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Challenge

Project 2 - Design following connections of a factory building

AArun A13 Feb 2023
Member end forces (Factored)
Shear Force Moment Axial force
Secondary beam 120 kN - 15 kN (C)
Primary beam 450 kN 280 kN-m -
Column 150 kN - 1250 kN (C)
Part 1 :
Secondary Beam to primary beam connection – Fin plate shear connection
Given Data
For Fe410 Grade of Steel,
Ultimate strength (fu)
 = 410 Mpa
The yield strength (fy )
 = 250 Mpa
Bolts
Grade of Bolts
 = 4.6 Grade
fub = 400 MPa
Factored Shear force
 = 120 kN
Assuiming dia. Of bolt
 = 20 mm
Stress area, Anb
 = 315 mm2
Minimum pitch, p = 2.5x dia.
 = 50 mm
Edge Distance, e
 = 44 mm
Diameter of bolt hole, do
 = 22 mm
Thickness of fin plate
 = 16 mm (Assumed )
Shear Strength of Bolt
Single Shear case
Since the web of ISMB 300 is connected to fin plate by bolts, the shear force on bolts acts on a single
plane on the bolts. Therefore it’s a single shear case
Shear Strength of Bolt
Design Shear strength of bolt is given by
Vnsb= Nominal shear capacity of boly
fub = Ultimate tensile stress of the bolt
nn = Number of shearing planes with threads within the shear plane
ns= Number of shear planes without threads intercepting the shear plane
Asb = Nominal shank area
Anb= Net tensile stress area of the bolt in each shear plane
γmb= Partial safety factor for bolt material
 = 1.25
Assuming the shear plane of the bolt lies in the shank of the bolt (i.e the threads excluded from the shear plane)
Refer sketch below:
fub = 800 MPa
nn = 0
ns= 1
Asb = N.A
Anb= 314.2 mm2
γmb= 1.25
Vdsb= 116083.2 N = 116.1 kN
Bearing Strength of Bolt
kb is small of following :
1 e/3do = 0.667
2
p/3do - 0.25 =
 0.758
3 fub / fu = 1.640
4 1 1.000
kb = 0.667
fub = Ultimate tensile stress of the bolt
fu = Ultimate tensile strength of plate
t = Aggregate thickness of connected plates
Vđť‘‘đť‘ťđť‘Ź= 174933.3333 N = 174.9 kN
Strength of bolt is the lower value of shear strength and bearing strength of bolt
Therefore, Strength of bolt,
Vs = 116.1 kN
Number of Bolts required = Factored Axial Tensile Force / Vs
Number of Bolts required =
 2
Number of Bolts
Providing 4 bolts in 2 columns
Weld Thickness
 = 8 mm
Strength of weld
Fin plate
Depth, Lw
 = 125 mm
Thickness, t
 = 12 mm
Thickness of weld
 = 8 mm
Effective throat thickness, tt = KS = 0.7 x t =
 5.6 mm
Design strength of the weld,
Pdw = 132716.763 N = 132.716763 kN > 120 kN Hence, OK
Therefore, provide weld of 6mm.
Provided length of weld is ok as capacity of weld is more than factored load.
Part 2:
Primary beam to column connection – Extended end plate connection
Property ISMB 500 ISMB 300
(All dimensions are in mm)
F1 Depth of section (h) 500 300
Width of flange (bf) 180 140
M
 Thickness of flange (tf) 17.2 12.4
Thickness of web (tw) 10.2 7.5
Gauge (g) 100 80
F2
Resolving moment for the lever arm from centre to centre of flanges
2(F1+F2)x(500-17.2)=280x10^3
Also, F1=F2=F
F= 144.99 kN
For Fe410 Grade of Steel,
Ultimate strength (fu)
 = 410 Mpa
The yield strength (fy )
 = 250 Mpa
Bolts
Grade of Bolts
 = 8.8 Grade
fub = 800 MPa
Assuiming dia. Of bolt
 = 24 mm
Stress area, Anb
 = 356 mm2
Tension Capacity of Bolt
Tnb = 0.9fub Anb /1.25
Tnb = 205056 N = 205.056 kN
Permissioble prying force Q = Tension Capacity - F
Q= 60.07 kN
Design of End Plate for Prying Action
Thicknes of End plate
 = 30 mm (Assume)
Edge distance
 = 50 mm (Assume)
Width of plate
 = 180 mm (Max = width of column)
Distance from centre line of bolt to fillet weld
Iy =
((180/2)-50)-10
Iy = 30 mm
Distance from centre line of bolt (Ie) to prying force is the minimum of following:
1. the end distance, n
 = 90 mm
3.
2
fo = 0.7 x440 = 560 Mpa
le= 69.84783461 mm
Therefore , le= 70 mm
Moment at toe of weld =
F x Iy- Q x le
= 153.977585 N/mm
Effective width of end plate for each bolt w= be / 2
= 90 mm
Moment Capacity =
= 4602.27 > 153.977585 N/mm Hence, OK
Minimum Prying force (IS 800, Cl. No.10.4.7) is as below:
n= 1.5
Q= 15.64 < 60.07
Hence, OK
Check for combined shear and tension (IS 800, Cl. No.10.4.6),
Vsf = 18.75 kN
Vdf= 60 kN
Tf= 160.63 kN
Tdf= 205.056 kN
= 0.71127169 < 1
Hence, OK
Part 3:
Column to base plate connection - Pinned connection
Given Data:
Properties of ISMB500 section
Flange width
 = 180 mm Property ISMB 500 ISMB 300
(All dimensions are in mm)
Overall depth
 = 500 mm Depth of section (h) 500 300
Flange thickness
 = 17.2 mm Width of flange (bf) 180 140
Web thickness
 = 10.2 mm Thickness of flange (tf) 17.2 12.4
Radius of gyration about major axis
 = 202.1 mm Thickness of web (tw) 10.2 7.5
Radius of gyration about minor axis
 = 35.2 mm Gauge (g) 100 80
Cross sectional area
 = 11074 mm2
Root radius
 = 17 mm
Axial compressive load
 = 1250 kN
Shear force
 = 150 kN
Given Data
For Fe410 Grade of Steel,
Ultimate strength (fu)
 = 410 Mpa
The yield strength (fy )
 = 250 Mpa
γmw= Partial safety factor for weld
 = 1.5
For M30 grade of concrete:
Bearing strength of concrete scbc= 0.45 fck = 0.45 × 30 =
 9 N/mm2
γm0= 1.1
γmw = 1.25 (for shop welding)
Area of Base Plate Requieed, Ar
Ar = P/ scbc
Ar = 138888.8889 mm2
Providing square base plate
L = 372.6779962 ~ 450
Providing square base plate of 450mmx450mm
Check for bearing pressure of concrete,
Bearing pressure,
w= P/A
w= 6.172839506 < 9 N/mm2
Hence, OK
Thickness of Base plate
a= 135 mm
b= -25 mm
ts= 34.99669297 mm Say 40 mm
ts= 40 mm
Thus providing 450mm x 450mm x 40mm base plate
Length available for welding around column profile,
La = 1630.8 mm
Size of weld
 = 8 mm
Total end returns
 = 12
Effective length of weld =
 1438.8 mm
Strength of weld per mm,
Strength of weld per mm
 = 884.8 N/mm
Required length of weld =
Shear Force / Strength of weld per mm
Required length of weld =
 169.5339721 mm < 1438.8 mm
Hence, OK
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