水工U形渡槽设计

工程概况：渡槽又称高架渠，是一组由桥梁，隧道或沟渠构成的输水系统。

用来把远处的水引到水量不足的城镇、农村以供饮用和灌溉。

分类：U型薄壳渡槽、矩形渡槽、薄腹梁渡槽、拱组合梁渡槽、下承式拱渡槽、T型刚构渡槽等世界上最早的渡槽诞生于中东和西亚地区。

公元前700余年，亚美尼亚已有渡槽。

公元前703年，亚述国一西拿基立下令建一条483公里长的渡槽引水到国都尼尼微。

渡槽建在石墙上，跨越泽温的山谷。

石墙宽21米，高9米，共用了200多万块石头。

渡槽下有5个小桥拱，让溪水流过。

古希腊的许多城市建有良好的渡槽，但古罗马人最为认真，把供水系统看作是公共卫生设施的重要部分。

罗马第一条供水渡槽是建于公元前312年的阿庇渡槽；第十条也是最后一条则是公元226年建成的阿历山大渡槽；最长最壮观的是建于公元前114年的马西亚渡槽，虽然水源离罗马仅37公里，但渡槽本身长达92公里。

这是因为渡槽要保持一定坡度，依地形蜿蜒曲折地修建。

漆水河渡槽水工U形渡槽槽身结构设计目录：1.水工U形渡槽槽身结构设计计算书。

3（1）原始资料。

2（2）内力分析。

4（3）结构计算。

6（4）配筋说明。

172.计算数据记录表。

53.内力分析图。

附图14.水工U形渡槽槽身结构配筋图。

附图2水工U形形渡槽槽身结构设计一、水工U形渡槽槽身结构设计计算书（一）原始资料（如图）1．横截面尺寸：槽壳内半径： 0R 1.60m = 垂直段总高： f =0.85 m 槽壳壁厚： t =0.12 m 侧 梁： 水平宽 a =0.30 m 外端高 b =0.15 m 斜边高 c =0.12 m 2．纵截面尺寸：槽壳总长： L =10.80 m横杆：高 0.20 m δ =宽 b 0.20' = 间距 e =1.50 m （近似值）3．端肋：沿水流方向厚度 d =0.30 m垂 直 边 高 度 1h =1.55 m 斜边垂直高度 2h =1.62 m 斜边水平投影 m =0.29 m 横流方向支座宽 1b =0.30 m 支座间距 n l =2.86 m 3．便桥：预制桥板：厚度 h 0.07 m ' = 宽度 B =0.60 m 栏杆：柱a a 0.12 1.2m m m ⨯⨯H =0.12 ⨯ ⨯间距为e 1.0m'≈ 扶手横档 0.080.10f f b h m m ⨯= ⨯ 便桥有效宽度：B =0.86 m ' 便桥人群荷载：2q 2 KN m /人＝ 4．材料：混凝土C 30：32.51025KN m γ⨯ /砼＝＝224215 1.50 3.010c t c f N mm f N mm E N mm =/ = / =⨯ /钢筋Ⅰ级： 252210 2.110y y yv s f f f N mm E N mm ='== / =⨯ / 5．建筑物等级：Ⅳ级。

查表知结构安全级别为Ⅲ级 00.9γ= 6．荷载组合：满槽水＋自重＋人群自重荷载（永久荷载）： 标准值222000022()2[2] 40.020.02 1.600.12 1.22(0.0630.070.620.080.101.5 1.01.60.850.12)25427.6725k f f b R R t R a Hg F h b h ft e e KN mππ2'δπ+-=∆++'B ++++γ'⨯⨯⨯ =⨯++⨯++⨯⨯π⨯(1.6+0.12)-⨯ +⨯+⨯ = /砼设计值0561.2905.16725.27=⨯=⨯=k G g r g 满槽水荷载（可变荷载）： 标准值2002[()]223.14 1.60[]10264.1920R q R f KN mδγ⨯=+2-⨯0.20 =+2⨯1.60⨯(0.85-)⨯2= /k 水水π设计值q 水 =61.701.11920.64=⨯=⨯水q Q γKN/m人群荷载（可变荷载）： 标准值2220.86 3.44q q B KN m ='=⨯⨯=/k 人人设计值q 人=m KN q Q /128.42.144.3=⨯=⨯人γ（二）内力分析用空间计算法程序2AQ 上机计算。

