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

maximum tensile load Pmax is 11.91 ksi

Step-by-step 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

`( \sigma y)/(FOS)  = (Pmax)/(area)` .............1

put here all these value we get Pmax in equation 1

`(55)/(3)  = (Pmax)/(0.65)`

Pmax = 11.91 ksi

so maximum tensile load Pmax is 11.91 ksi