Exemplo Metodo dos Deslocamentos Pórtico 1

Prévia do material em texto

EXEMPLO 2 - PÓRTICO ENGASTADO
≔b 15 cm ≔l1 5 m ≔P 30 kN
≔h 60 cm ≔l2 6 m ≔q2 20 ――
kN
m
≔q3 10 ――
kN
m
≔E 35 GPa ≔l3 7 m
≔I =――
⋅b h
3
12
270000 cm
4
≔A =⋅b h 900 cm
2
DESLOCABILIDADES
SISTEMA HIPERGEOMÉTRICO
CASO 0
Barra 1 Barra 2
≔β10 =−P −30 kN ≔β20 =――
⋅q2 l2
2
60 kN ≔β30 =―――
⋅q2 l2
2
12
60 ⋅m kN
≔β50 =――
⋅q2 l2
2
60 kN
Barra 3
≔β40 =―――
⋅−q3 l3
2
−35 kN ≔β60 =+―――
⋅−q2 l2
2
12
―――
⋅−q3 l3
2
12
−100.833 ⋅m kN
≔f =
β10
β20
β30
β40
β50
β60
⎡
⎢
⎢
⎢
⎢
⎢
⎢⎣
⎤
⎥
⎥
⎥
⎥
⎥
⎥⎦
−30
60
60 m
−35
60
−100.833 m
⎡
⎢
⎢
⎢
⎢
⎢
⎣
⎤
⎥
⎥
⎥
⎥
⎥
⎦
kN
CASO 1
≔k11 =⋅
⎛
⎜
⎝
+⋅E ―
A
l2
―――
⋅⋅12 E I
l1
3
⎞
⎟
⎠
1 m 534072 kN ≔k21 0 kN ≔k31 =⋅
⎛
⎜
⎝
―――
⋅⋅6 E I
l1
2
⎞
⎟
⎠
1 m 22680 ⋅m kN
≔k41 =⋅
⎛
⎜
⎝
⋅−E ―
A
l2
⎞
⎟
⎠
1 m −525000 kN ≔k51 0 kN ≔k61 ⋅0 kN m
CASO 2
≔k12 0 kN ≔k22 =⋅
⎛
⎜
⎝
+――
⋅E A
l1
―――
⋅⋅12 E I
l2
3
⎞
⎟
⎠
1 m 635250 kN
≔k32 =⋅
⎛
⎜
⎝
―――
⋅⋅6 E I
l2
2
⎞
⎟
⎠
1 m 15750 ⋅m kN ≔k42 0 kN
≔k52 =⋅
⎛
⎜
⎝
――――
⋅⋅−12 E I
l2
3
⎞
⎟
⎠
1 m −5250 kN ≔k62 =⋅―――
⋅⋅6 E I
l2
2
1 m 15750 ⋅m kN
CASO 3
≔k13 =
⎛
⎜
⎝
―――
⋅6 E I
l1
2
⎞
⎟
⎠
22680 kN ≔k23 =―――
⋅6 E I
l2
2
15750 kN
≔k33 =+―――
⋅⋅4 E I
l1
―――
⋅⋅4 E I
l2
138600 ⋅m kN ≔k43 0 kN
≔k53 =―――
⋅⋅−6 E I
l2
2
−15750 kN ≔k63 =―――
⋅⋅2 E I
l2
31500 ⋅m kN
CASO 4
≔k14 =⋅
⎛
⎜
⎝
―――
⋅−E A
l2
⎞
⎟
⎠
1 m −525000 kN ≔k24 0 kN
≔k34 ⋅0 kN m ≔k44 =⋅
⎛
⎜
⎝
+――
⋅E A
l2
―――
⋅⋅12 E I
l3
3
⎞
⎟
⎠
1 m 528306.122 kN
≔k54 0 kN ≔k64 =⋅
⎛
⎜
⎝
―――
⋅⋅6 E I
l3
2
⎞
⎟
⎠
1 m 11571.429 ⋅m kN
CASO 5
≔k15 0 kN ≔k25 =⋅
⎛
⎜
⎝
−―――
⋅⋅12 E I
l2
3
⎞
⎟
⎠
1 m −5250 kN
≔k35 =
⎛
⎜
⎝
−―――
⋅⋅6 E I
l2
2
⎞
⎟
⎠
1 m −15750 ⋅m kN ≔k45 0 kN
≔k55 =
⎛
⎜
⎝
+―――
⋅⋅12 E I
l2
3
――
⋅E A
l3
⎞
⎟
⎠
1 m 455250 kN ≔k65 =
⎛
⎜
⎝
―――
⋅⋅−6 E I
l2
2
⎞
⎟
⎠
1 m −15750 ⋅m kN
CASO 6
≔k16 0 kN ≔k26 =―――
⋅⋅6 E I
l2
2
15750 kN
≔k36 =―――
⋅⋅2 E I
l2
31500 ⋅m kN ≔k46 =―――
⋅⋅6 E I
l3
2
11571.429 kN
≔k56 =―――
⋅⋅−6 E I
l2
2
−15750 kN ≔k66 =+―――
⋅⋅4 E I
l2
―――
⋅⋅4 E I
l3
117000 ⋅m kN
MONTANDO A MATRIZ DE RIGIDEZ
