Transcript Document

FME201
Solid & Structural
Mechanics I
Dr. Hussein Jama
[email protected]
Office 414
Lecture: Mon 11am -1pm (CELT)
Tutorial Tue 12-1pm (E207)
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Lecture Outline
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This lecture is based on the book by R C
Hibbeler Chapter 5
5.1Torsional deformation of a circular shaft
5.2 The Torsion Formula
5.3 Power of Transmission
5.4 Angle of twist
5.5 Statically indeterminate Torque Loaded
Members*
5.8 Stress concentration
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Engineering Professor of the
Year
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Engineering Professor of the
Year
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Student complaints
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I'm currently studying for my mechanical
engineering exam on this Friday and I'd like
to hear stories of bad professors to lighten
my frustration. The professor for this class is
the least liked professor in the ME
department. The only preparation for what's
on the exam is basically the table of contents
that he typed out. He also noted that it's an
incomplete list so we have to figure out what
else to study. Roughly 30% of the class fails
and he doesn't care and has said so happily.
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Torsion
Torque is a moment that twists
a member about its longitudinal
axis.
External loads (T) produce
internal loads which
produce deformation, strain
and stress.
Application
• Design of shafts in machinery
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Introduction
Assumption:
•The angle of rotation is small
•The length of the shaft will remain unchanged
•The radius of the shaft will remain unchanged.
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Torsion
Before Torque
After Torque
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5.1 Torsional deformation of a
circular shaft
f(x) = angle of twist
(varies linearly along
the length, 0 at x = 0,
max at x = L)
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Torsional deformation cont..
Recall g = shear strain (rad)
f
df
g

x
dx
Notice, shear strain,
g varies linearly with
radial distance, ,
and is max on the
outer surface!!
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Torsional deformation cont..
Notice, shear strain, g
varies linearly with radial
distance, , and is max on
the outer surface!!
Distance from
center to point of
interest

g   g max
c
Distance from center
to outer fiver (i.e.
outer radius)
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Stress due to torsion
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Deformation = shear strain
Shear stress is proportional to shear strain
(Hooke’s Law)
  Gg
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If you can visualize deformation, you can
visualize stress
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Recap
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f = angle of twist varies from zero at fixed
support to max at end.
g = shear strain varies from zero at center to
max at outer fiber
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5.2 The Torsion formula
If linear elastic, Hooke’s
law applies, t = Gy
Therefore, stress follows
same profile as strain!!

    max
c
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Torsion formula
Derivation – simple Torque balance.
The torque produced by the stress
distribution over the entire cross
section must be equal to the resultant
internal torque, or:

T    dA      maxdA
A
c
A 
T
 max
c
2

 dA
A
This is simply polar moment of
inertia, J (an area property)
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Torsion formula
Torque (N-m, Nmm or lb-in, lbft, etc)
Tc
 max 
J
Outer radius of
shaft (m or in)
Polar moment of
inertia (m4 or in4)
Max shear stress
in shaft (MPa,
psi/ksi, etc.)
or
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T
 
J
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Polar moment of inertia
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This is a geometric property and is always
positive.
It has units of m^4 of mm^4
Solid shaft:
J 
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2
c
4
•Hollow shaft:
J 
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
c
2
4
o
 ci4

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Stress profile
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Stress profile – wood failure
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Example 5.3
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Solution Example 5.3
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Example 5.4
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Solution example 5.4
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5.3 Power transmission
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Nothing new, just calculate Torque, T,
from power equation:
Power (watts, ft-lb/s
or hp)
P = Tw
Angular velocity (rad/s or Hz)
Torque (N-m, lbft)
w  2f
f = Hz or rev/s
Careful with units!
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Power transmission cont..
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Shaft powered by 500 W electric motor spins
at 10 Hz, find Torque in shaft.
P
= Tw
10 Hz (2 rad/rev) = 62.83 rad/s
T = 500 N-m/s
62.83 rad/s
= 7.96 N-m
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Class example
Determine:
1. Torque throughout shaft
2. Stress throughout shaft
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Homework
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5.5
5.7
5.15
5.19
5.25
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Recommended Texts
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7/20/2015
Mechanics of Materials – 2nd Edition, Madhukar
Vable – available online FREE
Engineering Mechanics – Statics, R.C. Hibbler,
Engineering Mechanics – Statics, D.J. McGill & W.W.
King
Mechanics of Materials , J.M. Gere & S.P.
Timoshenko
Mechanics of solids, Abdul Mubeen, Pearson
Education Asia
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