Mecanica para Ingenieros Estática - Meriam - 6ed (Sol)

7/27/2019 Mecanica para Ingenieros Estática - Meriam - 6ed (Sol)

http://slidepdf.com/reader/full/mecanica-para-ingenieros-estatica-meriam-6ed-sol 1/2188

http://www.elsolucionario.blogspot .com

LIBROS UNIVERISTARIOS

Y SOLUCIONARIOS DE

MUCHOS DE ESTOS LIBROS

LOS SOLUCIONARIOS

CONTIENEN TODOS LOS

EJERCICIOS DEL LIBRO

RESUELTOS Y EXPLICADOS

DE FORMA CLARA

VISITANOS PARA

DESARGALOS GRATIS.



















































































































































































































































































































































































































































































































































































































































































4/10

As a whole:

trianglessimilar bym5.1: Note =CE

kN80.306)45cos5.13(3)45cos3(5:0

==−°++°=Σ

y

y A

D D M

Joint D:

°=°−

°= − 7.28

45cos36

45sin3tan 1θ

⎧ ==°−=Σ C CDCDF y

kN92.7,07.28sin80.3:0



⎩⎨ ==−°=Σ T DE DE F x kN94.6,07.28cos92.7:0

Joint C :

⎩⎨⎧

=−°+°−°=Σ

=°−°−°−=Σ

037.28sin92.745sin7.28sin:0

07.28cos92.745cos7.28cos:0

CE BC F

CE BC F

y

x

Solve simultaneously to obtain:)(kN74.2

)(kN70.5

C CE

C BC

−=

−=

Joint E :

°=°−°

= 5.672

45180 β

T BE

BE F y

kN10.2

045sin74.25.67sin:0

=

=°−°=Σ





















































4/124

As a whole:

By symmetry, Ay = Ey = 7.5 kips

0:0 ==Σ x x AF

Joint A:

°== − 7.3315

10tan 1

α

⎪⎩

⎪

⎨

⎧

==+°−=Σ

==+°−=Σ

C AB ABF

T AF AF F

y

x

kips52.13,05.77.33sin:0

kips25.11,07.33cos52.13:0



⎩ y

Joint B:

⎩⎨⎧

=−=−=Σ

=−−=Σ

05sin)52.13(:0

0cos)52.13(:0

α

α

BF BC F

BF BC F

y

x

Solve simultaneously to obtain:

BC = 9.01 kips C , BF = 4.51 kips C

Joint F :

By symmetry, EF = AF = 11.25 kips T

& DF = BF = 4.51 kips C

T CF CF F y kips5,07.33sin)51.4(2:0 ==°−=Σ

By symmetry, CD = 9.01 kips C , DE = 13.52 kips C , DF = 4.51 kips C













4/130

By symmetry, A y = G y = 6 kN

0:0 ==Σ x x AF



°==°=− 43.18

5.1

5.0tan,45 1 β α

m2

3= AC

0,0)2()5.4(3)3(5.1)3(6:0 ==−−−=Σ CH AC CH M I

C CDCDF y kN74.4or kN74.4,035.143.18sin6:0 −==−−°+=Σ

T AH AH F x

kN5.4,043.18cos74.4:0 ==+°−=Σ



4/132

By symmetry, A y = E y = 4 kN

Joint A:

°==°== −− 5.805.0

3tan,6.40

5.3

3tan 11 β α

⎨⎧ =+°+°=Σ 046.40cos5.80cos:0 AF ABF x



⎩⎨

=°+°=Σ 06.40sin5.80sin:0 AF ABF y

Solve to obtain: C AB kN73.4or kN73.4−=

Joint C :

By symmetry, BC = CD

C BC BC F y kN66.5,0845sin2:0 ==−°=Σ

Joint B:

045sin66.56.40sin5.80sin73.4:0 =°−°−°=Σ BF F y

T BF kN024.1=







4/136

We can begin at joint E without finding the external reactions.

