NCEES Sample questions- Breadth- Q126

[New2WR]

Dear All,

I hope all is going well with you preparing for the exam! I have a difficulty to understand problem# 126 solution (the alternate solution part) per the Breadth (morning) exam in the WR-ENV NCEES sample questions book. It solves the problem using the following equation which is the same as equation 79.47 per CERM 12 ed:

Y= Yvpc + g1X+[(g2-g1)/2L]X^2

Then it adjusts the left hand side to be the slope by subtracting Yvpc from both sides and dividing by X as shown below:

Y’ (slope) = g1 + [(g2-g1)/ L]

However I think there is a typo per the denominator it should be 2L instead of L. I checked the book's errata at the NCEES website but nothing there in this regard.

Your input guys is truly appreciated!

Thanks!

 

[ptatohed]

Which year NCEES Sample Q&Ss are you referring to? I checked my 2011 copy (Transpo depth but all breadth questions are the same) and #126 is a vertical curve problem.

 

[bradlelf]

Which year NCEES Sample Q&Ss are you referring to? I checked my 2011 copy (Transpo depth but all breadth questions are the same) and #126 is a vertical curve problem.

+1 give a year or post the problem for assistance

 

[K19]

This is indeed the vertical curve problem from the 2011 sample questions. @ NewtoWR, your second equation is missing an "X" at the end and should read:

Y’ (slope) = g1 + [(g2-g1)/ L] X

This is not an algebraic manipulation of the first equation (elevation Y as a function of X); rather it is taking the derivative of the first equation with respect to X to get a function for slope. The derivative of X^2 is 2X, and the 2 in the denominator is thus cancelled out. HTH

 

[ptatohed]

You're right K19. I guess New2WR never said #126 was not a vertical curve problem. I'm just not used to the term Y and Y' used in vertical curve problems so I assumed it was some sort of open channel flow problem.

I just checked my notes. Before I go any further, I'd like to state that it drives me nuts that the question and solution use the terms "VPC, VPI and VPT" instead of the correct terms "PVC, PVI and PVT". Anyway, moving on......

I solved this problem in (what I believe) to be less steps and less time:

r = (g2 - g1) / L = (3% - -2.3%) / 3 sta = 1.767% [the rate of change per station]

r (x) = 1.767% (2 sta) = +3.534% [the rate of change at Sta 14+00]

r14 + g1 = +3.534% + -2.3% = 1.23% [the tangent slope at Sta 14+00]

Answer (B)

 

[New2WR]

Thanks all for your inputs! Special thanks to K19 you nailed it man! yes that is exactly what was confusing me. I looked at it from the algebraic stand point and did not see it as a first derivative. Thanks again.