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# Resolução do Capítulo 6 - Análise de Estruturas

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```PROBLEM 6.1
Using the method of joints, determine the force in each member of the
truss shown. State whether each member is in tension or compression.

SOLUTION

FBD Truss:

Joint FBDs:
Joint B:

N

Joint C:
N

( ) ( )( )0: 6.25 m 4 m 315 N 0 240 NB y yM C\u3a3 = \u2212 = =C

0: 315 N 0 75 Ny y y yF B C\u3a3 = \u2212 + = =B

0: 0x xF\u3a3 = =B

75 N
5 4 3
AB BCF F
= =

125.0 N C ABF = W

100.0 N T BCF = W

By inspection:

260 N CACF = W

PROBLEM 6.2
Using the method of joints, determine the force in each member of the
truss shown. State whether each member is in tension or compression.

SOLUTION
FBD Truss:

Joint FBDs:
Joint C:

Joint A:

( ) ( )( )0: 14 ft 7.5 ft 5.6 kips 0 3 kipsA x xM C\u3a3 = \u2212 = =C

0: 0 3 kipsx x x xF A C\u3a3 = \u2212 + = =A

0: 5.6 kips 0 5.6 kipsy y yF A\u3a3 = \u2212 = =A

3 kips
5 4 3
BC ACF F
= =

5.00 kips C BCF = W

4.00 kips T ACF = W

1.6 kips
8.5 4
ABF
=

3.40 kips T ABF = W

PROBLEM 6.3
Using the method of joints, determine the force in each member of
the truss shown. State whether each member is in tension or
compression.

SOLUTION
FBD Truss:

Joint FBDs:
Joint C:

Joint B:

( )( ) ( )0: 6 ft 6 kips 9 ft 0 4 kipsB y yM C\u3a3 = \u2212 = =C

0: 6 kips 0 10 kipsy y y yF B C\u3a3 = \u2212 \u2212 = =B

0: 0x xF\u3a3 = =C

4 kips
17 15 8
AC BCF F
= =

8.50 kips T ACF = W

7.50 kips C BCF = W

By inspection:

12.50 kips C ABF = W

10 kips
5 4
ABF
=

PROBLEM 6.4
Using the method of joints, determine the force in each member of the
truss shown. State whether each member is in tension or compression.

SOLUTION
FBD Truss:

Joint FBDs:
Joint D:

Joint C:

Joint B:

( ) ( )( ) ( )0: 1.5 m 2 m 1.8 kN 3.6 m 2.4 kN 0B yM C\u3a3 = + \u2212 =

3.36 kNy =C

0: 3.36 kN 2.4 kN 0y yF B\u3a3 = + \u2212 =

0.96 kNy =B

20: 2.4 kN 0
F F\u3a3 = \u2212 =

3.48 kN T ADF = W

2.10: 0
F F F\u3a3 = \u2212 =

2.1 (3.48 kN)
2.9CD
F =

2.52 kN C CDF = W

By inspection: 3.36 kN C ACF = W

2.52 kN C BCF = W

40: 0.9 kN 0
5y AB
F F\u3a3 = \u2212 =

1.200 kN T ABF = W

PROBLEM 6.5
Using the method of joints, determine the force in each member of the
truss shown. State whether each member is in tension or compression.

SOLUTION
FBD Truss:

Joint FBDs:
Joint B:

Joint C:

Joint A:

0 :xF\u3a3 =

0x =C

By symmetry: 6 kNy y= =C D

10: 3 kN 0
5y AB
F F\u3a3 = \u2212 + =

3 5 6.71 kN T ABF = = W

20: 0
5x AB BC
F F F\u3a3 = \u2212 =

6.00 kN C BCF = W

30: 6 kN 0
5y AC
F F\u3a3 = \u2212 =

10.00 kN C ACF = W

40: 6 kN 0
5x AC CD
F F F\u3a3 = \u2212 + =

2.00 kN T CDF = W

1 30: 2 3 5 kN 2 10 kN 6 kN 0 check
55y
F \uf8eb \uf8f6 \uf8eb \uf8f6\u3a3 = \u2212 + \u2212 =\uf8ec \uf8f7 \uf8ec \uf8f7\uf8ed \uf8f8\uf8ed \uf8f8

By symmetry: 6.71 kN TAE ABF F= = W

10.00 kN C AD ACF F= = W

6.00 kN C DE BCF F= = W

PROBLEM 6.6
Using the method of joints, determine the force in each member of the
truss shown. State whether each member is in tension or compression.

