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Truck Bed Cover and Gas Spring Forces

  1. Aug 5, 2014 #1
    I am working on making an aluminum truck bed cover for a customer despite my attempts to pass it up, he insists that i make it for him. Fabricating it, is not a problem, but I am having a hard time figuring out the force and length of gas springs needed and the best mounting angles for them. I apologize if this is not a good place to ask this, but i saw another fella who asked a similar question years ago, although, he had a little more physics knowledge than I, so i figured it never hurts to ask.

    It will weigh approx 60lbs, possibly a tad heavier if he has it painted, and will be 52" long and needs to open up approx 30°, or 2 - 2 1/2 ft at the end. Center of Gravity will be very close to the center of the lid. Mounting position is very flexible, upper mount on lid approx 2" +/- from top, and if possible i would like to use something like these from McMaster http://www.mcmaster.com/#4138T63 or http://www.mcmaster.com/#9416K23, but am not limited to them if they will not work well, they are the longest decently priced ones from a supplier i use regularly.

    It would be nice if it self opened gently, but is not 100% necessary, as long as it opens fairly easily and holds it open.

    If anyone can help me figure this out, i would greatly appreciate it, as anything we have done in the past has just been a trial and error with forces and mounting positions. I would like to figure this one out properly, and learn how to for future needs.

    Thanks in Advance, I appreciate your knowledge!
    Brian
     
  2. jcsd
  3. Aug 5, 2014 #2

    Simon Bridge

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    What you need will depend on how the cover opens and where the gas springs are placed and how they are positioned - due to the lever action. Basically you have to draw a picture.
     
  4. Aug 5, 2014 #3
    OK, a few thoughts on how to go about it. I think it is trial and error.
    Firstly, draw your arrangement with the cover shown in both the open and closed position, in a side view.
    As a start choose a point on the side of the opening cover that is in line with the CG. Draw and arc, length equal to the open gas spring centred on the attachment point of the open cover. If you choose a point closer to the hinge the strut will need to provide a higher force but less strut travel can be used. A point further away from the hinge means a lower strut force but a strut with longer travel.
    Now draw an arc centred on the attachment point of the cover in the closed position, length equal to the closed spring, or maybe just a tad longer. Where those arcs cross gives a fixing for the other end of the gas strut.
    This arrangement now shows the angle of the strut in the open and closed positions, and the component of the force acting vertically up on the cover must be at least equal to the weight of the cover at that point (or half if using two struts).
    If this doesn't come out right you might have to choose another strut length and run through it again, but the answer should quickly converge.
    That's my initial thinking, but the explanation might be a bit clearer if I had to actually work through the exercise instead of do it as a thought experiment.

    Cheers,
    Terry
     
  5. Aug 5, 2014 #4
    Here is a quick drawing, spring is drawn with extended (33.94") and compressed (17.8", stroke is 16.14") lengths for mounting points. Is mounting at CG a must? I figured to make it need less force, i could mount it maybe 1/3 from end, and then mount it somewere close to the blue circle area. After drawing the springs that i referenced, it appears that if i do this, i wont need quite as long of a spring as i initially thought. I realize it may be somewhat trial and error, but i thought i could draw up the mounting, and calculate the forces from the angles that the shock is sitting in closed and open to get a pair of springs that will be enough, but not too much. Forgive me, as i am learning here.

    Edit: Another thought, is that i could change the springs around so that they mount in the back by the rear stake pocket near the tailgate, maybe at CG of the cover, and are perpendicular to the cover when open. That might actually be a better plan. What do you think? I think that would be easier to calculate also.

    Brian
     

    Attached Files:

    Last edited: Aug 5, 2014
  6. Aug 5, 2014 #5
    I don't think there is one 'right' solution, there are probably a few that will satisfy the constraints.
    One thing that comes to mind is that it may not be desirable to have excess strut force available in the closed position, i.e. if there is excess force when closed, unlocking the cover would cause it to fly up, so to speak, and could be dangerous. If the closed cover strut force is a just a little below the weight to be lifted it is easy for anyone to start opening the cover.
    With good geometry, as the cover opens the strut becomes more normal, or perpendicular, and the component of the strut force acting against the weight of the cover becomes progressively greater...but you don't want too much force in the open position or it will be hard to close the cover.
    A lot of compromises.
    I would probably do a number of trial arrangements with a few of the possible struts, and probably do it Sketchup or similar, so it is easy to draw the arc lengths, and measure angles to calculate force components.

    Cheers,
    Terry
     
  7. Aug 5, 2014 #6
    Terry, what your saying sounds like a good plan. What are the formulas or methods for figuring out the forces? I have a cad program, that is were that pic was drawn. I think i understand the weight that needs to be lifted, if connected at the CG, Horizontal would be 60lbs, 45deg would be 45lbs and vertical would be 0lbs. correct? How does the force exerted by the spring change with its angle? The same?

