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Chapter 6: Chapter 6: Structural AnalysisStructural Analysis

© 2007 Pearson Education South Asia Pte Ltd Teaching Slides – Free-Body DiagramsTeaching Slides – Free-Body Diagrams

Example 6.3

Determine the force in each member of the truss. Indicate whether the members are intension or compression.

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Example 6.3

)(10.4

0)30sin10.4(2;0

600

0)4(600)3(400)6(;0

600

0600;0

TkNF

FF

NA

mNmNmAM

NC

CNF

DB

DBy

y

yC

x

xx

A

B C

Cy

Cx

600N

Ay

400N

6m

3m

4m

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Example 6.3

A

B C

Cy

Cx

600N

Ay

400N

6m

3m

4m

600N

Cy

FAD

)(750

054

600

;0

)(450

0)750(53

;0

CNF

FN

F

TNF

NF

F

AB

AB

y

AD

AD

x

y

xA

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Example 6.3

A

B C

Cy

Cx

600N

Ay

400N

6m

3m

4m

FDB

600N

y

xD

450N

FDC

)(200

0)250(54

;0

)(250

250

060053

450

;0

CNF

NF

F

TNF

NF

NFN

F

DC

DC

y

DB

DB

DB

x

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Example 6.3

A

B C

Cy

Cx

600N

Ay

400N

6m

3m

4m

600NC

y

xFCB

200N

200N )(0200200

;0

)(600

0600

;0

checkNN

F

CNF

NF

F

y

CB

CB

x

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Example 6.4

Using the method of joints, determine all the zero-force members of the Fink roof truss. Assume all joints are pin connected.

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Example 6.4 5 kN

2 kN

A

B

C

H G F E

D

G

y

xFGH

FGCFGF

0;0 GCy FF

Joint G

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Example 6.4 5 kN

2 kN

A

B

C

H G F E

D

x

D

yFDF

FDC

FDE

Joint D

0;0 DFx FF

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Fx

y

Example 6.4 5 kN

2 kN

A

B

C

H G F E

D

Oθ

FFEFFG

FFC Joint F

0,90

0cos;0

FC

FCy

F

FF

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B

x

y

Example 6.4 5 kN

2 kN

A

B

C

H G F E

D

FBH

FBA

FBC

Joint B

)(2

02;0

CkNF

FkNF

BH

BHx

2 kN

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Example 6.7

Determine the force in member EB of the roof truss. Indicate whether the member are in tension or compression.

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Example 6.7 1000 N

A

F

E

B C

D

1000 N3000 N

1000 N

4000 N2000 N

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Example 6.7 1000 N

A

F

E

B C

D

1000 N3000 N

1000 N

4000 N2000 N

FED

FEB

FAB

FFB

FEDcos30°

FEDsin30°

)(2000 ;0

)(3000 ;0

)(3000 ;0

TNFF

CNFF

CNFM

EBy

EFx

EDB

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Example 6.8

Determine the forces acting in the members of the space truss. Indicate whether the members are in tension or compression.

© 2007 Pearson Education South Asia Pte Ltd Teaching Slides – Free-Body DiagramsTeaching Slides – Free-Body Diagrams

Example 6.8

Joint A

0577.0577.0577.04

0

;0

)577.0577.0577.0(

,,}4{

kFjFiFkFjFj

FFFP

F

kjiF

rr

FF

kFFjFFkNjP

AEAEAEACAB

AEACAB

AE

AE

AEAEAE

ACACABAB

z

y

x

A

FAB

FAC

FAE

P = 4 kN

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Example 6.8 z

y

x

B

RAB

FBD

FBEJoint B

0

show, To

)(2

)(66.5

0707.02;0

045sin4;0

0707.045cos;0

CEDCDE

BD

BEB

BEBDz

By

BEBx

FFF

CkNF

TkNFR

FFF

RF

FRF

FAB = 4kN

2 kN

45°

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Example 6.21

The 100kg block is held in equilibrium by means of the pulley and the continuous cable system. If the cable is attached to the pin at B, compute the forces which this pin exerts on each of its connecting members

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Example 6.21

B

By490.5 N

Bx

490.5 N

45°

Pulley B

NB

NNBF

NB

NBF

y

yy

x

xx

3.837

05.49045sin5.490;0

8.346

045cos5.490;0

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Example 6.21

B

By

490.5 N

Bx

Pin B

NB

NNBF

NB

NBF

y

yy

x

xx

3.837

05.49045sin5.490;0

8.346

045cos5.490;0

FAB

FCB