RESISTENCIA DE MATERIALES II
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PROBLEMAS RESUELTOS R2 RESITENCIA DE MATERIALES, MECANICA DE MATERIALES
Transcript of RESISTENCIA DE MATERIALES II
-
I. ALUMNO: Donaldo Josue HUANCA PONCE CDIGO: 122095
Para las siguientes vigas simplemente apoyadas. Determine la deflexin indicada por
los mtodos geomtricos y los mtodos energticos.
1.-
Solucin:
=0
*L= (
)
(
)+ (
)
(
)
=
AREA DE MOMENTOS:
=3
2
=3
4
=
4
=9
2
=9
4
=3
4
-
CASO AREA XC XD a
9
2
3
3
b
4
c
432
4
a 4
4
9
2
9
b
2
=
9
3
432
=2
=
=
, entonces: f =
=4
2
9
2
=34
3 4
Por lo tanto: 2= f - =
=
..Rpta:
2.- ALUMNO: Donaldo Josue HUANCA PONCE CDIGO: 122095
Solucin:
= .
2 =
=
=
=
=
=
=
-
AREA DE MOMENTOS:
CASO AREA XC XD A
3
B
-
cL = =
(
)
=
..Rpta:
3.- ALUMNO: Donaldo Josue HUANCA PONCE CDIGO: 122095
Solucin:
= .
=
=
Area =
2
=
=
2
0
2
0
=
=
=
=
2
=
4
-
AREA DE MOMENTOS:
CASO AREA XC XD a 4
2 3
4 3
4
4
4
b 2
32 2
2
3
3
a 4
2 3
4 3
4
4
4
a ( 4)
2 3
4 3
4
4
3
4
Diagrama de deformacin:
i) =2
3
32 2 (
3)
(
4) 3
3
2 3
4
3 4
4 3
Area = 4 2
2
4
3 =
4 3
2 3
=
=
Area =
=
=
Area = 4 2
2
4
3 =
4 3
2 3
=
=
= = (
4)
2 3
=
4 3
4
=
-
= 2 4 3 4
2 3 4
4=
= ( ( ) )
= 4 3
2 3
4 3
4
=
= =
4 2 2 4
= [ ( ) ]
Rpta:
ii) =2
3
32 2
(
4) 3
3
2 3
= 2
4 2 3
4=
=
= 4 3
2 3
= =
4
=
Rpta:
4.- ALUMNO: Donaldo Josue HUANCA PONCE CDIGO: 122095
Solucin:
-
AREA DE MOMENTOS:
CASO AREA XC XD a
2
2
3
3
b
4
c
432
4
a
4
4
b
29
94
9
3
9
Diagrama de deformacin:
=3
2
=3
4
=
4
=9
2
=9
4
=3
4
-
= =
2(2
3)
(4
)
432 4
2
3
=
= =
(
4 )
29 ( 94
9 )
432 4
=
. Rpta:
5.- ALUMNO: Donaldo Josue HUANCA PONCE CDIGO: 122095
Solucin:
AREA DE MOMENTOS:
CASO AREA XC XD a 2
2
2
b
2
2
3
3
L/2
=
=
=
Area =
2
=
=
Area =
2
=
=
-
a
2
a 3
9
= =
2(2
3)
2
(
2)
=
cL = =
(
)
=
..Rpta:
6.- ALUMNO: Donaldo Josue HUANCA PONCE CDIGO: 122095
Solucin:
AREA DE MOMENTOS:
=
=
=
2
=
4
Area = 4 2
2
4
3 =
4 3
2 3
=
=
Area =
=
=
Area = 4 2
2
4
3 =
4 3
2 3
=
=
-
CASO AREA XC XD a 4
2 3
4 3
4
4
4
b
2
2
3
3
a 4
2 3
4 3
4
4
4
a ( 4)
2 3
4 3
4
4
3
4
= =
2(2
3)
(
4) 3
3
2 3 (
4 3 4
4 3 )
=( (
4)
) (
4) 3
3
2 3 (
4( 3 4
4 3 ))
3 2
2
=
i) = =
(
)
(
)
=
3 4 2 2 3 4
2(2
3
4)
4
( 4 )
2 3
4 3
4
=
..Rpta:
i) = =
(
)
=
3 4 2 2 3 4
2
4
( 4 )
2 3
=
..Rpta:
= = (
4)
2 3
=
4 3
4
=
-
7.- ALUMNO: Donaldo Josue HUANCA PONCE CDIGO: 122095
Solucin:
AREA DE MOMENTOS:
=3
2
=3
4
=
4
=9
2
=9
4
=3
4
-
CASO AREA XC XD a
2
2
3
3
b
4
c
432
4
d
2
2
a
4
4
b
29
94
9
3
9
d 2
3
3
3
Del diagrama de deformacin tenemos:
= =
2(2
3)
3
(4
)
3
432(4
)
2
=
2
2
3 4
4 ..i)
= =
2 3
3
432
=
3 3
2 ..ii)
=
=
3 4
4
3 4
432
=23
, =
..^ =
Porl lo tanto:
= 2 =
2
(
4 )
3
29 ( 3
9 )
2
3
3
= 23
2
(
4 )
3
29 ( 3
9 )
2 2
3
3
= 3 9 4
49 Rpta:
8.- ALUMNO: Donaldo Josue HUANCA PONCE CDIGO: 122095
-
= .
