si
WJELDOX an
OXElOSUND
it
HMDOX
1 Read this first
WELDOX - EHS and HS steels in plate structurer
HARDOX - AR steel as protection against abrasion
Extra High-Strength (EHS) and Abrasion Resistant (AR) steels
are the steel industry's answer to the demands of markets,and
designers for steels that permit lighter, more durable and more
wear -resistant structures.
.
In order to permit lighter and more durable structures, the
steels, and especially the EHS steels, must have a high yield
stress.
Despite their great hardness, they permit satisfactory production economy. It is, for example, possible to drill in the AR steels
- a small number of holes with high-speed steel drills and larger
numbers with special high-speed drills (20% more expensive).
Yield stress, N/mm2
MS (Mild Steels)
e.g. S 235, BS 40, St 37 ................................ 200 - 300
HS (High Strength steels)
e.g. S 355, ASTM A 572-50, BS 50, St 52-3.... 300 - 500
EHS (Extra High Strength steels)
e.g. WELDOX 600, WELDOX 700,
WELDOX 900, WELDOX 960 ........................... 600 - 960
Demands for abrasion resistance have led to the development
of the AR (Abrasion-Resistant) steels, which are abrasion
resistant steels that can also be used as structural steels.
Brinell hardness HB
MS ................................................................. 100-150
HS .................................................................. 150 - 210
EHS ................................................................ 240 - 320
AR HARDOX 400, HARDOX 500 ........................ 360 - 560
The principal area of application for the EHS and AR steels is
within sectors where a premium is placed on reducing:
- the dead load of the structure
-wear
- the effects of impacts and shocks
- surface damage (by increasing the hardness of the component)
These steels are used very widely today within the transport
sector, on construction vehicles, mobile cranes, forestry machines,
within the process industry (cement and timber processing), in
the mining and minerals industry etc.
The Heavy Plate Division of SVENSKT STAL has sold more
than 700 000 metric tons of EHS and AR plate (1986).
Svenskt SteWs vast and long experience with EHS and AR
steels has been compiled in this design manual. The manual is
primarily intended to serve as a valuable aid to designers and,
hopefully, to encourage creative thinking and unconventional
design approaches.
EHS and AR steels have fundamentally the same analyses as
MS and HS steels and exhibit roughly the same weldability as
these steels. Their high strength and toughness have been
achieved through heat treatment.
The steels are very pure and therefore possess good bending
properties in the cold state.
WELDOX and HARDOX steels actually
involve nothing new in design work.
The usual formulae for stress and strain can be used and additional design data are given in this manual.
The design manual does not cover the entire field of welded
structures, concentrating instead on aspects of special interest
pertaining to the use of EHS and AR steels.
The purpose of the manual is to provide:
- extensive information on EHS and AR steels
- design data
- answers to questions that can arise when first working with
these steels
- a deeper understanding of important subjects such as:
fatigue, buckling, torsion, impact stress, wear etc, in
which instruction provided at present is inadequate
The contents of the manual are not bound to Swedish building codes and standards, in part because the work of codification and standardization is lagging behind engineering practice
and only partially covers EHS steels today, and in part because
EHS and AR steels are primarily used within areas where codes
and standards are not needed. Another reason is that the same
design data should be applicable also in export markets. Wherever portions of standards have been used, they have been
adapted to international practice and standards such as ISO,
ECCSand IIW.
This handbook has been produced at the initiative of the
Heavy Plate Division of Svenskt Stal in Oxel6sund, Sweden, in
cooperation with Dr. Arne Johnson Ingenj6rsbyra AB in Stockholm and SIKOB in Stockholm.
The principal author and chief editor of the d~sign manual has
been Lennart Bergqvist, M. Eng.
After this Design Manual was published in 1981,
developments in the production of heavy plate have forged
ahead, and so has effective utilization of this product in the
engineering industry. Steels with yield strengths of up to 1100
N/mm2 have been developed in Oxelosund, and they have also
found appropriate applications.
The WELDOX family has been expanded by the iclusion of
steels in the "intermediate range" with yield strengths of 420 •
500 N/mm2. The development of these steels has been ·guided
by the groving demands of the engineering industry for steels
that are favourable in production. Weldability and bendability
have been assigned priority in the development work.
HARDOX steels have developed further within the existing
framework, but are now being produced to closer tolerances on
hardness, etc.
This 4th edition has been relatively extensively revised by the
inclusion of new steels.
Claes Lowgren, M.Sc., the head of our Applications Development Group, has been responsible for the revision work.
We wish to thank those who have submitted valuable views on
various occasions concerning the contents of the manual.
These have led to certain amendments and additions.
Oxelosund, June 1991
SSAB OxeliSsund AB
, .,
Key to symbols used in this manual
As a rule, the SI system of units is used and N/mm2 is used for
mechanical stress.
A
area of cross section
R
stress ratio
a
. throat thickness for weld, crack length, length
s
standard deviation
dimension
critical crack size
Sf
T
safety factor
ac
b
temperature, time, transverse (shear) force
width
t
plate thickness
c
coefficient, length dimension
U
potential
Cv
impact strength
u
displacement
D
Palmgren-Miner cumulative damage sum, plate
V
voltage
stiffness
v
velocity
d
diameter
W
overall width
e
eccentricity, impact coefficient, shear centre
Wv
section modulus in torsion
ex, ey
distance from the centroid of a cross section to
an actual section
Ww
w
warping resistance
displacement
E
Young's modulus (modulus of elasticity), energy
x
position, displacement
F
fu nction of .....
y
position, displacement
G
shear modulus
z
position, displacement, utilization factor
H
horizontal force
a
angle, parameter for determination of permissible
h
height, depth (of section)
stress=~
radius of inertia
(u su = ReL )
uel
I
second moment of area, current
fJ
angle, parameter
J
J integral
cp
angle, function of a
K
stress intensity
~,o
deflection. COD value
kB
material factor
f
strain
Kc
Kf
fracture toughness
i:
strain rate
fatigue factor (fatigue strength reduction factor)
fA
parameter
kH
factor for Hertz surface pressure
A
d/W"lug"
kR
stress alternation factor
v
Poisson's ratio = 0.3
density, radius
ka ky.
buckling coefficients
(!
Kt
stress concentration factor
Us
upper yield point = ReH
Kv
section factor with respect to torsional stiffness
uO.2
proof stress (proportional limit)
Kw
section factor with respect to warping stiffness
UB
ultimate tensile strength = Rm
Kx
joint factor
ur
stress range
m
mass, mean value
u ru
endurance limit
stress intensity coefficient for weld
ua
stress amplitude
exponent in Norton's creep law
um
mean stress
N
normal force, load cycles
Uw
warping normal stress
Nd
nj
design load cycle number
ueL
Euler critical stress
load cycles, number of
uH
Hertz surface pressure
Nk
P
buckling load
Uj
initial stress
point force
Uj
equivalent stress
p
spectrum parameter, gas or liquid pressure, bearing
uk
critical stress
stress
up
necessary stress amplitude for propagation
shear stress due to torsion
Mk
n
Pm
maximum plastic deformation
Tv
Q
total load
Tw
shear stress due to warping
q
load per unit length
cp
angle
QB
probability of failure
!p
angle
radius
w
angle
1:2
2 What are WELDOX and
HARDOX steels?
Quenched and tempered steels.. ........ 2.1 •
Direct-cooled steels ........................... 2.2
Classification of the steels ................ ,. 2.3
Chemical analysis, strength values ...... 2.4
Comparison with ordinary steels .: ....... 2.5
Lean chemical composition for
excellent weldability ......................... 2.5.1
Strength ......................................... 2.5.2
Toughness ...................................... 2.5.3
Machinability .. ' ................................ 2.5.4
Economics ...................................... 2.5.5
2 What are WELDOX and HARDOX?
WELOOX - 'quenched and tempered' or 'direct-cooled' structural
steels.
HARDOX - 'quenched' or 'quenched and tempered" abrasionresistant steels.
2.1 Quenched and tempered steels
WELDOX quenched and tempered structural steels have very high
yield strengths and are known as Extra High Strength (EHSI'
steels. HARDOX is an Abrasion-Resistant (AA) steel.
High-strength steels are not difficult to produce. But refined
technology is needed if EHS and AR steels are to be made tough,
veldable and bendable. Modern EHS steels such as WELDOX and
AA steels such as HAADOX are imparted this combination of
strength and toughness by metallurgical purification and harde·
ning in a continuous roller-quenching plant.
The 'grade steps' shown in Figure 2.1 can be used to compare
these quenched and tempered steels with conventional structural
steels of lower strength.
2.1.1 Brief history - quenched and tempered steels
Quenched and tempered steel plate is nothing new. Armour plate
has been made in this way for decades. These steels had high
contents of carbon and alloying elements and were very difficult
to weld. It was not until the early 1950s that US Steel launched a
quenched and tempered steel with better weld ability. This was
known as T1 steel. The company then developed its static
quenching presses (see Figure 2.2a) and achieved better cooling
capacity.
Figure 2.2a
Figure 2.1
ReH
Nlmm'
,.....
)
I~'----------------.-------------
Static quenching plant
New research findings revealed that extremely small quantities
(0.002%) of boron (B) produced considerably increased
hardenability. The contents of carbon and alloying elements could
thus be greatly reduced, and steels with much better weldability
could be produced by the addition of boron.
The real "breakthrough came in the mid-1960s when Bethlehem
Steel, together with the Drever & Co. consultancy company,
developed a continuous quenching process in a 'roller quench"
plant. The principle is shown in Figure 2.2b.
5OO.~-----------------
Figure 2.2b
High-pressure zone
The classification of steels into EHS and AR steels conforms to
the practice that has long been accepted on the" market. In the
past, abrasion-resistant plate was regarded as plate intended for
chutes, troughs, buckets, etc. for which hardness was the only
requirement.
Modern abrasion·resistant plate has been developed so that
guaranteed high hardness can be combined with guaranteed
toughness and strength. In addition, most of these hard steels
are readily bendable in the cold state.
The boundary between structural steel and abrasion-resistant
steel plate is therefore becoming increasingly diffuse. Modern
abrasion-resistant steel plate should therefore be regarded as
structural steel plate which is also highly resistant to abrasive
wear.
The high strengths are put to use today for:
- designing structures with low deadweight
- designing more wear-resistant structures
- lowering the total cost by shortening the welding times
The mechanical properties of these steels are achieved principally by heating to approx. 900°C and then quenching to room
temperature (hardening). This determines the maximum yield
strength and hardness. During subsequent heating to 400 700°C (tempering), the yield strength and hardness gradually
decline to a predetermined level, but the toughness simultaneously increases.
Low-pressure zone
Continuous roller quench plant
The hot plate is quenched continuously as it is fed out of the
furnace."
.
In the roller quench plant, the plate is quenched directly and at
a cooling capacity which is far higher than that of the static quenching press.
In 1969, a roller quench plant was built in Oxel6sund with the
assistance of Drever, and production was in full swing in 1970.
Since that date, more than one million tonnes of EHS and AR
plate have been produced in Oxel6sund.
2:1
What are WELDOX and HARDOX steels?
2.2 Direct-cooled steels
A direct-cooling plant was completed in 1988. This is used for
cooling plate directly after rolling to achieve yield strength levels
between 420 and 500 MPa. The entire strength range between
high-strength, micro-alloyed, grainrefined steels such as S 355
(BS 50, St 52-3) and the quenched and tempered EHS steels are
covered by the directcooled WELDOX 420 and WELDOX 500
steels (see Figure 2.3).
2.3 Classification of the steels
Structural steels can be classified according to strength into
MS, HS and EHS steels. The strength is achieved by varying the
fundamental chemical composition and by employing different
manufacturing processes.
Yield stress, N/mm2
Figure 2.3
HS
R.H
H/mnl
(MP.)
1000
-
-_.--
MS (Mild Steels)
e.g. S 235 (BS 40, St 37)......................................
200 - 300
HS (High Strength) steels
e.g. S 355 (BS 50, St 52-3), WELDOX 420,
WELDOX 500 .. ........ ................... .......... ...... ..... .....
300 - 500
EHS (Extra High Strength) steels
e.g. WELDOX 600, WELDOX 700,
WELDOX 900, WELDOX 960 ................... ....... ........ 600 - 960
150
Modern AR steels are sometimes used as structural steels.
AR steels are classified principally according to hardness. These
steels also have a very high yield strength.
A controlled rolling process produces high toughness due to
the fine-grained structure achieved in this way. The strength is
achieved by cooling the steel directly after rolling (see Fig. 2.4).
The cooling procedure can be controlled within a wide range. The
new WELDOX 420 and WELDOX 500 steels have av microstructure which differs from that of the quenched and tempered steels
which have a martensitic hardened structure. WELDOX 420 and
WELDOX 500 are mainly ferritic and are thus related to the normalized, grain-refined steels.
Figure 2.4
Direct cooling
2:2
Hardness,
Brinell
MS ..................................... 100-150
HS ..................................... 150 - 210
EHS .................................... 240 - 320
AR HARDOX 400, HARDOX 500 ............... 360 - 560
What are WELD OX and HARDOX steels?
2.4 Chemical analysis
Table 2.b
Chemical analysis - typical values for WElDOX structural steel plate
NI
B
Mo
I
CE
Thickness mm
C
Si
Mn
WELDOX420
8 -50
(50)-80
.13
.14
.30
.30
14
14
.36
.38
WELDOX460
8-30
(30)-50
(50)-SO
.09
.09
.25
.15
.43
14
1.55
1.4
35
.36
41
8-50
(50)-80
.09
.15
.25
.43
1.55
1.4
37
41
WELOOX 600
6 -25
.13
45
1.4
WELOOX 700
6 -12
(12)-20
(20)-45
(45)-SO
.13
.15
.16
.17
.45
.45
.22
.22
lA
lA
lA
lA
.60
.25
.25
.50
WELOOX 900
6-60
1.4
.25
.50
6 -13
(13)-25
.17
.17
.22
WELOOX 960
.22
.22
1.4
1.2
.25
A5
1.0
.50
.50
Cr
NI
Mo
B
CE"
1.4
-
1.3
-
-
.9
.60
.25
.60
.002
.002
.002
.002
.37
.50
.56
.62
1.2
1.2
.9
.60
.50
.60
.002
.002
.002
.57
.59
.66
WELDOX 500
.17
.25
Cr
.10
.002
39
.002
.002
.002
.002
.37
41
.56
.56
n
i
\
.002.56
S6
.002
.64
Table 2.1b
Chemical analysis - typical values for HARDOX steel plate
-Thickness mm
C
Si
6-20
(20)-30
(32)-51
(51)-SO
.13
.16
.17
.24
.45
.32
.22
.25
6 -20
(20)-50
(50)-SO
.25
.25
.28
.50
.50
.25
HAROOX400
HAROOX 500
.) CE = C
Mn
lA
.50
.50
.35
-
-
-
.20
.35
.50
+ ~ + Cr + Mo + V + Ni + Cu
6
5·
15
Table 2.2a
WELDOX structural steel plate
Thickness
mm
ReH
N/mm'
Rm
N/mm'
A5
%
Impact
toughness
min J by
Hardness
Typical
values HB
-40°C
Bending recommendations
Min. internal radius
----
Perpendicular'
Parallel'
WELDOX420
S-16
(16)-60
(60)-SO
420
3S0
3S0
500- 650
4S0- 650
4S0- 650
19
19
19
40
40
40
190
180
ISO
1,0X!
1,OXt
1,OX!
1,5Xt
1,5XI
1,5Xt
WELOOX460
S-16
(16)-40
(40)-SO
460
440
420
530- 730
530- 720
510- 720
17
17
17
40
40
40
200
200
190
1,0Xt
l,OX!
I,OX!
1,5Xt
1.5Xt
1,5XI
WELDOX500
8-16
(16)-40
(40)-80
500
480
460
570- 720
570- 720
550- 720
16
16
16
40
40
40
210
210
200
I,OX!
l,OX!
1,0Xt
1,5Xt
1,5Xt
1.5XI
WElDOX 600
6-25
600
700- 850
14
40
240
1,5Xt
2,5Xt
WELOOX 700
6 -64
(64)-80
700
630
7S0- 930
690- 930
14
14
40
40
260
250
2Xt
2Xt
3Xt
3Xt
WELDOX900
6-50
(50)-60
900
830
940-1100
850-1100
12
12
40
40
310
300
3Xt
3X!
4Xt
4Xt
WELDOX 960
6 -25
960
980-1150
12
40
320
3Xt
4Xt
Table 2.2b
HARDOX steel plate
Hardness
HB
Re
N/mm'
Rm
N/mm'
A5
Toughness
%
Bending recommendations
Min. internal radiUS
Perpendicular'
Parallel'
HAROOX400
360-440
ca 1050
1250
ca 10%
ca30J, -40'C
3XI
4Xt
HAROOX 500
450-560
ca 1300
1550
caS%
ca 20JI, - 40 'C
-
-
*) Orientation of the bend line in relation to the direction of rolling.
2:3
What are WELDOX and HARDOX steels?
2.5 Comparison with ordinary steels
2.5.1 lean chemical composition for excellent weldability
A primary goal at Oxel6sund is that our steel must be readily
weldable, i.e. that it should have the lowest possible content of
alloying elements. On the other hand, a steel that is to be hardened must have a certain amount of alloying elements to ensure
through hardening.
So our philosophy is: Use as lean an analysis as possible and
quench rapidly. This produces a less expensive steel with better
weldability.
Table 2.3 shows an analysis comparison (typical analyses)
between WElDOX 700, HARD OX 400 and S 355 (SS 50), which
demonstrates that WElDOX 700 und HARDOX 400 are ordinary
steels. in terms of composition. In plate thicknesses below
20 mm, these steels therefore offer the same weldability as
ordinary steels.
For heavier plate, preheating of the workpiece is necessary for
welding, in the same way as for ordinary steels (see our welding
brochure).
2.5.2 Strength - High yield strength and high
yield/ultimate tensile strength ration.
WElDOX and HARDOX steels have high yield strengths. It may be
of interest to see the appearance of the tensile test curve as
compared to that for ordinary steels such as S 355 (SS 50,
St 52-3) This comparison is shown in Figure 2.5.
The tensile test curve shows that WElDOX and HARDOX steels
have a high yield strength, a smooth transition to a yield plateau
and low strain hardening. The vurves show that the yield/ultimate
strength ratio is higher and the rupture strain is lower than those
of ordinary steels.
Even though the rupture strain (-,) of WELDOX and HARDOX
steels is lower, the values are perfectly acceptable. This has
been clearly demonstrated by the experience gained from
practical operation of our steels in many demanding applications.
The strength values are shown in Table 2.2. The modulus of
elasticity (E) is the same as that of ordinary steels, i.e. 21· 10·
N/mm2.
Figure 2.5
Cl N/mm'
1000
Table 2.3
Typical analysis at t ~ 20 mm
800
Si
Mn
0.16 0.25
LlO
C
Cr
Mo
B
WELDOX 700 <45 mm
CE
600
S 235
WElDOX 500
0.34
S 355
0.16 0.43 1.36
0.38
WELDOX 500
0.09 0.25 1.55
0.37
WELDOX 700
0.15 0.45 1.40
0.10 0.002 0.41
HARDOX400
0.13 0.45 1.40
0.002 0.37
S 355
400
200
o
10
20
30
E%
"Carbon equivalents" (CE or El are the weightings of certain
alloying elements that have been obtained from experience. The
carbon equivalents describe the effects of various alloying
elements on the ability of the steel to harden during the fast
temperature changes that occur during welding. Various formulae
have been developed for this purpose, although the most common is the one from the International Institute of Welding (IIW):
CE = C + Mn + Cr + Mo + V + lli.±...CJ.!
6
5
15
If the CE is lower than 0.41 %, the steel is considered to be 'very
readily weldable". Our welding brochure includes further
information.
2:4
What are WELDOX and HARDOX steels?
Figure 2.7
2.5.3 Toughness
WELDOX and HARDOX steels are manufactured to very close tolerances on composition and also have very low inclusion
contents. Freedom from inclusions is particularly characteristic of
steels that must have a high guaranteed toughness and of steels
for which we give bending recommendations.
Sulphur (S) readily combines with manganese (Mn) to form
manganese sulphide (MnS). Manganese sulphide is soft during
the rolling process and is rolled out into long streaks, which
results in impaired toughness and bend ability of the steel. We
expect our WELDOX and HARDOX steels to have excellent toughness and bendability. To achieve this, we desulphurize them.
We carry out desulphurizing in our TN (silicon calcium, SiCa) or
our ASEA-SKF (cerium, Ce) plant. In these plants, the sulphur
combines with Ca to form CaS or with Ce to form CeS, and is
removed with the slag, i.e. the sulphur content drops radically.
The CaS or CeS residues that remain in the steel are hard during
rolling and retain their nodular shape. This shape is much more,
favourable from the toughness and bending viewpoints than the
long MnS streaks.
The sulphur contents of the steels for which we guarantee the
toughness and give bending recommendations is around 0.004%.
The bending recommendations for WELDOX and HARDOX
steels are summarized in Table 2.2, and a comparison with ordinary steels is made in Figure 2.6.
Figure 2.6
•
r----+_ _
w _-+-----,~
t
-t
Steel grade
r/t 1.
wit 1.
r/T /1
wit /1
S235
S 355
2.0
2.5
1.0
2.0
3.0
3.0
3.0
7.0
7.5
6.0
7.0
8.5
8.5
8.5
2.5
3.0
1.5
3.0
4.0
4.0
4.0
7.5
8.5
7.5
8.5
10.0
10.0
10.0
WELDOX500
WELDOX 700
WELDOX 900
WELDOX 960
HARDOX400
The steels are also fundamentally grain-refined, although actual
grain refinement takes place by the quenching and tempering process which results in high yield strength and high toughness.
Are hardened steels not brittle?
Materials technology textbooks often state that if a material is
made hard, it will also be brittle. But what this statement does not
take into account is that the hardness of martensites depends on
the carbon content (see Figure 2.7), and that low-carbon martensite is very ductile and lends itself readily to forming. The graph
applies to pure, untempered martensite and shows the hardness
test results for our WELDOX and HARDOX steels. The microstructure of our steels consists of virtually pure martensite.
We can therefore produce abrasion-resistant steels, particularly
HARDOX 400, which has a guaranteed hardness of at least 360
HB combined with a typical impact strength of 30 Joules at
-40°C.
For WELDOX 700, for instance, we can guarantee 40 J at
-60°C, if required.
Brinell
hardness
600
500
400
300
0,10
0,20
0,30
0,40
%C
2.5.4 Machinability
WELDOX and HARDOX steels can be gas-cut in the same way as
ordinary steels. Certain HARDOX steels should be gas-cut at
elevsted workpiece temperatures (150 - 200°C).
In machining operations, WELDOX structural steel plate can
very well be machined with high speed steel tools, although the
cutting data should be adjusted to suit.
High speed steel drills can be used for WELDOX and, on a
limited production scale, also on HARDOX 400. Special Coalloyed
high speed steel drills designated "HSS-E" (about 20% more
expensive than ordinary drills) are recommended for abrasionresistant steels. Cemented carbide tools can be used if the
machine tool is appropriately stable.
Machining costs can readily be maintained at a reasonable
level provided that the right machines, the right tools and the
right cutting data are used.
For more detailed information, please refer to our machining
brochures.
2.S.S Economics
WELDOX and HARDOX steels are more difficult to manufacture
than ordinary steels, and thereby have a greater technology
content. As a result, these steels are more expensive in terms of
price per tonne (see our price list).
.
.
Considering the price/unit of yield stress (Figure 2.8) or pnce/
unit of hardness (Figure 2.9), it will be obvious that WELDOX and
HARDOX steels are profitable provided that the strength and/or
hardness can be exploited.
In most WELDOX and HARDOX steel applications, the deadweight of the structure is extremely important to the end user and
thus also to the competitiveness of the manufacturer. In earthmoving vehicles, for instance, an expenditure of more than
1 GBP/kg can be justified for making the structure lighter.
The same reasoning can be applied to the hardness of
HARDOX steels. In most wear applications, increased hardness
results in reduced wear. A longer useful life reduces the number
of stoppages for replacement and repairs.
It can be said in summary that WELDOX and HARDOX steels
are profitable to use provided that the designer and user can
really exploit their advantages.
.
Some of the techniques for achieving this are explained in thiS
manual.
Some clarifying examples of applications for which WELDOX
and HARDOX steels have been put to use are given below (see
Figure 2.10).
:5
What are WELDOX and HARDOX steels?
Figure 2.8
Figure 2.9
GBP/guaranteed yield stress
GBP/typical Brinell hardness
The steels meet the same
toughness requirements
S 275 (SS 43)
30mm
M
S 355 (SS 50)
22mm
Plate thickness = 12 mm
'<t
WELDOX 7000
()')
!B
1.0
.....
11 mm
6
C'I
1.0
en
()')
!B
1.0
. 1.0
('/')
()')
0
0
0
.....
~
0
....J
IJ.J
~
0
0
'<t
X
0
0
Cl::
:f
0
0
1.0
X
0
0
«Cl::
:::c
2:6
What are WELDOX and HARDOX steels?
figure 2.10
\
Dumptrucks/Tippers
Suitable steel grades: HARD OX 400, WELDOX 700
in body, chassis, bumpers, protection plates, joint lugs.
Mining equipmentjloaders/Buckets
Suitable steel grades: WELDOX 700, HARDOX 400,
HARDOX 500 in lift arms, buckets and skips.
R
Crushers/Bins
Suitable steel gra~~s~~:rR~~~s:OO, HARDOX 500
~;
• • • • • • •EiiI.F~'1dNBI'.1:'lw.·iI.ltl'''lIi_~i
Mobile cranes
Suitable steel grades: WELDOX 700, WELD OX 900
in jibs, outriggers and chassis.
Cyclones/pipelines
Suitable steel grades: HARDOX 400, in certain
IWIIIIIII
cases WElDOX 700. in dosl colleclo". bark and
wood ship cyclones and dredging pipes ..
2:7
3
3 Factors of safety
Risks and probabilities, general.......... 3.1
Failure criteria ................................... 3.1
Safe-Life - Fail-Safe ............................ 3.1
Calculation of probability of failure
and coupling to factors of safety ......... 3.2
Choice of factors of safety .................. 3.2
Factors of safety for WELDOX ............. 3.2
Table of strength results ..................... 3.4
3 Factors of safety
Risks and probabilities, general
"Acceptable risks of failure" used in design
Enlightened people today know that 100% certainty doesn't
exist and cannot be expressed in mathematical terms. We
expose ourselves daily to risks of various kinds.
The concept of "probability of failure" is actually not particularly
associated with WELDOX and HARDOX steels. But because, the
steels are new to many and will often be used in completely new
structures with high stresses where experience may often be
lacking, it may be of benefit to introduce this approach.
Moreover, the market is becoming increasingly aware of
the importance of availability (= the percentage of the time a
product can be used) and will increasingly demand precise
guarantees of function from the manufacturer. Indeed, high
availability will become a very important selling argument (and
has already done so in some industrial sectors).
Below is a breakdown of the influence of various factors on
the availability of a product:
Objects
Product development and design
Improper use
Fabrication and quality control
Other factors
40%
30%
20%
10%
In other words, the design department plays an important role.
One measure of risk is probability of failure, i.e. the probability that a given structure will fail or "break" at a given point in a
given fasion.
In general, the entire system in a structure is included in the
concept of "Probability of failure" and this branch of science is
called "Reliability engineering" and is an interdisciplinary subject. In this Simple treatment, we must limit ourselves to studying a few details.
Readers who wish to find out more are referred to (2), which
contains many interesting references and is easy to read.
Government regulations for building structures often specify
permissible stresses, factors of safety or partial coefficients for
load and bearing capacity. These often stand in direct relation
to the desired probability of failure. Standards, codes and
government agencies often specify loads in the form of load
assumptions.
Some examples of probability of failure or risk of death (from
e.g. reference 1) are given in table 3.1.
Risk during period of service
Car components not vital to safety
Car's steering mechanism, suspension
Military airplanes
Passenger airplanes
Nuclear power plants (vital safety part)
5· 10-2
10-3
10-3
3· 10- 5
10- 10 - lO- 11 /hour
Load-bearing structures (experience and recommendations)
Risk during period
of service
Very limited material damage
Limited material damage
Bodily injuries or very extensive material damage
Probability of extensive bodily injuries
Regulations for Welded Steel Structures
StBK-N2 (fatigue)
10- 2
10- 3
10-4
10- 5
<10- 5
Failure criteria
Under former standards and codes, structures were designed
for the most dangerous load to which they could be expected to
be subjected.
The term "failure" in a wider sense means that the promised
function cannot be achieved.
Some typical failure criteria are, for example, when the structure reaches
a given stress
yield point
maximum deflection
ultimate strength
fracture toughness
greatest permissible crack length
fatigue fractu re
instability (overall or local buckling)
Safe-Life - Fail-Safe
Two principles of design can be worth mentioning in connection
with fatigue:
- Safe-life
- Fail-safe
Table 3.1
Reference levels
Disease, average
Natural disaster,
snakebite, etc
Risk of death
per hour
per year
W-o
10-2
High risk
10- 10
10-0
Negligible risk
Accident statistics
(resulting in death)
Risk
Risk during period
of service
Passenger car
(all causes)
10-6
4· 10-3/4000 hours
Passenger airplane
(all causes)
10-6
40· 10-3/40000 hours
Passenger airplane
(fatigue)
0.4· 10-6
8· 10-3 /20000 hours
The Safe-Life principle requires that the structure be simple
and isostatic. The risk of fatigue failure is kept extremely small
during the intended useful life of the structure. In order to
prolong the useful life of the structure, the component must be
replaced with a new one. If a fatigue crack should nevertheless
occur, it is catastrophic, since it will probably propagate and
lead to fracture before it is discovered. This method can be
suitable for inexpensive, easily replaceable components (for
example vital bolts).
The Fail-Safe principle requires that there be alternative
paths for the loads (hyperstatic structure). The structure is designed in such a manner that fatigue cracks are prevented from
propagating and do not lead to failure. Furthermore, periodic
inspection of the structure must be possible.
In such a structure, the stress level can be kept high and the
material can be better utilized, although not so high that cracks
appear before half of the useful life of the structure has been
reached (rule of thumb).
Before it is repaired, the residual static strength of a unit with
cracks must never be less than the maximum load (max. test
load).
:1
Fadors of sarety
Examples of Safe-Life and Fail-Safe principles:
Figure 3.1
Forged fork
Safe-life
laminated fork
o
Fail-safe
Solution:
m = 797 - 500 = 297
s=
V602 + 23 2 = 64
QB = 1 - 1>( ~ )
<i> ( ~ ) = <i> (
2il ) 1> (4.60)
=
The table 3.5 gives cP (4.60) = 0.9 5 7888
QB = 1 - 0.999997888 = 0.000002112
i.e. the probability of failure is 2· 10- 6
Our "factor of safety" Sf = m2/ml = 1.59.
Factors that influence overall probability
of failure
a.
b.
c.
d.
e.
Material strength (inci. surface treatment)
Fabrication (weld classes etc.)
External loads.
Accuracy of design calculation
Possibility of inspection
a, band c are statistically determined; we cannot analyze band
d here, but rather the individual designer must evaluate these
factors himself in the light of his own fabrication operation (or
that of sub-suppliers) and previous design experience.
Calculation of probability of failure
We shall only consider material strength and external loads
under the assumption that we have very good calculation
accuracy.
We know that loads are statistically distributed, both when it
comes to static loads (loading and unloading) and fatigue loads.
They are often normally distributed with a mean value m and
standard deviation s.
Material strength such as as, a B, fatigue strength etc. is also
statistically distributed.
Figure 3.2 shows the relationship between load, material
strength and probability of failure.
Figure 3.2
Frequency
e.g. tensile stress due to external load
We have made one error in this example. We have not taken
account of the fact that the steels have a guaranteed yield
strength. The steels are tested by means of tensile testing, and
if any plate is found to be below the guaranteed yield strength,
the plate is rejected and never reaches the customer, being
instead replaced in the consignment by a flawless plate.
In other words, the strength distribution of the steel plate in a
consignment to a customer is as shown in figure 3.3, giving an
even lower actual probability of failure.
Table 3.3 shows the scatter in material strength (static data).
What is difficult now is to determine the scatter in load.
Here we can only provide a few simple hints.
From experience, we know that load has greater scatter than
the material strength, and a rough rule of thumb is that the
standard deviation of the load is twice as great as that of the
material strength.
Try therefore to determine the various applications of the
structure and the scatter of the loads in these different applications.
In most cases, there are two load cases that the designer
must take into consideration in his calculations:
extreme load and
fatigue load
We have only dealt with extreme load, i.e. static load. A similar
probability of failure line of reasoning applies for fatigue load.
See the section entitled "Fatigue".
Figure 3.3
Frequency
This risk is the steel manufacturer's headache and never reaches the customer.
Data on our steels (m2, S2) are presented in table 3.3.
The dangerous region represents the probability of failure
(QB) in this case and is calculated as follows:
Choice of factor of safety
m = m2- ml
s=
Vs1 + S2
2
2
The function 1> (~ ) is obtained from table 3.5
Example 3.1
Assume a steel WELD OX 700 t = 10 mm with guaranteed yield
stress 700 N/mm2 and with a mean value m2 = 797 N/mm2 and
s2 = 23 N/mm2. See table 3.3.
The external load causes a tensile stress whose mean value
ml = 500 N/mm2 and s = 60 N/mm2. What is the probability
of failure?
3:2
The purpose of a factor of safety is to provide a safeguard
against an undesirable consequence. Despite the fact that
stresses and strains can be calculated and the steel has a
guaranteed yield strength, great uncertainty exists with respect
to factors such as scatter of loads, fabrication etc. Furthermore,
we cannot take all factors into account in design. In other
words, we do not feel secure about allowing the design stress to
be equal to the yield stress of the material.
In order to provide a margin of safety, we have introduced
factors of safety, which are often based on many years of experience.
There are different factors of safety in different industries, for
example in steel construction in Sweden the factor of safety is
Sf = 1.5 against as, for road vehicles Sf = 2 against as, for
vehicles on poor roads Sf = 3 against as.
In any case, it is understood that a probability of failure is
associated with the factor of safety.
Factors of safety
The factor of safety can be divided into four components:
Kj : Material strength, often determined via a guaranteed value,
i.e. = 1.0.
K2: Fabrication factor, which is dependent upon practice in
production, scope of inspection, inspection requirements
and production control (the right steel in the right place).
The more thorough the inspection, the higher the level to
which the material can be stressed.
Normal workshop practice
0.6
Extensive inspection
0.9
K3: Load factor 1.0 - 1.2 for static load, depending on how well
the conditions'are known.
For fatigue under known conditions
1.5
For fatigue under unknown conditions
2.0
K4 : Calculation accuracy
The accuracy in design calculations is dependent upon the
knowledge, experience and skill of the designer. In the case
of new structures where there is no experience to call. on, or
where measurements cannot be carried out, deviations of
30% are not unusual, i.e. factor = 1.3.
Even where experience is available and conditions are
favourable, the deviation can be 10%, i.e. factor = 1.10.
The factor of safety can be expressed as
It is essential in this connection for the designer to have a
feeling for all these factors. He must, for example, be intimately
familiar with his own fabrication operation.
Example 3.2
Estimate the factor of safety against as for a construction vehicle
that is manufactured in long production runs with small scope
of inspection, with varying load conditions (fatigue) under
known conditions, and with experience from previous manufacture and design calculation.
Solution:
Kj = 1
K3 ,", l.5
Sf == --,,1__.5~·_1._2--= 3.0 i.e. agrees well with previously
1 . 0.6
mentioned rules of thumb.
Factor of safety against what?
In most cases, a margin of safety is desired against bodily
injuries in connection with failures. These failures stem from
plastic deformation (yield), fatigue, buckling, brittle failure etc.
For the most part, we associate the factor of safety with safety
against plastic deformation, and this is right for many cases.
In this context, however, we would like to point out a possible
risk where a non-linear relationship exists between, for example,
moment and stress at a point in the structure, in other words
the factor of safety against an overload moment is not as great
as Sf indicates (see figure 3.4),
Figure 3.4
Moment
Ms-+--------------~
Mperm - + - - - - 7 f '
Factor of safety (static) for WELD OX steels
Of the more than one million tonnes of WELDOX and HARDOX
steel plate supplied thus far by Svenskt Stal, approximately 50%
is WELDOX. Most is used in highly stressed and advanced
structures that are not subject to national standards or
regulations. It has therefore been up to each fabricator to select
this own factors of safety in accordance with the principles
outlined above. As far as we know, WELDOX steel has not been
treated differently from HS (S 355, SS 50) in static design
calculations. We have not received any reports of failures with
WELDOX steel due to the fact that the same factors of safety
have been used as for ordinary steels.
Nevertheless, we often encounter doubt on the part of future
users of WELDOX steel, and especially on the part of national
authorities, when it comes to factors of safety.
Doubt is understandable, since the steels often have new and
different characteristics - higher stress, higher yield/ultimate
strength ratio and lower elongation - compared to ordinary
steels, and since the experience of the indiViduals concerned IS
based on C and CMn steels.
Previous design criteria were based on ultimate tensile
strength, yield strength or/and the yield strength to ultimate
tensile strength ratio.
This is associated with the fact that it was formerly (1940s
and 50s) believed that brittleness was related to the yield
strength to ultimate strength ratio. This view was supported by
the evidence from tests and use of C and CMn steels, namely
the fact that low toughness was associated with a high yield/ultimate strength ratio. Furthermore, it was known that the strain
and ageing of such steels results in a high yield/ultimate
strength ratio and reduced toughness.
In view of this, it is not surprising that certain designers and
authorities are negatively disposed towards quenched-and-tempered steels with a high yield/ultimate strength ratio (approx.
0.80 - 0.95, compared to 0.50 - 0.70 for ordinary steels)
We now know that this correlation does not generally hold.
In quenched-and-tempered steels, as the yield/ultimate
strength ratio increases with rising tempering temperature,
toughness increases as well.
The quenched-and-tempered steels exhibit excellent toughness,
despite their high strength (see chapter entitled "Toughnessbrittleness'). Good evidence is provided by applications with
WELDOX 700 for the handling of LPG (Liquid Petroleum Gas at
- 55°C).
The lower values of A5 (elongation) say nothing about the
ductility of the steel. For example, it is just as easy to bend
WELDOX 700 (os = 700 N/mm2, A5 = 14%) as S 355 (os =
350 N/mm2, A5 = 22%).
..
There seems to be only one case where As can be a Critical
factor. This is when it is necessary to have high rotational capacity in a plastic hinge in the event of e.g. the collapse of a
structure and the course of failure is dependent upon whether
or not the desired rotational capacity exists, provided that
buckling does not occur.
.
.'
Using the ultimate tensile strength as a deSign criterion
cannot be right in view of what we know today about fracture
mechanics. The ultimate tensile strength applies for a steel
without any stress-raisers, and as soon as we have a stressraiser, fracture mechanics should be used (more about this in
the chapter entitled "Toughness-brittleness" l.
It is therefore not so strange that factors of safety for WELDOX
steels vary between different government authorities and
countries.
.
The following are some examples of factors of safety against
yield stress in Sweden:
Table 3.2
Steel
grade
W500 P
W500
W 700
W900
L - - - - - L - - - - L - - i _ Stress
operm = as/Sf
Os
Always choose a factor of safety against the most critical
variable!
-Pressure
vessels
Storage
tanks
Building
structures
Cranes
1.5
1.5
-
-
-
1.7
-
1.65*
1.75*
1.66
-
1.7
1.72
'-----
* Preliminary
Equivalent German steels all have a factor of safety of 1.5 for
cranes and pressure vessels in relation too s'
::1
Factors of safety
Table 3.3
Example 3.3
Typical strength values for OX steels
What is the actual probability of failure for Sf = 1.5 for S 355 and
WELDOX 700?
