Vertical Stresses Below Applied Loads

Both theories assume an ideal elastic soil; real soil only approximates this.

Boussinesq's theory (point load on a half-space)

• The soil is a semi-infinite, homogeneous, isotropic, linearly elastic mass (stress is proportional to strain).
• The soil is weightless and continuous; the stress-strain relation does not depend on the stress level.
• The load acts at the surface of a horizontal half-space as a point load (areas are found by integration).
• Valid for the condition that stress is below failure and for relatively uniform soil. It gives results close to the observed for homogeneous deposits, and gives somewhat high stress for stratified soils.

Limitations: real soils are neither perfectly elastic nor homogeneous; layered soils, a stiff layer below, or a very soft layer limit the use. It also ignores the effect of foundation depth (the load is taken on the surface).

Westergaard's theory

• The soil is an elastic medium reinforced by thin, perfectly rigid horizontal sheets (e.g. sand or silt layers within clay) which allow no lateral strain.
• Poisson's ratio is taken as zero ($\nu = 0$).
• Suitable for stratified, anisotropic deposits (alternating sand and clay), and for soft clay with thin sand seams. It gives lower stress than Boussinesq (about 2/3 at the load axis).

Limitations: the assumption of perfectly rigid layers and $\nu = 0$ is ideal, and the actual stress is between the two theories.

Newmark's influence chart is a diagram made of concentric circles divided by radial lines into a number of small areas (elements), each of which causes the same vertical stress at the centre point at a given depth $z$ when loaded with a uniform pressure. It is based on Boussinesq's equation for a uniformly loaded circular area.

        \  |  /
      ---(   O   )---   circles: R/z = 0.27, 0.40,
        /  |  \          0.52, 0.64, 0.77, 0.92, 1.11,
                         1.39, 1.91, ∞  (influence 0.1 each)

The chart has 10 circles (stress ratio 0 to 1 in steps of 0.1) and 20 radial lines. Each element then has an influence value $I = \dfrac{1}{10\times20} = 0.005$.

Use

To find the vertical stress increase $\Delta\sigma_z$ at any point below a loaded area of irregular shape (any plan shape):

where $N$ is the number of elements covered by the plan of the loaded area (drawn to a scale such that the length OQ equals the depth $z$ and the point is placed at the centre of the chart), and $q$ is the uniform load intensity.

Approximate methods give a quick estimate of the vertical stress below a loaded area without using elastic theory. They assume that the load spreads out with depth over an increasing area.

2:1 method (2 vertical : 1 horizontal)

The load $Q$ is assumed to spread at a slope of 2 vertical to 1 horizontal (about 26.6° from the vertical) on all sides. At depth $z$, the loaded area increases by $z$ in each direction.

     |<-- B -->|
     ▇▇▇▇▇▇▇▇▇▇▇          ground
      \       /
       \     /     2V : 1H
        \   /
   z     \ /
      |<- B+z ->|

• Rectangular footing $B\times L$:

• Square footing ($B\times B$): $\Delta\sigma_z = \dfrac{Q}{(B+z)^2}$
• Strip footing (per unit length): $\Delta\sigma_z = \dfrac{Q}{B+z}$
• Circular footing of diameter $D$: $\Delta\sigma_z = \dfrac{Q}{\frac{\pi}{4}(D+z)^2}$

60° distribution (30° method)

The load is assumed to spread at an angle of $30^\circ$ with the vertical (60° with horizontal) on each side, so the width at depth $z$ is $B + 2z\tan30^\circ$:

Comments

• The result is the average stress over the width at depth $z$, not the maximum under the centre.
• The 2:1 method agrees reasonably with Boussinesq (under the centre) at depths of about 1 to 4 times the width of the footing. It is useful for quick settlement estimates of layered soils.

Given

Circular foundation $D = 10$ m ($R = 5$ m). Load $Q = 18000$ kN. Weak stratum at $z = 3$ m below the foundation level (assumed, as the depth is not printed), bearing capacity $q_a = 100$ kN/m².

