cantilever problem. 3

[HawkOkeoJr]

hallo everyone,
I have a cantilever with a point load at the far end e.g 10N and a spring of know stiffness e.g 100 N/m attached perpendicular to the beam at 1/3 of the beam from the fixed point. can somebody please help me know how i can calculate the deflection at the cantilever tip. I will appreciate
Thank you!

 

[rb1957]

this is a singly redundant problem ... review "unit force method".

or you can solve this iteratively ...
first solve the cantilever without the spring, calculate the delection at the spring point;
with this defelction you know the spring force,
solve this cantilever (with two forces, at the tip and at the spring)
calculate deflection at the spring point, and revise the spring force.

anticipating that the solution will oscillate (the cantilever alone will predict a large deflection at the spring, applying the spring force = to this deflection will probably produce an opposite deflection). if it does, i'd replace the spring force with 1/2 the first deflection, and see how that works out.

there are closed form solutions to the elastic support, but solving DEs is probably out of the question.

 

[GregLocock]

Another, non iterative approach, is to solve for the beam's stiffness at the spring's location, for a cantilever L/3 long. This will be k1. Then you know the total rate at that location is k1+100, so you know the deflection at the spring is 10/(k1+100). Then work out the angle of the beam at that point due to the force 10*k1/(k1+100), call it alpha.

Then work out the deflection of a cantilever 2L/3 long with a 10N load, call this d2

The total tip deflection is then 10/(k1+100)+2L/3*alpha+d2

Another way would be to do a work equation.

Cheers

Greg Locock


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[HawkOkeoJr]

@ GregLocock, why is the deflection due to the influence of the angle so high?. this part of the formula (2L/3*alpha)...lets say that my 2/3 of L is 25. its generating a very big deflection at the tip

 

[GregLocock]

Units?

Cheers

Greg Locock


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

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[HawkOkeoJr]

@rb1957 and GregLocock, or (anybody else who has an idea with the approach) i have another issue concerning deflections, lets say that the beam specified above is connected to another beam which lies on Z axis (Z axis, is the axis that is facing you) at an angle of 90 degrees. Meaning that both beams are on X-Z plane. so, my question is, how do i calculate the deflection Analytically at the connection point and also at the tip of the main beam which is lying parallel to the X-axis?......if its not clear enough, i would provide an attached drawing.

Thank you in advance, am really learning a lot from all this discussion.

 

[rb1957]

it sounds like you've changed the problem to an L beam, in the X-Z plane.

still a "basic" problem ...
draw a free body, it looks like you'll have three reactions (shear, moment, and torque) due to the applied load.
now you can solve the basic beam equations ("double integration") to get deflections of the beam generally (as equations);
get the deflection of the spring attach pt.
then set up a 2nd free body for the spring force, some unit value, 1?, applied as the spring force would act, opposite the deflection due to load;
calc the deflection of the spring pt.

now you can set up an equation to determine the spring force, F, ...
d1 = initial deflection of the pt (+ve), with only tip load,
du = deflection of the pt with a unit spring load (-ve),
ds = deflection of spring = ds = d1+F*du, ds = d1+(k*ds)*du

 

[HawkOkeoJr]

@ rb1957, Sir thanks for your constant reply, i really appreciate it a lot. The second question was without the spring.
I have attached a file, if you have some time, you can have a look at it and give me clue of the way forward.
N/B the material is Allum hence E= 70GPa. am looking forward to your reply.

 

[rb1957]

so you've replaced the spring with a cantilever ...
same deal ... you can calc the stiffness of the cantilever (i think someone did it here already, but try it for yourself first)

calc displacements of the loaded cantilever (double integration method). that'll give you displacements along the beam.
"all" you have to do is match the displacement at the common point ...

 

[IDS]

A simple hand calc is not going to give the same result as a computer analysis unless you relase the tortional and longitudinal connection in the computer analysis, so the transverse beam is only providing vertical restraint.

What is the length of the beams?

