PROBLEMA. RESOLVER LA SIGUIENTE VIGA POR EL METODO DE CORTES, Y POR EL METODO DE NEWTON RAPHSON CONOCER LAS DISTANCIAS EXACTAS DONDE EL MOMENTO FLECTOR DE LA VIGA ES IGUA A 0, SIN TENER EN CUANTA LOS EXTREMOS.
โ ๐น๐ = ๐
1 + ๐
2 โ 2(300) โ 480 โ ๐๐
1 = 300(1.5) + ๐
2(8) โ 480(4) โ 300(9.5) R2= 540 KN R1= 540 KN
ECUACIONES DE MOMENTO 0<X<3 = -50X2
3<X<7=-5X3+45X2+105X-1035
7<X<11= 5X3-165X2+1575X-4465
11<X<14= -50X2+1400X+9800
DIAGRAMA DE MOMENTOS
PROGRAMACION EN EXCEL RAIZ 1 F(x)= -5x^3+45x^2+105x-1035 Fยด(x)ยด= -15x^2+90x+105 ITERACION Xi F(xi) Fยด(Xi) Xi+1 Ea Et 1 4 -215 225 4.955555556 2 4.95555556 -18.0588203 182.637037 5.054433749 23.1951655 3 5.05443375 -0.29162259 176.6895296 5.056084229 3.36607806 4 5.05608423 -8.3969E-05 176.5877646 5.056084705 0.05762853 5 5.0560847 -7.0486E-12 176.5877353 5.056084705 1.661E-05 6 5.0560847 0 176.5877353 5.056084705 1.4164E-12 7 5.0560847 0 176.5877353 5.056084705 0
0.000005 0.000005 0.000005 0.000005 0.000005 0.000005
CONDICION SIGA SIGA SIGA SIGA SIGA PARE PARE
RAIZ 2 F(x)= 5x^3+165x^2+1575x-4465 F(x)ยด= 15x^2+330x+1575 ITERACION Xi F(xi) Fยด(Xi) Xi+1 Ea Et CONDICION 1 8.5 71.875 -146.25 8.991452991 SIGA 2 8.99145299 -8.46373312 -179.486084 8.944297619 18.5173597 0.000005 SIGA 3 8.94429762 -0.06751824 -176.611316 8.943915321 1.6277371 0.000005 SIGA 4 8.94391532 -4.507E-06 -176.587737 8.943915295 0.01335254 0.000005 SIGA 5 8.9439153 0 -176.587735 8.943915295 8.9151E-07 0.000005 PARE 6 8.9439153 0 -176.587735 8.943915295 0 0.000005 PARE 7 8.9439153 0 -176.587735 8.943915295 0 0.000005 PARE 8 8.9439153 0 -176.587735 8.943915295 0