楼板结构构件计算书

结构构件计算书一、构件编号: LB-1二、示意图三、依据规范 建筑结构荷载规范 GB50009-2001 混凝土结构设计规范 GB50010-2010四、计算信息 1.几何参数 计算跨度: Lx = 7200 mm; Ly = 4200 mm 板厚: h = 130 mm 2.材料信息 混凝土等级: C30 fc=14.3N/mm2 ft=1.43N/mm2 ftk=2.01N/mm2 Ec=3.00104N/mm2 钢筋种类: HRB335 fy = 300 N/mm2 Es = 2.0105 N/mm2 最小配筋率: = 0.214% 纵向受拉钢筋合力点至近边距离: as = 20mm 保护层厚度: c = 10mm 3.荷载信息(均布荷载) 永久荷载分项系数: G = 1.200 可变荷载分项系数: Q = 1.400 准永久值系数: q = 1.000 永久荷载标准值: gk = 7.000kN/m2 可变荷载标准值: qk = 4.000kN/m2 4.计算方法:弹性板 5.边界条件(上端/下端/左端/右端):固定/固定/固定/固定 6.设计参数 结构重要性系数: o = 1.00 泊松比: = 0.200五、计算参数: 1.计算板的跨度: Lo = 4200 mm 2.计算板的有效高度: ho = h-as=130-20=110 mm六、配筋计算(lx/ly=7200/4200=1.7142.000 所以按双向板计算): 1.X向底板钢筋 1) 确定X向板底弯矩 Mx = 表中系数(G*gk+Q*qk)*Lo2 = (0.0069+0.0373*0.200)*(1.200*7.000+1.400*4.000)*4.22 = 3.555 kN*m 2) 确定计算系数 s = o*Mx/(1*fc*b*ho*ho) = 1.00*3.555106/(1.00*14.3*1000*110*110) = 0.021 3) 计算相对受压区高度 = 1-sqrt(1-2*s) = 1-sqrt(1-2*0.021) = 0.021 4) 计算受拉钢筋面积 As = 1*fc*b*ho*/fy = 1.000*14.3*1000*110*0.021/300 = 109mm2 5) 验算最小配筋率 = As/(b*h) = 109/(1000*130) = 0.084% min = 0.214% 不满足最小配筋要求 所以取面积为As = min*b*h = 0.214%*1000*130 = 279 mm2 采取方案8180, 实配面积279 mm2 2.Y向底板钢筋 1) 确定Y向板底弯矩 My = 表中系数(G*gk+Q*qk)*Lo2 = (0.0373+0.0069*0.200)*(1.200*7.000+1.400*4.000)*4.22 = 9.554 kN*m 2) 确定计算系数 s = o*My/(1*fc*b*ho*ho) = 1.00*9.554106/(1.00*14.3*1000*110*110) = 0.055 3) 计算相对受压区高度 = 1-sqrt(1-2*s) = 1-sqrt(1-2*0.055) = 0.057 4) 计算受拉钢筋面积 As = 1*fc*b*ho*/fy = 1.000*14.3*1000*110*0.057/300 = 298mm2 5) 验算最小配筋率 = As/(b*h) = 298/(1000*130) = 0.229% min = 0.214% 满足最小配筋要求 采取方案8160, 实配面积314 mm2 3.X向支座左边钢筋 1) 确定左边支座弯矩 Mox = 表中系数(G*gk+Q*qk)*Lo2 = 0.0571*(1.200*7.000+1.400*4.000)*4.22 = 14.101 kN*m 2) 确定计算系数 s = o*Mox/(1*fc*b*ho*ho) = 1.00*14.101106/(1.00*14.3*1000*110*110) = 0.081 3) 计算相对受压区高度 = 1-sqrt(1-2*s) = 1-sqrt(1-2*0.081) = 0.085 4) 计算受拉钢筋面积 As = 1*fc*b*ho*/fy = 