1

2.5 Structural connections

All connections shall have a design resistance such that the structure remains effective

and is capable of satisfying all the basic design requirements. Connection design will be

based on the basis of realistic assumption of the distribution of internal forces provided

that:

(a) Assumed internal forces are in equilibrium with applied forces.

(b) Each element in the connection is capable of resisting the internal forces.

(c) Deformations are within the deformation capacity of the fasteners or welds and

of the connected parts.

Usually, members meeting at a joint should be arranged with their centroidal axes

meeting at a point. Where there is eccentricity at intersections the members and

connections should be designed to accommodate the moments which result. In the case

of bolted connections of angles and tees with at least two bolts per connection, the

setting out lines of the bolts may be adopted instead of centroidal axes.

Joints in simple construction should not develop significant moments adversely

affecting the structure. Joints between members in rigid construction should be capable

of transmitting the forces and moments calculated in design.

Bolted connections

The size of holes for all fasteners shall not exceed the dimension given in Table 2.5.1

Table 2.5.1 Maximum Dimensions of Holes

Bolt shank

diameter (mm)

Clearance hole

diameter (mm)

Oversize hole

diameter (mm)

Short slotted hole

dimensions (mm)

Long slotted hole

dimensions (mm)

14 d + 1 d + 4 d + 1 d + 4 d + 1 2.5d

14<d22 d + 2 d + 5 d + 2 d + 6 d + 2 2.5d

24 d + 2 d + 6 d + 2 d + 8 d + 2 2.5d

27 d + 3 d +8 d + 3 d +10 d + 3 2.5d

For edge distances and spacing of holes refer to table 2.1.1

Design shear rupture resistance

The design value of the effective resistance Veff,Rd for rupture along a block shear failure

path shall be determined from:

Veff,Rd = Mo

effvy Af

,60.0

2

Effective shear area, Av,eff = tLv + L1 + L2 - ndo

Where L1 = 2.5do but a1

L2 = 5.0 do but a2

n = number of fastener holes on the block shear failure path.

t = thickness of the web or bracket.

a

L

a

L

L

1

1

2

2

v

d0

Block shear

failure path

a

L

a

L

L

11

2

2

v

Block shear

failure path

a

L

a

L

L

11

22

v

Block shear

failure path

cope cope

cope

(a) plain end (b) coped end (c) doubled coped end

Figure 2.5.1 Net shear area – Block shear

Shear capacity of bolts:

The shear capacity per shear plane Fv,Rd of a bolt shall be taken as:

Fv,Rd = fv,dAs

Where fv,d = design shear strength.

fv,d = Mb

ubf

6.0 but

Mb

ybf

87.0

fub = nominal ultimate strength of the fastener.

fyb = nominal yield strength of the fastener.

As = shear area; for threads in shear plane As is taken at the bottom of the threads and if

threads do not occur in the shear plane As taken as the shank area.

Bearing capacity

The effective capacity of a bolt in bearing on any ply shall be taken as the Sesser of

the bearing capacity of the bolt and the bearing capacity of the connected ply.

The bearing capacity of the bolt Fbb,Rd will be taken as

Fbb,Rd = dtfbb,d

Where d = nominal diameter of the bolt.

t = thickness of the connected ply, or, if the bolts are countersunk, the thickness

of the ply mines half of the depth of counter sinking.

fbb,d = design bearing strength of the bolt.

3

fbb,d = M

ybub ff

)(9.0

The bearing capacity of the connected ply Fbp,Rd shall be taken as:

Fbp,Rd = dtfbp,d but 2

1 e1t fbp,d

Where d = nominal diameter of the bolt

t = thickness of the ply.

fbp,d = design bearing strength of the connected parts.

fbp,d = 2

)(8.0

M

yu ff

e1 = is the edge distance

Bolts subject to tension

The tension capacity Ft,Rd of a bolt (including countersunk bolts) shall be obtained

from: Ft,Rd = ft,d As

Where, ft,d = design tension strength

ft,d = M

yb

M

ubf

butf

0.17.0

As = tensile stress area.

Combined shear and Tension

For bolts subjected to both shear and tension the following additional relationship shall

be satisfied. 4.1,

,

,

,

Rdt

sdt

Rdv

sdv

F

F

F

F

The grade of a bolt is given by two figures separated by a point (table 2.5.3) the

first figure is 1% of the minimum ultimate strength in N/mm2 and the second is 1/10

th of

the percentage ration of the minimum yield strength to the minimum ultimate strength.

