1、孔径及净空净跨径L 0 =3m 净高h 0 = 2.5m2、设计安全等级三级结构重要性系数r 0 =0.93、汽车荷载荷载等级公路 —Ⅰ级4、填土情况涵顶填土高度H =0.8m 土的内摩擦角Φ =30°填土容重γ1 =18kN/m 3地基容许承载力[σ0] =200kPa5、建筑材料普通钢筋种类HRB335主钢筋直径12mm 钢筋抗拉强度设计值f sd =280MPa 涵身混凝土强度等级C 20涵身混凝土抗压强度设计值f cd =9.2MPa 涵身混凝土抗拉强度设计值f td = 1.06MPa 钢筋混凝土重力密度γ2 =25kN/m 3基础混凝土强度等级C 15混凝土重力密度γ3 =24kN/m 3(一)截面尺寸拟定 (见图L-01)顶板、底板厚度δ =0.3m C 1 =0.3m 侧墙厚度t =0.28m C 2 =0.5m 横梁计算跨径L P = L 0+t = 3.28m L = L 0+2t = 3.56m 侧墙计算高度h P = h 0+δ = 2.8m h = h 0+2δ =3.1m 基础襟边 c =0.2m 基础高度 d =0.4m 基础宽度 B =3.96m图 L-01(二)荷载计算1、恒载恒载竖向压力p 恒 = γ1H+γ2δ =21.90kN/m2恒载水平压力顶板处e P1 = γ1Htan 2(45°-φ/2) = 4.80kN/m 2底板处e P2 = γ1(H+h)tan 2(45°-φ/3) =23.40kN/m 22、活载钢 筋 混 凝 土 箱 涵 结 构 设 计一 、 设 计 资 料二 、 设 计 计 算汽车后轮着地宽度0.6m,由《公路桥涵设计通用规范》(JTG D60—2004)第4.3.4条规定,按30°角向下分布。
一个汽车后轮横向分布宽> 1.3/2 m < 1.8/2 m故横向分布宽度a = (0.6/2+Htan30°)³2+1.3 =2.824m同理,纵向,汽车后轮着地长度0.2m0.2/2+Htan30°=0.562 m < 1.4/2 m故b = (0.2/2+Htan30°)³2 =1.124m ∑G =140kN 车辆荷载垂直压力q 车 = ∑G /(a³b) =44.12kN/m 2车辆荷载水平压力e 车 = q 车tan 2(45°-φ/2) =14.71kN/m2(三)内力计算1、构件刚度比K = (I 1/I 2)³(h P /L P ) =1.052、节点弯矩和轴向力计算(1)a种荷载作用下 (图L-02)涵洞四角节点弯矩M aA = M aB = M aC = M aD =-1/(K+1)²pL P 2/12横梁内法向力N a1 = N a2 =0侧墙内法向力N a3 = N a4 =pL P /2恒载p = p 恒 =21.90kN/m 2M aA = M aB = M aC = M aD =-9.58kN ²m N a3 = N a4 =35.92kN 车辆荷载p = q 车 =44.12kN/m 2M aA = M aB = M aC = M aD =-19.30kN ²m 图 L-02N a3 = N a4 =72.36kN(2)b种荷载作用下 (图L-03)M bA = M bB = M bC = M bD =-K/(K+1)²ph P 2/12N b1 = N b2 =ph P /2N b3 = N b4 =0恒载p = e P1 =4.80kN/m 2M bA = M bB = M bC = M bD =-1.61kN ²m N b1 = N b2 =6.72kN(3)c 种荷载作用下 (图L-04)图 L-03M cA = M cD =-K(3K+8)/[(K+1)(K+3)]²ph P 2/60M cB = M cC =-K(2K+7)/[(K+1)(K+3)]²ph P 2/60N c1 =ph P /6+(M cA -M cB )/h P N c2 =ph P /3-(M cA -M cB )/h PN c3 = N c4 =0恒载p = e P2-e P1 =18.60kN/m2M cA = M cD =-3.43kN ²m M cB = M cC =-2.80kN ²m N c1 =8.45kN N c2 =17.59kN图 L-04(4)d 种荷载作用下 (图L-05)M dA =-[K(K+3)/6(K 2+4K+3)+(10K+2)/(15K+5)]²ph P 2/40.6/2+Htan30°=0.76 mM dB =-[K(K+3)/6(K2+4K+3)-(5K+3)/(15K+5)]²ph P2/4M dC =-[K(K+3)/6(K2+4K+3)+(5K+3)/(15K+5)]²ph P2/4M dD =-[K(K+3)/6(K2+4K+3)-(10K+2)/(15K+5)]²ph P2/4N d1 =(M dD-M dC)/h PN d3 = -N d4 =-(M dB-M dC)/L P车辆荷载p = e车 =14.71kN/m2M dA =-19.82kN²mM dB =9.00kN²mM dC =-13.92kN²mM dD =14.90kN²m图 L-05N d1 =10.29kNN d2 =30.88kNN d3 = -N d4 =-6.99kN(5)节点弯矩、轴力计算及荷载效应组合汇总表按《公路桥涵设计通用规范》(JTG D60—2004)第4.1.6条进行承载能力极限状态效应组合V x =ω1x+x2(ω2-ω1)/2L P-N3图 L-07=-27.13kN(3)左侧墙 (图L-08)ω1 =1.4e P1+1.4e 车=27.31kN/m 2ω2 =1.4e P2+1.4e 车53.35kN/m 2x =h P /2N x = N 3 =134.61kNM x =M B +N 1x-ω1²x 2/2-x 3(ω2-ω1)/6h P=-13.17kN ²mV x =ω1x+x 2(ω2-ω1)/2h P -N 1=11.69kN (4)右侧墙 (图L-09)ω1 = 1.4e P1 = 6.72kN/m 2ω2 = 1.4e P2 =32.76kN/m 2x =h P /2N x = N 4 =154.18kNM x =M C +N 1x-ω1²x 2/2-x 3(ω2-ω1)/6h P=-25.08kN ²m V x =ω1x+x 2(ω2-ω1)/2h P -N 1=-17.14kN(5)构件内力汇总表(四)截面设计1、顶板 (B-C)钢筋按左、右对称,用最不利荷载计算。
(1)跨中l 0 =3.28 m ,h =0.30 m , a =0.03 m ,h 0 =0.27 m ,b =1.00 m ,M d =70.29 kN ²m ,N d =35.66 kN , V d =9.78 kNe 0 = M d /N d =1.971m i =h/121/2=0.087m长细比l 0/i =37.87> 17.5由《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG D62—2004)第5.3.10条ξ1 = 0.2+2.7e 0/h 0 =19.913> 1.0 ,取ξ1 =1.00ξ2 = 1.15-0.01l 0/h =1.041> 1.0 ,取ξ2 =1.00η =1+(l 0/h)2ξ1ξ2h 0/1400e 0η =1.012由《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG D62—2004)第5.3.5条图 L-08图 L-09e = ηe0+h/2-a = 2.114mr0N d e =f cd bx(h0-x/2)67.85 =9200x(0.27-x/2)解得x =0.029 m≤ξb h0 =0.56³0.27 =0.151 m故为大偏心受压构件。
A s = (f cd bx-r0N d)/f sd =0.0008336m2= 833.6mm2μ = 100A s/(bh0) =0.31 %>0.2 %符合《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG D62—2004)第9.1.12条的要求。
选用 φ12 @130 mm,实际 A s =870.0mm20.51³10-3f cu,k1/2bh0 =615.8 kN>r0V d =8.8 kN故抗剪截面符合《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG D62—2004)第5.2.9条的要求。
由《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG D62—2004)第5.2.10条0.50³10-3α2f td bh0 =143.1 kN>r0V d =8.8 kN故可不进行斜截面抗剪承载力的验算,仅需按(JTG D62—2004)第9.3.13条构造要求配置箍筋。
(2)结点l0 =3.28 m ,h = δ+C1 =0.60 m , a =0.03 m ,h0 =0.57 m , b = M d =64.16 kN²m ,N d =35.66 kN, V d =154.18 kNe0 = M d/N d = 1.799mi =h/121/2 =0.173m长细比l0/i =18.94> 17.5由《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG D62—2004)第5.3.10条ξ1 = 0.2+2.7e0/h0 =8.723> 1.0 ,取ξ1 =1.00ξ2 = 1.15-0.01l0/h = 1.095> 1.0 ,取ξ2 =1.00η =1+(l0/h)2ξ1ξ2h0/1400e0η = 1.007由《公路钢筋混凝土及预应力混凝土桥涵设计规范》(JTG D62—2004)第5.3.5条e = ηe0+h/2-a = 2.082mr0N d e =f cd bx(h0-x/2)66.80 =9200x(0.57-x/2)解得x =0.013 m≤ξb h0 =0.56³0.57 =0.319 m故为大偏心受压构件。
A s = (f cd bx-r0N d)/f sd =0.0003087m2= 308.7mm2μ = 100A s/(bh0) =0.05 %<0.2 %应按最小配筋率配置受拉钢筋。