【正文】 88888888 單板抗拔穩(wěn)定系數(shù) （合格），條件困難時(shí)刻適當(dāng)減小。=，H=8m, ,得到，t=，考慮錨固需要及施工附加荷載的影響，當(dāng)墻高大于6m，取厚度為30cm。三到四個(gè)比較熟練的工人一天能完成20到40平方米的墻面施工（包含所有的施工工序）。致謝歷時(shí)將近兩個(gè)月的時(shí)間終于將這篇論文寫完，在論文的寫作過(guò)程中遇到了無(wú)數(shù)的困難和障礙，都在同學(xué)和老師的幫助下度過(guò)了。由于我的學(xué)術(shù)水平有限，所寫論文難免有不足之處，懇請(qǐng)各位老師和學(xué)友批評(píng)和指正！參考文獻(xiàn)[1] 50330—[S].北京:人民交通出版社,2002.[2]尉希成,[M].北京:中國(guó)建筑工業(yè)出版社,2004.[3]陳仲頤,周景星,[M].北京:清華大學(xué)出版社,1994.[4][M].北京:中國(guó)鐵道出版社,2002.[5][M].北京:人民交通出版社,2002.[6][M].北京：人民交通出版社.[7] [S].J1272001.[8][M].北京:中國(guó)鐵道出版社，2007.[9] [S].J447205.[10][M].北京:中國(guó)鐵道出版社,1992.[11]池書蘭,[M].北京:中國(guó)鐵道出版社,1007.[12]薛殿基,[M].北京:中國(guó)建筑工業(yè)出版社,2008.MODEL TESTSA useful tool in engineering is the analysis of the behavior of a structure by doing a model test, at a reduced scale. The purpose of the test may be just to investigate a phenomenon in a qualitative way, but more often its purpose is to obtain quantitative information. In that case the scale rules must be known. For a soil a special difficulty is that the mechanical properties often depend upon the state of stress, which is determined to a large extent by the weight of the soil itself. This means that in a scale model the soil properties are not well represented, because in the model the stresses are much smaller than in reality (the prototype).An ingenious way to simulate the stresses in a model is to increase gravity, by placing the scale model in a centrifuge, in which the model is rotated at high speed. The principles of this method are briefly presented in this chapter. Some attention is also paid to 1gtesting, the testing of a model without scaling gravity. It will appear that in some situations this can be useful method of model testing.The scale rules of a ceratin area of physics can be derived by considering the basic equations that fully describe a certain process, and then taking care that all relevant terms in each of the equations are scaled by the same factor. The equations describing the process may be partly symbolic, if a detailed description can not be given, but the character of the relations is known. It is essential that all important factors are taken into account. Less important factors may be disregarded, if their small influence can be demonstrated.Simple scale modelsOne of the most important properties of soils is that it may shear, possibly up to very large deformations, and that this shear is caused by the relative magnitude of the shear stress, pared to the normal stress. In Coulomb’s failure criterion ()this appears if the first term, the cohesion c, is very small. This is the case for sand. In that case one may write ()It appears that failure is determined only by a ratio of the stresses, not by their magnitude. This does not necessarily mean that the ratio of shear stress to normal stress determines the soil behavior throughout the entire range from zero deformation to failure. For very small deformations the behavior is more or less elastic, and it is not certain that in that range the ratio is the only parameter that governs the deformations. However, there is much evidence that the stiffness of soils increases with the stress level, both in shear as in pression (pare Terzaghi’s logarithmic pression formula). Thus, it is not unreasonable to assume, at least for sandy soils, that the deformations can be described by a formula of the character (48..3)where is an invariant of the stress tensor, say the isotropic stress. This means that the deformations are determined only by the ratio of the shear stressesand a characteristic normal stress, say the isotropic stress. For sands this is a useful approximation. It may be noted that in pression the deformation is also determined by a stress ratio, in this case the ratio of the stress to the initial stress. The assumption excludes effects as consolidation, creep and dilatancy. These must be small pared to shear and primary pression for the assumption () to be valid. Examples of problems for which the assumption is valid are a laterally loaded pile, or a caisson loaded by cyclic forces.Figure : Model test.If all the spatial dimensions are scaled down by a factor nL, . ()the equations of equilibrium, including the term representing the weight of the material, are satisfied if the scale factor for the stresses is also nL ()This can be verified by noting that the equations of equilibrium consist of terms of type and the gravity term . All these terms now are identical in the model and in the prototype.,If the relation between stresses and strain is of the form (), the deformations are represented at scale 1, ()Because the deformations are related to the displacements by derivatives with respect to the spatial coordinates (for example),the displacements are at the same scale as a length, ()In each of the relevant equations (equilibrium, patibility and constitutive equations) the ratio of all terms in the model is the same as the corresponding terms in the prototype. This means that it is indeed possible to study the behavior of the prototype in a scale model. The boundary values of stress and deformations must also be applied using the scale nL.A problem that can be studied in this way is a laterally loaded pile, see Figure . Compression is not important in this case, so that it is unlikely that pore water pressures will be generated. The determining factor for the deformations is the ratio of shear stress to normal stress. In the model these ratios will be the same as in the prototype if the material is the same. The deformations then are at scale 1. Similarly,problems of sheet pile walls, or retaining walls, can be studied by 1gmodels, if the material is noncoh