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[求助]基坑支护flac 3的模拟 [复制链接]

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只看楼主 倒序阅读 使用道具 楼主  发表于: 2013-09-28
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new X2xuwA  
gen zone brick p0 0 0 0 p1 80 0 0 p2 0 80 0 p3 0 0 30 size 40 40 15 Y&^P"Dw  
model  mohr Z xb_K  
prop bulk 9.2e10 shear 4.7e10 fric 50 $L2%u8}8:  
prop coh 1e10 tens 1e10 CY& hIh~S@  
set grav 0 0 -10 q;:6_Qr  
ini dens 2000 ;kJu$U  
fix x y range x -0.1 0.1 toaYsiIkzW  
fix x y range x  79.5 80.5 .y&QqxiE  
fix x y range y -0.1 0.1 n 8pt\i0  
fix x y range y  79.5 80.5 @y/wEBb  
fix x y z range z -0.1 0.1 ,\aUq|~  
rO:u6."_  
hist id=1 unbal \Fe_rh  
solve 0K-*WQ*#9  
save chushiyingli.save c0Dmq)HK?  
3~qR  
prop bulk 2.06e7 shear 1.01e7 fric 8 coh 15e3 A~dQ\M  
*E~VKx1  
set large >z`,ch6~  
ini xdis 0 ydis 0 zdis 0 AC.A'|"]i  
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&)/H?S;yN  
%K/G+  
S(YHwH":  
sel pile prop density=2500 V9);kD  
{&Gk.ODI7  
sel pile prop Emod=8.0e10 Nu=0.25 XCArea=0.1963  & v f/$`IJ  
              XCJ=6.13e-3 XCIy=3.07e-3  XCIz=3.07e-3  & } r\SP3  
              Per=1.57 & 8EVF<@{]  
              CS_sK=2e8 CS_sCoh=0    CS_sFric=0.0 & Uzm[e%/`  
              CS_nK=1.3e20 CS_nCoh=0 CS_nFric= 15.0 CS_nGap=off '44nk(hM69  
lMI ix0sSj  
model mech null range x 20 60  y 20 60 z 29 30 q#':aXcv"  
b{ubp  
K.sj"#D  
sel beam id 4 begin=(20,20,30)  end=(60,20,30)  nseg=20 Iu)(Huv  
sel beam id 4 begin=(60,20,30)  end=(60,60,30)  nseg=20 5i wikC=y  
sel beam id 4 begin=(20,60,30)  end=(60,60,30)  nseg=20 fQ?n(  
sel beam id 4 begin=(20,20,30)  end=(20,60,30)  nseg=20 a5Acqa  
vY TPZ@RL  
sel beam prop emod=2.0e11 nu=0.25 'AlSq:gZ  
sel beam prop XCArea=0.6 XCIz=0.0256 XCIy=0.0144 XCJ=0.03 9_CA5?y$:  
;zxlwdfcr'  
#uDBF  
SEL JOIN beam ID  4 pile ID 3 RANG Z 29.9 30.1 5IB:4zx^h  
SEL JOIN beam ID 4 beam ID 4 RANG Z 29.9 30.1 PY3Vu]zD  
Wcay'#K,  
hist id=2 unbal |SXMu_w  
step 5000 >t6'8g"T  
save gongkuang1.save ~QQ23k&  
vN6]6nUOiT  
S0o,)`ZB  
model mech null range x 20 60  y 20 60 z 22 29 ![."xHVeL  
PezWc18  
hist id=3 unbal 9@&Z`b_  
step 5000 |@n{tog+-  
save gongkuang2.save gQcr'[[a  
j~0ZE -e  
4) I/\  
model mech null range x 20 60  y 20 60 z 18 29 Y.hH fSp  
1Mhc1MU  
hist id=4 unbal NByN}e  
step 5000 y<LwrrJ>  
save gongkuang3.save HjY-b*B  
基坑尺寸40m×40m×12m,采用φ500的悬臂桩支护桩,请问一下各位老师为什么我模拟出的支护桩和桩后土向坑外移动?请老师们帮忙看看是不是桩的参数和冠梁参数取错了。 -gX2{dW  
这是基坑y向位移的一张云图。请各位老师帮我找下原因。
1条评分土币+3
沈建韩旭 土币 +3 - 2013-11-13
离线kunshan_zqs

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只看该作者 1楼 发表于: 2013-09-28
建议你仔细看看程序,你垂直方向取了30米,划分了15个单元格,每个单元格是2米,挖土阶段怎么会出现29米?
1条评分土币+3
须胡白 土币 +3 向您学习,牛人 2013-09-29
离线须胡白

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只看该作者 2楼 发表于: 2013-09-28
回 kunshan_zqs 的帖子
kunshan_zqs:建议你仔细看看程序,你垂直方向取了30米,划分了15个单元格,每个单元格是2米,挖土阶段怎么会出现29米? (2013-09-28 22:54)  qr;" K?NX  
Y }d>%i+  
首先谢谢您的解答。29m是施工冠梁的第一个工况!
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只看该作者 3楼 发表于: 2013-09-30
如果对半个单元的材料参数取为零,程序上是否允许,或者会出现什么状况?你是否可以考虑把上面28至30之间划为两层单元
离线须胡白

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只看该作者 4楼 发表于: 2013-09-30
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kunshan_zqs:如果对半个单元的材料参数取为零,程序上是否允许,或者会出现什么状况?你是否可以考虑把上面28至30之间划为两层单元 (2013-09-30 14:59)  IFe[3mB5  
>FO=ioNY  
我找到原因了,是桩的侧向刚度和粘结强度取值太大了,我调整了一下过去了。你的将28至30划分为两个单元,也是有理的,我试过如果将桩设置在网格中间,比如说20.5的位置处,就不能计算了,不知道这是什么原因……

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只看该作者 5楼 发表于: 2014-02-22
修改后的模拟效果可否给大家看一下呢
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