Let me try and explain this in more detail... because while this is a very simple problem to solve, it requires you to make a leap that to me isn't exactly intuitive.
The theory behind this question is straightforward. It requires you to recognize this is a lateral earth pressure question in an "at rest" situation. You know this because the basement wall is constrained from moving towards or away from the backfill because of the slab on grade and concrete floor. The problem states to use a triangular pressure distribution (which you should have known even if they didn't tell you... it's the most common situation and probably all you'd ever see on the exam), so you need to understand how pressure distribution works - it basically says the force varies linearly with depth. The total force is the simply the area under triangle (well, in front of because it's in the horizontal we're working on).
So far so good? You know to use the 45 pcf above and 35 pcf below. The only other question is what to do at the boundary of the water table...
And that's where I don't like this problem. With water you need to start thinking about effective pressure because some of the weight of the soil is floating on the water. The common equations you use to get to total horizontal force is to start with weight of soil (vertical pressure, σv) and subtract the weight of the water (pore pressure, u) to get the "effective" (sometimes called "effective overburden") vertical pressure (σ'v). This effective vertical pressure is converted to the horizontal pressure by multiplying by a coefficient of earth pressure (K, at rest, active, or passive). Then all you need do is add in the pore pressure (it acts the same in all directions) and you have the total (at rest, active, passive) earth pressure. This sounds easy, but given that the coefficient of earth pressure changes with different layers, it can make for a much more complex pressure distribution. I found the easiest way to solve these sort of problems was to create a table with depth going down the side (including the + and - limits at places where K changes), and total weight, water weight, effective pressure, horizontal soil, and horizontal total along the top. The values of horizontal total are used to draw the "shape" of the pressure distribution.
But... you aren't given K, so this is a pretty big hint that the "equivalent fluid densities" already account for the effective overburden pressure caused by the water... you couldn't account for pore pressure in the effective pressure even if you wanted to!
So then... you've got the top triangle and you calculate it's base using the straightforward equation σv = γsoil x h (45x4 in this case). Area is 1/2bh so you get the "1/2 x 45 x 4^2". The bottom layer is a little different... there's no discontinuity at h=4, so the pressure curve starts there - meaning there's a rectangle in the pressure diagram who's area is "45 x 4 x 9". The force at the bottom is the total weight of the soil (the 45x4 from above plus the 35x9 of the bottom layer) plus the 62.4x9 of the water in the bottom layer. But because you're just adding the areas together and you already accounted for the rectangle, you need only calculate the area of the bottom triangle "1/2 x (35 + 62.4)x9 x 9".
Again, this turns out to be a very easy problem but you need to recognize where the simplifications are made for it to be a useful study guide. Still... it's worth remembering: don't make the problems too hard! There are often lots of distracting (needless!) details given.
Really, I wish someone would start collaborating on
PE Notes - Geotechnical because this topic would do a lot of good to be explained in simple terms.
