A tricky question
A tricky question
(OP)
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
Any thoughts on this






RE: A tricky question
RE: A tricky question
RE: A tricky question
reaction A: X=11.25 kips, Y=1.125 kips
reaction B: X=-11.25 kips, Y=1.125 kips
load in strut = sqrt(11.25^2+1.125^2)=11.31 kips
delta strut = 16.48'*11.31/E/A
from delta strut, the deflection of hinge C can easily be derived.
Am I take it too simple?
RE: A tricky question
Yes, you COULD use SAP or STAAD or any FEA program of your choice, but a pencil and paper should be sufficient. A calculator will be handy, but in MY day, all we were allowed was a slide rule! (End "nostalgic old fogey" mode.)
RE: A tricky question
RE: A tricky question
Actually this is the standard designation for a hinged support. The only unusual part is to state 'hinged' and 'fixed' on the drawing. The triangle represents a rigid body whose base is fixed and top is hinged.
The rest is pretty straightforward. Did you get the job?
BA
RE: A tricky question
First, there's the buckling issue - I'd bet it would be hard to find a material with that E and a 0.155 sqin cross section that wouldn't buckle globally (if a rod) or locally (if some sort of tube).
The big kicker is that the axial deflection under that load is a smidgeon under an inch. Guess what happens when you subtract an inch from 16.48'? You get just less than 16.4'. The half span is 16.4', which means this thing flips into tension action pretty quickly.
Now, if they haven't buckled (or even if they have, as long as they didn't break), you can pretty quickly come up with where it ends up below the pins as a hangar with 2 or 3 iterations - I'll leave that one up to you to do...
If you "heard" it on the internet, it's guilty until proven innocent. - DCS
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RE: A tricky question
you got it. Rest obviously is very simple.
RE: A tricky question
BA
RE: A tricky question
RE: A tricky question
RE: A tricky question
with initial geometry, the load in the struts is 11306 lbs, which'll cause a 0.084ft change in length ... 16.48-0.084 < 16.4.
so now it'll deflect downwards more than 1.64' ('cause this'll reestablish the original length of the struts after the snap-thru). i figure (slow day) a downward displacement of 2.181ft (total displacement = 3.821ft), load in the struts = 8534 lbs.
RE: A tricky question
That would give it a stress of 8534/0.155 = 55,000 psi. Pretty hefty stress for a material with E = 14,503,000 psi. And what about the impact after it snaps through. Maybe the load ends up on the ground.
BA
RE: A tricky question
you wrote 16.48-0.084 < 16.4.
How it can be less than 16.4? less than minimum distance between two supports.
RE: A tricky question
RE: A tricky question
you are really blowing my mind off. With given data, howyou come up with fty = 72 ksi. Thanks.
RE: A tricky question
whilst i'm at it ... your earlier post 16.48-.084 is less than the length of the supports, therefore it can't happen ... the maximum load this structure will support before it snaps thru is about 1690 lbs.
RE: A tricky question
RE: A tricky question
1. Since members are stated to be struts we assume top center joint to be pinned.....also joints at base are stated to be pinned.....and length of each strut = 16.4818 ft. then
2. Vertical force component (Fy) at each support is 1.125 kips..then
Fx=1.125*16.4/1.64 = 11.25 kips and total force in each strut is F=11.306 kips
3. axial deflection = Fl/AE = .9947 inches...and vertical deflection is .9947*1.64/16.4818 = .0989 inches
4. deflection is symmetric therefore final position of center joint is at x=16.4 ft and y=(19.68-.0989)=19.5811 inches = 1.631758 ft.
5. final length is root(1.631758^2+16.4^2)=16.480978 ft. therefore no snap thru occurs...final stress is 11.3061/.155=72943 psi
No trick question and nothing exceptional.....If my undergrad Mechanics of Materials students could not provide a correct answer to this question they would receive and "F" for the course......
Ed.R.
RE: A tricky question
I've been very interested in snap-through buckling since reading Galambos' "Structural Stability of Steel". Do you have any good literature on snap-through buckling? Where did you come up with the 1690#?
Does anyone else have some good literature on snap-through buckling?
RE: A tricky question
I'm failing to see how this doesn't go into tension. You show the axial shortening correct, but the new axial length (after shortening) is less than the horizontal length from support to center, therefore it MUST snap through and go into tension.
Please explain how you see it differently.
Your step 3 assumes similar triangles such that the pre-deflected shape is the same as the post-deflected shape, which isn't true. If you take the new axial length of the strut as 16.4818'-0.0825' = 16.399', you can see how this is less than 16.4'.
