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9a-ex. 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 Su = 120
MPa.
During strenuous activity, the femur is subjected to a torque of 100 Nm.
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.
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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
-8m
4
so: t = TRo/J = 45.9 MPa
Reducing the sI equation:
and rearranging:
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For the bone to break, sI = Su = 120 MPa.Sincet = 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 tohave 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.
Sol'n:
Step 1. The Tresca Criterion is based on the maximum shear stress. The in-plane principal
stresses are:
Solving:
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.
Solving:
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:
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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
NOTES
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
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