Steel · Members & beams
Welded plate girder (flexure & shear) calculator
Section properties from plates, effective section modulus with flange/web slenderness, web shear buckling with optional intermediate stiffeners.
Standard: AS 4100 Cl 5.2, 5.10, 5.11
Inputs in the worked example
The live calculator starts with these values; change any of them and every result, check and diagram updates instantly.
| Input | Value | Unit |
|---|---|---|
| Overall depth | 1200 | mm |
| Flange width | 400 | mm |
| Flange thickness | 25 | mm |
| Web thickness | 10 | mm |
| Plate fy | 340 | MPa |
| Stiffener spacing (0 = unstiffened) | 1500 | mm |
| Design moment | 2800 | kNm |
| Design shear | 850 | kN |
Worked example — results
| Result | Value | Unit |
|---|---|---|
| d1 | 1,150 | mm |
| Ix | 8,171,562,500 | mm⁴ |
| Z | 13,619,271 | mm³ |
| S | 15,056,250 | mm³ |
| Flange λe (welded, λep 8, λey 14) | 9.10 | |
| Web λe (λep 82, λey 115) | 134.1 | |
| min(S, 1.5Z) | 15,056,250 | mm³ |
| Ze | 10,014,170 | mm³ |
| φMs | 3,064 | kNm |
| αv (unstiffened) = (82/λw)² | 0.374 | |
| αv stiffened: αv(0.75 + (d1/s)²) if s ≤ d1, else αv(1 + 0.75(d1/s)²) (buckling coefficient ratio; no tension field) | 0.539 | |
| φVu = 0.9·0.6 fy Aw αv | 1,137 | kN |
- ✓ Bending
- ✓ Shear
- ✓ Web d1/tw ≤ 200 unstiffened / 270 stiffened
Method, formulas and assumptions
- d1 = 1200-2*25 = 1150 mm
- Ix = (400*1200^3-(400-10)*1150^3)/12 = 8.172e+9 mm⁴
- Z = 2*8.172e+9/1200 = 1.362e+7 mm³
- S = 400*25*(1200-25)+10*1150^2/4 = 1.506e+7 mm³
- Flange λe (welded, λep 8, λey 14) = (400-10)/2/25*SQRT(340/250) = 9.0963
- Web λe (λep 82, λey 115) = 1150/10*SQRT(340/250) = 134.11
- min(S, 1.5Z) = MIN(1.506e+7,1.5*1.362e+7) = 1.506e+7 mm³
- Ze = IF(AND(9.0963<=8,134.11<=82),1.506e+7,IF(AND(9.0963<=14,134.11<=115),1.362e+7+MIN((14-9.0963)/6,(115-134.11)/33,1)*(1.506e+7-1.362e+7),1.362e+7*MIN(14/9.0963,(115/134.11)^2,1))) = 1.001e+7 mm³
- φMs = 0.9*340*1.001e+7/1E6 = 3064.3 kNm
- αv (unstiffened) = (82/λw)² = MIN(1,(82/134.11)^2) = 0.37385
- αv stiffened: αv(0.75 + (d1/s)²) if s ≤ d1, else αv(1 + 0.75(d1/s)²) (buckling coefficient ratio; no tension field) = IF(1500<=0,0.37385,MIN(1,0.37385*IF(1500/1150<=1,0.75+(1150/1500)^2,1+0.75*(1150/1500)^2))) = 0.53865
- φVu = 0.9·0.6 fy Aw αv = 0.9*0.6*340*1150*10*0.53865/1000 = 1137.3 kN
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For preliminary design and education. Check every result against the current standard and your own engineering judgement before use.