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Eligheor2

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Eligheor2

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Eligheor2

  1. 1.     Por  el  método  de  las  secciones:     Dibujaremos  el  diagrama  de  cuerpo  libre  para  hacer  el  calculo  de  las   reacciones  en  los  apoyos:     1!!"! 2!!"! !! 2!!"! 2!!"! !! 1!!"! !! 2,62!!! !! !! 0,46!!! COSMOS: Complete Online Solutions Manual Organization System COSMOS: Complete Online Solutions Manual Organization System !! !! !! !! !! !!! Chapter 4, Solution 19. 19. Chapter 4, Solution !!! !!! Free-Body Diagram: Free-Body Diagram: (a) (a) From free-body diagram of lever BCD free-body diagram of lever BCD2,4!!! 2,4!!! From 2,4!!! 2,4!!! ΣM C M C0:= 0:AB T50 (mmmm ) − 200 75(mmmm )0= 0 T ( AB 50 ) − 200 N ( N 75 ) = Σ= Aplicando  las  ecuaciones  de  equilibrio  obtenemos  que:   (b) (b) From free-body diagramlever BCD From free-body diagram of of lever BCD     ∴ TABTAB 300 ∴ = = 300 ΣFx Σ𝐹 x0:= 0:    𝑅N += + + x0.6 ( 300300 N )0= 0 = = 0: 200 N 0  C + 0.6 ( N ) = 200 Cx F ! ! ! ∴ C x = x−380380 N or or C x C x 380380 N = = N ∴C = −N   ΣFy Σ= y0:= 0: y + y + (0.8 ( 300 N )0= 0 C C 0.8 300 N ) = F 𝐹! = 0:     ∴ C y = −240240 N or or C y C y240240 N = = N ∴ Cy = − N   𝑅!! +  Then− 1𝑘𝑁  C = C 2− 2𝑘𝑁 2 (2𝑘𝑁 2− 1𝑘𝑁 )= =2 449.44 N 𝑅!! − 2𝑘𝑁 + 2 2 = − 380 ) + 2( 240 2 0   C Then and and C = x Cx + C y = y ⎛ Cy ⎞ C Cx ⎠ x ⎝ ⎝C ⎠ ( 380 ) + ( 240 ) = 449.44 N ⎛ − 240 ⎞ ⎝ − 380380 ⎠ ⎝− ⎠ ⎞ tan 1 ⎜ ⎜⎟ y ⎟ = 1 ⎜ ⎜ ⎟ = ⎞ = 32. ° θ = θ =−tan −1 ⎛⎟ = tan −tan −1 ⎛ − 240 32.276276° ⎟ ⎜ ⎜ ⎟ or C = 449449 N 32.3° ▹ ▹ or C = N 32.3°
  2. 2.   𝑅!! +   𝑅!! = 8  𝑘𝑁       2,16!!! 2,62!!! ! 0,46!!! 9,6!!!   Con  los  datos  del  dibujo  podemos  sacar:     2,16 𝛼 =   tan!! = 12,68°   9,6                                      
  3. 3. Si  se  conoce  la  inclinación  de  la  armadura,  podemos  encontrar  la  medida  que   poseen  verticalmente  los  elementos  DC,  FE  y  HG     !! !! !! !! !! !!     !! !! !! !! !! !! !! 𝐷𝐶 = 𝑦! +   𝐴𝐵   𝐹𝐸 = 𝑦! +   𝐴𝐵   𝐻𝐺 = 𝑦! +   𝐴𝐵     Donde  tenemos:                                                               𝑦! =   𝐴𝐶 tan 𝛼    ,                         𝑦! =   𝐴𝐸 tan 𝛼  ,                       𝑦! =   𝐴𝐺 tan 𝛼                                                                           𝐴 𝐶 =  2,4  𝑚    ,                         𝐴 𝐸 =  4,8  𝑚  ,                       𝐴 𝐺 =  7,2  𝑚                                                                                                           𝐴 𝐵 =  0,46  𝑚    ,                       𝛼 =  12,68°     Por  lo  tanto     𝐷𝐶 = 𝑦! +   𝐴𝐵 = 2,4 tan 12,68 + 0,46 = 1  𝑚     𝐹𝐸 = 𝑦! +   𝐴𝐵 = 4,8 tan 12,68 + 0,46 = 1,54  𝑚     𝐻𝐺 = 𝑦! +   𝐴𝐵 = 7,2 tan 12,68 + 0,46 = 2,08  𝑚      
  4. 4. COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 19. Free-Body Diagram: cuaciones  de  equilibrio  obtenemos  que:   Aplicando  las  e (a) From free-body diagram of lever BCD   ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0 𝑀! = 0:         ∴ TAB = 300 (b) From free-body diagram of lever BCD −2,4 2 −  4,8 2 − 7,2 Fx = 0: 200 N+ 9,6(𝑅0.6 ( 300 N ) = 0 Σ 2 − 9,6 1   + Cx + !! ) = 0       ∴ C x = −380 N or C x = 380 N 2,4 2 + 4,8 2ΣFy 7,2 2 C++9,6 (1 N ) = 0 + = 0: y 0.8 300 𝑅!! =   = 4  𝑘𝑁   9,6 ∴ C y = −240 N or C y = 240 N   Luego  tenemos  que:     C = Then 2 2 Cx + C y = ( 380 )2 + ( 240 )2 = 449.44 N ⎞ 𝑅!! = 8𝑘𝑁 − 𝑅!!