箱涵设计计算书

箱涵设计计算书 本文由pestjenny贡献 pdf文档可能在WAP端浏览体验不佳。建议您优先选择TXT，或下载源文件到本机查看。 一. 设计资料 地下通道净跨径 L0=6m,净高 h0=3.5m,箱顶填土厚为 3m,土的 内摩擦角φ为 30?,填土的密度γ1=20KN/m3.箱涵主体结构混凝 土强度等级为 C30,箱涵基础垫层混凝土强度等级为 C15,纵向受力 钢筋采用 HRB335 钢筋.地基为强风化砂岩.汽车荷载等级为城-A 级. 二. 设计计算 (一)尺寸拟定 顶板,底板厚度δ=50cm 侧墙厚度 t=50cm 故计算长度 l = L0 + t = 6 + 0.5 = 6.5m h = H 0 + δ = 3.5 + 0.5 = 4.0m (二)荷载计算 1.恒载 竖向恒载标准值 水平恒载标准值 qv = γ 1 H + γ 2 δ = 20 × 3 + 25 × 0.5 = 72.5 KN / m 2 顶板处 qh1 = tg 2 (45o ) γ 1 H = tg 2 (45o 2 φ 30o ) × 20 × 3 = 20 KN / m 2 2 底板处 φ 30o qh 2 = tg 2 (45o ) γ 1 ( H + h + δ ) = tg 2 (45o ) × 20 × (3 + 4.5) = 50 KN / m 2 2 2 2.活载 一个汽车后轮荷载横向扩散长度 0.6 1.8 ,故两辆 + 3 × tg 30o = 2.03 f 2 2 车相邻车轴由荷载重叠;一个汽车后轮荷载纵向扩散长度 f 3 .6 0.25 o 2 .按两辆车相邻计算车轴荷载扩散面积横向 + 3 × tg 30 = 1.86 2 p 6 .0 2 0.6 分布长 a = ( + 3 × tg 30o + 1.8) × 2 + 1.3 = 8.96m .纵向分布长分两种情况, 2 第一种情况考虑 1,2,3 轴荷载重叠,此时纵向分布长 0.25 + 3 × tg 30o ) × 2 + 3.6 + 1.2 = 8.52m ;第二种情况只考虑 4 轴荷载, 2 0.25 此时纵向分布长 b = ( + 3 × tg 30o ) × 2 = 3.72m .车辆荷载垂直压力,按 2 2 × (60 + 140 + 140) 纵向分布第一种情况计算, qv车 = = 8.91KN / m 2 ;按纵 8.96 × 8.52 2 × 200 向分布第二种情况计算, qv车 = = 12.0 KN / m 2 . 8.96 × 3.72 b=( 取车辆荷载垂直压力标准值 qv车 = 12.0 KN / m 2 . 车辆荷载水平压力标准值 φ 30o qh车 = qv车 tg 2 (45o ) = 12.0 × tg 2 (45o ) = 4.0KN/m 2 2 2 (三)内力计算 1.结点内力计算 K= h I2 4 = = 0.6154 l I1 6.5 ?竖向恒载作用下 计算简图如下: M A = M B = MC = M D = N1 = N 3 = qv l 2 72.5 × 6.52 = = 158.02 KN m 12( K + 1) 12 × (0.6154 + 1) N2 = N4 = 0 q l 72.5 × 6.5 = = 235.63KN 2 2 ?竖向活载作用下 qv车 l 2 12.0 × 6.52 = = 26.15 KN m M A = M B = MC = M D = 12( K + 1) 12 × (0.6154 + 1) N1 = N 3 = q l 12.0 × 6.5 = = 39.0 KN 2 2 N2 = N4 = 0 ?