A cylindrical 4340 steel bar is subjected to reversed rotating–bending stress cycling, which yielded the test results presented
1 answer:
Based on the maximum applied load, the factor of safety, and the distance between the loadbearing points, the minimum allowable bar diameter is <u>18.6 mm. </u>
<h3>What is the minimum allowable bar diameter?</h3>The diameter is included in the following formula:
Maximum stress / Factor of safety = (16 x Maximum applied load x distance between loadbearing points x 10⁻¹⁰) / (π x diameter³)
Solving gives:
(490 x 10⁶) / 2.25 = (16 x 5,000 x 55.0 x 10⁻¹⁰) / (π x diameter³)
217,777,777.78 = 0.00044 / (π x diameter³)
diameter = 18.6 mm
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Answer with Explanation:
The general equation of simple harmonic motion is
where,
A is the amplitude of motion
is the angular frequency of the motion
is known as initial phase
part 1)
Now by definition of velocity we have
part 2)
Now by definition of acceleration we have
part 3)
The angular frequency is related to Time period 'T' as
where
is the angular frequency of the motion of the particle.
Part 4) The acceleration and velocities are plotted below
since the maximum value that the sin(x) and cos(x) can achieve in their respective domains equals 1 thus the maximum value of acceleration and velocity is and
respectively.
Answer:
M = 281.25 lb*ft
Explanation:
Given
W<em>man</em> = 150 lb
Weight per linear foot of the boat: q = 3 lb/ft
L = 15.00 m
M<em>max</em> = ?
Initially, we have to calculate the Buoyant Force per linear foot (due to the water exerts a uniform distributed load upward on the bottom of the boat):
∑ Fy = 0 (+↑) ⇒ q'*L - W - q*L = 0
⇒ q' = (W + q*L) / L
⇒ q' = (150 lb + 3 lb/ft*15 ft) / 15 ft
⇒ q' = 13 lb/ft (+↑)
The free body diagram of the boat is shown in the pic.
Then, we apply the following equation
q(x) = (13 - 3) = 10 (+↑)
V(x) = ∫q(x) dx = ∫10 dx = 10x (0 ≤ x ≤ 7.5)
M(x) = ∫10x dx = 5x² (0 ≤ x ≤ 7.5)
The maximum internal bending moment occurs when x = 7.5 ft
then
M(7.5) = 5(7.5)² = 281.25 lb*ft
