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Analysis of a hyperstatic system by hand calculation
3

Analysis of a hyperstatic system by hand calculation

Analysis of a hyperstatic system by hand calculation

(OP)
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

RE: Analysis of a hyperstatic system by hand calculation

"hyperstatic" = "statically indeterminate" ? ... haven't heard it expressed that way before.

surprised you don't show V reactions at 1 and 7

your F load is applied mid-span which helps you solve the horizontal portion of you frame.

however, the vertical legs are also redundant, in that they can react the fixed end moment two ways ... as a moment and/or as a couple in H. no?

RE: Analysis of a hyperstatic system by hand calculation

Yea. There have to be downward shear forces at 1 and 7 (V/2 here) to counter V.

Mike McCann
MMC Engineering
http://mmcengineering.tripod.com

RE: Analysis of a hyperstatic system by hand calculation

The location of F2 has not been dimensioned. With four Degrees of Freedom, the problem is indeterminate to the first degree. You need an equation representing strain compatibility.

If F2 is at the midpoint of the beam and the structure is symmetrical about a vertical axis through F2, the problem can be simplified by considering only half the structure. By symmetry, the midpoint of the beam does not move horizontally and does not rotate.

If F2 is not at midspan or the structure is unsymmetrical, the problem becomes a little more difficult. If the stiffness of the columns are not equal, the structure is unsymmetrical. If the stiffness or length of the lower beams are unequal, the structure is unsymmetrical.

BA

RE: Analysis of a hyperstatic system by hand calculation

Okay, you told us that F2 is in the middle of the span. So if the structure is symmetrical, you can consider only half of it with a rotation and horizontal translation of zero at F2. The rotation at Point 1 is also zero.

As stated by rb1957 and m2, V1 and V7 are required for equilibrium.

BA

RE: Analysis of a hyperstatic system by hand calculation

does "e3.A" mean length e3 times Area ?

is this a student post ?

i think the problem is doubly redundant. i think trying to solve using a plane of symmetry will be easiest. you'll have three equations of equilibrium; vertical forces are easy, horizontal force is unknown (statically) and the moment at 1 also (once you know both of these you can determine the moment at 4, on the plane of symmetry). equally if you can find themoments at 1 and 4, you can determine the horizontal force ... which two reactions you choose to label as redundant is up to you (and doesn't change the solution, if it's done properly).

you'll have to read up on methods to solve redundant, or hyperstatic, problems ... unit force method is one.

RE: Analysis of a hyperstatic system by hand calculation

(OP)
thank you all for your quick replies.

@rb1957 as BAretired mentioned, it is called "indeterminate to the first degree". I didn't know the exact term of it, sorry.

@BAretired F2 is acting directly in the middle of the system, the symmetrical axis.

of course i do know that the vertical forces at 1 and 7 are required but check out the results from the attachment i've posted:

1) I don't really get, what K is good for
2) H1/H7 are calculated with K and also the term e.3A is in the formula but i don't get which force is creating this momentum with the lever arm e.3A

I think the vertical forces at 1 and 7 are "replaced" kind of by the term K. Otherwise, what is this term good for?

RE: Analysis of a hyperstatic system by hand calculation

(OP)
@rb1957

i think my answer was to quick :)


Quote (rb1957)

does "e3.A" mean length e3 times Area ?

Quote (rb1957)

no, it is called e3.A but it's just a length, consider it as e3 if you want
No it isn't. I have calculated this part in an FEM program and wanted to check the results by calculating it analytically. This is what I've found at the company for this kind of systems. It provides the same analysis as I need but I don't understand he steps of the calculation.

Quote (rb1957)


i think the problem is doubly redundant. i think trying to solve using a plane of symmetry will be easiest. you'll have three equations of equilibrium; vertical forces are easy, horizontal force is unknown (statically) and the moment at 1 also (once you know both of these you can determine the moment at 4, on the plane of symmetry). equally if you can find themoments at 1 and 4, you can determine the horizontal force ... which two reactions you choose to label as redundant is up to you (and doesn't change the solution, if it's done properly).

you'll have to read up on methods to solve redundant, or hyperstatic, problems ... unit force method is one.

I will try to calculate it with the symm. axis and will compare the results with the results of the attachment. Thank you.

RE: Analysis of a hyperstatic system by hand calculation

(OP)
oops sorry, wrong quotes at the post above..


Quote (rb1957)

is this a student post ?
No it isn't. I have calculated this part in an FEM program and wanted to check the results by calculating it analytically. This is what I've found at the company for this kind of systems. It provides the same analysis as I need but I don't understand he steps of the calculation.

RE: Analysis of a hyperstatic system by hand calculation

(OP)
so, I've tried it with the symm. axis. there is still the problem, that there are no horizontal forces. in the first attachment from my first post, you can see that the horizontal forces H1 and H7 do exist...

in this attachment are the "solution" with the symm. axis. I've compared the results with the results of the first attechment at point 3 and they aren't the same...

RE: Analysis of a hyperstatic system by hand calculation

I have to go out now, but I suggest the following:

1. Make the half structure determinate by considering a fixed support at point 4.
2. Apply H1 and V1 (V1 = F2/2 by symmetry)and determine the rotation at point 1 from these two forces.
3. Apply unit moment at point 1 and determine the rotation at point 1 due to unit moment.
4. Solve for M1, the moment required to change the rotation to zero. The other moments will follow automatically.

BA

RE: Analysis of a hyperstatic system by hand calculation

You forgot to show a horizontal force at F2.

EIT
www.HowToEngineer.com

RE: Analysis of a hyperstatic system by hand calculation

DerGeraet, your solution is wrong as you are assuming a zero moment at F2, which is not. The solution must include a slope or deflection equation to solve for the static indeterminacy.

prex
http://www.xcalcs.com : Online engineering calculations
http://www.megamag.it : Magnetic brakes and launchers for fun rides
http://www.levitans.com : Air bearing pads

RE: Analysis of a hyperstatic system by hand calculation

same as the two above ... you need H4 and M4.

and i thnk the problem is doubly redundant ... when you consider sum moments you have M1+M4+H*e3 = 0
consider if the frame is pinned at pt1 and pt4 then you can solve H = V/2*(e1/2+e2)/e3 (the frame becomes a two force member, so the forces are co-linear, acting from pt1 to pt4)
so to solve the problem you had to remove two reactions ... doubly redundant

RE: Analysis of a hyperstatic system by hand calculation

Just came on this. Are the reaction points bearings as stated or fixed as diagrammed? Either way, given the symmetricality, it is a fairly simple Moment Distribution or Slope Deflection problem.
Using the left half of the model, apply the same moment at each end of the beam; distribute it through the model; calculate the resulting unbalanced vertical force; proportion all moments up by the actual applied load/unbalanced force.

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

RE: Analysis of a hyperstatic system by hand calculation

DerGeraet,

Please confirm the type of support you have at points 1 and 7.

BA

RE: Analysis of a hyperstatic system by hand calculation

(OP)
Hi guys, I just came home. I will give you an answer as soon as possible, so within the next hour(s). thank you for the answers!

@BAretired
they are both "solid bearings" (so 2 DOF, one in horizontal and one in vertical direction). this is why the system is indeterminate to the first degree.

RE: Analysis of a hyperstatic system by hand calculation

"they are both "solid bearings" (so 2 DOF, one in horizontal and one in vertical direction)." ??

2 dof implies pinned, yes? ... but you show a moment at pt1 ??

or do you mean 2 rotational dof ??

RE: Analysis of a hyperstatic system by hand calculation

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.

BA

RE: Analysis of a hyperstatic system by hand calculation

To solve the problem with two hinges, make point 7 a horizontal roller and solve for Δ7 under load F2. Then calculate Δ7 with unit horizontal force applied at 7. Calculate H7 as the force required to bring point 7 back to its starting point.

BA

RE: Analysis of a hyperstatic system by hand calculation

(OP)
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/CourseOnlin... 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.

RE: Analysis of a hyperstatic system by hand calculation

(OP)

Quote (BAretired)

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.

RE: Analysis of a hyperstatic system by hand calculation

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).

RE: Analysis of a hyperstatic system by hand calculation

(OP)
my bad, sorry .. i will call it "fixing points" in the future :)

Quote (rb1957)

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"...

RE: Analysis of a hyperstatic system by hand calculation

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.

RE: Analysis of a hyperstatic system by hand calculation

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.

