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National University of Singapore
SESSION
2014
-
15
Semester
1
ME2113
Mechanics of Materials
I (Part 1)
Department of Mechanical Engineering
Applied Mechanics Group
NUS ME Dept Web Page
Office EA-05-13
Tel: 65162557
A/P Tay C. J.Tay Cho Jui
mailto:[email protected]:[email protected]8/9/2019 CH 1 Stress Strain 14 Ivle
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SESSION
2014
-
15
Semester 1
ME2113
Mechanics of Materials I
Modular Credits
: 3
Pre
-
Requisites
:
EG1109
-
Statics and Mechanics of Materials
Workload:
Lecture hr: 26 Tutorial hr: 5 Lab.
hr: 6 Two Lab sessions
Assessment:
Final Examination: 80%
Continuous assessment (
Consisting of Labs and quiz)
20% Course Lecturers
A/P C J Tay Part I, A/P
Vincent Tan Part II ;
Brief description of module: Part I
Analysis of Stress and Strain
Bending of Beams
Stresses in Loaded Beams
Deflection of Beams
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Basic Text:
A.C.
Ugural
,
Mechanics of Materials, McGraw
-
Hill,1993
(Chapters 4, 7, 9 & 10 for part I)
Supplementary Readings:
F. P. Beer and E. R. Johnston, Jr. and J.T.
DeWolf
,
Mechanics of Materials, McGraw
-
Hiil
, SI 3rd Ed., 2004. R. C.
Hibbeler,Mechanics
of Materials, Prentice Hall,
SI2nd Ed., 2005.
J. M.
Gere
, Mechanics of Materials, Thomson
Brooks/Cole Publishing Company, 6th ed., 2004.
R
. R. Craig, Jr., Mechanics of Materials, McGraw
-
Hill,2nd ed., 2000.
National University of Singapore
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NUS- Web Page IVLE
Module ME2113
Workbin
Lecture notes, Tutorial sheets, Lab manuals
FAQ
Tutorial Solutions, Past Exams Solutions
Discussion Forum
Announcement
National University of Singapore
Anonymous feedback via IVLE
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ME2113 Part I (6 weeks)
14 AugStresses and Strains
21 Aug
Stresses and Strains
28 Aug
Shear force and Bending Moment in Beams04 Sept
Stresses in Loaded Beams
11 Sep
Stresses in Loaded Beams/Deflection of Beams
18 SepDeflection of Beams
Mid-Semester Break
ME2113 Part II - Prof. Vincent Tan
LESSON PLAN
SESSION 2014-15
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ME2113 (part 1)
Tutorial Session
Day: Every Friday 9-10 am
Venue: Tutorial Room E3-06-02Timetable: http://me.nus.edu
.sg/
Consultation time for ME2113 (part 1)Day: Every Thursday
Time: 5.00-6.30 pm Venue: Office EA-05-13
http://me.nus.edu.sg/http://me.nus.edu.sg/http://me.nus.edu.sg/8/9/2019 CH 1 Stress Strain 14 Ivle
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ME 2113 Part I Assessment (on-line)
14/10/2014
(3rd week after mid-sem break)
multiple-choice question quiz
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Chapter 1
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ANALYSIS OF STRESS AND STRAIN
n is normal to dA
S1 , S2 are tangential
(in plane)
Apply general force dF on dA
Defn:
As dA 0, stress state is at the point P.
dA
dF
SheardA
dF
NormaldA
dF
S
dAS
S
dAS
n
dAn
2
02
1
01
0
lim
lim
lim
Note: Stress values depend on magnitude of dF and
also the direction of dF.
STRESSES
dA
P
s1
n
dF
s2
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y
x
z
yy
yx
yz
zx
zy
zz
xx
xy
xz
Stresses shown are
all positive on a
cube of 1 unit
length
For a small isolated element with planes perpendicular
to coordinate axes and surrounding a point P, there
exist 9 stress components.
They are
3 normal stresses
6 shear stresses
As size of parallelepiped reduces, in the limit, these 9 stress
components will define completely, the state of stress at the
point P.
