Seemingly simple free body diagram problem 6

[DerbyEngineer]

Hi all, hope you can help.

When drawing Free Body Diagrams (FBDs) is it necessary to consider the stiffness of elements within the structure? I never thought so, until I thought about a simple pin-jointed structure with two truss elements (see attachment). Before considering the relative stiffnesses, I would have immediately said that both truss elements equally share the load. Resolving forces confirms this. End of story.

However, if the truss elements are not of the same stiffness, I would then expect more of the load to be attracted to the stiffer element. But this would then contradict my FBD. What am I missing here? It seems such a simple problem, but I can't understand it!

Any information appreciated. Thanks,

Martin

 

[GregLocock]

Your reasoning in the top part of the document is correct. Your expectation and the conclusion in the bottom part is not.

Cheers

Greg Locock


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[rb1957]

ahh, "seemingly simple" ...

stiffness if relevent to statically indeterminate problems. statically determinate problems have only one solution (so stiffness isn't a factor) ...
now i'll open your link !

now, looking at you pic, the loads in the two structures (1" where the arms are the same section, and 2: where one arm is significantly heavier) are the same; assuming pinned joints. this is a classic three-force member problem, the three forces make a force polygon (no, not a dead parrot), with forces aligned to the driection of the members. a heavier member means "only" smaller stresses (and yes, smaller displacements, but you're (i'm ?) assuming small displacements (that the displacements have a negligible impact on force disribution inside the structure).

now, if the joints are fixed, so members can carry moments, then the solution is different, 'cause it is now redundant. then that the larger Fx (side) force in the heavier member is balanced by the same Fx in the smaller one (so that the force is no longer aligned to the member).

 

[electricpete]

(I also assume that the applied force is small and, therefore, the angles are unchanged.)

That sounds like a resonable assumption for most situations ... good to record your assumption.

However, what if the left member is (say) three times as stiff as the right member as shown. I would then expect it to take three times the load.

I can see how you come to this conclusion. You are imagining a small differential movement (stretching) when the force is applied. Further you are imagining that the differential movement is vertical. IF this differential motion was purely vertical, then you would be correct. In reality, the movement would not be purely vertical. It would be slightly in the horizontal direction of the stiffer sling so that the elastic strain distance of both slings is not the same. The three times as stiff sling will have approx 1/3 the strain as the other sling. The difference is very small in comparison to the lengths of the slings... has no significant effect on angles as per your previous assumption... the forces are practically the same as calculated the simple way without considering stiffness (as if the displacement is straight up).


=====================================
(2B)+(2B)' ?

 

[Cockroach]

The error is in your logic.

Geometry of the problem correctly concludes that the force through each leg of the truss must be identical. Again, the geometry dictates this because of the simularity in angles of the legs at the base. If the angles "theta" are different, then the force through the legs of the truss ARE NOT equal and one leg would carry more force than the other. You can easily see this from the mathematics of the free body diagram.

So "stiffness" which is not a function of geometry, does not affect load capacity through the truss. Your second statement is incorrect, three times the stiffness may mean a factor of safety of three, but the physical load due to geometrical similarity with the angle of the base means that the force through the members are equal. RB1957 is correct and correctly notes that in statically indeterminate structures, the additional equation to give independence is typically obtained from a material consideration.

For example, I have one member out of steel and the other out of timber. How would it be possible that in a symmetric application force through a leg would be unequal? Factor of safety or capacity to carry the load would not be, but that has nothing to do with geometry. You can proove this for yourself by simply going through the mathematics for unequal angles and obtaining a closed form solution set. Then substitute equal angles into the resulting equation and watch how quickly it reduces to the form where R1=R2=F/2.

I note that ElectricPete discussion about a Virtual Force, pertains to a method of solving theses problems, Method of Virtual Work. That assumes the structure would not change shape also for the work performed by a force over a virtual distance is zero. You're mixing up concepts.



