PREGUNTA 01
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Francisco Platero
PR081209 Mar 24 Jun 2008, 22:07
Xc=A1*X1+A2*X2…..An*Xn/At
Xc=(20)(11)+(20)(7)+(20)(1)/60= 6.33
Yc= A1*Y1+A2*Y2…..An*Yn/At
Yc=(20)(5)+(20)(11)+(20)(15)/60= 10.33
Centroide total: (6.33, 10.33)
secc A X Y Dx Dy
1 20 11 5 4.67 -5.33
2 20 7 11 0.67 0.67
3 20 1 15 -5.33 4.67
I1x= (2)(10) ³/12+((10)(4.67)²)
= 602.85
I2x= (10)(2) ³/12 ((20)(0.67)²)
= 15.64
I3X= (2)(10) ³/12((20)(-5.33)²)
= 734.84
Ixx=602.85+15.64+734.84= 1353.33in a la cuarta.
I1y= (10)(2) ³/12+((20)(-5.33)²)
= 571.51
I2y= (2)(10) ³/12+((20)(0.67)²)
= 175.65
I3y= (10)(2)³/12((20)(4.67)²)
= 442.84
Iyy= 571.51+175.65+442.84= 1190 in a la cuarta.
Xc=(20)(11)+(20)(7)+(20)(1)/60= 6.33
Yc= A1*Y1+A2*Y2…..An*Yn/At
Yc=(20)(5)+(20)(11)+(20)(15)/60= 10.33
Centroide total: (6.33, 10.33)
secc A X Y Dx Dy
1 20 11 5 4.67 -5.33
2 20 7 11 0.67 0.67
3 20 1 15 -5.33 4.67
I1x= (2)(10) ³/12+((10)(4.67)²)
= 602.85
I2x= (10)(2) ³/12 ((20)(0.67)²)
= 15.64
I3X= (2)(10) ³/12((20)(-5.33)²)
= 734.84
Ixx=602.85+15.64+734.84= 1353.33in a la cuarta.
I1y= (10)(2) ³/12+((20)(-5.33)²)
= 571.51
I2y= (2)(10) ³/12+((20)(0.67)²)
= 175.65
I3y= (10)(2)³/12((20)(4.67)²)
= 442.84
Iyy= 571.51+175.65+442.84= 1190 in a la cuarta.
PR081209- Invitado
Correccion de unidades (Francisco platero)
PR081209 Mar 24 Jun 2008, 22:11
Xc=A1*X1+A2*X2…..An*Xn/At
Xc=(20)(11)+(20)(7)+(20)(1)/60= 6.33
Yc= A1*Y1+A2*Y2…..An*Yn/At
Yc=(20)(5)+(20)(11)+(20)(15)/60= 10.33
Centroide total: (6.33, 10.33)
seccion A X Y Dx Dy
1 20 11 5 4.67 -5.33
2 20 7 11 0.67 0.67
3 20 1 15 -5.33 4.67
I1x= (2)(10) ³/12+((10)(4.67)²)
= 602.85
I2x= (10)(2) ³/12 ((20)(0.67)²)
= 15.64
I3X= (2)(10) ³/12((20)(-5.33)²)
= 734.84
Ixx=602.85+15.64+734.84= 1353.33 Cm a la cuarta.
I1y= (10)(2) ³/12+((20)(-5.33)²)
= 571.51
I2y= (2)(10) ³/12+((20)(0.67)²)
= 175.65
I3y= (10)(2)³/12((20)(4.67)²)
= 442.84
Iyy= 571.51+175.65+442.84= 1190 Cm a la cuarta.
