9aex. Failure Examples • Ex. 9a.1 • Ex. 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 S_{u} = 120 MPa. 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, 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 zx plane. The Maximum Principal Stress is: The contributing stresses are: 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. Using the Bending Stress formula (s=Mc/I), the Moment to break the bone is: 
Sol'n: Step 1. The Tresca Criterion is based on the maximum shear stress. The inplane principal stresses are:
Considering all directions, for plane stress, the maximum shear stress is:
The maximum shear stress is outofplane, t_{max} = 6.6 ksi

If the Equivalent Stress s_{o} exceeds the yield stress, the material is deemed to have yielded. Using either equation, solving for the equivalent stress gives: s_{o} = 12.2 ksi
NOTES

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