Does this mean you give yourself an "F"?
RE: A tricky question
I don't think I said it is too high. I said it was "pretty hefty", particularly when you consider dynamic loading. If the yield strength is exceeded, inelastic conditions will apply to your deflection calculation.
BA
RE: A tricky question
StructuralEIT ... assume a dispalcement of the apex (.1ft), calculate the load created in the struts, and so the load applied to the structure ... as i said before, "slow day".
RE: A tricky question
the strut would not shorten because both ends are restrained against translation (shortening) - draw a circle with radius=strut length about A, do the same for "B", there is only a single contact point between the two circles. If the strut has shortened, again draw the circles, which would not be in contact along the path of rotation.
RE: A tricky question
I wouldn't be so harsh as to give my students an F for not getting this question - I'd probably have given it as a bonus.
As StructuralEIT said, your problem is with your "vertical" deflection in step 3.
If you "heard" it on the internet, it's guilty until proven innocent. - DCS
http://www.eng-tips.com/supportus.cfm
RE: A tricky question
What? Of course the strut has to shorten. That's basic mechanics of materials. Both ends are restrained against translation, but the center hinge deflects down as the struts shorten.
RE: A tricky question
RE: A tricky question
RE: A tricky question
With the strut in compression, it's going to shorten. Approximately .0829', as has been calculated using statics and mechanics of materials. Does that part make sense?
RE: A tricky question
Use RobertEIT's P(strut) = 11.31 kips
L (Strut) = 197.7816"
E = 14503 ksi
A = 0.155"
S = PL/EA = 0.995"
L' = L-S = 197.7816 - 0.995" = 196.7866"
Half Span Length = 196.8" > L'
The pins are considered non-deformable rigid body, where is the 0.0134" (each side, total gap between strut = 0.0268") went?
RE: A tricky question
RE: A tricky question
RE: A tricky question
RE: A tricky question
RE: A tricky question
RE: A tricky question
I think we agree that the axial shortening of each strut is .9947 inches.....the horizontal and vertical components of such a shortening is then deltahoriz = delta * cos (theta) and deltavert = delta * sin (theta) and since sin(theta)= 1.64/16.4818 (for small displacements) and my step 3. is correct.....alternate using original vectors is .9947 [16.4,1.64,0]/16.4818 for horizontal and vertical components. Again I think I am correct......
Again I don't think this is a large displacement problem nor does snap-thru occur....also the length of the member after shortening is > than the horizontal length (16.4)....Note also that symmetry requires the final position to remain in the center......
Ed.R.
RE: A tricky question
Yep it would not be the first time I have been wrong and had to give myself an "F"
Ed.R.
RE: A tricky question
why do you think this isn't a large displacement problem ? just because the displacement is small compared to the geometry of the problem ? but the impact of this displcement is significant on the structure's ability to react load. Consider the displaced geometry, with the apex moved 1.63ft. What are the strut reactions now ?
RE: A tricky question
In your post you've already said that the axial shortening is .9947" or .0829' hence the struts final length is 16.3971
'. This is shorter than half the span length.
As far as finding the location of the end of one of the struts with the small displacement theory, that doesn't really matter because it's connected to end of the other strut. It seems that you've lost deformation compatibility at some point.
RE: A tricky question
Never, but never question engineer's judgement
RE: A tricky question
Really? How do you figure?
BA
RE: A tricky question
Never, but never question engineer's judgement
RE: A tricky question
BA
RE: A tricky question
RE: A tricky question
It was either a very good question, to ask engineers, or a very bad question, to ask in an exam.
Cheers
Greg Locock
SIG:Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.
RE: A tricky question
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.
RE: A tricky question
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'
RE: A tricky question
Please have a look to my calcs. Can u please explain why it will deflect extra .546 feet?
RE: A tricky question
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?
RE: A tricky question
BA
RE: A tricky question
?? 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.
RE: A tricky question
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.
RE: A tricky question
RE: A tricky question
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')?
RE: A tricky question
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.
RE: A tricky question
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?
RE: A tricky question
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.
RE: A tricky question
RE: A tricky question
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.
RE: A tricky question
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).
RE: A tricky question
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.
RE: A tricky question
Check this out (see attaced sketch).
RE: A tricky question
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:
Pcr=3.08EAno3
where
no=fo/L
fo=initial rise of C (500 mm)
L=distance of supports (10000 mm)
With the other data provided should be
Pcr=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 (
prex
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RE: A tricky question
"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?
