Ejercicio 6.68 SoluciΓ³n
ππ’π‘ = 58 πππ π Utilizando la ecuaciΓ³n 6-70
πβ²π = 0.506(76)π³π (1, 0.138) = 38.5 π³π(1, 0.138) πππ π Utilizando la tabla 6-10 se obtienen los valores para πΎπ : πΎπ = (14.5)(76)0.719 Γ ln(1,0.110) πΎπ =0.64Γ ln(1,0.110) ππ = 0.370(1.5) = 0.555 in Para hallar el factor de tamaΓ±o πΎπ , la ecuaciΓ³n 6-20
πΎπ = (
0.555 β0.107 ) 0.3
πΎπ = 0.935 De la ecuaciΓ³n de MarΓn 6-71 ππ = [0.64Γ ln(1,0.110)] Γ( 0.935) Γ [38.5 π³π(1, 0.138)] ππΜ
= (0.644)(0.936)(38.5) = 23,2 kpsi πΆππ = β(0.110)2 + (0.138)2 =0.176 ππ = 23,2 πΏπ(1,0.176)πππ π
Utilizando la tabla A-16, πβ = 0 π· 3 πβ = (16)β = 0.125 π· 1.5 A= 0.80 πΎπ‘ = 2.20 De las ecuaciones 6-78, 6-79 y la tabla 6-15
πΎπ‘ =
2.20πΏπ(1,0.10) 5 2(2.20 β 1) β76 1+ 2.20 β0.125
= 1.83πΏπ(1,0.10)
De la tabla A-16:
ππππ‘ = π = πΎπ
ππ΄π·3 π
32
ππππ‘
=
π(0.80)(1.5)3 32
= 0.265 ππ3
= 1.83πΏπ(1,0.10) (
1.5 0.265
) = 10.4 πΏπ(1,0.10)πππ π
πΜ
= 10.4 πππ π πΆπ = 0.10 Utilizando la ecuaciΓ³n 5-43: π§=
πΏπ[(23.2β10.4)β(1 + 0.102 )β(1 + 0.1762 )] βπΏπ[(1 + 0.1762 )(1 + 0.102 )]
Finalmente de la tabla A-10 ππ = 0.0000415 π
= 1 β ππ = 1 β 0.0000415 πΉ = π. πππππ Rta.
= β3.94
Ejercicio 6.69 Repita el problema 6-68, con la barra sometida a un momento torsional completamente reversible de 2000 lbf Β· pulg.
SoluciΓ³n