Statics/Method of Joints

The method of joints is a way to find unknown forces in a truss structure. The principle behind this method is that all forces acting on a joint must add to zero. If there were a net force, the joint would move.

Example 1Edit

QuestionEdit

StaticsTrussJointEx1.svg

Find the force in member BC of the truss pictured to the right.

AnswerEdit

Using the method of joints, the force could be found by isolating the joint at either end of the member (joint B or C). Neither joint can be solved without further analysis; however, joint B can be solved if the force in member  AB and  BH is found.

To find force  AB analyze joint A. This joint has an external vertical force of 300N which must be countered by the members attached to the joint. Member  AE cannot possibly support any vertical load, otherwise it would not be loaded axially and the entire structure would no longer be a truss. If  AE has no load then member  AB is in 300N of tension.

When joint H is analyzed it is found that the force in members  BH and  HC must be zero. The reason why neither member can carry any load is that member  BH can only take a vertical load and member  HC can only take a horizontal load. In a real world application this structure might be useful if there was a load applied at joint  H . Now joint  B can be analyzed.

StaticsTrussJointex1ans.svg

The picture to the left shows the forces affecting joint B.

 \sum F_y = 0 = BH - BA + BC \cos(50) - BE \cos(50)

 \sum F_x = 0 = BC \sin(50) + BE \sin(50)

SubstitutionEdit

From analysis of joint  A

 \ BA = 300N (Tension)

From analysis of joint  H

 \ BH = 0N

Put values for  BA and  BH into the equilibrium equations for joint B.

 \sum F_y = 0 = 0 - 300N + BC \cos(50) - BE \cos(50)

 \sum F_x = 0 = BC \sin(50) + BE \sin(50)

 \ -BC \sin(50) = BE \sin(50)

 \ -BC = BE

Now  -BC can be inserted in place of  BE in  sum F_y and  BC can be solved for.

Last modified on 23 March 2012, at 20:00