由于 r0 (N+1.27Qg)<1.6R ∴满足要求
二.桩身承载力验算 1) 桩身轴向力验算
r0 N=0.9×0.745=0.6705MN
2
ψc.fcA=0.8×14.3×0.25×3.14×0.8=5.75MN r0 N=0.6705< fcA=5.75MN ∴满足要求
2）桩身水平力验算
r0 H=0.9×0.01775=0.01598MN αhd(1+
2
n
0.5NG5)1.5d2?0.5d (JGJ94-94第4.1.1.2)
?mftA2
=40×10×0.8×(1+=0.0343MN
-3
0.5?0.74525)1.5?0.8?0.5?0.8
2?1.43?0.25?3.14?0.82r0 H=0.01598MN<αhd(1+
2
0.5NG5)1.5d2?0.5d=0.0343MN
?mftA∴满足要求
3）桩身承载力验算 （1） 桩身计算长度（以表5 5.3-2取） lc=0.7(lc+
4.0?)=0.7×(4.65+
4)=8.98m 0.489(2)桩身稳定性验算
lc8.98==11.226 d0.8查表得：?=0.935
0.9?( fcA+ fy′As′) (GB50010-2002第7.3.1)
=0.9×0.93×(14.3×0.25×л×0.8+300×0.00314)=6.805MN ∵Nmax=0.745MN<0.9?( fcA+ fy′As′)=6.805MN
∴满足要求 （2） 桩身正截面承载力验算(GB50010-2002第7.3.8-2～7.3.10-3) 非工作状况 m=
2
fyfcAs=300=20.979
14.310?0.022ρs==×100%=0.625% 20.8An=
N0.705==0.098 fcA14.3?0.503α=1+0.75mρs –0.52.25m2?s?3.5m?s?2(1?n) =1+0.75×20.979×0.625%
-0.52.25?20.9792?0.625%2?3.5?20.979?0.625%?2(1?0.098) =0.3398
αt=1.25-2α=1.25-2×0.3398=0.570
2sin???sin?t?2sin3??+ fArfcAryss?3?32sin(0.3398??) = ?14.3?0.503?0.4?3?+300?0.00314?0.340?sin0.3398??sin0.570?
?=0.368+0.036=0.404MN e0=
MN=
0.1800.745=0.242
ea=
d30=
0.830=0.026>0.02
∴取ea=0.02m
ei= e0+ ea=0.242+0.02=0.262m
0.5fcA0.5?14.3?0.503?1???4.827?1 ∴取?1=1
N0.745lc8.98??12.135 h0.4?0.34?2?1.15?0.01取?2?1 η=1?lc?1.15?0.01?11.226?1.038 hl( c)2?1?2 eh1400ih011( 12.135)2 0.27414000.7 =1? =1.269
?ei=1.269×0.274=0.348m N?ei=0.745×0.348=0.260MN.m
sin???sin?t?2sin3??∵ N?ei=0.260 MN.m 3??∴满足要求 （3）桩身斜截面承载力验算(GB50010-2002第7.5.12) 非工作状况下 ??M0.228??18.35?3 Vh00.01775?0.7∴取??3 0.3fcA=0.3×14.3×0.503=2.158>0.709MN取N=0.709 Vc?A1.75ft?(1.76r)?(1.6r)?fyv?sv?(1.76r)?0.07N ??1S1.752?5.027?10?5?(1.76?0.40)?1.43?(1.76?0.4)?(1.6?0.4)?210??0.07?0.745=0.2813?10.1509+0.099+0.050=0.3700MN ∵V=0.01775MN< Vc?A1.75ft?(1.76r)?(1.6r)?fyvsv?(1.76r)?0.07N?0.3700MN ??1S∴桩身斜载面承载力满足要求 五、塔吊承台基础验算 一、角桩对承台的冲切验算(GB50007-2002第8.5.17-5～7) 1、桩截面换算：b=0.8d=0.8×800=640mm 2、冲切力：Fl=N桩反力-G承台?0.745?0.025?5?5?1.35?1.2?0.542MN 44 3、角桩冲跨比：?1x??1y?a1x78.0??0.918 h08500.56?0.501 ?1y?0.2 4、角桩冲切系数：?1x??1y? 5、受冲切承载力截面高度影响系数：以插入法求得?hp=0.983 6、承台抗冲切力 F抗?