Planar truss determinacy – member count for a stable, statically determinate frame For a stable, statically determinate planar truss (simple frame), how many members m are required in terms of the number of joints j?

Introduction / Context:
Before analyzing a truss, engineers quickly check determinacy and stability using counting rules. A classic relation links members, joints, and reaction components for planar trusses.

Given Data / Assumptions:

• Planar truss, pin-jointed, loads applied at joints.
• Two reaction points providing three reaction components in total.
• No internal multi-member rigidities beyond pin connections.

Concept / Approach:
For a statically determinate and stable planar truss, the member count is governed by the relation m = 2j − 3. This arises by equating the number of unknown bar forces plus reactions to the available independent equilibrium equations (2 per joint, or 3 for the whole truss with method of joints approach).

Step-by-Step Solution:
Total unknowns: bar forces m + reaction components r (typically r = 3).Independent equations of static equilibrium for the whole truss do not suffice; instead use method of joints: each joint contributes 2 equations, giving 2j equations.Set unknowns equal to equations for determinacy: m + r = 2j → m + 3 = 2j → m = 2j − 3.

Verification / Alternative check:
For a basic triangular truss (j = 3), m = 2*3 − 3 = 3, which matches the three sides of a triangle.

Why Other Options Are Wrong:
3j − 3, 2j − 2, and 2j − 1 would render the truss indeterminate or unstable relative to the counting rule.

Common Pitfalls:
Confusing the planar truss relation with space frames (which have m = 3j − 6 for simple determinate cases); forgetting to count actual support reactions.

Final Answer:
m = 2j − 3