桥梁抗震算例

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抗震计算书4.18(内容清晰)

抗震计算书4.18(内容清晰)

抗震计算书4.18(内容清晰)十堰至天水高速公路桥墩抗震计算书一、项目概况徽县(大石碑)至天水高速公路是十堰至天水国家高速公路(G7011)甘肃境内路段,我院承担了该项目第STSJ2合同段的勘察设计工作。

路线起于西和县城南五里铺,终点位于天水市秦州区皂郊镇,路线全长81.625km。

本项目直接或间接影响区域均为四川汶川“5.12”大地震的受灾区。

地震动加速度峰值0.30g (抗震设防烈度为Ⅷ度),抗震设防措施等级为9度。

地震动反应谱特征周期0.4s。

由于本项目地震烈度较高,桥梁抗震计算显得非常重要。

二、计算内容(1)、地震作用本项目大部分桥梁均为20米、30米预制预应力混凝土连续箱梁桥,现选取几种典型结构及墩高组合计算抗震,为本项目桥梁抗震设计提供参考。

详细选取类型见下表:孔数(孔)墩高组合(米)-跨径(米)5X20 5+8+7+65X20 11+20+25+155X20 15+20+25+155X20 20+25+25+205X20 20+25+25+204X30 5+7+64X30 11+30+254X30 16+30+254X30 20+30+254X30 25+30+25注:墩高组合中“5+7+6”表示1号墩高5米,2号墩高7米,3号墩高6米。

以下类推。

根据公路桥梁抗震设计细则(JTG/T B02-01-2008),一般情况下,公路桥梁可只考虑水平向地震作用,直线桥可分别考虑顺桥向和横桥向的地震作用。

在顺桥向地震作用影响下,由于矮墩相对刚度较大,承担的力也相应较大。

因此,高低墩搭配情况下对矮墩更不利;横桥向地震作用下,高低墩搭配情况下对高墩更不利。

据此考虑,选取上述几种跨径和墩高组合进行抗震计算。

(2)桥梁结构概况1、跨径:5-20米、4-30米2、桥梁宽度:12.25米3、桥梁右偏角:90°4、墩台结构:柱式台、双柱式桥墩5、地震烈度:地震动加速度峰值0.30g(抗震设防烈度为Ⅷ度),抗震设防措施等级为9度。

简支空心板桥桥墩抗震计算书

简支空心板桥桥墩抗震计算书

简支空心板桥桥墩抗震计算书(一)设计资料1、上部构造:2孔20m连续桥面简支梁,20m先张法预应力混凝土简支宽幅空心板,计算跨径为19.32m,每跨(单幅)横向设8块板。

桥面现浇10cm50号混凝土,9cm沥青混凝土。

2、桥面宽度(单幅):0.5(防撞墙)+净11.5(行车道)+0.75m(波形护栏)=12.75m。

3、斜度:30°。

4、设计荷载:公路ⅰ级。

5、支座:墩顶每块板板端设gyz200×42mm板式橡胶支座2个。

6、地震动峰值加速度:0.20g。

7、下部构造:圆形双柱式墩,直径1.3m;钻孔桩直径1.5m,长40m。

墩柱为30号混凝土,桩基础为25号混凝土,hrb335钢筋。

桥墩通常结构如下(二)恒载计算1、上部恒载反力空心板:[(12.5+0.3)×6+(14.7+0.3)×2]×26=2776.8kn铰缝混凝土:2.22×7×26=404.0kn桥面铺装(包括50号混凝土和沥青混凝土):11.5×20×0.1×26+11.5×20×0.09×24=1094.8kn防撞墙:6×26=156kn波形护栏:5.6×26=145.6kn合计:2776.8+404.0+1094.8+156+145.6=4577.2kn2、下部恒载计算1)盖梁加防震挡块重力pg=28.8×25=720kn2)系梁重力px=8.1×25=202.5kn3)一个墩柱重力pd=?×1.3×5.6×25=185.8kn244)单桩自重力pz=?×1.5×40×25=1767.1kn24(三)水平地震力计算1、顺桥向水平地震力计算1)上部结构对板式橡胶支座顶面处产生的水平地震荷载eihs=kitp?ki?1nciczkh?1gspitp式中:ci=1.7,cz=0.3,kh=0.2根据地质资料分析,桥位所在地土层属ⅲ类场地,所以有β1=2.25×(0.45) t10.95对于板式橡胶支座的梁桥t1=2?1其中:ωg2gspgtp21=gtpk1?(k1?k2)gsp?{[gtpk1?(k1?k2)gsp]2?4gtpgspk1k2}1/2k1=?kisi?1n计算采用2孔×20m为一联,故n=1kis=?gdari?1ns?t其中:ns=2×16=32,gd=1200kn/m由橡胶支座计算知ar=?×0.2=0.0314m2224?t=0.042m∴kis=32×1200?0.0314=28708.6kn/m0.042k1=1×28708.6=28708.6kn/mk2=?kipi?1nkip=3i13e1li其中:墩柱使用30号混凝土,则ec=3.00×10mpae1=0.8×3.00×10×10=2.4×10kn/m按墩高h=7m控制设计,支座垫石+支座厚度=0.1+0.042=0.142mli=7+0.142=7.142m柱惯矩:i1=?×1.3=0.1402m4443724643?0.1402?2.4?107kip=7.1423×2=55418.0kn/mk2=1×55418.0=55418.0kn/mgsp=2×4577.2=9154.4kngtp=gcp+ηgp其中:gcp=720kngp=2×185.8=371.6knη=0.16(xf2+2xf12+xfxf1+xf1+1)222顺桥向作用于支座顶面的单位水平力在支座顶面处的水平位移为:xd=x0-φ0l0+xq其中:l0=li=7.142mxq=l03e1i137.1423==0.000036173?2.4?10?0.1402桩的排序宽度:b1=0.9(d+1)=0.9×(1.5+1)=2.25m桩在土中的变形系数:α=5m=10000kn/m其中:桩使用25号混凝土,则ec=2.80×10mpaei=0.8×2.8×10×?×1.5=5.567×1074644mb1ei64∴α=510000?2.25=0.332165.567?10桩长h=40m,∴αh=0.3321×40=13.284m>2.5m挑αh=4.0,故kh=0从而存有x0= b3d4?b4d3l0b3c4?b4c313eia3b4?a4b3?2eia3b4?a4b3。

抗震计算—3-30混凝土小箱梁抗震计算

抗震计算—3-30混凝土小箱梁抗震计算

一、工程概况K44+033.729/K44+035.763桥位于楚雄连汪坝至南华县城一级公路7合同牛凤龙段,为主线上跨楚大高速而设。

孔跨布置为左幅3×30+3×30+3×30 +(50+60+40) m,右幅4×30+4×30+(50+60+50)m先简支后连续预应力小箱梁和连续钢箱梁桥,错墩台设置。

本桥平面分别位于缓和曲线(起始桩号:K43+819.332,终止桩号:K43+925.604,参数A:312.65,右偏)、圆曲线(起始桩号:K43+925.604,终止桩号:K44+165.386,半径:425m,右偏)和缓和曲线(起始桩号:K44+165.386,终止桩号:K44+238.925,参数A:312.65,右偏)上,纵断面纵坡-1.81%;墩台径向布置。

采用3、4孔一联连续结构,按半幅计左幅桥设4联,右幅桥设3幅,全桥共设9道伸缩缝。

本计算为左幅第二联3×30先简支后连续预应力小箱梁。

根据《中国地震动参数区划图》(GB18306-2001)及《云南省地震动峰值加速度区划图》、《云南省地震动反应谱特征周期区划图》,桥位处中硬场地类型3区,地震动峰值加速度值为0.15g,地震动反应谱特征周期为0.45s,地震基本烈度值为Ⅶ度,分组为第二组。

图1.1 桥型布置图图1.2 剖面示意二、自振特性分析全桥有限元计算模型示于图2.1,从左到右依次是4号墩、5号墩,其自振周期及相应振型列于表2.1,示于图2.2。

图2.1 有限元模型模态号频率/Hz 周期/s 振型特征1 0.245193 4.078413 主梁纵飘2 0.915799 1.091943 5号墩顺桥向弯曲3 1.676969 0.596314 主梁横向对称弯曲4 2.775748 0.360263 5号墩扭转弯曲5 3.101696 0.322404 主梁竖向反对称弯曲6 3.189287 0.313550 主梁竖向对称弯曲第一阶振型主梁纵飘第二阶振型5号墩顺桥向弯曲第三阶振型主梁横向对称弯曲第四阶振型5号墩扭转弯曲第五阶振型主梁竖向反对称弯曲第六阶振型主梁竖向对称弯曲图2.2 振动模态三、地震输入E1、E2水准地震时,均按反应谱输入。

ch7 桥梁抗震设计示例

ch7 桥梁抗震设计示例
曲率延性系数 33.89
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图7.7 固定墩墩柱截面的弯矩-曲率关系 lp 0.5h
1 3 ( 1 ) 0 . l 5 h ( 1 0 . 2 h l ) 5 1 3 3 . 8 2 1 7 . 5 9 ( 1 0 . 2 1 7 . 5 5 ) 1 6 1
则,固定墩的设计地震力为:
P C i C z ( K h G s d G s ) 121.633 0. 4. 3kN ( 0 . 2 0 . 784 38995 0 . 02 28360 )
桥梁抗震
7.2.2固定墩的抗弯强度验算
固定墩一个墩柱的墩底弯矩为:
M P l 2163 .4 7 7571 .9 kN .m
2
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固定墩一个墩柱的恒载轴力为: N ( 35 .45 15 1 .5 1 .35 2 .5 7 ) 10 5671 .9 kN
根 据 《 公 路 钢 筋 混 凝 土 及 预 应 力 混 凝 土 桥 涵 设 计 规 范 》 规 定 的 计 算 公 式 计 算 , 图 7.3 所 示 的 截 面 在 5671.9kN 的 轴 力 作 用 下 , 所 能 承 受 的 弯 矩 为 :
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体系的自振周期为:
(I偏安全考虑,不折减)
反应谱值:
T 2
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桥梁抗震算例

