You are drawing separate FBDs for two frame members connected by an internal pin. You label the pin force on member A as components (Cx, Cy). What force components must appear on member B at that same pin?
A(Cx, Cy) — the same force acts on both members simultaneously
B(-Cx, -Cy) — Newton's third law requires the reaction to be equal and opposite
CNew independent unknowns (Dx, Dy), since each member has its own FBD
DZero, because internal pin forces cancel when the whole structure is in equilibrium
Newton's third law is the thread connecting separated FBDs: if A pushes on B with (Cx, Cy), then B pushes back on A with (-Cx, -Cy). Introducing independent unknowns (Dx, Dy) for each side of the pin — option C — is the classic mistake. It creates more unknowns than equations, making the system unsolvable. You introduce the pin force once, then negate it on the adjacent member.
Question 2 Multiple Choice
A frame member has pin connections at three points and carries a distributed load between two of them. How should it be classified and analyzed?
AAs a two-force member, since all forces must ultimately balance to maintain equilibrium
BAs a multi-force member — it carries bending and shear — requiring three equilibrium equations applied to its isolated FBD
CBy summing moments only, since axial and shear forces cancel in a symmetric member
DAs a truss member, provided the distributed load is replaced by equivalent point loads
A two-force member requires exactly two pin connections and no loads applied between them, so the force acts along the member's axis. Three connection points and an intermediate distributed load mean the member carries bending and shear — it is definitively a multi-force member. The two-force shortcut is invalid here. You must isolate it, draw a complete FBD, and apply ΣFx = 0, ΣFy = 0, ΣM = 0.
Question 3 True / False
Disassembling a frame at its internal pins, rather than analyzing the entire assembled structure, is necessary because the assembled FBD alone does not provide enough equations to find all internal forces.
TTrue
FFalse
Answer: True
The assembled structure gives at most 3 equilibrium equations (in 2D), which can only find the external support reactions. Internal pin forces between members are additional unknowns that only appear when you cut through the pins and draw individual member FBDs. Disassembly is what exposes those unknowns and provides the equations to solve them.
Question 4 True / False
A frame member connected at exactly two pins with no load applied between them can be treated as a two-force member, regardless of whether the overall frame is statically determinate.
TTrue
FFalse
Answer: True
The two-force member classification depends only on the member's own geometry and loading conditions — two pin connections, no intermediate loads — not on the overall structure's determinacy. Such a member carries only axial force along the line joining its pins. This simplification is valid whenever the member meets those conditions, and should be exploited to reduce unknowns.
Question 5 Short Answer
Why does violating Newton's third law when transferring pin forces between member FBDs make the system unsolvable?
Think about your answer, then reveal below.
Model answer: Each internal pin introduces two unknown force components. Newton's third law says those components appear with opposite signs on the two members sharing the pin — so there are only two unknowns per pin, not four. If you introduce independent unknowns on each side instead of negating, you double the unknowns without adding any equations, giving an underdetermined system with no unique solution. The law isn't just a physics requirement; it is the constraint that keeps the equation count equal to the unknown count.
This is why consistent FBD labeling is non-negotiable in frame analysis. Mistakes here don't produce a wrong number — they produce an unsolvable system, which is a signal to go back and check all pin force signs across every FBD before attempting to solve.