Ejercicios propuestos - Capítulo III

1. J=π2(0.04)4=1.28(10−6)π  m4J = \frac{\pi}{2} (0.04)^4 = 1.28(10^{-6})\pi\;m^4
τA=TcJ=5×103(0.04)1.28(10−6)π=49.7  MPa\tau_A = \frac{Tc}{J} = \frac{5\times 10^3(0.04)}{1.28(10^{-6})\pi} = 49.7\;MPa
τB=TρBJ=5×103(0.02)1.28(10−6)π=24.9  MPa\tau_B = \frac{T\rho_B}{J} = \frac{5\times 10^3(0.02)}{1.28(10^{-6})\pi} = 24.9\;MPa

3. 5.17  kN⋅m5.17\;kN\cdot m

5. τmaˊx=26.7MPa\tau_{máx} = 26.7 MPa

7. a) 8.35 ksi, b) 5.94 ksi

9. J=π2(0.034)=0.405(10−6)π  m4J = \frac{\pi}{2}(0.03^4)= 0.405(10^{-6})\pi\; m^4

ϕAC=−0.00838  rad=−0.480°\phi_{AC} = -0.00838\;rad = -0.480\degree