TORSION of

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.
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Net Torque Due to Internal Stresses

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

If t is the shear stress developed in the
element, then, the elementary resisting force is,
dF t dA
? Elementary resisting torsional moment is, dT
dF r t dA r
? Total resisting torsional moment is,
The internal forces develop to counteract the
torque.
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.
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 t dA
and dT dF r
where t is shearing stress
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ASSUMPTIONS IN THE THEORY OF PURE TORSION

1. The material is homogeneous, isotropic and obeys
   Hookes 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.

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TORSION FORMULA
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.

R
The line AB rotates by an angle F and the point
B shifts to point Bl, when the free end rotates
by an angle ? due to the applied torque T.
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If F is the shear strain (angle BABl) and ?
is the angle of twist in length L, then
tan(F) BBl / AB BBl / L
Since F is small, tan(F) F BBl/L
But BBl R?
gt L F R ? -----(1)
If t is the shear stress at the surface of the
shaft and G is the modulus of rigidity, then,
G t / F
gt F t / G
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Substituting in (1), we have, (R ?) (L
t) / G
Now for a given shaft subjected to a given torque
T, the values of G, ? 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.
Thus shear stress increases linearly from zero
at axis to maximum value of t at the surface.
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If B is a point at a distance r from centre
instead of on the surface, then
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Consider an elemental area dA at a distance
r from the axis of the shaft (or centre).
If tr is the shear stress developed in the
element, then, the elementary resisting force is,
dF tr dA
? Elementary resisting torsional moment is,
dT dF
r tr dA r
But from eq. (2), we have,
12
? Total resisting torsional moment is,
But ? r2 dA is nothing but polar moment of
inertia of the section, we have J ? r2 dA .
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From (A) and (B) , we get,
T Torsional moment
(N-m)
(m4)
J Polar moment of inertia
t Shear stress
(N/m2)
R Radius of the shaft
(m)
G Modulus of rigidity
(N/m2)
(radians)
? Angle of twist
L Length of the shaft
(m)
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POLAR MODULUS
Where ZP J/R polar modulus.
Where t is maximum shear stress (occurring at
surface) and R is extreme fibre distance from
centre.
Thus polar modulus is the ratio of polar moment
of inertia to extreme radial distance of the
fibre from the centre.
Unit m3
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TORSIONAL RIGIDITY (OR STIFFNESS)
Torsional stiffness is the amount of toque
required to produce unit twist.
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
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POLAR MODULUS
Solid Circular Section
RD/2
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Hollow Circular Section
R D1/2
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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,
Power, P Work done per second.
Unit N-m/s or Watt. 1H.P 736Watt 736
N-m/s
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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.
L 6m 6000mm T 12 kN-m 12 106 N-mm
G 82 103 N/mm2 ? 3º (3p/180) radians
12 106 6000 82 103 3p/180
J
16.7695 106 mm3
For solid circular shaft,
16.7695106 p.d4 32
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gt d 114.32mm
gt R d/257.16mm
gt t 40.903 N/mm2
gt t 40.903 MPa.
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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.
L 3000 mm T 25 kN-m 25 106 N-mm
G 85103 N/mm2 ?2.5º(2.5p/180)rad
t 90 N/mm2
and RD/2
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gt D 0.1456m
Taking D 0.1456m, we have,
d 0.125m
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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.
L 3000 mm P 150 kN-m/s 150106 N-mm/s
tmax 75 N/mm2 ?max 1º p/180 radians
G 82 109 N/mm2 N 200 rpm.
7161972.439 N-mm
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For maximum shear stress condition,
gt D 78.64 mm
For maximum twist of the shaft,
gt D 111.13mm.
Hence, the safe value of diameter which satisfies
both the conditions is,
D 111.13mm.
Note Relation between D and ? (or t) is
inversly proportional.
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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.
TS TH T tS,MAX tH,MAX tMAX
LS LH L
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Since torque and stress are same,
Considering the solid shaft
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Considering the hollow shaft
Equating (1) and (2),
gt D 0.93D1.
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Weight of solid shaft,
Weight of hollow shaft,

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Dividing (3) by (4), we have,
But D 0.93D1
1.557
Weight of hollow section 0.64 times weight of
solid section
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5. Prove that a hollow shaft is stronger and
stiffer than a solid shaft of same material,
length and weight.
To prove hollow shaft is stronger
Since material, weight and length are same,
gt D2 D12 D22 -----(1)
Weight of the solid shaft weight of the hollow
shaft
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For solid shaft, torque resisted is,
For hollow shaft, torque resisted is,
Substituting (1) in the above equation we have,
Since D (D12 D22)1/2
32
?
gt 1
Hence hollow shafts are stronger than solid
shafts of same material, length and weight.
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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
K G J
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Stiffness of hollow shaft is,
KH G JH G p(D14-D24)/32
Stiffness of solid shaft is,
KS G JS G pD4/32
gt KH gt KS
gt 1
Hence hollow shafts are stiffer than solid
shafts of same material, length and weight.
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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.
P 245 103 N-m/s 245 106 N-mm/s
N 240 rpm L 1000mm G 80 103 N/mm2
tmax 40N/mm2 ?max 1º p/180 radians Tmax
1.5T
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9748.24 103 N-mm
? Tmax 1.5T 14622.360 103 N-mm
a.) For solid shaft
Let D be the diameter of solid shaft.
gt D 123.02mm
gt D 101.6mm
Hence, the diameter to be provided is, D
123.02mm.
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b.) For hollow shaft
Let d1 be the external diameter. Then internal
diameter, D2 0.5D1
gt J 0.09204d14
gt D1 125.7mm
gt D1 103.21mm
Hence provide D1 125.7mm and D2 62.85mm
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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.
P 2.2 106 N-m/s 2.2 109 N-mm/s
t 85N/mm2 N 60 rpm ? (78009.81)N/m3
350140.874 103 N-mm
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For solid shaft
gt D 275.802mm
Weight of solid shaft,
59743.842 ? L
For hollow shaft
gt J 0.06711166DE4
gt DE 313.087mm
? DI 234.815mm
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Weight of hollow shaft,
33682.011 ? L
? Percentage saving in weight,
43.62
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EXERCISE PROBLEMS
T49

• 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.
• 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.

Contd..
42
T50
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 ? 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
Contd..
43
T51

• 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
• 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 20104N-mm produces an
  angular twist of 0.9º on a length of 0.25m. Find
  the Poissons ratio, modulus of elasticity and
  modulus of rigidity for the material.
• Ans µ 0.25 E 214.5GPa G 83GPa.

Contd..
44
T52
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
Contd..
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 tMAX 10.66 MPa