论坛首页    职业区    学术与写作    工程技术区    软件区    资料区    商务合作区    社区办公室
 您好! 欢迎 登录注册 最新帖子 邀请注册 活动聚焦 统计排行 社区服务 帮助
 
  • 帖子
  • 日志
  • 用户
  • 版块
  • 群组
帖子
  • 5031阅读
  • 16回复

[原创]讨论:沿用几十年的强度折减法正确吗? [复制链接]

上一主题 下一主题
离线shenyl0211
 

发帖
410
土币
3157
威望
549
原创币
0
只看楼主 倒序阅读 使用道具 楼主  发表于: 2009-09-04
        一个W0.2kN的正方形铁块,沿表面倾斜α30°的铅块滑动,如果视滑带为无穷小,两者接触面绝对光滑、无粘性(c0kPaφ),且铁块下部受到弹簧秤的足够支撑,则不计其它因素时,该铁块的稳定系数K为多少?请列出计算公式(与cφ有关的抗滑力用R表示、其它抗滑力用P表示、下滑力用T表示)。 )v!>U<eprD  
        首先,我们要承认铁块确实处于真正的极限平衡状态(K=1); W/O&(t  
        其次,我们也要承认它的稳定系数是可以计算的; qMAH~P0u  
        第三,我们可以得出R0TWsinα0.2×sin30°0.1kN,弹簧秤给铁块的PT0.1kN anj*a<C<  
    本人的解是铁道部、建设部、国土资源部规范的传递系数(超载)法:K=抗滑力/下滑力=(R+P)/T=(0+0.1)/0.1=1,符合物理现象,且与摩尔-库仑准则无关。 e[t+pnRh  
    其它边坡稳定性计算方法(强度折减法)基于摩尔-库仑准则,结果是:KR/(TP)0/(0.10.1)0/0=无解,不符合物理现象。 b 469  
    强度折减法有一个荒唐的、可悲的、致命的错误:没有把抗滑桩、挡墙等具有抗滑作用的建筑物看作是一种可以被滑体剪切破坏的高强度材料。也就是说,谁都知道这些抗滑建筑物也有粘聚力和内摩擦角,所以应该将这些抗滑建筑物看作是滑体的一部分,但实际上强度折减法却没有这样做。 lNSB "S  
    从力的性质来说,弹簧秤给铁块的力也应是抗滑力,但强度折减法对此不予承认。根据这个逻辑可以得出:如果c、φ不等于0,且上述的R=0.05kN,则P=0.05kN,此时K=1,但P的物理意义是把铁块切掉一半(下滑力减少0.05),但这样R<0.05kN,所以又错了 1mw<$'pm0  
    当多部规范采用传递系数(超载)法时,那么多岩土人却没有因此从错误的观点中醒悟过来,真是可怜。  '-F }(9M  
       
[ 此帖被shenyl0211在2009-09-05 09:41重新编辑 ]
离线ymcheng

发帖
1165
土币
1
威望
5358
原创币
0
只看该作者 1楼 发表于: 2009-09-04
Your analysis is failure controlled by the discontinuity. Strength reduction analysis is the failure within soil mass controlled by strength. We have mixed up the two concept.
离线shenyl0211

发帖
410
土币
3157
威望
549
原创币
0
只看该作者 2楼 发表于: 2009-09-05
引用第1楼ymcheng于2009-09-04 23:33发表的 : P)~PrTa%  
Your analysis is failure controlled by the discontinuity. Strength reduction analysis is the failure within soil mass controlled by strength. We have mixed up the two concept. iulM8"P  
{\G4YQ  
/ve8);cH\  
郑教授:请你不要混淆滑带和滑体的区别。强度指标为0的滑带也是滑带,坚硬的铁块也完全可以是滑体。
离线ymcheng

