ingenieria de materiales

claro 1 12 in modulo elastico 200 Gpa E=200Gpa(1*10ˆ9Pa/1Gpa)
claro2 10 in momento de inercia 0.049 inˆ4 E=2*10ˆ11Pa(1Psi/6.847Pa)
claro3 10 in FD -309.47 lb E=2.9*10ˆ10Psi
claro4 4 in Fc 477.47 lb
I=¼πrˆ4
I=.0490inˆ4
calculo de reacciones

∑MB=0 calculo de fuerzabanda en V
∑MB= Re(24) - 309.47(20)+477.47(10)=0
Re=58.9lb T=6300(12)/475rpm
T=1591.57lb-in
∑Fz FN=T/(D/2)=1591.57(10/2)
Re-Rd+Rc+Rb=0 FB=1.5FN= 1.5(318.3)
58.9-309.47+477.47-Rb=0 F en b V=477.47lb
Rb=226.9lb
calculo de fuerza banda plana

T=6300(3.5)/475rpm
D.C.L T=464.21lb-inFN=T/(D/2)=464.21(6/2)
FB=1.5FN= 2(154.75)
F en b plana=477.47lb



Obtencion de la ecuacion Diferencial de curva elastica
EId2y/dxˆ2=∑Mizq.=58.9x-309.47(x-4)+467.47(x-14)
Eidy/dx=58.9xˆ2/2 - 309.47(x-4)ˆ2/2 + 467.47(x-14)ˆ2/2+c1
Eiy=58.9xˆ3/6 - 309.47(x-4)ˆ3/6 + 467.47(x-14)ˆ3/6+c1x + c2Calculo de constantes C1 , C2
E y=0 x=o
Ei(0)=58.9(0)ˆ3/6 - 309.47((0)-4)ˆ3/6 + 467.47((0)-14)ˆ3/6+c1(0) + c2
C2=0

B y=0 x=24
Ei(0)=58.9(24)ˆ3/6 - 309.47((24)-4)ˆ3/6 + 467.47((24)-14)ˆ3/6+c1(24) + c2
C1=8292.05

obtener pendiente y deformacion a 12m
Eidy/dx=58.9(12)ˆ2/2 -309.47((12)-4)ˆ2/2 + 467.47((12)-14)ˆ2/2+8292.05
x=12 Eidy/dx=2629.81

dy/dx=2629.81/EI = 2629.8lb-inˆ2/(2.9*10ˆ10lb/inˆ2)(.0490inˆ2)
dy/dx= 1.85*10ˆ6mˆ2

deformacion a 12m
Eiy=58.9(12)ˆ3/6 - 309.47((12)-4)ˆ3/6 + 467.47((12)-14)ˆ3/6+8292.05
Eiy = -1152.85y= -1152.85lb/inˆ2/(2.9*10ˆ10lb/inˆ2)(.0490inˆ2)
y = 8*10ˆ-7inˆ2

Calcular localizacion de deformacion maxima

Eidy/dx=58.9xˆ2/2 - 309.47(x-4)ˆ2/2 + 467.47(x-14)ˆ2/2+829.05= 0





claro 1 12 in modulo elastico 200 Gpa E=200Gpa(1*10ˆ9Pa/1Gpa)
claro2 10 in momento deinercia 0.049 inˆ4 E=2*10ˆ11Pa(1Psi/6.847Pa)
claro3 10 in FD -309.47 lb E=2.9*10ˆ10Psi
claro4 4 in Fc 477.47 lb
I=¼πrˆ4
I=.0490inˆ4
calculo de reacciones

∑MB=0 calculo de fuerza banda en V
∑MB= Re(24) - 309.47(20)+477.47(10)=0
Re=58.9lb T=6300(12)/475rpm
T=1591.57lb-in
∑Fz FN=T/(D/2)=1591.57(10/2)Re-Rd+Rc+Rb=0 FB=1.5FN= 1.5(318.3)
58.9-309.47+477.47-Rb=0 F en b V=477.47lb
Rb=226.9lb
calculo de fuerza banda plana

T=6300(3.5)/475rpm
D.C.L T=464.21lb-in
FN=T/(D/2)=464.21(6/2)
FB=1.5FN= 2(154.75)
F en b plana=477.47lb



Obtencion de la ecuacion Diferencial decurva elastica
EId2y/dxˆ2=∑Mizq.=58.9x-309.47(x-4)+467.47(x-14)
Eidy/dx=58.9xˆ2/2 - 309.47(x-4)ˆ2/2 + 467.47(x-14)ˆ2/2+c1
Eiy=58.9xˆ3/6 - 309.47(x-4)ˆ3/6 + 467.47(x-14)ˆ3/6+c1x + c2


Calculo de constantes C1 , C2
E y=0 x=o
Ei(0)=58.9(0)ˆ3/6 - 309.47((0)-4)ˆ3/6 + 467.47((0)-14)ˆ3/6+c1(0) + c2
C2=0B y=0 x=24
Ei(0)=58.9(24)ˆ3/6 - 309.47((24)-4)ˆ3/6 + 467.47((24)-14)ˆ3/6+c1(24) + c2
C1=8292.05

obtener pendiente y deformacion a 12m
Eidy/dx=58.9(12)ˆ2/2 - 309.47((12)-4)ˆ2/2 + 467.47((12)-14)ˆ2/2+8292.05
x=12 Eidy/dx=2629.81

dy/dx=2629.81/EI =...