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

 

[Lion06]

rb-

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?

 

[Lion06]

Edr-

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"?

 

[BAretired]

rb1957,

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

 

[rb1957]

EdR ... i think your missing that if the struts apex drops 1.63ft then the geometry of the problem has changed significantly and the struts are much less effective in reacting the load, and so the axial load in the struts increases.

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".

 

[kslee1000]

I think RobertEIT got the job with minor comment:
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.

 

[swearingen]

EdR

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

 

[Lion06]

kslee-

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.

 

[kslee1000]

I think you can make a quick draw use autocad. Then shorten the two struts, see if you can make thir tips in touch, and at what position.

 

[Lion06]

Do you really need autocad for this? Really?

 

[PUEngineer]

kslee,

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?

 

[kslee1000]

Numerical Cal:

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?

 

[Lion06]

The struts are shortening. They just don't shorten below 16.4'. That is where they transition from compression to tension.

 

[PUEngineer]

YOU GOT IT! That's the reason for the discussion about snap through and the reason that the struts ultimately end up as tension memebers, with the middle joint ending up more than 1.64' below the pinned ends.

 

[kslee1000]

I concure.

 

[PUEngineer]

Thank you doctor...ha-ha. I the word concur and I think of that movie Catch Me If You Can.

 

[kslee1000]

How about "I agree", both you "PU" & "SEIT" were correct. Looks like I lost the job offering .

 

[EdR]

swearingen & structuralEIT:

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.

 

[EdR]

structuralEIT:

Yep it would not be the first time I have been wrong and had to give myself an "F"

Ed.R.

 

[rb1957]

EdR,
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 ?

 

[PUEngineer]

EdR,
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.