桥梁抗震算例

计算简图某城市互通立交匝道桥上部结构采用预应力混凝土连续梁桥体系,跨径布置为2×25m ,梁宽从10.972m 变化到15.873m ;桥墩和桥台上都设置板式橡胶支座。

以下为该桥采用《公路工程抗震设计规范》(JTJ004—89)的简化计算方法手算的计算步骤及计算结果:附2.1 顺桥向地震力计算该联支座全部采用板式橡胶支座,故地震力由两部分组成:上部结构对板式橡胶支座顶面处产生的水平地震荷载及桥墩地震荷载。

一、上部结构对板式橡胶支座顶面处产生的水平地震荷载上部结构对D6号墩板式橡胶支座顶面处产生的水平地震荷载按下式计算:zsp h z i ni itpitpihs G K C C KK E 10β∑==(附2-1)式中,3.1=i C ,2.0=z C ,1.0=h K 1、确定基本参数(1)全联上部结构总重力:2353.4825)86.527.518(⨯+⨯+=zsp G 255023.0⨯⨯⨯+kN 2.16155=(2)实体墩对支座顶面顺桥向换算质点重力:()pff tp ztp GX X G G ⎥⎦⎤⎢⎣⎡-+==2131由于不考虑地基变形,即0=f X故 ()p pff tp G GX X G 311312=⎥⎦⎤⎢⎣⎡-+= 而 kN G p 3.57525346.4295.5=⨯⨯= 得 kN G G G p tp ztp 8.1913/===(3)一联上部结构对应的全部板式橡胶支座顺桥向抗推刚度之和1K :m kN K /103915.23.5756244.2480)23(41⨯=⨯+⨯+=(4)设置板式橡胶支座的D6号桥墩顺桥向抗推刚度2K :8015.01=I 4m ,088.12=I 4m ,676.13=I 4m083.105.06.045.01321=-+=I I I I e 从而,得 49233.0m I e =m kN l EI K e D /1055.8746.49233.0103.3335373⨯=⨯⨯⨯== m kN K K D /1055.852⨯==∴ 2、计算桥梁顺桥向自振基本周期T 1[]{}ZspZtp Zsp Ztp ZspZtp Zsp Ztp G G K K G G G K K K G G K K K G g24)()(2121221121121-++-++=ω-24.11s 1=s T 673.1211==ωπ3、计算动力放大系数1β根据1T 及规范三类场地土动力放大系数函数,计算1β:646.045.025.295.01=⎪⎭⎫⎝⎛⨯=T β4、计算上部结构对D6号桥墩产生的水平地震力上部结构对D6号桥墩板式橡胶支座顶面处产生的顺桥向水平荷载按式(附2-1)计算:kN E E iihs hs 6.1302.16155646.01.02.03.1103915.23.575624=⨯⨯⨯⨯⨯⨯⨯==∑二、实体墩由墩身自重在墩身质点i 的顺桥向水平地震荷载实体墩由墩身自重在墩身质点i 的顺桥向水平地震荷载按下式计算:11hp i z h li i E C C K X G βγ=得 D6号墩kN E th 22.476.1910.10.18482.01.02.03.1=⨯⨯⨯⨯⨯⨯= 三、桥墩顺桥向地震剪力和弯矩第二联D6号桥墩墩底的顺桥向地震剪力和弯矩分别如下:kN Q D 82.13422.46.1306=+=()kN M D 93.585346.422.46.1306=⨯+=附2.2 横桥向地震力计算D6号桥墩横桥向水平地震荷载按下式计算(参见D6号墩计算简图):111i h p i z h i iE C C K X Gβγ= (附2-2)式中,3.1=i C ,2.0=z C ,1.0=h K 1、计算i X 1由于5031.14606.474<==B H 故取 ()fi f i X H H X X -⎪⎭⎫⎝⎛+=13/11不考虑地基变形时:0=f X故有 3/11⎪⎭⎫ ⎝⎛=H H X i i得 889.06.4744.3333/111=⎪⎭⎫⎝⎛=X ,621.06.4747.1133/112=⎪⎭⎫ ⎝⎛=X2、计算桥墩各质点重力i GkN G 6.80772/2.161550== kN G 4.32825146.2122.61=⨯⨯=kN G 61.247252.2502.42=⨯⨯= 3、计算横桥向基本振型参与系数1γ011.16.247621.04.328889.06.807716.247621.04.328889.06.80771220201=⨯+⨯+⨯⨯+⨯+⨯==∑∑==ni iini iiG XGX γ 4、计算D6号桥墩振动单元横桥向振动时的动力放大系数1β (1)计算横桥向柔度δ:934.11=I 4m ,700.32=I 4m ,254.103=I 4m32105.06.045.01I I I I e -+= 得 4569.2m I e =H 2H 1HD6号墩计算简图563731076.81/5.11419/10412.1646.5569.2103.333-⨯===+⋅=⨯=⨯⨯⨯==KmkN K K K Ks K m kN l EI K DS De D δ (2)计算桥墩横向振动的基本周期T 1s gG T t 72.122/11=⎪⎪⎭⎫ ⎝⎛=δπ(3)确定动力放大系数1β根据T 1及规范三类场地土动力放大系数函数,得629.045.025.295.01=⎪⎭⎫⎝⎛⨯=T β5、计算各质点的水平地震力根据公式(附2-2)计算作用于D6号桥墩各质点的横桥向水平地震力:kNE kN E kN E hp hp hp 40.26.247586.0011.1629.01.02.03.156.44.328839.0011.1629.01.02.03.155.1336.8077011.1629.01.02.03.1210=⨯⨯⨯⨯⨯⨯==⨯⨯⨯⨯⨯⨯==⨯⨯⨯⨯⨯= 6、计算横桥向地震剪力和弯矩D6号墩墩底的横桥向地震剪力和弯矩分别如下:kN Q D 51.14040.256.455.1336=++=m kN M D ⋅=⨯+⨯+⨯=34.598137.140.2334.356.4346.455.1336。

桥梁抗震算例

桥梁抗震算例

计算简图某城市互通立交匝道桥上部结构采用预应力混凝土连续梁桥体系,跨径布置为2×25m ,梁宽从10.972m 变化到15.873m ;桥墩和桥台上都设置板式橡胶支座。

以下为该桥采用《公路工程抗震设计规范》(004—89)的简化计算方法手算的计算步骤及计算结果:附2.1 顺桥向地震力计算该联支座全部采用板式橡胶支座,故地震力由两部分组成:上部结构对板式橡胶支座顶面处产生的水平地震荷载及桥墩地震荷载。

