JJ310 STRENGTH OF MATERIAL Chapter 6 Torsion

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STRENGTH OF MATERIAL

JJ 310

CHAPTER 6

TORSION

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Learning Outcomes:At the end of this lecture, student should be

able to;

i) Understand the definition of torsionii) Understand the general equation for

torsioniii) Define the polar second moment of

inertia for solid and hollow shaftiv) Understand the transmission of powerv) Understand the shafts in series and

parallelvi) Solve the problem related to solid and

hollow shaft

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INTRODUCTION TO TORSIONTorsion is the stress produced in an object

when it's twisted. The amount of force that twists the object is measured in torque.

Ways to put an object into torsion include holding one end of it still while you twist the other end, or twisting both ends in opposite directions (imagine wringing out a wet towel).

When an object is twisted beyond its limit, it breaks.

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Assumption in Torsion

The shaft has a uniform cross section

The shaft material is uniform throughout and the shear stress is proportional to the shear strain (Elastic region)

The shaft is straight and initially unstressed

The axis of twisting moment is the axis of the shaft

Plain transverse sections remain the same after twisting

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General Equation for Torsion

Where; τ = Shear stress (Pa) L = Length (m)R = Radius (m) T = Torque (Nm)G = Modulus of rigidity (Pa) J = Polar 2nd

moment θ = Angle of twist (Radian/Degree) of area (m4)

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Other formulas

Polar 2nd moment of area for solid and hollow shaft;

i) JSolid = πd4

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ii) JHollow = π (D4 - d4) D = outer diameter

32 d = inner diameter

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ExamplesExample 1:

A solid steel shaft (G = 80 GPa) of circular cross-section, length = 0.5 m, diameter = 20 mm, is twisted about its axis of symmetry by applying a torque of 72Nm. Calculate:

i) The maximum shear stressii) The angle of twist

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Solution to Example 1Given; G = 80 GPa, L = 0.5 m, d = 20 mm, T =

72 Nm

JSolid = πd4 i) Shear stress,

32 τ = TR

= π (0.02)4 J

32 = 72 x (0.01)

= 1.57 x 10-8 m4 1.57 x 10-8

= 45.9 MPa #

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ii) Angle of twist

θ = τL

RG

= (45.9 x 106) (0.5)

(0.01) (80 x 109)

= 22.95 x 106

800 x 106

= 0.03 rad #

Change radian to degree;x 180 π

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Example 2:

A shaft is made from tube 50mm outer diameter and 20mm inner diameter. The shaft is 0.7m long and subjected to a torque of 1200Nm. Given G = 90 GPa.

Calculate;

i) Maximum shear stressii) Angle of twist

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Solution to Example 2i) Max shear stress

JHollow = π (D4 - d4) τ = TR

32 J

= π (0.054-0.024) = 1200 x (0.025)

32 5.98 x 10-7

= 5.98 x 10-7 m4 = 50.2 MPa #

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ii) Angle of twist

θ = τL

RG

= (50.2 x 106) (0.7)

(0.025) (90 x 109)

= 35.14 x 106

2.25 x 109

= 0.02 rad #

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Transmission of PowerShafts are the medium to transmit power from the

point of power generation to the point of its application.

The power transmitted is the work done per second.

Thus, Power = Torque x Velocity of the shafts

P = Power (Watt)T = Torque (Nm)ω = Velocity of shaft in rad/sec

N = Velocity of shaft in rev/min (rpm) ω = 2πN / 60

P = Tω @

P = 2πNT / 60

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Example 1

A shaft is made from tube 25mm outer diameter and 20mm inner diameter. The shear stress must not exceed 150MPa. Calculate the maximum power that should be transmitted at 500 rev/min.

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Solution to Example 1 Given;

N = 500 rev/min d = 20 mm

D = 25 mm τ = 150 MPa

JHollow = π (D4 - d4) T = τJ

32 R

= π (0.0254 - 0.024) = (150 x106 ) x (2.26 x 10-8)

32 2.26 x 10-8

= 2.26 x 10-8 m4 = 271.2 Nm

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Thus the maximum power is;

P = 2πNT

60

= 2π(500) x 271.2

60

= 14.2 KW #

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Shafts in Series

When two shafts are joined in series and a single torque is applied, both shafts are subjected to the same torque .

T = T1 = T2

T1 T2 Bar 1 Bar 2

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Meanwhile the angle of twist is the sum of twist of each shaft.

Thus;θ = T1L1 + T2L2

G1J1 G2J2

θ = T L1 + L2

G1J1 G2J2

∆θ = θ1 + θ2

From the general torsion equation, we know;

θ = TL

GJ

Known;T = T1 = T2

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Shafts in Parallel

When two shafts are joined in parallel, torque applied is the sum of the torques on the two shafts.

T = T1 + T2

T1 T1Bar 1

Bar 2

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Meanwhile the angle of twist of each shaft are same.

Thus; θ = T1L1 = T2L2

G1J1 G2J2

∆θ = θ1 = θ2

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Exercise1. A shaft is made of solid round bar 30mm diameter and

0.5m long. The shear stress must not exceed 200MPa. Calculate; (Take G = 90 GPa)

i) The maximum torque (Answer: 1060 Nm)

ii) The angle of twist (Answer: 0.07 rad)

2. A steel shaft 5m long having diameter of 50 mm is to transmit power at a rotational speed of 600 rev/min. If the maximum shear stress is limited to 60MPa, determine; (Take G = 80 GPa)

i) The maximum power (Answer: 92.5 KW)

ii) The angle of twist (in degree) (Answer: 8.6°)