Do you think that individuals with the same disabilities have the same needs? If your answer is yes, how? /No, why?
1 answer:
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Answer:
B A and C
Explanation:
Given:
Specimen σ σ
A +450 -150
B +300 -300
C +500 -200
Solution:
Compute the mean stress
σ = (σ
+ σ
)/2
σ = (450 + (-150)) / 2
= (450 - 150) / 2
= 300/2
σ = 150 MPa
σ = (300 + (-300))/2
= (300 - 300) / 2
= 0/2
σ = 0 MPa
σ = (500 + (-200))/2
= (500 - 200) / 2
= 300/2
σ = 150 MPa
Compute stress amplitude:
σ = (σ
- σ
)/2
σ = (450 - (-150)) / 2
= (450 + 150) / 2
= 600/2
σ = 300 MPa
σ = (300- (-300)) / 2
= (300 + 300) / 2
= 600/2
σ = 300 MPa
σ = (500 - (-200))/2
= (500 + 200) / 2
= 700 / 2
σ = 350 MPa
From the above results it is concluded that the longest fatigue lifetime is of specimen B because it has the minimum mean stress.
Next, the specimen A has the fatigue lifetime which is shorter than B but longer than specimen C.
In the last comes specimen C which has the shortest fatigue lifetime because it has the higher mean stress and highest stress amplitude.
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.
Answer:
0.05%
Explanation:
From the question, we have;
The yield strength of the mild steel, = 43.5 ksi
Young's modulus of elasticity, ∈ = 29,000 ksi
The total strain, = 0.2% = 0.002
The inelatic strain is given as follows;
=
-
/∈
Therefore, we have;
= 0.002 - 43.5/(29,000) = 0.0005
Therefore, the inelastic strain, = 0.0005 = 0.05%
Taking the inelastic strain as the residual strain, we have;
The residual strain = 0.05%