CIV-340 · Lesson 9 of 12 · 20 min

Stresses under loaded areas

A footing pushes on the soil directly beneath it, but the stress it adds spreads out and decreases with depth, like the pressure under a hand pressed into a mattress. To estimate how much a clay layer several meters down will settle, the engineer needs the stress increase at the depth of the clay, not at the base of the footing. This lesson develops the elastic solutions that give those stress increases, compares them with the quick 2:1 method, and applies them to the garage footings designed in CIV-330, including the effect of the neighboring footings.

Why elasticity

Soil is not elastic. It yields, it creeps and its stiffness depends on stress. Nevertheless, for the stress increases below a footing that is well within its bearing capacity, the distribution of stress predicted by elastic theory turns out to be a good approximation. The reason is that stresses in a body depend mainly on equilibrium and compatibility, and much less on the stiffness of the material, as long as the material is reasonably uniform. Field measurements under footings and embankments have confirmed this for decades. The elastic stress solutions are therefore used even when the settlements themselves are calculated from consolidation theory, as in lesson 10.

Boussinesq's point load

In 1885, the French mathematician Joseph Boussinesq solved the problem of a point load \(P\) on the surface of a uniform, isotropic, linearly elastic half-space. The vertical stress increase at a depth \(z\) and a horizontal distance \(r\) from the load is

\[\Delta\sigma_z = \frac{3P}{2\pi z^2}\Big[\frac{1}{1 + (r/z)^2}\Big]^{5/2}\]

Directly below the load, at \(r = 0\), this reduces to \(\Delta\sigma_z = 3P/(2\pi z^2) = 0.477P/z^2\). The vertical stress does not depend on the elastic modulus or, remarkably, on Poisson's ratio, which is why the solution is so useful. Westergaard later solved the same problem for a soil reinforced by thin horizontal layers that prevent lateral strain, a model of varved clays; it gives somewhat lower stresses directly below the load and is used occasionally for such deposits.

Loaded areas

A footing spreads its load over an area. Integrating Boussinesq's solution over the loaded area gives the stress increase from any distributed load. Three cases are especially useful.

Under the center of a uniformly loaded circle of radius \(a\) carrying pressure \(q\), such as a storage tank,

\[\Delta\sigma_z = q\Big[1 - \Big(\frac{1}{1 + (a/z)^2}\Big)^{3/2}\Big]\]

Under the corner of a uniformly loaded rectangle of sides \(B\) and \(L\), the solution derived by Newmark in 1935 is written with \(m = B/z\) and \(n = L/z\) as \(\Delta\sigma_z = q I\), with an influence factor

\[I = \frac{1}{4\pi}\Big[\frac{2mn\sqrt{m^2 + n^2 + 1}}{m^2 + n^2 + m^2n^2 + 1}\cdot\frac{m^2 + n^2 + 2}{m^2 + n^2 + 1} + \tan^{-1}\frac{2mn\sqrt{m^2 + n^2 + 1}}{m^2 + n^2 + 1 - m^2n^2}\Big]\]

where the inverse tangent is taken in the range 0 to \(\pi\). Fadum published the factor as a chart, and tables of it appear in every soil mechanics text. For a strip footing of width \(B\) and infinite length, closed-form expressions in terms of angles subtended at the point are also available.

Superposition

Because the elastic solutions are linear, stresses from separate loads add. The stress under the center of a rectangle is four times the corner value for a rectangle a quarter of the size. The stress at a point outside a loaded rectangle is found by adding and subtracting rectangles that all have a corner at the point. The same principle gives the combined effect of neighboring footings, a fill next to a building or an excavation that removes load.

Left: contours of vertical stress increase beneath a uniformly loaded square footing of width B, plotted in a vertical section through its center, labeled 0.8q, 0.6q, 0.4q, 0.2q and 0.1q; the 0.1q contour, the pressure bulb, reaches about 2B below the footing. Right: the 2:1 method, with the load spreading downward at one horizontal to two vertical so that it acts on an area (B + z)(L + z) at depth z.
Stress increase below a square footing: elastic pressure bulbs and the 2:1 approximation.

