ANALISIS ESTRUCTURAL AUX. DOC.: EFRAIN SANTALLA ALEJO FORMULAS DE GIROS DE CADA APOYO PARA CARGAS PARTICULARES Nº
q1
CARGA
Giro derecho del apoyo izquierdo
Giro izquierdo del apoyo derecho
q 2 3
1
θ d d
θ d
θ i
=
L
q 2
θ d d
θ i
q
(8 ⋅ q + 7 ⋅ q ) 1
θ i
2
7⋅q⋅ L
=
θ d
=
θ i
360 ⋅ EI
=
L
360 ⋅ EI
(7 ⋅ q + 8 ⋅ q ) 1
2
q ⋅ L3 45 ⋅ EI
b
3
θ d d
360 ⋅ EI
3
L a
3
L
θ i
θ d
L
2 q ⋅ L3 b b = 1 + 7 − 3 L 360 ⋅ EI L
θ i
2 q ⋅ L3 a a = 1 + 7 − 3 L 360 ⋅ EI L
a
q
4
θ d d
θ i L
q ⋅ L ⋅ a2 a θ d = 2− L 24 ⋅ EI
2
θ i
q ⋅ L ⋅ a2
a = 2 − 24 ⋅ EI L
2
a
q
5
θ d d
q⋅a ⋅L
a
a θ d = 45 + 12 40 − 45 L 360 ⋅ EI L q ⋅ b2 ⋅ L
a
2
θ i L
2
θ i
2 q⋅a ⋅L
2 a = 5 − 3 90 ⋅ EI L
a 6
q θ
θ
θ d
=
20 − 15
a + 3 2
θ i
=
q ⋅ b2 ⋅ L
a − 10 3 2
ANALISIS ESTRUCTURAL AUX. DOC.: EFRAIN SANTALLA ALEJO FORMULAS DE GIROS DE CADA APOYO PARA CARGAS SIMETRICAS Nº
CARGA
Giro derecho del apoyo izquierdo q
θ d
11
θ i
θ d
=
L L/2 12
q
θ d
1
q⋅L
24
EI
3
θ d
θ i
a
θ d
=
θ d
θ i
=
L a
q θ d
θ i L
15
θ d
5 192
⋅
q ⋅ L3 EI
q ⋅ a2 ⋅ L 3 6 ⋅ EI
a
− 2 L
q ⋅ a2 ⋅ L
a θ d = 1 − 2 L 12 ⋅ EI ⋅ 3 q⋅L
2 a a θ d = 1 − 2 − L 24 ⋅ EI L
θ i
q
2º
2º
θ d
EI
5
q ⋅ L3
θ i
=
⋅
θ i
q⋅a ⋅L 3 a = − L 6 ⋅ EI 2
EI
192
2
θ i
=
q ⋅ a2 ⋅ L 12 ⋅ EI
1 −
a 2⋅ L
a
q
L
16
24
3
a
q
a
=
q⋅L
a
q
14
θ i
1
L/2
L
13
Giro izquierdo del apoyo derecho
θ i
θ d
=
q ⋅ L3 30
EI
θ i
3 q⋅L
2 a a = 1 − 2 − L 24 ⋅ EI L
q⋅L
3
θ i
=
30
EI
ANALISIS ESTRUCTURAL
Nº
AUX. DOC.: EFRAIN SANTALLA ALEJO FORMULAS GENERALES DE GIROS PARA DISTINTAS CARGAS
ESQUEMA DE LA CARGA 3 casos
a
b θ i
19
θ d
M
a
θ d =
M ⋅ L b EI ⋅ θ d = 1 − 3 ⋅ 6 L
EI ⋅θ i =
b
c
M ⋅ L
EI ⋅θ d = θ i
20
θ d
EI ⋅θ i =
m
a
b
EI ⋅ θ d =
c θ i
21
θ d
EI ⋅θ i =
m
20 casos a
b
EI ⋅ θ d =
c θ i
22
θ d
m1
FACULTAD DE INGENIERIA
m 2
Giro derecho del apoyo izquierdo ;
EI ⋅ θ i =
a 1 − 3 ⋅ L
6
2
2
θ i =
Giro izquierdo del apoyo derecho
2 m⋅b a + b 2 b + c ( a + b ) − c3 − b ⋅ L + c 2 2 ⋅ L 3 3 3
m⋅b
( a + b) − c + a + b − b ⋅ L b + a 3 3 3
2
2 ⋅ L
2
3
{bL ( c + b ) − ⋅ L 4
2
m ⋅b 24 ⋅ L
2
b 24
2
m ⋅b 24
c
2
b 24 ⋅ L
2
2
{bL ( a + b ) −
4
3
{bL c ( ⋅ L 2
4
2m1
(
a + b)
2
2 2
( 2c + L ) − 2c } 3
2
1
}
a + b ) − c ( 3L − 2c ) ( 3
2
}
+ m ) + b ( 3m + m ) − 2( m + m ) ( a + b) ( 2c + L) − 2c 2
2
1
{bL a ( m + m ) + b ( m + 4
2
5
1
U.M.S.A.
