meriam engineering mechanics statics 6th - solution manual

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Page 1: Meriam Engineering Mechanics Statics 6th - Solution Manual

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Phillip Molnar
Sticky Note
Ch 1
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Phillip Molnar
Sticky Note
Ch 2
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Phillip Molnar
Sticky Note
Ch. 3
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Phillip Molnar
Sticky Note
Ch 4
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4/10

As a whole:

trianglessimilarbym5.1:Note =CE

kN80.3

06)45cos5.13(3)45cos3(5:0

=

=−°++°=Σ

y

yA

D

DM

Joint D:

°=°−

°= − 7.28

45cos3645sin3tan 1θ

⎩⎨⎧

==−°=Σ

==°−=Σ

TDEDEF

CCDCDF

x

y

kN94.6,07.28cos92.7:0

kN92.7,07.28sin80.3:0

Joint C:

⎩⎨⎧

=−°+°−°=Σ=°−°−°−=Σ

037.28sin92.745sin7.28sin:007.28cos92.745cos7.28cos:0

CEBCFCEBCF

y

x

Solve simultaneously to obtain: )(kN74.2)(kN70.5

CCECBC

−=−=

Joint E:

°=

°−°= 5.67

245180β

TBE

BEFy

kN10.2

045sin74.25.67sin:0

=

=°−°=Σ

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Page 357: Meriam Engineering Mechanics Statics 6th - Solution Manual

4/124

As a whole: By symmetry, Ay = Ey = 7.5 kips

0:0 ==Σ xx AF

Joint A:

°== − 7.331510tan 1α

⎪⎩

⎪⎨⎧

==+°−=Σ

==+°−=Σ

CABABF

TAFAFF

y

x

kips52.13,05.77.33sin:0

kips25.11,07.33cos52.13:0

Joint B:

⎩⎨⎧

=−=−=Σ=−−=Σ

05sin)52.13(:00cos)52.13(:0

αα

BFBCFBFBCF

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

TCFCFFy 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

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4/130

By symmetry, Ay = Gy = 6 kN

0:0 ==Σ xx AF

°==°= − 43.185.15.0tan,45 1βα

m2

3=AC

0,0)2()5.4(3)3(5.1)3(6:0 ==−−−=Σ CHACCHM I

CCDCDFy kN74.4orkN74.4,035.143.18sin6:0 −==−−°+=Σ

TAHAHFx kN5.4,043.18cos74.4:0 ==+°−=Σ

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Page 366: Meriam Engineering Mechanics Statics 6th - Solution Manual

4/132

By symmetry, Ay = Ey = 4 kN

Joint A:

°==°== −− 5.805.0

3tan,6.405.3

3tan 11 βα

⎩⎨⎧

=°+°=Σ=+°+°=Σ

06.40sin5.80sin:0046.40cos5.80cos:0

AFABFAFABF

y

x

Solve to obtain: CAB kN73.4orkN73.4−= Joint C:

By symmetry, BC = CD

CBCBCFy kN66.5,0845sin2:0 ==−°=Σ

Joint B:

045sin66.56.40sin5.80sin73.4:0 =°−°−°=Σ BFFy

TBF kN024.1=

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Page 370: Meriam Engineering Mechanics Statics 6th - Solution Manual

4/136

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

°=°°−

= − 1530sin24

60sin2424tan 1θ

⎪⎩

⎪⎨⎧

==−°=Σ

==−°=Σ

CLDEDELF

CLEMLEMF

x

y

366.1,015cos932.1:0

932.1,0)15sin(2:0

Joint M:

°=°−°°−°

= − 5.3760cos2445cos2445sin2460sin24tan 1α

°=°−°

°−= − 9.32

60cos2445cos2460sin2424tan 1β

⎩⎨⎧

=°−°−°=Σ=°−°−°−=Σ

015sin932.15.37sin9.32sin:0015cos932.15.37cos9.32cos:0

MNDMFLMNDMF

y

x

Solve simultaneously to obtain: CLDM 785.0=

Joint D:

CLDN

DNLLFy

574.0

09.32sin785.0:0

=

=−−°=Σ

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Page 392: Meriam Engineering Mechanics Statics 6th - Solution Manual

4/15

By symmetry, Ay = Fy = 4 kN

Joint A: °== − 60

21cos 1θ

⎩⎨⎧

==°−=Σ

==°−=Σ

TAGAGF

CABABF

x

y

kN31.2,060cos62.4:0

kN62.4,060sinkN4:0

Joint G:

°=°

= − 9.402

60sin2tan 1α

TBGBGFy kN528.1,029.40sin2:0 ==−°=Σ

Joint B:

CBC

BCFy

kN46.3

09.40sin528.160sin60sin62.4:0

=

=°−°−°=Σ

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Page 398: Meriam Engineering Mechanics Statics 6th - Solution Manual

4/2

By symmetry, Ay = Cy = 1000 lb

Joint A:

°== − 3.30127tan 1θ

⎪⎩

⎪⎨⎧

==+°=Σ

−==+°=Σ

TABABADF

CADADF

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.

