Subido por Tobías Portocarrero

# Problema Slope Deflection

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```UNIVERSIDAD NACIONAL HERMILIO VALDIZ&Aacute;N
FACULTAD DE INGENIER&Iacute;A CIVIL Y ARQUITECTURA
E.A.P. INGENIER&Iacute;A CIVIL
PC N&ordm; 2 – SLOPE
DEFLECTION
ASIGNATURA
: AN&Aacute;LISIS ESTRUCTURAL I
DOCENTE
: ING. ANTONIO DOM&Iacute;NGUEZ MAGINO
INTEGRANTES
:
o
o
o
o
o
o
o
o
o
o
o
CICLO
ANCHILLO TIMOTEO, JACKELINE
&Aacute;NGEL VERTIZ, SUKKER
CABELLO PONCE, KEVING
CORNELIO ARGUMENDO, PEDRO ALFREDO
FALC&Oacute;N FABI&Aacute;N, LENYN JIROSHI
GER&Oacute;NIMO MARCOS JAIME
HU&Aacute;RAC ALBORNOZ, &Aacute;NGELA
ORTEGA PALACIOS, MILER
PORTOCARRERO DEL &Aacute;GUILA, TOB&Iacute;AS MART&Iacute;N
RODIL HUAM&Aacute;N, ROMEL
SULCA CORREA, HEISON
: VERANO
Hu&aacute;nuco – Per&uacute;
2018
FACULTAD DE INGENIER&Iacute;A CIVIL Y ARQUITECTURA
E.A.P. INGENIER&Iacute;A CIVIL
PROBLEMA:
P=6T
5
2
20&deg;C
3
0&deg;C
3m
20&deg;C
0&deg;C
0&deg;C
1
4
1m
1m
1m
DATOS:
EI = Cte.
Columna – Viga = (30 x 30)cm
f&acute;c = 210 Kg/cm2
α = 0.1 x 10-3 /&deg;C
G = E/2(1+u)
SOLUCI&Oacute;N:
Antes de iniciar, conviene definir las caracter&iacute;sticas de la estructura.
E
G
I
EI
20&deg;C
2173706.511928
724568.837309
0.000675
1467.251896
T/m2
T/m2
m4
T-m2
∆=2cm
FACULTAD DE INGENIER&Iacute;A CIVIL Y ARQUITECTURA
E.A.P. INGENIER&Iacute;A CIVIL
F&oacute;rmula:
𝑴𝒊𝒋 = 𝑴𝒊𝒋 +
𝟑∆𝒊𝒋
𝟐𝑬𝑰
𝜶𝑬𝑰∆𝑻
(𝟐𝜽𝒊 + 𝜽𝒋 −
)&plusmn;
𝑳
𝑳
𝒅
Entonces:
𝑀12 =
2𝐸𝐼
𝛼𝐸𝐼∆𝑇 &deg;
[2(0.0014) + 𝜃2 ] −
3
𝑑
𝑀21 =
𝑀25 =
𝑀52
2𝐸𝐼
3(𝛿5 )
𝛼𝐸𝐼∆𝑇 &deg;
[2𝜃2 + 𝜃5𝑖 −
]−
1
1
𝑑
2𝐸𝐼
3(𝛿5 )
𝛼𝐸𝐼∆𝑇 &deg;
[2𝜃5𝑖 + 𝜃2 −
]+
=
1
1
𝑑
𝑀53 =
𝑀35
2𝐸𝐼
𝛼𝐸𝐼∆𝑇 &deg;
[2𝜃2 + 0.0014] +
3
𝑑
2𝐸𝐼
∆3
𝛼𝐸𝐼∆𝑇 &deg;
[2𝜃5𝑑 + 𝜃3 − 3 ( )] −
2
2
𝑑
2𝐸𝐼
∆3
𝛼𝐸𝐼∆𝑇 &deg; 3𝑝𝑙
[2𝜃3 + 𝜃5𝑑 − 3 ( )] +
=
+
2
2
𝑑
16
𝑀34 =
2𝐸𝐼
𝑀43 =
2𝐸𝐼
3
3
(2𝜃3 + 𝜃4 ) −
(2𝜃4 + 𝜃3 ) +
𝛼𝐸𝐼∆𝑇 &deg;
𝑑
𝛼𝐸𝐼∆𝑇 &deg;
𝑑
FACULTAD DE INGENIER&Iacute;A CIVIL Y ARQUITECTURA
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2
3
δ5
δ5
t3/2
∆3
Ψ53
0,0014
1
4
∆=2cm
1 𝑀
2
 𝑡5 = 2 ( 𝐸𝐼25 ) 𝑥 3 =
2
 𝜃2 =
𝑀25
3𝐸𝐼
−𝑀25
3𝐸𝐼
+ 𝛿5
𝑀
 ∆= 𝛿5 + ∆3
25
𝛿5 = 𝜃2 − 3𝐸𝐼
𝑀
25
∆3 = 0.02 − 𝜃2 + 3𝐸𝐼
Resolviendo los momentos:
(*)
∆
FACULTAD DE INGENIER&Iacute;A CIVIL Y ARQUITECTURA
E.A.P. INGENIER&Iacute;A CIVIL
𝑀12 = 2.738870804 + 978.16793𝜃2 − 0.097816793 = 978.16793𝜃2 + 2.641053411
𝑀21 = 1956.33586𝜃2 + 1.369435102 + 0.097816793 = 1956.33586𝜃2 + 1.461251895