Steel
grade
Plate
thickness
mm
Us guaran-
teed
N/mm2
N/mm2
Standard
deviation
N/mm2
S 355
(SS 50)
8 -10
(10) - 20
(20) - 40
350
350
350
396
383
381
22
20
22
6 -12
(12) - 20
(20) - 40
(40) - 50
(50l- 70
700
700
700
700
700
797
768
779
774
780
23
29
38
41
47
6 -20
(20) - 40
(40) - 75
900
900
900
1150
1030
1056
50
51
43
WELDOX
700
HARDOX
400
Us mean
Assume normal distribution of load =
and s = 60 N/mm2 (standard deviation).
Table 3.4
Z
grade
Us
UB
Sf=~
.. U
perm
N/mm2
S 355
W700
W700
350 520
700 780
700 780
1.5
1.5
1.75
0' perm Us-Uperm UB -Uperm
233
467
400
117
233
300
287
320
380
Figure 3.5
S 355 (SS 50l
m=383 N/mm2
WELDOX 700
700
750 m=]68 N/mm2
It is important to note that steel with a yield strength below
the guaranteed level is never released to customers, but is
scrapped internally. The curves reflect the results at Svenskt
SteWs production unit in Oxelosund.
3:4
~;
Solution:
S 355 (BS 50)
ml = 350 = 233 N/mm2
1.5
SI = 60 N/mm 2
m2 = 383 N/mm 2
S2 =
20 N/mm 2
m = m2- ml = 150 N/mm2
s = vi 602 + 202 = 63.2
(~) = <I> (
s
150
63.2
) = <I> (2.37) (Table 3.5) =
= 0.991106 (0.92 11(5)
The probability of failure OB = 1- <I> = 0.009,
i.e. 9· 10-3
WELD OX 700
ml = 700 = 467 N/mm2
1.5
SI = 60 N/mm 2
m2 = 768 N/mm2
S2 = 29 N/mm 2
m = m2- ml = 301 N/mm2
It can also be of interest to study the appearance of the curves
of actual yield stress distribution obtained at Svenskt StAI, Oxelosund for S 355 and WELDOX 700, e.g. 12-20 mm (values
taken from table 3.3.). The curves are shown in figure 3.5.
350
=
We further assume that the strength values of both steels are
normally distributed.
<I>
Table 3.4 shows a comparison between different steels with
respect to yield stress, ultimate tensile strength, most commonly
occurring factors of safety, permissible stress and difference
between yield stress and permissible stress on the one hand
and ultimate tensile strength and permissible stress on the other
hand.
~
s = vi 292 + 602 = 66.6
<I> (
~~~ ) = <I> (4.52) = 0.956908
The probability of failure OB = 1-<1> = 0.0000030, i.e. 3 • 10"
We see that the relationship between the mean value and the
guaranteed value has great significance with regard to the probability of failure.
Nor do any grounds exist for burdening WELDOX steels with
higher factors of safety when it comes to instability (see chapter
on static strength and buckling of columns and plates).
In conclusion, the above shows that there is no reason to
demand higher factors of safety for WELDOX steels than for
ordinary steels.
Factors of safety
Table 3.5
Normalized normal distribution <fl
~)
x
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
.0
.4
.5000
.5398
.5793
.6179
.6554
.5040
.5438
.5832
.6217
.6591
.5080
.5478
.5871
.6255
.6628
.5120
.5517
.5910
.6293
.6664
.5160
.5557
.5948
.6331
.6700
.5199
.5596
.5987
.6368
.6736
.5239
.5636
.6026
.6406
.6772
.5279
.5675
.6064
.6443
.6808
.5319
.5714
.6103
.6480
.6844
.5359
.5753
.6141
.6517
.6879
.5
.6
.7
.8
.9
.6915
.7257
.7580
.7881
.8159
.6950
.7291
.7611
.7910
.8186
.6985
.7324
.7642
.7939
.8212
.7019
.7357
.7673
.7967
.8238
.7054
.7389
' .7703
.7995
.8264
.7088
.7422
.7734
.8023
.8289
.7123
.7454
.7764
.8051
.8315
.7157
.7486
.7794
.8078
.8340
.7190
.7517
.7823
.8106
.8365
.7224
.7549
.7852
.8133
.8389
1.0
1.1
.8413
.8643
.8849
.90320
.91924
.8438
.8665
.8869
.90490
.92073
.8461
.8686
.8888
.90658
.92220
.8485
.8708
.8907
.90824
.92364
.8508
.8729
.8924
.90988
.92507
.8531
.8749
.8944
.91149
.92647
.8554
.8770
.8962
.91309
.92785
.8577
.8790
.8980
.91466
.92922
.8599
.8810
.8997
.91621
.93056
.8621
.8830
.90147
.91774
.93189
1.9
.93319
.94520
.95543
.96407
.97128
.93448
.94630
.95637
.96485
.97193
.93574
.94738
.95728
.96562
.97257
.93699
.94845
.95918
.96638
.97320
.93822
.94950
.95907
.96712
.97381
.93943
.95053
.95994
.96784
.97441
.94062
.95154
.96080
.96856
.97500
.94179
.95254
.96164
.96926
.97558
.94295
.95352
.96246
.96995
.97615
.94408
.95449
.96327
.97062
.97670
2.0
2.1
2.2
2.3
2.4
.97725
.98214
.98610
.98928
.921802
.97778
.98257
.98645
.98956
.9 2 2024
.97831
.98300
.98679
.98983
.9 2 2240
.97882
.98341
.98713
.92 0097
.922451
.97932
.98382
.98745
.920358
.922651
.97982
.98422
.98778
.920613
.922857
.98030
.98461
.98809
.920863
.923053
.98077
.98500
.98840
.921106
.923244
.98124
.98537
.98870
.921344
.923431
.98169
.98574
.98899
.921576
.923613
2.5
2.6
2.7
2.8
2.9
.9 2 3790
.9 2 5339
.9 2 6533
.9 2 7445
.9 2 8134
.9 2 3963
.9 2 5473
.9 2 6636
.9 2 7523
.9 28193
.9 2 4132
.9 2 5604
.9 2 6736
.9 27599
.9 28250
.9 2 4297
.9 2 5731
.9 2 6833
.9 2 7673.
.928305
.9 2 4457
.9 2 5855
.9 2 6928
.9 2 7744
.9 28359
.9 2 4614
.9 2 5975
.9 2 7020
.9 2 7814
.9 2 8411
.9 2 4766
.9 2 6093
.9 2 7110
.9 2 7882
.9 2 8462
.9 2 4915
.9 2 6207
.9 2 7197
.9 2 7948
.928511
.9 2 5060
.9 2 6319
.9 2 7282
.9 2 8012
.9 2 8559
.9 2 5201
.9 2 6427
.9 2 7365
.9 2 8074
.9 28605
3.0
3.1
3.2
3.3
3.4
.9 2 8650
9 30324
.933129
93 5166
.93 6631
.9 2 8694
,930646
.9 3 3363
.9 3 5335
.93 6752
.9 2 8736
.9 30957
.9 3 3590
.9 3 5499
.9 3 6869
.9 2 8777
.9 3 1260
.93 3810
.9 3 5658
.9 3 6982
.9 2 8817
.9 3 1553
.93 4024
,935811
.9 3 7091
.9 2 8856
.9 3 1836
.93 4230
.93 5959
.93 7197
.9 2 8893
.93 2112
.93 4429
.93 6103
.9 3 7299
.9 2 8930
.9 3 2378
.9 3 4623
.9 3 6242
.9 3 7398
.928965
.9 3 2636
.9 3 4810
.9 3 6376
.9 3 7493
.9 2 8999
.9 3 2886
.9 3 4991
.9 3 6505
.9 3 7585
3.5
3.6
3.7
3.8
3.9
93 7674
.93 8409
.9 3 8922
.9 4 2765
.9 4 5190
.9 3 7759
.93 8469
9 3 8964
.9 4 3052
.9 4 5385
.9 3 7842
.9 38527
.9 4 0039
.9 4 3327
,945573
.93 7922
.93 8583
.9 4 0426
.94 3593
.94 5753
.9 3 7999
.93 8637
.94 0799
.94 3848
.94 5926
.93 8074
.93 8689
.9 4 1158
.9 4 4094
.94 6092
.93 8146
.93 8739
.9 4 1504
.9 4 4331
.94 6253
.9 3 8215
.9 38787
.94 1838
.94 4558
.9 4 6406
.9 3 8282
.9 3 8834
.94 2159
.9 4 4777
9 4 6554
.9 3 8347
.9 3 8879
.94 2468
.9 4 4988
.9 4 6696
4.0
4.1
4.2
4.3
4.4
.94 6833
.94 7934
.94 8665
.9 5 1460
.95 4587
.9 4 6964
,948022
.94 8723
.95 1837
.9 5 4831
.94 7090
.94 8106
.9 4 8778
.9 5 2199
.9 5 5065
.94 7211
.94 8186
.94 8832
.95 2545
.95 5288
.947327
.948263
.948882
.952876
.955502
.947439
.948338
.948931
.953193
.955706
.947546
.948409
.948978
.953497
.955902
.9 4 7649
.94 8477
.9 5 0226
.9 5 3788
.9 5 6089
.9 4 7748
.9 4 8542
.9 5 0655
.9 5 4066
.9 5 6268
.9 4 7843
.9 4 8605
.9 5 1066
.9 5 4332
.9 5 6439
4.5
4.6
4.7
4.8
4.9
.95 6602
.9 5 7888
.9 58699
.96 2067
.96 5208
.9 5 6759
.9 5 7987
.95 8761
.96 2453
.96 5446
.9 5 6908
.9 58081
.9 58821
.96 2822
.96 5673
.9 5 7051
.9 5 8172
.9 6 8877
.96 3173
.96 5889
.95 7187
.9 58258
.9 58931
.96 3508
.96 6094
9 5 7318
.9 5 8340
.9 5 8983
.96 3827
.96 6289
.9 5 7442
.9 5 8419
.96 0320
.96 4131
.96 6475
.9 5 7561
.9 58494
.96 0789
.9 6 4420
.9 6 6652
.9 5 7675
.9 5 8566
.96 1235
.96 4696
.96 6821
9 5 7784
.9 58634
.96 1661
.9 6 4958
.9 6 6981
.1
.2
.3
1.2
1.3
1.4
1.5
1.6
1.7
1.8
I
I
-----
(Example: <1> (3.57) = .93 8215 = 0.9998215.
3:5
4 Static strength
Load application ................................ 4:2
Lighter can be stiffer .......................... 4: 12
Buckling of columns ........................... 4:14
Buckling of plates .............................. 4:20
4
Torsion ............................................. 4:28
Location and strength of welds ........... 4:39
High-temperature strength .................. 4:47
4 Static strength
General
WELDOX and HARDOX steels possess high strength and can be
directly utilized to reduce the dead load or to increase the
maximum live load capacity of a structure.
Material cost in relation to strength (GBP/yield strength)
decreases with increasing strength. If we can exploit the high
strength of the WELDOX and HARDOX steels, it is clearly
economical to employ these steels.
Reduced dead load in most structures is of very great
importance for the end user.
For manufacturers who appreciate the marketing advantages
of lighter-weight products and structures and take this into
account when choosing a material, the WELDOX and HARDOX
steels present an inspiring challenge to exploit these
characteristics to full advantage.
Table 4.1 presents the guaranteed yield strengths of the most
common steels.
Table 4.1
Steel
minimum plate thickness in mm'
MS:
HS:
EHS:
AR:
S 235 (BS 40A,St 37-2)
S 355 (BS 50, St 52-3)
WELDOX 420
WELDOX 500
WELDOX 600
WELDOX 700
WELDOX 900
WELDOX 960
HARDOX 400
8
8
8
8
6
4
4
4
4
Guaranteed
yield strength
ReH (N/mm')
220
350
420
500
600
700
900
960
900
• If you wish to have other plate thicknesses, contact us!
For typical values and the scatter of strength values in our
production, see chapter 3.
High-temperature yield strengths are given in section 4.7.
The formulas for stress and strain of classical Materials Science
can be used in designing with WELDOX and HARDOX steels.
Permissible stresses have to be determined from case to case
with the aid of chapter 3, which shows that WELDOX and HARDOX
steels do not have to be treated differently from ordinary steels.
The modulus of elasticity for WELDOX and HARD OX steels is the
same as for ordinary steels (21 . 10' N/mm2). This means that if
the second moment of area (I) is not adjusted in proportion to
reduced plate thickness, greater elastic deformation will be
obtained.
Elasticity in design can have many advantages, as the word
"resilient' implies!
.
When WELDOX and HARDOX steels are used for the first time, a
number of questions naturally arise, especially when plate thickness is reduced. The following are the most common questions,
with reference to the pertinent section in the manual:
- How can load appliction best be arranged?
- Can stiffness be greater despite the choice
of thinner WELDOX steel?
- When can WELDOX steel be used in structures
susceptible to overall buckling?
- When does local buckling become critical?
- Twisting of open thin-walled sections gives rise
to large stresses that are added to bending
stresses. How are these stresses calculated?
- What is the strength of a welded joint in WELDOX
or HARDOX steel? Which welding electrode is
suitable?
4.1
4.2
4.3
4.4
4.5
4.6
The sections contain data and calculation methods for evaluating the situation at hand .
4: 1
4.1 Load application
Page No.
Some fundamental principles ." ......................................... 4:2
Normally·loaded and in-plane-loaded plates
A normally-loaded plate is a flat plate loaded by forces perpendicular to its plane, so that bending moments tend to deflect the
plate out of its plane, see figure 4.1.1.
Figure 4.1.1
Extreme load - Fatigue load .............................................. 4:4
Examples of load applications ........................................... 4:4
Lugs on panels - Calculation method .. ,............................. 4:5
Displacements caused by small deformations are perpendicular
to the plane of the plate.
The normal stresses in normally-loaded plates are caused by
the bending moment. They vary in the direction perpendicular
to the plane of the plate and can be very large - in most cases
larger than the shear stresses.
An in-plane-loaded plate is the oPPosite of a normally-loaded
plate. The forces act in the plane of the plate, and the moments
tend to rotate the plate in its plane, see figure 4.1.2.
Figure 4.1.2
OM
y
Some fundamental principles
A great deal of knowledge and experience is required to apply
loads .correctly in and between structural elements in such a
manner that the result is low local stresses and deformations.
The subject is difficult to treat generally. Few textbooks deal
with it, and the knowledge possessed by designers is usually
based on experience (usually bitter). An understanding of the
principles of load application is essential. As a rule, this is where
the greatest stresses are created.
The purpose of designing with WELDOX and HARDOX steel is
often to reduce plate thickness. The idea is to apply forces in such
a way that local stresses are as small as possible and are channeled in a manner that creates small stress concentrations (fatigue!).
On the other hand, we know that stresses at the points of load
application are very high. Since these stresses are often crucial in
determining whether or not we will have plastic deformation in the
structure, the WELDOX and HARDOX steels are the natural choice
due to their high yield strength!
If is difficult to cite any general rule for load application, but the
following is a broad philosophy of design:
Determine function
Calculate the size and direction of the forces
Channel the forces together largely as tension and compression
Avoid bending moments!
Us~ plate like skin on a load-bearing skeleton. Design intelligently to simplify production and minimize plate thickness.
In a properly designed structure, the final design will reflect the
play of forces. This results in a very attractive structure!
The normal stresses are constant in the direction perpendicular to the plane of the plate and are of the same order of
magnitude as the shear stresses. The displacements take place
in the plane of the plate.
A flat plate loaded with a general force system is simultaneously subjected to both in-plane and normal loads.
Why make this distinction?
The reason is that a plate loaded in its plane is a rigid structural
element than can take heavy loads with small deformations,
while the opposite applies to a normally-loaded plate (i.e. small
loads and large deformations).
When we design a structure - espe~ially a thin-walled one we must endeavour to make the components function as much
as possible as in-plane-loaded plates and as little as possible
as normally-loaded plates!
Compare:
Struts - in-plane-loaded plates
Beams - normally-loaded plates
Static strength - load application
Some examples
Figure 4.1.5
As an example, we can take a welded I beam acted upon by a
bending moment, see figure 4.l.3. The normal stress at the
lower edge of the flange is almost as great as at its upper edge.
The flange thus acts almost as an in-plane-loaded plate. The
web acts purely as an in-plane-loaded plate.
Figure 4.1.3
I beam
a
At the points of attachment, the stresses in the flanges of the
cross beam are small, so that their curtailment does not have
any great effect.
If, instead of a cross beam, we have a cantilever (see figure
4.1.6a), we cannot use the same technique, since the web will
then act as a normally-loaded plate. Curtailment of the flanges
reduces the load-bearing capacity and lateral stability of the
beam considerably. The technique illustrated in 4.1.6b should
be used instead. The web and flanges then act as in-planeloaded plates.
This stress configuration is disrupted close to pOints of load
application, such as a pOint force near the outer edge of the
flange, see figure 4.1.4.
Figure 4.1.4
Figure 4.1.6a
Point load on I beam flange
p
Figure 4.1.6b
B
.. I
Correct design
Here, the flange acts locally as a normally-loaded plate. The
force P produces large deformations and stresses.
The maximum bending stress at the web directly opposite the
force is, according to (27)
omax==0.5
and the deflection at P
b = 0.05'
p·o b2
LIB> 8
where 0 is the plate stiffness defined as
It is very common for forces from e.g. a hydraulic cylinder,
struts etc to be applied via a cantilever (lug) in a beam or panel
Figures 4.1.7a, b, c and d (taken in part from (26)) illustrate
how this can be done and how the maximum stress varies
between the different cases. Figure a shows the flanges and
web being used as in-plane-loaded plates. Figure b shows how
the web absorbs the vertical force by membrane action, while
the moment creates a couple perpendicular to the web, so that
the web acts both as an in-plane-loaded plate and a norrnallyloaded plate with large normal stresses.
Figure c does not give any improvement compared to a.
Finally, in figure d, the lug has been replaced by a tube that
gives slightly lower normal stresses than the lug in figure b.
This is because the load has been spread out over a larger
area and because the round shape of the tube produces a
smaller stress concentration.
AgU",@ I~bv
p
E· t3
o = ---=---=--=-2
12 (1 _v )
ax
v = 0.3 for steel
In other words, the point of load application on the flange
should be moved to a point directly above the web. The web
then functions as an in-plane-loaded plate with respect to load
application.
When a short rigid cross beam is fastened between two
torsionally flexible main beams (see figure 4.1.5), it is virtually
only transverse force (vertical force) which is transferred to the
main beam. This gives rise to in-plane stresses (membrane
stresses) in the web.
O
= 10 N/mm2
max = 30 N/mm2
Zmax
p
2
0Xmax = 350 N/mm 2
a Zmax -- 410 N/mm
11 I~dv
o Xmax = 5 N/mm2
o Zmax = 30 N/mm2
p
0Xmax = 150 N/mm2
o Zmax = 300 N/mm2
Static strength - Load application
What is critical, Extreme load or Fatigue load?
There are many excellent analyses that describe how loads
should be applied statically in such a way as to give rise to low
stresses. Such methods may also be mistakenly applied to the
design of components, frames etc. that are subjected to fatigue
load.
As explained in the section "Fatigue", welds cause very steep
stress concentrations in conjunction with fatigue, even when
they are not load-carrying!
A small weld bead laid on the flange of a beam that is
subjected to fatigue load reduces its fatigue strength by 50%!
From this, it is clear that the optimal design for static loading
is not necessarily the optimal design for fatigue loading.
As an example, we can take the lug in figure 4.1. 7a. This
solution is good for static loading. But if the channel beam is
subjected to bending fatigue (M br ), the weld at the lug will
completely determine its fatigue strength. The design shown in
figure 4.1. 7b is then preferable. If the force or, the lug is so
great that the stresses in the web exceed the permissible level,
stiffeners must be welded on, for example as shown in 4.1.8a.
If we are to have the same fatigue strength for the lug's weld
as for the flange when the beam is subjected to bending, C
must be "" OA . H if the flange is sheared or gas-cut. If the
edges of the flange are machined and rounded, C must be ""
0.25' H.
An alternative approach may be to put a plate underneath the
lug as shown in figure 4.1.8b.
Figure 4.1.9
a
b
/
bulkhead that
simultaneously
serves
as backing bar
bulkhead (cannot
be welded in RHS)
Figure 4.1.8a
Example of optimal design for both extreme static load and
fatigue load.
Pmax
...-_ ...-------
e and f are suitable in cases where many point loads must be
applied.
r::::=~
g
H
Figure 4.1.8b
/
Alternative
-~
deSignL:::====::::" •••••••• -••••••••
~
Cast component for application of load
due to two side-mounted hydraulic cylinders.
h
I
,
,
,
I
--.....,
I
I
I
I
I
I
:
,,
-----, "'::'
I
I
I
1. _____ .1
,
Box beams
Box beams made of plate, often bent, or RHS (rectangular hollow section) members have been used widely in frames, cranes,
excavator booms etc. A great deal of attention must be paid to
load application in all structures incorporating box beams,
especially when the beams consist of moving parts.
In the case of beams with moving parts, forces from hinge
joints and hydraulic cylinders must be applied. The same principle still applies, i.e. an endeavour must be made to channel
the forces into the webs, utilizing the webs as in-plane-loaded
plates.
There are many ways to in which this can be done. Some
examples are shown in figure 4.1.9.
4:4
~
-
- - t-.-!
.. ___ • .J
-
- ,
,
,I
,I
,
I
I
I
I
I
I
•I
;,,'
'"
One way to apply forces due to two side-mounted cylinders
when the forces act at a point where the stress in the boom is
high and the section is of small depth. The point of load application is moved by means of this solution.
Static strength - load application
Box beams for telescopic members
These telescopic members are used in crane jibs, frames
(variable-length vehicle frames), masts on fork-lift trucks etc.
The bending moment in the extension member must be
transmitted in the form of a couple to the member which the
extension member slides into. This can be done, for example,
by means of rollers or plastic blocks (see figure 4.l.1O), the
dimension L being of great importance.
Figure 4.1.10
Some crane manufacturers have chosen a cross-section as
shown in figure 4.1.13. In this way, the risk of flange buckling is
reduced (the width of the free panel is smaller), and at the
same time the flanges are utilized as in-plane-loaded plates for
load application into the webs. In order to prevent lateral buckling of the webs, a rigid frame is fitted around the outer section
at the point where load application takes place.
Naturally, compromises between the alternative designs are
possible.
Principle of telescopic action
L
Figure 4.1.13
Hexagonal beam
/,~
-..: ~
rollers or blocks
The precise shape of the cross-section is of essential
importance.
Where loads are moderate, it is possible to make use of the
very attractive section shown in figure 4.1.11, which consists of
two bent channels. Since the forces from the roller act a slight
distance away from the webs, the flanges will act as normallyloaded plates and will consequently be bent. The bending properties of the steel are of the utmost importance in minimizing
the radius at the corners. The larger the radius, the larger the
bending moment on the flange.
figure 4.1.11
Box beam made up of bent channel sections
The welds on the box beam must be located in the webs. If
they are located in the flanges, fatigue cracks will quickly be
initiated from the root of the weld.
The figure also illustrates the theoretical load distribution on
such a beam. It might be assumed during design that the flange
will be subjected to uniformly distributed contact pressure. This
is wrong. In actual fact, the flange will be subjected to much
higher pressures than we have assumed, and unless sufficiently
hard plate is used, such as EHS or AR plate, we will be in trouble.
figure 4.1.12
outer beam
inner beam (extension
member)
Load application in shells
Forces can be taken up easily and elegantly with !tie aid of shell
structures. Nature learned this prinCiple long ago and applied it
to skulls, bones, ball jOints, egg shells etc. We use shells ourselves in buildings, ships, motor vehicles, airplanes, pressure
vessels etc.
The stresses in an in-plane-loaded plate are a special case of
the membrane stresses in a shell. A shell is in the membrane
stress state if the stresses are not dependent upon the coordinate in the direction normal to the shell.
A great advantage will be gained if loads can be applied as
membrane stresses.
Example of a cylindrical shell. A force acts on a cylindrical
shell as shown in figure 4.1.14 left. In this case, the shell IS
subjected to flexure, and M = F1 . e.
With a longer lug, as shown in figure 4.1.14 'right, the
bending moment is avoided and the force is applied elegantly
as a membrane stress.
Figure 4.1.14
~xample of load application in cylindrical shel
Alternative to deSign shown in 4.1.11
t
+
tt
--------+----;
H~D
Another design for a cross-section as described above is
shown in figure 4.1.12. Here, better use is made of the material,
with thick flanges and thin webs. This means that the bending
stresses in the flange caused by the roller are reduced by about
1 and the def
'
"2
ormatlons
are reduced by about -
1
-3' and at
t
t
the same time the webs "support" the flanges better than in the
former case.
Lugs on panels
Demonstration of calculation methods for a large number of
different load application problems would require a special
book. For the purposes of this manual, we have therefore chosen a case, which we know is widespread and can giverise to
problems, namely load application in panels with welded-on
lugs.
Lugs on panels are widely used for such structural elements
as hinge joints, hydraulic cylinders, lifting lugs, coupling devices
~c.
.
4:5
Static strength - Load application
We have already discussed how loads should be applied and
how structural elements should be utilized as in-plane-loaded
plates. This is sometimes impossible, and the plate must be
subjected to normal loads.
In such cases, it is of great importance to know the local
stresses underneath the lug in the plate. The plate often carries
other stresses as well, and these stresses will be added to the
local stresses created by the load in the lug.
A calculation method is demonstrated below for determining
the maximum local stresses approximately at the ends of the lug
in the case when load is applied to a plate beam via one lug,
both when the longitudinal direction of the lug coincides with
that of the beam and when it is perpendicular to this.
b) Beam with transverse lug
The length of the panel is at least 10 times greater than its
width.
Figure 4.1.16
Symbols
a) Beam with longitudinal lug
The length of the panel is at least 10 times greater than its
width.
Figure 4.1.15
y
Symbols:
If
- -1'
I
I
I
I
I
~
B
I
T
I
I
I
I
I
',>
I-
(
.,
I
I
)
/--------- --------
y
(
'- j
-
I·
I
I
I
x
I
I
I
I
I
-I
a
Calculation method:
1. Divide the force into the components F and P.
2. Superimpose a coordinate system as shown in figure 4.1.16.
3. Calculate c = 112 (a - l)
4. Calculate cIa.
5. Obtain the following values from the graphs in figures
4.1.25-30 for the appropriate values of cIa and y/a:
Calculation method:
lOlog S; 1OIog Sy' lOlog S.;' lOlog 5." lOlog Sy"
1. Divide the force into the components F and P.
2. Superimpose a coordinate system as shown in figure 4.1.15.
lOlog Sxy"
6. Interpolate if necessary.
3. Determine the value of x for the desired point.
7. Calculate the above logarithms.
4. Calculate x/a.
8. Calculate the stresses for the appropriate value of y as follows:
5. Calculate d/a
6. Obtain the following values from the graphs in figures
4.1.19 -4.1.24 for the appropriate values of xla and d/a.
S'x S'y S'xy S"x S"y S"xy
(S' for forces parallel to the plate surface)
(S" for forces perpendicular to the plate surface)
F· H
a.t
ax=~'
S'
P
t
x+~'
S ..
x
7. Interpolate if necessary.
8. Calculate the stresses for the appropriate value of x as
follows:
ax=
ay =
r xy =
a. t
t
F. H
·S'+_P_· S "
x
t2
x
d . t2
F. H
d·e
F·H
d·e
Example 4.1.1
·S,+-p_· S "
y
t2
y
P Sxy"
. S'
xy+~'
t
This calculation method gives the maximum stress in the y
direction, which occurs near the corners, i.e. y = ±d. This
method gives an estimated accuracy of ± 10%. If greater
accuracy is required, FEM programs can be used.
4:6
S'
P
S ..
r XY = -F·H
- 2 - ' xy+--z-' xy
A box beam as shown in figure 4.1.17 is to take up the force
from a superstructure, and the designer plans to apply the load
via a lug on one web as per figure 4.1.29. Calculate the
maximum stresses at the lug.
Static strength - Load application
figure 4.1.17.
"x
-
ay
t
c
R = 50000 N (5 tonnes)
a = 35°
H= 100 mm
t = 12 mm
a = 400 mm
L == 160 mm
B= 20 mm
b = 180 mm
L
a
b
T xy
= 597 N/mm2
= 485 N/mm2
= 43 N/mm2
Of these, cry is of great interest, since it is added to the bending
stresses in the beam.
Assume that we have chosen S 355 steel with cr s == 350
N/mm2 and cr perm in the beam:
350
_ 233 NI,mm 2
-1.5
This stress is assumed to exist in the flange, and the bending
stress at the point x = c == 120 mm is
Gperm=233
ay bend = 233·
160
2
400 = 93 N/mm
Gytot{x =C } = 93 + 485 = 578 N/mm2 > a sl
As we see, it is very easy for high stresses to occur. In order to
guard against plastic deformation, steel with a high yield strength
should therefore be chosen. In this case, BS 50·D is not enough,
WELDOX 700 with O's = 700 N/mm2 is clearly superior.
It is evident from the formulas that 0' =
x
+
t
i.e., that if we can just get by with S 355, then for static loads, we
can use WELDOX 700 with a plate thickness of
Solution: Beam with transverse lug
Divide R into the components F and P
F = 50000· cos 35° = 40957 N
P = 5000 . sin 35° = 28678 N
Superimpose the coordinate system as shown in figure 4.1.16
Calculate c =
+
V
350 "" 0.7 of the plate thickness for S 355
700
In the case of fatigue, more information is required before the
size of the thickness reduction can be determined.
It is also evident from the graphs (figure 4.1.25-30) that wide
and elongated lugs reduce the stresses.
(a - U = 120 mm
Calculate cIa = 120/400 = 0.30
Obtain from graphs 4.1.25-30, for cIa = 0.3 and yla =
Example 4.1.2
8~2 = 41~ = 0.025
A hydraulic cylinder that operates the bucket is mounted on the
dipper arm of an excavator. The hydraulic cylinder is attached
to the dipper arm via a lug as shown in figure 4.1.30. Calculate
the maximum stresses in the dipper arm around the lug for the
extreme loading case when maximum load is lifted and cylinder
Cl holds the load under maximum pressure, i.e. R = 75000 N
(7.5 tonnes). Maximum bending stress in section A-A = 120
N/mm 2 .
1OIog Sx' = 0.68
Sx' = 10°·68 = 4.78
10log Sy' = 0.55
Sy' == 10°55 = 3.55
1OIog Sxy' = - 0.33
Sxy '=. 10 - 033 = 0.47
1OIog Sx" = 0.11
Sx" == 10°11 = 1.29
= 0.07
Sy" = 10°°7 = 1.17
10log Sy"
1OIog Sxy" == - 1.35
Sxy"= 10-1.35 = 0.05
Figure 4.1.18
The stresses are calculated as follows:
ax=_F_'"H,-. S'+ P . S"
a.t2
x T
x
a x = 40957· 100 . 4.78 + 28678.
400· 144
144 1.29
a x = 340 + 257 = 597 N/mm2
F. H
P
a Y= ' S '+--' S"
2
a. tY
t2
Y
a = 40957· 100 . 3.55
y
400· 144
+
28678. 1 17
144
.
ay = 252 + 233 = 485 N/mm2
rxy =
r xy =
F·H
r . t
--2-'
S'
P Sxy"
xy + -2-'
Section AA
t
40957· 100 . 047 + 28678. 005
400· 144
.
144'
rxy = 33 + 10 = 43 N/mm2
4:7
Static strength - load application
Solution: Beam with longitudinal lug
F· H
Oy=--" .
d·t~
1. Divide the force into F and P
F = 73860 N
p = 13023 N
2. S~perimpose the coordinate system as in figure 4.1.18. Even
though the lug is not symmetrical, we can place the origin in
the middle of the lug. This will give too Iowa value of the
maximum stress due to P, but as P is only 17% of F, the
error in a will be acceptable.
ay =
,
P
e
Sy+ - - ' S"
73860 . 60
100. 102
= 363 + 65
y
. 0.82 + 13023. 0.5 =
102
= 428 N/mm2
.xv= 0
3. Value of x = 100
i.e. maximum stress in section A - A
4. x/a = 0.5
Ox
ay
5, d/a = 0.5
6. The following values are obtained from graphs 4.1.19-24 for
x/a;" 0.5 and d/a = 0.5:
S'x = 1.06
S' = 0.82.
S.Yxy = 0
S"x = 1.1
S" = 0.5
S"Yxy -- 0
7. Interpolation is not necessary.
8. Calculate the stresses as follows
F· H
°x= (l:t2'
a x=
S'
x+
73860· 60
100· l(j-
P
S'
T'
x
. l.06 + 131~3. 1.1 =
= 470 + 143 = 613 N/mm2
4:8
v-
= 613 N/mm2
= 428 + 120 = 548 N/mm 2
Le. so high that plastic deformation occurs in the area when
S 355 is used, and WELD OX 700 provides good insurance against
this.
Static strength -, load application
figure 4.1.19
Figure 4.1.20
lug parallel, force parallel to plate
lug parallel, force parallel to plate
1.12
d/a
1.12
/0.15
//,020
/0.25
/0.30
1.04
1.04
0.96
~0.35
~0.40
0.96
0.88
__ 0.10
0.88
~0.45
080
-............0.05
0.50
0.80
0.72
><
0.72
0.64
In
0.64
>.
0.56
d/a
-Ul
0.50
0.45
0.40
0.35
0.30
0.25
0.20
0.15
0.10
0.05
0.00
0.56
0.48
0048
0.40
0040
0.32
0.32
0.24
0.24
0.16
0.16
0.08
0.08
0.00
U)
0
0
0
U)
0
ci
0
U)
N
N
0
0
0
LO
et)
ci
ro:
0
0
'<t
LO
ci
0
'<t
0
U)
U)
0
ci
x/a
U)
0.00
j
If)
0
0
Figure 4.1.21
0
ci
LO
N
0
If)
N
et)
et)
0
0
0
0
0
'<t
0
LO
I
0
t£l
'<t
lD
l!")
0
0
0
x/a
lug parallel, force perpendicular to piate
1.12
2.8
1.04
2.6
0.96
2.4
0.88
2.2
0.80
2.0
-0.10
1.8
-0.15
1.6
-0.20
lA
-0.25
-0.30
-0.35
0.72
d/a
{Wrt+\+H- O. 50
0.64
_?;:
ci
0
Figure 4.1.22
lug parallel, force parallel to plate
V>
LO
~tT+\+H-OA5
\\\+~+t--n
0.56
40
\\H,H+I-t\--() 35
d/a
-0.05
x
1;,
\\ \ \ Irtttll+--() 30
0.48
\\\\\.~r--0.25
1.2
20
.15
.10
.05
1.0
0.40
0.32
0.6
0.16
004
0.08
0.2
LO
0
ci
0
.....
LO
ci
ci
o
ci
N
lD
N
0
0
et)
ci
o
U)
ci
-0.50
0.8
0.24
0.00
-0040
-0.45
0.0
LO
lD
ci
x/a
lD
0
0
0
ci
lD
ci
0
If)
N
N
ci
ci
0
ro:
0
lD
0
'<t
l!")
'<t
0
l!")
et)
If)
lD
0
0
ci
0
ci
x/a
4:9
Static strength - load application
Figure 4.1.24
Figure 4,1,23
lug paraliel, force perpendicular to plate
lug parallel, force perpendicular to plate
2.8
0.28
2.6
0.26
2.4
0.24
2.2
0.22
2.0
0.20
d/a
-0.05
1.8
0.18
,.,x 0.16
,., 1.6
'in
'",
1.4
1.2
d/a
0.14
....."..---0.05
'0---''1---0.10
0.12
15
1.0
0.10
0.8
0.08
0.6
0.06
0.4
0.04
0.2
0.02
~-\--0.15
\+-\-+-0.20
-~-I---O. 25
, .......T\-\rI--- 0.30
\.\.\~cTtt-O. 35
\Wr't\+\-O .40
~t\m-0.45
0.0
ID
0
0
0
ID
0
0
o
ID
N
N
o
o
0
ID
("f)
("f)
0
o:t
0
o
0
o
ID
o
~ffit--O. 50
0.00
ID
ID
ID
0
0
0
0
.....
U'l
0
0
x/a
lug transverse, force parallel to plate
1.8
1.4
cia
Ol)
1.2
~0.45
0.40
0.35
0.30
1.0
~O.25
0.20
0.8
0.15
0.10
0.05
0.00
0.6
" 0.4
U'l
N
o
o
("f)
U"l
o
o
o
U'l
U'l
0
1.2
1.6
0
o
lug transverse, force parallel to plate
1.6
'",
N
Figure 4.1.26
2.0
x
o
x/a
Figure 4.1.25
1.4
......
1.0
0.8
Ol)
o
"
U"l
o:t
o
0
U'l
0
y/a
-0.6
-0.8
-1.0
-1.2
4·10
Static strength - load application
Figure 4.1.27
Figure 4.1.28
lug transverse, force paralle! to plate
>><
·111
OIl
0.40
0.7
0.35
0.6
0.30
0.5
0.25
0.4
0.20
0.3
0.15
0.2
0.10
0.1
x
1n 0.05
-00
•
0
0-
Lug transverse, force perpendicular to plate
0.8
o
lCl
0
lCl
lD
0
"<l"
lD
"<l"
0
ci
y/a
M
If)
ci
ci
ci
ci
ci
.q
-0.1
.q
0
y/a
-0.2
-0.3
cia
-0.4
-0.20
0.45
0.40
0.35
0.30
0.25
0.20
·-0.5
-0.6
cia
0.45
0.40
0.35
-0.25
-0.30
~ 0.25
0.20
0.20
0.15
0.15
0.10
0.100.050.00
0.05
Figure 4.1.30
Lug transverse, force perpendicular to plate
o
o
-0.6 ci
Figure 4.1.29
lCl
0
ci
0
0
N
ci
o
ci
I
i
lCl
N
0
I
0
lCl
(Y)
0
"<l"
lCl
(Y)
0
ci
ci
ci
I
I
I
"<l"
,
y/a
o
lCl
o
I
0.4:;
lug transverse, force perpendicular to plate
-0.7
-0.8
-0.9
-1.0
,:;:~~:::~~~~~~~~0.40
-0.35
0.30
0.25
0.20
0.15
",,0.10
,,0.05
0.00
-1.1
-1.2
>,x
<JI
-1.3
OIl
0
~ -1.4
-1.5
-1.6
-1.7
-1.8
-1.9
-2.0
';11
4.2 Lighter can be stiffer
Page No.
What is stiffness? ............................................................ 4:12
Different second moments of area can be obtAined
with the same or smaller amount of material...... ......... ..... 4: 12
Two examples ................................................................. 4: 12
The fact that different values of I can be achieved with the
same amount of material stems from the definition of the
second moment of area about an aXIs x, dI x = j.dA, where y is
the distance from the axis x to the area element dA.
in other words, all that is necessary to obtain a large value of
I is to move the material as far as possible outwards from the
x axis.
How far the material can be moved outwards depends upon
how much stress the material can take, which is where WELDOX
and HARDOX steels come in.
The bending stress at the point of attachment in figure 4.2.1
can be written
p. L· emax
ob max =
I
where emax is the largest distance
from the centroid of the section (the X axis) to the extreme fibre.
This means that the bending stress increases if the magnitude oB is constant (= same bending stiffness) but the
material is moved out, i.e. emax increases.
The examples below illustrate the relationships involved and
demonstrate the fact that it is possible to design both lighter and
stiffer structures with WELDOX and HARDOX steels.
Naturally, local buckling must be taken into account when
working with large, thin panels. Experience shows that buckling is
normally not a problem in machine structures, but more about
this in the section entitled 'Buckling of plates'.
Example 4.2.1
Which simple approximate relationships apply to the bending of a
beam when a change is made from milder steels to WElDOX,
loading conditions remaining the same as shown in figure 4.2.2?
It is quite possible to create both lighter and stiffer structures and
fabrications by using WELDOX and HARD OX steels instead of
ordinary steel.
Figure 4.2.2
p
L
What is stiffness?
