圆截面混凝土支护桩配筋的新算法.pdf

圆截面混凝土支护桩配筋的新算法李乾南1,熊宗海2(1.武汉科技大学中南分校,武汉 430200;2.中国地质大学,武汉 430074)摘 要:在沿整个圆周均匀配筋与局部均匀配筋相结合的条件下,提出一种钢筋混凝土支护桩配筋的新实用方法,构建了一个函数数表,同时提出修正值公式。通地四个实例的计算表明,新方法可节省钢筋14%31%,而仍保留仅沿整个圆周均匀配筋的优点。关键词:钢筋混凝土,支护柱,配筋中图分类号:TU473 文献标识码:B 文章编号:100423152(2011)04200782041 引言目前,混凝土支护桩的圆截面抗弯有两种配筋方式,一是沿整个周边均匀配置纵向钢筋,二是在截面受拉区和受压区周边局部均匀配置纵向钢筋或集中纵向钢筋,其受弯承载力计算,建筑基坑支护技术规程(J GJ120-99)1作了明确规定。全圆周均匀配筋的优缺点是24:形成刚劲钢筋笼,施工时需保证纵向钢筋直立,牢固不变位;同一桩体,诸截面弯矩变向时,容易照顾,方便考虑;无需另外设置纵向构造筋(一般约占纵向受力筋的15%25%),可由纵向受力筋承担;由于每根钢筋与中和轴的距离不等,距中和轴越近的钢筋,抗弯矩作用越小,不经济。而局部周边配筋的优缺点,恰恰与之相反,虽可节省钢筋,但施工与构造要求高,须复杂的设计。作者试图将此两种配筋结合在一起,即在截面整个周边除均匀配筋外,还在钢筋受拉屈服弧段内增加局部均匀钢筋,所增钢筋量为此段内原均匀配筋量的12倍,以达到扬长避短,互补完善的目的。理论上,可将 混凝土结构设计规范(GB50010-2002)5中的(7.3.8-1)、(7.3.8-2)两式,除令N=0,Nei=M(弯矩设计值)外,再各增加一项6,亦即:1fcA(1-sin2 2)+(-t)fyAs-nsfyAs=0(1)M=231fcA rsin3+fyAsrssin+sin t+fyAsrsnsin s(2)s=1cos-1-rrs+1.25(1+fy0.0033Es)1+r/rs1+rs/r(1-cos)(3)式中:s为圆截面受拉钢筋屈服区面积的圆心角与2的比值;为圆截面受压区面积的圆心角与2的比值;t=1.25-2,0.625时,取t=0;r为圆截面半径,rs为纵向钢筋所在圆周半径;n为钢筋受拉屈服区内所增钢筋的倍数,一般n=1或2,其余字符意义详见规范6。为避免利用上述(1)、(2)和(3)式进行配筋计算时求解超越方程的麻烦,引入两个无量纲参数、m,构建了一个函数s()、()和m()的数表,同时提供一对计算、s修正值的公式,以快速计算出所求数值。实例计算表明,新方法可节省钢筋14%31%,而仍保留仅沿整个圆周均匀配筋的优点。2 全域的无量纲参数s()、()和m()的函数表引入无量纲参数,即=fyAs1fcA=fyAsfcmA、m=M1fcA r=MfcmA r,考虑一般rrs=1.16以及热轧钢筋性质7,上述(1)、(2)和(3)式,可改写为:(,s)=-sin2 2-(1.25-3+ns)(4)m=0.2122sin3+0.2744sin+sin(1.25-2)+nsin s(5)收稿日期:2010210211 作者简介:李乾南,男,教授级高工。1994-2013 China Academic Journal Electronic Publishing House.All rights reserved.(,s)=2.080cos+cos s-0.9193(6)根据(4)、(5)和(6)式,可作出无量纲函数s()、()和m()的函数表1,供设计使用。表1 s()、()和m()函数表序号sn=1n=2=fyAs/fcmAm=M/fcmA r=fyAs/fcmAm=M/fcmA r10.19290.79030.030030.052360.019480.0510120.20290.77120.035880.061700.023210.0599030.21290.75250.042560.072150.027430.0697740.22290.73440.050190.083820.032220.0806950.23290.71640.058880.096780.037630.0927260.24290.69840.068770.11120.043750.106070.25290.68070.080010.12710.050630.120480.26290.66330.092680.14460.058320.136290.27290.64560.10710.16390.067010.1534100.28290.62810.12360.18520.076720.1720110.29290.61070.14200.20840.087620.1921120.30290.59290.16320.23420.099840.2140130.31290.57510.18740.26260.11360.2377140.32290.55720.21470.29370.12910.2632150.33290.53920.24630.32830.14660.2910160.34290.52110.28300.36690.16630