1．上机数据计算（准备两组数据，一组用于承载力计算，一组用于正常使用极限状态验算）00.121.60 1.6622t R R m =+=+= 0.200.850.752h f m δ=-=-=2 30(2)0.30(20.150.12)F 0.6322a b c m +⨯⨯+∆===10.800.3010.50l L d m =-=-= 33300.200.200.0000888881212 1.50b J m e 'δ⨯===⨯2000G G G {2}.6.81384998( 1.05 1.20)4.52366665 1.72 6.24366665( 1.0, 1.0)f f QG Q Q b R a H G F h B b h q B e e KN m KN m δ'=∆++'++γγ+'γ'/ γγ= =γ+γ= / γ= γ={ 人砼Q 承载力＝，正常0.750.4518072281.66h A R === 0.120.0722891561.66t B R ===0220.0630.022*******.66F C R ∆=== 10.50 6.3253102051.66l D R === 3300.1219.44019440.000088888t E J ===000226.8138499830.247272825(27.5666.2436666527.55610 1.660.226581022(27.556G G G F R =====γ⨯= {水承载力）正常使用）2．上机计算结果:承载力计算M 0 =-7.549942E-02 M 1 = .2115567 M 2 = .2214667 M 3 = .2072521 M 4 = .1726536 M 5 = .1223864 M 6 = 6.169657E-02 M 7 =-4.028062E-03 M 8 =-6.958687E-02 M 9 =-.1302293 M 10 =-.181855 M 11 =-.2211571 M 12 =-.2457184 M 13 =-.2540697S 0 =-300.5273 S 1 =-105.1067 S 2 =-49.97215 S 3 = 1.328807 S 4 = 47.78657 S 5 = 88.75562 S 6 = 123.7834 S 7 = 152.6342 S 8 = 175.5023 S 9 = 192.8914 S 10 = 205.4156 S 11 = 213.6268 S 12 = 218.1468 S 13 = 219.5523 Z= 79.68113N 0 =-.681385 N 1 =-.4693885 N 2 = .9664244 N 3 = 1.537593N 4 = 2.12781 N 5 = 2.71425 N 6 = 3.276332 N 7 = 3.796139N 8 = 4.258661 N 9 = 4.651817 N 10 = 4.9663 N 11 = 5.195339N 12 = 5.334429 N 13 = 5.381061 NG= .4858664T 0 = 0 T 1 = 92.71896 T 2 = 97.75652 T 3 = 99.33664T 4 = 97.74118 T 5 = 93.30577 T 6 = 86.40169 T 7 = 77.42258T 8 = 66.76346 T 9 = 54.79662 T 10 = 41.85808 T 11 = 28.24597T 12 = 14.22031 T 13 = 2.171492E-03 TM= 99.33664正常使用极限状态验算M 0 =-6.904241E-02 M 1 = .1919571 M 2 = .2009443 M 3 = .1879737M 4 = .1564427 M 5 = .1106412 M 6 = 5.534594E-02 M 7 =-4.535793E-03 M 8 =-6.426623E-02 M 9 =-.1195174 M 10 =-.1665533 M 11 =-.2023612M 12 =-.2247388 M 13 =-.2323477S 0 =-276.0692 S 1 =-96.50473 S 2 =-45.84194 S 3 = 1.29765S 4 = 43.94534 S 5 = 81.55782 S 6 = 113.7016 S 7 = 140.1754S 8 = 161.1712 S 9 = 177.1398 S 10 = 188.6404 S 11 = 196.2114S 12 = 200.3176 S 13 = 201.676 Z= 73.18598N 0 =-.6243666 N 1 =-.4275816 N 2 = .8805827 N 3 = 1.402606N 4 = 1.94208 N 5 = 2.478112 