Vertical Curve Length using AASHTO Standards

[maximus808]

Hi guys, can you guys help me understand this problem from Chelapati's Highway Book Volume 4.

The Question ask to find the length of Vertical Curve given a Design Speed. I would think you would use the Design Speed to find the SSD which the solution does. From going through the CERM, it tells you to use the table where you guess if SSD>L or SSD<L and if the if the L is what you assumed you're right. However in the solution, the SSD is 495'. And According the to the AASHTO Table for finding the length of Vert. Curves on Grades, using the equation that they used, the length of vertical curve is 264' which is less than the SSD. Thus, the equation we used would be wrong. Can someone help clarify?

Also, for part B, to find the low point station, couldn't I use the CERM method of x(lowpoint) = -G1/R where R = G2-G1/L. When I find x and add it to the STA of BVC. I don't get the result they had. Thanks for your help guys.

See attached.

Vertical_Curve_AASHTO_Problem.pdf

 

[Ambrug20]

Hi guys, can you guys help me understand this problem from Chelapati's Highway Book Volume 4.
The Question ask to find the length of Vertical Curve given a Design Speed. I would think you would use the Design Speed to find the SSD which the solution does. From going through the CERM, it tells you to use the table where you guess if SSD>L or SSD<L and if the if the L is what you assumed you're right. However in the solution, the SSD is 495'. And According the to the AASHTO Table for finding the length of Vert. Curves on Grades, using the equation that they used, the length of vertical curve is 264' which is less than the SSD. Thus, the equation we used would be wrong. Can someone help clarify?

Also, for part B, to find the low point station, couldn't I use the CERM method of x(lowpoint) = -G1/R where R = G2-G1/L. When I find x and add it to the STA of BVC. I don't get the result they had. Thanks for your help guys.

See attached.

Maximus808,

I did these problems without looking on the answer.

A. my result is the same as given.

I used the same formula as them CERM 78.58(B), based on the result L=149.6

Looked up table in the Green Book p.112 ex.3-1 SSD=495 and based on this data used table78.4 (CERM). Make sure to use sag curves formula for this problem. Since S>L, I had to use L=2S-(400+3.5S)/A L=381

B. when you calculate Low/High point see page 78-11 (CERM) fig. 78.10

There are different calculations for CREST & SAG. I thing that where you need to be careful.

Hope it helped.

 

[maximus808]

Thanks, I'll double check my answers

 

[maximus808]

Tanya,

In your answer for part A, how did you get L=381 using L=2S-(400+3.5S)/A? I get a result of 60 something? Don't we simply plug 495' for S and solve for L. Then we check if L is less than 495'?

S=495'

A=|G2-G1| = 2.3

L=63'

Can someone check my calcs? Thanks

 

[sac_engineer]

Tanya,
In your answer for part A, how did you get L=381 using L=2S-(400+3.5S)/A? I get a result of 60 something? Don't we simply plug 495' for S and solve for L. Then we check if L is less than 495'?

S=495'

A=|G2-G1| = 2.3

L=63'

Can someone check my calcs? Thanks

400

Technically, you're calculation is correct, but you need to check it with the Lmin value (AV^2/46.5: comfort criterion). This question is a bit tricky because the difference in slop transition is low, therefore the SSD will always be greater than the length, therefore using the maximum SSD would be favored.

It appears that the 300 feet approximation stated in the answer is calculated from the braking distance: 1.075x55^2 / (11.2 - .32 x 1.25)

I don't remember anything this tricky in the exam.

Good luck!

 

[maximus808]

Thanks. As long I knew that my calcs were ok. I thought it was an error in the solution. But Thanks.

 

[dneva]

Hi guys, can you guys help me understand this problem from Chelapati's Highway Book Volume 4.
The Question ask to find the length of Vertical Curve given a Design Speed. I would think you would use the Design Speed to find the SSD which the solution does. From going through the CERM, it tells you to use the table where you guess if SSD>L or SSD<L and if the if the L is what you assumed you're right. However in the solution, the SSD is 495'. And According the to the AASHTO Table for finding the length of Vert. Curves on Grades, using the equation that they used, the length of vertical curve is 264' which is less than the SSD. Thus, the equation we used would be wrong. Can someone help clarify?

Also, for part B, to find the low point station, couldn't I use the CERM method of x(lowpoint) = -G1/R where R = G2-G1/L. When I find x and add it to the STA of BVC. I don't get the result they had. Thanks for your help guys.

See attached.

Use y=-(g1^2*(L))/(2*(g2-g1)) +elvBVC to calculate the low point elevation.

elvBVC =.0125*200+105

This will give you the answer.

 

[maximus808]

Thanks, I reworked the problem using the CERM equation for low point STA and elev. and got the answer. Thanks for your help guys.

 

[Ambrug20]

Tanya,
In your answer for part A, how did you get L=381 using L=2S-(400+3.5S)/A? I get a result of 60 something? Don't we simply plug 495' for S and solve for L. Then we check if L is less than 495'?

S=495'

A=|G2-G1| = 2.3

L=63'

Can someone check my calcs? Thanks

L=2x495 - (400+3.5*495) / 2.3=63, you are right.

both methods as:

L=A*V2/46.5 = 150 (CERM eq.78-58)

and

using graph 78-15 K=115 for V=55 so L= K*A=115*2.3=265 You also may use Green Book p. 277

I am not sure whre did they have K>167 for flat curves and why do they use 300 as an answer.