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

 

[JAE]

Just use a simple free-body diagram?

 

[JStephen]

Calculate force in the struts using statics, calculate deflected length of struts, re-calculate the angles using that deflected length, and repeat if necessary.

 

[RobertEIT]

This isn't a tricky question, it's very simple.

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?

 

[jhardy1]

This is fundamental "Statics 101". It is a bit of a worry if engineering graduates are not being taught the basics any more!

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.)

 

[jhardy1]

Actually, what I found particularly interesting is that the problem was obviously originally presented in a metric source reference (E = 10^5 MPa; Span = 10 m; Rise = 500 mm; A = 100 mm^2; F = 10 kN; etc), but has been translated into (rounded) American conventional units for American candidates. No problem in that, and it might be argued that it adds the only "trickiness" to the problem, because the numbers don't divide out quite so nicely in American units, but it did leave me wondering whether the American tester lacked the confidence to re-present the question using nice rounded American units (e.g. E = 10^5 ksi; A = 0.1 in^2; Span = 20 feet; Rise = 1 foot; F = 1 kip; etc), and was unsure that they could re-calculate the correct answer themselves?

 

[BAretired]

Never saw anything like this. Same support is pinned and than fixed.

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

 

[swearingen]

It is a trick question - they just want to know how fast you realize (or if you realize) that it's not stable.

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

http://www.eng-tips.com/supportus.cfm

 

[dgkhan]

swearingen
you got it. Rest obviously is very simple.

 

[BAretired]

Should have checked that. Star for swearingen. Guess I wouldn't have got the job.

BA

 

[w12x26]

The system from a structural analysis point of view is stable but from a design view it is unstable because of the material and section properties. I would have gone through the determination of the reactions and loads in the strut and then the deflection before realizing it is unstable. It is indeed a tricky question. This is by the way how you design cathedral ceilings with no ridge beam. I guess I would not have gotten the job either.

 

[jhardy1]

Yep, I guess I would have missed the job too!

 

[rb1957]

running some numbers i think it'll snap thru ...
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.

 

[BAretired]

rb1957,

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

 

[dgkhan]

rb1957
you wrote 16.48-0.084 < 16.4.
How it can be less than 16.4? less than minimum distance between two supports.

 

[rb1957]

not really ... Al E = 10E6psi, fty = 72ksi ... all bets are off regarding dynamic loading on the snap-thru !

 

[dgkhan]

rb1957
you are really blowing my mind off. With given data, howyou come up with fty = 72 ksi. Thanks.

 

[rb1957]

7075T6 ... it doesn't relate to your problem, it was a reply to BAretired (who thought 55ksi was too high a stress for this material).

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.

 

[RobertEIT]

This kind of trick is more like a kid's game. Don't see much practical meaning. Hope NCEES will not bring such kind of trick into their exam.

 

[EdR]

Just to clarify some of the previous incorrect statements...

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.