水平恒载作用下 计算简图如下: MA = MB = Kqh1h 2 (q qh1 )h 2 K (3K + 8) h2 12( K + 1) 60( K 2 + 4 K + 3) 0.6154 × 20 × 42 (50 20) × 42 × 0.6154 × (3 × 0.6154 + 8) = = 18.46 KN m 12 × (0.6154 + 1) 60(0.61542 + 4 × 0.6154 + 3) MC = M D = = Kqh1h 2 (q qh1 )h 2 K (2 K + 7) h2 12( K + 1) 60( K 2 + 4 K + 3) 0.6154 × 20 × 42 (50 20) × 42 × 0.6154 × (2 × 0.6154 + 7) = 17.10 KN m 12 × (0.6154 + 1) 60(0.61542 + 4 × 0.6154 + 3) qh1 h (qh 2 qh1 )h M A M C 20 × 4 (50 20) × 4 18.46 + 17.10 + + = + + = 59.66 KN 2 6 h 2 6 4 q h (qh 2 qh1 )h M A M C 20 × 4 (50 20) × 4 18.46 + 17.10 N 4 = h1 + = + = 80.34 KN 2 3 h 2 3 4 N2 = N1 = N 3 = 0 ?水平活载作用下 计算简图如下: MA 10.18 K 5K + 1 0.6154 5 × 0.6154 + 1 = [ ? ]qh车 h 2 = [ ? ] × 4 × 42 = KN m MB 8.15 24( K + 1) 10(3K + 1) 24(0.6154 + 1) 10(3 × 0.6154 + 1) MC 5.82 K 0.6154 5 × 0.6154 + 3 5K + 1 KN m ? = [ ? ] × 4 × 42 = ]qh车 h 2 = [ MD 7.85 24(0.6154 + 1) 20(3 × 0.6154 + 1) 24( K + 1) 20(3K + 1) MC M D 5.82 + 7.85 = = 2.10 KN l 6.5 M M D 8.15 + 7.85 N2 = B = = 4 KN h 4 N1 = N 3 = N 4 = qh车 h MB MD 8.15 + 7.85 = 4× 4 = 12 KN h 4 ?节点弯矩和轴力计算汇总 表 关于同志近三年现实表现材料材料类招标技术评分表图表与交易pdf视力表打印pdf用图表说话 pdf 荷载种类 恒载 小计 活载 小计 A -158.02 -18.46 -176.48 -26.15 -10.18 -36.33 弯矩(KNm) B C -158.02 -158.02 -18.46 -17.10 -176.48 -175.12 -26.15 -26.15 8.15 5.82 -18.00 -20.33 D -158.02 -17.10 -175.12 -26.15 -7.85 -34.00 -257.74 -198.92 轴力(KN) 1 2 3 235.63 235.63 59.66 235.63 59.66 235.63 39.00 39.00 2.10 4.00 2.10 41.10 4.00 41.10 4 80.34 80.34 12.00 12.00 113.21 88.74 基本组合 -262.64 短期效应 -201.91 组合 -236.97 -189.08 -238.61 -189.35 340.30 77.19 340.30 264.40 62.46 264.40 注:基本组合 Sud = 1.2SGK + 1.4SQK ,短期效应组合 S sd = SGK + 0.7 SQK 2.跨中截 面内力计算 ?顶板 顶板均布荷载设计值 q = 1.2qv + 1.4qv车 = 1.2 × 72.5 + 1.4 × 12 = 103.8KN / m 2 跨中弯矩 M 中 = M C + N1 l ql 2 6.5 103.8 × 6.52 = 236.97 + 340.3 × = 320.81KN m 2 8 2 8 跨中剪力 V中 = ql 103.8 × 6.5 N1 = 340.3 = 2.95KN 2 2 ?底板 3qh车 h 2 3 × 4 × 42 = = 4.54 KN / m 2 q' = 2 2 6.5 l 底板非均布荷载设计值 w1 = 1.2qv + 1.4(qv车 q' ) = 1.2 × 72.5 + 1.4 × (12 4.54) = 97.44 KN / m 2 w2 = 1.2qv + 1.4(qv车 + q' ) = 1.2 × 72.5 + 1.4 × (12 + 4.54) = 110.16 KN / m 2 l w1l 2 ( w2 w1 )l 2 2 8 48 6.5 97.44 × 6.52 (110.16 97.44) × 6.52 = 262.64 + 340.3 × = 317.53KN m 2 8 48 M 中 = M A + N1 V中 = w1l ( w2 w1 )l 97.44 × 6.5 (110.16 97.44) × 6.5 + N1 = + 340.3 = 13.29 KN 2 8 2 8 ?