P
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y
x
z
yy
yx
yz
zx
zy
zz
xx
xy
xz
Take Moment about Z-axis
xy 1x1 -yx 1x1 = 0
i.e. xy = yx
The cube is stationary (in equilibrium)
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From equilibrium (i.e. taking moment about any axis), we can show
that:
zyyz
zxxz
yxxy
Number of unknown stresses reduced to 6.
i.e. xx ,yy , zz,
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yzxzxy ,,
y
x
z
yy
yx
yz
zx
zy
zz
xx
xy
xzP
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Stress Component on Arbitrary plane (2 dimensional case)
z
y
yx x
xy
x
y
xy
x
x
y
Arbitrary plane whose normal makes an angle with horizontal
Q. What are the values of x, xy in terms of x, y and xy ?
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Substitute for Tx and Ty
cossin)()sin(cos
22
'' xyxyyx
y can be found by substituting + /2 for
in expression for x,
i.e.
cossin2cossin 22 xyyxy
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Using the relations
cossin2sincos' 22 xyyxx
z
y
yx x
xy
x
y
xy
x
x
y
etc 2222
sin211cos2sincos2cos
cossin22 Sin
x
y x
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Rewriting,
(i)
(ii)
(iii)
2cos2sin2
)(,, xy
yx
yx
Element A
y
y
xy
xx
y
y'
xy
x x'
x
Stresses on element A
2sin2cos2
)(
2
)(xy
yxyx
x
2sin2cos2
)(
2
)(xy
yxyx
y
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When ,
)(
2tan
2
1
)(
22tan..
02cos2sin2
)(
1
yx
xy
yx
xy
xy
yx
or
ei
0,, yx
x' which is denoted as 1 is known as the maximum
principal stress
y' which is denoted as 2 is known as the minimum principalstress and
, which is denoted as is known as the principal angle
Eq. a
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y
x
z
yy
yxyz
zx
zy
zz
xx
xy
xz
Convention for denoting stress
1. Normal stress, ijiindicates the direction of a normal to the plane on
which the stress component acts;
jindicates the direction of the stress.
Usually denoted by ij , e.g. xx
For simplicity
xx is written as xyy is written as y
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2. Shear stress
iindicates the direction of a normal to the plane on
which the stress component acts;
jindicates the direction of the stress. y
x
z
yy
yx
yz
zx
zy
zz
xx
xy
xz
e.g. xy
ij
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x
y
x
y
Positive Shear Negative Shear
+xy -xy
Sign convention
Sign convention- Normal stresses
Stress is positive Tension (e.g. xx)
negative Compression (e.g. -xx)
Sign convention- Shear stresses
M h Ci l f 2 D St
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2sin2cos2
)(
2
)(xy
yxyx
x
(i)
22
2sin2cos2
)(
2
)(
xy
yxyx
x (a)
2sin2cos2
)(2
)(xy
yxyx
y (ii)
22
2sin2cos2
)(
2
)(
xy
yxyx
y (b)
2cos2sin2
)(,, xy
yx
yx
(iii)
22 2cos2sin
2
)(,,
xy
yx
yx(c)
We haveMohrs Circle for 2-D Stresses
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22
2sin2cos2
)(
2
)(
xy
yxyx
x (a)
22
2sin2cos2
)(
2
)(
xy
yxyx
y(b)
22 2cos2sin2
)(,,
xy
yx
yx(c)
Eqs. a) + c) gives
222
''
2
)2()2(xy
yx
yx
yx
x
Eqs. b) + c) gives
222
''
2 )2
()2
(xy
yx
yx
yx
y
(iv)
(v)
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Eqns (iv) and (v) represent a circle in the -planewith center
)0,2
( yx
2
2
2 xyyxR
and radius R given by:
National University of Singapore
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)0,2
( yx
2
2
2 xyyxR
x-axis represents normal stress
y-axis represents shear stress
1
2
max
min
222
''
2 )2
()2
(xy
yx
yx
yx
x
222)( Ryax
x
xyA point on the circumference is
given by itsx andy coordinates
ie it represents the stresses ona plane in the element
Q: What does a point on the
circumference of the circle represent?
https://www.google.com.sg/search?hl=en-GB&rlz=1T4GGNI_en-GBSG457SG457&biw=1920&bih=871&tbm=isch&q=puzzled+face+emoticon&revid=303442468https://www.google.com.sg/search?hl=en-GB&rlz=1T4GGNI_en-GBSG457SG457&biw=1920&bih=871&tbm=isch&q=puzzled+face+emoticon&revid=3034424688/9/2019 CH 1 Stress Strain 14 Ivle
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Y
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Hence the state of stress on Plane A or Plane B as shown
above can be represented by a point on the
circumference of a Mohrs circle
of center
and radius
)0,2
( yx
2
2
2 xy
yxR
xy
y
Y
X
xx
xy
y
Plane A
Plane B
A
B
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2
2
1
2
2
2 xy
yxR
X
Y
xy
x
yPlane B
Plane A
y
x
2
yx
CD direction of 1
(2 ) (1 )P
Q
2
yx
1 = max prin stress
2 = min prin stress
max = max shear
stress
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O C
xy
For a plane inclined at , x,
y,
xycan be obtained.