When you try to get someone to use WELDOX steel in a
structure, you often hear to following objection: 'There is no point
in using WELDOX. You get the same or greater elastic deformation and a flimsier structure."
The person recalls the formula for the elastic deflection of a
cantilever from his studies of Materials Science (see figure 4.2.1)
--- ----=:::
-- --
}
---~-::::~"':..":.:7
l"Jo
b = constant
If t < < band h, then
Figure 4.2.1
h
E, I
I==b·t·JL
2
W=b· t· h
p
L
-- -::.]}6.
::. -::. - - ==::: - - _
6. = PL3
3El
deflection
..........
Solution:
He also recalls that the modulus of elasticity (E) is roughly the
same for all steels.
His objection is only valid if the WELDOX beam has the same
overall dimensions and shape and the same or smaller plate
thickness as the ordinary steel beam.
Bending stiffness is the product of E . I, and different
second moments of area (I) can be achieved with the same
or lesser amount of material!
Analogously, torsional stiffness for "ordinary twisting" is the
product of G· Kv , where G is the shear modulus (for steel G
= ~,6 ) and Kv is the section factor with respect to torsional
stiffness.
Bending stresses:
MS 1
WELDOX 2
p. L
P. L
(4.2.1)
Static strength - lighter can be stiffer
Elastic deflection:
Example 4.2.2
Two beams, made of WELDOX 700 and S 355, have the following
dimensions:
3 . E·
h/
. o· t 1 · -2-
Figure 4.2.3
WELDOX 700
S 355 (SS 50-0)
as = 700 N/mm2
a, = 350 N/mm2
(4.2.2)
Combine equations 4.2.1 and 4.2.2
_L
1
oor
:er
a
,- - - - - - - -t-'
0
~
-
~
---_._-" r-'
(4.2.3)
if we have two given types of steel with different yield stresses
(permissible stresses), 02 > 01> deflection can be changed by
varying the depth of the beam.
~I.
280
~r-
~I.
280
~r
The thickness of the flanges is determined by
Beam data:
(4.2.1)
Ix
Wx = 16.2· 105 mm 3
Wx = 19.8' 105 mm3
A
Will the WElDOX beam be heavier?
= 397· 106 mm4
Ix = 422· 106 mm 4
= 10528 mm 2
A = 14848 md
The section area A= 2 . b . t (disregarding the thin web)
The WELDOX beam is:
(4.2.4)
Equation 4.2.3 gives
422
397 = 1.06
i.e. 6% stiffer
10528
14548 = 0.70
i.e. 30% lighter
1.5* . 700· 16.2 . 105
1.5* . 350· 19.8' 105 =: 1.64
i.e. 64% larger bending
moment
and substituted in equation 4.2.4
• Sf
= 1.5 for both steels.
With beam spans of 3000 mm between two supports, local
buckling is prevented when Sf = 1.5.
If we wish to have the same deflection
AwElDOX =: ~S'
(
01
2
-a;-)
e.g. a 1 =: 350 N/mm2
02 = 700 N/mm2
S 355
WElDOX 700
AwELDOX "" 0.26 . AMS
In other words, we can build much lighter beams of WElDOX than
of MS while keeping deflection the same!
Note that deflection is dependent upon both material and
shape!
In other words, for the same section area, different second
moments of area can be obtained!
4:13
4.3 Buckling of columns
Examples of imperfections are:
Page No.
Buckling of columns, general
4:14
Design procedure for column in axial
compression with regard to in-plane buckling .................. 4: 16
Influence of welding residual stresses .............................. 4: 15
Example calculation ........................................................ 4: 16
Are WELDOX columns economical? .................................... 4: 17
Simultaneous bending and buckling ................................ 4: 17
- initial curvature Wo
- initial eccentricity
- deviation in cross-sectional geometry
- scatter in material properties
- residual stresses (from the fabrication process)
For a column with wo> 0, a theoretical load deflection curve is
obtained as shown by the lower curve in figure 4.3.l.
Nk = the critical load is determined from
(4.3.1 )
Nk = ak' A
whereak = limit stress with respect to buckling
(critical stress)
A = section area
Ok is dependent upon the member's slenderness ratio, crosssectional shape, steel grade and residual stresses.
The following definitions are first necessary:
Elastic critical stress
("Euler critical stress")
Instability phenomena such as column buckling (overall buckling), plate buckling (local buckling) and interaction between
the two are becoming increasingly important as the trend goes
towards increasingly high material utilization and advanced
structural designs.
Columns occupy space, and an attempt is therefore made to
make them as slender as possible. Slender columns have a
smaller stress-absorbing area, and WELDOX steels become
interesting.
In many cases, it is economical to use WELDOX in columns,
see below.
When a column becomes critical with respect to buckling, the
cause is to be sought in the following factors: load, cross section, column length, restraint, yield strength and modulus of
elasticity.
It is difficult to specify general limits when it comes to the risk
of overall buckling. Long and slender columns must be checked. Limits are given below in mathematical form, and the
examples provide some information.
°el
(4.3.2)
I min = minimum second moment of area of the section (the
column buckles towards the side where it is weakest)
A = section area
E = modulus of elasticity
L = effective length as per table 4.3.1
Table 4.3.1
Euler's
buckling cases
IN
Buckling
configuration
it'
In~plane buckling, general
When a straight column (wo = 0) is acted upon by a compressive force N as shown in figure 4.3.1, no buckling wm takes
place until the theoretical load NE has been reached.
The perfectly straight column does not exist. In practice, all
columns are subject to imperfections.
2
1
L=
~
4
3
~
,
\
t
;
/
\
,,
-<
I
1
-~
nm~
2L
L
0.7 L
0.5 L
Table 4.3.2
Figure 4.3.1
N
N
Curve
ao
a
b
c
d
y
0.125
0.206
0.339
0.489
0.756
Wo = 0
Slenderness parameter a
a= ~=; .+0twhere i is the radius of gyration
L-------------------~Wm
N
4:14
'- - ~
I
-YA
in some literature, a is calledI
(4.3.3)
Static strength - Buckling of columns
The value of a determines how large a percentage of the yield
stress the critical stress Ok may amount to, see figure 4.3.2.
in figure 4.3.2, which is taken from the recommendations of
the ECCS (European Convention for Constructional Steelwork,
ref. 68), there are different curves aa, a, b, c and d, depending
upon the type of section, plate thickness, residual stresses etc.,
see table 4.3.3.
The curves in figure 4.3.2 can be expressed in mathematical
form.
whena':;; 0.2 ~= 1
Residual stresses
As is well known, welding gives rise to large residual tensile
stresses in the longitudinal direction of the weld. In order for
the cross-section to be in equilibrium, residual compressive
stresses therefore arise in other parts of the cross section.
These residual compressive stresses are added to the external
compressive stresses due to applied load, and Ok is reached
earlier than if there had been no welding residual stresses.
The distribution of the residual stresses for corner-welded box
sections is illustrated in figure 4.3.3.
as
J
whena> 0.2 :: = F -
F2 -
Figure 4.3.3
-;7
(4.3.4)
whereF= 1+1' (a-O 2l+a 2
2'a
z
(4.3.5)
I' is dependent upon which curve is to be used, and the value
ofI' for the different curves is given in table 4.3.2.
Figure 4.3.2
a) real
Ok
. &l
as
1.0+--_==-------
e
b
I;.
T
0.5
-:;-
I
:1L.J
b) idealized
o
2.0
a (X)
1.0
In addition to the five different buckling curves, the ECCS also
gives recommendations for utilization of the material's yield
strength. as" is therefore SUbstituted fora s in order to determinea and Ok. see table 4.3.3.
With knowledge of the welding parameters, it is possible to
calculate a/as for corner-welded box columns.
-;-~ = -'v-,'b-' 1-.- 0 - -;:-t
0.15' U'
(4.3.6)
1- 1
which is taken from ref. (64).
Table 4.3.3 Guidelines for choice of buckling curve and notional yield stress as· for EHS shapes
Cross-sectional
shape
Curve a 5*
[$1
.-_1_.
I
~
Rectangular and round tubes
.Bent in flanging machine
and welded
Roll formed and welded
v = welding rate m/s
I = current
A
U = voltage
volts
Typical welding parameters are given in table 4.3.4.
a
Table 4.3.4 Typical welding parameters for submerged-arc
welding, one pass.
----
f'
Welded box section
'-li,'Yy- BUC;i~g;abou'
High proportion of welding
residual stresses
0/02> 0,2
Welded I -section
Buckling about:
x - x Gas-cut flange
Rolled flange
y - y Gas-cut flange
Rolled flange
Stress-relieved box section
a
a
b
y-y
Thick-walled sections t ;;::: 40 mm
0.90 5
0.90 s
0.90 5
0.90 5
0.90 5
0.90 s
0.90 s
aa
as
aa
a
as
as
0,90 5
-
I
Amperes
V
m/h
4
6
7
34
34
35
800
850
875
50
35
25
a
6
8
10
35
36
36
700
850
925
35
30
27
<pa
5
6
7
35
35
35
35
450
500
600
650
46
42
38
35
~
~
""'.. ./
8
d
--
volts
70°
a
a
a
b
._-
a = throat
thickness
Joint type
Stress-relieved I-section
Buckling about
x -x
--- -
U
For manual arc welding, it can be assumed that
U ~ I in formula (4.3.6) = 5· 10- 7mIJ
4:15
Static strength - Buckling of columns
For box columns made of welded and bent channel sections,
it can be assumed thata;la 5 = 0.1.
Table 4.3.3 shows for welded box sections that, in order that
curve a may be used, it is necessary thata/a 5 < 0.2.
A rule of thumb states that a throat thickness < 6 mm generally givesa;los < 0.2 for most corner-welded columns.
It is desirable to minimize the area of the weld, and this can
easily be done if the columns are only to be acted upon by a
compressive force. The purpose of the weld will then be largely
to hold the plates together.
lf, however, transverse forces act on the column, the area of
the weld must be adjusted to this condition.
Calculation procedure
After all these definitions, it may be useful to outline the calculation procedure, the main purpose of which is to calculate what
load the column can take and to compare this with the actual
load.
Gel = ;1t2.
Gel =
467· lO6
21 . 104 - - - - - . . , , 18048· 30002
5958 N/mm2
-J
a -
0.9' 700
5958
a = 0,323
1. Determine cross section and curve (y) as well as 0 5 * from
tables 4.3.2 and 4.3.3.
4. F = 1 + Y (a - 0.2) + a 2
2a 2
Calculate:
2. 0 el from (4,3.2).
3, a from (4.3,3).
F=
4. F from (4.3,5) if a > 0,2.
5, Ok/Os from (4.3.4).
6,
1 + 0,339 (0.323 - 0.2) + 0.323 2
2· 0,323 2
F = 5.492
Nk (critical load) from (4,3.1),
7, The permissible compressive force is obtained by dividing
Nk oWk by a factor of safety Sf, which can be put equal to
1.5 for steel structures under ordinary load cases, and to 1.3
for exceptional load cases.
For rapid overload of short duration (max. 35), Sf= 1.1.
Otherwise, rely on experience and section 3.
Example 4.3.1
Adesigner has to design a box column of WELDOX 700 (as= 700
N/mm2) and chooses between two fabrication types:
a ) corner-welded box column
b) bent and welded box column
Assume the following dimensions:
l = 3000 mm (Euler "2") 0400 mm
Nk = 0.9' 700· 0.956' 18048
Nk ::: 1087 . 104 N ::: 1087 tonnes
t = 12 mm.
Bend radius = 3 . t = 36 mm,
Throat thickness of corner welds = 7 mm.
The section area and the second moment of area are as follows:
a) A = 18048 mm 2
b) A = 17511 mm 2
I = 467· 106 mm 4
I = 398· 106 mm 4
Alternative b)
1. Curve a ando s * = Os according to table 4.3.3
y = 0.206 according to table 4,3.2
Solution:
Alternative a
1. Before the curve and 0 s * can be determined, we have to
checkoilo s' According to table 4,3.4, throat thickness = 7
mm (automatic submerged arc welder) gives U = 35 volts,
I = 875 A, v = 25 m/h.
2. Oel=;r2. 21· Ht·
o el = 5234 N/mm
~ = ---::-::----:-c~__:_1:::---=-=-=--__
as
25· 400· 12· 700
3600· 0,15' 35· 875
~= 0,25> 0.2
Os
i,e. curve b and a 5 * = 0.9 Os according to table 4.3.3
y = 0,339 according to table 4.3.2
4, lE)
(4.3,6)
3. a =
_1700
y-523'4
a = 0.366
2
398· Id'
17511.30002
Static strength - Buckling of columns
4. F=
1 + 0.209 (0366 - 0.2) + 0.366 2
2· 0.3662
F = 4.362
There is an area in the lower part of the graphs in which the
wall thickness for WELDOX is less than 4 mm. According to ECCS
E.R.. the cross section has been given such a shape that the
condition
~:;;; 1.52
t
/
E
V0k'aJ
1/2
is satisfied. The columns are assumed to have the same overall
dimensions.
Figure 4.3.4 As welded
6. Nk = 690· 0.962' 17511
M~
Nk
/
S 355
(curve b)os= 0.94·o s
iWELDOX700 (curvea)os= 0.90'0 5
6
.. ' Alternative b should be chosen. Because bending is employed instead of welding. it is usually the cheapest alternative.
5
4
Example 4.3,2
3
If the column according to alternative a) in example 4.3.1 is
stress-relieved, how large a buckling load can it take then?
2
Solution:
According to table 4.3.3. a stress-relieved box section gives
curve aa. as * = as and y = 0.125 according to table 4.3.2.
Figure 4.3.5
3) a =
_(700
V
5958 = 0.340
4) F ==
1 + 0.125 (0.340 - 0.2) + 0.3402
49
2· 0.3402
= .
5) ok/as = 4.9 _
V
1492 _
O~__~__. -__.-~~-.__- .__- .__- .__- .__-.~L
7
8
9
10 fTl
6
3
4
2
5
MN
7.0
1
= 0.979
0.3402
6) Nk = 690 . 0.979' 18048 == 1220· 104N == 1220 tonnes
Stress-relieved
Nk
WELDOX 700
6.0
S355 (curvea)os' =0 5
5.0
4.0
Compared with 1071 tonnes
3.0
.. ' Residual stresses play a large role. Therefore. use the least
possible weld metal area.
2.0
1.0
L
Are WElDOX columns economical?
It is difficult to lay down set limits for the economic use WELDOX
steel in columns, since a large number of factors are involved.
Naturally, one important parameter is the price ratio between
WELDOX and competing materials. It is important to take all
functional requirements into account when making economic
assessments.
When it comes to buckling, the high strength of WELDOX steel
is applied to best advantage in columns with a low value of the
slenderness parameter «, i.e. in stiff columns.
In the case of slender columns, load-bearing capacity is primarily determined by the modulus of elasticity, which is independent
of steel grade.
Figures 4.3.4 (welded) and 4:3.5 (stress-relieved) present in
graphical form the results of a technical and economic study of
welded square box columns made of WELDOX 700 and S 355
(BS 50 D). The study is based on the 1978 ECCS recommendations. The price is given per kg of finished box column.
The figures are used as follows:
Given: Nk, L and price WELDOX/price S 355 finished box
column.
Problem: Is it economical to use WELDOX?
Solution: WELDOX is economically advantageous if the point
(N k, L) lies to the left of the curve for the given price ratio.
O~--r--'---r--.---'---.---r--'---.---r---
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0 10.0 fTl
Example 4.3.3
The price ratio for welded box columns made of WELD OX 700 in
relation to S 355 is about 1.50. The column length is 4 m af)d Nk
= 3 MN. Is WELDOX 700 economical, despite its higher steel
price?
Solution:
Use figure 4.3.4.
The point 3 MN, 4 m clearly lies to the left of the curve for 1.50.
.. ' It is very economical to use WELDOX 700.
Simultaneous compression and bending
The stress situation under simultaneous compression and bending is more complicated than under axial compression owing.
among other things. to second order effects. out-of-plane buckling and transverse load.
4:17
Static strength - Buckling of columns
If bending only occurs in one plane and the beam is braced
so that lateral torsional buckling cannot occur, the following
relationships hold true:
{3 . M + M + N . e*
a + --'
0
oE;a s
p. -1
W
p.
M
a + W oE; as (as a check)
(4.3.7)
(4.3.8)
An automatically welded box column of square cross section is
made of WELDOX 700 (as'" 700 N/mm2). The column is pinjointed at both ends and is acted upon by an axial force
N = 3 MN, see figure 4.3.6.
Afterwards it is stipulated that the column should also resist a
uniformly distributed transverse load q kN/m.
How large a value of q can be permitted?
l
Figur4.3.6
where:
a= -
Example 4.3.4
N· 3 MN
N
A
p. =ael
(4.3.9)
a
2
I/A
ael=jf . E 7
(4.3.2)
/'
(4.3.10)
Vq
-15
E
o.....
MJ/M 2 = ratio of bending moments at beam ends! Ml! ~! M2!
'15
....
M= M2
Mo = largest bending moment due to transverse load. In determining Mo, the member is regarded as being simply supported. If Mo and M2 have opposite signs and if
! Mo! ~ ! 2M2! ' Mo = 0 is sUbstituted.
W = section modulus
A = cross-sectional area
I = second moment of area
I-
0300
-I
IN = 3 MN
5°
(4.3.11)
For the general case (bending in two planes), see ECCS recommendations 1978.
Solution:
= jf2 . 21. 104 .
a
First, a/a s has to be calculated in order that the appropriate
buckling curve may be determined.
From table 4.3.4, we obtain the value for a fillet weld of throat
thickness 7 mm and multiply the welding rate v by 2 (half fillet
weld).
el
232.1· 106
:: 281 N/mm2
17100. 1OO~
a =
./630
V
28f = 1.497
F=
1 + y (a - 0.2) + a 2
2. a 2
u = 35 volts
F = 1 + 0.206 (1.497 - 0.2) + 1.4972
2· 1.4972
I = 875 A v = 2· 25 mlh
_a_i = _-:--1...,.-_
F = 0.783
v·b·t·a s _ 1
~0'""";'.1-;:-5-.-:-:U-·7--1
!i=
1
012< 02
as
-=2-'-:2=-=5-'-:3:'::0-=-0-'-=-1-:::5-'-=7:=00=- = .
.
3600· 0.15' 35· 875 -1
'.' curve a anda s*= 0.9a s according to table 4.3.3
y
= 0.206 according to table 4.3.2
as = 0.9'
I
ak = 0.375 • 630 = 236 N/mm2
700 = 630 N/mm2
= (3004 - 2704 )112 = 232.1' 106 mm4
1061150 =
W = 232.1'
1.548·
A = 3002 - 27~ = 17100 mm2
a el = jf2. E. 11 A
7
4:18
ak/as = 0.375
e*'=
106 mm3
(4.3.2)
(.!l
-1 ). (1-~)~
ak
a~
A
e* = (630 _ 1 ) .
236
e* = 24,2 mm
(4.3.11)
(1 _ 236 ). 1.548· 106
281
17100
Static strength - Buckling of columns
0=
N
-=
A
fl = °el
a
3· 106
17100 = 175.4 NJmm
175.4 + 21.5q + 129.;;;621
2
q = 14.7 N/mm = 14.7 kN/m
= ~ = 1.6
Equivalent value of q if the column had been made of S 355
(crs= 350 N/mm2)
175.4
M = 0 since the end moments are zero ..
Mo =
qss 50 D = 5.05 kN/m
~8 L2 (uniformly distributed load)
0 +fl
- - ' f3'M+M 0 +N·e* ';;;0s
fl -1
175.4 +
W
1.6
1.6 -1
i.e. considerably less.
4.3.7
q. 190002 + 3. 106 . 25
-----~~--
1.548. 106
.;;;621
4,4 Buckling of plates
Page No.
Buckling due to normal stress ......................................... 4:20
Buckling due to shear stress ........................................... 4:21
4:21
Simultaneous shear stress and normal stress
When must the influence of local buckling
be taken into account? .................................................... 4:21
Stiffeners ........................................................................ 4:22
Postcritical region .... ...... ...... ..... ..... .......... ........ .... ....... .... 4:22
Examples ........................................................................ 4:22
Interaction between local buckling and
overall buckling ............................................................... 4:27
The buckling of a plate is fundamentally different from the
buckling of a column. In this manual, a differentiation is made
between local buckling (i.e. plate buckling) and overall buckling
(column buckling).
Figure 4.4.1 schematically illustrates the relationship between
load (N) and deformation (ll) for a column Cl and for a panel b
supported along its edges so that these remain straight during
loading.
The graphs in figure 4.4.1 show that the panel can support a
load considerably in excess of the bifurcation load Ncr , unlike
the column, which cannot withstand higher loads than NE'
The dashed curves show what happens in actual practice.
- When the buckling stress is exceeded, a considerable reduction of the stiffness of the panel also takes place. This is of
importance in cases where the panel forms part of other
structural elements such as columns and beams etc.
- Conventional calculation methods for estimating load-bearing
capacity are generally related to the critical buckling stress.
In the following, instructions are first given for determining the
critical stress for plates (panels) for some common cases, after
which methods for determining load-bearing capacity are
discussed.
Buckling due to normal stress
The critical buckling stressoel is calculated as follows:
o el = k" . 0.905 E (Vb)2
where k" = the buckling coefficient, which is dependent upon
the stress distr.ibution, conditions of support and
the geometry of the plate. See table 4.4.1
E = modulus of elasticity
t = plate thickness
b = plate width
Table 4.4.1 The buckling coefficient ka for some common
cases. *)
Load
Figure 4.4.1a
N
N
a
I~
~I
r
-,
I
I
I
I
I
L _________ J
r-t:.
I
I
I
I
I
I
I
I
I
a/b > 1; - 1 ~ 1/1 ~ 1
NE :::..-=--:-,.-_-",~
N
L
\
\
I
\
\
k" = 4 + 2 (1 -l/d + 2(l-l/Jl
l/J =1~ka=4
l/J = -1 ~ ka = 24
ideal column
err--,.
I/~
actual column
I
I
a
al
N
r
ideal
panel
Figure 4.4.1b
;
"
/
..- ..-
I
I
I
I
."
;
".f \actual panel
1
I
I
J
V'a)
b
I
Free edge
I
I
1
--<l/J~ 1
3
'Pal
ka= 1 +43tp [(b/a}2 + 0.426]
!pal
a
lr
- The buckling stress is important in cases where freedom from
buckling is a requirement, since the amplitude of the buckles
grows considerably when the buckling stress is exceeded.
I
I
I
I
Free edge
j
a)
1/'a)
I
I
I
The extent of this postcriticalload-bearing capacity is chiefly
dependent upon the panel's slenderness, material properties
and imperfections such as initial buckles and residual stresses.
It may therefore appear to be relatively uninteresting to determine the buckling stress associated with the bifurcation load
(NE in figure 4.4.1). But this is not the case, as is evident from
the following points.
-I
..,
-
1
3<l/J~1
ka = _4_ [(b/a)2 + 0.426]
3+1/1
*) For other cases, see ref. No. (71) and (72).
0)
Static strength - Buckling of plates
The above expressions are based on the assumption that the
plate is simply supported. When the panel is incorporated in
section walls in beams and columns, there is generally some
restraint in adjacent panels. This means that the above table
4.4.1 usually gives values on the safe side in cases where there
is some form of restraint.
where
a eh l' el '" the buckling stress for the panel when acted upon
only by normal stress (a el) or shear stress (reil.
OJ =
V a/ + 31'2
s* = coefficient that expresses the combined influence of the
normal and shear stresses on the buckling load. See
figure 4.4.4.
Buckling due to shear stress
The critical shear stress rei is calculated as follows:
rei = kr . 0.905 E(tlb)2
(4.4.2)
where
s·
1,1
~ = the buckling coefficient under shear stress.
See figure 4.4.2
~ = 5.34 + 4 (b/a)2
b/a::;:; 1
Figure 4.4.4 Diagram for determining the coefficient s* under
combined loading.
(4.4.3)
Figure 4.4.2 Buckling due to shear stress
r
°1
to' i
°1
1
I
I
1,0
'Pal
-- -
I
I
-- -
-
tjJ0l
0,9
b
I.
When must the influence of local buckling be
a
pi
taken into account?
The expression for ~ given in equation 4.4.2 applies for a
simply supported panel. If the panel is clamped at all edges,
then
~ = 8.98 + 5.6 (b/at
(4.4.4)
Generally, the greater the width-thickness ratio bit, the greater
the risk of local buckling. In order to determine whether local
buckling should be taken into account, the value of the slenderness parameter a must be studied.
In the case of buckling due to normal stress
(b/a)::;:; 1
(4.4.7)
Simultaneous shear stress and normal stress
In general, this problem is very complicated. Approximate solutions have, however, been derived for certain less complicated
cases. Here, the treatment of this problem is restricted to a
single case: Shear stress and linearly varying normal stress. See
fig. 4.4.3.
Figure 4.4.3 Simultaneous shear stresses and normal stresses.
I~
a
r
-~I
"l
I
I
I
I
I
I
L
.....J
I
when <:l.. ~ ;:, 1
r
~
rr;:Vr;-
a = -
(4.4.8)
t
= a sf'V3
rei = critical buckling stress as per equation 4.4.2
b
t
In the case of simultaneous normal stresses and shear
stresses
(4.4.9)
In the treatment, it is assumed that the loading is proportional, i.e. thala and r increase to the same degree.
The approach in treating this case is to calculate a "critical
equivalent stress" = ajel' This is done in the following manner:
ajel = ael ~ s·
al
In the case of buckling due to shear stress
1's
I
I
as = yield stress of the material
ael = critical buckling stress as per equation 4.4.1
where
t
I
I
where
(4.4.5)
(4.4.6)
whereajel = critical equivalent stress as per equation (4.4.5) or
(4.4.6).
If the panel were free of imperfections, buckling would not
have any influence on load-bearing capacity when a ::;:; 1. In
view of the residual stresses and initial buckles that are present
in actual practice, however, the limiting value a 0 (when a ::;:; a 0
it is assumed that buckling does not have any influence) must
be put::;:; 1. Depending upon the extent of existing imperfections, a 0 can be estimated to lie in the interval 0.6::;:; a 0::;:; 0.8.
In the current Swedish Regulations for Steel Structures
StBK-Nl,a o is around 0.7 -0.8.
4:21
Static strength - Buckling of plates
What do we do when a > a o?
As in the case of overall (column) buckling, the conventional
method for taking into account the influence of local buckling
on strength is to reduce the design value of strength with reference to a buckling curve. The postcriticalload-bearing capacity
which is a characteristic of slender panels is thereby taken into
account to only a small extent.
This design method is simple to use, but leads in the case of
slender panels to an underestimate of load-bearing capacity. A
model that better describes the behaviour of the panel in the
postcritical region is described further on.
Table 4.4.2 contains design values - fundamentally the same
as in StBK-N 1 - for normal, shear and equivalent stress in the
case of a plate supported on all four sides.
Table 4.4.2 Design values of O,! and OJ with respect to local
buckling of plate supported along all four edges.
a
od/Os
0~a~0.67
0.67<a~0.80
0.80<a~ l.49
1
1
a> 1.49
wherea =
1.557 -O.696a
1.154/a 2
!d/Us
Ojd/Os
0.577
1.1
1.557-O.696a
0.577
0.9-0.403 a 1.557-O.696a
1. 154/a 2
0.666/a 2
y 0s/Oel or~' Y Us/Ojel
according 10 equations 4.4.7 - 9
Table 4.4.3 applies for panels with a free edge.
Table 4.4.3 Design values of 0 with respect to local buckling of
plate with free edge.
a =
V 0s/Oel
o
~a ~ 0.70
0.70 < a ~ l.36
a> 1.36
- A box section can, instead of one 90' bend, have two 45°
bends spaced close together. This reduces the flat surface on
both the web plate and the flange plate.
t
2
- Rememberoel -
(b)
Postcritical region
A detailed analysis of the behaviour of plates in the postcritical
region is highly complicated and is therefore avoided in practical design work. For this reason, simplified models have been
developed based on the concept of effective width in panels
subject to normal stress and tension field theories for panels
subject to shear stress. These concepts were originally developed in aeronautical engineering during the 1920s and 19305
and have since been applied within other fields as well.
Here, we will deal only with panels subject to normal stress
when l/! = 1 (i.e. uniformly distributed over the width of the
plate).
In the postcritical region, the panel does not have a uniformly
distributed normal stress over its width, even if the externally
applied stress is uniformly distributed.
The stress is greatest at the boundaries, and a common
failure criterion is when the stress along the edges is equal
to the yield stress. The maximum load-bearing capacity Nmax of
the panel in figure 4.4.1 can be written:
Nmax = 0mB . b· t
umB. which is the mean ultimate stress, can be expressed with
the aid of Winter's formula, which can be approximated as
follows:
0.78
(44.10)
0mB ~ Os
3~
wherea =
Od/Os
1
1.53 - 0.75a
0.938/a 2
V Us/Oel
Uel = 4· 0.905' E
t
2
(I))
Note that 0 mB is proportional to as213!
If a mB/Os according to (4.4.10) is compared with a diu s according to table 4.4.2 for different values of a, table 4.4.4 is
obtained.
Table 4.4.4
Stiffeners
When we encounter limitations due to local buckling or when
we wish to stiffen up our structure, we can make use of
stiffeners.
Here, we have a great deal to learn from sheet metal designers and others.
Hints:
Study household appliances, car bodies, ship hulls (30 m
wide plates of 20 mm thickness are actually sheet metal structures), airplanes, steel buildings etc.
Some examples:
- Bend the free edge of e.g. flanges.
- Longitudinal stiffeners on beam webs are often adequate and
more economical than transverse ones.
- Holes in the web of a beam can be strengthened with longitudinal stiffeners or transverse stiffeners, or with both (longitudinal on one side and transverse on the other side of the
web).
- The holes can also be strengthened with a cover plate so that
the panel is made thicker locally around the holes.
- The load-bearing capacity of a beam web with holes is particularly important if the transverse force is large. The influence of holes on the bending moment is relatively small.
- Eccentric location of holes in the vertical direction increases
the load-bearing capacity of the web under transverse force.
- The holes should not be spaced more closely than 1110 of
the web depth and should not be larger than 3/4 of the web
depth.
- Care should be taken when applying loads near holes!
4:22
a
0.8
1.0
1.2
1.4
1.6
1.8
2.0
amB/a s 0.905 0.780 0.691 0.623 0.570 0.527 0.491
1
Od/as
0.861
0.722 0.583 0.451
0.356 0.289
The table demonstrates what was mentioned earlier, namely
that use of table 4.4.2 underestimates the load-bearing capacity
of slender panels.
On the other hand, table 4.4.2 gives higher ultimate values at
a < 1.3.
Equation 4.4.10 should provide a better estimate of the loadbearing capacity of a panel subjected to uniformly distributed
compression.
If there are initial stresses in the plate, the value of 0 m8 is
reduced. The following expression - based on results from,
among others, Dwight (69) and Little (70) - can be used to
estimate the panel's load-bearing capacity:
amB = os' 0.78' (1-0.8' o;los)/~
where aj = initial compressive stress (cf. figure 4.4.3).
The expression fOrOmB also includes the yield stress of the
plate and mB - Os213.
a
Example 4.4.1
A simply supported beam with a span of 7 m is made of WELDOX
700 (cr s = 700 N/mm2). The beam is acted upon by a point load
(P d) at midspan. Determine Pd with respect to local buckling.
The beam is provided with stiffeners at the supports and the
load is applied in the middle of the beam.
The beam has the dimensions shown in figure 4.4.5.
Static strength - Buckling of plates
r ,I
figure 4.4.5
Pd = 0.966' 106 N with respect to flange buckling.
d
i
L-I
.I.
3500
Web buckling:
The web is acted upon by both normal stresses and shear
stresses, and is considered to be freely supported.
3500
a1 = -
ltf 12
=
1750 Pd
622.3' 106
t -- :--
tL = 6
I
0
0
~--+-
-
I.C)
11
::I:
M
(H/2 - tf)
I
T
T"" - -
2Aweb
(500/2 -12)= 669.3' 10-0· Pd
Pd
= ----"----
2· (500 - 2· 12)' 6
2· (H - 2 . tf )· tL
I
Owing to a symmetrical linearly varying bending stress, tp =
- 1, which gives ka = 24 according to table 4.4.1
I
kr according to equation (4.4.3)
I
I·
B = 360
kr = 5.34 + 4
Section I -I
~. ~ =
Critical section at midspan under the load Pd'
when
(3~060 )2 = 5.41
= 0.862
(4.4.5)
~. ~ < 1 willajel be according to equation (4.4.6)
I<u
T
Bending moment
M = ~ = Pd' 7000 = 1750 Pd
Shear force
P
-f
= 0.5 P
T=
4
a·
a Je1= T eI' ..::::.L.
at s'
4
Flange buckling: The top flange is acted upon by a uniformly
distributed load, therefore tp = 1, and according to table 4.4.1,
ka = 4
Gel= ka' 0.905· E·
t )2
(T
Gel = 4 . 0.905 . 21 . 104 • ( 360 ~22 . 6
yI
5.34 + 4
Figure 4.4.4 for tp = - 1 gives s* = 0.76
Solution:
a =
2=
669.3' 10-0 . 5.41
175· 10-0
24
I<u
T
(+ )
yl700 =
as =
ael
Y= 903 N/mm2
=
kr . 0.905' E·
T el =
5.41 . 0.905· 21 . 104
ajel = 163·
0.880
903
t )2
(T
T el =
a =
734.7· 10-6· Pd
175 . 10-6· Pd
yI as =
ajel
(4~6 Y= 163 N/mm2
yl700
520
.
0.76 =
520 N/mm2
= 1.16
According to table 4.4.2 when a = 0.880
according to table 4.4.2
adlaS = l.557 - 0.696 . a
ad = 700 (1.557 - 0.696· 0.874) = 663 N/mm2
The mean stress over the flange
ajd = 700· 1.557 - 0.696' 1.16 = 525 N/mm2
M
. (H/2 - t f /2)
I
a=-
aj
I = 360·
663 =
500
12
3
-
1750· Pd
622.3' 106
348· 476
12
3
= ajd give 734.7· 10-6· Pd = 525 N/mm2
= 6223· 106 mm4
(500/2 _ 12/2)
.
Pd = 0.714' 106 N with respect to flange buckling
. .' web buckling is crucial and max.
Pd =0.714· 10' N =72 tonnes
4:23
Static strength - Buckling of plates
Example 4.4.2
Figure 4.4.6
The main members of a vehicle frame consist of box beams
made of 8 mm S 355. It is desired to reduce the dead load of the
frame by using 6 mm EHS steel grade WELDOX 700. The same
overall dimensions are retained in order to obtain as high rigidity
as possible.
Is it possible to use WELDOX 700 in view of the risk of buckling
when the frame members are loaded as shown in the figure?
Stiffeners are provided at the points of load application. The
maximum forces are 2.5 times the static forces.
2 tonnes
~
6 tonnes
~
2 tonnes
~
~~~~~~.~1~OO~o___*~I.__~20~O~O___'4!_ti~lO~O~c~J
4.57 tonnes
5.43 tonnes
t:: 8
! = 54.6' 106 mm4
W = 390· 103 mm 3
A = 5.76· 103 mm 2
H = 280
-rJ-
t :: (;
l/R;=3' t
WElDOX 700 (cr s = 700 N/mm2)
I = 43.4· 106 mm 4
B = 120
W = 310· laJ mm3
A = 4.44· 103 mm 2
Solution:
Critical section directly underneath the force of 6 tonnes.
M = 2.5' 48.6' 106 N mm
Flange buckling:
The mean stress over the flange is calculated from
M
-1- (H/2-t/2)
T = 2.5' 3.43 tonnes = 2.5' 34.3' 103 N
al =
S 355 (BS 50)
WElDOX 700
al -
2.5' 48.6' 106
54.6' 106
. (280/2 _ 8/2)
al=
2.5' 48.6' 106
43.4· 106
. (280/2 _ 6/2)
a 1 = 302.5 N/mm2
ko = 4 according to table 4.4.1ljl = 1
ael =
ael
4·0.905·21· 104. (
8
120-16-48
)2
ael
= 15514 N/mm 2
= 4· 0.905' 21· 104. ( 120- ~2-36 )2
ael = 5279 N/mm2
a=
a =
vi
350
15514
= 0.15 < 0.8
a=
vi
700
5279
= 0.36 < 0.8
i.e. no risk of flange buckling
Web buckling:
The web is acted upon by both normal and shear stresses
M
a1= I
4:24
(H/2 - t - Ri) (only the flat surface)
Static strength - Buckling of plates
WELDOX 700
S 355 (BS 50)
2.5 . 48.6' 106 (280 _ 6 _ 18 )
2
43.4' 106
2.5' 48.6' 1<f (280 -8 -24 )
2
6
54.6' 10
al = 324 N/mm
T
r==~-=--
2· Aweb
r=
2.5' 34.3' 103
2.b.8
2.5 . 34.3' 103
2·b·6
r
b = 280 - 2·8 - 2· 24 = 216
b = 280 - 2 . 6 - 2 . 18 = 232
r = 24.8 N/mm2
r = 30.8 N/mm2
Symmetrical linearly varying bending stress 1/1 = - 1 gives ko = 24 according to table 4.4.1.
kr = 5.34+ 4'
(+ y
kr = 534
. + 4· (~)2
1000
kr = 5.34 + 4 ( :;~
kr = 5.52
kT = 5.55
.!:!....~ =
= 2.80
r
ko
y
383 . 5.55 = 2.88
30.8
24
Figure 4.4.4 for 1/1 = - 1 gives s*
s* = 0.94
s* = 0.94
aj . S*
a·leI = a. eI' --'al
4. ( 8 )2
216
ael = 24.0.905. 21· 10
ael = 24 ,0.905, 21· 104 .
ael = 6256 N/mm2
ael = 3050 N/mm2
aj =
ajel = 6256·
306
302.5 . 0.94 = 5746 NI mm2
a =
.1350
V
5746 = 0.25 < 0.67
V 3832 + 3 . 30.82 = 387 NI mm 2
ajel = 3050·
,
(2~2 )2
387
2
383 . 0.94 = 2896 N/mm
a--[Ci;
V 0;a =
.fTciO
V
2s96 = 0.49 < 0.67
No risk of buckling and WELDOX 700 can be exploited to the
full. The frame members are
2· (5100' (5.76-4.4)'103· 7800'10-9 )=
108 kg lighter/vehicle
4:25
Static strength - Buckling of plates
=500 Os = 700 N/mm2 Nd =2.46.10 N
Example 4.4.3
b
A stiffened 5 x 2 m panel consists of a number of smaller 500
x 1000 mm "free panels" of 14 mm SS 50 D steel. The panel
is subjected to a compressive load in its own plane and in the
longitudinal direction. The entire panel weighs 1.5 tonnes, of
which the plate (14 mm) weighs 1.12 tonnes. Dead load is a
great problem.
Is it possible to reduce the dead load by using WELDOX 700
without reducing the load-bearing capacity?
which gives t = 10,3 mm, Le. t = 12 mm
6
The entire panel will be (14 -12)·8· 10 = 160 kg lighter
(8 kglm2 • mm)
or"" 11% reduction
How much of the yield strength of WELDOX 700 is exploited?
Gel = 4·0.905· 21· 104
(5~
Y= 437 N/mm2
oo = 1.264
~
437
Solution:
a =
WELDOX 700 permits the use of thinner plate. Will buckling be
the limiting factor? Analysis of original free panel:
adlas = 1.557 - 0.696' 1.264 = 0.687
Figure 4.4.7
Le. about 70%
a = 1000
Thus, WELDOX gives a lighter structure, despite the fact that its
yield strength cannot be fully exploited.
b = 500
Example 4.4.4
-
- -__4 -____________________~__~_
S 355
t = 14 mm
ael = k,,' 0.905' E·
ka =
(+ Y
A panel made of S 355 in a structure subjected to a uniformly
distributed compressive load will be subjected to severe overload.
