◄
12.91
Design of welds
Welding is a process where a metal joint is made by fusing (melting ) the
components and filling in with molten metal from an appropr iate g~ade of
electrod e. Welding produces neat, strong and efficien t joints and with the
majority of applicat ions in connecti on design is superior to bolting.
In general engineer ing, electric arc ~elding is used, either manual or automatic.
This text consider s only manual welding using either E41XX electrod es (where
the ultimate tensile strength , fuw = 410 MPa) or E48XX electrod es, fuw =
480 MPa. Welding electrod es should preferab ly be of higher strength grade
than the parent metal, i.e., E48XX electrod es should be used with Grade 250
steel.
Th~ weld quality shall be either SP (special purpose) or GP (general purpose) .
When welds are subject to fatigue loading refer to Clause 11.1.5, AS4100.
The inspectio n requirem ents for welds are set down in AS 1554.1 and AS 1554.5.
The inspectio n procedur e on GP welds is the least stringen t.
Category SP(¢= 0.8) should be specifie d for statical ly loaded welds where
the design load on the weld is greater than 75% of the design capacity of
weld. Otherwise Category GP can be specifie d (for which¢= 0.6). Category
GP can also be nominated for welding other than structur al.
The two main types of welds
used are the butt (Fig. 86(a))
and fillet (Fig. 86(b)). Both
weld types have a number of
variation s.
I
7r
Butt welds ... Clause 9.7.2
Welds that have complete fusion
of the base metal and weld metal
throughout the joint are called
complete penetrat ion butt welds.
Where butt welds do not have the
above descripti on they are called
incomplete penetrati on butt welds.
Complete penetrati on butt weld
( a)
( b)
FIG. 86
When a complete penetrati on butt weld meets the requirem ents set down in
Claus: 9.7.2, AS4100, the design capacity of the weld equals the nominal
capacit~ of the weaker of the elements being joined multiplie d b the
appropri ate~ factor from Table No. 2.
y
Incomplet e penetrati on butt weld
~ese we:ds are designed as fillet welds having
a throat thickness as determined using Clause 9. 7. 2. 3' AS 4100.
Fillet welds
Clause 9.7.3, AS4100
The failure mode fore ual 1 f
.
plane) which is term dqth deg _illet welds is in
the minimum area (middle
of a fillet weld is e
_ef ' esign throat thickness of a weld, tt, The size
1eg 1ength tw,
speci .ied on shop detail d i
. reference
see Fig.
raw ngs with
to the
87
Clause 9. 7. 3' AS 4100 deal
.
maximum and minimum ~eld ~ with pref erred sizes of fillet welds· also
plates being joined.
sizes with r eference to the thickness es ~f the
12.92
I. .I
tw
Equal leg fillet weld
FIG. 87
FIG. 88
Stress distribu tion in fillet welds is very complex and design can only be
carried out if some simplifi cations are introduc ed. These are as follows:
(i) Weld strengt h~ weld size.
(ii) Weld strengt h~ weld metal strength .
(iii) Longitu dinal direct forces on a fillet weld are neglecte d as they do not
act on the failure plane.
(iv) Residua l stresses are ignored.
(v) Longitu dinal and transver se welds have the same strength .
(vi) Stress in a fillet weld can be consider ed as shear stress on the throat
of the weld regardle ss of the directio n of applied load. Shear st~~sse s
acting in differen t directio ns can be added by vectors.
Strength limit state for fillet weld
Clause 9.7.3.10 , AS4100
A fillet weld subject to a design force per unit length of weld (vt) shall
satisfy -
* <Pvw
vw:;;
where
~ =
0.8 for Category SP welds
= 0.6 for Category GP welqs
Vw
fuw
tt
kr
Table No. 2
Table No. 2
= 0.6fuwt tkr
= nominal tensile strength of weld metal
see Fig. 87
= design throat thicknes s
= a reductio n factor given in Table 9.7.3.10 (2), AS4100, to account
for the length of a welded lap connecti on (Lw), see Fig. 88. If
Lw :;; 1700 mm, kr = 1.00, otherwis e refer to the above mentione d
table. The kr factor allows for the non-unif ormity of stress
distribu tion in long lap welds. For all other connecti on types
kr = 1.0.
