1. The document discusses torsion of circular shafts, including pure torsion, assumptions in the theory of pure torsion, torsion formula, polar modulus, torsional rigidity, power transmitted by shafts, and numerical problems and solutions. 2. Key concepts covered include shear stress distribution in shafts under torsion, relationship between applied torque, shear stress, polar moment of inertia, and angle of twist. 3. Formulas are derived for calculating torque, shear stress, polar modulus, and torsional rigidity of solid and hollow circular shafts.

1. TORSION Of
CIRCulaR ShafTS

2. 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 of
shaft

3. Net Torque Due to Internal Stresses
• Net of the internal shearing stresses is
an internal torque, equal and opposite to
the applied torque.
( )∫ ∫== dARdFRT τr 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,

4. 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 of
shaft
dF3
dF2
dF1
Consider the section of shaft under pure torsion. The
internal forces develop to counteract the torque.

5. 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

6. 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.

7. TORSION FORMULA :
T
L
O
BA
B’
Φ
θ B
B’
O
θ
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.

8. 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 =

9. Substituting in (1), we have, (R × θ)= (L × τ) / G
τ
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.

10. 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=> =
B
r
R
R
τ = R × G×θ
L

11. r
dA
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).

12. ∴ 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)

13. From (A) and (B) , we get, T τ G × θ
J 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)

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

15. TORSIONAL RIGIDITY (OR STIFFNESS) :
T G × θ
J 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.

16. 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 :

17. Hollow Circular Section :
IXX = IYY = π(D1
4
–D2
4
)
64
x x
y
y
D1
D2
J = IXX + IYY = π(D1
4
– D2
4
)
32
R= D1/2
Polar modulus, ZP = J = π(D1
4
–D2
4
)
R 16D1

18. 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

19. 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×θ
J 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

20. ==> 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

21. 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

22. T τ G×θ
J 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×L
G×θ
d = 0.125m

23. 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 60P
2πN
(60 × 150×106
)
(2 × π × 200)
=> T = = 7161972.439 N-mm

24. 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 × θ
J L
=
T×L
G×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.

25. 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

26. 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)

27. D1
2D1/3
τMAXT
Considering the hollow shaft :
=
T J π.[D1
4
– (2D1/3)4
] /32 65π.D1
3
τmax R D1/2 1296
= = ----- (2)
Equating (1) and (2),
πD3
65πD1
3
16 1296
= => D = 0.93D1.

28. Weight of hollow shaft,
=WH π.[D1
2
– (2D1/3)2
] × L × ρ
4
=WH 5πD1
2
× L × ρ
36
----- (4)
Weight of solid shaft, =WS πD2
× L × ρ
4
----- (3)

29. Dividing (3) by (4), we have,
=WS
WH
π.D2
× L × ρ
4
5πD1
2
× 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

30. 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(D1
2
– D2
2
). L. ρ
4
= => D2
= D1
2
– D2
2
-----(1)
Weight of the solid shaft = weight of the hollow shaft

31. For solid shaft, torque resisted is,
JS ×τ
R
TS = =
(πD4
/32) τ
D/2
=> =TS
τ× πD3
16
For hollow shaft, torque resisted is,
JH ×τ
R
TH = =
[π(D1
4
–D2
4
)/32]×τ
D1/2
=> =TH
τ × π(D1
4
–D2
4
)
16D1
∴ TH D1
4
– D2
4
TS D3
× D1
=
Substituting (1) in the above equation we have,
∴ TH D1
4
– D2
4
TS (D1
2
– D2
2
)3/2
. D1
= [ Since D = (D1
2
– D2
2
)1/2
]

32. ∴
TH ( D1
2
– D2
2
) ( D1
2
+ D2
2
)
TS (D1
2
– D2
2
). (D1
2
– D2
2
)1/2
D1
=
> 1
( )
2
2
2
1
2
1
2
2
1
2
1
22
1
1
2
2
2
1
2
2
2
1
1
1
1
)(
1
1
1
)(






−














+
=






−














+
=
−
+
=
D
D
D
D
D
D
D
D
D
D
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.

33. 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
1×
=
G
J
K
K = G J

34. Hence hollow shafts are stiffer than solid shafts of
same material, length and weight.
∴ KH D1
4
– D2
4
KS D4
=
=
(D1
2
– D2
2
)(D1
2
+ D2
2
)
(D1
2
– D2
2
)2
=
D1
4
– D2
4
(D2
)2
=>
KH D1
2
+ D2
2
KS D1
2
– D2
2
= > 1 => KH > KS
Stiffness of hollow shaft is, KH = G × JH = G [π(D1
4
-D2
4
)/32]
Stiffness of solid shaft is, KS = G × JS = G [πD4
/32]

35. 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

36. P 2πNT
60
= = 9748.24 × 103
N-mm
60.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.

37. b.) For hollow shaft :
Let ‘d1’ be the external diameter. Then internal diameter,
D2 = 0.5D1
J =
π(D1
4
– D2
4
)
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

38. 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-mm
60P
2πN
=> T =

39. 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
– DI
4
)
32
=> J = 0.06711166DE
4
=> 350140874.8 85
0.067112DE
4
DE/2
= => DE = 313.087mm
∴ DI = 234.815mm
=WS πD2
× L × ρ
4

40. Weight of hollow shaft,
= 33682.011 ρ L
∴ Percentage saving in weight, =
WS – WH
WS
× 100
= 43.62%
*************************************************
=WH π.[DE
2
– DI
2
] × L × ρ
4

41. 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.]

42. 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%]

43. 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
20×104
N-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.]

44. 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]

45. 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]
*************************************************