Tensile Area
A threaded fastener with a joint coefficient of 0.5 is put under fatigue loading via an external force of 15 N which produces an alternating stress of 4 MPa. What is the tensile-stress area in 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.
$$$4\times 10^{6}\frac{N}{m^2}=\frac{(0.5)(15N)}{2A_t}$$$
$$$A_t=\frac{(7.5N)m^2}{8\times 10^{6}N}=9.375\times 10^{-7}m^2=0.94\:mm^2$$$
$$$0.94\:mm^2$$$
Time Analysis
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