Line method of fillet weld design
This method represen ts a convenie nt approach to the design of both simple and
complex weld groups.
The limiting shear stress for E48XX electrod es is 0.6fuw = 0.6 x 480 = 288 MPa.
The allowab le paralle l load per mm length of weld in a statical ly loaded fillet
weld is -
Q*
= ¢288tt
= cp X 288 X 0.707 X tw
= ¢200tw N/mm
12.93
Summary of formulas used in the line method of weld design
The weld or weld group s ar e trea ted as lines of no t hicknes s .
of welds and weld groups a re · given in Table No. 13.
The properties
Direc t sh ear
* N*
= T
vw
where
v! = design force per unit length of weld (or in Line Method terminology,
'stress' in N/mm)
N* = design force, N
t = total length of fillet weld, mm.
In-plane loading (torsion)
I
Refer to Fig. 89 and Table No. 13.
M*
I
I
X C
'
v* = _z__
w
Jw
where
* = 'stress', N/mm ·(at right angles to line 'c')
Vw
M*z
design moment (or torsion), Nmm
distance from centroid of weld group to
furthermost point, mm
J w = polar line moment of inertia (second
moment) of weld group, mm 3 •
=
C =
FIG. 89
Out-of-plane loading (bending)
Refer to Fig. 90 and Table No. 13.
I
Point A (transverse)
V*
W
I
I
~
~
=-
Zw
where
v: = 'stress', N/mm
Mx* = design moment, Nmm
Zw = llne modulus for bending,
d
rnm2
Point B (transverse)
v: = 0
FIG. 90
Resultant 'stress'
Where both primary and secondary forces are acting (or comb inations of
forces), the weld forces (or stresses) are added by means of vec t or s a nd
the resultant equated to the allowable 'stress' (ref er to p . 12.92) .
Worked example No. 29
Determine the required size of fillet
weld to join the plates shown in Fig . 91.
Use E48XX electr odes. Assume Category
SP welding .
Note: To prevent excessive joint rot at io~
upon load application, the minimum
lap length should be about f i ve t i.me-;
FIG. S;
ci1e thi ckneEs
of :h2 t ~i Ln2r 1la t e .
12.94
TABLE No. 13
Propertie s of weld treated as a line
Outline of welded
Joint
d = depth
b = width
Bending
about horizonta l axis X-X
,----~b _J
d
X-----X
Zw = bd
L__
z
2
2
_4bd+d _d (4b+dl
- 6(2b +di
6
wbottom
top
Iv b1
xjf.l
N
....!
Twisting
J _ I2b + dl3 _ b2(b + dl2
2
=_L_
Y 2b + d
Zw = bd
+
Q
w-
6
(2b+dl
12
Ny
w-
J =(b+2dl _d (b+d)
3(b+dl
bottom
-
3
top
w
J w2
_2bd+ l_d (2b+d-l
- 3(b+dl
3
wbottom
top
t- b-,
d2
Xf-·, ,_:____i
2(b+dl
t
N~7Nx=--
2
z
t- b-,
x-]E}
Zw = bd
+
a2
2
Z
w
z
d.
- :-
-x
~
D
= 2bd +
= 'Tld
w 4
2
g3
2
d
Zw = '!I._
2
12
w
J
2
+ 'll D
2
2
(b+2dl
3
= d (4b + di
w 6(b+d)
J
3
6
3
3
_4bd+ i = 4bd +d
6b + 3d
3
wbottom
top
3
J = (b+2dl _d (b+dl
.""-
2
lb+ 2dl
12
_ (b + d)
z
2
3
z _ 2 bd + d 2 _ d2 12 b + d I
3
2
_ b + 3bd + d
6
w2
3
3
J _ 2b + 6bd + d
w-
6
3
12.95
Answer
N*
100 x
lQ,:_ = 417 N/mrn
v; =T = 2 x120
eld = ~200tw
Design strength of w
P· 12,92
•••
Equating both values ~200tw = 417
0,8
200tw = 417
. tw = 2.6 mm (say 3 mm)
X
Worked example No. 30
ld
. d size of fillet we
Determine the require
. Fi 92 Use
h wn in g.
'
to join the plates so
SP welding
is
E48XX electrodes. Category
required.
Answer
*
N* _ 100 x 10
Vw =
T -
q*
~200tw N/mm
=
3
2 x 25
= 2000
N/mm
P· 12 •92
/
l
Bar 25 Wide
N. = 100 kN
FIG. 92
Equating both values ~200tw = 2000
0,8
200tw = 2000
X
t w = 12.5 mm (say 15 mm)
·
••
Worked example No. 31
Determine the required size of fillet weld to join
the plates shown in Fig. 93. Use E48XX electrodes
and design for Category SP welding.
t
60
Answer
*
Vw
=
N*
T
q*
=
~200tw N/mm
=
150 x 103
4 x 60
=
625 N/mm
p. 12.92
Equating both values ~200tw = 625
0,8
X
200tw
=
625
tw
=
3.9 mm (say 4 mm)
FIG. 93
Worked example No. 32
Determine the required size of fillet weld to join the plates shown in
Fig. 94(a). The applied dead load = 35 000 N. Use E48XX electrodes and
design for Category SP welding.
Answer
This i s an example of in-plane loading.
j
12.96
500
0
Ill
N
Weld line
hsooo
(a)
( b)
N -
FIG. 94
From TableI No. 13
J
w
= (2b
+ d) 3
12
2
b (b
+
2b +
d) 2
d
b = 125 mm
d = 250 mm
By substit ution Jw = 6.022 x 10 6 mm units
From Fig. 94(b) b2
Ny= 2b + d
Substi tution gives Ny= 31.25 mm
2
+ (125 - 31.25) 2 ]½ = 156.25 mm
C = [(250/2 )
Moment (torsio n) being applied to weld group 35 ooo x@5 00 + (125 - 31.25) J
6
= 25.98 x 10 Nmm
M~ =
Primary 'stress ',
35 000 X 1.25 = 87 •5 .N/mm
N*
*
500
vw = T =
Second ary 'stress ',
* ~ X C
Vw =
Jw
______-;;;J
. ts
Cr1·t·1caI po1n
of weld
25 • 98 X 10 6 X 156 • 25 6
6.022 X 10
= 674 N/mm
=
Fig. 95 shows the critica l points
of the weld group and the method
of determ ining the absolu te
'stress '. This foregoi ng procedu re
is in accorda nce with the rigid
plate analys is specifi ed in
Clause 9.8.1.1 , AS4100.
0
u,
N
\
E
E
Resultant
= 730 N/mm
zu,
c-.:
co
E
E
zu,
"
a:)
FIG. 95
12.98
EXERCISES
When answering the following questions allow for Category SP welding using
E48XX. electrodes.
,f,
16.5 Holes
....
0
16 (CS 1030)
I -++-++.-++--16
I
I
~
,P; ' A
L 1 .I
Gr 250 steel
90
+Load
8
10
FIG. 97
28.
,/
FIG. 98
Refer to Fig. 97.
A pipe is slotted and welded to the upper chord of a truss with a continuous 6mm fillet weld. Calculate the maximum permissible dead load which
can be applied to the pipe with respect to weld strength.
Ans. 809/1.25 = 647 kN
29.
Fig. 98 shows an anchor point for a guy. The guy tension from basic wind
speed= 10 kN (Wu= 1.5W). Allow for a wear condition by assuming slight
pin rotation is possible during operation.
(a) For the pin state:
(i) M* and ¢Mp
(ii)
(iii)
v1
vt
and ¢Vf
and ¢Vb (pin)
(iv) ~ and ¢Vb (ply). Assume sheared end.
(b) For the thread, state Ntf and <l>Ntf
(c) Calculate the size of fillet weld required to attach the links to the
support. Neglect the contribution of the welding across the 10mm
dimension of the links.
(d) (i) State N* for each link plate.
(ii) Determine <j>Nt for each plate for yield limit and fracture limit.
Ans. (a)(i) 105Nm; 136.5Nm (ii) 15kN; 49.9kN
(iii) 15kN; 35.8kN (iv) 15kN; 206.6kN
(b) 15 kN; 62.8 kN
(c) 0.24 (theoretically )
(d) 7.5 kN; 11 2 .3 kN and 98.8 kN
Note: The provis i ons of Claus e 9.1 . 4(b), AS4100, have not been considered
in this exercise.