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V{t àxÜ I
Transformations of
Stress & Strain
Ch 6- 1
1. Ferdinand P. Beer, E. Russell Johnston,Jr, John T. Dewolf, David F. Mazurek “ Mechanics of Materials” 5th Edition in SI units
2. R.C.Hibbeler “ Mechanics of Materials “ Seventh Edition
Materials for this chapter are taken from :
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The most general state of stress at a point may be
represented by 6 components,
Introduction
),, :(Note
stressesshearing,,
,,
xz zx zy yz yx xy
zx yz xy
z y x
τ τ τ τ τ τ
τ τ τ
===
Same state of stress is represented by a different set
of components if axes are rotated.
Ch 6- 2
the components of stress are transformed under arotation of the coordinate axes. The second part of
the chapter is devoted to a similar analysis of the
transformation of the components of strain.
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Plane Stress - state of stress in which two faces of the
cubic element are free of stress. For the illustrated
example, the state of stress is defined by
Plane Stress
.0,, and xy === zy zx z y x τ τ σ τ σ σ
State of plane stress occurs in a thin plate subjected to
forces acting in the midplane of the plate.
Ch 6- 3
State of plane stress also occurs on the free surface of a
structural element or machine component, i.e., at any
point of the surface not subjected to an external force.
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Consider the conditions for equilibrium of a prismatic
element with faces perpendicular to the x , y , and x’
axes.
Transformation of Plane Stress
( ) ( )
( ) ( )
( ) ( )
( ) ( ) θ θ τ θ θ σ
θ θ τ θ θ σ τ
θ θ τ θ θ σ
θ θ τ θ θ σ σ
sinsincossin
coscossincos0
cossinsinsin
sincoscoscos0
A A
A A A F
A A
A A A F
xy y
xy x y x y
xy y
xy x x x
∆+∆−
∆−∆+∆==∑
∆−∆−
∆−∆−∆==∑
′′′
′′
The equations may be rewritten to yield
Ch 6- 4
θ τ θ
σ σ
τ
θ τ θ σ σ σ σ
σ
θ τ θ σ
2cos2sin2
2sin2cos22
2sin2cos22
xy
y x
y x
xy y x y x
y
xy
y x y x
x
+
−
−=
−−
−+
=
+
−
+=
′′
′
′
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The previous equations are combined to yield
parametric equations for a circle,
222 τ σ σ =+− ′′′
Principal Stresses
22
22
where
xy y x y x
ave R τ σ σ σ σ
σ +
−=
+=
Principal stresses occur on the principal planes
of stress with zero shearing stresses.
Ch 6- 5
o
2minmax,
90 byseparatedanglestwodefines : Note
22tan
22
y x
xy p
xy y x y x
σ σ
τ θ
τ σ σ σ σ
σ
−=
+
−±
+=
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Maximum shearing stress occurs for ave x σ σ =′
Maximum Shearing Stress
45 byfromoffset
and90 byseparatedanglestwodefines: Note
2
2tan
2
o
o
22
max
p
xy
y x s
xy y x
R
θ
τ
σ σ θ
τ σ σ
τ
−−=
+
−==
Ch 6- 6
2 y xave σ σ σ σ
+==′
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For the state of plane stress
shown, determine (a) the
principal planes, (b) the principal
Find the element orientation for the principal
stresses from
Example 6.1
Solut ion
stresses, (c) the maximum
shearing stress and the
corresponding normal stress.
y x
xy p
σ σ
τ θ
−=
22tan
Determine the principal stresses from
22
minmax,22
xy y x y x
τ σ σ σ σ
σ +
−±+=
Calculate the maximum shearin stress with
Ch 6- 7
22
max2
xy y x
τ σ σ
τ +
−=
2
y x σ σ
σ
+
=′
F lt f M h i l E i i
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Find the element orientation for the principal
stresses from:
( )=
+== 333.1
40222tan
xyτ θ
Example 6.1
°°=−−−
1.233,1.5321050
p
y x
θ σ σ
°°= 6.116,6.26 pθ
Determine the principal stresses from:
2 −+ x σ σ σ σ
50MPa 40MPa
10MPa
x xy
y
σ τ
σ
= + = +
= −
Decl are !