    Brian
     
  8. Aug 5, 2014 #7
    Brian,
    I have knocked up a little diagram that might help, but blowed if I can see how one can attach an image...must be domestic blindness again!
    How can I do that?

    Cheers,
    Terry
     
  9. Aug 5, 2014 #8
    When posting, use "go advanced" button, then near bottom of page use the "manage attachments" button.
     
  10. Aug 6, 2014 #9
    OK...I had been using "quick reply", so now it all comes clear.

    You will see in the diagram attached, if my uploading works OK, I have two views, cover closed and cover open.
    The gas strut force is probably given as an axial force (I haven't looked at the strut specs), hence you need the component of that force that is in the direction of the cover weight at the attachment point.
    In each case the cover weight shown is W', which you need to calculate from the weight W and the location of the attachment point, noting that some weight will be carried at the hinge. You can calculate the weights (reactions) at the hinge and the attachment point using the following relationships, sum of vertical forces = zero, and sum of moments = zero. Sing out if that is foreign to you.
    The gas strut axial force is F, and the component you need is F'=F.sin(theta).
    As mentioned previously, play around with arc lengths equal to closed length + a bit (so it doesn't bind), and open length, from trial cover attachment points in open and closed positions.
    Hope that helps
     

    Attached Files:

  11. Aug 6, 2014 #10
    tduell, yes this is all foreign to me. I can build a CNC Plasma/Router/Scribe table from scratch, and make the aluminum cover no problem, but this i need to learn as it is greek to me. Maybe an example, would help. Lets say for ease of example, the weight of the cover is 60lbs, Length is 52", attachment point is dead center 26" from end, opened angle of cover is 30deg, and gas spring is 30deg while cover is closed with the vertical leg of triangle being 10" and the angle (or compressed spring length) being 20". If a spring force is needed for reference, lets use 100lbs x 2 springs. Thanks for being gentle with my lack of these particular math skills, i learned some in school, but haven't used or needed it since, until now.

    I am thinking that cover weight closed is 60lbs, and at 30deg would be 40lbs?
     
  12. Aug 6, 2014 #11
    At this stage I think it is worth you having a a good try at figuring it out yourself, with a bit of guidance.
    The attached gives an example of how to calculate the cover load at the strut attachment point, generalised a bit for an arbitrary attachment point. Have a think about this and see if you can use it for the case you have in mind. Just remember that a moment is force x distance, and in my examples we equate clockwise moments with anti-clockwise moments...ie they are equal if the system is in equilibrium. (sorry if this getting detailed).
    See how you go, and in the meantime I'll have a bit of look at the geometry you have described and see if I can arrive at conclusions.

    Cheers,
    Terry
     

    Attached Files:

  13. Aug 6, 2014 #12
    The cover weight is always 60 pounds no matter what orientation.
    Weight always acts in the vertical direction.

    What changes is the reaction forces at the pin at the one end of the cover, and the reaction forces at the pin where the cover and strut meet.
     
  14. Aug 6, 2014 #13
    when I said cover weight, I should have said force against the gas spring.
     
  15. Aug 6, 2014 #14
    Using that drawing and figuring the reaction force for both opened and closed, i get 60lbs for Rb, with the attachment point on Center of Gravity (center of lid). Not sure if im doing something wrong or missing something. The sum moments about hinge = 0, im understanding for this example it does not effect calculation, basically using the formulas in the boxes you made.

    When offseting the mounting point to be 1/3 from the open end, i am getting Rb=34.2 closed and Rb=45.4 open, which cant be right. ????

    Maybe i should tell him to just use a broomstick to hold it open.... ;)

    On another note, to figure the force required vertically to hold the lid in the closed and open positions (without regard to gas spring angle, still need to figure that), if i stuck a scale underneath with a rigid object attached to the upper mounting point, the scales reading (minus the weight of the rod) should give the same results as the formulas and provide the figure needed for this part, correct?
     
    Last edited: Aug 6, 2014
  16. Aug 6, 2014 #15
    What you have to calculate is based on the lever.
    220px-Lever_%28PSF%29.png

    Your box/ cover would act as the one in the middle, with the fulcrum at the hinged end, the resistance is the weight, W, of the cover, and effort, F, is that from the gas spring. W and F act in the vertical direction.

    With the force F at position A, W at position B, and the fulcrum at O, then from the sum of moments about O, we can say that
    F A = W B, or
    F = W B / A
    regardless if the cover is open or closed.

    One has to determine the distance A and B to find the mechanical advantage.

