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A uniform bar having a machined surface is subjected to an axial load varying from 400 kN to 150 kN. The material of the bar has Su= 630 MPa,kc = 0.7 and kf= 1.42 Using Goodman’s Criteria and factor of safety 1.5, The diameter of the rod is __________ mm, Where, Su = ultimate tensile strength,kc = surface finish factor, kf = fatigue st!ess concentration factor
  • a)
    48
  • b)
    50
Correct answer is option ''. Can you explain this answer?
Verified Answer
A uniform bar having a machined surface is subjected to an axial load...
Using Goodman’s equation with
Kc = 0.7, Kf = 1.42
Su = 630 MPa and Se = 315 MPa
Thus,
∴d = 48.69 mm
d ≈ 49 mm
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Most Upvoted Answer
A uniform bar having a machined surface is subjected to an axial load...
To determine the diameter of the rod using Goodman's Criteria, we can follow these steps:

1. Identify the given values:
- Ultimate tensile strength (Su) = 630 MPa
- Surface finish factor (kc) = 0.7
- Fatigue stress concentration factor (kf) = 1.42

2. Calculate the alternating stress (Sa):
The alternating stress is the difference between the maximum and minimum axial loads applied to the bar.
Sa = (Maximum load - Minimum load) / 2
= (400 kN - 150 kN) / 2
= 250 kN / 2
= 125 kN

3. Calculate the allowable alternating stress (Sa_allow):
The allowable alternating stress is determined using Goodman's Criteria:
Sa_allow = Su / (kf * (1 + kc))
= 630 MPa / (1.42 * (1 + 0.7))
= 630 MPa / (1.42 * 1.7)
= 630 MPa / 2.414
= 260.6 MPa

4. Calculate the diameter of the rod:
The diameter of the rod can be determined using the formula for axial stress:
σ = P / (π * d^2 / 4)
where σ is the axial stress, P is the axial load, and d is the diameter of the rod.

Since we know that the factor of safety (FS) is 1.5, we can write:
Sa_allow / FS = σ_allow = P / (π * d^2 / 4)

Solving for d:
d^2 = 4 * P / (π * σ_allow)
d = √(4 * P / (π * σ_allow))

Substituting the given values:
d = √(4 * 150 kN / (π * 260.6 MPa))
= √(600 kN / (π * 260.6 MPa))
= √(600 kN / (3.14159 * 260.6 N/mm^2))
= √(600 / (3.14159 * 260.6))
= √(600 / 819.948)
= √0.73171
≈ 0.8557

Converting the diameter to mm:
d_mm ≈ 0.8557 * 1000
≈ 855.7 mm

Rounding off to the nearest whole number, the diameter of the rod is approximately 856 mm.
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A uniform bar having a machined surface is subjected to an axial load varying from 400 kN to 150 kN. The material of the bar has Su= 630 MPa,kc = 0.7 and kf= 1.42 Using Goodman’s Criteria and factor of safety 1.5, The diameter of the rod is __________ mm, Where, Su = ultimate tensile strength,kc = surface finish factor, kf = fatigue st!ess concentration factora) 48b) 50Correct answer is option ''. Can you explain this answer?
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