x
y
2 1
)22cos('
ROCx
22
Plane B
Plane A
x
center )0,2
( yx
2
2
2 xy
yx
R
)22cos( R
PQ
Principal angle =2
2
(x, -xy)A
(y , xy)B
2
2
( x, xy - )D
( y xy , )E
xy x
y
x
Plane E
Plane D
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Stress components at any arbitrary plane can be determined from Mohrs
circle:
)22cos(
ROCx
2
yxOC
2
2
2 xy
yxR
On substituting and simplifying,
(cf. eqn(i))
2sin2cos2
)(
2
)(xy
yxyx
x
(i)
2sin2cos2
)(
2
)(xy
yxyx
x
2sin2sin2cos2cos)22cos(: Note
2sin2cos2
)(
2
)(xy
yxyx
y
and
2cos2sin2
)(,, xy
yx
yx
Similarly,
)(
2
2tanyx
xy
and
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O C
xy
x
y
2 1
Plane B
Plane A
x
)22cos( R
PQ
2
(x, -xy)A
(y , xy)B
2
2
( x, xy - )D
( y xy , )E
xy x
y
x
Plane E
Plane D
Note that rotation on Mohrs circle is twice that of element and sense of
direction of rotation of axes is the same for Mohrs circle and element.
Sign convention for stresses when constructing and analyzing Mohrs circle:
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2
2
1
2
2
2 xy
yxR
X
Y
xyx
yPlane B
Plane A
y
x
2
yx
CD direction of 1
(2 ) (1 )
Sign convention for stresses when constructing and analyzing Mohr s circle:
B?
A?
https://www.google.com.sg/search?hl=en-GB&rlz=1T4GGNI_en-GBSG457SG457&biw=1920&bih=871&tbm=isch&q=puzzled+face+emoticon&revid=303442468https://www.google.com.sg/search?hl=en-GB&rlz=1T4GGNI_en-GBSG457SG457&biw=1920&bih=871&tbm=isch&q=puzzled+face+emoticon&revid=303442468https://www.google.com.sg/search?hl=en-GB&rlz=1T4GGNI_en-GBSG457SG457&biw=1920&bih=871&tbm=isch&q=puzzled+face+emoticon&revid=303442468https://www.google.com.sg/search?hl=en-GB&rlz=1T4GGNI_en-GBSG457SG457&biw=1920&bih=871&tbm=isch&q=puzzled+face+emoticon&revid=3034424688/9/2019 CH 1 Stress Strain 14 Ivle
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Sign convention for stresses when constructing and analyzing Mohrs circle:
Shear stressesif the shear stresses on opposite faces of the element
produce forces that result in a clockwise couple, these stresses are taken as
positive.
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Plane A clockwise ,
+ve on Mohrs circle
Plane B anticlockwise,
-ve on Mohrs circle
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Sign convention for normal stresses
Positive is tensile and plotted along positive x-axis
- negative is compressive and plotted along negative x-axis
x
y
x
y
Positive Shear Negative Shear
+xy -xy
IMPORTANT:
Do not confuse Mohrs circle sign convention with that of an entire element
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Please be punctual
Lectures will start on time
not puncture!
ME 2113 Lectures will be webcast
Power point slides are available on webcast
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Applications of Mohrs circle
i) Spherical Pressure Vessel
Consider a spherical pressure
vessel with radius rand wall
thickness tsubjected to an
internal gage pressurep.
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P
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The normal stresses can berelated to the pressurep by
inspecting a free bodydiagram of the pressurevessel. To simplify theanalysis, we cut the vesselin half as illustrated.
From equilibrium, the stressaround the wall must have anet resultant to balance theinternal pressure across thecross-section.
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Applications of Mohrs circle
Stresses in SphericalPressure Vessel
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ii) Cylindrical
Pressure Vessel
Consider a cylindrical
pressure vessel with
radius rand wall
thickness tsubjected to
an internal gage
pressurep.
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P
1
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To determine the
longitudinal stress, we
make a cut across the
cylinder similar toanalyzing the spherical
pressure vessel.
