why do you use x/2 (a half of arc length CD) as the distance between D and shelf of Wx? Because the distance between D and shelf of Wx is smaller than x/2?
If this were a simple zipline, how would you work backwards to determine the load transferred to guys beyond A and B if you new the tension and expected load on the cable?
Hiền Trần That concept applies for catenary cables, where the weight acting is just the weight of the cable. In this case the wieght is external and the cable has no weight. There is nothing wrong in this concept
Thank uh for this vedio .... But can uh please explain me that If a heavy string attached at the ends at same horizontal level and when central Dip is very small approaches which type of curve and why ??? Please explain me
because this video is a case that outlines a uniformly distributed load along the cable. The situation you described is when its sags from self-weight, which is described by a different curve (catenary, next video in this series)
You are one of the best teachers in the field of mechanics...thank you for your contribution :)
Hai Loda kya bolke nikl gaya pate nhi chla
Thank you! Beautifully explained, perfectly clear and in less time than it took to find the video.
Glad it was helpful!
Civil engineering student here, best prof ever
Thank you. Glad you found our videos. 🙂
why do you use x/2 (a half of arc length CD) as the distance between D and shelf of Wx? Because the distance between D and shelf of Wx is smaller than x/2?
oh yeah, i got it
If this were a simple zipline, how would you work backwards to determine the load transferred to guys beyond A and B if you new the tension and expected load on the cable?
Great video. Thank you
Thank you. Glad you found our videos. 🙂
why does point C does not have Fy ?
Thank you.
thank u very much have enjoyed it
I think the shape behaves as hyperbolic curve. I have found a problem that the teacher use the length of CD as x is kind of wrong
Hiền Trần That concept applies for catenary cables, where the weight acting is just the weight of the cable. In this case the wieght is external and the cable has no weight. There is nothing wrong in this concept
@@vinayak1561 Thanks
how can we find the tension of the cable?
That is explained in the videos in this playlist. Keep watching.
How to compute the maximum force of cable?
The maximum force (tension) on the cable will be at the highest connection point, where it carries the most weight.
thank you
I was disappointed there wasn’t a lecture on a suspended cable with intermediate pulley supports
No we don;t have that type of lecture.
alguien sabe en que libro puedo encontrar el ejercicio
Thank uh for this vedio ....
But can uh please explain me that
If a heavy string attached at the ends at same horizontal level and when central Dip is very small approaches which type of curve and why ???
Please explain me
because this video is a case that outlines a uniformly distributed load along the cable. The situation you described is when its sags from self-weight, which is described by a different curve (catenary, next video in this series)
hello, can someone please recommend me to a channel i can learn the Calculation of deck arch bridges
Interesting
great and love from pakistan
Welcome to the channel.
Can you put the subtitles in Spanish please
The load is not uniform distributed
Wahhhhhhhh
Glad you liked it.
hi
Annoying
Why do you think so?