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Page 401: Meriam Engineering Mechanics Statics 6th - Solution Manual

4/22

Joint E:

°=°−°

= 752

30180θ

⎩⎨⎧

==+°−=Σ

==−°=Σ

TLEDEDLF

TLEFLEFF

x

y

268.0,075cos035.1:0

035.1,075sin:0

Joint D:

RDFRRRRDF 239.1,30cos)2)((24 222=°−+=

°==° 8.23,sin

239.130sin αα

RR

⎩⎨⎧

=°+°+−=Σ=°−°−−=Σ

075sin8.23sin:0075cos8.23cos268.0:0

CDDFLFCDDFLF

y

x

Solve simultaneously to obtain: CLDFTLCD 664.0,313.1 == Joint C:

⎩⎨⎧

=°−°−°=Σ=°+°−°−=Σ

075sin313.130sin45sin:0075cos313.130cos45cos:0

LCFBCFLCFBCF

y

x

Solve simultaneously to obtain: )(674.0 CLCF −=

Joint F: °=°+°+°= 8.83308.2330β

⎩⎨⎧

=−°+°−+°=Σ=+°+°++°−=Σ

0sin30sin75sinsin45sin:00cos30cos75coscos45cos:0

αβαβ

DFCFEFBFFGFDFCFEFBFFGF

y

x

Solve simultaneously to obtain: TLBF 814.1=

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Page 404: Meriam Engineering Mechanics Statics 6th - Solution Manual

4/25

Location of C: 22c 4 cyx −−=

°−=°−−=−++=

25sin4),25cos4()4()4( 222

ACyACxyxAC

cc

cc

Solve to obtain ft815.1=AC Joint B:

°=°−

°= − 04.18

25cos815.1425sin815.1tan 1θ

⎩⎨⎧

==−°=Σ

==−°=Σ

TABABF

CCBCBF

x

y

lb1535,004.18cos1615:0

lb1615,050004.18sin:0

Joint C:

°=°−=°=°−°

= 4525,702

40180 δβδ

⎩⎨⎧

=°−°−°−=Σ

=°−°−°=Σ

004.18cos161545cos25cos:0

004.18sin161545sin25sin:0

CDACF

CDACF

x

y

Solve simultaneously to obtain: CAC lb779orlb779−=

By symmetry across a line from the origin to point A, CACAD lb779==

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Page 408: Meriam Engineering Mechanics Statics 6th - Solution Manual

4/28

By symmetry, Ay = Iy = 5/2L, Ax = 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 xN = -8 m: yN2 = -82 + 162

yN = 13.86 m N = (-8, 13.86) m Location of A: yA

2= -122 + 162, yA = 10.58 m, A = (-12, 10.58) m Location of B: yB

2 = -122 + 182 =, yB = 13.42 m, B = (-12, 13.41) m Location of C: yC

2 = -82 + 182 =, yC = 16.12 m, C = (-8, 16.12) m Location of D: yD

2 = -42 + 182 =, yD = 17.55 m, C = (-4, 17.55) m Location of M: yM

2 = -42 + 162 =, yM = 15.49 m, C = (-4, 15.49) m Joint B:

°=+−−

= − 28.6128

42.1386.13tan 1α

°=+−−

= − 1.34128

42.1312.16tan 1β

⎩⎨⎧

=°+°+=Σ=°+°=Σ

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

BNBCLFBNBCF

y

x

Solve simultaneously to obtain: )(32.5,44.4 CLBCTLBN −== Joint N:

°=

+−−

= − 2.2284

86.1349.15tan 1γ

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⎩⎨⎧

==−+°+°−=Σ==°+°−=Σ

CLCNLCNLFTLNMNMLF

y

x

318.0,02.22sin76.428.6sin44.4:076.4,02.22cos28.6cos44.4:0

Joint C:

°=

+−−

= − 61.1984

12.1655.17tan 1δ

°=−−

= − 99.848

49.1512.16tan 1ε

⎩⎨⎧

=°−°++°=Σ=°+°+°=Σ

099.8sin61.19sin318.01.34sin32.5:0099.8cos61.19cos1.34cos32.5:0

CMCDLLFCMCDLF

y

x

Solve simultaneously to obtain: TLCM 41.3=

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4/3

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

⎪⎩

⎪⎨⎧

==°−=Σ

==−°=Σ

CADCDADF

TCDCDF

x

y

N3000,045cos:0

N4240,0300045sin:0

Joint C:

