1

6.0 Connections 6.1 Introduction Connections determine how the forces (moment and axial load) being transferred between two structural members. The terms rigid-connections (continuous design), semi-rigid connections (semi-continuous design) and simple connections (simple design) are therefore indicate the degree of moment transferred between members. The rigid connection and simple connection are the idealized assumptions that indicating full-moment transfer and zero-moment transfer; the semi-rigid connection is in actual condition which stand in between. The Figure 6-1 shows the difference between the effects of connection types to the force distribution in a structure.

Figure 6-1 Moment distribution in various type of connections

Recall the concept of effective length in previous chapter (see also Table 25 BS 5950), the single-bolt connecting two truss member is considered pin-jointed; double-bolts is consider semi-rigid. In this sense, the rigid connection may be gained if appropriate amount of bolt assigned to the connection. Normally welding is assumed rigid connection. The Figure 6-2 shows some example of the connection types mentioned above, and the Figure 6-3 shows the moment rotation characteristic of several types of connections.

Semi rigid connections

Pinned joint

Rigid joint

No moment at beam end

Moment due to eccentricity of beam-end force (based on BS 5950)

Beam end connection transfers a portion of moment to the columns

The least sagging moment

Sagging moment govern the beam size

Reduced sagging moment

Beam design using hogging moment.

Simple Construction

Semi-continuous Construction

Continuous Construction

2

v

Figure 6-2 Various types of connection

Web cleat (assumed to be simple construction) Fin plate (assumed to be

simple construction, could be semi-continuous)

Extended end plate (prior for semi-continuous)

Flush end plate (simple semi-continuous) Beam splice (mandatory to

be continuous design) Column splice (continuous)

Secondary beam-main beam connection (web cleat) (assumed as pin connection)

Web cleat and end bearing (simple construction)

Welded (continuous) Portal frame eave connection (continuous)

3

Figure 6-3 Moment rotation characteristic

In this chapter, we will focus on joints in simple design. In simple design, joints between members should be capable to transmit the calculated forces and should also capable of accepting the resulting rotation. No significant moment that adversely affects members of the structures taken into account. Bolting and welding are concerned as the major connection medium. Three common types of connections covered are:

- shear connection in which the connection is resisting shear deformation in longitudinal direction of both connected members (planar shear)

- tensile connection the connection is assigned to resist the uplift force, e.g. end plate connection.

- brackets connecting a horizontal member perpendicularly to a vertical member which inducing moment.

Figure 6-4 Various connections

Single shear

Double shear Lapped joint Face-connected joint

4

6.2 Bolting A bolt may be subjects to plane shear and tension. Some of the possible failure modes are shown in Figure 6-5 below.

Figure 6-5 Various failure mode of bolt

Significantly the failure may occur at the connected plate, bolt itself and the nut. Therefore, a bolt should be checked its resistance to all these kind of failure. A summary of items to be checked for bolt design is provided below: About the plate or connected part

- bolt spacing (maximum spacing and minimum spacing); - edge distance (maximum and minimum); - packing thickness and number of plies; - bolt holes effect on shear capacity; - block shear failure which occurs when a group of bolts are used; and - bearing capacity

About the bolt itself - shear capacity affected by the thickness and number of plies; - bearing capacity; and - tension capacity

Bearing Crushed Tore off

Plate failure Bolt yielded Frictional loss

5

6.2.1 Bolt Spacing and Edge Distances

Figure 6-6 Bolts spacing (adapted from cl. 6.2)

2.5d

Oversize hole

Slotted hole

Standard clearance hole

Fitted hole

d

d = nominal diameter of bolt

14t

t

t = thickness of the thinner plate *These plates are not stiffened by web or outstand

Distance between two lines of adjacent bolts

Stress direction

Connection exposed to corrosive influences:

The lesser of 16t and 200mm

t

*Any direction

Oversize hole

e

Standard clearance hole

Edge

D

e

e

Minimum edge or end distance (for unstiffened plates):

Quality of cut Edge and end distance For a rolled, machine flame cut, sawn or planed edge or end For a sheared or hand flame cut edge or end

1.25D 1.40D

NOTE D is the diameter of a standard clearance hole for a bolt of the relevant nominal diameter

