Hello, if you solve this example using the 'a' formula as given in NCEES PE/ FE handbook instead of 'R' formula, the answer comes to 118.03 which is close to 118.05. So the correct answer would be option 'd'. Both formulas are primarily the same but due to rounding off, the final answer changes which could throw you off if the options are within 0.1'. Please advise on more correct method to solve this. Thanks
The answer selections most likely won't be this close so it shouldn't be an issue. Always stick the the handbook for the equation to use would be our advice. We probably should have spaced these answers out a bit more and you could use either equation. Good call!
Hello, thanks for your videos, in the last step you are assuming that the 5.2425 stations distance are upwards, but I think the turning point is at 2.24 stations (from PVC to PVT), that means that elevation will go up until station 56.22 and then downwards. Let me know if I am confusing something. Thanks
Thanks for the video.The information given in the question is confusing. It clearly states "the vertical curve is 900 ft long." Then in the calculation it is used as the horizontal distance between PVC and PVT. Vertical curve lengths and horizontal distances are both important parameters in vertical curve problems. Vertical curve length and horizontal distance in this type of question need not to be interchanged because the results of the calculations would be different, and in an exam set up a student would take a lot of time to figure out whether vertical curve length = chord length. That would be a waste of time in the exam. Questions need to be explicit. Good explanation and thumbs up.
From what I understand the Vertical Curves Length is always the horizontal distance from the PVC to the PVT. This differs from Horizontal Curve where they have a chord and an arc length.
Question- How did you know the question was referring to the elevation on the chord rather than on the tangent or the curve itself? I've seen other videos and it refers to the elevation along tangent line or it would specify the curve. Thanks
Is X always L/2 in the case of a asymetrical vertical curve? I am studying for my FE Civil exam and this is crucial for me. The FE handbook on ncees does not have this. Also, can they give problems if they dont provide them in the FE handbook?
Danell Baptiste X is a point that can be anywhere on the curve. PVI is always L/2 for the vertical curve. The FE is really math and statistics heavy. They won't ask you anything you can't pull an equation out of the FE handbook if you're diving into civil questions.
Flightbyrd feet and stations are really the same thing. If I gave you a station of 1+00 that would equal 100 feet, or if I gave 250+00 that would equal 25000 feet. Hopefully that makes sense.
Sir , I would like to ask you a question since i am weak in calculations i am asking you this question.The summit curve length is of 50m. Initial tangent of Gradient 0.7% meeting negative gradient of -0.91%.where the chainage starts at 642.00 and ends at 692.00 and given k=29.083 at chainage 667.482(i.e maximim point of summit curve).I would like to know how to calculate formation road levels at center for every 2m intervals..FRL at ch 667.482 is given as 533.866 .FRL at chainage 642 is given as 533.796 and FRl at 692 is given as 533.743.EGL at ch 667.482 is given as 523.528
+Marie Kristin Baltazar 9 stations should be the total length of the curve. If you aren't familiar with stations just know that 900 feet is the same as 9 stations. 100 feet is 1 station, etc.
This vertical curve always confuses me. there's a guy on youtube making these transportation tutorials. in one clip for simple vertical curves, the PVI is L/2 (but not always). in another clip, he works out an example where PVI is not at L/2. I'm just so confuse since even the Ncess manual references the PVI is at L/2. could you please please help to confirm this?
WHY70122 Yes these can be confusing. I would always take the PVI at L/2 for vertical curves. See Example 79.6 in the CERM and see where they located their Vertex Point (PVI) even though the grades are different and it doesn't look completely symmetrical. The L is 400 and PVI is at 2+00 (have of L). So, yeah, I would take PVI at L/2. Maybe you can share a link of the video you had in question?? Hope that helps.