Ch 6- 8
( ) ( )22
minmax,
403020
22
+±=
+
= xyτ σ
MPa30
MPa70
min
max
−=
=
σ
σ
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Calculate the maximum shearing stress with
22
+
−
= y x
τ σ σ
τ
Example 6.1
( ) ( )22 4030 +=
MPa50max =τ
45−= p s θ θ
°°−= 6.71,4.18 sθ
50MPa 40MPa
10MPa
x xy
y
σ τ
σ = + = += −
Ch 6- 9
2
1050
2
−=
+==′ y x
ave
σ σ σ σ
The corresponding normal stress is
MPa20=′σ
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A plane stress system is as shown in the
diagram below. Determine the direct and
Example 6.2
SOLUTION: METHOD OF EQUILIBRIUM
=- .
10 MPa
8 MPa
15°
P
'
'
10 sin15cos75 8 sin15cos75
2 cos15cos15 8 cos15cos75 0
.
0;
10 sin15sin75 8 sin15sin15
x
y
A A A
A A
Ans
F
A A A
σ
σ
τ
∆ − ∆ + ∆
+ ∆ + ∆ =
=
Σ =
∆ + ∆ + ∆
-5.916MPa
տ
Ch 6- 10
2 MPa
P 2 cos15cos75 8 cos15cos15 0
.
A A
Ansτ + ∆ − ∆ == 3.93MPa
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With the physical significance of Mohr’s circle for
plane stress established, it may be applied with
simple geometric considerations. Critical values
Mohr’s Circle for Plane Stress
.
For a known state of plane stress
plot the points X and Y and construct the circle
centered at C .
22
22 xy
y x y xave R τ
σ σ σ σ σ +
−=
+=
Ch 6- 11
.
y x
xy p
ave R
σ σ
τ θ
σ σ
−=
±=
22tan
minmax,
The direction of rotation of Ox to Oa is the sameas CX to CA.
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y g g
Engineering Mechanics Centre of Studies
With Mohr’s circle uniquely defined, the state of
stress at other axes orientations may be depicted.
Mohr’s Circle for Plane Stress
For the state of stress at an angle q with respect to
the xy axes, construct a new diameter X’Y’ at an
angle 2q with respect to XY .
Normal and shear stresses are obtained from the
Ch 6- 12
coordinates X’Y’.
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Mohr’s circle for centric axial loading:
Mohr’s Circle for Plane Stress
0, === xy y x P
τ σ σ P
xy y x2
=== τ σ σ
Mohr’s circle for torsional loading:
Ch 6- 13
Tc xy y x === τ σ σ 0 0=== xy y x Tc τ σ σ
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Example 6.3
Solut ion
For the state of plane stress shown, (a)
’
Ch 6- 14
,
principal planes, (c) the principal stresses,
(d) the maximum shearing stress and the
corresponding normal stress.
Construction of Mohr’s circle
( ) ( )
( ) ( ) MPa504030
MPa40MPa302050
MPa202
1050
2
22 =+==
==−=
=−+
=+
=
CX R
FX CF
y xave
σ σ σ
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Principal planes and stresses
5020max +=+== CAOC OAσ
Example 6.3
MPa70max =σ
5020min −=−== BC OC OBσ
MPa30min −=σ
°=
==
1.532
30
402tan p
CP
FX
θ
θ
Ch 6- 15
°= 6.26 pθ
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Example 6.3
Ch 6- 16
Maximum shear stress
°+= 45 p s θ θ
°=6.71 sθ
R=maxτ
MPa50max =
τ
aveσ σ =′
MPa20=′
σ
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Example 6.4
Solut ion
For the state of stress shown,determine (a) the principal planes
and the principal stresses, (b) the
Ch 6- 17
stress components exerted on the
element obtained by rotating the
given element counterclockwise
through 30 degrees.
