chapter 5 torsion
TRANSCRIPT
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ll bb lRussell C. Hibbeler
Ch T i Chapter 5: Torsion
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• 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.
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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.
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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’
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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.
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ExampleExample
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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.
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Chapter 5: TorsionChapter 5: TorsionMechanics of Material 7Mechanics of Material 7thth EditionEdition
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Shaft Deformations• From observation, the angle of twist of theshaft is proportional to the applied torque andto the shaft length.
LT
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.
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Chapter 5: TorsionChapter 5: TorsionMechanics of Material 7Mechanics of Material 7thth EditionEdition
axisymmetric) shafts are distorted whensubjected to torsion.
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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
rc
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Chapter 5: TorsionChapter 5: TorsionMechanics of Material 7Mechanics of Material 7thth EditionEdition
maxmax and cr
Lc
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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,
Jc
dArc
dArT max2max
4444 dd
• The results are known as the elastic torsionformulas,
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Chapter 5: TorsionChapter 5: TorsionMechanics of Material 7Mechanics of Material 7thth EditionEdition
322
41
42
41
42 ddccJ
and max JTr
JTc
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TrTc and max JTr
JTc
Where
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.
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Angle of Twist in Elastic Rangef• Recall that the angle of twist and maximum
shearing strain are related,
c L
max
• In the elastic range, the shearing strain and shearare related by Hooke’s Law,
TJGTc
G max
max
• Equating the expressions for shearing strain andl i f th l f t i tsolving for the angle of twist,
JGTL
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
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i ii
ii
GJLT
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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.
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Example
TAB = +80N.m, TBC = -70N.m, TCD = -10 N.m
φA/D = (+80) LAB + (-70) LBC + (-10) LCDJG JGJG
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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.
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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.
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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
B
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r =15 mm
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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.
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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.
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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
fTP 2
For shaft design, the design or geometric parameter is
ll
TcJ
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Chapter 5: TorsionChapter 5: TorsionMechanics of Material 7Mechanics of Material 7thth EditionEdition
allowc
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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.
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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.