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Curved axis with initial plastic buckling load of compression members

Parts of the modulus. Southern equilibrium condition M, available N ===. X from (7), Shanley model based on available incomplete axial compression members of the critical force is 1) When the elastic buckling, E. A E. = E the critical load for N one; iv 2) When the plastic buckling, and the use of theoretical calculations of tangent modulus, E a E = E,, the critical load for N, one. 3) the amount of theoretical calculations using dual-mode, E a E, E a E, the critical load for N ===. Susan Chan 3N_ derived as a function of bending with the initial, post production at the beginning of zero bending load, in order to solve the ultimate load, assuming load to the tangent modulus load, the axial force N continue to increase, and that / z,[link widoczny dla zalogowanych], can be E. = E, and E. Depends on the convex side of the model is changing the strain, the use of notation: r first-class. (11) generation (7), have N one. Skill (4) generation (12), we get N-L4 ? e +(].( 13) due to bending deformation began in the zero load, eliminating e, so N A △ N-N. A N: one by one life 2 [z], L a. ... 'J' that N first-class [z]. ... United (13) and (14), we obtain a (a) / [(r_1) _ (r +1)]. ) to (15) generation (13), was No. 6 Wuxiang Guo, et al: a first curved plastic buckling axial load of 823 members * ~ a (2) / rhE a _El an E- -E,] 'with the dividing numerator and denominator on the type E, too [2 )/(+),/]. (16) l-El '~ where: EE, / E. (16) Shanley model is based on incomplete performance of axial buckling formula for compression members, but the actual components, E, is not constant, but decrease with the increase of compressive strain, the following function by solving the E-answers modified (16). 4E as a function of the plastic zone solution, a-E a (.) (17) for structural steel, when> the time, the tangent modulus for the E-or (or one a) E / Ecr ()]. (1 Cool

to q235 steel, for example, about 190N/ram, about 235N/mm. , Elastic modulus E a (200 ~ 206) × 10. N / mm, combined type (17), (1 Cool

, eliminate the availability of z (ay a) l / hE a 2 (+) (E +2 E,[link widoczny dla zalogowanych],) \. 4D \ (E ~ E) A 2 (+) (E + E,), / 4E, / hE a 2 (。+)( E +2 E), \ a D I_ II = dried noodles J,[link widoczny dla zalogowanych], 1. Divided on both sides of the same, have two!]: F the barrier! ± ± Mo 1.4 \ h (1 a E) a 2 (+) (1 + E) / / 4E / h-2 (+) ( 1 +2 E),, \ El \ h (1 a E) a 2 (+) (1 + E ),,'( 19) can be obtained from the above equation on the equation E: UE + UE ~ U2E +1- 0. (20) where fUl a Z (1). A Q (W1) l a zW2 +2 Q + +4 () l-z +8 ~ Q () 【eleven 4E () (21) where: Q a 'wl eleven 2, + w 2 a h ten ((,+), W. {a 4 (+). Finally, we can obtain E, a function: E-E, (0, h, l, E,, p,). (22) 5 Susan Chan function solution equation (22) into (16), by N-N (A, h, _】 _, l, E, 13 \Trace;. as the initial deflection amplitude; ∈ (O ,[link widoczny dla zalogowanych],+。。) is the displacement; l is the length of bar; E is Young's modulus; to the proportional limit; as the yield limit. by numerical calculation can be drawn a deformation of the load curve and solve the extreme value. This is the extreme bending of the axis with initial plastic buckling of compression members and the numerical buckling displacement of .6 of a platform structure bearing,[link widoczny dla zalogowanych], diameter 600mm, outer diameter of 1000mm, model parameters h = 1000mm, assume the existence of the amplitude of the initial bend. = 10mm, E a 2 × 10N/ram, proportional limit of a 190N/ram., yield limit of 13 \with the initial bending of the plastic buckling loads of axial compression members and the corresponding buckling displacement. MATIAB the use of sophisticated programming. were taken initial bending amplitude of a 5,1 O, 15film calculation, rendering the results are shown in Figure load deformation 2 shows, the critical buckling load and deformation as shown in Table 1. fur \ saddle blanket column Z0, saddle blanket column (a) = 5mm (b) 1 () mm

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