Resolução do Capítulo 6 - Análise de Estruturas
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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
2.9y AD
F F\u3a3 = \u2212 =
 
3.48 kN T ADF = W
 
 
2.10: 0
2.9x CD AD
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 =
 
 
same answers
 
 
 
 
 
( )17.5 40: 3 kips 3.70 kips 0
18.5 5x AD
F F\u3a3 = + \u2212 =
 
 
8.125 kipsADF =
 
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
5x AD
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 =