Job: to design a beam to column flange simple connection with ordinary bolts to a given shear force

Ved.

The column will be HEA and the beam an IPE section, with a specific steel quality for each student,

according to attached list. The joint configuration can be seen in the figure below.

Required to calculate:

$Number of bolts and bolt arrangement in the connection‐ ‐ Joint shear resistance according to EN 1993 1 8‐ ‐ Joint tying resistance according to EN 1993 1 8

Joint data:

Steel S275

Beam IPE 450

Column HEA 340

Plate 12mm

VEd=550kN

Column HEA340, S275

Depth

Width

Thickness of the web

Thickness of the flange

Fillet radius

Area

hc 330mm:=

bc 300mm:=

twc 9.5mm:=

tfc 16.5mm:=

r 27mm:=

Ac 133.5mm2

:=

Beam IPE450, S275

Depth

Width

Thickness of the web

Thickness of the flange

Fillet radius

Area

hb 450mm:=

bb 190mm:=

twb 9.4mm:=

tfb1 14.6mm:=

rb 21mm:=

Ab 98.8mm2

:=

Plate S275

tp 12mm:=Thickness

Yield s trength

Ultimate tensile strength

fy 275N

mm2

:=

fu 430N

mm2

:=

Bolts M20 8.8

Tensile stress area

Diameter of the shank

Diameter of the holes

Yield s trength

Ultimate tensile strength

As 245mm2

:=

d 20mm:=

do 22mm:=

fyb 640N

mm2

:=

fub 800N

mm2

:=

αv 0.6:=

Partial safety factors

γM2 1.25:= (for shear resistance at ULS)

γMu 1.1:= (for tying resistance at ULS)

γM0 1:=

Design shear force

VEd 550kN:=

Weld design(S275)

a 0.48 twb⋅ 4.512 mm⋅=:=

Choose as a=5mm welding.

Number of bolts

FvRd αv fub⋅

As

γM2

⋅ 94.08 kN⋅=:=

nVEd

FvRd

5.846=:=

I choose n 6:= bolts(3 on 2 plates).

Joint shear resistance

Shear resistance of the plates

VRd1 n FvRd⋅ 564.48 kN⋅=:=

VRd1 VEd>

Plates design

The plate height

hp hb 2 60⋅ mm− 330 mm⋅=:=

From Eurocode 1993-1-8:

Minimum

e1 1.2 do⋅ 26.4 mm⋅=:=

e2 1.2 do⋅ 26.4 mm⋅=:=

Maximum

e1 4 tp⋅ 40mm+ 88 mm⋅=:=

e2 4tp 40mm+ 88 mm⋅=:=

Choose e1 70mm:= ., e2 50mm:=

So,hp-e1-e1=190mm p1 2.2do 48.4 mm⋅=:=

p2 2.4do 52.8 mm⋅=:=

So i divide equally the transversal distance between bolts:

p1190

2mm 95 mm⋅=:= .

The plate base will have the dimension:bp e2 e2+ 100 mm⋅=:=

tp 12mm:=

We will have two plates 330x100x12 with 3 bolts on each.

Column's flange in shear

Av 0.9 hp⋅ tfc⋅ 4.901 103

× mm2

⋅=:= the shear area

VRd7 Av

fy

γM0 3⋅

⋅ 778.059 kN⋅=:=

Bearing resistance of a single bolt

Vertical bearing resistance of a single bolt•

e1

3do

1.061=p1

3 do⋅

1

4− 1.189=

fub

fu

1.86=

αb is the minimum of the above 3 relations and 1:

αb 1.061:=

2.8e2

do

⋅ 1.7− 4.664=

k1 is the minimum between above relation and 2.5

k1 2.5:=

FbRdv

k1 αb⋅ fu⋅ d⋅ tp⋅

γM2

218.99 kN⋅=:=

Horizontal bearing resistance of a single bolt•

e2

3do

0.758=fub

fu

1.86=

αb is the minimum of the above 3 relations and 1:

αb 0.758:=

2.8e1

do

⋅ 1.7− 7.209=

1.4p1

do

⋅ 1.7− 4.345=

k1 is the minimum between above relations and 2.5

k1 2.5:=

FbRdh

k1 αb⋅ fu⋅ d⋅ tp⋅

γM2

156.451 kN⋅=:=

The bearing resistance will be the minimum from horizontal and vertical bearing resistance:

FbRdh 156.451kN:=

Bearing resistance of the plates

VRd2 6 FbRdh⋅ 938.706 kN⋅=:=

Bearing resistance of the column's flange

1 bolt:

k1 2.5:=

αb 0.758:= (minimum from above relations)

FbRdf

k1 αb⋅ fu⋅ d⋅ tfc⋅

γM2

215.12 kN⋅=:=

6 bolts:

VRd8 6 FbRdf⋅ 1.291 103

× kN⋅=:=

Joint tying resistance

FvRdu αv fub⋅

As

γMu

⋅ 106.909 kN⋅=:=

NRdu1 n FvRdu⋅ 641.455 kN⋅=:=

Design load VEd 550 kN⋅=

VRd1 564.48 kN⋅=Shear resistance of the plate

VRd7 778.059 kN⋅=Shear resistance of the column's flange

VRd2 938.706 kN⋅=Bearing resistance of the plates

VRd8 1.291 103

× kN⋅=Bearing resistance of column's flange

NRdu1 641.455 kN⋅=Joint tying resistance