How Do Multiple Cantilever Beams with Rigid Links Affect Load Distribution?

  • Thread starter Thread starter DWSprings
  • Start date Start date
  • Tags Tags
    Beams Cantilever
AI Thread Summary
The discussion focuses on analyzing the load distribution in a system of three cantilever beams connected by rigid links. To determine the total spring constant, users are advised to apply Castigliano's theorem, calculating deflections at various points based on a single load applied at the end of the top beam. The rigidity of the links is debated, with suggestions that they may only transmit vertical loads rather than being truly rigid, complicating the analysis. For accurate results, a finite element model is recommended, especially if the rigid links introduce axial tension and compression. Theoretical methods are still being pursued alongside numerical simulations for validation.
DWSprings
Messages
2
Reaction score
0
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)
 

Attachments

  • IMG_2160.jpg
    IMG_2160.jpg
    20.2 KB · Views: 472
Engineering news on Phys.org
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.
 
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.
 
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
 
Posted June 2024 - 15 years after starting this class. I have learned a whole lot. To get to the short course on making your stock car, late model, hobby stock E-mod handle, look at the index below. Read all posts on Roll Center, Jacking effect and Why does car drive straight to the wall when I gas it? Also read You really have two race cars. This will cover 90% of problems you have. Simply put, the car pushes going in and is loose coming out. You do not have enuff downforce on the right...
Thread 'Physics of Stretch: What pressure does a band apply on a cylinder?'
Scenario 1 (figure 1) A continuous loop of elastic material is stretched around two metal bars. The top bar is attached to a load cell that reads force. The lower bar can be moved downwards to stretch the elastic material. The lower bar is moved downwards until the two bars are 1190mm apart, stretching the elastic material. The bars are 5mm thick, so the total internal loop length is 1200mm (1190mm + 5mm + 5mm). At this level of stretch, the load cell reads 45N tensile force. Key numbers...
I'm trying to decide what size and type of galvanized steel I need for 2 cantilever extensions. The cantilever is 5 ft. The space between the two cantilever arms is a 17 ft Gap the center 7 ft of the 17 ft Gap we'll need to Bear approximately 17,000 lb spread evenly from the front of the cantilever to the back of the cantilever over 5 ft. I will put support beams across these cantilever arms to support the load evenly
Back
Top