Maximum edge or end distance (for unstiffened plates):

e < 11te normally

e < 40mm + 4te where the parts are exposed to corrosive influences

Minimum bolts spacing in unstiffened plates

Maximum bolts spacing in unstiffened plates

Edge and End Distances

Normally:

* The standard dimensions of holes are listed in Table 33

6

6.2.2 Capacity of Plate or Connected Part

6.2.2.1 Shear Capacity No reduction of shear capacity if Av.net > 0.85Av/ Ke where Av.net is the net area after deducting bolt holes; Ke is the effective net area coefficient (see cl. 3.4.3) If Av.net < 0.85Av/ Ke, the shear capacity is 0.7pyKeAv.net

6.2.2.2 Block Shear Capacity of the Connected Part A group of bolt holes may induce shear failure which depicted in Figure 6-7:

Figure 6-7 Block shear failure of bolt group

To prevent it, checking must be done to ensure the reaction Fr at the connected part less than block shear capacity Pr where, Pr = 0.6pyt [Lv + Ke(Lt kDt)] where

Dt is the hole size for the tension face, generally the hole diameter, but for slotted hole the dimension perpendicular to the direction of load transfer should be used;

k is a coefficient with values as follows: - for a single line of bolts: k = 0.5; - for a two line of bolts: k = 2.5 Lt is the length of the tension face Lv is the length of shear face t is the thickness

Figure 6-8 Effective shear area (Figure 22)

Failure on the shear face of bolt group Tensile rupture on

tension face of bolt group

7

6.2.2.3 Bearing Capacity of the Connected Part (cl. 6.3.3.3) Bearing capacity of the plate or any connected part is indicated by: Pbs = kbsdtpbs but Pbs < 0.5kbsetpbs where

pbs is bearing strength of the connected part, (460N/mm2 for steel grade S 275, and 550N/mm2 for grade S 355)

kbs - for bolts in standard clearance holes : 1.0; - for bolts in oversized holes: 0.7; - for bolts in short slotted holes: 0.7; - for bolts in long slotted holes: 0.5; - for bolts in kidney-shaped slots: 0.5

6.2.2.4 Moment Capacity of the Connected Part (cl. 6.3.4.2) If the connected part is designed assuming double curvature bending, its moment capacity per unit width should be taken as Mc = pytp2/6 where

tp is the thickness of the connected part.

Figure 6-9 Double curvature bending at outstand

6.2.3 Bolts Capacity Two design methods are allowed in BS 5950-1:2000, they are simple method and more exact method. The simple method is commonly used, and it will be introduced in this chapter. To apply simple method, the cross-centre spacing of the bolt lines should not exceed 55% of the flange width or end plate width:

8

Figure 6-10 Maximum cross-centres of bolt lines for the simple method

6.2.3.1 Shear Capacity (cl. 6.3.2) The shear capacity of the whole connection is determined by the lesser of shear capacity of the bolt itself and the shear capacity of the connected members. The shear capacities of a bolt in different cases are depicted in the Figure 6-11 below:

Figure 6-11 Shear capacity of bolt in various conditions

6.2.3.2 Bearing Capacity (cl. 6.3.3) Since the failure of connection may occurs on bolts or the connected part, the capacity of concerned connection is taken as the lesser of the bolt plate bearing capacity and bolt bearing capacity Pbb Pbb = dtppbb where,

d/3 < tpa < 4d/3, Ps = psAs

pa389

dtdd

*The number of plies is preferably less than four

tpa < 4d/3

tpa < d/3 , Ps = psAs

A = At A = A

*At may be obtained from the appropriate bolt specification; if it is not defined, At is taken as the area at the bottom of the threads

Lj > 500mm, Ps = psAs

50005500 jL

Packing

Long joint

9

pbb is the bearing strength of the bolt, obtained from Table 31, (460N/mm2 for bolt grade 4.6; 1000N/mm2 for bolt grade 8.8 and 1300N/mm2 for bolt grade 10.9);

tp is the thickness of the connected part, or, if the bolts are countersunk, the thickness of the part minus half the depth of countersinking.

