Optimizing Cantilever Design: Solving for Minimum Mass and Maximum Strength

In summary, the student is trying to find the rod of the least mass that can still resist forces without yielding, but is running into an error. He has created formulas for axial, shear, torsional, and bending stresses, and has dragged and dropped them down different columns for different radius rods in order to find the combination of stresses that will be the least for the given rod. However, he is having trouble applying these stresses to a real-world situation. He has also forgotten about combined loads and is going to look into that.
  • #1
Vircona
4
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.
 
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  • #2
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.
 
  • #3
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!
 
  • #4
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.
 
  • #5
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!
 

1. What is cantilever failure?

Cantilever failure is a structural failure that occurs when a beam or structure supported at one end is unable to withstand the applied load and collapses or breaks at the unsupported end.

2. What are the common causes of cantilever failure?

The most common causes of cantilever failure include excessive loading, inadequate support or anchorage, and structural defects such as cracks or corrosion. Other factors such as dynamic loads, temperature changes, and improper design can also contribute to cantilever failure.

3. How can I prevent cantilever failure?

To prevent cantilever failure, it is important to conduct a thorough analysis and design of the structure using appropriate safety factors and load calculations. Regular inspections and maintenance can also help identify and address potential issues before they lead to failure.

4. What are the warning signs of cantilever failure?

Signs of potential cantilever failure may include visible cracks or deformations, excessive vibrations, and unusual noises. It is important to address these warning signs promptly and seek professional assistance if needed to prevent failure.

5. What should I do if a cantilever failure occurs?

If a cantilever failure occurs, it is important to evacuate the area and seek professional help immediately. It is also important to investigate the cause of the failure and make necessary repairs or modifications to prevent future failures.

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