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

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

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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.

As a whole:

trianglessimilar bym5.1: Note =CE

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

==−°++°=Σ

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

Solve simultaneously to obtain:)(kN74.2

)(kN70.5

Joint E :

°=°−°

= 5.672

45180 β

BE F y

kN10.2

045sin74.25.67sin:0

=°−°=Σ

As a whole:

By symmetry, Ay = Ey = 7.5 kips

0:0 ==Σ x x AF

Joint A:

°== − 7.3315

10tan 1

⎪⎩

==+°−=Σ

C AB ABF

T AF AF F

kips52.13,05.77.33sin:0

kips25.11,07.33cos52.13:0

Joint B:

⎩⎨⎧

=−=−=Σ

=−−=Σ

05sin)52.13(:0

0cos)52.13(:0

BF BC F

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

By symmetry, A y = G y = 6 kN

0:0 ==Σ x x AF

°==°=− 43.18

5.0tan,45 1 β α

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

By symmetry, A y = E y = 4 kN

Joint A:

°==°== −− 5.805.0

3tan,6.40

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=

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

Joint E :

°=°°−= − 15

30sin24

60sin2424tan 1θ

⎪⎩

⎪⎨⎧

==−°=Σ

C L DE DE LF

C L EM L EM F

366.1,015cos932.1:0

932.1,0)15sin(2:0

Joint M :

°°−°− 537

45sin2460sin24tan 1α

°=°−°

= 5.3760cos2445cos24

°=°−°

°−= − 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

Solve simultaneously to obtain: C L DM 785.0=

Joint D:

C L DN

DN L LF y

09.32sin785.0:0

=−−°=Σ

By symmetry, A y = F y = 4 kN

Joint A:

°== − 602

1cos 1

⎩⎨⎧

==°−=Σ

T AG AGF C AB ABF

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

Joint G:

= − 9.402

60sin2tan 1

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

Joint B:

BC F y

kN46.3

09.40sin528.160sin60sin62.4:0

=°−°−°=Σ

By symmetry, A y = C y = 1000 lb

Joint A:

°== − 3.3012

7tan 1

⎪⎩

⎪⎨⎧

==+°=Σ

−==+°=Σ

T AB AB ADF

C AD ADF

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.

Joint E :

°=°−°

30180θ

⎩⎨

==+°−=Σ

==−°=Σ

T L ED ED LF

T L EF L EF F

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

30sinα

⎩⎨⎧

=°+°+−=Σ

=°−°−−=Σ

075sin8.23sin:0

075cos8.23cos268.0:0

CD DF LF

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

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

Joint F :

°=°+°+°= 8.83308.2330 β

⎩⎨⎧

=−°+°−+°=Σ

=+°+°++°−=Σ

0sin30sin75sinsin45sin:0

0cos30cos75coscos45cos:0

DF CF EF BF FGF

Solve simultaneously to obtain: T L BF 814.1=

Location of C:22

c 4 c y x −−=

°−=°−−=

−++=

25sin4),25cos4(

)4()4( 222

AC y AC x

y x AC

Solve to obtain ft815.1= AC

Joint B:

°=°−

°= − 04.18

25cos815.14

25sin815.1tan 1θ

⎨⎧

==−°=Σ

T AB ABF

C CBCBF

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

Solve simultaneously to obtain: C AC lb779or lb779−=

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

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

For x N = -8 m: y N 2

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

Location of A: y A2= -12

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

Location of B: y B2

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

Location of C : yC 2

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

Location of D: y D2

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

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

Joint B:

°=+−

−= − 28.6

42.1386.13tan 1α

°=+−

−= − 1.34

42.1312.16tan 1 β

⎩⎨⎧

=°+°+=Σ=°+°=Σ

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

BN BC LF BN BC F

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

Joint N :

°=+−−= − 2.22

86.1349.15tan 1γ

⎨⎧

==−+°+°−=Σ

==°+°−=Σ

C LCN LCN LF

T L NM NM LF

318.0,02.22sin76.428.6sin44.4:0

76.4,02.22cos28.6cos44.4:0

Joint C :

°=+−

−= − 61.19

12.1655.17tan 1δ

°=−

−= − 99.8

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

Solve simultaneously to obtain: T LCM 41.3=

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

Joint D:

⎪⎩

⎪⎨⎧

==°−=Σ

==−°=Σ

C ADCD ADF

T CDCDF

N3000,045cos:0

N4240,0300045sin:0

Joint C :

⎪⎩

⎪⎨⎧

==°+−=Σ

==°−°=Σ

TBCBCF

C AC AC F

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

°==− 9.36

5.1tan 1

AE AE M G

lb1417

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

=°−°−+=Σ

T BC BC M F

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

= − 5.55

T L BE

0)5.55(sin23

=°⎟ ⎠

⎞⎜⎝

⎛ −⎟

⎞⎜⎝

⎛ =Σ

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

Joint C :

⎪⎩⎪⎨⎧

==−°=Σ

T CDCDF C BC BC F

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

AD BDF x

⎩⎨

=°−°−=Σ 01.55sin45sin:0 BD ADF y

Solve simultaneously to obtain:

N3590or N3590

Joint B:

N2950or N2950

045cos70701.55cos3590:0

=−°−°=Σ

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

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

Joint E :

T L EL

EL L LF y

0)30sin80.4(2:0

=−−°=Σ

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

1 ==⎟ ⎠

⎞⎜⎝

⎛ −−−=Σ −

By inspection: 0=CQ

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

°==°==−− 7.33

4tan,46.9

1tan 11 β α

( )[ ]

)(74.2

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

CD L L L L

=°−°−−++=Σ

T L JM JM L L

CD LF y 0901.0,07.33sin22

46.9sin3:0 ==°−−−°+=Σ

By symmetry, A y = H y = 3 L

0:0 ==Σ x x AF

Origin at center of arc

L ti f I2

B (9 23 3)

Location of I : y I 2

– 92, B = (9, 23.3) m

Location of J : y J 2

– 62, J = (6, 24.3) m

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

Location of K : yK

– 32

, K = (3, 24.8) m

°=−

=°=−

= −− 39.103

3.248.24tan,3.56

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

039.10cos3.56cos66.2:0

GK JK LF

JK GK LF

Solve simultaneously to obtain: T LGK 13.2=

By symmetry, A y = H y = 3 L

0:0 ==Σ x x AF

Origin at center of arc

Location of B: y B2

– (-15)2, B = (-15, 26.0) m

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

Location of C:2

( 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– (-3)

2, D = (-3, 29.8) m

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

°=−

=°=−

=−− 2.55

0.206.28tan,60.11

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

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(

°−°=== 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

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

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

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

kN75,cos0.96:0

kN0.96,0sin60:0

Joint B:

⎪⎩

⎪⎨⎧

==°+−=Σ

−==°+=Σ

T BH BH F

C BC BC F

kN60,03.51cos0.96:0

)(kN75,03.51sin0.96:0

Joint H :

⎪⎩

⎪⎨⎧

==−+=Σ

==+−=Σ

T GH GH F

C CH CH F

kN5.112,075cos0.48:0

kN0.48,030sin:0

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

Joint C :

°== −7.38

⎪⎩

⎪⎨⎧

==°−°−=Σ

==°+°−=Σ

T BC BC F

C AC AC F

kN45.6,07.38cos07.215cos5:0

kN07.2,07.38sin15sin5:0

Joint B:

°== −6.26

⎪⎩⎪⎨⎧

==+°−=Σ

==−°=Σ

C BD BDF

T AB ABF

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.

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.

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.

