Mechanical Engineering: Ch 10: Forces on Cables (9 of 33) Concentrated Load: Example (Part 2)

[@iftikharkhan8403]

Thank you so much sir love from india I have studied engineering for 4 years but no one taught this concept as simple as you have taught thanks a lot ...

[@MichelvanBiezen]

Glad it was helpful.

[@karimkhan1312]

u r the best -- there is no lecture like this - thank u

[@pranavamali05]

1 million soon 🎉

[@MichelvanBiezen]

We are getting close. 🙂

[@mmrafi2010]

Thank you for your video. However, the problem is not correctly posed. It appears that you are restricting both the sag of C and the sag of D. We can restrict only one of the three sags (B or C or D) for this problem. Letting y_AC (vertical distance between A and C) = 1m, y_AD = 1.167 (D is actually above point A). This can be easily verified by taking moment about D for the segment DE. In that case, T_CD would be 44.4 kN. If you choose to restrict both the sag of C and D, then you will have to make one of the applied forces unknown. Otherwise, there are more equations (10: 2 for each node) than unknowns (9: y_CB, T_AB, T_BC, T_CD, T_DE, Ax, Ay, Ex, Ey). Hope this helps.

[@kathakalimukherjee7685]

Exactly

[@nikhilshinde6546]

Keep up the good work👍👍

[@abuzarkhanafridi63]

the value of TCD =75914.33 N take jiont C SUM FY (TCDSIN(@)=24000 It gives 75914.33N sir

[@raquelarce7879]

Muy buena explicación. Gracias

[@MichelvanBiezen]

You're welcome.

[@mojetoj3620]

how abou an example for when catenary is pulled horizontal by an arm? i want to have a maximum angle of 14 degrees, how can i calculate this? thanks for all your help and videos

[@MichelvanBiezen]

Watch the rest of the videos in the playlist. We have a number of examples where some data is given and asking for the remainder in all the various combinations.

[@mojetoj3620]

@@MichelvanBiezen thank you sir

[@ashishtiwari1912]

Thank you sir..your videos were very informative..But I failed to understand why can we take moment about any point in the cable while ignoring one part of the section?

[@hamish518]

It's because the cable doesn't resist moments internally, and therefore, the internal moment in the cable at any point is zero.

[@kabelodale_]

They in equilibrium I think

[@brianhaygood183]

Think of it as doing "method of sections" We can state that everything left of C is not moving or rotating, and we know no torque is transmitted to that portion of the cable from things to the Right of C, so we can state the portion on the Left is in static equilibrium.

[@mostafamostafa-wt9jd]

THANK YOU VERY MUCH PLEASE IN REAL CASE I ONLY CAN HAVE HORIZONTAL DISTANCE AND FORCES HOW I CAN DETERMINE VERTICAL DISTANCE IN THIS EXAMBLE 1M THANK YOU AGAIN

[@rajeevkumarsam5499]

Sir what will be shape of cable due to uniformly distributed load parabola or catenary.i think catenary but in answer parabola is given.plz clarify it.

[@MichelvanBiezen]

You can find that here: MECHANICAL ENGINEERING 10: FORCES ON CABLES It should be a catenary.

[@ahmedyusuf4657]

What would happen if there is a load in Z axis, for example wind load. (you only know wind velocity). Can you please explain how you will solve in this condition

[@MichelvanBiezen]

You would need to know the diameter and shape of the cable, so you cold determine the force applied to the cable by the wind. Then you would treat it as any other distributed forces in that direction.

[@ahmedyusuf4657]

@@MichelvanBiezen do you have any plan to make a video about this. Would be nice

[@elijahkwemoi7590]

Thanks

[@tharjun9447]

Thanks sir

[@MichelvanBiezen]

You are welcome!

[@orlandorafaelriveroreyes6427]

34000/36000 is not 3/4..in cable DE..you are out

[@মোঃনাঈমহাসান-ছ৬থ]

I am bangladeshi