Is it possible to sustain the overload better by switching to
WELDOX, despite the fact thatlthere is already a risk of buckling
with" S 355?
The panel has the following dimensions:
Figure 4.4.8
.
4 according to table 4.4.11/1 = 1
ael = 4·0.905· 21· 104
a = 1600
.
(~)2 = 595 N/mm2
b = 980
t = 10
a
•
.
ra; = -)350
= 0.766
Va;
595
a = .
according to table 4.4.2, a dla s = 1 when a. < 0.8.
We can therefore fully exploit the yield strength of SS 50 D. The
total load-bearing capacity of the panel is then
Nd = ad' t· b = 350· 14· 500 = 2.46· 106 N
Solution:
First calculate a for S 355
WELDOX 700
In order to be able to use WELDOX 700 (os = 700 N/mm2), we
must make the panel more slender, i.e. a > 0.8.
Assume 0.8 < a 0::; 1.49, then adlas = 1.557 - 0.696' a.
according to table 4.4.2.
ka = 4 according to table 4.4.1 (1/1 = 1)
Gel =4' 0.905·21· 104
(91~0
Y= 79.1 N/mm2
r;;; = -) 350
= 2.1 (Le. fairly slender panel).
Va;
79
a. = .
The failure stress in the postcritical region G mB according to
(4.4.10) is
This is then substituted in the equation from table 4.4.2
GmB = G s '
2
Nd
=1.557-0.696'.jG S 'b
42
t'b'a s
4·0.905·21·1O·t
0.78
3y;;2
GmB = 350 0.78
3v2.i:2
_1_ ~ + 0.696' b - /
as
4
t (b'a s )
V 4·0.905·21·10
4:26
= 1.557
= 166 N/mm2
Nmax = amB . b· t = 166·980· 10 = 1.63· 106 N
Static strength - Buckling of plates
Assume that we choose WELDOX 900 (as'" 900 N/mm2) as a
suitable WELDOX steel.
_/900
V
79.f = 3.37
a =
If a comparison is made between corner-welded and bent box
columns with a slenderness parameter of aab "'" 0.6, the bent
columns have an approximately 10% higher load-bearing capacity than corner-welded columns of the same plate thickness
and overall dimensions.
Example 4.4.5
A column of WELDOX 700 (as = 700 N/mm') is to have a length
L =: 3000 mm and overall dimensions of 0400 mm t = 12 mm.
Nmax = 312·980· 10 = 3.07 . looN i.e. 88% greater load!
Bending radius R = 36 mm
A = 17511 mm 2 .
Calculate the maximum load with respect to local buckling l
Solution:
Interaction between !ocal buckling and overall
buckling
The interaction between local buckling and overall buckling is
relatively complicated.
When a column fabricated from welded plates undergoes
overall buckling, we suspect that the "buckling failure" must
start locally. Overall buckling is initiated by local buckling.
The buckling of e.g. a box column is a typical case of the
interaction between local buckling and overall buckling.
A simple calculation method for box columns based on
Ingvarsson's studies (63) of welded and bent columns of
WELDOX 700 (as = 700 N/mm') is given below.
alb= 0.5259'
~ [ ~ -(2-\12)· (~-+ 1)]
alb= 0.5259-
V
alb = 0.94
a/as = 0.1 bent corners t = 12 mm
aob=
v'6.
v'6
Ib = local buckling.
aob = -;;-'
2. Calculate or estimate the influence of welding residual
stresses a/a s.
aob = 0.34
3. Calculate aob ob == overall buckling.
4. Use one of the diagrams in figures 4.4.10 -11 in order to
determine Ok/as'
These diagrams are taken from (63) and apply for the EHS
steel WELDOX 700 (as'" 700 N/mm').
A
II
The calculation procedure is as follows:
1. Calculatealb
- ~ (3+ 1) ]
-700
- - [400
--(2-v2)·
21· 104
12
~~
w
V T
V
700
21· 10 4
3000
400
Figure 4.4.10 givesokios = 0.78
0s*
= as
= 700· 0.78 N/mm2
Ok
Nk = 700 . 0.78' 17511 = 9.55 . looN
5. Max. column load Nk = Ok' A, where A = the cross-sectional area.
Figure 4.4.10
Local buckiingalb is calculated from the formulas in figure
4.4.9 for the relevant section.
Ok/as
- r-
1.0
Figure 4.4.9
-.,
0 1
\
~
I
I
I
0.1 r-
-
as
r- .......
"
r- -. -......:: f:S ::.... '--- Euler
0.5
u,o ~ 0.5259'
/R
I~
~
"
w
I:
w
<
t
·1
"(j ~ ~ ~ "
~ ::--.. f::: ~ ~,
1,'/
alb =
y?j.!i
E
-"
O':f; L 1.0~ f ""f-1.4//~ §::: ::::-- r::::::::::: ~
o.~~
0.8 - l.\~ i - f-1.~1
l.8
0.9
1.2
0.0
0.0
0.5
1.0
"'=:
~
~ ....
-
-
"'"
2.0
1.5
Cl ob
Figure 4.4.11
-
r- -.,
The influence of welding residual stresses can be calculated as
described in the section "Buckling of columns" or estimated as
foHows:
For box columns with bent corners t max = 12 mm, a/as"'" 0.1
For box columns with welded corners with a maximum
throat thickness of 6 mm welded in one pass t max = 16 mm,
0.5
a/as ~ 0.2
alb = <
aob for a square box column is calculated from:
aob=
v'6.
II
V
where L = critical length
as .~
E
W
\
-=0.2 r-. ~
as
,
,
r- t-..
r- ~ ::---. ' / ' Euler
\
-
I
I
r-.....;
r-~
.... -..; F:: :::-...'
s: ~ ~ .
i""~ ~ r---; ~ ~
~ -l.4/,/
0.8 r-- l.qZ
LV r-- -1.~/
l.2
0.9
l.8
g:l~ ~
a:
-.
....
f:: :::::: b- ""-
0.0
0.0
0.5
l.0
1.5
2.0
aob
4:27
The warping u is heavily dependent upon the cross-sectional
shape of the beam - cL figure 4.5.1. In this connection, it is
common to speak of
4.5 Torsion
- warpfree cross-sections that consist of radially symmetriC
shapes
Page No.
General discussion of torsion ........................................... 4:28
- quasi-warpfree cross-sections, including solid box, Tand L-beams
Formulas and diagrams for calculation ............................ 4:28
Calculation examples
4:31
- warping cross-sections, including 1-, channel, C-, Z- and
hat sections.
Torsionally stiff - torsion ally flexible structures ...... ...... ..... 4:34
If free warping of the end cross-sections of the beam in figure
4.5.1 b) is prevented by means of some device such as an end
plate, this will give rise to normal stresses a w along the beam. In
the case of thin-walled sections, the warping normal stresso w is
calculated as follows:
0w = 0w (x,y,z) = EqJ "(x)w (y,z)
(4.51)
where
E
w
= Modulus of elasticity
= the warping function, which in this case is equal to
the sectorial coordinate (more about this below)
"Ordinary torsion" (Saint-Venant torsion) is well known to most
designers and does not normally cause any problems in design.
What are almost completely unknown to most designers are
the secondary stresses that arise in connection with torsion
especially in open sections, at holes in closed sections etc. '
These secondary stresses are often very high and frequently
crucial in determining the strength of e.g. a welded jOint in the
corner of a frame.
These high stresses call for the use of WElDOX steel in
order that plastic deformations in the structure may be
prevented.
This may seem at first glance to be a somewhat murky and
difficult subject, but with the aid of an analogy with 'ordinary"
beam bending, we have tried to simplify the analysis as much as
possible.
General discussion of torsion
If a beam is loaded with a twisting moment (torque), a rotation
Ijl = Ijl (x) of the cross-section of the beam occurs, see figure
4.5.1.a. In addition to this torsion, a displacement also takes
place perpendicular to the cross-section of the beam (except in
the case of radially symmetric cross-sections). This phenomenon
i~ called warping and is designated here by u = U (v,y,z), see
figure 4.5.l.b.
Figure 4.5.1
a)
The external torque MT is generally taken up as Saint-Venant
shea~ stress~s 'tv and as Vlasov shear stresses two The corresponding. portions of t~e torque MT are designated Mv and Mw,
respectively. If both kinds are present at the same time this is
called mixed torsion.
'
The shear stresses 'tv and 'tw behave in fundamentally different
ways, tv forming closed shear stress trajectories and 'tw open
ones - cf. figure 4.5.2
Figure 4.5.2
'Csv
y---Msv
Shear stresses in connection with
a. Saint-Venant torsion
b. Vlasov torsion
Cl' (xl
The following relationships apply for the partial torques Msv and
Mw:
, MSY= GKyqJ'
(4.5.2)
Mw = - (EKwp")'
(4.5.3)
where
b)
~;:::::==~--------------7~
G· = Shear modulus = E12Cl + v)
E = Modulus of elasticity
Kv = Section factor with respect to torsional stiffness
(see table 3.1)
Kw = Section factor with respect to warping stiffness
(see table 3.2.)
qJ
= Angle of twist
qJ', qJ ", qJ '" ':" dp/dx, etc.
The following relationship applies for mixed torsion:
MT = Mv + Mw = GKvqJ' - (EKwqJ")'
a) Radially symmetric closed cross-section - no warping (SaintVenantl.
b) Open thin-walled cross-section - considerable warping
(Vlasov).
4:28
(4.5.4)
In special cases, either My or Mw is almost completely dominant.
In such cases, the analysis can be performed on the basis of
equation (4.5.2) if Mv is dominant and equation (4.5.3) if Mwis
dominant.
Beams with a solid or closed thin-walled cross-section can be
assumed to undergo only Saint-Venant torsion, which means
that the analysis can be based on equation (4.5.2).
Static strength - Torsion
The value of the following parameter is studied for beams
with open thin-walled cross-sections:
VI<l50\l torsion:
(4.5.5)
Equation (4.5.3) gives the following equilibrium relationship cf. equation (4.5.6):
(4.5.10)
(EKwcp ")" = mT
Akesson (67) recommends the following limits for application
of the simpler relationships (4.5.2) and (4.5.3t
Equation (4.5.2) - only Saint-Venant torsion - can be assumed to apply when {J vL;;;. 15.
Equation (4.5.3) - only Vlasov torsion - can be assumed to
apply when fl yL < 0.7
When 0.7 < {J v L < 15, mixed torsion is assumed to exist and
equation (4.5.4) is used.
In figure 4.5.3, the ratio Mw/Mwfl v L = 0 is given as functions of {J v L, which can also be characterized by different crosssection types.
The symbol B = bimoment is used for the quantity - EKwlf'''.
The following relationship applies between Mw and B:
Mw = B'
Equation (4.5.10) is the same type of relationship that applies
for bent beams:
Figure 4.5.3
Mw
MwllvL=O
Warping with
warping restraint
(Vlasov)
1.0
Warping without
warping restraint
(Saint-Venant)
.......
0.8
0.6
o.1 0.2 0.4 0.8
--
0.6
Shape
C [
Vlasov torsion
An analogy exists between
Analogue "bent" beam
and
w
M
T
El
q
P
Deflection
Bending moment
Transverse force
Bending stiffness
Uniform distributed load
Point load
ti£.fff.f..&:.
4 6 810
2
(4.512)
Angle of twist
B
Bimoment
Mw Torque
EKw Warping stiffness
mT Uniform distributed torque
Mo Concentrated torque
mo
I'\.
04
0.2
(EIW")" = q
If'
~
"" ,
(4.511)
Mixed torsion
20
4060 100
80
I
o
In this case also, an analogous structure in the form of a bent
beam under tension with a tensile force = GK v, bending stiffness
EKw and transverse load = mr can be studied. For a
closer analysiS of this case, see Akesson (67). See also example
=
4.5.3.
The torque ratio Mw/Mwflv L = 0 as a function of{Jv L according to Kollbrunner, Basler 1969 (72).
Example 4.5.1
A C-beam is fixed at one end and has a clevis joint at the other
end. The beam is acted upon by a concentrated torque Mo at
midspan. Determine the maximum rotationcp if Mo = 100 kNIT!
Solution of the equations
Certain analogies with bent beams can be used to solve the
above equations.
Figure 4.5.4
Saint-Venant torsion:
A
My = GKvCP'
(4.5.2)
If the beam is acted upon by a distributed torque mT
D1T(x), the following equilibrium relationship applies:
1=4000
=
(4.5.6)
Solution
First determine whether a simplified analysis according to
equation (4.5.3) can be performed. In order for this to be
possible,fiv L~ 0.7.
which, together with .(4.5.2), gives
m1 = - (GKvcp'),
(4.5.7)
At constant torsional stiffness GK v, equation (4.5.7) can be
written
(4.5.8)
flv L = L
V GK/EK w
t3
103
Ky = + 300)+ 400] =
3 l II = -[2000
3
= 4· 105 mm4
The following equilibrium relationship applies for a bent beam:
M" =-q
(4.5.9)
The equations (4.5.8) and (4.5.9) are analogous, and for the
same boundary conditions, the angle of twist <p and the torque My
can be obtained as the bending moment M and the shear force T
in an analogous bent beam acted upon by a notional load q
mT/GKv and q = mT, respectively. The concentrated torque Mo is
replaced by notional point loads <p' MoIGKy and P = Mo,
respectively.
This analogy means that tables and diagrams for bent beams
given in handbooks can be used in the analysis.
=
=
Kw is determined with the aid of figure 4.5.20
~ = 100 = 0.25
a
400
~= 300 = 0.75
a
400
From the figure, we obtain
~
= 0.085
5
ta
Static strength - Torsion
Kw = 0.085' 10· 400 5 = 8.7· 1012 mm 6
VI .. so\, torsion for thin-walled open cross-section
G = E/2(1 + v) = E/2(1 + 0.3) = E/2.6
Both shear stresses Twand normal stresses a w arise in connection with Vlasov torsion as a result of the torque Mw and the
bimoment B = - EKwcp ", respectively.
flv L = Lv'GKy/EK w =
In general, the following applies:
= 4000 V4· 105 /2.6' 8.7· 1012 = 0.53
B(x)
a w (x,s) = -K- W (s)
When {:J v L < 0.7, the beam undergoes virtually only Vlasov
torsion, and a calculation can be based on equation (4.5.3).
Mw
= - EKw qJ '"
(4.5.14)
w
where
a w = the warping normal stress (assumed to be constant over
(EK w = Constant)
the wall thickness t (s) )
The following boundary conditions apply:
qJ
(0) = 0
• qJ' (0)
(l) = 0
qJ
=0
qJ" (l) =
0
= longitudinal coordinate of the beam
s
= curvilinear coordinate of the cross section
B = - EKwqJ "= the bimoment
The beam analogy is used to solve the problem. The boundary
conditions for the analogous beam are:
Kw = section factor with respect to warping stiffness
w
= standardized sectorial coordinate of the cross section
=0
w (0)= 0
w
w' (0)= 0
w" (l) = 0
(l)
x
The largest warping normal stress
which, expressed in verbal terms, means a beam which is
rigidly clamped at one end and simply supported at the other.
The maximum angle of twist is now obtained as the deflection
wmax of the beam with El = EKw which is loaded with a point
load P = Mo at midspan.
a \!]ax is calculated from:
max
(4.5.15)
Ow
where
Ww = Kw~ w I max
Figure 4.5.5
8
!
EI=EKw
I~
real beam
Values of Ww are given for some different cross sections in table
4.5.2.
P=M o
~
(4.516)
715};-
L
./
The warping shear stresses Tw caused by the torque Mw can be
calculated from:
TW(X,S)=
-
(4.5.17)
analogous beam
where
From a handbook we obtain:
Sw = the sectorial static moment.
PL3
w max = 107E I
->
qJ max =
Substituted numerical values give:
qJ max
=
Tw = the warping shear stress (assumed to be constant over
the wall thickness t (5) )
100· 106 . 40003
107· 2.1' 105 . 8.7· 1012
The sign convention for Tw is that Tw (s) is considered to be
positive when it is in the direction of the positive curvilinear
coordinate.
Expressions for the calculation of extreme values of Sw are
given in table 4.5.2. The directions of TW are also given in the
diagrams in table 4.5.2.
= 0.033 fad = 1.9°
Mixed torsion
Tv • T wand 0 ware calculated after M v' Mw and B have been
determined.
Calculation of stresses caused by a torque
Saint-Venant torsion
The maximum shear stress T ~ax occurs somewhere along the
boundary of the cross-section and is calculated from:
M
T max = _v_
v
Wv
Mv = torque
Wv = elastic section modulus in torsion of the crosssection
Table 4.5.1 gives values of Wy and the position ofT~ax for
some common cross-sections.
4:30
Summary of calculation procedure for beam
torsion
- Determine the type of loading and the end conditions
(boundary conditions).
- Determine the type of torsion involved - Saint-Venant, Vlasov
or mixed torsion.
- Determine the section forces (My, Mw etc) and deformation
(rp ).
- Determine the stresses (r y. Tw,Owl.
Static strength - Torsion
Example 4,5.2
Sw31 = 0.25(0.163 + 0.456)/2 = 0.0774
la
Determine the maximum shear stressT~ax and the maximum
warping normal stressa~ax for the beam in example 45.1.
SW32 = 0.0774 + [(0.75 -0.425)/2]0.163 = 0.1039
la
Solution:
The beam analogy gives the following torque and bimoment
distributions (Mw corresponds to the shear force T in the analogous beam and B corresponds to the bending moment M).
SW33
la
= 0.1039 - 0.425· 0.213/2 = 0.0586
The following is obtained from a handbook:
0.213 _ 0.0586 = _ 0.0054
SW4 =
ta 3
4
Figure 4.5.6
t
Equation (4.5.17) now gives
r=M o
tEl.",
L
1-------=-----<--1
11
Largest torque
1:6. 100· 106 . 0.1039' 10· 400= = 53 N/mm2
T max =
11
16' Mo
11
Mo
16
8.7 . 10 12 . 10
w
Mw (T)
Equations (4.5.15) and (4.5.16) give
3
a ~ax =
ill = 16". 100· 10 4000 = 629 N/mm2
6
.
119· 106
Ww
B (M)
3
.l.. MoL
I B Imax = 16 Mo' L
16
Example 4.5.3
Rectangular hollow section, with opening in wall, acted upon
by a twisting moment.
Figure 4.5.8
5
32 Mol
The following is obtained from table 4.5.2, case 4, and figure
4.5.19 (a = h):
--
~ = 0.425 e = 0.425· 400 = 170
Mo
a
I ..
Figure 4.5.7
b=300
,------
•I
Section A-A
I"
a
~~
e = 170
I.
~ = 0.425 = 0.213
i
2
(0.75-0.425)
2
=0.163
Wia = 0.163 + 0.25 (0.75 + 0.425) = 0.456
b
..
I
The part of the section that contains the opening can be
regarded as a beam of open cross section which is prevented
from warping at the point of transition to the parts of the beam
with a closed cross section. This is a simplified approach. since
adjacent parts of the beam are not completely rigid in practice
and therefore permit some warping. This simplified approach
causes the warping normal stresses to be overestimated while
the rotation is underestimated.
In applying the beam analogy, the case is equivalent to a
beam rigidly restrained at both ends, at one of the supports of
which there is settlement.
4:31
Static strength - Torsion
Figure 4.5.9
~ Mo
EI=EKw
.... __ ,
2
J ~2
In the case in question, N = GKv
A better value OfqJ2 is now obtained from
1
1+
which, after substitution in the above expression for the
bimoment B, gives
If,By L ~ 0.7 (cf. example 4.5.1), the case can be treated as
one of pure Vlasov torsion.
Owing to symmetry, the following is obtained directly:
Figure 4.5.10
For the stated limit,Bv l = 0.7, the above expression becomes
B = 0961 Mo' 11
.
B
2
i.e. the bimoment is overestimated in this case by 4 % if pure
Vlasov torsion is assumed.
The following applies for the torque:
Mw 11
I 11 I lfJ I I I 11
I
If mixed torsion is assumed to be involved, i.e. 0.7 <,Bv L < 15,
the case can be analyzed using the beam-column analogy, see
Akesson (67). In the beam-column analogy, the analogous
beam is also acted upon by an axial tensile force = GK y, and
changes in the geometry of the system are taken into account (a
problem of the second order). An approximate variant of this
analogy will be shown below.
SinceqJ' = 0 at the support, we therefore have a situation where
M~ax = Mo
M v reaches a maximum at midspan
qJ '3 can be determined from the relationship
qJ'3=8 1 +8 2
The left half of the beam is studied.
The following applies approximately:
Figure 4.5.11
8 1 + 8 2 "" B (1/2) + B (1/2) = ~= Mo 121
3EKw
6EKw
4EKw
8EKw
which gives
An equilibrium equation that takes into account the changes in
the geometry of the system gives
B = Moll!2 -GK vqJ2!2
In order for the bimoment B to be determined, qJ 2 must be
known.
If the axial load GK y is not taken into account, the following
relationship applies:
I1
qJz _ Mo (-2-)
3EKw
2-
3
_
-->qJ2-
Mo 113
12EKw
Owing to the axial tensile force GK v , however, the displacement
qJz is reduced. The following relationship can be used here:
which applies approximately to beams acted upon simultaneously by axial and transverse load. Wo is the deflection due to
transverse load alone, N is the normal force and NE is the
buckling load.
4:32
After B, Mw and Mv have been determined, the stresses can be
calculated in the same way as illustrated in example 4.5.2.
A numerical example of 'Cl case with a hole in the beam wall
(see example 4.5.4) is given below.
Static strength - Torsion
Example 4.5.4
When we have pure Vlasov torsion, example 4.5.3 shows
The frame of an off-road haulage vehicle consists of a closed
tube of dimensions given below.
A crane is mounted on the front of the frame and the axle
suspension on the rear.
The crane gives rise to a moment (15 tonne-metres) in the
frame, which is to be transmitted to the ground via the frame
and the axle suspension. The crane also causes a shear force of
two tonnes and the load 10 tonnes.
A hole has been made in the frame as shown in figure 4.5.12
in order to provide access to equipment located inside the
frame tube.
Calculate the stresses at the edge of the hole!
B = Mo' Ll
2
The warping shear stress is
Mw . SWmax
r max
w
Kw' t
The warping normal stress is
Figure 4.5.12
_ I BI
a max
W
-
WWmin
First calculate Sw max from table 4.5.2, case 4, and figure 4.5.19
(a = h).
E =
ela = 0.490
y = cia = 0.125
FtJO~m:~.
I..
.1
b
P = bla = 1.0
From table 4.5.2, case 4 (a = h):
a = 400 mm
~ = 0.490
i
2
b = 400 mm
= 0.245
2
50 mm
c=
1.0 - 0.49 = 0.255
2
Solution
Which type of torsion dominates?
Investigate the parameterpy' L = L1 ·
f
Kv = 3"" . I li =
W3 = w2 + y
V
GK y
EKw
(P + E) i
Wia = 0.255 + 0.125(1.0 + 0.49) = 0.441
83
y . a' t
""3 (50 + 400 + 400 + 400 + 50) =
2
= 2.219' 105 mm 4
-+
s.u
Kw is obtained from figure 4.5.21
cia = ~ = 0.125
400
bla = 1.0
ta
0.125
= - - (0.255 + 0.441) = 0.0435
2
s.u2 = ~ 1+ [ (f3 -E) a . tl2] . W 2
~2
- = 0.0435 +
ta 3
~
= 0.092
5
ta
(
1.0-0.49
) . 0.255 = 0.109
2
Kw = 0.092' 8· 4005 = 7.536' 10 12 mm6
s.u
E
G = 2.6 for steel
Pv' L = 400·
0.49
~ = 0.109 - - - . 0.245 = 0.0490
ta
2
2.219' 105
-2.-6-'-7-.5-36-'-1-0"12;- = 0.042 < 0.7
.: Pure Vlasov torsion.
~4
0.245
ta 3
4
- = - - -0.0490 =
0.0123
433
Static strength - Torsion
-~- = 0.109 is greatest and occurs in the middle of the
t· a3
horizontal parts, see figure table 4.5.2, case 4
w
Kw' t
w
3
WXA2 = ~ = 2.155' 106 mm3
15· 107 • 0.109'8.4003
T max
[t.
c
12+ t· c·
== 3.233' 108 mm 4
Mv' ~umax
T max
t . a3
a2
Ix"" ---r2+ 2· b· t 4+ 2
T- c
= 138 N/mm2
7.536' 10 12 • 8
M
abA2=
Sw 1 applies at points A 1 and A2
0.0435
TwAI, A2 = 138,
a max =
w
_
== 55 N/mm
0.109
atotA2 = 38.6 + 281 = 319 N/mm2
2
We see that the torsion makes the greatest contribution
tOOtotA2'
_I_BI_
°
Wwmin
Kw
Wwmin - w max
~ = 38,6 N/mm2
WXA2
If we had not included w, we would have made a large
underestimate ofa at point A2!
_ ~= 7.536'10 12
W3
0.441· 400"
Hint: do not locate any welds in sections through Al and A2.
Do as shown in figure 4.5.14, for example.
-
Figure 4.5.14
= 281 N/mm2
15' 107 . 400
ow
max =
1.068· 108 . 2
/
W3 applies at the point AI, A2
Large radii
.,' the following applies for the points Al and A2
o max = 281 N/mm2
w
The frame is simultaneously subjected to bending, and the
normal stresses can be added.
Our warping normal stresses will therefore be added to the
bending normal stresses!
Calculate the bending moment Mb in the section through Al
and A2
,
No transverse joints within this region
.\
Figure 4.5.13
5000
..
2200
2 ton
.
1500
1000
.
Torsionally stiff - torsionally flexible
Careful thought and planning is important when connecting or
anchoring beams subjected to torsion,
The example in figure 4,5.15 is taken from reI. (65).
"1400 300
10 tonnes
t
i
!
!
I
t
Front wheels
Al
A2
Bogie centre
3.28 tonnes
I I 8.72 tonnes
.,' The bending moment is greatest in the section through A2.
Calculate ob at the point A2
I x (section through A2)
4,3;1
Figure 4.5.15
Static strength - Torsion
Design a) appears natural, since the flanges are "available".
This design results in a beam that is alternately open and closed
with very high warping stressesa w at the points marked x.
This often results in fatigue cracks as shown in figure b. The
design has relatively high torsional stiffness.
Design c) permits warping and does not give at all the same
G w as design al.
Design c) is extremely torsionally flexible.
The designer should give careful thought to which solution is
best from a functional point of view, the torsional!y stiff or the
torsionally flexible alternative.
If torsionally stiff elements are to be used, the transitions
must be gradual and the stresses must be checked.
A transition between an open and a closed section should be
made as shown below.
Figure 4.5.18
The designer can choose between stiff and flexible alternatives
at points of attachment as well, as illustrated in the case below,
taken from ref. (66).
Figure 4.5.16
x
h
Figure 4.5.19
C - section of constant wall thickness
I.
b
..
I
In other words, welded all around.
In the alternative shown below, only the web and the wedges
have been welded to the end plate and the flanges have been
cut at an angle to permit warping.
a
Figure 4.5.17
=t
+-.----~I
23.5
1 . 5 , . - - - - - , - - - - - y - - - - , - - - - , - - - - , b/a
0.0524 rad
l _ _--,2.5
e
a
L_----12.0
I.
1.0-l----+~..:::::;.-+---_+--__!r_-__I
.. - 1 _ - - - ; 1. 5
lx
_~--"""Il. 0
c::::::::::+-=-_F=====l
0.5-+----b~.:::::::::=-b___
.9
.8
.7
.6
.5
.4
.3
.2
.1
The case with the flanges free gives approx. 50% lower
a max and an approx. 40% more torsionally flexible beam +
attachment.
0.0;-----11----+----1----+------1
0.0
0.1
0.2
0.3
0.4
c
0.5
a
4:35
Static strength - Torsion
Figure 4.5.20
figure 4.5.21
C - section of constant wall thickness
C - section of constant wall thickness
5.00
4.00
3.00
Kw
Tii5
.0 50 -t-~t_-<...c....,-+",c-+_-?"'I---lf_7"'-t--+:rL_-t---;
2.00
.040+---;;~~-/~~~-~~--_+~-t--+-~
1.50
.030+---~~r-~--;;~",c-+_--~-+---r~T-~
. 500+--+--::7'-F--+-:::".-qr--;jo~
4 OOI-F'--+_--lr""'-t--::o'f£--t-:7'''4- ,-J:,--t---;;>F--:;:j
.300-F'--+--b"""-t---l".-q-t-:ri--
.050+----""'-t---t£--t-''''----r--'''''t---+-'''---t--+'''--t--;
0.2
o
0.3
0.1
0.4 c
O.S
a
. OOOS'i--'--t--I-"---l---t---t---t--+--t---lr--l
o
0.1
0.2
0.3
04 c
O.S
11
4:36
Static strength - TO;SIon
Table 4.5.1
I
KVI Wv and Zv for beam cross sections Excerpt from reference 67.
I
I
Symbols
Kv
W,
lv
f max
t's
M
Mr
Mp
d<p1dx = M/GKy
= section factor of Saint-Venant torsional stiffness GK, Oength4 )
'" elastic section modulus in torsion (length 3 )
= plastic section modulus in torsion (Iengtth
'" maximum shear stress according to elastic theory (iorce/length 2 )
= shear stress at yield
= Saint-Venant twisting section moment (force' length)
'" twisting section moment at incipient yield
= yield moment for elastic - ideally - plastic material
Cross-section with the position of
'max = M/W,
Mf = W,Ts
Mp = l,.s
Z,
W,
K,
and description of
position ofT max
T rnax marked
1 Solid circle
d = 2r
0h=2,j
nd4 132 =.n:f 12
2 Solid rectangle
cl hf
Oi
\-!-l
3 Open cross section composed of
arbitrary number of thin-walled
parts, each of constant wail thickness t
:,~~
~
\l1.J
I,
4 Thin-walled circular ring of constant wall thickness t
.~
5 Thin-walled rectangle of constant
wall thickness t
~
.,:;"
7 Thin-walled rectangle of varying
wall thickness
g13
I1
I...-LlJ
;rd 3 1l2 = m~ 13
c2 hi'
In the middle of the long sides
(hi' 12)(1 - tl3h)
hit
1
1,5
2
4
8
20
'"
Cl
c2
c3
0.141
0.208
1.000
0.196
0.231
0.858
0.229
0.246
0.796
0.281
0.282
0.745
0.307
0.307
0.743
0.323
0.323
0.743
0.333
0.333
0.743
c2lhi t,2 12
cjlh j ti3 13
K,/t max
The factor cl corrects for the conditions at the connection between
parts of the cross-section. If the
parts of the cross-section are rectangular and more than two in number, the following is obtained for steelbeams of normal wall thickness:
Cl = 1.05 -1.15. For L-sections,
cl"'" LOO, for U-sections Cl "" 1.12
and for H-sections Cl "" 1.30
Stress concentration in concave
corners at the connections between
parts of the cross-section has been
ignored. Along the boundary on the
thickest part.
;rd3 tl4 = 2n~ t
.n:d2 tJ2 = 2nr2 t
Along the entire wall.
= W,
2a2b2 tI(a + b)
2abt
Stress concentration at the inside
corners has been ignored. Along the
entire wall.
= W,
Bredl's second formula:
4A2 1<1> (l/tlds
A is the area within the dashed midline
Bredt's first formula:
= W,
~3
I-L...l
6 Thin-walled ring of arbitrary
shape, of varying wall thickness t
.n:d3 116 =.n:~ 12
Along the entire circumference
4ib2
!(blt 1 + alt 2 + blt3 + a/(4)
c2 = 1,0 - 1,2
2Atmin
At the thinnest part of the wall
2abt min
= W,
Stress concentration at the inside
corners has been ignored. Along the
entire thinnest wall.
I
I
8 Symmetrical two-celled thin-wailed cross section with wall in
plane of symmetry
S~m
DOl
The cross section behaves in torsion as if the midwall did not exist. Use the formulas for cross sections 4,6 and 7.
4:37
Static strength - Torsion
Table 4.5.2
Kw, WW1 Zw, (jJ (s) and Sw (5) for open thin-walled beam cross sections Excerpt from ref. 67
Symbols
= section factor with respect to warping stiffness (length 5 )
= elastic warping resistance (length 4 )
= plastic warping resistance (length 4 )
'" standardized sectorial coordinate (length 2)
= sectorial static moment Oength4)
= centroid
= centre of torsion
w
Sw
TP
VC
Three end views of a positive cross-sectional surface are shown below for
each type of cross section. The geometry of the cross section is defined in
the first; thew (s) diagram is shown in the second and the Sw (5) diagram
is shown in the third. A positive bimoment B gives tensile stressesow (5)
where the w (5) diagram has been marked with plus signs and compressive
stresses where the diagram has been marked with minus signs. A positive
Vlasov twisting moment Mw gives shear stresses r w (s) which on the
positive cross-sectional surface are directed as shown by the arrows in the
5", d iagra m.
The diagrams are drawn for the case where the given algebraic expressions for the ordinatesw; and 5",;. i = 1.2 ... give a result with a positive
sign. The formulas for Kw apply even il these expressions give a result
with a negative sign.
Bisymmetric I-section
l
_
bh~:,aIbOla
_~,htbfl6
w(s)
_
Sw~s)
Kw = b3 h2t b 124
= I, (h/2t
Ww = b2 htb/6
= W2 h/2
Zw = b2htb 14
= Z, h/2
Kw =Ioh~+Iuhu2
= to ba 112
ha =[lu/(Ia+1u)Jh
= tu b u3 112
hu = h - ha
, bh/4
b,ht./16
2 Unisymmetric I-section
~~o J:
t'l~
:vvccY-; TIP hu h
tu
:,:r~:~la
_ b 2,hoto/B
Sw(s)
W(S)
Ww = minimum of 2Kw Iba ha and 2Kv/b u hu
h
, buhu/2
~
10
b'uhutu/B
Zw = minimum of hb/ tu f4 and hb u2 tu 14
3 Symmetrical channel section of constant wall thickness in each part
patrabOlaS.. 1
Wl +
+
e
W(S)
5",2
- W2 t e--
S.. (S):
e
Ww = K.,Iw2
= 3b2 tb 1(6btb + hth )
S",1
Zw = hb2 tb 12 and e p = 0 for 4bt h .;; hth
Zw = ep h2 t h /4+ !htb/21[ep2+ (b-epfjand
ep = (4btb - hth) 18tb for 4bt b > hth
= eh/2
5",1 = (b -
S",2
+ r)2 S,,,2
wl
Kw = b2 h2tb (2b-3e)/12
w2 =\b-e)h/2
er htb 14
5",2 = (b - 2e) bhtb 14
4 Symmetrical channel section with turned-in edge stiffeners and constant wall thickness (C-section)
E '
C=:~h
+ TP
V
t
Y
Parabola
Kw = (htl3)[(1 + 2,) Wj 2 + 2 ({3 + l' -dw22
'"
+ 21'w3 (W2 + (3)]
h ",(S)
l
e
12
-
=.8[21'(3-41'2)+ 3.81 f[l + 21'(41'2-61' + 3)+ 6{3]
Ww = Kw1w3
(1'1
~I
e=. b=/I h
5",1 = (1'htl2) (W2 + (3)
wl
= fh2 12
W2 = (.8 -e) h2 /2
W3
= W2 + l' (.8 + El h2
5",2 = 5",1+ [(.8 -Elhtl2] w2
5",3 = 5",2 - (ehtl2)w 1
5",4 = (htl4)Wl-~3
5 Polar-symmetric Z-section with turned-in edge stiffeners and constant wall thickness
Kw =(htl3[(3+2~)wI2+2!.8+1'-~)wl
+ 21'w 3 (W2 +(3)]
~
=.8[.8 + 21'(1 +1')]/[1 + 2(.8 +l'~
= ~h
Ww =Kw 1w 3
WI
= ~h2/2
5",1 = (1'htl2)(W2 + (3)
W2 = (.8 -~ lh2 12
5",2 = 5",1 + [(.8 -~)ht 12Jw 2
w3 =wz+1'.8h 2
5",3 = 5",2 - (~ht 12)w 1
Kw = b2 h2 tb .(2b - 31) 112
= b2 tI(2bt b + hth)
Ww = K.,Iw2
6 Polar-symmetric Z-section 01 constant wall thickness in each part
~~
Y-VC=TPr:
t~h!2\
I'b I b I
4:38
If'2
Zw =hb2 tb 12 and Ip.= 0 for 2btb ';;ht h
Zw = Iph 2 t h/2+ !htb/21[fp2+ (b-fptl and
Ip
= (2bt b - hthl 14tb lor 2bt b > hth
wl
= Ih 12
5", I = (b - It htb 14
W2
= (b-flh/2
5",2 = (b - 21) bhtb 14
Splicing of beams with different web depths
4.6 Location and strength
of welds
Figure 4.6.3
Page No.
General discussion of weld location and weld design ....... 4:39
Static strength of welds ....................................... :........... 4:40
Calculation of stresses in welds ... .................................... 4:40
Soft zones ...................... .............. ............................ ....... 4:41
Filler material for WELDOX and HARDOX steels................... 4:42
Tensile test results with different welded joints, filler
materials, heat inputs and plate thicknesses for
WELDOX and HARDOX steels...................... ..............
4:43
Formulas for different types of welded joints .................... 4:45
Dynamic normal load
Predominantly
static normal load
Design and location of welds
What is said below is applicable to all weldable structural steels,
but jf WELDOX and HARDOX steels are to be utilized to the
greatest advantage, even more thought must be given to weld
location and weld design than in the case of ordinary steels. This
is mainly because of the problem of fatigue at welds, the possibility of using welding electrodes intended for softer steels (undermatching) and the fact that location of welds in areas with lower
stresses reduces the danger associated with weld defects.
The structure should be designed with smooth, gradual transitions for the flow of forces, and the welds should be located in
regions of low stress and designed to give a low notch effect.
The structure should be designed so that it is easy to determine the pathways for the flow of forces, which often results in an
attractive appearance and also facilitates calculations.
Access must be provided for both welding and the inspection
of welds during the service life of the structure in order to permit
the early detection of fatigue cracks.
Some examples of the above:
Predominantly static moment
Dynamic moment
Figure 4.6.1
Figure 4.6.4
-,---
Welding defects resulting from welding in inaccessible positions.
Figure 4.6.2
Example of favourable welding conditions
r
/\
Better weld design
Ordinary weld design
Static strength - location and strength of welds
If excessive deflection is a problem, this qm be counteracted by
cambering, for example by means of a suitable welding sequence. Good information on welding sequences and weld deformations is provided in (23).
Figure 4.6.5
unloaded
p
~Ioaded
~
t
15
There are many reasons for this:
- The designer has succeeded in locating the welds in regions
of low stress.
- Softer filler materials are less susceptible la cracking during
welding and do not require preheating to the same extent.
- Filler materials with lower yield strengths are cheaper per kg
of weld metal, and the assortment is wider, especially for
submerged-arc welding.
- Welding shops are familiar with electrodes for ordinary steels,
and their assortment of electrodes in stock does not have to
be broadened just because they start to use WELDOX and
HARDOX steels.
When quenched and tempered or hardened steels are welded,
a certain zone adjacent to the weld will be tempered (softened)
by the heat of welding. Many designers have therefore asked
whether it is really possible to utilize the high strength of the
parent metal in the welded joint. These questions are dealt with
in this chapter.
Figure 4.6.6
Weld close to the neutral layer on beams in flexure. See the
chapter on load application and the chapter on fatigue
(examples 5.5.5,5.3.7,5.112).
Calculation of stresses in welds in general
)
Stresses in welds can be calculated in two principal ways:
a) For butt welds and complete penetration fillet welds see
figure 4.6.8.
b) For fiUet welds, see figure 4.6.9.
Figure 4.6.8
Figure 4.6.7
Note that welds that do not transmit any appreciable force can
nevertheless cause fatigue failure for example in the beam on
which they are placed.