.3209170.35290.50260.32550.40990.18890.3536180.36290.48380.37530.45830.21450.3889190.37290.46470.43440.51350.24400.4277200.38290.44520.50520.57680.27840.4705210.39290.42530.59140.65080.31860.5179220.40290.46480.69880.73900.36640.5710230.41290.38370.83560.84710.42390.6317240.42290.36191.01600.98410.49470.7020250.43290.33911.26601.16700.58420.7855260.44290.31531.63601.42800.70120.8885270.45290.29012.24101.84200.86211.0220280.46290.26313.42602.63301.10001.2090290.47290.23376.86004.88401.49301.5020300.48290.2008221.8000144.32.29602.0750310.48320.19972.33402.1010320.49290.16225.07603.9910330.50000.128865.7945.24340.5005注:(1)fcm=1fc;(2)对C 20、C 25、C 30混凝土,圆截面素混凝土的m=0.0495,故从0.1929起;(3)n=1时,变化区间为0.1929,0.4832;(4)n=2时,变化区间为0.1929,0.5005。3、s的修正值公式当已知参数值与表1中数值不相符时,可先按表1作线性插值,而后按以下公式(7)、(8),对、s进行修正,即hi+1=(i,si)sin si-(i,si)n(1-cos2 i+3)sin si+2.080nsin i(7a)ki+1=(i,si)(1-cos2 i+3)+(i,si)2.080sin i(1-cos2 i+3)sin si+2.080sin i(7b)i+1=i+hi+1(8a)si+1=si+ki+1(8b)公式(7)、(8)由Newton2Raphson法导出,由于初值i、si取自表1由线性插值得出,已相当接近真值,故仅需修正13次,即可达到足够准确的程度。4 应用举例4.1 算例一圆截面混凝土支护桩,直径D=1200 mm,混凝土强度等级C30,fcm=16.5 MPa,级钢筋fy=310 MPa。已知弯矩设计值M=2074 kNm。试用本文方法配置纵向钢筋(n取1)。解:m=MfcmA r=207410616.5(600)3=0.1852。按n=1,查表1得=0.2829,s=0.6281,=0.1236。所以2 s=360 0.6281=226.116,As=fcmAfy=0.123616.5310(600)2=7440 mm2。故配筋如图1所示,1525+1925=7364+图1 算例一 配筋图4418=11782 mm2,(文献8,仅周边均匀配筋97 第4期 李乾南等:圆截面混凝土支护桩配筋的新算法 1994-2013 China Academic Journal Electronic Publishing House.All rights reserved.2825=13745 mm2)节省钢筋14%。4.2 算例二圆截面混凝土支护桩,直径D=600 mm,混凝土强度等级C30,fcm=16.5 MPa,II级钢筋fy=310 MPa,已知弯矩设计值M=658.5 kNm。试按本文方法配置纵向钢筋(n取2)。解:m=MfcmA r=688.510616.5(300)3=0.4705按n=2,查表1得=0.3829,s=0.4452,=0.2784,所以2 s=360 0.4452=280.15,As=fcmAfy=0.278416.5310(300)2=4190 mm2 故配筋如图2所示,1122+1022=4181+3801=7982 mm2(文献 8 ,仅周边均匀配筋2822=10643mm2)节省钢筋25%。图2 算例二 配筋图4.3 算例三某基坑支护桩,直径D=500 mm,混凝土强度等级C30,fcm=16.5 MPa,II级钢筋fy=310 MPa,已知沿周边均匀配有825钢筋,试按本文方法作配筋图并算出其能承受的最大弯矩设计值。解:=fyAsfcmA=3108490.916.5(250)2=0.3758 按n=1,查表1得=0.3629,s=0.4838,m=0.4583,所以2 s=360 0.4838=287.08。故配筋如图3所示,825+325=3927+1473=5400 mm2,所求最大弯矩设计值M=mfcmA r=0.458316.5(250)3=371.2 kNm。(文献8,仅周边均匀配筋1625=7854 mm2,M=393kNm)节省钢筋31%。图3 算例三 配筋图4.4 算例四某基坑支护桩,直径D=800 mm,混凝土强度等级C25,fcm=11.9 MPa,钢筋II级,fy=300MPa,已知沿周边均匀配有618钢筋,试按本文方法作配筋图并算出其能承受的最大弯矩设计值。