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请私信作者！）团队成果，侵权必究！（温馨提示，本文档没有计算功能，请在作者个人中心中下载对应的Excel计算表格，填入基本参数后，Excel表格会计算出各分项结果，并显示计算过程！）tanα1′=H×0.18+B11H1=1.8×0.18+0.991.8=0.73α1′=36.13°β=33.69°h′′=H1sinβ(cotβ+tanα1′)=1.8×sin33.69°×(1.5+0.73)=2.23mh0=0时的破裂角R=cotφcot(2φ+β)+h′′2cos(φ+β)P2(H1+a)2sinφsin(2φ+β){1+(H1+a)2h′′2tan(φ+β)∙[2h′′(H1+a)sinβ−cotβ−h′′2(h1+a)2cotβ]−2P(H1+a)cosφh′′cos(φ+β)}=cot35°cot(70°+33.69°)+ 2.232cos (35°+33.69°)3.82sin35°sin (70°+33.69°)×{1+ 3.822.232×tan(35°+33.69°)×[2×2.233.8sin33.69°−1.5−1.52.2323.82]−2×3.8cos35°2.23×cos (35°+33.69°)}=−1.686 Q=h′′(H1+a)sin (2φ+β)−cot (2φ+β)= 2.233.8×sin (70°+33.69°)−cot (70°+33.69°)=0.848计算第一破裂面倾角：tanβi=−Q+√Q2−R=−0.848+√0.8482+1.686=0.703βi=35°06′04′′=35.10°计算第二破裂面倾角：tan(αi−β)=cot(φ+β)−P(H1+a)cosφh′′sin(φ+β)(1−tanφtanβi)=cot (35°+33.69°)−3.8×cos35°2.23×sin (35°+33.69°)×(1−tan35°×0.703)=−0.371αi −β=−20°20′21′′=−20.34°αi =33.69°−20.34°=13.35°α1′>αi ,验证出现第二破裂面假定条件是否成立：L 0=(H 1+α)tanβi +H 1tanα1′−b =3.8×0.703+1.8×0.73−1.5×2=0.985(m)所以假定条件成立。

b.土压力计算因此可用表3-2-2中出现第二破裂角时的第六类公式计算土压力。

B =5.6+(1.8+2×2)tan30°=8.95(m )(取B =10m)L 0=0.985−0.75−(0.5−0.3)=0.04m在“B ×L 0”范围内布置的车轮重∑G ：∑G =(1500.6)×0.04=10（KN ） h 0=∑G γBL 0=1010×10×0.985=0.06(m)根据求得的h 0重新求第一、第二破裂面倾角：P =√1+2h 0H 1+a =√1+2×0.063.8=1.02Q=h′′P(H1+a)sin (2φ+β)−cot (2φ+β)= 2.231.02×3.8sin (70°+33.69°)=0.836R=cotφcot(2φ+β)+h′′2cos (φ+β)P2(H1+a)2sinφsin (2φ+β){1+(H1+a)2h′′2tan(φ+β)∙[2h′′(H1+a)sinβ−cotβ−h′′2(h1+a)2cotβ]−2P(H1+a)cosφh′′cos (φ+β)}=−1.707tanβi=−Q+√Q2−R=−0.836+√0.8362+1.707=0.7151第一破裂面倾角：βi=35°34′06′′=35.57°tan(αi−β)=cot(φ+β)−P(H1+a)cosφh′′sin(φ+β)(1−tanφtanβi)=cot(35°+33.69°)−1.02×3.8cos35°2.23sin(35°+33.69°)(1−tan35°×0.7151)=−0.373αi−β=−20°27′12′′=−20.45°第二破裂面倾角：αi=33.69°−20.45°=13.24°土压力系数：K=cos(βi+φ)sin(βi+αi+2φ)(tanβi+tanαi)=cos(35.57°+35°)sin(35.57°+13.24°+70°)×(0.715+tan13.24°)=0.361H1′=H1×1+tanα1′tanβ1+tanαi