For a planar truss with j joints and m members, identify the condition for a simple (statically determinate and stable) truss.

Introduction / Context:
Truss classification (simple, compound, complex) and determinacy are foundational in structural analysis. A simple plane truss can be built by starting with a basic triangle (j = 3, m = 3) and adding two members and one joint at a time while maintaining determinacy and stability.

Given Data / Assumptions:

• Planar truss; pin joints; straight members carrying only axial forces.
• External reactions are determinate (typically three for a planar pin-supported truss).

Concept / Approach:
The necessary (though not always sufficient) count condition for a simple, statically determinate plane truss is m = 2j − 3. This arises from equilibrium equations: each joint contributes two scalar equations; three overall are consumed by support reactions, leaving 2j − 3 independent member force unknowns for determinacy.

Step-by-Step Solution:
1) Begin with a triangle: j = 3, m = 3 → m = 2j − 3 holds (3 = 6 − 3).2) Add a new joint with two members to existing joints → j → j+1, m → m+2 → relation preserved.3) By induction, the relation holds for any simple truss formed this way.

Verification / Alternative check:
Compare with method-of-joints determinacy: unknown member forces = m; available independent equations = 2j − r (r = 3 external reaction components). For a internally determinate truss with determinate supports, m = 2j − 3.

Why Other Options Are Wrong:

• j = 2m − 3, m = 3j − 2, j = 3m − 2: Do not match the equilibrium/determinacy logic for plane trusses.

Common Pitfalls:

• Assuming m = 2j − 3 guarantees stability in every geometry; special mechanisms can exist if members align unfavorably.

Final Answer:
m = 2j - 3.