水池计算书水池概况：有顶盖地上式矩形水池，平面尺寸5m ×8.2m ，池壁净高6.0m ，设计水深5.7m ，材料采用C40混凝土和HRB400级钢筋，工艺提供池内水容重311/w r kN m =,池壁保护层c=35mm.材料及相关条件：混凝土强度等级：C40 219.1/c f N mm = 21.71/t f N m m= 22.39/tk f N mm = 226.8/ck f N mm = 423.2510/c E N mm =⨯钢筋 HRB400级 2360/y f N m m= 522.010/s E N mm =⨯ 荷载计算：池内水压力:011 5.762.7(/)wk w p r H kN m ==⨯=1.27 1.2762.779.63(/)wk p p kN m ==⨯=水内CB1轴水池池壁计算 1.池壁厚度估算57000.6958200H l == , 按三边固定，顶部自由计算，查表得0.0276α=-, 0220.027679.63 5.771.41(./)y M pl kN m m α==-⨯⨯=-假设配筋率00.5%s A bh ρ==3600.0050.09419.1y Cf f ξρ==⨯= 221(1)1(10.094)0.09022s a ξ----===0.5(10.5(10.94s γ==⨯=0217.4()h mm === 参照此值，池壁厚度为取h=300mm 。

2. 池壁内力计算池壁内力计算（三边固定，顶部自由），h=300mm池壁竖向计算高度取 5.7y l m =水平向计算长度8.2x l m = / 5.7/8.20.695y x l l == 池壁外侧内力计算: 内侧水平向固端弯矩：0220.020479.63 5.752.78(./)x M pl kN m m α==-⨯⨯=-竖向底端固端弯矩：0220.027679.63 5.771.41(./)y M pl kN m m α==-⨯⨯=-竖向跨中弯矩：220.006279.63 5.716.04(./)y M pl kN m m α==⨯⨯= 水平跨中弯矩：220.006779.63 5.717.33(./)x M pl kN m m α==⨯⨯= 3. 池壁配筋计算(1)内侧水平钢筋按受弯构件计算052.78./xM kN m m = 62252.78100.03919.11000265s c o M f bh α⨯===⨯⨯ 相应的ξ=0.039<ξb=0.614。

配筋率ρ=ξ*fc/fy=0.039*19.1/360=0.207% >ρmin=0.2%,需要的钢筋截面积为2min 00.002071000265548.6()S A bh mm ρ==⨯⨯=选配212@150,754S A mm =裂缝宽度验算52.7841.56(./)1.27q M kN m m ==裂缝截面的钢筋应力为：62041.5610239.08(/)0.870.87265754qsq S M N mm h A σ⨯===⨯⨯按有效受拉区计算的受拉钢筋配率：7540.0050.50.51000300S te A bh ρ===⨯⨯,小于0.01,取0.01te ρ=钢筋应变不均匀系数:20.650.65 2.391.1 1.10.450.01239.081tkte sq f ψρσα⨯=-=-=⨯⨯,大于0.4，取0.45ψ=最大裂缝宽度：max 11.8(1.50.11)(1)sqstedc E σωψαυρ=++239.08121.80.45(1.5350.11)(10)0.72000000.01=⨯⨯⨯⨯+⨯⨯+⨯0.1250.2()mm mm =<符合要求 (2)竖向底端钢筋按受弯构件计算071.41./yM kN m m = 62271.41100.05319.11000265s c o M f bh α⨯===⨯⨯ 相应的ξ=0.053<ξb=0.614。

配筋率ρ=ξ*fc/fy=0.053*19.1/360=0.28% >ρmin=0.2%,需要的钢筋截面积为2min 00.00281000265742()S A bh mm ρ==⨯⨯=选配214@150,1026S A mm =裂缝宽度验算71.4156.23(./)1.27q M kN m m ==裂缝截面的钢筋应力为：62056.2310237.7(/)0.870.872651026qsq S M N mm h A σ⨯===⨯⨯按有效受拉区计算的受拉钢筋配率：10260.00680.50.51000300S te A bh ρ===⨯⨯,小于0.01,取0.01te ρ=钢筋应变不均匀系数:20.650.65 2.391.1 1.10.450.01237.71tkte sq f ψρσα⨯=-=-=⨯⨯,大于0.4，取0.45ψ=最大裂缝宽度：max 11.8(1.50.11)(1)sqstedc E σωψαυρ=++237.7141.80.45(1.5350.11)(10)0.72000000.01=⨯⨯⨯⨯+⨯⨯+⨯0.140.2()mm mm =<符合要求(3)竖向跨中钢筋按受弯构件计算16.04./y M kN m m = 62216.04100.01219.11000265s c o M f bh α⨯===⨯⨯ 相应的ξ=0.012<ξb=0.614。