Joint E :

°=°°−= − 15

30sin24

60sin2424tan 1θ

⎪⎩

⎪⎨⎧

==−°=Σ

==−°=Σ

C L DE DE LF

C L EM L EM F

x

y

366.1,015cos932.1:0

932.1,0)15sin(2:0

Joint M :

°°−°− 537

45sin2460sin24tan 1α



°=°−°

= 5.3760cos2445cos24

tanα

°=°−°

°−= − 9.32

60cos2445cos24

60sin2424tan 1 β

⎩⎨⎧

=°−°−°=Σ

=°−°−°−=Σ

015sin932.15.37sin9.32sin:0

015cos932.15.37cos9.32cos:0

MN DM F

L MN DM F

y

x

Solve simultaneously to obtain: C L DM 785.0=

Joint D:

C L DN

DN L LF y

574.0

09.32sin785.0:0

=

=−−°=Σ











































4/15

By symmetry, A y = F y = 4 kN

Joint A:

°== − 602

1cos 1

θ

⎩⎨⎧

==°−=Σ

==°−=Σ

T AG AGF C AB ABF

x

y

kN31.2,060cos62.4:0kN62.4,060sinkN4:0



Joint G:

°=°

= − 9.402

60sin2tan 1

α

T BG BGF y kN528.1,029.40sin2:0 ==−°=Σ

Joint B:

C BC

BC F y

kN46.3

09.40sin528.160sin60sin62.4:0

=

=°−°−°=Σ











4/2

By symmetry, A y = C y = 1000 lb

Joint A:

°== − 3.3012

7tan 1

θ



⎪⎩

⎪⎨⎧

==+°=Σ

−==+°=Σ

T AB AB ADF

C AD ADF

x

y

lb1714,03.30cos:0

)(lb1985,010003.30sin:0

By symmetry: CD = AC = 1985 lb C

BC = AB = 1714 lb T

By inspection of joint B, BD is a zero-force member.





4/22

Joint E :

°=°−°

= 752

30180θ

⎩⎨

⎧

==+°−=Σ

==−°=Σ

T L ED ED LF

T L EF L EF F

x

y

268.0,075cos035.1:0

035.1,075sin:0

Joint D:

R DF R R R R DF 239.1,30cos)2)((24 222

=°−+=

°==°

8.23,sin

239.1

30sinα

α

R R

⎩⎨⎧

=°+°+−=Σ

=°−°−−=Σ

075sin8.23sin:0

075cos8.23cos268.0:0

CD DF LF

CD DF LF

y

x

Solve simultaneously to obtain: CLDFTLCD 66403131 ==



Solve simultaneously to obtain: C L DF T LCD 664.0,313.1 ==

Joint C :

⎩⎨⎧

=°−°−°=Σ

=°+°−°−=Σ

075sin313.130sin45sin:0

075cos313.130cos45cos:0

LCF BC F

LCF BC F

y

x

Solve simultaneously to obtain: )(674.0 C LCF −=

Joint F :

°=°+°+°= 8.83308.2330 β

⎩⎨⎧

=−°+°−+°=Σ

=+°+°++°−=Σ

0sin30sin75sinsin45sin:0

0cos30cos75coscos45cos:0

α β

α β

DF CF EF BF FGF

DF CF EF BF FGF

y

x

Solve simultaneously to obtain: T L BF 814.1=





4/25

Location of C:22

c 4 c y x −−=

°−=°−−=

−++=

25sin4),25cos4(

)4()4( 222

AC y AC x

y x AC

cc

cc

Solve to obtain ft815.1= AC

Joint B:

°=°−

°= − 04.18

25cos815.14

25sin815.1tan 1θ

⎩

⎨⎧

==−°=Σ

==−°=Σ

T AB ABF

C CBCBF

x

y

lb1535,004.18cos1615:0

lb1615,050004.18sin:0

Joint C :



°=°−=°=°−°

= 4525,702

40180δ β δ

⎩⎨⎧

=°−°−°−=Σ

=°−°−°=Σ

004.18cos161545cos25cos:0

004.18sin161545sin25sin:0

CD AC F

CD AC F

x

y

Solve simultaneously to obtain: C AC lb779or lb779−=

By symmetry across a line from the origin to point A, C AC AD lb779==







X4/28

By symmetry, A y = I y = 5/2 L, A x = 0

By inspection of point A , AN = 0

Coordinate origin is at the center of

the two concentric arcs.

