Module 3: Analysis of Statically Determinate Structures

Every equilibrium solution starts with a complete free-body diagram. Replace supports by reaction components, replace distributed loads by equivalent resultants only when appropriate for the equilibrium calculation, preserve distances and directions, and state a sign convention before writing equations.

A cut through a planar beam or frame generally exposes axial force $N$, shear force $V$, and bending moment $M$. Their signs must follow one consistent convention throughout the lesson and the plotted diagrams.

The change in shear between two locations equals the signed area under $-w(x)$, while the change in moment equals the signed area under $V(x)$. Concentrated forces create jumps in the shear diagram; concentrated couples create jumps in the moment diagram.

An internal hinge transmits force but not bending moment, so $M=0$ at the hinge. Separating the structure at the hinge provides additional free bodies and equilibrium equations. The hinge forces on the two adjacent pieces are equal and opposite.

• At an unloaded joint with only two non-collinear members, both are zero-force members.
• At an unloaded joint with three members where two are collinear, the non-collinear member is zero-force. These inspection rules do not apply when an external load or support reaction acts at the joint in a way that changes equilibrium.

Isolate a truss joint as a concurrent force system and apply $\sum F_x=0$ and $\sum F_y=0$. Begin at a joint with no more than two unknown member forces. Assume unknown members are in tension; a negative result then indicates compression.

Pass an imaginary cut through the members of interest and isolate one side of the truss. For a planar truss, a section cutting no more than three independent unknown member forces can usually be solved directly using the three planar equilibrium equations. Taking moments about the intersection of two unknown force lines can isolate the third.

A cable under a load uniformly distributed per unit horizontal projection forms a parabola under the idealized model. A cable carrying its own uniform weight per unit cable length forms a catenary. The two curves are similar for shallow sag but arise from different loading definitions and should not be interchanged without justification.

First solve global force and moment equilibrium. Then use $M_C=0$ at the internal hinge $C$ on either isolated half to determine the remaining horizontal thrust. Once the reactions are known, cut the arch at the point of interest and resolve the section force into axial, shear, and moment components relative to the local arch tangent.

• R. C. Hibbeler, Structural Analysis, 11th ed., Pearson, 2024