From equilibrium,
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dx
rP
h
h
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To determine the hoop
stress, we make a cut
along the longitudinal
axis.
From equilibrium, the
hoop stress yields,
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NOTE:
The hoop stress is twice as much as the longitudinal stress for thecylindrical pressure vessel.
The above formulas are good for thin-walled pressure vessels. ieradius ris larger than 5 times its wall thickness t(r> 5 t).
When a pressure vessel is subjected to externalpressure, thestresses are negative since the wall is now in compression insteadof tension.
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Q: What would happen to the sausages if they were
overcooked?
A: Cracks along the longitudinal direction appear first.
= 2
This is because the internal pressure generated by the
steam inside the sausage causes the skin to fail from hoop
stress since the hoop stress is twice as much as the
longitudinal stress in the sausage (cylindrical pressurevessel .
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EXAMPLE 1
The state of plane stress at a point (in a pressurevessel) is represented by the figure shown.Determine the stresses on an element oriented at300 counterclockwise from the position shown.Illustrate your answer on a diagram.
6 MPa
y
x
12 MPa
8 MPa 300
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From Eqns (i) to (iii) we have
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From Eqns. (i) to (iii), we have
(i)
(ii)
(iii)
Substituting = 300 , we have
To construct the Mohrs circle,
2sin2cos2
)(
2
)(' xy
yxyx
x
2sin2cos2
)(
2
)(
' xy
yxyx
y
2cos2sin2
)(,,
xy
yx
yx
MPax 2.8'
MPay 2.12'
MPayx 66.5,,
circleofcenterMPayxavg 22128
2
circleofradiusR xyyx
66.116102
222
2
Question: Where should point A be ?
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2
R = 11.66
8
6 MPa
y
x
12 MPa
8 MPa
Plane A
Plane B
Point A ?
Point A ?
x
Clockwise,
+ve shear
6
600
x
y
12
y
6
300
X'
y
Q p
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From the Mohrs circle,
MPax
2.8'
MPay 2.12'
MPayx 66.5,,
5.66 MPa
y'
x
12.2 MPa
8.2 MPa
300
300
National University of Singapore
2
R = 11.66
xClockwise,
+ve shear 6
600
x
12
y
6
-ve shear stressat this point
y
8
Hence
anticlockwise
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Deformation and Strain
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Deformation and Strain
Deformation is a physical phenomenon it can be measured.
Strain is a mathematical concept.
Basic modes of deformation (displacement)
Rigid body and Non rigid body
Rigid body Translation, Rotation
Non rigid body Elongation, Angular Distortion
O Angular Distortion
A
A'
B B'
P P'A
B'
A'
B
P
ORotation
OElongation
B
B
PP A A'
O
B
AP
P' A'
B'
Translation
Definition of Strain
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Definition of Strain
Line element (direct strain)
Engineering strain of line element PQ.
O
P
Q
Q'
P'
PQ
PQQP
,,
Rotation between two line elements (shearing strain)
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Rotation between two line elements (shearing strain)
Shearing strain
When is small
,
)tan( ,,
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1.2.2 StrainDisplacement Relationships
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Normal Strains
P
x
y
z
x
u
x
y
v
y
z
w
z
where u,v and w represent the displacements in thex, y and
zdirections respectively.
1.2.2 StrainDisplacement Relationships
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Shear Strains
(ii)
y
u
x
vxy
y
w
z
v
yz
x
w
z
u
xz
Assumptions
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Assumptions
A
A' B B'
D
D'
C
C'
ii indicates the direction in which the elongation or contraction is required. (the
sides are of an undeformed element)
Positive Strain for elongation (e.g. x)
Negative Strain for contraction (e.g. y)
1. Deformations are infinitesimally small
2. Displacement of a point on the element is continuous, i.e.
no cracks, overlapping, slippage, etc. and also body from
which element is isolated, is continuous throughout.3. Element is small, i.e. surroundings within close
neighbourhood of point P.
Convention for Strains
Normal Strains
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Shear Strains
A A' B
B'
C
C'
D D'
x
y
iji, j indicate directions of two mutually perpendicular sides of an
undeformed element whose extent of angular deformation is required
(e.g. xy
)
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Strains at a point
The strains x, y, z, xy, yz and xz are those of a cube elementsurrounding the point of interest, P(x, y, z). If the size of the cube is
allowed to become infinitesimally small, the strains can be regarded as
being the strains at point P(x, y, z) within a body.
