Plane pin-jointed trusses and frames – redundancy check using member–joint count Let n be the number of members and j be the number of joints in a planar pin-jointed frame (truss). The frame contains redundant (statistically indeterminate) members when which of the following relations holds true?

Introduction / Context:For planar pin-jointed trusses, a quick determinacy check compares the number of members and joints. This avoids lengthy calculations and flags whether additional compatibility equations (beyond static equilibrium) are required.

Given Data / Assumptions:

• Planar (2D) truss with ideal pin joints and two-force members.
• Proper external support reactions so the overall structure is stable.
• No internal mechanisms (i.e., geometry prevents collapse under small loads).

Concept / Approach:The basic relation for a statically determinate, stable planar truss is n = 2j − 3. If the actual number of members n exceeds this value, there are more unknown member forces than can be solved by the three global equilibrium equations and joint-by-joint equilibrium; such a truss is statically indeterminate (redundant). If n is less, the truss is a mechanism (unstable) unless additional bracing exists.

Step-by-Step Solution:Determinacy criterion (planar truss): n_determinant = 2j − 3.Redundancy condition: if n − (2j − 3) > 0 → extra (redundant) members.Therefore, redundancy ⇔ n > 2j − 3.

Verification / Alternative check:Try a triangular truss: j = 3, 2j − 3 = 3; a triangle with n = 4 is indeterminate, aligning with the rule.

Why Other Options Are Wrong:

• n = 2j − 3: determinate (no redundancy).
• n = 3j − 2: unrelated to 2D pin-jointed determinacy.
• n < 2j − 3 and n < j − 2: imply a mechanism/instability, not redundancy.

Common Pitfalls:Forgetting that stability also depends on geometry; a truss may satisfy n = 2j − 3 yet be unstable if poorly arranged.

Final Answer:n > 2j − 3