Example 9.1.1 - Femur Failure

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.

Solution: Since the bone is brittle, it follows the Maximum Normal Stress criteria. If the maximum Principal Stress is greater than the material Tensile Strength, s_I > S_u, 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:

Solving the geometric terms:

so:    t = TR_o/J = 45.9 MPa

Reducing the s_I equation:

and rearranging:

For the bone to break, s_I = S_u = 120 MPa.
Since t = 45.9 MPa, we can solve for s_x : s_x = 102.44 MPa.

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

Example 9.1.2

Given: A Plane Stress element in a part made of the 6061-T6 is found to have the following stress:
s_x = 5.6 ksi; s_y = 9.9 ksi, and t_xy = 5.0 ksi.
The Axial Yield Strength, S_Y, of 6061-T6 aluminum is 35 ksi, and its Shear Yield Stress, t_Y, is 24 ksi.

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

Solution:

Step 1: The Tresca Condition is based on the Maximum Shear Stress. The Max. In-Plane Shear Stress is given by:

Absolute Value: |tmax,in-plane| = 5.4 ksi

To find the Out-of-Plane Shear Stresses, we need to find the In-Plane Principal Stresses:

Absolute Values: |sI| = 13.2 ksi        |sII| = 2.3 ksi

Considering all directions, for Plane Stress, the Maximum Shear Stress, t_max is:

Solving: tmax = 6.6 ksi

Thus:

Step 2: The von Mises Criterion is based on Maximum Distortion Energy. The von Mises (or Equivalent) Stress for a Plane-Stress element is:

From Principal Stresses	From General Stresses

If the Equivalent Stress, s_o, exceeds the uniaxial Yield Stress, the material is deemed to have yielded.

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

• Note, the ratio S_Y:t_Y for design values of 6061-T6 Aluminum is 35:20 = 1.75. Note, these values are rounded.
• From the von-Mises Criterion, if only shear stress, t, acts at a point (s_x=s_y=0), then at yielding, s_o = 1.732t = 1.732t_Y = S_{Y; or SY:tY=1.732.}
• From Tresca, if we calculate the Maximum Shear Stress, t_max, developed in a uniaxial tension test at yielding (s_x=S_Y; s_y=0), we would find that t_max = s_x/2 = S_Y/2 and predict the Shear Yield Stress, t_Y = S_Y/2 = 17, or _SY:tY=2.
• Thus, the von Mises Model is the better model for 6061-T6 Aluminum.