problema 3 inciso 2 3 4 5 faltan imagenes
Post on 17-Feb-2016
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2
Ʈ max ¿ R=¿√(σ ' z−σ ' x2
)2
+Ʈ xz2¿
Ʈ min ¿ R=¿−√(σ ' z−σ ' x2
)2
+Ʈ xz2 ¿
3.
σ ' 3+Rcos (2θ−2α )=σ ' z+σ ' x2
Ddα [Rcos (2θ ) cos (2α )+Rsen (2θ ) sen (2α ) ]= D
dα [ σ ' z+σ 'x2−σ '3]
Para obtener la dirección de los planos principales
−2 Rcos (2θ ) sen (2α )+2 Rsen (2θ ) cos (2α )=0
sen (2θ )cos (2α )=cos (2θ ) sen (2α )
tg (2α )=tg (2θ )=2Ʈ xzσ ' z−σ
'x
α=12arctg ( 2Ʈ xzσ ' z−σ
'x)
4.
2 β=2θ+90
cos (2 β)=cos (2θ+90)
cos (2 β )=cos (2θ )cos (90 )−sen (2θ ) sen (90 )
cos (2 β )=−sen (2θ )( I )
2 β=2θ+90
sen(2 β)=sen (2θ+90)
sen (2β )=sen (2θ ) cos (90 )+cos(2θ)sen(90)
sen (2β )=cos (2θ )(II)
de I y II
tg (2β )= −1tg (2θ )
β=12arctg ( σ
'x−σ
'z
2Ʈ xz )5.
σ 'α+Rcos (180−2α )=σ ' z+σ
'x
2
σ 'α=σ 'z+σ
'x
2−Rcos (2α )
σ 'α=σ 'z+σ
'x
2−√( σ 'z−σ 'x2 )
2
+4Ʈ xz2 .cos (2α )
Ʈ α=Rsen (180−2α )
Ʈ α=Rsen (2α )
Ʈ α=√( σ ' z−σ ' x2 )2
+4Ʈ xz2 . sen (2α )
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