Question

A 15-kN tensile load will be applied to a 50-m length of steel wire with E = 200 GPa. Determine the smallest diameter wire that can be used, knowing that the normal stress must not exceed 150 MPa and that the increase in length of the wire must not exceed 25 mm

Answer 1

Answer:

d=13.81 mm

Explanation:

Given that

P = 15 KN  ,L = 50 m

E= 200 GPa

ΔL = 25 mm

σ  = 150 MPa

Lets take d=Diameter

There are we have two criteria to find out the diameter of the wire

Case I :

According to Stress ,σ  = 150 MPa

P = σ  A

$A=\dfrac{P}{\sigma}$

$d=\sqrt{\dfrac{4P}{\pi \sigma}}$

By putting the values

$d=\sqrt{\dfrac{4\times 15000}{\pi \times 150}}$

d= 11.28 mm

Case II:

According to elongation ,ΔL = 25 mm

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

$A=\dfrac{PL}{E\Delta L}$

$A=\dfrac{4PL}{\pi E\Delta L}$

$d=\sqrt{\dfrac{4\times 15000\times 50000}{\pi \times 200\times 1000\times 25}}$

d=13.81 mm

Therefore the answer will be 13.81 mm .Because it satisfy both the conditions.