Contact pressure:

Stress at the top of the weak stratum (below the centre)

Boussinesq's equation for a uniformly loaded circular area:

(A quick check by the 2:1 method: $Q/[\frac{\pi}{4}(10+3)^2] = 135.6$ kN/m², also above 100.)

Check

$\Delta\sigma_z = 198\ \text{kN/m}^2 \approx 19.8\ \text{t/m}^2 > 100\ \text{kN/m}^2\ (10\ \text{t/m}^2)$

Answer: The stress at the top of the weak stratum is about 198 kN/m², which exceeds the bearing capacity of the stratum (100 kN/m²). It is not safe to construct the water tower with this foundation size.

The foundation must be enlarged (a diameter of about 15 m gives about 97 kN/m²) or the load shifted to a deeper stratum (piles).

The basic difference is in the assumed behaviour of the soil mass.

• Boussinesq: the soil is a homogeneous, isotropic, elastic half-space. It can deform laterally freely; Poisson's ratio does not enter the vertical stress formula.
• Westergaard: the soil is an elastic medium reinforced by thin, rigid, horizontal layers (like alternating sand/clay sheets) which prevent lateral strain; Poisson's ratio is taken as zero.

Westergaard's solution gives smaller stress directly under the load.

Scaling in Newmark's method makes the chart consistent with the depth $z$ at which the stress is required.

• The chart has a scale line OQ. The length OQ represents the depth $z$ (the circles are drawn for radius ratios $R/z$, with OQ as the unit).
• Draw the plan of the loaded area on tracing paper to a scale in which the depth $z$ is equal to the length OQ of the chart. Thus the scale is $= OQ/z$.
• Place the point at which the stress is required over the centre of the chart. Count the elements covered by the plan.

For a different depth, redraw the plan to a new scale (same chart). The stress is then $\Delta\sigma_z = I\,N\,q$.

Methods used to find the increase in vertical stress below a loaded area:

1. Boussinesq's equation (point load, and its integrals for line, strip, circular and rectangular loads)
2. Westergaard's equation (for stratified soil)
3. Newmark's influence chart (for irregular areas)
4. Fadum's chart / influence factor charts (for rectangular areas)
5. Approximate 2:1 (or 60°) load spread method
6. Pressure bulb (isobar) diagrams and Newmark's influence method for embankments (e.g. Osterberg's chart for triangular loads).

Significant depth is the depth below a foundation up to which the increase in vertical stress due to the load is large enough to cause significant settlement. Below it the stress increase is negligible. It is commonly taken as the depth at which the vertical stress increase falls to 10% of the contact pressure ($0.1\,q$), or 20% of $q$ in some practice (or where $\Delta\sigma_z = 0.1\,\sigma'_{v0}$ for the overburden).

Importance:

• It fixes the depth of exploration (boreholes should reach at least this depth).
• It fixes the soil thickness to be included in settlement and consolidation calculations. Layers below it can be neglected.
• For footings, the depth is approximately $1.5B$ to $2B$ for a square footing and $4B$ to $6B$ for a strip footing.

Construction of an isobar for significant depth

An isobar is a contour of equal vertical stress.

1. Select the stress level, e.g. $\Delta\sigma_z = 0.1\,q$.
2. For a chosen set of depths $z$ below the footing, compute $\Delta\sigma_z$ at various horizontal distances $x$ from the centre using Boussinesq or the strip/circular/rectangular footing equations (or charts).
3. At each depth, mark the points where $\Delta\sigma_z = 0.1\,q$ on both sides of the axis.
4. Join the points by a smooth curve to get a bulb-shaped isobar.
5. The depth of the lowest point of the 0.1q isobar, on the axis, is the significant depth.

   |<---- B ---->|
   ▇▇▇▇▇▇▇▇▇▇▇▇▇▇▇  q
    (   0.5 q   )
     (  0.2 q  )
      ( 0.1 q )  <-- isobar
        \___/    significant depth below this

An isobar diagram (pressure bulb) is a plot of contours (isobars) joining points below a loaded area that have the same vertical stress increase, usually expressed as a fraction of the contact pressure $q$ (e.g. 0.8q, 0.5q, 0.2q, 0.1q). The contours form bulb-shaped curves under the foundation and show how stress spreads and dies out with depth.