Xc=(20)(11)+(20)(7)+(20)(1)/60= 6.33
Yc= A1*Y1+A2*Y2…..An*Yn/At
Yc=(20)(5)+(20)(11)+(20)(15)/60= 10.33
Centroide total: (6.33, 10.33)
seccion A X Y Dx Dy
1 20 11 5 4.67 -5.33
2 20 7 11 0.67 0.67
3 20 1 15 -5.33 4.67
I1x= (2)(10) ³/12+((10)(4.67)²)
= 602.85
I2x= (10)(2) ³/12 ((20)(0.67)²)
= 15.64
I3X= (2)(10) ³/12((20)(-5.33)²)
= 734.84
Ixx=602.85+15.64+734.84= 1353.33 Cm a la cuarta.
I1y= (10)(2) ³/12+((20)(-5.33)²)
= 571.51
I2y= (2)(10) ³/12+((20)(0.67)²)
= 175.65
I3y= (10)(2)³/12((20)(4.67)²)
= 442.84
Iyy= 571.51+175.65+442.84= 1190 Cm a la cuarta.
PR081209- Invitado
respuesta 1
GR040641 Mar 24 Jun 2008, 22:51
Jaime Omar Guzman Ramirez
Xc = A1X1 + A2X2 + A3X3 / A total
Xc = (20cm² )(1cm) + (20cm²)(7cm) + (20cm²)(11cm) / (20 cm2) + (20 cm2) +(20 cm2)
Xc = (20cm³) + (140cm³) + (220cm³) / (60cm²)
Xc = 6.33cm
Yc = A1Y1 + A2Y2 +A3Y3 / A total
Yc = (20cm²)(15cm) + (20cm²)(11) + (20cm²)(5cm) / (60cm²)
Yc = (300cm³) +(220cm³) + (100cm³) / (60cm²)
Yc = 10.33cm
MOMENTO DE INERCIA
Respecto con el eje Xc
I tot = I tot 1 + I tot 2 + I tot 3
I tot 1 = IoX1 + A1d12
I tot 1 = bh3 / 12 + A1d12
I tot 1 = (2cm)(10cm)3 / 12 + (20cm²)(4.67cm)2
I tot 1 = 602.845 cm4
I tot 2 = IoX2+ A2d22
I tot 2 = bh3 / 12 + A2d22
I tot 2 = (10cm)(2cm)3 / 12 + (20cm²)(0.67)2
I tot 2 = 15.645 cm4
I tot 3 = IoX3 + A3d32
I tot 3 = bh3 / 12 + A3d32
I tot 3 = (2cm)(10cm)3 / 12 + (20cm²)(5.33)2
I tot 3 = 734.845 cm4
I tot = I tot 1 + I tot 2 + I tot 3
I tot = 602.845cm4 + 15.645cm4 + 734.845cm4
I tot = 1353.34cm4
Respecto con el eje Yc
I tot = I tot 1 + I tot 2 + I tot 3
I tot 1 = IoY1 + A1d12
I tot 1 = bh3 / 12 + A1d12
I tot 1 = (10cm)(2cm)3 / 12 + (20cm²)(5.33cm)2
I tot 1 = 574.845 cm4
I tot 2 = IoY2+ A2d22
I tot 2 = bh3 / 12 + A2d22
I tot 2 = (2cm)(10cm)3 / 12 + (20cm²)(0.67cm)2
I tot 2 = 175.645 cm4
I tot 3 = IoY3 + A3d32
I tot 3 = bh3 / 12 + A3d32
I tot 3 = (10cm)(2cm)3 / 12 + (20cm²)(4.67cm)2
I tot 3 = 442.845 cm4
I tot = I tot 1 + I tot 2 + I tot 3
I tot = 574.845 cm4 + 175.645 cm4 + 442.845 cm4