Location of N : x2

+ y2

= 162

For x N = -8 m: y N 2

= -82

+ 162

y N = 13.86 m N = (-8, 13.86) m

Location of A: y A2= -12

2+ 16

2, y A = 10.58 m, A = (-12, 10.58) m

Location of B: y B2

= -12

2+ 18

2=, y B = 13.42 m, B = (-12, 13.41) m

Location of C : yC 2

= -8

2+ 18

2=, yC = 16.12 m, C = (-8, 16.12) m

Location of D: y D2

=

-42

+ 182

=, y D = 17.55 m, C = (-4, 17.55) mLocation of M : y M

2=

-4

2+ 16

2=, y M = 15.49 m, C = (-4, 15.49) m

Joint B:



°=+−

−= − 28.6

128

42.1386.13tan 1α

°=+−

−= − 1.34

128

42.1312.16tan 1 β

⎩⎨⎧

=°+°+=Σ=°+°=Σ

028.6sin1.34sin)2/5(:0028.6cos1.34cos:0

BN BC LF BN BC F

y

x

Solve simultaneously to obtain: )(32.5,44.4 C L BC T L BN −==

Joint N :

°=+−−= − 2.22

84

86.1349.15tan 1γ

⎩

⎨⎧

==−+°+°−=Σ

==°+°−=Σ

C LCN LCN LF

T L NM NM LF

y

x

318.0,02.22sin76.428.6sin44.4:0

76.4,02.22cos28.6cos44.4:0

Joint C :

°=+−

−= − 61.19

84

12.1655.17tan 1δ

°=−

−= − 99.8

48

49.1512.16tan 1ε

⎩⎨⎧

=°−°++°=Σ

=°+°+°=Σ

099.8sin61.19sin318.01.34sin32.5:0

099.8cos61.19cos1.34cos32.5:0

CM CD L LF

CM CD LF

y

x

Solve simultaneously to obtain: T LCM 41.3=





4/3

We can begin at joint D without finding the external reactions.

Joint D:

⎪⎩

⎪⎨⎧

==°−=Σ

==−°=Σ

C ADCD ADF

T CDCDF

x

y

N3000,045cos:0

N4240,0300045sin:0

Joint C :

⎪⎩

⎪⎨⎧

==°+−=Σ

==°−°=Σ

TBCBCF

C AC AC F

x

y

N6000,0)45cos4240(2:0

N4240,045sin424045sin:0



⎪⎩ +Σ T BC BC F x N6000,0)45cos4240(2:0

Joint B:

From equilibrium of entire truss

T ABF y

N3000,0From ==Σ

4/30

°==− 9.36

2

5.1tan 1

α



T AE

AE AE M G

lb1417

0)6(9.36sin)5.1(9.36cos)6(600)4(800:0

=

=°−°−+=Σ

4/31

T BC BC M F

kips81.4,0)5.2(4)5.0(5)60sin3(:0 ==−−°=Σ







4/34

°=

°

= − 5.55

6

20tan

6

4

tan 1

d

d

α



T L BE

d BE

d L M

D

809.0

0)5.55(sin23

:0

=

=°⎟ ⎠

⎞⎜⎝

⎛ −⎟

⎠

⎞⎜⎝

⎛ =Σ











4/4

We can begin at joint C without finding the external reactions.

