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楼主 |
发表于 2010-3-8 15:12:32
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来自 江苏徐州
谢谢了!
那有的后面没有写时,def……end是怎么运行的?
如:
;-----------------------------------------------------------
; 剑桥模型各向同性压缩实验例题
;-----------------------------------------------------------
new
gen zone brick p0 0 0 0 p1 1 0 0 p2 0 1 0 p3 0 0 1 size 1 1 1
title
Isotropic compression test for normally consolidated soil
; --- 材料参数 ---
model cam-clay
prop shear 250. bulk_bound 10000.
prop mm 1.02 lambda 0.2 kappa 0.05
prop mpc 5. mp1 1. mv_l 3.32
; --- 边界及初始条件 ---
fix x y z
ini sxx -5. syy -5. szz -5.
ini xvel -0.5e-4 range x 0.9 1.1
ini yvel -0.5e-4 range y 0.9 1.1
ini zvel -0.5e-4 range z 0.9 1.1
; --- 定义fish 函数 --- (numerical values for p, q, v)
def camclay_ini_p
p_z = zone_head
loop while p_z # null
mean_p = -(z_sxx(p_z) + z_syy(p_z) + z_szz(p_z))/3.0 - z_pp(p_z)
z_prop(p_z,'cam_cp') = mean_p
p_z = z_next(p_z)
endloop
end
camclay_ini_p ;这一块可以理解!!
def path
p_z = zone_head
sp = z_prop(p_z,'cam_cp')
sq = z_prop(p_z,'cq')
sqcr= sp*z_prop(p_z,'mm')
if sp = 0.0 then
sp = 1.
endif
lnp = ln(sp)
svol = z_prop(p_z,'cv')
mk = z_prop(p_z,'bulk')
mg = z_prop(p_z,'shear')
cpc = z_prop(p_z,'mpc')
end
;这里没有path,path这一块通过什么命令运行的?我的理解:是通过后面的 his path 采样记录,再通过trip段,得以运行。对吗?多谢你的指点……
; ... 装载-卸载练习 ...
def trip
loop i (1,5)
command
ini xvel -0.5e-4 range x 0.9 1.1
ini yvel -0.5e-4 range y 0.9 1.1
ini zvel -0.5e-4 range z 0.9 1.1
step 300
ini xvel mul -.1 yvel mul -.1 zvel mul -.1
step 1000
ini xvel mul -1. yvel mul -1. zvel mul -1.
step 1000
end_command
end_loop
end
; --- 采样记录 ---
his nstep 20
his unbal
his path
his sp
his lnp
his sq
his svol
his mk
his mg
his gp zdisp 0 0 1
; --- 试验 ---
trip ;
;结果
plot his 3 vs -9
plothis6 vs 4
plothis7 8 vs -9
save aaaaa |
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