A part has been tested to have Sut = 530 MPa, f = 0.9, and a fully corrected Se = 210 MPa. The design requirements call for the
part to be loaded at three different fully-reversed loads and cycled at each one for a set number of cycles. First, it will be loaded at ±350 MPa for 5,000 cycles. Then, it will be loaded at ±260 MPa for 50,000 cycles. Finally, it will be loaded at ±225 MPa until it fails. How many cycles do we expect the part to last at the final loading? Use Miner's method. :g
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Answer:
Δx = 25 ft.
Explanation:
Assuming that the person on the ground starts running at the same time as the egg is dropped, we have two simultaneous trajectories:
1 ) Egg falling:
If the egg is dropped, and we neglect the air resistance, we can use the kinematic equation that relates the distance and fall time, as follows:
yf-y₀ = 1/2* g* t²
If we take the up direction as positive, we can solve for t as follows:
0-100 ft = 1/2* (-32.15 ft/s²)* t²
⇒ t = = 2.5 sec.
2) Person on the ground running away:
In order to be able to run away, and then return to catch the egg, running at constant speed, he must run during exactly the half of the time that the egg is falling, i.e., 1.25 sec.
We can get the distance at which he can reach, applying the definition of velocity:
v = (xf-x₀) / (tfi-t₀)
If we choose t₀=0 and x₀ = 0 , we can solve for xf, as follows:
xf = v*t = 20 ft/sec*1.25 sec = 25 ft.