[?1x(c2?a1ya)??1y(c1?1x)]?hpfth0) 222 =2×[0.501×(0.3+0.78)]×0.983×1.43×0.85 =0.826MN ∵Fl=0.542 二、塔吊基础对承台的冲切验算(GB50007-2002第8.5.17-1～4) 1、冲跨比 ?1x??1y?a1x78.0??0.918 h0850 2、冲跨系数 ?0x??0y? 3、抗冲切力 0.840.84??0.751 ?1y?0.20.918?0.2 F抗?2?[?0x(bc?a0y)??0y(hc?a0x)]?hpfth0 =2×[0.751×(1.6+6.103)] ×2×0.983×1.43×0.85 =9.702MN 4、冲切力（以附着最大荷重对应状态考虑） Fl=N塔吊+N标准节=0.628×1.2=0.754MN ∵ Fl=0.754MN 三、斜截面承载力验算(GB50007-2002第8.5.18-1～2) 1、剪跨比 ?x??y? 2、剪切系数 ??700?0.824 8501.751.75??0.959 ??1.00.824?1 3、受剪切承载力截面高度影响系数?hs8008004?(0)4?()?0.985 1200h11 4、斜截面抗剪承载力 V抗=βhsβftbh0=0.985×0.959×1.43×5.2×0.85=5.970MN V<2Nmax=2×0.745=1.49MN ∵ V<2Nmax=1.49< V抗=5.970MN ∴满足要求 四、正截面承载力验算 1、控制内力Mimax<2Nmaxｘi=2×0.745×(0.4+0.7)=1.639MN.m 2、抗弯承载力 根据塔吊说明书要求配置上下纵横各配置27Ф20 则： 2-6 M抗=fyAsrsh0=300×27×0.25×?×20×0.9×0.85×10=1.947MN.m ∵M=1.639MN.m ∴满足要求 五、正常使用极限状态验算 （一）、裂缝控制验算:使用环境按一类考虑，按荷载效应的标准组合计算的弯矩 Mk=2×Nmax.Xi 0.529?0.0139)0.2025=2?[(?]?1.1 1.21.4=1.314 裂缝限值[wlim]=0.3mm (根据GB50010-2002 3.3.4条确定) GB50010-2002 8.1.2条: deg Wmax=acrψbsk(1.9c?0.08)Es?te 式中：acr—构件受力特征系数取2.1 ψ—裂缝间受拉钢筋应变不均匀系数：（有效要求混凝土截面面积取桩距，纵向受拉钢筋按桩距内根距确定） ?=1.1-0.65 =1.1-0.65 tk ??tesktkA?Asp.?skAteff =1.1-0.65× sk0.5?3800?1000其中 1.314?103?1062k?sk???267.087N/mm0.87hA0.87?900?6283.190sM2.016283.19?? ?sk?267.087N/mm2?[?sk]?300N/mm2满足配筋强度要求 As?20?10???6283.19mm2 则?=1.1-0.65× 2.01??0.379 267.0870.5?3800?1000C—最外层纵向受控制钢筋外进缘至受拉区底边的距离： C=100mm>65mm,则取c=65mm 52 Es—钢筋弹性模量 Es=2.0×10N/mmdep?nidi20?202???20mm ?ni?idi20?1.0?202?tc?As?ApAte?6283.19?0.331% 0.5?3800?1000则： Wmax?2.1?0.2?267.08720?(1.9?65?0.08?) 0.331%2.0?105 =0.340mm>[wlim]=0.300mm 为此，提高钢筋强度等级或减少间距，本例拟采用Ф20@150双层双向可满足裂缝控制要求（计算略） （二）、受弯构件挠度验算(GB50010-2002 8.2.2～5)条 短期刚度Bs= EsAsh01.154?0.2?26aE?1?3.5rf' 式中：Es—钢筋弹性模量Es?2.0?10N/mm 2 As=受拉区纵向非预应力钢筋截面面积As?7853.99mm（同裂缝验算数值） 52
塔吊施工方案(完整版)
