Static Force Analysis of a Seat

[davidmandis]

Hi,
I have a seat which is about to be pull tested. The load in this test is applied as shown in the image attached. The applied load is transmitted to the suspension though 4 bolts which attach the seat frame to the suspension. I want to transfer the forces from the three locations that the force is applied on the seat to the suspension and find the reactions on the pins in the suspension. The final goal is to find the diameter of the pins to withstand the loads.
My plan is to,
1) divide the loads in half (since I am considering only half the seat in my free body diagram)
2) transfer the forces as a combination of a force and a moment onto the suspension upper plate (Fig 2)
3) analyze the suspension as a frame and find the reactions on the pins (Fig 3)
The problem is after I transfer the forces to the suspension upper plate, it becomes very difficult to analyze the suspension as a single frame since it becomes statically indeterminate. Second option is to just consider the suspension upper plate as a cantilever beam and design the left uppermost pin by dividing the reaction force at the pin into half. The third approach is to carry out an FEA analysis. But not considering the FEA analysis, do you guys think my approach towards the static analysis is correct? Would like to hear any comments or suggestions about the problem.

Thanks,
David

 

[rb1957]

your bolt loads are easy to get, ok?

the suspension plate looks problematical ... the way you're supporting the suspension plate, the only thing stopping it from pivoting is the links taking transverse shear.

i think a simple free body diagram (of 1/2 a seat) can show you how the thing will react the loads. sure there is a little confusion with the fixed end moment (which in this case is really differential loading on the two links, since they look like they're attached by single bolts) but you can run a couple of scenarios thru the hand calcs to see how much things can change (it'll some where between a pinned joint and a fully fixed joint). you can run a simple FEA, or more complicated calcs (for redundant structures).

good luck, i think you'll need it ... the seat looks pretty unstable

 

[davidmandis]

Yes the bolt loads are easy to get, but the problem is figuring out the reactions on the pins. If I consider the plate as a cantilever with a fixed end, the load would be distributed equally amongst the two pins at the end, but if I do not consider it as a fixed end then I cannot treat it as a fixed end cantilever since there is no resistance to moment and will have to consider the entire frame. Basically the problem is confusing the hell outta me.

David

 

[rb1957]

the pins connect the links to the suspension plate (immediately under the seat, and to the base plate (the rest of the world, the reaction at the bottom.)

forget the pins for a minute.

the links have to react the applied force. that tells you they have significant transverse shear. that tells you that the links (either individually or together) react moment at the ends.

the only redundancy you have is if the links are cantilevered (built-in) at both ends. start assuming that the links are pinned to the suspension plate ... that's statically solvable (i think). then assume a moment at this end. play around a little and i think you shhould see a trend. i suspect that the moment you calculate with one end pinned is in reality shared (equally) between the two ends.

i still think you structure is pretty unstable (and inefficient) ... but presumably you have your reasons.

 

[TomFin]

You can reduce the number of unknowns once you realize that the links in your parallel linkage are 2 force members. Being two force members you can deduce the sense (direction) and solve for magnitudes.

Failure is a prerequisite of successful design

 

[TomFin]

If you simplify the suspension upper plate as part of the parallel linkage, this is the Force analysis result...
See attached.

This pic looks like it is from MGA's FMVSS Test Fixture

Failure is a prerequisite of successful design

 

[rb1957]

i don't think the links are two force members, due to their mis-alignment to the applied force ...

two force member assumes pin ends, which would make this a mechanism

there has to be moment reaction, 'cause the links have to react transverse force

 

[davidmandis]

hmmm you seemed to have solved it like a truss problem. But, will it be the same since the joints in a truss are either welded or riveted while we have a pin joint here which will not resist a moment?
and rb1957 how will it be statically solvable since even if I use the method of sections or joints, the number of unknowns are more than the number of equations? remember each pin will have two reactions and there are 4 pins if I consider half the seat. I guess it could be statically determinate if I consider only one pin at the end and solve for half the load. what do you think?
BTW thanks to both of you for your help so far!

