Structville...

A hub for civil engineering related designs, analysis, discussion, information, and knowledge.....

Trending Posts:

Thursday, March 29, 2018

Structural Design of Steel Fin Plate Connection


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 tp/zp ≥ 0.15
Long tp/zp < 0.15

Where zp 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
A secondary beam UKB 305 x 165 x 40  is supported on a primary beam of  UKB 533 x 210 x 101 with an ultimate shear force of 100.25 kN. We are to design a fin plate connection for this joint using grade 8.8 non preloaded bolts.

Initial Considerations
Since the depth of beam is less than 610mm, let us adopt fin plate size of 100 x 10mm
Gap gh = 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 VEd ≤ VRd

VRd = nFv,Rd/sqrt[(1 + αn)2 + (βn)2]

Fv,Rd = (αvfubA)/γm2

For M20 8.8 bolts;
Fv,Rd = (0.6 × 800 × 245)/1.25 = 94080 N = 94.08 kN

For a single vertical line of bolt (n2 = 1, and n = n1); α = 0

β = 6z/[n1(n1 + 1)p1]
β = (6 × 30)/[3 × (3 + 1) × 70] = 0.214

Therefore;
VRd = (3 × 94.08)/sqrt[(1 + 0 × 3)2 + (0.214 × 3)2)] = 237.506 kN

VEd = 100.25 kN < 237.506 kN
Therefore bolt group is ok for shear




(ii) Fin plate in bearing
Basic requirement VEd ≤ VRd

VRd = n/sqrt{[(1 + αn)/(Fb,ver,Rd )]2 + [βn/(Fb,hor,Rd]2}

α = 0; β = 0.214

The vertical bearing resistance of a single bolt
Fb,ver,Rd = (k1αbfupdtp)/γm2

k= min[2.8(e2/d0) - 1.7; 2.5) =  min[2.8(50/22) - 1.7; 2.5)
= min[4.66;2.5] = 2.5

α= min[e1/3d0; (p1/3d0 - 1/4); fub/fup; 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

Fb,ver,Rd = (2.5 × 0.61 × 410 × 20 ×10)/1.25 = 100040 N = 100.04 kN

The horizontal bearing resistance of a single bolt
Fb,hor,Rd = (k1αbfupdtp)/γm2

k= min[2.8(e1/d0) - 1.7; 1.4(p1/d0) - 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= min[e2/3d0; fub/fup; 1.0]
 = min[50/(3 × 22); 800/410; 1.0]
= min[0.76; 1.95; 1.0] = 0.76

Fb,hor,Rd = (2.5 × 0.76 × 410 × 20 ×10)/1.25 = 124640 N = 124.64 kN

VRd = (3)/sqrt{[(1 + 0 × 3)/100]2 + [(0.214 × 3)/124.64]2) = 288.218 kN

VEd = 100.25 kN < 288.218 kN Ok

(iii) Supported Beam - Fin Plate

Shear
Basic requirement VEd ≤ VRd,min

VRd,min = min(VRd,g;VRd,n;VRd,n)

Gross section
VRd,g = (hp.tp.ffy,p)/[1.27 × sqrt(3) × γm0]

VRd,g = [(220 × 10 × 275)/(1.27 × √3 × 1.0)] × 10-3 = 275.034 kN

Net Section
VRd,n = Av,net × [fu,p / (sqrt(3) × γm2)]

Net Area = Av,net =  tp.(h- n1.d0)
Av,net =  10(220 - 3 × 22) = 1540 mm2

VRd,n = 1540  × [410 / (√3 × 1.1)] × 10-3 = 331.399 kN

Block Tearing
VRd,b = [(0.5fu,p Antm2] +  [fy,p Anv / (sqrt(3) × γm0)]

Net area subject to tension = Ant = tp[e2 -  0.5d0]
Ant = 10[e2 -  0.5d0]
Ant = 10[50  -  0.5 × 22] = 390 mm2

Net area subject to shear = Anv = tp[h- e1 -  (n- 0.5)d0]
Anv = 10[20 - 40 -  (3 - 0.5) × 22] = 1250 mm2

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

VRd,min = min(275.034; 331.399; 271.145 ) = 271.145 kN

VRd,min = 100.25 kN < 271.145 kN Ok

Lateral Torsional Buckling of fin plates
Basic requirement VEd ≤ VRd
tp/zp = 220/50 = 4.4 > 0.15

Therefore fin plate is short;
VRd = [(Wel,p/z) × (fy,p / γm0)]

Wel,p = (tphp2)/6 = (10 × 2202)/6 = 73333.333 mm3

VRd = [(73333.333/50) × (275 / 1.0)] = 403333.333 N = 403.333 kN
VEd = 100.25 kN < 403.33 kN Ok

Supported Beam in Shear
Gross Section
VRd,g = [Av,web × (fy,b1  / (sqrt(3) × γm0))]

Gross Area Av  = ATee - btf,b1 + (tw,b1 + 2rb1× 0.5tf,b1

ATee =  (265.2 - 10.2) × 6 + (165 × 10.2) = 1530 + 1683 = 3213 mm2
Av  = 3213 - (165 × 10.2) +  (6 + 2 × 8.9) × 0.5 × 10.5 = 3213 - 1683 + 124.95 = 1654.95 mm2

VRd,g = 1654.95 × 275 / (sqrt(3) × 1.0) = 262758.603 N = 262.758 kN




Net Section
VRd,n = [Av,net × (fu,b1  / (sqrt(3) × γm2))]

Net area; Av,net = A- n1d0tw,b1 1654.95 - (3 × 22 × 6) = 1258.95 mm2

VRd,n = 1258.95 × 410 / (sqrt(3) × 1.1) = 270918.727 N = 270.918 kN

Block Tearing
VRd,b = [(0.5fu,b1 Antm2] +  [fy,b1 Anv / (sqrt(3) × γm0)]

Net area subject to tension = Ant = tw,b1[e2,b -  0.5d0]
Ant = 6[40  -  0.5 × 22] = 174 mm2

Net area subject to shear = Anv = tw,b1[e1,b + (n- 1)p- (n- 0.5)d0]
Anv = 6[40 + (3 - 1)70 -  (3 - 0.5)22] = 750 mm2

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

VRd,min = min(262.758; 270.918; 154.955) = 154.955 kN

VRd,min = 100.25 kN < 154.955 kN Ok

Check for welding (supporting beam)
For a beam in S275 steel;

Basic requirement; a ≥ 0.5tp
0.5tp = 0.5 × 10 = 5mm
a = 5.7mm > 0.5tp

Thank you for visiting Structville today. God bless you.

No comments:

Post a Comment