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Ejercicios n7

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Ejercicios n7

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Ejercicios n7

  1. 1. Ejercicio Práctico Único: Para la viga y condiciones de carga mostradas en la figura: a) Diagrama de cuerpo libre de la viga con sus reacciones; b) Cálculo de las reacciones en los apoyos; c) Diagramas de fuerza cortante y momento flexionante o flector; d) Momento de Inercia de la viga; e) Ubicación del eje neutro de la viga; y f) determínense los esfuerzos máximos de tensión y de compresión de la viga. 25 mm 25 mm
  2. 2. Free-Body Diagram: Chapter 4, Solution 19. b) and 0,6!!! 𝑃𝑜𝑟  𝑠𝑖𝑚𝑒𝑡𝑟𝑖𝑎  𝑒𝑛  𝑒𝑙  𝑠𝑖𝑠𝑡𝑒𝑚𝑎    𝑅! = 𝑅! 𝐹! = 0      , 𝑅! = ( 380 )2 + ( 240 )2 = 449.44 N C y = 240 N 32.3° ▹ Fuerza Homogénea 12   𝑘𝑁 𝑚 0,9!!! or C = 449 N ⎛ Cy ⎞ − 240 ⎞ ⎟ = tan −1 ⎛ ⎟ = 32.276° ⎜ ⎟ ⎝ − 380 ⎠ ⎝ Cx ⎠ 2 2 Cx + C y = 0,3! θ = tan −1 ⎜ ⎜ C = or !! ∴ C y = −240 N C x = 380 N ∴ TAB = 300 !! Then or ΣFy = 0: C y + 0.8 ( 300 N ) = 0 ∴ C x = −380 N ΣFx = 0: 200 N + Cx + 0.6 ( 300 N ) = 0 (b) From free-body diagram of lever BCD ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0 (a) From free-body diagram of lever BCD COSMOS: Complete Online Solutions Manual Organization System a) 1,8  𝑚 = 21,6  𝑘𝑁 21,6!!"! 24!!"! 24!!"! !! 0,9!!! 1,8!!! !! 0,3! 𝑅! + 𝑅! = 69,6  𝑘𝑁 69,6  𝑘𝑁 = 34,8  𝑘𝑁 2 𝑅! = 𝑅! = 34,8  𝑘𝑁 0,6!!! 𝑅! + 𝑅! − 24 + 21,6 + 24 𝑘𝑁 = 0
  3. 3. (b) From free-body diagram of lever BCD ΣM C = 0: TAB ( 50 mm ) TAB ∴ −2 ΣFx = 0: 200 N + Cx + 0.6 ( 300 N ) = 0 ΣFx = 0: 200(b) +From free-body diagram of lever BCD N Cx + 0.6 ( 300 N ) = 0 ∴ C x = −380 N or C x = 380 N ∴ C x = −380 N or C = 380 NC + 0.6 ( 300 N ) = 0 ΣFx = 0: x 200(b) +From free-body diagram of lever BC N x ΣFy = 0: C y + 0.8 ( 300 N ) = 0 (a) From free-body diagram of lever BCD ΣFy = 0: C y + 0.8 ( 300 N ) = 0 ∴ C = −380 N ΣFx = 0: x 200 N + Cx + 0. or C = 380 N x ∴ C y = −240 N or C y = 240 N ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0 ∴ C y = −240 N or y = 0: y C y + 0.8 ( 300 N ) = 0 ∴ C x = −380 N C = 240 N ΣF 2 2 2 2 C = C x + C y = ∴(TAB )= + ( 240 ) = 449.44 N 380 300 Then 2 2 ΣF 2 2 or y = 0: y C y + 0.8 ( 300 N C = 240 N C = C x + C y = ( 380 ) + ( 240 ) ∴ C y = −240 N = 449.44 N Then (b) From free-body diagram of lever BCD Cy ⎞ ⎛ ⎛ − 240 ⎞ 2 2 ∴ C y = −240 N and θ = tan −1 ⎜ ⎟ = tan −1 ⎜ 2 ⎟ = 32.276° C Cx + 2 ΣFx = 0: 200 N + Cx + 0.6 ( 300 N ) = ⎜ C x ⎟ 0 ⎠ ⎝ − 380 ⎠= tan −1 ⎛ C y ⎞ Then −1 ⎛ − 240 ⎞ = 32.276°C y = ( 380 ) + ( 240 ) = 449.44 N ⎝ ⎟ ⎜ and θ ⎜ C ⎟ = tan ⎜ − 380 ⎟ = 2 2 2 ⎠ C ⎠ ⎝ ∴ C x = −380 N or C x = 380 N or C =x 449 N ⎝32.3° ▹ ⎛ C y ⎞ Then −1 ⎛ − 240 ⎞ = C x + C y = ( 380 ) −1 ⎟ ⎜ and θ = tan C = 449 tan ⎜32.3° ▹= 32.276° or ⎜ C ⎟ = N ⎝ − 380 ⎟ ⎠ ⎝ x⎠ ⎛ Cy ⎞ ΣFy = 0: C y + 0.8 ( 300 N ) = 0 ⎛ − 24 and θ = tan −1 ⎜ ⎟ = tan −1 ⎜ ⎜ = ⎟ − 38 or C C x 449 N ⎝32.3 ⎠ ⎝ ∴ C y = −240 N or C y = 240 N Free-Body Diagram: (a) From free-body diagram of lever BCD Chapter 4, Solution 19. (a) From 200 ΣM C = Free-Body Diagram: N ( 75 mm ) = of lever BCD 0: TAB ( 50 mm ) −free-body diagram 0 ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0 (a) From AB ∴ T free-body Free-Body Diagram:= 300 diagram of lever BCD ∴ − 200 N ( 75 ΣM = 0: T ( 50 mm ) TAB = 300 diagram of lever BC (b) From free-body diagram of lever BCD (a) From free-body mm ) = 0 : Complete Online Solutions Manual Organization System c) Se consideran 5 secciones en la viga: COSMOS: Complete Online Solutions Manual Organization System er 4, Solution 19. AB Body Diagram: 1. 𝐹! = 0      ,        34,8  𝑘𝑁 − 𝑉! = 0       ⇒       𝑉! = 34,8  𝑘𝑁 (a) From free-body diagram of lever BCD ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0 𝑀! = 0      ,          34,8  𝑘𝑁 0  𝑚 + 𝑀! = 0         ⇒       𝑀! = 0 ion 19. am: C ine Solutions Manual Organization System (b) From free-body diagram of lever BCD ΣFx = 0: 200 N + Cx + 0.6 ( 300 N ) = 0 2. ∴ C = −380 N C = 380 N or x 𝐹! = 0      ,        x 34,8 − 24  𝑘𝑁 − 𝑉! = 0       ⇒       𝑉! = 10,8  𝑘𝑁 ΣFy = 0: C y + 0.8 ( 300 N ) = 0 (a) From free-body diagram of lever BCD ∴C −240 N or ΣM C = 0: TAB ( 50 mm ) y−=200 N ( 75 mm ) = 0 C y = 240 N 𝑀! = 0      ,           34,8 0,6  𝑘𝑁 ∙ 𝑚 + 𝑀! = 0         ⇒       𝑀! = −20,88    𝑘𝑁 ∙ 𝑚 C AB 𝑀! = 0,         34,8 3 − 24 2,4 − 21,6 1,2  𝑘𝑁 ∙ 𝑚 + 𝑀! = 0                               ( 2 + ( 240 ) = )2−20,88  𝑘𝑁 449.44 N ∙ 𝑚 C = C x + C y = 380 Then OSMOS: Complete Online Solutions Manual Organization System = ⇒       𝑀! 2 2 ∴ TAB = 300 , Solution 19. and C = θ = tan −1 ⎜ ⎜ Then and ΣM C =2 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0 ( 380𝑀! + (0      ,          𝑃𝑜𝑟  𝑠𝑖𝑚𝑒𝑡𝑟𝑖𝑎       ⇒       𝑀! = 0 ) = 240 )2 = 449.44 N ∴ TAB = 300 ⎛ Cy ⎞ ⎛ − 240 ⎞ −1 ⎟ free-body diagram.276° ⎜ θ = tan(b) From tan −1 ⎜ ⎟ = 32 of lever BCD ⎜C ⎟= ⎝ − 380 ⎠ ⎝ x⎠ ΣFx = 0: 200 N + Cx + 0.6 ( 300 N ) = 0 or C = 449 N 32.3° ▹ ∴ C x = −380 N or C x = 380 N C = Then 2 2 Cx + C y = y Diagram: mplete Online Solutions Manual Organization System (b) From free-body diagram of lever BCD ⎛ Cy ⎞ 1 ⎛ − 240 ⎞ and θ = tan −1 ⎜0: ⎟ = tan −+⎜Cx + 0.6= 300276°= 0 ΣFx = ⎜ 200 ⎟ N ⎝ − 380 ⎟ ( 32. N ) ⎠ ⎝ Cx ⎠ hapter 4, Solution 19.5. ∴ C x = −380 N or C = 380 N or C = x449 N 32.3° ▹ ΣFy = 0: C y + 0.8 ( 300 N ) = 0 𝐹 = 0         ⇒       𝑉! = 0 ! Free-Body Diagram: (a) C y = −free-body diagram C ylever BCD of = 240 N ∴ From 240 N or Free-Body Diagram: or C = 449 N 2 2 2 2 ∴T C = C x + C y = ( 380 ) + ( 240 ) = 449.44 N AB = 300 Then (b) From free-body diagram of lever BCD ⎛ Cy ⎞ ⎛ − 240 ⎞ −1 ⎜ and 4.ΣFx θ = tan200 N ⎟ = tan −10.6 ( 300⎟N )32.276° = 0: + ⎜ C ⎟ Cx + ⎜ − 380 ⎠ = = 0 ⎝ ⎝ x⎠ ∴ C x = −380 N or C x = 380 N or − 449 N 𝐹! = 0      ,         34,8 − 24 − 21,6C =24  𝑘𝑁 −32.3=▹ ⇒       𝑉! = −34,8  𝑘𝑁 𝑉! ° 0       ΣFy = 0: C y + 0.8 ( 300 N ) = 0 (a) From free-body diagram of lever BCD ΣM =∴ C yT= −240mm ) − or N ( C y mm ) =N 0: 200 75 = 240 0 ( 50 N ⎛ Cy ⎞ − 240 ⎞ ⎟ = tan −1 ⎛ ⎟ = 32.276° ⎜ ⎟ ⎝ − 380 ⎠ ⎝ Cx ⎠ m: 𝑀! = 0, 34,8 1,8 − 24 1,2  𝑘𝑁 ∙ 𝑚 + 𝑀! = 0     ⇒   𝑀! = −33,84  𝑘𝑁 ∙ 𝑚 ( 380 )2 + ( 240 )2 Chapter 4, Solution 19. n 19. AB = 449.44 N C Solutions Manual Organization System 32.3° ▹ 2 2 ∴ T 2 2 AB = 300 C = C x + C y = ( 380 ) + ( 240 ) = 449.44 N Then (b) From free-body diagram of lever BCD ⎛ Cy ⎞ ⎛ − 240 ⎞ −1 = tan −1 ⎜ ⎟ 3. and Fx = 0: θ = tan + ⎜Cx +⎟ 0.6 ( 300 N ) = 0 = 32.276° Σ 200 N ⎜ Cx ⎟ ⎝ − 380 ⎠ ⎠ ⎝ ∴ C x = −380 N or C x = 380 N or = 449 N 32.3 ▹ 𝐹! = 0      ,         34,8 − 24 − 21,6  𝑘𝑁 −C 𝑉! = 0       ⇒       𝑉!°= −10,8  𝑘𝑁 ΣFy = 0: C + 0.8 ( 300 N ) = 0 (a) From free-body diagram ofy lever BCD ΣM = 0: T ∴ (C y mm240 N N ( 75 mmC = = 240 N 50 = − ) − 200 or )y 0 2 2 Cx + C y = Chapter 4, Solution 19. olutions Manual Organization System 19. ∴ TAB = 300 ( 300 N ) = 0 chanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., senberg, William E. Clausen, David Mazurek, Phillip J. Cornwell ΣFy = 0: C y + 0.8 he McGraw-Hill Companies.