配筋率ρ=ξ*fc/fy=0.012*19.1/360=0.064%<ρmin=0.2% 需要的钢筋截面积为200.0021000265530()S A bh mm ρ==⨯⨯=选配212@200,565S A mm =(4).水平跨中钢筋计算: 水平钢筋按偏心受拉构件计算17.33(./)x M kN m m = 62217.33100.01319.11000265s c o M f bh α⨯===⨯⨯ 相应的ξ=0.013<ξb=0.614。

配筋率ρ=ξ*fc/fy=0.013*19.1/360=0.069%<ρmin=0.2% 需要的钢筋截面积为200.0021000265530()S A bh mm ρ==⨯⨯=选配212@200,565S A mm =CB2轴水池池壁计算荷载计算：池内水压力:011 4.751.7(/)wk w p r H kN m ==⨯=1.27 1.2751.765.66(/)wk p p kN m ==⨯=水内1.池壁厚度估算47000.756300H l == , 按三边固定，顶部自由计算，查表得0.0291α=-, 0220.029165.66 4.742.21(./)y M pl kN m m α==-⨯⨯=-假设配筋率00.5%s A bh ρ==3600.0050.09419.1y Cf f ξρ==⨯=221(1)1(10.094)0.09022s a ξ----===0.5(10.5(10.94s γ==⨯=0167.1()h mm === 参照此值，池壁厚度为取h=300mm 。

2. 池壁内力计算池壁内力计算（三边固定，顶部自由），h=300mm池壁竖向计算高度取 4.7y l m =水平向计算长度 6.3x l m = / 4.7/6.30.75y x l l == 池壁外侧内力计算: 内侧水平向固端弯矩：0220.022265.66 4.732.2(./)x M pl kN m m α==-⨯⨯=-竖向底端固端弯矩：0220.029165.66 4.742.21(./)y M pl kN m m α==-⨯⨯=-竖向跨中弯矩：220.006565.66 4.79.43(./)y M pl kN m m α==⨯⨯= 水平跨中弯矩：220.007665.66 4.711.02(./)x M pl kN m m α==⨯⨯= 3. 池壁配筋计算(1)内侧水平钢筋按受弯构件计算32.2./xM kN m m = 62232.2100.02419.11000265s c o M f bh α⨯===⨯⨯ 相应的ξ=0.024<ξb=0.614。

配筋率ρ=ξ*fc/fy=0.024*19.1/360=0.127% <ρmin=0.2%,需要的钢筋截面积为2min 00.0021000265530()S A bh mm ρ==⨯⨯=选配212@200,565S A mm =裂缝宽度验算32.225.35(./)1.27q M kN m m ==裂缝截面的钢筋应力为：62025.3510194.6(/)0.870.87265565qsq S M N mm h A σ⨯===⨯⨯按有效受拉区计算的受拉钢筋配率：5650.0040.50.51000300S te A bh ρ===⨯⨯,小于0.01,取0.01te ρ=钢筋应变不均匀系数:20.650.65 2.391.1 1.10.30.01194.61tkte sq f ψρσα⨯=-=-=⨯⨯,小于0.4，取0.4ψ=最大裂缝宽度：max 11.8(1.50.11)(1)sqstedc E σωψαυρ=++194.6121.80.4(1.5350.11)(10)0.72000000.01=⨯⨯⨯⨯+⨯⨯+⨯0.090.2()mm mm =<符合要求 (2)竖向底端钢筋按受弯构件计算042.21./yM kN m m = 62242.21100.03119.11000265s c o M f bh α⨯===⨯⨯ 相应的ξ=0.031<ξb=0.614。

配筋率ρ=ξ*fc/fy=0.031*19.1/360=0.164% <ρmin=0.2%,需要的钢筋截面积为2min 00.0021000265530()S A bh mm ρ==⨯⨯=选配212@200,565S A mm =裂缝宽度验算42.2133.24(./)1.27q M kN m m ==裂缝截面的钢筋应力为：62033.2410255.2(/)0.870.87265565qsq S M N mm h A σ⨯===⨯⨯按有效受拉区计算的受拉钢筋配率：5650.0040.50.51000300S te A bh ρ===⨯⨯,小于0.01,取0.01te ρ=钢筋应变不均匀系数:20.650.65 2.391.1 1.10.490.01255.21tkte sq f ψρσα⨯=-=-=⨯⨯,大于0.4，取0.49ψ=最大裂缝宽度：max 11.8(1.50.11)(1)sqstedc E σωψαυρ=++255.2121.80.49(1.5350.11)(10)0.72000000.01=⨯⨯⨯⨯+⨯⨯+⨯0.1450.2()mm mm =<符合要求(3)竖向跨中钢筋按受弯构件计算9.43./y M kN m m = 6229.43100.00719.11000265s c o M f bh α⨯===⨯⨯ 相应的ξ=0.007<ξb=0.614。