2-lectures lec 16 distortion energy theory
TRANSCRIPT
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Dr. A. Aziz Bazoune
King Fahd University of Petroleum & MineralsKing Fahd University of Petroleum & MineralsKing Fahd University of Petroleum & MineralsKing Fahd University of Petroleum & MineralsMechanical Engineering Department
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FAILURE CRITERIA
Failure Criteria (Static Load +Ductile Material )
6-5- Distortion Energy Theory ( DE), Von-Mises or Von-Mises-HenckyTheory
(Hueber, Poland: 1904, von-Mises, Germany+US, 1913and Hencky, Germany+US, 1925)
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The Distortion Energy Theory (DE) originated from the
observation that ductile materials stressed
hydrostatically exhibited yield strengths greatly in
excess of the values given by the simple tension test.Therefore, it was postulated that yielding was not a
simple tensile test or compressive phenomenon at all,
but, rather that it was related somehow to the angular
distortion of the stressed element.
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The total strain energy of an element of material could be dividedinto two parts :
Due to change in volume (Hydrostatic) Due to change in shape (Distortion)
where : total strain energy: strain energy due to change in volume
: strain energy due to change in shape
V S U U U = +
U V U
S U
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The stress state shown in Figure 6-16b is one of hydrostatic
tension due to the stress acting in each of the same principaldirections as shown in Figure 6-16a. The formula for is givenby
1 2 3
3av
+ +
=
av
av
(a)
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Figure 6-16(a) Element with triaxial stresses; this element undergoes both volume change and angulardistortion. (b) Element under hydrostatic tension undergoes only volume change. (c) Elementhas angular distortion without volume change .
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The distortion energy theory is found by:
Computing the total strain energy
Subtracting the energy due to volumetric change
Notice that when a material extends in one direction, it contracts
in another direction. Refer to Table 4-2 Page 124, the principal
strains are given by
( )( )
( )( )
( )( )
1 1 2 3
2 2 1 3
3 3 1 2
1
1
1
E
E
E
= +
= +
= +
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The total strain energy per unit volume subjected to threeprincipal stresses is :
( )
( )
1 1 2 2 3 3
2 2 21 2 3 1 2 2 3 3 1
1
21
22
u
E
= + +
= + + + + (b)
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T e stra n energy or pro uc ng on y vo ume c ange can eobtained by substituting for and into Eq. (b). Theresult is
vuav 1 2, , 3
( )
( )
2 2 2 2 2 2
2
12
2
31 2
2
v av av av av av av
av
u E
E
= + + + +
= (c)
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Substituting Eq.(a) into Eq. (c) and simplifying we get
The distortion energy is obtained by subtracting Eq.(6-7) from Eq.(b). This gives
(6-7)( )2 2 21 2 3 1 2 2 3 3 11 2
26v
u E
= + + + + +
( ) ( ) ( )2 22
1 2 2 3 3 11 + + +
= = 6-8
10
3 2v
E Notice that the distortion energy is zero if 1 2 3 0. = = =
For the simple tensile test, at yield , and andfrom Eq. (6-8) the distortion energy is
1 yS = 2 3 0, = =
21
3d yu S
E
+=
(6-9)
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Failure Criterion:Failure occurs when the distortion energy of the material reachesor exceeds the strain energy of the tensile test
( ) ( ) ( )2 22
1 2 2 3 3 1 21 1
3 2 3 yS
E E
+ + + +
11
(6-10)
Therefore,
( ) ( ) ( ) 1 22 221 2 2 3 3 12 y
S + +
Strain EnergyDistorsion Energy for Tensile Test
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For a simple case of tension , then yield would occur whenThus, the left of Eq. (6-10) can be thought as a single , equivalent ,or effective stress for the entire general state of stress given by
and .
This effective Stress is usually called von Mises stress and is givenby
yS
1 2,
3
'
12
(6-12)( ) ( ) ( ) 1 22 221 2 2 3 3 1'2
+ + =
Eq. (6-10) , for yield, can be written as
' yS (6-11)
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(6-14)( ) ( ) ( ) ( )1 22 2 2 2 2 21' 6
2 x y y z z x xy yz zx
= + + + + +
Using xyz components of the three dimensional stress, the von Misesstress can be written as
Plane stress1 22 2 2' 3 = + + (6-15)
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The distortion shear-energy theory is also called The von Mises or von Mises Hencky theory The shear energy Theory
The Octahedral-shear-stress theory
Simple tension and shear
( )1 22 2' 3 x xy = + (6-15a)
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(6-13)
( )1 22 2
' A A B B = +
For plane stress, let and be the two nonzero principal stressstresses. Then from Eq. (6-12), we get
A
B
Eq. (6-13) represents a rotated ellipse in the plane as shownin Figure (6.17) with .
, A B ' yS =
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(6-13)
( )1 22 2
' A A B B = +
For plane stress, let and be the two nonzero principal stressstresses. Then from Eq. (6-12), we get
A
B
Eq. (6-13) represents a rotated ellipse in the plane as shownin Figure (6.17) with . The dotted lines in the figure
, A B ' yS =
15
, ,hence, more conservative.
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Octahedral Shear Stresses
Consider an isolated element in whichthe normal stresses on each surface areequal to the hydrostatic stress .
There are 8 surfaces symmetric to therinci al directions that contain that
av
19
stress. This form an octahedron asshown in Figure 6-18.
Figure 6-18 Octahedral SurfaceThe shear stresses on these surfacesare equal and are called the octahedral
shear stresses .
( ) ( ) ( )1 22 22
1 2 2 3 3 1
1
3oct = + + (6-16)
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which is identical to Eq.(6-10) the verifying that the maximumoctahedral shear stress theory is equivalent to the distortion energytheory.
The model for the MSS ignores the contribution of the normalstresses on the 45 o surfaces of the tensile specimen. However,theses stresses are P/2A , and NOT the hydrostatic stresses, whichare P/3A. Herein lies the difference between the MSS and DEtheories.
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Eq.(6-11) can be expressed as a design equation by
' yS
n = (6-19)
The DE theory predicts no failure under hydrostatic stress andagrees well with all data for ductile behavior. Hence it is the mostwidely used theory for ductile materials and is recommended fordesign problems unless otherwise specified
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Consider a case for pure shear where for plane stressEq. (6-15) with (6-11) for yields gives
Thus, the shear yield strength predicted by the distortion-energytheory is
2 3 0. = = xy
23 xy yS
= or3
y
xy
S = (6-20)
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0.577sy yS S = (6-21)
Which as stated earlier, is about 15% greater than the predicted bythe MMS theory (see the pure shear load line in Fig-6-17).