Answer:
79.74*10^6 Pa
Explanation:
Based on the parameters provided, we have:
ε[tex]_{t}[/tex] = ln([tex]l_{f}/l_{i}[/tex])
Where initial gauge length = 3 cm and the final gauge length is 3.5 cm. Therefore:
ε[tex]_{t}[/tex] = ln(3.5/3) = ln(1.167) = 0.154
Similarly,
σ[tex]_{E}[/tex] = F/[3.142*(di^2)/4]
Where σ[tex]_{E}[/tex] = 120*10^6 Pa and di = 1 cm = 0.01 m
Therefore,
F = 120*10^6 * 3.142*(0.01^2)/4 = 9426 N
σ[tex]_{t}[/tex] = F/[3.142*(df^2)/4 = 9426/[3.142*(0.00926^2)/4 = 9426/6.74*10^-5 = 139.95*10^6 Pa
σ[tex]_{t}[/tex] = k*ε[tex]_{t} ^{0.5}[/tex] = 139.95*10^6
k = 139.95*10^6/(0.154)^0.5 = 356.63*10^6 Pa
Therefore, when ε[tex]_{t}[/tex] = 0.05 cm/cm
σ[tex]_{t}[/tex] = 356.63*10^6 (0.05)^0.5 = 79.74*10^6 Pa