David

 

[TomFin]

First of all the system is unstable as it looks. I'm sure there is some sort of actuator that attaches to the parallel arms or Suspension Upper Plate which needs to be included as an external force/reaction in the analysis.

Observing the OPs pics there is indeed 4 "Pins" which he has identified.Truss analysis assumes pinned connections, even though they are bolted or welded in practice. The OPs concern is to properly size the pins which requires an FBD of the pin including all the forces from each member.

The definition of a 2 force member can be paraphrased as a member that is restrained at the endpoints and has no external forces between the restraints. The forces according to Newton's 3rd law will be equal and opposite as shown in the FBD.
Using the concept of 2 force members reduces the unknowns and complexity of the problem. Remember, for each link FBD, 3 Eqns can be applied employing the moment at the pin with the most unknowns yielding the reactions.






Failure is a prerequisite of successful design

 

[rb1957]

david,

just to put us on the same page, we're considering 1/2 the seat (symmetry).

i'm not solving the structure as a truss. you've got an applied force vector, which has to be reacted by the link. This reaction is sensibly resolved into axial and transverse components on the link. As the applied force induces a moment about where you attach to the "rest of the world", you have to have a moment reaction there; in your case i'd consider this as differential axial loads in the two links (if each one has only one pin, a single link cannot react moment at the end). thinking about it this way, i think you can determine the reactions between the link and the suspension plate statically (there'll be a force and a moment). still (like Tom) i think you seat support is dubious.

Tom, i'll accept that definition of a two force member. the example i usually use for a two force member is a strut, pin ended, carrying only axial (equal and opposite) loads.

 

[GregLocock]

From memory there's a spring/damper somewhere in the system, this is an important load path.

Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[vonlueke]

davidmandis: Your structure, as currently drawn, is a mechanism and is therefore unstable. If you apply your test force to it, the linkage rods rotate upward, and the suspension upper plate rotates downward, which is a dynamic problem. In order for this structure to be static, you must fix at least one of your four pins. If you fix one pin, then the structure is stable and statically determinate. If you fix two or more pins, the structure is statically indeterminate. Therefore, in the following answer, I assume the right-hand end of the lower linkage rod (point 1 in TomFin's diagram) is fixed, and points 2, 3, and 4 are pins, free to rotate.

Only a rod that has a pin at each end is a truss member or two-force member. Therefore, using my above assumption, linkage rod 1-3 is not a truss member, but linkage rod 2-4 is, which means the force in rod 2-4 is required to be axial, whereas the force in rod 1-3 is not.

Let's say your applied test load is Fo, and the angle between vector Fo and the horizontal is phi. Therefore, from statics, you can sum forces at pins 3 and 4, and sum moments about pin 4. You end up with the following end forces on the linkage rods.

H3 = (Fo/a3)[a1*sin(phi) - a2*cos(phi)],
F4 = [H3 - Fo*cos(phi)]/cos(theta),
V3 = Fo*sin(phi) - F4*sin(theta),

where Fo = applied test load, phi = angle between vector Fo and horizontal, theta = angle between linkage rod axial centerline and horizontal, a1 = horizontal distance from point 4 to Fo, a2 = vertical distance from point 4 to Fo, a3 = vertical distance from point 3 to point 4, F4 = axial force in upper linkage rod (positive is tension), H3 = horizontal force on linkage rod end 3 (positive to the right), V3 = vertical force on linkage rod end 3 (positive upward).

Now that you have the end forces on the upper end of both linkage rods, you therefore have equal and opposite end forces on the lower end of the linkage rods. And you can also solve for the moment on linkage rod end 1, again using statics. The shear force on pins 2 and 4 is F4.

By the way, if you have more than one applied test load (Fo), you can compute the above equations once for each applied test load, then add each H3, V3, and F4 result together to get the total linkage rod end forces.

 

[vonlueke]

Correction: My second sentence should read, "If you apply your test force to it, the linkage rods rotate upward (and the suspension upper plate stays level), which is a dynamic problem."