一、上部结构对板式橡胶支座顶面处产生的水平地震荷载上部结构对D6号墩板式橡胶支座顶面处产生的水平地震荷载按下式计算:zsp h z i ni itpitpihs G K C C KK E 10β∑==(附2-1)式中,3.1=i C ,2.0=z C ,1.0=h K 1、确定基本参数(1)全联上部结构总重力:2353.4825)86.527.518(⨯+⨯+=zsp G 255023.0⨯⨯⨯+kN 2.16155=(2)实体墩对支座顶面顺桥向换算质点重力:()pff tp ztp GX X G G ⎥⎦⎤⎢⎣⎡-+==2131由于不考虑地基变形,即0=f X故 ()p pff tp G GX X G 311312=⎥⎦⎤⎢⎣⎡-+= 而 kN G p 3.57525346.4295.5=⨯⨯= 得 kN G G G p tp ztp 8.1913/===(3)一联上部结构对应的全部板式橡胶支座顺桥向抗推刚度之和1K :m kN K /103915.23.5756244.2480)23(41⨯=⨯+⨯+=(4)设置板式橡胶支座的D6号桥墩顺桥向抗推刚度2K :8015.01=I 4m ,088.12=I 4m ,676.13=I 4m083.105.06.045.01321=-+=I I I I e 从而,得 49233.0m I e =m kN l EI K e D /1055.8746.49233.0103.3335373⨯=⨯⨯⨯== m kN K K D /1055.852⨯==∴2、计算桥梁顺桥向自振基本周期T 1[]{}ZspZtp Zsp Ztp ZspZtp Zsp Ztp G G K K G G G K K K G G K K K G g24)()(2121221121121-++-++=ω-24.11s 1= s T 673.1211==ωπ3、计算动力放大系数1β根据1T 及规范三类场地土动力放大系数函数,计算1β:646.045.025.295.01=⎪⎭⎫⎝⎛⨯=T β4、计算上部结构对D6号桥墩产生的水平地震力上部结构对D6号桥墩板式橡胶支座顶面处产生的顺桥向水平荷载按式(附2-1)计算:kN E E iihs hs 6.1302.16155646.01.02.03.1103915.23.575624=⨯⨯⨯⨯⨯⨯⨯==∑二、实体墩由墩身自重在墩身质点i 的顺桥向水平地震荷载实体墩由墩身自重在墩身质点i 的顺桥向水平地震荷载按下式计算:11hp i z h li i E C C K X G βγ=得 D6号墩kN E th 22.476.1910.10.18482.01.02.03.1=⨯⨯⨯⨯⨯⨯= 三、桥墩顺桥向地震剪力和弯矩第二联D6号桥墩墩底的顺桥向地震剪力和弯矩分别如下:kN Q D 82.13422.46.1306=+=()kN M D 93.585346.422.46.1306=⨯+=附2.2 横桥向地震力计算D6号桥墩横桥向水平地震荷载按下式计算(参见D6号墩计算简图):111i h p i z h iiE C C K X G βγ= (附2-2)式中,3.1=i C ,2.0=z C ,1.0=h K 1、计算i X 1由于5031.14606.474<==B H 故取 ()fi f i X H H X X -⎪⎭⎫⎝⎛+=13/11不考虑地基变形时:0=f X故有 3/11⎪⎭⎫ ⎝⎛=H H X i i得 889.06.4744.3333/111=⎪⎭⎫⎝⎛=X ,621.06.4747.1133/112=⎪⎭⎫ ⎝⎛=X2、计算桥墩各质点重力i GkN G 6.80772/2.161550==kN G 4.32825146.2122.61=⨯⨯=kN G 61.247252.2502.42=⨯⨯=3、计算横桥向基本振型参与系数1γ011.16.247621.04.328889.06.807716.247621.04.328889.06.80771220201=⨯+⨯+⨯⨯+⨯+⨯==∑∑==ni iini iiG XGX γ 4、计算D6号桥墩振动单元横桥向振动时的动力放大系数1β (1)计算横桥向柔度δ:934.11=I 4m ,700.32=I 4m ,254.103=I 4m 32105.06.045.01I I I I e -+= 得 4569.2m I e =H 2H 1HD6号墩计算简图563731076.81/5.11419/10412.1646.5569.2103.333-⨯===+⋅=⨯=⨯⨯⨯==KmkN K K K Ks K m kN l EI K DS De D δ (2)计算桥墩横向振动的基本周期T 1s gG T t 72.122/11=⎪⎪⎭⎫ ⎝⎛=δπ(3)确定动力放大系数1β根据T 1及规范三类场地土动力放大系数函数,得629.045.025.295.01=⎪⎭⎫⎝⎛⨯=T β5、计算各质点的水平地震力根据公式(附2-2)计算作用于D6号桥墩各质点的横桥向水平地震力:kNE kN E kN E hp hp hp 40.26.247586.0011.1629.01.02.03.156.44.328839.0011.1629.01.02.03.155.1336.8077011.1629.01.02.03.1210=⨯⨯⨯⨯⨯⨯==⨯⨯⨯⨯⨯⨯==⨯⨯⨯⨯⨯= 6、计算横桥向地震剪力和弯矩D6号墩墩底的横桥向地震剪力和弯矩分别如下:kN Q D 51.14040.256.455.1336=++=m kN M D ⋅=⨯+⨯+⨯=34.598137.140.2334.356.4346.455.1336。

桥梁抗震设计示例

桥梁抗震设计示例

M R 5998 kN.m M
可见,墩柱截面的抗弯强度不满足要求,必须增加截面纵筋。在截面上下缘各增加一
排 1025 钢筋,如图 7.5 所示。根据计算,该截面的抗弯强度为:
M R 7641 kN.m M
图7.5 截面配筋修改图
桥梁抗震
7.2.3 固定墩的延性能力检算
1 固定墩的延性需求确定 Cz 0.3
7.2.1 设计地震力计算
ms
P CiCz (KhGs id Ri )
K
图7.4 自振特性计算简图
桥梁抗震
K
2
3EI l3
3 3.0 1.35 1.53
2
12 73
1.99 105 kN / m
(I 偏安全考虑,不折减)
体系的自振周期为:
反应谱值:
T 2
ms 2 K
3899 .5 1.99 105
设计
• 减隔震概念设计 • 两种对策比较
桥梁抗震
7.1 基本设计资料
图7.1 某一联高架桥立面图(单位:cm)
中墩每一立柱顶设置一个固 定盆式支座,其它立柱顶设 置单向活动盆式支座。
桥梁上部结构的质量为:
图7.2 某一联高架桥横断面图(单位:cm)
ms (0.618 2.5 0.13 18 2.5 2.6) 110 35.45 110 3899 .5t
0.00427

f’l2 /f’பைடு நூலகம்
co
对于矩形截面,有效约束系数K e 可取为 0.75,则:
f 'lx / f 'c K e x f yh / f 'c 0.75 0.00427 340 /(30 0.85) 0.0427

