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NCEES Afternoon Q# 538 Retaining Wall Problem


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#1 pleasepass

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Posted 10 April 2012 - 04:42 PM

This problem has been bugging me for a month now... anyhow, for those of you who are familiar with this retaining wall problem, when calculating for F2 = (45)(4)(9)... I'm not understanding why we would include the 4 in this equation, which would be the height of the soil above the water table... and then it's multiplied by the height of soil below the water table??? Previous iterations of these problems that I've seen don't seem to jive with this rationale. Can anyone help to explain this? Thanks!

#2 Dano_PE

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Posted 10 April 2012 - 10:02 PM

4 forces on the wall.

Distribution 1 - (.5)(45)(4^2) = Resultant force of Triangle Distribution #1. To get the value of the bottom of the triangle do (45)(4) not squared. By doing squared you are basically finding the area in front of the distribution.
Distribution 2 - (4)(45)(9) = bottom of triangle distribution. Multiply this by the depth from the water table to the footing bottom (9) as shown bolded. This distribution is a rectangle and is the force of the soil above the water tables effect of foundation below. Again you are finding area in essence.
Distribution 3-(.5)(35)(9)^2 - This is a triangle distribution and is the effect of the saturated soil below the water table.
Distribution 4- (.5)(62.4)(9^2) - This is a triangle distribution of the water "pressure" so to say. For this step you usually do (Ka-1) not just Ka.

Sum these and there is your answer.

Dis 1 = 350 lb / ft
Dis 2 = 1620 lb/ ft
Dis 3 = 1417.5 lb/ft
Dis 4 = 2527.20 lb/ft

Sum = 5924.70lb/ft

Hope that helps and anyone correct me if I am wrong because I did this very fast and without checking it over too much. NCEES just lumps my Dis 3 and Dis 4 together.
Good luck.

#3 maverickmauresmo

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Posted 16 October 2012 - 11:25 PM

Why do we use At rest pressure? What type of scenario would we use Active pressure instead of At rest pressure?



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