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 370 mm (14.57 in.), what must be the diameter to allow an elongation of 0.53 mm (0.02087 in.)?

Respuesta :

Answer:

d= 7.32 mm

Explanation:

Given that

E= 110 GPa

σ = 240 MPa

P= 6640 N

L= 370 mm

ΔL = 0.53

Area A= πr²

We know that  elongation due to load given as

[tex]\Delta L=\dfrac{PL}{AE}[/tex]

[tex]A=\dfrac{PL}{\Delta LE}[/tex]

[tex]A=\dfrac{6640\times 370}{0.53\times 110\times 10^3}[/tex]

A= 42.14 mm²

πr² = 42.14 mm²

r=3.66 mm

diameter ,d= 2r

d= 7.32 mm

Answer:

[tex]d=7.32\ mm[/tex]

Explanation:

Given:

  • Young's modulus, [tex]E=110\times 10^3\ MPa[/tex]
  • yield strength, [tex]\sigma_y=240\ MPa[/tex]
  • load applied, [tex]F=6640\ N[/tex]
  • initial length of rod, [tex]l=370\ mm[/tex]
  • elongation allowed, [tex]\Delta l=0.53[/tex]

We know,

Stress:

[tex]\sigma=\frac{F}{A}[/tex]

where: A = cross sectional area

Strain:

[tex]\epsilon = \frac{\Delta l}{l}[/tex]

& by Hooke's Law within the elastic limits:

[tex]E=\frac{\sigma}{\epsilon}[/tex]

[tex]\therefore 110\times 10^3=\frac{F}{A}\div \frac{\Delta l}{l}[/tex]

[tex]\therefore 110\times 10^3=\frac{6640\times 4}{\pi.d^2}\div \frac{0.53}{370}[/tex]

where: d = diameter of the copper rod

[tex]d=7.32\ mm[/tex]

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