1.000*14.3*1000*110*0.085/300 = 446mm2 5) 验算最小配筋率 = As/(b*h) = 446/(1000*130) = 0.343% min = 0.214% 满足最小配筋要求 采取方案8110, 实配面积457 mm2 4.X向支座右边钢筋 1) 确定右边支座弯矩 Mox = 表中系数(G*gk+Q*qk)*Lo2 = 0.0571*(1.200*7.000+1.400*4.000)*4.22 = 14.101 kN*m 2) 确定计算系数 s = o*Mox/(1*fc*b*ho*ho) = 1.00*14.101106/(1.00*14.3*1000*110*110) = 0.081 3) 计算相对受压区高度 = 1-sqrt(1-2*s) = 1-sqrt(1-2*0.081) = 0.085 4) 计算受拉钢筋面积 As = 1*fc*b*ho*/fy = 1.000*14.3*1000*110*0.085/300 = 446mm2 5) 验算最小配筋率 = As/(b*h) = 446/(1000*130) = 0.343% min = 0.214% 满足最小配筋要求 采取方案8110, 实配面积457 mm2 5.Y向上边支座钢筋 1) 确定上边支座弯矩 Moy = 表中系数(G*gk+Q*qk)*Lo2 = 0.0800*(1.200*7.000+1.400*4.000)*4.22 = 19.757 kN*m 2) 确定计算系数 s = o*Moy/(1*fc*b*ho*ho) = 1.00*19.757106/(1.00*14.3*1000*110*110) = 0.114 3) 计算相对受压区高度 = 1-sqrt(1-2*s) = 1-sqrt(1-2*0.114) = 0.122 4) 计算受拉钢筋面积 As = 1*fc*b*ho*/fy = 1.000*14.3*1000*110*0.122/300 = 637mm2 5) 验算最小配筋率 = As/(b*h) = 637/(1000*130) = 0.490% min = 0.214% 满足最小配筋要求 采取方案10120, 实配面积654 mm2 6.Y向下边支座钢筋 1) 确定下边支座弯矩 Moy = 表中系数(G*gk+Q*qk)*Lo2 = 0.0800*(1.200*7.000+1.400*4.000)*4.22 = 19.757 kN*m 2) 确定计算系数 s = o*Moy/(1*fc*b*ho*ho) = 1.00*19.757106/(1.00*14.3*1000*110*110) = 0.114 3) 计算相对受压区高度 = 1-sqrt(1-2*s) = 1-sqrt(1-2*0.114) = 0.122 4) 计算受拉钢筋面积 As = 1*fc*b*ho*/fy = 1.000*14.3*1000*110*0.122/300 = 637mm2 5) 验算最小配筋率 = As/(b*h) = 637/(1000*130) = 0.490% min = 0.214% 满足最小配筋要求 采取方案10120, 实配面积654 mm2七、跨中挠度计算： Mk - 按荷载效应的标准组合计算的弯矩值 Mq - 按荷载效应的准永久组合计算的弯矩值 1.计算荷载效应 Mk = Mgk + Mqk = (0.0373+0.0069*0.200)*(7.000+4.000)*4.22 = 7.507 kN*m Mq = Mgk+q*Mqk = (0.0373+0.0069*0.200)*(7.000+1.0*4.000)*4.22 = 7.507 kN*m 2.