solucionario ANALISIS
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-
PROBLEMA N 1:
RESOLVER:
DATO DEL PROBLEMA:
- P=10 ton
- L = 4m
- E = 10 000 kips/2 = 2 x 106kg/2 = 2 x 107 ton/2
- =0.7 x 105/ c
- 1 = 80 4 ; 1 = 40
2
- 3 = 80 4 ; 3 = 40
2
- 2 = 150 4 ; 2 = 5
2
- b = L/20 y h = L/10
HALLAR:
a. {AD}
b. {ADL}
c. {ADT}
d. {ADR}
e. {ADS}
f. {K}
g. {D}
h. {AMD}
i. DMF
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CALCULO De E ; (cada elemento)
- 1 = (0.025)(2 x 107) = 516128 ton.
- 2 = (0.025)(2 x 107) = 516128 ton.
- 3 = (3.22 x 103)(2 x 107) =64516 ton.
- 1 = (3.33 x 105)(2 x 107) =665.9703 ton- 2.
- 2 = (6.243 x 105)(2 x 107) =1248.6943 ton- 2
- 3 = (3.33 x 105)(2 x 107) =665.9703 ton- 2
1. DETERMINACION DEL GRADO DE LIBERTAD:
2. PARA IMPEDIR EL DESPLAZAMIENTO EMPOTRAMOS LA VIGA:
3. DETERMINAMOS AD:
a).
AD = {-8 } Ton-m
-
4. DETERMINAMOS ADs:
4.1 DETERMINAMOS {ADL}:
AD1 = M12 + M13 = Pa2
2 +
w2
12 = -
10(4)(4)2
82 +
2.5(12)2
12 = -10 + 30 =20
b)
4.2 DETERMINAMOS {ADT}:
{ADT} = -{(1)(12)
}= {-
0.7 x 105(665.9703 )(1030)
0.8} = {0.1748}
c)
{ADL} = { AD1} = {20}
{ADT} = { 0.1748}
-
4.3 DETERMINAMOS {ADR}:
- {ADR} = {21 ()
} =
2(665.9703)(0.004)
8 = -0.666
d)
5. DETERMINAMOS {ADS}
{ADS} = {ADL} + {ADT} + {ADR}
{ADS} = {20} + {0.1748} + {-0.666} e)
6. DETERMINAMOS {K}
D1 = 1 1 =0
{ADR} = { -0.666}
{ADS} = {19.5088}
-
11 = 4(665.9703)
8 +
4(1248.6943)
12 = 749.2166
f)
7. DETERMINAMOS {D}
{AD} = {ADS} + {K} {D}
{D} = {}1 {{AD} {ADS}}
{D} = {749.2166}1 {{-8} {19.5088}}
g)
8. DETERMINAMOS {AM}
{AM} = {AMS} + {AMD}{D}
Sabemos que:
{AMS} = {AML} + {AMT} + {AMR}
{K} = {749.2166}
{D} = {-0.0367}
-
- {AML}
{AML} ={
21121331
} = {
1010 3030
}
- {AMT}
{AMT} ={
2112
00
} = {
0.17480.1748
00
}
-
- {AMR}
{AMR} ={
2112
00
} = {
1.3320.666
0 0
}
DETERMINAMOS {AMS}
{AMS} = {AML} + {AMT} + {AMR}
{AMS} = {
1010 3030
} + {
0.17480.1748
00
} + {
1.3320.666
0 0
}
{AMS}= {
8.493210.4912
3030
}
-
DETERMINAMOS {AMD}
h)
{AM} = {AMS} + {AMD}{D}
{AM} = {
8.493210.4912
3030
} + {
166.492575332.98515 416.2314208.1157
}{-0.0367}
{AM} = {
2.382922522.711755 14.724307637.6378462
}
{AMD} = {
166.492575332.98515 416.2314208.1157
}
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I) DMF
