EJERCICIO 1 DATOS ππβ ππ2 ππ ππ¦ = 2800 β 2 ππ ππ£ππ 2.22 π = β β πππ β πππ π‘ β = 55 β 4 β 0.95 β = 48.94 ππ 2 2 πΒ΄π = 280
Views 86 Downloads 2 File size 622KB
EJERCICIO 1
DATOS ππβ ππ2 ππ ππ¦ = 2800 β 2 ππ ππ£ππ 2.22 π = β β πππ β πππ π‘ β = 55 β 4 β 0.95 β = 48.94 ππ 2 2 πΒ΄π = 280
π΄π = 2#7 = (2)(3.88) = 7.76 ππ2
π½1 = 0.85 para un πΒ΄π = 280
ππβ ππ2
Equilibrio de Fuerzas πΆπ = ππ 0.85πΒ΄π ππ = ππ¦ π΄π π=
ππ¦ π΄π 2800 Γ 7.76 = = 3.65 ππ 0.85πΒ΄π π 0.85 Γ 280 Γ 25 π=
π 3.65 = = 4.30 ππ π½1 0.85
DeformaciΓ³n del acero ππ =
(π β π)πππ’ 0.003(48.94 β 4.30) = = 0.03114 π 4.30 ππ > 0.005 πΆπ²
ππ‘ππππ§ππ β
= 0.9
Haciendo momento en el acero a tensiΓ³n ππ = πΆπ (π β πβ2) ππ = 0.85πΒ΄π ππ(π β πβ2) ππ = 0.85(280)(3.65)(25)(48.94 β 3.65β2) ππ = 1023220.013 ππ. ππ = 10.23 πππ. π
Calculando momento ultimo ππ’ = β
ππ ππ’ = 0.9(10.13) π΄π = π. ππ π»ππ. π
EJERCICIO 2
DATOS ππβ ππ2 ππ ππ¦ = 2800 β 2 ππ πΒ΄π = 210
π = β β πππ β πππ π‘ β ππ£ππ = 55 β 4 β 0.95 β 2.87 = 47.18 ππ π΄π = 6πΒ°9 = (6)(6.45) = 38.70 ππ2 Porcentaje de refuerzo: ππ = 0.0371 π=
π΄π 38.70 = = 0.03281 β π < ππ πΉππππ πππ π‘πππ πΓ³π ππ 25π₯47.18
Entonces: ππ = ππ¦ π¦ ππ =
(π β π)πππ’ 2,800 ; ππ¦ = = 0.0013 π 2.03π₯106
Equilibrio: πΆπ = ππ 0.85πΒ΄π ππ = ππ¦ π΄π
π=
π=
ππ¦ π΄π 0.85πΒ΄π π
2800 Γ 38.70 = 24.28 ππ 0.85 Γ 210 Γ 25
π=
π 24.28 = = 28.56 ππ π½1 0.85
Entonces: ππ =
0.003(47.18 β 28.56) = 0.00196 β ππ¦ < ππ < ππ = 0.005 28.56
ππππ ππ π‘ππππ πππΓ³π β π = 0.65 + (ππ β ππ¦ ) (
π = 0.65 + (0.00196 β 0.00138) (
0.25 ) 0.005 β ππ¦
0.25 ) = 0.69 0.005 β 0.00138
Momento nominal, haciendo sumatoria de momentos en acero a tensiΓ³n: ππ = πΆπ (π β πβ2) = 0.85πΒ΄π ππ(π β πβ2) ππ = 0.85(210)(24.28)(25)(47.18 β 24.18β2 ) ππ = 37.97 πππ. π Momento ultimo: ππ’ = β
ππ = 0.69(37.97) π΄π = ππ. ππ π»ππ. π
EJERCICIO 3
DATOS: ππβ ππ2 ππ ππ¦ = 3500 β 2 ππ ππ£ππ 2.22 π = β β πππ β πππ π‘ β = 55 β 4 β 0.95 β = 48.49 ππ 2 2 ππ£ππ 1.90 πΒ΄ = πππ + πππ π‘ + = 4 + 0.95 + = 5.90 ππ 2 2 πΒ΄π = 420
π΄π = 3#7 = (3)(3.88) = 11.64 ππ2 π΄Β΄π = 2#6 = (2)(2.85) = 5.70 ππ2 π½1 = 0.75 para un πΒ΄π = 420
ππβ ππ2
Relaciones geomΓ©tricas πππ’ πΒ΄π ππ = = π π β πβ² π β π ππ =
πππ’ (πβπ) π
πΒ΄π =
πππ’ (πβπΒ΄) π
Equilibrio de Fuerzas πΆπ + πΆπ = ππ 0.85πΒ΄π ππ + π΄Β΄π πΒ΄π = ππ¦ π΄π 0.85πΒ΄π ππ +