Regards,
Cockroach

 

[electricpete]

I note that ElectricPete discussion about a Virtual Force, pertains to a method of solving theses problems, Method of Virtual Work. That assumes the structure would not change shape also for the work performed by a force over a virtual distance is zero. You're mixing up concepts.

Nope, not mixed up about the concepts at all (unless your picking on the terminology). With real materials (non-infinite stiffness), there is displacement when load is applied. The drawing presumes a symmetry. I assume the symmetry applies in the non-loaded condition. In the case where there is a non-infinite stiffness, that symmetry will be destroyed when load is applied. The amount of movement is so small as to be negligible in terms of angles (per assumption) and therefore negligible in terms of forces. However, if you assumed (incorrectly) that symmetry applies in both conditions, that would lead you to wonder how it is that both slings expand the same amount with the same deflection even though they have different stiffnesses. The answer is that exact symmetry cannot apply in both unloaded and loaded conditions if the stiffnesses are not the same. (and again, the deviation is not significant to predicting angles and therefore loads, but does play a role in reconciling the behavior as above).


=====================================
(2B)+(2B)' ?

 

[electricpete]

Correction

that would lead you to wonder how it is that both slings expand the same amount with the same deflection even though they have different stiffnesses

should be:

that would lead you to wonder how it is that both slings expand the same amount with the same force even though they have different stiffnesses

=====================================
(2B)+(2B)' ?

 

[MiketheEngineer]

Try RISA or something equal??

 

[electricpete]

Let me restate the purpose of my coments.
First of all, in case it's not obvious, I agree with Greg: The method on top of original attachment is correct, method on bottom is wrong. No controversy there, I trust.

The purpose of my comments was to address the question: why would someone think as stated on the bottom?

The logic on bottom is obviously trying to reconcile the relationships:
T1 = K1*deltaL1
T2 = (3*K1)*deltaL2
Where T is tension, deltaL is stretch distance, K is stiffness, K2=3*K1 postulated

If we assumed that deltaL1 = deltaL2, we would conclude T2 = 3*T1, which is in conflict with the correct conclusion offered on top of the page.

The resolution of the conflict comes in realizing that the assumptiondeltaL1=deltaL2 would be incorrect.
The two slings (or whatever) will stretch by different amounts. The difference in stretch is significant and necessary to consider when looking at the equation T = K*deltaL. But that difference in detlaL is small in comparison with all other dimensions of the problem by assumption. So there is no siginficant change in the angles and from logic on top no significant change in the tensions.

Also I have assumed the applied force is applied directly upward at the junction of those slings, wherever it ends up in equilibrium... i.e. the applied force is free to float horizontally or vertically so that it remains at the junction of those slings. I consider this assumption reasonable because the alternative assumption is not: The alternative assumption would be a force constrained to act in the original horizontally-centered location.... that scenario could not develop equilibrium in presence of unequal stiffness slings without a horizontal component of applied force.

=====================================
(2B)+(2B)' ?

 

[GregLocock]

Ah, I think the simple explanation is that in practice the bottom one is more commonly experienced - replace the pin joints by welded joints (or pins with friction) and lo and behold, the stiff member takes more load.



Cheers

Greg Locock


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[zekeman]

"New PostGregLocock (Automotive)
8 Jan 13 0:56
Ah, I think the simple explanation is that in practice the bottom one is more commonly experienced - replace the pin joints by welded joints (or pins with friction) and lo and behold, the stiff member takes more load."

Yes, you have made a determinate problem indeterminate and now elasticity has a fundamental role.

But the OP is talking pin joints and as everybody in the universe knows a single element pin connected on both ends has only 2 forces, equal colinear and opposite( and at the pin joints including partial buckling of the element ), period.Elasticity can play only the 2nd order role of changing overall geometry, but is as pointed out negligible for the case posed.

So the proper way to analyze this problem for low elasticty materials is first do a static analysis, find the deflections including some buckling,etc,and after the dust settles,find where the top pin has settled, and then from the new geometry do the successive iterations of static equilibrium to get a result.

 

[Cockroach]

This pretty much summarizes the points raised in this discussion. As per attachment.

Regards,
Cockroach