6.2.3.3 Tension Capacity The tension capacity of a bolt may be determined by either simple method or more exact method. The simple method provides: Pnom = 0.8ptAt where pt is the tension strength of the bolt (240N/mm2 for bolt grade 4.6; 560N/mm2 for bolt grade 8.8 and 700N/mm2 for bolt grade 10.9)

6.2.4 Bracket Brackets may be connected to either the web or the flange of the column using bolts, welds or a combination of the two. It may be fabricated from offcuts of rolled sections, or from plates appropriately shaped and welded together. The connection may be subjects to:

- the moment acts out-of-plane producing tension in the bolts; - the moment is in the plane of the connection resulting in a shear effect in the bolts

Figure 6-12

6.2.4.1 Bracket Subjects to Out-of-plane Moment A single bolt in a bracket subjects to out-of-plane moment (faced connected bracket) is necessary to be checked its adequacy in resisting the direct shear stress and tensile stress due to moment. The direct shear stress Fs of the bolt may be obtained by: Fs = P/n where P is the acted force (see Figure 6-9); n is the number of bolt in the bracket The shear stress due to moment: Ft = Peymax/ 2y where

a) Face connected-moment (out of plane) b) Lapped-torsional moment (in plane)

10

ymax is the distance from the zero-moment point to the centre of furthest bolt, which labeled as d in Figure 6-9 a);

y is the distance from the zero-moment point to centre of each bolt. The effect due combination of shear and tension also should be checked through satisfying the following expression (for simple method) (cl. 6.3.4.4):

4.1nom

t

s

s PF

PF

6.2.5 Worked Example for Face-Connected Bracket Check the adequacy of the connection depicted below in Figure 6.13:

Figure 6-13

Try 8 no. 20mm bolts (grade 8.8)

Direct Shear Force Fs = P/n = 462/8 = 57.8kN

Shear Capacity of Connection For 20mm bolts (grade 8.8), ps = 375N/mm2 and As = 245mm2 Ps = psAs = 375 245 10-3 = 91.9kN > Fs = 57.8kN Ok

Shear Capacity of the Connected Part Assume standard clearance hole used, therefore, kbs = 1.0 and d = 22 Pbs = kbsdtpbs = 1.0 22 20 460 10-3 = 202.4kN > Fs = 57.8kN Ok

cl. 6.3.2.1 Table 30

Tension Force Due to Out-of-plane Moment Determine the zero-moment point, overall depth d = 480, d/7 = 68.57mm from bottom, use 70mm

cl. 6.3.3.3

200 45

7

480

3 @

100

20

154.3

Gk = 210kN Qk = 105kN

Design force

P = 1.4Gk + 1.6Qk

= 1.4 210 + 1.6 105

= 462kN

Out-of-plane moment

M = 462 0.2

= 92.4 kNm

11

Figure 6-14 Tension Capacity Pnom = 0.8ptAt = 0.8 560 10-3 303 = 135.74kN > Ft = 98.9kN Ok

Combination Check

nom

t

s

s

PF

PF =

74.1359.98

1148.57 = 1.23 < 1.4 Ok

cl. 6.3.4.2

Comments Bolts satisfactory

cl. 6.3.4.4

Bracket Subjects to In-plane Moment Similar to the design of face-connected bracket, the bolts in a bracket subject to in-plane moment (lapped bracket) is required to check its adequacy on resisting direct shear. The moment due to eccentricity is also converted into shear onto the bolt. Therefore, these shear vectors are sum together in force analysis.

Figure 6-14 The vector sum of shear stress

Formula of Fs is similar to the calculation of lapped bracket. For shear stress due to moment, Ft = Permax/( 2x + 2y ) where

Tension force

Ft = Peymax/ 2y = 232022012020 103202.0462 2222

3

= 89.3 kN

100

100

100

20 70

Bol

t ten

sion

o

Fs

Ft

FR

rmax

A

o

APe

y1 y2

x1

12

rmax = 2max2

max yx , is the distance from the centroid of bolt group to the centre of outset bolt, which subject to greatest stress

The vector sum of shear force is gained by using trigonometry formula:

FR = cos2 ts2t2s FFFF

6.2.6 Worked Example for Lapped Bracket Reconsider the previous case, now replace the connection to be the gusset plate on two faces of the column flanges, which as depicted in Figure 6-16 below. Check the adequacy of the connection below:

Figure 6-15

Shear Force Due to In-plane Moment

rmax = 2max2

max yx = 22 15075 = 167.71mm Ft = Permax/( 2x + 2y ) = 462 550 167.71 / [(8 752) 2 + (4 502 + 4 1502) 2] = 147.0kN