I tot = 1193.34cm4
Xc = A1X1 + A2X2 + A3X3 / A total
Xc = (20cm² )(1cm) + (20cm²)(7cm) + (20cm²)(11cm) / (20 cm2) + (20 cm2) +(20 cm2)
Xc = (20cm³) + (140cm³) + (220cm³) / (60cm²)
Xc = 6.33cm
Yc = A1Y1 + A2Y2 +A3Y3 / A total
Yc = (20cm²)(15cm) + (20cm²)(11) + (20cm²)(5cm) / (60cm²)
Yc = (300cm³) +(220cm³) + (100cm³) / (60cm²)
Yc = 10.33cm
MOMENTO DE INERCIA
Respecto con el eje Xc
I tot = I tot 1 + I tot 2 + I tot 3
I tot 1 = IoX1 + A1d12
I tot 1 = bh3 / 12 + A1d12
I tot 1 = (2cm)(10cm)3 / 12 + (20cm²)(4.67cm)2
I tot 1 = 602.845 cm4
I tot 2 = IoX2+ A2d22
I tot 2 = bh3 / 12 + A2d22
I tot 2 = (10cm)(2cm)3 / 12 + (20cm²)(0.67)2
I tot 2 = 15.645 cm4
I tot 3 = IoX3 + A3d32
I tot 3 = bh3 / 12 + A3d32
I tot 3 = (2cm)(10cm)3 / 12 + (20cm²)(5.33)2
I tot 3 = 734.845 cm4
I tot = I tot 1 + I tot 2 + I tot 3
I tot = 602.845cm4 + 15.645cm4 + 734.845cm4
I tot = 1353.34cm4
Respecto con el eje Yc
I tot = I tot 1 + I tot 2 + I tot 3
I tot 1 = IoY1 + A1d12
I tot 1 = bh3 / 12 + A1d12
I tot 1 = (10cm)(2cm)3 / 12 + (20cm²)(5.33cm)2
I tot 1 = 574.845 cm4
I tot 2 = IoY2+ A2d22
I tot 2 = bh3 / 12 + A2d22
I tot 2 = (2cm)(10cm)3 / 12 + (20cm²)(0.67cm)2
I tot 2 = 175.645 cm4
I tot 3 = IoY3 + A3d32
I tot 3 = bh3 / 12 + A3d32
I tot 3 = (10cm)(2cm)3 / 12 + (20cm²)(4.67cm)2
I tot 3 = 442.845 cm4
I tot = I tot 1 + I tot 2 + I tot 3
I tot = 574.845 cm4 + 175.645 cm4 + 442.845 cm4
I tot = 1193.34cm4
GR040641- Invitado
julio cesar rodriguez barbero
RB080774 Mar 24 Jun 2008, 23:09
Aca estan mis respuestas del ejercicio 1
RB080774- Invitado
DANIEL AMAYA
AA070857 Mar 24 Jun 2008, 23:26
DANIEL OCTAVIO AMAYA ARÁUZ AA070857
RESOLUCION PROBLEMA #1
Xc=(A1 X1+A2 X2+A3X3)/Atotal
Xc=((2)(10)(1)+ (8 )(2)(6)+ (2)(12)(11))/60
Xc=(20+96+264)/60
Xc=380/60
Xc=6.333
Yc=(A1 Y1+A2 Y2+A3 Y3)/Atotal
Yc=((2)(10)(15)+ (8 )(2)(11)+ (2)(12)(6))/60
Yc=(300+176+144)/60
Yc=620/60
Yc=10.333
Ix1=〖(b)(h)〗^3/12+ (A)(d)²
Ix1= 〖(2)(10)〗^3/12+ (20)(15-10.33)²
Ix1=602.8446 cm⁴
Ix2=〖(b)(h)〗^3/12+ (A)(d)²
Ix2=〖(8 )(2)〗^3/12+ (16)(11-10.33)²
