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Elenna [48]
3 years ago
9

A cylinder of aluminum-magnesium alloy 0.5 m long is subjected to an elastic tensile stress of 10.2 MPa. The measured elastic el

ongation is 0.34 mm. Determine the Young’s modulus of this material in MPa. (answer format X)
Engineering
1 answer:
elena55 [62]3 years ago
8 0

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

\varepsilon =\dfrac{\Delta L}{L}

\varepsilon =\dfrac{0.34}{0.5\times 1000}

\varepsilon =0.00068

We know that

\sigma = \varepsilon  E\\\\E=\dfrac{10.2}{0.00068}\\E= 15000\ MPa\\E=15\ GPa

Therefore the young's modulus will be 15 GPa.

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*3–32. The rubber block is subjected to an elongation of 0.03 in. along the x axis, and its vertical faces are given a tilt so t
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