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I want to use a 6061 aluminum alloy for a tie rod with a circular cross section. The yield strength of this alloy is 240 MPa, and the elastic modulus is 70 GPa.
1. If this tie rod has to hold a load of 20,000N, what is the minimum diameter of the tie rod needed to avoid yielding. Enter the answer in meters.

Respuesta :

Answer:

[tex]d \geq 0.0103 m[/tex]

Explanation:

Given data:

yield strength = 240 MPa

E = 70 GPa

P = 20,000 N

we know that

TO prevent yielding yield strength [tex]\leq 240 MPa[/tex]

[tex]\sigma_Y = frac{P}{area\ of\ cross\ section}[/tex]

taking [tex]\sigma_y = 240 MPa[/tex]

[tex]240 \times 10^6 = \frac{P}{\frac[pi}{4} \times d^2}[/tex]

solving for diameter

[tex]d^2 = \frac{20,000 \times 4}{240 \times 10^6 \times \pi}[/tex]

[tex]d^2 \geq 10.3 mm[/tex]

[tex]d \geq 0.0103 m[/tex]

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