Vector sum of shear force

FR = cos2 ts2t2s FFFF = 71.167/7595.1469.28295.1469.28 22 = 162.0kN

Shear Capacity of Connection For 24mm bolts (grade 8.8), ps = 375N/mm2 and As = 353mm2 Ps = psAs = 375 353 10-3 = 132.4kN > Fs = 162.0kN Ok

cl. 6.3.2.1 Table 30

Shear Capacity of the Connected Part Assume standard clearance hole used, therefore, kbs = 1.0 and d = 26 Pbs = kbsdtpbs = 1.0 26 20 460 10-3

cl. 6.3.3.3

Use 8 no. 24mm bolts (grade 8.8) on

each face

Direct Shear Force

Fs = P/n = 462/(8 2)

= 28.9kN

462kN 550

100

75

100

100

75

13

= 239.2kN > Fs = 162.0kN Ok Comments Bolts satisfactory

6.2.6 Worked Example of Web Cleats Design a beam-to-column connection using web cleats. The beam is 457 191 90UB S275 and the column is 203 203 86UC S275. Factored reaction at the beam end is 402.85kN.

Figure 6-16

Sizing Gauge, g = 134.4mm Try angle cleat 90 90 10L Length of cleat, lc > 0.6Db = 0.6 467.4 = 280.44mm, Adopt lc = 290mm

Bolt Adopt M20 grade 8.8 bolts in standard clearance holes db = 20mm Db = 22mm Minimum bolt spacing = 2.5db = 2.5 20 = 50mm Maximum bolt spacing = 14tmin = 14 10 = 140mm Minimum edge distance = 1.25Dh = 1.25 22 = 27.5mm (for rolled edge) = 1.40Dh = 1.40 22 = 30.8mm (for cut edge) Maximum edge distance = 11tmin = 11 10 = 110mm = 1.0 since thickness of both web splice and flange splice are less than 16mm Adopt Bolts spacing = 50mm Edge distance = 30mm for the rolled edge and 35mm for the cut edge

Table 33 cl. 6.2.1.1 cl. 6.2.1.2 Table 29 cl. 6.2.2.5 Table 9 Table 12

Shear capacity of bolt group connecting cleats to web of supported beam

Fv = 402.85kN (factored)

14

Basic requirement: FR < 2Ps due to bolts are double shear Force on the outermost bolt Shear force per bolt, Fs = Fv / (number of bolts) = 402.85 / 6 = 67.14kN Shear force per bolt due to eccentric moment, Ft = Permax/( 2x + 2y ) rmax = 2max

2max yx = 22 12560

= 138.65mm Ft = 402.85 60 138.65 / [2(3 602 + (252 + 752 + 1252))] = 51.28kN Vector sum

FR = cos2 ts2t2s FFFF = 65.138/6028.5114.67228.5114.67 22 = 100.58kN Bolt shear capacity As = At = 245mm2, ps = 375N/mm2 Capacity of a single bolt = 2Ps = 2psAs = 2 375 245 10-3 = 183.75kN 2Ps > FR = 100.58kN OK Shear and Bearing Capacity of Cleat Connected to Supported Beam For shear Force to be resisted by each cleat F = Fv / 2 = 402.85 / 2 = 201.43kN Shear capacity of a single angle cleat, Av = 0.9Anet = 0.9 tc (lc 6Dh) = 0.9 10 (320 6 22) = 1692mm2 Pv = 0.6pyAv = 0.6 275 1692 10-3 = 279.18kN > F OK For bearing Force to be resisted by each cleat F = FR / 2 = 100.58 / 2 = 50.29kN Bearing capacity, Pbs kbs = 1.0, db = 22mm and e = 30mm Pbs = kbsdbtcpbs = 1.0 22 10 460 10-3 = 101.2kN 0.5 kbsetcpbs = 0.5 1.0 30 10 460 10-3 = 69kN < Pbs Therefore Pbs = 69kN > F = 50.29kN OK

cl. 4.2.3 cl. 6.3.3.3

Shear and Bearing Capacity of the Supported Beam For shear (Block shear capacity) Ke = 1.2 for steel grade S275 k = 0.5 for single line of bolts Pr = 0.6pytb [Lv + Ke(Lt kDt)] = 0.6 275 11.4 [250 + 1.2 (60 0.5 22)] 10-3 = 580.85kN > Fv = 402.85kN OK

