Chapter 5 Torsion

ll bb lRussell C. Hibbeler

Ch T i Chapter 5: Torsion

• The effects of applying a torsional loading to a long

Objectives

• The effects of applying a torsional loading to a long straight member such as shaft or tube.

• To determine both stress distribution within the member and the angle of twist when the material behaves in a linear elastic manner and also when it isbehaves in a linear-elastic manner and also when it is inelastic.

Due to axial distribution of shearstress shafts made from woodstress, shafts made from woodtend to split along the axial planewhen subjected to excessivetorque

This tubular drive shaft for a truck wassubjected to an overload resulting in failurecaused by yielding of the material.

Torsional Loads on Circular ShaftsTorsional Loads on Circular Shafts• Stresses and strains of circular

shafts subjected to twisting couplesor torques*

T bi t t T th

• Shaft transmits the torque to the

• Turbine exerts torque T on theshaft

• Shaft transmits the torque to thegenerator

• Generator creates an equal andopposite torque T’

Chapter 5: TorsionChapter 5: TorsionMechanics of Material 7Mechanics of Material 7thth EditionEdition* Momen that tends to twist a member about its longitudinal axis. Its effect is of primary concern in the design of axles or drive shafts used in vehicle and machinery.

ExampleExample

Net Torque Due to Internal Stressesq• Net of the internal shearing stresses is an internaltorque, equal and opposite to the applied torque,

dArdFrT

• Although the net torque due to the shearing• Although the net torque due to the shearingstresses is known, the distribution of the stresses isnot

• Distribution of shearing stresses is staticallyindeterminate – must consider shaft deformations

• Unlike the normal stress due to axial loads, thedistribution of shearing stresses due to torsionalloads can not be assumed uniform.

Chapter 5: TorsionChapter 5: TorsionMechanics of Material 7Mechanics of Material 7thth EditionEdition

Shaft Deformations• From observation, the angle of twist of theshaft is proportional to the applied torque andto the shaft length.

L• When subjected to torsion, every cross‐sectionof a circular shaft remains plane andpundistorted.

• Cross‐sections for hollow and solid circularshafts remain plain and undistorted because a

• Cross‐sections of noncircular (non‐i t i ) h ft di t t d h

pcircular shaft is axisymmetric.

axisymmetric) shafts are distorted whensubjected to torsion.

Shearing Strain• Consider an interior section of the shaft. Asa torsional load is applied, an element on theinterior cylinder deforms into a rhombusinterior cylinder deforms into a rhombus.

• Since the ends of the element remain planar,the shear strain is equal to angle of twist

• It follows that

the shear strain is equal to angle of twist.

LrrL or

• Shear strain is proportional to angle of twistand radius

maxmax and cr

Stresses in Elastic RangeM l i l i h i i b h h• Multiplying the previous equation by the shearmodulus,

max GcrG

44 dcJ maxcr

From Hooke’s Law, G , so shear stress

• Recall that the sum of the moments from the

322J The shearing stress varies linearly with the radial

position in the section.

• Recall that the sum of the moments from theinternal stress distribution is equal to the torqueon the shaft at the section,

dArT max2max

4444 dd

• The results are known as the elastic torsionformulas,

42 ddccJ

and max JTr

TrTc and max JTr

τmax = the maximum shear stress in the shaft, which occurs at the

outer surface

T = the resultant internal torque acting at the cross section.

Its value is determine from the method of sections and

the equation of moment equilibrium applied about the

h ft’ l it di l ishaft’s longitudinal axis.

J = the polar moment of inertia of the cross‐sectional area

c = the outer radius of the shaftc = the outer radius of the shaft.

Angle of Twist in Elastic Rangef• Recall that the angle of twist and maximum

shearing strain are related,

• In the elastic range, the shearing strain and shearare related by Hooke’s Law,

• Equating the expressions for shearing strain andl i f th l f t i tsolving for the angle of twist,

If th t i l l di h ft ti• If the torsional loading or shaft cross‐sectionchanges along the length, the angle of rotation isfound as the sum of segment rotations

Sign Con entionSign Convention• Sign convention for thei t l t d thinternal torque and theangle of twist of one endof the shaft with respect tothe other endthe other end.

• Use right hand rule: bothtorque and angle will be

iti id d thpositive, provided thethumb is directed outwardfrom the shaft when thefingers curl to give thefingers curl to give thetendency for rotation.

Example

TAB = +80N.m, TBC = -70N.m, TCD = -10 N.m

φA/D = (+80) LAB + (-70) LBC + (-10) LCDJG JGJG

Example 1Determine the maximum shearing stress caused by a torque of magnitude T = 800 N.m.

Example 1

torque of magnitude T 800 N.m.

Example 2Knowing that the internal diameter of the hollow shaft shownis d = 23 mm, determine the maximum shearing stress caused

Example 2

is d 23 mm, determine the maximum shearing stress causedby a torque of magnitude T = 1.0 kN.m.

Example 3 The shaft is supported by two bearings and is subjected to three torques. Determine the shear stress developed at points A and B, located at section a–a of the shaft.

A C =75 mm

r =15 mm

Example 4Under normal operating conditions, the electric motor exerts a torque of2.4 kN.m at A. Knowing that each shaft is solid, determine the maximum

Example 4

shearing stress (a) in shaft AB, (b) in shaft BC, (c) in shaft CD.

E l 5The allowable stress is 104 MPa in the 38 mm diameter rod AB and 55 MPain the 46 mm diameter rod BC Neglecting the effect of stress

Example 5

in the 46 mm diameter rod BC. Neglecting the effect of stressconcentrations, determine the largest torque that may be applied at A.

Power TransmissionPower Transmission Power is defined as the work performed per unit of

timetime. For a rotating shaft with a torque, the power is

Since the power equation is

dtdTP / locity,angular veshaft where

f2rad2cycle1 Since , the power equation is

For shaft design the design or geometric parameter

f2rad2cycle 1

For shaft design, the design or geometric parameter is

allowc

Example 6A solid steel shaft AB as shown in the figure is to be used to transmit 3750W from the motor M to which it is attached. If the shaft rotates at N = 175

d h l h ll bl h f 100 MP

Example 6

rpm and the steel has an allowable shear stress of tallow = 100 MPa,determine the required diameter of the shaft to the nearest mm.

Example 7Example 7The motor delivers 30 kW to the shaft while it rotates at 20 Hz. The shaft issupported on smooth bearings at A and B, which allow free rotation of theshaft. The gear C and D fixed to the shaft removed 18 kW and 12 kWrespectively. Determine the diameter of the shaft to the nearest mm if theallowable shear stress is τallow = 56 MPa and the allowable angle of twist of C

D i 0 20o G 76 GPrespect to D is 0.20o. G = 76 GPa.

Chapter 5 Torsion

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