Both stresses tend towards zero for z → ∞, of course, but the horizontal normal stress appears to tend towards zero much faster than the vertical normal stress.. It also appears that at
Trang 1In 1892 Flamant obtained the solution for a vertical line load on a homogeneous isotropic linear elastic half space, see Figure 30.1 This is the two
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F
σxz
σzx
Figure 30.1: Flamant’s Problem.
dimensional equivalent of Boussinesq’s basic problem It can be considered as the superposition of an infinite number of point loads, uniformly distributed along the y-axis A derivation is given in Appendix B.
In this case the stresses in the x, z-plane are
π
π
π
force per unit length, so that F/r has the dimension of a stress.
Expressions for the displacements are also known, but these contain singular terms, with a factor ln r This factor is infinitely large in the origin and at infinity Therefore these expressions are not so useful.
On the basis of Flamant’s solution several other solutions may be obtained using the principle of superposition An example is the case of a uniform load of magnitude p on a strip of width 2a, see Figure 30.2 In this case the stresses are
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x z p r1 r2 θ1 θ2 Figure 30.2: Strip load In the center of the plane, for x = 0, θ 2 = −θ 1 Then the stresses are x = 0 : σzz = 2p π [(θ1 + sin θ1 cos θ1], (30.7) x = 0 : σxx = 2p π [(θ1 − sin θ1 cos θ1], (30.8) x = 0 : σxz = 0 (30.9) That the shear stress σ xz = 0 for x = 0 is a consequence of the symmetry of this case The stresses σxx and σzz are shown in Figure 30.3, as functions of the depth z Both stresses tend towards zero for z → ∞, of course, but the horizontal normal stress appears to tend towards zero much faster than the vertical normal stress It also appears that at the surface the horizontal stress is equal to the vertical stress At the surface this vertical stress is equal to the load p, of course, because that is a boundary condition of the problem Actually, in every point of the surface below the strip load the normal stresses are σxx = σzz = p. . .
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σzz/p σxx/p
z/a Figure 30.3: Stresses for x = 0.
the axis of symmetry x = 0 in the case of a strip load, see Figure 30.2 It can be expected
axis x = 0 This means that this solution can also be used as the solution of the problem that is obtained by considering the right half of the strip problem only, see Figure 30.4.
In this problem the quarter plane x > 0, z > 0 is supposed to be loaded by a strip load
of width a on the surface z = 0, and the boundary conditions on the boundary x = 0
smooth and rigid vertical wall The wall is supposed to extend to an infinite depth, which
is impractical For a smooth rigid wall of finite depth the solution may be considered as
a first approximation.
The formulas (30.7) and (30.8) can also be written as
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z p Figure 30.4: Strip load next to a smooth rigid wall Integration of the horizontal stress σ xx from z = 0 to z = h gives the total force on a wall of height h, Q = 2 π ph arctan( a h ). (30.12) For a very deep wall (h a) this becomes, because then arctan(a/h) ≈ a/h, h → ∞ : Q = 2 π pa = 0.637 pa. (30.13) The quantity pa is the total vertical load F (per unit length perpendicular to the plane of the drawing) It appears that the horizontal reaction in an elastic material is 0.637 F For a very shallow wall (h a) the total lateral force will be, because then arctan(a/h) ≈ π/2, h → 0 : Q = ph (30.14) This is in agreement with the observation made earlier that the value of the horizontal stress at the surface, just below the load, is σxx = p For a very short wall the horizontal force will be that horizontal stress, multiplied by the length of the wall Another interesting application of Flamant’s solution is shown in Figure 30.5 In the case of a surface load by two parallel line loads it can
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x z F F
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x z F
Figure 30.5: Line load next to a smooth rigid wall.
Trang 4be expected that at the axis of symmetry (x = 0) the horizontal displacement and the shear stress will be zero, because of symmetry This means that the solution of that problem can also be used as the solution of the problem of a line load at a certain distance from a smooth rigid
half of Figure 30.5, these conditions are satisfied by the solution of that problem.
The horizontal stresses against the wall are given by equation (30.2) for Flamant’s basic problem, multiplied by 2, because there are two line loads and each gives the same stress In this formula the value of x should be taken as x = a, where a is the distance of the force to the wall The horizontal stress against the wall in this case is
π
The distribution of the horizontal stresses against the wall are also shown in Figure 30.5 The maximum value occurs for z = 0.577 a, and that maximum stress is
a The total force on a wall of depth h can be found again by integration of the horizontal stress over the depth of the wall This gives
π
F
If a = 0 this is Q = 0.637 F If a increases the value of Q will gradually become smaller A force at a larger distance from the wall will give smaller stresses against the wall.
Problems
30.1 Is it a coincidence that in each of the formulas (30.12) and (30.16) the same factor 2/π appears?
30.2 Transform the stresses in Flamant’s solution into polar coordinates, or in other words, derive expressions for the stress components σrr, σθθ and
σrθ Note that the result is somewhat peculiar.