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Answer:
a) the temperature to which the pin must be cooled for assembly is
b) the radial pressure at room temperature after assembly is
c) the safety factor in the resulting assembly = 6.4
Explanation:
Coefficient of thermal expansion
Yield strength = 400 MPa
Modulus of elasticity (E) = 209 GPa
Room Temperature = 20°C
outer diameter of the collar
inner diameter of the collar
pin diameter =
Clearance c = 0.06 mm
a)
The temperature to which the pin must be cooled for assembly can be calculated by using the formula:
-0.09 =
-0.09 =
=
−0.07523262 =
b)
To determine the radial pressure at room temperature after assembly ;we have:
c) the safety factor of the resulting assembly is calculated as:
safety factor =
safety factor =
safety factor = 6.4
Thus, the safety factor in the resulting assembly = 6.4
Answer:
The stress level at which fracture will occur for a critical internal crack length of 6.2mm is 135.78MPa
Explanation:
Given data;
Let,
critical stress required for initiating crack propagation Cc = 112MPa
plain strain fracture toughness = 27.0MPa
surface length of the crack = a
dimensionless parameter = Y.
Half length of the internal crack, a = length of surface crack/2 = 8.8/2 = 4.4mm = 4.4*10-³m
Also for 6.2mm length of surface crack;
Half length of the internal crack = length of surface crack/2 = 6.2/2 = 3.1mm = 3.1*10-³m
The dimensionless parameter
Cc = Kic/(Y*√pia*a)
Y = Kic/(Cc*√pia*a)
Y = 27/(112*√pia*4.4*10-³)
Y = 2.05
Now,
Cc = Kic/(Y*√pia*a)
Cc = 27/(2.05*√pia*3.1*10-³)
Cc = 135.78MPa
The stress level at which fracture will occur for a critical internal crack length of 6.2mm is 135.78MPa
For more understanding, I have provided an attachment to the solution.