θ = tan −1 ⎛ −y 4𝑘𝑁 =1 ⎛ − 240 ⎞ = 32.276° 4  𝑘𝑁   ⎟ ⎟ = tan − ⎜ ⎜ and = 8  𝑘𝑁 C ⎜C ⎟ ⎝ x⎠ − 380 ⎠ ⎝   or C = ual,   Calcularemos  el  valor  de  cada  componente  de  las  reacciones,  por  lo  c449 Nse   32.3° ▹ seccionará  la  armadura  en  los  elementos  CE,  DE  y  DF     !! 2!!"! !!" !! 1!!"! ! !! !! !! !!" !! !!" 4!!"! 2,4!!!   2,4!!! 2,4 𝛽 =   tan!! Beer, = Russell Johnston, Jr., 67,38°   Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P.1 E. Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.  
  5. 5. COSMOS: Complete Online Solutions Manual Organization System Chapter 4, Solution 19. Free-Body Diagram:ecuaciones  de  equilibrio  obtenemos  que:   Aplicando  las   (a) From free-body diagram of lever BCD   ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0 𝑀! = 0:         ∴ TAB = 300 (b) From free-body diagram of lever BCD COSMOS: Complete Online Solutions Manual Organization System 1 𝐹!"   −  2,4 4𝑘𝑁 =+  2,4 1𝑘𝑁 ΣFx 0: 200 N + C= + 0.6 ( 300 N ) = 0 x 0     ∴ C x = −380 N or C x = 380 N   Chapter 4, Solution 19. ΣFy = 0: C y + 0.8 ( 300 N ) = 0 2,4 4𝑘𝑁 −  2,4 1𝑘𝑁 𝐹!" =   = 7,2  𝑘𝑁        𝑻    𝑒𝑙  𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑜  𝐶𝐸  𝑒𝑠𝑡𝑎  𝑒𝑛  𝑡𝑟𝑎𝑐𝑐𝑖ó𝑛   1 ∴ C y = −240 N or C y = 240 N Free-Body Diagram:   2 2 (a)   C = C 2 + C 2 of ( 380 ) + ThenFrom free-body diagram = lever BCD ( 240 ) = 449.44 N x y ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0 𝑀! = C y ⎞ ⎛ 0:     ⎛ − 240 ⎞ and θ = tan −1 ⎜ ⎟ = tan −1 ⎜ ⎟ = 32.276° ∴ TAB = 300 ⎜C ⎟ ⎝ − 380 ⎠   ⎝ x⎠ (b) From free-body diagram of lever BCD   or C = 449 N 32.3° ▹ −4,8 4𝑘𝑁 + 4,8 1𝑘𝑁 + 2,4 2𝑘𝑁 − 1Fcos 12,68 (𝐹!"Cx− 2,4 (sin 12,680 (𝐹!" ) = 0   Σ x = 0: 200 N + ) + 0.6 300 N ) =   𝐹!" ∴ C = −380 N or C x = 380 N x 4,8 4𝑘𝑁 − 4,8 1𝑘𝑁 − 2,4 2𝑘𝑁 =   = −6,34  𝑘𝑁   Fy 0: C y + 1 cos 12,68 +Σ  2,4= sin 12,680.8 ( 300 N ) = 0 COSMOS: Complete Online Solutions Manual Organization System   ∴ C y = −240 N or C y = 240 N Chapter 4, Solution 19.   2 2 2 2 ∴ 𝐹!" =  6,34  𝑘𝑁        𝑪      𝑒𝑙  𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑜  𝐷𝐹  𝑒𝑠𝑡𝑎  𝑒𝑛  𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑖ó𝑛  449.44 N C = C x + C y = ( 380 ) + ( 240 ) = Then Free-Body Diagram:   ⎛ Cy ⎞ ⎛ − 240 ⎞ (a)   and From free-body −diagram of lever BCD ⎟ = 32.276° θ = tan 1 ⎜ ⎟ = tan −1 ⎜       ⎟ ⎜ ⎝ 380 ⎠ ΣM C = 0:⎝ C xAB ( 50 mm )−− 200 N ( 75 mm ) = 0 T ⎠ 𝑀! = 0:     or C = 449 N ∴32.3° = 300 TAB ▹ (b) From free-body diagram of lever BCD 2,4 1𝑘𝑁 − 2,4 4𝑘𝑁 + 1 634𝑘𝑁 − 1 N + 67,38 (𝐹300 N ) 0   0 ΣFx = 0: 200 sin Cx + 0.6 ( !" ) = = ∴ C = −380 N                 C = 380 N ∴ C y = −240 N 𝐹!" or or C y = 240 N x −2,4 1𝑘𝑁 + 2,4 4𝑘𝑁 − 1 x634𝑘𝑁 =   = N) = 0 −0,93  𝑘𝑁   ΣFy = 1 sin 12,68 0: C y + 0.8 ( 300 2 2 2 2 ∴ 𝐹!" =  0,93  𝑘𝑁        𝑪      𝑒𝑙  𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑜  𝐷𝐸  𝑒𝑠𝑡𝑎  𝑒𝑛  𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑖ó𝑛  449.44 N C = C x + C y = ( 380 ) + ( 240 ) = Then and ⎛ Cy ⎞ − 240 ⎞ ⎟ = tan −1 ⎛ ⎟ = 32.276° ⎜ ⎟ ⎝ − 380 ⎠ ⎝ Cx ⎠ θ = tan −1 ⎜ ⎜ Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies. or C = 449 N 32.3° ▹

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