Answer:
maximum length of the specimen before deformation = 200 mm
Explanation:
Hi!
If we have a cylinder with length L₀ , and it is elasticaly deformed ΔL (so the final length is L₀ + ΔL), the strain is defined as:
[tex]\epsilon =\frac{\Delta L}{L_0}[/tex]
And the tensile stress is:
[tex]\sigma = \frac{F}{A}\\F = \text{tensile load}\\A = \text{ cross section area}[/tex]
Elastic modulus E is defined as:
[tex]E = \frac{\sigma}{\epsilon }[/tex]
In this case ΔL = 0.45 mm and we must find maximum L₀. We know that A=π*r², r=(3.3/2) mm. Then:
[tex]\sigma=\frac{2340N}{\pi (1.65 \;mm)^2}=273.68 MPa = [/tex]
[tex]E=123\;GPa=\frac{L_0 \;(273.68MPa)}{0.45\;mm } \\ L_0 = 200\; mm[/tex]
