Một trong những vấn đề đơn giản giới hạn thấp hơn giới hạn trên có thể được xác định là trường hợp của một tải dải dài vô tận trên một lớp vật liệu gắn kết đồng nhất (φ = 0), xem hình 40,1. Trọng lượng của vật liệu sẽ được bỏ qua, ít nhất là trong chương này p. Điều đó có nghĩa rằng nó là giả định rằng γ = 0. Các x . . | Chapter 40 STRIP FOOTING One of the simplest problems for which lower limits and upper limits can be determined is the case of an infinitely long strip load on a layer of homogeneous cohesive material 0 see Figure . The weight of the material will be disregarded at least in this chapter. That means that it is assumed that Y 0. The Ị Ị Ị . Ị Ị Ị Ị Ị Ị Ị 1 . _ problem is a first schematization of the shallow foundation of a structure using a long strip foundation made of concrete for instance. It will first be attempted to obtain a lower bound for the failure load using an equilibrium system. Such a system should consist of a field of stresses that satisfies the conditions of equilibrium in all points of the field that agrees with the given stress distribution on the soil surface and that does not violate the yield condition in any point. Lower bound Figure Strip footing. An elementary solution of the conditions of equilibrium in a certain region is that the stresses in that region are constant because then all conditions are indeed satisfied. In a two-dimensional field these equilibrium conditions are in the absence of gravity 0 dx dz ff 0 dx dz f xz fzx- The main difficulty is to satisfy the boundary condition because the normal stress fzz is discontinuous along the surface see Figure . This difficulty can be surmounted by noting that in a statically admissible field of stresses an equilibrium system not all stresses need be continuous. 227 Arnold Verruijt Soil Mechanics 40. STRIP FOOTING 228 . . . . . . . . .Formally this can be recognized by inspection of the equations of equilibrium eqs. - . All partial derivatives in these equations must exist which . I 1 1 I 1- I 1 1 I . . means that the stresses must at least be continuous in the directions in which they . have to be differentiated. It follows that the shear stress ơxz must be continuous .1111111111 in both directions that the normal stress ơxx must be continuous in