计算受弯构件的短期刚度 Bs 1) 计算按荷载荷载效应的两种组合作用下，构件纵向受拉钢筋应力 sk = Mk/(0.87*ho*As) 混规(7.1.4-3) = 7.507106/(0.87*110*314) = 249.811 N/mm sq = Mq/(0.87*ho*As) 混规(7.1.4-3) = 7.507106/(0.87*110*314) = 249.811 N/mm 2) 计算按有效受拉混凝土截面面积计算的纵向受拉钢筋配筋率 矩形截面积: Ate = 0.5*b*h = 0.5*1000*130= 65000mm2 te = As/Ate 混规(7.1.2-4) = 314/65000 = 0.483% 3) 计算裂缝间纵向受拉钢筋应变不均匀系数 k = 1.1-0.65*ftk/(te*sk) 混规(7.1.2-2) = 1.1-0.65*2.01/(0.483%*249.811) = 0.017 因为不能小于最小值0.2，所以取k = 0.2 q = 1.1-0.65*ftk/(te*sq) 混规(7.1.2-2) = 1.1-0.65*2.01/(0.483%*249.811) = 0.017 因为不能小于最小值0.2，所以取q = 0.2 4) 计算钢筋弹性模量与混凝土模量的比值 E E = Es/Ec = 2.0105/3.00104 = 6.667 5) 计算受压翼缘面积与腹板有效面积的比值 f 矩形截面，f=0 6) 计算纵向受拉钢筋配筋率 = As/(b*ho)= 314/(1000*110) = 0.285% 7) 计算受弯构件的短期刚度 Bs Bsk = Es*As*ho2/1.15k+0.2+6*E*/(1+ 3.5f)(混规(7.2.3-1) = 2.0105*314*1102/1.15*0.017+0.2+6*6.667*0.285%/(1+3.5*0.0) = 1.396103 kN*m2 Bsq = Es*As*ho2/1.15q+0.2+6*E*/(1+ 3.5f)(混规(7.2.3-1) = 2.0105*314*1102/1.15*0.017+0.2+6*6.667*0.285%/(1+3.5*0.0) = 1.396103 kN*m2 3.计算受弯构件的长期刚度B 1) 确定考虑荷载长期效应组合对挠度影响增大影响系数 当=0时，=2.0 混规(7.2.5) 2) 计算受弯构件的长期刚度 B Bk = Mk/(Mq*(-1)+Mk)*Bs (混规(7.2.2-1) = 7.507/(7.507*(2.0-1)+7.507)*1.396103 = 6.982102 kN*m2 Bq = Bsq/ (混规(7.2.2-2) = 1.396103/2.0 = 6.982102 kN*m2 B = min(Bk,Bq) = min(698.186,698.186) = 698.186 4.计算受弯构件挠度 fmax = f*(qgk+qqk)*Lo4/B = 0.00239*(7.000+4.000)*4.24/6.982102 = 11.733mm 5.验算挠度 挠度限值fo=Lo/200=4200/200=21.000mm fmax=11.733mmfo=21.000mm，满足规范要求!八、裂缝宽度验算: 1.跨中X方向裂缝 1) 计算荷载效应 Mx = 表中系数(gk+qk)*Lo2 = (0.0069+0.0373*0.200)*(7.000+1.00*4.000)*4.22 = 2.793 kN*m 2) 光面钢筋,所以取值vi=0.7 3) 因为C 20，所以取C = 20 4) 计算按荷载效应的准永久组合作用下，构件纵向受拉钢筋应力 sq=Mq/(0.87*ho*As) 混规(7.1.4-3) =2.793106/(0.87*110*279) =104.601N/mm 5) 计算按有效受拉混凝土截面面积计算的纵向受拉钢筋配筋率 矩形截面积，Ate=0.5*b*h=0.5*1000*130=65000 mm2 te=As/Ate 混规(7.1.2-4) =279/65000 = 0.0043 因为te=0.0043 0.01,所以让te=0.01 6) 计算裂缝间纵向受拉钢筋应变不均匀系数 =1.1-0.65*ftk/(te*sq) 混规(7.1.2-2) =1.1-0.65*2.010/(0.0100*104.601) =-0.149 因为=-0.149 0.2,所以让=0.2 7) 计算单位面积钢筋根数n n=1000/dist = 1000/180 =5 8) 计算受拉区纵向钢筋的等效直径deq deq= (ni*di2)/(ni*vi*di) =5*8*8/(5*0.7*8)=11 9) 计算最大裂缝宽度 max=cr*sq/Es*(1.9*C+0.08*Deq/te) (混规(7.1.2-1) =1.9*0.200*104.601/2.0105*(1.9*20+0.08*11/0.0100) =0.0257mm 0.30, 满足规范要求 2.