Method of Joints

Method of Joints | Analysis of Simple Trusses Method of Joints The free-body diagram of any joint is a concurrent force system in which the summation of moment will be of no help. Recall that only two equilibrium equations can be written

and

This means that to solve completely for the forces acting on a joint, we must select a joint with no more than two unknown forces involved. This can be started by selecting a joint acted on by only two members. We can assume any unknown member to be either tension or compression. If negative value is obtained, this means that the force is opposite in action to that of the assumed direction. Once the forces in one joint are determined, their effects on adjacent joints are known. We then continue solving on successive joints until all members have been found. Problem 001-mj | Method of Joints

At joint A

Problem Find the force acting in all member s of the truss shown in Figure T-01.

At joint B

Solution

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At joint D

At joint E

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Summary

At joint F

At joint C

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Problem 002-mj | Method of Joints

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Problem 002-mj The structure in Fig. T-02 is a truss which is pinned to the floor at point A, and supported by a roller at point D. Determine the force to all members of the truss.

At joint G

Solution 002-mj

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At joint B

At joint A

At joint F

At joint D

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Summary At joint C

FAB = 8.73 kN tension FAG = 21.82 kN compression FBC = 15.71 kN tension FBF = 8.73 kN compression FBG = 8.73 kN tension FCD = 5.24 kN tension FCE = 13.09 kN tension FCF = 13.09 kN compression FDE = 13.09 kN compression FEF = 10.48 kN compression FFG = 12.22 kN compression

At joint E

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Problem 003-mj Find the force in each member of the truss shown in Fig. T-04.

At joint B

Solution 003-mj

HideClick here to show or hide the solution At joint C

At joint E

At joint D At joint A

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Summary

At joint F

Checking

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Top chords FDE = 64 kN tension FEF = 176 kN tension

Bottom chords FAB = 120 kN compression FBC = 32 kN compression

Web members FAF = 140 kN tension FAE = 150.78 kN compression FBE = 150.78 kN tension FBD = 86.16 kN compression FCD = 86.16 kN tension - See more at: http://www.mathalino.com/reviewer/engineeringmechanics/problem-003-mj-method joints#sthash.dDWmMfac.dpuf

Problem 004-mj | Method of Joints

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Problem 004-mj The truss pinned to the floor at D, and supported by a roller at point A is loaded as shown in Fig. T06. Determine the force in member CG.

Solution 004-mj

At joint A

At joint B

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At joint G

answer 

At joint F

Another Solution to 004-mj

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At joint C

At joint F answer 

Problem 005-mj | Method of Joints

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At joint D

Problem 005-mj Compute the force in all members of the truss shown in Fig. T-08.

Solution 005-mj

HideClick here to show or hide the solution At joint E

At joint C

At joint A

At joint E

At joint B

At joint D

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At joint F

Solution 404

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Summary

Problem 404 Roof Truss - Method of Joints

SPONSORED LINKS Problem 404 Determine the forces in the members of the roof truss shown in Fig. P-404. At Joint A

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At Joint D

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At Joint C

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Summary AB = 450 N compression AC = 389.71 N tension BC = 450 N tension BD = 900 N compression CD = 389.71 N tension

At Joint B