Building planning for services
Planning means deciding where each part goes before building. For services (water, drainage, air-conditioning, electricity) we plan:
- Shafts: vertical spaces from bottom to top for pipes, ducts and cables. Stacking toilets and kitchens above each other shortens the pipes.
- Plant rooms: rooms for pumps, tanks, lifts and air units, placed where noise and weight are easy to handle.
- Floor height: leave extra height under beams and above ceilings for ducts and pipes.
- Access: panels so workers can repair pipes later.
Structure of buildings
A frame building has these parts.
- Foundation: spreads the building weight over the soil.
- Columns: vertical members that carry load down.
- Beams: horizontal members that carry the floor to the columns.
- Slab: the floor plate.
- Walls: may carry load (wall structures) or only divide space (frame structures).
Common materials: reinforced concrete (concrete is strong in squeezing, steel bars take the pulling), steel and timber.
Structural mechanics: loads and forces
Loads are the forces on a building.
- Dead load: the building's own weight.
- Live load: people, furniture, goods.
- Wind and earthquake load: sideways forces.
For a simple beam of span L with an even load w (kN per metre): total load = w × L, each support carries half of it (w × L ÷ 2), and the biggest bending moment is at the middle, M = w × L² ÷ 8. A column carries the load of the floor area it serves, summed over all floors. Pressure on soil = load ÷ footing area.
Bending of beams and holes for services
A beam under load bends. For the same load and span, the sag is about proportional to 1 ÷ depth³. If the depth doubles, the sag becomes one eighth. So deep beams are very stiff.
Pipes often need to cross a beam. Rules of thumb: make the hole round, small, at mid-depth, and in the middle third of the span. Do not cut holes near the supports (shear is highest there) or near the top or bottom edge. Plan such holes with the structural engineer before concreting.
Key formulas and definitions
- Total load on beam = w × L (w in kN/m, L in m)
- Support reaction (even load) = w × L ÷ 2
- Maximum bending moment = w × L² ÷ 8
- Sag ∝ load ÷ depth³ (same span and material)
- Soil pressure = load ÷ footing area
Worked examples
1. A beam of span 6 m carries an even load of 10 kN/m. Find the total load and each support reaction.
Total = 10 × 6 = 60 kN. Each support = 60 ÷ 2 = 30 kN.
2. For the same beam find the maximum bending moment.
M = w × L² ÷ 8 = 10 × 36 ÷ 8 = 45 kN·m.
3. A beam is made twice as deep. By what factor does its sag change?
Sag ∝ 1 ÷ depth³, so the factor is 1 ÷ 2³ = 1/8. It sags eight times less.
4. A column serves 20 m² of floor on each of 3 floors. Floor load is 5 kN/m². Find the column load.
Per floor = 20 × 5 = 100 kN. Three floors = 300 kN.
5. That column rests on a square footing of 1.5 m². Find the soil pressure.
Pressure = 300 ÷ 1.5 = 200 kN/m² (200 kPa).
6. A 6 m beam needs a pipe hole. Which middle-third range of the span can be used?
Middle third = from 2 m to 4 m from one support. Put the round hole at mid-depth within this range.
Common mistakes
- Making beams wide and flat. Depth, not width, gives stiffness (depth is cubed).
- Drilling holes near supports or at the bottom of a beam. Cracks start there.
- Planning services after the structure is cast. Plan shafts and sleeves early.
- Forgetting the weight of the building itself (dead load) when adding loads.