Construct Mohr’s circle
( ) ( ) ( ) ( ) MPa524820
MPa802
60100
2
2222 =+=+=
=+
=+
=
FX CF R
y xave
σ σ σ
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Example 6.4
Ch 6- 18
Principal planes and stresses
°=
===
4.672
4.220
482tan
p
pCF
XF
θ
θ
clockwise7.33 °= pθ
5280
max
+=
+== CAOC OAσ
5280
max
−=
−== BC OC OAσ
MPa132max +=σ MPa28min +=σ
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g g
Example 6.4
180 60 67.4 52.6
80 52cos52.6
80 52cos52.6
x
y
OK OC KC
OL OC CL
ϕ
σ
σ
′
′
= ° − °− ° = °
= = − = − °
= = + = + °Stress components after rotation by 30o
Ch 6- 19
s n . x yτ ′ ′ = = −
Points X’ and Y’ on Mohr’s circle thatcorrespond to stress components on the
rotated element are obtained by rotating XY
counterclockwise through °= 602θ
48.4MPa
111.6MPa
41.3MPa
x
y
x y
σ
σ
τ
′
′
′ ′
= +
= +
= −
S l P bl 6 1
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Supplementary Problems 6.1
1) At a point in the structural member, the stresses
are represented as in figure below. Find;
a) the magnitude and orientation of theprincipal stresses and,
[96.1MPa;23.94 MPa;28.15o]
a) the magnitude and orientation of the
maximum shearing stresses andassociated normal stresses.
[36.08MPa; -16.85o ; 60 MPa]
80 MPa
40 MPa
30 MPa
y
10 MPa
Ch 6 - 20
2) For the state of plane stress shown, determinethe normal and shearing stresses on the planes
whose normal are at +60o and + 150o with the x-
axis. Show stresses on a sketch of the element.
[for +60: σ N = 7.32MPa; τ N = -10MPa]
[for +150: σ N = -27.32MPa; τ N = 10MPa]
x
20 MPa
10 MPa
S l P bl 6 1
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Supplementary Problems 6.1
3) A structural member is subjected to two set of loadings. Each separately produce
stress conditions at a point A as shown in Figure 3(a) and Figure 3(b).
x, y xy .
[37.5 MPa; 22.5 MPa; -14.33 MPa]
a) By continuing the effect of both stress conditions Q3(a), determine the
principal stresses and their orientations with respect to the x -axis and also the
magnitude of the maximum shearing stress.
10 MPa
20 MPa30 MPa
Ch 6 - 21
Figure 3(a) Figure 3(b)
AA
15 MPa
5 MPa
30o
x
+
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State of stress at Q defined by: zx yz xy z y x τ τ τ σ σ σ ,,,,,
Consider the general 3D state of stress at a point and thetransformation of stress from element rotation
General State of Stress
Consider tetrahedron with face perpendicular to the line QN
with direction cosines: z y x λ λ λ ,,
The requirement leads to,∑ = 0n F
x z zx z y yz y x y
z z y y x xn
λ λ τ λ λ τ λ λ τ
λ σ λ σ λ σ σ
222
222
+++
++=
Ch 6- 22
Form of equation guarantees that an element orientationcan be found such that
222ccbbaan λ σ λ σ λ σ σ ++=
These are the principal axes and principal planes and the
normal stresses are the principal stresses.
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Application of Mohr’s Circle to the Three- Dimensional Analysis of Stress
Transformation of stress for an element The three circles represent the normal
Ch 6- 23
ro a e aroun a pr nc pa ax s may e
represented by Mohr’s circle.
an s ear ng s resses or ro a on aroun
each principal axis.
Points A, B, and C represent the principal
stresses on the principal planes (shearing
stress is zero)minmaxmax
21 σ σ τ −=
Radius of the largest circle yields the
maximum shearing stress.
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In the case of plane stress, the axis
perpendicular to the plane of stress is a principal
axis (shearing stress equal zero).
Application of Mohr’s Circle to the Three-Dimensional Analysis of Stress
a) the corresponding principal stresses arethe maximum and minimum normal
stresses for the element
If the points A and B (representing the principal
planes) are on opposite sides of the origin, then
Ch 6- 24
b) the maximum shearing stress for the
element is equal to the maximum “in-plane”shearing stress
c) planes of maximum shearing stress are at
45o to the principal planes.
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Application of Mohr’s Circle to the Three-Dimensional Analysis of Stress
If A and B are on the same side of the origin (i.e.,
have the same sign), then
b) maximum shearing stress for the element is
equal to half of the maximum stress
a) the circle defining smax, smin, andt max for the element is not the circle
corresponding to transformations within
the plane of stress
Ch 6- 25
c) planes of maximum shearing stress are at
45 degrees to the plane of stress
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Transformation of Plane Strain
Plane strain - deformations of the material takeplace in parallel planes and are the same in
each of those planes.