左侧墙 w1 = 1.2qh1 + 1.4qh车 = 1.2 × 20 + 1.4 × 4 = 29.6 KN / m 2 w1 = 1.2qh 2 + 1.4qh车 = 1.2 × 50 + 1.4 × 4 = 65.6 KN / m 2 h w1h 2 ( w2 w1 )h 2 M中 = M C + N2 2 8 48 2 4 29.6 × 4 (65.6 29.6) × 4 2 = 238.61 + 77.19 × = 155.43KN m 2 8 48 V中 = w1h ( w2 w1 )h 29.6 × 4 (65.6 29.6) × 4 + N2 = + 77.19 = 0.01KN 2 8 2 8 ?右侧墙 w1 = 1.2qh1 = 1.2 × 20 = 24 KN / m 2 w1 = 1.2qh 2 = 1.2 × 50 = 60 KN / m 2 h w1h 2 ( w2 w1 )h 2 M中 = M D + N2 2 8 48 2 4 24 × 4 (60 24) × 4 2 = 257.74 + 77.19 × = 163.36 KN m 2 8 48 V中 = w1h ( w2 w1 )h 20 × 4 (60 20) × 4 + N2 = + 77.19 = 17.19 KN 2 8 2 8 3.作用基本组合构件内力汇总表 构件 M -238.61 -262.64 -262.64 -236.97 C-D A-B A-C B-D N V M C 77.19 340.3 320.81 A 113.21 340.3 317.53 A 340.3 113.21 -155.43 B 340.3 77.19 -163.36 N V M N V 跨中 D 77.19 -2.95 -257.74 77.19 -340.3 跨中 B 113.21 -13.29 -236.97 113.21 -340.3 跨中 C 340.3 0.01 -238.61 340.3 -77.19 跨中 D 340.3 -17.19 -257.74 340.3 -113.21 4.作用短期组合构件内力汇总表 作用短期组合构件内力计算方法及步骤同基本作用, 汇总表如下: 构件 Ms C -189.35 A -201.91 A -201.91 B -189.08 Ns 62.46 88.74 264.4 264.4 Ms 跨中 Ns 62.46 跨中 Ms D -198.92 B -189.08 C -189.35 D -198.92 Ns 62.46 88.74 264.4 264.4 C-D A-B A-C B-D 242.70 241.33 跨中 88.74 264.4 跨中 -120.03 -124.00 264.4 (四)截面配筋计算 结构左右和上下对称配筋,采用最不利荷载计算 ?顶板跨中 ?配筋计算 计算跨度 l0 = 6.5m ,截面高度 h = 500mm ,截面宽度 b = 1000mm ,钢 筋保护层厚度 a = 40mm e0 = i= M 320.81 = = 4.156m N 77.19 bh 2 = 12 0.52 = 0.1443 12 l0 6.5 = = 45.03 > 17.5 i 0.1443 根据《公路钢筋混凝土和预应力钢筋混凝土桥涵设计规范》 (以下 简称桥涵规范)JTG D62-2004 第 5.3.10 条 ζ 1 = 0.2 + 2.7 e0 4156 = 0.2 + 2.7 × = 25.42 > 1.0 ,取ζ1 = 1.0 h0 445 ζ 2 = 1.15 0.01l 0.01 × 6500 = 1.15 = 1.02 ,取ζ2 = 1.0 h 500 η = 1+ 1 1 6500 2 l ( 0 ) 2 ζ 1ζ 2 = 1 + ×( ) × 1.0 × 1.0 = 1.013 1400e0 / h0 h 1400 × 4156 / 445 500 按桥涵规范 (5.3.5-3) 公式, 轴向压力作用点至纵向受拉钢筋的合 力点的距离: 500 h a = 1.013 × 4156 + 55 = 4405mm 2 2 x Ne = f cd bx(h0 ) 2 x 77.19 × 1000 × 4405 = 13.8 × 1000 × x × (445 ) 2 e = ηe0 + 解得 x = 59.3mm < ξb h0 = 0.56 × 445 = 249.2 ,为大偏心受压构件 As = f cd bx N 13.8 × 1000 × 59.3 77.19 × 1000 = = 2647mm 2 fy 280 实配 25@125, As = 3927mm2 跨中剪力值较小,未做斜截面抗剪承载力验算 ?正常使用极限状态计算 根据桥涵规范 6.4.4 条 γ f '= (b f 'b)h f ' bh0 =0 l0 / h = 6500 / 500 = 13 < 14 取 η s = 1.0 ys = 500 h as = 55 = 195mm 2 2 es = η s e0 + ys = 1.0 × 4156 + 195 = 4351mm z = [087 0.12(1 γ s )( h0 2 445 2 ) ]h0 = [0.87 0.12 × (1 0) × ( ) ] × 445 = 386.6mm es 4351 σ ss = N s (e s z ) 62.46 × (4351 386.6) = = 163.10 N / mm 2 As z 3927 × 386.6 根据桥涵规范 6.4.3 条,验算裂缝宽度 对带肋钢筋 C1 = 1.0 作用长期影响系数 C2 = 1 + 0.5 C3 = 0.9 61.26 Nl = 1 + 0.5 × = 1.4904 62.46 Ns 裂缝宽度 Wtk = C1C2C3 σ ss Es ( 30 + d ) 0.28 + 10 ρ 163.10 30 + 25 ( ) = 0.168mm < 0.2mm 5 2.0 × 10 0.28 + 10 × 0.007854 == 1 × 1.4904 × 0.9 × 满足要求 ?顶板节点 计算步骤同上. e0 = M 257.74 = = 3.339m 77.19 N l 1 1 6500 2 ( 0 ) 2 ζ 1ζ 2 = 1 + ×( ) × 1.0 × 1.0 = 1.016 1400e0 / h0 h 1400 × 3339 / 445 500 η =1+ e = ηe0 + 500 h a = 1.016 × 3339 + 55 = 3588mm 2 2 x Ne = f cd bx(h0 ) 2 x 77.19 × 1000 × 3588 = 13.8 × 1000 × x × (445 ) 2 解得 x = 47.7 < ξb h0 = 0.56 × 445 = 249.2 ,为大偏心受压构件 As = f cd bx N 13.8 × 1000 × 47.7 77.19 × 1000 = = 2075mm 2 f sd 280 实配 22@125, As = 3041mm2 根据《公路钢筋混凝土和预应力钢筋混凝土桥涵设计规范》 (以下 简称桥涵规范)JTG D62-2004 中 5.2.9 条 0.51× 10 3 f cu , k bh0 = 0.51× 10 3 × 30 × 1000 × 445 = 1243.06 KN > γ 0V = 1.0 × 340.3 = 340.3KN 抗剪截面尺寸符合要求 根据桥涵规范 5.2.10 条 0.5 × 10 3 × 1.25 × α 2 f td bh0 = 0.5 × 10 3 × 1.25 × 1.0 × 1.39 × 1000 × 445 = 386.59 > γ 0Vd = 1.0 × 340.3 = 340.3KN 式中 1.25 为板式受弯构件提高系数,抗剪配筋按构造设置 ?