# I need some help about mutiple cantilever beams.

1. Dec 12, 2013

### DWSprings

Hi guys! I need some help mutiple cantilever beams.
There are three cantilever beams. Single load is applied at the end of an above beam only.
Each beam has own spring constant(k1, k2, k3), thickness(t1,t2,t3) and same width, B. They are connected with blue rigid links like on figure.
they are just applied for connected condition of beams and transfer forces to the other beams. Then they can be ignored structurally.

1. How can get total spring constant?

2. How can get load deviationon each section of beams?
(ab, bc, de, ef, gh)

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2. Dec 13, 2013

### timthereaper

You'll have to draw 3 separate free-body diagrams and use Castigliano's theorem (CT) a lot. To get the total spring constant, you'll have to assume the deflection is x at the end of the top beam (point C) and then use CT to get the deflection at point B. Then, if those really are rigid links, then the deflection at B is the same as at point F. Do CT again to get the deflection at F. Then assume the deflection at F is the same as at point H. Then, since you have the deflections of all the beam ends in terms of x, then the equation becomes F = k_1*d_c + k_2*d_f + k_3*d_h, where d_z is the deflection at point z. Rearrange to get the form F = k*x and then k becomes the total spring constant.

The deviation in the beams can be solved for by Castigliano's theorem. I'll leave you to figure that out.

3. Dec 13, 2013

### AlephZero

If this is a textbook-type question, the "rigid links" are probably meant to transmit only loads in the vertical direction - i.e. they are actiully pinned to the beams, not rigidily connected. In that case, what timthereaper said.

If the links really are rigid, so the rotation of the beam at each end must be the same, and when a finite length of link rotates the ends also move horizontally (so the beams have axial tension and compression as well as bending), this is getting too tough for a sane person to want to solve it by hand. Making a finite element model would be much more practical.

4. Dec 16, 2013

### DWSprings

Actually the attached figure is a simplified model of laminated parabolic leaf spring.
then i drew the rigid links just for expressing the contact points of spring.
the FEA was aleady preformed but i want to get results with the theoretical method too.
i will try to get the constant with Casigliano's theorem~
timthereaper, AlephZero, Thanks for your help!! :D