Analysis of a hyperstatic system by hand calculation 3

[DerGeraet]

Hi,

I'm currently trying to solve the problem, attached to the pdf-file. It's a hyperstatic system due to the fact that it's connected by two solid bearings (2 DOF*2). The forces in the horizontal direction, H1 and H7, are calculated but the vertical forces are kind of replaced? by K. Force F2 is in the middle of e1.A. I have really no idea, how to get the same results as listed on the file. I would be very thankful for every tip.

regards

 

[DerGeraet]

I hope my question and the problem is clear in the new attachement.

I found this file for the displacement method

http://www.sut.ac.th/engineering/Civil/CourseOnline/430332/pdf/03_SlopeDeflection.pdf

The example is almost the same.

But what I don't understand is the way of analyzing this part like the guy did with K=e3/e1. He never uses any vertical reaction forces at position 1 or position 7. Instead of, the momentums in the structure are calculated only by the horizontal force (H1 or H7), K and F.

Thank you all anyway for your support and sorry for my bad expression regarding bearing etc.

 

[DerGeraet]

If the supports at 1 and 7 are hinges, I misinterpreted the situation before. I thought the supports prevented vertical movement and rotation but permitted horizontal.

The supports at 1 and 7 prevent translation in horizontal and vertical direction. Normaly, the momentum at these supports has to be 0 but as you can see in the "torque path", it isn't. Also in the FEM-analysis the momentums were not 0 at the supports. The disabled DOF's in the FEM where translational movements in horizontal and vertical directions.

 

[rb1957]

well that's quite a bit different ...

don't use terms like "bearing" ... i immediately pictured a ball bearing of some type (was wondering what stops this frame from tipping over).

i'd suggest solving it from scratch as a redundant (or hyperstatic) problem. picking up someone else's solution and trying to figure out why they did it that way is going to be much easier if you work through the solution. there may be a perfectly obvious reason why he uses the length ratios in the solution, maybe it's just a short form to simplify the results.

the difficulty of studying on your own is that it is hard to figure out the assumptions he's making. M1 and M7 look like "prying moments". and if this is a four fastener brkt, then there'd be a 2nd horizontal force (out-of-plane).

i believe (still) that the problem is doubly redundant, more clearly seen if you look at the 1/2 frame (cut on the axis of symmetry).

 

[DerGeraet]

my bad, sorry .. i will call it "fixing points" in the future

i believe (still) that the problem is doubly redundant, more clearly seen if you look at the 1/2 frame (cut on the axis of symmetry).

i can't figure out what you mean with "doubly reduntant"...

 

[rb1957]

maybe one way to look at the problem is to assume the legs (e3) are rigid; then the reactions at 1 and 7 are as though the frame is a beam from 1 to 7; the moment in the legs is constant. if the legs aren't rigid, then the moment varies along the length, then M2 < M3 and the difference is reacted as a couple H*e3 and M1 and M7 are similarly reduced.

 

[paddingtongreen]

Since the model is symmetrical, tackle only half of it. The center of the beam remains horizontal but moves vertically. Apply an arbitrary moment (ignoring the load) at the center point and the same direction and magnitude at the end of the beam. Use moment distribution to run the moments through the half structure. From the moments calculate the resulting force at the load point and multiply the moments and reactions by the load divided by the resulting force.

Michael.
Timing has a lot to do with the outcome of a rain dance.

 

[BAretired]

Looking at the sketch of the bracket and assuming that the lower tabs of the bracket bear on a rigid body, the supports at 1 and 7 are closer to fixed than hinged, i.e. rotation and translation are both prevented.

If the direction of force were to be reversed, there would be an additional vertical support left of point 2 and another right of point 6. They are not at points 2 and 6 because of the rounded corners. Thus the bracket behaves differently according to the direction of the force F.

For the direction of F shown with fixity at 1 and 7, the structure is indeterminate to the third degree. For F reversed, it is indeterminate to the fifth degree.

The diagram to the right of the bracket shows a dimension of e[sub]2A[/sub]/2 between points 3 and 4. That should be e[sub]1A[/sub]/2.



BA

 

[rb1957]

"i can't figure out what you mean with "doubly reduntant"..." ... a statically indeterminate (= hyperstatic ?) structure can't be solved by equations of equilibrium 'cause there are too many reactions. statically indeterminate structures are also called redundant structures 'cause thay have redundant reactions (reactions that can be removed and you still have a structure and not a pile of rubble). the advantage of using this term, redundant, is that you can describe how redundant the strcuture is.

for example, a cantilever is statically determinate. a propped cantilever (with a pinned support at the other end) is singly redundant ... you can remove one the reactions and you'll still have a structure. a double cantilever (fixed at both ends) is doubly redundant ... you have to remove two reactions to get a determinate structure.

in your case if points 1 and 7 are fixed then the problem is doubly redundant (like a double cantilever).

 

[BAretired]

But what I don't understand is the way of analyzing this part like the guy did with K=e3/e1. He never uses any vertical reaction forces at position 1 or position 7. Instead of, the momentums in the structure are calculated only by the horizontal force (H1 or H7), K and F.

Let's revise the symbols for simplicity. I prefer L, a, h corresponding to e1, e2, e3.

Make the structure determinate by making joints 1 and 7 horizontal rollers. Place a vertical roller at 4. Calculate θ1 and δ1 for F, H and M. Assume relative EI = 1 for all members.

