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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 CΣ = − = =C

0: 315 N 0 75 Ny y y yF B CΣ = − + = =B

0: 0x xFΣ = =B

75 N

5 4 3AB BCF F= = 125.0 N C ABF =

100.0 N T BCF =

By inspection: 260 N CACF =

PROBLEM 6.2


SOLUTION

FBD Truss:

Joint FBDs:

Joint C:

Joint A:

( ) ( )( )0: 14 ft 7.5 ft 5.6 kips 0 3 kipsA x xM CΣ = − = =C

0: 0 3 kipsx x x xF A CΣ = − + = =A

0: 5.6 kips 0 5.6 kipsy y yF AΣ = − = =A

3 kips5 4 3BC ACF F= =

5.00 kips C BCF =

4.00 kips T ACF =

1.6 kips8.5 4

ABF =

3.40 kips T ABF =

PROBLEM 6.3


SOLUTION

FBD Truss:

Joint FBDs:

Joint C:

Joint B:

( )( ) ( )0: 6 ft 6 kips 9 ft 0 4 kipsB y yM CΣ = − = =C

0: 6 kips 0 10 kipsy y y yF B CΣ = − − = =B

0: 0x xFΣ = =C

4 kips17 15 8AC BCF F= =

8.50 kips T ACF =

7.50 kips C BCF =

By inspection: 12.50 kips C ABF =

10 kips5 4ABF =

PROBLEM 6.4


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 CΣ = + − =

3.36 kNy =C

0: 3.36 kN 2.4 kN 0y yF BΣ = + − =

0.96 kNy =B

20: 2.4 kN 0

2.9y ADF FΣ = − = 3.48 kN T ADF =

2.10: 02.9x CD ADF F FΣ = − =

2.1 (3.48 kN)2.9CDF = 2.52 kN C CDF =

By inspection: 3.36 kN C ACF =

2.52 kN C BCF =

40: 0.9 kN 05y ABF FΣ = − = 1.200 kN T ABF =

PROBLEM 6.5


SOLUTION

FBD Truss:

Joint FBDs:

Joint B:

Joint C:

Joint A:

0 :xFΣ = 0x =C

By symmetry: 6 kNy y= =C D

10: 3 kN 05y ABF FΣ = − + =

3 5 6.71 kN T ABF = =

20: 05x AB BCF F FΣ = − = 6.00 kN C BCF =

30: 6 kN 05y ACF FΣ = − = 10.00 kN C ACF =

40: 6 kN 05x AC CDF F FΣ = − + = 2.00 kN T CDF =

1 30: 2 3 5 kN 2 10 kN 6 kN 0 check

55yF Σ = − + − =

By symmetry: 6.71 kN TAE ABF F= =

10.00 kN C AD ACF F= =

6.00 kN C DE BCF F= =

PROBLEM 6.6


SOLUTION

FBD Truss:

Joint FBDs:

Joint C:

Joint D:

( ) ( )( ) ( )( )0: 25.5 ft 6 ft 3 kips 8 ft 9.9 kips 0A yM CΣ = + − =

2.4 kipsy =C

0: 2.4 kips 9.9 kips 0y yF AΣ = + − =

7.4 kipsy =A

0: 3 kips 0x xF AΣ = − + =

3 kipsx =A

2.4 kips12 18.5 18.5

CD BCF F= =

3.70 kips T CDF =

3.70 kips C BCF =

or: 0:x BC CDF F FΣ = = 60: 2.4 kips 2 0

18.5y BCF FΣ = − =

same answers

( )17.5 40: 3 kips 3.70 kips 018.5 5x ADF FΣ = + − =

8.125 kipsADF = 8.13 kips T ADF =

( ) ( )6 30: 3.7 kips 8.125 kips 018.5 5y BDF FΣ = + − =

6.075 kipsBDF = 6.08 kips C BDF =

Joint A:

PROBLEM 6.6 CONTINUED

( )4 40: 3 kips 8.125 kips 05 5x ABF FΣ = − + − =

4.375 kipsABF = 4.38 kips C ABF =

PROBLEM 6.7


SOLUTION

FBD Truss:

Joint FBDs:

Joint A:

Joint B:

0: 480 N 0 480 Ny y yF AΣ = − = =A

( )0: 6 m 0 0A x xM DΣ = = =D

0: 0 0x x xF AΣ = − = =A

480 N

6 2.5 6.5AB ACF F= = 200 N C ABF =

520 N T ACF =

200 N

2.5 6 6.5BE BCF F= = 480 N C BEF =

520 N T BCF =

Joint C:

Joint D:


By inspection: 520 N T CD CEF F= =

( )2.50: 520 N 06.5x DEF FΣ = − = 200 N C DEF =

PROBLEM 6.8


SOLUTION

FBD Truss:

Joint FBDs:

Joint F:

Joint C:

( ) ( )( ) ( )( )0: 9 ft 6.75 ft 4 kips 13.5 ft 4 kips 0E yM FΣ = − − =

9 kipsy =F

0: 9 kips 0 9 kipsy y yF EΣ = − + = =E

0: 4 kips 4 kips 0 8 kipsx x xF EΣ = − + + = =E

By inspection of joint :E 9.00 kips T ECF =

8.00 kips T EFF =

By inspection of joint :B 0ABF =

0BDF =

40: 8 kips 05x CFF FΣ = − = 10.00 kips C CFF =

30: (10 kips) 05y DFF FΣ = − = 6.00 kips T DFF =

( )40: 4 kips 10 kips 05x CDF FΣ = − + =

4.00 kips T CDF =

( )30: 9 kips 10 kips 05y ACF FΣ = − + =

3.00 kips T ACF =

Joint A:


40: 4 kips 05x ADF FΣ = − = 5.00 kips C ADF =

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