A question about Example 7-1 of CT Trenching and shoring manual

[ysna]

Hi, I am trying to design a braced soldier pile wall with the help of CT manual, but I seem to have problem finding the required depth of embedment of piles. According to the manual I need to take moments about brace location on the pile to calculate embedment depth using a factor of safety of 1.3. Doing that leads to a third degree equation which gives only negative roots for D. I checked the example 7-1 of the manual (page 7-14) and put M_R=1.3M_D on page 7-18 which in turn gave me this equation

D^3+25.66D^2+94.96D+185.71=0

which also gives negative values for D and off coarse totally different from what the manual suggested which is

D^3+18.7D^2-114.53D-223.98=0

I checked the procedure before equation above and they all seem to be right but putting M_R=1.3M_D just doesn't give D^3+18.7D^2-114.53D-223.98=0.
I just wonder if there is something wrong with the procedure suggested by the CT manual or I'm just too stupid to do basic calculation

I found a link to mentioned example here (

http://files.engineering.com/downlo...952-4374-a46d-a51c80be834c&file=Tie_Back.pdf)

would you please take a look at page 7-18 and help me with that example. Thanks.

 

[Panars]

What the example in the manual does not make clear is that when you set M_R equal to 1.3 M_D, the sign on M_R is negative. This is because the sum of the moments equals zero (M_R + M_D = 0) for the stable condition. When you move M_R to the other side of the equation, it is negative. So when you solver for D you are actual solving the equation -M_R=1.3 M_D. When you do that, you get the results in the example.

 

[ysna]

Thank you Panars, that solved my problem... Seems that I have forgotten simple equilibrium equations