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

 

[rb1957]

"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?

 

[msquared48]

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

 

[BAretired]

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

 

[BAretired]

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 m[sup]2[/sup], V1 and V7 are required for equilibrium.

BA

 

[rb1957]

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.

 

[DerGeraet]

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?

 

[DerGeraet]

@rb1957

i think my answer was to quick

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

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.

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.

 

[DerGeraet]

oops sorry, wrong quotes at the post above..

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.

 

[DerGeraet]

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...

 

[BAretired]

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

 

[RFreund]

You forgot to show a horizontal force at F2.

EIT

http://www.HowToEngineer.com

 

[prex]

DerGeraet, your solution is wrong as you are assuming a zero moment at F[sub]2[/sub], which is not. The solution must include a slope or deflection equation to solve for the static indeterminacy.


prex
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[rb1957]

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

 

[paddingtongreen]

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.

 

[BAretired]

DerGeraet,

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

BA

 

[DerGeraet]

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.

 

[rb1957]

"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 ??

 

[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.

BA

 

[BAretired]

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

BA