By G. Fowles, G. Cassiday
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Additional info for Analytical Mechanics [SOLUTIONS MANUAL]
A function that does this is: ςφ Ι(z) = 1 2 a / z2 . 19) In the following the boundary conditions are checked: 1) If y = 0 it follows that z = x and thus: ςφ Ι(z) = Ι(x) = a 2 1♣ ♥ /x 2 = i· 1 •≠ ςφ 2 a / x2 . e. Re Ι(z) = 0. Consequently: (z) = 0 . e. 19) becomes Ι(z) = ς φ. Because now Im Ι(z) = 0, it follows that: ςx = ςy = Re Ι(z) = ς φ . 3) For z = ±a we obtain Ι(z) ο φ. So at the crack tips (y = 0, thus z = x) we get: ςy = Re Ι(z) ο φ. The Mode I Stress Intensity Factor It will prove convenient to translate the origin of the co-ordinate system to the crack tip at z = +a by introducing the variable Κ= z a.
16. Then, by performing a fatigue crack propagation test with a standard specimen over a certain crack length range, da/dn- K data must be generated. Using this, da/dn can simply be eliminated leading to a K-a relation. KI for the unknown configuration can now be determined knowing both the crack length and the load. The Weight Function Method From expressions for the energy release rate, G, and the relation G = K2/E (see chapter 4) it can be shown that the stress intensity factor for a particular stress distribution can be calculated if the stress intensity factor is known for another stress distribution on the same configuration of specimen and crack geometries.
2. The Elastic Stress Field Approach 35 2) The case of a plate uniaxially loaded in the y direction cannot be solved using the complex Westergaard stress function, since then it is required that ςx = ςy for y = 0, cf. 18). a), see reference 3. 19, the boundary conditions for the uniaxial case are satisfied. Thus the stress field in a uniaxially loaded plate is identical to that in a biaxially loaded plate with the exception of ςx, which is reduced by the remote stress ς. 24), this correction is usually omitted because near the crack tip ςx is much larger than ς.