⎪⎩

⎪⎨⎧

==°+−=Σ

==°−°=Σ

TBCBCF

CACACF

x

y

N6000,0)45cos4240(2:0

N4240,045sin424045sin:0

Joint B: From equilibrium of entire truss TABFy N3000,0From ==Σ

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4/30

°== − 9.3625.1tan 1α

TAE

AEAEM G

lb1417

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

=

=°−°−+=Σ

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4/31

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

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4/34

°=°

= − 5.55

6

20tan64

tan 1

d

TLBE

dBEdLM D

809.0

0)5.55(sin23

:0

=

=°⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛=Σ

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Page 422: Meriam Engineering Mechanics Statics 6th - Solution Manual

4/4

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

⎪⎩

⎪⎨⎧

==−°=Σ

==−°=Σ

TCDCDF

CBCBCF

x

y

N5000,045cos7070:0

N7070,0500045sin:0

Joint D:

m31.4,45cos)6)(5(265 222=°−+= DBDB

°=⇒°

= 1.5531.445sin

5sin θθ

⎩⎨⎧

=°−°−=Σ=°−°+=Σ

01.55sin45sin:0045cos1.55cos5000:0

BDADFADBDF

y

x

Solve simultaneously to obtain: TAD

CBD

N4170

N3590orN3590

=

−=

Joint B:

CAB

ABFx

N2950orN2950

045cos70701.55cos3590:0

−=

=−°−°=Σ

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Page 426: Meriam Engineering Mechanics Statics 6th - Solution Manual

4/43

By symmetry, Ay = Iy = 3L

0:0 ==Σ xx AF

°=°−°+

= − 8.268

25tan2430tan166tan 1θ

CLLDE

DELLLLM L

80.4or80.4

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

)16()8(:0

−=

=°+°−°−−++=Σ

⎪⎩

⎪⎨⎧

−−+°+°−°−=Σ

=°+°+°−=Σ

22325sin8.26sin30sin80.4:0

025cos8.26cos30cos80.4:0LLLLMDLLF

LMDLLF

y

x

Solve simultaneously to obtain: TLLMTLDL 54.4,0446.0 == Joint E:

TLEL

ELLLFy

80.3

0)30sin80.4(2:0

=

=−−°=Σ

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4/44

By symmetry, Ay = Gy = 3L

0:0 ==Σ xx AF

TLDQDQLLLFy 576.0,0

25.3tansin

223:0 1 ==⎟

⎠⎞

⎜⎝⎛−−−=Σ −

By inspection: 0=CQ

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Page 430: Meriam Engineering Mechanics Statics 6th - Solution Manual

4/47

By symmetry, Ay = Gy = 3L 0:0 ==Σ xx AF

°==°== −− 7.3364tan,46.9

61tan 11 βα

( )[ ])(74.2

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

:0

CLCD

CDLLLLM J

−=

=°−°−−++=Σ

TLJMJMLLCDLFy 0901.0,07.33sin22

46.9sin3:0 ==°−−−°+=Σ

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Page 432: Meriam Engineering Mechanics Statics 6th - Solution Manual

4/49

By symmetry, Ay = Hy = 3L

0:0 ==Σ xx AF

Origin at center of arc

Location of I: yI2 = 252 – 92, B = (9, 23.3) m

Location of J: yJ2 = 252 – 62, J = (6, 24.3) m

Location of G: yG = yI + 6, G = (6, 29.3) m Location of K: yK

2 = 252 – 32, K = (3, 24.8) m

°=−

=°=−

= −− 39.103

3.248.24tan,3.563

8.243.29tan 11 βα

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

)6(3:0 CLFGLyyFGLLM KGK −==−−+−=Σ

⎪⎩

⎪⎨⎧

°−°+=Σ

=°−°−=Σ

3.56sin39.10sin23:0

039.10cos3.56cos66.2:0

GKJKLF

JKGKLF

y

x

Solve simultaneously to obtain: TLGK 13.2=

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4/50

By symmetry, Ay = Hy = 3L

0:0 ==Σ xx AF

Origin at center of arc

Location of B: yB2 = 302 – (-15)2, B = (-15, 26.0) m

Location of A: yA = yB – 6, A = (-18, 20.0) m Location of C: yC

2 = 302 – (-9)2, C = (-9, 28.6) m Location of M: yM = yA, M = (-9, 20.0) m Location of D: yD

2 = 302 – (-3)2, D = (-3, 29.8) m Location of L: yL = yA, L = (-3, 20.0) m

°=−

=°=−

= −− 2.556

0.206.28tan,60.116

6.288.29tan 11 βα

CLCDLCDCDLLLM L

79.2or79.20)60.1190sin()0.208.29()12()6()15(3:0

=−==°−°−−++−=Σ

TLCLCLLLLLFy 534.0,02.55sin60.11sin79.23:0 ==°−°−−−=Σ

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4/53

N5660,0)81.9(100060sin2:0 ==−°=Σ TTFy

From 0=Σ xF , Ax + Bx = 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=

°−°=== DEEFFG

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

yB

A

TTTTAM

N90023,0)81.9(100007014:0 ==−+−=Σ yyy BBF

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Page 438: Meriam Engineering Mechanics Statics 6th - Solution Manual