Figure 4.6.9
Fatigue cracks
a = throat thickness
-'--_ _ _ _-+_--':: G,
Static strength of welds
A load is' considered to be static in accordance with most
standards and codes when the number of load cycles is less
than 103 - 104 . Information on the static strength values of
welds is therefore needed.
The yield strength of the mixed weld metal obtained when
welding C and CMn steels is often around 500 N/mm 2 .
This makes it easy to meet the requirement of many standards that "The weld metal shall have a strength at least equal
to that of the parent metal", since the yield stresses of the
standardized structural steels are < 500 N/mm 2 •
The situation is different when WELDOX and HARDOX steels
are welded, since these steels can readily be welded with filler
material intended for C, CMn and grain-refined steels as well as
with electrodes that give a yield strength of 700 N/mm2 and, in
som cases, higher.
Basic rule: Don't choose a filler material with a higher
strength than is absolutely necessary.
4:40
Figure 4.6.10
Static strength - Location and strength of welds
Thus, according to a) the stress in joints with complete weld
penetration which are in tension is equal to the nomiilal stress
in a structural cross section.
In the case of fillet welds, the stress components have to be
added together according to some formula to arrive at an equivalent stress or similar.
2
crj
_.?
=Ou<+0J..--ou·0J.. + 3TH
2
+ 3TJ..
2
ref8
the stipulation being that
0i :::; a perm and, at the same time, no individual stress components may exceed a perm'
Formulas for stress calculation for the most common types of
welded joints are presented in table 4.6.1, which is taken from
(65).
When it comes to the welding of WELDOX and HARDOX steels,
and when it is absolutely necessary to have the same strength
in the weld as in the parent metal, filler material can be selected
with the aid of figure 4.6.15, which is taken from our welding
brochure.
Figure 4.6.13 WELDOX 700, 20 mm
Submerged arc NiCrMo 2.5, {3 4 mm OK 10.61
h
CA:l1
1.7 kJ/mm
4 passes
350~------~--------~------------
OB = 880 N/rnrn2
10 mm
(weld reinforcement left on)
Since a certain relationship exists between hardness (HB)
and ultimate tensile strength
Soft zones
As was mentioned above, it is possible in many cases to use
filler metal of lower strength than the parent metal (undermatching). These undermatched weld metals, together with the
heat-affected zone created by the welding process, form soft
zones in the welded joint, which bring to mind the concept of
the "weakest link". What strength do we get when we weld
hardened and quenched-and-tempered steels, and what
strength is it possible to achieve with undermatching weld metal
in WELDOX and HARDOX steels?
A soft zone can be compared to a piece of rubber glued between two pieces of steel, as shown in figure 4.6.11 a. Under a
tensile load, the piece of rubber assumes a shape as shown in
4.6.11 b. It has basically the same shape in cross section as well.
Figure 4.6.11
the hardness curve (HVlO) also reflects the strength.
Tensile tests with the weld reinforcement left on gave OB =
880 N/mm2, which well covers the guaranteed value for
WELDOX 700 (780 N/mm2).
The strength requirement is met even at a higher heat input of
2.5 kJ/mm, as is evident from figure 4.6.14. At higher heat inputs, the toughness of the welded joint is lower (see also section
entitled "Toughness-brittleness"). As was mentioned above, there
can be certain advantages to using undermatched filler material.
As has also been shown, smaller plate thicknesses give softer
and wider heat-affected zones.
Figure 4.6.14 WELDOX 700
Submerged arc NiCrMo 2.5, {3 4 mm OK 10.61
h
~'
____
~~L
~t
_ _ _ ___
-_
-P
We can see that the state of stress is triaxial, and sense
instinctively that the pieces of steel help to prevent excessive
constriction of the piece of rubber.
We also understand that if the soft zone is narrow in relation
to the plate thickness, the joint can sustain a greater load than if
it is wide.
Thus, many factors are involved in determining the strength of
the soft zone. These factors are enumerated in figure 4.6.12.
Figure 4.6.12
LO-J
lOmm
2.5kJ/mm
OB = 805 N/mm2
(weld reinforcement left on)
OL----------------------------------Approximately 400 tensile tests have been carried out with
varying plate thickness, filler metal, heat input and joint type on
WELDOX 700 and HARDOX 400 (no preheating) to provide a body
of data to serve as a basis for design work.
- The ultimate tensile strength of the parent metal a Bp'
- The ratio of the ultimate tensile strength of the soft zone to
that of the parent metal, a Bsla Bp'
- The size of the soft zone in relation to the other dimensions,
hit and h/w.
- The capacity of the soft zone to undergo strain hardening
a - Efl.
The strength and size of the heat-affected zone is influenced by
the plate thickness and the heat input (kJ/mm).
The thinner the plate, the wider and softer the zone. Figure
4.6.13 shows that the heat-affected zone is about 3 mm wide for
the submerged-arc welding of 20 mm WELDOX 700, at a rate of
1. 7 kJ/mm, In 4 passes.
Static strength - location and strength of welds
Figure 4.6.15
Figure 4.6.1501
GAS METAL MC WElD!NG Wire electrode! (MiG-MAGl
Filler materials for
WELDOX and
HARDOX.
Note! Only when filler
material and parent
metal must be of
same strength.
... ...
.,
~~~
...
...
Approximate yield atrenqth (MPa)
500
ERaOS-X
ER70$.X
Mmnuteetul1H'
100
EA12(JS..X
!:A110S·X
EA100s
ERIOS-X
BOHlEA
EMK8
2.5Ni-!G
NiCrMo2,5-\G
X70-!G
EtGA
ElgamaliC 100
EJgaml!llie130
E1oamatk: 135
ESAB
OK Autrod 12.51
OK Autrod 12.64
OK Autrod 13.00
OK Autrod 13.12
OKAutrod 13.13
FILARC
Fit.re PZ eooo-S
FH.re PZ eooo
FlIare PZ 6042
MVAEX
Soslrand aWl
Bostrand 20
LINCOLN'
LNM-25
NiMo1·!G
X90-IG
OKAlrtrod 13.29
Bo-strsnd41
~tr.ndLWl
BoS1rand42
SMITWELO
lNM-28
lNM·12
LNM-H!1
OEAUKON
C.rbofll 100
CarbolifMo
SAF
NK: 70S
NlC 701..
CllrbofilNIMoCr
Carbofi! NiMo-1
N!C ..
N!C"
Figure 4.6.1Sb
GAS METAL ARC WELDING Flux cored electrodes (fCAW)
..
el...
~:
A..
ManufaeluRr
..
TI52:·FD
ElGA
ESAB
DWA55E
DWA55L
OK Tubrod 15.00
OK Tubrod 15.17
OK Tubrod 15.11
OK Tubrod 15.14
OKTubrod 15.25
7..
E1oxr-x
E9XT-X
eOHLEA
E12XT-X
E11XT·X
OK Tubrod 15.27
OK Tubrod 15.26
FILA~C
Filarc PZ 6125
Fil.rc PZ 6138
Filan: PZ 6130
MUREX
Coroflt955
CCfOftt R56
Fitere PZ 6145
Corofil NO 1
Corof!! B65
LINCOLN!
SMITWElD
Outershield 71 C-H
0ut9f$hl9ld 81 N11-H
OutershleJd T55-H
Oul9rshi~ld 91 K2-H
lrtn~/ekJ NR 203 Nil
OERUKON
Fluxofll30
AUJl.ofil:31
S8.ldulll 1tSA
Saldual100
Flu){QflI40
Flu){ofil20
SAF
...
ApptOJl!m.ta yield Ilrengih IMP.)
5..
EaXT·X
Enr-x
Filarc PZ 6147
Fllare PZ6148
FnarePZ8149
Fluxofil41
Fluxofl142
FJuxofll4S
Safdua1128
Satdual121
Figure 4.6.1Sc
GAS METAL ARC WELDING Metal powder cored electrodes (MCW)
cl...
~~S.
..
...
ERXT·X
...
£IXT-X
Manufacturer
...
Approximate yteld llrength (MP.)
7..
EHIXT-X
E11XT·)(
£12XT·)(
E9XT·X
:8~~~=
Hl52·FO
HL 5O-FO
ElGA
MXA100
MXA55T
ESAB
OK Tubrod 1•. 00
OK Tubrod 14.12
OK Tubrod 1-4.02
FILARC
Filarc PI 6102
MUREX
COl'Omlg57
LINCOLN!
SM!TWElO
OI1l&rshield MC-.7t().H
OERLlKON
SAF
AurofllM10
Seldu!l! 206
OK Tubrod 14.03
Fiflrc PZ6103
Coromlg NI 2,5
Coromig Mo 0,5
Safdua12S5
Safdua1270
FluxofllM42
figure 4.6.1Sd
SHIELDED METAL ARC WELDING (SAW)
.. ...
cia..
~~
..
Approximate yield strength. (MP.)
500
F7A(s4)-EM12
100
Fl0A( s4-)EXXX.X
FIA{s4)-EXXX.X
SOHlEA
B825!EMS2
BB25INi2-UP
BB25!3NICrMo2.5-UP
ESAB
OK Flul( 10.711
OK Aux 10.B2I
OK Autrod 12.24
OK Autrod 12.34
OKAux 10.62/
OK Aulrod 13.40
OK Aulrod 12.20
OK Autrod 12.22
LlNCOLNI
SM!TWELD
F11A(s4)-EXXX·X
...
F12A(s4rEXXX.X
F9A(s5)-EXXX·X
"enu'.etur"r
MUREX
CoromlgN1Mo
S3tinare BX3001
8os!umdWeQ
Satll'l3rc 8X4OO1
SaUnare BX400l
Bo!Ilrand S3Mo
BostTand S4Mo
lNS 129-P230
OK Flux 10.621
OK Autrod 13.43
L 61-Uncoln Weld 960
LNS 140A-P230
LA 9O-P240
LA 100-P240
OERLlKON
OP41 TTfRuxocord 31
OP121IOE-SD3
OP121TT/Oe·S2Mo
OP41TT/Fluxoooni 40
OP41nfFtuxocord 42 OP41TIIFtuxocord 45
OP41TIIFIUXOCOfd 41
OP121TTfOe-S3N1Mo1 Of'121TTfFluxocoro 42 OP121TIlAuxOCQfd 45
SAF
ASS89IAS 36-
AS 5891AS40
AS sag/AS 81
LNS 164·P240
lNS 166-P230
AS S891SAFCORE 6501
Figure 4.6.15e
MANUAL METAL ARC WELDiNG (MMA)
i~~~
el••a
Manufacturer
BOHlER
ELGA
ES~B
Approximate yield etrength (MPa)
390-
,90-
4so.S50
530-820
800-890
740--130
E7G1S
E7016
E'TtI28
EIOt8
EP01.
Etot8-M
E10018-M
E11018-M
E1201l-M
FOX EV SO
FOX ev 55
FOX HL 180 Kb
FOXEV60
FOX EV 63
FOX EV 70
FOXEV75
FQXEV85
Max~!.21
Ne
P48K
P70
P62MA
Maxel.22
P110MR
Mave1a 110
OK 48.00
OK Femu: 38.48
OK Femax 38.85
OK~.68
OK 74".78
OK 75_75
OK 76.92
FHerc C6
Fllarc75S
F\larc98
Fitarc 108
FilllfC SSS
FHarc 765
Fi!arcS1H
FilarcCSH
FHare C6HH
Fllerc B8S
Fllllrc 27P
Fortr8X 7019
Fttrromax
Fortrel( 8018C1
OK48.08
FILAAC
MUAEX
Filare 35
Filare 36S
Filllre98
FlIarc 118
Flb,.-e 108MP
Fortrs" 9018
Fortrel( N02
FortrexNQ1
Ferex 7016
4:42
OK 75.76
LINCOLN'
SMITWELO
Conare 49C
Baso 100
ConareV 180
K')'03
Conarc 600
C011tl1'C 70 G
E 100 18-M
E11018-1;(
C011are 80
e 120 13·M
OeALIKON
Extra
T4!'naclto
Fabaclto 160S
Tenacllo 10
Tenacllo 708
Ten,elloeS
Tenacllo7S
SAF
Safer N 48
Safer NF 510A
Sa!er NF 58
Safer NF 59A
Safer NF 52
SalerMD 56
SIl'erND65
SaferNF 59
Conatc 85
Tenaefto 100
Tenaelto90
Safer NO eo
SalerNO 100
Static strength - Location and strength of welds
Butt welds
Only ultimate tensile strengths are given, and it may be assumed
that the yield strength is about 0.95 of the ultimate strength.
As can be seen, it is advisable not to exceed 1.0 kJ/mm for 8
mm WELDOX 700 if the guaranteed values of the parent metal
are to be maintained.
2.4 kJ/mm can be permitted for 16 mm WELDOX 700.
It is perhaps surprising that undermatched filler materials such
as PhC6 and OK 12.51, which are intended for much softer
steels, have such high strength in welded joints with WELDOX
700 and HARDOX 400.
For example, Ph C6 in 16 mm WELDOX 700 in a double-V butt
joint gave CJ = 765 N/mm2, and the same filler material in
HARDOX 400 (single-V butt joint) gave 797 N/mm2!
Figure 4.6.16 shows clearly that it is often possible to produce
satisfactory joints with filler material of lower strength.
-I
Figure 4.6.16
~
Butt weld with weld reinforcement left on
Joint
uB(Rm)
uB(Rm)
Parent mtrl
N/mm2
B
mm
Electrode
8
8
8
8
50
50
50
50
Ph 118 (EllO 18-M)
Ph 118 (E110 18-M)
Ph C6 (E7028)
OK 12.51 (E70S6) + SK 203
6 x 3.25
1 x 3.25+ 2 x 4
5 x 3.25
2 x 1.0
V
V
V
V
0.8
1.6
0.9
1.3
812
760
763
598
827
827
827
WELDOX 700
16
16
16
16
16
16
16
16
100
100
100
100
100
100
100
100
Ph 118 (E110l8-M)
Ph 118 (E11018-M)
Ph 118 (EH018-M)
Ph 118 (EH018-M)
Ph 98 (E9018-M)
Ph 98 (E9018-M)
Ph C6 (E7028)
OK 12.51 (E70S6) + SK 203
2 x 3.25 + 5 x 4
4 x 3.25 + 2 x 4
2 x 3.25 + 3 x 4
2 x 3.25 + 3 x 5
2 x 3.25 + 5 x 4
4 x 3.25 + 2 x 4
2 x 3.25 + 2 x 4
3 x 1.2
V
X·
V
V
V
X
X
V
1.7
1.4
2.3
2.4
1.7
1.5
1.5
2.6
835
829
789
778
795
822
765
643
828
828
817
817
828
828
828
817
HARDOX400
8
8
8
8
8
50
50
50
50
50
Ph 118 (EH018-M)
Ph 118 (El1018-M)
Ph 118 (Ell018-M)
Ph C6 (E7028)
OK 12.51 (E70S6)+ SK 203
6 x 3.25
2 x 3.25 + 2 x 4
2 x 3.25 + 1 x 5
5 x 3.25
3 x 1.0
V
V
V
V
V
0.8
1.2
1.4
0.9
1.2
856
777
766
797
828
1324
1308
1308
1324
1324
HARDOX400
16
16
16
100
100
100
Ph 118 (Ell018-M)
Ph 118 (El1018-M)
OK 12.51 (E70S6) + SK 203
2 x 3.25 + 3 x 4
2 x 3.25 + 3 x 5
3 x 1.2
V
V
V
2.3
2.5
2.5
850
854
694
1276
1276
1276
WELDOX 700
Number of
passes
HI
kJ/mm
t
mm
Steel
N/mm2
BOl
• X = double -V butt joint
Non-load-carrying fillet weld
i.e. a bar welded across the test rod with two fillet welds, figure
4.6.17. With this type of joint, the filler material makes no difference. only the heat input is important. In this case also, 'some
caution should be observed, and the heat input for 8 mm
WELDOX 700 should be max. 1.3 kJ/mm if the guaranteed values
are to be met. With 16 mm WELDOX 700, 3.4 kJ/mm may be
used without adverse effects.
Il
Figure 4.6.17
~CI====~~====::::JI---
Non-load-carrying fillet weld across the test rod (bar left in place)
Steel
t
mm
B
mm
Electrode
Number of
passes l )
throat
thickness
mm
HI
kJ/mm
u.2 (Rel)
u.2
uB (Rm)
N/mm2
Parent mtrl
N/mm2
N/mm2
uB
Parent mlrl
N/mm2
WELDOX 700
8
8
8
50
50
50
Ph 118 (E11018-M)
Ph 118 (E11018-M)
OK 12.51 (E70S6) + SK 203
4 x 4.0
2 x 5.0
2 x 1.2
5
5
5
1.3
2.0
1.7
715
664
727
770
770
770
803
791
813
B01
B01
B01
WELDOX 700
16
16
16
100
100
100
Ph 118 (E11018-M)
Ph 118 (E110l8-M
OK 12.51 (E70S6) + SK 203
12 x 4.0
4x 5.0
4 x 1.2
10
10
10
1.2
3.4
2.4
786
795
766
775
775
775
825
832
832
817
817
817
HARDOX400
8
8
50
50
OK 12.51 (E70S6) + SK 203
OK 12.51 (E70S6) + SK 203
4 x 1.2
2 x 1.2
5
5
0.9
1.7
853
790
972
972
987
936
1308
1308
16
16
100
100
OK 12.51 (E70S6) + SK 203
OK 12.51 (E70S6) + SK 203
12 x 1.2
4 x 1.2
10
1.0
2.4
1053
1096
1096
1284
1203
1276
1276
HARDOX400
10
') Two fillet welds.
4:4
Static strength - Location and strength of welds
Non-load-carrying fillet weld along the test rod
This is where heat input is of the greatest significance, 8 mm
WELDOX 700 and HARDOX 400 being most affected.
The test rods here are relatively narrow, and with wider joints,
the soft zone is naturally of less importance. Plate thickness
above 16 mm causes no problems up to 2.5 kJ/mm.
Cf. figure 4.6.18
Figure 4.6.18
Non-load-carrying fillet weld along the test rod (bar left in place)
Number of
passes!)
aB (RmY)
a.22 )(ReU
Throat
HI
0'.2
aB
Parent mtrl.
thickkJ/mm
Parent mtrl.
ness mm
N/mm2
N/mm2
N/mm2
N/mm2
Steel
t
mm
B
mm
Electrode
WELDOX 700
8
8
8
50
50
50
Ph 118 (Ell0l8-M)
Ph 118 (Ell018-M)
OK 12.51 (E70S6) + SK 203
4 x 4.0
2 x 5.0
2 x 1.2
5
5
5
1.2
2.5
2.1
578
560
590
776
776
776
760
741
746
803
803
803
WELDOX 700
16
16
16
90
90
90
Ph 118 (E110l8-M)
Ph 118 (E11018-M)
OK 12.51 (E70S6) + SK 203
12 x 4.0
4 x 5.0
4 x 1.2
10
10
10
1.1
2.5
2.5
703
721
703
767
767
767
839
808
796
805
805
805
HARDOX400
8
8
50
50
OK 12.51 (E70S6) + SK 203
OK 12.51 (E70S6) + SK 203
4 x 1.2
2 x 1.2
5
5
1.1
2.3
716
571
937
937
839
749
1315
1315
HARDOX400
16
16
70
70
OK 12.51 (E70S6) + SK 203
OK 12.51 (E70S6) + SK 203
12 x 1.2
4 x 1.2
10
10
1.3
2.6
918
787
972
972
1059
897
1274
1274
1) Two fillet welds
2) Stress
= force
area
(Le. bending is not included). The actual maximum stress will be considerably higher.
Load-carrying fillet weld
Figure 4.6.19 shows that for a throat thickness"" 0.7 . tin
WELDOX 700, there is no need for filler material and parent metal
to be of the same strength. The load-carrying capacity of the welded joint is determined by the throat thickness and the
strength of the weld metal.
I.e., the softer the filler material, the greater the throat thickness must be.
But even this has a marginal effect if the throat thickness ""
0.7' t.
A heat input of up to 2.1 kJ/mm is quite permissible for 8 mm
WELDOX 700.
.
The corresponding, data for HARDOX 400 are presented in
figure 4.6.20.
Figure 4.6.19
Load-carrying fillet weld - OX 812
Steel
WELDOX 700
WELDOX 700
WELDOX 700
t
mm
Electrode
mm
Number of
passes l )
Throat
thick
ness
mm
0'.2 (Rel)
HI
kJ/mm
0'.2
Parent
mtrl
aB (Rm)
N/mm2
N/mm2
N/mm2
8
8
8
100
100
100
Ph 118 (E11018-M)
Ph 118 (E110l8-M)
Ph PZ 6132 (Ell018-M)
12x3.25
4x4.0
4x 1.6
6
6
6
1.2
2.1
1.8
737
784
775
766
800
798
8
8
8
100
100
100
Ph PZ 6130 (E70 T-5)
Ph C6H (E7028)
Ph C6H (E7028)
4x 1.6
4x4.0
12x4.0
6
6
9
1.8
2.1
2.0
781
756
765
797
768
788
12
12
12
100
100
100
Ph 118 (ll018-M)
Ph 118 01018-M)
Ph PZ 6132
24x3.25 ,
12x5.0
12x 1.6
8.5
8.5
8:5
1.2
2.0
1.9
813
815
823
843
846
850
12
12
12
100
100
100
Ph PZ 6130 (E70 T-5)
Ph C6H (E7028)
Ph C6H (E7028)
i2x 1.6
12x4.0
12x4.0
8.5
8.5
12
1.9
1.75
2.2
810
811
818
837
844
848
16
16
16
100
100
100
Ph 118 (Ell018-M)
Ph C6H (E7028)
Ph C6H (E7028)
24x5.0
12x4.0
40x4.0
11.5
11.5
18
1.8
2.1
2.1
772
1) Four fillet welds
4:44
B
..
780
812
744
807
Static strength - Location and strength of welds
Figure 4.6.20
Load-carrying fillet weld - HARDOX 400
HI
Os (Rm)
Throat
thickness N/mm2
mm
Electrode
Number of
passes ll
100
100
100
100
Ph 118 (Ell018-M)
Ph C6H (E7028)
Ph PZ 6130 (E70 T-5)
Ph PZ 6130 (E70 T-5)
12x3.25
4x4.0
4x 1.6
12x 1.6
6
6
6
9
1.2
2.1
1.8
2.0
1070
849
752
1036
12
12
12
12
12
100
100
100
100
100
Ph 118 (EH018-M)
Ph 118 (E11018-M)
Ph C6H (E7028)
Ph PZ 6130 (E70 T-5)
Ph PZ 6130 (E70T-5)
24x3.25
12x5.0
12x4.0
12x 1.6
12x 1.6
8.5
8.5
8.5
8.5
12
1.2
2.0
1.75
1.8
2.0
1100
920
777
893
928
16
16
16
16
100
100
100
100
Ph 118 (E11018-M)
Ph C6H (E7028)
Ph C6H (E7028)
Ph C6H (E7028)
24x5.0
12x4.0
40x4.0
24x5.0
11.5
11.5
18
18
1.8
2.1
2.1
2.4
1023
812
1053
988
B
t
mm
mm
HARDOX400
8
8
8
8
HARDOX400
HARDOX400
Steel
1) Total number of passes for four fillet welds
In summary:
- Filler material of the same strength as the parent material is
only needed in butt welds where the guaranteed values of the
parent material must be met.
- Be careful with heat input at smaller plate thicknesses.
- It is best to use tried-and-tested filler materials for softer
steels. They give the best production economy.
TabeI4.6.1
Formulas for different types of welded joints.
Excerpt from reference 65.
a...c:,........
S....,
TenSion.
compression
a - f-
Loadonidia&<am
I
r -
A
2
r
0b-
•
"
M,
W,
"
8
7
6
5
4
f.- -
DII·~
Mbo;'X
Notes
M,
M..
M"
!!.
3
Torsion
Benchna
Q
f
My
;'
L/t=¥ ~ M.
~
,?\
~
I'
q
2
F-c-
~
~fr-F
I~
X
Yj
~
~
tFq
M
r~2 r-X
/iy
,--
aZ
4
,--
5
,--
6
r--
.E~'7tJ)·
,X
a -f-
r ... ~
.,1
all-
flQ-~
A"
6a~~~
."1
f
ill' 1', + 12'1'2
f-
f,
Ob- MbI'Xlor~
a,'11 + az'l'z
1..
1.&
Ob-~or~
1..
looy
iI,·
ill
1',< I> 1'1
Incomplete weld
penelr ahon akW8
enllte ItOilh
f
a--
r.~
a---ta, +
r.~
2· ,-I'
a,·
i2
1',- 1'2- I'
2'0'1"
~
F-f-
§
§§
;:
==
t
Fq
1
r--
a_
3
6'M ba
f,
a -FA"
f
,f-F
.I'
Xy
i2)' 1
(a,+ a2)-1
Ob-~
(al + '2'- 1
o.~
•
I ..
Incomplele .eld
pentllahon
a __
f _
2, 1
a·
,,+a2- 1
(1-~
2·,·1
ab -
3,·.~r
(1t1-
(1b-
;~1'
(1.,-
~L
I ..
Iy
a-f
.,1
r _ !a..' I
6a~~r
"1" il2 - a - l
4:45
Static strength - Location and strength of welds
Bas.it rcrmull}
She;y
TenSiOn,
Bending
compres~"
C
loading dIagram
1
0'"
!.....
A
2
9
~F
!-11
t,
12
13
14
15
Ob'"
r"'~
(1b '"
62·.~r
",2 ~
a -,
,.
f,
(a,+ a2 + 1'3'" c.!I1 1
Ob'" M'.l~· )(1., MI;n" KZ
,.---'~
(a,+3]+2'a]) 1
Ob'" Mtp:·X,.Me.·Xz
1.
1.
cb'
v. _ _
' __
2· la, + .121 I
1'''~
Ob"
",
o ...-~-
r .. ~
{al -+ a.. 1 1
ab-
o•
FQY
t I Iy
~4
a4
Structural
cross section
16
FQ,
,
F
(31+ 32+ 2 a)l
1
lal + al)- I
0
___
' __
r .. ~
2- a\, 1
2- "I 1
c. _ _
' __
2· al' I
r.~
O"'~
f ..
2· a)· 1
18
.---'-2-a(H+Bl
a
6·M!1x
.
6·M!?l
11'11+"2+ 2 -<13)
<lJ"" 34
l'
3· ~'c.
al ,.., <li
~-+;-<\7
"1 ..
0, • ~
c, •
:3. Mbr
l'
a] ,.. 3;
qb"'~
Oh""
'I -1
"
0 _ _' _
4· a H
Connecbon
cross se<:!;on
61·'~Y'
"
Structural
cross sedan
21
T.~
o~- ~'.~~
0,'
6'H~b;1
Qb'S
,.4-
2· Am' a
B- H
M,
c, _
0,-
!7"'-W
', __M_,_
2' a· tH+al-tB-+al
f""~
S-H
. iN,M,
','
2· B' a
B- "
,
3'M!!t:'{S+2-a)
a'S 3 +a J 'H-+3 a-B-1B+ar
~
2· H· a
c._'-
20
-
0
011"" dZ = G
i'l]'" a2
ob'
3· MM' (H + 2, a)
:=
11) = 3'~
"
Ob"'~
33' ]L
3· M!»'tl
" - I·fal + 3· tll-a}J
1!~
a~ '" ]J~ '" 0
{IJ\ + .121- l~
1.
a'Hj'a3oB+3'~
I
-IS\- f-!i-l
!--
~~. MI,a" X?
3 Mtl."H!+2'311
0, -
rDEbJ~
19
MbJ;' X
" 1·1a1'+3·itl~
0,'
Single fdlet
. .Id
la! + 32+ "3+ i!4)'1:!
I.
c·
8
5;.,?11t
c,~
I.
I.
7
My
Fqy
-
.£.L
1",
~F
xi yl
!--
1.
2 la! + all· I
1 I)
~~'
II
,- 1
(al -+ "l+ a3+ a~l'l
c-
g.
!-!--
IFq'IY'"
L
W,
6t.~J~
o ... ~ _
a· 1
!i
5
5
0t.""
'-I
~M
1-
10
t"l"'-
1,
!a..
t""
I-I
My
Ub""~
I,
Not",
M,
M",
Mbx' X
4
OIE~_
{IO!>!; !-eCtion
!--
Ob""
3
Structural
8
!S.
A
r'"
Torsion
M..
Q
3· M.' (8 -+ 2· ~l
a· H3 + a'· H + 3· a· H· (H -+
Qb"'"
a-f
3· M!?I· (B -+ 2· a)
,s. H3+ a)' 8 + 3'3' B· tB-+ af
M,
',' 2· A,.,., - a
B""H
~~'1
1','""
!i
W,
S-H
H
~
r"O_~'Hj
e~H
fI?l ~x
My
--~F
~~
22
1', ... 2L2 - Am -.
,
0"-O,lt" Cl
1'''''~
0,'
iT:-
(J1l-
O'lI"' Cl
16· Mm: (0+ 2· a)
tro({D+ 2 oaf_oi J
Ob'
32· MbJ,
O)'n
o!) ...
',m __2_-",,_ _
16'Mpz'(O+2'4)
rr"{{D+ 2· )l_OJl)
ro+a~·a·1f
D'
Connection
cross section
a
Y2
27
4:46
Tension. compressiOn
0, -
y~
xli.
P--I
32" Mby
Shear stress due to
transverse forces
ft"
~
It?·.~t
Equivalent stres~
Mbl · YI .. Mbo ' Yz
1,
I,
ILl
rr
~m i()
1 YI
!--
f -
under bendmg
25
26
° '" or;-
Beam
2'
-
.,
Structural
cross section
23
hH
0, "
Me.' y'! .. Meo" y'2
I,
I,
T"~"'~
1~' 2·
a
11 " 2· a
o~"'''s,2+ 2-r Z "OdM
Slatic l02d
(u::. )~+ (k-)2~
Dynamic load
w
ab'" Mbi:" y'\ .. MI)I:" Y2
1.
l~
t,..~
,- h
1.1
o~- "0/ + 2f2"'Ol!~W
Static !Q3O
(~)'+
(!...}'
a.!,n
r,,. .. 11
Dynamic IMd
4.7 High-temperature strength
Page No.
Strength data at high temperatures ................................. 4:47
Creep data ..................................................................... 4:47
WELDOX and HARDOX steels are not intended for use at temperatures above 400°C. They can, however, be used to advantage at
temperatures up to 400°C, owing to their high yield strengths.
Note that at temperatures above 350°C, creep must be taken
into account.
The yield strength of a steel decreases with increasing temperature, and at a limiting temperature Tg that is = 0.4' the melting
temperature (oK), a viscous deformation also takes place, which
is of considerable importance ab~ve Tg.
During this viscous deformation, the steel deforms continuo~
usly under a constant state of stress. The steel is said to creep.
The region above Tg is called the creep region, and in this
region there is no lower limit of the stress state below which
creep does not occur.
Of our WELDOX and HARDOX steels, the WELDOX steels are
delivered in the quenched (hardened) and tempered state with
tempering temperature = 600°C, while the HARDOX steels are
delivered in either the hardened or the quenched and tempered
state. We know through testing that creep first appears at about
350-400°C for these steels, which does not entail any additional
tempering for WELDOX steels, but bear in mind that HARDOX
steels lose both in hardness and in yield strength at these temperatures. Contact us in special cases for HARDOX steels.
WELDOX and HARDOX steels are not heat-resistant in the
normal sense but can well be used at temperatures up to Tg,
owing to their high yield strength. See table 4.7.1, which shows
yield strength at high temperatures (design values).
In the case of hot forming or in other special cases, it can be
useful to know yield strengths at even higher temperatures.
These typical values are shown in table 4.7.2. Note that creep
must be taken into account at temperatures above 350°C.
Creep can be described and classified in accordance with
figure 4.7.l.
If a steel e)(hibits a pronounced secondary period when the
creep rate £ is constant (a = constant), the steel is said to be
stabie with regard to creep. This secondary period is of the
greatest interest, since it can have a duration of several years.
Data of interest in this respect are the creep modulus and the
creep rupture limit. The creep modulus a c7 (N/mm') is defined as the stress that gives rise to a creep rate of E = 10-7
l/hour, i.e. the stress that gives rise to 1% creep in 100.000
hours. Similarly, 0c6 is the stress that gives rise to 1% strain in
10.000 hours. The exponent n in Norton's creep law is also of
interest. Some values are given in table 4.7.3.
The creep rupture limit cB m (N/mm') is the stress per
original area that gives rise to rupture after creep for 10'" hours
at a given temperature. Values for WELD OX steels are given in
table 4.7.3. This table shows that WELDOX 700 t>45 mm is the
steel that has the best creep properties.
WELDOX and HARDOX steels have found their widest use at
high temperatures below Tg, by virtue of their high hardness
and yield strength, in applications such as flue gas fans,
cyclone cleaners and pipelines where highly abrasive particles
occur in combination with temperatures below Tg.
°
E strain
rupture
Figure 4.7.1
secondary
tertiary
time
Table 4.7.1
°0.2 Design values N/mm2
Steel
20"C
50
75
100
125
150
175
200
250
300
350
400
OX 520 P 355
WELDOX 700
350
343
338
329
319
309
299
289
270
240
216
196
Data sheei
690
643
620
600
595
590
585
580
570
560
550
540
Ref44
Table 4.7.2
Typical 0.2% proof stress a 0.2 at temperatures> 400°C. Creep must be taken into account.
Steel
450"C
500"C
550"C
600"C
WELDOX 700 t> 45
WELDOX 700
620
550
580
470
530
350
410
230
Table 4.7.3
Creep data for Extra High Strength (EHS) steels
Steel
WELDOX 700
t
mm
temp
°C
O~B
6-20
400
450
500
400
450
500
400
450
500
640
330
205
640
450
300
645
520
365
21-45
46-75
3477
°cB
610
300
180
615
410
250
615
485
315
4
o cB
555
375
220
600
460
275
5
o cB
404
263
132
nl )
°c6
°cl
450
205
400
128
4.9
445
320
180
395
220
80
6.1
2.8
242
90
1) Estimated values
4:47
5 Dynamic strength
Fatigue. general ................................. 5:2
Practical design
against fatigue failure ......................... 5: 15
Examples - Fatigue ............................ 5:24
Something about crack
propagation ....................................... 5:31
Design of panels against
impact ............................................... 5:37
5
5 Dynamic strength
The sections in this chapter can be used independently of each
other with the exception of "Examples - fatigue" which must be
used together with "Practical design against fatigue failure."
Section 5.1 "Fatigue, general" has been included to provide
explanations of this complex phenomenon - fatigue - and
thereby, hopefully, fill a knowledge gap.
This section is recommended as it gives the user a better
understanding of the factors, which are of importance for fatigue
strength, especially of welded joints.
Section 5.4 "Something about crack propagation" provides a
simplified presentation of a method by means of which the
propagation of cracks can be calculated so that the life of
structures containing cracks or flaws (e.g. root defects) can be
estimated.
Section 5.5. "Design of panels against impact" presents a
simple design method and data to permit the design of panels
against impact effects due to pieces of rock being dropped on
them from a height, etc.
51
b. The subject is very complex and interdisciplinary (covers the
entire chain of design - manufacture - use).
5.1 Fatigue, general
c. It can be difficult for a designer to know enough about the
service loads on the structure.
Page No.
5:2
Why is it so important to understand fatigue?
d. Weld defects and stress-raisers of various types cause the
most problems, and the welder is not aware of this.
Initiation - propagation .................................................. .. 5:3
What is special about fatigue failure is that the phenomenon is
very local. A fatigue crack can form at a single point! It is
therefore necessary to know which point or points will be critical
for a given load case. Before this can be answered, it is necessary to have a drawing of the part in question.
Furthermore, design against fatigue extends to all stages in
the design and production of the product:
Welds - a problem of geometry ...................................... . 5:4
Fatigue - WELD OX and HARDOX steels .......................... . 5:5
Influence of welding residual stresses ............................. . 5:6
Notch effect, tables etc. .. ............................................... . 5:7
Advantages of WELDOX and HARDOX steels ..................... . 5:8
a. Selection and purchase of the "right material"
Importance of the load spectrum ..................................... 5:8
b. Correct structural design (e.g. location of welds)
Methods of improvement ....... ................ ............ ............. 5: 11
c. Acceptable analytical model
Some special cases: ....................................................... 5:12
Axial misalignment, Angular errors, Stress relieving, Throat
thickness In fillet welds, Influence of plate thickness, Gascut edges, Practical conversions, Multiaxial state of stress
strain cycle fatigue
'
d. Correct information to workshop
e. Choice of shop methods (welding method, filler material etc.)
and production parameters (current, feed, flow etc.)
f.
The individual operator (the welder)!
g. Surface treatment (not just for the sake of appearances l )
By fatigue failure we mean a type of fracture (rupture) that
can occur far below the static ultimate tensile strength of the
material if the loading consists of repeated applications of a
stress a sufficient number of times.
Example: You wish to cut a piece of wire, but you have no
wire clippers. If you bend the wire back and forth, a crack will
form, and after a sufficient number of bendings, the wire will
break. This is a fatigue failure.
It is known that machines and other kinds of structures are
subjected to fatigue loads and that 80-90% of all fractures that
occur are fatigue fractures.
However, it is easy to understand why this happens, since
the structure is usually designed against plastic deformation
(yielding) and not against fatigue!
h. Inspection and quality control
Inspection routines and design drawings are all well and
good, but strength control is not complete until the person
who has carried out the fatigue calculation has inspected and
approved the part!
What is said above demonstrates that everyone who participates in the development and production of a product has a
direct influence on the fatigue strength of the product.
Designing against fatigue - is it science or engineering art?
The precise fatigue diagrams and load data that are needed
are almost never available, and it is almost always necessary to
extrapolate or interpolate a bit. Extrapolating or interpolating
correctly is an engineering art!
This demonstrates once again that we are dealing with a
complex process and that it is extremely important that designers and engineers should be able to visualise the stresses in
their structures.
There are several reasons for this:
a. Instruction concerning the design of welded structures to
withstand fatigue is often inadequate in engineering schools.
Figure 5.1.1
infinite life
----".
C ----v'
iinite life
==========~>
1 m s=<
~
u
100 mm
'"E
10mm
'"
U
o
U
to be detected
by N.D.T.
starting from defect
~efect does not grow
1 mm
-'"
u
~
u
E
\\O~\
s\l}\\\(\'&
0
U
10mm
\'1>c,'<-
~
~
u
-
o'/~'
- --
t
-
---} non-- }
propagating
cracks
----
-;;;in sizes
~S
1000 A
0
,::;\ ~
,<-0<;
1 mm
c
-
\\)S\o~
.~(,
100mm
\\0
. ,,'l,
~~
c}.'1>
100 A
:J
C
10 A
--------
----------
no crack nucleation
atomic distance
1A
20
40
-
52
60
80
percentage fatigue life
100%
~~~~~:~~
Dynamic strength - Fatigue, general
Fatigue general
The total number of cycles to failure can be written
N = Ni + Np
= number of cycles to initiate a crack
= number of cycles for the crack to propagate (grow) until it
reaches critical size, at which the final failure occurs.
We know that crack initiation occurs readily in the local stress
concentrations found on surfaces, such as scratches, pores,
radii, welds etc., but surface defects also arise in very pure and
smooth materials as a result of microstructural dislocations
(movements of lattice defects) after a certain number of load
alternations.
Or N/mm2
Blast-cleaned and primed
x = fracture
x/= survivor
1000
700
:>.
I""
~
400
300
Figure 5.l.1 shows the different starting points of a microcrack,
and demonstrates that the crack can result in either finite or
infinite life.
Whether life will be finite or infinite depends upon the stress
concentration factor Kt
t
WELDOX 7008 mm
amin = 30 N/mm2
500
In the vast majority of cases, we will initiate a crack. How the
crack will propagate and possibly lead to fatigue failure is illustrated by figure 5.l.1, taken from (5).
local stress
nominal stress
Figure 5.1.3
r--..
NOTE! The prerequisite for crack initiation: Surface defect! (including internal surfaces, pores, etc.)