桥梁抗震计算书学习

桥梁抗震计算书学习

计算书计算 : XXX校核 : XXX审核 : XXX二零零九年一二月1. 设计规范1.1. 公路工程技术标准(JTG B01-2003)1.2. 公路桥涵设计通用规范 (JTG D60-2004)1.3. 公路钢筋混凝土及预应力混凝土桥涵设计规范 (JTG D62-2004) 1.4. 公路桥涵地基与基础设计规范(JTG D63-2007)1.5. 公路桥梁抗震设计细则(JTG/TB 02-01-2008)2. 设计资料2.1. 使用程序 : MIDAS/Civil, Civil 2006 ( Release No. 1 )2.2. 截面设计内力 : 3D2.3. 构件类型 : 普通混凝土桥梁2.4. 地震作用等级 : E1作用3. 主要材料指标3.1. 混凝土3.2. 普通钢筋4. 模型简介4.1. 单元数量 : 单元 42 个4.2. 节点数量 : 112 个4.3. 边界条件数量 : 6 个4.4. 施工阶段 : 1 个5. 荷载组合说明5.1. 荷载工况说明5.1.1. 静力荷载工况5.1.2. 反应谱荷载工况5.2. 荷载组合说明5.2.1. 荷载工况名称5.2.2. 荷载组合6. 验算结果表格6.1. 桥墩单元强度验算6.2. 盖梁强度-抗弯验算6.3. 盖梁强度-抗剪验算6.4. 基础强度验算验算6.5. 支座厚度(板式橡胶)验算6.6. 支座抗滑稳定性(板式橡胶支座)验算6.7. 支座水平承载力(固定盆式支座)验算7. 抗震详细计算过程7.1. 桥墩单元强度验算:44单元i截面使用阶段正截面轴心抗压承载能力验算:截面偏心矩为0,做轴心抗压承载力验算:γ0*Nd = 826885.72 NNn = 0.90φ(fcdA+fsd'As')=0.90*1.00*(13.80*1767145.87+280.00*12568.00) = 25115087.68 Nγ0*Nd ≤ 0.90φ(fcdA+fsd'As')[5.3.1],轴心受压满足要求. OK--------------------------------------------------------------------------44单元i截面Fx最小时(My)的偏心受压验算:e0 = Md/Nd = 24577517.71/826885.72 = 29.72 mme = ηe0+h/2-as = 1.00*29.72+1500.00/2-150.00 = 629.72 mme' = ηe0+as'-h/2 = 1.00*29.72+150.00-1500.00/2 = -570.28 mmNd = 826885.72 N, γ0Nd = 826885.72 N.γ0Nde = 520708950.58 N.mm,γ0Nde' = -471553915.15 N.mmA= 3.10; B= 0.04; C=2.85; D=0.26Nn1 = Ar^2fcd+Cρr^2fsd [5.3.9-1]'= 3.10*750.00*750.00*13.80+2.85*0.003556*750.00*750.00*280.00 =25657645.04 NNn2 = (Br^3fcd+Dρgr^3fsd')/e0/η [5.3.9-2]= (0.04*750.00^3*13.80+0.26*0.003556*0.80*750.00^3*280.00)/29.72/1.00 = 10764877.58 Nγ0Nd ≤ Nn, 偏心受压满足验算要求. OK.--------------------------------------------------------------------------44单元i截面My最大时的偏心受压验算:e0 = Md/Nd = 24577517.71/826885.72 = 29.72 mme = ηe0+h/2-as = 1.00*29.72+1500.00/2-150.00 = 629.72 mme' = ηe0+as'-h/2 = 1.00*29.72+150.00-1500.00/2 = -570.28 mmNd = 826885.72 N, γ0Nd = 826885.72 N.γ0Nde = 520708950.58 N.mm,γ0Nde' = -471553915.15 N.mmA= 3.10; B= 0.04; C=2.85; D=0.26Nn1 = Ar^2fcd+Cρr^2fsd [5.3.9-1]'= 3.10*750.00*750.00*13.80+2.85*0.003556*750.00*750.00*280.00 = 25657645.04 NNn2 = (Br^3fcd+Dρgr^3fsd')/e0/η [5.3.9-2]= (0.04*750.00^3*13.80+0.26*0.003556*0.80*750.00^3*280.00)/29.72/1.00 = 10764877.58 Nγ0Nd ≤ Nn, 偏心受压满足验算要求. OK.--------------------------------------------------------------------------44单元i截面My最小时的偏心受压验算:e0 = Md/Nd = 24358340.95/1266564.36 = 19.23 mme = ηe0+h/2-as = 1.00*19.23+1500.00/2-150.00 = 619.23 mme' = ηe0+as'-h/2 = 1.00*19.23+150.00-1500.00/2 = -580.77 mmNd = 1266564.36 N, γ0Nd = 1266564.36 N.γ0Nde = 784296959.88 N.mm,γ0Nde' = -735580277.99 N.mmA= 3.10; B= 0.04; C=2.85; D=0.26Nn1 = Ar^2fcd+Cρr^2fsd [5.3.9-1]'= 3.10*750.00*750.00*13.80+2.85*0.003556*750.00*750.00*280.00 = 25657645.04 NNn2 = (Br^3fcd+Dρgr^3fsd')/e0/η [5.3.9-2]= (0.04*750.00^3*13.80+0.26*0.003556*0.80*750.00^3*280.00)/19.23/1.00 = 16637236.11 Nγ0Nd ≤ Nn, 偏心受压满足验算要求. OK.--------------------------------------------------------------------------44单元j截面使用阶段正截面轴心抗压承载能力验算:截面偏心矩为0,做轴心抗压承载力验算:γ0*Nd = 826885.72 NNn = 0.90φ(fcdA+fsd'As')=0.90*1.00*(13.80*1767145.87+280.00*12568.00) = 25115087.68 Nγ0*Nd ≤ 0.90φ(fcdA+fsd'As')[5.3.1],轴心受压满足要求. OK--------------------------------------------------------------------------44单元j截面Fx最小时(My)的偏心受压验算:e0 = Md/Nd = 43359333.77/826885.72 = 52.44 mme = ηe0+h/2-as = 1.00*52.44+1500.00/2-150.00 = 652.44 mme' = ηe0+as'-h/2 = 1.00*52.44+150.00-1500.00/2 = -547.56 mmNd = 826885.72 N, γ0Nd = 826885.72 N.γ0Nde = 539490766.63 N.mm,γ0Nde' = -452772099.09 N.mmA= 2.93; B= 0.18; C=2.69; D=0.39Nn1 = Ar^2fcd+Cρr^2fsd [5.3.9-1]'= 2.93*750.00*750.00*13.80+2.69*0.003556*750.00*750.00*280.00 = 24232887.73 NNn2 = (Br^3fcd+Dρgr^3fsd')/e0/η [5.3.9-2]= (0.18*750.00^3*13.80+0.39*0.003556*0.80*750.00^3*280.00)/52.44/1.00 = 22837544.68 Nγ0Nd ≤ Nn, 偏心受压满足验算要求. OK.--------------------------------------------------------------------------44单元j截面My最大时的偏心受压验算:e0 = Md/Nd = 43359333.77/826885.72 = 52.44 mme = ηe0+h/2-as = 1.00*52.44+1500.00/2-150.00 = 652.44 mme' = ηe0+as'-h/2 = 1.00*52.44+150.00-1500.00/2 = -547.56 mmNd = 826885.72 N, γ0Nd = 826885.72 N.γ0Nde = 539490766.63 N.mm,γ0Nde' = -452772099.09 N.mmA= 2.93; B= 0.18; C=2.69; D=0.39Nn1 = Ar^2fcd+Cρr^2fsd [5.3.9-1]'= 2.93*750.00*750.00*13.80+2.69*0.003556*750.00*750.00*280.00 = 24232887.73 NNn2 = (Br^3fcd+Dρgr^3fsd')/e0/η [5.3.9-2]= (0.18*750.00^3*13.80+0.39*0.003556*0.80*750.00^3*280.00)/52.44/1.00 = 22837544.68 Nγ0Nd ≤ Nn, 偏心受压满足验算要求. OK.--------------------------------------------------------------------------44单元j截面My最小时的偏心受压验算:e0 = Md/Nd = 42994039.16/1266564.36 = 33.95 mme = ηe0+h/2-as = 1.00*33.95+1500.00/2-150.00 = 633.95 mme' = ηe0+as'-h/2 = 1.00*33.95+150.00-1500.00/2 = -566.05 mmNd = 1266564.36 N, γ0Nd = 1266564.36 N.γ0Nde = 802932658.10 N.mm,γ0Nde' = -716944579.77 N.mmA= 2.93; B= 0.18; C=2.69; D=0.39Nn1 = Ar^2fcd+Cρr^2fsd [5.3.9-1]'= 2.93*750.00*750.00*13.80+2.69*0.003556*750.00*750.00*280.00 = 24232887.73 NNn2 = (Br^3fcd+Dρgr^3fsd')/e0/η [5.3.9-2]= (0.18*750.00^3*13.80+0.39*0.003556*0.80*750.00^3*280.00)/33.95/1.00 = 35278128.28 Nγ0Nd ≤ Nn, 偏心受压满足验算要求. OK.--------------------------------------------------------------------------43单元i截面使用阶段正截面轴心抗压承载能力验算:截面偏心矩为0,做轴心抗压承载力验算:γ0*Nd = 826811.74 NNn = 0.90φ(fcdA+fsd'As')=0.90*1.00*(13.80*1767145.87+280.00*12568.00) = 25115087.68 Nγ0*Nd ≤ 0.90φ(fcdA+fsd'As')[5.3.1],轴心受压满足要求. OK--------------------------------------------------------------------------43单元i截面Fx最小时(My)的偏心受压验算:e0 = Md/Nd = 43376544.70/826811.74 = 52.46 mme = ηe0+h/2-as = 1.00*52.46+1500.00/2-150.00 = 652.46 mme' = ηe0+as'-h/2 = 1.00*52.46+150.00-1500.00/2 = -547.54 mmNd = 826811.74 N, γ0Nd = 826811.74 N.γ0Nde = 539463590.40 N.mm,γ0Nde' = -452710501.01 N.mmA= 2.93; B= 0.18; C=2.69; D=0.39Nn1 = Ar^2fcd+Cρr^2fsd [5.3.9-1]'= 2.93*750.00*750.00*13.80+2.69*0.003556*750.00*750.00*280.00 = 24232887.73 NNn2 = (Br^3fcd+Dρgr^3fsd')/e0/η [5.3.9-2]= (0.18*750.00^3*13.80+0.39*0.003556*0.80*750.00^3*280.00)/52.46/1.00 = 22826440.83 Nγ0Nd ≤ Nn, 偏心受压满足验算要求. OK.--------------------------------------------------------------------------43单元i截面My最大时的偏心受压验算:e0 = Md/Nd = 43376544.70/826811.74 = 52.46 mme = ηe0+h/2-as = 1.00*52.46+1500.00/2-150.00 = 652.46 mme' = ηe0+as'-h/2 = 1.00*52.46+150.00-1500.00/2 = -547.54 mmNd = 826811.74 N, γ0Nd = 826811.74 N.γ0Nde = 539463590.40 N.mm,γ0Nde' = -452710501.01 N.mmA= 2.93; B= 0.18; C=2.69; D=0.39Nn1 = Ar^2fcd+Cρr^2fsd [5.3.9-1]'= 2.93*750.00*750.00*13.80+2.69*0.003556*750.00*750.00*280.00 = 24232887.73 NNn2 = (Br^3fcd+Dρgr^3fsd')/e0/η [5.3.9-2]= (0.18*750.00^3*13.80+0.39*0.003556*0.80*750.00^3*280.00)/52.46/1.00 = 22826440.83 Nγ0Nd ≤ Nn, 偏心受压满足验算要求. OK.