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4.1.-Calcular el incremento de longitud que tendrá un pilar de hormigón de 50 x 50 cm sección y de 3 m de longitud, que se encuentra apoyado en su base inferior, debido a su propio peso. 3 Datos: E= 25 GPa , γ ( peso específico del hormigón)= 24 KN/m γ(peso
γ = 24 E =
L
∆ L = ∫ 0
N .dx
kN m
3
= 24.10 3
N m
3
25 GPa = 25.10 9 Pa = 25.10 9
N m
2
siendo N = f ( x )
E . A
x
3m
Peso
x
-
Pesox
N
RA
18000
RA
=0 R = Peso → R = γ .V = 24.10 .(0,5.0,5.3) = 18000 N ∑ F semester Master your with Scribd Read Free For 30this Days Sign up to vote title 3 on 0 − x − 3 N = − R + Peso = − R + γ .V = −18000 + 24.10 .(0,5.0,5. x) & The New York Times Useful Not useful N = −18000 + 6000. x 3
A
A
A
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x
x
A
= 0 → N = −18000 N
x
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x = 3 → N = 0
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4.2.-Una barra de sección variable y peso despreciable está empotrada en su extremo superior y sometida a las cargas que se indican en la figura. Se pide determinar: 1) Diagramas de fuerzas normales. 2) Diagramas de desplazamientos. 3) Tensión máxima, indicando donde se dará, e incremento de longitud de la barra. Datos: E = 210000 N/mm 2 RA
A1 = 4 cm2
2m 2m
20000 N
2m
A2 = 2 cm 2
4m
A3 = 1 cm2 10000 N
Cálculo de las reacciones:
∑ F = 0
R A
+ 10000 = 20000 → R A = 10000 N
1) Diagramas de fuerzas normales You're Reading a Preview
0 − x − 2 4 − x − 6
Unlock full access with = −10000 2 − xa free − 4trial. N = − R A = −10000 6 − x − 10 N = − R A + 20000 = 10000 N = − R A + 20000 = 10000
N = − R A
Download With Free Trial − 10000. x N . L 2) Diagramas de desplazamientos: = u = ∆ L( x) = E . A 210000.10 6.4.10 −4
0-x-2
x = 0 → u
=0
x = 2 → u
= −2,38.10 − 4 m
RA = 10000
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2-x-4
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4-x-6 RA = 10000 2
u
4 cm
1
x
2 cm
2
2
= ∆ L ( x) = ∑ =
20000
N i . Li E . Ai
= ∆ L1 + ∆ L2 + ∆ L3 =
− 10000 .2 − 10000 .2 10000 .( + + 10 −4 10 −4 21.10 .4.10 21.10 .2.10 21.1010.
3
x
= 4 → u = −7,14.10 − 4 m
x
=6→u =−
6-x-10 RA = 10000 2
4 cm
1
x 2 2
2 cm 20000
2
u
= ∆ L ( x) = ∑
N i . Li E . Ai
=∆ L1 + ∆ L2 + ∆ L3 + ∆ L4 =
− 10000 .2 − 10000 .2 10000 .2 + + + 21.1010.4.10 − 4 21.1010.2.10 − 4 21.1010.2.10 − 4
3 1 cm
4
x
= 6 → u = −2,38.10 − 4 m
x
= 10 → u = 16,67
You're Reading a Preview 10000
N (N) Unlock full access with a freeutrial. (m)
Download With Free Trial
2,38.10-4
10000
20000
7,14.10 10000 16,67.10-4
Master your semester with Scribd x 10000 & The New York Times Special offer for students: 3) σMAXOnly , �$4.99/month. L
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-4
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4.7.-La figura representa una barra rígida AB que está soportada por un pasador si fricción en A y por los alambres CD y EF. Cada alambre tiene una sección de 62,5 mm una longitud de 2 m, siendo el alambre CD de una aleación de aluminio y el EF de acero Determinar el valor de la carga P que hará que se rompa primero alguno de los dos cable 2 2 Datos: cable EF de acero: f u = 410 N/mm , E = 210000 N/mm cable CD de aluminio: f u = 310 N/mm 2 , E = 70000 N/mm 2
D 2m
F
al
ac
P E
C
A
B
1m
1m
2m
Ecuaciones de equilibrio de la barra rígida AB:
Fal
RA
Faca Preview P You're Reading Unlock full access with a free trial.
1m
2 m Download With Free 1 m Trial
R + F + F = P (1) ∑ F = 0 ∑ M = 0 P.4 = F .3 + F .1 (2) Master your semester with Scribd Es un caso hiperestático, se busca una ecuación de deformación: Read Free Foron 30this Days Sign up to vote title & The New York Times Useful Not useful A
A
Special offer for students: Only $4.99/month. 1m
al
ac
ac
al
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2m
1m
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desarrollemos la ecuación (3): F ac . Lac E ac . Aac
= 3.
