Answer:
1.435
Explanation:
The stress (σ) on an object is equivalent to the ratio of pressure applied (P) and cross sectional area (A). Thus:
σ = P/A = P/(πd²/4) = 4P/πd²
The highest stress on the cylindrical part is:
σ[tex]_{max}[/tex] = 4(1+0.1)P/π[(1-0.05)d]² = 1.22*4P/πd² = 1.22σ
Similarly, the minimum strength can be estimated as:
[tex]S_{min} = (1-0.15)S[/tex] = 0.85*S
Failure occurs when the maximum stress is equivalent to the minimum strength. Therefore:
0.85*S = 1.22σ
The design factor = S/σ = 1.22/0.85 = 1.435
