Steel · Connections: bolts & welds
Pin connection calculator
Pin in double shear, pin bearing, pin bending between plates and the lug net tension across the pin hole.
Standard: AS 4100 Cl 9.4 (pins) · Cl 7 lug net section
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 |
|---|---|---|
| Design force P* | 300 | kN |
| Pin diameter | 50 | mm |
| Pin fy | 350 | MPa |
| Central lug thickness | 25 | mm |
| Each clevis plate thickness | 16 | mm |
| Clearance each side | 2 | mm |
| Lug width | 140 | mm |
| Lug fu | 430 | MPa |
Worked example — results
| Result | Value | Unit |
|---|---|---|
| Pin area | 1,963 | mm² |
| φVf = 0.8·0.62 fyp·2·Ap | 681.7 | kN |
| φVb = 0.9·1.4 fyp dp tm | 551.3 | kN |
| Pin moment (P/2)(tc/2 + gap + tm/4) | 2.44 | kNm |
| φMp = 0.9 fyp dp³/6 | 6.56 | kNm |
| Lug net tension 0.9·0.85 fu (bl − dp − 1) tm | 731.9 | kN |
- ✓ Pin shear
- ✓ Pin bearing
- ✓ Pin bending
- ✓ Lug net section
Method, formulas and assumptions
- Pin area = PI()*50^2/4 = 1963.5 mm²
- φVf = 0.8·0.62 fyp·2·Ap = 0.8*0.62*350*2*1963.5/1000 = 681.73 kN
- φVb = 0.9·1.4 fyp dp tm = 0.9*1.4*350*50*25/1000 = 551.25 kN
- Pin moment (P/2)(tc/2 + gap + tm/4) = 300/2*(16/2+2+25/4)/1000 = 2.4375 kNm
- φMp = 0.9 fyp dp³/6 = 0.9*350*50^3/6/1E6 = 6.5625 kNm
- Lug net tension 0.9·0.85 fu (bl − dp − 1) tm = 0.9*0.85*430*(140-50-1)*25/1000 = 731.91 kN
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For preliminary design and education. Check every result against the current standard and your own engineering judgement before use.