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Physics
A cylindrical rod of copper (E = 110 GPa, 16 × 106 psi) having a yield strength of 240 MPa (35,000 psi) is to be subjected to a load of 6640 N (1493 lbf). If the length of the rod is 380 mm (14.96 in.), what must be the diameter to allow an elongation of 0.50 mm (0.01969 in.)?
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Answer:

The diameter is 7.64 mm.

Explanation:

Given that,

Force = 6640 N

Length = 380 mm

Elongation = 0.50 mm

Modulus of elasticity =  110 GPa

We need to calculate the area

Using formula of stress

[tex]\sigma=\dfrac{F}{A}[/tex]....(I)

[tex]\sigma=\dfrac{E\Delta l}{l}[/tex]....(II)

From equation (I) and (II)

[tex]\dfrac{F}{A}=\dfrac{E\Delta l}{l}[/tex]

[tex]A=\dfrac{Fl}{E\Delta l}[/tex]

Put the value into the formula

[tex]A=\dfrac{6640\times380\times10^{-3}}{110\times10^{9}\times0.50\times10^{-3}}[/tex]

[tex]A=4.587\times10^{-5}\ m^2[/tex]

We need to calculate the diameter

[tex]A=\dfrac{\pi d^2}{4}[/tex]

[tex]4.587\times10^{-5}=\dfrac{\pi d^2}{4}[/tex]

[tex]d=\sqrt{\dfrac{4\times A}{\pi}}[/tex]

[tex]d=\sqrt{\dfrac{4\times4.587\times10^{-5}}{\pi}}[/tex]

[tex]d=7.64\ mm[/tex]

Hence, The diameter is 7.64 mm.

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