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

a) Fuerza Homogénea

12  𝑘𝑁𝑚 1,8  𝑚

= 21,6  𝑘𝑁

b)

𝑃𝑜𝑟  𝑠𝑖𝑚𝑒𝑡𝑟𝑖𝑎  𝑒𝑛  𝑒𝑙  𝑠𝑖𝑠𝑡𝑒𝑚𝑎    𝑅! = 𝑅!

𝐹! = 0      , 𝑅! + 𝑅! − 24+ 21,6+ 24 𝑘𝑁 = 0

𝑅! + 𝑅! = 69,6  𝑘𝑁

𝑅! =69,62  𝑘𝑁 = 34,8  𝑘𝑁

𝑅! = 𝑅! = 34,8  𝑘𝑁

!!

!!

!!

!! 24!!"! 24!!"! 21,6!!"!

0,6!!! 1,8!!!

0,9!!! 0,9!!!

0,3! 0,3! 0,6!!!

CO

SM

OS

: C

om

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izat

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Sy

stem

V

ecto

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s fo

r E

ng

inee

rs:

Sta

tics

and

Dyn

am

ics,

8/e

, F

erd

inan

d P

. B

eer,

E.

Ru

ssel

l Jo

hn

sto

n,

Jr.,

Ell

iot

R.

Eis

enb

erg

, W

illi

am E

. C

lau

sen

, D

avid

Maz

ure

k,

Ph

illi

p J

. C

orn

wel

l ©

20

07

Th

e M

cGra

w-H

ill

Co

mp

anie

s.

Ch

ap

ter

4, S

olu

tion

19.

Fre

e-B

od

y D

iag

ram

:

(a)

Fro

m f

ree-

bo

dy

dia

gra

m o

f le

ver

BCD

()

()

0:

50

mm

200

N75 m

m0

CAB

MT

Σ=

−=

300

AB

T∴

=(b

) F

rom

fre

e-b

od

y d

iag

ram

of

lev

er BCD

()

0:

200

N0.6

300

N0

xx

FC

Σ=

++

=

3

80

N

o

r3

80

Nx

xC

∴=

−=

C

()

0:

0.8

30

0 N

0y

yF

CΣ

=+

=

N

240

or

N 240

=−

=∴

yy

CC

Th

en

()

()

22

22

38

02

40

449

.44

Nx

yC

CC

=+

=+

=

and

°

=⎟ ⎠⎞

⎜ ⎝⎛

−−=

⎟⎟ ⎠⎞⎜⎜ ⎝⎛

=−

−276

.32

380

240

tan

tan

11

xy

CCθ

or

4

49

N=

C32.3

°▹

c) Se consideran 5 secciones en la viga:

1.

𝐹! = 0      ,        34,8  𝑘𝑁 − 𝑉! = 0       ⇒      𝑉! = 34,8  𝑘𝑁

𝑀! = 0      ,          34,8  𝑘𝑁 0  𝑚 +𝑀! = 0         ⇒      𝑀! = 0

2.

𝐹! = 0      ,         34,8− 24  𝑘𝑁 − 𝑉! = 0       ⇒      𝑉! = 10,8  𝑘𝑁

𝑀! = 0      ,           34,8 0,6  𝑘𝑁 ∙𝑚 +𝑀! = 0         ⇒      𝑀! = −20,88    𝑘𝑁 ∙𝑚

3.

𝐹! = 0      ,         34,8− 24− 21,6  𝑘𝑁 − 𝑉! = 0       ⇒      𝑉! = −10,8  𝑘𝑁

𝑀! = 0, 34,8 1,8 − 24 1,2  𝑘𝑁 ∙𝑚 +𝑀! = 0     ⇒  𝑀! = −33,84  𝑘𝑁 ∙𝑚

4.

𝐹! = 0      ,         34,8− 24− 21,6− 24  𝑘𝑁 − 𝑉! = 0       ⇒      𝑉! = −34,8  𝑘𝑁

𝑀! = 0,         34,8 3 − 24 2,4 − 21,6 1,2  𝑘𝑁 ∙𝑚 +𝑀! = 0                              ⇒      𝑀! = −20,88  𝑘𝑁 ∙𝑚

5.