π΄Β΄π πΈπ πππ’ (π β πΒ΄) = ππ¦ π΄π π
(5.70)(2.03π₯106 )(0.003)(π β 5.90) = 11.64(3500) π
0.85(420)(0.75)(25)π +
6693.75π +
34713(π β 5.90) = 40740 π
34713π β 204806.7 = 40740π β 6693.75π 2 6693.75π 2 β 6027π β 204806.7 = 0 Resolviendo ecuaciΓ³n cuadrΓ‘tica se obtiene π1 = 5.99 ππ π2 = β5.01 ππ
π = π½1 π = 0.75(5.99) = 4.49 ππ
Comprobando deformaciones ππ¦ =
πΒ΄π =
πππ’ (πβπΒ΄) π
ππ =
=
πππ’ (πβπ) π
3500 = 0.001724 2.03π₯106
0.003(5.99β5.90) 5.99
=
= 0.0000451
0.003(48.94β5.99) 5.99
ππ > 0.005
= 0.021511
ππ‘ππππ§ππ β
= 0.9
πΒ΄π < ππ¦ ππΎ ππ > ππ¦
Calculando esfuerzo del acero πΒ΄π = πΒ΄π πΈπ = (0.0000451)(2.03π₯106 ) = 91.50
ππ ππ2
Haciendo momento en el acero a tensiΓ³n
ππ = πΆπ (π β πβ2) + πΆπ (π β πΒ΄) ππ = 0.85πΒ΄π ππ(π β πβ2) + π΄Β΄π πΒ΄π (π β πΒ΄) ππ = 0.85(420)(4.49)(25)(48.94 β 4.49β2) + (5.70)(91.50) (48.94 β 5.90) ππ = 18.98 πππ. π
Calculando momento ultimo ππ’ = β
ππ ππ’ = 0.9(18.75) π΄π = ππ. ππ π»ππ. π
EJERCICIO 4
DATOS π β² π = 350 ππ/ππ2 ππ¦ = 2800 ππ/ππ2 π΄π 1 = π΄π 2 = π΄π 3 = π΄π 4 = 2(1.98) = 3.96ππ2 ππ¦ =
ππ¦ 2800 = = 0.001379 πΈ 2.03π₯106 π = π½1 π
Como f'c = 350 kg/cm2;
Ξ²= 0.8
Calculando distancias entre hasta las varillas π1 = π1 =
β
πΒΊ8 + β
ππ π‘ + π
ππ 2
1.59 + 0.95 + 4.0 = 5.75 ππ 2
π4 = β β π4 = 55 β
β
πΒΊ8 β β
ππ π‘ β π
ππ 2
1.59 β 0.95 β 4.0 = 49.26 ππ 2
Para d2 y d3 se calcula la separaciΓ³n vertical S π=
π4 β π1 49.26 β 5.75 = = 14.5 ππ 3 3
Calculando π2 = π1 + π = 5.75 + 14.5 = 20.25 ππ π3 = π2 + π = 20.25 + 14.5 = 34.75 ππ
Relaciones GeomΓ©tricas (Se asume que todas las secciones de acero estΓ‘n a tensiΓ³n y que las secciones 2, 3 y 4 ya han entrado a fluencia. πππ’ 0.003 ππ 1 ππ 2 ππ 3 ππ 4 = = = = = π π π1 β π π2 β π π3 β π π4 β π
ππ 1 =
0.003(π1 β π) < ππ¦ π
ππ 2 =
0.003(π2 β π) > ππ¦ π
ππ 3 =
0.003(π3 β π) > ππ¦ π
ππ 4 =
0.003(π4 β π) > ππ¦ π
Equilibrio Se tiene que ππ 2 = ππ 3 = ππ 4 = ππ¦ πΆπ = ππ 0.85 π β² π ππ = π΄π 1 ππ¦ + π΄π 2 ππ¦ + π΄π 3 ππ¦ + π΄π 4 ππ¦
0.85 (350)(0.8)(25)π = 3.96 (
0.003(5.75 β π) ) (2.03π₯106 ) + 3(3.96)(2800) π
Simplificando la ecuaciΓ³n se obtiene: 5950 π 2 β 9147.6 π β 139071.24 = 0 Factorizando se llega a que: πΆ = 5.66 ππ Comprobando con las deformaciones: ππ 1 =
0.003(5.75 β 5.66) = 0.000048 < ππ¦ 5.66
ππ!
