Foundation settlement in layered soil

[pelelo]

Hello,

I am working on a foundation project were I need to compute the predicted settlements.

My soil profile is composed of 10' of Sandy silt (MLS), 20 of Silty sand (SM), 15 of silty gravel (GM) and another 15' layer of MLS.

I don;t have much of experience dealing with foundation settlements but I know there are bunch of methods available for computing elastic settlements. At least the most common I have seen are Theory of Elasticity and Schmertmann's method. To my knowledge Schmertmann's method is better than Theory of Elasticity because it takes into account several Elasticity Modulus values for layered soils, such as my case.

For the ones who like to compute settlements using the Theory of Elasticity, I have seen few examples were it is required to compute settlements for several layers and I have noticed the same formula (S = q * B * (1-u2)/E) is applied for each layer. What I noticed is those authors don't take into account the change (decrease) of stress (instead they used q), using boussinesq's chart, as the soil profile goes deeper.

Also I am not sure if the same footing width "B" I am supposed to keep it when analysing each layer. Probably the deeper the soil profile is, some reduction factor of the Base should be taken into account.

Can anyone give me some imput about this?

On the other hand, can anyone let me know what are the limitations of Schmertmann's method?, Or is this one simply the ideal for solving multilayered profiles?

Thanks a lot.

 

[BigH]

YannisChasiotis -- But not AASHTO (and therefore USA bridge designs).

 

[pelelo]

Mccoy,

Sorry for the late response.

You said you got about 1.5 cm of settlement, did you use B&B´s approach or any other?. You assumed D = 1 m, L = 10 m. Wouldn´t a 10 m long footing be too much? (it is more a mat than a footing).

In my case, B = 2m and Depth of influence Zi = 1.7 m (2^1.7).

Still using you Ic value (1.81) I continue getting low settlement results.

Again,

Using B&B´s approach:

q = 200 Kpa,
Nave = 25
B = 2 m

Sc = B^075 * ((1.7/(NSPT^1.4)) * q

Sc = (2^0.75) * ((1.7/(25^1.4)) * 200 = 6.31 mm

Please let me know if I am using a wrong value,

Sorry for the inconvience.