Bearing Capacity in Clay

Bearing Capacity in Clay

In clay soils and plastic silts, the cohesion term plays a major role in bearing capacity. The friction angle is considered to be zero for clays and plastic silts.

Design Example

This example explores a column footing in a homogeneous clay layer. Find the ultimate bearing capacity of a 1.2 m x 1.2 m square footing placed in a clay layer. The density of the soil is found to be 17.7 kN/m 3 and the cohesion was found to be 20 kPa. See Fig.

Column footing in a homogeneous clay layer

Solution

STEP 1: Find the Terzaghi bearing capacity factors from Table Below.

For clay soils, the friction angle (~) is considered to be zero.

Nr = 5.7

Nq= 1.0

Ny = 0.0

STEP 2: Find the shape factors using Table

For a square footing Sc = 1.3 and Sy = 0.8.

STEP 3" Find the surcharge (q).

q = F x d - 17.7 x 1.2 = 21.2 kPa

STEP 4" Apply the Terzaghi bearing capacity equation.

qult = 20 x 5.7 x 1.3 + 21.2 x 1.0 + 0

qult = 169.4 kPa

allowable bearing capacity(qallowable) -- qult/F.O.S. - qult/3.0 = 56.5 kPa

The total load (Qallowable) that could safely be placed on the footing is found as

Qallowable = qallowable x area of the footing

Q allowable = qallowable x (1.5 x 1 . 5 ) - 1 2 7 . 1 kN

Permeability Test

Transport of water through soil media depends on the pressure head, velocity head, and the potential head due to elevation. In most cases, the most important parameter is the potential head due to elevation.
See Fig. 1.43. Water travels from point A to point B due to high potential head.

The velocity of the traveling water is given by the Darcy equation.

v = k x i

where

v - velocity
k = coefficient of permeability (cm/sec or in./sec)
i= hydraulic gradient = h/L
L = length of soil
Q -- A x v = volume of water flow
A -- area
v- velocity

Design Example
Find the volume of water flowing in the pipe shown in Fig. 1.44. The soil Permeability is 10 -5 crn/sec. The area of the pipe is 5 cm 2. The length of the soil plug is 50 cm.

Solution:

Apply the Darcy equation

v = k x i
v = k x (h/L)
v - 10 -5 x 20/50 - 4 x 10 -6 cm/sec
volume of water flow- A x v - 5 x 4 x 10 -6 cm3/sec
= 2 x 10 - s cm3/sec

Liqued limit and Plastic limit

Water flowing through soil

Water flowing through soil

Water flowing due to gravity

Seepage Rate:

Water movement in soil occurs through the voids within the soil fabric. The more voids there are in a soil, the more water can flow through. This seepage path can be seen in Fig.

Seepage Path

The velocity of water seepage through a soil mass is dependant upon the void ratio or porosity of the soil mass. This seepage velocity can be seen in Fig. 1.46. The volume of water traveling through the soil is

Seepage Velocity

where

A = area

v = velocity

The porosity (n) of a soil is defined as

n= Vv / V

where

Vv - volume of voids = L x Av

Av - area of voids

V = total volume -- L x A

L -- length

A --total cross-sectional area

Hence

n -- (L x Av)/(L x A) = Av/A

Av--n x A

Q=v x A

The velocity of water traveling through voids (v~) is known as the seepage velocity.

Q=vs x Av

Q = v x A = vs x Av = vs X (n x a)

v x a - vs X (n x a)

Vs - v/ n

SPT-CPT CORRELATIONS

In the United States, the Standard Penetration Test (SPT) is used extensively. On the other hand, the Cone Penetration Test (CPT) is popular in Europe. A standard cone has a base area of 10sq cm and an apex angle of 60 ~ . European countries have developed many geotechnical engineering correlations using CPT data. Fig. 1.25 shows a standard CPT device.

Standard CPT device

The correlation between SPT and CPT shown in Table can be used to convert CPT values to SPT numbers or vice versa. Qc = CPT value measured in bars.

where

1 bar = 100 kPa

N = SPT value

D50 = size of the sieve that would pass 50% of the soil

Design Example 

SPT tests were done on a sandy silt with a D50 value of 0.05 mm. The average SPT (N) value for this soil is 12. Find the CPT value.

Solution

From Table 1.4, for sandy silt with a D50 value of 0.05 mm

Qc/N =3.0

N=12

Hance

Qc = 3 x 12 = 36 bars = 3,600 kPa