question about a cut fill problem

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ketanco

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in a cut and fill question here is the info:

sta 1+00

cut 150sf

fill 0

sta 2+00

cut 50sf

fill 0

sta 3+00

cut 0

fill 40sf

in the solution, between station 2 and 3, it takes full 100 feet for calculation of for each of both cut AND fill (for each of them)

but shouldnt the cut PLUS fill distance equal 100? because from 2 to 3 it is transitioning from cut to fill. so shouldnt the cut distance and fill distance add to 100 but not each of them 100?

also, it is calculating the area of cut by (50+0) / 2 but shouldnt it divide by 3 and not 2? because look at cerm 13th edition, page 80-5 formula 80.10, if one of the area is 0, (which is 0 where it transitions from cut to fill) then we must divide by 3 and not 2 am i wrong?

this is the problem 101 in NCEES sample problems book for water resources and environmental, under morning questions.

 
Ketanco,

I don’t have the problem either, but it sounds like a really good one. It includes the major concepts of cut/fill problems.

  1. Figuring out total net cut/fill between stations when there is only a cut or fill area.
  2. Figuring out total net cut/fill between stations when there is both a cut and fill area.
  3. Also it makes you think about which method to use when estimating volumes.

So if you understand this problem, you will be able to figure out most cut/fill problems. With that said let me address your question;

Question #1. “shouldn’t the cut distance and fill distance add to 100 but not each of them 100”.

- The answer is no, the way they did it sounds correct. For estimating you usually find the volume of the cut and then find the volume of the fill. Then subtract the two volumes to get the total cut/fill for that length. I will answer this better by the solution below.

Question #2. “ It is calculating the area of cut by (50+0)/2 but shouldn’t it divide by 3 and not 2”

- I would say yes to this, It really depends on how they asked the question. If the question tells you to use the Average end area method to solve then they are right, but I think if they left it up to you to decide then the Pyramid method would be the most appropriate.

Solution: This is how I would have solved it:

Step 1: Find Cut between sta 1+00 – 2+00:

(150 sf + 50 sf)/2 * 100 ft = 10,000cf of cut

Step 2: Find Cut/Fill between sta 2+00 – 3+00: (I agree with you and will use the pyramid method)

(50+0)/3 * 100 ft = 1666 cf of cut

(0+40)/3 * 100 ft = 1333 cf of fill

1666 cf – 1333 cf = 333 cf of cut

- or you could do step 2 like this which sounds like it would make more intuitive sense to you. Be careful doing it like this when cut/fill areas are in both stations, you still can but you have to account for all the cut and fill area.

(50-40)/3 * 100 ft = 333 cf of cut

Step 3: Find total cut:

10,000 cf + 333 cf = 10,333 cf They usually like it in cy so 10,333 cf / (27 cy/cf) = 383 cy of cut

I hope that helps,

www.learncivilengineering.com

 
desantmf said:
Ketanco,

I don’t have the problem either, but it sounds like a really good one. It includes the major concepts of cut/fill problems.

  1. Figuring out total net cut/fill between stations when there is only a cut or fill area.

Figuring out total net cut/fill between stations when there is both a cut and fill area.
Also it makes you think about which method to use when estimating volumes.


So if you understand this problem, you will be able to figure out most cut/fill problems. With that said let me address your question;

Question #1. “shouldn’t the cut distance and fill distance add to 100 but not each of them 100”.

- The answer is no, the way they did it sounds correct. For estimating you usually find the volume of the cut and then find the volume of the fill. Then subtract the two volumes to get the total cut/fill for that length. I will answer this better by the solution below.

Question #2. “ It is calculating the area of cut by (50+0)/2 but shouldn’t it divide by 3 and not 2”

- I would say yes to this, It really depends on how they asked the question. If the question tells you to use the Average end area method to solve then they are right, but I think if they left it up to you to decide then the Pyramid method would be the most appropriate.

Solution: This is how I would have solved it:

Step 1: Find Cut between sta 1+00 – 2+00:

(150 sf + 50 sf)/2 * 100 ft = 10,000cf of cut

Step 2: Find Cut/Fill between sta 2+00 – 3+00: (I agree with you and will use the pyramid method)

(50+0)/3 * 100 ft = 1666 cf of cut

(0+40)/3 * 100 ft = 1333 cf of fill

1666 cf – 1333 cf = 333 cf of cut

- or you could do step 2 like this which sounds like it would make more intuitive sense to you. Be careful doing it like this when cut/fill areas are in both stations, you still can but you have to account for all the cut and fill area.

(50-40)/3 * 100 ft = 333 cf of cut

Step 3: Find total cut:

10,000 cf + 333 cf = 10,333 cf They usually like it in cy so 10,333 cf / (27 cy/cf) = 383 cy of cut

I hope that helps,

www.learncivilengineering.com
Click to expand...
thanks a lot but for 1 i am still not convinced. because at station +2, you have a cut and at station +3 you have a fill. so it means, somewhere in between 2 and 3, along that 100', you transitioned from cut to fill. this means, the length of cut section and the length of fill section adds up to 100. but they are not EACH 100. why are you and they taking 100 for each? I am still not clear on it.

