@@user-bj8zl5dp2j ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-fU0V7BoAu_k.htmlsi=xTWB5jMd7zOjZC39 This video may help you in determining length and bearing of unknown side in a close traverse.
a vertical curve joins a -1.2% grade to a + 0.8% grade The P.I of the vertical curve is at station 75 m elevation 50.9 m above sea level The center line of the roadway must clear a pipe located at the station 75 + 40 by 0.8 m The elevation of the top of the pipe is 51.1 meter above sea level what is the minimum length of the vertical curve that can be used
Thanks for your explanation. I apricate it, if you could solve another problem for compound curve. If tangent points and bearing of T1 and T3 are fixed, how can I find the two radii? Thank you!
Chainage is the distance from one destination to another. If you want to find the chainage at PI then calculations steps depends upon the point of known chainage. I hope it clears your doubt.
I'm a bit unclear about your question, but based on what I understood, here is my answer: "I don't believe there is a need for chainage to calculate the deflection angle of the last chord. It solely depends on the length of the curve." Please let me know if this addresses your query.
sir i have a doubt since stdia is fixed and in first case the midpoint of upper and lower stdia gives the middle stdia and in second case it fails. why??
I think you are talking about the load obtained by the externally applied uniformly varying load? If yes. The following discussions may clear your doubts. The value of 576 will be obtained if we had a uniformly distributed load UDL (12*48=576) but in this case, we have a uniformly varying load (UVL), hence (1/2*12*48=288). It is the same like the area of a rectangle (UDL) is b*h but the area of a triangle (UVL) is 1/2*b*h.
Sir, can you please prepare a video regarding the application of straight lines in graphics. Or if you have any video links regarding that application of straight line topic please attach them to my comment. I hope you will help me 🙏🏼🙏🏼🙏🏼
The cross-section is exactly perpendicular to the QR member. However, the x-x section is not perpendicular to the QR member. Cross-sectional dimensions are given. Therefore, to convert this cross-sectional dimension to that on the x-x section we need to use a right-angled triangle. That's what it is done in this video.
For uniformly varying load(UVL), the load will act at 1/3 of the total length of UVL from the larger side. 12 x 1/3 = 4 ft, larger side is B, hence 4ft from point B.
