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nadya68 [22]
2 years ago
9

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.)?
Physics
2 answers:
Ksenya-84 [330]2 years ago
5 0

Answer:

d=7.32\ mm

Explanation:

Given:

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

We know,

Stress:

\sigma=\frac{F}{A}

where: A = cross sectional area

Strain:

\epsilon = \frac{\Delta l}{l}

& by <u>Hooke's Law within the elastic limits:</u>

E=\frac{\sigma}{\epsilon}

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

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

where: d = diameter of the copper rod

d=7.32\ mm

Fynjy0 [20]2 years ago
4 0

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

\Delta L=\dfrac{PL}{AE}

A=\dfrac{PL}{\Delta LE}

A=\dfrac{6640\times 370}{0.53\times 110\times 10^3}

A= 42.14 mm²

πr² = 42.14 mm²

r=3.66 mm

diameter ,d= 2r

d= 7.32 mm

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