--------------------------------------------------------------------------43单元i截面My最小时的偏心受压验算:e0 = Md/Nd = 43011250.09/1266638.34 = 33.96 mme = ηe0+h/2-as = 1.00*33.96+1500.00/2-150.00 = 633.96 mme' = ηe0+as'-h/2 = 1.00*33.96+150.00-1500.00/2 = -566.04 mmNd = 1266638.34 N, γ0Nd = 1266638.34 N.γ0Nde = 802994256.18 N.mm,γ0Nde' = -716971756.01 N.mmA= 2.93; B= 0.18; C=2.69; D=0.39Nn1 = Ar^2fcd+Cρr^2fsd [5.3.9-1]'= 2.93*750.00*750.00*13.80+2.69*0.003556*750.00*750.00*280.00 = 24232887.73 NNn2 = (Br^3fcd+Dρgr^3fsd')/e0/η [5.3.9-2]= (0.18*750.00^3*13.80+0.39*0.003556*0.80*750.00^3*280.00)/33.96/1.00 = 35266071.49 Nγ0Nd ≤ Nn, 偏心受压满足验算要求. OK.--------------------------------------------------------------------------43单元j截面使用阶段正截面轴心抗压承载能力验算:截面偏心矩为0,做轴心抗压承载力验算:γ0*Nd = 826811.74 NNn = 0.90φ(fcdA+fsd'As')=0.90*1.00*(13.80*1767145.87+280.00*12568.00) = 25115087.68 Nγ0*Nd ≤ 0.90φ(fcdA+fsd'As')[5.3.1],轴心受压满足要求. OK--------------------------------------------------------------------------43单元j截面Fx最小时(My)的偏心受压验算:e0 = Md/Nd = 62759235.47/826811.74 = 75.91 mme = ηe0+h/2-as = 1.00*75.91+1500.00/2-150.00 = 675.91 mme' = ηe0+as'-h/2 = 1.00*75.91+150.00-1500.00/2 = -524.09 mmNd = 826811.74 N, γ0Nd = 826811.74 N.γ0Nde = 558846281.17 N.mm,γ0Nde' = -433327810.24 N.mmA= 2.69; B= 0.34; C=2.46; D=0.56Nn1 = Ar^2fcd+Cρr^2fsd [5.3.9-1]'= 2.69*750.00*750.00*13.80+2.46*0.003556*750.00*750.00*280.00 = 22293535.84 NNn2 = (Br^3fcd+Dρgr^3fsd')/e0/η [5.3.9-2]= (0.34*750.00^3*13.80+0.56*0.003556*0.80*750.00^3*280.00)/75.91/1.00 = 28672632.86 Nγ0Nd ≤ Nn, 偏心受压满足验算要求. OK.--------------------------------------------------------------------------43单元j截面My最大时的偏心受压验算:e0 = Md/Nd = 62759235.47/826811.74 = 75.91 mme = ηe0+h/2-as = 1.00*75.91+1500.00/2-150.00 = 675.91 mme' = ηe0+as'-h/2 = 1.00*75.91+150.00-1500.00/2 = -524.09 mm Nd = 826811.74 N, γ0Nd = 826811.74 N.γ0Nde = 558846281.17 N.mm,γ0Nde' = -433327810.24 N.mmA= 2.69; B= 0.34; C=2.46; D=0.56Nn1 = Ar^2fcd+Cρr^2fsd [5.3.9-1]'= 2.69*750.00*750.00*13.80+2.46*0.003556*750.00*750.00*280.00 = 22293535.84 NNn2 = (Br^3fcd+Dρgr^3fsd')/e0/η [5.3.9-2]= (0.34*750.00^3*13.80+0.56*0.003556*0.80*750.00^3*280.00)/75.91/1.00 = 28672632.86 Nγ0Nd ≤ Nn, 偏心受压满足验算要求. OK.--------------------------------------------------------------------------43单元j截面My最小时的偏心受压验算:e0 = Md/Nd = 62247823.02/1266638.34 = 49.14 mme = ηe0+h/2-as = 1.00*49.14+1500.00/2-150.00 = 649.14 mme' = ηe0+as'-h/2 = 1.00*49.14+150.00-1500.00/2 = -550.86 mmNd = 1266638.34 N, γ0Nd = 1266638.34 N.γ0Nde = 822230829.11 N.mm,γ0Nde' = -697735183.08 N.mmA= 2.93; B= 0.18; C=2.69; D=0.39Nn1 = Ar^2fcd+Cρr^2fsd [5.3.9-1]'= 2.93*750.00*750.00*13.80+2.69*0.003556*750.00*750.00*280.00 = 24232887.73 NNn2 = (Br^3fcd+Dρgr^3fsd')/e0/η [5.3.9-2]= (0.18*750.00^3*13.80+0.39*0.003556*0.80*750.00^3*280.00)/49.14/1.00 = 24367724.80 Nγ0Nd ≤ Nn, 偏心受压满足验算要求. OK.--------------------------------------------------------------------------42单元i截面使用阶段正截面轴心抗压承载能力验算:截面偏心矩为0,做轴心抗压承载力验算:γ0*Nd = 826743.24 NNn = 0.90φ(fcdA+fsd'As')=0.90*1.00*(13.80*1767145.87+280.00*12568.00) = 25115087.68 Nγ0*Nd ≤ 0.90φ(fcdA+fsd'As')[5.3.1],轴心受压满足要求. OK--------------------------------------------------------------------------42单元i截面Fx最小时(My)的偏心受压验算:e0 = Md/Nd = 62775441.91/826743.24 = 75.93 mme = ηe0+h/2-as = 1.00*75.93+1500.00/2-150.00 = 675.93 mme' = ηe0+as'-h/2 = 1.00*75.93+150.00-1500.00/2 = -524.07 mmNd = 826743.24 N, γ0Nd = 826743.24 N.γ0Nde = 558821385.89 N.mm,γ0Nde' = -433270502.07 N.mmA= 2.69; B= 0.34; C=2.46; D=0.56Nn1 = Ar^2fcd+Cρr^2fsd [5.3.9-1]'= 2.69*750.00*750.00*13.80+2.46*0.003556*750.00*750.00*280.00 = 22293535.84 NNn2 = (Br^3fcd+Dρgr^3fsd')/e0/η [5.3.9-2]= (0.34*750.00^3*13.80+0.56*0.003556*0.80*750.00^3*280.00)/75.93/1.00 = 28662855.61 Nγ0Nd ≤ Nn, 偏心受压满足验算要求. OK.--------------------------------------------------------------------------42单元i截面My最大时的偏心受压验算:e0 = Md/Nd = 62775441.91/826743.24 = 75.93 mme = ηe0+h/2-as = 1.00*75.93+1500.00/2-150.00 = 675.93 mme' = ηe0+as'-h/2 = 1.00*75.93+150.00-1500.00/2 = -524.07 mmNd = 826743.24 N, γ0Nd = 826743.24 N.γ0Nde = 558821385.89 N.mm,γ0Nde' = -433270502.07 N.mmA= 2.69; B= 0.34; C=2.46; D=0.56Nn1 = Ar^2fcd+Cρr^2fsd [5.3.9-1]'= 2.69*750.00*750.00*13.80+2.46*0.003556*750.00*750.00*280.00 = 22293535.84 NNn2 = (Br^3fcd+Dρgr^3fsd')/e0/η [5.3.9-2]= (0.34*750.00^3*13.80+0.56*0.003556*0.80*750.00^3*280.00)/75.93/1.00 = 28662855.61 Nγ0Nd ≤ Nn, 偏心受压满足验算要求. OK.--------------------------------------------------------------------------42单元i截面My最小时的偏心受压验算:e0 = Md/Nd = 62264029.46/1266706.85 = 49.15 mme = ηe0+h/2-as = 1.00*49.15+1500.00/2-150.00 = 649.15 mme' = ηe0+as'-h/2 = 1.00*49.15+150.00-1500.00/2 = -550.85 mmNd = 1266706.85 N, γ0Nd = 1266706.85 N.γ0Nde = 822288137.28 N.mm,γ0Nde' = -697760078.36 N.mmA= 2.93; B= 0.18; C=2.69; D=0.39Nn1 = Ar^2fcd+Cρr^2fsd [5.3.9-1]'= 2.93*750.00*750.00*13.80+2.69*0.003556*750.00*750.00*280.00 = 24232887.73 NNn2 = (Br^3fcd+Dρgr^3fsd')/e0/η [5.3.9-2]= (0.18*750.00^3*13.80+0.39*0.003556*0.80*750.00^3*280.00)/49.15/1.00 = 24362699.75 Nγ0Nd ≤ Nn, 偏心受压满足验算要求. OK.--------------------------------------------------------------------------42单元j截面使用阶段正截面轴心抗压承载能力验算:截面偏心矩为0,做轴心抗压承载力验算:γ0*Nd = 826743.24 NNn = 0.90φ(fcdA+fsd'As')=0.90*1.00*(13.80*1767145.87+280.00*12568.00) = 25115087.68 Nγ0*Nd ≤ 0.90φ(fcdA+fsd'As')[5.3.1],轴心受压满足要求. OK--------------------------------------------------------------------------42单元j截面Fx最小时(My)的偏心受压验算:e0 = Md/Nd = 82519462.77/826743.24 = 99.81 mme = ηe0+h/2-as = 1.00*99.81+1500.00/2-150.00 = 699.81 mme' = ηe0+as'-h/2 = 1.00*99.81+150.00-1500.00/2 = -500.19 mmNd = 826743.24 N, γ0Nd = 826743.24 N.γ0Nde = 578565406.75 N.mm,γ0Nde' = -413526481.21 N.mmA= 2.69; B= 0.34; C=2.46; D=0.56Nn1 = Ar^2fcd+Cρr^2fsd [5.3.9-1]'= 2.69*750.00*750.00*13.80+2.46*0.003556*750.00*750.00*280.00 = 22293535.84 NNn2 = (Br^3fcd+Dρgr^3fsd')/e0/η [5.3.9-2]= (0.34*750.00^3*13.80+0.56*0.003556*0.80*750.00^3*280.00)/99.81/1.00 = 21804836.91 Nγ0Nd ≤ Nn, 偏心受压满足验算要求. OK.--------------------------------------------------------------------------42单元j截面My最大时的偏心受压验算:e0 = Md/Nd = 82519462.77/826743.24 = 99.81 mme = ηe0+h/2-as = 1.00*99.81+1500.00/2-150.00 = 699.81 mme' = ηe0+as'-h/2 = 1.00*99.81+150.00-1500.00/2 = -500.19 mmNd = 826743.24 N, γ0Nd = 826743.24 N.