Thus 5.6 grade means that the minimum ultimate strength is 500 N/mm2 and the yield

strength is 60% of this strength which is 300 N/m2. The nominal values of the yield

strength fyb and the ultimate strength fub to be adapted as characteristic values in

calculations are given below:

Table 2.5.3

Bolt grade 4.6 4.8 5.6 5.8 6.8 8.8 10.9

Fyb(N/mm2) 240 320 300 400 480 640 900

Fub (N/mm2 400 400 500 500 600 800 1000

4

Tensile stress Areas for Bolts

Tensile stress area for bolts as determined by ISO standards shank and tensile areas are

tabulated below (Table 2.5.2)

Table 2.5.2

Bolt diameter (mm) Tensile area (mm2) Shank area (mm

2)

12 84 113

16 157 201

20 245 314

22 303 380

24 353 452

27 459 572

30 561 707

Example 2.5.1

A machine prepared connection shown in the figure below is

subjected to a design tensile force of 240kN. All data regarding the

member and connection are shown in the figure. The steel grade is Fe

430, the bolt grade is 8.8 and its diameter is 20 mm. check that the

connection is adequate (all dimensions are in mm).

2L 70 x 7

50 80 50

10

0

t = 1570 15 70

70

30

Solution

Geometry, material and load:

Angle: 2L70 x 7, Fe430 fy = 275MPa

Cross-sectional area, Ag = 2 x 940 = 1880mm2

5

Gusset plate: 100mm x 15mm, Fe 430 fy = 275MPa.

Bolts: Dia = 20mm, grade 8.8 fyb = 640MPa

fub = 800MPa, Area at the bottom of thread: As =

245mm2

(i) Holes diameter and spacing:

Hole dia. do= d + 2 = 22mm (table 2.5.1)

Min. edge distance, e2 = 1.25do = 1.25 * 22 =

27.5mm<30mm Ok

Minimum hole distance, P1 = 2.5do = 2.5*22 = 55mm<

80mmOk

Max.edge distance, e1 = 12t = 12 * 7 = 84mm>50mmOk

Max hole distance, P1 = 14t = 14*7 = 98mm>80mmOk

(ii) Capacity of connected members:

The gross sectional area resistance capacity of the angle:

Npl,Rd = OkkNkNAf

M

y

2404701.1

10*275*1880 3

1

The gross sectional area resistance capacity of the gusset plate:

Npl,Rd = OkkNkNAf

M

y

2403751.1

10*275*15*100 3

1

The net section area of the angle, Anet = 27867*222

1880mm

Angle is connected by a single row of bolts in one leg:

Spacing of holes: P1 = 80mm = 3.64do

Since P1 is b/n 2.5do and 5do interpolation must be used:

Nu,Rd = 2

4.0

M

unet fA

for P1 = 2.5do

And Nu,Rd = 2

7.0

M

unet fA

for P1 = 5do

Observing the coefficients 0.4 and 0.7

Interpolating for the coefficient:

Coef. = 0.4 + )4.07.0(*5.25

5.264.3

= 0.5368

6

Hence the design ultimate resistance of the net section

Nu,Rd = 25.1

10*430*)786*2(*5368.05368.0 3

2

M

unet fA

= 290.3kN> 240.0 kN

Ok

Net area of the gusset plate:

Anet = 100*15 - 22*15 = 1170mm2

Des. Ultimate resistance of the net section:

Nu,Rd = 25.1

10*430*1170*9.09.0 3

2

M

ueff fA

= 362.2 kN > 240.0 kN Ok

(iii) Shear capacity of bolts:

Assumption threads are in shear plane i.e. As = 245mm2

The bolts are subjected to double shear (two shear planes each at the

interface b/n the gusset plate and angle)

Fv,Rd = fvdAs =

OkkNAf

OkkNAfAf

AfF

Mb

syb

Mb

syb

Mb

subsvdRdv

18821825.1

10*245*640*87.0*287.0

1202

240188

25.1

10*245*800*6.0*287.0)2(6.0

3

3

,

(iv) Bearing capacity of members and bolts:

The gusset plate is not the critical member

Since t = 15mm > 2*7 = 14mm

The bearing capacity of the gusset plate

Fbp,Rd = dtfbp,d 2

1e1tfbp,d

Fbp,Rd =

2

)(8.0

M

ybub ffdt

=

25.1

10*)275430(*8.0*15*20 3

Fbp,Rd = 135.36kN > OkkN 0.1202

240

OkkNkN

4.1352.16925.1

10*)275430(8.0*15*50*

2

1 3

7

The bearing capacity of one angle:

Fbp,Rd =

dbp

M

ybubtfe

ffdt,

2

,2

1(8.0

= OkkN

602*2

2402.63

25.1

10*)275430(*8.0*7*20 3

OkkN

2.6396.7825.1

10*)275430(*8.0*7*50*

2

1 3

The bearing capacity of the bolt

Fbb,Rd = dtfbb,d = Mb

ybub ffdt

)(9.0(

= OkkN

2403.29025.1

10*)640800(*9.0*14*20 3

Therefore the connection is adequate to resist the design tensile load.