If you are referring to the final answer it looks like I get 117.91 ft and I just checked again to get the same thing. I hope that answers your question. Let me know if you see problems or have more questions. Thanks!
hi, how can you draw the shear and moment diagrams of beam having four(4) span from A to E, a settlement at B=1" and at C=2" with both end fixed. Span AB has a UDL of 3.33k/f with Length 24'-0"; Span BC has a point load at midspan with length 16'-0"; Span CD has no loading with length14'-0" and Span DE has a UDL of 1k/f with length 30'-33"
What does the settlement have to do with this question? Also is there any connection on points B, C, D? This type of problem is not something you would ever have to worry about on the PE breadth exam. It would be much simpler. In order to solve this you would first need to solve for all the reactions along the beam. Because you have 3 equations to solve this and more than 3 unknowns this is an indeterminate structure and you would need to use something like Castigliano's method to get your reactions. From there we could draw it up. Good luck with this one!
You can also use feet, instead of stations if you use decimal precents....use the numbers as they actually are.
Hello, if you solve this example using the 'a' formula as given in NCEES PE/ FE handbook instead of 'R' formula, the answer comes to 118.03 which is close to 118.05. So the correct answer would be option 'd'. Both formulas are primarily the same but due to rounding off, the final answer changes which could throw you off if the options are within 0.1'. Please advise on more correct method to solve this.
Thanks
The answer selections most likely won't be this close so it shouldn't be an issue. Always stick the the handbook for the equation to use would be our advice. We probably should have spaced these answers out a bit more and you could use either equation. Good call!
If PVI = 58+26.90, elevation at Sta 59+00.00 = 117.94. That's not much of a difference.
Hello, thanks for your videos, in the last step you are assuming that the 5.2425 stations distance are upwards, but I think the turning point is at 2.24 stations (from PVC to PVT), that means that elevation will go up until station 56.22 and then downwards. Let me know if I am confusing something. Thanks
The offset Rxx/2 corrects the vertical distance from the tangent to the curve at any point.
Thank you very much for PE courses
Thanks for the video.The information given in the question is confusing. It clearly states "the vertical curve is 900 ft long." Then in the calculation it is used as the horizontal distance between PVC and PVT. Vertical curve lengths and horizontal distances are both important parameters in vertical curve problems. Vertical curve length and horizontal distance in this type of question need not to be interchanged because the results of the calculations would be different, and in an exam set up a student would take a lot of time to figure out whether vertical curve length = chord length. That would be a waste of time in the exam. Questions need to be explicit. Good explanation and thumbs up.
From what I understand the Vertical Curves Length is always the horizontal distance from the PVC to the PVT. This differs from Horizontal Curve where they have a chord and an arc length.
How did you get the "5" digit in the 10-thousandths place of 5.2425 Sta if only dividing 524.2 ft by 100 ft/Sta?
Question- How did you know the question was referring to the elevation on the chord rather than on the tangent or the curve itself? I've seen other videos and it refers to the elevation along tangent line or it would specify the curve. Thanks
+Jerry Kong I would say that if they don't explicitly say what to solve for then you would find it on the curve because these are curve problems.
How is the elevation at PVC, 119.64' higher than the elevation at the 59+00 station, 117.91ft?
Is X always L/2 in the case of a asymetrical vertical curve? I am studying for my FE Civil exam and this is crucial for me. The FE handbook on ncees does not have this. Also, can they give problems if they dont provide them in the FE handbook?
Danell Baptiste X is a point that can be anywhere on the curve. PVI is always L/2 for the vertical curve. The FE is really math and statistics heavy. They won't ask you anything you can't pull an equation out of the FE handbook if you're diving into civil questions.
@@CivilEngAcademy agree here. Took and passed FE.
great problem. I need to focus on remembering that finding y at PVC = y@PVI+/- g1(L/2). That plus or minus is key.
you dont need to remember this, use tangent elevation from the reference, at PVI x= L/2
How is it that in the calculation for the elevation at PVC you can add ft and stations?