°=°+

°= − 8.28

15sin5.125.215cos5.1tan 1β

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

⎩⎨⎧

=°+°+°++−=Σ=°+°+°=Σ

01.25sin8.28cos15sin420199002307014:001.25cosCKsin28.8cos1542019:0

CDCKFCDF

y

x

Solve simultaneously to obtain: CCKCK N9290orN9290 =−=

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4/8

As a whole: 00 =⇒=Σ xx AF

kN60== EAy by 0=Σ yF and symmetry.

Joint A: ( )( )°== − 7.385/4tan 1θ

⎪⎩

⎪⎨⎧

=−=Σ

==−=Σ

TAHAHF

CABABF

x

y

kN75,cos0.96:0

kN0.96,0sin60:0

θ

θ

Joint B:

⎪⎩

⎪⎨⎧

==°+−=Σ

−==°+=Σ

TBHBHF

CBCBCF

y

x

kN60,03.51cos0.96:0

)(kN75,03.51sin0.96:0

Joint H:

⎪⎩

⎪⎨⎧

==−+=Σ

==+−=Σ

TGHGHF

CCHCHF

x

y

kN5.112,075cos0.48:0

kN0.48,030sin:0

θ

θ

Joint G:

TCGFy 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

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4/9

Joint C:

°== − 7.3854tan 1θ

⎪⎩

⎪⎨⎧

==°−°−=Σ

==°+°−=Σ

TBCBCF

CACACF

y

x

kN45.6,07.38cos07.215cos5:0

kN07.2,07.38sin15sin5:0

Joint B:

°== − 6.2642tan 1β

⎪⎩

⎪⎨⎧

==+°−=Σ

==−°=Σ

CBDBDF

TABABF

x

y

kN89.12,06.26cos42.14:0

kN42.14,045.66.26sin:0

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Phillip Molnar
Sticky Note
Ch 5
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Page 1052: Meriam Engineering Mechanics Statics 6th - Solution Manual

Text Illustrations

To Accompany

Meriam/Kraige – Statics 6e

Appendix A: Area Moments of Inertia.

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A 01

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A 02

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A_03

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A 04

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A 05

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A 06

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A 07

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P A 01

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P A 02

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P_A_03

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P_A_04

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P_A_05

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P A 06

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P A 07

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P A 08

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P A 09

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P_A_10

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P A 11

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P A 12

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P_A_13

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P A 14

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P A 15

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P A 16

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P A 17

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P A 18

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P A 19

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P A 20

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P A 21

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P A 22

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P A 23

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P A 24

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P A 25

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P A 27

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P A 28

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P A 29

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P_A_30

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P A 31

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P A 32

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P_A_33

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P A 34

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P A 35

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P A 36

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P A 37

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P A 38

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P A 39

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P_A_40

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P_A_41

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P A 42

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P A 43

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P A 44

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P_A_45

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P A 46

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P_A_47

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P A 48

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P A 49

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P_A_50

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P A 51

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P A 52

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P_A_53

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P A 54

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P_A_55

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P A 56

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P A 58

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P A 59

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P A 60

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P A 61

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P_A_62

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P A 63

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P A 64

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P A 65

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P A 66

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P_A_67

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P A 72

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P A 87

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Text Illustrations

To Accompany

Meriam/Kraige – Statics 6e

Appendix B: Summary of Mass Moments of Inertia.

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Text Illustrations

To Accompany

Meriam/Kraige – Statics 6e

Appendix D: Useful Tables.

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D 01tbl

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D 02tbl

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D 03tbl

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D 03tbl cont

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D 04 tbl cont2

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D 04 tbl cont3

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D 04tbl

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D 04tbl cont

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Text Illustrations

To Accompany

Meriam/Kraige – Statics 6e

Chapter 1 : Introduction to Statics.

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01_08

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P_01_09

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Text Illustrations

To Accompany

Meriam/Kraige – Statics 6e

Chapter 2 : Force Systems.

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02 02

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02_19

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02_27

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Text Illustrations

To Accompany

Meriam/Kraige – Statics 6e

Chapter 3 : Equilibrium.

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Text Illustrations

To Accompany

Meriam/Kraige – Statics 6e

Chapter 4 : Structures.

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Text Illustrations

To Accompany

Meriam/Kraige – Statics 6e

Chapter 5 : Distributed Forces.

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Text Illustrations

To Accompany

Meriam/Kraige – Statics 6e

Chapter 6 : Friction.

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Text Illustrations

To Accompany

Meriam/Kraige – Statics 6e

Chapter 7 : Virtual Work.

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