Example 2.5.2

Check that the secondary girder to primary girder connection by

means of angles shown in the figure below is adequate. All data

required are provided in the figure.

Main girder: Fe 430, fy = 275N/mm2

Secondary girder: Fe430, fy = 275N/mm2

Angles: 2L 90 X 9, Fe430, fy = 275N/mm2

Bolts: Diameter = 22mm; Grade 8.8

fyb = 640 N/mm2, fub = 800N/mm2

8 @ 60

28

50

50

90

900

155

300

800

t = 15

18.5 40

35

90

9 40

8

Bolt area at the bottom of thread: As = 303mm2

Applied load: shear force V = 890kN (at the center line of the web of

the main girder)

Solution

Diameter of holes:

Do = d + 2 = 22 + 2 = 24mm. (Table 2.5.1)

Min.edge distance, e1 =1.25do = 1.25*24 = 30mm<40mmOk

Min,hole distance, P1 = 2.5do = 2.5*24 = 60mmOk

Max.edge distance, e1 = 12t = 12*9 = 108mm>40mmOk

Max.hole distance, P1 = 14t = 14 X 9 = 126mm> 60mmOk

Shear capacity of bolts

From the figure one shear area per bolt is obtained.

We assume that threads are in shear plane.

Shear capacity of a bolt:

Fv,Rd = fvd As = Mb

syb

Mb

subAfAf

87.06.0

= .4.499*2

8904.116

25.1

10*303*800*6.0 3

kN

OkkN

4.1160.13525.1

10*303*640*87.0 3

Capacity of connection: Main Girder and connection Angle

Bearing capacity of bolts:

Since the web thickness of the beam tw = 18.5mm is greater than the

angle leg thickness ta = 9.0mm, the angle is the critical member.

The bearing capacity is:

Fbb,Rd = dtfbb,d =

Mb

ybub ffdt

(9.0

= kN4.499*2

8903.205

25.1

10*)640800(*9.0*9*22 3

9

Bearing capacity of angle:

Fbp,Rd =

dbp

Mb

ybubtfe

ffdt,1

2

1)(8.0

= kN4.499*2

8903.89

25.1

10*)275430(*8.0*9*22 3

25.1

10*)275430(8.0*9*40*

2

1 3 = 81.2 < 89.3kN but > 49.4kN.

Capacity of connection secondary girder and connection

Angle.(two shear area per bolt)

Design moment:

Applied load V = 890kN

Eccentricity, e = 50 + 18.5/2 = 59.25mm.

The moment caused by the eccentricity of the group of bolts is:

Msd = 890 * 59.25 * 10-3 = 52.7kNm.

For one shear area per bolt, Msd = kNm35.262

7.52

Forces in the bolts:

The internal forces are distributed to the bolts according to their

distance from the center of gravity of group of bolts.

Σr2 = y2 + z2

= 2(602 + 1202 + 1802 + 2402)

= 2.16 * 105 mm2

Horizontal force: (at far end bolt)

Msdr / Σr2 = (52.7*106*240) / (2.16*105)

= 5.86 * 104 N = 58.6kN.

For one shear area per bolt Hsd = kN3.292

6.58

Similarly, vertical force: Vsd = kN44.499*2

890

Resultant force Fsd = 22 44.493.29 = 57.50kN. Ok

Shear capacity of bolts (per one shear area)

10

!35.11600.13525.1

1030364087.0

!50.5735.11625.1

103038006.0

87.06.0

3

3

,

OKkN

OKkN

AfAfAfF

Mb

syb

Mb

sub

svdRdV

Bearing capacity of bolts:

Web thickness of the secondary girder is tw =15.0mm < 2 * 9.0 =

18.0mm. Therefore, the web thickness of the secondary girder is

critical. Max force is on far end bolt equivalent to two shear surfaces.

Fsd = 2*57.5 =115 kN.

