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Design of Various Structures: Timber, Reinforced Concrete and Steel

Timber is light, renewable and strong along the grain but weak across it and sensitive to moisture and rot. Concrete is strong in compression and weak in tension, so steel bars (rebar) are placed where the beam is stretched: reinforced concrete (RC). Steel is strong in both push and pull, ductile and light for its strength; slender steel members can buckle and steel loses strength in fire. In all three the load follows one path: slab, beam, column, foundation, ground.

🎬 Step-by-step story

  1. Timber: light and easy to work. It is strong along the grain, like a bundle of straws. Add a load and it sags a little but holds.
  2. Plain concrete beam. Concrete likes being squeezed but hates being stretched. Watch the bottom: cracks open and the beam fails.
  3. Reinforced concrete: steel bars are placed at the bottom, where the beam is pulled. Now only hairline cracks show and the beam holds the same load.
  4. Steel I-beam: the top and bottom flanges carry push and pull, the thin web keeps them apart. Strong, light and good for long spans.
  5. Every structure sends its load down a path: beam, column, foundation, ground. Any weak link in this chain is the failure point.
  6. Free play: choose a material, raise the load and switch the bars on and off. Which one holds? Which one fails first?

Tip: drag the 3D scene to turn it. Use two fingers to zoom.

🤔 Common doubts, cleared

Why does timber hold well along the grain?

Timber is made of long fibres, like a bundle of straws, so pulling or pushing along them is strongest. Across them it splits easily.

Why does the concrete beam crack at the bottom?

When it sags, the bottom is stretched. Concrete is about ten times weaker in tension than in compression, so cracks open there.

Why are there still cracks in the RC beam?

Tiny hairline cracks are normal: the steel is stretching a little and sharing the load. They stay very thin and harmless.

Why is the middle of the I-beam so thin?

Near the middle the bending stress is almost zero, so a thin web is enough to keep the flanges apart and carry shear. Material is saved.

What happens if one part of the load path is weak?

The chain breaks at its weakest link: a good beam on a weak column or a sinking foundation still fails.

Timber structures

Timber is renewable and light (about 6 kN/m³). Strong along the grain, weak across it; strength drops with knots, cracks and damp. Used for roof trusses, floor joists, posts and beams, and modern engineered wood (glulam, CLT) for larger buildings.

Reinforced concrete structures

Concrete (cement, sand, stone, water) is strong in compression (typically 20–40 MPa) but only about one tenth as strong in tension. Reinforced concrete puts steel bars (yield about 500 MPa) where tension acts: at the bottom of a sagging beam, at the top over supports.

Fine hairline cracks at the bottom of an RC beam are normal: they show that the steel is working.

Steel structures

Steel has high strength (yield 250–355 MPa), is ductile and stiff (E = 200 GPa), and is equally good in tension and compression. Rolled shapes: I, H, channel, angle, pipe. The I-shape puts material far from the neutral axis where it is most useful.

Choosing a material

Weight per cubic metre: timber about 6 kN, concrete 25 kN, steel 78.5 kN. Steel is heavy per volume but needs very little volume, so a steel frame is light overall. Pick by span, load, fire rules, cost, speed, local supply and look. Whatever the material, check the load path: slab, beam, column, foundation, soil.

Key formulas and definitions

Worked examples

1. The tension at the bottom of an RC beam is 60 kN. Allowable steel stress is 150 MPa. How many 12 mm bars (113 mm² each)?

A_s = 60 000 / 150 = 400 mm². 400 / 113 = 3.5, so use 4 bars (452 mm²).

2. A timber joist must carry M = 2 kN·m with allowable stress 10 MPa. The joist is 50 mm wide. Find the depth.

M = 2 × 10⁶ N·mm. Z = 2 × 10⁶ / 10 = 0.2 × 10⁶ mm³. Z = bh²/6, so h² = 6 × 0.2 × 10⁶ / 50 = 24 000 and h = 155 mm. Use 50 × 160 mm.

3. A steel column, pinned both ends, L = 3 m, I = 2 × 10⁶ mm⁴, E = 200 000 N/mm². Find the Euler load.

P_cr = π² E I / L² = 9.87 × 200 000 × 2 × 10⁶ / 3000² = 4.39 × 10⁵ N = 439 kN.

4. Compare the weights of 0.1 m³ of timber, concrete and steel.

Timber 6 × 0.1 = 0.6 kN; concrete 25 × 0.1 = 2.5 kN; steel 78.5 × 0.1 = 7.85 kN.

5. A 100 × 100 mm timber post is exposed to fire on all four sides for 30 minutes (char rate 0.65 mm/min). What section is left?

Char depth = 0.65 × 30 = 19.5 mm per side. Left: 100 − 2 × 19.5 = 61 mm. So 61 × 61 mm.

6. A 20 mm steel bar (314 mm²) has yield stress 250 MPa. Find the yield load and the safe load with a safety factor of 1.5.

Yield load = 250 × 314 = 78 500 N = 78.5 kN. Safe load = 78.5 / 1.5 = 52.3 kN.

Common mistakes

Practice quiz

1. In a simply supported RC beam the main bars go:
2. Concrete is weak in:
3. Timber is strongest:
4. The Euler load of a column falls when its length:
5. Which material is the lightest per cubic metre?

Practice: answer these yourself

Type or choose your answer, then press Check. Use a hint if you are stuck; the full solution appears after you answer.

Frequently asked questions

Why is steel added to concrete?

Concrete cracks in tension. Steel bars are strong in tension, so together they cover each other's weakness. Concrete also protects the steel from rust and fire.

Is timber a good structural material?

Yes, if kept dry and protected. It is light, renewable and strong along the grain. Engineered timber is now used even in tall buildings.

Why are steel beams I-shaped?

Bending stress is highest at the top and bottom edges. The I-shape puts most material there, and a thin web joins them, giving high strength for little weight.

Where this is taught

Japan高校(専門学科)1〜3年Structural Design of Buildings

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