# 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 1

### Question

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

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 ${\displaystyle AB}$  and ${\displaystyle BH}$  is found.

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

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

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

${\displaystyle \sum F_{y}=0=BH-BA+BC\cos(50)-BE\cos(50)}$

${\displaystyle \sum F_{x}=0=BC\sin(50)+BE\sin(50)}$

#### Substitution

From analysis of joint ${\displaystyle A}$

${\displaystyle \ BA=300N(Tension)}$

From analysis of joint ${\displaystyle H}$

${\displaystyle \ BH=0N}$

Put values for ${\displaystyle BA}$  and ${\displaystyle BH}$  into the equilibrium equations for joint B.

${\displaystyle \sum F_{y}=0=0-300N+BC\cos(50)-BE\cos(50)}$

${\displaystyle \sum F_{x}=0=BC\sin(50)+BE\sin(50)}$

${\displaystyle \ BC\sin(50)=-BE\sin(50)}$

${\displaystyle \ BC=-BE}$

Now ${\displaystyle BC}$  can be inserted in place of ${\displaystyle -BE}$  in ${\displaystyle \sum F_{y}}$ , which gives:

${\displaystyle 2BC\cos(50)-300N=0}$

Finally, ${\displaystyle BC}$  can be solved for as follows:

${\displaystyle BC=233.4N}$