ππ 2 =
0.003(20.25 β 5.66) = 0.007733 > ππ¦ 5.66
ππ!
ππ 3 =
0.003(34.75 β 5.66) = 0.015419 > ππ¦ 5.66
ππ!
ππ 4 =
0.003(49.26 β 5.66) = 0.023109 > ππ¦ 5.66
ππ!
TambiΓ©n: π = π½1 π π = (0.8)(5.66) = 4.53 ππ
Calculando βM en el eje que pasa por d1 π ππ = πΆπ (π1 β ) + ππ ((π2 β π1 ) + (π3 β π1 ) + (π4 β π1 )) 2 ππ = 33677 (5.75 β
4.53 ) + 11088((20.25 β 5.75) + (34.75 β 5.75) + (49.26 β 5.75)) 2 ππ = 1083131.225 πΎπ. ππ ππ = 10.82 πππ. π
Calculando Momento Γltimo. Como ππ 4 =
0.003(49.26 β 5.66) = 0.023109 > ππ = 0.005 5.66
Entonces β
= 0.9 ππ’ = β
ππ ππ’ = (0.9)(10.82) π΄π = π. ππ π»ππ. π
EJERCICIO 5
DATOS π β² π = 280
πΎπ β 2 ππ
ππ¦ = 2800
πΎπβ ππ2
π = 50 β 4 = 44 ππ π΄π = 2π5 = (2)(1.98) = 3.96 ππ2 Γrea a compresiΓ³n: π΄π =
ππ β² 2
De relaciones geomΓ©tricas: 50 π = 25 π β² πβ² =
25 π 50
πβ² =
π ; 2
Entonces: π΄π =
π π π2 ( ) β π΄π = 2 2 4
Equilibrio: (fs=fy) πΆπ = ππ 0.85π β² π π΄π = π΄π ππ¦ (0.85)(π β² π) (
π2 ) = π΄π ππ¦ 4
4(π΄π ππ¦) π=β 0.85π β² π (4)(2800)(3.96) π=β (0.85)(280) π = 13.65 ππ TambiΓ©n: π=
π 13.65 = 0.85 0.85
π = 16.06 ππ
Comprobando con deformaciones: ππ¦ =
ππ¦ 2800 = β ππ¦ = 0.001379 πΈπ 2.03 π₯ 106
De Relaciones GeomΓ©tricas πππ’ ππ = π πβπ ππ =
(π β π)πππ’ (44 β 16.06)(0.003) = π 16.06 ππ = 0.00522
πΆπππ ππ > ππ¦ π¦ π¦π ππ’π ππ > 0.005,
π = 0.9
Haciendo Momento en el acero π = 0.85π β² π π΄π (π β
π = 0.85π β² π
π = (0.85)(280) (
2π ); 3
π΄π =
π2 4
π2 2π (π β ) 4 3
13.652 2 β 13.65 ) ) (44 β 4 3
ππ = 386907.99 ππ. ππ Momento ΓΊltimo: ππ’ = πππ = (0.9)(38907.99) = 348217.19ππ. ππ β π΄π = π. ππ πππ. π
EJERCICIO 6
DATOS π β² π = 210 ππ/ππ2 ππ¦ = 3500 ππ/ππ2 π΄π = 2(3.88) = 7.76ππ2 ππ¦ =
ππ¦ 3500 = = 0.001724 πΈ 2.03π₯106
π = π½1 π
Ξ² = 0.85
π = 50 β 7.5 = 42.5 ππ Relaciones GeomΓ©tricas 50 50 β π = 25 πβ² πβ² =
50 β π 2
TambiΓ©n: ππ =
πππ’ (π β π) > ππ¦ π
Equilibrio: πΆπ = ππ 0.85 π β² π (
0.85 π β² π (
0.85 (210) (
625 β (
