Post on 19-May-2018
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Ishik University / Sulaimani
Civil Engineering Department
Mechanics of Materials
CE 211
CHAPTER -5-
TORSION
Part -2-
CHAPTER -5- TORSION
Outlines of this chapter:
1.Chapter Objectives
2.Introduction
3.Torsional Deformation of a Circular Shaft
4.The Torsion Formula
5.Power Transmission
6.Angle of Twist
7.Statically Indeterminate Torque-Loaded Members
8.*Solid Noncircular Shafts
9.*Thin-Walled Tubes Having Closed Cross Sections
10.Stress Concentration
11.*Inelastic Torsion
12.*Residual Stress
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6. Angle of Twist
Angle of twist is important when analyzing reactions on
statically indeterminate shafts.
=
T(x) dx
J(x) G ∫0 L
= angle of twist, in radians.
T(x) = internal torque at arbitrary position x, found from
method of sections and equation of moment
equilibrium applied about shaft’s axis.
J(x) = polar moment of inertia as a function of x.
G = shear modulus of elasticity for material.
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Constant torque and x-sectional area
=
TL
JG
If shaft is subjected to several different torques, or x-
sectional area or shear modulus changes suddenly from
one region of the shaft to the next, then apply the above
equation to each segment before vectorially adding each
segment’s angle of twist: =
TL
JG
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For Multiple-Section Shafts:
SHAFTS WITH A VARIABLE CIRCULAR CROSS SECTION
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Sign convention
Use right-hand rule: torque and angle of twist are positive
when thumb is directed outward from the shaft
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Example,
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Internal Torque
Use method of sections and equation of moment
equilibrium applied along shaft’s axis.
If torque varies along shaft’s length, section made at
arbitrary position x along shaft is represented as T(x).
If several constant external torques act on shaft between
its ends, internal torque in each segment must be
determined and shown as a torque diagram.
Procedure for analysis
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Angle of twist
When circular x-sectional area varies along shaft’s axis,
polar moment of inertia expressed as a function of its
position x along its axis, J(x).
If J or internal torque suddenly changes between ends of
shaft, = ∫ (T(x)/J(x)G) dx or = TL/JG must be applied
to each segment for which J, T and G are continuous
or constant.
Use consistent sign convention for internal torque and
also the set of units.
Procedure for analysis
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Example 5.4
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Solution ;
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Solution ;
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Solution ;
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A torsionally loaded shaft is statically indeterminate if moment equation of equilibrium, applied about axis of shaft, is not enough to determine unknown torques acting on the shaft
7. Statically Indeterminate Torque-Loaded Members
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From free-body diagram, reactive torques at supports A and B are unknown, Thus,
Since problem is statically indeterminate, formulate the condition of compatibility; end supports are fixed, thus angle of twist of both ends should sum to zero
Mx = 0; T TA TB = 0
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Assume linear-elastic behavior, and using load-displacement relationship, = TL/JG, thus compatibility equation can be written as
Solving the equations simultaneously, and realizing that L = LAC + LBC, we get
TA LAC
JG
TB LBC
JG = 0
TA = T LBC
L ( ) TB = T LAC
L ( )
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Procedure for analysis
Equilibrium
Draw a free-body diagram.
Write equations of equilibrium about axis of shaft.
Compatibility
Express compatibility conditions in terms of rotational displacement caused by reactive torques.
Use torque-displacement relationship, such as
= TL/JG
Solve equilibrium and compatibility equations for unknown torques.
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Example 5.5
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Solution ;
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Solution ;
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Solution ;
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Example 5.6
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Solution ;
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8. Solid Noncircular Shafts
Shafts with noncircular x-sections are not axisymmetric, as such, their x-sections will bulge or warp when it is twisted.
Torsional analysis is complicated and thus is not considered for this text.
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Results of analysis for square, triangular and elliptical x-sections are shown in table
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Example 5.7
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Solution ;
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Thin-walled tubes of noncircular cross section are often used to
construct light-weight frameworks such as those used in aircraft. In
some applications, they may be subjected to a torsional loading. In
this section we will analyze the effects of applying a torque to a thin-
walled tube having a closed cross section, that is, a tube that does not
have any breaks or slits along its length. Such a tube, having a
constant yet arbitrary cross-sectional shape, and variable thickness t,
is shown in Fig. 5–28a. Since the walls are thin, we will obtain the
average shear stress by assuming that this stress is uniformly
distributed across the thickness of the tube at any given point.
9. Thin-Walled Tubes Having Closed Cross Sections
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Example 5.8
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Solution ;
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Solution ;
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Solution ;
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Example 5.9
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Solution ;
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Assignment 5.1
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Assignment 5.2
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Assignment 5.3