Ix2=12.516 cm⁴
Ix3=〖(b)(h)〗^3/12+ (A)(d)²
Ix3=〖(2)(12)〗^3/12+ (24)(10.33-6)²
Ix3=737.9736 cm⁴
Ixx= Ix1+Ix2+Ix3
Ixx= 602.8446+12.516+737.9736
Ixx=1353.3342 cm⁴
Iy1=〖(b)(h)〗^3/12+ (A)(d)²
Iy1=〖(10)(2)〗^3/12+ (20)(4.67)²
Iy1= 442.8446
Iy2=〖(b)(h)〗^3/12+ (A)(d)²
Iy2=〖(2)(8 )〗^3/12+ (16)(0.67)²
Iy2=91.887
Iy3=〖(b)(h)〗^3/12+ (A)(d)²
Iy3=〖(12)(2)〗^3/12+ (24)(4.33)²
Iy3=457.973
Iyy= Iy1+Iy2+Iy3
Iyy= 442.845+91.887+457.973
Iyy= 992.7052 cm⁴
RESOLUCION PROBLEMA #1
Xc=(A1 X1+A2 X2+A3X3)/Atotal
Xc=((2)(10)(1)+ (8 )(2)(6)+ (2)(12)(11))/60
Xc=(20+96+264)/60
Xc=380/60
Xc=6.333
Yc=(A1 Y1+A2 Y2+A3 Y3)/Atotal
Yc=((2)(10)(15)+ (8 )(2)(11)+ (2)(12)(6))/60
Yc=(300+176+144)/60
Yc=620/60
Yc=10.333
Ix1=〖(b)(h)〗^3/12+ (A)(d)²
Ix1= 〖(2)(10)〗^3/12+ (20)(15-10.33)²
Ix1=602.8446 cm⁴
Ix2=〖(b)(h)〗^3/12+ (A)(d)²
Ix2=〖(8 )(2)〗^3/12+ (16)(11-10.33)²
Ix2=12.516 cm⁴
Ix3=〖(b)(h)〗^3/12+ (A)(d)²
Ix3=〖(2)(12)〗^3/12+ (24)(10.33-6)²
Ix3=737.9736 cm⁴
Ixx= Ix1+Ix2+Ix3
Ixx= 602.8446+12.516+737.9736
Ixx=1353.3342 cm⁴
Iy1=〖(b)(h)〗^3/12+ (A)(d)²
Iy1=〖(10)(2)〗^3/12+ (20)(4.67)²
Iy1= 442.8446
Iy2=〖(b)(h)〗^3/12+ (A)(d)²
Iy2=〖(2)(8 )〗^3/12+ (16)(0.67)²
Iy2=91.887
Iy3=〖(b)(h)〗^3/12+ (A)(d)²
Iy3=〖(12)(2)〗^3/12+ (24)(4.33)²
Iy3=457.973
Iyy= Iy1+Iy2+Iy3
Iyy= 442.845+91.887+457.973
Iyy= 992.7052 cm⁴
AA070857- Invitado
Oswaldo Antonio Berrios Chavarria
BC060610 Miér 25 Jun 2008, 01:09
Mi Respuesta Problema 1
Primero encontramos los respectivos centroides tanto en "X" como en "Y":
Xc= ((20cm^2)(1cm) + (20cm^2)(7cm) + (20cm^2)(11cm)) / (60cm^2)
Xc= (20cm^3 + 140^3 + 220^3) / 60cm^2
Xc= (380cm^3)/(60cm^2) = 6.33cm
Yc= ((20cm^2)(15cm)+(11cm^2)(7cm)+(20cm^2)(5cm))/(60cm^2)
Yc= (300cm^3 + 77cm^3 + 100cm^3) / 60cm^2
Yc= (620cm^3)/(60cm^2) = 10.33cm
Ahora que conocemos sus centroides encontramos los momentos de inercia de la siguiente manera:
Ixtot= Iototx1 + Iox2 + Iox3
Iox=bh^3/12 +Ad^2
Iox1= bh^3/12 +Ad^2
Iox1= ((2cm)(10cm)^3 / 12) + (20cm^2)(-4.67cm)^2
Iox1= (2000cm^4 / 12) + 436.178cm^4
Iox1= 166.67cm^4 + 436.178cm^4
