considering the volumetric strains due toswamidas/engi5312-classnotes1-word... · web...
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Plane Strain
Each strain is acting independent of one another
Due to normal strain x Due to normal strain y
x =
Due to shear strain
1
dxdu1
dv1
dv2
x
y
du2
dv3
x
du3
dy
x
Negligible
y
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General Equations of Plane Strain Transformation
Sign Conventions
(1) Normal strains are + ve, if they cause elongations along x and y axes, respectivelty.
(2) Normal strains are - ve, if they cause shortening along x and y axes, respectively
(3) Shear strains are + ve, if the interior angle AOB becomes less than 900.
(4) Shear strains are – ve, if the interior angle AOB becomes greater than 900.
2
x
y
dx
dy
du = x dx
dv = y dy
O A
B
x
yx
y
dy
dy
dx
dy dx
O
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From the figure
Problem
Using the above orientations of axes, determine the strains along xoy axes due to
, defined w.r.t. xoy axes.
Effect of normal strain x, along x axis
Effect of normal strain y, along x axis
3
x
yx
y
dy
dy
dx
dydx
dv1du1
du
du=xdx
x
yx
y
dy
dx
dy dx
du1 =dv sin = x dysin
dv1 = y dycos
dv = ydy
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Effect of shear strain xy, along x - axis
[Assume that dx remains fixed in position, and the shear strain xy is represented by the
change in angle of dy]
+du1
=
(I)
+dv1
=
= (the angle of shear distortion along x axis)
4
x
x
y
dy
dx
dxxy
dv1=xydysindu1=xydycos
xy dy
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By rotating the angle through 900, in the clockwise direction, the rotation of elemental
length dy can be obtained.
(-) = rotation of the right angle xoy
(II)
From Equation I,
(III)
From Equation II,
(IV)
can be obtained by introducing (90+) for in
5
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=
=
=
To find the principle strain
i.e.,
6
2p
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=
=
7
= p
= p
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Similarity Between Stress and Strain Transformation EquationsStresses at a point Strains at a point
9
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Principle Stresses Principle Strains
for p1,
Maximum in-plane shear stress
10
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Principal stress plane and maximum
shear stress planes are inclined at 450.
Consequently,
twice the values of these
angles will be inclined at 900.
As a result
Maximum in-plane Shear Strain
Principal strain plane and maximum (in-
plane) shear stress planes are inclined at
450 to one another.
Consequently, twice the values of these
angles will be inclined at 900.
As a result
11
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Mohr’s Circle
For Plane Stress For Plane Strain
2
1
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Material Property Relationships
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When only shear stresses are acting
i.e.,
i.e., (A)
When a body is subjected to normal stresses the body under goes only
change in volume.
Volume change =
(volume change)/unit volume =
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=
p/e = bulk modulus = K =
=
Theories of FailureIn the design of structural members, it becomes important to place an upper limit on the
state of stress that defines the material's failure.
Ductile Materials Brittle Materials
Stress Stress
Y - Yield stress (steel)
ult – Ultimate stress ult – Ultimate stress
- Not used since strain is very
high at this stress level
0.1% Proof stress
(stress at o.1% elongation)
(Aluminum)
Strain Strain
y – Yield strain (0.15 to 0.2 ult – Ultimate strain (0.2% to
for mild steel) 0.3% for CAST IRON)
ult – Ultimate strain (20 to 25%
for mild steel)
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The material behaviour – either ductile or brittle – does not remain a constant one for
any material. It is dependent on:
Temperature
Rate of loading
Chemical environment
Forming/shaping methods
In order to apply the theories of failure:
(i) The state of stress in a structure, at a point where the maximum stresses
are expected - - are determined first.
(ii) Thereafter, the principal stresses and maximum shear stresses are
determined -
Failure Theories
For ductile materials For brittle materials
1. Maximum shear stress theory 1. Maximum normal stress theory
Proposed by Tresca
2. Maximum distortion energy 2. Mohr’s failure criterion
theory – Proposed first by Huber - Proposed by Otto Mohr
and refined later by Von Mises and
Hencky
- Huber–Mises–Hencky theory
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1. Maximum Shear Stress Theory
“Failure (by yielding) will occur in a material (at a point) when the maximum shear
stress in the material is equal to the maximum shear stress that will occur when the
material is subjected to an axial tensile test’.
