1

5. Design of welds in connections

5.1 Introduction

Clause 13.13 in SANS 10162-1 controls the design of welded joints. The fact that this

clause is so short is an indication of the relative simplicity of the design of welded

connections as compared to bolted ones.

We assume in what follows that ‘matching electrodes’ will be used for the welds,

meaning electrodes with an ultimate tensile strength at least equal to that of the steel

(‘parent metal’). If the code for the electrode starts with D7, E7 or F7 it can be

regarded as ‘matching’ for the structural steels commonly used in South Africa,

including Grade S355.

5.2 Complete and partial joint penetration groove welds

Figure 5.1 (a) shows a complete joint penetration groove (CPJP) weld and (b) a partial

joint penetration groove (PJPG) weld.

For a weld loaded in shear as shown in Figure 5.1 (c) Clause 13.13.2.1 states that the

resistance rV will be the smallest of:

umwr fAV 67,0 (5.1)

uwwr xAV 67,0 (5.2)

where w = 0,67

uf = tensile strength of steel

ux = tensile strength of weld metal

wA = area of the effective weld throat

mA = shear area of the effective fusion face

2

Figure 5.1 – Complete and partial joint penetration welds

For a PJPG weld the following will apply:

1baAw (5.3)

2baAm (5.4)

Defining mA for a CJPG weld is more complex, but if uu fx we can simply say that for

a CJPG weld the shear resistance is given by rV :

uwr tbfV 67,0 (5.5)

while for a PJPG weld:

uwr bfaV 167,0 (5.6)

For a weld loaded in tension as shown in Figure 5.1 (d), Clause 13.13.3 states effectively

that a CJPG weld should be treated as though the base metal is not interrupted by the

weld. For a PJPG weld the following applies:

ygunwr fAfAT (5.7)

where baAn 1

tbAg

3

A joint subject to combined tension and shear can be regarded as satisfactory if

ru TT and ru VV .

5.3 Fillet welds

Fillet welds are the most popular welds in structural engineering. As shown in Figure 5.2,

the load on a fillet weld can be either in the longitudinal direction of the weld uvf , or

at right angles to it upf , or at any other angle uf . (We define here uf = force on

weld per unit length, and vrf , rpf and rf the resistances per unit length corresponding

to uvf , upf and uf )

Figure 5.2 – Loads on fillet weld

It has been amply demonstrated that rvrp ff . In fact, the following expression in

Clause 13.13.2.2 defines the resistance per unit length of the weld:

utwwr xaf 67,0 ( 5,01 Sin 5,1 ) (5.8)

where twa =throat dimension of weld.

The rules for determining twa for various shapes of the fillet weld are shown in Figure 5.3.

In South African practice the size of a normal fillet weld as depicted in Figure 5.3 (a) is

given as wa , the leg length of the weld. The throat is then:

wtw aa 7071,0 (5.9)

The is no reason to have the geometry shown in (b) – it’s a waste of material and one

might as well assume that wtw aa 7071,0 . A concave weld as shown in (c) is not what

we strive for, except where this shape is required for aesthetic reasons. The comforting

4

thought is that it is relatively easy to measure the dimension twa directly. The situation in

(d) occurs where the angle between the two steel surfaces is less than 90°, and another

situation occurs where the angle is more than 90°. AWS D1.1 contains extensive

information on the subject and SAISC recommends that the designer and the steelwork

contractor spend time together discussing what can be reliably achieved, how this will

be done and checked, and what the strength will be.

A similar situation occurs when the process used by the fabricator (for example

submerged arc welding) allows him to achieve penetration while doing a fillet weld,

even though no preparation was made, as illustrated in Figure 5.3 (e). Aware of the

fact that it is difficult to have full confidence in the penetration, even with all the

requisite quality assurance, the SAISC recommends the following expression for the

throat thickness:

'8,0 twtwtw aaa

(5.10)

Figure5.3 – Rules for determining throat dimension of a fillet weld

Clause 13.13.2.2 states that, irrespective of the direction of the load, the resistance of a

fillet weld as determined by the base metal is:

uwwr faf 67,0 (5.11)

The object of this expression is to prevent failure in the base metal along the leg of the

weld. It has since been proven that such a limit is not required unless ux is significantly

higher than uf . However, there is a different reason for accepting the limit until such

time as SANS 10162-1 is reviewed again: in a joint with both horizontal and transverse

welds the fact that the transverse ones are stiffer than the longitudinal ones will cause

the force not to be distributed among the welds proportional to their resistances. Thus a

slight degree if conservatism is appropriate.