The isobar of $0.1Q$ is the contour of points in the soil below a point load $Q$ where the vertical stress equals $0.1Q$ (per m², with $Q$ in kN and distances in m). Use Boussinesq's equation:

with $I_B = 0.4775\,[1+(r/z)^2]^{-5/2}$.

Computation

Setting $\Delta\sigma_z = 0.1\,Q$: $z^2 = I_B/0.1$, so $z = \sqrt{10\,I_B}$ and $r = (r/z)\,z$.

Diagram

Plot the points $(\pm r, z)$ with $z$ downward and join them by a smooth curve.

        Q (load)
  ------●-------------  surface
     .  |  .       r max = 0.94 m at z = 1.18 m
    .   |   .
    .   |   .
     .  |  .
      . | .
        ●    z max = 2.19 m on the axis (r = 0)

The isobar is a closed bulb that begins and ends at the load point on the surface and extends to $z = 2.19$ m on the axis. Inside the bulb the stress is greater than $0.1Q$.

Boussinesq's equation for the vertical stress below a point load on the surface is based on these assumptions:

1. The soil mass is a semi-infinite half-space bounded by a horizontal surface.
2. The soil is homogeneous, with the same properties at all points.
3. The soil is isotropic, with the same properties in all directions.
4. The soil is linearly elastic (stress proportional to strain, following Hooke's law).
5. The soil is weightless; the stresses considered are only those due to the applied load.
6. The load acts as a concentrated point load on the surface, vertical, and the soil is initially free of stress.
7. The soil is a continuous medium, and the stress distribution is independent of the elastic constants (the vertical stress does not depend on $E$ or $\nu$).

Assumptions

Uniform pressure $q$ on a flexible circular area of radius $R$ on the ground surface; the soil is homogeneous, isotropic and elastic (Boussinesq). The stress is required at depth $z$ below the centre.

Derivation

Boussinesq's equation for a point load $dQ$ at horizontal distance $r$:

An elemental ring of radius $r$ and width $dr$ carries $dQ = q\,(2\pi r\,dr)$. All points of the ring are at the same distance $r$ from the axis, so:

Let $t = 1 + r^2/z^2$, so $dt = 2r\,dr/z^2$:

Example

Take $q = 100$ kPa and $R = 2$ m.

The stress decreases with depth, and is nearly $q$ close to the surface.

Assumptions

Each column and footing transmits one-third of the total load as a point load: $P = 10^6/3 = 3.333\times10^5$ kN (as printed, $10^6$ kN). The columns are at the corners of an equilateral triangle of side 10 m. Boussinesq's point load equation is used, with $z = 5$ m:

        C1
       /  \
   10 m    10 m          centre O at 5.774 m from
     /  O   \            each footing
   C2 ------- C3
        10 m

Distance from the centre O to each column: $r = \dfrac{10}{\sqrt3} = 5.774$ m.

(a) At the centre of the tank (z = 5 m)

$r/z = 1.155$, $1 + (r/z)^2 = 2.333$

For one column: $\Delta\sigma = \dfrac{3\times333333}{2\pi\times25}\,(2.333)^{-2.5} = 6366.2\times0.1202 = 765.5$ kPa

For three columns:

(b) Below each footing (z = 5 m)

• Own column ($r = 0$): $\dfrac{3P}{2\pi z^2}\times1 = 6366.2$ kPa
• Each of the other two columns ($r = 10$ m, $r/z = 2$, $(1+4)^{-2.5} = 0.01789$): $6366.2\times0.01789 = 113.9$ kPa

Answer: $\Delta\sigma_z \approx 2296$ kPa at the centre and $\approx 6594$ kPa below each footing, 5 m below the foundation level.

