TORSION ofCircular shafts

A bar subjected to moment in a plane perpendicular to the longitudinal axis (i.e., in the plane of cross – section of the member) is said to be in ‘TORSION’.

This moment is called ‘Twisting Moment’ or ‘Torque’.

Torque = T Unit : N-m, kN-m, etc.

T

T

Axis ofshaft

Net Torque Due to Internal Stresses

• Net of the internal shearing stresses is an internal torque, equal and opposite to the applied torque.

dARdFRT r r

The internal forces develop to counteract the torque.

If ‘τ’ is the shear stress developed in the element, then, the elementary resisting force is, dF = τ × dA

∴ Elementary resisting torsional moment is, dT = dF × r = τ × dA × r

∴ Total resisting torsional moment is,

PURE TORSION :

A member is said to be in ‘Pure torsion’, when its cross sections are subjected to only torsional moments (or torque) and not accompanied by axial forces and bending moment.

T

Axis ofshaft

dF3

dF2

dF1

Consider the section of shaft under pure torsion. The internal forces develop to counteract the torque.

At any element, the force dF developed is in the direction normal to radial direction.

This force is obviously shearing force and thus the elements are in pure shear.

If dA is the area of the element at a distance ‘r’ from the axis of the shaft, then

dF = τ × dA and dT = dF × r

where τ is shearing stress

ASSUMPTIONS IN THE THEORY OF PURE TORSION:

1. The material is homogeneous, isotropic and obeys Hooke’s law (stresses are within elastic limit, i.e., shear stress is proportional to shear strain).

2. Cross sections which are plane before applying twisting moment remain plane even after the application of twisting moment, i.e., no warping takes place.

3. Radial lines remain radial even after applying torque, i.e., circular sections remain circular.

4. The twist along the shaft is uniform.5. Shaft is subjected to pure torsion.

TORSION FORMULA :

T

L

OBA

B’

Φθ B

B’

Oθ

The line AB rotates by an angle Φ and the point B shifts to point Bl, when the free end rotates by an angle ‘θ’ due to the applied torque ‘T’.

R

When a circular shaft is subjected to torsion, shear stresses are set up in the material of the shaft. To determine the magnitude of shear stress at any point on the shaft, Consider a circular shaft of length ‘L’, fixed at one end and subjected to a torque ‘T’ at the other end as shown.

If ‘Φ’ is the shear strain (angle BABl) and ‘θ’ is the angle of twist in length ‘L’, then

tan(Φ) = BBl / AB = BBl / L

Since Φ is small, tan(Φ) = Φ = BBl/L

But BBl = Rθ

=> L × Φ = R × θ -----(1)

If ‘τ’ is the shear stress at the surface of the shaft and ‘G’ is the modulus of rigidity, then,

G = τ / Φ => Φ = τ / G

R × θ L

∴ Φ =

Substituting in (1), we have, (R × θ)= (L × τ) / Gτ R

G × θ L

==> ---------(A)

Thus shear stress increases linearly from zero at axis

to maximum value of τ at the surface.

Now for a given shaft subjected to a given torque T, the values of G, θ and L are constant. Hence shear stress produced is proportional to the radius R

From equation A, it is clear that intensity of shear stress at any point in the cross – section of the shaft subjected to pure torsion is directly proportional to its distance from the centre.

If B is a point at a distance ‘r’ from centre instead of on the surface, then

τr r

G × θ L

=τr ----------(2) r

τ R=> =

Br

R

R

τ = R × G×θ L

rdA

dF

R R

τ

r τr

If ‘τr’ is the shear stress developed in the element,

then, the elementary resisting force is, dF = τr × dA

∴ Elementary resisting torsional moment is,

dT = dF × r = τr × dA × r

But from eq. (2), we have, τr τ × r R

=

∴ dT = τ × dA × r2 R

Consider an elemental area ‘dA’ at a distance ‘r’ from the axis of the shaft (or centre).

∴ Total resisting torsional moment is,

But ∫ r2 × dA is nothing but polar moment of inertia of the section, we have J = ∫ r2 × dA .

∴ T = τ × J R

τ × r2 × dA τ ∫ r2 × dA R R∫T = => T =

∴ T τ τr

J R r= = ----------(B)

From (A) and (B) , we get, T τ G × θJ R L

= =

T = Torsional moment

J = Polar moment of inertia

τ = Shear stress

R = Radius of the shaft

G = Modulus of rigidity

θ = Angle of twist

L = Length of the shaft

(N-m)

(m4)

(N/m2)

(m)

(N/m2)

(radians)

(m)

POLAR MODULUS :

T τ × J τ × ZP

R==

Where ZP = J/R = polar modulus.

Thus polar modulus is the ratio of polar moment of inertia to extreme radial distance of the fibre from the centre.

