Presented by: Civil Engineering Academy

Foundation Settlement Presented by: Civil Engineering Academy

Compaction vs. Consolidation: ◦ These are 2 terms that are confusing to

people. Compaction increases the density

of a soil by reducing the volume of air.

Consolidation is a time processes of

increasing the density by draining out

water.

Immediate settlement occurs within hours or

days of a load being applied and is a small

portion of the total settlement.

Simmediate = IpqB

1−𝜇

𝐸

Simmediate = immediate settlement

Ip = influence factor (use a table)

q = applied pressure (P/A)

B = width or diameter of applied load

𝝁 = Piosson’s ratio of soil (use a table)

E = modulus of elasticity of soil (use a table)

Settlement of foundations is an extension of consolidation (covered previously). Because of consolidation you get foundation settlement.

Assuming that the stress increase at the center of the clay layer due to the mat foundation is 200 psf, the consolidation settlement (inches) of the clay layer is?

Solution:

H = 10 ft p’ = pressure at setter of clay

Cc = 0.4 p’ = 100pcf(5)+(120pcf-62.4)(5)+(125-62.4)(5’) e = 0.4 p’ = 1101 psf

𝝙p’ = 200 psf

S=10𝑓𝑡(0.4)

1+0.4log[(1101+200)/1101] = 0.207 ft = 2.49 in

half clay layer =10/2

You have to talk about pressure distribution

when talking about foundation settlement.

2:1 Method

Boussinesq Method

To find the increase in vertical stress under the

center of a rectangular load you can use this

approximate method. P

B

B Z/2 Z/2

2V:1H Z

Footing dimensions (BxL)

Stress at depth Z = 𝑃

(𝐵+𝑍)(𝐿+𝑍)

B+Z

Can tell you what the distribution of stresses

are for a point load applied.

Form uniformly loaded circular and

rectangular areas you need to use an

influence factor from tables. The vertical

stress is determined by:

q= applied bearing pressure (q = load/area)

I = influence factor from tables (see next slides)

A flexible rectangular area measures 10x20 ft in plan view. It supports a pressure of 3,000 psf. Determine the vertical stress increase due to the applied load at a depth of 20 ft below the corner of the foundation. Use the Newmark method.

Solution: m = B/z = 10/20=0.5, n=L/z = 20/20 = 1.0 see chart I = 0.12

∆𝜎 = 𝐼𝑞 = 0.12 3000 𝑝𝑠𝑓 = 360 𝑝𝑠𝑓

A rectangular area measures 40’x40’ in plan view. It supports a pressure of 6,000 psf. Determine the increase in stress at a depth of 20 ft below the center of the footing.

Solution: Use Boussineq stress chart:

B=40 ft

Horizontal distance from center of footing: x/B

= 0/40 = 0

Vertical distance from center of footing:

z/B = 20/40 = 0.5B

From the chart I= 0.5p = 0.7(6000psf) = 4200 psf

More example problems!!

Slope stability of soils. Good stuff.