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 Force required to bend a tube suspension link... Post Reply Forum
 Posted by: astroracer ® 03/01/2004, 14:31:18 Author Profile Mail author Edit I have a problem I am working on. I want to determine the amount of force a 1 1/2 o.d. x 1/4 wall D.O.M tube can withstand before bending. The tube is the lower link in a four link suspension on the back of a 3300 lb truck. It is 36" long with the front pivot located to the frame with a rod end. The rear is attached to the rearend with a rod end. The truck is lifted with airbags which sit between this lower bar and the frame. The rear most bag mount (welded to the bar) is 20" from the rearend (lever arm). I figure the weight distribution is pretty close to 50/50 so I am using 1650lbs to calculate the loads. I've included a sketch which will hopefully clarify the problem. I am worried about the bar bending during hard suspension hits and /or work hardening and breaking. Are my fears justified and what kind of forces will this bar withstand before it loses it's elasticity and bends perminently or stresses and breaks. Thanks in advance for any help.

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 Is this really all that difficult? Re: Force required to bend a tube suspension link... -- astroracer Post Reply Top of thread Forum
 Posted by: astroracer ® 03/04/2004, 13:31:34 Author Profile Mail author Edit I know this will take some serious caculations. I am not a mechanical engineer, I am a designer. I can see some inherent strength problems with this design but would like to have some real world numbers to prove it to a friend... Any help would be greatly appreciated and I can help if more info is needed. Thanks Mark

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 Re: Is this really all that difficult? Re: Is this really all that difficult? -- astroracer Post Reply Top of thread Forum
 Posted by: ReubenAG ® 03/10/2004, 16:24:37 Author Profile Mail author Edit The calcs for this are not too difficult - the link can be approximated as a simply supported beam. I converted all your imperial units to metric - the force applied creates a bending moment of approx 2000 MPa (about 290 ksi) for a tube of the given dimensions. I don't know what steel you are using, but ordinary mild steel yields (deforms permanently) at anywhere between 200 and 250 Mpa (35 ksi). The drawing directions are also not clear - is this link a trailing arm type link or a link in a double wishbone (a-arm) setup. Normal convention for vehicles is to put x positive forwards, y positive left and z positive up. Also some clarity - when you talk about a 3300 lb truck do you mean a truck carrying a 3300 lb load, or the truck and load has a mass of 3300 lb?

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 Re: Is this really all that difficult? Re: Re: Is this really all that difficult? -- ReubenAG Post Reply Top of thread Forum
 Posted by: astroracer ® 03/11/2004, 13:42:39 Author Profile Mail author Edit Hey Rueben, Thanks for responding. The link is a trailing arm. This is a 4 link style rear suspension with two lower links and two upper links all mounted with rod ends. The mat'l is 1 1/2" O.D. x .25 Wall D.O.M. steel tubing (mild steel). The pic I provided has "X" positive to the left BUT... I work in GM on Unigraphics V18 and UG has the Coordinates all backwards as it was an airplane design program at one time. X+ is rearward. Y+ is out the passenger side door and Z+ is upward. Anyway... the front of the truck is to the right in the pic... The total mass of the truck (it won't carry any kind of load) is approximately 3300-3400 lbs. I figure 50% of the weight will be on the lower link with the bags aired up. This is the 1650lbs. I have indicated in the pic. This is a static load... not taking into account the loads induced when hitting a pothole or speed bump at speed. Thank you for helping. Mark

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 Re: Is this really all that difficult? Re: Re: Is this really all that difficult? -- astroracer Post Reply Top of thread Forum
 Posted by: ReubenAG ® 03/20/2004, 15:09:30 Author Profile Mail author Edit Even if you assume that the 1650 lbf carried by the rear axle is spread over two rear links, the yield strength of the links would be exceeded. In any situation such as this, it is a good idea to place the air bag, spring or whatever is used to support the load as close as possible to the point of application of the load. This minimises the bending moment (torque) experienced by the link. Sometimes vehicle designers have to attach the spring element some distance from where the load comes in due to space considerations - however, I am struggling to see why one would want to mount the spring as far from the axle as in your sketch.

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 Bending loads Re: Re: Is this really all that difficult? -- ReubenAG Post Reply Top of thread Forum
 Posted by: astroracer ® 03/22/2004, 12:44:04 Author Profile Mail author Edit Reuben, This is not my design... I am trying to explain to a young man why this is not a good idea. The reason for the excessive distance from the load is so the bag will provide more "height", or lift, when it is aired up. This is not a requirement for the suspension to work but a vanity thing. I guess you are only as cool as the height of your truck when aired up....  I know this is not a good design but explaining it to some of these young "builders" is tough if I don't know how you got the numbers you came up with. If you could possibly run through the calculations for me I would really appreciate it. Thank You Mark

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 Re: Bending loads Re: Bending loads -- astroracer Post Reply Top of thread Forum
 Posted by: ReubenAG ® 03/25/2004, 09:40:11 Author Profile Mail author Edit You had already shown how to find to bending moment (torque to some) at the point where the air bag attaches to the strut. Given a bending moment M, the maximum stress in any beamStress = concentration factor x Bending moment x distance from centre of area to top of beam / second moment of areaSome of the terms in this are a little obscure -this is the general form for calculating stresses in a beam which is not necessarily circular or even symmetrical. For the simplified case of a beam of circular cross section and thickness t, the following is true:Distance from centre of area to top of beam = Outer radius Second moment of area = pi *(outer diameter ^4-inner diameter^4)/64. BTW, I only work in metric units so I have no idea what conversion factors are used if the inputs are pounds and inches. The concentration factor is used if there are any sharp corners, shoulders, holes etc which may cause stress concentrations.

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