Iox1= 602.85 cm^4
Iox2= bh^3/12 +Ad^2
Iox2= ((10cm)(2cm)^3 / 12) + (20cm^2)(0.67cm)^2
Iox2= (80cm^4 / 12) + 8.978cm^4
Iox2= 6.67cm^4 + 8.978cm^4
Iox2= 15.65 cm^4
Iox3= bh^3/12 +Ad^2
Iox3= ((2cm)(10cm)^3 / 12) + (20cm^2)(5.33cm)^2
Iox3= (2000cm^4 / 12) + 568.178cm^4
Iox3= 166.67cm^4 + 568.178cm^4
Iox3= 734.85 cm^4
Sacando valor Ixtot nos queda:
Ixtot= Iox1 + Iox2 + Iox3
Ixtot= 602.85cm^4 + 15.65 cm^4 + 734.85 cm^4
Ixtot= 1353.35 cm^4
Ahora vamos con el momento de inercia en "Y":
Iytot= Ioy1 + Ioy2 + Ioy3
Ioy=bh^3/12 +Ad^2
Ioy1= bh^3/12 +Ad^2
Ioy1= ((10cm)(2cm)^3 / 12) + (20cm^2)(5.33cm)^2
Ioy1= (80cm^4 / 12) + 968.178 cm^4
Ioy1= 6.67cm^4 + 568.178cm^4
Ioy1= 574.85 cm^4
Ioy2= bh^3/12 +Ad^2
Ioy2= ((2cm)(10cm)^3 / 12) + (20cm^2)(0.67cm)^2
Ioy2= (2000cm^4 / 12) + 8.978cm^4
Ioy2= 166.67cm^4 + 8.978cm^4
Ioy2= 175.65 cm^4
Ioy3= bh^3/12 +Ad^2
Ioy3= ((10cm)(2cm)^3 / 12) + (20cm^2)(4.67cm)^2
Ioy3= (80cm^4 / 12) + 436.178cm^4
Ioy3= 6.67cm^4 + 436.178cm^4
Ioy3= 442.85 cm^4
Sacando valor Iytot nos queda:
Iytot= Ioy1 + Ioy2 + Ioy3
Iytot= 574.85 cm^4 + 175.65 cm^4 + 442.85 cm^4
Iytot= 1193.35 cm^4
Primero encontramos los respectivos centroides tanto en "X" como en "Y":
Xc= ((20cm^2)(1cm) + (20cm^2)(7cm) + (20cm^2)(11cm)) / (60cm^2)
Xc= (20cm^3 + 140^3 + 220^3) / 60cm^2
Xc= (380cm^3)/(60cm^2) = 6.33cm
Yc= ((20cm^2)(15cm)+(11cm^2)(7cm)+(20cm^2)(5cm))/(60cm^2)
Yc= (300cm^3 + 77cm^3 + 100cm^3) / 60cm^2
Yc= (620cm^3)/(60cm^2) = 10.33cm
Ahora que conocemos sus centroides encontramos los momentos de inercia de la siguiente manera:
Ixtot= Iototx1 + Iox2 + Iox3
Iox=bh^3/12 +Ad^2
Iox1= bh^3/12 +Ad^2
Iox1= ((2cm)(10cm)^3 / 12) + (20cm^2)(-4.67cm)^2
Iox1= (2000cm^4 / 12) + 436.178cm^4
Iox1= 166.67cm^4 + 436.178cm^4
Iox1= 602.85 cm^4
Iox2= bh^3/12 +Ad^2
Iox2= ((10cm)(2cm)^3 / 12) + (20cm^2)(0.67cm)^2
Iox2= (80cm^4 / 12) + 8.978cm^4
Iox2= 6.67cm^4 + 8.978cm^4
Iox2= 15.65 cm^4
Iox3= bh^3/12 +Ad^2
Iox3= ((2cm)(10cm)^3 / 12) + (20cm^2)(5.33cm)^2
Iox3= (2000cm^4 / 12) + 568.178cm^4