2 =
=
=
=
AREA DE MOMENTOS:
CASO AREA XC XD a
3
b
-
c
2
4
4
Del diagrama de deformacin:
= =
2
= (
)(
2
) =
4 =
=
=
=
Area =
2
=
=
2
0
2
0
=
-
= = 3
3
3
2
2
4
= = 3
3
3
2
2 2
4 3
2
4
= 4 2 4 24 3 3 2
9 4..Rpta:
9.- ALUMNO: Donaldo Josue HUANCA PONCE CDIGO: 122095
Solucin:
AREA DE MOMENTOS:
= =
=
2
=
=
=
2
2
=
=
=
2
= 2
3 , =
3
-
CASO AREA XC XD a
2
2
3
3
b
3
4
4
c 4
2
2 2
3 2
2 3
3 4
2
3 4
d
2
2
e
2 2 3
3
3 4
3 4
a 2
9
4
9
2
9
b 4
2
d 2
3
3
3
Del diagrama de deformacin tenemos:
= =
2(2
3)
(3
4)
4
2 (
2 2
3 2 ) (
2 3
3 4 ) (
2)
2 2 3 (2
3 3
3 4 )
=
2
2
..i)
= =
2
4
2 (
2 2
3 2 )
2 2 3
=
..ii)
=
=
= 3 2 2 4 3 2 4
4 2 3 4 ,
= 92 2 3 9 4
2 3 4
Porl lo tanto:
= 2 =2
2
9(2
9)
4 3
(
)
4 3
2 2 2
3 2
2
3 4
2
3
3
-
=2(
3 2 4 2 4 2 3 4
)
9(2
9)
4
(
)
4
2 (
2 2
3 2 ) (
2
3 4 )
2 (
92 9 2 3 4
)
3(
3)
=
.Rpta:
III. ALUMNO: Donaldo Josue HUANCA PONCE CDIGO: 122095
Solucin:
=
2= 3
2= 4
= 4 2 4
= 29
= = 2
= 324
=
=
Momentos de empotramiento perfecto:
= 2
= 2
=
= 2
=
= 4
-
Ecuaciones de giros y desplazamientos:
2 3 { = 2 2 4 2 = 2 2 2 4
}
3 4 { = 2 4 2 = 2 2 2 4
}
4 { = 324 4 2
= 4 324 2 4 }
Ecuaciones de continuidad:
= = = 4 4 432
= = = 2 432 2 432
= = = 44 432 2
[ = = =
] = [ 4 432 432 2 432 432 2
] [
] [
4 2
44 ]
= 4 2 3 4 = 3
= 2 23 93 = 9
= 3 9 2 42 49 = 2
Por lo tanto:
2 3 { =
= 2 9 2 }
3 4 { = 2 9 2 = 3
}
4 { = 3 = 3
}
-
= =
Diagrama de deformacin:
IV. ALUMNO: Donaldo Josue HUANCA PONCE CDIGO: 122095
Solucin:
= 2
= 32
= =3 2
3
=
=
-
Momentos de empotramiento perfecto:
=
= 2
=
=
= 2
=
Ecuaciones de giros y desplazamientos:
2{ =
4 2
= 2
2 4
}
2 3{ = 4 2
= 2 4 }
3 4{ = 2
4 2
=
2 4
}
Ecuaciones de continuidad:
= = = 4 4 432
= = = 2 432 2 432
= -> =
= -> =
Por lo tanto:
=
2 2
= 32
2 2
= 2 2
= 2 2
4
[
] =
[ 22
3
32
3 2
3 2
3 2
3 2
2 2 2 2
2
3 ]
[
] [
2 2
4
]
-
= 4 3 49
= 9 43 9 2 4
= 32 3 4
= 22222222
V. ALUMNO: Donaldo Josue HUANCA PONCE CDIGO: 122095
Solucin (TRABAJO VIRTUAL)
=
(
9 2
4 2
9 9
4 9
9 2 9 9 9 2 9 9
4 2 4 9 4 2 4 9
9 2 3 3
4 2 2 3
9 9 9 9
4 9 4 9 3 3 9 9
2 3 4 9 )
-
=
(
4 2
4 2
2 3
3 )
=
(
3
4 233 2 29 2 4
3 4 233 2 29 2 3 3333333
33 33 2 29 2
2 )
-> =
=
(
9 4 2222222 4444444 2 2 3333333 2 2 2 9 3 34 9 24 9 3 3333333 4 4 3333333 4 4 3 42 2 43 42 2 4 42 2 4 4 4 924 2 4 94 924 2 449 293 329 449 39 2 94 9 39 2 942 42 2 4 3 9 4 943 9 42 943 4 4 3 4 2 4 4
4 4 3 42 2 43 42 2 4 42 2 42 3 29 2 3 93 32 3 32
943 9 4 9 9 4 4444444
4 4 4 2 9 4 9 9 4 )
-
Deformaciones!!
-
VI. ALUMNO: Donaldo Josue HUANCA PONCE CDIGO: 122095
2.
Solucin:
= ,4 = 2 = 2
=
= (2
)
=
3 , =
=
=
= (2
) (2 )
( )
=
2(2 ) 2 2 , =
= (
)
=
=
Rpta
