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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?

Respuesta :

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

[tex]\frac{ \sigma y}{FOS}  = \frac{Pmax}{area}[/tex]  .............1

put here all these value we get Pmax in equation 1

[tex]\frac{55}{3}  = \frac{Pmax}{0.65}[/tex]

Pmax = 11.91 ksi

so maximum tensile load Pmax is 11.91 ksi

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