𝐹! = 0         ⇒      𝑉! = 0

𝑀! = 0      ,          𝑃𝑜𝑟  𝑠𝑖𝑚𝑒𝑡𝑟𝑖𝑎       ⇒      𝑀! = 0

CO

SM

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izat

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Sy

stem

V

ecto

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ng

inee

rs:

Sta

tics

and

Dyn

am

ics,

8/e

, F

erd

inan

d P

. B

eer,

E.

Ru

ssel

l Jo

hn

sto

n,

Jr.,

Ell

iot

R.

Eis

enb

erg

, W

illi

am E

. C

lau

sen

, D

avid

Maz

ure

k,

Ph

illi

p J

. C

orn

wel

l ©

20

07

Th

e M

cGra

w-H

ill

Co

mp

anie

s.

Ch

ap

ter

4, S

olu

tion

19.

Fre

e-B

od

y D

iag

ram

:

(a)

Fro

m f

ree-

bo

dy

dia

gra

m o

f le

ver

BCD

()

()

0:

50

mm

200

N75 m

m0

CAB

MT

Σ=

−=

300

AB

T∴

=(b

) F

rom

fre

e-b

od

y d

iag

ram

of

lev

er BCD

()

0:

200

N0.6

300

N0

xx

FC

Σ=

++

=

3

80

N

o

r3

80

Nx

xC

∴=

−=

C

()

0:

0.8

30

0 N

0y

yF

CΣ

=+

=

N

240

or

N 240

=−

=∴

yy

CC

Th

en

()

()

22

22

38

02

40

449

.44

Nx

yC

CC

=+

=+

=

and

°

=⎟ ⎠⎞

⎜ ⎝⎛

−−=

⎟⎟ ⎠⎞⎜⎜ ⎝⎛

=−

−276

.32

380

240

tan

tan

11

xy

CCθ

or

4

49

N=

C32.3

°▹

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°▹

CO

SM

OS

: C

om

ple

te O

nli

ne

So

luti

on

s M

anu

al O

rgan

izat

ion

Sy

stem

V

ecto

r M

echa

nic

s fo

r E

ng

inee

rs:

Sta

tics

and

Dyn

am

ics,

8/e

, F

erd

inan

d P

. B

eer,

E.

Ru

ssel

l Jo

hn

sto

n,

Jr.,

Ell

iot

R.

Eis

enb

erg

, W

illi

am E

. C

lau

sen

, D

avid

Maz

ure

k,

Ph

illi

p J

. C

orn

wel

l ©

20

07

Th

e M

cGra

w-H

ill

Co

mp

anie

s.

Ch

ap

ter

4, S

olu

tion

19.

Fre

e-B

od

y D

iag

ram

:

(a)

Fro

m f

ree-

bo

dy

dia

gra

m o

f le

ver

BCD

()

()

0:

50

mm

200

N75 m

m0

CAB

MT

Σ=

−=

300

AB

T∴

=(b

) F

rom

fre

e-b

od

y d

iag

ram

of

lev

er BCD

()

0:

200

N0.6

300

N0

xx

FC

Σ=

++

=

3

80

N

o

r3

80

Nx

xC

∴=

−=

C

()

0:

0.8

30

0 N

0y

yF

CΣ

=+

=

N

240

or

N 240

=−

=∴

yy

CC

Th

en

()

()

22

22

38

02

40

449

.44

Nx

yC

CC

=+

=+

=

and

°

=⎟ ⎠⎞

⎜ ⎝⎛

−−=

⎟⎟ ⎠⎞⎜⎜ ⎝⎛

=−

−276

.32

380

240

tan

tan

11

xy

CCθ

or

4

49

N=

C32.3

°▹

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°▹

CO

SM

OS

: C

om

ple

te O

nli

ne

So

luti

on

s M

anu

al O

rgan

izat

ion

Sy

stem

V

ecto

r M

echa

nic

s fo

r E

ng

inee

rs:

Sta

tics

and

Dyn

am

ics,

8/e

, F

erd

inan

d P

. B

eer,

E.

Ru

ssel

l Jo

hn

sto

n,

Jr.,

Ell

iot

R.

Eis

enb

erg

, W

illi

am E

. C

lau

sen

, D

avid

Maz

ure

k,

Ph

illi

p J

. C

orn

wel

l ©

20

07

Th

e M

cGra

w-H

ill

Co

mp

anie

s.

Ch

ap

ter

4, S

olu

tion

19.

Fre

e-B

od

y D

iag

ram

:

(a)

Fro

m f

ree-

bo

dy

dia

gra

m o

f le

ver

BCD

()

()

0:

50

mm

200

N75 m

m0

CAB

MT

Σ=

−=

300

AB

T∴

=(b

) F

rom

fre

e-b

od

y d

iag

ram

of

lev

er BCD

()

0:

200

N0.6

300

N0

xx

FC

Σ=

++

=

3

80

N

o

r3

80

Nx

xC

∴=

−=

C

()

0:

0.8

30

0 N

0y

yF

CΣ

=+

=

N

240

or

N 240

=−

=∴

yy

CC

Th

en

()

()

22

22

38

02

40

449

.44

Nx

yC

CC

=+

=+

=

and

°

=⎟ ⎠⎞

⎜ ⎝⎛

−−=

⎟⎟ ⎠⎞⎜⎜ ⎝⎛

=−

−276

.32

380

240

tan

tan

11

xy

CCθ

or

4

49

N=

C32.3

°▹

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°▹

CO

SM

OS

: C

om

ple

te O

nli

ne

So

luti

on

s M

anu

al O

rgan

izat

ion

Sy

stem

V

ecto

r M

echa

nic

s fo

r E

ng

inee

rs:

Sta

tics

and

Dyn

am

ics,

8/e

, F

erd

inan

d P

. B

eer,

E.

Ru

ssel

l Jo

hn

sto

n,

Jr.,

Ell

iot

R.

Eis

enb

erg

, W

illi

am E

. C

lau

sen

, D

avid

Maz

ure

k,

Ph

illi

p J

. C

orn

wel

l ©

20

07

Th

e M

cGra

w-H

ill

Co

mp

anie

s.

Ch

ap

ter

4, S

olu

tion

19.

Fre

e-B

od

y D

iag

ram

:

(a)

Fro

m f

ree-

bo

dy

dia

gra

m o

f le

ver

BCD

()

()

0:

50

mm

200

N75 m

m0

CAB

MT

Σ=

−=

300

AB

T∴

=(b

) F

rom

fre

e-b

od

y d

iag

ram

of

lev

er BCD

()

0:

200

N0.6

300

N0

xx

FC

Σ=

++

=

3

80

N

o

r3

80

Nx

xC

∴=

−=

C

()

0:

0.8

30

0 N

0y

yF

CΣ

=+

=

N

240

or

N 240

=−

=∴

yy

CC

Th

en

()

()

22

22

38

02

40

449

.44

Nx

yC

CC

=+

=+

=

and

°

=⎟ ⎠⎞

⎜ ⎝⎛

−−=

⎟⎟ ⎠⎞⎜⎜ ⎝⎛

=−

−276

.32

380

240

tan

tan

11

xy

CCθ

or

4

49

N=

C32.3

°▹

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°▹

CO

SM

OS

: C

om

ple

te O

nli

ne

So

luti

on

s M

anu

al O

rgan

izat

ion

Sy

stem

V

ecto

r M

echa

nic

s fo

r E

ng

inee

rs:

Sta

tics

and

Dyn

am

ics,

8/e

, F

erd

inan

d P

. B

eer,

E.

Ru

ssel

l Jo

hn

sto

n,

Jr.,

Ell

iot

R.

Eis

enb

erg

, W

illi

am E

. C

lau

sen

, D

avid

Maz

ure

k,

Ph

illi

p J

. C

orn

wel

l ©

20

07

Th

e M

cGra

w-H

ill

Co

mp

anie

s.

Ch

ap

ter

4, S

olu

tion

19.