解:=fyAsfcmA=3006254.511.9(400)2=0.076585按n=1,查表1并作线性插值,得0=0.2429+0.2529-0.24290.08001-0.06877(0.076585-0.06877)=0.2499s0=0.6984+0.6807-0.69840.08001-0.06877(0.076585-0.06877)=0.6861按公式(4)、公式(6):(0,s0)=0.2499-sin20.24992-0.076585(1.25-30.2499+0.6861)=-0.00016(0,s0)=2.080cos0.2499+cos0.6861-0.9193=-0.0002按公式(7)、公式(8):h1=(-0.00016 sin0.6861-(-0.0002)10.076585)(1-cos20.2499+30.0076585)sin0.6861+2.0800.076585sin0.2499=0.0001130k1=-0.0002(1-cos20.2499+30.076585)+(-0.00016)2.08 sin0.24993.574=-0.00027561=0+h1=0.2499+0.000113=0.2500s1=s0+k1=0.6861-0.0002756=0.6858 将1、s1代入式(4)、式(6)、式(7)和式(8),重复以上计算过程可得:2=1+h2=0.2500+0.00002017=0.2500s2=s1+k2=0.6858-0.0001934=0.6860所以2 s=360 0.6860=2123.5。故配08土 工 基 础 2011 1994-2013 China Academic Journal Electronic Publishing House.All rights reserved.筋如图4所示,618+418=1527+1017=2544 mm2根据式(5),可得m=0.2122sin0.2500+0.27440.076585sin0.2500+sin(1.25-20.2500)+sin0.6860=0.1223 所求最大弯矩设计值M=mfcmA r=0.122311.9(400)3=292.6 kNm(文献8,仅周边均匀配筋822=3041 mm2,M=300 kNm)节省钢筋16%。图4 算例四 配筋图参考文献1 建筑基坑支护技术规程(J GJ120-99)S.19992 唐永祥.圆截面挡土灌注桩配筋的探讨J.建筑技术,19933 施文华.深基坑桩锚支护设计中的几个问题J.建筑技术,19934 孙宝俊.殷惠光.圆形截面单侧配筋预应力钢筋混凝土支护桩的配筋计算J.建筑结构,19985 混凝土结构设计规范(GB50010-2002)S.20026滕智明.环形及圆形截面钢筋混凝土构件正截面强度计算R.清华大学科研报告,19857 滕智明,朱金铨.混凝土结构及砌体结构(上册)M.北京:中国建筑工业出版社,20028 武汉市城乡建设管理委员会.武汉地区深基坑工程技术指南(WBJ-7-95)S.1995A New Method to Calculate Reinforcement inCircular Section Concrete Retaining PilesLI Qian nan1,XION G Zong hai2(1.Zhongnan Branch,Wuhan University of Science and Technology,Wuhan430200;2.China University of Geosciences,Wuhan430074,China)AbstractBased on the view point of the uniform reinforcement along entire circular section combined with partial uniform re2inforcement.There is a new method to calculate reinforcement for R.C.Retaining Piles with circular section,in which thefunction table and the correction formula are included.The results shows that the application of new method the steel bar issaved about 1431%with the advantage of uniform reinforcement alone entire circle.Key wordsreinforced concrete,retaining pile,reinforcement18 第4期 李乾南等:圆截面混凝土支护桩配筋的新算法 1994-2013 China Academic Journal Electronic Publishing House.All rights reserved.