(K =
There are different ways of depicting the results of a fatigue
test graphically, but for welded joints (which are of greatest
interest to us), the S-N diagram is the most suitable.
When the crack has been initiated and the stress amplitude
and stress concentration factor are sufficiently large, all that
remains is propagation until the crack is large enough for the
component to break.
200
100
103
)
10 5
10 6
107
Number of load cycles
and the stress amplitude a , figure 5.1.2 (6).
The manner in which propagation takes place is illustrated by
figure 5.1.4. Each time the crack opens up, it propagates a bit,
and when the stress at the tip of the crack is high, a plasticization also takes place, resulting in a rounding of the crack tip.
Figure 5.1.2
Figure 5.1.4
a
- - a- - I
~I
1
1
::::>
I
All cracks propagate to complete fracture
:=j
===b
Q)
-0
:2 -01
0.
E
1!1
u
gl
'" :31
~
Critical alternati ng
stress amplitude required to
propagate a crack
oil
III
(/)
Q.
tl
I
I
~
I
~~
I
I
I
1
I
I
I
I
Ktcrit
I
I
2
4
6
8
10
12
14
16
18
20
Kt
I.e, we can "live with" fatigue cracks as long as Kt and/or
a a < a p are sufficiently low (a).
The phenomenon of infinite and finite life is also illustrated by
an S-N curve from a fatigue test, figure 5.l.3. The endurance
limit is usually plotted in the graph horizontally at the level
where none of the speCimens go to failure. The sloping part of
the curve is the one that shows finite life (fatigue strength) and
represents the "mean fracture curve", i.e. 50% probability of
fracture.
When the crack closes. the crack front folds and we obtain a
very sharp notch. The crack has then propagated a small distance Ll a for one load cycle. The figure shows how the characterisitic striations arise. These striations can only be seen in an
electron microscope. The striations that are often seen with the
naked eye on a fracture surface are called "oystershell" or
"beach" markings and are formed when the stress amplitude
changes. A fatigue fracture at constant amplitude exhibits a
completely smooth surface.
The size of the final fracture in relation to the entire crosssectional area shows whether the load at the time of the fracture
was high or low. (Large final fracture = high load.)
A fatigue crack always grows perpendicular to the direction of
the largest principal stress.
It is possible to observe how the crack is oriented at a fatigue
fracture and to compare this with computations and assumptions.
.
5:3
Dynamic strength - Fatigue, general
Fatigue fractures are not desirable in any way, but when they
do occur, we should try to learn from them - they do provide a
sort of test record, and give us a chance to learn from our
mistakes!
As always, progress often proceeds by trial and error, and,
unfortunately, nothing teaches us as much as our failures.
Figure 5.1.7
Welds - a problem of geometry
It took a relatively long time before it was established that the
geometry of the weld at the transition between the parent metal
and the weld metal (the weld toe) is the primary factor that
determines the fatigue strength of the weld.
As late as the mid-1960s, it was assumed that fatigue in
welded joints was primarily a metallurgical phenomenon.
When a weld is examined under a microscope, the weld toe
can been seen to contain microcracks originating from cold
flows or hydrogen. Non-metallic inclusions may also be present
just below the surface, see figure 5.l.5.
Figure 5.1.5
i ----'~-I'------,---<.\JLF--r-Figure 5.1.8
Figure 5.1.9
Figure 5.1.10
These inclusions are slag particles that have not had time to
float up to the surface of the weld pool during the welding
procedure before being "frozen fast" due to rapid heat dissipation through the plates.
The inclusions have a size of 10-200 microns and are located at a depth of about 200 microns below the undercut.
These slags are created in most welding methods, since the
impurities originate from the parent material.
Naturally. a welded joint does not exhibit this apperance
along the entire length of the weld, but the presence of such
stress raisers cannot be completely disregarded.
It has been clearly proven that the microgeometric stress
raiser at the transition between weld metal and parent metal
constitutes the crack initiation. The interaction between microgeometric and macrogeometric stress raisers will ultimately
determine the fatigue strength of the welded joint. This in turn
demonstrates who has the greatest influence on the result: the
welder!
Figure 5.1.11
Figure 5.1.12
The welder or production engineer is usually unaware of this!
It is therefore not so strange that "bad welds" account for the
majority of fatigue failures. as was mentioned above.
The welds are often poorly executed from the viewpoint of
fatigue because the welding personnel lack training.
A great deal would be gained ifthe welder could be taught to
understand fig. 5.1.5. It is important to understand why and
how the fatigue strength of welded joints can be improved.
This improvement can be brought about by welding in a
manner that produces smaller undercuts or by using the
improvement methods that are available, e.g. TIG dressing or
grinding. In addition, it is of the utmost importance to choose
suitable places for electrode changes or welding stops.
At corners (external and internal), transitions etc., for
example, it is advisable to weld the most difficult part without
stopping and to make the stop further on.
Figures 5.1.6-14 show the positions of the critical points with
respect to fatigue fracture for different welded joints. Note that
welds that are not load-carrying are nevertheless just as critical for fatigue fracture in the component on which they are
located if this component is load-carrying. (Figures 5.1.8-12.)
Instruction of welding personnel in these matters does pay, as
shown by experience from numerous companies.
5:4
Figure 5.1.13
(
Figure 5.1.14
Dynamic strength - Fatigue, general
The macrogeometric notch effect is dependent upon the shape
of the weld reinforcement and the type of welded joint. This is
shown by fig. 5.1.15, which applies to a butt weld. The higher
the reinforcement, the poorer the fatigue strength of the weld.
It is thus the interaction of microgeometric and macrogeometric stress raisers that determines the fatigue strength of
the welded jOint.
We can therefore conclude that crack initiation in welded
joints is already predetermined. The life of the joint therefore
consists solely of propagation to critical size.
This is why in-service inspection is so important.
Figure 5.1.15
Some important definitions
Figure 5.1.16
o
Fatigue strength at 2· 106 cycles
N/mm2
time
G r = G max -0' min stress range
Gm =
mean stress
0' a =
stress amplitude
R = 0' min stress ratio
G max
In order to be able to define the stress situation in connection
with fatigue, we need to define the stress variation and the
absolute stress leve\.
This can be done in a number of ways. Here are the most
common ones:
G max and G min
a r andO'min
50
O'r and R (and sometimesam1nl
amaxand R
0~--------+--------4----
lOO'
120'
140'
aa anda m
We will usea p R
____~________~
160'
e
180'
Why is knowledge of fatigue so important in connection
with WElDOX and HARDOX steels?
There are three reasons:
1. Fatigue cracks propagate at roughly the same rate in all steels,
and since the life of welded joints is dependent upon crack
propagation, welded WELDOX and HARDOX steels exhibit the
same fatigue strength at around 2 . 106 load cycles as ordinary
welded steels.
2. The reason for choosing WELDOX instead of HS steels is to
reduce plate thickness. When this is done, the stresses in the
steel - both static and fatigue stresses - will naturally increase
for a given load case. This means that design against fatigue is
more important when WELDOX steel is used in welded structures, since fatigue strength does not increase at the same
rate as static strength.
3. WELDOX and HARDOX steels are often used in structures
subjected to high fatigue loads.
The above may give the impression that WELDOX and HARDOX
steels are not at all suitable for use in structures subject to
fatigue loads, but this is not the case!
We need only look around at the applications in which WELDOX
and HARDOX steels are used to find that these are indeed subject
to high fatigue load and that they are clearly functioning satisfactorily!
Why?
The answer to this question will be given later, but first we have
to introduce certain concepts and show the importance of the
welding residual stresses.
Figure 5.1.17 The importance of different values of R R = a mm
O'max
+0
3~--------------------------------,,~~
2~--------------~~n-~",~---+rH~~
0~----~~--+H~--4+~~~~~--------~
_l~~~~UL~~---Y~~~
____________~
-2~WW~------------------------------4
-3~~~------------------------------~
R=3
-0
Observation on the use of 0 r instead of a max
R = - 1 and a max is the "worst case".
When it comes toa r = 0max - amino the worst case occurs
when a, has the greatest effect on crack propagation. 0 r has the
greatest effect when there is no compressive stress during the
cycle, i.e. when the load consists of pulsating tension. The
crack does not propagate under compression.
Figure 5.1.18
R= -1
R= 0
Ifa q = 0"2' thena'2 (R = 0) is worse from the viewpoint of
fatigue than a,) (R = -1 1
5:5
Dynamic strength - Fatigue, general
Influence of welding residual stresses in
connection with fatigue
°
Consider a welded I -beam. After welding, the residual stresses
are on a level with the yield strength of the weld metal (which is
often on a level with that of the parent metal).
r~s
s
Assume that the parent metal and the weld metal have the
following tensile test curve.
o
U'L
The stress in the two welds will vary from the yield stress
downwards byo 1, i.e. the only load variation felt by the welds is
a 1, and this stress variation is called the stress range r'
As is evident from the earlier line of reasoning, fatigue cracks
can very well be initiated in welds on the "compression side",
which often seems surprising, but is quite natural owing to the
welding residual stresses.
This means that it is the stress range or that is the significant variable in conjunction with the fatigue of welded structures!
The ° r philosophy has been verified by the testing of beams
(about 2000 as of this time) and has been generally accepted
internationally. with a few exceptions.
This represents a considerable simplification in design, insofar as knowledge of r is sufficient when it comes to welded
non-stress-relieved structures.
As far as small specimens or short welded joints are concerned, the residual stresses are often small, with some influence
of the mean stress, i.e. the stress ratio R.
This can be illustrated by means of a Haig diagram, which
applies for a certain number of load cycles and shows the stress
amplitude as a function of the mean stress.
°
f
Assume further that the beam is loaded in flexure with a fatigue
load that gives the following stress distribution.
Figure 5.1.19
Amplitude
kvww
0. N/mm2
-y-
160
b=~==========:::;r:d+0 1--A+
120
p
-01
Fillet welds small specimens
2· 106 load cycles
01
External and internal stresses can be added. This has the following consequences:
Top flange
Bottom flange
The weld against the top
flange will be unloaded
The weld against the bottom
flange will be overloaded and
will yield slightly
(°5+°1=°5)
see tensile test curve
At 1st loading
(I
(1,
-rio.,----------
11,
r+-....- - - - - -
(),
'"
80
.....
I
40
11
0::
Mean
O+-_ _ _
R,=_+_l_ _...-_ _- r_ _---._ _ _-r-_ stress
o
40
80
120
160
200 am N/mm2
Figures 5.1.19 and 5.1.20 are taken from (7) and figure
5.1.19 shows a dependence on R for small specimens, while
figure 5.1.20, which applies for beams, does not show any
dependence on R.
Usingo r consistently means working on the safe side. With
good knowledge of the mean stress (external + internal
stresses), higher stresses can be allowed.
At unloading
Figure 5.1.20
Amplitude
Oa N/mm 2
Fillet welds beams
2.10 6 load cycles
160
At 2nd loading
120
80
and after unloading
40
.....
I
11
0::
\)
qj
R= +1
40
5:6
80
120
160
200
Mean
stress
am N/mm2
Dynamic strengtll - Fatigue, general
Figure 5.1.21, taken from ref. (8), gives an idea of how much
the stresses can be increased. The figure shows a factor kR that
is multiplied by a (perm' This is assuming that the stress ratio R
(external + internal stresses) has been clearly established and
determined. The result is better material utilization for components subjected to external stress ratios < 0 and low residual
stresses e.g. "short welds", rolled beams, stress-relieved structural elements etc.
whereo lJ3 , (2' 106 ) is the median stress range for fatigue failure
at 2· 10 load cycles and R = O.
Ref. {8l sets out a number of classes for K x, see table 5.1. I.
Table 5.1.1
Kx: 1.3, 1.5, 1.7, 2.0, 2.3, 2.6,3.0, 3.5,4.0 and 5.0
=
=
e.g. mill scale Kx
1.3, butt weld
2.3, fillet weld"" 4.0. See
table 5.2.3.
In table 5.2.3, r is the nominal stress range and is calculated
in a section without the influence of stress concentration
(approx. 0.2 . plate thickness from the weld reinforcement).
The weld causes diHerent stress concentrations in relation to
the stress direction over the weld, Kx -L perpendicular to the
weld and Kx 11 when the stress runs parallel to the direction of
the weld.
With figure 5.1.5 in mind, we understand that the quality of
the weld means a great deal for the fatigue strength of the jOint.
Now comes the difficult part; classification of a finished weld.
There are a number of different international weld classification systems and even more applied in the internal standards of
diHerent manufacturers.
We feel that it would be wrong to use the internal standard of
a given company, so we base our approach on StBK-N2 Regulations for Welded Steel Structures (8).
°
Figure 5.1.21
1.3
I--'
I--' i-'
V V
1.2
V
V
V
/'
1.1
/
/
See table 5.1.2.
1.0
V
o
-0.5
-1.0
R
-l.~
-2.0
Representation of notch effect
Geometric stress raisers (notches) give rise to stress concentrations. Such a stress raiser is defined by a stress concentration
factor Kt.
K _
local stress
t-
nominal stress
Kt is based on elastic theory, often obtained through calculation,
photostress analysis and experimental verification. Good and
practical tables are provided in references (3) and (10).
Note that Kt is determined only by geometry and that material
properties have no bearing. Welded jOints can be classified with
respect to their notch effect, and figures 5.1.6 - 14 show where
this is greatest.
A traditional method of specifying the reduction of fatigue
strength due to the notch effect is by means of a fatigue factor
K f (also known as the fatigue-strength reduction factor or the
factor of stress concentration in fatigue), which is defined as the
ratio of the fatigue strength of a plain, unnotched specimen to
the fatigue strength of a specimen containing a stress concentration at a given number of load cycles.
However, it is difficult to determine fatigue strength if Kt is
used. A value of Kf would be needed for a given material, Rvalue, a r' number of load cycles and stress raiser! .
This would make it difficult and complicated to design against
fatigue.
It would be very difficult to find any particular case in reference tables. Approximate methods would have to be resorted to
Table 5.1.2 Quality requirements for weld classes in accordance with StBK-N2.
Weld
class
Requirement concerning surface, shape and hornogeneity.
SvO
The surface of the weld shall not contain cavities which
render cleaning, painting or galvanization difficult.
Svl
The surface of the weld shall be free from cavities. surface
pores, intrusions and overlaps. In Construction Class 2,
large weld reinforcernent. root defects, penetration beads or
sharp discontinuities shall not occur.
The homogeneity of the weld shall substantially satisfy the
requirements applicable to radiographic grading
No 3 (green" l.
SvOT
SvlT
The weld shaH be fluidtight. In other respects, requlrernents
for SvO and Svl respectively shall apply. In the case of SvO,
there is the additional requirement that the surface shall be
free from surface cavities.
Sv2
Fillet and T-butt welds shall preferably Ilave a concave or
flat surface. A slightly convex surface Will however be
approved.
A butt weld shall have a flat or convex surface at both the
top and the root. The height of the weld reinforcernent shall
not exceed 15% of the width. The height of the penetration
bead shall not exceed 30% of the width of the penetration
bead, but shall not in any case exceed 2 mm.
The transition between weld and parent material shall be
smooth and free from sharp discontinuities such as undercuts. Craters shall be filled and irregularities removed.
Apart from these requirements, the appropriate parts of the
requirements for Svl, concerning surface finish and shape,
shall apply.
(1)
The homogeneity of the weld shall substantially satisfy the
requirements applicable to radiographic grading No 4
(blue"). However, subsurface pores and non-contiguous
slag entrapments corresponding to radiographic grading
No 3 (green") will be approved.
Considerable simplifications are possible when working with
welded jOints.
In practice, it is quite adequate to proceed on the basis of a
single steel.
a mm ) can be disregarded owing to
°max
the welding residual stresses, and fatigue cracks propagate at a
largely equal rate in all types of steel.
When the R-value is known, we can multiply the permissible
stresses by the factor kR in accordance with the above line of
reasoning.
The R-value (stress ratio
Sv3
Fillet and T-butt welds shall have a concave or flat surface.
The transition between weld and parent material at fillet
and T-butt welds shall be dressed to a smooth and rounded
shape.
In the case of butt welds, the top and bottom reinforcernent
shall be dressed flush.
Kx-value
Apart from these requirements, the appropriate parts of
the requirements for Svl and Sv2, concerning surface
finish and shape, shall apply.
It has been found convenient to classify the notch effect of
welds by means of a material-independent joint factor Kx.
The homogeneity of the weld shall satisfy the requirements
for radiographic grading No 4 (blue").
referred to in (8) and defined as
Kx = 315 (N/mm:l
OrB' (2,10)
a See Collection of Reference Radiographs of Welds in Steel,
published by the IIW.
5:7
Dynamic strength - Fatigue, general
One possibility is, of course, to determine Kx for a weld by
means of fatigue testing and to use the definition of Kx.
It should also be borne in mind that most test results that are
available come from different laboratories where the weld has
been made in the most favourable position under favourable
conditions (no piecework system, good working premises etc.).
The welding position, for example, is especially important.
Table 5.2.3 contains Kx factors, and more detailed design
instructions are provided in section 5.2 "Practical design against
fatigue failure".
WElDOX and HARDOX steels in structures subjected to
fatigue loading
Despite everything that has been said previously, there are five
areas where WELDOX and HARD OX steels can be used with
success in structures subjected to fatigue loading:
1. Parent material unaffected by weld
2. High stress levels
3. Low load cycle numbers
4. Suitable load spectra
5. When it is possible and desirable to increase the fatigue
strength of the welds.
l.
Parent material unaffected by weld (sharp notches)
If the designer has succeeded in locating welds in areas of low
stress in the structure., the fatigue conditions for the parent
material and for the welds are somewhat different. Here, the
crack initiation phase must be dealt with first, which results in
an increase of the fatigue strength of the structure.
The fatigue strength of the parent metal is proportional
to the ultimate tensile strength of the steel. This is valid for a
polished test specimen.
a rB = k· aB k = OA - 0.6, see figure 5.1.22, which applies for
mill scale.
Figure 5.1.23 is an S-N diagram for three different steels with
blast-cleaned and primed surfaces.
2.
High stress levels
There are many structures where the load consists of a high
static load and a smaller fatigue load, for example stands, certain frames, bridges, penstocks etc. Here, it is easy to exceed
the permissible static stresses or the yield strength of ordinary
steels.
WELDOX and HARDOX steels can therefore be utilized to great
advantage and we can permit the same 0" r at maximum
high stress levels as at low ones, which has been proved by
the O"r - philosophy for welds and experimentally for WELDOX and
HARDOX steels (11, 12)
3.
low load cycle numbers
The region for the permissible stress range is limited by the S-N
curve and by the yield stress (or permissible static stress) of the
steel.
In other words, WELDOX and HARDOX steels are advantageous
when the number of load cycles is less than 10', there is a full
load spectrum (constant amplitude) and R = O.
This is illustrated by figure 5.1.24 (13.14J, which applies for
butt welds.
Figure 5.1.24
Or N/mm'
Butt weld
1000
700
WELDOX 7
.....
500
Figure 5.1.22 Fatigue strength at 2· 106 cycles, R =0
Mill scale
400
N/mm'
300
Or
S 355
S 275·
600
200
450
300
100
102
150
150
300
450
800
950 N/mm2
Ultimate tensile strength
Mill scale is a very vague definition of a surface, which is why
we prefer to test blast-cleaned and primed surfaces, since they
are more precisely defined and because this is the mostcomman service condition of the plate surface.
N/mm'
750
Cv0"
Figure 5.1.23
fir
600
I)).' •
Blast-cleaned affd primed surface
R=O p= 1
1000
700
500
""-
~ 400
~,
//
~
"
WELDOX ]00
S ~52. (BS 50)
--300
-
S 275 (BS 40) 200
100
10'
5:8
10'
10'
107
10·
10'
10'
10'
10·
The diagram shows that below a certain load cycle number
Ng, it is only the yield stress (static permissible stress) which
imposes a limitation.
For R = - 1, Ng decreases buta r is doubled.
4.
Suitable load spectra
This area is perhaps the most important one.
The actual fatigue load on a structure is usually not of
constant amplitude. For example, the work cycle of a haulage
vehicle consists of: loading - off -road driving with load - onroad driving with load - dumping - off-road driving without
load - on-road driving without load - off-road driving without
load - loading etc.
If a fatigue test is carried out for a butt weld first with constant
load amplitude (ka) and then with a load spectrum (spJ, e.g.
p = 112 (more about this later), we will get two different S-N
curves. The S-N curve for the load spectrum is displaced towards a load cycle' number which is more than an order of
magnitude higher! See figures 5.1.25, 26 and 27, which are
taken from (14).
In other words, it is absolutely wrong to design with constant
amplitude data if the load is of variable amplitude.
Describing load spectra is a science in itself. Trying here to
penetrate more deeply into the subject would only complicate
matters unnecessarily.
This section may be regarded as an introduction to design
against fatigue using load spectra instead of constant load
amplitude. Some simplifications are therefore necessary.
Dynamic strength - Fatigue, general
Figure 5.1.25
Or
N/mm'
R=O
Mill scale
1000
700 r- -
~LQOJ<; TQO ca
r WELDO~ ~~O sp
l'ol
'"
500
400 ;:..::
/
"
~S 355 sp
Figure 5.1.28
l~~~~~~::::============~~~P=l
-
- S 355 ca
300
III
III
Point 3 will give us something called the spectrum parameter p.
A histogram is then plotted with declining values of a,la rref
along a base of values of log N.
A straight line can now be drawn from (a r/a rr f = 1, N t = 100),
in such a way that it is tangential to the tops of the histogram,
and the spectrum parameter can be read off at the right of the
diagram, see figure 5.1.28.
5/6
200
ca p
sp p
100
10'
10·
105
1
Or
1/2
1/2
Or rei
10'
107
1/3
Figure 5.1.26
Or
N/mm 2
1/6
R=O
Butt weld
1000
O~
700
500
400
WELDOX
-
I-
~
The choice of straight lines is practical. and it is also found that
most of the load spectra of interest to us are close to being
straight lines.
Some standard regulations, for example Swedish bridge regulations and crane regulations (74), give us values of p for different service classes. The standardized stress spectra in the
crane regulations are not linear, but follow a gaussian normal
distribution, see figure 5.1.29. These spectra are based on the
German crane standard DIN 15018, which is in turn based on
experiences from LBF (Laboratorium fur Betriebsfestigkeit in
Darmstadt, Germany), where it was believed in the 1950s that
crane spectra closely followed a gaussian distribution.
S 355 sp
300
200
S 355 ca
ca p
sp p
100
10'
ID·
1
1/2
107
10'
i'TO-()"_m
r
i"""iO::::==:::::c:::--,---,---rc--,----,3/3
S3
R=O
Fillet weld
bo-am
Figure 5.1.29
Figure 5.1.27
° N/mm2
~O
1
log N
log Nt
700 sp
WELDOX 700 ca ......
--
______________________________
1000
700
WELDOX 700 sp
500
"'--,
400
-
III
III
III
en
S 355 sp
1
300
200
WELDOX 700 ca
100
10'
0
I;; 'I;;
ca p sp p
S 355 ca
I
9'o 19'
~
1
1/2
i!
III
III
10"
107
10·
For example: - Disregard the mean stress; it is primarily the
welds that are of interest.
- Use the Palmgren - Miner cumulative damage
rule.
- Pay no attention to the sequence of the loads.
In order to describe in a simple manner how "heavy" (the
heavier, the closer to constant amplitude) our load spectrum is,
we choose the same method as that used in the Swedish
Regulations for Welded Steel Structures (8).
It is then necessary to know:
1. The largest stress range in the spectrum, a rref'
2. The total number of load cycles Nt .
3. The way the different stress ranges a r are distributed among
different load cycle numbers.
~_-..l.
o
_ _-L-_-:---L-_ _L -__--L_----lI 0/3
1/6
2/6
3/6
4/6
5/6
6/6
~
I!
l
IgN _ _ ____._
Ig N
Despite more than 100 years of fatigue testing, most design
data derive from constant amplitude testing, while our structures are subjected to variable amplitude (spectrum) loads.
This has its rools both in inadequate testing equipment - it is
considerably more expensive to carry out a spectrum test - and
in the fact that investigators have often contented themselves
with comparative testing, therefore considering constant amplitude to be adequate. Furthermore, uncertainty has prevailed
with regard to the testing procedure (i.e. how actual load spectra should be simulated, actual spectra being in turn difficult to
measure).
What is therefore needed is a theory that makes possible the
use of the results of constant amplitude test for estimation of the
life of a structure under variable amplitude load.
5:9
J
I
1
Dynamic strength - Fatigue, general
Such a theory is the linear cumulative damage rule originally
proposed by Palmgren in 1924 and restated by Miner, usually
known as the Palmgren - Miner cumulative damage rule. It is
the simplest and least data-demanding cumulative damage rule
and is no less accurate than the others.
If we assume that one stress cycle with 0 ri is independent
of the changes in the material caused by the preceding load
cycles, then it causes a partial damage of
6.D=_l_
Ni
where Ni = expected life at constant amplitude foro w
If a structure is subjected during a service time To to a mixture
of stress cycles of varying size:
There are now two ways to proceed:
a. Start with the load spectrum and use the Palmgren - Miner
cumulative damage rule for each case, using permissible
stress ranges at constant amplitude.
b. Compute new SoN curves for permissible stress ranges for
each Kx and p-value with the aid of the Palmgren - Miner
cumulative damage rule. The latter is the most convenient
method, but if it is difficult to obtain a load spectrum sufficiently accurate to carry out a cumulative damage calculation, we will have to settle for making an estimate of the
spectrum parameter p.
Method b is used in section 5.2 "Practical design against
fatigue failure".
Something about the shortcomings of the Palmgren - Miner
cumUlative damage rule
nl cycles ofo q , which at constant amplitude gave NI'
n2 cycles of 0 r2' which at constant amplitude gave N2.
Figure 5.1.31 (1) shows what happens to crack propagation
when the stress range changes. When the change is "Low ~
High", crack growth increases immediately, whereas when the
change is "High ~ Low", crack growth stops suddenly, only to
start up again after a certain delay at a rate corresponding to the
new stress range!
ni cycles of 0 ri' which at constant amplitude gave Ni'
then the total life of the structure
j ni
D = I6.D = I i = 1 Ni
Figure 5.1.31
Crack length
will be expended during the time To.
The life of the structure is then exhausted when
D= I
~ reaches unity.
To
The estimated life will be T = I ~
N
See also figure 5.1.30.
Time
Crack length
Figure 5.1.30
"r
Time
Or
Time
(Number of load cycles)
10gN
The sequence in which the different levels come is therefore
of the utmost importance.
The reason why a crack retardation occurs in the High ~
Low case is that when the crack is propagating (growing). 0 s is
always reached in the crack tip and in a small region in front of
Table 5.1.3
Fatigue life in 103 cycles 0 max overload = 0 min + 1.67· 0 r
N/mm2
Constant
amplitude
One overload
144
144
144
177
177
177
220
220
220
1676
1020
3780
727
1062
640
223
441
325
10000*
10000
10000
1294
719
616
670
1050
410
Or
*Survivor
5:10
Number of cycles between overloads 6. N
105
10000*
-
4692
919
1962
1604
1014
435
104
7800*
10000*
6200*
1333
5049
1418
718
799
685
1~
1~
4215
1126
1673
1339
1367
1741
623
326
628
3260
1586
1130
590
410
685
238
503
251
10
850
1332
980
391
260
422
150
142
163
Dynamic strength - Fatigue, general
this. When the load is removed, this region will contain compressive stresses. Both the size of the region and the magnitude
of the compressive stresses depend on the size of the preceding
stress range.
Strain hardening also occurs in this region. This means that if
a smaller stress range is then imposed, its influence on crack
propagation will decrease by the compressive stress caused by
the preceding higher stress range. The duration of this effect
will depend upon the size of the compressive stress region.
From this it follows that even occasional high a r have a favourable effect.
If, on the other hand, an external compressive stress is imposed, the effect is reduced sharply.
The same effects have also been found with regard to crack
initiation in notches with radii of between 1 and 5 mm (16).
The effect of the delay for welded jOints (non-load-carrying
cruciform weld) under overload is described in (17) and applies
for R "" O. Table 5.1.3 shows the results.
As can be seen, if the number of cycles between overloads is
~ Hi, there is a reduction of fatigue life, while for> Hj cycles
there is a clear increase.
Fora r = 144 N/mm2 and L\N > 103 , none of the speCimens
failed.
Thus, the Palmgren - Miner cumulative damage rule does
not take into consideration High ~ Low or Low ~ High and the
distance between these events.
Load cycles that are smaller than the endurance limit (ar5 in
figure 5.1.30) make no contribution to I n/N, since according to
the definition N ~ 00. In practice, these small a r also shorten
the fatigue life of the structure by accelerating the growth of
previously formed cracks.
This situation can be rectified by modifying the S-N curves
either by disregarding the endurance limit and extrapolating the
S-N curve downwards or by introducing a new level of the
endurance limit that corresponds to a threshold value (L\ Kth ) at
which a certain .size of the crack does not propagate.
It is also possible to give the S-N curve another slope below
the endurance limit (k' = 2k - 1).
Stress ranges with Iowa min (compressive stresses) are more
damaging than what corresponds to ~. This unfavourable
Ni
effect will be greater if such events occur periodically
throughout the entire service time.
It would seem as if the Palmgren - Miner cumulative damage
rule is a good approximation and gives values on the "safe side"
for R ~ 0, but can give highly uncertain results if R <0, i.e.
cycles with compressive stress.
The best way to check a component is to record an actual
load spectrum in service and then mount the component in a
rig and feed the testing equipment with the spectrum in question. This is known as service simulation testing.
The reader who wishes to find out more about load spectra is
referred to (1).
The practical part of the calculation work will be dealt with in
the section "Practical design against fatigue failure".
We have two basic options:
a. Improve the design of the joint so that we get larger radii and
smoother transitions between the weld and the parent material (undercut).
b. Introduce compressive stresses in the critical region (a crack
can only propagate under tensile stresses).
There are a number of methods where these mechanisms are
involved either singly or in combination. Among these, three
stand out as being the most practical:
Peening, grinding and TIG dressing.
Peening is a cold working process in which a pneumatic tool
(slag hammer) with a rounded tip is used to work harden the
transition between the weld and the parent material. In this
manner, compressive stresses are introduced and crack initiation is delayed. This works excellently until the welded joint is
overloaded, whereupon the compressive stress state is eliminated.
The results of the peening process are nearly impossible to
check and considerable environmental problems are associated
with peening (noise).
Figure 5.1.32 As - welded
Figure 5.1.33 Ground
5. When it is possible and desirable to increase the fatigue
strength of welds
When it has been found that technical/economic advantages can
be gained by using WELDOX and HARDOX steel plate in a structure, it is often discovered during design that a few welded joints
are completely critical with respect to fatigue in determining
whether or not WELDOX and HARDOX steel plate can be used.
In such cases, it can be well worthwhile to increase the fatigue
strength of such critical welded jOints in some way.
The section "Fatigue, general" clearly shows that the problem
~f fatigue in welded joints is a problem of geometry, and from
figure 5.1.5 we can see that we must in some way prevent or
prolong the initiation phase and "suppress" crack propagation ~n ord~r to be able to increase the fatigue strength of the
welded JOint; In other words we must increase the influence of the
term Ni in
Figure 5.1.34 TIG-dressed
Dynamic strength - Fatigue, general
Grinding is the most widely used method at present, but it is
often used at the wrong place.
Correct grinding involves grinding of the transition between
the weld and the parent material (the weld toe) so that microcracks and slag inclusions are removed. At the same time, a
smoother transition with a large radius is created.
Important! The undercut must be ground down to about 0.5
mm below the surface of the plate in order to make sure that all
stress raisers have been removed, see figure 5.1.5! Care should
also be taken in grinding to make sure that the score marks
caused by the grinding process are not oriented across the flow
of stress. If they are, they will act as new stress raisers.
In prattice, grinding imposes high demands on the operator.
Bringing about a measurable improvement in fillet welds with
only a grinding disc is very difficult. It is often necessary to use a
burr.
Grinding of the top or root weld reinforcements is meaningless and only arouses suspicions as to the quality of the weld.
See figures 5.1.32 and 5.1.33.
TIG dressing is a relatively new method that is rapidly gaining
more and more adherents.
A TIG torch is used to re melt the weld metal in the weld toe
without the use of filler material. A TIG torch consists of a
tungsten electrode surrounded by argon shielding gas.
In this manner, a very smooth and fine transition between the
weld metal and the parent metal is obtained, and at the same
time the entrapped slag inclusions are released and can float up
to the surface. See figure 5.1.34.
Properly executed weld dressing using the above methods
greatly extends fatigue life at approx. 1()6 load cycles, for example by a factor of 2 - 3 for butt welds of types Sv 2 and 3 and
by a factor of 3 - 4 for fillet welds of type Sv 2; alternatively, a r
can be increased by about 20 and 40%, respectively.
TIG dressing is the preferable method of the two with respect
to environmental hygiene and ease of execution and inspection.
The environmental advantages of TIG welding over grinding
are obvious.
Accessibility is very good! In principle, TIG dressing can be
performed everywhere it has been possible to weld. Its degree
of difficulty is equivalent to that of ordinary gas welding.
The results are very easy to check by means of visual inspection. The evaluation method is the same as for welds. Experience shows that TIG dressing is the method that gives the
most reliable results. TIG dressing is about 3 times faster than
grinding.
Important!
Note that only those parts of the welded joint that are critical for
fatigue failure should be dressed. Figures. 5.1.6 - 14, and above
all practical experience, should serve as a guide.
For design I(urposes, an Sv 2 class weld can be counted as
an Sv 3 class weld, provided that dressing has been carried out
as described above, see table 5.2.3. The improvement can
sometimes be considerably greater, but a thorough analysis is
necessary in order to be sure of this.
If the risk of fatigue is reduced in this manner at one point,
this means that there will be an increase of the fatigue risk at
another place. The way in which a possible failure situation is
altered must, of course, be carefully analyzed.
Some special cases and results from experience
Gas-cut free edges can be dealt with in the same manner as
welds (a r) with Kx = 1. 5 for the best cutting class Sk3 as per
StBK-N2 (8) and Kx = 1.7 for Sk2, see table 5.1.4. .
It is very important that the transition between the as-rolled
surface and the gas-cut surface along the edge be chamfered,
since the gas-cut surface is critical, and experience shows that
crack initiation often occurs in this transition. Cutting defects
reduce fatigue life by 50% and should be ground off with very
smooth transitions. All chamfering and grinding shall be done
in such a way that the score marks are aligned along the edge.
See also ref. 18.
5:12
Table 5.1.4
Quality requirements for cutting classes as per StBK-N2
Cutting Application
class
Dressing of
surface
Quality requirement
Dressing
Surface
roughness of free
mm
edge
~I
•
\
The surface shall be
dressed so that the
necessary corrosion
protection can be
applied (see StBK·N4)
SkO
Structural
element
not designed for
transmission of
force (non·load·
carrying)
Ski
Construction
a.; 1.0
class I in struc·
tures subjected
to small number
of load cycles
or load cycles of
small magnitude
Sk2
Construction
classes 1 and 2
a..;; 0.3
Edges shall be
deburred
Sk3
Construction
a..;; 0.2
classes 1 and 2
Edges shall
be chamfered
tos;!< 2mm
Stag and spatter As for SkO
shall be removed
from edges
Cutting defects. surface
cracks and other surface
, defects shall be removed
by grinding. In construc·
tion class 1. a surface
defect may be repaired by
welding and subsequently
ground. Apart from these
requirements. the specifi·
cations given for SkO
shall apply
As for Sk2
Influence of some defects of geometry
1. Axial misalignment in butt welds has been studied in (19)
with pulsating tension and bending, see table 5.1. 5.
- - - - - - ~---t+~
+,
The stress concentration factor for a butt weld with axial
misalignment and under pulsating tension can be approximated
as follows:
e
Kt = 1 + 3 t
Table 5.1. 5 Pulsating tension
25
50
75
100%
38
46
65
72%
Grinding does not have much effect
50
25
elt
11
increase ofa r at 106
10
75
100%
28%
elt
Reduction of a r at
le! cycles
13
Deposition of additional filler material in order to improve the
geometry of the weld does not bring about any appreciable
improvement.
In pulsating bending, axial misalignment has very little
negative effect.
2. Axial misalignment in load-carrying fillet welds
Kockums has investigated this in (20) and found that axial
misalignments of 5 mm in incomplete penetration jOints did not
lead to any reduction of fatigue strength at 106 load cycles (t =
35 mm). Axial misalignments of 10 mm and greater caused a
marked reduction. This could be compensated for to some extent by increasing the root gap!
Dynamic strength - Fatigue, general
Figure 5.1,36 B
it was also shown - and this has also been found by other
investigators - that the same fatigue strength can be achieved
for a fillet-welded joint as for a complete penetration joint if the
throat thickness is correct. See below, optimum throat thickness.
Tp
H/Tp
1.4-.-----~--,.----._--._-__,
50
Angular misalignm~nt in butt welds
Figure 5.1.35 shows what a great influence an angular
misalignment has on the stress range.
1.2
25
19
Figure 5.1.35
12.5
1.0
8
L = 350 mm
Or
6.25
= 20mm
= 200 N/mm2
0.8
0.6
Fracture in A
0.4
ar
0.2
1
a (0)
.16
3
.5
Increase afar ('Ye)
14
43
Simply supported
h (mm)
5
8
.8
1.3
72
116
O+-----.-----r----.-----.----~
Stress relieving of the welded joint for the purpose of increasing
its fatigue strength is only worthwhile for those welded joints that
are subjected to an external a r which includes compressive
stress (i.e. R < O!) some time during the stress cycle. The
increase in a r corresponds to kR in fig. 5.l.21 and is verified by
(21 and 22). The increase is greatest when N > 106 load
cycles. Up to 25 % increase in allowable stress range is possible.
Influence of test load (overload) has been dealt with in the
section on spectrum load and has been shown to have a
favourable effect.
a. Angular misalignments are straightened out
b. Crack tips are rounded
c. After unloading, compressive stress is obtained around the
cracks that have been subjected to tensile stress.
It has been demonstrated (23) that an optimum throat thickness in statically loaded structures does not give an optimum
result under fatigue loading. This is especially true at large
throat thicknesses. To resist fatigue, the welded jOint should be
made uniformly strong so that there is an equally great risk of
initiation from the root and from the undercut. From the viewpoint of inspection, initiation from the undercut is easier to
detect.
Figure 5.1.36 A and B shows how throat thickness should
be chosen for different plate thicknesses and penetrations. In
actual practice, throat thicknesses are usually too small! This is
also shown by actual fatigue failures, where the fracture usually
initiates at the root l
0.4
0.6
0.8
1.0
2a,/T-p
Influence of plate thickness
In the case of smooth components, the greater the material
volume, the higher the probability of failure. The critical material volume for welds is very small. The influence of e.g. plate
thickness is statistically negligible.
For fillet welds, on the other hand, it has been shown that the
thicker the plate, the larger the region of high stress where
fatigue cracks initiate.
In practice, fatigue strength is reduced by about 10% be·
tween 10 mm and 30 mm plate thickness.
Practical conversions at 2· 106 load cycles
a r (R = 0) "" 0.8· Or (R = - 1)
a rtension (R = 0) = 0.8' a rflexure (R = 0)
Multiaxial states of stress are very common, and research
results here are almost non-existent.
A multiaxial state of stress is defined by three principal stresses a), a2, a3 and their corresponding directions ~.rl and ~.
In the general case, all six components vary with time and in
a way that need not be at all regular, nor need the components
be interrelated.
This entails an enormous complexityl
It is therefore understandable why there is no simple, or even
complex, theory that accurately describes the situation.
The following steps are recommended, in the order given, in
order that reasonable progress may be made in design.
Figure 5.1.36 A
-
0.2
- - - - - Increased penetration
Prototypes versus mass-produced specimens
Prototypes made by hand-picked personnel often exhibit a longer life than mass-produced specimens. The difference can be a
factor of 2.
