Trusses, Frames, Centroids, and Friction

Choose the structural model

FE Mechanical statics is not one problem type. A truss, a frame, a machine, and a beam can all be drawn with pins and loads, but they do not use the same assumptions. A truss is modeled as pin-connected two-force members loaded only at joints. A frame contains at least one multi-force member intended to support loads. A machine transmits or modifies forces through connected members.

That classification controls your free-body diagrams. In a simple truss, member forces act along member centerlines and are tension or compression. In a frame or machine, pins can transmit two components, and members may carry moments internally even if the pin itself does not transmit a couple.

Truss method selection

For a planar truss, check determinacy with m + r = 2j. Here m is member count, r is external reaction count, and j is joint count. Equality suggests a stable, statically determinate truss if the geometry is also stable. A smaller value suggests a mechanism; a larger value suggests static indeterminacy.

With the method of joints, start at a joint with no more than two unknown member forces. Assume all unknowns are tension pulling away from the joint; negative answers indicate compression. With sections, cut through no more than three unknown members and take moments about a point that eliminates two of them.

Centroids and area moments

Centroid questions are bookkeeping problems. Break the area into simple shapes, pick a reference origin, and compute x_bar = sum A_i x_i / sum A_i and y_bar = sum A_i y_i / sum A_i. Holes count as negative area. If the problem gives a curve, use the centroid integrals from the handbook or convert the shape to a known triangle, rectangle, semicircle, or composite area.

Area moment of inertia requires the correct axis. The handbook lists centroidal formulas such as rectangle I_x = bh^3/12. If the requested axis is offset, use the parallel-axis theorem I = I_c + Ad^2. Many FE misses come from using bh^3/12 about a base axis, where bh^3/3 would apply for a rectangle about its own base.

Friction model

Dry friction is a force limit, not always an equality. If no slipping is imminent, static friction is whatever value equilibrium requires up to f_s <= mu_s N. At impending motion, f_s = mu_s N and points opposite the impending relative motion. Once sliding occurs, kinetic friction is f_k = mu_k N.

For blocks on inclines, draw weight as mg downward, not perpendicular to the plane. Resolve into components parallel and normal to the surface: W sin theta along the slope and W cos theta normal to the slope. Use consistent units. If mass is given in kg, weight is mg in newtons; if weight is given in lb, it is already a force for statics.