Fbb,Rd = dtfbb,d=

mb

ybub ffdt

9.0

=25.1

10*)640800(*9.0*15*22 3=342.1 kN > 115 kN ------OK!

Bearing capacity of the secondary girder:

Fbp,Rd = dtfeffdt

bp

m

ybub,

2

1)(8.01

2

kN1159.14825.1

10)275430(15228.0 3

=25.1

10*)275430(8.0*15*40*

2

1 3=135.4<148.9 but > 115 kN-----OK!

Shear capacity of angles

Two angles are used to connect main girder and secondary girder.

The shear capacity of one angle is

(Av,net (fy/fu) Avholes need not be considered)

Av,net = 560*9 - 9*24*9=3096mm2

Avu

y

f

f= 560*9*

430275 = 3223mm2

An effective area may be assumed. Thus,

Av,eff=Av,net y

u

f

f= 3096*

275430 =4841mm2 <5040mm2=Av.

11

The resistance to shear can be determined as

Vp,Rd = mo

vA

3

yf=

3*4.1

275*4841 *10-3 = 698.74 kN > 2

890 = 445 kN.

Capacity of secondary girder at section A-A

The section is class 3.

Designs shear force,

Vsd = 890 kN

Design moment (due to eccentricity of group of bolts).

Msd = 890* (115+18.5/2) *10-3 = 146.2 KNm

cross. Sectional properties at section A.A

AA = 300* 28+712*15= 19080mm2

Centroid: ZI = 19080

14*28*300)282

712(*15*712 = 221.1mm

IZI = 12712*15 3

+ 712*15*(740 - 221.1 - 2

712 )2 + 12

28*300 3

+ 300 *

28 * (221.1 -14)2

= 1095417 *103 mm4

WZI, el = )1.221740(

10*1095417 3

= 2111.0* 103 mm3.

AZI,V =712*15 = 10680mm2

The resistance to bending:

Mel,Rd = mo

yelZI fW

, =

1.1275*10*0.2111 3

*10-6

207.1 + 14 = 221.1

Centroid ZI

28

300

A

155

A

740

15

12

= 527.76 kNm > 146.2 kNm OK!

The resistance to shear:

0.5 VRd = 0.5 Mo

yv fA

3= 0.5 * 310*

1.1*3

275*10680 = 770.76kN < 890kN

Hence, interaction of bending and shear is required:

Mel,v,Rd = Mv,Rd = mo

yf

w

v

elZl t

AW

4

2

,

= (2Vsd/ VRd - 1)2 = ( 2)176.770*2

890*2 = 0.024

Mel,v,Rd = kNmkNm 2.1463.5161015*410680*024.010*0.2111

1.1275 6

23

Shear rupture resistance:

The design shear ruptures resistance:

Veff,Rd = mo

effvy Af

,6.0

Av,eff = t (Lv + L1 +L2 - ndo)

Lv = 8x60 = 480mm

L1 = 2.5do a1 = 2.5 * 24 = 60mm < 130mm

L2 = 5do a2 = 5*24 = 120 mm < 130mm

Aeff = 15* (480+ 60 + 120 – 9*24 ) = 66.6* 102mm2

Veff,Rd = 1.1

10*6.66*275*6.0 2

* 10-3 = 999.0 kN > 890 kN OK

155

130

8060

130

41

13

Theoretical background of Eccentrically loaded

bolted connections

There are six bolts shown in the joint. P is the eccentrically applied

force at a distance e from the centroid of the bolt group. The forces

in any bolt can be due to the direct action of the applied force P and

due to the moment due to eccentricity.

Direct force in one bolt F1 = Force P / No. of bolts = P/n

Due to moments each bolts get torsional shear. Forces due to

moment in any bolt act perpendicular to radius vector ‘r’.

Shear stress (f2) due to moment in a bolt is proportional to the

radius vector ‘r’

(f2) α r or (f2) = kr

Force due to moment in one bolt = F2 = A.(f2) = Akr

Moment of resistance of one bolt = F2. r = Akr2

Total Moment of resistance of one bolt = Σ Akr2 = Pe = M

k = 2Ar

Pe F2 =

2Ar

PeAr =

2r

Mr

F2 may have two components along vertical and horizontal directions

based on its line of action (ie. F2 cos θ and F2 sin θ, where θ is the

angle between F1 and F2).

