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A cylinder of aluminum-magnesium alloy 0.5 m long is subjected to an elastic tensile stress of 10.2 MPa. The measured elastic elongation is 0.34 mm. Determine the Young’s modulus of this material in MPa. (answer format X)

Respuesta :

Answer:

E= 15 GPa.

Explanation:

Given that

Length ,L = 0.5 m

Tensile stress ,σ = 10.2 MPa

Elongation ,ΔL = 0.34 mm

lets take young modulus  = E

We know that strain ε given as

[tex]\varepsilon =\dfrac{\Delta L}{L}[/tex]

[tex]\varepsilon =\dfrac{0.34}{0.5\times 1000}[/tex]

[tex]\varepsilon =0.00068[/tex]

We know that

[tex]\sigma = \varepsilon E\\\\E=\dfrac{10.2}{0.00068}\\E= 15000\ MPa\\E=15\ GPa[/tex]

Therefore the young's modulus will be 15 GPa.

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