Basics of design
A structure must carry loads: its own weight (dead load), people, grain, snow, wind (live load). The load travels through beams, columns and foundations to the ground.
- Find the loads.
- Choose the material and shape.
- Find the capacity (how much each part can carry).
- Check safety factor = capacity ÷ load. Keep it above 1; designers often use 1.5 or more because loads and materials are not perfectly known.
Too small a safety factor is unsafe. Too large wastes money.
Beams
A beam lies across a gap and rests on supports. A load makes it bend: the top is squeezed and the bottom stretched. For a point load P in the middle of a span L the biggest bending moment is M = P × L ÷ 4, in the middle. At the supports it is zero.
Each support carries half the load: P ÷ 2.
Stiffness grows with the cube of the depth: double the depth, and the sag is 8 times less. So tall thin sections (like a plank on its edge) beat flat wide ones.
Trusses
A truss is a frame of straight bars joined into triangles. A triangle cannot change shape without changing the length of a side, so it is rigid.
With a load in the middle: the top chord is in compression (squeezed), the bottom chord is in tension (stretched), and the diagonals share both. Every bar only pushes or pulls, never bends, so it can be thin. This makes trusses light for long roofs and bridges.
Rigid frames
A rigid frame (portal frame) joins columns and a beam with stiff corner joints that cannot rotate. The corners pass bending between column and beam, so the frame keeps its shape under side push such as wind, and the beam bends less than it would on simple supports.
With pinned (hinged) corners, the same frame would fold sideways unless a diagonal brace is added. Poly-houses, sheds and grain stores are often portal frames.
Try it
In the 3D: in the last step choose the beam, load 8 kN and size 1. Is it safe? Raise the size until the safety factor passes 1.5. Then do the same with the truss and compare the sizes.
At home: rest a ruler on two books and press the middle with a coin stack. Turn the ruler on its edge and try again. Which sags less?
Key formulas and definitions
- Safety factor = capacity ÷ load
- Beam, point load P at the middle of span L: M = P × L ÷ 4; each support carries P ÷ 2
- Sag of a beam ∝ P × L³ ÷ (E × b × h³): double the depth h → 1/8 of the sag
- Truss: top chord compression, bottom chord tension (load in the middle)
- Key terms: load, capacity, bending moment, compression, tension, pinned, rigid
Worked examples
1. A post can carry 15 kN and the load is 5 kN. Find the safety factor.
15 ÷ 5 = 3.
2. A simply supported beam has a span of 4 m and a 10 kN load at the middle. Find the maximum bending moment and the support reactions.
M = P × L ÷ 4 = 10 × 4 ÷ 4 = 10 kN·m. Each support carries 10 ÷ 2 = 5 kN.
3. A beam sags 16 mm. If its depth is doubled (same width, load, span, material), what is the new sag?
Sag goes with 1/h³, so it becomes 16 ÷ 8 = 2 mm.
Common mistakes
- Thinking a safety factor of 1 is fine. It means no reserve at all.
- Putting a plank flat instead of on edge. Depth matters much more than width.
- Mixing up compression and tension: for a beam under load the top is squeezed and the bottom is stretched.
- Thinking a rigid frame has no bending. Its corners carry bending.