N 6 = 2.991876 N 7 = 3.466993N 8 = 3.88974 N 9 = 4.249081 N 10 = 4.536512 N 11 = 4.745846N 12 = 4.872967 N 13 = 4.915588 NG= .4417494T 0 = 0 T 1 = 85.16774 T 2 = 89.7917 T 3 = 91.23868T 4 = 89.76901 T 5 = 85.69219 T 6 = 79.34942 T 7 = 71.10252T 8 = 61.31363 T 9 = 50.32399 T 10 = 38.44205 T 11 = 25.94059T 12 = 13.05981 T 13 = 1.520416E-03 TM= 91.238683．绘内力（应力）图①承载力计算内力图：②正常使用极限状态验算内力图：4．找出最大纵向正应力m σ和最大剪应力m τ①承载力计算22993.7466 =2168.978 m m KN m KN m τ=/σ/②正常使用极限状态验算 22=912.7137 =1991.615 m m KN m KN m τ/ σ/ （三）结构计算所用参数：01.20.90.85d ct γ= γ= ψ=0.95 α=1．正截面承载力计算320s 1.20.90.95797.236510A 3895.069210d y Z mm f γγψ⨯⨯⨯⨯===选配钢筋：2s (A 3895.069) 1212}mm φφ = 内层1416外层2．正截面抗裂验算查表1.35m γ'=修正值：30300) 1.35(0.7)(0.85 1.60.12)101.2029617831.203m m f R t 300γ=γ'(0.7+=⨯+++++⨯ = ≈222168.978 2.168978m KN m N mm σ=/=/ 00.9 2.168978 1.95209802m γσ=⨯=0.85 1.203 2.25 2.301ct m tk f αγ=⨯⨯=mctm tkf αγσ<γ∴满足正截面抗裂要求3．斜截面承载力计算220.120.24240b t m mm ==⨯==000.85 1.6 2.452450h f R m mm =+=+==该梁均布荷载设计值：G q Q q q g =γ+γ+γk 人k 水水砼=70.61+4.128+29.0561 =103.7941KN/m剪力 01()0.90.950.5103.7941 2.86128.23501572n V q l K N ψ=γ=⨯⨯⨯⨯= 则3000.07 1.2128.2350157100.0724*******01.25 1.252102450d c yv V bh f f h γ-⨯⨯-⨯⨯⨯=<⨯⨯所以按结构配筋4．斜截面抗裂验算2993.7466m KN m τ= /222221987.4932 1.9874932tp m m KN m N mm σ=τ=⨯993.7466 KN/= /= / 2200.9 1.8247736 1.9874932tp N mm N mm γσ=⨯ /= /20.85 2.25 1.9125ct tk f N mm α=⨯= /tpct tk f γσα<∴满足斜截面抗裂5．跨中截面横向承载力计算1000120b mm h t mm = == 25a a mm ='=001202595952570h h a mm h a mm =-=-= -'=-=6025352ha mm -=-= （1）槽底截面横向承载力计算13133.13515153.85122M M KN m N N KN ==⋅ ==0 3.135150.058218736583553.851222M h e m mm a mm N ==== >-= 故 属于大偏心受拉构件，内侧受拉04735122he e a mm =-+=-= 设 0b x h ξ= 查0.426sb α=23200 1.253.851221012150.4261000950)21070d c sb s y Nef bh A f h a αγ-⨯⨯⨯-⨯⨯⨯'==<'(-'⨯选配26@130(21)s A m m m φ '= 7 / 30220()1.253.85122101221021770015100095d y s s c Nef A h f bh ααγ-''-'⨯⨯⨯-⨯⨯==<⨯⨯ 故按2x α<' 计算 s A012025478222h e a e mm '=-'+=-+= 320 1.253.851221082360.4734()21070d s y Ne A mmf h a γ'⨯⨯⨯===-'⨯配筋：22min 08@130(360.47340.15%100095142.5)s A mm m bh mm m ρφ =/>=⨯⨯=/（2）槽侧横向承载力计算221.8337049.864072M