≔k
k11 k12 k13 k14 k15 k16
k21 k22 k23 k24 k25 k26
k31 k32 k33 k34 k35 k36
k41 k42 k43 k44 k45 k46
k51 k52 k53 k54 k55 k56
k61 k62 k63 k64 k65 k66
⎡
⎢
⎢
⎢
⎢
⎢
⎢⎣
⎤
⎥
⎥
⎥
⎥
⎥
⎥⎦
ENCONTRANDO OS DESLOCAMENTOS
＝−f ⋅k d
≔d =⋅k
−1
−f
0.008263
−0.000062
−0.001923
0.008265
−0.00018
0.000546
⎡
⎢
⎢
⎢
⎢
⎢
⎣
⎤
⎥
⎥
⎥
⎥
⎥
⎦
ENCONTRANDO AS REAÇÕES
CASO 0
≔Ma0 ⋅0 kN m ≔Mb0 =―――
⋅q3 l3
2
12
40.833 ⋅m kN
≔Va0 0 kN ≔Vb0 0 kN
≔Ha0 0 kN ≔Hb0 =―――
⋅−q3 l3
2
−35 kN
CASO 1
≔Ma1 =―――
⋅⋅6 E I
l1
2
22680 kN ≔Mb1 0 kN
≔Vb1 0 ――
kN
m≔Va1 0 ――
kN
m
≔Ha1 =――――
⋅⋅−12 E I
l1
3
−9072 ⋅―
1
m
kN ≔Hb1 ⋅0 ――
kN
m
CASO 2
≔Ma2 0 kN ≔Mb2 0 kN
≔Vb2 0 ――
kN
m≔Va2 =―――
⋅−E A
l1
−630000 ⋅―
1
m
kN
≔Hb2 ⋅0 ――
kN
m≔Ha2 0 ――
kN
m
CASO 3
≔Ma3 =―――
⋅⋅2 E I
l1
37800 ⋅m kN ≔Mb3 ⋅0 kN m
≔Vb3 0 kN
≔Va3 0 kN
≔Hb3 ⋅0 kN
≔Ha3 =―――
⋅⋅−6 E I
l1
2
−22680 kN
CASO 4
≔Ma4 0 kN ≔Mb4 =―――
⋅⋅6 E I
l3
2
11571.429 kN
≔Va4 0 ――
kN
m
≔Vb4 0 ――
kN
m
≔Ha4 ⋅0 ――
kN
m
≔Hb4 =――――
⋅⋅−12 E I
l3
3
−3306.122 ⋅―
1
m
kN
CASO 5
≔Ma5 0 kN ≔Mb5 0 kN
≔Va5 0 ――
kN
m ≔Vb5 =―――
⋅−E A
l3
−450000 ⋅―
1
m
kN
≔Ha5 ⋅0 ――
kN
m ≔Hb5 0 ――
kN
m
CASO 6
≔Mb6 =―――
⋅⋅2 E I
l3
27000 ⋅m kN ≔Ma6 ⋅0 kN m
≔Va6 0 kN
≔Vb6 0 kN
≔Ha6 ⋅0 kN
≔Hb6 =―――
⋅⋅−6 E I
l3
2
−11571.429 kN
CALCULANDO AS REAÇÕES
≔D1 0.0082625833 m
≔D2 −0.0000618129 m
≔d =⋅k
−1
−f
0.0082625833
−0.0000618129
−0.0019225093
0.0082651655
−0.0001801287
0.0005460609
⎡
⎢
⎢
⎢
⎢
⎢
⎣
⎤
⎥
⎥
⎥
⎥
⎥
⎦
≔D3 −0.0019225093
≔D4 0.0082651655 m
≔D5 −0.0001801287 m
≔D6 0.0005460609
≔Vb =++++++Vb0 ⋅Vb1 D1 ⋅Vb2 D2 ⋅Vb3 D3 ⋅Vb4 D4 ⋅Vb5 D5 ⋅Vb6 D6 81.058 kN
≔Hb =++++++Hb0 ⋅Hb1 D1 ⋅Hb2 D2 ⋅Hb3 D3 ⋅Hb4 D4 ⋅Hb5 D5 ⋅Hb6 D6 −68.644 kN
≔Mb =++++++Mb0 ⋅Mb1 D1 ⋅Mb2 D2 ⋅Mb3 D3 ⋅Mb4 D4 ⋅Mb5 D5 ⋅Mb6 D6 151.217 ⋅m kN
≔Va =++++++Va0 ⋅Va1 D1 ⋅Va2 D2 ⋅Va3 D3 ⋅Va4 D4 ⋅Va5 D5 ⋅Va6 D6 38.942 kN
≔Ha =++++++Ha0 ⋅Ha1 D1 ⋅Ha2 D2 ⋅Ha3 D3 ⋅Ha4 D4 ⋅Ha5 D5 ⋅Ha6 D6 −31.356 kN
≔Ma =++++++Ma0 ⋅Ma1 D1 ⋅Ma2 D2 ⋅Ma3 D3 ⋅Ma4 D4 ⋅Ma5 D5 ⋅Ma6 D6 114.725 ⋅m kN
MONTANDO OS DIAGRAMAS
DMF (kN.m):
DEC (kN):
DEN (kN):