发帖
1165
土币
1
威望
5358
原创币
0
只看该作者 3楼 发表于: 2009-09-05
K=抗滑力/下滑力=(R+P)/T ]* -9zo0  
K=R/(T-P) 8}m] XO  
F(/^??<5  
Both definitions are used for slope stability analysis based on Mohr Coulomb. If you use program SLIDE, you are allowed to choose which definition you prefer. It is not the problem of strenght reduction, but definition of factor of safety. a  1bu  
W&y%fd\&3  
In case there are external loads, these two definitions give different factors of safety. This problem has been discussed in Geotechique and Journal of Institution of Structural Engineers more than 10 years ago. It is well known to many people, and may be you should consult these references. Such difference in the detinition also apply to analysis and design of sheet pile structures. zw^jIg$  
li\hHd5  
This is not something new. Just look up the old papers and journals. I think one of the paper is by Burland (not very sure). The other one, I can't remember now. u2'xM0nQ  
 5I5~GH  
Your view is not new, and it is clear to many engineers outside China. Look up the active and passive option in SLIDE.
离线ymcheng

发帖
1165
土币
1
威望
5358
原创币
0
只看该作者 4楼 发表于: 2009-09-05
There different ways to define factor of safety. For classical strength reduction method, only the shear strenght is used. For your case, shear strenght is 0 and base shear is also zero, so 0/0 is correct for strength reduction, and it is indeterminate based on this definition. It is not wrong, as there is no shear strength and base shear. ciI;U/V  
2#5SI  
For your second definition K=1.0, it is a factor of safety based o the system, not just the shear strength. :*TfGV  
zLue j'  
So both factors of safety are correct, but they are different factors of safety. You have assumed them to have the same meaning ? vWI9ocl`W  
ws4a(1  
Both factors of safety are also used, depending on your view. There are even engineers who compute both factors of safety and choose the smaller value. This issue has been well discussed in the last 20 years, so no one disucss it now. May be you can refer to these old references to see the arguments (I think 1 report by Institution of Structural Engineers 20-25 years ago, the other reference is in Geotechnique, also more than 10 years ago) gPY2Bnw;l  
j&) +qTV  
You are correct that very few engineers in China notice about two different definitions of safety. Also look up the user guide of program SLIDE, it is also documented there. May be you should alert the engineers about the existence of at least two definitions (there are even more definitions). However, strength reduction method cannot be taken as wrong. Without external load, it is the same as the other definition. With external load, engineers should pick up his own definition FOS to shear strength or FOS to system. These two definitions are different.
离线shenyl0211

发帖
410
土币
3157
威望
549
原创币
0
只看该作者 5楼 发表于: 2009-09-05
回 3楼、4楼(ymcheng) 的帖子
        1、我从来没有说过我的观点是新的,你也不用多次指出。传递系数法早在1954年就已提出,之后被列入铁路规范,继而建设部、国土资源部相继采纳,交通部在推力计算时也放弃原来的强度折减法,说明传递系数法是正确的,我现在只不过是重申传递系数法的观点,而不象你的软件slope2000及我国的不少院士、教授、博士后、博士、硕士等有丰富知识的人在各种刊物、讲座歪曲传递系数法的正确原理。 `TKe+oS)  
        2、不论从什么角度出发考虑问题,铁块的稳定系数只有1个固定值,不可能无解(无穷多个值)。这是一个很简单的物理现象和物理问题。 $d?<(n  
        3、不管是你们对摩尔-库仑准则的误解,还是摩尔-库仑准则本身的问题或不完善,都说明强度折减法是错误的——在公式推导过程中数学上是成立的,但无法解释T-P的物理意义是什么?当采用残余强度的时候强度还能再折减吗(如c的最小值是20kPa,如取安全系数为1.25,则c能取16kPa吗)?显然不能,则此时得出的显然不是基于摩尔-库仑准则的稳定系数。所以强度折减法根本经不起任何推敲。 Ra[{K@  
        4、传递系数法已经用了50多年了,即使这样还有人胡乱给其取名。该法在铁道部用了几十年,才被建设部、国土资源部、交通部逐渐接受。可见,国人的脑子不够用,容易受权威人士特别是外国人的影响,而出现那么多稳定系数的定义更说明了只看数学表达方式、不顾物理意义的人比比皆是。 L~SM#?z:ue  
        5、从我国院士们对传递系数法的几十年误解(甚至至今也没有明白)可以证明,我们不应该轻易相信任何权威,我们应该用自己的头脑解决有争论的问题。所以,你不要跟我说权威人士的观点。对传递系数法的认识上,越是权威的可能越是错误的,这是中国的国情——迷信造成的。 jR\&2;T  
        6、当然,当稳定系数趋于1、安全系数接近于1、R与T相近、滑体不是很大、地震力不控制时,折减法和传递系数法的稳定系数和推力是相近的,对工程不会有多大影响,反之则不然。 Q&^ti)vB  
   
