1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Cantilever Failure: Help?

  1. Apr 19, 2015 #1
    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.
  2. jcsd
  3. Apr 19, 2015 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    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.
  4. Apr 19, 2015 #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!
  5. Apr 19, 2015 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    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.
  6. Apr 19, 2015 #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!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Cantilever Failure: Help?
  1. Cantilever cable (Replies: 8)

  2. Cantilever Help (Replies: 6)

  3. Cantilever experiment (Replies: 2)

  4. Cantilever Design (Replies: 3)

  5. Cantilever Experiment (Replies: 0)