Resultant force in the bolt = R = 2

2

2

21 )sin()cos( FFF

P e

P M = Pe

Forces due to Moment Direct Forces

14

Tension and compression zones for the bolts subjected to axial forces

Example 2.5.3

Verify the adequacy of the connection with respect to the capacity of

bolts against the combined action of shear and tension, in the bracket

connection shown in the figure below; bolt grade used is grade 8.8.

Shear force = 462 kN

Out of plane moment = 462(0.2415) = 112 kN m

Shear per bolt = 462/8 = 57.8 kN

Shear capacity of one bolt =

!35.11600.13525.1

1030364087.0

!80.5735.11625.1

103038006.0

87.06.0

3

3

,

OKkN

OKkN

AfAfAfF

Mb

syb

Mb

sub

svdRdV

3 x 100

8 no. of 22 mm

dia bolts - 4 on

either side

T section acting

as bracket

241.5 80 mm

80 mm

462 kN

I section acting

as Column

All dimensions are in mm

dc

dc/7

Load

Tension

Compression

15

Bearing capacity of one bolt =

Fbb,Rd = dtfbb,d=

mb

ybub ffdt

9.0

=25.1

10*)640800(*9.0*10*22 3=228.09 kN > 57.8 kN ----OK!

dc/7 = 460/7 = 65.7 = 66 mm

Ft,Rd = ft,d As

Tension capacity = Ft,Rd = ft,d As

Where, ft,d = design tension strength

ft,d = M

yb

M

ubf

butf

0.17.0

As = tensile stress area.

okf

f

M

yb

M

ub

44851225.1

640*0.10.1

44825.1

800*7.07.0

Ft,Rd = ft,d As = 448 * 303 * 1000 = 135.74 kN > 111.6 kN

--------- Safe

Combined shear and Tension

4.1,

,

,

,

Rdt

sdt

Rdv

sdv

F

F

F

F

Ie. Ok 4.132.174.135

6.111

35.116

8.57

66 mm

460 mm

Maximum tensile force will be in the top bolt; Radius vector to the first bolt = (300+(80 – 66))

= 314 mm

Ft = 2r

Mr

=

kN6.1112*]31421411414[

314*1000*1122222

;

16

Welded connections

The particular advantage of welding is that it produces a more rigid

connection than can be obtained by the use of bolts. Welding does

require, however, greater skill and more supervision, and is therefore

more expensive.

Welds shall generally be classified as: (EBCS.3)

a) Fillet welds.

b) Butt welds ( with full or partial penetration)

c) Slot welds.

d) Plug welds.

e) Flare groove welds.

Table 2.5.3 common types of welded Joints.

Types of

weld

Types of joint

But joint Tee-butt joint Lap joint

Fillet weld

Slot weld

Full

penetration

butt weld*

Double J

Single J

Double u

hole

Single bevel

Double bevel

Double v

Single v

Single u

17

Partial

penetration

butt weld.

Plug weld

Fillet welds

Butt welds are used to lengthen plates in the end-on position, and are

not often used in structural connections because plate sizes in the

connections are relatively small. The fillet weld most commonly used

in structural connections has equal leg lengths and has a flat or

convex face. The maximum size of fillet weld from a single run metal

arc manual process is 80mm, but the maximum preferred to

guarantee quality is 6mm. fillet welds are generally continuous,

although they may be intermittent provided that they satisfy code

requirements.

Provision of EBCS-3

Fillet welds may be used for connecting parts where the fusion

faces form an angle of between 600 and 1200

Fillet welds terminating at the ends or sides of parts should be

returned continuously around the corners for a distance of not less

than twice the leg length s of the weld unless access or the

configuration renders this impracticable. This detail is particularly

important for fillet welds on the tension side of parts carrying a

bending load.

In lap joint the minimum lap shall be not less than 4t where t is

the thickness of the thinner part joined. Single fillet welds should

Double v

Double bevel

Double u

18

only be used where the parts are restrained to prevent opening of

the joint.

Fillet welds may be continuous or intermittent.

Design of a fillet weld

Effective Length

The effective length of a fillet weld shall be taken as the overall

length less one leg width s for each end which does not continue

at least twice the leg width s round a corner. The effective length

should not be lese than 40mm or 6 times the throat thickness.

Provided that the weld is a full size throughout one, no reduction

in effective length need be made for either the start or the

termination of the wild.

Throat thickness

The effective throat size a of a fillet weld shall be taken as the

perpendicular distance from the root of the weld to a straight line

joining the fusion faces which lies within the cross section of the

weld . It should not, however, be taken greater than 0.707 times

the effective leg width s.

