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

 

[prex]

I'll have to insist, especially with swearingen, rb1957 and miecz: it is not sufficient, for determining whether a snap through will occur, to calculate the shortening of the struts using the initial geometry. To be convinced of that, try with half the imposed load: I expect your conclusion would be that it doesn't, but as the load would still be over the critical load, the snap through is assured. However you all are of course correct in stating that a snap through will occur with the load specified.
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

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: Air bearing pads

 

[Tomfh]

Obviously we are supposed to ignore strut buckling. That's why they didn't include "I" in the section properties.

But regardless of that, why do you assume the struts will buckle? Perhaps "I" is very large.

 

[marks625]

My first three questions were:

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

 

[COEngineeer]

If want to be accurate, you have to put delta theta in there and do some integration lol.

Never, but never question engineer's judgement

 

[BAretired]

If the strut is a 10M bar, it has an area of 100mm^2. It could not hold itself up let alone a 10kN load at point C. So it might as well be a cable and point C is being held up by a skyhook.

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

 

[JStephen]

One thing to note in connection with the snap-through buckling is that the stresses are not proportional to the load. So when you calculate a critical load for snap-through, a safety factor should be applied to the load, rather than to the stresses.

 

[kslee1000]

I think the trick lies in the position change in struct needs to match both the geometry & mechanic restrains. There are 3 inter-related unknowns - y (vertical deflection), H (horizontal force) & T (force in strut).
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.

 

[JStephen]

Here's a problem that seems related to me. This was from a dynamics textbook from long ago. The problem is flawed, fatally flawed. Anyway, this was in a section of the book dealing with energy theory, and presumably that is how it was to be worked. The question was: Given the mass, spring constant, and dimensions as shown, it is desired to swing the mass/pendulum down at the right velocity so it stops at the position marked A. So what initial velocity is necessary to do this?

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.

 

[Tomfh]

Yes, position A is on the downhill. It can't stop there.

 

[BAretired]

It can't stop there because the only forces acting on the mass are (a) its own weight and (b) a horizontal force from the spring which is transferred directly to the hinge. The mass would continue to a position where the vertical component of the spring force balances the gravitational force of the mass.

BA

 

[civeng80]

Yes definitely a trick question here, but unrealistic in an engineering design office in my opinion. Just wondering what sort of job dgkhan was applying for.

By the way I would have lost the job too as I would have solved it by the earlier posts.

 

[jhardy1]

Having realised that I made a fundamental blunder in my earlier posts (in believing this was a "simple" statics problem), I decided to check for myself.

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!

 

[kslee1000]

julian:

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?

 

[BAretired]

kslee,

In your equations for e and e', was the original length 16.4' or the diagonal length of 16.48179602'?



BA

 

[kslee1000]

BA:

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'.

 

[BAretired]

kslee,

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

 

[kslee1000]

BA:

I failed to see why is that. Will take some time to think about it tomorrow.

 

[kslee1000]

Ok, let me try to verify my solution posted on May 23, 17:16.

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.

 

[kslee1000]

BA:

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.

 

[civeng80]

Why not state after doing a few statics calculations that the structure under the applied load is unstable ? That is it is a mechanism and does not act as a structure at least for the initial external form.

Remember in the structural mechanics classes when some frames were not structures but mechanisms and we had to pick them out?