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1m
3m
300 kgf/m

1m
3m
300 kgf/m
Problema 707

L1
L2

L1
L2
Método Tres momentos

2M2L1+L2+M3L2+6A1a1L1+6A2b2L2=0
Ecuación 1
A1a1L1=qL324
2M24+3M3+630024+6300*3324=08M2+3M3=-2100 ……..1

L2

L3

L2

L3

Ecuación 2
M2L2+2M3L2+6A2a2L2=0
A2a2L2=qL324
3M2+6M3+6300*3324=0
3M2+6M3=-2025 …..2
Por suma y resta resolvemos el sistema de ecuaciones-2*(8M2+3M3=-2100)
3M2+6M3=-2025
--------------------------------------------------
-16M2-6M3=4200
3M2+6M3=-2025-------------------------------------------------------
-13M2=2175
M2=-217513
De ……..1 sacamos el valor de M3
8-217513+3M3=-2100
M3= -330013
De L1 se hace sumatoria demomentos a la izquierda
Mizq=M2
-217513=RA-qL22
RA= -217513+150
RA= -22513
Mizq=M3
-330013=RAL1+L2+RBL2-qL22
-330013=4-22513+RB3-300*162
RB=960013
Fy=0
Fy=RA+RB+RC-300*4=0-22513+960013+RC-1200=0
RC=622513

1m
3m

1m
3m

1m
3m

1m
3m
Rotación en el apoyo simple en x= 0


θA=θ'A+θ''A+θ'''A
θA=-qL36-22513422+960013322
θA=-300*436-180013+4320013
θA=-20013Método Cross
FD | 1 | 3/4 | 1/4 | 0 |
MEP | 25 | -25 | 225 | -225 |
1 D | -25 | -150 | -50 | 0 |
TRANS | -150/2 | -25/2 | 0 | -25 |
2 D | 150/2 | 75/8 | 25/8 | 0 |TRANS | 75/16 | 150/4 | 0 | 25/16 |
3 D | -75/16 | -225/8 | -75/8 | 0 |
TRANS | -225/16 | -75/32 | 0 | -75/16 |
4 D | 225/16 | 225/128 | 75/128 | 0 |
TRANS | 225/256 |225/32 | 0 | 75/256 |
5 D | -225/256 | -675/128 | -225/128 | 0 |
TRANS | -675/256 | -225/512 | 0 | -225/256 |
6 D | 675/256 | 675/2048 | 225/2048 | 0 |
TRANS | 675/4096 |675/512 | 0 | 225/1096 |
7 D | -675/4096 | -2025/2048 | -675/2048 | 0 |
SUMATORIA | 0 | -167.358 | +167.358 | -253.656 |

MIzq.=-167.358
RA1-qL22=-167.358
RA1-3002=-167.358...