METODO DE DISTRIBUCION DE MOMENTOS · PDF file3. Determine carry-over factors The carry-over...
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METODO DE DISTRIBUCION DE MOMENTOS
Transcript of METODO DE DISTRIBUCION DE MOMENTOS · PDF file3. Determine carry-over factors The carry-over...
METODO DE DISTRIBUCION DE
MOMENTOS
Rigidez en un extremo apoyado de una barra.
MA
Rigidez = KAB = MA / fA
Flexibilidad = 1/KAB = fA / MA
fA =0= MA ·L ·L/2EIz -
RA·L3/3EIz
dva = 0
dvb = 0
dhb = 0
f Flb = 0
B
fA =MA·L/EIz - RA·L2/2EIz
MA·3/2·L = RA
KAB = MA / fA= 4·E·Iz /
L
fA =MA·L/EIz – 3/2·MA·L/2EIz
Coeficiente de transmisión.
MA
KAB = MA / fA=
4·E·Iz/L
B
dva = 0
dvb = 0
dhb = 0
f Flb = 0
MB MA B
fA = MA·L/3EIz - MB·L/6EIz
fB = 0 = - MBL/3EIz + MAL/6EIz
MA =
2·MB
=>
CtAB = MB/MA= 1/2
dva = 0
dvb = 0
dhb = 0
fA = MA·L/3EIz
CtAB = MB/MA= 0
MA B
KAB = MA / fA= 3·E·Iz/L = 0,75·
4·E·Iz/L
Coeficiente de transmisión.
Coeficiente de transmisión.
MA B
CtAB = MB/MA= 0
KAB = MA / fA= 3·E·Iz/L
MA B
KAB = MA / fA= 4·E·Iz/L
CtAB = MB/MA= 1/2
MA B
CtAB = MB/MA= 0
KAB = MA / fA= 0
Rigidez de un nudo. Coeficientes de reparto o
factores de distribución.
C
B
D
E
MA= MAB + MAC + MAD + MAE
MAB
MB
MB
MAC
MAD
MAE
MA
MAB
MAC
MAD
MAE
MA
Rigidez de un nudo. Coeficientes de reparto o
factores de distribución.
MAC C
CtAC = MC/MAC= 0
KAC = MAC/ fA=
3·E·Iz/L
MAB B
KAB = MAB / fA=
4·E·Iz/L
CtAB = MB/MAB= 1/2
MAD D
CtAD= MD/MAD= 0
KAD = MAD/ fA= 0
MAE E
CtAE = ME/MAE= 0
KAE = MAE/ fA= 3·E·Iz/L
Rigidez de un nudo. Coeficientes de reparto o
factores de distribución.
C
B
D
E
MA
CtAC = MC/MAC= 0
KAC = MAC / fA=
3·E·Iz/L
KAB = MAB / fA=
4·E·Iz/L CtAB = MB/MAB= 1/2
CtAD = MD/MAD= 0
KAD = MAD / fA=
0
CtAE = MC/MAE= 0
KAE = MAE / fA= 3·E·Iz/L
KA = KAB + KAC + KAD + KAE
MA= MAB + MAC + MAD + MAE
KA = MA / fA = 4·E·Iz/L + 3·E·Iz/L + 0 + 3·E·Iz/L = 10·E·Iz/L
= (4/10)·KA
= (3/10)·KA
= (0/10)·KA
= (3/10)·KA
MAB= (4/10)·MA MB = (2/10)·MA
MAC= (3/10)·MA
MAD= (0/10)·MA
MAE= (3/10)·MA
Momentos de empotramiento perfecto (no
admiten giro)
B A
L MA MB
fB = 0 =q·L3/24EIz - MBL/3EIz - MAL/6EIz
| MB | = | MA | = M =>
q·L3/24EIz = M·L/2EIz
M = q·L2/12
MA = + q·L2/12 MB = - q·L2/12
Momentos de empotramiento perfecto (no
admiten giro)
dva = 0
dvb = 0
dhb = 0
f Flb = 0
MB B P
fA = 0 = (P·L/2·L/2·1/2·(2/3·L/2+L/2)-RA·L·L·1/2·2/3·L)/EIz = (5/48·P·L3- 1/3·RAL3)/EIz
5/16·P = RA MB = -1/2·P·L + RAL = -3/16·P·L
MA = 0
B P A
b·RB
a·RA
Momentos de empotramiento perfecto (no
admiten giro)
fB = 0 = (RA·a2/2·(b+1/3·a) + RB·b3/3 - MBL2/6 - MAL2/3)/EIz
=>
SMA = 0 = MA + M - MB + RB·L
B A
L a b
C
SMC = 0 = MA + M - MB + RB·b - RA·a
fA = 0 = (RB·b2/2·(a+1/3·b) + RA·a3/3 - MAL2/6 - MBL2/3)/EIz
fB = 0 = (RA·a2/2 + RB·b2/2 - MBL/2 - MAL/2)/EIz
MA MB
RA RB RA
a·RA
b·RB R’B R’A
Momentos de empotramiento perfecto (no
admiten giro) Tipo de carga y Ligaduras MA MB
+ q·L2/12 - q·L2/12
0 - q·L2/8
+ q/L2·[L2·1/2·((a+c)2-a2) - 2/3·L·((a+c)3-a3) + 1/4·(a+c)4-a4)] - q/L2·[ 1/3·L·((a+c)3-a3) - 1/4·(a+c)4-a4)]
0 - q/8L2·[a4 -(a+c)4 + 2·L2·c(2·a+c)]
+ q·L2/30 - q·L2/20
0 - q·L2/15
0 - 7·q·L2/120
+ 5/96 · q·L2 - 5/96 · q·L2
ab
c
ab
c
q
q
q
q
Momentos de empotramiento perfecto (no admiten
giro)
Tipo de carga y Ligaduras MA MB
0 - 5/64 · q·L2
+ P·a·b2/L2 - P·b·a2/L2
0 - P·a·b(2·a+b) / 2·L2
+ P·a·(a+c)/L - P·a·(a+c)/L
+ P/L2·(a·b2 - a2·b) - P/L2·(a·b2 - a2·b)
+ M/L3·[a·b·(2a+b) - b3) + M/L3·[a·b·(a+2b) - a3)
BAa b
P
BAa b
P
BAa c
P
b
P
BA
ba
BAa
bP
aP
b
M
Procedimiento
1. Restrain all possible displacements.
2. Calculate Distribution Factors:
The distribution factor DFi of a member connected to any joint J is
where S is the rotational stiffness , and is given by
3. Determine carry-over factors
The carry-over factor to a fixed end is always 0.5, otherwise it is 0.0.
4. Calculate Fixed End Moments. (Table 3.1).
These could be due to in-span loads, temperature variation and/or
• relative displacement between the ends of a member.
5. Do distribution cycles for all joints simultaneously
Each cycle consists of two steps:
1. Distribution of out of balance moments Mo,
2.Calculation of the carry over moment at the far end of each member.
The procedure is stopped when, at all joints, the out of balance moment is a
negligible value. In this case, the joints should be balanced and no carry-over
moments are calculated.
6. Calculate the final moment at either end of each
member.
This is the sum of all moments (including FEM) computed
during the distribution cycles.
Example
Stiffness-Factor Modification
Example
ESTRUCTURAS SIMETRICAS
ESTRUCTURAS SIN
DESPLAZAMIENTO
ESTRUCTURAS CON
DESPLAZAMIENTO
Moment Distribution for frames:
sidesway
