Design methods for steel structures
Steel has a yield strength fy (about 235 MPa for mild steel). Beyond it, the steel stretches for good. Design keeps loads far below it. There are two main methods.
Allowable stress method: the stress from the real loads must stay below fy ÷ safety factor (for example 235 ÷ 1.5 = 157 MPa).
Limit state method: multiply the loads by load factors (for example 1.35 for dead load and 1.5 for live load) and compare with the member strength, which is reduced by a resistance factor. We also check serviceability: the beam must not sag too much (deflection limit about span ÷ 300).
Always check three things: bending, shear and deflection. Long, slim beams may also twist sideways (lateral buckling), so they need side supports.
Design of H-beams
An H-beam (or I-beam) has two flanges joined by a web. The flanges are far from the middle line, so they give most of the bending strength.
The bending strength depends on the section modulus Z = I / (d/2), where I is the second moment of area and d is the depth. For a plain rectangle, Z = b d² / 6. Notice d is squared: double the depth gives four times the Z.
Design steps: (1) find M = wL²/8 and V = wL/2; (2) required Z = M ÷ allowable stress; (3) pick an H-beam with at least that Z from the steel tables; (4) check shear τ = V ÷ (d × tw); (5) check deflection δ = 5wL⁴ / (384 E I) is under the limit.
Design of plate girders
When the span or load is too big for rolled H-beams, engineers weld plates into a plate girder: two thick flange plates and a tall, thin web plate. They can make it as deep as needed.
The flanges take the moment. A quick estimate of the force in one flange is F ≈ M / h, where h is the distance between the flange centres. The flange area needed is F ÷ allowable stress. The web takes the shear.
A tall thin web can buckle (wrinkle) like a thin sheet of tin. To stop this, transverse stiffeners (vertical plates) are welded to the web, and bearing stiffeners are placed at the supports and heavy loads. Closer stiffeners allow a thinner web.
Try it
At home: take a sheet of paper and lay it across two books. It sags. Fold it into a letter "V" or an accordion shape and it holds a coin. Same paper, deeper shape.
In the 3D: set the load to 40 kN/m and find the smallest depth that keeps the bar green.
Key formulas and definitions
- Allowable stress = f_y ÷ safety factor (235 ÷ 1.5 ≈ 157 MPa)
- Factored load = 1.35 × dead + 1.5 × live (typical)
- M = wL²/8, V = wL/2 (uniform load, simple span)
- σ = M / Z; required Z = M ÷ allowable stress
- Rectangle: Z = b d² / 6
- Shear: τ = V / (d × t_w)
- Deflection: δ = 5 w L⁴ / (384 E I)
- Plate girder flange force F ≈ M / h
Worked examples
1. A simply supported steel beam, span 6 m, carries 20 kN/m. Find M and V.
M = wL²/8 = 20 × 36 / 8 = 90 kNm. V = wL/2 = 20 × 6 / 2 = 60 kN.
2. Allowable stress is 150 MPa. Find the required section modulus for M = 90 kNm.
Z = M / σ = 90 × 10⁶ / 150 = 600 × 10³ mm³ = 600 cm³. Choose an H-beam with Z of at least 600 cm³.
3. The beam has depth 300 mm and web thickness 8 mm. Find the shear stress for V = 60 kN.
Web area = 300 × 8 = 2400 mm². τ = 60 000 / 2400 = 25 MPa. This is low, so shear is fine.
4. Dead load 40 kN and live load 20 kN. Find the factored load (1.35 and 1.5).
1.35 × 40 + 1.5 × 20 = 54 + 30 = 84 kN.
5. A rectangular bar is 100 mm wide and 200 mm deep. Find Z. What if the depth is doubled?
Z = b d² / 6 = 100 × 200² / 6 = 666 667 mm³ ≈ 667 cm³. At d = 400: Z = 100 × 160 000 / 6 = 2 666 667 mm³, four times bigger.
6. A plate girder carries M = 400 kNm. The flange centres are 0.8 m apart. Find the flange force and area (allowable stress 150 MPa).
F = M / h = 400 / 0.8 = 500 kN. Area = 500 000 / 150 = 3333 mm² (for example 250 mm × 14 mm).
Common mistakes
- Using the whole depth for the flange force. The lever arm is the distance between flange centres.
- Choosing a section by weight alone. Check Z, shear and deflection too.
- Forgetting side supports. A deep, slim beam can twist sideways before it yields.
- Mixing kNm and N·mm. 1 kNm = 10⁶ N·mm.