Resolución de Integrales

Resolución de integrales
Caso A
M 1 ( x) = −
x
l
M 0 ( x) =
ql
q
x − x2
2
2
q
1
1
l
1
l
ql
2
ql
2
x
x
M
max
1
=1
M
l
δ10 = ∫ M 1 ( x)κ 0 ( x)dx =
0
I10
EI
max
0
ql 2
=
8
l
I10 = ∫ M 1 ( x) M 0 ( x)dx
0
l
q l
q l
q x3
q x4
+
I10 = − ∫ x 2 dx + ∫ x 3 dx = −
2 0
2l 0
2 3 0 2l 4
l
0
ql
⎛1 1⎞
= − ql 3 ⎜ − ⎟ = −
24
⎝6 8⎠
1
= − M 1max M 0max l
(fórmula de tablas)
3
3
Caso B
M 1 ( x) = −
x≤l 2
⎧P 2 ⋅ x
M 0 ( x) = ⎨
⎩P 2 ⋅ x − P ( x − l 2) l 2 ≤ x ≤ l
x
l
P
1
1
l
1
l
P
2
P
2
x
x
M
max
1
=1
M 0max =
l
δ10 = ∫ M 1 ( x)κ 0 ( x)dx =
0
I10
EI
Pl
4
l
I10 = ∫ M 1 ( x) M 0 ( x)dx
0
l
P l
P l
xl
P x3
P x 3 x 2l
I10 = − ∫ x 2 dx + ∫ ( x 2 − )dx = −
+
−
l l2
2l 0
2
2l 3 0 l 3
4
l
l 2
( 8 − 16 + 12 + 2 − 3) = − 3 Pl 2
⎛1 1 1 1 1 ⎞
= − Pl 2 ⎜ − + + − ⎟ = − Pl 2
48
48
⎝ 6 3 4 24 16 ⎠
1
= − 1.5 ⋅ M 1max M 0max l
(fórmula de tablas)
6
Caso C
x
l
M 1 ( x) = −
κ 0 ( x) = α
1
1
l
(∆T S − ∆T I )
= −κ 0
h
∆T S
∆T I
1
l
x
x
M 1max = 1
l
κ0
0
l
δ10 = ∫ M 1 ( x)κ 0 ( x)dx =
=
1 max max
M1 κ 0 l
2
∫
l
0
xdx =
κ 0max = κ 0
κ 0 x2
l
2
l
=
0
κ 0l
2
(fórmula de tablas)
Caso D
x
l
M 1 ( x) = −
M 0 ( x) = − M 0
x
l
1
1
l
M0
1
l
M0
l
M0
l
x
x
M 0max = M 0
M 1max = 1
l
δ10 = ∫ M 1 ( x)κ 0 ( x)dx =
0
I10 =
M0
l2
∫
l
0
x 2 dx =
1
= M 1max M 0max l
3
I10
EI
3 l
M0 x
l2 3
0
l
I10 = ∫ M 1 ( x) M 0 ( x)dx
0
=
M 0l
3
(fórmula de tablas)
Caso E
M 1 ( x) = −
⎛ x⎞
M 0 ( x) = − M 0 ⎜1 − ⎟
⎝ l⎠
x
l
1
1
l
M0
1
l
M0
l
x
x
M 0max = M 0
M 1max = 1
l
δ10 = ∫ M 1 ( x)κ 0 ( x)dx =
0
M
I10 = 0
l
=
I10
EI
l
I10 = ∫ M 1 ( x) M 0 ( x)dx
0
l
⎛
M 0 x 2 x3
x2 ⎞
⎛1 1⎞ M l
x
dx
−
=
−
= M 0l ⎜ − ⎟ = 0
∫0 ⎜⎝ l ⎟⎠
l 2 3l 0
6
⎝ 2 3⎠
l
1 max max
M1 M 0 l
6
(fórmula de tablas)
M0
l