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Clase de dinamica cls # 6

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  • 1. CLASE DE DINAMICA REALIZADO POR: ING. ROMEL VALENZUELA ING. FERNANDO LEIVA Clase 6
  • 2. Ejemplo Datos: L=100pulg EI=10x10^8lbs*pulg² W=3000lbs k=2000lbs/pulg Xo=1 pulg Vo=20pulg/seg Calcular el desplazamiento y la velocidad después de 1seg = + 2 = 3 = 3 + 2 = 3 + 2 = = = 3000 32.20 ∗ 12 = 7.76 ∗ = = 3 ∗ 10 ∗ 10 + 2 ∗ 2000 ∗ 100 7.76 ∗ 100 = 23.54
  • 3. La ecuación e movimiento es: − = 0 = ( + ∅) v = − ( + ∅) 0 = 1 = (23.54(0) + ∅) 1 = (∅) v 0 = 20 = − 23.54 (23.54(0) + ∅) 20 = − 23.54 (∅) = −20 23.54 ∅ 1 = (∅) 1 = cos(∅) 1 −20 23.54 ∅ = cos(∅) 23.54 ∅ −20 = cos(∅) tan ∅ = −20 23.54
  • 4. ∅ = tan −20 23.54 ∅ = −0.7043 = −20 23.54 ∅ = −20 23.54 −0.7043 = 1.312 = ( + ∅) = 1.312 (23.54 − 0.7043) 1 = 1.312 (23.54 1 − 0.7043) 1 = −0.8713 ∗ v = −1.312(23.54) (23.54 − 0.7043) v = − ( + ∅) v 1 = −1.312(23.54) (23.54 1 − 0.7043) v 1 = 23.09
  • 5. Ejemplo Calcular la frecuencia para el desplazamiento horizontal del marco de acero. Se supone que la viga es infinitamente rígida, se desprecia la masa de las columnas, determine la variación del momento flexionarte y cortante de las columnas, para las condiciones iniciales Xo, Vo Solución Del AISC 10WF33 Ixx=171 pulg^4 8WF24 Ixx=82.50pulg^4 E 29000Ksi = 29,000,000 ∗ 171 = 4.959 9 ∗ = 29,000,000 ∗ 82.50 = 2.3925 9 ∗
  • 6. = 12 Para columnas rígidas en sus dos extremos = 3 Para columnas rígidas en uno de sus extremos y articulado en el otro k10 12 EI10 L 3 := k10 1.993 10 4  lbf in = w 50000lbf:= m w g := m 129.504 lbf s 2  in = k8 3 EI8 L 3 := k8 2.404 10 3  lbf in = ke 2k8 k10+:= ke 2.474 10 4  lbf in = ω ke m := ω 13.821 1 s =
  • 7. T 2π ω 0.455 s=:= f 1 T 2.2 1 s =:= x t( ) Acos ωt ϕ+( )= x0 20in:= v0 0 in s := Condiciones iniciales: x 0( ) x0= Acos ω 0( ) ϕ+( )= x0 Acos ϕ( )= v t( ) Aωsen ω t ϕ+( )-= v 0( ) Aωsen ω 0( ) ϕ+( )-= 0= Aωsen ϕ( )- 0= sen ϕ( ) 0= ϕ 0= x0 Acos ϕ( )= A x0:= A 20in= = 20cos(13.821 ) = −20(13.821)sen(13.821t) = −276.42sen(13.821t)
  • 8. = 6 ( ) = 6 4.959 9 12 ∗ 12 20cos(13.821 ) = 28,697,916.67cos(13.83 ) = 12 ( ) = 12 4.959 9 12 ∗ 12 20cos(13.821 ) = 398,582.18cos(13.821 ) = 3 ( ) = 3 2.3925 9 12 ∗ 12 20cos(13.821 ) = 6,922,743.06cos(13.83 ) = 3 ( ) = 3 2.3925 9 12 ∗ 12 20cos(13.821 ) = 48,074.61cos(13.821 )
  • 9. Ejemplo Momento en los apoyos ∆= 192 = 192 = = 192 = = 192 2 = 2 = = = 96 ( ) = 8 = 24 ( ) El desplazamiento máximos de una viga doblemente empotrada con carga puntual
  • 10. Si L=120in, EI=10^9 lb*in^2, w=5000lbs, Xo=0.50in, Vo=15in/s, Determine x,v,a para t=2seg Además de las reacciones en los extremos = = 192 = 92.66 Solución 0 = 92.66(0) + ∅ = 0.50 0 = − 92.66 92.66 0 + ∅ = 15 −92.66 ∅ = 15 = − 0.1619 ∅ 92.66(0) + ∅ = 0.50 ∅ = 0.50 = 0.50 cos ∅ = 0.50 cos ∅ = − 0.1619 ∅ tan ∅ = − 0.1619 0.50 ∅ = tan − 0.1619 0.50 = −0.3132 = 0.50 cos −0.3132 = 0.5255
  • 11. = 0.5255cos(92.66 − 0.3132) = −0.5255 92.66 (92.66 − 0.3132) = −48.69 (92.66 − 0.3132) = 96 ( ) = 96 0.5255 cos 92.66 − 0.3132 2 = 274,527.19 = 24 ( ) = 24 (0.5255cos(92.66 − 0.3132)) 2 = 8,235,815.73 ∗ 2 = 0.5255cos(92.66(2) − 0.3132) 2 = −0.4941 2 = −48.69 (92.66(2) − 0.3132) 2 = −16.57 = − ( + ∅) 2 = 353.56