Results 1 to 10 of 10

Thread: Force Required to Lift Door

  1. #1
    Associate Engineer
    Join Date
    May 2017
    Posts
    6

    Force Required to Lift Door

    I have a relatively simple mechanics problem that I am stuck on at work. I need to calculate the force required to lift a bifold door using lift cables.

    Background:

    Bifold doors are a type of door where the top of the top section is hinged to the header and the bottom is hinged to the bottom half of the door. Lift straps or cables are used to pull the bottom of the door upwards. There are rollers on the bottom of the door that roll up the side of the building as the door opens. I have included a picture pulled from Pinterest for use as a visual reference.



    The weight of the door is known (4000 lbs overall) and the height of the door is 20 feet. I've drawn the FBD but am having a difficult time remembering how to calculate the force required to lift based on this information. I've done a summation of forces and my moment equations but what I am coming up with does not make sense. Any assistance would be greatly appreciated. Thanks in advance!

    Attached Images Attached Images

  2. #2
    Associate Engineer
    Join Date
    May 2017
    Posts
    6
    I have attached a picture of what I have completed. I am still unsure how to calculate the hp and torque required to lift the door. If anyone has any assistance they could give I would greatly appreciate it.

    Attached Images Attached Images

  3. #3
    Lead Engineer Cake of Doom's Avatar
    Join Date
    Aug 2012
    Location
    UK
    Posts
    433
    Work out the perpendicular distance for your lever arm then multiply the applied force by that distance. Thats the simple approach so you may want to throw some safety factors in there. Personally, I like a safety factor of 3 for mechanical lifting but check your local codes for whats allowable.

  4. #4
    Technical Fellow Kelly_Bramble's Avatar
    Join Date
    Feb 2011
    Location
    Bold Springs, GA
    Posts
    2,611
    Your equilibrium equation are incomplete.. You are looking for Rby and you are summing about a point - not a distributed load.

    Here are my Modeled numbers:


    Attached Images Attached Images
    Tell me and I forget. Teach me and I remember. Involve me and I learn.

  5. #5
    Associate Engineer
    Join Date
    May 2017
    Posts
    6
    There would not be a Rby since it is a roller, not a hinged connection. I just used resultant forces for the weight of each section of the door to simplify things. Ultimately I am looking for the force required lift the door and I am unsure of how to calculate it.

  6. #6
    Technical Fellow Kelly_Bramble's Avatar
    Join Date
    Feb 2011
    Location
    Bold Springs, GA
    Posts
    2,611
    Quote Originally Posted by oilmanjr View Post
    There would not be a Rby since it is a roller, not a hinged connection. I just used resultant forces for the weight of each section of the door to simplify things. Ultimately I am looking for the force required lift the door and I am unsure of how to calculate it.
    So, how does the bi-fold shutter lift?


    Yes, there is a Rby Force... that is the force used to lift it.
    Tell me and I forget. Teach me and I remember. Involve me and I learn.

  7. #7
    Associate Engineer
    Join Date
    May 2017
    Posts
    6
    Yes I misunderstood what you were saying. The Rby is what I'm looking for (it is denoted as "L" in my diagram).

    It is my understanding that we are looking at this entire system in equilibrium, so the force "L" or "Rby" that is found is only the force that is required to hold the door in the position that it is in. So would it just take any amount of force greater than "L" to lift the door?

  8. #8
    Technical Fellow Kelly_Bramble's Avatar
    Join Date
    Feb 2011
    Location
    Bold Springs, GA
    Posts
    2,611
    Quote Originally Posted by oilmanjr View Post
    Yes I misunderstood what you were saying. The Rby is what I'm looking for (it is denoted as "L" in my diagram).

    It is my understanding that we are looking at this entire system in equilibrium, so the force "L" or "Rby" that is found is only the force that is required to hold the door in the position that it is in. So would it just take any amount of force greater than "L" to lift the door?
    Yes, plus whatever friction there is...

  9. #9
    Associate Engineer
    Join Date
    May 2017
    Posts
    6
    Since the weight of the door is 4 kips and the Rby that you found for both situations was 2 kips is it safe to assume that the force required to lift is always 1/2*W? Or will this change depending upon different stages of the door opening? i.e. Will it be more difficult to lift the door when it starts to open or when it gets closer to full open, or will both be the same?

  10. #10
    Technical Fellow Kelly_Bramble's Avatar
    Join Date
    Feb 2011
    Location
    Bold Springs, GA
    Posts
    2,611
    Treat the bottom node like it is a static pin or pinned connection. Gravity is the loading and it only goes one way.

    is it safe to assume that the force required to lift is always 1/2*W?

    Run the numbers at several levels of lift (bottom, middle, highest lift angle) and look for the maximum lift force required. Then pick an adequate FOS for your end design.

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •