Por  el  método  de  las  secciones:    Dibujaremos  el  diagrama  de  cuerpo  libre  para  hacer  el  calculo  de  las  reacciones  en  los  apoyos:    

 Aplicando  las  ecuaciones  de  equilibrio  obtenemos  que:    

𝐹! = 0:    𝑅!! = 0    

𝐹! = 0:      

𝑅!! +  𝑅!! − 1𝑘𝑁  − 2𝑘𝑁 − 2𝑘𝑁 − 2𝑘𝑁 − 1𝑘𝑁 = 0  

1!!"!

2!!"!

2!!"! 2!!"!

1!!"!

!!! !!!

!!!

!!

!! !! !! !! !!

!!

!! !!

!!

2,4!!! 2,4!!! 2,4!!! 2,4!!!

2,62!!!

0,46!!! COSMOS: Complete Online Solutions Manual Organization System

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.

Chapter 4, Solution 19.

Free-Body Diagram:

(a) From free-body diagram of lever BCD

( ) ( )0: 50 mm 200 N 75 mm 0C AB

M TΣ = − =

300AB

T∴ =(b) From free-body diagram of lever BCD

( )0: 200 N 0.6 300 N 0x x

F CΣ = + + =

380 N or 380 Nx x

C∴ = − =C

( )0: 0.8 300 N 0y y

F CΣ = + =

N 240or N 240 =−=∴yy

C C

Then ( ) ( )2 22 2380 240 449.44 N

x yC C C= + = + =

and °=⎟⎠

⎞⎜⎝

⎛

−−=⎟⎟

⎠

⎞⎜⎜⎝

⎛= −−

276.32380

240tantan

11

x

y

C

Cθ

or 449 N=C 32.3°▹

COSMOS: Complete Online Solutions Manual Organization System

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.

Chapter 4, Solution 19.

Free-Body Diagram:

(a) From free-body diagram of lever BCD

( ) ( )0: 50 mm 200 N 75 mm 0C AB

M TΣ = − =

300AB

T∴ =(b) From free-body diagram of lever BCD

( )0: 200 N 0.6 300 N 0x x

F CΣ = + + =

380 N or 380 Nx x

C∴ = − =C

( )0: 0.8 300 N 0y y

F CΣ = + =

N 240or N 240 =−=∴yy

C C

Then ( ) ( )2 22 2380 240 449.44 N

x yC C C= + = + =

and °=⎟⎠

⎞⎜⎝

⎛

−−=⎟⎟

⎠

⎞⎜⎜⎝

⎛= −−

276.32380

240tantan

11

x

y

C

Cθ

or 449 N=C 32.3°▹

 𝑅!! +  𝑅!! = 8  𝑘𝑁  

   

   Con  los  datos  del  dibujo  podemos  sacar:    

𝛼 =   tan!!2,169,6 = 12,68°  

                                   

2,16!!!

0,46!!!

9,6!!!

2,62!!! !

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 tan𝛼    ,                        𝑦! =  𝐴𝐸 tan𝛼  ,                      𝑦! =  𝐴𝐺 tan𝛼                                                                          𝐴𝐶 =  2,4  𝑚    ,                        𝐴𝐸 =  4,8  𝑚  ,                      𝐴𝐺 =  7,2  𝑚                                                                                                          𝐴𝐵 =  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    

!! !! !! !! !!

!!

!! !!

!! !!

!!

!!

!!

Aplicando  las  ecuaciones  de  equilibrio  obtenemos  que:    

𝑀! = 0:        

−2,4 2 −  4,8 2 − 7,2 2 − 9,6 1  + 9,6(𝑅!!) = 0      

𝑅!! =  2,4 2 + 4,8 2 + 7,2 2 + 9,6 1

9,6 = 4  𝑘𝑁  

 Luego  tenemos  que:    

𝑅!! = 8𝑘𝑁 − 𝑅!! = 8  𝑘𝑁 − 4𝑘𝑁 = 4  𝑘𝑁    Calcularemos  el  valor  de  cada  componente  de  las  reacciones,  por  lo  cual,  se  seccionará  la  armadura  en  los  elementos  CE,  DE  y  DF    

   

𝛽 =   tan!!2,41 = 67,38°  

COSMOS: Complete Online Solutions Manual Organization System

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.

Chapter 4, Solution 19.

Free-Body Diagram:

(a) From free-body diagram of lever BCD

( ) ( )0: 50 mm 200 N 75 mm 0C AB

M TΣ = − =

300AB

T∴ =(b) From free-body diagram of lever BCD

( )0: 200 N 0.6 300 N 0x x

F CΣ = + + =

380 N or 380 Nx x

C∴ = − =C

( )0: 0.8 300 N 0y y

F CΣ = + =

N 240or N 240 =−=∴yy

C C

Then ( ) ( )2 22 2380 240 449.44 N

x yC C C= + = + =

and °=⎟⎠

⎞⎜⎝

⎛

−−=⎟⎟

⎠

⎞⎜⎜⎝

⎛= −−

276.32380

240tantan

11

x

y

C

Cθ

or 449 N=C 32.3°▹

!! !! !!

!!

!!

!!

4!!"!

2!!"!

1!!"!

!

2,4!!! 2,4!!!

!!"

!!"

!!"

Aplicando  las  ecuaciones  de  equilibrio  obtenemos  que:    

𝑀! = 0:        

1 𝐹!"  −  2,4 4𝑘𝑁 +  2,4 1𝑘𝑁 = 0      

𝐹!" =  2,4 4𝑘𝑁 −  2,4 1𝑘𝑁

1 = 7,2  𝑘𝑁        𝑻    𝑒𝑙  𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑜  𝐶𝐸  𝑒𝑠𝑡𝑎  𝑒𝑛  𝑡𝑟𝑎𝑐𝑐𝑖ó𝑛      

𝑀! = 0:        −4,8 4𝑘𝑁 + 4,8 1𝑘𝑁 + 2,4 2𝑘𝑁 − 1 cos 12,68 (𝐹!")− 2,4 sin 12,68 (𝐹!") = 0    

𝐹!" =  4,8 4𝑘𝑁 − 4,8 1𝑘𝑁 − 2,4 2𝑘𝑁1 cos 12,68 +  2,4 sin 12,68 = −6,34  𝑘𝑁  

   

∴ 𝐹!" =  6,34  𝑘𝑁        𝑪      𝑒𝑙  𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑜  𝐷𝐹  𝑒𝑠𝑡𝑎  𝑒𝑛  𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑖ó𝑛      

𝑀! = 0:        

2,4 1𝑘𝑁 − 2,4 4𝑘𝑁 + 1 634𝑘𝑁 − 1 sin 67,38 (𝐹!") = 0    

𝐹!" =  −2,4 1𝑘𝑁 + 2,4 4𝑘𝑁 − 1 634𝑘𝑁

1 sin 12,68 = −0,93  𝑘𝑁  

   

∴ 𝐹!" =  0,93  𝑘𝑁        𝑪      𝑒𝑙  𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑜  𝐷𝐸  𝑒𝑠𝑡𝑎  𝑒𝑛  𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑖ó𝑛              

COSMOS: Complete Online Solutions Manual Organization System

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.

Chapter 4, Solution 19.

Free-Body Diagram:

(a) From free-body diagram of lever BCD

( ) ( )0: 50 mm 200 N 75 mm 0C AB

M TΣ = − =

300AB

T∴ =(b) From free-body diagram of lever BCD

( )0: 200 N 0.6 300 N 0x x

F CΣ = + + =

380 N or 380 Nx x

C∴ = − =C

( )0: 0.8 300 N 0y y

F CΣ = + =

N 240or N 240 =−=∴yy

C C

Then ( ) ( )2 22 2380 240 449.44 N

x yC C C= + = + =

and °=⎟⎠

⎞⎜⎝

⎛

−−=⎟⎟

⎠

⎞⎜⎜⎝

⎛= −−

276.32380

240tantan

11

x

y

C

Cθ

or 449 N=C 32.3°▹

COSMOS: Complete Online Solutions Manual Organization System

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.

Chapter 4, Solution 19.

Free-Body Diagram:

(a) From free-body diagram of lever BCD

( ) ( )0: 50 mm 200 N 75 mm 0C AB

M TΣ = − =

300AB

T∴ =(b) From free-body diagram of lever BCD

( )0: 200 N 0.6 300 N 0x x

F CΣ = + + =

380 N or 380 Nx x

C∴ = − =C

( )0: 0.8 300 N 0y y

F CΣ = + =

N 240or N 240 =−=∴yy

C C

Then ( ) ( )2 22 2380 240 449.44 N

x yC C C= + = + =

and °=⎟⎠

⎞⎜⎝

⎛

−−=⎟⎟

⎠

⎞⎜⎜⎝

⎛= −−

276.32380

240tantan

11

x

y

C

Cθ

or 449 N=C 32.3°▹

COSMOS: Complete Online Solutions Manual Organization System

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.

Chapter 4, Solution 19.

Free-Body Diagram:

(a) From free-body diagram of lever BCD

( ) ( )0: 50 mm 200 N 75 mm 0C AB

M TΣ = − =

300AB

T∴ =(b) From free-body diagram of lever BCD

( )0: 200 N 0.6 300 N 0x x

F CΣ = + + =

380 N or 380 Nx x

C∴ = − =C

( )0: 0.8 300 N 0y y

F CΣ = + =

N 240or N 240 =−=∴yy

C C

Then ( ) ( )2 22 2380 240 449.44 N

x yC C C= + = + =

and °=⎟⎠

⎞⎜⎝

⎛

−−=⎟⎟

⎠

⎞⎜⎜⎝

⎛= −−

276.32380

240tantan

11

x

y

C

Cθ

or 449 N=C 32.3°▹