混凝土外文參考資料及譯文

1、畢 業(yè) 設(shè) 計(jì)（論 文）外 文 參 考 資 料 及 譯 文譯文題目：Tensile Strength of Members 學(xué)生姓名：學(xué)號(hào)：專業(yè)：所在學(xué)院：指導(dǎo)教師：職稱：2013 年1月6 日外文參考資料：Tensile Strength of Members1.INTRODUCTIONWhen the section of a member is under the action of an axial tension force N,the section is said to be an axial tension member section. Just as the column

2、section,when there is a moment M acting simultaneously with the axial tensile force,the section is an eccentric tension member with eccentricity e0 =M/N on the tensile side of the member.Though tensile resistance is a shortcoming of concrete, there are still many members with tension used in enginee

3、ring practice. Tension members may be found in the lower chord of a pre-cast truss, in one of the legs of a twin-legged column, in the wall of a vessel under internal pressure , in the tie rod of a tied arch and etc. Fig. 8.1 shows some reinforced concrete members with axial tension and with eccentr

4、ic tension used in civil engineering.in truss rectangular reservoirCircular reservoir bunker latticed columnFig.1some reinforced concrete members with axial tension and with eccentric tensionDesign of tension members is often based on serviceability limit state. The design value of tension steel rei

5、nforcement is controlled by the allowable crack width. The Chinese code stipulates that in reinforced concrete tension members, the design value of tensile reinforcement should be no more than 300Mpa.2.STENGTH OF MEMBERS WITH AXIAL TENSIONTest results on reinforced concrete members with axial tensio

6、n show that the tension may be taken by both concrete and reinforcement before concrete is cracked, and the tension is taken by the reinforcement alone after concrete is cracked. Fig.8.2 shows an axial tension section at the ultimate state. The equation may be written as (1)Where , N isdesign value

7、of axial tension , fy is design value of the yielding strength of reinforcement.Fig.2 an axial tension section3. STRENTH OF MEMBERS WITH ECCENTRIC TENSIONA reinforced concrete eccentric tension member section may be divided into two types,i.e. ,a full tension section (or small-eccentricity tension s

8、ection ) and a partial tension section (or large-eccentricity tension section ). Fig.3 shows a section with reinforcement As on one side and As on the other side. If the eccentric tension force N is acting within the space be-tween As and As fig. 3(a), then once the section is cracked by the tension

9、 stress on the side of As , it can be deduced by equilibrium that a compression zone cannot be maintained in the section and the section will be cracked through. This is the case of a full tension section. It must be noticed that before the section cracks, there may be a compression zone in the sect

10、ion. It is only after the concrete has cracked and the tension on the section has shifted to As that the compression zone can no longer be maintained and the section is cracked through.If the eccentric tension force is acting on the outside of the space between As and As Fig.3(b), then after the sec

11、tion is cracked , part of the section must be in compression to maintain equilibrium and the section will be cracked through, this is the case of a partial tension section.3.1 Full Tension Section (Small-eccentricity Tension sections )Test results show that all the steel of As and As , in full tensi

12、on section are in tension and they will yield at the ultimate state if the ratio of As to As is adjusted to suit the eccentricity of the load. Fig.4 shows a full tension section at the ultimate state, by the equilibrium of force and moments we have(a)full tension section (b)partial tension sectionFi

13、g.3 eccentric tension section Fig.4 full tension section(2)(3)(4)where, e is the lever arm measured from the eccentric tension to As, and e is the lever arm to As ; ho is the depth of As from the top of the section; e =-eo as and e=+eo-as.If the section is symmetric and as=as , putting e =(ho-as)/2-

14、eo, e=(ho-as)/2 +eo and M = Neo , the Eqs.(3) and (4) may be written as (5)(6)These two equations may be used to design the required reinforcement for a given section under given the load N and M. The first term shows the reinforcement required by the axial tension N, the second term shows the effec

15、t of the moment M on the reinforcement required. The existence of Mincreases the amount of As and reduces the amount of As. thus, if the section is to be designed for combined loads, then As should be designed for the combination of the largest axial tension N with the largest moment M, and As shoul

