Soil Stockpil height Question

[Mike_NC]

Problem:

You excavate 207 BCY of soil from the south end of your project site and stockpile at the north end (flat end) of the site to be used later. If the angle of repose for the stockpiled soil is 36 degrees, the maximum height of the stock piled soile is most nearly:

A). 5 ft

B). 14 ft

C). 29 ft

D). 37ft

I found 79-3 of Lindenburg, heading 12 (Volumes of piles), however without knowing the shape or size of the base, I can't seem to figure out how to utilize the equations. Any pointers would be great.

Mike

 

[Fluvial]

Simplest would be to assume a circular base. Then it would be a question of "what height cone will contain this volume with this angle of repose". I don't know if that's ok to assume or not.

 

[MA_PE]

Based on the problem statement, I'd assume the the shape is a right circular cone with an angle between the base and the side = to rhe angle of repose (36 degrees). You know the volume = 207 BCY or 5589 cubic feet.

the volume of a right circular cone is V = 1/3 * R^2 * h

tan(36 degrees) = h/R therefore R = h / tan(36 deg)

substituting for R and solving for h, I get 14.12 ft.

edit: Fluvial I think you need to assume that and there's no reason not to.

 

[Mike_NC]

Thanks for the help.

 

[Fluvial]

edit: Fluvial I think you need to assume that and there's no reason not to.

It's been a while since I dealt with soil. Water doesn't have much of an angle of repose, yanno.

 

[EnvEngineer]

I the real world the loader would flatten the pile to about 5', but the test is not the real world.

 

[PE2BII]

Based on the problem statement, I'd assume the the shape is a right circular cone with an angle between the base and the side = to rhe angle of repose (36 degrees). You know the volume = 207 BCY or 5589 cubic feet.
the volume of a right circular cone is V = 1/3 * R^2 * h

tan(36 degrees) = h/R therefore R = h / tan(36 deg)

substituting for R and solving for h, I get 14.12 ft.

edit: Fluvial I think you need to assume that and there's no reason not to.

 

[PE2BII]

Hello,

I have an addendum to the solution.

The volume of a right circular cone is V=(1/3) x 3.14159 x R^2 x h

tan (36) =0.726, so 0.726= h/R; h=0.726R

substituting in the eqn 5589 ft^3= (1/3) x 3.14159 x R^2 x 0.73 R: h = 14.05 ft

The best

 

[smart]

Based on the problem statement, I'd assume the the shape is a right circular cone with an angle between the base and the side = to rhe angle of repose (36 degrees). You know the volume = 207 BCY or 5589 cubic feet.
the volume of a right circular cone is V = 1/3 * R^2 * h

tan(36 degrees) = h/R therefore R = h / tan(36 deg)

substituting for R and solving for h, I get 14.12 ft.

edit: Fluvial I think you need to assume that and there's no reason not to.

What is 207 BCY " mean?

 

[Mike_NC]

BCY is 'Bank Cubic Yard'... The soil in its natural state in the ground.

There are typical swell factors (Loose Cubic Yard 'LCY'), but since none is given, I did not assume any. I think the CERM has a list of typical swell and shrinkage factors for certain types of materials. I feel on the real exam, if they expect you to use a factor, they will either give it to you or give you enough information on the soil to make a judgement call on what the factor is. Somone else who may know better can confirm or reject that...

Mike

 

[PE_2009]

Based on the problem statement, I'd assume the the shape is a right circular cone with an angle between the base and the side = to rhe angle of repose (36 degrees). You know the volume = 207 BCY or 5589 cubic feet.
the volume of a right circular cone is V = 1/3 * R^2 * h

tan(36 degrees) = h/R therefore R = h / tan(36 deg)

substituting for R and solving for h, I get 14.12 ft.

edit: Fluvial I think you need to assume that and there's no reason not to.

Volume of right circular cone = Pie*r*r*h/3

Why didn't you use pie in solution?

 

[MA_PE]

Volume of right circular cone = Pie*r*r*h/3
Why didn't you use pie in solution?

because I forgot to type pi when I typed the equation.

 

[sgsmith2_PE]

I'm sure this is the same thing but here are some formulas that I used to solve this:

D = ((7.64*V)/(tan R))^1/3

D is diameter in feet

V is Volume in cubic feet

R is angle of repose

Once you have diamater, H=0.5D(tan R)

H will be in feet.

Obviously it's for conical piles only, but it just seemed easier than all that substituting junk...lol