Iox3= 166.67cm^4 + 568.178cm^4
Iox3= 734.85 cm^4
Sacando valor Ixtot nos queda:
Ixtot= Iox1 + Iox2 + Iox3
Ixtot= 602.85cm^4 + 15.65 cm^4 + 734.85 cm^4
Ixtot= 1353.35 cm^4
Ahora vamos con el momento de inercia en "Y":
Iytot= Ioy1 + Ioy2 + Ioy3
Ioy=bh^3/12 +Ad^2
Ioy1= bh^3/12 +Ad^2
Ioy1= ((10cm)(2cm)^3 / 12) + (20cm^2)(5.33cm)^2
Ioy1= (80cm^4 / 12) + 968.178 cm^4
Ioy1= 6.67cm^4 + 568.178cm^4
Ioy1= 574.85 cm^4
Ioy2= bh^3/12 +Ad^2
Ioy2= ((2cm)(10cm)^3 / 12) + (20cm^2)(0.67cm)^2
Ioy2= (2000cm^4 / 12) + 8.978cm^4
Ioy2= 166.67cm^4 + 8.978cm^4
Ioy2= 175.65 cm^4
Ioy3= bh^3/12 +Ad^2
Ioy3= ((10cm)(2cm)^3 / 12) + (20cm^2)(4.67cm)^2
Ioy3= (80cm^4 / 12) + 436.178cm^4
Ioy3= 6.67cm^4 + 436.178cm^4
Ioy3= 442.85 cm^4
Sacando valor Iytot nos queda:
Iytot= Ioy1 + Ioy2 + Ioy3
Iytot= 574.85 cm^4 + 175.65 cm^4 + 442.85 cm^4
Iytot= 1193.35 cm^4
BC060610- Invitado
vicente- Invitado
RESPUESTA1
MS080756 Miér 25 Jun 2008, 12:47
JOSÉ FERNANDO MARTÍNEZ SARMIENTO MS080756
PRIMERO DETERMINAMOS LAS AREAS:
AREA1 = B*H = 2CM*10CM = 20CM^2
AREA2 = B*H = 10CM*2CM = 20CM^2
AREA3 = B*H = 2CM*10CM = 20CM^2
AHORA PASAMOS A LOS EJES CENTROIDALES:
Xc = (A1*X1 + A2*X2 + A3*X3)/ ATOTAL
Xc= (20CM^2)(1CM)+ (20CM^2)(7CM) + (20CM^2)(11) / 60 CM^2
Xc = 380 CM^3 /60 CM^2
Xc= 6.33 CM
Yc = (A1*Y1 + A2*Y2 + A3*Y3)/ ATOTAL
Yc = (20CM^2)(15CM) + (20CM^2)(11CM) + (20CM^2)(5CM) / 60CM^2
Yc =620 CM^3 / 60CM^2
Yc =10.33 CM
AHORA PASAMOS A LOS MOMENTOS DE INERCIA:
EN X:
Iox1= (2CM)((10CM)^3)/12) + (20CM^2)(4.67CM)^2
Iox1= 602.84CM^4
Iox2= (10CM)((2CM)^3)/12) + (20CM^2)(0.67CM)^2
Iox2= 15.64CM^4
Iox3= (2cm)((10CM)^3)/12) + (20CM^2)(5.33CM)^2
Iox3= 734.84CM^4
EN Y:
Ioy1= (10CM)((2CM)^3)/12) + (20CM^2)(5.33CM)^2
Ioy1= 574.84CM^4
Ioy2= (((2CM)(10CM)^3)/12) + (20CM^2)(0.67CM)^2
Ioy2= 175.64cm^4
Ioy3= (10CM(2CM^3)/12) + 20CM^2(4.67CM)^2
Ioy3= 442.84 CM^4
AHORA BIEN YA QUE HEMOS OBTENIDO LOS DOFERENTES MOMENTOS DE INERCIA LO QUE HACEMOS ES SUMARLOS TODOS PARA OBTENER LOS RESPECTIVOS MOMENTOS DE INERCIA TOTALES TANTO EN X COMO EN Y.
