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1 C:\W\whit\Classes\304_01_ver_3\_Notes\6_FailureAnalysis\1_failureMechanics_intro.doc p. 1 of 1 Failure Mechanics Performance can be affected by temperature (high and low), corrosion, light, moisture, contact with fuels, radiation We will concentrate mainly on the effects of stress. What is failure? Separation Yield Loss of stiffness when the structure can no longer perform the required function Why do structures fail? Excessive loads o Static failure o Plasticity o Buckling o Aeroelastic effects Creep Corrosion Fatigue Fracture Two basic design philosophies Safe life design (landing gear, hinge points, ) Fail Safe design (damage tolerant) Often lighter, since easier to be redundant than to guarantee life Definitions Limit load: The maximum static or dynamic load an aero-space vehicle or its structural elements are expected to experience at least once during its service life. ( Factor of safety= (allowable stress, etc.)/(actual stress, etc.) Design allowable A limiting value for a material property that can be used to design a structural or mechanical system to a specified level of success with 95% statistical confidence. B-basis allowable: material property exceeds the design allowable 90 times out 100. A-basis allowable: material property exceeds the design allowable 99 times out of 100.
2 C:\W\whit\Classes\304_01_ver_3\_Notes\6_FailureAnalysis\_staticFailure.doc p. 1 of 4 Static Failure without Cracks Basic challenge: there are 6 stress components. There are an infinite number of combinations that can exist in structures. How can we characterize the material so that we can make predictions for any combination? There are many criteria. Here are a few. Criteria Maximum stress: failure when stress component exceeds allowable Maximum strain: failure when strain component exceeds allowable Yielding of Metals o In the following, σ i are the principle stresses o Tresca Maximum resolved shear stress Yielding governed by largest of σ1 σ σ1 σ3 σ σ3 o Von Mises distortional energy for isotropic material see C. T. Sun book, p σm = ( σ1 σ) + ( σ1 σ3) + ( σ σ3) σ M is the equivalent uniaxial stress based on the Von Mises criterion Various forms of the von Mises Criterion (
3 C:\W\whit\Classes\304_01_ver_3\_Notes\6_FailureAnalysis\_staticFailure.doc p. of 4 Comparison of yield criteria Assume plane stress conditions. Plot the yield surface for the Tresca and von Mises criteria in the σ1 σstress space, where σ1and σare the inplane principal stresses.
4 C:\W\whit\Classes\304_01_ver_3\_Notes\6_FailureAnalysis\_staticFailure.doc p. 3 of 4
5 C:\W\whit\Classes\304_01_ver_3\_Notes\6_FailureAnalysis\_staticFailure.doc p. 4 of 4 What do the Tresca and von Mises criteria give you for Uniaxial extension in the x-direction? (other stresses =0) Pure shearσ xy? (other stresses = 0)
6 C:\W\whit\Classes\304_01_ver_3\_Notes\6_FailureAnalysis\b_example_yield_1.doc For general case: First step in using Tresca criterion is to determine the principal stresses.
7 3_3._A&H.mw Determine the torque Mx that will cause yielding based on the Tresca and Von Mises yield criteria. Assume the internal pressure is 100 psi. restart : currentdir ; Digits d 15 : sigmayield d.0e4 : allowableshear d sigmayield ; "C:\W\whit\Classes\304\Notes\6_FailureAnalysis" r d 10 : t d.1 : p d 100 : sigma x, x d 0: σθ, θ d p r t J d evalf $Pi$r 3 $t ; sigma x, theta d Mx$r ; J Mx Allowable torque based on Tresca criterion Mohr's circle At zero rotation coordinates in stress space (x-theta plane) p0 d sigma theta, theta, sigma x, theta ; , Mx At 90 rotation in physical space, but 180 degrees in stress space p90 d σ x, x, Kσ x, θ 0, K Mx p0 Cp90 center d evalm ; (1) () (3) (4) (5) (6)
8 radius d sqrt center 1 Kp0 1 C center Kp0 ; C Mx σ p1 d center 1 K radius; σ p d center 1 C radius; σ p3 d 0.0 : (7) K C Mx C C Mx (8) max1 d max d max3 d σ p1 Kσ p σ p1 Kσ p3 σ p Kσ p3 ; ; ; K C Mx K C C Mx C Mx (9) solve abs max1 = allowableshear, Mx ; solve abs max = allowableshear, Mx ; solve abs max3 = allowableshear, Mx ; K , , K I, I , K , K Allowable torque based on von Mises cirterion This uses the principle stresses 1 vm d sqrt $sqrt σ p1 Kσ p C σp1 Kσ p3 C σp Kσ p3 ; vm d simplify vm ; C Mx C (10) K C Mx C C C Mx 1/
9 C Mx solve vm = sigmayield, Mx ; K , (11) (1) Now work it using the formula that does not require the principle stresses xx rr thetatheta xr xtheta rtheta s d 0, 0, sigma theta, theta, 0, sigma x, theta, 0 ; 0, 0, , 0, Mx, 0 1 vm d sqrt $ sqrt s 1 Ks C s 1 Ks 3 C s Ks 3 C6$ s 4 Cs 5 Cs 6 ; solve vm = sigmayield, Mx ; C Mx , K (13) (14)
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