16、d be designed for the combination of the largest axial tension N with the smallest moment M.It is a very common practice to place the reinforcement symmetrically in an eccentric tension section so that As=As. In such case, the steel As will not yield and we may calculate As(=As) from Eq.(6) .3.2 Par

17、tial Tension Sections (Large-eccentricity Tension Sections ) The failure mode of a partial tension section is similar to a under-reinforced section under flexure or to a compression column section with failure initiated by the yielding of the tension reinforcement As followed by the extension of cra

18、cks and the crushing of concrete. However, a failure initiated by the crushing of concrete without the tension yielding of steel cannot be excluded if the section is over-reinforced.By replacing the equivalent stress block for the actual stress distribution in the same way as for a compression colum

19、n section (Fig.5), the equilibrium equations may be Fig.5 Partial tension sectionestablished as(7) (8)where, e=eo-+as. To ensure the tension steel can reach its yielding strength fy, and the compression steel can reach its yielding strength fy at the ultimate load, the following conditions should be

20、 satisfied(9)If it is found that 2as , then restart the analysis assuming =2as, the equation may be expressed by the approximate formula (10)where ,e= eo+-as. when the reinforcements are symmetrically placed so that as = as, As=As, the equations can be written as (11)The design of a section and the

21、calculation of ultimate loads for the partial tension section are similar to the compression column section.【example 1】Given a rectangular section of b=350mm and h=400mm ,the concrete is of grade C25 and reinforcement is of HRB335. Design the required reinforcements for an axial tension N=500kN and

22、a moment M=40kN·m. Data: C25 concrete fc = 11.9N/mm2, 1=1.0HRB335 steelfy = 300MPa, b=0.55ho =400-35=365mm Minimum steel ratio min = 0.2%>45ft/fy = 0.16% Solution:so the section is a full tension section (small-eccentricity tension section)By formula (5) and (6)Provide 4 20 with the steel ar

23、ea of 1256 mm2 for As, and 4 12 with the steel area of 452 mm2 for As.【example 2】The wall of a water tank has a thickness of 200 mm.The concrete is of grade C20 and reinforcement is of HRB235. Under internal water pressure, there are an axial tension N = 80kN and a moment M= 70kN·m acting on ea

24、ch meter length of the wall. Design the required reinforcements As and As. Data: C20 concrete fc = 9.6N/mm2, 1=1.0 HRB235 steelfy =fy= 210MPa, b=0.55Effective depth ho =h-as=200-25=175mm, b=1000mmMinimum steel ratiomin = 0.2% Solution:so the section is a partial tension section(large-eccentricity te

25、nsion section).Assuming =0.3Let As = minbh = 0.2%×1000×200 = 400 mm2Provide 10 190 with the area of As = 413 mm2Provide 18100 with the area of As = 2545mm24. SHEAR STRENGTH OF MEMBERS WITH TENSIONTests show that the shear strength of tension member is less than that of the same member with

26、out axial tensile force. It is because the tensile force in tension members will induce the development of inclined crack due to the decrease of the shear compression zone of the section and then the decrease in shear strength of the member.The Chinese Code suggests a simple and conservative linear

27、relationship to evaluate the shear bearing capacity of a tension member with rectangular section, “T” section or “I” section, and is expressed as (12)Where, N is the design value of axial tension force corresponds to shear force V, is shear span ratio of the calculated section, the value of is taken

28、 as section 7.5.If the value calculated from the right hand side of Eq.(12) is less than fyv, then V is taken as equal to fyv, and the value of fyvcannot be less than 0.36ftbho.譯文：1. 引言當(dāng)構(gòu)件的截面上作用有軸心拉力N時(shí)，稱該構(gòu)件為軸心受拉構(gòu)件；當(dāng)構(gòu)件的截面上既作用有軸心拉力又作用有彎矩M時(shí)，或作用有一偏心距為 e0 =M/N 的拉力時(shí)，稱該構(gòu)件為偏心受拉構(gòu)件。盡管混凝土的抗拉強(qiáng)度較低，但是工程實(shí)踐中仍有許多受拉構(gòu)