cl. 3.4.3 cl. 6.2.4

15

For bearing kbs = 1.0, db = 22mm and e = 50mm Pbs = kbsdbtbpbs = 1.0 22 11.4 460 10-3 = 115.37kN 0.5 kbsetbpbs = 0.5 1.0 50 11.4 460 10-3 = 131.1kN < Pbs Therefore Pbs = 115.37kN > FR = 100.58kN OK

cl. 6.3.3.3

Shear Capacity of Bolt Group Connecting Cleats to Supporting Column Force to be resisted for each top row bolt, F = 402.85 / 6 = 67.14kN Shear capacity of a single bolt for the top row bolts Due to bolt shank failure, Ps = ps As = 375 245 10-3 = 91.9kN > F = 67.14kN OK Due to end tearing Pbs = 0.5 kbsetcpbs = 0.5 1.0 30 10 460 10-3 = 69.0kN > F = 67.14kN OK

cl. 6.3.2.1 cl. 6.3.3.3

Shear Bearing Capacity of Cleats Connected to Supporting Column For shear Force to be resisted by each cleat, F = Fv / 2 = 201.43kN Shear capacity of a single angle cleat, Av = 0.9Anet = 0.9 tc (lc 4Dh) = 0.9 10 (320 4 22) = 2088mm2 Pv = 0.6pyAv = 0.6 275 2088 10-3 = 344.52kN > F = 201.43kN OK

cl. 6.3.2.1

For Bearing Force to be resisted by each bolt, F = Fv / 12 = 402.85/12 = 33.57kN Pbs = kbsdbtcpbs = 1.0 22 10 460 10-3 = 101.2kN 0.5 kbsetcpbs = 0.5 1.0 30 10 460 10-3 = 69.0kN < Pbs Therefore Pbs = 69.0kN > F = 33.57kN OK

cl. 6.3.3.3

Shear Bearing Capacity of Connected Column Flange It need not be checked since the thickness is more than the web cleats.

6.3 Welding A weld is produced by passing a current (between 50 and 400 amperes), through an electrode or filler wire to produce an arc which complete the path from the power source through the specimen to earth. Therefore the generated heat (from 2800 to 16700C) will melt both the electrode and the parent metal and the plates being welded fuse together on cooling. The Figure 6-18 shows some types of welding.

Figure 6-17 Various types of weld

Butt weld Fillet weld Deep penetration fillet weld

16

There are two design methods allowed which are the simple method and directional method. The simple method is to be introduced since it involves shorter procedure. Generally, the design of weld involves only the calculation of vector sums the design stresses as in the bolt design, assigning appropriate amount of welding material and fulfilling some rules stated in BS 5950.

6.3.1 Weld Strength The longitudinal shear capacity PL per unit length of weld is taken as: PL = pwa where pw is the design strength of fillet welds (Table 37), which often taken as

220N/mm2 a is the throat size of a fillet weld (see Figure 6-19 below)

Figure 6-18 Effective throat size of a plain fillet weld

Since the effective throat size is not allowed to be more that 0.7s, (where s is the smaller leg for a plain fillet weld or the smaller fusion face for any other case), and the design strength pw = 220N/mm2 for steel S275 in all electrode classes, the formula of shear capacity PL per unit length could be rearranged, so: PL = pwa = 220 0.7s = 0.154s kN/mm

17

6.3.2 Rules in Weld Design

6.3.2.1 Effective Length (cl. 6.8.2)

Figure 6-19 Effective length

Figure 6-20 At the end of member

Figure 6-21 Le for a lapped joint

6.3.2.2 Angle of Intersection of a Fillet Weld (cl. 6.8.1)

Figure 6-22 Limitation of intersection angle for a fillet weld

Le = L 2s

s

s

s

L

* Le is taken as the length over which the fillet is full size

* If Le < 4s or

40mm, the weld should not be used to carry load

* In case where the end return (which is taken as minimum 2s) exist, the Le is taken as

Le = L s * L > Tw

L > 4t

t = thickness of the thinner part jointed

120 > > 60

Fusion faces

18

6.3.3 Simple Design

6.3.3.1 For Welds Subject to Axial Force Only For welds which are parallel to force direction. The weld length may be divided in proportion of distance from weld to the centroid, see Figure 6-24

Figure 6-23 Proportion of weld length in parallel of force direction

For welds which are perpendicular to the direction of force, the stress is assumed distributed uniformly over the weld length.