Joint C :

⎪⎩⎪⎨⎧

==−°=Σ

==−°=Σ

T CDCDF C BC BC F

x

y

N5000,045cos7070:0 N7070,0500045sin:0

Joint D:

m31.4,45cos)6)(5(265 222

=°−+= DB DB

°=⇒°= 1.5531.445sin

5sin θ θ

⎩⎨⎧

°°Σ

=°−°+=Σ

0155i45i0

045cos1.55cos5000:0

BDADF

AD BDF x



⎩⎨

=°−°−=Σ 01.55sin45sin:0 BD ADF y

Solve simultaneously to obtain:

T AD

C BD

N4170

N3590or N3590

=

−=

Joint B:

C AB

ABF x

N2950or N2950

045cos70701.55cos3590:0

−=

=−°−°=Σ







4/43

By symmetry, A y = I y = 3 L

0:0 ==Σ x x AF

°=°−°+

= − 8.268

25tan2430tan166tan 1θ



C L L DE

DE L L

L L M L

80.4or 80.4

060sin)625tan2430tan24()24(3)24(2

)16()8(:0

−=

=°+°−°−−++=Σ

⎪⎩

⎪⎨

⎧

−−+°+°−°−=Σ

=°+°+°−=Σ

22325sin8.26sin30sin80.4:0

025cos8.26cos30cos80.4:0

L L L LM DL LF

LM DL LF

y

x

Solve simultaneously to obtain: T L LM T L DL 54.4,0446.0 ==

Joint E :

T L EL

EL L LF y

80.3

0)30sin80.4(2:0

=

=−−°=Σ

4/44

By symmetry, A y = G y = 3 L

0:0 ==Σ x x AF



T L DQ DQ L

L LF y

576.0,02

5.3tansin

223:0

1 ==⎟ ⎠

⎞⎜⎝

⎛ −−−=Σ −

By inspection: 0=CQ





4/47

By symmetry, A y = G y = 3 L 0:0 ==Σ x x AF



°==°==−− 7.33

6

4tan,46.9

6

1tan 11 β α

( )[ ]

)(74.2

046.990sin)5()9(3)3()6()9(2

:0

C LCD

CD L L L L

M J

−=

=°−°−−++=Σ

T L JM JM L L

CD LF y 0901.0,07.33sin22

46.9sin3:0 ==°−−−°+=Σ



4/49

By symmetry, A y = H y = 3 L

0:0 ==Σ x x AF

Origin at center of arc

L ti f I2

252

92

B (9 23 3)



Location of I : y I 2

= 252

– 92, B = (9, 23.3) m

Location of J : y J 2

= 252

– 62, J = (6, 24.3) m

Location of G: yG = y I + 6 , G = (6, 29.3) m

Location of K : yK

2

= 252

– 32

, K = (3, 24.8) m

°=−

=°=−

= −− 39.103

3.248.24tan,3.56

3

8.243.29tan 11 β α

)(66.2,0)3()()6(2

)6(3:0 C LFG L y yFG L

L M K GK −==−−+−=Σ

⎪⎩

⎪⎨

⎧

°−°+=Σ

=°−°−=Σ

3.56sin39.10sin2

3:0

039.10cos3.56cos66.2:0

GK JK LF

JK GK LF

y

x

Solve simultaneously to obtain: T LGK 13.2=



4/50

By symmetry, A y = H y = 3 L

0:0 ==Σ x x AF

Origin at center of arc

Location of B: y B2

= 302

– (-15)2, B = (-15, 26.0) m

Location of A: y A = y B – 6, A = (-18, 20.0) m

Location of C:2

302

( 9)2C ( 9 28 6) m



Location of C : yC = 30 – (-9) , C = (-9, 28.6) m

Location of M : y M = y A, M = (-9, 20.0) mLocation of D: y D

2= 30

2– (-3)

