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MECHANICS OF
MATERIALS
Seventh Edition
Ferdinand P. Beer
E. Russell Johnston Jr.
John T. !e"ol#!avid F. Ma$ure%
Le&ture Notes'
Bro&% E. Barr(
).S. Militar( A&ade*(
CHAPTER
Copyright 2015 McGraw-Hill Education. Permiion re!uired "or reproducti
3Torsion
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MECHANICS OF MATERIALSSeventh
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Torsional Loads on Circular Shafts
, -
Stresses and strains in members of
circular cross-section are subjected
to twisting couples or torques
Generator creates an equal and
opposite torque T
Shaft transmits the torque to thegenerator
Turbine exerts torque Ton the shaft
Fig. 3.2 (a) A generator provides power at aconstant revolution per minute to a turbinethrough shaft AB. (b) Free body diagram of shaftAB along with the driving and reaction torques onthe generator and turbine, respectively.
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MECHANICS OF MATERIALSSeventh
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Net Torque Due to Internal Stresses
, - ,
( ) == dAdFT
Net of the internal shearing stresses is aninternal torque equal and opposite to the
applied torque
!lthough the net torque due to the shearing
stresses is "nown the distribution of the
stresses is not#
$nli"e the normal stress due to axial loads the
distribution of shearing stresses due to torsional
loads cannot be assumed uniform#
%istribution of shearing stresses is staticall&
indeterminate ' must consider shaft
deformations#
Fig. 3.24 (a)Free body diagram ofsection BC with torque at Crepresented by the representablecontributions of small elements of areacarrying forces dF a radius from thesection center. (b) Freebody diagram
of section BC having all the small areaelements summed resulting in torque
Fig. 3.3 "haft sub#ect to torquesand a section plane at C.
S
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MECHANICS OF MATERIALSSeventh
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Axial Shear Components
, - /
Torque applied to shaft produces shearing
stresses on the faces perpendicular to the
axis#
(onditions of equilibrium require the
existence of equal stresses on the faces of the
two planes containing the axis of the shaft#
The slats slide with respect to each other
when equal and opposite torques are applied
to the ends of the shaft#
The existence of the axial shear components is
demonstrated b& considering a shaft made up
of slats pinned at both ends to dis"s#
Fig. 3.6 $odel of shearing in shaft(a) undeformed% (b) loaded and
deformed.
Fig. 3.5 "mall element in shaftshowing how shear stress componentsact.
MECHANICS OF MATERIA SS
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MECHANICS OF MATERIALSSeventh
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Shaft Deformations
, - 0
)rom obser*ation the angle of twist of theshaft is proportional to the applied torque and
to the shaft length#
L
T
+hen subjected to torsion e*er& cross-section
of a circular shaft remains plane and
undistorted#
(ross-sections for hollow and solid circular
shafts remain plain and undistorted because acircular shaft is axis&mmetric#
(ross-sections of noncircular ,non-
axis&mmetric shafts are distorted when
subjected to torsion#Fig. 3.8 Comparison of deformations
in circular (a) and square (b) shafts.
Fig. 3.7 "haft with &'ed support andline AB drawn showing deformationunder torsion loading (a) unloaded%(b) loaded.
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MECHANICS OF MATERIALSSeventh
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Shearin Strain
, - 1
(onsider an interior section of the shaft# !s a
torsional load is applied an element on the
interior c&linder deforms into a rhombus#
Shear strain is proportional to twist and radius
maxmax and
cL
c==
LL
== or
.t follows that
Since the ends of the element remain planar
the shear strain is equal to angle of twist#
Fig. 3.13 "hearing "train inematicde&nitions for torsion deformation. (a)
!he angle of twist (b) *ndeformedportion of shaft of radius with (c)+eformed portion of the shaft having
same angle of twist, and strain, anglesof twist per unit length, .
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MECHANICS OF MATERIALSSeventh
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Stresses in Elastic Rane
, - 2
Jc
dAc
dAT max/max
= ==
0ecall that the sum of the moments of the
elementar& forces exerted on an& cross
section of the shaft must be equal to the
magnitude T of the torque1
andmax J
T
J
Tc
==
The results are "nown as the elastic torsion
formulas
2ultipl&ing the pre*ious equation b& the
shear modulus
max Gc
G =
max
c
=
)rom oo"es 4aw G= so
The shearing stress *aries linearl& with thedistance from the axis of the shaft#
Fig. 3.14 +istribution of shearingstresses in a torqued shaft% (a) "olidshaft, (b) hollow shaft.
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MECHANICS OF MATERIALSSeventh
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Normal Stresses
,- 3
Note that all stresses for elements aand cha*e
the same magnitude#
5lement cis subjected to a tensile stress ontwo faces and compressi*e stress on the other
two#
5lements with faces parallel and perpendicular
to the shaft axis are subjected to shear stressesonl Normal stresses shearing stresses or a
combination of both ma& be found for other
orientations#
( )
max6
6max78
6max6max
/
/
/78cos/
o
===
==
A
A
A
F
AAF
(onsider an element at 78oto the shaft axis
5lement ais in pure shear#
Fig. 3.17 Circular shaft with stresselements at dierent orientations.
Fig. 3.18 Forces on faces at -/ to shafta'is.
Fig. 3.19 "haft elements with onlyshear stresses or normal stresses.
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MECHANICS OF MATERIALSSeventh
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Torsional !ailure "odes
, - 4
%uctile materials generall& fail in shear# 9rittle materials are wea"er in
tension than shear#
+hen subjected to torsion a ductile specimen brea"s along a plane of
maximum shear i#e# a plane perpendicular to the shaft axis#
+hen subjected to torsion a brittle specimen brea"s along planes
perpendicular to the direction in which tension is a maximum i#e# along
surfaces at 78oto the shaft axis#
Photo 3.2 "hear failure of shaft sub#ect totorque.
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MECHANICS OF MATERIALSSeventh
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Sample Pro#lem $%&
, - 56
ShaftBCis hollow with inner and outer
diameters of :6 mm and ;/6 mm
respecti*el ShaftsABand CDare solid
and of diameter d# )or the loading showndetermine ,a the minimum and maximum
shearing stress in shaftBC ,b the
required diameter dof shaftsABand CD
if the allowable shearing stress in these
shafts is 4$T.>N1
(ut sections through shaftsAB
andBCand perform static
equilibrium anal&ses to find
torque loadings#
Gi*en allowable shearing stress
and applied torque in*ert the
elastic torsion formula to find therequired diameter#
!ppl& elastic torsion formulas tofind minimum and maximum
stress on shaftBC.
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MECHANICS OF MATERIALSSeventh
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, - 55
Sample Pro#lem $%&
S>4$T.>N1 (ut sections through shaftsABandBC
and perform static equilibrium anal&sisto find torque loadings#
( )
CDAB
ABx
TT
TM
==
==
m"N