Calculating reactions from center of gravity

[NTME76249]

Hi Guys
Sorry for posting something so basic. i think i slept through this part of statics

The attached image shows my problem.

The intended use of this is do determine loading on a skid from a know CG and weight from a SolidWorks model.
I know how many cross members i have and i need a way to determine my loading on each cross member.
Sorry if this has been asked before. but i was unable to find it.
Thanks

Michael McMillan

http://cadtechie.blogspot.com/

 

[drawoh]

NTME76249,

I take it your points a, b, c and d are your reaction points?

Your first diagram is statically indeterminate. The reaction forces depend partially on the stiffness of the structure. You have to solve this algebraically, then work out boundary conditions. This should be explained somewhere in your mechanics of materials text.

It looks like you solved the second diagram.

Just as a style note: I would not factor your lengths the way you did. If you write your moment equation as...

Σ(a) = (84in-12in)(100lb) - (96in-12in)d...

It will be easier to compare your equation to your diagram.

--
JHG

 

[rb1957]

pls clarify your sketch ...
i think you have 4 cross-braces supported by two side-braces ?

so in the first pic, i think you'd have a+e where you have a, b+f where you have b, etc
and in the 2nd a+b+c+d where you have a and e+f+g+h where you have b ...

i think the 2nd pic shows that the cg is towards one side (ie not in the middle).

you show be able to figure out the total reaction on one side (a+b+c+d) and the other (e+f+g+h) (hint equations of equilibrium ... sum forces, sum moments).

then maybe you can figure out what to do with a, b, c, and d ... you know the sum of the forces, and you know where this acts. with only two equations availiable you need to express the 4 loads with two variables (i'd suggest using a linear variation).

clear as mud ?

 

[GregLocock]

As drawoh says, can't be done without more detail or making some assumptions.

Quite likely would be c~d and a is -ve





Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

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[rb1957]

personally i doubt the force at the far end is -ve since it'd only be a contact force ... i think zero is it's minimum.

 

[Twoballcane]

I'm not allowed to open files, but in any case may I suggest (if not already using it) the cut and sum method coupled with method of sections if it is a determinate truss problem. These are two great techniques that you can use to find the loads in each beam.



Tobalcane
"If you avoid failure, you also avoid success."
“Luck is where preparation meets opportunity”

 

[rb1957]

looking again, i see how horribly messed your first calc is !

the second calc, that is determinate with only two reactions, looks ok

but the first one ?? 6*100 is the moment due to the 100 lbs load 6 ft from a, ok ... but 4*100+1*100 ??
and then all the different "d" terms ??
if you've got four unknowns and only two equations, then you need to make assumptions. a rational assumption would be to assume linear distribution of reactions. i'd express the reactions as a, a+2/7b, a+5/7b, a+b ... two unknowns a and b
now you can sum forces 4a+2b = 100
and moments (about a)(a+2/7b)*2+(a+5/7b)*5+(a+b)*7 = 6*100
so that 14a+78/7b = 600

now if a is -ve i think you should say a = 0 and redo with three reactions.

clear as mud ?

 

[drawoh]

rb1957,

I am interpreting the thing as a cantilever beam, solvable by double integration method. The OP can assume that the deflection is zero at the reaction points, and use the boundary conditions to solve the integration constants and reactions.

Alternately, he can assume that the beam is very rigid and that it is sitting on soft anti-vibration mounts, with a known spring rate.

Either way, there is lots of fun math.

--
JHG

 

[rb1957]

you gotta get out more ...

it could be a beam on many rigid supports,
a linear assumption means you're assuming plane section remain plane (usually a good assumption).

many ways to skin cats ...