某工程双孔箱涵设计计算书(按新规范计算)(精)

园中路双孔箱涵计算书一、设计资料箱涵净跨径L。=2×4m，净高H。=3.6m，箱涵顶面铺装沥青砼0.05m+C40细石砼层0.2m（平均），两端填土r=18KN/m3,Φ=30°，箱涵主体结构砼强度等级为C30，箱涵基础垫层采用C10砼，受力钢筋采用HRB335钢筋，地基为粉质粘土，汽车荷载为城-B。二、设计依据《公路钢筋砼及预应力砼桥涵设计规划》（JTGD62-2004）三、内力计算1、荷载计算1）恒载恒载竖向压力p1=r1·H+r2·δ=24×(0.05+0.2+25×0.4=16KN/m2恒载水平压力：顶板处：p2=245(tan21ψγ-⋅⋅H=1.5KN/m2底板处：p3=245(tan(21ψγ-⋅+⋅hH=27.87KN/m22）活载a1=a2+2H=0.25+2×0.25×tan30°=0.54mb1=b2+2H=0.6+2×0.25×tan30°=0.89m车辆荷载垂直压力q车=11baG⨯=89.054.060⨯=124.84KN/m2车辆荷载水平压力e车=q车·tan2(45°-Ψ/2=124.84×0.333=41.61KN/m23作用于底板垂直均布荷载总和q1=1.2q恒1+1.4q车1q恒1=p1++BddHr2(43+⨯⨯=16+9.83.03.02(6.325+⨯⨯⨯=25KN/mq车1=124.84KN/mq1=1.2q恒1+1.4q车1=1.2×25+1.4×124.84=204.78KN/m4作用于顶板垂直均布荷载总和q2=1.2q恒2+1.4q车2q恒2=16KN/mq车2=124.84KN/mq2=1.2q恒2+1.4q车2=193.98KN/m5作用于侧墙顶部的水平均布荷载总和q3=1.2q恒3+1.4q车3q恒3=1.5KN/mq车3=41.61KN/mq3=1.2q恒3+1.4q车3=60.05KN/m6作用于侧墙底部的水平均布荷载总和q4=1.2q恒4+1.4q车4q恒4=27.87KN/mq车4=41.61KN/mq4=1.2q恒4+1.4q车4=91.7KN/m2、恒载固端弯矩计算mKNLqMFAC⋅-=⨯-=⨯-=65.24123.416122212恒恒mKNMMFACFCA⋅=-=65.24恒恒mKNLqMFBD⋅=⨯=⨯-=52.38123.425122211恒恒mKNMMFBDFDB⋅-=-=52.38恒恒mKNLqqLqMFAB⋅=⨯-+⨯=⨯-+⨯=06.163045.187.27(1245.130(12222234223恒恒恒恒mKNLqqLqMFBA⋅-=⨯--⨯-=⨯--⨯-=10.232045.187.27(1245.120(12222234223恒恒恒恒3、活载固端弯矩计算mKNLqMFAC⋅-=⨯-=⨯-=36.192123.484.124122212车车mKNMMFACFCA⋅=-=36.192车车mKNLqMFBD⋅=⨯=⨯=36.192123.484.124122211车车mKNMMFBDFDB⋅-=-=36.192车车mKNLqqLqMFAB⋅=⨯-+⨯=⨯-+⨯=48.5530461.4161.41(12461.4130(12222234223车车车车mKNLqqLqMFBA⋅-=⨯--⨯-=48.5520(122234223车车车车3、抗弯劲度计算005.03.4124.04124313=⨯⨯=⨯=LdKAC顶005.03.4124.04124313=⨯⨯=⨯=LdKBD底00225.04123.04124323=⨯⨯=⨯==LdKKBAAB侧4、杆端弯矩的分配系数计算69.000225.0005.0005.0=+=+=ABACACACKKKμ31.000225.0005.000225.0=+=+=ABACABABKKKμ31.0005.000225.000225.0=+=+=BDBABABAKKKμ69.0005.000225.0005.0=+=+=BDBABDBDKKKμ5、杆端弯矩的传递系数各杆件向远端的传递系数均为0.56、结点弯矩分配计算恒载弯矩分配计算表注：弯矩符号以绕杆端顺时针旋转为正。