I was working through some of the six minute solutions problems, and I noticed that on 2 questions, they used different methods to calculate the answer. For one problem, when calculating the density of the clay soil below the water table they used the dry density*(1+w), for the second problem, same circumstance, they calculated the saturated density. I guess, i'm just confused as to why they used the the wet density in the first problem for clay below the water table. I was under the impression that clay below the water table was assumed to be saturated. Anyone else notice this? I wish i could give the problem numbers but I am at work

# 6 min solutions

Started by
teammike
, Oct 18 2010 02:29 PM

1 reply to this topic

### #1

Posted 18 October 2010 - 02:29 PM

### #2

Posted 18 October 2010 - 04:38 PM

QUOTE (teammike @ Oct 18 2010, 07:29 AM) <{POST_SNAPBACK}>

I was working through some of the six minute solutions problems, and I noticed that on 2 questions, they used different methods to calculate the answer. For one problem, when calculating the density of the clay soil below the water table they used the dry density*(1+w), for the second problem, same circumstance, they calculated the saturated density. I guess, i'm just confused as to why they used the the wet density in the first problem for clay below the water table. I was under the impression that clay below the water table was assumed to be saturated. Anyone else notice this? I wish i could give the problem numbers but I am at work

This is a situation where you need to read the question carefully. I would assume that soil below the water table is at saturated conditions, but if the dry unit weight and moisture content are provided, then all you can do is solve for the moist unit weight of the soil.

Saturated soil just means that there's no air volume. But if the moisture content, void ratio or porosity are provided, then calculating the total (moist) unit weight is the only option.

Good luck!

#### 0 user(s) are reading this topic

0 members, 0 guests, 0 anonymous users