 
After rethinking this, yes you should use the average end area like the book has.

Also to help you with number 1, you are not over thinking it you are thinking about what you are solving wrong.

There is no doubt that we are assuming that the profile view goes from 50 sf cut at 2+00 then in a straight line goes to 40 sf fill at station 3+00. If you draw it out, it crosses the x-axis a little before 2+56. (As pointed out earlier these are assumptions, we don't know what really is between there.) - now that is what you are thinking about we should be getting our numbers from, which we could if we wanted to spend a lot of time figuring out where it crosses the x-axis and then find the area of the triangle. And subtract those numbers.

But all we care about is how much cut or fill is left total between those two stations(mass diagram). So to do that you can simply say

50/2 * 100 - 40/2 * 100= 500cf --- what you are doing is subtracting the cut by fill over the whole distance. If you would plot both of these lines under the profile view, you would see the fill line crosses the cut line at the same distance. So at station 2+56 you are have more fill area then cut area, so the amount of cut quantity will start going down at that point

I know kind of confusing written out, Let me know if you get that or want me to include the diagrams?

 
ketanco said:
desantmf said:
Ketanco,

I don’t have the problem either, but it sounds like a really good one. It includes the major concepts of cut/fill problems.

  1. Figuring out total net cut/fill between stations when there is only a cut or fill area.

Figuring out total net cut/fill between stations when there is both a cut and fill area.
Also it makes you think about which method to use when estimating volumes.


So if you understand this problem, you will be able to figure out most cut/fill problems. With that said let me address your question;

Question #1. “shouldn’t the cut distance and fill distance add to 100 but not each of them 100”.

- The answer is no, the way they did it sounds correct. For estimating you usually find the volume of the cut and then find the volume of the fill. Then subtract the two volumes to get the total cut/fill for that length. I will answer this better by the solution below.

Question #2. “ It is calculating the area of cut by (50+0)/2 but shouldn’t it divide by 3 and not 2”

- I would say yes to this, It really depends on how they asked the question. If the question tells you to use the Average end area method to solve then they are right, but I think if they left it up to you to decide then the Pyramid method would be the most appropriate.

Solution: This is how I would have solved it:

Step 1: Find Cut between sta 1+00 – 2+00:

(150 sf + 50 sf)/2 * 100 ft = 10,000cf of cut

Step 2: Find Cut/Fill between sta 2+00 – 3+00: (I agree with you and will use the pyramid method)

(50+0)/3 * 100 ft = 1666 cf of cut

(0+40)/3 * 100 ft = 1333 cf of fill

1666 cf – 1333 cf = 333 cf of cut

- or you could do step 2 like this which sounds like it would make more intuitive sense to you. Be careful doing it like this when cut/fill areas are in both stations, you still can but you have to account for all the cut and fill area.

(50-40)/3 * 100 ft = 333 cf of cut

Step 3: Find total cut:

10,000 cf + 333 cf = 10,333 cf They usually like it in cy so 10,333 cf / (27 cy/cf) = 383 cy of cut

I hope that helps,

www.learncivilengineering.com
Click to expand...
thanks a lot but for 1 i am still not convinced. because at station +2, you have a cut and at station +3 you have a fill. so it means, somewhere in between 2 and 3, along that 100', you transitioned from cut to fill. this means, the length of cut section and the length of fill section adds up to 100. but they are not EACH 100. why are you and they taking 100 for each? I am still not clear on it.
Click to expand...
Not necessarily. You can be cutting on the left side of the road and filling on the right. There are plenty of times you have both cut and fill in the same cross section.

 
I pulled out my solution to this problem which I did 2-3 years ago. The scan I am attaching is the exact work I did 2-3 years ago, no modifications. Although my memory is foggy (!), it appears that I was thinking along the same lines as ketanco - that is, the cut and fill distance between stations 2+00 and 3+00 should add up to 100', not be 100' each.

I weighted the distance based on the amount of cut/fill. So, since there is 50 SF of fill (and 40 SF of cut), I multiplied the distance of fill by [50/(50+40)] x 100' =~ 55'. I multiplied the cut by (100'-55'=) 45'

In hindsight, I agree with what everyone is saying here. It's totally fine to use 100'. The "extra" length of cut will nearly be a wash with the "extra" length of fill. For instance, I got an answer of 388 CY and the solution shows 389 CY (Answer C = 390). Close enough. :)

ketanco, as for your question about using the pyramid formula as opposed to the average end area method when one end is zero, I think you are right. But, from what I noticed on these types of problems, and from what you can see with this problem, is the answers have a large enough difference between them that you'll still get close enough to the correct answer choice with whatever method you choose. Also, if I am not mistaken, aren't you supposed to use the pyramid method only when one end comes to a point? One of the ends in this problem comes to zero, but I don't think it comes to a point necessarily, right? (ex. It's more of a "wedge" than a "pyramid").

20130520124504.jpg

 
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