Fre

e-B

od

y D

iag

ram

:

(a)

Fro

m f

ree-

bo

dy

dia

gra

m o

f le

ver

BCD

()

()

0:

50

mm

200

N75 m

m0

CAB

MT

Σ=

−=

300

AB

T∴

=(b

) F

rom

fre

e-b

od

y d

iag

ram

of

lev

er BCD

()

0:

200

N0.6

300

N0

xx

FC

Σ=

++

=

3

80

N

o

r3

80

Nx

xC

∴=

−=

C

()

0:

0.8

30

0 N

0y

yF

CΣ

=+

=

N

240

or

N 240

=−

=∴

yy

CC

Th

en

()

()

22

22

38

02

40

449

.44

Nx

yC

CC

=+

=+

=

and

°

=⎟ ⎠⎞

⎜ ⎝⎛

−−=

⎟⎟ ⎠⎞⎜⎜ ⎝⎛

=−

−276

.32

380

240

tan

tan

11

xy

CCθ

or

4

49

N=

C32.3

°▹

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°▹

0,6!!! 0,3! 0,9!!! 0,9!!! 0,3! 0,6!!!

34,8!!"!

−34,8!!"!

−33,84!!" ∙!!

1

2 3

4

5

d)

𝑆𝑖          𝑟 = 25  𝑚𝑚

𝑏 = 25  𝑥  2 = 50  𝑚𝑚

Áreas

𝐴! =𝜋𝑟!

2 =𝜋 25  𝑚𝑚 !

2 = 981,74  𝑚𝑚!

𝐴! = 𝑏ℎ = 50  𝑚𝑚  𝑥  25  𝑚𝑚 = 1250  𝑚𝑚! Centroide áreas

𝑦! =4𝑟3𝜋 =

4  𝑥  25  𝑚𝑚3𝜋 = 10,61  𝑚𝑚

𝑦! = −ℎ2 =

25  𝑚𝑚2 = −12,5  𝑚𝑚

Centroide de la figura

𝑦 =𝐴!𝑦! + 𝐴!𝑦!𝐴! + 𝐴!

=981,74 10,61 + 1250 −125  𝑚𝑚!

981,74+ 1250  𝑚𝑚! = −2,334  𝑚𝑚

1

2

!!!

!!!

!! 25!!!!

25!!!!

Esto responde la pregunta e) ubicación del eje neutro que se encuentra en el centroide de la figura.

𝐼! = 𝐼𝑥! − 𝐴!𝑦!! =𝜋𝑟!

8 − 𝐴!𝑦!! =𝜋 25  𝑚𝑚 !

8 − 981,74  𝑚𝑚! 10,61  𝑚𝑚 !

𝐼! = 42881,54  𝑚𝑚!  

 𝑑! = 𝑦! − 𝑦 = 10,61  𝑚𝑚 + 2,334  𝑚𝑚 = 12,944  𝑚𝑚 𝑰𝟏 = 𝐼! + 𝐴!𝑑!

! = 42881,54  𝑚𝑚! + 981,74  𝑚𝑚! 12,944 ! = 207369,26  𝑚𝑚!

𝐼! =𝑏ℎ!

12 =50  𝑚𝑚 25  𝑚𝑚 !

12 = 65104,16  𝑚𝑚!

𝑑! = 𝑦! − 𝑦 = −12,5  𝑚𝑚 + 2,334  𝑚𝑚 = 10,166  𝑚𝑚

𝑰𝟐 = 𝐼! + 𝐴!𝑑!! = 65104,16  𝑚𝑚! + 1250  𝑚𝑚! 10,166 ! = 194288,6  𝑚𝑚!

El momento de Inercia d) es

𝐼 = 𝑰𝟏 + 𝑰𝟐 = 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! = −𝑀𝑦!𝐼 = −

−33,84  𝑘𝑁 ∙𝑚 0,027334  𝑚401,65  𝑥  10!!  𝑚! = 2,3  𝐺𝑃𝑎

𝜎! = −𝑀𝑦!𝐼 = −

−33,84  𝑘𝑁 ∙𝑚 −0,022666  𝑚401,65  𝑥  10!!  𝑚! = −1,9  𝐺𝑃𝑎