βπ (β β π)π β² β ) = π΄π ππ¦ 2 2
βπ (β β π) (50 β π) β ) = π΄π ππ¦ 2 2 2
(50)(25) (50 β π)2 β ) = 7.76(3500) 2 2
502 β 100π + π2 7.76(3500) )= 4 0.85 (210)
Simplificando: π2 β 100π + 608.64 = 0 Factorizando se obtiene: π = 6.51 ππ Entonces: π=
π 6.51 = = 7.66 ππ π½ 0.85
Comprobando con Relaciones geomΓ©tricas: ππ =
0.003(42.5 β 7.66) = 0.013645 > ππ¦ 7.66
Por lo que: πβ² =
50 β 6.51 = 21.75 ππ 2
CΓ‘lculo de Momento: π π ππ = πΆπ1 (π β ) + πΆπ2 (π β ) 2 3
ππ!
(25 β π β² )π π π ππ = 0.85 π β² π π β² π (π β ) + 0.85 π β² π ( ) (π β ) 2 2 3 ππ = 0.85(210)(21.75)(6.51) (42.5 β + 0.85(210) (
6.51 ) 2
(25 β 21.75)(6.51) 6.51 ) ) (42.5 β 2 3
ππ = 1068043.799 ππ. ππ = 10.68 πππ. π
Debido a que ππ =
0.003(42.5 β 7.66) = 0.013645 > ππ = 0.005 7.66
Momento ΓΊltimo: ππ’ = β
ππ ππ’ = (0.9)10.68 π΄π = π. ππ π»ππ. π
β
= 0.9
EJERCICIO 7
DATOS π β² π = 210
πΎπβ ππ2
ππ¦ = 3500
πΎπ β 2 ππ
π = 50 β 4 = 44 ππ π΄π = 2π6 = (2)(2.85) = 5.70 ππ2
Equilibrio: Asumiendo a=7.5 cm, y acero en fluencia (fs=fy) πΆπ = ππ 0.85π β² π ππ = π΄π ππ¦ (0.85)(210)(7.5)(25) = (5.7)(3500) 33468.75 = 19950
El esfuerzo del concreto cuando a = 7.5 cm mayor al del acero, βaβ debe ser menor, entonces se considera una secciΓ³n rectangular de ancho 25 cm. Equilibrio suponiendo fluencia en el acero: πΆπ = ππ 0.85π β² π ππ = π΄π ππ¦ (0.85)(210)(π)(25) = (5.7)(3500) π = 4.47ππ < 7.5ππ Solo una parte del patΓn estΓ‘ trabajando. Entonces: π=
π 4.47 = = 5.26ππ 0.85 0.85
Comprobando fluencia en el acero: ππ¦ =
ππ¦ 3500 = β ππ¦ = 0.00152 πΈπ 2.03 π₯ 106
De relaciones geomΓ©tricas: πππ’ ππ (π β π)πππ’ (44 β 5.26)(0.003) = β ππ = β β ππ = 0.02209 π πβπ π 5.26 πΆπππ ππ > ππ¦ π¦ π¦π ππ’π ππ > 0.005,
π = 0.9
Momento en el acero: π π = (0.85π β² π ππ) (π β ) 2 π = (0.85)(210)(4.47)(25) (44 β
4.47 ) 2
ππ = 833102.12 ππ. ππ Momento ΓΊltimo: ππ’ = πππ = (0.9)(833102.12) = 749791.91ππ. ππ β π΄π = π. ππ πππ. π
EJERCICIO 8