2
3m 2
1
3
2
) − 2( m + m ) 2( a + b ) − c ( 3L − 2c ) } 3
1
2
2
ANALISIS ESTRUCTURAL Nº
ESQUEMA DE LA CARGA P a b
3 casos
AUX. DOC.: EFRAIN SANTALLA ALEJO FORMULAS GENERALES DE GIROS PARA DISTINTAS CARGAS θ d = Giro derecho del apoyo izquierdo ; θ i = Giro izquierdo del apoyo derecho
P ⋅ a ⋅b
EI ⋅ θ d =
θ d
θ i
q
a
P ⋅ a ⋅b
EI ⋅θ i =
6 ⋅ L
q
b
a
EI ⋅θ d =
q ⋅b b 2 a +b b b 3 + c (a + b) + c + 2 ⋅ c 3 + a − b 2 L + c 2 6 ⋅ L 2 3 2 4
θ i
EI ⋅θ i =
q ⋅b b 3 b c 3 + c ( a + b ) + 3c 2 + a + a + b − b 2L b + a 2 2 6 ⋅ L 2 2 3 4
c
EI ⋅θ d =
2 q ⋅b b 2 a +b b L b 3 2 c a b c c b a + + + + ⋅ + − 3 2 ) ( + c 2 12 ⋅ L 3 3 3 2 5
θ i
EI ⋅θ i =
c
EI ⋅θ d =
θ i
EI ⋅θ i
c
EI ⋅ θ d =
θ i
EI ⋅ θ i
25
θ d
20 casos
b
a
q 2
q1
26
θ d
4 casos
a
b
q nº
27
θ d
FACULTAD DE INGENIERIA
(a + L)
c
24
θ d
(b + L)
6 ⋅ L
23
2 q ⋅b b 3 c b L4 2 2 c a b c b a a b 2 + + + 3 + + + − ( ) b + a 2 12 ⋅ L 3 3 3 2 5 3 L
5 4 b ( 3c + b) ( b + c ) − c 5 c q + ⋅ + − 10 15 3 2 2 2 4 L L b⋅L 360 L 5 5 4 4 b ( 3a + b ) ( a + b) − a 5 ( a + b) − a 5 L3 b ( 3a + 2b ) a+b a q q = ⋅ 10 − 15 + 3 + ⋅ 10 + 15 − 3 1 2 L L L2 b ⋅ L4 L2 b ⋅ L4 360
q1 ⋅ 10
b ( 3c + 2b )
4
b + c (a + b ) 3 6 ( n + 1) L n+2 b q ⋅b = + c (a + b ) 2 6 ( n + 1) L n+2 q ⋅b
5
6 ⋅b ⋅ L a +b n +1 b b +a − + c + 2⋅c + c 3 n+2 ( n + 2)( n + 3) n + 4 2
2
2
3
2
5
(b + c) − c b+c − 15 +3 4 b⋅ L L
3
6 ⋅b ⋅ L n +1 c n +3 b + a + a + b − b + a + 3⋅c n+2 3 ( n + 2)( n + 3) n + 4 2
