塔吊扶墙附着计算书

NN次附着附着点高h1(m)附着点净h01(m)风压等效高度变化zμ工作状态风荷载体sμ非工作状μs”工作状态风振系数z非工作状态风振系ββ”z工作状态非工作状风压等效态风压等均布线荷效均布线载标准值荷载标准1次附着计算依据：1、《塔式起重机混凝土根底工程技术规程》JGJ/T187-20232、《钢构造设计标准》GB50017-2023一、塔机附着杆参数塔机型号塔机型号塔身桁架构造类型方钢管H(m)121.5B(m)1.6l1(m)起重臂与平衡臂截面计算高度h(m)57l2(m)12.91.06工作状态时回转惯性力产生的扭矩标准值Tk1(kN·m)270工作状态倾覆力矩标准值Mk(kN·m)1039Mk”(kN*m)1668附着杆数附着杆数四杆附着附墙杆类型Ⅰ类附墙杆截面类型格构柱C(m)1.8二、风荷载及附着参数附着次数附着次数N31a1(m)71b1(m)10.92a2(m)72b2(m)9.13a3(m)7.53b3(m)9.14a4(m)ω0(kN/m2)α07.50.24b4(m)ω0”(kN/m2)10.910.35qskqsk”30300.7741.951.951.791.9450.291.57855250.9221.951.951.7551.9120.3391.84880251.0611.951.951.7121.8770.3812.088110301.2061.951.951.6951.8630.4292.355着着悬臂端塔机附着立面图三、工作状态下附墙杆内力计算1、在平衡臂、起重臂高度处的风荷载标准值qkqk=0.8βzμzμsω0α0h=0.8×1.695×1.206×1.95×0.2×0.35×1.06=0.237kN/m2TkTk2Tk2=1/2qkl12-1/2qkl22=1/2×0.237×572-1/2×0.237×12.92=365.287kN·m0.9〕Tk=0.9(Tk1Tk2)=0.9×(270+365.287)=571.758kN·m3、附着支座反力计算计算简图

剪力图在工作状态下，塔机起重臂位置的不确定性以及风向的随机性，在计算支座4处锚固环截面内力时需考虑塔身承受双向的风荷载和倾覆力矩及扭矩。4、附墙杆内力计算4处的附墙杆担当Nw=20.5RE=110.273kN。计算简图：α1=arctan(b1/a1)=57.291°α2=arctan(b2/a2)=52.431°α3=arctan(b3/a3)=50.505°α4=arctan(b4/a4)=55.469°β1=arctan((b1-c/2)/(a1+c/2))=51.691°β2=arctan((b2+c/2)/(a2+c/2))=51.691°β3=arctan((b3+c/2)/(a3+c/2))=49.97°β4=arctan((b4-c/2)/(a4+c/2))=49.97°T4杆并代以相应多余未知。δ11×X1+Δ1p=0X1=1时，各杆件轴力计算：T11×sin(α1-β1)×(b1-c/2)/sinβ1+T21×sin(α2-β2)×(b2+c/2)/sinβ2-T31×sin(α3-β3)×(b3+c/2)/sinβ3-1×sin(α4-β4)×(b4-c/2)/sinβ4=0T11×cosα1×c-T31×sinα3×c-1×cosα4×c-1×sinα4×c=0、Tk同时存在时，θ0～360°循环，各杆件轴力计算：T1p×sin(α1-β1)×(b1-c/2)/sinβ1+T2p×sin(α2-β2)×(b2+c/2)/sinβ2-T3p×sin(α3-β3)×(b3+c/2)/sinβ3-Tk=0T1p×cosα1×c-T3p×sinα3×c-Nw×sinθ×c/2+Nw×cosθ×c/2-Tk=0T2p×cosα2×c-T3p×sinα3×c+T3p×cosα3×c-Nw×sinθ×c/2-Nw×cosθ×c/2-Tk=0δ11=Σ(T2L/(EA))=T

2(a/cosα

)/(EA)+T 2(a/cosα

)/(EA)+T 2(a/cosα

)/(EA)+12(a14/cosα4)/(EA)

