Ejercicio 6 68.docx

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Ejercicio 6.68 SoluciΓ³n

𝑆𝑒𝑑 = 58 π‘˜π‘π‘ π‘– Utilizando la ecuaciΓ³n 6-70

𝑆′𝑒 = 0.506(76)𝑳𝒏 (1, 0.138) = 38.5 𝑳𝒏(1, 0.138) π‘˜π‘π‘ π‘– Utilizando la tabla 6-10 se obtienen los valores para πΎπ‘Ž : πΎπ‘Ž = (14.5)(76)0.719 Γ— ln(1,0.110) πΎπ‘Ž =0.64Γ— ln(1,0.110) 𝑑𝑒 = 0.370(1.5) = 0.555 in Para hallar el factor de tamaΓ±o 𝐾𝑏 , la ecuaciΓ³n 6-20

𝐾𝑏 = (

0.555 βˆ’0.107 ) 0.3

𝐾𝑏 = 0.935 De la ecuaciΓ³n de MarΓ­n 6-71 𝑆𝑒 = [0.64Γ— ln(1,0.110)] Γ—( 0.935) Γ— [38.5 𝑳𝒏(1, 0.138)] 𝑆𝑒̅ = (0.644)(0.936)(38.5) = 23,2 kpsi 𝐢𝑆𝑒 = √(0.110)2 + (0.138)2 =0.176 𝑆𝑒 = 23,2 𝐿𝑛(1,0.176)π‘˜π‘π‘ π‘–

Utilizando la tabla A-16, 𝑑⁄ = 0 𝐷 3 π‘Žβ„ = (16)⁄ = 0.125 𝐷 1.5 A= 0.80 𝐾𝑑 = 2.20 De las ecuaciones 6-78, 6-79 y la tabla 6-15

𝐾𝑑 =

2.20𝐿𝑛(1,0.10) 5 2(2.20 βˆ’ 1) ⁄76 1+ 2.20 √0.125

= 1.83𝐿𝑛(1,0.10)

De la tabla A-16:

𝑍𝑛𝑒𝑑 = 𝜎 = 𝐾𝑓

πœ‹π΄π·3 𝑀

32

𝑍𝑛𝑒𝑑

=

πœ‹(0.80)(1.5)3 32

= 0.265 𝑖𝑛3

= 1.83𝐿𝑛(1,0.10) (

1.5 0.265

) = 10.4 𝐿𝑛(1,0.10)π‘˜π‘π‘ π‘–

πœŽΜ… = 10.4 π‘˜π‘π‘ π‘– 𝐢𝜎 = 0.10 Utilizando la ecuaciΓ³n 5-43: 𝑧=

𝐿𝑛[(23.2⁄10.4)√(1 + 0.102 )⁄(1 + 0.1762 )] βˆšπΏπ‘›[(1 + 0.1762 )(1 + 0.102 )]

Finalmente de la tabla A-10 𝑃𝑓 = 0.0000415 𝑅 = 1 βˆ’ 𝑃𝑓 = 1 βˆ’ 0.0000415 𝑹 = 𝟎. πŸ—πŸ—πŸ—πŸ—πŸ” Rta.

= βˆ’3.94

Ejercicio 6.69 Repita el problema 6-68, con la barra sometida a un momento torsional completamente reversible de 2000 lbf Β· pulg.

SoluciΓ³n

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