Optimizing Cantilever Design: Solving for Minimum Mass and Maximum Strength

AI Thread Summary
The discussion revolves around a project involving a 1.2m cantilever beam subjected to axial and torque/shear forces, requiring the selection of a rod with minimal mass that can withstand these forces without yielding. The user initially focused on axial stress calculations for a 2-cm radius aluminum rod, concluding that it could support a force of 120 kN, which raised concerns about the accuracy of their calculations. Feedback highlighted the importance of considering combined stresses from axial, bending, torsional, and shear forces, rather than relying solely on axial stress. The user acknowledged the oversight regarding combined loadings and plans to revise their approach accordingly. The conversation emphasizes the complexity of cantilever design and the need for a comprehensive analysis of all acting forces.
Vircona
Messages
4
Reaction score
0
I'm in a sophomore-level general engineering class (mechanics of materials), and I was assigned a project that I had a good idea on how to tackle, but I'm running into an error.

The project: I'm given a 1.2m long cylindrical cantilever with forces acting on it; one axial tensile force, and two forces producing torque/shear along the outside edge of the beam. I can choose either a solid, hollow, or composite rod, and can choose from steel, aluminum, or titanium. I can also vary my outer radius (and inner/boundary radii in the cases of hollow or composite rods). I need to find the rod of the least mass that can still resist the forces without YIELDING (not failing) given a 1.5 factor of safety.

My approach: I wrote up the equations for axial, shear, torsional, and bending stresses in excel. I just dragged and dropped the formula for each down the columns with varying radii in the first column (only for solid rods so far; once I get past this hiccup I'll be working on composites).

My problem: Using a type of aluminum with a ~250 MPa tensile yield strength and 2.7 g/cm^3 density, my equations show that a 2-cm radius aluminum rod can resist a force of 120 kN (diagram shows 80 kN --> 120 kN accounting for the FoS). This definitely sounds wrong, obviously, and the equations for just the tensile force is exceedingly simple. I still, for some reason, can't figure out where I've gone wrong.

My Calculations: Without going into all of the different radii I used, I'll show what I did for the 2-cm radius rod. Axial stress = F/A; the force is 120 kN. Area is (obviously) pi*r^2 --> pi*(0.02)^2 --> 0.001256637. F/A = 120,000/0.001256637 = 95.5 x 10^6 N/m^2 = 95.5 MPa. So the rod is experiencing 95.5 MPa of stress, but its tensile yield strength is 250 MPa, so this 2-cm radius aluminum rod can hold 120 kN (or roughly 27,000 lbs) of force (according to my surely-wrong calculations).

I know this should be very easy, but I honestly can't figure out where I'm going wrong. Any help would be greatly appreciated.
 
Engineering news on Phys.org
It's not clear why you think this obviously simple tensile stress calculation must be wrong. (Note: it isn't)

Since the bar must withstand not only axial, but a combination of axial, bending, torsional, and shear stresses, it is unlikely that applying a purely axial load is going to be the limiting factor in the selection of the size of the rod or the material from which the rod is fabricated.

Apply several different loads to your bar and see what the combined stresses are. If you haven't covered theories of failure (besides loading things up to the tensile yield point), then these theories are likely going to come into play once you have bending stresses combined with shear stresses.
 
Sorry, I should have specified in the original post. I was only concerned with axial forces for now because it seems strange that an aluminum rod 2-cm in radius could hold 27,000 lbf. However, I do have the calculations written for shear, bending, and torsional stresses as well.

I did forget about combined loadings, though. We just went over them, and I'd started this project before combined loading was discussed in class, so I hadn't really put the two together yet. I'll take a look at that and should be able to figure it out using that. Thanks!
 
Vircona said:
Sorry, I should have specified in the original post. I was only concerned with axial forces for now because it seems strange that an aluminum rod 2-cm in diameter could hold 27,000 lbf. However, I do have the calculations written for shear, bending, and torsional stresses as well.

I did forget about combined loadings, though. We just went over them, and I'd started this project before combined loading was discussed in class, so I hadn't really put the two together yet. I'll take a look at that and should be able to figure it out using that. Thanks!

According to your OP, the rod is not 2 cm in diameter; the radius of the rod is 2 cm. The diameter of the rod is ... ?

In any event, tensile loads are some of the easiest to support. Compressive loads, and the possibility of structural instability (i.e., buckling), are much more interesting (and complicated) to analyze.
 
Edited my post; I meant radius, typed diameter. Oops.

Luckily, there's not a compressive loading on the beam. I guess I'm just surprised to find that such a small rod can hold that much force, so I assumed my calculations had to have been off.

Anyway, I'll start analyzing the combined loadings and see if I can tinker with my formulas to get them to work for me. Thanks for the help!
 
Hi all, I have a question. So from the derivation of the Isentropic process relationship PV^gamma = constant, there is a step dW = PdV, which can only be said for quasi-equilibrium (or reversible) processes. As such I believe PV^gamma = constant (and the family of equations) should not be applicable to just adiabatic processes? Ie, it should be applicable only for adiabatic + reversible = isentropic processes? However, I've seen couple of online notes/books, and...
Thread 'How can I find the cleanout for my building drain?'
I am a long distance truck driver, but I recently completed a plumbing program with Stratford Career Institute. In the chapter of my textbook Repairing DWV Systems, the author says that if there is a clog in the building drain, one can clear out the clog by using a snake augur or maybe some other type of tool into the cleanout for the building drain. The author said that the cleanout for the building drain is usually near the stack. I live in a duplex townhouse. Just out of curiosity, I...
I have an engine that uses a dry sump oiling system. The oil collection pan has three AN fittings to use for scavenging. Two of the fittings are approximately on the same level, the third is about 1/2 to 3/4 inch higher than the other two. The system ran for years with no problem using a three stage pump (one pressure and two scavenge stages). The two scavenge stages were connected at times to any two of the three AN fittings on the tank. Recently I tried an upgrade to a four stage pump...
Back
Top