Optimum throat thickness in load-carrying fillet welds
0_
o
l. Try to calculate or estimate the prinCipal stress in combina-
-
a
tion with a stress raiser that is most critical for fatigue failure
at a given pOint. Assume a uniaxial state of stress. (Remember! A fatigue crack always propagates perpendicular to the
largest prinCipal stress.)
2. Use an approximate method, for example the one given in
StBK-N2 (8).
3. Consult the special literature and start with (1) part 2,
special methods.
5:13
Dynamic strength - Fatigue, general
Strain cycle fatigue can be involved in two main cases:
1. When the component is subjected to loads that are controlled more by strains than stresses, for example a flexible
component between two rigid components subjected to thermal variations etc.
2. When the component is subjected to an extremely small
number of high loads N < 1000 - 10000 (Iow cycle fatigue).
In this life range, the S-N curve is nearly horizontal, so stress
is a poor parameter (cf. the tensile test curve at (J s)'
Strain is therefore a more sensitive measure of the state of
the material in this region.
This subject is a very extensive one, so we shall confine ourselves to a few approximate formulas only.
For unnotched steel Kt = 1, the total strain amplitudeE:ta can
be written in accordance with Coffin-Manson:
5:14
eta = 0.5' 0°.6 • N-O·6 + 1.75
(a: ). N -0.12
where 0 = el og _1_
I-tjI
N
aB
E
tp
= number of load cycles
= ultimate tensile strength of the steel
= modulus of elasticity of the steel
reduction of area at rupture in tensile test
=
The notch effect in connection with strain cycle fatigue can, as
a first approximation, be put equal to the stress concentration
factor Kt, and in many cases gives a result on the safe side.
Those who wish to find out more are referred to (1 ) part two,
special methods, and ASME Ill.
5.2 Practical design against
fatigue failure.
7. Compareo rmax withorperm
Iformax~arperm OK.
If a fmax > 0rperm' change dimensions or welded joint and
carry out new calculation. (Note that the position of the critical point may be changed!)
life?
Page No.
Calculation procedure ..................................................... 5:15
Typical load spectra ........................................................ 5:15
Comments and hints on the calculation metrlod .............. 5:16
Load analysis .................................................................. 5:16
Evaluation of load spectra ............................................... 5: 17
If it is instead desired to determine the expected life N, proceed
according to points I, 2 and 3, and 0 rmax gives N from the
appropriate S-N curve.
If either of points 5 or 6 is met, correct the SoN curve.
P.S. Check that the finished structure conforms to the designer's intentions with regard to stress concentrations etc.
ProbaQility of failure QB .. ................... ..... ........ ................. 5: 17
Table 5.2.1
Typical spectrum parameters lip values"
Efficient and convenient calculation methods for designing
welded jOints have long been lacking, but with the introduction
of the stress range philosophy (0 r) in 1974, simplifications were
possible so that designers had a practical calculation method to
work with.
The Swedish Regulations for Welded Steel Structures StBKN2 (8) took a first step in this direction. The method used here
is largely based on these regulations.
The differences lie, first and foremost, in the permissible
stresses. Since the publication of StBK-N2, extensive reviews and
supplementary tests have been carried out. In our tables, we
have taken these newer results and international viewpoints into
account. For the most part, the failure curves agree with StBK-N2
at 2· 106 load cycles, but the slope has been adjusted to
WELDOX and HARDOX steels and to the slopes that have been
adopted internationally.
As has been illustrated earlier, in order that the fatigue strength
of a component may be calculated, the design of the component
must already have been largely determined, so that this type of
calculation is actually more of a check calculation.
The reader will undoubtedly find the calculation method simple
after having gone through a few examples in section 5.3.
Calculation procedure (see also comments below)
Given: Statically designed structure, i.e. steel and dimensions
given, welded joints positioned and designed in the most
favourable manner.
Required service life Nd (load cycles) given.
Determine the maximum permissible stress range Or perm'I
Nd
P
2. 106
1/3R
Bridges
1/2R
Steel structures
2.105
ON
MObile crane, hook operation
1/2 R
Mobile crane, hook operation
Overhead travelling crane
2/3 -IN
at steel mill
2/3 N
Container crane
108
OR
Ship bottom
Excavators (boom, dipper
4 - 5 years
112 - 2/3 N
arm)
Vehicle frames:
2· 106 4 - 5 years
2/3 R
Dumptrucks, tippers
4 - 5 years
1/2 - 2/3 R
Forestry machines
0- 112 R
Mobile crane chassis
Forks on forklift trucks
at sawmills
Lift arms on wheel loaders,
hard duty, regular cycle
Lift arms on wheel loaders,
normal varying duty
2· 106
1/3 R
114 N
0- 116 R
Nd == Design load cycle number
= Linear load spectrum, see figure 5.1.28
== Normally distributed load spectrum, see figure 5.1.29
R
N
1. Choice of point for analysis.
Determination of weld class etc., stress direction.
Determination of K., table 5.2.3.
Or
2. Load analysis
N/mm2
Measure, calculate or estimate the load spectrum and calculate 0 rmax' Determine the spectrum parameter p, table 5.2.l.
3. Determine the required probability of failure QB
Kx, P and QB give the appropriate SoN curve, table 5.2.5 a
or b (depending upon which value of QB is desired).
4. With the aid of the SoN curve, Nd gives 0 r erm' figures 5.2.1
- 5 or table 5.2.5 a or b.
Figure 5.2.1
1000
...... III!i;>.l.
500
300
200
6. If the point in question is not a gas-cut edge or heat-affected
material, 0 rperm can be multiplied by ks, depending upon
the steel, see table 5.2.2.
......
.........
~
~R'
Kx
1.3
P
5. If the structure does not contain any residual stresses, or if it
can be
contains known residual stresses and R < 0,0 r
multiplied by kR' depending upon the R value, ~i~1igure
5.l.2l.
Assumed SoN curves
(failure curves)
~
100
50
20
10 3
1.5
1.7
2.0
2.3
2.6
3.0
3.5
4.0
5.0
10 4
10 5
106
10 7
5:15
Dynamic strength - Practical design against fatigue failure
Figure 5,2.2
Comments and hints on the calculation proce~
dure
a,
N/mm'
1000
......
""
500
.:...;
200
Kx
100
1.7
~~
1.3
1.5 iZ
20
10 3
1
~
300
50
Qs 10- 2
v( ~rm
-p
t!oo.
2.
2.3
2.6
3.0
3.5
4.0
5.0
•
5.l.6-14_
iO'
Figure 5.2.3
N/mm'
1000
OB .10- 3
p=l
;:0.,;
500 :'"
~
300
:-;;
200
:--...
?
it
100
'?
~.
~.
50
'i
Watch out for fillet welds, even if they are not load-carryingl
The joint factor Kx is determined fr.om table 5.2.3 on the
basis of weld class etc. and principal stress direction. Note
where the design sections are located (marked in the table l.
A combined stress concentration effect from holes and
welds is not considered to exist if the distance between the hole
and the weld is greater than the diameter of the hole.
If the distance is less, the nominal stress range is multiplied
by Y KXhole and the Kx factor of the welded joint is used.
In the case of intersecting welds, the largest Kx value increased by one step from the series 1.3, l.5, 1.7,2.0,2.3,2.6,3.0,
3.5, 4.0 and 5.0 is used. If more than two welds interact, the
largest Kx value is increased by two steps_
If Kx is greater than 5.0, a higher weld class is chosen or the
joint is redesigned.
If the desired Kx value is not found in table 5_2.3, test results
can be used or an attempt can be made to estimate the Kx
value, some appropriat~ factor from the Kx series being used.
to-
"
20
10 3
L Choice of points
The choice of point(s) for analysis is the most important part.
The wrong point will yield a meaningless calculation. The secret
here is to find the most severely stressed region ((J r, principal
stress direction, stress gradient) in combination with existing
stress raisers (K x), which together give the most criticql point
with respect to fatigue failure_
As a safeguard, preliminary calculations should be carried out
for a number of likely points before it is possible to determine
which is the most critical.
Studies of failures and personal experience are very valuable.
Some idea of which points are critical is provided by figures
5.0
10 4
315 N/mm2
107
10 5
(definition)
Figure 5.2.4
<1,
Nlmm'
UUU
"Pl"m
p
500 ~
08=10--'
Certain welded joints can be improved from Sv 2 to Sv 3 if the
weld is TIG-dressed.
1
~
300
2.
200
IP
11.
100
I~'
50
§~
11
I?·
.0
-
')
liQ
20
10 7
10'
Figure 5_2.5
0,
Nlmm'
1000
500
OB 10 5
~ ~.
300
=
.::--.
200
100
n
50
:?
1
:?
20
103
5:16
III
~.C
10 4
105
106
-
10 7
load analysis
Load analysis is the weak link in the calculation chain and the
most difficult to get a grip on.
Load~ are calculated with the aid of the laws of mechanics to
start with. In many cases, this is all that can be done.
It can sometimes be very valuable to carry out some simple
measurement as a check.
Complex measurements are difficult to evaluate. Experience
shows that the effort is seldom thought to be worthwhile_
Comparative calculations performed on satisfactorily functioning, similar structures can be very valuable.
It is even more difficult to deal with the load spectrum. The
best method is to record a spectrum covering several operation
cycles on a tape recorder and have a computer evaluate it.
This is expensive, and few designers have access to such
equipment.
A simpler method is to use a pen recorder and evaluate
manually. Here, one must often settle for a single operation
cycle .
Typical spectrum parameters p ana Nd exist for certain structures, see table 5.2.1.
These p values should be used with circumspection. In the
case of cranes, for example, the Swedish crane regulations (74)
should be observed.
Experience has shown that severe (above-normal) service
gives a fuller, often normally distributed spectrum, while normal
use gives a linear spectrum. Furthermore, a normal spectrum
becomes linear on a vehicle frame during haulage driving.
Load spectra from manoeuvring, steering, cranes and excavators are often normally distributed_
Dynamic strength - Practical design agaill:st fatigue failure
A recorded spectrum can be evaluated in a number of ways.
In the case of welded structures, Range-Pair Exceedance
Count, often called simply Range-Pair, seems to be the most
popular method. This is because the minimum stress can be
disregarded for welded jOints (except in the case of stressrelieved jOints).
Some caution should be observed in using this method in the
case of Signals that resemble a pure sinusoidal Signal, see ref.
(l).
A spectrum is evaluated in the following manner:
l. Choose a number of suitablea r1' a r2' .... a rj levels.
2. Count the number of Nj for each a r by going through the
spectrum for each a fj' tin practice, la template with an
opening corresponding to each a fj is used.)
3. Plot the histogram. The 100 cycles with the highest a r may
be disregarded (favourable), but shall be checked to make
sure that they are less than a 5 ora permstat"
4. Determine the p value and place greater emphasis on the
curve having a good fit at high N/Nd values.
Figure 5.2.6 Evaluation of load spectrum
a
important! Do not confuse the load spectrum diagram in
fig. 5.2.6 with an SoN curve!
In other words, a load spectrum with a given p value, Nd , Kx
and a rmax is represented by one point in the SoN diagram for
the same Kx and p value.
3. Probability of failure GB
When the Kx and p values have been determined, the next step
is to determine,the probabiiity of failure, i.e. the desired factor of
safety.
As in the case of static load, discussed in the chapter "Factors of safety", there is also a scatter in the load and strength
data in connection with fatigue.
The SoN curve (the failure curve) represents a median curve
for 50% probability of failure in fatigue testing (where there is a
very small scatter in load).
If a curve for a lower probability of failure is desired, e.g. 10-3 ,
it is necessary to move the SoN curve along the N axis a sufficient number of standard deviations so as to obtain a 10-3
probability of failure. The distance the curve has to be moved is
thus dependent upon the standard deviation for the curve. This
varies with different Kx values. The scatter is often greater at low
Kx values and less at high ones. As a practical value for this
type of joint - mill scale, gas-cut, welded joint etc. - it can be
assumed that SN = 0.20 in 10 log N.
Scatter in spectrum testing is often less than in constant
amplitude testing (especially at low Kx).
Symmetrical spectra give considerably greater scatter than
asymmetrical spectra (1).
As far as the choice of probability of failure is concerned, 10-3
is usually obtained with the use of a safety factor along the a r
axis of Sf a = 1. 5-2.0, or along the N axis of SfN "'" 4.
f
It is often very difficult to know the scatter of the load, and it
is common to disregard it and to choose a probability of failure
of 10-4 _ 10.5 .
For our recommended values of a fperm' we have chosen the
following probabilities of failure:
10-3
Total number of operation cycles = 103
0"
0,
0,
N
10-4
10-5
N,..
corresponding to 3.10 standard deviations.
corresponding to 3.72 standard deviations
corresponding to 4.27 standard deviations
N'mm' Ollfla.l
0,4
0,3
0,
°1
400
300
200
100
I
075
05
025
Many factors influence the choice of probability of failure, but
here are some rules of thumb:
IO l - 100' 900
3.9· 10l
3· 10 3
5· 10 3
89·lO l
8.10 3
169· IO l
Very limited bodily injuries
Limited bodily injuries
Bodily injuries or very extensive material damages
Risk of extensive bodily injuries
• The 100 cycles with the largest Or may be
disrega,'ded, but must be "" a 5or 0 perm.lal'
10-2
10-3
10-4
10-5
In the Swedish Regulations for Welded Steel Structures
StBK-N2 (8) a probability of failure of < 10-5 is used.
Assumed SoN curves and curves for permissible stress range
GrlOrmax
a rperm for different probabilities of failure are presented in
1.0....,,------,..------------,
figures 5.2.1-5 for p = 1, i.e. constant amplitude.
For p < 1, see tables 5.2.5 a and b for linear spectra and
different probabilities of failure.
For parent material unaffected by welding, gas-cutting or very
sharp notches, a rperfT) may be multiplied by a material factor of
kB' as given in the taole below.
0.5
Table 5.2.2
Steel
p=1/6
103
104
log N
S 235 (BS 40, St 37-2)
S 355 (BS 50, St 52-3)
WELDOX 500
WELDOX 600
WELDOX 700
WELDOX 900
HARDOX400
1.0
1.2
1.3
lA
1.5
1.5
1.5
!
5:17
Dynamic strength - Practical design against fatigue failure
Table 5.2.3
No Constructional detail
In the figures, - - - marks the areas affected by the constructional detail
for which the stated values of K, are applicable. The arrows indicate the
direction of the stress and not the type of stress (normal stress - shear
stress, tension - compression).
lJ
Butt weld in single V joint
Weld K"
class
K'l
Svl
3.0
2.0
1.5
Sv2
Sv3
a) Where Ihe root of the weld is nol given a sealing run, the value of K,
shall be increased by one step for K"" and by two steps for K, l' in
the series of K, values in table 5.1.1 page 5:7.
2.3
1.7
1.5
a~all
b) The value of K, for Sv3 may, in this case, be applied also to TIG
dressed welds in Sv2.
OU""'-
--01
Butt weld in single V jOint
Sv1
Sv2
Remarks
With root with a sealing
run, alternatively
welded against a
backing strip which
is removed
cl The connection is capable of transmitting shear lorce.
Remarks
No Constructional detail Quality elc
K,
O! Parent material,
n.O)
12
ground surface
02
1.5 Rough rolled surface
of finish corresponding to a surface
rougher than Cutting
Class Sk3, see table
5.1.4
1.3 Surface finish corresponding to Sk3, burrs
and surface defects
removed
Parent material.
rolled surface
Parent material,
shot blasted
surface
04
Parent material,
hot-dip galvanised
surface
1.7
Parent material,
1.3 Surface irregularities
removed. Edges chamferecl according to Sk3
05
sawn surface
13
011__
--01
Butt weld in single V joint with
backing strip left in position
Svl
Sv2
011
14 Butt weld with incomplete
penetration
Svl
Sv2
OB
Thermally cut
surface
Sk2
Sk3
1.7
I.S
Open circular
holes
1.5d< c< 3d
2.6 Stress range may be
calculated over the
gross area. For
reamed holes with
chamfered edges, the
value of K, may be
reduced by one
step.
2.3 For values of K, for
bolted connections,
see StBK-N3.
~(J
~
.
---.
(1--
c> 3d
2.6
2.0
1.7
3.5
2.6
2.0
Root with sealing
run a)
01.-~DJ.
__--~-a1.
01
.,;;; 1:2
16 Butt weld at change of plate
thickness
2.3
1.7
1.5
3.0
2.3
1.7
Root with sealing
run a)
,;:; 1:3
,;:; 1:4
01.
Sv1
Sv2
Sv3
-~ __--~-(Jl
(JJ. 0J.
,;:; 1:3
Blitt weld at change of plate
width
Svl
Sv2
Sv3
-
3.0
2.3
1.7
-
OJ.-r::li:\
Root with sealing
run a)
Butt weld in double V joinl
K,"
2.3
1.7
1.5
~Oll
-01
K,J.
3.0
2.0
1.5b )
18 Butt weld at g'rder splice
__ (JJ.
~
1:3
OJ.--Svl
Sv2
Sv3
5:18
Svl
Sv2
Sv3
___...JJJ.
~,;:; 1:3
~(J
011---
-
R "" 6b
0---
10
-
.,;;; 1:2
3.0 Automatic flash welding
Composite connections
(studsf
2.6
2.0
011""'-
17
09
4.0
3.S
~Oll
,;:; 1:3
07
2_6
2.0
-01
1.3 See No OS
Parent material,
sheared surface
No sealing run on root.
but quality reQuirement applies to the
root side also
~all
15 Butt weld al change of plate
thickness
06
4.0
2.6
~all
See Nos 02 and 04-07
03
2.6
2.0
,;:;
Svl
Sv2
Sv3
-
nCIJn Ht::r:ln
3.0
2.3
1.7
Butt weld with sealing
run on root a)
Rolled or welded girder.
The K, value for Sv3,
however. applies only
to a welded girder.
On a welded girder,
other sections, see
e.g. No 30, shall also
be checked.
Dynamic strength - Pf'lu:tica! design against fatigue failure
No Constructional detail
Weld K'n
class
19 Butt weld at girder splice
Svl
Sv2
Sv3
(rolled girder)
-
K.l.
Remarks
No Constructional detail
3.0
2.3
2.0
Root with sealing run a).
Drilled or ground hole.
For Sv3, edges of hole
must also be dressed
(see No 08) (See Nos 25
and 33 ior welded
girders)
27
uCI:Jll
20
Continuous single V T-butt weld Svl
at attachment of circular or
Sv2
rectangular hollow section
Sv3
to stiff plate
Svl
Sv2
Sv3
5.0
4.0
-
Root with sealing run
21
28
5.0
4.0
Svl
Sv2
Sv3
a1
t
I
I--T
t-----tau ,
-I
all
~all
4.0
3.5
2.6
5.0
4.0
3.5
all~---a1
Root with sealing run
29
T-butt weld at edge of, or
parallel to, stressed plate
Svl
Sv2
Sv3
Single V T -butt weld
-I
an'
Svl
2.3
3.5
Sv2
Sv3
1.7
2.6
1.5
2.0b)
all-"'-"-
30
a1
Svl
Sv2
2.6
2.0
4.0
3.0
t
Sv3
1.7
2.3
'--T
I
-
25
Svl
Sv2
Sv3
3.0
2.6
23
-
T -butt weld, e.g. at beamcolumn JOint
a1t -,
A-A
rBtb~~8
!b . a a1
~tA-l
B-~·t
-+
-
Manual weld
4.0
3.5
3.0
Automatic weld
!~!
,all
I
I
Fillet weld
Svl
Sv2
Sv3
2.3
2.0
1.7
-I
32
Svl
Sv2
Sv3
3.0
2.6
2.3
-
-
-
Section b-b
Svl 2.3 3.5
Sv2 1.7 2.6
Sv3 1.5 2.0
Force at Section b-b
may be assumed dispersed over 45' as in figure
(applies also to rolled
column of I section. in
which case the values of
K, for Section b-b may
be put equal to 2.01
I--T
Fillet weld on one side only
Svl
Sv2
5v3
3.0
2.6
2.3
-
3.5
3.0
2.3
-
3.5
3.0
2.3
-
-
-
1--1
--IL ___: __ -I-all
Drilled or ground hole
K, values apply to ends
of T -butt weld
all
33
Section a-a
Svl
3.5
Sv2 2.6
Sv3 2.0
'
all-i- __ _~---i all
Penetration shall be
equal to at least half
the plate thickness
(bottom plate in the
figure 1
OiIEfj
+- -.-l-JfrjOUT
---+
26
1.7
4.0
3.5
3.0
tal
I
T -butt weld at girder splice
(welded girder)
2.6
2.3
+------t-
31
-I.+- __ '___ -f"
I--T
all
3.5
tal
t
Incomplete penetration weld
2.6
/~4~a1
Svl
Sv2
Sv3
a1
24
5.0
4.0
_~ ~all
--~
Fillet weld
--I
all'
Root with sealing run,
Symmetrical cross
section.
I-girder or box girder
'all
--t------+
4.0
3.5
tal
a1
23
5.0
4.0
Sv1
Sv2
T -butt weld at edge of, or
parallel to, stressed plate
all~al
T-butt weld
5.0
4.0
I~ 250 mm
tal
___ all
22
5.0
4.0
3.5
Sv3
-a1
Butt weld at edge of stressed
plate
-
Remarks
-~~~~~:~~~~~~- a
~all
all---
-
K,l.
I
a
Butt weld at edge of stressed
plate
Weld K"ll
class
Fillet weld at girder splice
Svl
Sv2
Sv3
-
-
OiImJrj~'T
-1-- -+ --134
Intermittent fillet weld between
flange and web in I girder
--I
I,all
-1-------+
all' 0-I
--
Svl
Sv2
Sv3
Drilled or ground hole.
Stated K, values apply
to ends of fillet weld.
Other sections shall
also be checked. see
e.g. No 12
-
t
5:19
Dynamic strength - Practical design against fatigue failure
Weld K,"
class
No Constructional
detail
35
-
Svl
Sv2
Sv3
Fillet weld at edge of. or
parallel to. stressed plate
5.0
4.0
tal
K,J.
Remarks
No Constructional
detail
-
Stated K, values also
apply to section through
the weld metal
44
5.0
4.0
.....-an
-
all36
~
Svl
Sv2
Sv3
Fillet weld at edge of. or
parallel to. stressed plate
4.0
3.5
2.6
5.0
4.0
3.5
~~
Stated K, values also
apply to section
through the weld metal
45
Box girder with stiffeners
ii
""
- / 450--. al
Fillet welded longitudinal cleatc)
~~
f-4
..~ h= !I'&:II
-a'
I ' 11:
. I':.1 '-la
r---It--l;-;
t
Svl
Sv2
Sv3
For I.. 100 mm. values of K,
5.0 may be reduced by one step
4.0
3.0b)
Svl
Sv2
Sv3
U
Iia ~
i
I -4+.I,
-I
a.
ttl I
5.0 If width of cleat is less
4.0 than half the plate width.
3.0b) K, values may be reduced by
one step. If I > lOO mm.
No 48 shall be applied
3.0
2.6
2.0
Manual weld
Svl
Sv2
Sv3
2.6
2.3
2.0
Automatic weld
Svl
Sv2
Sv3
-
It~ I
I
.
47 Girder with longitudinal
web stiffeners
I
1
I~ 100 mm
HI I
Svl
Sv2
Sv3
Fillet weld at transverse
attachment c)
5.0
3.5
2.3
i
Weld not returned at the ends.
Stated K, values may also be
applied to T-butt weld
~a
!~,
I
48 Girder with cover plate
aSvl
Sv2
Sv3
40 Fillet weld at transverse
attachment c)
4.0 Weld returned at the ends.
3.0 Stated K, values may also be
2.0b) applied to T-butt weld
I
~a
.
49 Girder with cover plate
Fillet weld at longitudinal
attachment c)
Svl
Sv2
Sv3
5.0
4.0
3.0
For I .. 100 mm. values of K,
may be reduced by one step.
Weld not deSigned for transmission of force
;::::::::::::::...
a-
~ ~
0.~
Svl
Sv2
Sv3
4.0
3.5
2.6
~(J
Good contact between rail
and top flange. Rail and
girder are assumed to interact when fll is determined.
but not when"lI is determined
(fm~mlilIl
-
With transverse fillet weld.
5.0 For Sv3. transverse weld and
4.0b) at least 50 mm of longitudinal welds nearest the
corners shall be dressed.
.
50 Girder with cover plate
Sv3
3.0
~o
Transverse fillet weld shalt
be dressed to taper of 1:3
or less. At least 50 mm of
longitudinal fillet welds
nearest the corners shall
be dressed to Sv3
a---
aContinuous beam with stiffeners Svl
over intermediate supports (the Sv2
figure shows three alternatives) Sv3
Svl
Sv2
Sv3
~a
~
-!::-~~(J
~~
--~~
Intermittent fillet weld between
crane rail and crane girder
With or without transverse
5.0 fillet weld to Sv2.
4.ob) For Sv3. transverse weld and
at teast 50 mm of longitudinal welds nearest the
corners shall be dressed.
a-
~a
a-
5:20
Svl
Sv2
Sv3
I!)
46 Girder with longitudinal
web stiffeners
I
38 Fillet welded transverse cleatC)
43
Weld returned at the ends.
Stated values of K, also
apply to single stiffeners
~
all---
42
3.5
3.0
2.6
1
1
.~
41
Svl
Sv2
Sv3
Il1
,D
ttl
_all
~= ....~
39
'" (f, iff, < 0.6",)
shall be calculated at edge
of stiffener. Weld returned
at the ends. Stated values
of K, also apply to single
stiffeners
-al
tal
37
Remarks
3.5
2.6
2.0
Svl
Beam with web stiffeners in
the span and over end supports Sv2
(see also No 43)
Sv3
ffWI)mJIl!
----~>II'Ii!'
Weld K,
class
4.0
3.0
2.3
", (f, iff, > 0.6",)
shall be calculated at edge
of stiffener. Weld returned
at the ends
51
Girder with cover plate
Svl
Sv2
Sv3
2.3
2.0
1.7
~a
a-
Refer to section at least
one flange width from the
end of the flange plate
Dynamic strength - Practical design against fatigue failure
No
Constructional
detail
Weld K.
class
52
Fillet welded connection
of member
Svl
5v2
Sv3
'
, --I
+~
! :. '-a
!
JI
+-53
5.0
4.0
3.5
Remarks
Kx values for pressure vessel nozzles under pulsating internal
pressure, from ret (43).
Table 5.2.4
Type
Joint
Fatigue strength
N/mm at
Kx
N = 2· 104 105
Svl
5v2
5v3
fillet welded connection
of member
2· 106
5.0
1
~
3.5
140
102
49
2
rlEh
2.5
152
120
71
3.0
180
177
58
4.0
! • ·--a fa
+~
+-
54 Fillet welded symmetrical
Svl
Sv2
Sv3
lap Joint
a-I
~
: i
I
55
11
: i-a
11
11
I
:
,
~,
5.0
4.0
Sv3
3.5
i_a
~
I!
I !
at attachment of circular or
rectangular hollow section to
stiff plate
a
Sv2
11
56 Contir,uQus single fillet weld
Svl
5v2
5v3
-
n
4
AA
3.0
167
116
50
5
~
1.9
160
112
52
6
~
2.2
195
140
77
I
-;::1::::~--1I~ a
~ 250 mm
3
!
Svl
Fillet welded symmetrical
lap joint
a-I
4.0
3.5
3.0bl
5.0
4.0
The arrow indicates the area for the kx value.
5:21
Dynamic strength - Practical design against fatigue failure
Table 5.2.5 a
Permissible stress rangeu r N/mm2 at probability of failure QB = lo-l
Kx
N
p
10'
104
105
6· 105
106
2· 106
900
636
357
228
201
169
1.5
900
598
321
197
172
143
p = 5/6
104
105
6· 105
106
2.106
107
755
425
271
239
201
135
711
382
235
204
170
110
664
344
206
178
146
92
628
306
175
149
120
73
535
248
137
115
92
62
472
219
121
102
81
55
430
200
110
93
74
45
389
181
99
84
67
39
361
168
92
78
62
36
305
142
78
66
52
31
p = 2/3
104
105
6· 105
2.106
107
lOS
900
522
334
294
247
165
874
469
289
251
209
135
817
423
254
219
180
114
772
376
215
183
148
89
658
306
168
142
112
70
581
270
149
125
99
61
529
246
135
114
91
55
479
222
122
103
82
48
444
206
114
96
76
44
376
175
96
81
64
38
p = 1/2
104
105
6· 105
106
2.106
107
900
643
471
428
387
308
900
607
347
326
271
175
900
548
328
284
233
147
900
488
279
238
191
116
854
396
218
184
146
95
754
350
193
162
129
82
687
319
176
148
118
75
621
288
159
134
106
69
576
268
147
124
99
58
488
226
125
105
83
49
p = 113
104
105
6· 105
106
2· Hr
107
900
793
582
555
496
415
900
735
512
480
439
358
900
713
469
416
354
243
900
689
393
335
270
163
900
560
308
260
206
l35
900
494
272
230
182
118
900
451
248
209
166
109
877
407
224
189
150
93
815
378
208
176
l39
81
689
320
176
149
118
69
p = 1/6
104
105
6· 105
106
2· 106
107
900
900
804
753
688
559
900
900
711
662
602
481
900
900
700
636
558
412
900
900
635
563
479
328
900
777
483
422
351
240
900
813
448
378
300
205
900
741
408
344
273
185
900
670
369
311
247
144
900
622
342
289
229
l34
900
526
290
244
194
113
104
105
6.105
106
2· 106
107
900
900
900
900
861
690
900
900
900
900
839
645
900
900
900
900
816
594
900
900
900
900
772
521
900
900
876
758
623
395
900
900
789
683
561
356
900
900
811
684
543
318
900
900
679
581
470
288
900
900
634
543
440
269
900
1.3
p
=1
p=o
5:22
1.7
559
289
176
149
123
2.0
900
529
258
147
125
101
2.3
900
450
209
115
97
77
2.6
856
398
184
101
86
68
3.0
781
362
168
93
78
62
3.5
705
327
152
84
71
56
4.0
655
304
141
78
66
52
5.0
554
257
119
66
55
44
900
900
560
473
375
220
Dynamic strength - Practical design against fatigue failure
Table 5.2.5 b
Permissible stress rangeo r N/mm2 at probability of failure Ga = 10-5
P
N
900
481
249
150
129
106
2,0
900
450
219
125
107
86
Kx
2,3
/81
363
168
93
78
62
2.6
692
322
149
82
69
55
3.0
642
298
138
76
64
51
3,5
592
275
128
70
59
47
4,0
554
257
119
66
55
44
5,0
491
228
106
58
49
39
p= 1
103
104
105
6· 105
106
2· 106
900
557
312
199
176
148
1.5
900
514
276
170
148
123
p = 5/6
104
105
6· 105
106
2· 106
107
661
372
238
209
176
118
612
328
202
176
146
95
572
296
178
154
126
80
535
261
149
127
102
62
431
200
110
93
76
55
382
177
98
82
69
44
354
164
91
76
61
38
327
152
83
70
56
33
306
142
78
66
52
31
271
126
69
58
46
27
p = 2/3
104
105
6· 105
106
2· 106
107
813
457
292
257
216
145
752
404
249
217
180
ll6
704
365
218
189
155
98
658
321
183
156
126
76
530
246
135
114
93
63
470
218
120
101
84
56
436
202
94
75
48
402
186
103
87
69
40
376
175
96
81
64
38
333
155
85
72
57
33
p = 1/2
104
105
6· 105
106
2· 106
107
900
592
378
333
280
187
900
522
322
281
233
151
900
472
283
245
201
127
853
416
237
202
163
99
687
319
176
148
118
83
609
283
156
131
108
73
565
262
144
122
97
64
520
241
133
112
89
55
488
226
125
105
83
49
432
201
llO
93
74
43
P = 1/3
104
105
6· 105
106
2· 106
107
900
710
50G
470
433
358
900
680
430
397
356
278
900
620
398
344
282
178
900
586
335
286
230
139
900
450
248
209
186
119
861
400
220
186
149
106
799
370
204
172
137
93
736
342
188
159
126
80
689
320
176
149
118
69
610
284
156
132
105
61
p = 1/6
104
105
6· 105
106
2. 106
107
900
900
673
630
577
469
900
892
665
611
545
419
900
850
579
528
466
348
900
831
536
473
399
269
900
741
408
344
273
205
900
658
362
305
242
180
900
610
336
283
225
150
900
562
309
261
207
131
900
526
290
244
194
116
900
466
257
217
172
101
p=o
104
105
6· 105
106
2· 106
107
900
900
900
900
833
644
900
900
900
900
789
554
900
900
900
893
760
522
900
900
900
794
665
439
900
900
811
684
543
333
900
900
668
572
462
298
900
900
623
533
432
270
900
900
599
505
401
234
900
900
561
473
375
220
900
900
497
419
333
195
1.3
1.7
III
5:23
5.3
Example 5.3.2
Examples - Fatigue
A welded joint in a crane jib is loaded as shown in figure 5.3.2
Page No.
Example
5.3.1 ....... ........... ... ...... ........ ...... ...... ............ ... 5:24
Example
5.3.2 ............................................................... 5:24
Example
5.3.3
Example
5.3.4
5:25
Example
5.3.5
5:26
Example
5.3.6
5:26
Example
5.3.7
5:27
Example
5.3.8
5:27
Example
5.3.9
5:28
Example
5.3.10
5:29
Example
5.3.11
5:29
Example
5.3.12
5:29
Figure 5.3.2
5:25
and designed with the following data:
a r = 125 N/mm2
p = 112
Nd = 2· 106 load cycles
Weld class Sv2
Probability of failure Qs = 10-5
Example 5.3.1
A frame is spliced with a butt weld as shown below. Weld class
Sv 2.
The joint is designed to withstand 2· 106 load cycles. Full
spectrum, i.e. p = 1.
During fabrication, an additional request is made: to put a light
on the jib.
The fabricator, who wishes to oblige the customer, drills a
hole as shown in the figure. What happens to the joint now?
What stress range can be. permitted when the probability of
failure Qs is:
a. 10-2
b. 10-5
Solution:
Hole*
Weld
Figure 5.3.1
case 08
case 13
table 5.2.3
table 5.2.3
Kx = 2.6
Kx = 2.0
'The distance between the weld and the hole is less than the hole diameter. so
the two stress-raisers are considered to interact.
To adjust for a hole, the nominal stress range shall be
multipliE'd by
V Kxhole
.: a r = 125· v'2.6 = 201 N/mm2
. a rperm in accordance with table 5.2.5b (10-5 )
Solution:
Stresses: Mainly bending stresses
Critical point: The corner A. Note! Intersecting welds i.e. the
highest Kx value is increased by one step in the Kx series
Kx.L = 3.5 (Case 13, table 5.2.3) } -+ KXA = 4.0
KXII = 2.0 (12)
5:24
Qs
Figure
a rperm
N/mm2
a) 10-2
b) 10-5
5.2.2
5.2.5
57
43
Kx = 2.0
p
112
Nd = 2· loG
=
I
a rperm = 163 N/mm2
. : a r > a rperm
What is the probability of failure now?
If we compare with table 5.2.5a, which applies for Qs = 10-3 ,
we getarJ?e!11J. = 191 N/mm2, so it can be said that QB lies
between 1()" and lQ-J
Dynamic strength - Examples - fatigue
Example 5.3.3
A welded joint for a tipping cylinder lug is to be designed
against fatigue, and the lug is to be welded to a frame as shown
in figure 5.3.3. What throat thickness should be chosen, and
will the joint hold?
Figure 5.3.3
p
= 1
Nd = 6· 105
Weld class Sv2
Q B == 10-3
W,
10· 106 mm 3
Pr = 40 tonnes
=
The frame is naturally subjected to other loads as well, so this
point must be checked for the total load spectrum (including
the tipping cycle) that acts on the frame.
Example 5.3.4
Assume that someone wishes to load more on his vehicle than
the frame permits. The frame must then be reinforced with a
cover plate on the flanges.
How thick and how long should the reinforcement be?
QB = 10-5
Weld class Sv2
p = 2/3
Nd == 2· 106 load cycles
3000
Figure 5.3.4
50 "'I
Present design:
2000
IPr = 12 tonnes
t 2000
A-o
~
N
Pr = 40 tonnes
t = ?-l---I------r
Solution:
Section modulus W = 1.26' 106 mm 3
In the case of a load-carrying fillet weld, cracks can initiate
either at the root or in section A-A.
Desired design:
We check that the lug will hold in section A-A
a rl =
KXl =
P
Pr
350· t
40· 104
350. 20
2
= 57 N/mm
4.0
=1
Nd = 6· 105
QB = 10-3
~ according to table 5.2.5a a rperm = 78 NI mm 2
OKa r < arperm
The throat thickness is selected from figure 5.l.36, which gives
optimum throat thickness. Assume a certain penetration. Take
2a)/T p = 0.8
Tp = 20 mm~ H/Tp = 0.98~ H = 19.6
throat thickness (a)=
~ = ~ = 13.8
V2
V2
Will the frame hold under the stress concentration effect of the
lug?
KXII = 2.0 case 51 table 5.2.3
QB = 10-5
Nd
= 2/3
= 2· 106
W
=
new
The bending stress in the frame member is:
->
(according to table 5.2.5b)
a rperm = 126 NI mm 2
18· 104 . 2000
2. 126
= 1.43. 106 mm 3
= 60 N/mm2
The most critical point is at the ends of the weld jOining the lug
to the frame member
K'II
p
Nd
QB
There are two critical points for fatigue here:
a. In the middle of the beam along the weld on the cover plate
b. At the end welds on the cover plate
Point a. determines the required thickness t.
P
i.e. throat thickness = 14 mm
4
Pr· L
a rbend =2.
- - = 40· 10 • 3000
W
2· 10· 106
Solution:
t:. W "'" (t· 100)'
(130 + _t_ ) 2. 2· _1_ = 0.17' 106 mm 3
2
130
t"= 6 mm i.e. 6· 100 mm
(case 36 table 5.2.3) = 3.5
= 1
~ according to !able 5.2.5a
= 6· 105
= 10-3
a perm = 84 N/mm2
a r bend < a rperm OK at least
for tip loading.
End weld (case 48) Kx
= 5.0
10-5
p = 2/3
Nd = 2· 106
QB
=
~ arperm = 57 N/mm2
5:25
Dynamic strength - Examples - fatigue
The design section for the cover plate is such that we must
calculate where a r = 57 N/mm2 is located for the desired load,
but with original W.
Example 5,3.6
The fabricator is unfortunately forced to weld a socket to a
frame flange with dimensions as shown in figure 5.3.7.
a. How much must the stress be reduced?
Figure 5.3.5
b. If we TIG-dress the weld, how much must we then reduce
the stress?
OB = 10-3
Weld class Sv 2
p = 2/3
Figure 5.3.7
For end weld:
Mmax = arperm' W = 57·1.26· 106 Nmm
57· 1.26' 106 . 2
.
Distance from support L =
4
18· 10
Nd =
2· 106
)
o
co
90
78
= 798"'" 800 mm
12
1.. 100 .1
i.e. the length of the cover plate = 4000 - 2 . 800 = 2400 mm
Solution:
The critical point is the transition between the flange and the
weld for the socket.
Example 5.3.5
A socket for a hydraulic tube is to be welded onto a vehicle
frame.
If the frame has the dimensions shown in figure 5.3.6, how
high can the socket be made if it is welded at the neutral layer
and the frame is designed against fatigue?
Os = 10-5
Figure 5.3.6
Weld class Sv 2
p = 1/3
Nd = 2· 106
- ~C::::;;:::::;;:J
[} )
The following applies for the longitudinal weld on the beam:
KXII
= 2.0 case 31
table 5.2.3
OB
= 10-3
= 2/3
= 2.106
P
Nd
-> arperm (table 5.2.5bl
= 148 N/mm2
For the top surface of the flange
90
2
armaxperm= 148· 78 = 171 N/mm
a. But when we weld the socket on, Kx = 3.0 (case 40) and
0rmaxperm = 91 N/mm 2 , i.e. 47% reduction!
o
N
N
20
Thus, this socket, whir::h is non-load-carrying, completely determines the fatigue strength of the beam!
b. TIG dressing, whose purpose is to improve the geometry of
the weld, can be employed to great advantage here
Figure 5.3.8
Solution:
TIG dressing
The height of the socket shall match a rperm for the longitudinal
weld.
