Respuesta :
Answer:
[tex]E_c=110716.332\ MPa[/tex]
Explanation:
Given:
- young's modulus of aluminium, [tex]E_a=6.9\times 10^4\ MPa[/tex]
- young's modulus of tungsten, [tex]E_t=2.8\times 10^5\ MPa[/tex]
When the rods are joined in series the the force applied on each rod is equal to the end force on the composite series rod.
Now, equating the forces on the two rods:
[tex]F_a=F_t[/tex]
[tex]\sigma_a.A=\sigma_t.A[/tex]
[tex]\sigma_a=\sigma_t=\sigma \rm (say)[/tex]
Now, young's modulus for the composite rod:
[tex]E_c=\frac{\sigma}{\epsilon_c}[/tex]
[tex]E_c=\frac{\sigma}{\frac{\Delta L_c}{2L} }[/tex]
[tex]E_c=\frac{\sigma}{(\frac{\Delta L_a+\Delta L_t}{2L}) }[/tex]
[tex]E_c=\frac{\sigma}{\frac{\epsilon_a}{2}+\frac{\epsilon_t}{2}}[/tex]
[tex]E_c=\frac{2\sigma}{\epsilon_a+\epsilon_t}[/tex]
[tex]E_c=\frac{2\sigma}{\frac{\sigma}{E_a} +\frac{\sigma}{E_t} }[/tex]
[tex]E_c=\frac{2}{\frac{1}{E_a} +\frac{1}{E_t} }[/tex]
[tex]E_c=\frac{2}{\frac{1}{6.9\times 10^5} +\frac{1}{2.8\times 10^5} }[/tex]
[tex]E_c=110716.332\ MPa[/tex]