M KN m N N KN == ⋅ ==0 1.8337040.185897185.897359.8640722M h e m m a mm N ====>-= 故 属于大偏心受拉构件，外侧受拉022*******he e a mm =-+=-= 设 0b x h ξ= 查得0.426sb α== 23200 1.29.86407210194150.4261000950()21070d c sb s y Nef bh A f h a γ-α⨯⨯⨯-⨯⨯⨯=-<'-'⨯选配 26@130(217)s A m m m φ =/3020()1.29.8640724100d y s s c Ne f A h a f bh 2γ-''-'⨯⨯⨯194-210⨯193⨯70α==<15⨯1000⨯95故按 2x a <' 计算 s A01202522926422h e a e mm '=-'+=-+= 320 1.29.86407210264212.58017()21070d s y Ne A mmf h a γ'⨯⨯⨯===-'⨯配筋：22min 06@1302170.15%1000142.5s mm m bh mm m ρ2φ (A =/>=⨯⨯95=/6．横向抗裂验算010*********b mm h mm a mm h mm = = '= =3003000.70.7 1.11201.551.11.705mh+=+>γ=⨯=∴542.11073.010s E c E E ⨯α===⨯ （1）槽底13133.13515153.85122s s M M KN m N N KN == ⋅ ==02()1000E s s A bh A A mm=+α+'=⨯120+7⨯(387+217) =124228200021000738795721725212422860.33527E s E s bh A h A a y A mm2+α+α''=⨯120+⨯⨯+⨯⨯ == 3322000000324()()()33100060.335277387339560.33527)7217148919980.3E s E s by b h y I A h y A y a mm 32-=++α-+α'-'⨯1000⨯(120-60.33527) =++⨯ ⨯(-+⨯⨯(60.33527-25) =3000148919980.32468839.78212060.33527I W mm h y ===-- 330023.13515110 1.70553.85122102468839.782124026.66670.741564944s m s M N W A N mm γ⨯⨯⨯+=+= / 21.7050.85 2.25 3.2608125m ct tk f N mm γα=⨯⨯= /则00s m sm ct tk M N f W A γ+<γα 故 满足抗裂要求（2）槽侧221.8337049.864072s s M M KN m N N KN == ⋅ ==02()10001207(217217)E s S A bh A A mm=+α+'=⨯+⨯+ =122893.33332000210007217957217252122893.333360E s E s bh A h A a y A mm2+α+α''=⨯120+⨯⨯+⨯⨯ == 3322000000324()()()331000607217339560)7217147544333.3E s E s by b h y I A h y A y a mm 32-=++α-+α'-'⨯1000⨯(120-60) =++⨯ ⨯(-+⨯⨯(60-25) = 3000147544333.32459072.22212060I W mm h y ===--330021.83370410 1.7059.864072102459072.222122893.33330.137598s m s M N W A N mm γ⨯⨯⨯+=+=/ 21.7050.85 2.25 3.2608125m ct tk f N mm γα=⨯⨯= /则00s m sm ct tk M N f W A γ+<γα 故 满足抗裂要求7．