3d-forces-moments 3

[amorphous]

Hello All,
I am trying to resolve a force in to four supports but could not do it. Its been a while I studied engineering statics, I guess I forgot the basics of it. I know its a very simple problem. I would really appreciate if any one can help solve this thing.

Thanx in advance.

 

[hydtools]

You need to get into a deflection calculation or virtual work method to solve what is a statically indeterminate problem.

http://www.public.iastate.edu/~fanous/ce332/force/cdmoment.html

Ted

 

[IDS]

You need to get into a deflection calculation or virtual work method to solve what is a statically indeterminate problem.

If the four supports were not exactly in the same plane and/or the supports were not rigid it would be indeterminate and you would have to know the stiffness of the plank and supports, but if they are in the same plane and the supports are assumed to be rigid you can just replace the 100 lb load with a 70 lb load on line AD and a 30 lb load on line BC and distribute those loads to the supports at the end of each line (ie in the ratio 15:85). That will give you four forces in equilibrium with the applied load.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[amorphous]

Thanks guys fot the prompt response.

@hydtools: Initillay I did not think it was a indeterminate problem and I thought I was missing something. It took sometime for me to understand the link you sent me and now I'll try solve my problems using those concepts. I am not soure I will succeed but I'll try.

@IDS: I totally understand your way of solving this problem and I can tell the loads on the supports by ratios of distances as it is a simple problem. But it becomes very difficult when you have multiple loads acting all over the plank. So I was more or less looking for a general method of solving the problem.

Does anyone know of a software (not going into FEA) out these to solve these kind of problems like Frame2D. Any help is much appreciated.

Thanks again

 

[IDS]

Amorphous - I don't see your problem, if you have multiple loads you just apply the same simple method to each load. How can it be any simpler than that? It would be easy to set up on a spreadsheet.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[kingnero]

@ IDS: the approach you suggested gives the exact same equations as amorphous stated in his attached pdf file.

I still cannot solve for either reaction force (A, B, C or D) with the given equations.

Matrix is as follows:


1 1 1 1 -100
0 1 1 0 -30
1 0 0 1 -70
0 0 1 1 -15
1 1 0 0 -85

but as you can see they all come down to the same equation, ie. Eq. #1

 

[IDS]

You are making it way too difficult; you don't need to solve any simultaneous equations.

You can replace the 100 kN load with a 70 kN load on AD and a 30 kN load on BC.

Each of these loads is distributed 85% (i.e. 17/20) to the nearer support and 15% (i.e. 3/20) to the further support. That gives you:

At A 0.85 x 70 = 59.5
At D 0.15 x 70 = 10.5
At B 0.85 x 30 = 25.5
At C 0.15 x 30 = 4.5

Total reaction = 100 OK
Moment about AD = (25.5 + 4.5) *10 = 300 OK
Moment about AB = (10.5 + 4.5) *20 = 300 OK

Same simple calculation for a load anywhere else.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[kingnero]

Ofcourse... I didn't quite got that from the first time...

 

[rb1957]

i'm suspicious when two user's have the same problem ... sound like school work ...

 

[chicopee]

This is the way that I was taught when I took Statics and Dynamics in the early '60's. See attachment which has a little more details for clarity purposes. The plate and supports are assumed rigid.

 

[rb1957]

nice looking method. alternative is to say AD reacts 70% and BC 30% (or AB 85% and CD 15%) which will put a small +ve reaction at C (unlike the -ve calc'd above). this is a result of solving an indeterminant problem by assumptions. to solve it "properly" remove on corner support (C) and see what the displaced shape tells you

 

[hydtools]

chicopee, nice solution.
As you have solved, force at C is a downward reaction not an upward reaction. Weight W is to the left of a line connecting reactions B and D, so reaction at C must be downward to create a moment opposing that of W and adding to moment of reaction A acting around line BD, assuming the connections at all supports can react either up or down.

If connection at C could not react downward, then reaction at C would be 0 and only A, B, and D would react to W.

Ted

 

[IDS]

The solution provided by chicopee gives a different answer to mine because it assumes elastic supports and a rigid plate.

My solution assumes rigid supports and a flexible plate. Both are correct for the assumed support conditions.

As said before, if you want to take the support displacements into account then you need to know the stiffness of the plate and the supports, and you can then use a method similar to that in the link provide by hydtools; except that was for a beam and it becomes considerably more complicated for a slab supported on its corners. You could probably find something reasonably straightforward in Roarke though.

The easiest and quickest way would be to plug it into an FEA program.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[hydtools]

IDS,
Your solution is not correct. You moved forces without including associated moments. In order for the displaced force to be equivalent to the original force, you must include a moment equal to the force times the displaced distance. That is what chicopee's solution shows when he moved W to the center of the plate. His is correct.

Ted

 

[IDS]

Your solution is not correct. You moved forces without including associated moments. In order for the displaced force to be equivalent to the original force, you must include a moment equal to the force times the displaced distance. That is what chicopee's solution shows when he moved W to the center of the plate. His is correct.

How come the reactions are in moment equilibrium with the applied force then? If you analyse it in a FEA program with rigid supports you will get the same answer.

Chicopee's solution is different because it incorporates different assumptions about the support and plate stiffness. Both are correct for the chosen assumptions.



Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[emonje]

"IDS,
Your solution is not correct. You moved forces without including associated moments. In order for the displaced force to be equivalent to the original force, you must include a moment equal to the force times the displaced distance. "

IDS has actually added the moments, just that the moment needed to move the 70 lbs by 3 ft cancels the moment needed to move 30 lbs by 7ft.

Had he, for example, moved the entire 100 lbs force to AD, he would need to add a (100 x 3) 300 lb-ft (assuming dimensions are in feet) moment clockwise looking from A to D.

 

[rb1957]

yeah, he replaced the original load with an equivalent set ... do the math Mx about A 100*3 = (70+30)*3, My about A 100*3 = 30*10 ...