Answer:
a) 2.452
b) 1.256
Explanation:
Stress due to dead weight. = 14 Ksi
Stress due to fully loaded tractor-trailer = 45Ksi
ultimate tensile strength of beam = 76 Ksi
yield strength = 50 Ksi
endurance limit = 38 Ksi
Determine the safety factor for an infinite fatigue life
a) If mean stress on fatigue strength is ignored
β = ( 45 - 14 ) / 2
= 15.5 Ksi
hence FOS ( factor of safety ) = endurance limit / β
= 38 / 15.5 = 2.452
b) When mean stress on fatigue strength is considered
β2 = 45 + 14 / 2
= 29.5 Ksi
Ratio = β / β2 = 15.5 / 29.5 = 0.5254
Next step: applying Goodman method
Sa = [ ( 0.5254 * 38 *76 ) / ( 0.5254*76 + 38 ) ]
= 19.47 Ksi
hence the FOS ( factor of safety ) = Sa / β
= 19.47 / 15.5 = 1.256
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Answer:
The energy, that is dissipated in the resistor during this time interval is 153.6 mJ
Explanation:
Given;
number of turns, N = 179
radius of the circular coil, r = 3.95 cm = 0.0395 m
resistance, R = 10.1 Ω
time, t = 0.163 s
magnetic field strength, B = 0.573 T
Induced emf is given as;
where;
ΔФ is change in magnetic flux
ΔФ = BA = B x πr²
ΔФ = 0.573 x π(0.0395)² = 0.002809 T.m²
According to ohm's law;
V = IR
I = V / R
I = 3.0848 / 10.1
I = 0.3054 A
Energy = I²Rt
Energy = (0.3054)² x 10.1 x 0.163
Energy = 0.1536 J
Energy = 153.6 mJ
Therefore, the energy, that is dissipated in the resistor during this time interval is 153.6 mJ
Answer:
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