DATOS ππβ ππ2 ππ ππ¦ = 2800 β 2 ππ πΒ΄π = 210
π = 50 β 3.75 = 46.25 ππ π΄π = 2πΒ°6 = (2)(2.85) = 5.70 ππ2 ππ¦ = 0.00138 Asumiendo a ππ = 0.005 πΉππππ π π‘πππ πΓ³π π¦ β
= 0.90
Momento nominal, haciendo sumatoria de momentos en acero a tensiΓ³n: ππ = πΆπ (π β πβ2) = 0.85πΒ΄π ππ(π β πβ2) ππ = 0.85(210)(8.94)(10)(46.25 β 8.94β2 ) ππ = 6.67 πππ. π Momento ultimo: ππ’ = β
ππ = 0.90(6.67) π΄π = π. ππ π»ππ. π
EJERCICIO 9:
DATOS ππβ² = 210
ππ ππ , ππ¦ = 4200 2 ππ ππ2
ππ Β° 7 = 2.22 ππ, π΄πΒ°7 = 3.88 ππ2 π΄π = 4(3.88) = 15.52 ππ2 π = 45 β 7.5 = 37.5 ππ Relaciones geomΓ©tricas: πππ’ ππ πππ’ (π β π) = ; ππ = π ( π β π) π Porcentaje de acero balanceado: ππ = ππ€π + ππ ππ€π = 0.0214 ( π‘ππππ π΄. 7 ) ππ =
π΄π π 0.85 ππβ² βπ (π β ππ€ ) 0.85(210)(5)(25 β 10) ; π΄π π = = = 3.19 ππ2 ππ€ π ππ¦ 4200 ππ =
π΄π π 3.19 = = 0.0085 ππ€ π (10)(37.5)
ππ = 0.0214 + 0.0085 = 0.0299
Porcentaje de acero a tensiΓ³n:
π=
π΄π 15.52 = = 0.0414 ; π > ππ ππ€ π 10(37.5)
πππππ π ππππππππΓ³π
ππ < ππ¦ β ππ = ππ πΈπ Equilibrio: πΆπ€π + πΆπ = ππ 0.85πβ²π π ππ€ + π΄π π ππ = π΄π ππ πππ’ (π β π) πππ’ (π β π) 0.85πβ²π π ππ€ + π΄π π ( ) πΈπ = π΄π ( ) πΈπ π π π=
π ππ ; πβ²π = 210 < 280 β π½1 = 0.85 π½1 ππ2
π 0.003 (37.5 β ) 0.85 0.85(210)(10)π + 3.19 ( ) (2.03 Γ 106 ) π 0.85 π 0.003 (37.5 β ) 0.85 = 15.52 ( ) (2.03 Γ 106 ) π 0.85 Despejando a: π = 21.20 ππ ;
π > βπ = 5 ππ πΆπ²
Entonces:
π=
π 21.20 = = 24.94 ππ π½1 0.85
Comprobando deformaciΓ³n del acero: πππ’ (π β π) 0.003(37.5 β 24.94) = = 0.00151083 π 24.94 ππ¦ 4200 ππ¦ = = = 0.00206897 ; πΊπ < πΊπ πΆπ² πΈπ 2.03 Γ 106 ππ =
ππ = ππ πΈπ = 0.00151083(2.03 Γ 106 ) = 3066.98
ππ < ππ¦ ππ2
Momento en el acero a TensiΓ³n: ππ = πΆπ€π (π β
βπ π ) + πΆπ (π β ) 2 2
ππ = 0.85πβ²π π ππ€ (π β ππ = 0.85(210)(21.20)(10) (37.5 β
βπ π ) + π΄π π ππ (π β ) 2 2
21.20 5 ) + 3.19(3066.98) (37.5 β ) 2 2
ππ = 13.60 π. π
ππ’ = β
ππ
ππ < ππ¦ β β
= 0.65
ππ’ = 0.65(13.60) π΄π = π. ππ π». π