Flightbyrd feet and stations are really the same thing. If I gave you a station of 1+00 that would equal 100 feet, or if I gave 250+00 that would equal 25000 feet. Hopefully that makes sense.
is PVI always at x=L/2?? thanks
Andres Bonilla yes vertical curves are always symmetrical about PVI, even if they don’t look symmetrical
117.94'
Sir , I would like to ask you a question since i am weak in calculations i am asking you this question.The summit curve length is of 50m. Initial tangent of Gradient 0.7% meeting negative gradient of -0.91%.where the chainage starts at 642.00 and ends at 692.00 and given k=29.083 at chainage 667.482(i.e maximim point of summit curve).I would like to know how to calculate formation road levels at center for every 2m intervals..FRL at ch 667.482 is given as 533.866 .FRL at chainage 642 is given as 533.796 and FRl at 692 is given as 533.743.EGL at ch 667.482 is given as 523.528
karadu sainathreddy if you could email me the problem by through my contact page I can take a look at this. www.civilengineeringacademy.com/contact
117.94 when PVI=58+26.90
If the station at PVI is 58+26.90, elevation at station 59+00 is 117.94.
where did you get 9 sta?? i'm confuse
+Marie Kristin Baltazar 9 stations should be the total length of the curve. If you aren't familiar with stations just know that 900 feet is the same as 9 stations. 100 feet is 1 station, etc.
u teach very clearly but handwriting is difficult to understand,thnx
Thanks for the feedback! I'll try to improve. My handwriting has never been that good :) Maybe it's the engineer in me.
elev. 117.94 ft.
Elev.@59+00 = 117.94 ft when PVI sta. 58+26.90
got 119.63 ft for Sta. 58+26.90.
Elevation of 59+00 IS 120.9788 , when PVI is 58+26.90
Answer Using PE Handbook 5.3.1 (pg 286) Curve elevations = Ypvc + g1x + ax^2 -> NO EQUATION GIVEN, THIS IS JUST MATH: Ypvi = Ypvc + g1 (L/2) -> Ypvc = Ypvi - g1 (L/2) = 119.64 -> g1 = given -> a = (g2 - g1)/ (2L) = -0.000044 -> X IS THE TRICKY PART x = STAx - STApvi + (L/2) = 524.2 CORRECT ANSWER = 117.91'
I got elevation of 117.91 at station 59+00.00
This vertical curve always confuses me. there's a guy on youtube making these transportation tutorials. in one clip for simple vertical curves, the PVI is L/2 (but not always). in another clip, he works out an example where PVI is not at L/2. I'm just so confuse since even the Ncess manual references the PVI is at L/2. could you please please help to confirm this?
WHY70122 Yes these can be confusing. I would always take the PVI at L/2 for vertical curves. See Example 79.6 in the CERM and see where they located their Vertex Point (PVI) even though the grades are different and it doesn't look completely symmetrical. The L is 400 and PVI is at 2+00 (have of L). So, yeah, I would take PVI at L/2. Maybe you can share a link of the video you had in question?? Hope that helps.
is the elevation 117.94 ft?
If you are referring to the final answer it looks like I get 117.91 ft and I just checked again to get the same thing. I hope that answers your question. Let me know if you see problems or have more questions. Thanks!
I agree with Corbyn Cools
For the question to the audience, using PVI Sta 58+26.90
The elevation is 117.94 ft at Sta 59+00
got station, 53+76.9 and elev = 117.93 ft
Elev is 117.911
hi, how can you draw the shear and moment diagrams of beam having four(4) span from A to E, a settlement at B=1" and at C=2" with both end fixed. Span AB has a UDL of 3.33k/f with Length 24'-0"; Span BC has a point load at midspan with length 16'-0"; Span CD has no loading with length14'-0" and Span DE has a UDL of 1k/f with length 30'-33"
What is the point load at the midspan of span BD and we can start working on it!?
Civil Engineering Academy The point load at Span BC is 35kips
Lasana Fofana Hey! i did not hear from you again.
What does the settlement have to do with this question? Also is there any connection on points B, C, D? This type of problem is not something you would ever have to worry about on the PE breadth exam. It would be much simpler. In order to solve this you would first need to solve for all the reactions along the beam. Because you have 3 equations to solve this and more than 3 unknowns this is an indeterminate structure and you would need to use something like Castigliano's method to get your reactions. From there we could draw it up. Good luck with this one!
fOR ELEVATION at PVI you + g1 not subtract. This video is just wrong