P 02 01

P 02 02

P 02 03

P 02 04

P 02 05

P 02 06

P_02_07

P 02 08

P 02 09

P 02 100

P 02 101

P 02 102

P 02 103

P 02 104

P 02 105

P 02 107

P 02 108

P 02 109

P_02_11

P 02 110

P 02 111

P 02 112

P 02 113

P_02_114

P 02 115

P 02 116

P 02 117

P 02 118

P_02_119

P 02 12

P 02 120

P 02 121

P 02 122

P 02 123

P_02_124

P 02 125

P 02 126

P 02 127

P_02_128

P 02 129

P 02 13

P 02 130

P 02 131

P 02 132

P_02_133

P_02_134

P 02 135

P 02 136

P 02 137

P 02 138

P 02 139

P_02_14

P 02 140

P 02 141

P 02 142

P 02 143

P_02_144

P 02 145

P 02 146

P_02_147

P 02 148

P 02 149

P 02 15

P 02 150

P 02 151

P 02 152

P 02 153

P 02 154

P_02_155

P 02 156

P 02 157

P 02 159

P_02_16

P 02 160

P 02 161

P_02_162

P 02 163

P 02 164

P 02 165

P 02 166

P_02_167

P 02 168

P 02 169

P 02 17

P 02 170

P 02 171

P 02 172

P 02 173

P 02 174

P 02 175

P 02 176

P 02 177

P_02_178

P 02 179

P 02 18

P 02 180

P_02_181

P 02 182

P 02 183

P 02 184

P 02 19

P 02 20

P 02 21

P 02 22

P 02 24

P 02 25

P 02 26

P 02 27

P 02 28

P 02 29

P 02 30

P 02 31

P 02 32

P 02 33

P_02_34

P 02 35

P_02_36

P 02 37

P_02_38

P 02 39

P 02 40

P_02_41

P 02 42

P_02_43

P 02 45

P_02_46

P 02 47

P 02 48

P 02 49

P 02 50

P 02 51

P 02 52

P 02 53

P 02 54

P 02 55

P 02 6

P_02_60

P_02_64

P_02_65

P_02_68

P_02_70

P_02_74

P 02 78

P_02_79

P 02 80

P 02 81

P_02_82

P_02_83

P 02 84

P_02_85

P 02 86

P_02_87

P_02_88

P 02 89

P 02 90

P_02_91

P 02 92

P 02 93

P_02_94

P_02_95

P 02 96

P_02_97

P_02_98

P 02 99

Text Illustrations

To Accompany

Chapter 3 : Equilibrium.

03 01 cont

P 03 01

P_03_02

P 03 03

P 03 04

P 03 05

P_03_06

P 03 07

P 03 08

P 03 09

P 03 10

P_03_100

P_03_101

P 03 102

P 03 103

P 03 104

P 03 105

P 03 106

P 03 107

P 03 108

P 03 109

P 03 11

P 03 110

P 03 111

P 03 112

P 03 113

P 03 114

P 03 115

P 03 116

P 03 117

P_03_118

P_03_119

P 03 12

P 03 120

P 03 121

P 03 122

P 03 123

P 03 124

P 03 125

P 03 13

P 03 14

P 03 15

P 03 16

P_03_17

P 03 18

P 03 19

P 03 20

P 03 21

P_03_22

P 03 23

P_03_24

P 03 25

P_03_26

P 03 27

P_03_28

P_03_29

P 03 30

P 03 31

P 03 32

P 03 34

P 03 35

P 03 36

P 03 37

P 03 38

P_03_39

P 03 40

P 03 41

P_03_41

P 03 42

P_03_43

P 03 44

P_03_45

P 03 46

P_03_46

P 03 47

P 03 48

P_03_49

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Chapter 4 : Structures.

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Chapter 5 : Distributed Forces.

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Chapter 6 : Friction

Chapter 6 : Friction.

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To Accompany

Chapter 7 : Virtual Work.

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

Documents

Dinámica Meriam Kraige 3 Edicion

Instrumentacion Industrial Creus 6ed

Solucionario Dinamica Meriam 2th Edicion PDF

Matemáticas Discretas 6ed

Mecanica para Ingenieros Estática - Meriam - 3ed

Manual CTO 6ed - Fisiología

Manual CTO 6ed Dermatología

Solucionario Dinamica Meriam 3th Edicion

solucionario dinamica meriam 3th edicion.pdf

Ejercicio 4 de la práctica meriam

Manual amir Hematología 6ed

Mecánica para ingenieros estática 4ta edicion j. meriam, l. g. kraige

2- Sistemas de Fuerzas - Aula Virtual de Aprendizajetic.uis.edu.co/avacopiaIIsemestre2013/pluginfile.php/77967/mod... · Modificado de [Meriam, 5ª Edición ] Estática- Sistemas

Manual CTO 6ed - Endocrinología

Resumen APA 6ed

Termodinamica Faires Simmang 6ed

Solucion Estatica Meriam 4 Ed.pdf

W40K 6ed Bolsillo eBook

Manual cto 6ed oftalmología

Manual cto 6ed anatomía