**1.0 Introduction**

The fin plate connection is very popular in the construction industry due to its ease of erection, and the absence of shared bolts in two sided connections. It consists of a length of plate welded to the supporting member (eg column or primary beam), and the supported member bolted to the fin plate. Technically, fin plates derive their rotational capacity from shear deformation of the bolts, and hole distortion in the fin plate and/or the beam web.

**2.0 Practical Considerations and Recommended Geometry**

With 10mm thick fin plate in S275 steel, 8mm fillet weld to the supporting member will guard against any possibility of weld failure (CSI, 2011). Fin plates may be classified as short or long as follows;

Short t_{p}/z_{p} ≥ 0.15

Long t_{p}/z_{p} < 0.15

Where z_{p} is the distance between the face of the support and the first line of bolts. For short fin plates, erection on site is usually more difficult, but with long fin plates, care must be taken against lateral torsional buckling, especially if the beam is laterally unrestrained.

According to SCI (2011), when detailing the joint, the following recommendations should be followed;

(1) Full strength fillet welds are provided

(2) The fin plate is positioned close to the top flange in order to provide positional restraint

(3) The depth of the fin plate is at least 0.6 times the supported beam depth in order to provide the beam with adequate torsional restraint

(4) The thickness of the fin plate or the beam web is ≤ 0.50d (for S275 steel)

(5) Property class 8.8 bolts non-preloaded are used

(6) All end and edge distances on the plate and the beam web are at least 2d

**3.0 Solved Example**

**Initial Considerations**

Since the depth of beam is less than 610mm, let us adopt fin plate size of 100 x 10mm

Gap g_{h} = 10mm

Bolt rows = 3

Length of plate = 0.6 × 303.4 = 182.04mm (provide plate length = 220mm)

**(i) Supported beam Bolt Shear Check**

Basic requirement V_{Ed} ≤ V_{Rd}

_{
}V_{Rd} = nF_{v,Rd}/sqrt[(1 + αn)^{2} + (βn)^{2}]

F_{v,Rd} = (α_{v}f_{ub}A)/γ_{m2}

For M20 8.8 bolts;

F_{v,Rd} = (0.6 × 800 × 245)/1.25 = 94080 N = 94.08 kN

For a single vertical line of bolt (n_{2} = 1, and n = n_{1}); α = 0

β = 6z/[n_{1}(n_{1} + 1)p_{1}]

β = (6 × 30)/[3 × (3 + 1) × 70] = 0.214

Therefore;

V_{Rd} = (3 × 94.08)/sqrt[(1 + 0 × 3)^{2} + (0.214 × 3)^{2})] = 237.506 kN

V_{Ed} = 100.25 kN < 237.506 kN

Therefore bolt group is ok for shear

**(ii) Fin plate in bearing**

Basic requirement V_{Ed} ≤ V_{Rd}

_{
}V_{Rd} = n/sqrt{[(1 + αn)/(F_{b,ver,Rd} )]^{2} + [βn/(F_{b,hor,Rd}]^{2}}

α = 0; β = 0.214

*The vertical bearing resistance of a single bolt*

F_{b,ver,Rd} = (k_{1}α_{b}f_{up}dt_{p})/γ_{m2}

k_{1 }= min[2.8(e_{2}/d_{0}) – 1.7; 2.5) = min[2.8(50/22) – 1.7; 2.5)

= min[4.66;2.5] = 2.5

α_{b }= min[e_{1}/3d_{0}; (p_{1}/3d_{0} – 1/4); f_{ub}/f_{up}; 1.0]

= min[40/(3 × 22); (70/(3 × 22) – 0.25); 800/410; 1.0]

= min[0.61; 0.81; 1.95; 1.0] = 0.61

F_{b,ver,Rd} = (2.5 × 0.61 × 410 × 20 ×10)/1.25 = 100040 N = 100.04 kN

*The horizontal bearing resistance of a single bolt*

F_{b,hor,Rd} = (k_{1}α_{b}f_{up}dt_{p})/γ_{m2}

k_{1 }= min[2.8(e_{1}/d_{0}) – 1.7; 1.4(p_{1}/d_{0}) – 1.7; 2.5)

= min[2.8(40/22) – 1.7; 1.4(70/22) – 1.7;2.5)

= min[3.39; 2.75; 2.5] = 2.5

α_{b }= min[e_{2}/3d_{0}; f_{ub}/f_{up}; 1.0]

= min[50/(3 × 22); 800/410; 1.0]

= min[0.76; 1.95; 1.0] = 0.76

F_{b,hor,Rd} = (2.5 × 0.76 × 410 × 20 ×10)/1.25 = 124640 N = 124.64 kN

V_{Rd} = (3)/sqrt{[(1 + 0 × 3)/100]^{2} + [(0.214 × 3)/124.64]^{2}) = 288.218 kN

V_{Ed} = 100.25 kN < 288.218 kN Ok

**(iii) Supported Beam – Fin Plate**

**Shear**

Basic requirement V_{Ed} ≤ V_{Rd,min}

V_{Rd,min} = min(V_{Rd,g};V_{Rd,n};V_{Rd,n})