6.3.4 Worked Example for Welds Subject to Axial Load

6.3.4.1 Welds Parallel to Force Direction Determine the weld lengths if only side welding applied on the connection depicted below:

Figure 6-24

Loading P = 1.4Gk + 1.6Qk = 1.4 50 + 1.6 60 = 166kN

Overall Sizing Try 6mm fillet weld, s = 6mm PL = 0.154s = 0.154 6 = 0.924kN/mm Le = P/PL = 166 / 0.924 = 180mm

Detail Sizing

bab

LLL

2e1e1e

651.21

1801e L

Assume P = Le1 + Le2 Moment at point A: Pb = Le1 (a + b) Therefore,

2e1e

1e1e

LLL

PL

bab

; and

2e1e

2e

LLL

baa

ab

Le1

Le2

P

A

65 50 8L

43.9 21.1

Le1

Le2

Gk = 50kN Qk = 60kN

65

19

Le1 = 58.43mm Le2 = 180 - Le1 = 180 58.43 = 121.57mm Applied Weld Length Since no end turning, L1 = Le1 + 2s = 58.43 + 2 6 = 70.43mm Apply 75mm L2 = Le2 + 2s = 121.57 + 2 6 = 133.57mm Apply 135mm

6.3.4.2 Welds in Both Parallel and Perpendicular to Force Direction Determine the weld lengths if only side welding applied on the connection depicted below:

Figure 6-25

Detail Sizing

bab

LLTL w

2e1e

1e 2/ , where Tw is the overall depth of the angle section

651.21

1802/651e L

Le1 = 27.2mm Le2 = 180 Tw - Le1 = 180 65 - 27.2 = 87.8mm

Applied Weld Length Since no end turning, L1 = Le1 + s = 27.2 + 6 = 33.2mm Since 33.2mm is less than Tw, applied L1 should be at least equal to Tw, therefore L1 = 65mm Apply 65mm L2 = Le2 + s = 87.8 + 6 = 93.8mm Apply 95mm

65 50 8L65

43.9 21.1

Le1

Le2

Gk = 50kN Qk = 60kN

20

6.3.4.3 For Welds Subject to In-plane Moment and Out-of-plane Moment Similar to bolt design, the stresses to be checked are the direct stress and stress due to eccentric moment. Therefore, the general formulas used in both cases are: Fs = P/ eL Ft = Permax/( xI + yI ) FR = cos2 ts2t2s FFFF

Figure 6-26 The vector summation of shear stress

o

Fs

Ft

FR

rmax

A

x

y

y

e

o

A

rmax

P

x

21

6.3.5 Worked Example for Lapped Joint (In-plane Moment) Determine the appropriate weld size for the connection shown below:

Figure 6-27

Direct Shear Force Fs = P/ eL = 200 / [2 (300 + 160)] = 0.217kN/mm

Shear Force Due to In-plane Moment Assume the thickness of weld is 1mm to ease the calculation

Ix = 2

2

3

15016012

300 = 11.7 106mm3

Iy = 2

2

3

8030012

160 = 4.52 106mm3

rmax = 22 15080 = 170mm Ft = Permax/( xI + yI ) = 200 150 170 / [(11.7 + 4.52) 106] = 0.314kN/mm

Vector Sum FR = cos2 ts2t2s FFFF = 170/80314.0217.02314.0217.0 22 = 0.458kN/mm

Sizing PL = FR = 0.458kN/mm 0.458 = 0.154s s = 0.458 / 0.154 = 2.97mm Apply 4mm fillet weld , Grade 35

300

150

o

A

rmax

200kN

160

x x o

rmax

x

Fs

Ft

FR x

y

yWelded at back face

22

6.3.6 Worked Example (In-plane Moment) (2 Flanges and 3 Welded Lines) Determine the appropriate weld size for the connection shown below:

Figure 6-28

Direct Shear Force Consider one plate only Fs = P/ eL = 200 / (300 + 2 160) = 0.323kN/mm

