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