The throat thickeners of a fillet weld should not be less than 3mm.

Long Joints

In lap joints longer than 150a the design resistances of a fillet

weld should be reduced by a reduction factor Lw given by,

Lw,1 = 1.2 – 0.2Lj / (150a) but Lw,1 1.0

45°

45°

s

s

a

h h = sssss 414.122 222

a = ½ (h) = ½ (1.414 s) = 0.707 s

19

Where Lj = overall length of the lap in the direction of the force

transfer.

For fillet welds longer than 1.7 meters connecting transverse

stiffeners in plated members, the reduction factor Lw may be

taken as:

Lw,2 = 1.1 - Lw/ 17 but 0.6 Lw,2 1.0

Where Lw = length of the weld in meter.

Example 1:

A connection shown in the figure below is subjected to a design

tensile force of 168kN. All data regarding the member and

connection are shown in figure. The steel grade is Fe 360. Check

that the connection is adequate.

50

50

3

55

t = 10

5

50

3

3

2L 50 * 5

55

55

50

20

Geometry, material and loading:

Plate: t = 10 mm Fe 360 fy = 235 MPa

Angle: 2L 50 *5 Fe 360 fy = 235 MPa

Throat thickness: a = 3 mm

Length of weld l = 4 * 55 + 2 * 50 = 320 mm

Applied tensile force = Nsd = 168 kN

Solution:

According to section 6-5-10, in angles connected by one leg, the

eccentricity of welded lap joint connection may be allowed for by

adopting an effective cross sectional area and then treating the

member as concentrically loaded.

For an equal angle or unequal angle connected by its larger leg, the

effective area may be taken as equal to the gross area.

Welds: The minimum weld length as per 6-5-5-1: 40 mm or 6(a)

6 (a) = 6 * 3 = 18 mm

Weld lengths provided are 50 mm and 55 mm > 40 mm –ok

As per section 6-5-5-3, check for long joints:

150 (a) = 150 * 3 = 450 mm

The length of the weld in our problem = 55 mm <450 mm

Hence, no reduction required for the design resistance.

The design strength per unit length of the weld

Fw,Rd = fvw,d (a)

fvw,d = Mw

uf

*65.0 = MPa2.187

25.1

360*65.0

Fw,Rd = 187.2 * 3 * 10-3 = 0.561 kN per Unit length

Design strength based on entire length of the weld

= 0.561 * 320 mm = 179.7 kN > 168 kN - Hence Safe

Check for tearing out:

As per the section 4-6-1-2 and 4-4, the design tension resistance

capacity is NRd =

kNkNf

Af

AM

y

t

M

y

v 1685.24210*}1.1

23510*50

)1.1(3

235*10*55*2

32{ 3

00

21

Therefore, the connection is adequate to resist the design tensile

load.

Example 2:

A tie bar consisting of a single angle 60mm * 60mm * 10mm is to be

welded to a gusset plate. The tie bar carries a load of 150 kN axially.

Design the joint if both the side fillets and end fillet has to be

provided. Let the material of the angle be Fe 430.

Solution:

For Fe 430, fu = 430 MPa

Design strength of weld fvw,d = Mw

uf

*65.0 = MPa6.223

25.1

430*65.0

Let the leg length of the weld be assumed to be = 10mm,

Then, throat thickness = 0.707 * 10 = 7 mm

Strength of the weld per mm length = 7 * 223.6 = 1565.2 N

Maximum length of end fillet = 60 mm

Strength of end fillet = 1565.2 * 60 * 310 = 93.912 kN

Strength to be taken by the side fillets = 150 – 93.912 = 56.088kN

Strength to be taken by one side fillet = 56.088/2 = 28.044 kN

Length of one side fillet = 28044/1565.2 = 17.92 mm = 18 mm

Effective length of the side fillets = 18 mm

Overall length of the side fillets = 18 + 1 leg length on either end

= 18 + 10 + 10 = 38 mm

Also effective length should not be less than 40 mm or 6 (a)

6 (throat) = 6 * 7 = 42

Overall length of the side fillet = 42 mm + 10 + 10 = 62 mm

22

Determine the design resistance of two 15 mm thick and 120mm

wide flats of Fe430 steel which are butt welded. Weld type is full

penetration butt weld.

Yield strength of Fe430: fy = 275 MPa

Design strength = fy/1.1 = 250 MPa

Weld area = 15 * 120 = 1800 mm2

Design resistance = area * design strength = 1800 * 250 *10-3

= 450 kN