RE: Analysis of a hyperstatic system by hand calculation

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 e2A/2 between points 3 and 4. That should be e1A/2.

BA

RE: Analysis of a hyperstatic system by hand calculation

"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).

RE: Analysis of a hyperstatic system by hand calculation

Quote (DerGaraet)

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 = FL2/8.
θ1 = θ2 = θ3 = FL2/16
δ1 = δ2 = θ3*h = hFL2/16

(2) Force H acting outward at Joints 1 and 7
M1 = M2 = 0
M3 = M4 = H*h
θ1 = Hh(L+h)/2
δ1 = Hh2(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

RE: Analysis of a hyperstatic system by hand calculation

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)

?

RE: Analysis of a hyperstatic system by hand calculation

(OP)
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!

RE: Analysis of a hyperstatic system by hand calculation

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 ?

RE: Analysis of a hyperstatic system by hand calculation

(OP)
unfortunately not :D 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...

RE: Analysis of a hyperstatic system by hand calculation

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 = -FL2/16 (counterclockwise)
δ1 = δ2 = θ3*-h = hFL2/16 (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 = -Hh2(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

RE: Analysis of a hyperstatic system by hand calculation

hi BA,
sorry not seeing it ...
now it is ...
-FL2/16 + Hh(L+h)/2 + M(L/2+h+a) = 0
hFL2/16 + Hh2(L/2+h/3) + Mh(L+h)/2 = 0 ... divide by -h ...
-FL2/16 - Hh(L/2+h/3) - M(L+h)/2 = 0 ... so that ...
Hh(L+h)/2 + M(L/2+h+a) = FL2/16 = -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.

RE: Analysis of a hyperstatic system by hand calculation

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

RE: Analysis of a hyperstatic system by hand calculation

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 = Hh2/(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

RE: Analysis of a hyperstatic system by hand calculation

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

BA

RE: Analysis of a hyperstatic system by hand calculation

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".

RE: Analysis of a hyperstatic system by hand calculation

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

RE: Analysis of a hyperstatic system by hand calculation

Had another look at this file. My previous expressions for rotation and deflection at Joint 1 with F only acting were wrong. They should be as follows: (see attachment)

(1) Force F acting upward at Joint 4
Simple span moment = F(L+2a)2/4.
θ1 = -F(a2+2ah+aL+L2/4) (counterclockwise)
δ1 = Fh(aL+L2/4+ah)/4 (left to right)

The expressions for H and M were correct before:
(2) Force H acting outward at Joints 1 and 7
M1 = M2 = 0
M3 = M4 = Hh
θ1 = +Hh(L+h)/2 (clockwise)
δ1 = -Hh2(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)


1) -F(a2+2ah+aL+L2/4) + Hh(L+h)/2 + M(L/2+h+a) = 0
2) Fh(aL+L2/4+ah)/4 - Hh2(L/2+h/3) - Mh(L+h)/2 = 0

Solving these two equations for M and H got too messy in terms of the variable names L,a and h, so I tried the values of 20, 5 and 10 for L, a and h.

The result was:
H = 0.3174F
M = 1.346F
which seems to pass the sanity check, so I think they are probably okay but I wouldn't guarantee it.

BA

RE: Analysis of a hyperstatic system by hand calculation

While it would be relatively easy to set up a spreadsheet to calculate the elastic solution of H and M for various values of L, a and h, it would be considerably easier to consider a plastic hinge forming at each corner of the bracket.

Mp = φZ*Fy = φwt2Fy/4
where φ = 0.9
and Z, w and t are the plastic modulus, width and thickness of the bent plate.

Then FuL/8 = Mp or Fu/2*a/2 = Mp
where Fu is the ultimate value of F.

So Fu = 8Mp/L or 4Mp/a, whichever is smaller.
If L>2a, 8Mp/L will govern.

Depending on which code one is using, F = Fu/LF where LF is the load factor (probably between 1.5 and 4, depending on the application).

BA

RE: Analysis of a hyperstatic system by hand calculation

To the first paragraph, I should have added "and at the load point". If L>2a, the failure mode would be a plastic hinge at Joints 3, 4 and 5. If L<2a, the failure mode would be a plastic hinge at Joints 1 and 2.

BA

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