A tricky question 10

[dgkhan]

This is not an assignment. I went for job interview and I am given this attached problem to solve it. I can simply model it in SAP or STAAD but how to model for the end supports. Never saw anything like this. Same support is pinned and than fixed.
Any thoughts on this

 

[COEngineeer]

It will move 3.28 to the bottom, and then it is stable and you will have axial elongation! Final answer!

Never, but never question engineer's judgement

 

[BAretired]

COEngineer,

Really? How do you figure?

BA

 

[COEngineeer]

it is unstable so it will rotate then the strut become in tension

Never, but never question engineer's judgement

 

[BAretired]

I know that, but how do you come up with 3.28? Please provide us with your substantiating calculations.

BA

 

[Tomfh]

1.64 + 1.64 = 3.28

 

[GregLocock]

I thought that 3.28 +a bit was one of the better answers, although I liked the buckling one even though it made assumptions about the section shape.

It was either a very good question, to ask engineers, or a very bad question, to ask in an exam.






Cheers

Greg Locock

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

 

[kslee1000]

P = 11.31 kips
A = 0.155 in2
P/A = 73 ksi

I think it would have stretched beyond yield, the final position of the load is on the ground.

 

[rb1957]

actually it's got to be more than 3.28 (= 2*1.64) as this "just" restores the struts to their undeformed lengths (the struts lengths are the same if the apex is 1.64' above the ends or if it 1.64' below).

i figure it's final position is .546' below the flipped original geometry ... displacement = 2*1.64+.546 = 3.826' and in this geometry the strut load is reduced to 8514lbs, the strain in the struts is 0.0038, the extension is 0.063'

 

[dgkhan]

rb1957
Please have a look to my calcs. Can u please explain why it will deflect extra .546 feet?

 

[kslee1000]

rb:

Just to discuss.
I guess you reached the stated position through several iterations. However, unless the axial force in the struct approaces zero, or it is negaligible small (need criteria), won't the struct continue to lengthen?

 

[BAretired]

I agree with rb1957 that it has to be more than 2*1.64' (1 m) because, once it has snapped through, the "struts" are going to elongate which will increase the deflection of point C. How much is a moot point because of the dynamic force involved in falling through.

BA

 

[rb1957]

"unless the axial force in the struct approaces zero, or it is negaligible small (need criteria), won't the struct continue to lengthen?"

?? the strut stretches under the stress of the load. in more typical structures, we can get away with saying the impact of displacements on the loadpaths in the structure is small ("samll displacement theory"). this structure isn't, as the effect of displacements, even after the snap-thru is quite significant. the original geometry snapped-thru (apex 1.64' below ends) gives the loads you've calc'd. but these loads cause the legs to strain, extend which increases the vertical companent of the strut load calc'd ... therefore force balance requires that the strut load falls. you're right, i did it by iteration, if i was smarter, i'd be able to do it with algerbra.

 

[kslee1000]

rb:

I was not argue the notion "large vs small displacement", which I think the former is the proper description for this case.

My question was, at the position you stated, there is still a tension of 8.4474 kipn pulling the struct, then wouldn't be the struct continue to lengthen until 1) beyond ultimate breaking strength, 2) T = 0?

I might have missed something.

 

[RockEngineer]

This has been a fun little exercise with lots of lively discussion. I come only did 3 iterations and come up with a vertical deflection of about 3.82 ft and horizontal reactions of 8.46K toward the center and vertical reactions of 1.25K up. I've given it to about a dozen engineers here at work and other places and it is surprising who gets it and who doesn't.

 

[kslee1000]

Rock:

As long as horizontal reaction exists, with vertical reaction being constant, there would be force (tension) in the struct. What was the reason it would stop to yield (elongate) when reaches the deflection you have stated (3.82')?

 

[Lion06]

kslee-

For the tension side, you actually work backwards. You have an initial elongation, and tie force (It's a tie now, not a strut), based on the original geometry. When it deflects further down, however, the horizontal reaction is LESS than you assumed in your first run, so your tension force and elongation get less until you converge on an answer, not more.

 

[kslee1000]

SEIT:

That's the question, when is convergence to occur in this case? Mathematically it (elongation = TL/EA) can get to infinitive, because, ignore selfweight, for struct with T (struct) = 0, it has to be in vertical position, which is impossible without setting criteria for breaking strength of the strut, otherwise, it would keep on going, wouldn't it be?

 

[dgkhan]

Kslee1000,
I too want to know this. When the apex has fallen 1.64' down. why it stops after further falling .58" whcih some of us are assuming.

 

[kslee1000]

dg: similarly simple mind, welcome on board

 

[EdR]

rb1957:

I think it is not a large displacement problem because of work/energy arguments.....something about the displacements that satisfy equilibrium make the energy a minimum (also known as the Theorem of Minimum Potential Energy). Assuming these laws have not been repealed (which may happen any day in the US) then the structure must first pass through my solution before going on to any large displacement solution....Since my solution is the first equilibrium position encountered it must be the correct solution based on minimum work/energy requirements....to get a large displacement (or snap thru) solution the applied load must be increased to cause added vertical (and axial) deformation.....

For those making arguments about the final and initial lengths of the members I would point out that:

Initial member length (inches) = 197.7816
final member length (my solution) = 197.7717
position of center = 196.8

therefore the member is still longer than the distance to the center.......

Ed.R.