9a-ex. Failure Examples
Ex. 9a.1Ex. 9a.2
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Example 9a.1 - Femur Failure [see also book]

Background: The shaft of a femur (thigh bone) can be approximated as a hollow cylinder. The significant loads that it carries are torques and bending moments.

Given: The femur shaft has an outside diameter of 24 mm and an inside diameter of 16 mm. The tensile strength of bone is taken to be Su = 120 MPa.
During strenuous activity, the femur is subjected to a torque of 100 N·m.

Req'd: The effective moment the bone can take without failure. Consider only torsion and bending loads, and take the bone to be a brittle material.

Image: eSkeletons Project

Sol'n: Since the bone is brittle, it follows the Maximum Normal Stress criteria. If the maximum Principal Stress is greater than the material Tensile Strength, sI > Su, the bone fails (fractures).

The element is taken on the left side of the femur as we look at it; the element is in the z-x plane. The Maximum Principal Stress is:

The contributing stresses are:

Solving the geometric terms:

I = 1.3069 x 10-8 m4;   J = 2.6138 x10-8 m4

so:    t = TRo/J = 45.9 MPa

Reducing the sI equation:

and rearranging:

For the bone to break, sI = Su = 120 MPa.
Since t = 45.9 MPa, we can solve for sx : sx = 102.44 MPa.

Using the Bending Stress formula (s=Mc/I), the Moment to break the bone is:

Example 9a.2

Given: A plane stress element in a part made of the 6061-T6 is found to have the following stress:
        sx = 5.6 ksi; sy  = 9.9 ksi;  txy = 5.0 ksi
The axial yield strengthof 6061-T6 aluminum is 35 ksi, and its shear yield stress is ty = 20 ksi.

Req'd: Determine the Factor of Safety:
  (a) using the Tresca Criterion.
  (b) using von Mises Criterion.


Step 1. The Tresca Criterion is based on the maximum shear stress. The in-plane principal stresses are:

  sI = 13.2 ksi        sII = 2.3 ksi 

Considering all directions, for plane stress, the maximum shear stress is:

The first term in the square brackets is the maximum in-plane shear stress.

  tmax,in-plane = 5.4 ksi

  tmax,out = 6.6 ksi

The maximum shear stress is out-of-plane, tmax = 6.6 ksi

F.S.(Tresca) = 20/tmax = 3.0

Step 2. The von Mises Criterion is based on a maximum distortion energy. The von Mises ( Equivalent) Stress for a plane-stress element is:

If the Equivalent Stress so exceeds the yield stress, the material is deemed to have yielded.

Using either equation, solving for the equivalent stress gives: so = 12.2 ksi

F.S.(von Mises) = 35/so = 2.9

  • The ratio Sy:ty for design values of 6061-T6 Aluminum is 35:20 = 1.75.
  • From the von-Mises Criterion, Sy:ty=1.732.
  • From Tresca,ty = Sy/2 = 17.5 ksi, or Sy:ty=2.
  • Thus, the von Mises Model is the better model for 6061-T6 Aluminum.

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Updated: 05/23/09 DJD