29、件的應(yīng)用實(shí)例，例如：預(yù)制桁架的下弦桿、雙肢柱的肢桿、承受內(nèi)壓力的管壁和拱的拉桿等，如圖1所示，給出了一些常見的鋼筋混凝土軸心受拉和偏心受拉構(gòu)件。 桁架 矩形容器圓形容器 儲(chǔ)倉格構(gòu)柱圖.1 一些鋼筋混凝土軸心受拉和偏心受拉構(gòu)件受拉構(gòu)件通常是按照正常使用極限狀態(tài)來設(shè)計(jì)的，因此，受拉鋼筋強(qiáng)度的設(shè)計(jì)值的大小對(duì)控制裂縫起著決定性的作用。規(guī)范中規(guī)定，鋼筋混凝土受拉構(gòu)件中受拉鋼筋強(qiáng)度的設(shè)計(jì)值不應(yīng)超過300MPa。2. 軸心受拉構(gòu)件對(duì)于軸心受拉構(gòu)件，在混凝土開裂前，混凝土與鋼筋共同承受拉力，開裂以后，裂縫截面的全部拉力只有鋼筋承受，如圖2所示，其受拉承載力可按公式（1）計(jì)算而得。(1)圖.2 軸心拉截面3.偏

30、心受拉構(gòu)件鋼筋混凝土偏心受拉構(gòu)件的截面可分為兩大類，一類是全部受拉截面（或小偏心受拉截面），另一類是部分受拉截面（或大偏心受拉截面）。設(shè)矩形截面上距軸向力N較近一側(cè)的縱向鋼筋為As，較遠(yuǎn)一側(cè)的縱向鋼筋為As ，如圖3所示。當(dāng)軸力N作用于As和As之間時(shí)，混凝土開裂后，由力矩平衡關(guān)系可知，截面上沒有保留受壓區(qū)，截面全部貫通，這種情況稱為小偏心受拉。值得注意的是：混凝土開裂前，截面上還可能保留有受壓區(qū)，但混凝土開裂后，鋼筋A(yù)s受拉，截面上不再保留受壓區(qū)，形成貫通整個(gè)截面的通縫。當(dāng)N作用于鋼筋A(yù)s與As間距以上時(shí)，當(dāng)截面開裂以后，截面部分受壓以保持平衡，截面不會(huì)全部貫通，這種情況稱為大偏心受拉。3.

31、1 全部受拉截面（小偏心受拉截面）試驗(yàn)結(jié)果表明，小偏心受拉構(gòu)件破壞時(shí)，拉力完全由鋼筋承受，鋼筋A(yù)s 和（a）小偏心受拉 （b） 大偏心受拉圖.3 偏心受拉截面 圖.4小偏心受拉截面 (2)(3) (4)As 的拉應(yīng)力均達(dá)到屈服強(qiáng)度。根據(jù)平衡條件，可寫出小偏心受拉構(gòu)件的計(jì)算公式。可由式（5）和 （6）求得給定截面在給定荷載（N、M）作用下兩側(cè)的受 (5)(6)拉鋼筋。第一項(xiàng)代表軸向力N所需的鋼筋，第二項(xiàng)代表彎矩M所需的鋼筋。M增加了As的用量而降低了As的用量。因此設(shè)計(jì)中有幾組不同荷載組合（N、M）時(shí)，應(yīng)按最大N和最大M求As，按最大N和最小M求As。對(duì)稱配筋時(shí)，As不屈服，可按公式（6）計(jì)算 As As。3.2 部分受拉截面（大偏心受拉截面）大偏心受拉構(gòu)件的破壞形態(tài)類似于受彎構(gòu)件或受壓構(gòu)件，破壞始自于受拉鋼筋A(yù)s的屈服，隨后裂縫展開，直到混凝土被壓碎。但若為超筋截面，則破壞始自于混凝土的壓碎，此時(shí)受拉鋼筋A(yù)s有可能并未屈服。大偏心受拉構(gòu)件的承載力可按式（7）、（8）計(jì)算，適用條件為式（9）。當(dāng)圖