Respuesta :
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
[tex]A=\dfrac{P}{\sigma}[/tex]
[tex]d=\sqrt{\dfrac{4P}{\pi \sigma}}[/tex]
By putting the values
[tex]d=\sqrt{\dfrac{4\times 15000}{\pi \times 150}}[/tex]
d= 11.28 mm
Case II:
According to elongation ,ΔL = 25 mm
[tex]\Delta L=\dfrac{PL}{AE}[/tex]
[tex]A=\dfrac{PL}{E\Delta L}[/tex]
[tex]A=\dfrac{4PL}{\pi E\Delta L}[/tex]
[tex]d=\sqrt{\dfrac{4\times 15000\times 50000}{\pi \times 200\times 1000\times 25}}[/tex]
d=13.81 mm
Therefore the answer will be 13.81 mm .Because it satisfy both the conditions.
The diameter is 13.81 mm
Details required to determine the diameter:
A 15-kN tensile load will be applied to a 50-m length of steel wire with E = 200 GPa.
Calculation of the diameter:
As per the Stress ,σ = 150 MPa
P = σ A
[tex]d = \sqrt{\frac{4\times 15000}{\pi \times 150} }[/tex]
= 11.28mm
Now
According to elongation ,ΔL = 25 mm
[tex]= \sqrt{\frac{4\times 15000 \times 50000}{\pi \times 200\times 1000 \times 25} }[/tex]
= 13.81 mm
For determining the wire of the smallest diameter, the above formulas should be used.
Therefore, we can conclude that the diameter is 13.81 mm.
Find out more information about the wire here:
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