Case 1: Pin-Pin
Does not apply. There is only one pin support.
Case 2: Two Member Joint
Joints 6 and 10 have two non-parallel members and no applied force or support reaction. Therefore, members I, E, Q, and H are zero force members.
With the elimination of member I, joint 1 is only shared between two non-parallel members: A and J. Since there is no applied force or support reaction at that joint, members A and J are also zero force members.
With the elimination of member A, joint 2 is only shared between two members: B and K. Since there is a force parallel to member K, the other non-parallel member, B, is a zero force member.
Case 3: Three Member Joint
Three members are connected at joints 4 and 8, and two of the members are parallel. Since no applied force or support reaction is present at those joints, the third non-parallel members, M and O, are zero force members.
Simplified Truss
Knowing that the internal forces in members A, B, E, H, I, J, M, O and Q are zero, we can eliminate them from the picture to simplify truss analysis.
StructNotes 