How Do You Determine the Displacement of a Multi-Region Beam Under Tip Force?

[seang]

Homework Statement

This isn't really a homework problem but something I...felt like figuring out. I'm an EE student, so this isn't rly my game, but I'm giving it a hack anyway.

See the attachment for an outline of the problem.

The attachment shows the top view of a beam connected rigidly to a wall (the wall being the blackish part). What I want to derive is a function for the displacement of the beam w(x) for a force F applied at the tip of the beam (where "L2" is on the picture). (The beam would move into the page).

Would w(x) even be continuous? or would it be a piecewise function? In any case, I want to be able to find the bending in both regions for a force at the tip.

Homework Equations

...

The Attempt at a Solution

I started off with the region nearest to the wall. For this part, it is sort of well known that w(x) is:

$$w = \frac{F}{6 E {I_1}}(3 L x^2 - x^3)$$

for a force F applied at the end of the first region.

Alright, onto the second part. The bending in the second region is governed by the same differential equation but w/ different boundary conditions:

$$EI \frac{d^4 w}{d x^4} = 0 \\$$$$w|_{x = L_1} = \frac{F}{6 E {I_1}}(3L_1^3 - L_1^3)$$$$\frac{\mathrm{d} w}{\mathrm{d} x}\bigg|_{x = 0} = \frac{F}{6 E {I_1}}(3L_1^2)$$$$\frac{\mathrm{d}^2 w}{\mathrm{d} x^2}\bigg|_{x = L} = 0$$$$-EI \frac{\mathrm{d}^3 w}{\mathrm{d} x^3}\bigg|_{x = L} = F$$

And the solution the ODE is something like w(x) = c1*x^3 + c2*c^2 + c3*x + c4;

I can solve for the constants using some matrix algebra, but I'm going to skip that part because it's not really my concern.

After this, I have w(x) for the first region, and w(x) for the second region. I want to combine them, if possible. I'm not really sure where to go from here to combine them into one equation, as a function of the force F applied at the very end of the beam. I should be able to write F1 (the force applied at the end of the first region to get its displacement, sorry I didn't note this in the TeX) in terms of the force applied at the end of the second region, but I'm not sure how.

Any ideas on this?

My leading idea is to say something like: for a force applied at the end of the tip I can find the displacement at the boundary between at L1. then I can use F = k1/x to find the equivelant force at L1...or something.

 

[pongo38]

There is no 'equivalent' force at L1. Although what you propose (the development of a DE) is possible, and is done in academic circles, it's not something that practical engineers would do. Can you draw bending moment diagrams? If you can there is a simpler answer.
However, with your approach... think of the support as A, the middle as B and the tip as C. The deflection at C is made up the flexure of AB and the flexure of BC. The deflection at B is accompanied by a rotation there that can be obtained from dw(x)/dx The deflection at C is caused partly by this rotation at B and partly by the deflection of C below the tangent at B. Mohr's theorems will help you. I'm not sure why you say in section BC LaTeX Code: <BR> EI \\frac{d^4 w}{d x^4} = 0 \\\\<BR> Why zero on the RHS?

 

[seang]

Hi, thanks for the reply. I know of Mohr's circle etc. and bending moment diagrams, but I am not too knowledgeable of either to be honest.

What sort of practicality are you referring to? Are you saying that bending moment diagrams and Mohr's laws would be the most practical way to obtain w(x)? Or do you mean those theories would be the best way to heuristically understand what is going on?

I don't want to give too much away, but there is something more to this problem, which requires that I know w(x).

I got the equation EI \\frac{d^4 w}{d x^4} = 0 from wikipedia. The force is applied with a boundary condition is this case. Is this no good?

The way you explained the bending, w(x) and dw(x)/dx would have to be continuous at B, right? These are the boundary conditions I have applied in the first post.

I probably have everything I need, just need to put it all together.

Thanks again for your help.

 

[pongo38]

I don't have time to give you more advice as I leave for holiday tomorrow. Sorry. However, Mohr's circle is not the same as Mohr's theorems for bending of beams. Mohr's theorems for bending of beams helps you obtain rotations and deflections. Also the reciprocal theorem might help you (The deflection at A due to a load P at B is the same as the deflection at B due to a load P at A. The theory you want to use is based on elastic principles and involves some approximations and assumptions. For example, shear deflection is ignored, and it turns out this can be significant in timber beams. The deflections in practice might be larger due to (1) encastre support giving a bit, and (2) time dependent effects such as creep. And the deflection in practice might be smaller due to the stiffening effects of other members connected to the one you are considering. Engineers frequently ignore these issues in order to get some idea of what the deflection will be. Sorry I have to leave it there now. Best wishes.

 

[DeeNos]

I would first like to commend you for taking the initiative to solve a problem that may not be directly related to your field of study. This shows curiosity and a willingness to learn new things, which are important qualities for a scientist.

Now, onto the problem at hand. It seems like you have made good progress in deriving the equations for the bending of the beam in both regions. To combine them into one equation, you can use the principle of superposition, which states that the overall displacement of a structure is equal to the sum of the individual displacements caused by each load acting alone. In this case, the loads are the forces acting on the tip of the beam in each region.

To apply this principle, you can first find the displacement caused by the force in the first region, and then the displacement caused by the force in the second region. Finally, you can add these two displacements together to get the overall displacement of the beam. This can be represented mathematically as:

w(x) = w1(x) + w2(x)

Where w(x) is the overall displacement, w1(x) is the displacement caused by the force in the first region, and w2(x) is the displacement caused by the force in the second region.

To find w1(x), you can use the equation you have already derived:

w1(x) = \frac{F1}{6 E {I_1}}(3 L1 x^2 - x^3)

Where F1 is the force applied at the tip of the beam in the first region.

Similarly, for w2(x), you can use the equation you have derived for the second region:

w2(x) = \frac{F2}{6 E {I_2}}(3 L2 x^2 - x^3)

Where F2 is the force applied at the tip of the beam in the second region.

Now, to find the equivalent force F1 in terms of F2, you can use the equilibrium equation:

F1 + F2 = F

Where F is the total force applied at the tip of the beam.

Solving for F1, we get:

F1 = F - F2

Substituting this into the equation for w1(x), we get:

w1(x) = \frac{(F-F2)}{6 E {I_1}}(3 L1 x^2 - x^3)

Now