P = (Y) A
For a two-dimensional stress system,
(I)
Under simple tension test,
(II)
Using in Eqn. I
450
Thin mild steel strip Luder’s lines
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Governing criteria (III)
Considering a three-dimensional element (with two-dimensional stress state)
Arranging the stresses in the order of decreasing magnitudes,
(i) Case (a):
Hence
Failure will occur first in the plane.
i.e., (IV)
Failure in shear will occur, when the maximum principal stress is equal to Y.
(ii) Case (b):
1
3
2
1
3 = 0
2
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V (a)
Failure will occur in the plane containing stresses
Generalizing this for a plane-stress failure wherein act along x-y axes and 3 acts
along z-axis, (zero stress), one can rewrite Equation (V (a)) as
V (b)
Failure envelope or Yield loci
2. Maximum Distortion Energy Theory
Y
Y
Y
Y
2
1
1, -2
+ve -ve
-1, +2
-ve +ve
Failure occurs in (1, 2) plane
Failure occurs in (1, 2) plane
(+ve)
(-ve)
(-ve)
(+ve)
2 = Y, 3 =0-Failure in this plane (1, 3)
1 = Y (3 =0)-Failure in this plane (1 , 3)
-Failure occurs in plane (1, 3)
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“ Failure (by yielding) will occur when the shear or distortion energy in the material
(at a point) reaches the equivalent value that will occur when a material is subjected to
uniaxial tensile test”.
Let us say that the principal stresses in an element, at a point, is given by
Total strain energy stored in the given system = Total volumetric strain energy + Total
distortion strain energy
ut = uv + ud (VI)
=
+
1
3
2
1
2
3
1
3
2
1
3
11
21
31
2
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(VII)
Also
= 0 (VIII)
Using the earlier stress-strain relationships
Considering the volumetric strains due to
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= 0 (IX) [Since according to Eqn. (VIII)]
Equation (IX) states that no volumetric change occurs in the material due to the stresses
(but it does produce a change of shape). Due to the three stresses
,
(Total strain energy)
Hence strain energy per unit volume
=
Considering only [the mean stresses and strains due to ( ) and (
)],
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(X)
Considering the normal (or principal) stresses and strains,
(XI)
Since ,
(XII)
When the specimen is under uniaxial tension,
From eqn. (XII),
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(XIII)
For a general state of stress,
= (XIV)
From Eqns. (XIII) and (XIV), equating the distortional energies due to an uniaxial state
of stress and that due to a multiaxial state of stresses,
(XV)
For a two-dimensional state of stresses,
Hence equation reduce to
i.e.,
(XVI)
This is an equation to an inclined ellipse.
Y
2
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Plot of Eqn. (XVI) gives the failure envelope or yield loci for a system subjected to a
two-dimensional state of stress.
Brittle Materials
Applicable to cast iron that tends to fail suddenly by fracture, without any warning.
1. Maximum normal stress Theory:
In a tension (or compression) test, brittle fracture occurs when the normal stress reaches
the ultimate stress ult.
Y Y
Y
1
Maximum distortion-energy theory Maximum
shear stress theory
450
450
450
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In a torsion test, brittle fracture occurs due to a maximum tensile stress (in a plane 450 to
the shear direction) when it reaches the ultimate stress ult.
Failure criteria or failure loci:
Statement
When the maximum principal stress in the material reaches a limiting value
that is equal to the ultimate normal stress the material can sustain, failure by fracture
occurs.
CompressionTensionTorsional shear
ult
ult
1
2
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-Eg. Chalk: under tension, under bending and under torsion.
2. Mohr’s Failure Criterion
For materials (brittle) that have different fracture properties in tension and compression,
this criterion holds good.
-Specially for metals
-For nonmetals like concrete (Rock, concrete, soils) another theory
is applicable (we will briefly deal with this later)
Three tests done to determine failure criteria –
- Tension test that gives (ult)t
- Compression test that gives (ult)c
- Torsion test that gives ult
Circle B1 = ult
2 = 03 = -ult
Circle C1 = (ult)t
2 = 03 = 0
(ult)c
A
BC
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Mohr’s circle for each test
Mohr’s failure criteria
Failure occurs when the absolute value of either one of the principal stresses reaches a
value greater than (ult)t or (ult)c or in general, if the stress at a point is defined by the
stress coordinate (1, 2), which is plotted on the boundary or outside the shaded area.
Circle A1 = 02 = 03 = -(ult)c
ult
Failure envelope
1
2
(ult)t
(ult)t
(ult)c
(ult)c
x (1, 2) material has failed
Material is under limiting condition
(11, 2
1)