跨中Y方向裂缝 1) 计算荷载效应 My = 表中系数(gk+qk)*Lo2 = (0.0373+0.0069*0.200)*(7.000+1.00*4.000)*4.22 = 7.507 kN*m 2) 光面钢筋,所以取值vi=0.7 3) 因为C 20，所以取C = 20 4) 计算按荷载效应的准永久组合作用下，构件纵向受拉钢筋应力 sq=Mq/(0.87*ho*As) 混规(7.1.4-3) =7.507106/(0.87*110*314) =249.811N/mm 5) 计算按有效受拉混凝土截面面积计算的纵向受拉钢筋配筋率 矩形截面积，Ate=0.5*b*h=0.5*1000*130=65000 mm2 te=As/Ate 混规(7.1.2-4) =314/65000 = 0.0048 因为te=0.0048 0.01,所以让te=0.01 6) 计算裂缝间纵向受拉钢筋应变不均匀系数 =1.1-0.65*ftk/(te*sq) 混规(7.1.2-2) =1.1-0.65*2.010/(0.0100*249.811) =0.577 7) 计算单位面积钢筋根数n n=1000/dist = 1000/160 =6 8) 计算受拉区纵向钢筋的等效直径deq deq= (ni*di2)/(ni*vi*di) =6*8*8/(6*0.7*8)=11 9) 计算最大裂缝宽度 max=cr*sq/Es*(1.9*C+0.08*Deq/te) (混规(7.1.2-1) =1.9*0.577*249.811/2.0105*(1.9*20+0.08*11/0.0100) =0.1772mm 0.30, 满足规范要求 3.支座上方向裂缝 1) 计算荷载效应 Moy = 表中系数(gk+qk)*Lo2) = 0.0800*(7.000+1.00*4.000)*4.22 = 15.523 kN*m 2) 光面钢筋,所以取值vi=0.7 3) 因为C 20，所以取C = 20 4) 计算按荷载效应的准永久组合作用下，构件纵向受拉钢筋应力 sq=Mq/(0.87*ho*As) 混规(7.1.4-3) =15.523106/(0.87*110*654) =248.023N/mm 5) 计算按有效受拉混凝土截面面积计算的纵向受拉钢筋配筋率 矩形截面积，Ate=0.5*b*h=0.5*1000*130=65000 mm2 te=As/Ate 混规(7.1.2-4) =654/65000 = 0.0101 6) 计算裂缝间纵向受拉钢筋应变不均匀系数 =1.1-0.65*ftk/(te*sq) 混规(7.1.2-2) =1.1-0.65*2.010/(0.0101*248.023) =0.576 7) 计算单位面积钢筋根数n n=1000/dist = 1000/120 =8 8) 计算受拉区纵向钢筋的等效直径deq deq= (ni*di2)/(ni*vi*di) =8*10*10/(8*0.7*10)=14 9) 计算最大裂缝宽度 max=cr*sq/Es*(1.9*C+0.08*Deq/te) (混规(7.1.2-1) =1.9*0.576*248.023/2.0105*(1.9*20+0.08*14/0.0101) =0.2059mm 0.30, 满足规范要求 4.支座下方向裂缝 1) 计算荷载效应 Moy = 表中系数(gk+qk)*Lo2 = 0.0800*(7.000+1.00*4.000)*4.22 = 15.523 kN*m 2) 光面钢筋,所以取值vi=0.7 3) 因为C 20，所以取C = 20 4) 计算按荷载效应的准永久组合作用下，构件纵向受拉钢筋应力 sq=Mq/(0.87*ho*As) 混规(7.1.4-3) =15.523106/(0.87*110*654) =248.023N/mm 5) 计算按有效受拉混凝土截面面积计算的纵向受拉钢筋配筋率 矩形截面积，Ate=0.5*b*h=0.5*1000*130=65000 