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关于国外规范中混凝土结构最小配筋率如何确定的?

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麻酱面 发表于 2009-3-6 16:52:07 | 显示全部楼层 |阅读模式
最近老师让我写一篇关于各国最小配筋率之间的比较。额,美国规范倒还好,能找到最小配筋率公式的出处。可是欧洲规范和英国规范就不知上哪查资料了。。。
欧洲规范上也就只短短几行说明了下最小配筋的取值,可是这咋来的呢???
国内发表的文章很少有提及关于欧洲规范的,好不容易找到一篇,也就提到是根据经验确定的,可这怎么个经验法也没说。国外的文章又不知道应该从何下手....

哎,被折磨的相当郁闷那~~~
我想知道,大家有没有对此有了解的,或者关于欧洲规范,有没有相关的参考书目,像是条文说明什么的(偶不才,实在在网上寻觅不到)。同样存在疑问的还有英国规范BS8110-1
哪位大牛能指条明路,谢过谢过

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wangjianping 发表于 2009-11-11 22:36:32 | 显示全部楼层
Let us first define the criteria for the minimum reinforcement. The minimum reinforcement is defined
as a bending reinforcement concentrated at the bottom of the beam which must assure: 1. Stable beam
response after Mfi is reached (Mfi corresponds to the bending moment at which the tensile strength of
concrete at the beam bottom is reached) and 2. the ultimate bending moment Mu must be
approximately equal or larger than Mfi. In addition to the above criteria, the minimum reinforcement
ratio should not be to high since otherwise the beam may fail in diagonal shear or concrete
compression rather than in bending.
Let us first consider the minimum reinforcement ratio for RC beams without stirrups and with any
distributed reinforcement, which is the most extreme case occurring in practice. As aforementioned,
the minimum reinforcement requirement has recently been extensively investigated by Carpinteri and
co-workers. According to their work the minimum reinforcement ratio should decrease when the beam
depth increases. However, from the theory [11,12] and experiments [2] we know that large plain
concrete beams generally exhibit more brittle (explosive) behavior after cracks in concrete occur than
the small beams. Comparing the response of a small plain concrete beam with an extremely large one,
it can be observed that the small beam exhibits a relatively ductile response, i.e. even without any
reinforcement the beam response is relatively stable. Consequently, for small beams the energy
equilibrium criterion is not relevant and therefore for the minimum reinforcement requirement we
propose the stress equilibrium criterion in the form:
M,=M^ + M„„ > M,, (3)
were Mcon stays for the contribution of concrete to the ultimate resistance. Note that there is an
important difference between Eq. (1) and Eq. (2) proposed by Carpinteri and Eq. (3). According to (1)
My > Men which leads to the increase of the minimum reinforcement when the beam size increases.
Since the minimum reinforcement should prevent brittle failure, there is no much sense to increase it if
the structural response anyhow tends to be more ductile by decreasing the beam size. This is the main
reason why we propose a stress equilibrium criterion different than that proposed by Eq. (1).
From the stress equilibrium point of view, Eq. (3), the most critical case occurs when the contribution
of concrete to the ultimate resistance is small, i.e. Mcon = 0 (for instance the concrete fracture energy
GF -^ 0, as for example in pre-cracked concrete or in very large beams). Therefore, from Eq. (3) it
follows:
M^>M^^ => P m i n ^ ^ (4)
WQTQfy = yield limit of the reinforcement, z = lever arm between tensile and compressive forces of the
cross-section. The above criterion is independent of the beam size. It is the lowest practical minimum
reinforcement ratio because the energy criterion becomes relevant when a brittle cracking occurs, i.e.
for larger beams. Theoretically, forh=>0 one needs no minimum reinforcement since the concrete is
for such a case perfectly elasto-plastic already without any reinforcement (for h ^> 0, (dU/ da) / GF ^>
0\ with U = total accumulated energy, a = crack length).
In contrast to small beams, a large plain concrete beam exhibits an elasto-brittle response and fails
immediately after the peak load (tensile strength of concrete) is reached. Therefore, intuitively, one

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wangjianping 发表于 2009-11-11 22:42:38 | 显示全部楼层
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lichengcaili 发表于 2010-3-12 10:18:53 | 显示全部楼层
那外高手总结一下
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wind715 发表于 2010-4-11 16:40:23 | 显示全部楼层
有很多相关的资料
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eastinent 发表于 2010-7-1 22:18:26 | 显示全部楼层
为了保持结构的延性,混凝土梁的开裂弯矩至少要不大于受拉钢筋屈服时所对应的弯矩,否则很容易造成突然的破坏。请回忆一下,我们在计算一个梁截面的弯矩--曲率关系的时候,通常取三个点,分别是开裂弯矩,屈服弯矩 (受压混凝土还处于弹性状态),以及最后的极限弯矩(受压混凝土已经处于塑性状态),同时求出这个状态的曲率。
在W.H.Mosley的教科书(3rd edition)中 (P159),针对英国规范,提到As/(bh) should not be less than 0.24% for mild steel and 0.13% for high-yield steel.我自己没有计算过,你可以试着算下。

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scloss84 发表于 2010-7-27 09:16:05 | 显示全部楼层
In Australian Standard, there are two requirements for minimum reinforcement.
1)        Minimum strength – this is when the beam suddenly crack and loses its tensile capacity, the steel must have an equal amount of tensile capacity. This can be calculated using concrete tensile strength and cross sectional area.
2)        Crack control – explained below, but even the supplementary material did not provide calculations.

