Fundamental period of a layered soil profile by the density-weighted Rayleigh quotient

Usage

estimateWeightedRayleighPeriod(VSm, hs, zm, gsat)

Arguments

VSm: Positive layer shear-wave velocities in m/s.
hs: Positive layer thicknesses in metres, aligned with VSm.
zm: Layer midpoint depths from the surface in metres, aligned; they order the layers, the overburden is accumulated from hs.
gsat: Positive layer unit weights in kN/m3, aligned; any unit cancels, only the ratios between layers matter.

Value

One estimate of the fundamental period in seconds, unrounded.

Mathematics

For a shear beam whose density $\rho(z)$ and velocity $V(z)$ vary with depth, Dobry, Oweis and Urzua (1976) write the Rayleigh quotient $$\omega^2 = \frac{\int_0^H \rho V^2 (dX/dz)^2\, dz}{\int_0^H \rho X^2\, dz}$$ and obtain the first-iteration mode shape from a unit initial shape, $$X(z) = \int_0^z \frac{1}{\rho(z_a) V^2(z_a)} \int_{z_a}^H \rho(z_b)\, dz_b\, dz_a,$$ with $z$ measured from the base (hydrated copy, equations eq-rayleigh-frequency and eq-rayleigh-mode-iteration). The slope of that shape is the overburden mass above the point divided by $\rho V^2$, so the numerator integrates by parts into $\int_0^H \rho X\, dz$. Evaluated at layer midpoints with layer-wise constant properties, and with $g$ cancelling between $\rho$ and the unit weight $\gamma$, $$T = 2\pi \sqrt{\frac{\sum_k \gamma_k h_k X_k^2}{\sum_k \gamma_k h_k X_k}}, \qquad X_k' = \frac{\sigma_k}{\gamma_k V_k^2},$$ where $\sigma_k$ is the total overburden at the midpoint of layer $k$, accumulated from the thicknesses and unit weights of the layers above it, and $X_k$ the deflection accumulated from the base. With a uniform unit weight the expression reduces to estimateRayleighPeriod(), the constant-density form the paper calls Method 7. The result is a conditional evaluation of the supplied profile and carries no dispersion.

Examples

estimateWeightedRayleighPeriod(VSm = c(150, 300), hs = c(10, 10), zm = c(5, 15),
                               gsat = c(16, 20))
#> [1] 0.2863489