    With A, 1/3 from the end, or 2/3 from the fulcrum O, and B at the midway point, and the length of the cover L, we can see that
    F = W (1/2 ) L / (2/3) L = W (3/4) --> the length cancels out.
    For W = 60 pounds, F = 45 pounds vertically, whether open or closed

    This is not the force that the spring has to push on the cover. The spring can push only along its length, so we have to somehow convert that push into a force to match the vertical force.

    The spring will be at an angle from the horiziontal. We can break the spring force, Fs, into its horizontal force and its vertical force. The horizontal force, Fh, is the spring force, Fs, multiplied by the cosine of the angle. The vertical force, Fv, is the spring force, Fs, mutiplied by the sine of the angle. One can see that these forces form a triangle with Fh on the bottom, Fv is vertical, and Fs completes the triangle.

    One has to compute Fv from the spring when it is at a different angle whether open or closed as being equal to the lever force F from above. We can then compute the spring force Fs that we need.

    I think that is the simple way to do.
     
  17. Aug 6, 2014 #16
    Because your attachment is at the centre of gravity the reaction will always be 60 lbs regardless of angle.
    Using the figures you provided for your possible arrangement with strut attachment at centre of cover, I arrive at the following...
    Vertical lift at attachment on cover, cover closed = 0.5 * axial strut force (closed strut is at 30 degrees to horizontal)
    Vertical list at attachment on cover, cover open = 0.85 * axial strut force (open strut is at 59 deg to horiz.)
    Length of open strut = 26.8 inches.

    I may have misinterpreted your arrangement.
    Do any of those angles and lengths clock up with your CAD layout?

    Yes, you could measure the cover weight by simply supporting the cover on scales, as described.

    Cheers,
    Terry
     
  18. Aug 6, 2014 #17
    I think i was over complicating things and getting confused with all the terminology that i am not used to. Thank you for describing it differently. I will look at it again probably after supper and see what i come up with and see how clear things are.
     
  19. Aug 6, 2014 #18
    The spring lower mount is yet to be determined, just a general area, but could change. I read that if possible they should be kept at 30deg or more at the place the rest the most (closed) to insure that the oil keeps the seal lubricated fully.

    I have attached another picture with a theoretical lower mounting point now for the sake of conversation and understanding.

    Without regard to the formulas, my thought would be that the spring being 100% vertical would mean full spring force is lifitng the cover, and if 100% Horizontal, 0% of the force is lifting the cover (it is just pushing away from the hinge, so if it is at 45deg, then 50% of the force is against the lid.

    So a 60lb cover weight with CG mounting and springs at 30deg would mean 180lbs force is needed, divided by 2 springs = 90lbs each, which would make the lid virtually weightless to lift from closed.

    In my attached drawing, the angles are 32deg closed (need 168.75lbs spring force) and 54deg open (need 100lbs spring force) divide by 2 springs and need 84.375lbs force per spring. Use 80lb springs for slight lifting needed, or 90lb for self opening or close to it. 80lb springs would mean 60lbs force needed to close from open, and 90lb would mean 80lb force to close. Maybe too much? Not sure. Does this seem right?

    After i post this, i will try to decipher the formulas using this latest picture, and see if i get the same thing.

    Brian
     

    Attached Files:

  20. Aug 6, 2014 #19
    at 45 deg the vertical force is 70% of axial force (sin 45).
    For your arrangement as shown, at 32 deg the vertical force from struts will be 0.53 * axial force (sin 32)
    At 54 deg the vertical force will be 0.81 * axial force (sin 54).
    Note that if the strut force is more than the weight when open, you don't necessarily have to provide all the excess force needed at the attachment to pull the cover down. At the end of the cover furthest from hinge the force required is much less, and can be calculated by using moments about hinge.
    Looks like you are getting close to having it figured out.

    Cheers,
    Terry
     
  21. Aug 6, 2014 #20
    Ok, another picture with the lengths and dealing with the sin = opposite/hyp or 14.5537/26.9133 = .54 of Axial Spring force.

    Vertical lift at attachment on cover, cover closed = 0.54 * axial strut force, OR

    Weight of 60 / .54 = 111.111 lbs needed of spring Force for 60lbs Vertical Force (to make lid weightless) at 32deg spring mounting when lid is closed? Not the same as my last figure, and still dont know if im doing it right, the way i did it seams right to me and i understand how i came to that with the % of force with different spring angles. When i use the sin formula, i cannot relate to why how those figures give me a logical answer. There is also a small margin of error, as the hinge and spring mount and not 100% on the some horizontal plane, but im not adding that in right now, and its probably small enough that it can be negligible as we will have a range of force that will provide acceptable results, and the mounting can be slightly adjusted to fine tune things also.

    This is so aggravating, it seems so simple but I cannot make your equations equal what seams to make sense to me.
     

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