Question

Given the bar & loading of problem [1]: estimate & explain.

(a) if f ty of the part is 20 ksi, is this maximum stress valid?
(b) if fty of the part is 60 ksi, is this maximum stress valid?

Answer 1

The maximum stress of 6.00 × 10²6N/m² is valid when compared to the given tensile yield strengths of 20 ksi and 60 ksi, as it is lower than both converted values of the yield strength in N/m².

Step-by-step explanation:

The question refers to evaluating if the maximum stress experienced by a component is valid given its tensile yield strength (f ty). The maximum stress value mentioned is 6.00 × 106N/m², and it is important to compare this with the provided yield strengths of 20 ksi and 60 ksi.

One needs to convert ksi to the same units as the maximum stress (N/m²) to make a valid comparison. The conversion factor is 1 ksi = 6.895 x 106 N/m². Thus, 20 ksi = 137.9 × 106 N/m² and 60 ksi = 413.7 × 106 N/m².

(a) If f ty of the part is 20 ksi (137.9 × 106 N/m²), the maximum stress of 6.00 × 106N/m² is valid because it is less than the tensile yield strength.
(b) If f ty of the part is 60 ksi (413.7 × 106 N/m²), the maximum stress is still valid for the same reason.