Plane strain occurs in a plate subjected along
its edges to a uniformly distributed load and
restrained from expanding or contracting
laterally by smooth, rigid and fixed supports
( )0 :strainof components
x === zy zx z xy y γ γ ε γ ε ε
Ch 6- 26
Example: Consider a long bar subjected to
uniformly distributed transverse loads. State
of plane stress exists in any transverse
section not located too close to the ends of
the bar.
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Transformation of Plane Strain
State of strain at the point Q results indifferent strain components with respect to
the xy and x’y’ reference frames.
22
( ) ( )
( ) y xOB xy
xy y xOB
y y x
ε ε ε γ
γ ε ε ε ε
γ ε ε ε
+−=
++=°=
=
2
45
coss ns ncos
21
Applying the trigonometric relations used
for the transformation of stress,
Ch 6- 27
cos2 sin 22 2 2
cos2 sin 22 2 2
sin 2 cos22 2 2
x y x y xy
x
x y x y xy
y
x y x y xy
ε ε ε ε γ
ε θ θ
ε ε ε ε γ ε θ θ
γ ε ε γ θ θ
′
′
′ ′
+ −
= + ++ −
= − −
−= − +
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Mohr’s Circle for Plane Strain
The equations for the transformation of plane
strain are of the same form as the equations
for the transformation of plane stress - Mohr’s
circle techni ues a l .
Abscissa for the center C and radius R ,
22
222
+
−=
+=
xy y x y xave R
γ ε ε ε ε ε
Principal axes of strain and principal strains,
Ch 6- 28
R R aveave
y x
xy
p
−=+=
−=
ε ε ε ε
ε ε
γ
θ
minmax
2tan
( ) 22max 2 xy y x R γ ε ε γ +−==
Maximum in-plane shearing strain,
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Example 6.5
The strain at a point in a body are as
follows:
800 200 600ε ε = = =
Solut ion
Using Mohr’s circle
800 xε µ =
Determine:
a) The principal strains and directions,
and
b) The strain εa in the direction of 60º
with the x -axis and the strain εbperpendicular to εa and the shearing
x y xy
Setting the coordinates
( )
( )
800, 300
200,300
X
Y
µ
µ
−
2001
3002
y
xy
ε µ
γ µ
=
=
Ch 6- 29
strain γab.
Centre OC800 200
5002
OC +
= =
Radius, R
2 2300 300 424.3 R µ = + =
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Absolute Maximum Shear Strain
Absolute maximum shear strain is found from the circle having the largest radius.
It occurs on the element oriented 45° about the axis from the element shown in its
original position.
minmax
minmaxmaxabs
ε ε ε
ε ε γ
+=
−=
Ch 6- 31
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Absolute Maximum Shear Strain
For plane strain we have,
This value represents the absolute maximum shear strain for the material.
Ch 6- 32
( ) maxmax''maxabs ε γ γ ==
z x minmaxmax''maxabs ε ε γ γ −==
y x
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Example 6.6
The state of plane strain at a point is represented by the strain components,
( ) ( ) ( )6 6 6400 10 , 200 10 , 150 10 x y xyε ε γ − − −= − = =
e erm ne e max mum n-p ane s ear s ra n an e a so u e max mum s ear s ra n.
From the strain components, the centre of the circle is on the ε axis at:
( ) ( )66 10100102
200400 −− −−=+−
=avg ε
−γ −−
Solut ion
Ch 6- 33
nce , e re erence po n as coor na es
2
= ,−
Thus the radius of the circle is
( ) ( ) ( )9622 103091075100400 −− =
+−= R
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Example 6.6
Computing the in-plane principal strains, we have
( )66
66
max 1020910309100
−−
−−
−=−−=
=+−=ε
From the circle, the maximum in-plane shear strain is
m n
( )[ ] (Ans) 1061810409209 66minmax planeinmax−− =−−=−= ε ε γ
10409,0, 10209 6minint6
max
−− −=== ε ε ε From the above results, we have
Thus the Mohr’s circle is as follows,
Ch 6- 34
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Three-Dimensional Analysis of Strain
Previously demonstrated that three principal
axes exist such that the perpendicular
element faces are free of shearing stresses.
By Hooke’s Law, it follows that the shearing
strains are zero as well and that the principal
planes of stress are also the principal planes
of strain.
Ch 6- 35
represented by Mohr’s circles.
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Three-Dimensional Analysis of Strain
For the case of plane strain where the x and y
axes are in the plane of strain,
the z axis is also a principal axis
the corres ondin rinci al normal
strain is represented by the point Z = 0 or the origin.