2, D = (-3, 29.8) m

Location of L: y L = y A, L = (-3, 20.0) m

°=−

=°=−

=−− 2.55

6

0.206.28tan,60.11

6

6.288.29tan 11 β α

C LCD LCD

CD L L L M L

79.2or 79.2

0)60.1190sin()0.208.29()12()6()15(3:0

=−=

=°−°−−++−=Σ

T LCLCL L L L LF y 534.0,02.55sin60.11sin79.23:0 ==°−°−−−=Σ





4/53

N5660,0)81.9(100060sin2:0 ==−°=Σ T T F y

From 0=ΣxF , A x + B x = 0

C = (0, 2.25) m

L = (0, 4.5) m



H = (6cos15°, 4.5 + 6sin15°) = (5.80, 6.05) m

G = (6cos15° + 1cos75°, 4.5 + 6sin15° - 1sin75°) = (6.05, 5.09) m

°=−= − 1.2505.6

25.209.5tan 1α

m524.1)151.25cos(

5.1=

°−°=== DE EF FG

E = (6cos15° + 1cos75°-2(1.524cos25.1°), 4.5 + 6sin15° - 1sin75° -

2(1.524sin25.1°)) = (3.30, 3.79) m

N07014

0)09.5(60cos)05.6(60sin)79.3(60cos)30.3(60sin)3(:0

−=

=°+°−°−°−−=Σ

y

y B

A

T T T T A M

N90023,0)81.9(100007014:0 ==−+−=Σ y y y B BF

°=°+

°= − 8.28

15sin5.125.2

15cos5.1tan 1 β

N42019,0)75)(sin25.2()3(07014:0 ==°−=Σ LK LK M C

⎩⎨⎧

=°+°+°++−=Σ

=°+°+°=Σ

01.25sin8.28cos15sin420199002307014:0

01.25cosCKsin28.8cos1542019:0

CDCK F

CDF

y

x

Solve simultaneously to obtain: C CK CK N9290or N9290 =−=





























































4/8

As a whole: 00 =⇒=Σ x x AF

kN60== E A y by

0=Σ yF and symmetry.

Joint A: ( )( )°== − 7.385/4tan 1θ

⎪⎩

⎪⎨⎧

=−=Σ

==−=Σ

T AH AH F

C AB ABF

x

y

kN75,cos0.96:0

kN0.96,0sin60:0

θ

θ

Joint B:

⎪⎩

⎪⎨⎧

==°+−=Σ

−==°+=Σ

T BH BH F

C BC BC F

y

x

kN60,03.51cos0.96:0

)(kN75,03.51sin0.96:0



Joint H :

⎪⎩

⎪⎨⎧

==−+=Σ

==+−=Σ

T GH GH F

C CH CH F

x

y

kN5.112,075cos0.48:0

kN0.48,030sin:0

θ

θ

Joint G:

T CGF y kN600 =⇒=Σ

By symmetry:

FG = 112.5 kN T , CF = 48.0 kN C

CD = 75 kN C , DF = 60 kN T

EF = 75 kN T , DE = 96.0 kN C





















4/9

Joint C :

°== −7.38

5

4tan

1θ

⎪⎩

⎪⎨⎧

==°−°−=Σ

==°+°−=Σ

T BC BC F

C AC AC F

y

x

kN45.6,07.38cos07.215cos5:0

kN07.2,07.38sin15sin5:0

Joint B:

°== −6.26

4

2tan

1 β



⎪⎩⎪⎨⎧

==+°−=Σ

==−°=Σ

C BD BDF

T AB ABF

x

y

kN89.12,06.26cos42.14:0

kN42.14,045.66.26sin:0























































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































































Text Illustrations

To Accompany



Meriam/Kraige – Statics 6e

Appendix A: Area Moments of Inertia.



