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

 

[hokie66]

Julian has presented an elegant solution to what was a trick question. Final answer. Lock it in.

 

[BAretired]

After snap-thru, Point C moves down to -500 and the tie makes an angle x with the horizontal. So x = arctan(500/5000) = 5.71^o.

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 - 1/cos(x))

By Hooke's Law, Delta(L) = TL/AE = P/2sin * 5000/cos(x)* 1/AE

Equating these two expressions, sin(cos(x)/cos -1) = P/2AE = 10*1000/(100*10^5) = 0.0005

Angle y is approximately 7.575^o.

5000tan = 665mm, 165mm below the unstressed, snap-through position.





BA

 

[jhardy1]

For all of those who were concerned about the "unreal" aspects of the problem (struts would buckle; assembly can't support its own weight; etc), try solving exactly the same problem, but scale the frame dimensions by a factor of 1:50 (i.e. half-span = 100 mm; rise = 10 mm), but keep the material properties, section properties and loads the same. We can now see that the struts would not buckle (e.g. Euler Load for a 10 mm x 10 mm bar x 100.5 mm long is about 81 kN, which is greater than the axial load of 50.2 kN which is obtained from "small displacements" analysis). Granted, the axial stress of 502 MPa is pretty high, but not outrageous.

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.

 

[civeng80]

Great effort Julian, a very well thought of solution to a fairly vexing problem. I may file this solution for future reference.

 

[jhardy1]

civeng80,

Yes, you never know when someone might ask you to design a very flat 3-pin arch / truss!

Cheers!

 

[JStephen]

You can ignore deflection and calculate the forces involved by statics only. But then you need to calculate deflections to see if this is a reasonable answer. Turns out it isn't.

But by the same reasoning, you can't ignore deflection of the supports unless you check them as well.

 

[petertor]

We investigated a 3D version of this problem with distributed loads instead of one point load, on a wagon wheel roof truss that had sagged down into an inverted shallow cone. After discounting fabrication errors and contractor hole hogging in the heavy timber truss connectors, we realized the structural engineer had unwittingly applied a (+) sign to the snow load instead of a (-) (computer model thought snow load was acting in up +Z direction), and this was in October, with snows due any day. Everyone from peer review to code review had blindly stamped the calcs, and passed them higher. So they shored the entire roof structure while the white hats and lawyers kicked the can down the road for about a year, then everyone agreed to tear it down. "You'll never work in this town again!" %)

 

[dgkhan]

iv63

I used simple excel. File attached for your review. Though my calcs are wrong.