If the points A and B lie on opposite sides of
the origin, the maximum shearing strain is themaximum in-plane shearing strain, D and E .
Ch 6- 36
If the points A and B lie on the same side of theorigin, the maximum shearing strain is out of the
plane of strain and is represented by the points
D’ and E’ .
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Th D l A l f S
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Three-Dimensional Analysis of Strain
Corresponding normal strains,
Consider the case of plane stress,
0=== z b ya x σ σ σ σ σ
( ) ( )babac
bab
baa
E
E E
E E
ε ε ν
ν σ σ
ν ε
σ σ ν ε
σ ν σ ε
+−
−=+−=
+−=
−=
1
Ch 6- 37
If B is located between A and C on the Mohr-
circle diagram, the maximum shearing strain is
equal to the diameter CA.
not zero.
Supplementary Problems 6.2
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1) For the given state of plane strain, determine (a) the orientation and magnitude of the
principal strains, (b) the maximum in-plane strain, and (c) the absolute maximum
shearing strain.a 37.53o -32.4e-6 -187.6e-6 b 155.2e-6 c 187.6e-6
ε x = - 90 µ ; ε y = - 130 µ ; γ xy = 150 µ
2) For the given state of plane strain, determine the state of strain associated with axes
x’ and y’ rotated via the angle θ .[-653e-6; 303e-6; -829e-6]
ε x = - 800 µ ; ε y = 450 µ ; γ xy = 200 µ ; θ = -25o
Ch 6 - 38
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G li d H k ’ L
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For a triaxial state of stress, the general form for Hooke’s law is as follow:
( ) ( ) ( ) x z x z x x vvv σ σ σ ε σ σ σ ε σ σ σ ε +−=+−=+−=1
, 1
, 1
Generalized Hooke’s Law
They are valid only for a linear–elastic materials.
Hooke’s law for shear stress and shear strain is written as
And relationship between G, E and v is;
1 1 1 xy xy yz yz xz xz
G G Gγ τ γ τ γ τ = = =
Ch 6- 39
( )v E G+= 12
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Generalized Hooke’s Law
Stresses in terms of strains
2 E
v or G eσ ε ε ε ε σ ε λ = + + − = +
( ) ( ) ( )( )
( ) ( ) ( )( )
1 1 2
21 1 2
21 1 2
: ;
y y x z y y y
z z x y z z z
x y z
v v
E v or G e
v v
E
or G ev v
vE ote and e
σ ε ε ε ε σ ε λ
σ ε υ ε ε ε σ ε λ
λ ε ε ε
+ −
= + + − = ++ −
= + + − = ++ −
= = + +
Ch 6- 40
−
( ) ( ) ( ); ;
2 1 2 1 2 1 xy xz yz y xz xz
E E E
v v vτ τ τ γ γ γ = = =
+ + +
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Example 6.7
The copper bar is subjected to a uniform loading along its edges. If it has a = 300 mm, b = 50mm, and t = 20 mm before load is applied, find its new length, width, and thickness after
application of the load. Take E CU = 120 GPa, v = 0.34.
From the loading we have
The associated normal strains are determined from the generalized Hooke’s law,
800 MPa , 500 MPa , 0 , 0 x y xy z
σ σ τ σ = = − = =
Ch 6- 41
( )
( )
( )
0.00808
0.00643
0.000850
x x y z
y
y x z
z z x y
v E E
v
E E
v E E
σ ε σ σ
σ ε σ σ
σ ε σ σ
= − + =
= − + = −
= − + = −
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Example 6.8
At a point in a stresses body, the strains related to the coordinate set xyz are
given by:
Determine the complete stress components for this body. Assume Young
Modulus, E = 100 GPa and Poisson’s ratio, v = 0.33.
6300 100 500 10
200 500 400
− − × −
Ch 6- 43
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Example 6 8
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Example 6.8
( )( ) xx xx yy zz xx E
σ ε υ ε ε ε = + + −−
Normal stresses on each axis:
( ) ( )( ) ( )( )
( )
( ) ( ) ( )( )
6
6 6 6 6
11
11
100 10200 10 0.33 100 10 400 10 200 10
1 0.33 1 2 0.33
2.2114 10 0.000031 .
