Analysis and Mechanics 1978
DOI: 10.1016/b978-0-08-022142-7.50032-6
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Green's Function for Thru-Crack Emanating From Fastener Holes

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Cited by 9 publications
(7 citation statements)
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“…Thus a local weighted average of the stress intensity factor can be obtained from separate nodal perturbations. Similarly, perturbations in nodal displacements associated with each crack advance can again be determined using (10). Now consider this loading, stress intensity factor distribution and set of displacement perturbations (one nodal vector 6u for each crack front perturbation Sl(s) in equation (12)) to be the 'starred' fields of equation (3).…”
Section: Three-dimensional Applicationsmentioning
confidence: 99%
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“…Thus a local weighted average of the stress intensity factor can be obtained from separate nodal perturbations. Similarly, perturbations in nodal displacements associated with each crack advance can again be determined using (10). Now consider this loading, stress intensity factor distribution and set of displacement perturbations (one nodal vector 6u for each crack front perturbation Sl(s) in equation (12)) to be the 'starred' fields of equation (3).…”
Section: Three-dimensional Applicationsmentioning
confidence: 99%
“…The stiffness would be K + SK and the nodal displacement vector would be u f Su. Assuming that SF = 0, the system of equations to be solved is then (9) Now, by using the fact that in the original solution, Ku = F, and ignoring the higher order term K~u = -SKU (10) This is extremely useful, for it shows that the perturbation in nodal displacements can be obtained from a second back-substitution through the already-factored initial stiffness matrix. The right-hand side is obtained from the initial displacements u and the perturbation of the stiffness matrix SK, which was also already calculated at the element level as a part of the determination of K I in equation (7).…”
Section: '($)+Sx'($)mentioning
confidence: 99%
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“…-m2r4 sin (26 -2 a ) -3mr' sin 2a + sin (2a + 26) C = m3rb cos 29 -m2r4 cos 49 + 3mr' cos 29 -2m2r4 -1 D = m3r6 sin 28 -m2r4 sin 46 + my2 sin 28 E = m cos 28 + m2r2 -cos (28 + 21%) -mr' cos 2a F = m sin 28 -sin (28 + 2 a ) -my2 sin 2a G = 2mr4 cos 2a -m2r4 -2r' cos (21% + 28) + 1 H = m2r4 -2mr' cos 28 + 1 (7) If the plate is subjected to uniaxial stress S, (Fig. 3a), the stress distribution along the x axis, oyl (x), can be obtained from equation (7) Fig.…”
Section: Stress Distributionsmentioning
confidence: 99%
“…Hsu and Rudd [7], and Cartwright and Rooke [8] discussed the principle, which can be illustrated schematically from Fig. 1.…”
Section: Introductionmentioning
confidence: 98%