The pressure bulb

Plotting contours of equal stress increase beneath a footing gives the familiar pressure bulbs. Beneath a square footing, the stress falls to about half the applied pressure at a depth of one footing width and to about 10 % at two widths. Beneath a strip footing, the bulbs extend about twice as deep, because the load is spread over a long line. This is the basis of the rule in lesson 1 that boreholes should reach at least one and a half to two footing widths below a footing, and deeper where footings are close together or for a raft, whose width is the width of the whole building.

The 2:1 method

For quick estimates, the load can be assumed to spread with depth at a slope of one horizontal to two vertical. A rectangular footing \(B \times L\) carrying a total load \(Q\) then produces at depth \(z\) an average stress increase

\[\Delta\sigma_z = \frac{Q}{(B + z)(L + z)}\]

This is the average over the spread area, so it underestimates the stress below the center at shallow depth and overestimates it at large depth. Between about one and four footing widths it agrees with the elastic center value within about 30 %, which is adequate for preliminary work and for checking a computer result.

The garage footings

The CIV-330 footings are 10 ft square with their bases 5 ft below grade. For consolidation, the load that matters is the sustained load: the dead load plus the part of the live load present for long periods, here 430 kips per column. The footing itself weighs about 37.5 kips and the backfill over it about 28.8 kips. The net pressure on the soil is the gross pressure minus the weight of the soil removed:

\[q_{net} = \frac{(430 + 37.5 + 28.8)(1000)}{10^2} - 115(5) = 4963 - 575 = 4388\ \text{psf}\]

Only the net pressure adds effective stress below the footing, because the soil excavated for the footing had already been carrying its own weight. A basement or a deeply embedded footing can reduce the net pressure almost to zero; this is the principle of a floating, or compensated, foundation.

Worked problem

Stress at the top of the clay

Find the vertical stress increase at the top of the clay, 8 ft below the footing base, under the center of an isolated footing, by the elastic solution and by the 2:1 method.

Show solution
  1. Divide the footing into four 5 × 5 ft squares meeting at the center. For each, \(m = n = 5/8 = 0.625\):
  2. \[I = \frac{1}{4\pi}\Big[\frac{2(0.3906)\sqrt{1.781}}{1.781 + 0.153}\cdot\frac{2.781}{1.781} + \tan^{-1}\frac{2(0.3906)\sqrt{1.781}}{1.781 - 0.153}\Big] = 0.1123\]
  3. \[\Delta\sigma_z = 4(0.1123)(4388) = 1971\ \text{psf}\]
  4. By the 2:1 method, with \(Q = 4388(100) = 438{,}800\) lb:
  5. \[\Delta\sigma_z = \frac{438{,}800}{(10 + 8)^2} = 1354\ \text{psf}\]

Answer: At this shallow depth, 0.8 of the footing width, the 2:1 average is a third lower than the elastic center value. Lesson 10 uses the elastic values.

Sofia's script computes the stress increase beneath the center and the corner of a footing at several depths through the clay, compares it with the 2:1 method and adds the effect of the four nearest neighboring footings on the column grid.

Vertical stress increase beneath the garage footings by Boussinesq superposition, the 2:1 method and the effect of neighboring footings.
# Vertical stress increase under the garage footings (Boussinesq, elastic half-space)
from math import pi, sqrt, atan2, copysign

def corner(a, b, z):
    """Influence factor at depth z below the corner of a uniformly loaded a x b rectangle."""
    if a == 0 or b == 0: return 0.0
    m, n = abs(a) / z, abs(b) / z
    s = m * m + n * n + 1
    I = (2 * m * n * sqrt(s) / (s + m * m * n * n) * (s + 1) / s + atan2(2 * m * n * sqrt(s), s - m * m * n * n)) / (4 * pi)
    return copysign(1, a) * copysign(1, b) * I

def rect(q, x1, x2, y1, y2, x, y, z):
    """Stress increase at (x, y, z) from load q on the rectangle x1..x2, y1..y2 (superposition of corners)."""
    return q * (corner(x2 - x, y2 - y, z) - corner(x1 - x, y2 - y, z) - corner(x2 - x, y1 - y, z) + corner(x1 - x, y1 - y, z))