In Australian Standard AS3600 Supp1 – 1994 C9.4.3.4. Reinforcement in the secondary direction in restrained slabs

In exposure classifications A1 and A2, the amount of reinforcement required is defined in terms of the degree of crack control required. For most applications a minor degree of crack control is adequate for durability. However, other considerations may dictate a greater degree of crack control.

The degree of control should be interpreted as follows:
a)        minor control – Where cracks are aesthetically inconsequential or hidden from view
b)        moderate control – Where cracks will be seen but can be tolerated
c)        strong control – Where the appearance of obvious cracks is unacceptable, or where cracks may reflect through finishes.

Hope this helps^^

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woshishuishiwo 发表于 2010-10-3 18:04:22 | 显示全部楼层
本帖最后由 woshishuishiwo 于 2010-10-3 18:05 编辑

这个问题有人写过文章了,呵呵,

01钢筋混凝土构件纵向钢筋最小配筋率的功能与取值.pdf

194.69 KB, 下载次数: 89, 下载积分: 金币 -2 金币

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主编徐有邻还是很牛的!  详情 回复 发表于 2017-1-13 15:16

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woshishuishiwo 发表于 2010-11-14 02:02:53 | 显示全部楼层
这么好的文章,居然没人下载,唉!
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xxy1216 发表于 2010-11-14 04:31:43 | 显示全部楼层
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ljmtidilgw 发表于 2010-11-14 11:08:27 | 显示全部楼层
woshishuishiwo 发表于 2010-11-14 02:02
这么好的文章,居然没人下载,唉!

是有段时间没有统计下载次数

现在好了
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ljmtidilgw 发表于 2011-1-5 21:39:56 | 显示全部楼层
看来不是简单的一个数,涉及结构本身、荷载级别等因素。
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paul1130 发表于 2011-1-14 17:11:23 | 显示全部楼层
不错的东西,正好可以借鉴用到
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ljmtidilgw 发表于 2011-4-27 12:07:58 | 显示全部楼层
Page 338, Practical Foundation Engineering - R W Brown - Mgh

1.JPG
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kengawk 发表于 2012-3-18 11:50:14 | 显示全部楼层
woshishuishiwo 发表于 2010-10-3 18:04
这个问题有人写过文章了,呵呵,

这个吧?
                                                                                          《建筑结构》                                                                        2003年08期                                                                       
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          钢筋混凝土构件纵向钢筋最小配筋率的功能与取值——《混凝土结构设计规范》(GB50010—2002)修订背景介绍        

        
        
          【摘要】:介绍了此次《混凝土结构设计规范》修订中纵向钢筋最小配筋率有关条文的修订背景。其中着重说明了构件中受拉和受压纵向钢筋最小配筋率在非抗震结构及抗震结构中所起的作用及其取值思路 ;在对国外规范相应规定进行对比分析的基础上 ,对我国修订后规范最小配筋率取值水平作了判断 ;并对一些遗留问题进行了讨论 ,给出了相应的建议

        
          【作者单位】:                                                                重庆大学土木工程学院  中国建筑科学研究院结构研究所  重庆大学土木工程学院  
【关键词】:                                                        钢筋混凝土  配筋构造  纵向钢筋最小配筋率  
【分类号】:TU375
【正文快照】:         
我国混凝土结构设计规范对各类构件中受拉和受压纵向钢筋最小配筋率的规定最早引自原苏联规范 ,取值偏低。规范GBJ1 0— 89(以下称原规范 )在此基础上只稍许提高了受拉纵向钢筋的最小配筋率。此次修订规范 ,从受力机理角度进一步探讨了纵向钢筋最小配筋率的功能 ,对比分析了

        



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ljmtidilgw  在2012-11-3 01:36  送朵鲜花  并说:我非常同意你的观点,送朵鲜花鼓励一下
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n900 发表于 2012-3-18 11:56:01 | 显示全部楼层
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CHJ1997-2008 发表于 2012-3-18 14:27:34 | 显示全部楼层
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eagleto 发表于 2012-8-11 04:39:13 | 显示全部楼层
英国规范本来就是欧洲规范的雏形,只不过现在的欧洲规范在某些方面改进了。不像美国规范那样,讲解公式等是如何来的,欧洲规范侧重使用方面,不少东西没有讲解的。
所以BS8110更侧重于经验公式等。而En1992等加入了一些结构改进的方法。基本原理各国都差不多,只不过针对当地的具体情况有些小差异而已。

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ypac 发表于 2012-10-26 15:37:12 | 显示全部楼层
好文章下了学习一下!
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