SOLU1
3
SOLUCION ∆ AM +∆ MN + ∆NP RA ( L 4 ) AE + RA −2 P ( L 4 ) AE + RA −3 P ( L 2 ) AE =∆ RAl 4 + ( Ra−2 p ) L 4 +( RA −3 P) L 2 = ∆E∆ RA 4 + ( Ra−2 p ) 4 + ( RA −3 P 2 ) = ∆E ( L 2 ) ∆ =10 −4 L ∆ L =10 −4 RA 4 + RA −2 P 4 + RA −3 P 2 =25 x 10 −4 x 2 x 10 12 x 10 −4
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Transcript of SOLU1
SOLUCION
∆ AM+∆MN+∆ NPRA ( L4 )AE
+RA−2P( L4 )
AE+RA−3 P( L2 )
AE=∆
RAl4
+(Ra−2 p ) L
4+(RA−3P ) L
2=∆ E∆
RA4
+(Ra−2 p )
4+( RA−3 P2 )=∆ E (L2 )
∆=10−4 L
∆L=10−4
RA4
+ RA−2P4
+RA−3 P2
=25 x10−4 x2 x1012 x10−4
RA=2P+500000
FUERZAS NORMALES
N AM=2 P+500000
NMN=RA−2 P=500000
NMN=RA−3 P=2 P+500000−3P=500000−P
COMO NB ESTA EN COMPRESION
N NB=−(P−500000)
σconpresion=NN baa
4 x106= p−50000025 x 10−4
P=600000N
P=600N
B) Si el espacio A se aumenta 2 veces
∆=3 x10−4 l
∆l=3 x10−4
Reemplazando en (1)
RA4
+ RA−2P4
+RA−3 P2
=25 x10−4 x 5 x1012 x 3x 10−4
ra=2 p+1.5x 106
Fuezasnormales :
N AM=2 P+1.5 x106
NMN=RA−2 P=1.5 x106
N NB=RA−3 P=1.5 x 106−p=−¿
σcompresion=N NA
A
4 X 106=PP−1.5 X 106
25 x10−4
p−1.6 x 106N
P=1600KN