Trusses, Frames, Centroids, and Friction

Trusses: method of joints vs method of sections

A truss is a pin-connected assembly of straight two-force members loaded only at joints, so every member carries pure tension or compression along its axis. First find the support reactions by treating the whole truss as a rigid body (ΣF_x, ΣF_y, ΣM = 0). Then choose a member-force method.

Method of joints isolates one pin at a time. Each joint gives two scalar equations (ΣF_x = 0, ΣF_y = 0), so start at a joint with at most two unknown members. Draw unknown member forces pointing away from the joint (assume tension); a negative result means compression. Work joint to joint until you reach the member you need.

Method of sections cuts an imaginary line through the truss, including the member of interest, and applies the three equilibrium equations to one cut-off portion. This is faster when only one or two specific member forces are wanted deep inside the truss, because a single moment equation can isolate one member directly.

Zero-force members (exam shortcut)

Identifying zero-force members before computing saves time and is a favorite FE question:

• At a joint with two non-collinear members and no external load or support, both members are zero-force.
• At a joint with three members where two are collinear and no external load, the third (non-collinear) member is zero-force.

A zero-force member is not useless — it prevents buckling and stabilizes geometry — but it carries no load in the given configuration. After removing them, the remaining solution simplifies.

Frames and machines

Unlike trusses, frames contain at least one multi-force member (loaded at more than two points or carrying a couple), so members can bend, not just stretch. Dismember the structure: draw separate free-body diagrams for each member, applying Newton's third law (equal-and-opposite) at connecting pins. A two-force member inside a frame still carries axial load only, which fixes the direction of that pin force and reduces unknowns. Machines transmit and modify forces and contain moving parts; solve them the same way — dismember and apply equilibrium piece by piece.

Centroids and moments of inertia

The centroid of a composite area uses the first-moment formula from the Handbook:

$\bar{x} = \frac{\sum A_i x_i}{\sum A_i}, \qquad \bar{y} = \frac{\sum A_i y_i}{\sum A_i}$

Break the shape into rectangles, triangles, circles, and semicircles whose individual centroids you know; treat holes as negative areas. For a rectangle b×h, the centroidal area moment of inertia about its horizontal axis is I = bh³/12; for a circle of diameter d, I = πd⁴/64.

To shift an inertia to a parallel non-centroidal axis, use the parallel-axis theorem:

$I = I_c + A d^2$

where d is the distance between the axes. Worked example: a 100 mm × 200 mm rectangle (tall) about its base. I_c = (100)(200)³/12 = 6.67×10⁷ mm⁴; d = 100 mm to the base; A = 20 000 mm². I_base = 6.67×10⁷ + 20 000(100)² = 6.67×10⁷ + 2.0×10⁸ = 2.67×10⁸ mm⁴. The base inertia is four times the centroidal value — exactly the I_base = bh³/3 result.

Dry friction

Coulomb friction resists impending or actual sliding. The maximum static friction is

$F_{max} = \mu_s N$

Friction equals μ_s N only at impending motion; below that it takes whatever value equilibrium requires, up to the maximum. Once sliding, kinetic friction F = μ_k N applies, with μ_k < μ_s, so a block lurches forward when it breaks loose.

Tipping vs sliding: a block on a plane either slides (when the pushing force reaches μ_s N) or tips about its leading edge (when the resultant of weight and applied force falls outside the base). Compare the two thresholds — whichever occurs at the lower force governs. On an incline, a block is on the verge of sliding when tan θ = μ_s, the angle of repose: a 0.30 static coefficient gives θ = arctan(0.30) = 16.7°.

Worked truss example and friction applications

Consider a triangular truss with a horizontal bottom chord AB, a top joint C above the midpoint, supports at pin A and roller B, and a 10 kN downward load at C. By symmetry the reactions are R_A = R_B = 5 kN up. At joint A the two members are AC (the diagonal) and AB (the bottom chord). 07 kN. Horizontal equilibrium gives AB = AC·cos45° = 5 kN tension. The diagonals carry compression and the bottom chord carries tension — the classic load path the method of joints reveals two equations at a time.

Friction in machine elements

FE friction questions extend beyond blocks on planes. A wedge uses small angles to amplify force, analyzed by drawing a free body of the wedge and of the loaded block, including friction on every contact surface (two or three surfaces, each with its own μN opposing impending motion). A belt or rope over a fixed drum obeys the capstan (belt-friction) equation T₁/T₂ = e^(μβ), where β is the wrap angle in radians and T₁ > T₂ is the tight-side tension — a small wrap and modest μ multiply tension dramatically, which is why a few turns around a post hold a heavy load.

A rope with μ = 0.3 wrapped a half turn (β = π) holds T₁/T₂ = e^(0.3·π) = e^0.942 = 2.57 — one person resisting 257 N can hold a 100 N pull on the other side.