mm2 te=As/Ate 混规(7.1.2-4) =654/65000 = 0.0101 6) 计算裂缝间纵向受拉钢筋应变不均匀系数 =1.1-0.65*ftk/(te*sq) 混规(7.1.2-2) =1.1-0.65*2.010/(0.0101*248.023) =0.576 7) 计算单位面积钢筋根数n n=1000/dist = 1000/120 =8 8) 计算受拉区纵向钢筋的等效直径deq deq= (ni*di2)/(ni*vi*di) =8*10*10/(8*0.7*10)=14 9) 计算最大裂缝宽度 max=cr*sq/Es*(1.9*C+0.08*Deq/te) (混规(7.1.2-1) =1.9*0.576*248.023/2.0105*(1.9*20+0.08*14/0.0101) =0.2059mm 0.30, 满足规范要求 5.支座左方向裂缝 1) 计算荷载效应 Mox = 表中系数(gk+qk)*Lo2 = 0.0571*(7.000+1.00*4.000)*4.22 = 11.080 kN*m 2) 光面钢筋,所以取值vi=0.7 3) 因为C 20，所以取C = 20 4) 计算按荷载效应的准永久组合作用下，构件纵向受拉钢筋应力 sq=Mq/(0.87*ho*As) 混规(7.1.4-3) =11.080106/(0.87*110*457) =253.337N/mm 5) 计算按有效受拉混凝土截面面积计算的纵向受拉钢筋配筋率 矩形截面积，Ate=0.5*b*h=0.5*1000*130=65000 mm2 te=As/Ate 混规(7.1.2-4) =457/65000 = 0.0070 因为te=0.0070 0.01,所以让te=0.01 6) 计算裂缝间纵向受拉钢筋应变不均匀系数 =1.1-0.65*ftk/(te*sq) 混规(7.1.2-2) =1.1-0.65*2.010/(0.0100*253.337) =0.584 7) 计算单位面积钢筋根数n n=1000/dist = 1000/110 =9 8) 计算受拉区纵向钢筋的等效直径deq deq= (ni*di2)/(ni*vi*di) =9*8*8/(9*0.7*8)=11 9) 计算最大裂缝宽度 max=cr*sq/Es*(1.9*C+0.08*Deq/te) (混规(7.1.2-1) =1.9*0.584*253.337/2.0105*(1.9*20+0.08*11/0.0100) =0.1820mm 0.30, 满足规范要求 6.支座右方向裂缝 1) 计算荷载效应 Mox = 表中系数(gk+qk)*Lo2 = 0.0571*(7.000+1.00*4.000)*4.22 = 11.080 kN*m 2) 光面钢筋,所以取值vi=0.7 3) 因为C 20，所以取C = 20 4) 计算按荷载效应的准永久组合作用下，构件纵向受拉钢筋应力 sq=Mq/(0.87*ho*As) 混规(7.1.4-3) =11.080106/(0.87*110*457) =253.337N/mm 5) 计算按有效受拉混凝土截面面积计算的纵向受拉钢筋配筋率 矩形截面积，Ate=0.5*b*h=0.5*1000*130=65000 mm2 te=As/Ate 混规(7.1.2-4) =457/65000 = 0.0070 因为te=0.0070 0.01,所以让te=0.01 6) 计算裂缝间纵向受拉钢筋应变不均匀系数 =1.1-0.65*ftk/(te*sq) 混规(7.1.2-2) =1.1-0.65*2.010/(0.0100*253.337) =0.584 7) 计算单位面积钢筋根数n n=1000/dist = 1000/110 =9 8) 计算受拉区纵向钢筋的等效直径deq deq= (ni*di2)/(ni*vi*di) =9*8*8/(9*0.7*8)=11 9) 计算最大裂缝宽度 max=cr*sq/Es*(1.9*C+0.08*Deq/te) (混规(7.1.2-1) =1.9*0.584*253.337/2.0105*(1.9*20+0.08*11/0.0100) =0.1820mm 0.30, 满足规范要求第9页，共9页