Truck Bed Cover and Gas Spring Forces

[Metal Man]

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

 

[Simon Bridge]

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.

 

[tduell]

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

 

[Metal Man]

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 won't 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

 

[tduell]

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

 

[Metal Man]

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

 

[tduell]

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

 

[skills4u]

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

When posting, use "go advanced" button, then near bottom of page use the "manage attachments" button.

 

[tduell]

When posting, use "go advanced" button, then near bottom of page use the "manage attachments" button.

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

 

[Metal Man]

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. Let's 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, let's 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?

 

[tduell]

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. Let's 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, let's 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?

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

 

[256bits]

I am thinking that cover weight closed is 60lbs, and at 30deg would be 40lbs?

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.

 

[Metal Man]

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.

when I said cover weight, I should have said force against the gas spring.

 

[Metal Man]

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 I am doing something wrong or missing something. The sum moments about hinge = 0, I am 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 can't 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?

 

[256bits]

What you have to calculate is based on the lever.

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.

 

[tduell]

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 I am doing something wrong or missing something. The sum moments about hinge = 0, I am 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 can't 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?

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

 

[Metal Man]

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.

 

[Metal Man]

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?

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

 

[tduell]

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

 

[Metal Man]

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 don't know if I am 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 I am 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.

 

[Metal Man]

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

Ok, this helps me understand a little better. Using the angle and sin on the calculator to get the figure we need. I did fail to think about the leverage closing it. One more thing to grind my brain with. I am going to have to shut down the thinker here before too long and get ready for another day so i still have some brain cells functioning in the morning. :)

Edit: So i am thinking with this layout and the 32sin and 54sin figures, that i need 113lbs or 56.5lbs per spring at closed and and 74lbs or 37lbs per spring at open. So if i use 60lb springs, i would have 7lbs lift when closed, and 46lbs excess Force when open...so divide by 2? for leverage since mounting in the center, and it takes 23lbs to close. Does this sound better? If so, I think i may have this simple version figured out without moving the mount off of CG, and i think this would be an acceptable amount of force. Or i could bump down to the 50lb springs and require lifting to open, or i could move my mount slightly to make a little less lift when closed.

Now if i can just remember to use the sin on the angles and divide cover weight by it and absorb all this info, ill be in good shape.

 

[Metal Man]

So for further confirmation and understanding here is another picture with a different lower mount with 9° closed and 40° open. Needed spring force here at closed to make cover weightless is 384, or divide by 2 springs = 192lbs each. Probably not a good plan and puts unneeded stress on the hinge and spring mounts. This is were i would want to move the upper mount out for more mechanical advantage as long as i had access to long enough gas springs.

If i have this part down, maybe i need to verify i can figure the leverage correctly if moving the upper mount, as the theoretical mounting position may not be an option, and mounting the top mount further away from the hinge may prove to be better, but ill wait for some verification that i am correct, and call it a night for now.

Thanks abunch guys i greatly appreciate your knowledge and help, and hopefully someone else will get something out of this conversation sometime as well.

Brian

 

[Metal Man]

Ok, still not sleeping. Here is a link to something i have seen in the past that is dealing with a similar principle, and i think it may actually be wrong according to what we have discussed here.

http://www.rodandcustommagazine.com/techarticles/0812rc_coilover_spring_rates/

On their correction factor chart, 45deg mentions .5 correction factor. Is this wrong or am i missing something, shouldn't it be 45sin or .71?

 

[tduell]

The latest picture didn't turn up here.
Have another try at posting the picture as it will help ensure I don't misinterpret, and I'll check out the numbers.

Cheers,
Terry

 

[Metal Man]

Picture added...

 

[tduell]

Ok, still not sleeping. Here is a link to something i have seen in the past that is dealing with a similar principle, and i think it may actually be wrong according to what we have discussed here.

http://www.rodandcustommagazine.com/techarticles/0812rc_coilover_spring_rates/

On their correction factor chart, 45deg mentions .5 correction factor. Is this wrong or am i missing something, shouldn't it be 45sin or .71?