Thus, 6 components of strains are required to define completely the state of
strain at point.
Transformation of Axes.
Often the strain components at a point referred to a set of axes are different
from the original axes.
For 2- D analysis, if x, y and xy are strains in x-y plane, what are theequivalent strains x, y and xy referred to x and y axes that make an
angle with the x-y axes.
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The three strain components referred to x-y axes which are at an angle to
x-y axes:
(d)
(e)
(f)
2sin2
2cos22
'xyyxyx
x
2sin2
2cos22
'xyyxyx
y
2cos2
2sin22
'' xyyxyx
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Principal Strains
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There will be a plane in the element which does not experience any shear
strains,
i.e. xy=0 .
From eqn(f)
0
2cos2sin''
xyyxyx
i.e.
)(2tan yx
xy
or
(g)
For this value of ,
(h)
(i)
)(
tan2
1 1
yx
xy
22
'222
)(
xyyxyx
x
22
' 222
)(
xyyxyx
y
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and are called the maximum and minimum principal strains
respectively.
(Usually denoted by 1 and 2)
The particular angle x or y denotes the direction of the principal axes, and
is denoted by . The rotation is defined as positive for counter-clockwise
rotation.
'x 'y
National University of Singapore
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(i)
(ii)
(iii)
2sin2cos2
)(
2
)(' xy
yxyx
x
2sin2cos2
)(2
)(' xyyxyxy
2cos2sin2
)(,, xy
yx
yx
The three strain components referred to x-y axes which are at an angle to
x-y axes:
(d)
(e)
(f)
2sin2
2cos22
'
xyyxyx
x
2sin2
2cos22
'xyyxyx
y
2cos2
2sin22
'' xyyxyx
Mohrs Circle of Strain
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Re-examining eqns (d), (e) & (f) and compare
with eqns (i), (ii) & (iii) for x , y & xy
The equations are similar in form.
Therefore a Mohrs circle for strain can similarly be constructed
with center at in the x , plane. The radius of the
circle is
0,
2
yx
2
xy
22
22
xyyxR
However for Mohrs strain circle,
x axis normal strainyaxis half shear stain
2
.
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y
x
y
x
+ve Shear stresses +ve Shear strains
Gi th t i t t
Mohrs Circle of Strain
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Given the strain state
x , y , xy
the corresponding Mohrs strain circle can be plotted.
Hence the strain components x , y and xy at any orientation can be determined.
x
y
x
xy
x
y
xy
2
2,
''
'
yx
y
2,
''
'
yx
x
2
2
2
2,
xy
y
y
2,
xy
x
2
yx
22
22
xyyxR
x 1
StressStrain Relationships
St d t i l t d th h th i i ti f th t i l
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Stress and strain are related through the engineering properties of the material
of the body.
Assumptions
All the stresses / strains are within the elastic range of the material
Material is homogeneous (i.e. properties uniform throughout)
Material is isotropic (i.e. properties independent of direction)
Hookes law
(Uniaxial tensile test)E
xx
E Youngs modulus
Lateral contraction in y direction =
- Poissons ratio.
Application of yproduces strains
E
x
)(
)(
directionxE
directionyE
y
yy
Application of and simultaneously
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EE
EE
xyy
yxx
)(1
)(1
)(1
yxzz
zxyy
zyxx
E
E
E
GGG
xzxz
yzyz
xyxy
;;
(i)(a)
(b)
(a) (a)
Application of x and y simultaneously,
For threedimensions,
For shear strains,
G shear modulus of elasticity.
Eqns (ii) and (iii) represent the generalized Hookes law
(for isotropic, homogeneous materials).
(b)
(c)
(ii)
(iii)
The Hookes law is applicable to any orthogonal stress system
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The Hooke s law is applicable to any orthogonal stress system.
e.g. )(1
zrE
)(1 3211 E
Substituting the values of y in Eq. (i)b into Eq. (i)a and express the stress
in terms of strains, we have
yxx E
2
1
Likewise we can obtain
xyyE
21
EE
EE
xyy
yxx
(i)(a)
(b)
EXAMPLE 2
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EXAMPLE 2
The strain components at a point in a machinemember are given by
x = 900 , y = -100 , xy = 600 .
Using Mohrs circle, determine the principal strainsand the maximum shearing strains.
Centre of circle: (x+ y )/2 = (900 -100)/2 = 400
Radius of circle
5832
600
2
100900
22
2222
xyyxR
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