Centroid Assuming weld thickness is 1mm, consider only one plate A = 300 + 2 160 = 620mm2 y = (2 160 1 80) / 620 = 41.29mm from left

Shear Force Due to In-plane Moment Consider only one plate,

Ix =

2

3

150160212

300 = 9.45 106mm3

Iy =

2

3

29.4130012

1602 = 1.19 106mm3

rmax = 22 15029.41160 = 191.29mm Ft = Permax/( xI + yI ) = 200 (150 + 80 41.29) 191.29 / [(9.45 + 1.19) 106]

300

150

A 400kN

160

o

rmax

x

Fs

Ft

FR x

y

yNo welding at back face

Plan view

o

rmax

x x

y

y

Plates on both flanges

C.L.

23

= 0.679kN/mm Vector Sum FR = cos2 ts2t2s FFFF = 29.191/29.41160679.0323.02679.0323.0 22 = 0.915kN/mm

Sizing PL = FR = 0.915kN/mm 0.915 = 0.154s s = 0.915 / 0.154 = 5.94mm Apply 6mm fillet weld , Grade 35

24

6.3.7 Worked Example for Face Joint (Out-of-plane Moment) Determine whether the connection shown below is safe to be used.

Figure 6-29

Loading P = 1.4Gk + 1.6Qk = 1.4 80 + 1.6 110 = 288kN Total Weld Length Le, f = 173.2 2 8 = 157.2mm Le, w = 280 2 8 = 264mm Le = 2 [157.2 + 264] = 842.4mm

Direct Shear Force Fs = P/ eL = 288 / 842.4 = 0.342kN/mm

Shear Force Due to Out-of-plane Moment Assume the thickness of weld is 1mm to ease the calculation

Ix = 2

2

3

2/3642.15712

264 = 13.48 106mm3

rmax = 364 / 2 = 182mm Ft = Permax/( xI + yI ) = 288 250 182 / 13.48 106] = 0.972kN/mm

Vector Sum FR = cos2 ts2t2s FFFF = 0972.0342.0 22 = 1.031kN/mm

250

364

Gk = 80kN Qk = 110kN

Offcut of 356 171 67UB

173.2

280 xx

8mm grade 35 welding

25

Sizing PL = FR = 1.031 kN/mm 1.031 = 0.154s s = 1.031 / 0.154 = 6.691mm < 8mm 8mm fillet weld is adequate!

6.3.8 Worked Example for Connection Combined Welding and Bolting (Flexible End Plate)

Design a beam-to-beam connection using flexible end plate. The supported beam is 305 102 33UB S275 and the supporting beam is 457 191 90UB S275. Factored reaction at the supported beam end is 79.37kN.

Figure 6-30

Sizing Gauge, g adopted as 100mm; plate thickness, tp = 8mm; and, plate length, lp > 0.6Db = 0.6 312.7 = 187.62mm Adopt lp = 220mm Use M20 grade 8.8 bolts in standard clearance holes db = 20mm Db = 22mm Minimum bolt spacing = 2.5db = 2.5 20 = 50mm Maximum bolt spacing = 14tp = 14 8 = 112mm Minimum edge distance = 1.25Dh = 1.25 22 = 27.5mm Maximum edge distance = 11tp = 11 8 1.0 = 88mm Adopt Bolts spacing = 75mm and edge distances = 35mm

Table 33 cl. 6.2.1.1 cl. 6.2.1.2 Table 29 cl. 6.2.2.5

Shear Capacity of Bolt Group Shear force to be resisted = Fv = 79.37kN

Fv = 79.37kN (factored) (see worked example in 3.4)

e1 =

e2 =

26

Bolt shear capacity Capacity of a single bolt itself, As = At = 245mm2, ps = 375N/mm2 Ps = psAs = 375 245 10-3 = 91.88kN 6Ps = 551.25kN > FR = 79.37kN The top pair of bolts may subject bearing failure Pbs = 0.5 kbsetppbs = 0.5 1.0 35 8 460 10-3 = 64.4kN > Fv / 6 = 13.23kN OK

cl. 6.3.2.1 cl. 6.3.3.3

Shear and Bearing Capacity of End Plate For shear Force to be resisted, F = 79.37kN Shear capacity of a single angle cleat, Av = 0.9Anet = 0.9 tp (lp 3Dh) = 0.9 8 (220 3 22) = 1108.8mm2 Pv = 0.6pyAv = 0.6 275 1108.8 10-3 = 182.95kN > Fv = 79.37kN OK