SOLUTION
FBD Truss:

Joint FBDs:
Joint C:

Joint D:

( ) ( )( ) ( )( )0: 25.5 ft 6 ft 3 kips 8 ft 9.9 kips 0A yM C\u3a3 = + \u2212 =

2.4 kipsy =C

0: 2.4 kips 9.9 kips 0y yF A\u3a3 = + \u2212 =

7.4 kipsy =A

0: 3 kips 0x xF A\u3a3 = \u2212 + =

3 kipsx =A

2.4 kips
12 18.5 18.5
CD BCF F
= =

3.70 kips T CDF = W

3.70 kips C BCF = W

or: 0:x BC CDF F F\u3a3 = =

60: 2.4 kips 2 0
18.5y BC
F F\u3a3 = \u2212 =

( )17.5 40: 3 kips 3.70 kips 0
F F\u3a3 = + \u2212 =

8.13 kips T ADF = W

( ) ( )6 30: 3.7 kips 8.125 kips 0
18.5 5y BD
F F\u3a3 = + \u2212 =

6.075 kipsBDF =

6.08 kips C BDF = W

Joint A:

PROBLEM 6.6 CONTINUED

( )4 40: 3 kips 8.125 kips 0
5 5x AB
F F\u3a3 = \u2212 + \u2212 =

4.375 kipsABF =

4.38 kips C ABF = W

PROBLEM 6.7
Using the method of joints, determine the force in each member of the
truss shown. State whether each member is in tension or compression.

SOLUTION
FBD Truss:

Joint FBDs:
Joint A:

Joint B:

0: 480 N 0 480 Ny y yF A\u3a3 = \u2212 = =A

( )0: 6 m 0 0A x xM D\u3a3 = = =D

0: 0 0x x xF A\u3a3 = \u2212 = =A

480 N
6 2.5 6.5
AB ACF F
= =

200 N C ABF = W

520 N T ACF = W

200 N
2.5 6 6.5
BE BCF F
= =

480 N C BEF = W

520 N T BCF = W

Joint C:

Joint D:

PROBLEM 6.7 CONTINUED

By inspection: 520 N T CD CEF F= = W

( )2.50: 520 N 0
6.5x DE
F F\u3a3 = \u2212 =

200 N C DEF = W

PROBLEM 6.8
Using the method of joints, determine the force in each member of the
truss shown. State whether each member is in tension or compression.

SOLUTION

FBD Truss:

Joint FBDs:
Joint F:

Joint C:

( ) ( )( ) ( )( )0: 9 ft 6.75 ft 4 kips 13.5 ft 4 kips 0E yM F\u3a3 = \u2212 \u2212 =

9 kipsy =F

0: 9 kips 0 9 kipsy y yF E\u3a3 = \u2212 + = =E

0: 4 kips 4 kips 0 8 kipsx x xF E\u3a3 = \u2212 + + = =E

By inspection of joint :E

9.00 kips T ECF = W

8.00 kips T EFF = W

By inspection of joint :B

0ABF = W

0BDF = W

40: 8 kips 0
5x CF
F F\u3a3 = \u2212 =

10.00 kips C CFF = W

30: (10 kips) 0
5y DF
F F\u3a3 = \u2212 =

6.00 kips T DFF = W

( )40: 4 kips 10 kips 0
5x CD
F F\u3a3 = \u2212 + =

4.00 kips T CDF = W

( )30: 9 kips 10 kips 0
5y AC
F F\u3a3 = \u2212 + =

3.00 kips T ACF = W

Joint A:

PROBLEM 6.8 CONTINUED

40: 4 kips 0
F F\u3a3 = \u2212 =

5.00 kips C ADF = W

PROBLEM 6.9
Determine the force in each member of the Gambrel roof truss shown.
State whether each member is in tension or compression.

SOLUTION
FBD Truss:

Joint A:

Joint B:

Joint D:

0: 0x xF\u3a3 = =H

By symmetry: 12 kNy= =A H

By inspection of joints and ,C G

and 0CE AC BCF F F= = W

and 0EG GH FGF F F= = W

30: 12 kN 3 kN 0
5y AB
F F\u3a3 = \u2212 \u2212 =

15.00 kN C ABF = W

( )40: 15 kN 0
5x AC
F F\u3a3 = \u2212 =

12.00 kN T ACF = W

( )4 10 40: 15 kN 0
5 10.44 5x BD BE
F F F\u3a3 = \u2212 \u2212 =

( )3 3 30: 15 kN 6 kN 0
5 10.44 5y BD BE
F F F\u3a3 = \u2212 \u2212 + =

Solving yields 11.93 kN C BDF = W

0.714 kN C BEF = W

0: by symmetryxF\u3a3 =```