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Answer:
diameter is 14 mm
Explanation:
given data
power = 15 kW
rotation N = 1750 rpm
factor of safety = 3
to find out
minimum diameter
solution
we will apply here power formula to find T that is
power = 2π×N×T / 60 .................1
put here value
15 × = 2π×1750×T / 60
so
T = 81.84 Nm
and
torsion = T / Z ..........2
here Z is section modulus i.e = πd³/ 16
so from equation 2
torsion = 81.84 / πd³/ 16
so torsion = 416.75 / / d³ .................3
so from shear stress theory
torsion = σy / factor of safety
so here σy = 530 for 1020 steel
so
torsion = σy / factor of safety
416.75 / d³ = 530 × / 3
so d = 0.0133 m
so diameter is 14 mm
Answer:
Machine Safeguards must meet these minimum general requirements: Prevent contact: The safeguard must prevent hands, arms, or any other part of a worker's body from making contact with dangerous moving parts. Be secure: Workers should not be able to easily remove or tamper with the safeguard.
Answer:
endurance length is 236.64 MPa
Explanation:
data given:
d = 37.5 mm
Sut = 760MPa
endurance limit is
Se = 0.5 Sut
= 0.5*760 = 380 MPa
surface factor is
Ka = a*Sut^b
where
Sut is ultimate strength
for AISI 1040 STEEL
a = 4.51, b = -0.265
Ka = 4.51*380^{-0.265}
Ka = 0.93
size factor is given as
Kb =1.29 d^{-0.17}
Kb = 0.669
Se = Sut *Ka*Kb
= 380*0.669*0.93
Se = 236.64 MPa
therefore endurance length is 236.64 MPa
Explanation:
150 divide by 150 and that how you do the is you what to divide together 15/ 150 you welcome have a good day is you need something else