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16 Nov 2006: 2.89 3.13 1.76 1.99 0.62 0.841 4.46 4.76 3.35 3.64 2.24 2.51 1.11 1.37. ... 2.72 22.450.5 9.62 9.72 6.39 6.53 3.15 3.34 0.11 0.140.75 12.10 12.24 8.88 9.06

20 Mar 2006: Hence substituting Equation 23. into Equation 24, one obtains. (T1T ′̄AT3TB̄T1TC̄ T3TD̄) θA. ... 23. 7 Definition of hinge angle θA. 24. 8 Deployable square frame proposed by Hedgepeth.

2 Feb 2006: 24. 35 dq. dĈ. " #C/:. ð4:3ÞTo find equilibrium configurations of the structure, we look for points where. ... K̂ Zv2Û=vq2 v2Û=vq vĈ. v2Û=vq vĈ v2Û=vĈ2. 24. 35; ð4:5Þ. is positive definite.

6 Jun 2006: PDF-1.4 %âãÏÓ 87 0 obj> endobj xref 87 57 0000000016 00000 n 0000001904 00000 n 0000001436 00000 n 0000001984 00000 n 0000002171 00000 n 0000002465 00000 n 0000002909 00000 n 0000003487 00000 n 0000004162 00000 n 0000004894 00000 n 0000005360

18 Jan 2006: 2.24 1532 0.73 36% 42.5 13.1 FD 0.28 1551 0.71 92% 52 25 FM 0.28 1386 0.91 30% 39.9 9.9. ... Varying the mean particle size between 0.28 mm and 2.24 mm had a negligible effect on the peak uplift resistance and mobilisation distance.

11 May 2006: n 16-48 64-96 112-145Elements along tendons 100 116 134Elements at equator 30 24 20Number of elements 3460 3216 3072Number of nodes 1831 1725 1671. ... Figures 24(a) and 24(b) show this type ofplot respectively for nCA118-10-290 and nCR118-10-290 balloons

16 Nov 2006: At the end of this step of calculation the equivalent tunnel diameterfrom equation (9) will be 4.24 m. ... The energy calculation is repeatedin the further steps as requested, with VL 2% and D 4.24 m, etc.

17 Feb 2006: A T300/LTM45 plain weave CFRP, 3-ply laminate (0/45/0) with uniform thickness wasadopted.Rmin was set at 24 mm, which provides a margin of 2.3 on material

12 Oct 2006: 24. 24 2. 4. 4. Fig. 6. Idealisation of pile in load-transfer analysis (Randolph,2003a).

23 Feb 2006: 12(1 ν2)π 4 A2. 4λ4. (24). The stretching strain energy per unit area can be obtained, neglecting stretchingin the η-direction, from. ... U =(1 ν)E t γ. 2. (π 2t 2. 24(1 ν2)λ2+.

12 Oct 2006: The undrainedstrength is given by: cu = co c1z = 0.21 8.19z 24.5 kPa, wherecoundrained shear strength at the ground surface, c1rate ofvariation of undrained shear strength with depth, and ... Geotech.Geoenviron. Eng., 1281, 13–24. Ikuta, Y., Maruoka, M.

15 Aug 2006: 24 and 25. Fig. 12. Volume strain with moderate cyclic shear loading of drysand or silt. ... 24 and 25. This is called “cyclic mobility” Castro 1975. ndrained monotonic shear.

18 Jan 2006: Figure 24 shows different representativeshapes of stress–strain curve for samples extracted from thesame in situ conditions with K0 = 1.0. ... 24. Different shapes of stress–strain curve (K0 = 1.0). Curve A n1 m1 B n2 m2 b2.