F al . Lal
F ac .2
E al . Aal
210000.106.62,5.10−6
= 3.
F al .2
70000.106.62,5.10−6
(3)
Hipótesis: “El cable de acero es el primero que alcanza la rotura” σ ac =
F ac Aac
= f u ( ac) = 410 N / mm 2 → F ac = 410 Aac = 410.62,5 = 25625 N
= 19930 se comprobará ahora para esta hipotesis el estado de tensiones del cable de alu mi F 2847,2 = 45,6 N / mm 2 < f u (al ) = 310 N / mm 2 → la hipotesis es corr σ al = al = 62,5 Aal si se lleva este valor a las ecuaciones
P
(1), (2) y (3) : F al = 2847,2 N
P
= 19,93 kN el cable de acero se romperá
Nota: si se hubiese tomado como hipótesis la contraria, es decir, que el cable de aluminio entr en fluencia: σ al =
F al Aal
= f u (al ) = 310 N / mm 2 → F al = 310 Aal = 310.62,5 = 19375 N You're Reading a Preview
si se lleva este valor a las ecuaciones (1), ( 2) y (3) : F Unlock full access with a free trial.ac
= 174375 N
P
= 13562
se comprobará ahora para esta hipotesis el estado de tensiones del cable de acero
σ ac =
F ac Aac
=
Download With Free Trial 174375 = 2790 N / mm 2 > f u ( ac) = 410 N / mm 2 → la hipotesis no es 62,5
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4.8.-Una placa rígida de acero se sostiene mediante tres soportes de hormigón de alta resistencia. Cada soporte tiene una sección transversal cuadrada de 20x20 cm 2 y una longitud de 2 m. Antes de aplicar la carga P se observa que el soporte central es 1 mm má corto que los otros dos. Determinar la carga máxima P que podrá aplicarse al conjunto si se sabe que la tensión máxima a la que podrá estar sometido el hormigón es de 18 MPa. Datos: E ( hormigón ) = 30 GPa
P
1 mm
3
1
2m
2
Se verá en un principio el valor de las tensiones en los pilares 2 y 3 cuando acortan 1 mm:
You're Reading a Preview
F2 �L2=1
mm
Unlock full access with a free trial.
F 2 . L2 = 1 mm Download With Free Trial E 2 . A 2
∆ L2 =
F 2 .2
2
= 0,001 → F 2 = 600000 N 30.10 9.20 2.10 −4 F 600000 σ 2 = 2 = 2 − 4 = 15000000 Pa = 15 MPa < 18 M A2 20 .10
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Cancel anytime. luego cuando el pilar 2 se acorta 1 mm, aun no alcanza la tensión de 18 MPa, así pues entrará a Special offer for students: Only $4.99/month. trabajar el pilar central 1 Useful
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P
1 mm
�L1
3
1
Desarrollando dicha ecuación:
30.10 9.20 2.10 −4
+ 0,001 =
2
2m
∆ L1 + 1 mm = ∆L2 (3)
Ecuación de deformación:
F 1 .1,999
�L2
F 1 . L1 E 1 . A1
+ 0,001 m =
F 2 .2
30.10 9.20 2.10 −4
F 2 . L2 E 2 . A2
(3)
You're Reading a Preview
Por último se impone la condición de que la tensión en algún pilar alcance el valor de 18 MPa. Ésto ocurrirá en los pilares 2 y 3 que sonfull losaccess que van tener un mayor acortamiento y por tanto Unlock with a free trial. estarán sometidos a mayores tensiones que el pilar 1. Así pues: σ 2 =
F 2 A2
Download With Free Trial
= 18 MPa = 18000000 Pa → F 2 = 18000000. A2 = 18000000.20 2.10 − 4 = 7200
y llevando este valor a las ecuaciones (1), (2) y (3): F 3 = F 2 = 720000 N F 1 = 120000 P = 1560060 Master your semester with Scribd N = 1560 kN Read Free For 30 Days Sign up to vote on this title & The New York Times Useful Not useful Special offer for students: Only $4.99/month.
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4.13.-La barra de sección circular, de radio R, mostrada en la figura, está empotrada en s extremo izquierdo. Al aplicarla las cargas indicadas se pide: 1) Dimensionamiento a resistencia de la barra empleando un margen de seguridad de 35 % 2) Para la sección de la barra obtenida del apartado anterior, calcular su alargamiento. 3)Dimensionar la barra a rigidez con la condición: ∆L = ≤ 0,15 mm Datos: tensión límite elástico f y = 275 N/mm2 , coef. seguridad material γ γM = 1,05, 2 E = 210000 N/mm
20 KN
40 kN
30 cm
30 cm
1) Dimensionamiento a resistencia: Cálculo de las reacciones: Diagramas de esfuerzos:
∑F = 0
R A
= 40 + 20 = 60 kN
You're Reading a Preview
0 − x − 300 N = 60 kN
300 − x − 600 N = 20 kN
Unlock full access with a free trial.