RE: A tricky question
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
RE: A tricky question
RE: A tricky question
prex
http://www.xcalcs.com : Online engineering calculations
http://www.megamag.it : Magnetic brakes for fun rides
http://www.levitans.com : Air bearing pads
RE: A tricky question
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
RE: A tricky question
I don't understand your printout. How can the elongation at 45 degrees be 0.1548"? Please explain.
BA
RE: A tricky question
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 ?
RE: A tricky question
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.
RE: A tricky question
No problem.
BA
RE: A tricky question
RE: A tricky question
RE: A tricky question
...
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.
RE: A tricky question
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.
RE: A tricky question
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.
RE: A tricky question
Just curious - what application did you use to generate these nice-looking sketches and cals? Word and AutoCAD?
Regards,
IV
RE: A tricky question
It snaps thru 3.28 feet (1000mm), then deflects 0.54 (165mm) feet further before reaching equilibrium.
RE: A tricky question
Anyway this problem stays meaningless, as, with the data provided, there is no way, as correctly stated by swearingen the first, to prevent the buckling of the struts (even supposing node C to be restrained laterally). And of course, if this was a real world problem, one would start by checkng global buckling as the first thing.
prex
http://www.xcalcs.com : Online engineering calculations
http://www.megamag.it : Magnetic brakes for fun rides
http://www.levitans.com : Air bearing pads
RE: A tricky question
But regardless of that, why do you assume the struts will buckle? Perhaps "I" is very large.
RE: A tricky question
1. What's I?
2. What's the cross-section shape?
3. What's the weight of the material?
Then I realized that #1 and #2 don't matter because if "I" was large enough to keep it from buckling the section would be so thin that it would crush. So the center hinge translates, the struts are now in tension and since the weight of the material wasn't given it is to be ignored.
From there I see four possible conditions:
1. The hinges are at ground level and the struts will be in compression in the x direction but probably bowed;
2. The supports are above ground level the struts will be below the supports and still in compression;
3. The supports are exactly 1.64' plus the y component of the stretched lengths of the struts above the ground so the load is sitting on the ground and there are no reaction forces;
4. The supports are more than 1.64' plus the y component of the stretched lengths of the struts above the ground and the struts are in tension.
Then I would have chosen #3 and turned in my answer.
Mark Schroder, P.E.
Crystal River, Florida
RE: A tricky question
Never, but never question engineer's judgement
RE: A tricky question
When the skyhook is suddenly released, the load falls freely for a distance of one meter whereupon it starts to stretch the cable.
Falling through that height would create an impact force which would likely be sufficient to break the cable or at least stretch it beyond its elastic limit.
So, the answer cannot be determined precisely.
BA
RE: A tricky question
RE: A tricky question
We can write two equations to get rid of one unknown.
Let
Lo = Horizontal span length, 16.4'
e = T*Lo/EA, EA = E*A
L = Lo+e
y = (L^2 - Lo^2)^0.5
y/Lo = (L^2 - Lo^2)^0.5/Lo (1)
Then from force equilibrium, we get
y/yo = V/H (2)
Let (1)=(2), after some operation, we get
[2T/EA + T^2/(EA)^2}^0.5 = V/H
Differentiate the equation above can get the exact solutions (dt/dH), but I am too far from calculus, thus can only get the approximate solutions using spreadsheet:
T = 11.27 kips
H = 11.21 kips
y = 1.644 ft below the ref datum (supports A & B)
It is interesting to see the curves, each representing the equation on either side of the equal sign.
RE: A tricky question
You can in fact work the problem using energy theory and get an answer. However, what immediately struck me is that when it gets in position A, it has to be accelerating downward at that point in time (due to gravity) and consequently can't very well stop at that position. The problem in this case is that you equate the potential energy from two different positions, but it can't get from one to the other without passing through a state of higher energy first. So at the "right" setting, it would flip over partway, and bounce back. At any higher initial velocity, it would keep going past the horizontal.
A related problem that illustrates the issue is if you are asked to roll a ball such that it stops on top of a hill. In theory at least, this can be done. But, if there is a bigger hill between you and the hill in question, then there is no solution, and that is the case with the spring-pendulum shown here.
I was in a class of maybe 20 or 30 people. None of the rest noticed this. The textbook writer didn't. I could not persuade the professor that anything was wrong here, either.
RE: A tricky question
RE: A tricky question
BA
RE: A tricky question
By the way I would have lost the job too as I would have solved it by the earlier posts.
RE: A tricky question
Firstly, I think we can all agree that if this was a "real world" problem, using real-world materials, it would fail in any practical engineering sense. However, treat this as a philosophical exercises - nobody said this was on Earth (so we can ignore gravity), nobody told us what the inertia and lateral effective length is (it could be fully restrained laterally, and could actually be a very thin, very deep section so it won't buckle), and it could be immensely strong (gotta get me some of that Unobtainium!).
