Vigas Con Cargas Triangulares Resuelto #1
7
1.00m 189kg/ 3.00m 3.00m 3.00m 3.00m CALCULO DE REACCIONES. 1- + ∑ FH =0 ;EH =0 2- +∑ FV =0 ; AV + EV −208.5−150−180−193.50−75=0 AV+EV = 807Kg ecuación 1 3- + ∑ ME =0 ;−( 75∗2 )−( 193.5 ∗6) −( 180∗5.5) −( 150∗8.5) −( 208.5∗8) + ( 10 AV )=0 AV = 524.40Kg Sustituir en ecuación 1 AV+EV = 807Kg ecuación 1 AV+EV =807 AV = 807 - 524.40 AV = 282.60Kg CALCULO DE CORTANTE Y MOMENTO FLECTOR POR INTEGRACIÓN TRAMO A-B W = 189−50 3 X – 189 = 46.33X – 189 V(X) = ∫ 46.33 X−189 75kg 193.50 60kg/m 208.50 50kg/m 50kg/m 180k 150k
description
ejercicios de vigas con cargas trapezoidales y triangulares por método de integración
Transcript of Vigas Con Cargas Triangulares Resuelto #1
1.00m
189kg/m
3.00m 3.00m 3.00m3.00m
CALCULO DE REACCIONES.1- + ∑ FH=0 ;EH=0
2- +∑ FV=0 ; AV +EV−208.5−150−180−193.50−75=0
AV+EV = 807Kg ecuación 13- + ∑ME=0;−(75∗2 )−(193.5∗6 )−(180∗5.5 )−(150∗8.5 )− (208.5∗8 )+ (10 AV )=0
AV = 524.40KgSustituir en ecuación 1AV+EV = 807Kg ecuación 1AV+EV =807AV = 807 - 524.40AV = 282.60KgCALCULO DE CORTANTE Y MOMENTO FLECTOR POR INTEGRACIÓN TRAMO A-B W = 189−50
3 X – 189 = 46.33X – 189
V(X) = ∫ 46.33 X−189
V(X) = 46.33 X22
-189X +524.40V(0) = 524.40V(3) = 165.89
75kg193.50kg 60kg/208.50kg/m 50kg/50kg/m180kg150kg
M(X) = ∫ 46.33 X22
-189X +524.40M(X) = 46.33 X3
3 - 189 X2
2 + 524.40X
M(0) = 0 M(3) = 1139.67TRAMO B-C W = 60−189
3 X – 60 = - 43X – 60
V(X) = ∫−43 X−60
V(X) = - 43 X22
−¿60X + 165.89V( 3) = -207.61M(X) = ∫−43 X2
2−¿¿60X + 165.89
M(X) =- 43 X33
- 60 X22
+ 165.89X + 1139.67M(3) = 980.34TRAMO C-DW = 0V(X) = -207.61V(1) = -207.61M(X) =∫−207.61
M(X) = -207.61X +980.34M(1) =772.73TRAMO D-EW = 50 X
3−¿ 50
V(X) = ∫16.67 X – 50
V(X) = 8.34 X2– 50X -207.61V(3) = - 282.55 V(3) = - 282.55 + 282.60V(3) =0.00M(X) = ∫8.34 X2- 50X – 207.61 M(X) = 8.34 X3
3 - 50 X2
2 – 207.61X + 772.73
M(3) = 0.00CALCULO DE MOMENTO MAXIMOV (X) −43 X2
2 – 60X +165.89 = 0
- 81.50X2 +165.89 -X2 165.89−81.50
X = √ 165.89−81.50
X = 1.44M(X) =- 43 X3
3 - 60 X2
2 + 165.89X + 1139.67
M (1.44) = 1273.54
INERCIA
TC:? Tt : ? (máx.)Caculo de áreas A1 = 360.00 cm2A2 = 312. 00 cm2A3 = 360.00 cm2.∑ A=1032
D1 = 44 cmD2 = 25 cmD3 = 6 cmEN = A1∗D 1+A 2∗D 2+A3∗D 3∑ A
EN = (360∗44 )+ (312∗25 )+(360∗6)1032
EN= 25cm2
D3=6 D2=25
D1=44 A12cm26cm12cm
18cm18cm 12cmA1
A3
M máx.= ( kgcm )∗Ψ t ;cI
T =( kgcm 2 )I1=( 30∗12312 ) = (360*192) = 134280.00cm4I2=( 12∗26312 ) = (312*02) = 17576.00cm4I3=( 30∗12312 ) = (360*192) = 134280.00cm4. ∑ ¿286136.00 cm4
Ttracción = (1273.54kg /m )∗(100cm )∗(25cm)286136cm4
=11.13 kgcm2
Tcompresión = (1273.54kg /m )∗(100cm )∗(25cm)286136cm4
=¿11.13 kgcm2
Ψt =
Ψc = T
C
189kg/m
3.00m 3.00m 1.00m 3.00m
524.4
165.89
207.61 207.61 582.60
772.7
L||
60kg/50kg/ 50kg/
