Fatigue Loading
With a joint coefficient of 0.25, a threaded fastener is put under fatigue loading via an external force of 10 N. Calculate the alternating stress (MPa) if the tensile-stress area is 0.5 mm^2.
Expand Hint
Fatigue loading is the observed material changes under stress due to cyclic loading (which produces a range of stress levels). The mean stress is the arithmetic mean of both the max and min stresses. The alternating stress is the difference between the peak stresses and the mean stress.
Hint 2
If an externally applied load varies between zero and
$$P$$
, the alternating stress is:
$$$\sigma_a=\frac{CP}{2A_t}$$$
where
$$C$$
is the joint coefficient,
$$P$$
is the externally applied load, and
$$A_t$$
is the tensile-stress area.
Fatigue loading is the observed material changes under stress due to cyclic loading (which produces a range of stress levels). The mean stress is the arithmetic mean of both the max and min stresses. The alternating stress is the difference between the peak stresses and the mean stress.
If an externally applied load varies between zero and
$$P$$
, the alternating stress is:
$$$\sigma_a=\frac{CP}{2A_t}$$$
where
$$C$$
is the joint coefficient,
$$P$$
is the externally applied load, and
$$A_t$$
is the tensile-stress area.
$$$\sigma_a=\frac{(0.25)(10N)}{2(0.5mm^2)}=\frac{2.5N}{1\cdot 10^{-6}m^2}=2.5\:MPa$$$
2.5 MPa
Time Analysis
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