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
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
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
Mike McCann
MMC Engineering
http://mmcengineering.tripod.com
RE: Analysis of a hyperstatic system by hand calculation
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
As stated by rb1957 and m2, V1 and V7 are required for equilibrium.
BA
RE: Analysis of a hyperstatic system by hand calculation
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
@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
i think my answer was to quick :)
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 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
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
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
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
EIT
www.HowToEngineer.com
RE: Analysis of a hyperstatic system by hand calculation
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
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
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
Please confirm the type of support you have at points 1 and 7.
BA
RE: Analysis of a hyperstatic system by hand calculation
@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
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
BA
RE: Analysis of a hyperstatic system by hand calculation
BA
RE: Analysis of a hyperstatic system by hand calculation
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
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
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
i can't figure out what you mean with "doubly reduntant"...
RE: Analysis of a hyperstatic system by hand calculation
RE: Analysis of a hyperstatic system by hand calculation
Michael.
Timing has a lot to do with the outcome of a rain dance.
RE: Analysis of a hyperstatic system by hand calculation
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
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
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
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
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
otherwise it might just be a bunch of numbers ?
RE: Analysis of a hyperstatic system by hand calculation
RE: Analysis of a hyperstatic system by hand calculation
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
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
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
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
BA
RE: Analysis of a hyperstatic system by hand calculation
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
BA
RE: Analysis of a hyperstatic system by hand calculation
(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
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
BA
RE: Analysis of a hyperstatic system by hand calculation
This is a link to a short spreadsheet I wrote for your bracket using OpenOffice.calc. It might be interesting to compare the results with the formula you provided in the original post.
BA