The values are very large because the load is taken as $10^6$ kN; for $10^6$ N the stresses are 1000 times smaller (2.30 kPa and 6.59 kPa).

Principle

Excavation removes soil weight, so it acts like a negative uniform load on the base of the excavation:

The reduction in vertical stress is obtained by the rectangular area (Fadum / Newmark) corner formula $I(m,n)$ with $m = B/z$, $n = L/z$, where $z$ is the depth below the excavation base. Depth required = 6 m from the ground, so $z = 6 - 2.5 = 3.5$ m.

(i) Below the centre

Divide the 3 m × 6 m area into four rectangles of 1.5 m × 3 m (corner at the centre).

(ii) At a point 6 m from the centre (along the 6 m side direction)

The point is 3 m beyond the end of the excavation. Use a rectangle from 3 m to 9 m (length) and 1.5 m (half-width), doubled for the two halves:

(Along the 3 m side direction the same method gives 0.11 kN/m².)

Answer: The effective vertical stress at 6 m depth is reduced by about 2.0 kN/m² below the centre and about 0.2 kN/m² at 6 m from the centre.

If the intended unit weight is 20 kN/m³ ($q = 50$ kN/m²), all reductions are 10 times larger: 20.3 kN/m² and 2.0 kN/m². The reduction is small, so excavation has little effect on points far away.

Given

Ring diameters 4 m and 6 m (as printed: taken as outer $D = 6$ m, inner $d = 4$ m), so $R = 3$ m and $r = 2$ m. Pressure $q = 100$ kN/m².

Method

The stress under the centre of a ring equals that of the full circle of radius $R$ minus that of the inner circle (radius $r$) of the same pressure:

Calculation

Distribution curve

 depth z (m)   stress (kPa)
   0 |*   0
 0.5 |*  1
   1 |* 5.8
   2 |---* 18.3
   4 |----* 20.4   (maximum)
   8 |--* 9.2

The stress along the axis is near zero at the surface (the centre is unloaded), increases to a maximum of about 20 kPa near $z \approx 3$ to $4$ m, and then decreases.

Answer: $\Delta\sigma_z = 0.98,\ 5.78,\ 18.29,\ 20.35,\ 9.22$ kN/m² at 0.5, 1, 2, 4 and 8 m.

Given

Strip footing $B = 2$ m, load $Q = 500$ kN/m. Contact pressure $q = 500/2 = 250$ kN/m². Depth $z = 5$ m, below the centre (where stress is maximum).

Boussinesq (strip load)

Below the centre, $\delta = 0$, so $\Delta\sigma_z = \dfrac{q}{\pi}\left(\alpha + \sin\alpha\right)$ where $\alpha = 2\tan^{-1}\left(\dfrac{B/2}{z}\right) = 2\tan^{-1}(0.2) = 0.3948$ rad (22.62°).

$\sin\alpha = 0.3846$

2:1 method

Comparison

Answer: Maximum stress at 5 m depth $= 62.0$ kN/m² (Boussinesq); the 2:1 method gives 71.4 kN/m², about 15% higher (conservative).

Given

Outer $D = 3.60$ m ($R = 1.80$ m), inner $d = 2.40$ m ($r = 1.20$ m), $q = 135$ kN/m², $z = 1.80$ m, point on the axis.

Stress on the axis of a ring = stress of full circle (radius $R$) minus stress of the inner circle (radius $r$):

Calculation

• Outer circle: $R/z = 1.0$, $(1+1)^{3/2} = 2.828$, factor $= 1 - 0.3536 = 0.6464$
• Inner circle: $r/z = 0.667$, $(1+0.444)^{3/2} = 1.736$, factor $= 1 - 0.576 = 0.4240$

Answer: Vertical stress at 1.80 m depth below the centre $\approx 30.0$ kN/m².