Unit : m3

T τJ R

=

Where τ is maximum shear stress (occurring at surface)and R is extreme fibre distance from centre.

TORSIONAL RIGIDITY (OR STIFFNESS) :

T G × θJ L

= T G × J × θ L

=

When unit angle of twist is produced in unit length, we have, T = G × J × (1/1) = GJ.

Thus the term ‘GJ’ may be looked as torque required to produce unit angle of twist in unit length and is called as, ‘torsional rigidity’ or ‘stiffness’ of shaft.

Unit : N-mm2

Torsional stiffness is the amount of toque required to produce unit twist.

Solid Circular Section :

x x

y

y

D

IXX = IYY = πD4

64

J = IXX + IYY = πD4

32 R=D/2

Polar modulus, ZP = J = πD3

R 16

POLAR MODULUS :

Hollow Circular Section :

IXX = IYY = π(D14 –D2

4) 64

x x

y

yD1

D2

J = IXX + IYY = π(D14 – D2

4) 32

R= D1/2

Polar modulus, ZP = J = π(D14 –D2

4) R 16D1

POWER TRANSMITTED BY SHAFTS :

Consider a shaft subjected to torque ‘T’ and rotating at ‘N’ revolutions per minute (rpm). Taking second as the unit of time, we have,

Angle through which shaft moves = N × 2π 60

Power, P = T × N × 2π = 2π NT 60 60

Power, P = Work done per second.

Unit : N-m/s or Watt. 1H.P = 736Watt = 736 N-m/s

NUMERICAL PROBLEMS AND SOLUTIONS

1.What is the maximum diameter of a solid shaft which will not twist more than 3º in a length of 6m when subjected to a torque of 12 kN-m? What is the maximum shear stress induced in the shaft ? Take G = 82 GPa.

T τ G×θJ R L

= =

Solution : L = 6m = 6000mm; T = 12 kN-m = 12 × 106 N-mm

G = 82 × 103 N/mm2; θ = 3º = (3π/180) radians

J = T×L G×θ

= 16.7695 × 106 mm3

J = π.d4 32

For solid circular shaft,

J = 12× 106 ×6000 82 × 103 ×3π/180

16.7695×106 = π.d4 32

==> τ T.R J

T τ J R

=

==> τ 12 × 103 × 0.5716 16.7695 × 106

=> τ = 40.903 N/mm2

=> τ = 40.903 MPa.

=> d = 114.32mm => R = d/2=57.16mm

2. A hollow shaft 3m long transmitted a torque of 25kN-m. The total angle of twist in this length is 2.5º and the corresponding maximum shear stress is 90MPa. Determine the external and internal diameter of the shaft if G = 85 GPa.

Solution :

d

D/2

D/2

L = 3000 mm; T = 25 kN-m = 25 × 106 N-mm

G = 85×103 N/mm2; θ=2.5º=(2.5π/180)rad

τ = 90 N/mm2

and R=D/2J = π.(D4 - d4) 32

T τ G×θJ R L

= =

τ R

G × θ L

=Taking, R ==> τ×L G×θ

∴ R D (90 × 106 ) × (3) 2 (85 × 109) ×(2.5π/180)

= = => D = 0.1456m

=J T×L G×θ

Also,

Taking D = 0.1456m, we have,

=> π(D4 - d4) 32

= T×LG×θ

d = 0.125m

3. A solid circular shaft has to transmit 150kW of power at 200 rpm. If the allowable shear stress is 75 MPa and permissible twist is 1º in a length of 3m, find the diameter of the shaft. Take G = 82 GPa.

Solution : L = 3000 mm; P = 150 kN-m/s = 150×106 N-mm/s

G = 82 × 109 N/mm2; N = 200 rpm.

τmax = 75 N/mm2; θmax = 1º = π/180 radians;

Power, P = 2π NT 60

==> T 60P2πN

(60 × 150×106) (2 × π × 200)

=> T = = 7161972.439 N-mm

For maximum shear stress condition,

T τ J R

= => πD4 32

7161972.4 75 D/2

= => D = 78.64 mm

For maximum twist of the shaft,

Hence, the safe value of diameter which satisfies both the conditions is,

T G × θJ L

=T×LG×θ

=> J = => =πD4 32

7161972.4 × 3000 82 × 103 × (π/180)

=> D = 111.13mm.

D = 111.13mm.

Note: Relation between ‘D’ and ‘θ’ (or τ) is inversly proportional.

4. Two circular shafts of same material are subjected to same torque producing the same maximum shear stress. If the first shaft is of solid section and the second shaft is of hollow section, whose internal diameter is 2/3 of the external diameter, compare the weights of the two shafts. Both the shafts are of equal length.