7、各部位剪力、轴向力及控制截面弯矩计算。（1）杆件AC（顶板）及杆件BD（底板）剪力计算KNLMMLqQCAACAC69.313.453.2889.1623.4162112=+--⨯=+-⨯=恒恒恒恒KNLMMLqQCAACAC41.2293.425.24858.8023.484.1242112=+--⨯=+-⨯=车车车车QAC=1.2QAC恒+1.4QAC车=359.2KNKNLMMLqQCAACCA11.373.453.2889.1623.4162112-=+--⨯-=+-⨯-=恒恒恒恒KNLMMLqQACCACA40.3073.425.24858.8023.484.1242112-=+--⨯-=+-⨯-=车车车车QCA=1.2QCA恒+1.4QCA车=-474.89KNKNLMMLqQBDDBDB88.573.468.2644.4423.4252111=+--⨯=+-⨯=恒恒恒恒KNLMMLqQBDDBDB39.3073.459.8024.24823.484.1242111=+--⨯=+-⨯=车车车车QDB=1.2QDB恒+1.4QDB车=499.80KNKNLMMLqQBDDBBD62.493.468.2644.4423.4252111-=+--⨯-=+-⨯-=恒恒恒恒KNLMMLqQBDDBBD42.2293.459.8024.24823.484.1242111-=+--⨯-=+-⨯-=车车车车QBD=1.2QBD恒+1.4QBD车=-380.73KN（2）杆件AB（侧墙）剪力计算KNLMMLqqLqQBAABAB13.18468.2689.16345.187.27(2487.273(2223424-=--⨯-+⨯-=+-⨯-+⨯-=恒恒恒恒恒恒KNLMMLqqLqQBAABAB22.83459.8058.803461.4161.41(2461.413(2223424-=--⨯-+⨯-=+-⨯-+⨯-=车车车车车车QAB=1.2QAB恒+1.4QAB车=-138.26KNKNLMMLqqLqQBAABBA61.40468.2689.16645.187.27(2487.276(2223424=--⨯--⨯=+-⨯--⨯=恒恒恒恒恒恒KNLMMLqqLqQBAABBA22.83459.8058.806461.4161.41(2461.416(2223424=--⨯--⨯=+-⨯--⨯=车车车车车车QBA=1.2QBA恒+1.4QBA车=165.24KN（3）各加腋起点截面弯矩及跨间最大弯矩计算加腋尺寸为0.5×0.3m则底板左加腋起点截面跨结点D的距离X1左=L1-d1/2-0.5=3.65m底板右加腋起点截面跨结点D的距离X1右=0.5+d1/2=0.65m顶板左加腋起点截面跨结点A的距离X2左=X1右=0.5+d1/2=0.65m顶板右加腋起点截面跨结点A的距离X2右=X1左=L1-d1/2-0.5=3.65m侧墙上加腋起点截面跨结点B的距离X上=L2-d2/2-0.3=3.5m侧墙下加腋起点截面跨结点B的距离X下=0.3+d2/2=0.5m各加腋起点截面的弯矩分别为：mKNXqXQMMDBDB⋅=⨯-⨯+-=⨯-+=32.59265.378.20465.38.49986.400222111左左底左mKNXqXQMMDBDB⋅-=⨯-⨯+-=⨯-+=25.119265.078.20465.08.49986.400222111右右底右mKNXqXQMMACAC⋅=⨯-⨯+-=⨯-+=42.59265.098.19365.02.35908.133222222左左顶左mKNXqXQMMACAC⋅-=⨯-⨯+-=⨯-+=15.114265.398.19365.32.35908.133222222右右顶右mKNLXqqXqXQMMBABA⋅-=⨯⨯-+⨯-⨯+-=⨯-+⨯-+=45.73465.005.607.91(25.07.915.024.16577.1446(232233424下下下侧下mKNLXqqXqXQMMBABA⋅-=⨯⨯-+⨯-⨯+-=⨯-+⨯-+=55.71465.305.607.91(25.37.915.324.16577.1446(232233424上上上侧上底板（BD）、顶板（AC）、侧墙（AB）跨间最大弯矩截面位置1X（距结点D的距离）、2X（距结点A的距离）、3X（距结点B的距离）mqQXDB44.278.20480.49911===mqQXAC85.198.1932.35922===mLqqLqqQqqXBA97.1405.607.91(405.607.91(24.16527.917.91((222342342443=--⨯⨯--=--⨯⨯--=底板（BD）、顶板（AC）、侧墙（AB）跨间最大弯矩mKNXqXQMMDBDB⋅=⨯-⨯+-=⨯-+=06.209244.278.20444.28.49986.4002221111mKNXqXQMMACAC⋅=⨯-⨯+-=⨯-+=49.199285.198.19385.12.35908.1332222222mKNLXqqXqXQMMBABA⋅=⨯⨯-+⨯-⨯+-=⨯-+⨯-+=83.124697.105.607.91(297.17.9197.124.16584.1446(23223334234338、轴向力计算根据力的平衡原理，顶板轴向力等于侧墙上端剪力；底板轴向力等于侧墙下端剪力；侧墙轴向力等于顶板及底板板端剪力。9、内力计算成果注：弯矩符号以洞壁内侧受拉为正，外侧受拉为负；轴向力以压力为正，拉为负。四、截面设计（1）顶板（A-C）钢筋按左、右对称，用最不利荷载计算。a跨中：l0=4.3mh=0.4ma=0.04mh0=0.36mmNMedd443.126.13849.1990===115.0124.01222===bhi5.1739.37115.03.40==il由《公路钢筋混凝土及预应力砼桥涵设计规范》（JTGD62-2004）第5.3.10条0.102.1136.0443.17.22.07.22.01取=⨯+=+=heξ0.104.14.03.401.015.1.015.12取=⨯-=-=hlξ021.1114.03.4(443.1140036.01(140012212=⨯⨯⨯⨯+=⨯⨯+=ξξηhleh由《公路钢筋混凝土及预应力砼桥涵设计规范》（JTGD62-2004）第5.3.5条mahee633.104.024.0443.1021.12=-+⨯=-+=η2360(10008.1316331026.1389.02(30xxxhxbfeNcdd-⨯⨯⨯=⨯⨯⨯-⋅⋅⋅=⋅⋅γ解得mmhmmxb6.20136056.05.43=⨯=⋅=ξ为大偏心受压构件。2317002801026.1389.05.4310008.13mmfNbxfAsbdcds=⨯⨯-⨯⨯=-=γ用Φ20@100mm，实际As=3141mm2，偏安全。2.087.036010003141100100=⨯⨯==bhAsμ满足《公路钢筋混凝土及预应力砼桥涵设计规范》（JTGD62-2004）第9.1.12条规定。KNQKNbhfdkcu31051.3449.06.10053601000301051.01051.03,3=⨯==⨯⨯⨯⨯=⨯--γ抗剪配筋按构造设置。b角点：l0=4.3mh=0.7ma=0.04mh0=0.66mmNMedd963.026.13808.1330===202.0127.01222===bhi5.1729.21202.03.40==il由《公路钢筋混凝土及预应力砼桥涵设计规范》（JTGD62-2004）第5.3.10条0.114.466.0963.07.22.07.22.01取=⨯+=+=heξ0.109.17.03.401.015.1.015