A 01



A 02



A_03



A 04



A 05



A 06



A 07



P A 0 1



P A 0 2



P_A_03



P_A_04



P_A_05



P A 0 6



P A 0 7



P A 0 8



P A 0 9



P_A_10



P A 1 1



P A 1 2



P_A_13



P A 1 4



P A 1 5



P A 1 6



P A 1 7



P A 1 8



P A 1 9



P A 2 0



P A 2 1



P A 2 2



P A 2 3



P A 2 4



P A 2 5



P A 2 7



P A 2 8



P A 2 9



P_A_30



P A 3 1



P A 3 2



P_A_33



P A 3 4



P A 3 5



P A 3 6



P A 3 7



P A 3 8



P A 3 9



P_A_40



P A 4 1



P A 4 2



P A 4 3



P A 4 4



P_A_45



P A 4 6



P_A_47



P A 4 8



P A 4 9



P_A_50



P A 5 1



P A 5 2



P_A_53



P A 5 4



P_A_55



P A 5 6



P A 5 8



P A 5 9



P A 6 0



P A 6 1



P_A_62



P A 6 3



P A 6 4



P A 6 5



P A 6 6



P_A_67



P A 6 8



P A 6 9



P A 7 0



P A 7 1



P A 7 2



P A 7 3



P A 7 4



P A 7 5



P A 7 6



P A 7 7



P A 7 8



P_A_79



P A 8 0



P_A_81



P A 8 2



P A 8 4



P A 8 5



P A 8 6



P A 8 7



P A 8 8



P A 8 9



P A 9 0



P A 9 1



P A 9 2

Text Illustrations

To Accompany


A di B S f M M f I i



Appendix B: Summary of Mass Moments of Inertia.



B 01



P 04 01



P_04_02



P 04 03



P_04_05



P 04 09



P 04 10



P 04 11



P_04_12



P 04 15



P_04_16

Text Illustrations

To Accompany


Appendix D: Useful Tables



Appendix D: Useful Tables.



D 01tbl



D 02tbl



D 03tbl



D 03t b l c on t



D 04 t bl co nt 2



D 04 t bl co nt 3



D 04tbl



D 04t bl c on t

Text Illustrations

To Accompany


Chapter 1 : Introduction to Statics.



Chapter 1 : Introduction to Statics.



01_01



01_02



01_03



01_04



01 05



01 06





















02 04



02 05



02 06



02 07



02 08



02_09



02 10



02_11



02_12



02 13



02 14



02 15



02 16



02 17



02 18



02_19



02 20



02 21



02 22



02 23



02 24



02 25



02_26



02_27



02_28



02_29



02 30



P 02 01



P 02 02



P 02 03



P 02 04



P 02 05



P 02 06



P_02_07



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P_02_11



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P_02_38



P 02 39



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P_02_95



P 02 96



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Text Illustrations

To Accompany


Chapter 3 : Equilibrium.





03 01



03 01 cont



03 02



03 03



03 04



03 05



03 06



03 07



03 08



03 09



03 10



03 A



03 B



03 C



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P_03_02



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P_03_06



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P_03_22



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P_03_24



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P_03_26



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P 03 40

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P_03_50



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Text Illustrations

To Accompany


Chapter 4 : Structures.





04_01



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P_04_123



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http://slidepdf.com/reader/full/mecanica-para-ingenieros-estatica-meriam-6ed-sol 1598/2188P 04 144















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P 04 26



P 04 27



P_04_28



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P 04 31



P_04_32



P 04 33



P_04_34



P 04 35



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P_04_37



P 04 38



P_04_39



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P_04_41



P_04_42



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P 04 45



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P_04_47



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P_04_50



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P 04 54



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P_04_70



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P_04_77



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P_04_84



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P_04_86



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P 04 94



P_04_95



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T t Ill t ti



Text Illustrations

To Accompany


Chapter 5 : Distributed Forces.



