Solving an equation
Solving an equation
(OP)
I was not real sure where to post it but here it goes.
I am trying to solve the depth of sheet pile needed and the equation I am solving for is:
g=gamma
D^4-[8p/g(Kp-Ka]D^2-[12PL/g(Kp-Ka)]D-[2P/g(Kp-Ka)]^2=0
this reduces to D^4-54.54D^2-1227.1D-13.63^2=0
My problem is how to solve for D, I know the answere is +/-12.5.
The above problem is out of a book but is does not show how to solve for D. I need to know how to solve it for another problem. I must be getting old because I cannot remember how to do it.
I am trying to solve the depth of sheet pile needed and the equation I am solving for is:
g=gamma
D^4-[8p/g(Kp-Ka]D^2-[12PL/g(Kp-Ka)]D-[2P/g(Kp-Ka)]^2=0
this reduces to D^4-54.54D^2-1227.1D-13.63^2=0
My problem is how to solve for D, I know the answere is +/-12.5.
The above problem is out of a book but is does not show how to solve for D. I need to know how to solve it for another problem. I must be getting old because I cannot remember how to do it.





RE: Solving an equation
RE: Solving an equation
RE: Solving an equation
Stop by when you get a chance.
RE: Solving an equation
If you start with the "reduced formula" and substitute + or - 12.5 in the formula it doesn't =0. Therefore, 12.5 is not the correct answer to the value of "D"
RE: Solving an equation
RE: Solving an equation
Again thanks everyone for the help.
RE: Solving an equation
RE: Solving an equation
RE: Solving an equation
RE: Solving an equation
Hydrology, Drainage Analysis, Flood Studies, and Complex Stormwater Litigation for Atlanta and the South East - http://www.campbellcivil.com
RE: Solving an equation
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RE: Solving an equation
RE: Solving an equation
X^4=54.54X^2+1227.1X+185.8
x= f(x) = {54.54x^2+1227.1X + 185.8} ^0.25
now plot x-f(x), and I find it crosses zero at x=12.45 or thereabouts.
RE: Solving an equation
www.PeirceEngineering.com
RE: Solving an equation
Thank you to everyone for your help.
RE: Solving an equation
www.PeirceEngineering.com
RE: Solving an equation
I derived a cubic moment equilibrium equation. After solving, I got a different embedment depth than Das but got the same moment. I also used CivilTech's Shoring program which gave the same solution as my hand calcs.
See attached PDF.
10.08' is the equilibrium depth, not 12.45'. It is probably just a coincidence that 12.45' is about equal to D x 120% = 12.1'.
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RE: Solving an equation
www.PeirceEngineering.com
RE: Solving an equation