²

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)

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

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

T_eq = (M2 +T2)^0.5

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

Therefore ratio σ/z = 2 M_eq/T_eq