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

 

[rb1957]

EdR,

i thought you calc'd a displacement something like 1" ... 197.78-1 = 196.78" but it's more than that IMHO. i think it is a large displacement problem, that the strut goemetry changes significantly as the struts approach the snap-thru point, you'd agree (i think) that if the struts have an apex 0.164' above their ends that the loading in the struts is much higher (as a ratio of the applied vertical load).

 

[Lion06]

It must snap-through. I don't believe that's even open for debate.


kslee-

The reason it doesn't "keep elongating" is that the first order analysis you do is the maximum it could be. Doing iterations will only make it smaller. The reason is this:
For the initial geometry, you get a tension in the tie and a vertical displacement. The new geometry with the vertical displacement makes the tension in the rod less (likely significantly less on the first iteration), which means there is less axial elongation.

You can try this to prove it to yourself. Put a problem of similar geometry into an analysis program (with the load such that the ties are in tension). Run it with a first order analysis and note the displacement and axial forces. Next run it with a second order analysis and you will notice that the axial forces and the vertical displacement are LESS than the first order analysis.

 

[kslee1000]

EDR:

Check this out (see attaced sketch).

 

[prex]

This kind of problem is treated in books on theory of elasticity, as it represents a type of elastic instability different from the more common column buckling and the like, where a new configuration (lateral deflection of a straight column) becomes possible at a certain load because of energy considerations.
In this case, instead, the instability is because, in some specific configurations, the internal force in the structure during the deformation may rise faster than the external load: the instability occurs when the internal force may increase alone, without a change in the external load.
The critical load is calculated, for the geometry of this problem as:
P_cr=3.08EAn_o³
where
n_o=f_o/L
f_o=initial rise of C (500 mm)
L=distance of supports (10000 mm)
With the other data provided should be
P_cr=3850 N = 0.865 kips
So the applied load is well above the critical load, and those who voted for C going snap though please ask dgkhan for their prizes.
And dgkhan, I'm afraid that your solution is not correct at least because of these two points:
1)the shortening of the struts without consideration of the change in geometry (side note for EdR: this is a problem where the effect of the change in geometry cannot be neglected, so it is in its essence a problem of large displacements) is not a sufficient criterion to determine if there is instability. In fact, if my numbers above are correct, the instability would occur also with half the load, but your reasoning wouldn't catch that (didn't check myself your numbers though)
2)There is an extra displacement of C in the underside position, due to the deformation of the struts now acting as ties: if I'm not in error, this extra displacement would be of the order of 250 mm!
Of course this a badly posed problem, because the properties of the struts (A and E) are unrealistic: the struts are much more like rubber than any suitable structural material. I'm sure (I hope in fact) the problem has not been prepared by an engineer (), but likely by a physicist ().

prex

http://www.xcalcs.com

: Online engineering calculations

http://www.megamag.it

: Magnetic brakes for fun rides

http://www.levitans.com

: Air bearing pads

 

[kslee1000]

prex:

"2)There is an extra displacement of C in the underside position, due to the deformation of the struts now acting as ties: if I'm not in error, this extra displacement would be of the order of 250 mm!"

No, the extra is 300 mm! Why not?

 

[prex]

And of course there is one more problem: the critical buckling load of those struts is ridiculously low.
I think the only sensible answer to the examiner is (perhaps what the examiner expects BTW):
The problem has no answer because the proposed data are unrealistic and such an arrangement cannot exist on earth.

prex

http://www.xcalcs.com

: Online engineering calculations

http://www.megamag.it

: Magnetic brakes for fun rides

http://www.levitans.com

: Air bearing pads

 

[rb1957]

prex ... the material isn't That unreasonable ... it's quite like aluminium ... E = 10E6psi, Fty = 72ksi

 

[prex]

Sorry, I had a problem in reading the units for E, I see now that E=100 GPa = 100000 MPa, so it's better than aluminum. However my numbers were correct and the critical load of those struts is some 2 million (!) times lower that the starting applied load.

prex

http://www.xcalcs.com

: Online engineering calculations

http://www.megamag.it

: Magnetic brakes for fun rides

http://www.levitans.com

: Air bearing pads

 

[kslee1000]

A quick cal based on original length = 16.4' at horizontal position.

