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Eligheor

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Eligheor

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Eligheor

  1. 1. Usaremos el metodo de las secciones debido a que calcularemos las fuerzas de un pequeño numero de barras de la armadura Primero dibujamos el diagrama de cuerpo libre: !! 20!!"! 20!!"! !! 36!!"! 2,4!!! !! !! !! omplete Online Solutions Manual Organization System !! B ! 4, Solution 19. 4,5!!! 4,5!!! 4,5!!! COSMOS: Complete Online Solutions Manual Organization System dy Diagram: Ahora aplicando las ecuaciones de equilibrio tenemos: OSMOS: Complete Online Solutions (a) From free-body diagram Manual Organization System Chapter 4, Solution 19. of lever BCD ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0 𝑀! = 0:          36 2,4 − 𝐵 13,5 + 20 9 +  20 4,5 = 0 hapter 4, Solution 19. Free-Body Diagram: (b) From free-body diagram of lever BCD of lever BCD (a) From free-body diagram Free-Body Diagram: Then and ∴ TAB = 300 2,4 9 50 N − ΣFx = 0:36 ΣM C +=C20 TAB (+  20 )4,50 N ( 75 mm ) = 0 200 N   +0: + 0.6 300mm )= 200 x 𝐵 =   = 26,4  𝑘𝑁   ↑ (a) From free-body diagram 13,5 BCD of lever ∴ C x = −380 N or C x = 380 N ∴ TAB = 300 ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0 Σ(b) =From C y + 0.8 ( 300 N ) = of lever BCD Fy 0: free-body diagram 0 ∴ TAB = 300 ΣFx = 0: 200 N C36 +   300 0 x ∴ C y = −240 N 0:                  +−C y+=0.6 (𝐾! =N ) = 0 240 N 𝐹! = or (b) From free-body diagram of lever BCD ∴ C x = −380 N or C x = 380 N 2 2 2 2 C = CΣFx C y 0: 200 N)+ + x 240 ) ( 300 N ) = N C ( + 0.6 = 449.44 0 x + = = ( 380 ΣFy =  𝐾! = y  36  𝑘𝑁   → N ) = 0 0: C + 0.8 ( 300 or C x = 380 N C y ⎞ ∴ C x−1= − 380 ⎞ ⎛ ⎛ −240 N θ = tan −1 ⎜ ⎟ = tan ⎜∴ C y = =240276° or C y = 240 N ⎟ − 32. N Σ⎜ C x=⎟ 0: C y ⎝ − 380 ⎠ N ) = 0 F + 0.8 ( 300 ⎝y ⎠ 𝐹! = 0:                  26,4 − 20 −2 20   +  𝐾! = 0 2 2 2 C =C C x −+ C yN= or or = 449240240= 449.44 N ( 380 ) +C( y N ) 32.3° ▹ C ∴ y = 240 = N Then 2 2 ⎛ − 240 = 2 2 ⎛C ⎞ C x = tany−1 = y( 380tan+1 ( 240 ) ⎞ =449.44 N θ + C ⎜ ⎟= ) − ⎜ ⎟ 32.276° ⎜C ⎟ ⎝ − 380 ⎠ ⎝ x⎠ ⎛ Cy ⎞ ⎛ − 240 ⎞ θ = tan −1 ⎜ ⎟ = tan −1 ⎜ ⎟ = 32.276° or C = 449 N ⎜C ⎟ − 380 Then and C = and 32.3° ▹
  2. 2.  𝐾! = 40 − 26,4 = 13,6  𝑘𝑁   ↑ Como ya se calculo el valor en cada componente de las reacciones, seccionaremos la armadura en los elementos AD, CD y CE: !!" !!" 36!!"! !! !! ! 17 15! 1,2!!! !! 8! 15! 17 ! 1,2!!! 8! !! !! OS: Complete Online Solutions Manual Organization System 2,25!!! !!" 26,4!!"! ter 4, Solution 19. 4,5!!! Ahora aplicando nuevamente las ecuaciones de equilibrio tenemos: -Body Diagram: (a) From free-body diagram of lever BCD ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0 𝑀! = COSMOS: Complete Online Solutions Manual Organization System 0:          36 1,2 − 26,4 2,25 − 𝐹!" 1,2 = 0 ∴ TAB = 300 (b) From free-body diagram of lever BCD ΣFx = 0: 36 1,2+ Cx 26,4 ( 300 N ) = 0 200 N − + 0.6 2,25 𝐹!" =   Chapter 4, Solution 19. Free-Body Diagram: ∴ C x = −3801,2 N = −13,5  𝑘𝑁   or C x = 380 N ΣFy = 0: 𝒆𝒍  𝒆𝒍𝒆𝒎𝒆𝒏𝒕𝒐  𝒔𝒆  𝒆𝒏𝒄𝒖𝒆𝒏𝒕𝒓𝒂  𝒆𝒏  𝒄𝒐𝒎𝒑𝒓𝒆𝒔𝒊𝒐𝒏 𝐹!" = 13,5  𝑘𝑁            𝑪   C y + 0.8 ( 300 N ) = 0 Then and (a) From free-body diagram of lever BCD ∴ C y = −240 N or C y = 240 N ΣM C = 0: TAB ( 50 mm8 − 200 N ( 75 mm ) = 0 ) 2 𝑀!2 = ( 380 )2 2 C = C x + C y = 0:                     + ( 240 ) 𝐹!" 449.44 N 0 = 4,5 = 17 ∴ TAB = 300 ⎛ Cy ⎞ (b) From free-body diagram− 240 ⎞  0 BCD ⎛ of lever θ = tan −1 ⎜ ⎟ = tan −1 ⎜𝐹!" = ⎟ = 32.276° ⎜C ⎟ ⎝N+C ⎠ ⎝ ΣFx x = 0: 200 − 380 ⎠x + 0.6 ( 300 N ) = 0 32.3° ∴ C x = −380 N or C = 449 N = 380 N ▹ or Cx ΣFy = 0: C y + 0.8 ( 300 N ) = 0 ∴ C y = −240 N Then C = 2 2 Cx + C y = or ( 380 )2 + ( 240 )2 C y = 240 N = 449.44 N
  3. 3. ter 4, Solution 19. -Body Diagram: (a) From free-body diagram of lever BCD 15 ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0 𝑀! = 0:                     17 𝐹!" 2,4   − 26,4 4,5 = 0 ∴ TAB = 300 (b) From free-body diagram of lever BCD ΣFx = 0: 200 17 Cx + 0.64,5 N ) = 0 N + 26,4 ( 300 𝐹!" =   15 ∴ C x = −380 N 2,4 or = 56,1  𝑘𝑁   C x = 380 N ΣFy = 0: C y + 0.8 ( 300 N ) = 0 𝐹!" =  56,1  𝑘𝑁            𝑻   𝒆𝒍  𝒆𝒍𝒆𝒎𝒆𝒏𝒕𝒐  𝒔𝒆  𝒆𝒏𝒄𝒖𝒆𝒏𝒕𝒓𝒂  𝒆𝒏  𝒕𝒓𝒂𝒄𝒄𝒊𝒐𝒏 ∴ C = −240 N or C = 240 N y Then C = 2 2 Cx + C y = and y ( 380 )2 + ( 240 )2 = 449.44 N θ = tan −1 ⎜ ⎜ ⎛ Cy ⎞ − 240 ⎞ ⎟ = tan −1 ⎛ ⎟ = 32.276° ⎜ ⎟ ⎝ − 380 ⎠ ⎝ Cx ⎠ Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell The McGraw-Hill Companies. or C = 449 N 32.3° ▹

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