Given

Outer $D = 8$ m ($R = 4$ m), inner $d = 4$ m ($r = 2$ m), $q = 150$ kN/m². Depth of interest is 8 m below the base of the footing, so $z = 8$ m. The effect of excavation is neglected.

Calculation

• Outer: $R/z = 0.5$, $(1.25)^{1.5} = 1.3975$, factor $= 1 - 0.7155 = 0.2845$
• Inner: $r/z = 0.25$, $(1.0625)^{1.5} = 1.0952$, factor $= 1 - 0.9131 = 0.0869$

Answer: Vertical stress at 8 m below the footing base along the axis $\approx 29.6$ kN/m².

Given

Outer $D = 10$ m ($R = 5$ m), inner $d = 7.5$ m ($r = 3.75$ m), $q = 160$ kN/m², $z = 4$ m below the foundation. The maximum stress at this depth is taken on the axis, below the centre, where the effect of the circular loaded area is calculated by Boussinesq's circle equation.

Calculation

• Outer: $R/z = 1.25$, $(1+1.5625)^{3/2} = 4.102$, factor $= 1 - 0.2438 = 0.7562$
• Inner: $r/z = 0.9375$, $(1+0.8789)^{3/2} = 2.575$, factor $= 1 - 0.3883 = 0.6117$

Answer: Vertical stress at 4 m below the foundation (on the axis) $\approx 23.1$ kN/m².

Method

Divide the 4 m × 5 m rectangle into four rectangles that have P as a common corner. The stress under a corner of a rectangle is $\Delta\sigma_z = q\,I$ where $I = f(m, n)$, $m = B/z$ and $n = L/z$. Here $z = 2.5$ m and $q = 200$ kN/m².

   2 m       3 m
 +-------+-----------+
 | 2x2   |   3x2     |  2 m
 |   (1) |   (2)     |
 +-------P-----------+
 |   (3) |   (4)     |  2 m
 | 2x2   |   3x2     |
 +-------+-----------+

For $(m,n) = (0.8, 1.2)$ the table has no column for 1.2, so $I$ is interpolated between $n = 1.0$ (0.1598) and $n = 2$ (0.1812): $I = 0.1598 + 0.2\times(0.1812 - 0.1598) = 0.1641$.

Stress at P

Answer: $\Delta\sigma_z \approx 122$ kN/m² at 2.5 m below P.

Using the exact formula for the influence factor (which gives $I = 0.1461$ for 0.8 × 0.8 and $0.1684$ for 0.8 × 1.2), the result is about 126 kN/m², so the table-based answer is correct to about 3%.

Method

Divide the T-shaped loaded area into rectangles that have P as a common corner and use the corner stress $\Delta\sigma_z = q\,I_N$ for each. Depth $z = 5$ m, $q = 120$ kPa.

 +--------+---------------------+
 |  (1)   |        (2)          |  3 m
 |  3x3   |        6x3          |
 +--------P-----+---------------+
          | (3)  |
          | 3x1.5|              1.5 m
          +------+

P lies at the left corner where the stem meets the flange (on the lower edge of the flange).

• Rectangle (1): left flange, 3 m × 3 m: $m = 0.6$, $n = 0.6$, $I_N = 0.1069$
• Rectangle (2): right flange, 6 m × 3 m: $m = 0.6$, $n = 1.2$ (given as ~1.0), $I_N = 0.1431$
• Rectangle (3): stem, 3 m × 1.5 m: $m = 0.6$, $n = 0.3$, $I_N = 0.0629$

(The given values are the same as the exact influence factors for these shapes.)

Stress at P

Answer: The vertical stress at P at 5 m depth $\approx 37.5$ kPa.