11 1

21 2

31 3 3Δ1p=Σ(T1×TpL/(EA))=T11×T1p(a1/cosα1)/(EA)+T21×T2p(a2/cosα2)/(EA)+T31×T3p(a3/cosα3)/(EA)X1=-Δ1p/δ11各杆轴力计算公式如下：T1=T11×X1+T1p，T2=T21×X1+T2p，T3=T31×X1+T3p，T4=X1按图上方向设置时求解各杆最大轴拉力和轴压力：T1=272.152kN，T2=0kN，T3=350.549kN，T4=0kNT1=0kN，T2=312.799kN，T3=0kN，T4=304.904kN按图上反方向设置时求解各杆最大轴拉力和轴压力：T1=0kN，T2=312.799kN，T3=0kN，T4=304.904kNT1=272.152kN，T2=0kN，T3=350.548kN，T4=0kN四、非工作状态下附墙杆内力计算载产生的扭转力矩。1、附着支座反力计算计算简图剪力图得：RE=223.29kN2、附墙杆内力计算支座4处锚固环的水平内力Nw=RE=223.29kN。依据工作状态方程组Tk=0，θ由0～360°循环，求解各杆最大轴拉力和轴压力：最大轴压力T1=65.355kN，T2=206.929kN，T3=219.383kN，T4=71.665kNT1=65.354kN，T2=206.93kN，T3=219.382kN，T4=71.665kN五、附墙杆强度验算格构柱参数格构柱参数格构柱截面类型四肢格构柱缀件形式缀条l01(mm)400a(mm)500格构柱分肢参数格构柱分肢材料L63X5iy0(cm)1.25A0(cm2)I0(cm4)fy(N/mm2)6.1423.17Z0(cm)1.74345分肢材料抗拉、压强度设计值f(N/mm2)345格构柱缀件参数缀条材料Az(cm2)L50X54.8inim(cm)0.98附图如下:塔机附着格构柱截面1杆件轴心受拉强度验算A=4A0=4×6.14×100=2456mm2σ=N/A=350549/2456=142.732N/mm2≤[f]=345N/mm2满足要求！2格构式钢柱换算长细比验算1的计算长度：L0=(a12+b12)0.5=12954.15mm整个格构柱截面对X、Y轴惯性矩：Ix=4[I0+A0(a/2-Z0)2]=4×[23.17+6.14×(50/2-1.74)2]=13380.318cm4整个构件长细比：λx=λy=L0/(Ix/(4A0))0.5=1295.415/(13380.318/(4×6.14))0.5=55.5分肢长细比：λ1=l01/iy0=40/1.25=32分肢毛截面积之和：A=4A0=4×6.14×100=2456mm2A1x=2Az=2×480=960mm2格构式钢柱绕两主轴的换算长细比：x λ0max=(λ2+40A/A )0.5=(55.52+40×2456/960)0.5x 附墙杆1长细比：λ01max=56.414≤[λ]=100，查标准表得：φ1=0.761满足要求！附墙杆2长细比：λ02max=50.217≤[λ]=100，查标准表得：φ2=0.803满足要求！附墙杆3长细比：λ03max=51.525≤[λ]=100，查标准表得：φ3=0.794满足要求！附墙杆4长细比：λ04max=57.582≤[λ]=100，查标准表得：φ4=0.752满足要求！1轴心受压稳定系数：σ1＝N1/(φ1A)=272152/(0.761×2456)=145.612N/mm2≤[f]=345N/mm2满足要求！2轴心受压稳定系数：σ2＝N2/(φ2A)=312799/(0.803×2456)=158.607N/mm2≤[f]=345N/mm2满足要求！3轴心受压稳定系数：σ3＝N3/(φ3A)=350548/(0.794×2456)=179.762N/mm2≤[f]=345N/mm2满足要求！4轴心受压稳定系数：σ4＝N4/(φ4A)=304904/(0.752×2456)=165.089N/mm2≤[f]=345N/mm2满足要求！3、格构式钢柱分肢的长细比验算1钢柱分肢的长细比：λ1=32≤0.7λ01max=0.7×56.414=39.49满足要求！2钢柱分肢的长细比：λ2=32≤0.7λ02max=0.7×50.217=35.152满足要求！3钢柱分肢的长细比：λ3=32≤0.7λ03max=0.7×51.525=36.067满足要求！4钢柱分肢的长细比：λ4=32≤0.7λ04max=0.7×57.582=40.307满足要求！六、附着杆与构造连接节点验算附着杆与建筑物连接方式附着杆与建筑物连接方式铰接dt(mm)16连接钢板强度等级Q345建筑物混凝土强度等级C35连接板固定方式连接板固定方式锚固螺栓连接板耳板排数24锚固螺栓类型地脚螺栓3052826fv(N/mm)b214017011地脚螺栓承载力计算下：7.062kN;0.717kN;2.95kN;2.835kN;Nvb=nvπd2fvb/4=1×3.14×282×140×10-3/4=86.162kNNtb=nvπde2ftb/4=1×3.14×262×170×10-3/4=90.212kNNcb=d×dt×fcb=28×16×305×10-3=136.64kN1：NV=V1/n=147.062/4=36.766kNNt=F1/n=228.997/4=57.249kN[(NV/Nvb)2+(Nt/Ntb)2]0.5=[(36.766/86.162)2+(57.249/90.212)2]0.5=0.765≤1NV=36.766kN≤Ncb=136.64kN2：NV=V2/n=190.717/4=47.679kNNt=F2/n=247.932/4=61.983kNv t /Nb)2]0.5=[(47.679/86.162)2+(61.983/90.212)2]0.5v t NV=47.679kN≤Ncb=136.64kN3：NV=V3/n=222.95/4=55.738kNNt=F3/n=270.513/4=67.628kNv t /Nb)2]0.5=[(55.738/86.162)2+(67.628/90.212)2]0.5v t NV=55.738kN≤Ncb=136.64kN4：NV=V4/n=172.835/4=43.209kNNt=F4/n=251.186/4=62.797kNv t /Nb)2]0.5=[(43.209/86.162)2+(62.797/90.212)2]0.5v t NV=43.209kN≤Ncb=136.64kN承压承载力满足要求。D(mm)50[σ](N/mmD(mm)50[σ](N/mm2)205[τ](N/mm2)