Respuesta :
Answer:
[tex]\tau_a = 22.7 ksi[/tex]
Explanation:
Given data:
[tex]\alpha = 12.2 \times 10^{-6} [/tex]degree F
Diameter is d = 0.4375
[tex]L_1 = 19 inch[/tex]
[tex]A_1 = 0.50 in^2[/tex]
[tex]L_2 = 27 inch[/tex]
[tex]A_2 = 0.65 in^2[/tex]
force in titanium and bronze will be equal
from equilibrium condition we have
F_T = F_B = p
From the information given as bar is tightened from ends thus net deformation is assummed to be zero
so we have
[tex]\Delta_T +\Delta_B = \Delta = 0[/tex]
[tex]\frac{PL}{AE}_T + \alpha \Delta T L = \frac{PL}{AE}_B + (\alpha \Delta T L)_B[/tex]
[tex]\frac{19\times P}{0.5\times 16500} + 19 \times 5.3\times 10^{-6} \times 40 = \frac{27P}{0.65 \times 15200} + 27\times 12.2\times 10^{-6} \times 40 = 0 [/tex]
solving for P we get
P = 341 kips
average shear at B
[tex]\tau_a = \frac{P}{A}[/tex]
[tex]\tau_a = \frac{341}{\frac{\pi}{4} \times 0.4375^2}[/tex]
[tex]\tau_a = 22.7 ksi[/tex]