(1) Force F acting upward at Joint 4
Simple span moment = FL/4.
Area under M/EI diagram is (FL/4)L/2 = FL[sup]2[/sup]/8.
θ1 = θ2 = θ3 = FL[sup]2[/sup]/16
δ1 = δ2 = θ3*h = hFL[sup]2[/sup]/16

(2) Force H acting outward at Joints 1 and 7
M1 = M2 = 0
M3 = M4 = H*h
θ1 = Hh(L+h)/2
δ1 = Hh[sup]2[/sup](L/2 + h/3)

(3) Moment M at Joint 1 and -M at Joint 7
θ1 = M(L/2 + h + a)
δ1 = Mh(L+h)/2

The sum of θ1 due to F, H and M = 0
The sum of δ1 due to F, H and M = 0

Solve for H and M (2 equations, 2 unknowns)
V1 and V7 do not enter into the calculation.

BA

 

[rb1957]

maybe i'm not summing properly but i get ...
for theta1 ... FL^2/16 + Hh(L+h)/2 + M(L/2+h+a) = 0 ....... (1)
and delta1 ... hFL^2/16 + Hh^2(L/2+h/3) + Mh(L+h)/2 = 0 ... (2)
divide (2) by h ... FL^2/16 + Hh(L/2+h/3) + M(L+h)/2 = 0 .. (3)

but (1) and (3) look like they dissolve into one equation ...

Hh(L+h)/2 + M(L/2+h+a) = -FL^2/16 = Hh(L/2+h/3) + M(L+h)/2
Hh(h/6) = -M(h/2+a)

?

 

[DerGeraet]

hi guys,

sorry for my late reply, just came home...

I could talk to the guy who has calculatet this part with the strange results et voila, he used a table for that... unfortunately I've left the table at the company but I promise that I will post it tomorrow.

and thank you for the answers, I will go trough them this evening. great forum with very helpful people!

 

[rb1957]

a table should have a derivation to go with it ... i'd look for that too. hopefully it's still around and didn't leave when "johnny" left, or didn't get tossed in a clean-up, or left behind in a move ...

otherwise it might just be a bunch of numbers ?

 

[DerGeraet]

unfortunately not just the results of the equations.. I'm trying to get the book and go trough the derivation of such examples, so I'm goint to order it soon. it's a german book FYI...

 

[BAretired]

rb1957,

If we use the sign convention that clockwise rotations are positive and deflections from left to right are positive then:

(1) Force F acting upward at Joint 4
Simple span moment = FL/4.
Area under M/EI diagram is (FL/4)L/2 = FL2/8.
θ1 = θ2 = θ3 = -FL[sup]2[/sup]/16 (counterclockwise)
δ1 = δ2 = θ3*-h = hFL[sup]2[/sup]/16 (left to right)

(2) Force H acting outward at Joints 1 and 7
M1 = M2 = 0
M3 = M4 = Hh
θ1 = +Hh(L+h)/2 (clockwise)
δ1 = -Hh[sup]2[/sup](L/2 + h/3) (right to left)

(3) Moment M at Joint 1 and -M at Joint 7
θ1 = +M(L/2 + h + a) (clockwise)
δ1 = -Mh(L+h)/2 (right to left)

The sum of θ1 due to F, H and M = 0
The sum of δ1 due to F, H and M = 0

This should result in the correct solution.

BA

 

[rb1957]

hi BA,
sorry not seeing it ...
now it is ...
-FL[sup]2[/sup]/16 + Hh(L+h)/2 + M(L/2+h+a) = 0
hFL[sup]2[/sup]/16 + Hh[sup]2[/sup](L/2+h/3) + Mh(L+h)/2 = 0 ... divide by -h ...
-FL[sup]2[/sup]/16 - Hh(L/2+h/3) - M(L+h)/2 = 0 ... so that ...
Hh(L+h)/2 + M(L/2+h+a) = FL[sup]2[/sup]/16 = -Hh(L/2+h/3) - M(L+h)/2 ... simplify ...
Hh(L+5h/6) = -M(L+3h/2+a) ... ie your two equations appear to be related.

i think you need to consider the boundary conditions at pt4.

 

[BAretired]

rb1957,
I did consider the boundary conditions at Point 4 when I set up the deflections and rotations.

I don't see what your problem is. Solve for M in terms of H from your last equation, then substitute that in the first equation and solve for H in terms of F. You will then know H and M in terms of F which is what we want.


BA

 

[BAretired]

hi rb,
Some signs were wrong in your equations:

1) -FL2/16 + Hh(L+h)/2 + M(L/2+h+a) = 0
2) hFL2/16 - Hh2(L/2+h/3) - Mh(L+h)/2 = 0 ... divide by h ...
3) FL2/16 - Hh(L/2+h/3) - M(L+h)/2 = 0 ... so that ...
4) Hh(h/6) + M(h/2+a) = 0

M = Hh[sup]2[/sup]/(3h+6a)

Let's try values of 5, 20, 10 for a, L, h

M = -H*100/30+30) = -1.6667H

From 1) -250F +H 150-41.6666) = 0
so H = 2.308F
and M = -3.846F

check 1)...-250F + 2.308F(10)(15) - 3.846F(25) = 0
check 3)...-250F - 307.7 + 57.6 = 0



BA

 

[BAretired]

Something wrong there. My solution does not pass the sanity check.

BA

 

[rb1957]

yeah, i didn't notice the change in theta1 directions ...

i thought i'd posted a reply "oops, i was looking at F as though it was a variable".

 

[BAretired]

Well, there is still something wrong. I would have expected M1 to be clockwise and would not have expected H to be as high as 2.3F. The results do not make sense, so cannot be accepted. The equations need to be reviewed.

BA