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[rb1957]

sure ...
hand calcs have the advantage that the engineer is frameing the question ... neglecting things that are IHO negligible.
hand calc have the disadvantage that because they are framed by the engineer, he may not include a feature that is relevent.

computer solutions have the advantage that they have defaults so the engineer doesn't have to think much (in fact more thought is spent on getting the damned thing to work)
computer solutions have the disadvantage that because they don't require much thought, generally not much thought is given.

pick your poison !
(it'll kill you either way)

 

[HawkOkeoJr]

the x-axis beam is 40cm while the z-axis one is 15cm. i think the lengths of the beams should not be that relevant, we can choose them arbitrarily BUT sure, to generate the displacement value that i gave, you need them.

 

[IDS]

the x-axis beam is 40cm while the z-axis one is 15cm. i think the lengths of the beams should not be that relevant, we can choose them arbitrarily BUT sure, to generate the displacement value that i gave, you need them.

I was interested in how much difference it made having a moment connection between the two beams.

With a moment connection I get much the same result as you (0.136 m) and releasing all rotational restraint where the transverse beam joins the longitudinal beam that increases to 0.173 m, which is what I'd expect you to get with a hand calculation (just done it, that's what I get).

By the way, the simplest way to check the computer output is to take the forces from the computer and apply them to two separate cantilevers, and see if the deflections at the connection point are equal.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[rb1957]

sorry, but of course the lengthsd are relevent ... a longer cantilever is goping to deflect more.

now maybe (maybe !) if the joint is 15cm from the fixed end of the 40cm (so that the two fixed supports are equally 15cm from the common point), then possibly (i'd expect) that the two beam will deflect the same, and the inner portion of the longer beam is deflecting at though it has double the inertia (assuming both beams have the same section). but that'd be a special solution and i think it's better to understand the general solution.

 

[HawkOkeoJr]

@ rb1957 and IDS, Sir, so, i guess a general Analytic solution is hard to come by?....cause have really tried it in and out but still not working out. the beams surely have the same section.
my other idea is to try and find the force acting at the junction but the problem is, the force acting at that point is shared by the two arms of the beam.
any ideas, will still be welcomed cause its my most crucial point to proceed.
Thank you!

 

[ishvaaag]

You can use the "force method" where you find the "force" redundants or the "displacement method"; at the contact point you will have (except that you dismiss some) 3 force and 3 torque redundants or 3 linear displacement and 3 rotation unknowns following the case, that you -for the benefit of an analytical solution- try to reduce longhand, what is prone to error; this advocates forfeiting an analytical solution by hand and using some FEM method program where such problem has the input properly ordered and the solution readily got. I am saying more or less the same that IDS first said.

 

[rb1957]

"a general Analytic solution is hard to come by" ... no in my opinion.

(1) solve the long cantilever, reactions (simple equilibrium equations) and displacements along the length (double integration method).

(2) now that you've solved a cantilever, you can solve the short cantilever with a load P at the tip

(3) now you can solve the long cantilever with two forces applied, the tip load and a load P at the join of the two cantilevers,
like (1) reactions and displacements (ok, they're generalised 'cause you don't have a value for P)

(4) now guess a value for P, say 1/2 the tip load. is the displacement of the end of the short cantilever (2) the same as the long cantilever with two loads and the common point ?

(5) adjust your guess accordingly.

this isn't that hard (IMHO), any text book should cover it.

 

[BAretired]

The problem is not difficult, but you first have to decide how the beams are attached to each other. If the attachment is deemed to be a hinge acting at the centroid of the two beams, then there is only one unknown vertical force acting at the connection point, say F.

Force F acts upward on Beam A and downward on Beam B. The solution can be found using strain compatibility. Beam A deflects downward an amount delta due to force P at the end combined with force F at the connection point. Beam B deflects downward an amount delta due to Force F acting at the connection point. One equation is required to solve for F.

If you consider the members connected rigidly together, then the torsional rotation of each beam must be considered as well. At the connection point, there will be an unknown vertical force F as before, but there will also be two unknown moments Mx and Mz. These can be solved directly using compatibility equations, which is what your computer program presumably did. You will need three equations to solve for three unknowns, namely delta, alpha_x and alpha_z representing the vertical deflection, and the rotation about the x and z axes respectively.



BA