EJERCICIO 10:
DATOS ππβ² = 280
ππ ππ , ππ¦ = 4200 , π
= 22.5 ππ 2 ππ ππ2
ππ Β° 7 = 2.22 ππ, π΄πΒ°7 = 3.88 ππ2 π΄π = 2(3.88) = 7.76 ππ2 π = 45 β 7.5 = 37.5 ππ Relaciones geomΓ©tricas: πππ’ ππ πππ’ + ππ πππ’ (π β π) = = ; ππ = π ( π β π) π π π 2
π = π
(1 β cos ) π = π
β2 β 2 cos π Γrea de segmento circular π΄ππ = π
2 ( π΄ππ π ππ π‘π ππ ππππππππ 1
Γrea del sector triangular:
π΄1 = 2 πβ
Γrea del sector circular:
π΄ π = π΄1 + π΄ππ
π π π β sin cos ) 2 2 2
π
2 ( π β sin π) = 2
Equilibrio: πΆπ = ππ 0.85ππβ² π΄ππ = π΄π ππ¦ 0.85(280)π΄ππ = 7.76(4200) π΄ππ = 136.94 ππ2 (Γ‘rea del segmento circular)
π΄ππ =
π
2 ( π β sin π) 2
136.94 =
136.94 =
π
2 ( π β sin π) 2
22.52 ( π β sin π) 2
0.54099753 = π β sin π
Resolviendo para ΞΈ: π = 1.540539837 Entonces: π 1.540539837 π = π
(1 β cos ) = 22.5 (1 β cos ) = 6.35 ππ 2 2
π=
π ππ ; πβ²π = 280 β π½1 = 0.85 π½1 ππ2 π=
6.35 = 7.47 ππ 0.85
Evaluando deformaciΓ³n del acero a tensiΓ³n: ππ =
πππ’ (π β π) 0.003(37.5 β 7.47) = = 0.01206024 π 7.47
ππ > 0.005
β
= 0.9 πππππ π πππππΓ³π
Γrea del sector triangular: π = π
β2 β 2 cos π = 22.5β2 β 2 cos 1.540539837 = 31.33 ππ 1 1 π΄1 = πβ = (31.33)(22.5 β 6.35) = 252.99 ππ2 2 2 2 2 π¦1 = β = (22.5 β 6.35) = 10.77 ππ (πππ ππ ππ ππππ‘ππ πππ πππππ’ππ) Μ
Μ
Μ
3 3 Γrea del sector circular: π΄ π = π΄1 + π΄ππ = 252.99 + 136.94 = 389.93 ππ2
πΜ
=
π 1.540539837 2π
sin 2 2(22.5) sin 2 = = 13.56 ππ (πππ ππ ππ ππππ‘ππ πππ πππππ’ππ) π 1.540539837 3( ) 3 ( ) 2 2
Recordando EstΓ‘tica: β π΄ πΜ
= β π΄π π¦Μ
π π΄ π πΜ
= π΄1 Μ
Μ
Μ
π¦1 + π΄ππ Μ
Μ
Μ
Μ
Μ
π¦π΄ππ 389.93(13.56) = 252.99(10.77) + 136.94 Μ
Μ
Μ
Μ
Μ
π¦π΄ππ Μ
Μ
Μ
Μ
Μ
π¦π΄ππ = 18.71 ππ (ππππππ πππ ππ ππ ππππ‘ππ πππ πππππ’ππ)
Momento en el acero a tensiΓ³n: ππ = πΆπ π = 0.85ππβ² π΄ππ (π β (π
β Μ
Μ
Μ
Μ
Μ
)) π¦π΄ππ ππ = 0.85(280)(136.94)(37.5 β (22.5 β 18.71)) ππ = 10.987 π. π ππ’ = β
ππ
ππ > 0.005 β β
= 0.90
ππ’ = 0.90(10.987) π΄π = π. ππππ π». π