A student analyzes a two-member frame with an internal pin at joint B. For member AB, they assume the pin at B exerts a force of 10 N to the right on member AB. What force should they show at joint B on member BC's free-body diagram?
A10 N to the right — both members experience the same force from the pin
B10 N to the left — Newton's third law requires equal and opposite reaction
CUnknown direction — the direction on BC must be solved independently
DZero — the pin force on AB is already accounted for in the overall structure FBD
Newton's third law is non-negotiable at every internal pin: if member AB experiences 10 N rightward from the pin at B, then member BC experiences exactly 10 N leftward. The pin force doesn't have an independent direction — once you assign a direction on one member, the other member gets the exact opposite. Violating this (drawing the same direction on both) is physically wrong and produces incorrect results even when equations appear to balance numerically.
Question 2 Multiple Choice
A structural member has pin connections at both ends and also has a distributed load applied along its length. How should this member be analyzed?
AAs a two-force member, since it has only two pin connections
BAs a multi-force member, since the distributed load creates loading at more than two points and introduces bending and shear
CAs a truss element, since the pins at both ends allow rotation
DEither method works — the choice only affects calculation complexity, not correctness
A two-force member requires forces applied at exactly two points with no moment between them — this allows the collinearity argument that forces must be axial. A distributed load violates this condition: it acts at infinitely many points along the member, producing bending and shear in addition to axial load. Treating such a member as two-force gives wrong answers because it ignores those internal components. The number of pin connections doesn't determine the analysis type — the number of loading locations does.
Question 3 True / False
The correct procedure for solving a multi-force frame is to analyze each member individually first, then combine the results to find the external reactions on the overall structure.
TTrue
FFalse
Answer: False
The correct procedure is the opposite: first analyze the entire structure as a single rigid body to find external support reactions, then disassemble at internal pins and analyze each member. Starting with individual members leaves you with more unknowns than equations (you don't know the external reactions yet), making the system unsolvable or requiring large simultaneous systems. The whole-structure FBD uses three equations to solve for external reactions cleanly before introducing internal pin forces.
Question 4 True / False
When shared pin forces between two members are labeled with assumed directions on one member's free-body diagram, they must be drawn in the exact opposite direction on the other member's free-body diagram.
TTrue
FFalse
Answer: True
This is Newton's third law applied to every internal joint. The pin exerts a force on member A and an equal-and-opposite force on member B. If you draw Bx pointing right on member A's FBD, you must draw Bx pointing left on member B's FBD. If your algebra gives a negative value, it means the actual force is opposite to your assumed direction — but as long as the pairing is consistent, the signs handle themselves correctly.
Question 5 Short Answer
Why can't a structural member with pin connections at both ends and a transverse load applied between them be analyzed as a two-force member? What goes wrong if you try?
Think about your answer, then reveal below.
Model answer: A two-force member relies on the collinearity argument: with forces only at two points and equilibrium to satisfy, those forces must be equal, opposite, and directed along the line between the two points. The moment you add a transverse load at an intermediate point, moment equilibrium about either pin is non-zero, so collinearity fails. The reaction forces at the pins are no longer axial — they have transverse components to balance the transverse load. Treating it as two-force ignores these components, gives wrong pin reactions, and makes the member appear to carry only axial load when it is actually bending.
The practical consequence is that the internal shear and bending moment distribution — which determines whether the member might fail — would be completely missed. Multi-force member analysis reveals the full loading state; two-force shortcutting hides it. The presence of the transverse load doesn't just change the magnitude of the answer; it changes the type of loading the member experiences.