Challenge

Project 1 - Design a simple double cleat angle shear connection

Project 1- Challenge
Part 1:
Considering 4.6 Grade Bolts in double shear
Given Data
For Fe410(E250) Grade of Steel ,
Ultimate strength (fu)
= 410 Mpa
The yield strength (fy )
= 250 Mpa
Bolts
Grade of Bolts
= 4.6 Grade
fub = 400 MPa
Assuiming dia. Of bolt
= 16 mm
Stress area, Anb
= 202 mm2
Anb = 0.78x Asb
ssd
Minimum pitch, p = 2.5x dia.
= 40 mm ~ 40 mm (Say)
Edge Distance, e
= 44 mm ~ 45 mm (Say)
Diameter of bolt hole, do
= 18 mm
Considering angle section of L 80x80x8
Thickness of Connecting Angle
= 10 mm
Shear Strength of Bolt
Double Shear case
Since the gusset is connected on angle section on eiter side of gusset plate , the shear force on bolts acts on a double
plane on the bolts. Therefore it’s a double 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 = 1
ns= 1
Asb = 157.8 mm2
(Anb = 0.78x Asb)
Anb= 201.1 mm2
γmb= 1.25
Vdsb= 132613.4 N = 132.6 kN
Bearing Strength of Bolt
kb is small of following :
1 e/3do = 0.815
2
p/3do - 0.25 =
0.741
3 fub / fu = 1.640
4 1 1.000
kb = 0.741
fub = Ultimate tensile stress of the bolt
fu = Ultimate tensile strength of plate
t = Aggregate thickness of connected plates
V𝑑𝑝𝑏= 97185.18519 N = 97.2 kN
Strength of bolt is the lower value of shear strength and bearing strength of bolt
Therefore, Strength of bolt,
Vs = 97.2 kN
Reaction from secondary beam =
150 kN
No. of Bolts required
= 1.5 Nos
Providing 2 Nos of bolts of 16mm dia.
ISMB 400 Main Beam
ISMB200 Beam
40
45
40
Total Length of angle cleat =
125 mm
Part 2:
Considering 4.6 Grade Bolts in single shear
Given Data
For Fe410(E250) Grade of Steel ,
Ultimate strength (fu)
= 410 Mpa
The yield strength (fy )
= 250 Mpa
Bolts
Grade of Bolts
= 4.6 Grade
fub = 400 MPa
Assuiming dia. Of bolt
= 16 mm
Stress area, Anb
= 202 mm2
Anb = 0.78x Asb
ssd
Minimum pitch, p = 2.5x dia.
= 40 mm ~ 40 mm (Say)
Edge Distance, e
= 44 mm ~ 45 mm (Say)
Diameter of bolt hole, do
= 18 mm
Considering angle section of L 80x80x8
Thickness of Connecting Angle
= 10 mm
Shear Strength of Bolt
Single Shear case
Since the gusset is connected on angle section on only one side of gusset plate , 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:
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