Question
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.)?
Solve problem with AI

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

$$\sigma=\dfrac{F}{A}$$....(I)

$$\sigma=\dfrac{E\Delta l}{l}$$....(II)

From equation (I) and (II)

$$\dfrac{F}{A}=\dfrac{E\Delta l}{l}$$

$$A=\dfrac{Fl}{E\Delta l}$$

Put the value into the formula

$$A=\dfrac{6640\times380\times10^{-3}}{110\times10^{9}\times0.50\times10^{-3}}$$

$$A=4.587\times10^{-5}\ m^2$$

We need to calculate the diameter

$$A=\dfrac{\pi d^2}{4}$$

$$4.587\times10^{-5}=\dfrac{\pi d^2}{4}$$

$$d=\sqrt{\dfrac{4\times A}{\pi}}$$

$$d=\sqrt{\dfrac{4\times4.587\times10^{-5}}{\pi}}$$

$$d=7.64\ mm$$

Hence, The diameter is 7.64 mm.

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