Solution : TS = TH = T ; τS,MAX= τH,MAX= τMAX ; LS = LH = L

τMAX

D

T

D1

2D1/3

τMAXT

τMAX

D

T

Considering the solid shaft :

T τ J R

=T J

τ R==>

Since torque and stress are same,

TS TH T

τS,max τH,max τmax

= = = Constant.

=T J πD4/32 πD3

τmax R D/2 16 = = ----- (1)

D1

2D1/3

τMAXT

Considering the hollow shaft :

=T J π.[D1

4 – (2D1/3)4] /32 65π.D13

τmax R D1/2 1296 = = ----- (2)

Equating (1) and (2),

πD3 65πD13

16 1296= => D = 0.93D1.

Weight of hollow shaft,

=WH π.[D12 – (2D1/3)2] × L × ρ

4

=WH 5πD12 × L × ρ

36----- (4)

Weight of solid shaft,

=WS πD2 × L × ρ 4

----- (3)

Dividing (3) by (4), we have,

=WS

WH

π.D2 × L × ρ 4

5πD12 × L × ρ

36

Weight of hollow section = 0.64 times weight of solid section

= 1.5579 D 2

5 D1

=

=>WH

WS

= 0.64 = 64%

But D = 0.93D1

5. Prove that a hollow shaft is stronger and stiffer than a solid shaft of same material, length and weight.

Solution : To prove hollow shaft is stronger:

τ

D

TS

D1

D2

τTH

Since material, weight and length are same,

πD2 . L. ρ 4

π(D12 – D2

2). L. ρ 4

= => D2 = D12 – D2

2 -----(1)

Weight of the solid shaft = weight of the hollow shaft

For solid shaft, torque resisted is,

JS ×τ R

TS = =(πD4/32) τ D/2

=> =TSτ× πD3

16

For hollow shaft, torque resisted is,

JH ×τ R

TH = =[π(D1

4–D24)/32]×τ

D1/2=> =TH

τ × π(D14–D2

4) 16D1

∴ TH D14 – D2

4

TS D3 × D1

=

Substituting (1) in the above equation we have,

∴ TH D14 – D2

4

TS (D12 – D2

2)3/2 . D1

= [ Since D = (D12 – D2

2)1/2 ]

∴TH ( D1

2 – D22 ) ( D1

2 + D22)

TS (D12 – D2

2) . (D12 – D2

2)1/2 D1

=

> 1

2

2

2

1

2

1

2

21

2

1

221

12

22

1

22

21

11

1

)(1

1

1

)(

DD

DD

DDD

D

DD

D

T

T

DDD

DD

T

T

S

H

S

H

Hence hollow shafts are stronger than solid shafts of same material, length and weight.

To prove hollow shaft is stiffer:

Stiffness of shaft may be defined as torque required to produce unit rotation in unit length. Let this be denoted by K. Then from torsion formula

1

1G

J

KK = G J

Hence hollow shafts are stiffer than solid shafts of same material, length and weight.

∴ KH D14 – D2

4

KS D4 =

=(D1

2 – D22)(D1

2 + D22)

(D12 – D2

2)2

=D1

4 – D24

(D2)2

=>KH D1

2 + D22

KS D12 – D2

2 = > 1 => KH > KS

Stiffness of hollow shaft is, KH = G × JH = G [π(D14-D2

4)/32]

Stiffness of solid shaft is, KS = G × JS = G [πD4/32]

6. A shaft is required to transmit 245kW power at 240 rpm. The maximum torque is 50% more than the mean torque. The shear stress in the shaft is not to exceed 40N/mm2 and the twist 1º per meter length. Taking G = 80kN/mm2, determine the diameter required if,a.) the shaft is solid.b.) the shaft is hollow with external diameter twice the internal diameter.

Solution : P = 245 × 103 N-m/s = 245 × 106 N-mm/s

N = 240 rpm; L = 1000mm; G = 80 × 103 N/mm2

τmax = 40N/mm2; θmax = 1º = π/180 radians; Tmax = 1.5T

P 2πNT 60= = 9748.24 × 103 N-mm60.P

2πN=> T =

∴ Tmax = 1.5T = 14622.360 × 103 N-mm

a.) For solid shaft : Let ‘D’ be the diameter of solid shaft.