tanβ=1.8×1+0.73×tan33.69°1+tan13.24°tan33.69°=2.31m a′=(H1+a)−H1′=3.8−2.31=1.49b′=a′cotβ=1.49×1.5=2.24mh3= b′−a′tanβitanβi+tanαi=2.24−1.49×0.71510.7151+tan13.24°=1.24m h4=H1′− h3=2.31−1.24=1.07m K1=1+2a′H1(1−h32H1′)+2h0h4H1′2=1+2×1.492.31×(1−1.242×2.31)+2×0.06×1.072.312=1.97E1=12γH1′2KK1=12×18×2.312×0.361×1.97=34.15kNE1x=E1cos(αi+φ)=34.15×cos(13.24°+35°)=22.74kNE1y=E1sin(αi+φ)=34.15×sin(13.24°+35°)=25.47kNE IH=E1cos(αi+φ+i)=34.15×cos(13.24°+35°+11.31°)=17.31kNE IV=E1sin(αi+φ+i)=34.15×sin(13.24°+35°+11.31°)=29.44kN作用于实际墙背上的土压力：E1x′=E1x=22.74kNE1y′=E1x tanα1=22.74×0.18=4.09kN土压力对验算截面的弯矩：Z1x=H1′3+a′(H1′− h3)2+h0h4(3h4−2H1′)3H1′2K1=2.313+1.49×(2.31−1.24)2+0.06×1.07×(3×1.07−2×2.31)3×2.312×1.97=0.82mZ1y=B1−Z1x tanα1=0.91−0.82×0.18=0.76mM E1=E1y′Z1y−E1x′Z1x=4.09×0.76−22.74×0.82=−15.54kN∙mc.上墙自重及弯矩计算W1=12(0.5+0.91)×1.8×23=29.19kNZ1=(0.52+0.5×0.91+0.912)+（1+0.91）×0.05×1.83×（0.5+0.91）=0.40m M w1=29.19×0.40=11.67kN∙mP149 截面应力验算∑N1=E′1y+W1=4.09+29.19=33.28 KN∑H1=E′1X=22.74 KN∑M1=M E1+M W1=−15.54+11.67=−3.87 KN.me1=B12−∑M1∑N1=0.912−−3.8733.28=0.571 m>B14σmin=∑N1B1(1−6e1B1)=33.280.91×(1−6×0.5710.91)=1−101.1 KPa>−110 KPa直剪应力：τ1=∑H1−0.40∑N1B1=22.74−0.4×33.280.91=10.1 KPa<80 KPa斜剪应力：A=∑H1−0.5γK B12+tani∑N1+f(∑H1tani−∑N1)∑H1tani−∑N1−f(∑H1−0.5γK B12+tani∑N1)∑H1−0.5γK B12+tani∑N1=22.74−0.5×23×0.912+0.05×33.28 =14.88∑H1tani−∑N1=22.74×0.05−33.28=−32.14A=14.88+0.4×(−32.14)−32.14−0.4×14.88=−0.053tanθ=A+√A2+1=−0.053+√0.0532+1=0.948θ=43.48°=43°28′59″τ2=cos2θB1[∑H1(1+ftanθ)(1−tanj tanθ)+∑N1(tanθ−f)×(1−tanj tanθ)+12⁄γH B12tanθ(tanθ−f)]=cos243.48°0.91×[22.74×(1+0.4×0.948)(1−0.05×0.948)+33.28×(0.948−0.4)(1−0.05×0.948)+12⁄×23×0.912×0.948×(0.948−0.4)]=0.5786×[29.876+17.373+4.947]=30.2 KPa<80 KPa基底截面强度及稳定性验算上墙及强身计算同前破裂角假设上墙第一破裂面交于荷载内，下墙破裂面亦交于荷载内，查公路设计手册《路基》（第二版）表3-2-3第三类公式得：ψ=φ+α2+δ2=38.46°A=−tanα2=0.25tanβ3=−tanψ+√(cotφ+tanψ)(tanψ+A)=0.72915β3=36.10°=36°05′52″假设条件成立。