底板跨中 e0 = M 317.53 = = 2.805m N 113.21 1 1 6500 2 l ( 0 ) 2 ζ 1ζ 2 = 1 + ×( ) × 1.0 × 1.0 = 1.019 1400e0 / h0 h 1400 × 2805 / 445 500 η = 1+ h 500 a = 1.019 × 2805 + 55 = 3054mm 2 2 x Ne = f cd bx(h0 ) 2 x 113.21 × 1000 × 3054 = 13.8 × 1000 × x × (445 ) 2 e = ηe0 + 解得 x = 60.4 < ξb h0 = 0.55 × 445 = 244.75 ,为大偏心受压构件 As = f cd bx N 13.8 × 1000 × 60.4 113.21 × 1000 = = 2573mm 2 f sd 280 实配 25@125, As = 3927mm2 ?底板节点 e0 = M 262.64 = = 2.32m N 113.21 1 1 6500 2 l ( 0 ) 2 ζ 1ζ 2 = 1 + ×( ) × 1.0 × 1.0 = 1.023 1400e0 / h0 h 1400 × 2320 / 445 500 η = 1+ h 500 a = 1.023 × 2320 + 55 = 2569mm 2 2 x Ne = f cd bx(h0 ) 2 x 113.21 × 1000 × 2569 = 13.8 × 1000 × x × (445 ) 2 e = ηe0 + 解得 x = 50.2 < ξb h0 = 0.56 × 445 = 249.2 ,为大偏心受压构件 As = f cd bx N 13.8 × 1000 × 50.2 113.21 × 1000 = = 2070mm 2 f sd 280 实配 22@125, As = 3041mm2 ?左右侧墙跨中 计算跨度 l = 4.0m ,截面高度 h = 500mm ,截面宽度 b = 1000mm ,钢 筋保护层厚度 a = 40mm 计算步骤同上. e0 = M 163.36 = = 0.48m 340.3 N e0 480 = 0.2 + 2.7 × = 3.11 > 1.0 ,取ζ1 = 1.0 h0 445 ζ 1 = 0.2 + 2.7 ζ 2 = 1.15 η = 1+ 0.01l 0.01 × 4000 = 1.15 = 1.07 ,取ζ2 = 1.0 h 500 1 1 4000 2 l ( 0 ) 2 ζ 1ζ 2 = 1 + ×( ) × 1.0 × 1.0 = 1.042 1400e0 / h0 h 1400 × 500 / 445 500 h 500 a = 1.042 × 500 + 55 = 695.4mm 2 2 x Ne = f cd bx(h0 ) 2 x 340.3 × 1000 × 695.4 = 13.8 × 1000 × x × (445 ) 2 e = ηe0 + 解得 x = 40.4 < ξb h0 = 0.56 × 445 = 249.2 ,为大偏心受压构件 As = f cd bx N 13.8 × 1000 × 40.4 340.3 × 1000 = = 776mm 2 f sd 280 实配 22@125, As = 3041mm2 ?左右侧墙节点 计算步骤同上. e0 = M 262.64 = = 0.772m 340.3 N 1 1 4000 2 l ( 0 ) 2 ζ 1ζ 2 = 1 + ×( ) × 1.0 × 1.0 = 1.026 1400e0 / h0 h 1400 × 772 / 445 500 500 h a = 1.026 × 772 + 55 = 987mm 2 2 η = 1+ e = ηe0 + x Ne = f cd bx(h0 ) 2 x 340.3 × 1000 × 987 = 13.8 × 1000 × x × (445 ) 2 解得 x = 58.5 < ξb h0 = 0.56 × 445 = 249.2 ,为大偏心受压构件 As = f cd bx N 13.8 × 1000 × 58.5 340.3 × 1000 = = 1668mm 2 f sd 280 实配 22@125, As = 3041mm2 本TXT由“文库宝”下载: