Can this be solved by statics?

[IH1980]

I am tying myself up in knots with this one. I have a system like the picture to assess. There are a variety of geometric configurations so i was planning to plug it all into a spreadsheet, but I am not happy that I am programming it right. So, I modelled one scenario and I cannot mirror the results by hand. Nodes 1-3 is 1.57m, 3-4 is 1.83m and 3-4 vertical is 2m. The rotated roller at 4 represents a slope, so it will provide a restraint but would slip down the slope on its own - member 2 is there to stop this happening.

I can't help feel i'm making a meal of this, but something is wrong somewhere!

 

[IH1980]

Self weight is off! I'm glad that at least I have the analysis principle right now. Being out by a kN is not a big deal but i'd be worried if I was dealing with forces an order of magnitude greater.

And the appreciation for not believing the computer is appreciated!

 

[rb1957]

i suspect your "error" is round off

another day in paradise, or is paradise one day closer ?

 

[IDS]

btw, you have implicitly assumed that the members are rods (axial load only, no bending). if the members are beams then the problem becomes redundant (as noted above) 'cause there can be an internal moment at node 3, balanced by a couple at nodes 1 and 3; the problem becomes a beam on three supports.

No, because as noted above, assuming rigid supports, the force in the left hand beam is horizontal, so there will be no moment in either beam,

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[rb1957]

not the way i see it ... the LH end looks like an anchor point to me, fully pinned, not like the roller at node 3.

if a moement was applied at node 3, the structure could support it.

so it follows that if the members can support bending, and if the joint at node 3 allows it, then an internal moment would develop at node 3.

another day in paradise, or is paradise one day closer ?

 

[haynewp]

Indeterminate if beam members, one too many unknowns and can get a vertical Rx at left pin due to deformation.

 

[Lion06]

OP said later on that it's a hinge between the two members and no internal bending moment can exist at the hinge between the two members. Therefore, it is determinate and easily solved by statics.

 

[IDS]

rb1957 - OK, for it to be completely statically determinant the horizontal member would have to be completely rigid, as well as the supports, but for any realistic member stiffness the horizontal displacement at the bottom of the sloping member will be very small, and the moment will be negligible.

I agree with the calculated force in the sloping member of 16.41 kN, and so does Strand7 (both with a hinge at the bottom, and a moment connection).

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[haynewp]

Then yes, takes away an unknown and I didn't see the hinge comment.

 

[IDS]

I thought I'd have a look at how much difference a moment connection at Node 3 makes (see attached spreadsheet).

The conclusions are:

Including the moment at Node 3 makes less than 0.1% difference to the reaction values.
Spreadsheet calculations agree with FEA results (Strand7) almost exactly.
Doing the calculation on a spreadsheet isn't that hard (once you have got it set up).

The remaining question is why the FEA results found in the OP are significantly different. At the moment I can't see why.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[IH1980]

Doug,

I have only just noticed your reply and thought I would acknowledge your effort in writing the spreadsheet to check it out for me - really appreciated, thanks. I've downloaded it and will be giving it a once over, I've wanted to pick up a few tips on programming the stiffness method so it will be really helpful.

 

[IDS]

IH980 - Glad to help. Solving simple problems on a spreadsheet is a great way to get a better understanding of what the big compiled programs are doing under the hood, which is one of the main reasons I do it.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[medeek]

What is the software you are using?

A confused student is a good student.
Nathaniel P. Wilkerson, PE

http://www.medeek.com

 

[IH1980]

The software used in the screenshots I posted is Robot Structural Analysis from Autodesk.

 

[Tomfh]

Just split the force into its vector components perpendicular to the roller and the member. Wash-rinse-repeat at the next node.