Structural systems applied

Simplify it to members and joints, add the real loads, follow each load to the ground, name what each member does, and say where it would fail first.

Triangulation for a mechanism; depth for bending, since stiffness rises with the cube of depth; a shorter span for deflection, which falls with the fourth power of span.

Strength and stiffness are different properties, and a stronger grade usually has almost the same stiffness. Geometry changes deflection enormously; material hardly at all.

At a joint, a fixing, or the most slender compression member — rarely in the middle of a solid part.

σ = F ÷ A using the original area; ε = ΔL ÷ L using the original length; E = σ ÷ ε taken on the straight part of the graph.

It is a length divided by a length, so the units cancel. A 200 mm bar stretched 0.4 mm has a strain of 0.002.

From the straight part only, before the yield point. Past yield the line curves and a gradient taken there is not the modulus.

Steel about 200 GPa, aluminium about 70 GPa, timber 10 to 15 GPa, polymers 1 to 3 GPa.

Overloading, material choice, size and shape — usually more than one at once, and size and shape are the cheapest to fix.

Stress concentrates there, and the sharper the corner the higher the local stress. A fillet radius spreads it, which makes a radius a structural feature.

The highest stress in that model, not necessarily a failure. Compare the value on the scale with the material's yield strength before concluding anything.

The loads, constraints and material data were all assumed. Analyse the wrong load case and the plot is precise, colourful and useless.

The forces balance — up equals down — and the moments about any point balance. Both must hold.

Take moments about one support so its reaction drops out, solve for the other, then use up-equals-down to find the first.

Arrows pointing away from each other along a member mean tension; arrows pointing towards each other mean compression.

Buckling, not just crushing. A slender member goes unstable sideways well below its crushing strength, and effective length decides it.

SF = ultimate ÷ allowable. Rearranged: allowable = ultimate ÷ SF, and required ultimate = allowable × SF.

The heaviest plausible user, plus anything carried, plus the dynamic peak from jumping or swinging, plus realistic misuse — never the average user.

Required ultimate load = working load × SF. Required area = that load ÷ the material's ultimate stress. Then round up to a standard stocked size.

It costs material, mass, money and embodied impact, can make the product worse to use, and can hide sloppy analysis. Reducing uncertainty by testing is often better.