Does a uniform loading of an elastic half space result in a uniaxial stress state or a uniaxial strain state?

Question

Suppose for instance a soil is loaded by a building over an area of length $L$ (load is in the $z$ direction). In the neighborhood of a point at depth $h$, $h \ll,L$, in the soil under the loaded area, do we expect a uniform uniaxial $\sigma_{zz}$ stress state (all other terms of the stress tensor vanish) or a uniform $\epsilon_{zz}$ strain state (all other terms of the strain tensor vanish)?

Answer 1

Strictly neither, but approximately both.

The exact stress distribution can be calculated by integrating the Green tensor of elasticity for a vertical point load. (This is called the Boussinesq problem; see Chapter IX in Love and Farrell's "Deformation of the Earth by Surface Loads". The point-load solutions are given in many soil mechanics and elasticity texts, e.g., Eqs. (8-2) here.) The strain distribution can then be obtained from generalized Hooke's law. Various normal and shear stresses and strains arise from the influence of the material freedom outside the load.

But perhaps we wish to look at limiting cases, e.g., $h\ll L$, as you specified. Is an approximation available?

Directly at the surface, we have a single external stress $\sigma_{zz}$ within the loaded region and no other external loading. For very small depths, far inside the edge of the loading, the stress state can be idealized as that of a half-space with uniform loading $\sigma_{zz}$. (The justification is that the material doesn't "know" that there's any other loading type over the half-space other than that directly above and near it; the exact solution shows that the influence of distant loads disappears quickly as an inverse power of increasing distance.) It follows that the radial strain (symmetric $\varepsilon_{xx}=\varepsilon_{yy}$) could be assumed to be constrained as zero, as a uniformly loaded half-space—or semi-infinite medium—has nowhere to expand into or contract from.

We apply generalized Hooke's Law for an isotropic material, with Poisson's ratio $\nu$, Young's modulus $E$, $\delta$ as the Kronecker delta, and repeated indices indicating summation:

$$\varepsilon_{ij}=\frac{1+\nu}{E}\sigma_{ij}-\frac{\nu}{E}\sigma_{ii}\delta_{ij};\tag{Generalized Hooke's Law}$$

$$0=\varepsilon_{xx}(=\varepsilon_{yy})=\frac{1}{E}\sigma_{xx}-\frac{\nu}{E}\sigma_{yy}-\frac{\nu}{E}\sigma_{zz};\tag{Radial strain}$$

$$\sigma_{xx}=\sigma_{yy}=\frac{\nu}{1-\nu}\sigma_{zz};\tag{Solve for radial stress}$$

$$\varepsilon_{zz}=\frac{(1+\nu)(1-2\nu)}{E(1-\nu)}\sigma_{zz};\tag{Solve for vertical strain}$$

If the soil is porous, for example, then Poisson's ratio is small, and $\sigma_{zz}$ could be taken as $\varepsilon_{zz}E$, the same answer we get for uniaxial compression of a rod or bar with free sides—the basis of defining Young's modulus $E$.

The idealization also results in zero shear stress and strain, which is not strictly the case for finite $L$.

Note that the same derivation provides us with the P-wave modulus of acoustics; here, also, the material is essentially fixed in place laterally when considering a propagating elastic wave.

To summarize: only uniaxial $\sigma_{zz}$ is applied at the surface, but lateral stresses arise underneath when Poisson effects are constrained. A reasonable assumption is uniaxial $\varepsilon_{zz}$ at depths $h\ll L$ underneath vertical loading of lateral dimension $L$. If Poisson's ratio is small, this corresponds to approximately uniaxial $\sigma_{zz}$.

Please let me know if anything's unclear.

Answer 2

If we compare with an uniaxial compression test, where the probe has no friction with the plates of the machine, it is clear that the example of the building is not uniaxial stress.

The vertical compression produces a lateral spreading, according to the Poisson coefficient of the material. If that spreading is restricted, what happens by the soil around the building, there is also lateral compression.

The result is a triaxial state of stress. If the restriction is complete (no lateral spreading allowed), it is also a uniaxial strain state, because there is no lateral strain, only vertical.