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!

 

[canwesteng]

4 unknowns, three equations. So it's indeterminate, and can't be solved by statics

 

[IH1980]

I was thinking that too, but you can change the pin at 1 for an X-axis only restraint without changing the results of the computer analysis, leaving three unknowns - Hz at 1 and Vt at 3 and 4? (Have only just done this since the OP by the way!)

 

[FishTheStructure]

Starting at Node 4 and summing the forces: 2 unknowns (force in Member 2 and the reaction force at Support 4) and 2 equations (sum of Fx = 0 and sum of Fy = 0) lets you solve for the force in Member 2 and the reaction force at Support 4
Then summing the forces at Node 3: 2 unknowns (force in Member 1 and the reaction force at Support 3) and 2 equations (sum of Fx = 0 and sum of Fy = 0) lets you solve for the force in Member 1 and the reaction force at Support 3

As I do not know the geometry of your members and supports, I cannot solve it exactly but if I assume that Member 2 and Support 4 are both angled at 45 degrees, i get a force of 11.12 at Node 1 which is only 1.6% different than the value in the image you posted.

Great spirits have always encountered violent opposition from mediocre minds - Albert Einstein

 

[canwesteng]

Are member 1 and 2 different members? I assume it is one continuous member, just separated into two finite elements.

 

[rb1957]

i read node4 reaction as normal to member 2, in which case you can calc the load in member 2

at node 3, the reaction is normal to element 1, which is a component of the load in member 2

at node 1, the reaction is the other component of the load in member 2.

i think your solution works.

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

 

[IH1980]

1 and 2 are two different members. They will be connected at Node 3.

In my time zone its late so I will go through the responses you all have very kindly given me in the morning and respond.

 

[canwesteng]

If it's a hinge connection between 1 and 2, you have six unknowns and six equations so you're fine. If it's a moment connection you're back to one more unknown than equation

 

[IH1980]

Correct, its a hinge. And now I am definitely off to bed...

 

[IDS]

For the given loading applied at the support, member 2 only has an axial load, so there is no rotation at Node 3, so it doesn't make any difference if the member is continuous or not. The force in member 1 is equal to the horizontal component of the force in member 2.



Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[canwesteng]

Assuming infinitely stiff supports that's the case. It's still statically indeterminate though, it's just easy to visually apply deflection compatibility with only axial forces in the member

 

[IH1980]

Ok, I am not to proud to admit I am being an idiot. I have attached how I would go about this and perhaps someone can point out the obvious!

I have drawn out the geometry in the top half. At the bottom, I have started with N4 and immediately get it wrong - I know it wrong because I have a vertical reaction equal to the applied load. This can't be right as then if I have a vertical at N3 I lose equilibrium. Surely my triangle of forces at N4 is wrong but I can't see it any other way as this arrangement gives the applied force as a resultant (if you see what I mean).

I think i'd better go back to school after this one...

 

[canwesteng]

First, sum moments about point 3 to solve for the angled reaction. Then use sum of forces in x to solve F2 and sum of forces in Y to solve the reaction at 3. Alternatively, method of nodes works because it's a truss, I don't actually see the error in your sketch, you have 16.4 kN calculated vs 16.2 kN from software

 

[IH1980]

True, perhaps I am a bit hasty in calling it wrong. It is after all 99% of the software analysis which is more than close enough! Also when I work out F2 using 16.4 x cos 47.54 i get 11.08kN which again is 99%.

My 'problem' with it is that I think the vertical component of reaction at 4 is 22.24kN as drawn on my sketch which does not match the output and cannot satisfy equilibrium if there is a vertical reaction at N3 - is this where I am fundamentally wrong with something?

 

[canwesteng]

Yes. The vertical component of reaction 4 is smaller than reaction 4 - sum of squares of the vertical and horizontal reaction at 4 should equal the reaction

 

[IH1980]

OK, so I thought I had cracked it. Attached is my work through of the example posted previous - the results come out pretty close. To be extra sure I ran a second check and I got different answers to 'the machine'! I will put the screen grab from Robot in the next post as I can only add one file per post. I also did a 3050x2000 (not attached) and that was even further away!

I should add that I am not blindly assuming the computer is right here, and I don't need to get results that match 100%, but I need to be right!

The help offered so far is really appreciated.

 

[IH1980]

Second screen grab attached.

 

[canwesteng]

Are you including member self weight in the computer analysis? That will prevent results from matching. Otherwise your methodology is fine, just make sure you're doing the geometry/trig correctly.

 

[rb1957]

well, IMO, well done for not believing the computer, for working it out for yourself, and for checking your own work.



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

 

[rb1957]

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.

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