B = 10.0                                  # square footing (ft), base 5 ft below grade
P, W_ftg, W_fill = 430.0, 37.5, 28.75     # sustained column load, footing and backfill weights (kips)
q_net = (P + W_ftg + W_fill) * 1000 / B**2 - 115.0 * 5.0
print(f"Net bearing pressure q_net = {q_net:.0f} psf")
spacing = 25.0                            # column grid (ft)
print(" z below base   center   corner   2:1     center with 4 neighbors")
for z in (2.0, 4.0, 8.0, 12.0, 16.0, 19.5, 25.0, 31.0):
    c = rect(q_net, -B/2, B/2, -B/2, B/2, 0, 0, z)
    k = rect(q_net, -B/2, B/2, -B/2, B/2, B/2, B/2, z)
    two = q_net * B * B / (B + z) ** 2
    nb = c + sum(rect(q_net, dx - B/2, dx + B/2, dy - B/2, dy + B/2, 0, 0, z)
                 for dx, dy in ((spacing, 0), (-spacing, 0), (0, spacing), (0, -spacing)))
    print(f"{z:8.1f} ft {c:8.0f} {k:8.0f} {two:7.0f} {nb:12.0f}   psf")

# Point load check: the whole footing load as a point at depth 6 m
z = 20.0
print(f"\nPoint load approximation at z = {z} ft: {3 * q_net * B * B / (2 * pi * z * z):.0f} psf")
Output
Net bearing pressure q_net = 4388 psf
 z below base   center   corner   2:1     center with 4 neighbors
     2.0 ft     4214     1091    3047         4215   psf
     4.0 ft     3509     1053    2239         3515   psf
     8.0 ft     1971      877    1354         2010   psf
    12.0 ft     1127      665     907         1224   psf
    16.0 ft      703      493     649          863   psf
    19.5 ft      496      382     504          700   psf
    25.0 ft      314      264     358          557   psf
    31.0 ft      209      186     261          460   psf

Point load approximation at z = 20.0 ft: 524 psf

Reading the results

Under the center of an isolated footing, the stress increase falls from 4,214 psf just below the base to 1,971 psf at the top of the clay, about 500 psf at mid-clay (19.5 ft below the base) and 209 psf at the bottom of the clay. The 2:1 method gives similar values in the middle of the layer and higher values near its base. Under the corner, the stress is only a quarter of the center value near the surface but nearly the same at depth, as the bulbs merge.

The neighboring footings, at 25 ft centers, add little near the top of the clay but contribute about 200 psf at mid-depth and 250 psf at the base, more than doubling the stress increase at the base. Deep below a group of footings, the stress approaches the average pressure of the whole building spread over its plan area. This is why interior footings settle more than corner footings, and why a building on a thick clay settles in a dish shape.

Large loaded areas

An area fill or a raft that is wide compared with the depth of the compressible layer produces a stress increase that hardly decreases with depth: the pressure bulb is wider than the layer is deep. For a fill many times wider than the clay is thick, the stress increase is simply the fill pressure, \(\Delta\sigma_z = \gamma_{fill} H_{fill}\), at every depth. This one-dimensional condition is exactly the condition of the laboratory consolidation test, which is why settlements under wide fills are predicted more reliably than under small footings.

The yard at the operations center will be raised by 4 ft of granular fill over an area about 260 ft square, a load of about 4 × 125 = 500 psf. Because the area is so much wider than the clay is deep, Owen takes the full 500 psf through the clay.