[ 此帖被shenyl0211在2009-09-05 22:54重新编辑 ]
离线ymcheng

发帖
1165
土币
1
威望
5358
原创币
0
只看该作者 6楼 发表于: 2009-09-06
摩尔-库仑准则的误解,还是摩尔-库仑准则本身的问题或不完善, this is not a problem of Mohr Coulomb. You can use any yield criterion, and the argument is still there. G/\t<>O8o  
6",1JH,;p  
FOS depends on definition. 铁块的稳定系数只有1个固定值 is your own interpretation. Different definition has different FOS. 5Y)*-JY1g  
g5lf- }?  
For bridge abutment design in France, they use another definition for the bridge engineers. Not the c, phi or the above two definitions, but the ultimate load on abutment against applied pressure. This is the third definition. They also use this for 100 years. 3D[IZ^%VtM  
{ :'#Ts<  
I think you cannot accept that different definition has different FOS. ^k]OQc7q'  
8-UlbO6  
K=抗滑力/下滑力 - Why the definition cannot be defined as Ultimate moment/mobilized moment instead of force ? This definition is also used in the world. You insist on only 1 definition but fail to appreciate that different people look at different issue, so different FOS exist in the world. ^aN;M\  
:84fd\It4  
A very simple argument is FOS can be defined by force or moment K=抗滑力/下滑力  or K=抗滑moment /下滑moment , so there are at least two. There are more. Your definition K=抗滑力/下滑力 is correct, but other definitions are also correct, depends on what you are looking at. Just be open to accept that there are other definitions. w_q{C>- cR  
DzYi> E:*  
There are sufficient argument, let the readers to think, and it ends.
离线shenyl0211

发帖
410
土币
3157
威望
549
原创币
0
只看该作者 7楼 发表于: 2009-09-07
回 6楼(ymcheng) 的帖子
1、“K=抗滑力矩/下滑力矩”是错误的,所以瑞典法被淘汰了。 ?<bByxa  
2、上述公式不是来自稳定系数定义,而是来自力矩平衡条件和摩尔-库仑准则。根据表现形式下定义是错误的。传递系数法、简化简布法、罗厄法、摩根斯顿-普莱斯法等的稳定系数定义又应怎样定义呢?当你给别人递钱的时候(这是个现象),原因应该有很多种吧? h7f&7v  
3、毕肖普法的稳定性计算公式也能理解为“K=抗滑力矩/下滑力矩”吗?貌似正确,其实不然。 s"?Z jV)`  
4、滑坡、边坡稳定性分析是个物理问题,应该尊重客观实际,一旦上升到错误的纯数学问题后,结果往往就不准确了。
离线baijinjie

发帖
382
土币
2653
威望
2018
原创币
0
只看该作者 8楼 发表于: 2009-09-08
有点高深了 俺不敢接话了 只能学习了
离线koostreet

发帖
332
土币
113
威望
2140
原创币
0
只看该作者 9楼 发表于: 2009-10-22
回 5楼(shenyl0211) 的帖子
” K=抗滑力/下滑力“不知楼主抗滑计算是怎么计算的?不也用摩尔库伦准则了吗?不平衡推力法计算和其他几个条分法一样进行这样那样的假设,应该有限使用。基于前述安全系数定义时,两种方法是一样的。个人愚见,仅供参考!
离线lw4051581

发帖
133
土币
0
威望
1272
原创币
0
只看该作者 10楼 发表于: 2009-10-23
              
离线xjywgy

发帖
6850
土币
856
威望
67214
原创币
0
只看该作者 11楼 发表于: 2010-02-20
谢谢楼主,顶
鲜花(xjyxr) 谢条评分
快速回复
限100 字节
温馨提示:欢迎交流讨论,请勿纯表情、纯引用!
 
上一个 下一个

      https://beian.mps.gov.cn/ 粤公网安备 44010602012919号 广州半山岩土网络科技有限公司 粤ICP备2024274469号

      工业和信息化部备案管理系统网站