Table 5.1 near the end of this chapter lists resistances for fillet welds of different sizes

(defined by the leg length wa ) and with a variable direction of force . The table is

based on Equations 5.8 and 5.9

5.4 PJPG weld reinforced by a fillet weld

5

Clause 13.13.3.3 is intended to deal with the situation depicted in Figure 5.4, where a

PJPG weld is reinforced by a fillet weld. The joint is in tension.

(Note wa may be bigger than 1a )

Figure 5.4 – PJPG weld reinforced with a fillet weld

The resistance of this weld is given by:

22

uwunwr xAfAT (5.12)

where, for a unit length:

ww aA 7071,0 (5.13)

1aAn

The similarity with the situation depicted in Figure 5.3 (e) is interesting. Equation 5.10 with

Equation 5.8 yields slightly more conservative results than Equation 5.12.

5.5 Resistance of a weld subject to a combination of forces

Figure 5.5 – Straight welded joint under combined forces

6

Figure 5.5 (a) shows a welded joint under the action of a longitudinal force uV , a

perpendicular force uF and a moment uM , all acting at its centroid. Figure 5.5 (b)

shows a weld with a force uP acting on it along the line AB. This case can be handled

by resolving uP into its components uF and uV acting at C , a distance e away from

the centroid. The problem can then be expressed in exactly the same terms as in (a),

i.e. with the forces acting at the centroid, by letting uu eFM .

If the weld is a CJPG weld we can assume that the stresses in it (force per unit length)

will be the same as in the parent metal and check that it can resist the tensile and

shear stresses at each point along its length.

For a PJPG weld or fillet welds we cannot always assume that the weld or the steel it

connects will be ductile enough to permit redistribution of stresses and it would be

prudent to assume a linear elastic stress distribution for the bending moment uM .

Noting that each of a PJPG and a fillet weld has a different strength depending on the

direction of the applied force, we proceed to discuss the fillet weld situation, on the

assumption that it will be clear how to handle a PJPG weld.

The maximum perpendicular force per unit length of weld is given by:

2

6

w

u

w

uup

MFf

(5.14)

and the longitudinal force per unit length by

w

uuv

Vf

(5.15)

The resultant force uf per unit length is thus:

22

uvupu fff (5.16)

In the case of a fillet weld we also need to know the direction of the resultant force,

which is given by

uv

up

f

f1tan (5.17)

The resistance of the weld can then be obtained from Equation 5.8 or Table 5.1.

If a weld is subjected to forces in three dimensions, i.e. a shear force per unit length uvf ,

a perpendicular force in the z ’ direction upzf and one in the ‘ y ’ direction upyf , the

resultant force is given by:

2

.

22

vuupyupzu ffff (5.18)

And the angle by:

7

uv

upyupz

f

ff 22

1tan (5.19)

These calculations are all rather cumbersome for a simple welded joint and it is

advisable to try to avoid them where possible by making the joint concentric to the

forces. In Figure 5.6 the whole problem with a moment acting on the weld was solved

by working with two ‘virtual’ gussets (the shaded areas), each of which has to resist the

force in the diagonal attached to it.

Figure 5.6 – ‘Virtual’ gussets to simplify calculation

8

9

5.7 Fatigue in welded joints

Metal fatigue can happen in a straight, smooth piece of steel, but only after so many

cycles that it is unlikely to be of interest to the structural engineer. As a rule, a fatigue

crack originates at a sharp discontinuity, and in normal steelwork practice such

continuities typically happen at welds, assuming that the designer, detailer and

fabricator will all avoid sharp corners and notches in other places. No wonder then

that, looking at the sketches in Clause 26 of SANS 10162-1, it is striking how welded joints

predominate.

Our intent here is not to explain the concept of fatigue, but rather to present a step-by-

step procedure for assessing whether a particular detail subjected to a large number of

stress cycles will be satisfactory in accordance with Clause 26.

Step 1 – Find a similar example

Find the number of an example in Figure 2 in Clause 26 that is similar to the detail under

consideration.

Step 2 – Determine the category

Find the number of the example in Column 4 in Table 10, and also read the description

in Column 2 to determine which detail category in Column 3 applies.

Step 3 – Obtain constants

From Table 9, read the values of , srtf , 'nN and

' for the specific detail category.

Step 4 – Determine the stress ranges and numbers of cycles

The stress range srf is the difference between the maximum and minimum stress in the

detail in question during one application of a load, i.e. from before the load is applied

until it is removed again. Note that, according to Clause 26.3.1 we need not worry

about fatigue if the stress in the detail remains compressive during the whole process,

but if the stress varies from compression to tension it is prudent to work with the whole

stress range. Welds are typically in a state of tensile residual stress, consequently

applying compressive stress amounts to changing the tensile stress in the detail.

N is the number of times the load is applied during the life of the structure (i.e. the

number of cycles for the stress range srf ).