Limits of the elastic solutions

The Boussinesq solutions assume a uniform half-space, a flexible load and small stresses. Real ground is layered. A stiff layer over a soft one spreads the load more widely, reducing stresses in the soft layer; a soft layer over a rigid base concentrates them. Elastic solutions for two- and three-layer systems, developed by Burmister for pavements, and finite element analyses capture these effects. A rigid footing, which settles uniformly, has a contact pressure that is higher at the edges than the center on clay and higher at the center on sand, but at depths greater than about half the footing width the stress distribution is the same as for a flexible footing with the same total load. For most foundation settlement work, the Boussinesq solution applied to the net pressure is the accepted standard.

Practical points

  • Use the net pressure, not the gross pressure, for settlement calculations.
  • Use the sustained load for consolidation of clay and the full service load for immediate settlement.
  • Include neighboring footings when the stress is needed at depths greater than about the footing spacing.
  • Remember that the stresses from wide fills hardly decrease with depth.
  • Check computer results against a hand calculation at one depth.

Common mistakes

  • MistakeApplying the corner influence factor directly at the center of a footing.

    InsteadDivide the footing into four rectangles meeting at the center and add their contributions.

  • MistakeUsing the 2:1 method for precise calculations at shallow depth.

    InsteadUse the elastic solution; the 2:1 method gives an average and underestimates the center value near the surface.

  • MistakeIgnoring the load from neighboring footings deep in the clay.

    InsteadSuperpose the contributions of nearby footings and area loads.

  • MistakeUsing the gross bearing pressure as the stress increase.

    InsteadSubtract the weight of the soil removed to get the net pressure.

Summary and key terms

  • Boussinesq's point load: \(\Delta\sigma_z = (3P/2\pi z^2)[1 + (r/z)^2]^{-5/2}\), independent of the elastic constants.
  • Uniformly loaded rectangles: \(\Delta\sigma_z = qI(m, n)\) under a corner; superposition gives any point.
  • Pressure bulbs under a square footing reach about 10 % of q at a depth of 2B.
  • The 2:1 method gives \(Q/[(B + z)(L + z)]\), a useful average.
  • Under the garage footing center, \(q_{net} = 4388\) psf gives 1,971 psf at the top of the clay and about 500 psf at mid-clay; neighbors add about 200 psf there.
  • Wide fills impose nearly the full fill pressure throughout the clay.
Boussinesq solution
The elastic solution for the stresses in a half-space under a surface point load.
Influence factor
The ratio of stress increase to applied pressure at a point below a loaded area.
Superposition
Adding the effects of separate loads, valid for linear elastic solutions.
Pressure bulb
A contour of equal vertical stress increase below a loaded area.
2:1 method
An approximation in which load spreads at one horizontal to two vertical with depth.
Net pressure
The pressure added to the soil at founding level, gross pressure minus the overburden removed.
Compensated foundation
A foundation whose excavated soil weight offsets much of the building load.
Sustained load
The part of the load present long enough to cause consolidation.

Practice problems

Work each problem on paper first, then reveal the answer to check.

  1. A 100 kip point load acts on the surface. Find Δσz at 6 ft depth directly below and at r = 6 ft.

    Show answer

    Below: \(0.477(100{,}000)/36 = 1325\) psf; at r = z: \(1325(0.5)^{2.5} = 234\) psf.

  2. A 6 ft by 10 ft footing carries a net pressure of 3,000 psf. Estimate Δσz at 10 ft below its base by the 2:1 method.

    Show answer

    \(3000(60)/[(16)(20)] = 563\) psf.

  3. Why do interior footings of a building on deep clay settle more than corner footings with the same load?

    Show answer

    Their stress bulbs overlap with more neighbors, so the stress increase deep in the clay is larger.

  4. A storage tank 40 ft in diameter applies 2,000 psf. Find Δσz at 20 ft below its center.

    Show answer

    \(a/z = 1\): \(2000[1 - (0.5)^{1.5}] = 1293\) psf.

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