MIT - 16.003/16.
004
Spring, 2009
Unit M5.3 Yield (and Failure) Criteria
Readings: CDL 5.11, 5.13, 6.9
16.003/004 -- Unified Engineering Department of Aeronautics and Astronautics Massachusetts Institute of Technology
Paul A. Lagace 2008
MIT - 16.003/16.004
Spring, 2009
LEARNING OBJECTIVES FOR UNIT M5.3
Through participation in the lectures, recitations, and work associated with Unit M5.3, it is intended that you will be able to
.explain why shearing is a key mechanism in material failure (yielding) in many cases .describe typical failure/yield criteria, their origin, and the importance of hydrostatic stress .use these criteria in assessing the failure for cases with multiaxial stress fields
Paul A. Lagace 2008
Unit M5-3 p. 2
MIT - 16.003/16.004
Spring, 2009
Thus far we have talked about the manifestation of yielding in the overall stress-strain response and the mechanisms/origins of yielding. We would like to move forward and be able to predict yielding (and failure) in structures under general (multiaxial) Load/Stress states.
Before we look at two (classic) criteria which have been devised to do this, let us consider two key facts. First the..
Maximum Shear Plane
Crystals (grains) slip along certain planes. In a material with many crystals and thus randomly oriented grains, overall slip occurs along a more or less oriented plane Thus, the material property is actually yield (shear yield stress)
Paul A. Lagace 2008
Unit M5-3 p. 3
MIT - 16.003/16.004
Spring, 2009
This yields the question.. How is this related to yield? --> Consider the uniaxial tensile test
Figure 5.3-1
Coupon under uniaxial tension
x1 x2
P Stress state: 11 = P/A 22 = 0 33 = 0
Recall stress transformation (Mohrs circle)
Paul A. Lagace 2008
Unit M5-3 p. 4
MIT - 16.003/16.004
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Figure 5.3-2
General in-plane transformation of uniaxial stress state
x1 x2
11
x2
x1
~
Figure 5.3-3
12
22
and maximum shear stresses! Mohrs circle and maximum shear stress for in-plane uniaxial stress state
shear
11
max
90 extensional
11
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Unit M5-3 p. 5
MIT - 16.003/16.004
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1 1 = y Thus, yield occurs when: 2
yield = 2 yield
What angle does max occur at? 45 to principal stress direction 11
slip/yield should occur along 45 lines
Figure 5.3-4
45 slip line in coupon under uniaxial stress P
(shows in failure modes) P
Paul A. Lagace 2008
Unit M5-3 p. 6
MIT - 16.003/16.004
Spring, 2009
The second fact deals with
The Importance of Hydrostatic Stress
Hydrostatic stress is a state of stress such that: 11 = 22 = 33 = p (principal stresses all equal no shear) Experimental data shows no yielding under hydrostatic stress
Figure 5.3-5
Unit cube under state of hydrostatic stress p
Mohrs circle collapses to a point!
Paul A. Lagace 2008
Unit M5-3 p. 7
MIT - 16.003/16.004
Spring, 2009
Conclusion: Any yield criterion must not allow yielding under hydrostatic stress/pressure There are two classic criteria we will consider devised for isotropic materials (yield is one value) The first is the
Tresca Criterion
(1868) Material yields if the maximum shear stress exceeds yield Generalizing this to three dimensions gives:
Paul A. Lagace 2008
Unit M5-3 p. 8
MIT - 16.003/16.004
Spring, 2009
= yield
or or or
= yield = yield for yield
= yield
or
= yield = yield
Recall y = 2y Check case of hydrostatic stress:
= = = C no yield
--> Look at the case of plane stress ( = 0):
Paul A. Lagace 2008
Unit M5-3 p. 9
MIT - 16.003/16.004
Spring, 2009
= y
= y
= y
Plotting the failure envelope..
Figure 5.3-6
Tresca failure envelope for case of plane stress II | II | y | I - II | | I | - y y
| I | - y | II |
| I - II |
A second criterion is the
Paul A. Lagace 2008
Unit M5-3 p. 10
MIT - 16.003/16.004
Spring, 2009
von Mises Criterion
Very similar to Tresca but not discontinuous
) + ( ) + ( ) = 2 2 y
at yield Still the differences of the principal stresses Now sum up the effects Still no yielding for case of hydrostatic stress
--> get a rounded-off Tresca --> Look at case of plane stress ( = 0):
) + 2 + 2 = 2 2 y
2
Paul A. Lagace 2008
+ 2 = 2 y
at yield
Unit M5-3 p. 11
MIT - 16.003/16.004
Spring, 2009
(Note: often write as:
2 2 1 1 1 1 2 2 + 2 2 + 3 1 2 = y 2
by using transformations to non-principal axes) Compare to Tresca.
Figure 5.3-7
Comparison of Tresca and von Mises failure criteria for case of plane strain II y
- y
y
- y
Paul A. Lagace 2008
I von Mises Tresca
Unit M5-3 p. 12
MIT - 16.003/16.004
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Finally, consider
Using Yield Criteria
The steps for a general structure are: 1. Analyze structure to obtain stresses (ij) and principal stresses (, , ) 2. Obtain yields/ultimates via handbook (e.g., MIL HDBK 5, 17) or experimentation 3. Choose yield/failure criterion 4. Utilize calculated stresses in failure/yield criteria with associated material yields/ultimates NOTE: Failure criteria get far more complex for inhomogeneous, nonisotropic material
Thus far weve concentrated on material failure by yielding. We next look at the phenomenon of fracture.
Paul A. Lagace 2008
Unit M5-3 p. 13
MIT - 16.003/16.004
Spring, 2009
Unit 5.3 (New) Nomenclature yield (y) -- shear yield stress yield -- yield stress , , -- principal stresses
Paul A. Lagace 2008
Unit M5-3 p. 14