Tmax τmax J R

= => 14622.36×103 40 πD4/32 D/2

= => D =123.02mm

Tmax G×θmax

J L=

=> D = 101.6mm

=> 14622.36×103 (80×103)×(π/180) πD4/32 1000

=

Hence, the diameter to be provided is, D =123.02mm.

b.) For hollow shaft :

Let ‘d1’ be the external diameter. Then internal diameter, D2 = 0.5D1

J = π(D1

4 – D24)

32=> J = 0.09204d1

4

Tmax τmax J R

= => 14622.36×103 40 0.09204D1

4 D1/2= => D1 =125.7mm

Tmax G× θmax

J L=

=> D1 = 103.21mm

=> 14622.36×103 (80×103)×(π/180) 0.09204D1

4 1000=

Hence provide D1 = 125.7mm and D2 = 62.85mm

J = π(D1

4 – (0.5D1)4) 32

7. A power of 2.2MW has to be transmitted at 60 r.p.m. If the allowable stress in the material of the shaft is 85MPa, find the required diameter of the shaft, if it is solid. If instead, a hollow shaft is used with 3DE = 4DI , calculate the percentage saving in weight per meter length of the shaft. Density of the shaft material is 7800kg/m3.

Solution : P = 2.2 × 106 N-m/s = 2.2 × 109 N-mm/s

τ = 85N/mm2; N = 60 rpm; ρ = (7800×9.81)N/m3

P 2πNT 60= = 350140.874 × 103 N-mm60P

2πN=> T =

For solid shaft :

T τ J R

= => 350140.874 ×103 85 πD4/32 D/2

= => D =275.802mm

Weight of solid shaft, = 59743.842 ρ L

For hollow shaft :

T τ J R

=

J = π(DE

4 – DI4)

32=> J = 0.06711166DE

4

=> 350140874.8 850.067112DE

4 DE/2= => DE = 313.087mm

∴ DI = 234.815mm

=WS πD2 × L × ρ 4

Weight of hollow shaft,

= 33682.011 ρ L

∴ Percentage saving in weight,

=WS – WH

WS × 100

= 43.62%

*************************************************

=WH π.[DE2 – DI

2] × L × ρ 4

T49

Contd..

EXERCISE PROBLEMS :

1. A hollow shaft 75mm outside diameter and 50mm inside diameter has a maximum allowable shear stress of 90N/mm2. What is the maximum power that can be transmitted at 500 rpm ?[Ans : 312.93 kW or 417 H.P.]

2. Determine the diameter of a solid circular shaft which has to transmit a power of 90 H.P at 210 rpm. The maximum shear stress is not to exceed 50MPa and the angle of twist must not be more than 1º in a length of 3m. Take G = 80GPa.[Ans : D = 119 mm.]

T50

Contd..

3. Find the diameter of the shaft required to transmit 12kW at 300 rpm if the maximum torque is likely to exceed the mean torque by 25%. The maximum permissible shear stress is 60N/mm2. Taking G = 0.84 × 105 N/mm2, find the angle of twist for a length of 2m.[Ans : θ = 4.76º.

A solid shaft of circular cross-section transmits 1200kW at 100 rpm. If the allowable stress in the material of the shaft is 80MPa, find the diameter of the shaft.

If the hollow section of the same material, with its inner diameter (5/8)th of its external diameter is adopted, calculate the economy achieved.[Ans : D = 194mm; % Saving in material = 31.88%]

T51

Contd..

5. A hollow shaft of diameter ratio 0.6 is required to transmit 600 kW at 110 rpm. The maximum torque being 12% more than the mean. The shearing stress is not to exceed 60MPa and the twist in the length of 3m not to exceed 1º. Calculate the maximum external diameter of the shaft. G = 80GPa.[Ans : 190.3mm]

6. During test on sample of steel bar 25mm in diameter, it is found that the pull of 50kN produces a extension of 0.095mm on the length of 200mm and a torque of 20×104N-mm produces an angular twist of 0.9º on a length of 0.25m. Find the Poisson’s ratio, modulus of elasticity and modulus of rigidity for the material.[Ans : µ = 0.25; E =214.5GPa; G = 83GPa.]

T52

Contd..

7. A solid aluminium shaft 1m long and 60mm diameter is to be replaced by a tubular steel shaft of same length and same outside diameter (60mm) such that each of the two shaft could have same angle of twist per unit torsional moment over the total length. What must be the inner diameter of the tubular shaft if GS = 4GA.[Ans : Di = 45.18mm]

8. A hollow shaft has diameters DE = 200mm and DI = 150 mm. If angle of twist should not exceed 0.5º in 2m and maximum shear stress is not to exceed 50MPA, find the maximum power that can be transmitted at 200 rpm. Take G = 84GPa.[Ans : 823kW]

T53

9. A hollow marine propeller shaft turning at 110 rpm is required to propel a vessel at 12m/s for the expenditure of 6220kW, the efficiency of the propeller being 68 percent. The diameter ratio of the shaft is to be (2/3) and the direct stress due to thrust is not to exceed 8MPa. Calculate (a) the shaft diameters(b) the maximum shearing stress due to torque.

[Ans : DI = 212mm; DE = 318mm; τMAX = 10.66 MPa]

*************************************************