Question

For a rod of annealed AISI 1018 steel with a cross sectional area of 0.65 in^2?; what is the maximum tensile load Pmax that should be placed on it given a design factor of 3 to avoid yielding?

Answer 1

Answer:

maximum tensile load Pmax is 11.91 ksi

Explanation:

given data

area = 0.65 in²

design factor of safety = 3

to find out

what is the maximum tensile load Pmax

solution

we know here area is 0.65 in² and FOS = 3

so by steel table for rod of annealed AISI 1018 steel table we know σy = 55 ksi

so

we use here design factor formula that is

$\frac{ \sigma y}{FOS} = \frac{Pmax}{area}$  .............1

put here all these value we get Pmax in equation 1

$\frac{55}{3} = \frac{Pmax}{0.65}$

Pmax = 11.91 ksi

so maximum tensile load Pmax is 11.91 ksi