*Gross section*

V_{Rd,g} = (h_{p}.t_{p}.f_{fy,p})/[1.27 × sqrt(3) × γ_{m0}]

V_{Rd,g} = [(220 × 10 × 275)/(1.27 × √3 × 1.0)] × 10^{-3} = 275.034 kN

*Net Section*

V_{Rd,n} = A_{v,net }× [f_{u,p} / (sqrt(3) × γ_{m2})]

Net Area = A_{v,net }= t_{p}.(h_{p }– n_{1}.d_{0})

A_{v,net }= 10(220 – 3 × 22) = 1540 mm^{2}

V_{Rd,n} = 1540 _{ }× [410 / (√3 × 1.1)] × 10^{-3} = 331.399 kN

*Block Tearing*

V_{Rd,b} = [(0.5f_{u,p }A_{nt})γ_{m2}] + [f_{y,p} A_{nv }/ (sqrt(3) × γ_{m0})]

Net area subject to tension = A_{nt }= t_{p}[e_{2} – 0.5d_{0}]

A_{nt }= 10[e_{2} – 0.5d_{0}]

A_{nt }= 10[50 – 0.5 × 22] = 390 mm^{2}

Net area subject to shear = A_{nv }= t_{p}[h_{p }– e_{1} – (n_{1 }– 0.5)d_{0}]

A_{nv }= 10[20 – 40 – (3_{ }– 0.5) × 22] = 1250 mm^{2}

V_{Rd,b} = [(0.5 × 410 × 390)1.1] + [(275 × 1250)/ (sqrt(3) × 1.0)] = 72681.818 + 198464.155 = 271145.973 = 271.145 kN

V_{Rd,min} = min(275.034; 331.399; 271.145 ) = 271.145 kN

V_{Rd,min} = 100.25 kN < 271.145 kN Ok

*Lateral Torsional Buckling of fin plates*

Basic requirement V_{Ed} ≤ V_{Rd}

t_{p}/z_{p} = 220/50 = 4.4 > 0.15

Therefore fin plate is short;

V_{Rd} = [(W_{el,p}/z) × (f_{y,p} / γ_{m0})]

W_{el,p }= (t_{p}h_{p}^{2})/6 = (10 × 220^{2})/6 = 73333.333 mm^{3}

V_{Rd} = [(73333.333/50) × (275 / 1.0)] = 403333.333 N = 403.333 kN

V_{Ed} = 100.25 kN < 403.33 kN Ok

**Supported Beam in Shear**

*Gross Section*

V_{Rd,g} = [A_{v,web }× (f_{y,b1} _{ }/ (sqrt(3) × γ_{m0}))]

Gross Area A_{v }= A_{Tee }– bt_{f,b1} + (t_{w,b1} + 2r_{b1}) × 0.5t_{f,b1}

A_{Tee }= (265.2 – 10.2) × 6 + (165 × 10.2) = 1530 + 1683 = 3213 mm^{2}

A_{v }= 3213 – (165 × 10.2) + (6 + 2 × 8.9) × 0.5 × 10.5 = 3213 – 1683 + 124.95 = 1654.95 mm^{2}

V_{Rd,g} = 1654.95 × 275 / (sqrt(3) × 1.0) = 262758.603 N = 262.758 kN

*Net Section*

V_{Rd,n} = [A_{v,net }× (f_{u,b1} _{ }/ (sqrt(3) × γ_{m2}))]

Net area; A_{v,net }= A_{v }– n_{1}d_{0}t_{w,b1} = 1654.95 – (3 × 22 × 6) = 1258.95 mm^{2}

V_{Rd,n} = 1258.95 × 410 / (sqrt(3) × 1.1) = 270918.727 N = 270.918 kN

*Block Tearing*

V_{Rd,b} = [(0.5f_{u,b1 }A_{nt})γ_{m2}] + [f_{y,b1} A_{nv }/ (sqrt(3) × γ_{m0})]

Net area subject to tension = A_{nt }= t_{w,b1}[e_{2,b} – 0.5d_{0}]

A_{nt }= 6[40 – 0.5 × 22] = 174 mm^{2}

Net area subject to shear = A_{nv }= t_{w,b1}[e_{1,b }+ (n_{1 }– 1)p_{1 }– (n_{1 }– 0.5)d_{0}]

A_{nv }= 6[40 + (3 – 1)70 – (3_{ }– 0.5)22] = 750 mm^{2}

V_{Rd,b} = [(0.5 × 410 × 175)1.1] + [(275 × 750)/ (sqrt(3) × 1.0)] = 35873.9 + 119081.986 = 154955.886 = 154.955 kN

V_{Rd,min} = min(262.758; 270.918; 154.955) = 154.955 kN

V_{Rd,min} = 100.25 kN < 154.955 kN Ok

*Check for welding (supporting beam)*

For a beam in S275 steel;

Basic requirement; a ≥ 0.5t_{p}

0.5t_{p} = 0.5 × 10 = 5mm

a = 5.7mm > 0.5t_{p}

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