横杆设计22 1.66 3.32l R m ==⨯=022 1.6 3.2n l R m ==⨯=0.20.225 1.05 1.05G g b KN m ='δγγ=⨯⨯⨯= /砼G e 2 1.5 1.20.07 1.525 1.05 6.35625KN m Q q q h e =γ+'γγ⨯⨯+⨯⨯⨯ /人砼＝＝00.8835231M KN m =-⋅则22002211()8210.90.95 1.05 3.32 6.356250.60.8835231)22.970559076G M gl qB M KN mψ=γ+-1 =⨯⨯(⨯⨯+⨯⨯+8 = ⋅01[()]0.90.950.5(1.05 6.35625) 3.2210.13175n V q g l KN ψ=γ+=⨯⨯⨯+⨯ = 4.595488G N KN =0.22000.220025b b m mm h m mm a a mm ='== =δ== ='=002002517517525150200257522h h a h a mm h a =-=-= -'=-=-=-=（1）按,G G N M 计算跨中截面的纵向受力筋0 2.970559076639.879611()754.5954882G G M he mm a N ===>-= 故按大偏心受拉计算0639.87961175564.87961152he e a mm =-+=-= 设 0b x h ξ= 查得0.426sb α== 23200 1.2 4.59548810564.8796115150.4262001750()210150d G c sb s y Nef bh A f h a γ-α⨯⨯⨯-⨯⨯⨯=-<'-'⨯选配 210(157)sA m m2φ = 3020()1.2 4.595488100d G y s s c Nef A h a f bh 2γ-''-'⨯⨯⨯564.8796115-210⨯157⨯150α==<15⨯200⨯175故按 2x a <' 计算 s A02002554562022h e a e mm '=-'+=-+= 320 1.2 4.595488101085.410499()210150d G s y Ne A mmf h a γ'⨯⨯⨯620===-'⨯配筋：22min 01570.20%20017570s mm m bh mm ρ22φ10 (A =/>=⨯⨯=（2）按0,G N M 计算两端截面的纵向受力筋000.8835231192.25882()754.5954882G M he mm a N ===>-= 故按大偏心受拉计算015575802he e a mm =-+=-= 设 0b x h ξ= 查得0.426sb α== 23200 1.2 4.85866410150.4262001750()210150d G c sb s y Nef bh A f h a γ-α⨯⨯⨯80-⨯⨯⨯=-<'-'⨯选配 210(157)sA m m2φ = 3020()1.2 4.858664100d y s s c Ne f A h a f bh 2γ-''-'⨯⨯⨯80-210⨯157⨯150α==<15⨯200⨯175故按 2x a <' 计算 s A02002515523022h e a e mm '=-'+=-+= 320 1.2 4.8586641042.6()210150d G s y Ne A mmf h a γ'⨯⨯⨯230===-'⨯配筋（与跨中同步配筋）：22min 01570.20%20017570s mm m bh mm ρ22φ10 (A =/>=⨯⨯=（3）按V 计算横向钢筋31.210.131751012158.1d V N γ=⨯⨯= 00.070.07152001752450c f bh N =⨯⨯⨯=则 00.07d c V f bh γ< 故 按结构设计配筋 210@140(78.5)sv A mm φ = （单肢箍筋） 8．端肋设计1 2.860.3 3.16n l l b m =+=+= 3.72 3.160.2822L l k m --===201220)][][2]}2]0.30}25241.96821053n R l h h d h m R f d KNπ2{+++-++⋅γ {⨯++⨯+⨯π⨯1.6⨯+⨯⨯+⨯⨯ = 1砼[端肋自重]＝[2(m b ＝[2(0.290.3) 2.86][1.55 1.62]0.30-[0.29 1.622 1.60.8500(2)2][]220.12(2 1.60.12)20.850.1220.063][5.40.3]252t R t Lft F d KNπ+++∆⋅-⋅γπ⨯⨯⨯+ ⨯⨯++⨯⨯-⨯ =126.4551702 砼[槽壳自重]＝[2＝[2224)220.20.2 1.6 1.2=(20.070.640.080.10) 5.4251.5 1.031.8456f f a H Lh B b h e e KN2'δ=+'++⋅⋅γ'⨯⨯⨯2⨯0.12⨯ +⨯⨯+⨯⨯+⨯⨯ = 0砼[便桥自重](2b R[]=[=41.96821053126.455170231.8456=200.2689807KN++ 则总自重端肋]＋[槽壳]＋[便桥]B 220.86 5.418.5762Lq KN '=⨯⨯⨯= 人[人群]=2 2)67.74219691 5.4365.807863322Rb LhR KN e π'δ+-⋅⋅γ⨯2水[水]R =(＝＝所以承载力计算中：[][][]2200.2689807 1.0518.576 1.2365.8078633 