Solving an equation 1

[wettpaint]

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.

 

[MiketheEngineer]

Write a quick formula in Excel and iterate to a solution??

 

[wettpaint]

I tried that but, I guess I don't know how to do that either. Because I did not come up with the answer they did.

 

[JedClampett]

My HP 48GX has a polynomial solver.
Stop by when you get a chance.

 

[OldPaperMaker]

wetpaint,
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"

 

[cvg]

or the reduced formula is incorrect...

 

[wettpaint]

Thanks everyone for the help. After thinking about it last night, I plugged the 12.5 back into the equation and could not get it to come out correct. I am following an example problem in a text book and there must be an error in the rounding of their answer and it makes a big difference. I entered sever numbers before coming up with 12.4287374 or about 12.5.

Again thanks everyone for the help.

 

[chicopee]

Go on the internet as there are quartic equation solvers and solutions.

 

[BridgeEI]

i think excel would be the easiest thing to use. Set up one cell for your "D" value and another cell that has the equation in it. Excel has a solver function and all you have to do is open it up and tell it to iterate the cell for "D" and your done. Less than a minute to do everything.

 

[MiketheEngineer]

BTW - Many textbooks have many errors. I have found a few myself!!

 

[beej67]

Learn the "Goalseek" function in Excel.

Hydrology, Drainage Analysis, Flood Studies, and Complex Stormwater Litigation for Atlanta and the South East -

http://www.campbellcivil.com

 

[PEinc]

Whenever I do a hand calculations for sheet pile embedment, I get a cubic equation for embedment depth, not a fourth power equation. Check your load diagrams and equation derivation.

http://www.PeirceEngineering.com

 

[normm]

No doubt my response is a bit late. But I attach a Mathcad graph of the equation. Looks the value is close to zero when the depth is approx 7.5.

 

[davefitz]

One way to write an iteration scheme is to ensure the order of the f(x) is less than 1.0, as follows:

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.

 

[PEinc]

I think too much effort is being spent on trying to solve the equation as it is written. As I mentioned above, the equation should not be a 4th power equation. I have designed sheet pile embedments for many, many walls using hand calculations and the moment equilibrium equation has always been to the 3rd power, i.e., D^3. See the attached sample embedment calc for a sheet pile wall. wettpaint should check his original equation.

www.PeirceEngineering.com

 

[wettpaint]

Here is where I got the formula.

Thank you to everyone for your help.

 

[PEinc]

Rather than copying a formula from a book, you should derive the formula yourself. That's what I gave you, but without the top line load. Take the example in the Das book. Separate the active and passive pressures. Add the top line load. Then, sum the moments about the bottom tip of the sheet piling, assuming the embedment is D. Then see if you get the same equation.

http://www.PeirceEngineering.com

 

[PEinc]

I solved Das' Example Problem 8.3. L = 15' with no unbalanced soil load behind the wall, an applied 2000 plf load, and the given soil properties.

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'.

www.PeirceEngineering.com

 

[PEinc]

P.S. The equation will change if there is retained soil behind the wall. You will need to use the previous sample solution I gave earlier and add any other applied loads. If the wall is not a cantilevered wall, the whole method will change.

http://www.PeirceEngineering.com

 

[csd72]

If all else fails just do trial and error start with numbers that you know are definately too large and too small and go from there by continually taking the average between the two closest numbers.