If the detail will be subject to various magnitudes of stress range, we can group the

stress ranges of different magnitudes together and assess the number of cycles

corresponding to each stress range. The result will be a number of stress ranges, each

with a range srif and a number of cycles iN .

Step 5 – Assessment for a single stress range

This applies only where we only have only one stress range.

If srtsr ff fatigue can be ignored.

10

Otherwise, srf may not exceed frf , where

333,0

Nf fr

If this is not satisfied, the detail or structure must be changed.

Check the result against Figure 1 in Clause 25.

Step 6 – Assessment for multiple stress ranges

For each stress range i with srtsri ff , calculate fiN from:

3

sri

fif

N

(5.20)

For each stress range i with srtsri ff , calculated fiN from

5

'

sri

fif

N

(5.21)

The detail will not fail in fatigue if:

0,11

n

i fi

i

N

N (5.22)

where n = number stress ranges.

Note that, according to Clause 26.3.6, we need not worry about fatigue if N is less than

either 20 000 or 3

srf

for the particular detail.

5.8 Guidance for designers of welded structures

The designer of a connection is the person who decides how strong a weld must be to

do a particular job. He/she may also other required characteristics, such as the furface

shape, penetration and degree of freedom from defects. The designer should specify

what evidence is required regarding achieving the desired quality standards. An

engineer may specify how the welding should be done, but in most instances the steel

fabricator and his welding engineer will, or at least should be more knowledgeable

about welding processes.

This leads us to suggesting the following procedure:

o The engineer specifies the weld, typically by putting a symbol or note next to a

joint. Figure 5.6 shows examples of a number of basic bits of information for

welded joints.

(a) and (d) are ways of specifying 6 mm fillet welds on both sides of the upstand.

(b) and (e) require CJPG welds, without providing any further information. (c)

and (f) are ways of specifying PJPG welds with an effective thoat of 8 mm.

11

Figure 5.7 – Minimum information on welds

o The steel fabricator and his welding engineer can either accept the instructions

of the engineer on face value – for example that a 6mm fillet weld must be

provided - or they can propose how the same result can be achieved in other

ways (for example by deep penetration SAW). The steelwork contractor has the

responsibility to submit to the engineer all the requisite welding procedures and

a quality plan that will assure the latter that welds that satisfy the design intent

will be achieved consistently and reliably.

o Both the designer and the steelwork contractor have to do quality assurance

during the course of the project to the point where they are both satisfied that

the intent of the designer was achieved.

Practical guidance on all aspects of welding of interest to steel designers can be

obtained from Chapter 4 above. Here we only want to add the following advice:

o Specify the smallest weld size that meets design requirements – trust the theory!

o Except where fatigue controls or where there are other good reasons to avoid

them, use intermittend welds.

12

5.9 Examples

Example 5.1

Check whether the following connection is strong enough:

13,0190xM u = 24,7 kN.m

uV =190 kN

Equation 5.14 (with a weld on each side of the plate):

2

6

2

1

w

u

w

u

up

MFf

250

107,246

250

0

2

1 6xxfup = 1,19 kN/mm

Equations 5.15:

2502

190

x

Vf

w

uuv

= 0,38 kN/mm

Equation 5.16:

2222 38,019,1 uvupu fff =1,25 kN/mm

13

Equation 5.17:

38,0

25,1tantan 11

uv

up

f

f =73°

From Table 5.1 :

rf =1,29 kN/mm for a 6 mm weld > 1,25 kN/mm OK

Check shear in plate:

Clause 13.4.2:

1225066,0.66,0 xxfV yr = 417 kN> 190kN OK

Example 5.2

What is the resistance rT of the following welds?

Equation 5.12:

222 uwunwr xAfAT

127071,07071,0 xaA ww =8,49 mm2/mm

1aAn = 10 mm2/mm

2248049,84701067,02 xxxTr = 8,34 kN/mm.

14

5.10 Resistance tables

Table 5.1 shows the resistances of fillet welds involving Grade S355 steel and E70XX

electrodes. The table is based on Equations 5.8 and 5.9, which assumes a shape of fillet

assumes a shape of fillet weld as depicted in Figure 5.3 (a). For welds with other cross

sections twa can be determined as discussed in 5.3 above.

The basis of Tables 5.2 to 5.9 is discussed in 5.6 above. The use of the tables can be

explained in the following steps:

i) Choose the weld group with the dimension to be called .

ii) Identify the dimension to be called b and calculate /b .

iii) Where applicable, read the value of /x depending on /b and calculate x .

iv) Determine the value of e and calculate /e .

v) With /b and /e known, read the value of C from the table. Interpolate where

necessary.

vi) Calculate the required leg length wa of the fillet weld:

C

Pa u

w (mm)

15

Table 5.1 – Factored shear resistance of fillet welds

16

17

18

19

20