1.1 634.9622794 G Q GKN=⋅γ+⋅γ+⋅γ⨯+⨯+⨯水自重人群水＝＝ 634.96227942170.68878483.72Gg KN m L === /则计算得max max 206.3627408269.68828M KN m V KN= ⋅ =内力图如下正常使用极限状态验算中：[][][]2200.2689807 1.018.576 1.0365.8078633 1.0 584.652844 G Q GKN=⋅γ+⋅γ+⋅γ⨯+⨯+⨯水自重人群水＝＝ 584.6528442157.1647433.72Gg KN m L === /计算得max max 190.0121743248.3202939M KN m V KN= ⋅ =内力图如下120()()(1.55 1.62)(0.85 1.600.12)0.6600h h h f R t m mm=+-++=+-++ ==0.330050b m mm a mm == =060050550h h a mm =-=-=（1）正截面承载力计算6s 220 1.2206.3627408100.18191753115300550d c M f bh αγ⨯⨯===⨯⨯110.202378535ξ==20150.2023785353005502385.175602210c s y f bh A mm f ξ⨯⨯⨯=== 选配 2820(2513)sA m m φ = ，分两排，净距 30e mm =min 025131.52%0.20%300550s A bh ρ===>ρ=⨯ （2）斜截面承载力计算31.2269.6882810323625.936d V N γ=⨯⨯=00.250.2515300550618750323625.936c d f bh N V N =⨯⨯⨯=>γ=故截面尺寸满足抗剪要求00.070.0715323625.936c c d V f bh V N ==⨯⨯300⨯550=173250N <γ=应由计算确定腹筋首先选配 12@150φ 双肢箍筋 2226sv A mm =max 150200s mm s mm =<= 可以02261.25 1.25210550217525150sv sv yvA V f h N s ==⨯⨯⨯= 323625.936217525541150.936323625.936cs c sv d V V V N V N =+=+=>γ=min 2260.5%0.12%300150sv sv A bs ρ===>ρ=⨯ 故所选箍筋满足要求（3）正截面抗裂验算3003000.70.7 1.2 1.1600h+=+=>1.551.11.705m γ=⨯=∴0230007E s A bh A mm=+α=⨯60+⨯3041 =200273.333320023007304155022000273.3333325.3E s bh A h y A mm2+α=⨯600+⨯⨯ == 33200000324()()33300325.37325.3070504)336538818613E s by b h y I A h y mm 3-=++α-⨯300⨯(600-325.3) =++⨯3041⨯(550- =3000653881861323804103.22600325.3I W mm h y ===--0 1.7050.85 2.2533804103.2277620717.3577.62071735192.0121743m ct tk f W N mm KN m M KN mγα=⨯⨯⨯= ⋅ =⋅<= ⋅按最大裂缝宽度验算1231.0 1.0 1.55050221587c mm ea c d mm ααα= = = ==++=++=22228730052200te A ab mm ==⨯⨯= 30410.05825670432200s te te A A ρ=== 620190.012174310140.00011410.870.875133041s ss s M N mm h A σ⨯===/⨯⨯max 1235max (30.1)140.0001141221.0 1.0 1.5(3500.1)0.0582567042.1100.19[]0.3ssstedw c E mm w mm σααα=+ρ =⨯⨯⨯⨯⨯+⨯⨯ =<= 故满足裂缝宽度要求（4）斜截面抗裂验算32max 248.32023410 1.379557186300600m V N mm bh ⨯τ===/⨯21.379557186tp m N mm σ=τ= /2200.9 1.379557186 1.24tp N mm N mm γσ=⨯ /= /20.85 2.25 1.9125ct tk f N mm α=⨯= /tpct tk f γσα<∴满足斜截面抗裂要求（四）配筋说明1．U 型渡槽薄壳混凝土保护层取25m m ，端肋取50m m 。