Steel · Connections: bolts & welds
Lifting lug (plate with pin hole) calculator
Plate lifting lug loaded through a pin at a sling angle in its plane: net tension, tear-out above the hole, pin bearing and the fillet weld group at the base under combined direct, shear and moment.
Standard: AS 4100 tension (7.2), block shear (9.1.9), pin bearing (9.6), fillet welds (9.7.3) · design load γ·DLF·SWL (verify against AS 4991 / project rules)
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 |
|---|---|---|
| Safe working load per lug | 50 | kN |
| Dynamic load factor | 1.25 | |
| Load factor | 1.5 | |
| Sling angle from vertical (in plane) | 30 | ° |
| Plate thickness | 20 | mm |
| Plate width at base | 160 | mm |
| Plate width across the hole | 120 | mm |
| Hole diameter | 40 | mm |
| Pin diameter | 37 | mm |
| Hole centre to top edge | 70 | mm |
| Hole centre height above base | 110 | mm |
| Plate fy | 280 | MPa |
| Plate fu | 430 | MPa |
| Fillet weld leg (both faces) | 8 | mm |
| Weld metal fuw (E48XX) | 480 | MPa |
Worked example — results
| Result | Value | Unit |
|---|---|---|
| Design load N* = γ·DLF·SWL | 93.75 | kN |
| Vertical component | 81.19 | kN |
| In-plane horizontal component | 46.87 | kN |
| Tension across hole φmin(Ag·fy, 0.85An·fu) | 526.3 | kN |
| Tear-out φ0.6fu·2(a − dh/2)t (φ 0.75) | 387.0 | kN |
| Pin bearing φ1.4fy·dp·t | 261.1 | kN |
| Weld throat w/√2 | 5.66 | mm |
| Weld capacity φ0.6fuw·tt (φ 0.8) | 1.30 | kN/mm |
| Weld vertical force/mm | 0.858 | kN/mm |
| Weld horizontal force/mm | 0.146 | kN/mm |
| Weld resultant force | 0.870 | kN/mm |
| Maximum utilisation | 0.668 |
- ✓ N* ≤ tension capacity
- ✓ N* ≤ tear-out capacity
- ✓ N* ≤ pin bearing
- ✓ Weld stress ≤ capacity
- ✓ Hole clearance ≤ 10 % of pin diameter
Method, formulas and assumptions
- Design load N* = γ·DLF·SWL = 1.5*1.25*50 = 93.75 kN
- Vertical component = 93.75*COS(RADIANS(30)) = 81.19 kN
- In-plane horizontal component = 93.75*SIN(RADIANS(30)) = 46.875 kN
- Tension across hole φmin(Ag·fy, 0.85An·fu) = 0.9*MIN(120*20*280,0.85*(120-40)*20*430)/1000 = 526.32 kN
- Tear-out φ0.6fu·2(a − dh/2)t (φ 0.75) = 0.75*0.6*430*2*(70-40/2)*20/1000 = 387 kN
- Pin bearing φ1.4fy·dp·t = 0.9*1.4*280*37*20/1000 = 261.07 kN
- Weld throat w/√2 = 8/SQRT(2) = 5.6569 mm
- Weld capacity φ0.6fuw·tt (φ 0.8) = 0.8*0.6*480*5.6569/1000 = 1.3033 kN/mm
- Weld vertical force/mm = 81.19/(2*160)+6*46.875*110/(2*160^2) = 0.85797 kN/mm
- Weld horizontal force/mm = 46.875/(2*160) = 0.14648 kN/mm
- Weld resultant force = SQRT(0.85797^2+0.14648^2) = 0.87038 kN/mm
- Maximum utilisation = MAX(93.75/526.32,93.75/387,93.75/261.07,0.87038/1.3033) = 0.66781
- Check the shackle WLL, out-of-plane loading (avoid or design for it), and the parent member under the lug separately.
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For preliminary design and education. Check every result against the current standard and your own engineering judgement before use.