N (kN)
60
Download With Free Trial +
20
x (mm)
Master your semester with Scribd Fórmula para el dimensionamiento a resistencia de una sección a Tracción: (según Normativa Read Free Foron 30this Days Sign up to vote title CTE-DB-SE-A): * & The New York Times Useful Not useful Comprobación a realizar: N ≤ N , = A. f Special offer for students: Only $4.99/month.
pl d
yd
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2) Alargamiento de la barra: (Las solicitaciones se emplean sin mayorar)
60.103.300 20.103.300 ∆ L = ∑ = + ≤ 0,37 mm E.Ai 210000.π .9,922 210000.π .9,922 Ni .Li
3) Dimensionamiento a rigidez con la condición: ∆L ≤ 0,15 mm
60.103.300 20.103.300 ∆ L = ∑ = + ≤ 0,15 → E. Ai 210000..π .R 2 210000..π .R 2 Ni .Li
R ≥ 15,57 mm
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4.14.-La estructura articulada de la figura está formada por dos barras de sección circula de acero. Si la estructura ha de soportar una carga de 30 kN en el nudo C, se pide: 1) Calcular las tensiones en ambas barras 2) Calcular el desplazamiento del nudo C. 3) Calcular el valor de la resistencia plástica de la barra AC 4) Calcular el valor de P que haría que la barra AC entrase en plasticidad 2 Datos: barra AC: : R = 1 cm; barra BC: R = 1,2 cm.; E = 210000 N/mm ; f y = 275 N/mm γ γM = 1,05 A
1m
C
B
P = 30 kN 1,5 m
1) Esfuerzos a los que estarán sometidas las barras
tagα =
You're Reading a Preview 1 = 0,666 → α = 33,7 º BC 1,5 Unlock full access with a free trial. ∑ F x = 0 Fac .cos 33,7º = F bc AB
=
∑
Equilibrio del nudo C:
.sen33,7º F y = 0With FacFree Download Trial = 30 resolviendo :
Fac 33,7º Fbc
o
C
= 54 kN (tracción y ) Fbc = 45 kN (compresión)
Fac
x
Master your semester30with Scribd & The New York Times Special offer for students: Only $4.99/month. barra AC:
σ ac =
F ac
=
54 .10
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3
2
= 171,9
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N 2
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2) Desplazamiento nudo C A y
α
C2
B
C ∆Lbc
x
∆Lac
C3
C1 α
C´
54.103.1,8.103 ∆ Lac = CC1 = = = 1,47 mm 210000.π .10 2 Eac . Aac Fac .Lac
= AC = AB 2 + BC 2 = 1,52 + 12 = 1,8 m = 1,8.103 mm 45.103.1,5.103 Fbc .Lbc ∆ Lbc = CC2 = = = 0,71 mm 210000.π .12 2 Ebc . Abc siendo Lac
You're Reading a Preview
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Desplazamientos de C :
δ x = CC2 = ∆Lbc = 0,71 mm
Download With Free Trial
δ y = C2 C ´= C 2C3 + C3 C´= CC1 . senα +
= 1, 47.sen33,7º +
C1C3 tagα
∆Lac .cos α + ∆ Lbc tagα
=
1,47.cos33,7º +0,71 = 3,71 mm tag 33,7º
Master your semester with Scribd δ = 0,71 mm ← & The New York Times δ = 3,71 mm ↓ x
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= ∆ Lac . senα +
y
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4.15.-Un tanque cilíndrico que contiene aire comprimido, tiene un espesor de pared de 7 mm y un radio medio de 25 cm. Las tensiones en la pared del tanque que actuan sobre un elemento girado tienen los valores mostrados en la figura. ¿Cuál será la presión del aire el tanque?. y
90 N/mm2
σ2
130 N/mm2
30 N/mm2 x
σ1
σ1
σ2
El círculo de Mohr correspondiente al estado de tensiones dado será: τ
B 30 C 130 σ1 σ2 N You're90Reading a Preview M
O
σ
30
Unlock full access with a free trial.
A
Download With Free Trial
Centro : OC =
σ x1 + σ y1
2
=
130 + 90 = 110 2 2
2 σ x1 − σ y1 130 − 90 2 + τ x1 y1 = Radio : CA = + 30 2 = 36 2 Scribd 2 your semester with
Master Pr incipales : & The NewTensiones York Times
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= Centro + Radio = 110 + 36 = 146 N / mm 2 2 ON OC CN Centro Radio 110 36 74 /
σ 2 =Only = OC + CM OM $4.99/month. Special offer for students:
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