𝑀25 = 8694826.048𝜃2 + 4347413.024𝜃52 − 13042239.07𝜃2 + 2962.462963𝑀25
− 0.09781679
𝑀25 =
4347413.022𝜃2 − 4347413.024𝜃52 + 0.097816793
2961.962963
𝑀25 = 1467.74726𝜃2 − 1467.74726𝜃52 + 3.308431334𝑥10−5
T-m
𝑀52 = 8694826.048𝜃52 + 4347413.024𝜃2 − 13042239.07𝜃2 + 2962.968963𝑀25
+ 0.097816793
𝑀52 = 8694826.048𝜃52 + 4347413.024𝜃2 − 13042239.07𝜃2
+ 2962.968963(1467.74726𝜃2 − 1467.747226𝜃52 + 3.302431334𝑥10−5 )
+ 0.097816793
𝑀52 = −4345945.276𝜃2 + 4345945.278𝜃52 + 0.1956666103
T-m
𝑀53 = 2934.503792𝜃53 + 1467.251896𝜃3 + 4401.755688𝜃2 − 88.03511376 − 𝑀25
− 0,097816793
𝑀53 = 2934.503792𝜃53 + 1467.251896𝜃3 + 4401.755688𝜃2 − 88.03511376
− (1467.74726𝜃2 − 1467.747226𝜃52 + 3.302431334𝑥10−5 ) − 0,097816793
𝑀53 = 2934.008428𝜃2 + 1467.74726𝜃52 + 1467.251896𝜃3 + 2934.503792𝜃53
− 0,09784981731
T-m
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E.A.P. INGENIER&Iacute;A CIVIL
𝑀35 = 2934.503792𝜃3 + 1467.74726𝜃53 − 88.03511376 + 4401755688𝜃2 − 𝑀25
+ 0,097816793
𝑀35 = 2934.503792𝜃3 + 1467.74726𝜃53 + 4401755688𝜃2 − 88.03511376 − 1467.74726𝜃2
+ 1467.74726𝜃52 + 3.302431334𝑥10−5 + 0,097816793
𝑀35 = 2934.008428𝜃2 + 1467.74726𝜃52 + 2934.503792𝜃3 + 1467.74726𝜃3 − 87.93733999
T-m
𝑀34 = 1956.33586𝜃3 − 0.097816743
𝑀43 = 978.16793𝜃3 − 0.097816743
T-m
Ahora: ecuaciones de equilibrio
 𝑀21 + 𝑀25 = 0
0 = 1956.33586𝜃2 + 146725.1895 + 1467.74726𝜃2 − 1467.74726𝜃52 + 3.302431334𝑥10−5
3424.08312𝜃2 − 1467.74726𝜃52 = −1.467284919
(1)
 𝑀52 + 𝑀53 = 0
4345995.276𝜃2 + 434594.278𝜃52 + 0.1956666103 + 2934.008428𝜃2 − 1467.74726𝜃52
+ 1467.251896𝜃3 + 2934.503792𝜃53 − 0.09784981731 = 0
4348879.284𝜃2 + 4347413.025𝜃52 + 1467.251896𝜃3 = −0.09784981731
 𝑀35 + 𝑀34 = 0
(2)
FACULTAD DE INGENIER&Iacute;A CIVIL Y ARQUITECTURA
E.A.P. INGENIER&Iacute;A CIVIL
2934.008428𝜃2 + 1467.74726𝜃52 + 2934.503792𝜃3 + 1467.7426𝜃53 − 87.93732999
+ 1956.33586𝜃3 − 0.097816793 = 0
2934.008428𝜃2 + 1467.74726𝜃52 + 4890.839652𝜃3 + 1467.7426𝜃53 = 88.03514678
 𝑀12 + 𝑀21 + 𝑀43 + 𝑀34 = 0
978.16793𝜃2 + 2.641053411 + 1456.33586𝜃2 + 1.467251895
= 978.16793𝜃3 + 0.097816793 + 1956.33586𝜃3 − 0. 097816793
2934.50379𝜃2 − 2934.50379𝜃3 = −4.10835723
Entonces resolviendo el sistema tenemos:
 𝜃2 = 0.001119 𝑟𝑎𝑑
 𝜃5𝑖 = −0.00252 𝑟𝑎𝑑
 𝜃5𝑑 = −0.0030316 𝑟𝑎𝑑
 𝜃3 = −0.00418 𝑟𝑎𝑑
(4)
(3)
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Diagrama de momento flector