  4. 4. 1 34,8!!"! 5 2 3 4 −34,8!!"! 0,6!!! 0,3! 0,9!!! 0,9!!! 0,3! 0,6!!! −33,84!!" ∙ !!
  5. 5. d) 𝑆𝑖          𝑟 = 25  𝑚𝑚 𝑏 = 25  𝑥  2 = 50  𝑚𝑚 !! ! 25!!!! !! 1 !! ! 25!!!! 2 Áreas 𝐴! = 𝜋𝑟 ! 𝜋 25  𝑚𝑚 = 2 2 ! = 981,74  𝑚𝑚! 𝐴! = 𝑏ℎ = 50  𝑚𝑚  𝑥  25  𝑚𝑚 = 1250  𝑚𝑚! Centroide áreas 𝑦! = 4𝑟 4  𝑥  25  𝑚𝑚 = = 10,61  𝑚𝑚 3𝜋 3𝜋 𝑦! = − ℎ 25  𝑚𝑚 = = −12,5  𝑚𝑚 2 2 Centroide de la figura 𝑦= 𝐴! 𝑦! + 𝐴! 𝑦! = 𝐴! + 𝐴! 981,74 10,61 + 1250 −125  𝑚𝑚! = −2,334  𝑚𝑚 981,74 + 1250  𝑚𝑚!
  6. 6. Esto responde la pregunta e) ubicación del eje neutro que se encuentra en el centroide de la figura. 𝐼! = 𝐼𝑥! − 𝐴! 𝑦! ! = 𝜋𝑟 ! 𝜋 25  𝑚𝑚 − 𝐴! 𝑦! ! = 8 8 ! − 981,74  𝑚𝑚! 10,61  𝑚𝑚 ! 𝐼! = 42881,54  𝑚𝑚!    𝑑! = 𝑦! − 𝑦 = 10,61  𝑚𝑚 + 2,334  𝑚𝑚 = 12,944  𝑚𝑚 𝑰 𝟏 = 𝐼! + 𝐴! 𝑑! ! = 42881,54  𝑚𝑚! + 981,74  𝑚𝑚! 12,944 𝐼! = 𝑏ℎ! 50  𝑚𝑚 25  𝑚𝑚 = 12 12 ! ! = 207369,26  𝑚𝑚! = 65104,16  𝑚𝑚! 𝑑! = 𝑦! − 𝑦 = −12,5  𝑚𝑚 + 2,334  𝑚𝑚 = 10,166  𝑚𝑚 𝑰 𝟐 = 𝐼! + 𝐴! 𝑑! ! = 65104,16  𝑚𝑚! + 1250  𝑚𝑚! 10,166 ! = 194288,6  𝑚𝑚! El momento de Inercia d) es 𝐼 = 𝑰 𝟏 + 𝑰 𝟐 = 207369,26  𝑚𝑚! + 194288,6  𝑚𝑚! = 401657,86  𝑚𝑚! 𝐼 = 401,65  𝑥  10!!  𝑚𝑚! f) Esfuerzo máximo de tensión y compresión 𝑦! = 25  𝑚𝑚 + 2,334  𝑚𝑚 = 27,334  𝑚𝑚 = 0,027334  𝑚 𝑦! = −25  𝑚𝑚 + 2,334  𝑚𝑚 = −22,666  𝑚𝑚 = 0,022666  𝑚 𝜎! = − 𝜎! = − 𝑀𝑦! −33,84  𝑘𝑁 ∙ 𝑚 0,027334  𝑚 =− = 2,3  𝐺𝑃𝑎 𝐼 401,65  𝑥  10!!  𝑚! 𝑀𝑦! −33,84  𝑘𝑁 ∙ 𝑚 −0,022666  𝑚 =− = −1,9  𝐺𝑃𝑎 𝐼 401,65  𝑥  10!!  𝑚!

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