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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.)?

2 Answers

6 votes

Answer:


d=7.32\ mm

Step-by-step explanation:

Given:

  • Young's modulus,
    E=110* 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=(F)/(A)

where: A = cross sectional area

Strain:


\epsilon = (\Delta l)/(l)

& by Hooke's Law within the elastic limits:


E=(\sigma)/(\epsilon)


\therefore 110* 10^3=(F)/(A)/ (\Delta l)/(l)


\therefore 110* 10^3=(6640* 4)/(\pi.d^2)/ (0.53)/(370)

where: d = diameter of the copper rod


d=7.32\ mm

User Prontto
by
8.9k points
6 votes

Answer:

d= 7.32 mm

Step-by-step 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=(PL)/(AE)


A=(PL)/(\Delta LE)


A=(6640* 370)/(0.53* 110* 10^3)

A= 42.14 mm²

πr² = 42.14 mm²

r=3.66 mm

diameter ,d= 2r

d= 7.32 mm

User Julien Bourdic
by
8.0k points