ITOALX= 602.84CM^4 + 15.64CM^4 + 734.84CM^4
ITOTALX= 1353.3CM^4
ITOTALY= 574.84CM^4 + 175.64CM^4 + 442.84 CM^4
ITOTALY= 1193.3CM^4
PRIMERO DETERMINAMOS LAS AREAS:
AREA1 = B*H = 2CM*10CM = 20CM^2
AREA2 = B*H = 10CM*2CM = 20CM^2
AREA3 = B*H = 2CM*10CM = 20CM^2
AHORA PASAMOS A LOS EJES CENTROIDALES:
Xc = (A1*X1 + A2*X2 + A3*X3)/ ATOTAL
Xc= (20CM^2)(1CM)+ (20CM^2)(7CM) + (20CM^2)(11) / 60 CM^2
Xc = 380 CM^3 /60 CM^2
Xc= 6.33 CM
Yc = (A1*Y1 + A2*Y2 + A3*Y3)/ ATOTAL
Yc = (20CM^2)(15CM) + (20CM^2)(11CM) + (20CM^2)(5CM) / 60CM^2
Yc =620 CM^3 / 60CM^2
Yc =10.33 CM
AHORA PASAMOS A LOS MOMENTOS DE INERCIA:
EN X:
Iox1= (2CM)((10CM)^3)/12) + (20CM^2)(4.67CM)^2
Iox1= 602.84CM^4
Iox2= (10CM)((2CM)^3)/12) + (20CM^2)(0.67CM)^2
Iox2= 15.64CM^4
Iox3= (2cm)((10CM)^3)/12) + (20CM^2)(5.33CM)^2
Iox3= 734.84CM^4
EN Y:
Ioy1= (10CM)((2CM)^3)/12) + (20CM^2)(5.33CM)^2
Ioy1= 574.84CM^4
Ioy2= (((2CM)(10CM)^3)/12) + (20CM^2)(0.67CM)^2
Ioy2= 175.64cm^4
Ioy3= (10CM(2CM^3)/12) + 20CM^2(4.67CM)^2
Ioy3= 442.84 CM^4
AHORA BIEN YA QUE HEMOS OBTENIDO LOS DOFERENTES MOMENTOS DE INERCIA LO QUE HACEMOS ES SUMARLOS TODOS PARA OBTENER LOS RESPECTIVOS MOMENTOS DE INERCIA TOTALES TANTO EN X COMO EN Y.