05 01



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05 03



05 04



05 05



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05 08



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05 10



05_11



05_12



05 13



05 14



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P 05 03



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P_05_05



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P_05_07



P 05 08



P 05 09



P 05 10



P_05_100



P_05_101



P_05_102



P_05_103



P_05_104



P_05_105



P 05 106



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P 05 108



P_05_109



P 05 11



P_05_110



P 05 111



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P 05 119



P 05 12



P_05_120



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P_05_129



P 05 13



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P 05 135



P_05_136



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P 05 139



P 05 14



P_05_140



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P_05_142



P_05_143



P 05 144



P_05_146



P 05 147



P_05_148



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P_05_15



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P_05_157



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P_05_159



P 05 16



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P_05_163



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P 05 17



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P_05_176



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P 05 178



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P 05 18



P_05_180



P 05 181



P 05 182



P 05 183



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P 05 185



P 05 186



P 05 187



P_05_188



P 05 189



P 05 19



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P_05_191



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P 05 194



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P 05 196



P 05 197



P_05_198



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P 05 20



P_05_200



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P 05 202



P 05 203



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P_05_21



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P_05_215



P 05 216



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P 05 219



P 05 22



P 05 220



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P 05 222



P 05 224



P 05 225



P_05_226



P_05_227



P 05 228



P_05_229



P 05 23



P_05_230



P 05 231



P_05_232



P_05_233



P_05_234



P 05 24



P 05 25



P 05 26



P 05 27



P 05 28



P 05 29



P 05 30



P 05 31



P 05 32



P 05 33



P_05_34



P 05 35



P 05 36



P 05 37



P_05_38



P 05 39



P 05 40



P_05_41



P 05 42



P 05 43



P 05 44



P 05 45



P 05 46



P 05 47



P 05 48



P 05 49



P 05 50



P 05 51



P 05 52



P_05_53



P 05 54



P 05 55



P 05 56



P 05 57



P 05 58



P 05 59



P 05 60



P 05 61



P 05 62



P 05 63



P 05 64



P 05 66



P 05 67



P 05 68



P 05 69



P 05 70



P_05_71



P_05_72



P 05 73



P 05 74



P 05 75



P 05 76



P 05 77



P_05_78



P_05_79



P 05 80



P 05 81



P_05_82



P 05 83



P_05_84



P_05_85



P_05_86



P_05_87



P_05_88



P 05 89



P_05_90



P 05 91



P_05_92



P 05 93



P 05 94



P_05_95



P_05_96



P_05_97



P_05_98



P_05_99

Text Illustrations

To Accompany


Chapter 6 : Friction



Chapter 6 : Friction.



06_01



06_02



06_03



06_04



06_05



06_06



06_07



06_08



06_09



06_10



06_11



06_12



P_06_01



P_06_02



P_06_03



P_06_04



P_06_05



P_06_06



P_06_07



P_06_08



P_06_09



P_06_10



P_06_100



P_06_101



P_06_102



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P_06_12



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P_06_122



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P_06_124



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P_06_13



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P_06_137



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P_06_14



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P_06_144



P_06_145



P_06_15



P_06_16



P_06_17



P_06_18



P_06_19



P_06_20



P_06_21



P_06_22



P_06_23



P_06_24



P_06_25



P_06_26



P_06_27



P_06_30



P_06_31



P_06_32



P_06_33



P_06_34



P_06_35



P_06_36



P_06_37



P_06_38



P_06_39



P_06_40



P_06_41



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P_06_67



P 06 68



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P 06 73



P 06 74



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P 06 78



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P 06 81



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P_06_83



P 06 84



P 06 85



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P_06_87



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P_06_90



P 06 91



P 06 92



P 06 93



P 06 94



P 06 95



P 06 97



P 06 98



P_06_99

Text Illustrations

To Accompany


Chapter 7 : Virtual Work.





07 01



07_02



07 03



07 04



07 05



07 06



07 07



07_08



07 09



07_10



07 11



07 12



07 13



07 14



07_15



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P 07 02



P 07 03



P 07 04



P 07 05



P 07 06



P_07_07



P_07_08



P_07_09



P_07_10



P_07_11



P_07_12



P_07_13



P_07_14



P_07_15



P_07_16



P_07_17



P_07_19



P_07_20



P_07_21



P_07_22



P_07_23



P_07_24



P_07_25



P_07_26



P_07_27



P_07_28



P_07_29



P_07_30



P_07_31



P_07_32



P_07_33



P_07_34



P_07_36



P_07_37



P_07_38



P_07_39



P_07_40



P_07_41



P_07_42



P_07_43



P_07_44



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P_07_52



P_07_53



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P_07_59



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P_07_61



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P_07_68



P_07_69



P_07_70



P_07_71



P_07_72



P_07_73



P_07_75



P_07_76



P_07_77

Documents

Mecanica para Ingenieros Estática - Meriam - 6ed (Sol)