Accepting the premise, we can demonstrate that the assembly will unconditionally "snap through" at a load significantly less than 10 kN. We can also demonstrate easily that it is truly a "large displacements" problem.
I believe the correct solution is:
Final deflection = 1164.8 mm = 3.821 ft
Tension in "strut" = 37.94 kN = 8.53 kip
Vertical Reaction at each side = 5.00 kN = 1.125 kip (as expected)
Horizontal reaction at each side = 37.61 kN = 8.45 kip inwards (as "struts" are in tension)
My working is attached.
Thanks and a star to dgkhan for posting the problem, and for all the lively debate this has generated. And its great to know that even we "grey hairs" can still get fooled form time to time!
RE: A tricky question
Quite impressive. Just a little question here.
I believe the final deflection, y = 3.821'-1.64' = 2.181'.
Check force-geometry slope compability:
V = 1.125
H = 8.45
V/H = 0.133 &
y = 2.181
L = 16.4
y/L = 0.133 = V/H, check, so far so good.
Check material-geometry compability:
y = 2.181
L = 16.4
e = (2.181^2+16.4^2)^0.5 - 16.4 = 0.1444
T = 8.53
L = 16.4
E = 14503
A = 0.155
e' = TL/EA = 8.53*16.4/(14503*.155) = 0.0622 < e = 0.144
----> e' needs to equal e so a stable trangle could form, in this case, it wouldn't.
Can you check and explain if I have missed something?
RE: A tricky question
In your equations for e and e', was the original length 16.4' or the diagonal length of 16.48179602'?
BA
RE: A tricky question
I used L = 16.4' as the base.
e = Sqr(y^2 + L^2)- L
e' = TL/EA - L
These two quantities need to be equal to satify both material properties (P,L,E,A) and trig.
Please I have verified the vertical displacement, y=2.181' from datum w/r to supports A & B, in the opening. Thus the struct length is (2.181^2+16.462^2)^0.5 = 16.544'.
RE: A tricky question
The original length of the strut, and after fall through, of the tie, was the slope length of 16.4 horizontal and 1.64 vertical, i.e. 16.482'.
Your equation:
e = (2.181^2+16.4^2)^0.5 - 16.4 = 0.1444
should be:
e = (2.181^2+16.4^2)^0.5 - 16.482 = 0.06259
BA
RE: A tricky question
I failed to see why is that. Will take some time to think about it tomorrow.
RE: A tricky question
V = 1.125
H = 11.21
V/H = 1.125/11.21 = 0.1 &
y = 1.644
L = 16.4
y/L = 1.644/16.4 = 0.1 = V/H, check.
L' = sqr(y62+L^2) = (1.644^2+16.4^2)^0.5 = 16.4822
e = L'-L = 16.4822-16.4 = 0.0822
T = 11.27
L = 16.4
E = 14503
A = 0.155
e' = TL/EA = 11.27*16.4/(14503*0.155) = 0.0822 = e, check
Thus, the system is in complete equilibrium under the loads and deformations indicated.
RE: A tricky question
Please draw the triangle with dutum set at the level of supports A & B, you might figure out what I was thinking, assuming, and mistakes if any.
RE: A tricky question
Remember in the structural mechanics classes when some frames were not structures but mechanisms and we had to pick them out?
RE: A tricky question
RE: A tricky question
Neglecting impact forces, if the 10kN load is placed gently on the structure, the tie moves to an angle y with the horizontal.
Delta(L) = 5000(1/cos(y) - 1/cos(x))
By Hooke's Law, Delta(L) = TL/AE = P/2sin(y) * 5000/cos(x)* 1/AE
Equating these two expressions, sin(y)(cos(x)/cos(y) -1) = P/2AE = 10*1000/(100*10^5) = 0.0005
Angle y is approximately 7.575o.
5000tan(y) = 665mm, 165mm below the unstressed, snap-through position.
BA
RE: A tricky question
The same conclusions arise - the assembly will unconditionally "snap through" and go into tension, and the final displacement is EXACTLY 1:50 of what I calculated for the original problem. Bar forces and reactions are the same as for the previous analysis.
Revised working attached.
RE: A tricky question
RE: A tricky question
Yes, you never know when someone might ask you to design a very flat 3-pin arch / truss!
Cheers!
RE: A tricky question
But by the same reasoning, you can't ignore deflection of the supports unless you check them as well.
RE: A tricky question
RE: A tricky question
I used simple excel. File attached for your review. Though my calcs are wrong.