Universidad Tecnológica de Panamá Facultad de Ingeniería Civil Estructuras III- Asignación I
lobaheridaApuntes29 de Agosto de 2017
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Universidad Tecnológica de Panamá Facultad de Ingeniería Civil Estructuras III- Asignación I
José Perurena 8-902-866
Para la siguiente estructura, que puede estar cargada uniformemente, como también con una carga puntual; inserte los valores de L, I, E, w y a (donde a es la distancia desde el extremo izquiero del claro hasta la posición de la carga puntual) para cada claro y el grado de precisión (n) que requiere para el cálculo de los momentos internos.
LAB ≔ 10 ft
4
IAB ≔ 1 in
EAB ≔ 1
kip
2
in
LBC ≔ 12 ft
4
IBC ≔ 1 in
EBC ≔ 1
kip
2
in
LCD ≔ 10 ft
4
ICD ≔ 1 in
ECD ≔ 1
kip
2
in
wAB ≔ 9
kip ft
PAB ≔ 0 kip
aAB ≔ 0 ft
wBC ≔ 0
kip ft
PBC ≔ 20 kip
aBC ≔ 6 ft
wCD ≔ 9
kip ft
PCD ≔ 0 kip
aCD ≔ 0 ft
[pic 2][pic 3]
n ≔ 0.001 kip ⋅ ft
, .
bAB ≔ LAB − aAB
bBC ≔ LBC − aBC
bCD ≔ LCD − aCD
KAB ≔
EAB ⋅ IAB LAB
KBC ≔
EBC ⋅ IBC LBC
KCD ≔
ECD ⋅ ICD LCD
KB ≔ KAB + KBC KC ≔ KBC + KCD
FDDBA ≔
KAB KB
= 0.545
FDDBC ≔
KBC KB
= 0.455
FDDCB ≔
KBC KC
= 0.455
FDDCD ≔
KCD KC
= 0.545
⎛ wAB ⋅ LAB MEAB ≔ −⎜[pic 4]
2
+ PAB ⋅ bAB
2
⋅ aAB ⎞
⎟ = −75 kip ⋅ ft
⎝ 12
2
LAB ⎠
2
⎛ wAB ⋅ LAB MEBA ≔ ⎜
+ PAB ⋅ aAB
2
⋅ bAB ⎞
⎟ = 75 kip ⋅ ft
⎝ 12
⎛ wBC ⋅ LBC[pic 5]
LAB
2
PBC ⋅ bBC
⎠
⋅ aBC ⎞
MEBC ≔ −⎜ +
⎝ 12
2
LBC
⎟ = −30 kip ⋅ ft
⎠
⎛ wBC ⋅ LBC MECB ≔ ⎜[pic 6]
2
+ PBC ⋅ bBC ⋅ aBC
2
⎞
⎟ = 30 kip ⋅ ft
⎝ 12
⎛ wCD ⋅ LCD[pic 7]
LBC
2
PCD ⋅ bCD
⎠
⋅ aCD ⎞
MECD ≔ −⎜ +
⎝ 12
2
2
LCD
⎟ = −75 kip ⋅ ft
⎠
2
⎛ wCD ⋅ LCD MEDC ≔ ⎜
+ PCD ⋅ bCD ⋅ aCD
2
⎞
⎟ = 75 kip ⋅ ft
⎝ 12
LCD ⎠
Valores iniciales
MAB ≔ 0[pic 8][pic 9]
MBA ≔ 0
MBC ≔ 0
MCB ≔ 0
MCD ≔ 0
MDC ≔ 0
Proceso iterativo para el cálculo de los momentos
MAB ≔ ‖ MDDB ← MEBA + MEBC
‖
- MDDC ← MECB + MECD
- while |MDDB| > n
= −90.882 kip ⋅ ft
- ‖ MDEBA ←−FDDBA ⋅ MDDB
‖ ‖
- ‖ MDEBC ←−FDDBC ⋅ MDDB
- ‖ MTBA ← MDEBA ⋅ 0.5
- ‖ MTBC ← MDEBC ⋅ 0.5
- ‖ MDDC ← MDDC + MTBC
‖ ‖
- ‖ MDECB ←−FDDCB ⋅ MDDC
- ‖ MDECD ←−FDDCD ⋅ MDDC
- ‖ MTCB ← MDECB ⋅ 0.5
- ‖ MTCD ← MDECD ⋅ 0.5
‖ ‖
- ‖ MDDB ← MTCB
- ‖ MDDC ← 0
- ‖ MAB ← MAB + MTBA
‖
- MAB ← MAB + MEAB
MBA ≔ ‖ MDDB ← MEBA + MEBC
‖
- MDDC ← MECB + MECD
- while |MDDB| > n
= 43.235 kip ⋅ ft
- ‖ MDEBA ←−FDDBA ⋅ MDDB[pic 10][pic 11]
‖ ‖
- ‖ MDEBC ←−FDDBC ⋅ MDDB
- ‖ MTBA ← MDEBA ⋅ 0.5
- ‖ MTBC ← MDEBC ⋅ 0.5
- ‖ MDDC ← MDDC + MTBC
‖ ‖
- ‖ MDECB ←−FDDCB ⋅ MDDC
- ‖ MDECD ←−FDDCD ⋅ MDDC
- ‖ MTCB ← MDECB ⋅ 0.5
- ‖ MTCD ← MDECD ⋅ 0.5
‖ ‖
- ‖ MDDB ← MTCB
- ‖ MDDC ← 0
- ‖ MBA ← MBA + MDEBA
‖
- MBA ← MBA + MEBA
MBC ≔ ‖ MDDB ← MEBA + MEBC
‖
- MDDC ← MECB + MECD
- while |MDDB| > n
= −43.235 kip ⋅ ft
- ‖ MDEBA ←−FDDBA ⋅ MDDB
‖ ‖
- ‖ MDEBC ←−FDDBC ⋅ MDDB
- ‖ MTBA ← MDEBA ⋅ 0.5
- ‖ MTBC ← MDEBC ⋅ 0.5
- ‖ MDDC ← MDDC + MTBC
‖ ‖
- ‖ MDECB ←−FDDCB ⋅ MDDC
- ‖ MDECD ←−FDDCD ⋅ MDDC
- ‖ MTCB ← MDECB ⋅ 0.5
- ‖ MTCD ← MDECD ⋅ 0.5
‖ ‖
- ‖ MDDB ← MTCB
- ‖ MDDC ← 0
- ‖ MBC ← MBC + MDEBC + MTCB
‖
- MBC ← MBC + MEBC
MCB ≔ ‖ MDDB ← MEBA + MEBC
‖
- MDDC ← MECB + MECD
- while |MDDB| > n
= 43.235 kip ⋅ ft
- ‖ MDEBA ←−FDDBA ⋅ MDDB[pic 12][pic 13]
‖ ‖
- ‖ MDEBC ←−FDDBC ⋅ MDDB
- ‖ MTBA ← MDEBA ⋅ 0.5
- ‖ MTBC ← MDEBC ⋅ 0.5
- ‖ MDDC ← MDDC + MTBC
‖ ‖
- ‖ MDECB ←−FDDCB ⋅ MDDC
- ‖ MDECD ←−FDDCD ⋅ MDDC
- ‖ MTCB ← MDECB ⋅ 0.5
- ‖ MTCD ← MDECD ⋅ 0.5
‖ ‖
- ‖ MDDB ← MTCB
- ‖ MDDC ← 0
- ‖ MCB ← MCB + MDECB + MTBC
‖
- MCB ← MCB + MECB
MCD ≔ ‖ MDDB ← MEBA + MEBC
...