cl. 6.3.2.1

For Bearing Force to be resisted by each bolt, F = Fv / 6 = 13.23kN Pbs = kbsdbtPpbs = 1.0 22 8 460 10-3 = 80.96kN 0.5 kbsetppbs = 0.5 1.0 35 8 460 10-3 = 64.4kN < Pbs Therefore Pbs = 64.4kN > F = 13.23kN OK

cl. 6.3.3.3

Shear Capacity of the Supported Beam Web at the End Plate Force to be resisted = Fv = 79.37kN Pvb = 0.6pyAv = 0.6pytblp = 0.6 275 6.6 220 10-3 = 239.58kN > Fv = 79.37kN OK

Capacity of Fillet welds Connecting End Plate to Supported Beam Web Try 6mm fillet welds, s = 6mm Design strength (assumed electrode classification 35), pw = 220N/mm2 PL = pwa = 220 0.7s = 220 0.7 6 10-3 = 0.924N/mm Effective length of the weld (both side of the beam web), lw = 2(lp 2s) = 2 (220 2 6) = 416mm Pweld = PL lw = 0.924 416 = 384.38kN > Fv = 79.37kN OK

27

Problems 1. Two pieces of steel plates which subjects to ultimate axial load 200kN is to be connected

to each other (as shown in Figure 6-32). Determine the number of bolt, bolt size, edge distances of the plate and spacing between bolts.

Figure 6-31

2. A double-C-channel member which resisting 820kN ultimate tensile force is composed of

two sections which are connected end-to-end (see Figure 6-33). Both top and bottom faces of the channel are covered by a piece of 10 162 steel plate. The gap within the double-C is filled by another piece of 10 198 plate. a. Determine the number of bolts needed for this system. b. Check the adequacy of the C-channels in resisting the bolt bearing. c. Check the adequacy of the plates.

Figure 6-32

210m

m

?

8mm thick grade S 275 steel plates

10 198

10 162

203 76 23.82 C

P = 820kN

28

3. Determine bolt size for the case below.

Figure 6-33

4. Determine bolt size for the case below.

Figure 6-34

5. Design a splice for 533 210 82UB. Ultimate moment and shear force at the point of

connection are 30kNm and 175kN. The formula Ft = Permax/( 2x + 2y ) could be replaced by Ft = Mrmax/( 2x + 2y ) in this case, where M is the design moment.

Figure 6-35

6. Determine the leg size of the connection which shown in Figure 6-37 below.

250

380

P = 300kN

5 @

60

12 no. 24 bolts

20mm thick gusset plates at each flanges

300

100

50

100

100

50

120kN

M

V

?

29

Figure 6-36

7. Determine the leg size of the connection which shown in Figure 6-38 below.

Figure 6-37

8. Determine the leg size of the connection which shown in Figure 6-39 below.

Figure 6-38

9. Design the fin plate which depicted in the Figure 6-40 below:

250

364

P = 350kN

Offcut of 533 210 82UB

208.7

348.

3

300

300

A 500kN

125

x x

Welded at back face125

2 no. 20mm thick plates

Elevation view Plan view

20

65

P = 90kN

60 60 30 6L

12mm thick gusset plate

30

Figure 6-39 Fin plate

10. Design a beam-to-column connection using web cleats. The beam is 533 210 92UB

S275 and the column is 254 254 89UC S275. Factored reaction at the beam end is 700kN.

11. Design a beam-to-beam connection using flexible end plate. The supported beam is 533

210 92UB S275 and the supporting beam is 610 229 140UB S275. Factored reaction at the supported beam end is 600kN.

References 1. L. J. Morris, D. R. Plum (1988), Structural Steelwork Design to BS 5950, Longman

Scientific & Technical, UK. 2. BSI (2000), BS 5950-1:2000 Guide to Amendments, SCI, UK. 3. D. A. Nerthercot (1991), Limit States Design of Structural Steelwork (Second

Edition), Chapman & Hall, London. 4. SCI/BCSA, Joints in Simple Construction Volume 1: Design Methods Second

Edition

Fv = 450kN (factored)