兰州交通大学毕业设计（论文）
7.6＜17.698（与假设公式不符） ﹙2﹚ 设破裂面交于荷载内B点
=0.5×(3.7＋10) ×3.7＋10＋2x3) =134.945
=0.5×3.7×5.55＋(5.55＋0.3＋2.14) ×3+0.5×10×(2×3+2×3.7＋10) ×tan11.3°
=9.7455
=﹣1.0247+ =0.5653
7.7446＜15.574（与假设公式不符） ﹙3﹚ 设破裂面交于荷载外C点
=0.5×(3.7+10) 2 =93.845
=0.5×3.7×5.55－3.7×3＋0.5×10×(2×3.7＋10) ×tan11.3°
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兰州交通大学毕业设计（论文）
=16.5675
=﹣1.0247+ =0.6392
8.757＜11.724（与假设公式不符） ﹙4﹚ 设破裂面交于荷载内E点
=0.5×(3.7+10) ×(3.7+10+2×3) =134.945
=0.5×3.7×5.55+(5.55+2.14) ×3+0.5×10×(2×3+2×3.7+10) ×tan11.3° =56.6895
=﹣1.0247+ =0.9128
12.5＞9.724（与假设公式不符） ﹙5﹚ 设破裂面交于荷载外D点
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兰州交通大学毕业设计（论文）
=0.5×(3.7+10) 2 =93.845
=0.5×3.7×5.55+0.5×10×(2×3.7+10) ×tan11.3° =27.6675
=﹣1.0247+ =0.6588
9.02556＞7.874（与假设公式不符） ﹙6﹚ 设破裂面交于荷载内F点
=0.5×(3.7+10) ×(3.7+10+2×3) =134.945
=0.5×3.7×5.55+(5.55+2.14) ×3+0.5×10×(2×3+2×3.7+10) ×tan11.3° =56.6895
=12.33/13.7
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兰州交通大学毕业设计（论文）
=0.9 θ=42° 土压力系数
=(0.9－0.1998) ×0.174/0.999 =0.1218
=20×(134.945×0.9－56.6895) ×0.174/0.999 =96.6485 kN/m
压力强度图各段高度
=(5.55－3.7×0.9) ÷(0.9－0.1998) =3.15m
=2.124÷(0.9－0.1998) =3.049m
=10－3.15－3.049 =3.801m 土压力作用点距墙趾的高度和水平距离
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兰州交通大学毕业设计（论文）
=[103+3.7×(3×102－3×3.15×10+3.152) ] ÷[3×(102+2×3.7×10－3.7×
3.15+2×3×3.801) ]
=4.67m
=3+4.67×0.1998 =3.934m 各段的压力强度
=20×3×0.1218 =7.9084 kPa
=20×3.7×0.1218 =3.8247 kPa
=20×10×0.1218 =24.3614 kPa 压力强度图形面积
=0.5×(3.15+3.049+3.801) ×24.3614+(3.049+3.801) ×10.7152+3.801×8.688 =95.777 kN/m =墙身自重
=0.5×23×10×(3+1)
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