ITOALX= 602.84CM^4 + 15.64CM^4 + 734.84CM^4
ITOTALX= 1353.3CM^4
ITOTALY= 574.84CM^4 + 175.64CM^4 + 442.84 CM^4
ITOTALY= 1193.3CM^4
MS080756- Invitado
REPUESTA No 1
TR080758 Sáb 19 Jul 2008, 06:44
JORGE ALBERTO TEOS
Xc= (20cm2)(1cm)+ (20cm2)(7cm) + (20cm2)(11) / 60 cm2
Xc = 380 cm3 /60 cm2
Xc= 6.33 cm
Yc = (20cm2)(15cm) + (20cm2)(11cm) + (20cm2)(5cm) / 60cm2
Yc =620 cm3 / 60cm2
Yc =10.33 cm
Con respecto a X
Iox1 = B x H3 / 12 + (20cm2)(4.67cm)2
Iox1 = (2cm) x (10cm)3 /12 + 436.2 cm4
Iox1 = 166.7 cm4 + 436.2 cm4
Iox1 = 602.84 cm4
Iox2 = B x H3 /12 + (20cm2)(0.67cm)2
Iox2 = (10cm) x (2cm)3 /12 + 8.978 cm4
Iox2 = 6.667 cm4 + 8.978 cm4
Iox2 = 15.64 cm4
Iox3 = B x H3 /12 + (20cm2)(5.33cm)2
Iox3 = (2cm) x (10cm)3 / 12 + 568.2 cm4
Iox3 = 166.7cm4 + 568.2 cm4
Iox3 = 734.84 cm4
ITOTX = Iox1+Iox2+Iox3
ITOTX= 602.84 cm4 +15.64 cm4 +734.84 cm4
ITOTX=1353.32
Con respecto a Y
Ioy1 = B3 x H /12 + (20cm2)(5.33cm)2
Ioy1 = (2cm)3 x (10cm) /12 + 568.2 cm4
Ioy1 = 6.667 cm4 + 568.2 cm4
Ioy2 = B3 x H /12 + (20cm2)(0.67cm)2
Ioy2 = (10cm)3 x (2cm) /12 + 8.978 cm4
Ioy2 = 166.7 cm4 + 8.978 cm4
Ioy2 = 175.7 cm4
Ioy3= B3 x H /12 + (20cm2)(4.67cm2)
Ioy3 = (2cm)3 x (10cm) /12 + 436.2 cm4
Ioy3 = 6.667 cm4 + 436.2 cm4
Ioy3 = 442.8 cm4
ITOTY = IoY1+IoY2+IoY3
ITOTY= 574.8 cm4 + 175.7 cm4 + 442.8 cm4
ITOTY= 1193 cm4
Xc= (20cm2)(1cm)+ (20cm2)(7cm) + (20cm2)(11) / 60 cm2
Xc = 380 cm3 /60 cm2
Xc= 6.33 cm
Yc = (20cm2)(15cm) + (20cm2)(11cm) + (20cm2)(5cm) / 60cm2
Yc =620 cm3 / 60cm2
Yc =10.33 cm
Con respecto a X
Iox1 = B x H3 / 12 + (20cm2)(4.67cm)2
Iox1 = (2cm) x (10cm)3 /12 + 436.2 cm4
Iox1 = 166.7 cm4 + 436.2 cm4
Iox1 = 602.84 cm4
Iox2 = B x H3 /12 + (20cm2)(0.67cm)2
Iox2 = (10cm) x (2cm)3 /12 + 8.978 cm4
Iox2 = 6.667 cm4 + 8.978 cm4
Iox2 = 15.64 cm4
Iox3 = B x H3 /12 + (20cm2)(5.33cm)2
Iox3 = (2cm) x (10cm)3 / 12 + 568.2 cm4
Iox3 = 166.7cm4 + 568.2 cm4
Iox3 = 734.84 cm4
ITOTX = Iox1+Iox2+Iox3
ITOTX= 602.84 cm4 +15.64 cm4 +734.84 cm4
ITOTX=1353.32
Con respecto a Y
Ioy1 = B3 x H /12 + (20cm2)(5.33cm)2
Ioy1 = (2cm)3 x (10cm) /12 + 568.2 cm4
Ioy1 = 6.667 cm4 + 568.2 cm4
Ioy2 = B3 x H /12 + (20cm2)(0.67cm)2
Ioy2 = (10cm)3 x (2cm) /12 + 8.978 cm4
Ioy2 = 166.7 cm4 + 8.978 cm4
Ioy2 = 175.7 cm4
Ioy3= B3 x H /12 + (20cm2)(4.67cm2)
Ioy3 = (2cm)3 x (10cm) /12 + 436.2 cm4
Ioy3 = 6.667 cm4 + 436.2 cm4
Ioy3 = 442.8 cm4
ITOTY = IoY1+IoY2+IoY3
ITOTY= 574.8 cm4 + 175.7 cm4 + 442.8 cm4
ITOTY= 1193 cm4
TR080758- Invitado
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