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A 10m long AISI 1040 steel rod is subjected to a tensile load of 15 kN. If the allowable total elongation is not exceeded 8mm and the yield stress is 345 MPa. and a factor of safety of 2.5 based on the yield stress., compute the required rod diameter. The proportional limit of the steel is 240 MPa. Young's modulus of the steel is 207,000MPa.

Respuesta :

Answer:

diameter = 11 mm

Explanation:

given data

long = 10m

tensile load = 15 kN

total elongation = 8mm  

yield stress is 345 MPa

factor of safety = 2.5

proportional limit of the steel = 240 MPa

solution

we get here diameter base on elongation

\delta l = [tex]\frac{PL}{AE}[/tex]   ...............1

put here value and we will get

0.008 = [tex]\frac{15\times 10^3\times 10}{\frac{\pi }{4}d^2\times 207000}[/tex]    

solve it we get

d = 10.74 mm  = 11 mm

and

now we get here allowable diameter that is

[tex]\sigma max = \sigma au[/tex]

[tex]\frac{P}{A} = \frac{\sigma u }{FOS}[/tex]    

[tex]\frac{15\times 10^3\times }{\frac{\pi }{4}d^2} = \frac{345\times 10^6}{2.5}[/tex]  

solve it we get

d = 11.76 mm  = 12 mm

so for safety purpose

diameter = 11 mm

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