Without looking too deeply into it, I suspect this is a different situation. They are talking about spring rate, which is force per unit deflection. You are only concerned with vertical force, not rate...or maybe I've lost the plot as well :-)

Cheers,
Terry

 

[tduell]

So for further confirmation and understanding here is another picture with a different lower mount with 9° closed and 40° open. Needed spring force here at closed to make cover weightless is 384, or divide by 2 springs = 192lbs each. Probably not a good plan and puts unneeded stress on the hinge and spring mounts. This is were i would want to move the upper mount out for more mechanical advantage as long as i had access to long enough gas springs.
Brian

OK, looking at this arrangement, when closed, the angle between the strut axis and the vertical is 81 deg, and the vertical component of the strut force is 0.15 (cos 81), hence total strut force required is 60/0.15=400, or 200 lbf per strut.
When open the angle between strut axis and vertical is 50 deg, and vertical component of axial force is 0.64 (cos 50), hence total strut force required is 60/0.64=93.75, or 46.8 lbf per strut.
I think you are right to think it may not be a good plan due to the high forces when closed. It would be a better arrangement to get the strut closer to vertical in both conditions if that can be achieved. Your previous arrangement did seem to be closer to this ideal.

Cheers,
Terry

 

[256bits]

One force we have not dealt with is he horizontal force from the spring on the cover, and does that effect the opening and closing.
At the location on the cover where the spring acts, the spring force Fs can be resolved into the vertical force Fv and the horizontal force Fh. Drawing these force components we can see that Fv pushes up attempting to open the cover, and Fh acts to attempt to close the cover. If the spring is more vertical than horizontal Fv > Fh and Fh we can neglect, and the cover does not open too much, hence the simpler previous analysis.

But if the spring is more horizontal, we should include that in the calculation as it can be significant to how much spring force we need.

The horizontal force gives a moment Fh multiplied by the distance of a horizontal line from the cover fulcrum. This distance is the sine of the angle of how much the cover is open. We can see that with cover closed the angle is zero, the horizontal force acts along the cover so there is no moment. After opening of the cover the value of the moment may become significant.

The complete moment equation then becomes, summing arounf O,

∑M = F$_{v}$ a - W b - F$_{h}$ d

where,
F$_{s}$ is the spring force, acting at point A
F$_{v}$ is the vertical force of Fs
F$_{h}$ is the horizontal force of Fs
W is the weight of the cover acting at point B
a is the horizontal distance from O to A, where Fs acts
b is the horizontal distanc from O to B, where the weight acts
d is the vertical distance from O to a horizontal line drawn from A

d varies as the sine of the angle of the cover from the horizontal

A little bit more complicated.

 

[256bits]

∑Unfortunately, the moment equation has 2 unknowns in it: F$_{v}$ and F$_{h}$

Thus we also have to do a calculation for the sum the forces in the vertical and horizontal directions to solve.

 

[tduell]

One force we have not dealt with is he horizontal force from the spring on the cover, and does that effect the opening and closing.

You are quite right, well spotted as they say. That had slipped by me.
It is a real good reason to try to mount the struts close to vertical.
If one was sufficiently motivated, one could write an analysis package in Octave, or whatever, with a little database of strut specs, and run a trial and error analysis to arrive at possible solutions, and calculate all the forces and moments that apply...but that would require a fair bit of motivation :-)

Cheers,
Terry

 

[Metal Man]

Just as i think i might be starting to understand, things complicate further. :(

I think i have somewhat of an idea anyhow, so i will continue with making the cover, and finding the possible mounting locations, calculate what i think i have learned to be close, and probably post up again with actual figures and maybe with a little help, I can find atleast a range that i should try, and then just try it. I don't have to be spot on, but close would be helpful. It will probably be atleast a week or so before i get all the parts and materials and get it made.

The other option that i know i could handle (non smart version, less brain and more physical) is making the cover and upper mount, and then a makeshift lower mount that i could attach a scale to, and a solid rod in place of the spring, and see what the force or lbs is on the scale when just starting to lift, and at open position. Probably could have done that already by now. :)

Brian