1 1 2
2.2114 10 0.000133 .
yy yy xx zz yy
Ans
E
ns
σ ε υ ε ε ε υ υ
−
− − − −×= × + − × − × − ×
+ −
= × − =
= + + −+ −
= × − =
-6.86MPa
-29.4MPa
Ch 6- 44
( ) ( )1 1 2 zz z
E σ ε υ υ
=+ − ( )( )
( )112.2114 10 0.000235 .
z xx yy zz
Ans
υ ε ε ε + + −
= × − = -51.97MPa
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Example 6.7
Continue.
( )9
6100 10300 10 .
1 1 0.33 xy xy
E ns
vτ ε
−×= = × =
+ +22.6MPa
( )
( )
9
6
9
6
100 10 200 10 .1 1 0.33
100 10500 10 .
1 1 0.33
xz xz
yz yz
E nsv
E ns
v
τ ε
τ ε
−
−
×= = × =
+ +
×= = × =
+ +
15.04MPa
37.6MPa
Thus, complete stress component is:
−
Ch 6- 45
. . .
22.6 29.4 37.6 .
15.04 37.6 51.97
Pa Ans
−
= −−
σɶ
Supplementary Problems 6.3
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1) Determine the complete stress component of this material. Use E = 200 GPa and G
= 80 GPa. 200 100 0
100 300 400 µ
2) A square ABCD of 20 mm side is scribed on the surface
of a thin plate while the plate is unloaded, as shown in
Figure Q2. After the plate is loaded, the lengths of sides
AB and AD are observed to have increased by 4 x10-6 m and 6 x10-6 m, respectively while angle DAB is
observed to have decreased by 160 x10-6 rad.
Determine;
D C
y
Ch 6 - 46
i. the principal strains and its orientation,[344.34µ,155.66µ,29o]
ii. the maximum in-plane shear strain, and [188.68µ]
iii. the average normal strain. [250µ]
iv. The principal stresses using E = 200 GPa and v =
0.3. [85.94 MPa,56.92 MPa]
A B
x
Figure Q2
Previous Exam Questions
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OCT2009/MEC411/KJM454 Q2(b)
An element is subjected to two mutually perpendicular strains, εx= 400µ and εy= 200µ,
together with an unknown shear strain γxy. If the maximum principal strain in the
i. the magnitude of the shear strain, [346.4µ]
ii. the other principal strain and the direction of the principal strain, [100µ,30o]
iii. the strain εa in a direction of 30° with the x-axis, and [500µ]
iv. If E = 200 GPa and v = 0.03, determine the principal stresses.
APR2011/MEC411 Q5(b)
Ch 6 - 47
e s a e o p ane s ra n a a po n n a e orma e o y s e ne y εx = – µ, εy =
– 140 µ, and γxy = 150 µ. Calculate :
i. the orientation and magnitude of the principal strains, [37.53o,-42.38µ,-197.62µ]
ii. the maximum in-plane strain, and [150 µ]
iii. the new state of εx , i.e., εx’, when the element is rotated + 30o with respect to the
x-axis. [-45.05 µ]
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JUN2011/MEC411 Q5(d)
The principal stresses (at a point on a certain material) are given as 90 MPa and 30
MPa, both is in tension. Determine the normal and shear stresses on a plane making an
angle of tan-10.25 (ccw) with the principal directions. [86.47 MPa,14.1 MPa]
JAN2012/MEC411 Q5(b)
A state of plane stress consists of a tensile stress 60 MPa exerted on vertical surfaces
and of unknown shearing stresses. If the largest normal stress is 100 MPa, calculate;
i. the average normal stress, [30 MPa]
Ch 6 - 48
ii. the magnitude of the shearing stress, [63.25 MPa]
iii. the maximum shearing stress, and [70 MPa]iv. the minimum principal stresses. [-40 MPa]
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JUN2012/MEC411 Q5(a)
A steel bar is subjected to an axial force of P . The state of stress at a point on the bar
caused by the axial loading is shown in Figure Q5(a). Knowing that the stresses on
plane a-a are σy
’ = -100 MPa and τx
’y
’ = 35 MPa, determine using the Mohr’s circle;
i) the normal stress, σX’ of the stress component on plane a-a, [-12.25 MPa]
ii) The angle α that plane a-a forms with the horizontal, and [19.3o]
iii) The maximum compressive stress in the bar. [-112.25 MPa]
P
σy
Ch 6 - 49
State of stress
P
α
a
a
σy
y
x
z
Figure Q5 (a)