γ0Nde = 578565406.75 N.mm,γ0Nde' = -413526481.21 N.mmA= 2.69; B= 0.34; C=2.46; D=0.56Nn1 = Ar^2fcd+Cρr^2fsd [5.3.9-1]'= 2.69*750.00*750.00*13.80+2.46*0.003556*750.00*750.00*280.00 = 22293535.84 NNn2 = (Br^3fcd+Dρgr^3fsd')/e0/η [5.3.9-2]= (0.34*750.00^3*13.80+0.56*0.003556*0.80*750.00^3*280.00)/99.81/1.00 = 21804836.91 Nγ0Nd ≤ Nn, 偏心受压满足验算要求. OK.--------------------------------------------------------------------------42单元j截面My最小时的偏心受压验算:e0 = Md/Nd = 81861932.47/1266706.85 = 64.63 mme = ηe0+h/2-as = 1.00*64.63+1500.00/2-150.00 = 664.63 mme' = ηe0+as'-h/2 = 1.00*64.63+150.00-1500.00/2 = -535.37 mmNd = 1266706.85 N, γ0Nd = 1266706.85 N.γ0Nde = 841886040.29 N.mm,γ0Nde' = -678162175.35 N.mmA= 2.93; B= 0.18; C=2.69; D=0.39Nn1 = Ar^2fcd+Cρr^2fsd [5.3.9-1]'= 2.93*750.00*750.00*13.80+2.69*0.003556*750.00*750.00*280.00 = 24232887.73 NNn2 = (Br^3fcd+Dρgr^3fsd')/e0/η [5.3.9-2]= (0.18*750.00^3*13.80+0.39*0.003556*0.80*750.00^3*280.00)/64.63/1.00 = 18530222.89 Nγ0Nd ≤ Nn, 偏心受压满足验算要求. OK.--------------------------------------------------------------------------41单元i截面使用阶段正截面轴心抗压承载能力验算:截面偏心矩为0,做轴心抗压承载力验算:γ0*Nd = 826680.22 NNn = 0.90φ(fcdA+fsd'As')=0.90*1.00*(13.80*1767145.87+280.00*12568.00) = 25115087.68 Nγ0*Nd ≤ 0.90φ(fcdA+fsd'As')[5.3.1],轴心受压满足要求. OK--------------------------------------------------------------------------41单元i截面Fx最小时(My)的偏心受压验算:e0 = Md/Nd = 82533710.52/826680.22 = 99.84 mme = ηe0+h/2-as = 1.00*99.84+1500.00/2-150.00 = 699.84 mme' = ηe0+as'-h/2 = 1.00*99.84+150.00-1500.00/2 = -500.16 mmNd = 826680.22 N, γ0Nd = 826680.22 N.γ0Nde = 578541841.83 N.mm,γ0Nde' = -413474420.79 N.mmA= 2.69; B= 0.34; C=2.46; D=0.56Nn1 = Ar^2fcd+Cρr^2fsd [5.3.9-1]'= 2.69*750.00*750.00*13.80+2.46*0.003556*750.00*750.00*280.00 = 22293535.84 NNn2 = (Br^3fcd+Dρgr^3fsd')/e0/η [5.3.9-2]= (0.34*750.00^3*13.80+0.56*0.003556*0.80*750.00^3*280.00)/99.84/1.00 = 21799410.90 Nγ0Nd ≤ Nn, 偏心受压满足验算要求. OK.--------------------------------------------------------------------------41单元i截面My最大时的偏心受压验算:e0 = Md/Nd = 82533710.52/826680.22 = 99.84 mme = ηe0+h/2-as = 1.00*99.84+1500.00/2-150.00 = 699.84 mme' = ηe0+as'-h/2 = 1.00*99.84+150.00-1500.00/2 = -500.16 mmNd = 826680.22 N, γ0Nd = 826680.22 N.γ0Nde = 578541841.83 N.mm,γ0Nde' = -413474420.79 N.mmA= 2.69; B= 0.34; C=2.46; D=0.56Nn1 = Ar^2fcd+Cρr^2fsd [5.3.9-1]'= 2.69*750.00*750.00*13.80+2.46*0.003556*750.00*750.00*280.00 = 22293535.84 NNn2 = (Br^3fcd+Dρgr^3fsd')/e0/η [5.3.9-2]= (0.34*750.00^3*13.80+0.56*0.003556*0.80*750.00^3*280.00)/99.84/1.00 = 21799410.90 Nγ0Nd ≤ Nn, 偏心受压满足验算要求. OK.--------------------------------------------------------------------------41单元i截面My最小时的偏心受压验算:e0 = Md/Nd = 81876180.23/1266769.87 = 64.63 mme = ηe0+h/2-as = 1.00*64.63+1500.00/2-150.00 = 664.63 mme' = ηe0+as'-h/2 = 1.00*64.63+150.00-1500.00/2 = -535.37 mmNd = 1266769.87 N, γ0Nd = 1266769.87 N.γ0Nde = 841938100.72 N.mm,γ0Nde' = -678185740.26 N.mmA= 2.93; B= 0.18; C=2.69; D=0.39Nn1 = Ar^2fcd+Cρr^2fsd [5.3.9-1]'= 2.93*750.00*750.00*13.80+2.69*0.003556*750.00*750.00*280.00 = 24232887.73 NNn2 = (Br^3fcd+Dρgr^3fsd')/e0/η [5.3.9-2]= (0.18*750.00^3*13.80+0.39*0.003556*0.80*750.00^3*280.00)/64.63/1.00 = 18527920.09 Nγ0Nd ≤ Nn, 偏心受压满足验算要求. OK.--------------------------------------------------------------------------41单元j截面使用阶段正截面轴心抗压承载能力验算:截面偏心矩为0,做轴心抗压承载力验算:γ0*Nd = 826680.22 NNn = 0.90φ(fcdA+fsd'As')=0.90*1.00*(13.80*1767145.87+280.00*12568.00) = 25115087.68 Nγ0*Nd ≤ 0.90φ(fcdA+fsd'As')[5.3.1],轴心受压满足要求. OK--------------------------------------------------------------------------41单元j截面Fx最小时(My)的偏心受压验算:e0 = Md/Nd = 102544893.63/826680.22 = 124.04 mme = ηe0+h/2-as = 1.00*124.04+1500.00/2-150.00 = 724.04 mme' = ηe0+as'-h/2 = 1.00*124.04+150.00-1500.00/2 = -475.96 mmNd = 826680.22 N, γ0Nd = 826680.22 N.γ0Nde = 598553024.94 N.mm,γ0Nde' = -393463237.67 N.mmA= 2.69; B= 0.34; C=2.46; D=0.56Nn1 = Ar^2fcd+Cρr^2fsd [5.3.9-1]'= 2.69*750.00*750.00*13.80+2.46*0.003556*750.00*750.00*280.00 = 22293535.84 NNn2 = (Br^3fcd+Dρgr^3fsd')/e0/η [5.3.9-2]= (0.34*750.00^3*13.80+0.56*0.003556*0.80*750.00^3*280.00)/124.04/1.00 = 17545352.14 Nγ0Nd ≤ Nn, 偏心受压满足验算要求. OK.--------------------------------------------------------------------------41单元j截面My最大时的偏心受压验算:e0 = Md/Nd = 102544893.63/826680.22 = 124.04 mme = ηe0+h/2-as = 1.00*124.04+1500.00/2-150.00 = 724.04 mme' = ηe0+as'-h/2 = 1.00*124.04+150.00-1500.00/2 = -475.96 mmNd = 826680.22 N, γ0Nd = 826680.22 N.γ0Nde = 598553024.94N.mm,γ0Nde' = -393463237.67 N.mmA= 2.69; B= 0.34; C=2.46; D=0.56Nn1 = Ar^2fcd+Cρr^2fsd [5.3.9-1]'= 2.69*750.00*750.00*13.80+2.46*0.003556*750.00*750.00*280.00 = 22293535.84 NNn2 = (Br^3fcd+Dρgr^3fsd')/e0/η [5.3.9-2]= (0.34*750.00^3*13.80+0.56*0.003556*0.80*750.00^3*280.00)/124.04/1.00 = 17545352.14 Nγ0Nd ≤ Nn, 偏心受压满足验算要求. OK.--------------------------------------------------------------------------41单元j截面My最小时的偏心受压验算:e0 = Md/Nd = 101741245.50/1266769.87 = 80.32 mme = ηe0+h/2-as = 1.00*80.32+1500.00/2-150.00 = 680.32 mme' = ηe0+as'-h/2 = 1.00*80.32+150.00-1500.00/2 = -519.68 mmNd = 1266769.87 N, γ0Nd = 1266769.87 N.γ0Nde = 861803165.99 N.mm,γ0Nde' = -658320675.00 N.mmA= 2.69; B= 0.34; C=2.46; D=0.56Nn1 = Ar^2fcd+Cρr^2fsd [5.3.9-1]'= 2.69*750.00*750.00*13.80+2.46*0.003556*750.00*750.00*280.00 = 22293535.84 NNn2 = (Br^3fcd+Dρgr^3fsd')/e0/η [5.3.9-2]= (0.34*750.00^3*13.80+0.56*0.003556*0.80*750.00^3*280.00)/80.32/1.00 = 27098125.92 Nγ0Nd ≤ Nn, 偏心受压满足验算要求. OK.--------------------------------------------------------------------------40单元i截面使用阶段正截面轴心抗压承载能力验算:截面偏心矩为0,做轴心抗压承载力验算:γ0*Nd = 826622.69 NNn = 0.90φ(fcdA+fsd'As')=0.90*1.00*(13.80*1767145.87+280.00*12568.00) = 25115087.68 Nγ0*Nd ≤ 0.90φ(fcdA+fsd'As')[5.3.1],轴心受压满足要求. OK--------------------------------------------------------------------------40单元i截面Fx最小时(My)的偏心受压验算:e0 = Md/Nd = 102556851.27/826622.69 = 124.07 mme = ηe0+h/2-as = 1.00*124.07+1500.00/2-150.00 = 724.07 mme' = ηe0+as'-h/2 = 1.00*124.07+150.00-1500.00/2 = -475.93 mmNd = 826622.69 N, γ0Nd = 826622.69 N.γ0Nde = 598530462.30 N.mm,γ0Nde' = -393416759.77 N.mmA= 2.69; B= 0.34; C=2.46; D=0.56Nn1 = Ar^2fcd+Cρr^2fsd [5.3.9-1]'= 2.69*750.00*750.00*13.80+2.46*0.003556*750.00*750.00*280.00 = 22293535.84 NNn2 = (Br^3fcd+Dρgr^3fsd')/e0/η [5.3.9-2]= (0.34*750.00^3*13.80+0.56*0.003556*0.80*750.00^3*280.00)/124.07/1.00 = 17542085.48 Nγ0Nd ≤ Nn, 偏心受压满足验算要求. OK.