Angle Tension Elong V-Defl
(Deg) (Kips) (inch) (inch) (MM)
1 71.6234 6.2703 3.5441 90.0201
2 35.8171 3.1356 6.9777 177.2336
3 23.8842 2.091 10.4092 264.3937
4 17.9195 1.5688 13.8375 351.4725
5 14.3421 1.2556 17.2617 438.4472
10 7.1985 0.6302 34.2834 870.7984
15 4.8296 0.4228 51.045 1296.543
20 3.6548 0.32 67.419 1712.4426
25 2.9578 0.2589 83.2807 2115.3298
30 2.5 0.2189 98.5095 2502.1413
35 2.1793 0.1908 112.989 2869.9282
40 1.9447 0.1703 126.6101 3215.8965
45 1.7678 0.1548 139.2681 3537.4097
50 1.6318 0.1429 150.867 3832.0218
55 1.526 0.1336 161.3186 4097.4924
60 1.4434 0.1264 170.5433 4331.7998
65 1.3792 0.1207 178.470 4533.1583
70 1.3302 0.1165 185.041 4700.0414
75 1.2941 0.1133 190.2036 4831.1714
80 1.2693 0.1111 193.9196 4925.5578
85 1.2548 0.1099 196.1606 4982.4792
86 1.2531 0.1097 196.43 4989.322
87 1.2517 0.1096 196.6397 4994.6484
88 1.2508 0.1095 196.7895 4998.4533
89 1.2502 0.1094 196.8794 5000.7368

 

[BAretired]

kslee,

I don't understand your printout. How can the elongation at 45 degrees be 0.1548"? Please explain.

BA

 

[rb1957]

prex ... the critical load is 2E-6 of the applied load ?

kslee ... don't get your point ??
small error in tension P/2 = 1.125kip (not 1.25kip)
elongation is based on tension,
not sure what MM is ...
but you don't close the loop at what angle is the vertical displacement consistent with the assumed angle ?

 

[kslee1000]

BA:

Thanks for pointing out. I was using simple geometric method to demostrate the elongation in continuation. However, this method obviously is invalid, my apology for posting before checking.

 

[BAretired]

kslee,

No problem.

BA

 

[rb1957]

yeah, it's a little like the hare and the tortise racing one another ... if the hare gives the tortise a head start, a simplistic (and flawed) argrument says that he can never over-take him.

 

[miecz]

I get the same answer as rb1957. The attached sheet shows the algebra, to a point. In the end, I resorted to a MathCAD solve routine.

 

[racookpe1978]

The problem has no answer because the proposed data are unrealistic and such an arrangement cannot exist on earth.

...

Had a problem (er, "solution" or "answer") like that on my nuke PE exam many years ago: Answered the rest of the test, came back to the problem with a few hours left available on the exam and tried a few different other ways.

None worked, none were reasonable, and all required that the "inside" region between two walls be below freezing to transfer the heat load that the problem stated. So I wrote it up that way, showed my work, and stated my conclusion that either the data given in the problem statement was wrong, my assumptions and derived values for the material coefficients were wrong (but showed how I got them from the tables and flows), or that there was an incredible amount of ice inside the concrete.

Was I right? Don't know - but I passed and got the license.

 

[RockEngineer]

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')?

kslee1000
You have to look at your possible bounds. The highest tension in your tie (when below) would be when you ignore any stretch and your geometry is the same as originally given. (shallowest angle gives highest horizontal component) This is also the shortest the tie can be on the tension side.
When you apply the highest tension to PL/AE the tie will also be the longest it could stretch but at the longest possible stretch you have a steeper angle which lowers the load. Longest possible length gives lowest boundary for horizonal component and thus lowest tension.

Pmax=((1.125K)^2+(12.25K)^2)^1/2 = 12.30 K upper bound
Lmin=((16.4')^2+(1.64')^2)^1/2= 16.48' lower bound length

Lmax= Lmin+Pmax/AE = 16.48'+.083'=16.563' Upper bound length
Lower bound of tension is at upper bound of length
Hmax=((16.563')^2-(16.4')^2)^1/2 = 2.318' Upper bound of deflection below horizontal

So your overall deflection down will be between 1.64' and 2.318'. A few interations and your problem comes to equalibrium.

There are real world problems like this but usually more complications. I just worked on one which required some of the same thinking. It was a horizontal life line. The more sag to begin with the more deflection when loaded and bigger angle when holding a fallen person and the lower the loads on the structure. The less sag the smaller the angle and if tight enough you can fail your building members or line when the person falls on it. This installed line is now interfering with a crane. How much can you tighten up the sag without causing problems with your building members and attachments. When loaded by a man hanging at the center it is our tie problem but calculating the initial length is harder due to the cantanary.

Isn't engineering fun.

 

[EdR]

OK....I finally do get it....I did not recognise just how small a load it takes before the geometry becomes important..

The attached file shows the deformed shape, using both bar and beam elements and small displacement assumptions, and it is clear that at some load < 1 kip (< 1/2 the stated load) the geometry becomes important...Also, even though the displacement never reaches the 19.68 inch height I have little doubt that the stiffness decreases so much that snap thru probably does occur......

Thats what you get when you make assumptions.......

Ed.R.

 

[iv63]

dgkhan:
Just curious - what application did you use to generate these nice-looking sketches and cals? Word and AutoCAD?
Regards,
IV

 

[Tomfh]

I agree with 3.82 feet (1165mm).

It snaps thru 3.28 feet (1000mm), then deflects 0.54 (165mm) feet further before reaching equilibrium.