Steel · Connections: bolts & welds
Bracing gusset (Whitmore, Thornton, block shear) calculator
Whitmore effective width, gusset tension yield and fracture, compression buckling over the Thornton length, and block shear around two bolt lines.
Standard: AS 4100 Cl 7, 6, 9.1.9 · Whitmore section · Thornton buckling
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 |
|---|---|---|
| Brace force (tension or compression) | 400 | kN |
| Bolt rows along brace | 3 | |
| Pitch | 70 | mm |
| Gauge between bolt lines (2 lines) | 60 | mm |
| End distance | 40 | mm |
| Bolt size | M20 | |
| Bolt category | 8.8/S | |
| Gusset thickness | 12 | mm |
| Gusset grade | Grade 250 plate | |
| Thornton length (Whitmore to support) | 180 | mm |
| Effective length factor | 0.65 |
Worked example — results
| Result | Value | Unit |
|---|---|---|
| Bolt diameter df | 20.00 | mm |
| Bolt tensile strength fuf | 830.0 | MPa |
| Tensile stress area As | 245.0 | mm² |
| Core area Ac (threads) | 225.0 | mm² |
| Shank area Ao | 314.0 | mm² |
| Plate yield stress fy (AS/NZS 3678) | 260.0 | MPa |
| Plate tensile strength fu | 410.0 | MPa |
| Joint length (n − 1)p | 140.0 | mm |
| Whitmore width g + 2 Lj tan30° | 221.7 | mm |
| Hole | 22.00 | mm |
| Whitmore gross area | 2,660 | mm² |
| Whitmore net area | 2,132 | mm² |
| φNt = 0.9 min(Ag fy, 0.85 An fu) | 622.4 | kN |
| r = t/√12 | 3.46 | mm |
| λn = (le/r)√kf √(fy/250) | 34.44 | |
| αa | 16.23 | |
| λ = λn + αa·αb | 42.56 | |
| η | 0.095 | |
| ξ | 2.95 | |
| Slenderness reduction αc | 0.894 | |
| φNc = 0.9 αc Ag fy | 556.5 | kN |
| Block shear gross shear area 2(Lj + e)t | 4,320 | mm² |
| Net shear area | 3,000 | mm² |
| Net tension area (g − dh)t | 456.0 | mm² |
| φRbs = 0.75 (min(0.6fu Anv, 0.6fy Agv) + fu Ant) | 645.7 | kN |
| Bolts (2nr, single shear, bearing) | 555.8 | kN |
- ✓ Whitmore tension
- ✓ Gusset buckling (compression)
- ✓ Block shear
- ✓ Bolts
Method, formulas and assumptions
- Bolt diameter df = CHOOSE(3,12,16,20,24,30,36) = 20 mm
- Bolt tensile strength fuf = IF(2=1,400,830) = 830 MPa
- Tensile stress area As = CHOOSE(3,84.3,157,245,353,561,817) = 245 mm²
- Core area Ac (threads) = CHOOSE(3,76.2,144,225,324,519,759) = 225 mm²
- Shank area Ao = CHOOSE(3,113,201,314,452,707,1018) = 314 mm²
- Plate yield stress fy (AS/NZS 3678) = CHOOSE(1,IF(12<=8,280,IF(12<=12,260,250)),IF(12<=8,320,IF(12<=12,310,IF(12<=20,300,280))),IF(12<=12,360,IF(12<=20,350,340))) = 260 MPa
- Plate tensile strength fu = CHOOSE(1,410,430,450) = 410 MPa
- Joint length (n − 1)p = (3-1)*70 = 140 mm
- Whitmore width g + 2 Lj tan30° = 60+2*140*TAN(RADIANS(30)) = 221.66 mm
- Hole = 20+2 = 22 mm
- Whitmore gross area = 221.66*12 = 2659.9 mm²
- Whitmore net area = (221.66-2*22)*12 = 2131.9 mm²
- φNt = 0.9 min(Ag fy, 0.85 An fu) = 0.9*MIN(2659.9*260,0.85*2131.9*410)/1000 = 622.42 kN
- r = t/√12 = 12/SQRT(12) = 3.4641 mm
- λn = (le/r)√kf √(fy/250) = 0.65*180/3.4641*SQRT(1)*SQRT(260/250) = 34.444
- αa = 2100*(34.444-13.5)/(34.444^2-15.3*34.444+2050) = 16.233
- λ = λn + αa·αb = 34.444+16.233*0.5 = 42.56
- η = MAX(0,0.00326*(42.56-13.5)) = 0.094737
- ξ = ((42.56/90)^2+1+0.094737)/(2*(42.56/90)^2) = 2.9477
- Slenderness reduction αc = MIN(1,2.9477*(1-SQRT(1-(90/(2.9477*42.56))^2))) = 0.89412
- φNc = 0.9 αc Ag fy = 0.9*0.89412*2659.9*260/1000 = 556.52 kN
- Block shear gross shear area 2(Lj + e)t = 2*(140+40)*12 = 4320 mm²
- Net shear area = 2*(140+40-(3-0.5)*22)*12 = 3000 mm²
- Net tension area (g − dh)t = (60-22)*12 = 456 mm²
- φRbs = 0.75 (min(0.6fu Anv, 0.6fy Agv) + fu Ant) = 0.75*(MIN(0.6*410*3000,0.6*260*4320)+410*456)/1000 = 645.66 kN
- Bolts (2nr, single shear, bearing) = 2*3*MIN(0.8*0.62*830*225/1000,0.9*3.2*20*12*410/1000) = 555.77 kN
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For preliminary design and education. Check every result against the current standard and your own engineering judgement before use.