Answer:
[tex]\dot L_{steel} = 3.448\times 10^{-4}\,\frac{in}{min}[/tex]
Explanation:
The Young's module is:
[tex]E = \frac{\sigma}{\frac{\Delta L}{L_{o}} }[/tex]
[tex]E = \frac{\sigma\cdot L_{o}}{\dot L \cdot \Delta t}[/tex]
Let assume that both specimens have the same geometry and load rate. Then:
[tex]E_{aluminium} \cdot \dot L_{aluminium} = E_{steel} \cdot \dot L_{steel}[/tex]
The displacement rate for steel is:
[tex]\dot L_{steel} = \frac{E_{aluminium}}{E_{steel}}\cdot \dot L_{aluminium}[/tex]
[tex]\dot L_{steel} = \left(\frac{10000\,ksi}{29000\,ksi}\right)\cdot (0.001\,\frac{in}{min} )[/tex]
[tex]\dot L_{steel} = 3.448\times 10^{-4}\,\frac{in}{min}[/tex]
Answer:
vsteel = 0.00034483 inch/min
Explanation:
Given
vAl = 0.001 inch/min
EAl = 10,000 ksi
Esteel = 29,000 ksi
P is the same value for the dogbone specimens
A is the same value for the dogbone specimens
L is the same value for the dogbone specimens
We can apply
ΔL = P*L/(A*E)
⇒ ΔL*E = P*L/A
then
ΔLAl*EAl = ΔLsteel*Esteel
ΔLAl*10,000 ksi = ΔLsteel*29,000 ksi
⇒ ΔLAl = 2.9*ΔLsteel
If ΔLAl = 2.9*ΔLsteel we have
vAl = 2.9*vsteel
0.001 inch/min = 2.9*vsteel
⇒ vsteel = 0.00034483 inch/min