--------------------------------------------------------------------------40单元i截面My最大时的偏心受压验算:e0 = Md/Nd = 102556851.27/826622.69 = 124.07 mme = ηe0+h/2-as = 1.00*124.07+1500.00/2-150.00 = 724.07 mme' = ηe0+as'-h/2 = 1.00*124.07+150.00-1500.00/2 = -475.93 mmNd = 826622.69 N, γ0Nd = 826622.69 N.γ0Nde = 598530462.30 N.mm,γ0Nde' = -393416759.77 N.mmA= 2.69; B= 0.34; C=2.46; D=0.56Nn1 = A r^2fcd+Cρr^2fsd [5.3.9-1]'= 2.69*750.00*750.00*13.80+2.46*0.003556*750.00*750.00*280.00 = 22293535.84 NNn2 = (Br^3fcd+Dρgr^3fsd')/e0/η [5.3.9-2]= (0.34*750.00^3*13.80+0.56*0.003556*0.80*750.00^3*280.00)/124.07/1.00 = 17542085.48 Nγ0Nd ≤ Nn, 偏心受压满足验算要求. OK.--------------------------------------------------------------------------40单元i截面My最小时的偏心受压验算:e0 = Md/Nd = 101753203.13/1266827.40 = 80.32 mme = ηe0+h/2-as = 1.00*80.32+1500.00/2-150.00 = 680.32 mme' = ηe0+as'-h/2 = 1.00*80.32+150.00-1500.00/2 = -519.68 mmNd = 1266827.40 N, γ0Nd = 1266827.40 N.γ0Nde = 861849643.90 N.mm,γ0Nde' = -658343237.64 N.mmA= 2.69; B= 0.34; C=2.46; D=0.56Nn1 = Ar^2fcd+Cρr^2fsd [5.3.9-1]'= 2.69*750.00*750.00*13.80+2.46*0.003556*750.00*750.00*280.00 = 22293535.84 NNn2 = (Br^3fcd+Dρgr^3fsd')/e0/η [5.3.9-2]= (0.34*750.00^3*13.80+0.56*0.003556*0.80*750.00^3*280.00)/80.32/1.00 = 27096172.05 Nγ0Nd ≤ Nn, 偏心受压满足验算要求. OK.--------------------------------------------------------------------------40单元j截面使用阶段正截面轴心抗压承载能力验算:截面偏心矩为0,做轴心抗压承载力验算:γ0*Nd = 826622.69 NNn = 0.90φ(fcdA+fsd'As')=0.90*1.00*(13.80*1767145.87+280.00*12568.00) = 25115087.68 Nγ0*Nd ≤ 0.90φ(f cdA+fsd'As')[5.3.1],轴心受压满足要求. OK--------------------------------------------------------------------------40单元j截面Fx最小时(My)的偏心受压验算:e0 = Md/Nd = 122784694.60/826622.69 = 148.54 mme = ηe0+h/2-as = 1.00*148.54+1500.00/2-150.00 = 748.54 mme' = ηe0+as'-h/2 = 1.00*148.54+150.00-1500.00/2 = -451.46 mmNd = 826622.69 N, γ0Nd = 826622.69 N.γ0Nde = 618758305.63 N.mm,γ0Nde' = -373188916.43 N.mmA= 2.42; B= 0.48; C=2.17; D=0.78Nn1 = Ar^2fcd+Cρr^2fsd [5.3.9-1]'= 2.42*750.00*750.00*13.80+2.17*0.003556*750.00*750.00*280.00 = 20011066.94 NNn2 = (Br^3fcd+Dρgr^3fsd')/e0/η [5.3.9-2]= (0.48*750.00^3*13.80+0.78*0.003556*0.80*750.00^3*280.00)/148.54/1.00 = 20675473.17 Nγ0Nd ≤ Nn, 偏心受压满足验算要求. OK.--------------------------------------------------------------------------40单元j截面My最大时的偏心受压验算:e0 = Md/Nd = 122784694.60/826622.69 = 148.54 mme = ηe0+h/2-as = 1.00*148.54+1500.00/2-150.00 = 748.54 mme' = ηe0+as'-h/2 = 1.00*148.54+150.00-1500.00/2 = -451.46 mmNd = 826622.69 N, γ0Nd = 826622.69 N.γ0Nde = 618758305.63 N.mm,γ0Nde' = -373188916.43 N.mmA= 2.42; B= 0.48; C=2.17; D=0.78Nn1 = Ar^2fcd+Cρr^2fsd [5.3.9-1]'= 2.42*750.00*750.00*13.80+2.17*0.003556*750.00*750.00*280.00 = 20011066.94 NNn2 = (Br^3fcd+Dρgr^3fsd')/e0/η [5.3.9-2]= (0.48*750.00^3*13.80+0.78*0.003556*0.80*750.00^3*280.00)/148.54/1.00 = 20675473.17 Nγ0Nd ≤ Nn, 偏心受压满足验算要求. OK.--------------------------------------------------------------------------40单元j截面My最小时的偏心受压验算:e0 = Md/Nd = 121834928.62/1266827.40 = 96.17 mme = ηe0+h/2-as = 1.00*96.17+1500.00/2-150.00 = 696.17 mme' = ηe0+as'-h/2 = 1.00*96.17+150.00-1500.00/2 = -503.83 mmNd = 1266827.40 N, γ0Nd = 1266827.40 N.γ0Nde = 881931369.38 N.mm,γ0Nde' = -638261512.15 N.mmA= 2.69; B= 0.34; C=2.46; D=0.56Nn1 = Ar^2fcd+Cρr^2fsd [5.3.9-1]'= 2.69*750.00*750.00*13.80+2.46*0.003556*750.00*750.00*280.00 = 22293535.84 NNn2 = (Br^3fcd+Dρgr^3fsd')/e0/η [5.3.9-2]= (0.34*750.00^3*13.80+0.56*0.003556*0.80*750.00^3*280.00)/96.17/1.00 = 22629982.47 Nγ0Nd ≤ Nn, 偏心受压满足验算要求. OK.--------------------------------------------------------------------------39单元i截面使用阶段正截面轴心抗压承载能力验算:截面偏心矩为0,做轴心抗压承载力验算:γ0*Nd = 826570.64 NNn = 0.90φ(fcdA+fsd'As')=0.90*1.00*(13.80*1767145.87+280.00*12568.00) = 25115087.68 Nγ0*Nd ≤ 0.90φ(fcdA+fsd'As')[5.3.1],轴心受压满足要求. OK--------------------------------------------------------------------------39单元i截面Fx最小时(My)的偏心受压验算:e0 = Md/Nd = 122794418.63/826570.64 = 148.56 mme = ηe0+h/2-as = 1.00*148.56+1500.00/2-150.00 = 748.56 mme' = ηe0+as'-h/2 = 1.00*148.56+150.00-1500.00/2 = -451.44 mmNd = 826570.64 N, γ0Nd = 826570.64 N.γ0Nde = 618736804.84 N.mm,γ0Nde' = -373147967.59 N.mmA= 2.42; B= 0.48; C=2.17; D=0.78Nn1 = Ar^2fcd+Cρr^2fsd [5.3.9-1]'= 2.42*750.00*750.00*13.80+2.17*0.003556*750.00*750.00*280.00 = 20011066.94 NNn2 = (Br^3fcd+Dρgr^3fsd')/e0/η [5.3.9-2]= (0.48*750.00^3*13.80+0.78*0.003556*0.80*750.00^3*280.00)/148.56/1.00 = 20672534.34 Nγ0Nd ≤ Nn, 偏心受压满足验算要求. OK.--------------------------------------------------------------------------39单元i截面My最大时的偏心受压验算:e0 = Md/Nd = 122794418.63/826570.64 = 148.56 mme = ηe0+h/2-as = 1.00*148.56+1500.00/2-150.00 = 748.56 mme' = ηe0+as'-h/2 = 1.00*148.56+150.00-1500.00/2 = -451.44 mmNd = 826570.64 N, γ0Nd = 826570.64 N.γ0Nde = 618736804.84 N.mm,γ0Nde' = -373147967.59 N.mmA= 2.42; B= 0.48; C=2.17; D=0.78Nn1 = Ar^2fcd+Cρr^2fsd [5.3.9-1]'= 2.42*750.00*750.00*13.80+2.17*0.003556*750.00*750.00*280.00 = 20011066.94 NNn2 = (Br^3fcd+Dρgr^3fsd')/e0/η [5.3.9-2]= (0.48*750.00^3*13.80+0.78*0.003556*0.80*750.00^3*280.00)/148.56/1.00 = 20672534.34 Nγ0Nd ≤ Nn, 偏心受压满足验算要求. OK.--------------------------------------------------------------------------39单元i截面My最小时的偏心受压验算:e0 = Md/Nd = 121844652.65/1266879.44 = 96.18 mme = ηe0+h/2-as = 1.00*96.18+1500.00/2-150.00 = 696.18 mme' = ηe0+as'-h/2 = 1.00*96.18+150.00-1500.00/2 = -503.82 mmNd = 1266879.44 N, γ0Nd = 1266879.44 N.γ0Nde = 881972318.23 N.mm,γ0Nde' = -638283012.94 N.mmA= 2.69; B= 0.34; C=2.46; D=0.56Nn1 = Ar^2fcd+Cρr^2fsd [5.3.9-1]'= 2.69*750.00*750.00*13.80+2.46*0.003556*750.00*750.00*280.00 = 22293535.84 NNn2 = (Br^3fcd+Dρgr^3fsd')/e0/η [5.3.9-2]= (0.34*750.00^3*13.80+0.56*0.003556*0.80*750.00^3*280.00)/96.18/1.00 = 22629106.01 Nγ0Nd ≤ Nn, 偏心受压满足验算要求. OK.--------------------------------------------------------------------------39单元j截面使用阶段正截面轴心抗压承载能力验算:截面偏心矩为0,做轴心抗压承载力验算:γ0*Nd = 826570.64 NNn = 0.90φ(fcdA+fs d'As')=0.90*1.00*(13.80*1767145.87+280.00*12568.00) = 25115087.68 Nγ0*Nd ≤ 0.90φ(fcdA+fsd'As')[5.3.1],轴心受压满足要求. OK--------------------------------------------------------------------------39单元j截面Fx最小时(My)的偏心受压验算:e0 = Md/Nd = 143210868.75/826570.64 = 173.26 mme = ηe0+h/2-as = 1.00*173.26+1500.00/2-150.00 = 773.26 mme' = ηe0+as'-h/2 = 1.00*173.26+150.00-1500.00/2 = -426.74 mmNd = 826570.64 N, γ0Nd = 826570.64 N.γ0Nde = 639153254.96 N.mm,γ0Nde' = -352731517.47 N.mmA= 2.42; B= 0.48; C=2.17; D=0.78Nn1 = Ar^2fcd+Cρr^2fsd [5.3.9-1]'。

