Beam bending problem, calculate deflection of a beam

[Robin91]

Homework Statement

The problem is to determine the maximum deflection when a person is standing on the beam in the attachment.

E= 206.8 GPa, v=0.3

Homework Equations

(although I am not completely sure if this one is relevant)

The Attempt at a Solution

I've thought a bit about the problem, but I don't know exactly where to start. The beam is asymmetical, which implies that it will most likely bend into the negative z direction and twist a bit.

If you could steer me into the right direction it will be most appreciated!

Thanks,
Robin

 

[Robin91]

I've thought a bit more about it, but I'd like to know if I'm thinking in the right direction.. I think the person should stand on the spot of the picture, exactly in the middle (seen longtitudinal perspective) of the beam for maximum deflection. Then the beam will rotate and bend. Am I supposed to translate the force to one colinear with the shear center and then have a force and a moment? And what is next then?

Thanks in advance,
Robin

 

[pongo38]

Suppose you put the person at the shear centre and obtain that deflection (without twist). If you then move the person to another position in the cross-section, how will that change the result you already have? I can't see the relevance of the stress equation, btw.

 

[Robin91]

I guess the beam then bends with

\delta_{max}=\frac{Pl^3}{48EI}

Is that correct? But how can I calculate the I, what is the neutral axis? Is that just the horizontal line through the centroid (because I use the deflection with a force in the shear center)..? And how do I use the Poisson ratio in this problem?

Thanks in advance,
Robin

 

[pongo38]

There are two centroidal axes, x-x and y-y, origin at the centroid, not corresponding to the shear centre, When the load is passing through the shear centre, the neutral axis is the x-x axis and the I is Ixx obtained using the parallel axis theorem. If take account of the load not being at the shear centre, then you need to add a twisting moment, the angular twist requiring a formula similar to delta max you quote but involving the torgue T, and J instead of I. If you don't know what I am talking about, you need to do some background reading on torsion of open sections, and then it should become clearer. I can't see how Poisson's ratio is involved.

 

[Robin91]

Ok, thanks, that should get me started. I only wondered about the poisson's ratio because it is given in the exercise... (\nu=0.3)

 

[diflection1]

general idea to understand:did you have any idea how shear and moment diagram look like?if so it will be a great help!SAYING:ONE PICTURE'S WORTH A THOUSAND WORD,LEADING TO THE WAY I think,thank you, I 'll get back----

 

[Robin91]

Hi Diflection,

Indeed that's a good thing to remember. However, in the few months that passed by since I posted it, I already passed the course :).

Thanks for your help,
Robin