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)

qq

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