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计算简图
某城市互通立交匝道桥上部结构采用预应力混凝土连续梁桥体系,跨径布置为2×25m ,梁宽从10.972m 变化到15.873m ;桥墩和桥台上都设置板式橡胶支座。

以下为该桥采用《公路工程抗震设计规范》(JTJ004—89)的简化计算方法手算的计算步骤及计算结果:
附2.1 顺桥向地震力计算
该联支座全部采用板式橡胶支座,故地震力由两部分组成:上部结构对板式橡胶支座顶面处产生的水平地震荷载及桥墩地震荷载。

一、上部结构对板式橡胶支座顶面处产生的水平地震荷载
上部结构对D6号墩板式橡胶支座顶面处产生的水平地震荷载按下式计算:
zsp h z i n
i itp
itp
ihs G K C C K
K E 10
β∑==
(附2-1)
式中,3.1=i C ,2.0=z C ,1.0=h K 1、确定基本参数
(1)全联上部结构总重力:
2353.4825)86.527.518(⨯+⨯+=zsp G 255023.0⨯⨯⨯+kN 2.16155=
(2)实体墩对支座顶面顺桥向换算质点重力:
()p
f
f tp ztp G
X X G G ⎥⎦
⎤⎢⎣

-+==2
131
由于不考虑地基变形,即0=f X
故 ()p p
f
f tp G G
X X G 3
1
1312
=⎥⎦
⎤⎢⎣

-+= 而 kN G p 3.57525346.4295.5=⨯⨯= 得 kN G G G p tp ztp 8.1913/===
(3)一联上部结构对应的全部板式橡胶支座顺桥向抗推刚度之和1K :
m kN K /103915.23.5756244.2480)23(41⨯=⨯+⨯+=
(4)设置板式橡胶支座的D6号桥墩顺桥向抗推刚度2K :
8015.01=I 4m ,088.12=I 4m ,676.13=I 4m
083.105.06.045.013
21=-+=I I I I e 从而,得 49233.0m I e =
m kN l EI K e D /1055.8746.49233.0103.33353
73⨯=⨯⨯⨯== m kN K K D /1055.852⨯==∴
2、计算桥梁顺桥向自振基本周期T 1
[
]
{}
Zsp
Ztp Zsp Ztp Zsp
Ztp Zsp Ztp G G K K G G G K K K G G K K K G g
24)()(2
1
2
12
21121121-++-++=ω
-24.11s 1= s T 673.121
1==
ωπ
3、计算动力放大系数1β
根据1T 及规范三类场地土动力放大系数函数,计算1β:
646.045.025.295
.01=⎪


⎝⎛⨯=T β
4、计算上部结构对D6号桥墩产生的水平地震力
上部结构对D6号桥墩板式橡胶支座顶面处产生的顺桥向水平荷载按式(附2-1)计算:
kN E E i
ihs hs 6.1302.16155646.01.02.03.1103915.23
.575624
=⨯⨯⨯⨯⨯⨯⨯=
=∑
二、实体墩由墩身自重在墩身质点i 的顺桥向水平地震荷载
实体墩由墩身自重在墩身质点i 的顺桥向水平地震荷载按下式计算:
11hp i z h li i E C C K X G βγ=
得 D6号墩kN E th 22.476.1910.10.18482.01.02.03.1=⨯⨯⨯⨯⨯⨯= 三、桥墩顺桥向地震剪力和弯矩
第二联D6号桥墩墩底的顺桥向地震剪力和弯矩分别如下:
kN Q D 82.13422.46.1306=+=
()kN M D 93.585346.422.46.1306=⨯+=
附2.2 横桥向地震力计算
D6号桥墩横桥向水平地震荷载按下式计算(参见D6号墩计算简图):
111ihp i z h i i E C C K X G βγ=
(附2-2)
式中,3.1=i C ,2.0=z C ,1.0=h K 1、计算i X 1
由于
5031.1460
6
.474<==B H 故取 ()f
i f i X H H X X -⎪


⎝⎛+=13
/11
不考虑地基变形时:0=f X
故有 3
/11⎪

⎫ ⎝⎛=H H X i i
得 889.06.4744.3333
/111=⎪


⎝⎛=X ,621.06.4747.1133
/112
=⎪⎭
⎫ ⎝⎛=X
2、计算桥墩各质点重力i G
kN G 6.80772/2.161550==
kN G 4.32825146.2122.61=⨯⨯=
kN G 61.247252.2502.42=⨯⨯=
3、计算横桥向基本振型参与系数1γ
011.16
.247621.04.328889.06.807716
.247621.04.328889.06.807712
20
201=⨯+⨯+⨯⨯+⨯+⨯=
=
∑∑==n
i i
i
n
i i
i
G X
G
X γ 4、计算D6号桥墩振动单元横桥向振动时的动力放大系数1β (1)计算横桥向柔度δ:
934.11=I 4m ,700.32=I 4m ,254.103=I 4m 3
2105.06.045.01I I I I e -+= 得 4569.2m I e =
H 2
H 1
H
D6号墩计算简图
56
3
731076.81
/5.11419/10412.1646.5569.2103.333-⨯==
=+⋅=⨯=⨯⨯⨯==K
m
kN K K K Ks K m kN l EI K D
S D
e D δ (2)计算桥墩横向振动的基本周期T 1
s g
G T t 72.122
/11=⎪⎪⎭
⎫ ⎝
⎛=δπ
(3)确定动力放大系数1β
根据T 1及规范三类场地土动力放大系数函数,得
629.045.025.295
.01=⎪


⎝⎛⨯=T β
5、计算各质点的水平地震力
根据公式(附2-2)计算作用于D6号桥墩各质点的横桥向水平地震力:
kN
E kN E kN E hp hp hp 40.26.247586.0011.1629.01.02.03.156.44.328839.0011.1629.01.02.03.155.1336.8077011.1629.01.02.03.1210=⨯⨯⨯⨯⨯⨯==⨯⨯⨯⨯⨯⨯==⨯⨯⨯⨯⨯= 6、计算横桥向地震剪力和弯矩
D6号墩墩底的横桥向地震剪力和弯矩分别如下:
kN Q D 51.14040.256.455.1336=++=
m kN M D ⋅=⨯+⨯+⨯=34.598137.140.2334.356.4346.455.1336。

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