I
Case 32 table 5.2.3 gives:
KXII = 2.6 5
OB = 10p
= 1/3
Nd
= 2.106
-> according to table 5.2.5b
a rperm = 149 N/mm2
The socket gives Kx case 39 = 3.5 andorperm = 126 N/mm2
h is calculated from
220/2 - 20
149
h/2
126
- - - - - = - - -> h = 152 mm
A TIG-dressed weld of class Sv 2 may, for certain joints (where
TIG dressing brings about an improvement), be counted as Sv 3.
Case 40 Sv 3 Kx = 2.0
Owing to the stress gradient,
a rmax perm = 148 NI mm2
armax perm =
°rmax
148
= 0.86
148. 90
78
Le. 14% reduction with
TIG dressing compared to
47% without TIG dressing'
In order to obtain the same strength in the beam after welding
of the socket, it is necessary to increase the thickness of the
flange to 23 mm in case a. and 14 mm in b.
526
Dynamic strength - Examples - fatigue
Example 5.3.7
A skilled designer has located his welds where the stresses are
low in the structure. Accordingly, the maximum stress is located
where the material is unaffected.
How much lighter can the structure (the beam) be made of
WELOOX 700 than of S 355 (BS 50) with the same overall
dimensions?
Figure 5.3.9
o
Mr(
(((!(((((((((((l(((((((((({
)Mf
N
no
=
WELDOX 700 as = 700 N/mm2
What is the difference in weight between the two steel grades?
3
Moment of inertia I "" 2· [ t'100
12
+ 30· t· 452 ] =
= t· 2.88' 105 (mm4)
__
t= 10 mm
CV)
-------
Two steels are usually considered:
as 350 N/mm2 < arperm!
S 355
Required section modulus W = ~
orperm
-W§QJ,
1.8· 104 . 1650 = 8.49. 104 mm3
350
Ws 355
Solution:
w"", 2· t' 6320' + 2· _1_ ·160· t· 160'zt'85335
1.8· 104 . 1650
439
W WELDOX 700 =
= 6.77' 104 mm3
160
W= _1_
50
Mr = 0rperm' W
For material unaffected by welding, a rperm = kB . a rperm with k8
obtained from table 5.2.2
Kx mill scale case 02 table 5.2.3 = 1.3
5
OB = 10-
=1
Nd = 2· 10°
p
S 355
1 a
fperm = 148 N/mm2
(table 5.2.5b)
Kx = 1.3
kB = 1.2
WX 700
Kx = 1.3
kB
= 1.5
t\'IX 700' k 1 · 222
t=
W' 50
2.88.105
t BS 50 D =
8.49· 104 . 50
2.88' 105
= 14.7, i.e. 16 mm
tWELDOX 700 = 11.7 mm i.e. 12 mm
What is the difference in weight for two forks?
a rperm = 148· 1.2 = 178 N/mm2
a rperm = 148· 1.5 = 222 N/mm2
= 10· k 1 · 178-> tWX700 = 8 mm
The cross-sectional areas
m
= Ltot ' 2 . 260· t . 7800· 10-9 [ kg]
mBS 50 0 = 3300·2· 260· 16· 7800· 10-9 = 214 kg
mox 812 = 160 kg
t.m
"wx 700 (640 + 320) . 8
= 54 kg or 25%!
ABS 50 D (640 + 320) . 10
i.e.
Awx 700
ABS 50 0
= 0.8
i.e. 20% lighter!
Figure 5.3.10
A
Example 5.3.8
A designer has to choose a material for a log grapple mounted
on a forklift truck. The log grapple must be able to withstand 3
tonnes (1.8 tonnes on each of the two tines, since the load may
be unbalanced). The profile of the grapple shall be as shown in
figure 5.3.10. The fabricator can weld to class Sv 2, and a
probability ot failure of OB = 10-3 is considered adequate.
Ltot = 3300 mm
3 ton
K, 11 = 2.0 /
Sv 2 -
__
Gas-cut, cutting class 3
Sk3-K x = 1.5
/
Solution:
First, we have to know the load spectrum and the required
service life.
=
2· 106 for
Use e.g. table 5:2.1, which gives p = 1/3 and Nd
forks on forklift trucks at sawmills. The critical point is the gascut surface in section A-A.
Kx = 1.5 case 07 table 5.2.31
OB = 10-3
P = 1/3
Nd = 2· 106
o
o
:c
:c
o
o
0"1
~t
* gives equal fatigue
strength when
Kx gas-cut = 1.5 and
KXII welded = 2.0
-> a rpe(m =
439 N/mm2
(table 5.2.5a)
5:27
Dynamic strength - Examples - fatigue
Example 5.3.9
A stress spectrum on the frame (S 355) of an industrial truck
(prototype) has been measured as shown below during 10 hours
of typical operation at a point where a risk of fatigue failure is
considered to exist. (After some time, the frame broke here.)
The frame has the dimensions specified in the figure and
must be spliced due to limitations in fabrication technology.
Have the right steel and plate thickness been chosen to
achieve a life of 104 hours (4 years) with a probability of failure
QB =
a fmax E; as
1O-3?
Nd ". 1.019· 106 load cycles and the value of p lies between 0
and 1/6. choose p = 116. since the largest contribution to the
cumulative damage takes place at N > 105 .
The value of Kx at point Dis then determined.
Sv 2, intersecting welds!
Kxll "'" 2.0 case 12 table 5.2,3 (can be compared to longitudinal
butt weld)
Kx.l = 3.5 case 13
Kx tot { Kx increased by one step} = 4.0
Figure 5.3.11
=
= 4.0
10-3 1
P = 1/6
Kx
a rperm = 289·N/mm 2 (table 5.2.5a)
QB
Nd "" 106
We havea r ax = 500 N/mm2 and 10 mm plate thickness. So it
wasn't so sFrange that the frame broke at this particular point.
Increase the plate thickness to 500. 10 = 17.3 mm
289
Take 18 mm plate thickness a rperm < O's for S 355 (BS 50)
.,' S 355 is adequate here.
We also check point E.
Only longitudinal weld Sv 2, Kx 11 = 2.0
Kxll = 2.0
I
~B : ~?: -+according to t~ble 5.2.5ao rperm
Nd
B-B
Weld class Sv 2 at E. D, S, A
Sv 1 at C
= 106
The maximum stress range at point E after the increase in
plate thickness = 289· 2540
= 489 N/mm 2. (It was even
1500
larger before!)
Measured stress spectrum
ar
N/mm2 aria rmax
500
450
400
350
300
250
200
150
100
50
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
N
10 hours
Ntot
N
104 hours
1· 163
1
2
3
6
25
52
75
183
235
437
This means that the actualarmax = 489 N/mm2 is less than
arperm = 563 N/mm 2 .
1· 163
2
3
6
25
52
75
183
235,
437
3
6
12
37
89
',' 18 mm WELD OX 100
164
Now we also check the weld for the buckling stiffener at point C,
a - (section depthf approximately
347
582
Nd = 1019
a r maxc = (650)2. 600 . 289 = 76.3 N/mm2
800
1500
Sv 1-+ Kx = 5.0 case 39 (Note! not intersecting welds)
First we determine the value of p from the measured spectrum
by plotting the histogram.
Note that we may disregard the 100 largest a r-values.
~~ ~
~ "-. 1-"'"
0.5
i"-.. ~ 1 -........
i'l
""" '"
5:28
Nd =
-+table 5.2.5aa rperm = 244 N/mm2
106
.,' Steel WElDOX 700, plate thickness 18 mm!
~
~ r-....l
.......
~~
102
~~ : !~-31
OK sincearmaxC= 76.3 <a rperm = 244 N/mm2
Figure 5.3.12
Ur/ur max
o
OK. Bur larger than 0' 5S355 = 350 N/mm'. Not enough!
WElDOX 700 must be used here!
Solution:
1.0
= 563 N/mm
103
104
r-...
105
p= 1/3
............
~
L 1/6
~
106
Dynamic strength - Examples - fatigue
Figure 5.3.13
Example 5.3.10
or/omax
A critical point in a welded structure has the following data:
Kx = 3.0
QB = 10-3
P == 2/3
Nd = 107
R = a min
--> table 5.2.5aorperm =
53 N/mm2
= _ 0.6
°max
a min = - 20 N/mm2
i.e. we have an external stress which becomes compressive
some time during the load cycle.
Question: Is it possible to increaseorperm by means of stress
relieving so that we meet the conditionormax = 65 N/mm 2 ?
oL---~----~--~~===d
10 3
102
104
10 6
10 5
Nd = 10 5
Which steel can be used when QB = 10 -5?
Solution:
Use figure 5.1.21, which gives the factor kR = 1.13 for R = -0.6
a rpermstress.rel. = kR . a rperm
Solution:
Table 5.2.5b
p = 0 and 105 givesorperm = 900 N/mm2
a rpermstress-rel. = 1.13' 53 = 60 NI mm 2 < a rmax
i.e. WELDOX 900 up to and including Kx = 5.0!
Answer: No
(Note: This is an actual application today!)
Example 5.3.11
Example 5.3.12
A mobile crane only utilizes its maximum capacity about 250
times during its life. It frequently utilizes its lifting height, how-
The frame for a forwarder (an articulated forestry machine for
hauling felled timber cross-country) is considered to be too heavy.
The material that has been used in the frame is S 355. It has not
been thought possible to utilize a higher yield strength, owing to the
stress-raising effect of the welded joints. The rear frame is now to
be redesigned.
ever.
A typical load spectrum is shown in figure 5.3.13.
Figure 5.3.14
Mk
~
I
P210ad
Present design
Log bunk
Crane moment Mk
~
Crane mount
3.5
5.0
Bogie mount
Moment and
Transverse force
from front section
O
KXll=1.7
t=10
320
A-A
o Marks critical points for fatigue failure
The numbers indicate Kx values in Sv 2
120
Proposed
design
Frame length 4300 mm
o
Section 0-0
Section E-E
5:29
Dynamic strength - Examples - fatigue
The loads come for the most part from the front section, the
crane, the log bunks and the bogie (the 'rear axle').
Experience shows that extreme loads can easily arise in connection with e.g. loading and driving over very rough terrain.
It has been necessary to utilize S 355 up to its yield stress
(350 N/mm2) for such cases. A probability of failure Qs 10.3 ,
a load spectrum p 1/3 and a life Nd 2 . 106 are assumed.
The fabricator can weld to class Sv 2, but does not accept
grinding.
How much lighter can the frame be made? Can steel with a
higher yield strength be utilized?
The present design of the frame is shown in figure 5.3.14.
=
=
=
Now we only have the longitudinal weld on the frame beams as
the limiting Kx. It would also be possible to move this weld to the
web and utilize the ks factor and be able to use steel with a higher yield strength. In this situation, all welds must be moved even
more towards the middle of the web and be located on an area of
0.2 . H (beam depth) in order that e.g. WELDOX 700 may be
utilized to its limit. If we can do this, we have come a long way!
Usually, it proves difficult in practice to come below 0.4 . H,
and then we should be able to permit stresses as shown in
figure 5.3.15.
Figure 5.3.15
Solution:
Mark the most probable positions of the critical points and indicate the Kx values.
Calculations, testing or experience will show where the most
critical points are. This analysis is not carried out here.
The result is that we have largely the same high fatigue
stresses along the entire frame on the top and bottom flanges. It
is then relatively easy to identify the most critical points, those
with the highest Kx values.
The bogie mount, Kx = 5.0, determines the maxa rperm
which, according to table 5.2.5a, is:
Kx = 5.0
p = 1/3
Nd = 2· 106
orperm = 118 N/mm2
This was, .of course, known, and a plate thickness in the frame
beams of up to lD mm was therefore chosen.
We redesign the frame and try to move the welds to areas of
low stress and simultaneously reduce the Kx values.
A proposed design is shown at the bottom of the figure (fig.
5.3.14).
Section B-B: The crane mount has been welded to the vertical
plates instead of directly to the beams.
The reinforcement wedge has been pulled up
on the beam web and the vertical weld against
the beam web goes only so far down that it is of
the same strength as the longitudinal weld on the
beam.
Note also the vertical plate's weld against the
beam.
Section C-C: Mount for bottom and top protective plate is combined and plug-welded to the beam web.
Section D-D: No weld on beam flanges!
Section E-E: The log bunks are fastened with bolted jOints and
the tubes are positioned so that the Kx value and
o rperm match Kx and 0 rperm on the flanges of the
frame beam.
5:30
0 , " {,;"-_-; K. = 3.5- 0, = 15.0 N/mm2
Orperm= 375 N/mm2
Assume that Kxflange = 1.7 is limiting, thenorperm = 354 N/mm2.
Compare this to the original permissible stress range of
118 N/mm 2.
We can then reduce the thickness of the plate in the beams to
118
- - 'lD= 3mm!
354
This is an unrealistic plate thickness in practice.
Choose therefore t = 6 mm!
There is a risk here now! In the original proposed design, the
yield stress 350 N/mm2 had been utilized for extreme loads.
The stress from these loads now increases to:
10
"'6
. 350 = 583 N/mm 2
Therefore choose WELDOX 700 (as = 700 N/mm2) t = 6
mm, then the frame can withstand both fatigue and extreme
10ads.WELDOX 700 is also better able to withstand
unforeseen extreme loads in fatigue.
When the frame is 4300 mm long, the weight reduction is then:
dm = M· L· (2B' 2H)' 7800· lD-9
dm = (10 -6)' 4300 (2·120 + 2· 320)' 7800· lD-9 = 118 kg
i.e. about 100 kg lighter, which means when the frame weighs a
total of about 800 kg, about 12% lighter!
5,4
Something about crack
propagation
Page No.
Calculation method for crack propagation under fatigue load
If we examine figure 5.1.4, which shows how a crack propagates
a distance A a during a load cycle I:!. N, we can see that there
must be some relationship between the steel, a r' A a and I:!. N
in order that the way in which a crack grows may be described.
Such a relationship is the Paris fatigue crack propagation
equation:
Calculation method for crack propagation
under fatigue load .............................................. ,', .. ,....... 5:31
Example ................................ ,.... ', ........... " .. ,........ ,.... " ... 5:33
Design data ....................... " ........................................... 5:35
~~ - crack propagation per load cycle in mm/cycle
This section is not specific to WELDOX and HARDOX steels, but is
nevertheless of great interest in dealing with fatigue and in evaluating how rapidly a crack propagates under fatigue load.
Crack propagation is one of the areas of Material Scienc(i! in
which large advances have been made recently.
C and m are material constants
I:!.K == variation in stress intensity N/mm 3/2 , MN/m3/2
What is meant by stress intensity?
If we are to describe the stress of a sharp notch using elastic
theory, the maximum stress is:
G max
= anominal
(1 + 2 ~ )
where a is the depth of the notch
r is the radius of the notch
Introduction
So far, our treatment of fatigue has been aimed mainly at determining what happens at the end of a component's life: Failure
or no failure, This is not enough in some cases; it is also necessary to be able to determine how a crack initiates and how
rapidly it propagates,
Different phases during fatigue are of interest in connection
with different structures.
1. Automobile engines: "Cracks must not form." Infinite life is
aimed at. Crack propagation is of little interest here. Design
and production eliminate all factors th-at promote the initiation of cracks,
2, Welded structures, e,g. bridges, offshore drilling rigs, reactor
vessels: "It is unrealistic to assume that welded structures
are free of defects." We must therefore assume that cracks
already exist and the initiation phase is of no interest. Crack
propagation is, however, of great interest. How long is the
safe life of the structure?
3, Airplanes: "Finite life is accepted," Both initiation and propagation of the crack are of interest.
Where our steels are involved, it is usually point 2 that attracts
the greatest interest.
In fatigue cracks, r is extremely small, so that a max ---> 00, The
state of stress must therefore be described in some other way,
for which purpose the concept of stress intensity is excellent.
K==a nom
== stress intenSity MN/m 3/2 , N/mm 3/2
a nom = nominal stress in undisturbed region M NI m2 == NI mm 2
a == crack depth m, mm
f ( ~, 8 , ,. ) = geometry function that describes the size
w
of the crack in relation to other
dimensions, location etc.
Stress intensity can be described in this form for all cracks, It is
only
f(:,8 ... )
that is more or less complicated. For a central crack in a very
large plate as shown in figure 5.4.1, for example
t(
Figure 5.4.1
(Jnom
~LLlLl_ULltlJ~
I
I
I
I
I
I
I
:,8 ... ) 1
==
L\. K is then the variation of the stress intensity when the stress
varies by L\. a
Since we are talking about crack propagation and a crack only
propagates under tensile stress, we must impose a restriction
when it comes to L\. a,
l:J.a = a max -a min. a > 0
I
I
!
~'f (:,8 ... )
I·
2,
·1
i
I
If we take the log of the Paris equation
da - C (!:J. K
dN
r, we have
da
log = m' log L\. K + log c
dN
I
I
I
I
T1T1TfTlTfTrnr
anom
K = a nom . ~ for a through crack
which is the equation of the straight line in a log-log diagram,
where m is the slope, This is also the line of regression obtained
in crack propagation experiments, see figure 5.4,2, What is
interesting here is the coupling to the SoN diagrams of welds. As
was pOinted out previously, the fatigue of welded jOints consists
mostly of crack propagation, If the slope of an S-N curve (n) for
a welded joint is compared with the slope of the line of regression in crack propagation experiments (m), we find that
n = - m. See figure 5.4,3.
5:31
Dynamic strength - Something about crack propagation
5.4.2
We know from section 5, L that below Cl given notch effect in
combination with low stress, a crack cannot propagate. An endurance limit exists here.
This is represented in the crack propagation diagram in the
form of an asymptote. Below about 10~ mm/cycle, the curve is
almost vertical. At very small da/dN, we can determine the
threshold value 6. Kth .
11 K < 11 Kth thus results in no crack growth. 11 Kfu for WELDOX
700 8.9 MN/m3/2 and for HARDOX 400 8.7 MN/m3/2, perpendicular to the direction of rolling, and R = O. The endurance
limit (1ru can be obtained from the definition of 11 K:
log~
dN
mm/cycle
=
=
_
a ru -
6.K th
~o'f (~, e .. )
Note that this equation contains ao - the actual crack length.
Thus, an S-N diagram can, in principle, be plotted on the basis
of crack propagation, see figure 5.4.4.
1O-5+----~
figure 5.4.4
log Or
\
Omax =
Rm
log 11 K
N/mm 312
N =_1_
k or"
n
Figure 5.4.3
logo,
n
o
2
3
4
5
6
7
8
log N
S-N curve derived from theories of fracture mechanics.
log N
log~
dN
m =- n
Near the threshold value, the stress ratio R = 0 min/O max exercises an influence which can be very great in some cases. This
is disregarded in this general treatment.
At large values of 6. K, we will approach a critical value 6. K-->
Kc (fracture toughness) when the crack propagates unstably
and a "final fracture" is obtained.
Handbooks exist with tables of different values of K for different geometries, for example in ref. (29), and a collection is
provided in ref. (30). (Cf. fig. 5.4.10-17).
When it comes to welded joints, the geometry is special and
the crack lengths are small (for most of the life of the joint).
Maddox (Welding Institute) has proposed that the geometry
function f (~ ... ) be multiplied by the factor Mk.
When it comes to very short cracks, f (~ ... ) for most
crack geometries "'" 1 and can thus be replaced with this value.
For welds:
6.K = Mk '6.0' \Ga
log I!J. K
5:32
Mk is dependent on weld geometry and relative crack depth
and is a function of a. In most cases, we can put Mk equal to a
constant.
Dynamic strength - Something about crack propagation
Figure 5.4.5
when ao «
Mk
ai, we can leave out the term containing at
l-~
Approximate values of Mk (a/t '" 0.01, a = crack depth, t = plate thickness)
Butt weld
Nf =
2
._ao~___
C (M k . do . film
~_ 1
for m,* 2
0>0
2
The constants C and m are dependent upon microstructure,
environment and, to some extent, R value. The variation of C
has the greatest influence. According to Gurney, ref. (31), the
approximate values for structural steels and their welds are:
Load-carrying fillet weld
C;: 1.832· 10-13 and m = 3.0
Mk = 2.5-3.5
N-mm system
These values have also been verified for WELDOX and HARDOX
steels.
The initial size of the crack (defect) is obtained from nondestructive testing. When it comes to welds, it can be assumed that
the notch at the weld toe gives a typical average value of ao =
0.15 mm and a typical max. value of ao '" 0.40 mm.
Calculation procedure for estimation of life
1. Determine the size of the initial crack. If this has to be
Non-load-carrying fillet weld'
l-
Mk = 2.0-2.8
estimated, choose one that is sufficiently large to be improbable.
2. Calculate do
3. Select Mk
4. Select C and m if these are not known from similar tests, or
assume as above.
5. Use the Paris fatigue crack propagation equation·
6. Fatigue life factor >3 (usually 4, depending on scope of
inspection) in view of scatter in data.
7. Compare with desired life.
*If we have several levels of do, the calculation can be carried
out in steps.
-
Mk = 2.0-2.8
Example 5.4.1
A connecting rod made of WELDOX 700 for a compacting
machine that compacts refuse has been deeply scored (score
depth 2.2 mm). See figure 5.4.6.
Figure 5.4.6
Approximate values are given in figure 5.4.5.
The Paris equation can now be written:
~ = C (M k ' do' "I/liar
dN
And this can be integrated in accordance with the following
formula from an initial crack ao to critical crack size.
And since C, Mk , do and m are constants, the following is
obtained:
l-~
2
1
l-~
2
N, = -----=,.--. _af'--_ _-_a-"!o___ for m
C (M k ' /la.
fir
1_~
2
'* 2
The customer doesn't dare operate the machine, since he
thinks there is a risk of fatigue failure, which would lead to very
extensive damage.
It takes 6 weeks to get a new connecting rod, and a mountain
of refuse quickly starts to grow.
Question: Can he run the machine without risking fatigue
failure during the 6 weeks while he is waiting for the new part,
during which time he compacts 2000 times/day a.nd a max =
180 N/mm2 and a min = - 30 N/mm2 in the cross section in
question? The plate is 20 mm thick.
Solution:
The initial crack is known to be 2.2 mm deep
do = 180 N/mm2
There is no weld here.
5:33
Dynamic strength - Something about crack propagation
The function f
(~ ) can be tal<en from figure 5.4.17, and
f
(! ) ~
C "" 1.832· 10-13
m= 3.0
1
_ 18-0·5 + 8.5-05
1.832· 10-13 n.5· 120· v'iYo
0.5
1.5
for an elliptical surface crack gives
K = 1.12'0 v;ra
1.832· 10-13
C=
N f = 3.71 . 104 cycles
m= 3.0
NI = ----------------------~13
1.832· 10- 0.12' 180· Vn to
2.2
3
1-2
With a fatigue life factor of 3, we may expect the life to be
1.24, 104 cycles.
3
--1
2
NI =
3.18' 105 cycles
With a fatigue life factor of 3, we can permit"'" 105 cycles 2000
compactions/day in 6 weeks = 2000· 7· 6 = 8.4· 104 cycles
'.' we can use the machine for 6 weeks and then replace the
connecting rod.
Example 5.4.2
Example 5.4.3
A weld on a manhole cover has dimensions according to figure
5.4.9
Figure 5.4.9
20
Load-carrying fillet welds have been made with insufficient
throat thickness, and there is no S-N diagram for this case.
-H-
~
What is the approximate fatigue life?
am., = 120 N/mm2
Omln = 0
A-~
0max= 150N/mm2
0min = 0
Kx { section A-A} = 4.0
What life is obtained with
a) Crack propagation theory
b) S-N diagram
Solution:
The crack initiates from the root and propagates through the
weld metal. Here, 2a o "" 2af, so the value of af must be estimated.
Solution:
What throat thickness is needed to ensureo max = 120 N/mm2
to prevent plastic yielding?
/),,0 =
°max . t = 2· athroat thickness' as' Y2
120·20
- - - == athroat thickness' 500 . Y2
2
athroat thickness = 1.69 mm"" 2 mm
af = 10 + 8· v'2 - 2 . v'2 == 10 + 6· Y2 = 18.5 mm
f ( ~ ) :Central crack in finite in-plane-loaded plate, fig. 5.4.10
a/w =
ao
10 + 8Y2
== ~ "" 0.4
Figure 5.4.8
C = 1.832· 10-13
Nr =
m = 3.0
0.15-05
1
1.832·
10-13 (2.5' 150·
v'ir"t
0.5
NI = 9.6' 104 "" 105
b) S-N diagram figure 5.2.1 (fracture curves) for p = 1
~ N f = 3· 105 cycles
= 4.0
= 150 N/mm2
Agrees fairly well with a)
21.3
~ = 0.7 as an average value during propagation
w
a) Assume initial crack a o == 0.15 mm
150 N/mm2
Mk == 2.5 "well made" weld (little stress concentration)
Example 5.4.4
If, in example 5.4.3, we also require a residual strength that
gives a maximum permissible crack depth = 3 mm, how many
load cycles will the structure withstand then?
Solution:
The S-N diagram, which only gives the number of cycles tu
failure, cannot solve this problem.
ao = o.i 5 mm
co
+
_3-05 + 0.15-05
1
o
1.832·
5:34
af = 3 mm
With the same data as above, we obtain:
10-13 (2.5' 150·
fit
0.5
NI = 3.72' 104 load cycles. Compare with example 5.4.3.
NI = 3· 105
.: Most of the fatigue life is expended at the start of crack
propagation.
Dynamic strength - Something about crack propagation
Figure 5.4.10
Figure 5.4.11
00
i
i
Kr= 0 0' Via fl (W'~)
t
==
~
•
I~
I1 (W·~)
l
ao
-I
2W
Mo
14
2h
K1= 00' vJia· fi6
0.4
(W)
6 Mo
3.5
oo=~
h
W
f l 6 (~)
3.0
1.15
V
0.7
1.10
2.0 1----+--+----l---+--;---cF---;<-+-7"~.." 0.8
1.0
1.05
1. 5 t---+---t--7f---:J'£--:;;./""-:;~7f""-----::?f---7I 00
1.0
OLIIIIIiiii~0~.1~~0~.2~==oE.3:::::==0[.4=--oL.5--0L.6--.-J0. 7
1.00
a
V
/
;/
o
0.1
0.2
0.3
0.4
a 0.5
W
W
Figure 5.4.13
Figure 5.4.12
2a
a
L-
..c
I--
N
14 2W
·1
3.0
00
15
(t· if)
2.5
1.15
W
11
2.0
LlO
W
11
1.05
1.00
16
4
2
1
0
0.5
1.0
1.5
c
11
2.0
1.5
\\
'--2
t--
4
6
8
a
10
r
5:35
5.0
1.4
h/W=y
1.3
1.2
-------
1.1
0.1
0.2
,/""
./
:/
/'
<
~
4.0
//
l.---"" W - 3
0.3
0.4
0.5
/
3.0
2.0
0.6 a 0.7
1.0
o
0.1
W
--
~
0.2
V
0.3
/
0.5
0.4
-M
Surface crack:
6M
(c)
I nternal crack:
( a)
Kr = (10 v'lTii' f 10 C
TW2
(10 =
a
KI= 1.12 (1o·Vi'a·!lQ
a
t = thickness
J
2.5
/
2.0
1.5
--5:36
0.1
0.2
--
:/
-------
0.3
0.4
1.0
/
0.5
- I - -r--
0.5
0.6
a
W
0.7
0.2
0.4
0.6
a
W
Figure 5.4.17
Figure 5.4.16
/
/
----
0.6
0.8
a
c
1.0
0.7
5.5
Design of panels against
impact.
Page No.
Table 5.5.1
Yield strength
N/mm2
guaranteed typical value
Steel
Mild Steel (MS)
S 235 (BS 40, St 37-2)
220
250
High-Strength Steel (HS)
S 355 (BS 50, St 52-3)
350
380
Extra High-Strength Steel
WELDOX 700
700
750
Abrasion-Resistant Steel (AR)
HARDOX400
900
1050
Problem description and consequences .................. ........ 5:37
Mechanics .... ....... ...... .............. .......... .............. ............... 5:37
When elastic theory can be used ..................................... 5:38
Plastic theory must be used for extreme loads ................ , 5:39
Design data and formulas for extreme loads .................... 5:39
Examples ....... .... ........ ...... ............ ........ ........................ ... 5:38
Problem description
When panels are subjected to abrasive wear, impact stresses
also frequently arise, for example in bins and hoppers, truck
bodies, feed tables, pioe bends, underbody protection plates,
loader buckets etc.
As we all know, impact stresses cause undesirable dents in
the panels.
A higher yield strength indirectly means that the steel is harder,
which is extremely important in most cases of abrasion. When
new structures are to be designed or a better material is to be
chosen, it is necessary to proceed by trial and error.
This costs a great deal of time and money.
There is a need for:
- Comparison between different steels under impact
- Design philosophy
- Design data
- Analytical models
Figure 5.5.1
A simple approach for the low-speed « 10 m/sI impact
phenomenon is given below.
Mechanics
As a result:
- The abrasion will be concentrated primarily to the peaks on
the panel, leading to rapid wear.
- The transported material will stick more readily to the panel.
- Friction increases.
- The structure gives an impression of weakness.
The concentration of abrasion to the peaks is disastrous for
stiffened panels (such as dumptruck bodies), since the plate
will rapidly wear out over the welded joint. Sticky material will
collect in the "valleys" ,(the dents) and thereby provide a substrate for additional sticking.
When the material slides along the panel (e.g. during dumping) this will give rise to very high friction (which in practice
often requires a steeper tipping angle).
Even without material adhesion, friction will increase owing to
the wavy surface, and this is often a very great disadvantage, for
example on underbody protection plates (on off-road vehicles).
In general, a structure with pents and heavily deformed panels looks weak in the eyes of the end user! The dents are most
obvious on the side of the plate that is not subjected to impact
and abrasion. This side is usually painted, and the paint is
brittle, so that the impact marks appear as ugly "sunbursts".
After some time, the cracks rust and the "sunbursts" stand out
even more.
In order to be able to tackle the problem correctly, it is important to be familiar with the mechanics of impact.
If, for example, we study the loading of a vehicle, the entire
system can be regarded as a "mass-spring system". See figure
5.5.2.
Often, a certain mass m (kg) is given, for example a stone
falling from a specified height h (m). The energy input to the
system is
Etot = m' g' h (Nm), (g = 9.81 m/s2).
Figure 5.5.2
h
"mx+kx=F"
mXl-klX ...
m2x2+ klXj-...
x
m3 3+ .. .
m4 x4+ .. .
m/e.g. rock
The dents often irritate the end user more than wear!
The designer must therefore try to minimize the plastic deformations to a level that can be accepted by the end user.
We know that plastic deformation can be reduced if we
choose a steel with a high yield strength. It is also known that
the thicker the plate, the less plastic deformation there will be.
Thick plate weighs more and costs more!
The best technical-economic solution is to use a plate with a
high yield strength! A comparison of the yield strengths of different steels is given in table 5.5.1.
kl spring constant of panel
m2 mass of body
k2 spring constant of body mounts
m3 mass of frame
k3 axle spring constant
m4 mass of axles and wheels
k4 tyre spring constant
5:37
Dynamic strength - Design of panels against impact
Such a system can be described by force equations
Solution:
and the result is a system of differential equations that is solved
with the aid of numerical methods and a computer.
It is the contact force (PI) between the stone and the panel,
i.e. the force in spring kl' which is of the greatest interest during
the impact. along with the parameters which influence its magnitude.
Usually m and m2 are of r.oughly the same order of magnitude, in which case almost all of the energy of m will be transmitted to m2 before m2 can start to move, at which point the
maximum force (P1maxl in spring kl has already been reached.
Due to the inertia of m;t. Pmax in spring kl is largely independent of the resilient mounting k2 of m2. which can be proven by
calculations.
While a softer spring k2 spares the frame and underlying
parts in the system, it does not appreciably affect PImax '
Interest should therefore be concentrated on the parameters
m, h, m2 and k} when studying the force Plan the panel.
The influence of the size and fall height of the falling mass is
obvious.
Reducing m2 reduces P1max slightly, but for practical reasons, it is only possible to reduce m2 by about 20%, in which
case the reduction of Pimax will only be marginal.
A weaker spring kl could, however, lead to an appreciable
reduction of P1max' If kl (the body) is made too weak, the risk of
plastic deformation will increase. Thanks to the high yield
strengths of the WElDOX and HARDOX steels, kl can be made
fairly small without a risk of plastic deformation (but not for extreme loads!), e.g. HARDOX 400 in the plating and WElDOX 700 or
WElDOX 900 in the ribs.
Experience shows that the deformation of the plate will be circular as vieyved from above, which means that we can approximate the panel as being circular and rigidly clamped.
(mx + kx = P (xl l,
We further assume the propagation of the load to be
2· ro= 30 mm.
The stress will be greatest directly underneath the load, and
the stressor can be calculated from (3) case 7, page 218:
Or =
~
[(m + 1) log ~+ (m + 1) r
2n m Iro
4a
When designing panels against impact, we assume the most
stringent load case that the structure is guaranteed to withstand.
An elastic approach involves the assumption that the given
energy (E = mgh) is to be absorbed elastically.
This is represented 'by the area underneath the curve in a
load displacement diagram (P -!:J.) in figure 5.5.3.
p
Ps+-------f
p+-----Jf
0:
]
3 . p. (m 2 _ 1) . a2
Deformation !:J. =
4 . n . E· m2 .
e
For the elastic case
2 ro = 30 mm
m
The elastic case
Figure 5.5.3
300
.
ra dIUS a = - 2
a
k
1
v
1
0.3
300
-2- = 150 mm
= 210000 N/mm2
lOmm
E
P
!:J.
4n E m2 ~
3 (m 2 -1) a2
k
p2
4.2961 . 104 N/mm
p2
4288· 1aJ· 4.2961 . 104 • 2
P
6.06' 105 N i.e. 60 tonnes!
Eel' 2 k
With substituted values, 0 = 8670 N/mm2. This is greater than
Os = 700 N/mm2 (for WELD OX 700), which means that the panel
will be plastically deformed.
155 kg falling from 2.82 m on a panel made of 10 mm 700
N/mm2 steel is not a particularly high requirement.
How large a mass will such a panel just be able to withstand if
the mass is released from about 3 m?
p.!:J.
Eel = -2-
i.e.
(v = v'2gh~ v = 7.67 m/s).
k!:J.2_ p2
2k
= -2- -
0rmax=os= 690N/mm2~ Pmax= 3.3346·105 N
p2 = Eel' 2· k
. Eel =
Ps corresponds to 0 = 0 s
p2
2k
(3.3346' lOS?
2· 4:2961' 104
= 1.294. 104 N ~ m
= 43.9 kg!
Example 5.5.1:
A panel of 10 mm WElDOX 700 (os = 700 N/mm2) must withstand a mass of 155 kg falling from a height of 2.82 m.
Figure 5.5.4
u
1)
I·
300
Calculate the maximum stress in the plate.
5:38
·1
With knowledge of realistic plate thicknesses and conditions of
service, it can be said directly that design on the basis of elastic
theory is not feasible. The result would be far too heavy structures.
Such a result may be correct if no plastic deformation whatsoever is the requirement.
The elastic theory does not appear feasible in practical design
work.
Dynamic strength - Design of panels against impact
This proves not to be a limitation, since very large values of !J.. Pm
are usually of no interest.
The formula has been checked by means of fall tests (see
ret. 62).
80 fall tests were performed in Oxelosund (62) on rigidly
clamped 600· 600 mm panels.
The falling body (radius 50 mm) had a mass of 155, 300,
500 and BOO kg and was released from a height of 2.B2 m. The
maximum plastic deformation !J.. Pm was measured.
Three steel grades were included in the study:
Design philosophy
From the above, it would appear difficult in practice to design
panels with the aid of elastic theory to withstand borderline
loads.
It should be possible to permit some plastic deformation,
but it is our job to keep it within the framework of what the
end user (or other specifying party) can accept.
In other words, we must use plastic theory.
S 355 (BS 50, St 52-3)
WELDOX 700 (StE 690, RQT 700)
HARDOX400
The nominal plate thicknesses were 6, B, 10, 12 and 16 mm.
The results are presented in figures 5.5.7-10, which can be
used as design guides.
The plastic case
We can examine the tensile test curve for the steels in question
as shown in figure 5.5.5 which is elastic - perfectly plastic.
Just as in the elastic case with the P-!J.. curve, the energy is
represented by the area underneath the U -f curve.
Figure 5.5.5
Figure 5.5.7
a
Plate thickness in mm
15
Epl
E
We consider once again the problem of the falling mass and the
given input energy Einput that is to be absorbed by the plate. Let
us see how two different steels - WELDOX 700 and HARDOX 400
- the same plate thickness sustain this load. See figure 5.5.6.
\ "\
10
~DOX400~ ~
HARDOX400
ar900 N/mm2
6Pm
Plate thickness in mm
=Einput
1\ \
15
j
Epl.WX 700 E
plastic HX 400 =
40mm
30
20
10
Figure 5.5.8
I
I
j
I~
(>
a
I
I'
5
o
f
'\
m=155kg
h=2.82 m
Figure 5.5.6
WELDOX 700
a
,WElDOX 700
HARDOX400\
E
700 . f plasticwx 700 = 0.77 . f plastic wX700
900
10
The result is approx. 25% less plastic deformation if
HARDOX 400 is used instead of WELDOX 700!
The corresponding figure for S 355 (BS 50) and HARDOX 400 is
about 60%!
Usually, when it comes to calculating extreme loads, Ep.I»
Eel, and the elastic contribution can therefore be neglectea.
.~
I\WElDOX7~
1\ ~
\
\
m=300 kg
h=2.82 m
S 355
(BS 50. St 52·3)
""'""-
~
1\
~
'\
5
6Pm
"(
o
10
20
40mm
30
Figure 5.5.9
Plate thickness in mm
Calculation of required plate thickness
A formula has been derived (62) from reference (4) for calcula-
ting the required plate thickness t (mm) when the following are
known:
Einput = Input energy (mgh)
N mm
Us
= Yield strength of the steel
NI mm2
!J.. Pm = Maximum plastic deformation
mm
-~ ~
15
HARDOX400
~lDOX 700' "
~
"
~
'" "'- '"
I'...
10
"'-
I Einput -~!J..p 2,!J..p < 1.3 (EuinpsuI )113
V If·Us·!J..Pm 36 m m
t= -
The formula is sufficiently accurate when
~
~S355
(BS 50, St 52·3)
m=500 kg
h=2.82 m
5
6Pm
(>
o
10
20
30
40 mm
5:39
Dynamic strength - Design of panels against impact
Solution:
Figure 5.5.10
HAROOX400
WElDOX700
Plate thickness in mm
Figure 5.5.7 gives
:\
15
"\.
HARDOX4oo\
' \WElDOX 700
Of the fall height had been only 2 m, what would the plate
thicknesses have been then?
Solution:
m=800 kg
h=2.82 m
5
(>
LlPm
o
40 mm
10
20
30
As these diagrams show, Ll Pm decreases with increasing
strength, and especially upon changing from WELDOX 700 to
HARDOX 400.
t = 10 mm,
t = 9.8 mm
t:: 10 mm
i.e.
(slightly too thin, but the next thickness
in the stock list is 16 mm)
1\ '\
10
I
t = 13 mm
t = 12 mm
i.e.
34% for m = 155 kg
29%
m = 300 kg
37%
m = 500 kg
18%
m = 800 kg
The decrease is roughly the same (25%) as was obtained by
a comparison of the tensile test curves.
Figure 5.5.11 shows a comparison of the deformation of different steels of 16 mm plate thickness subjected to 500 kg
falling from 2.82 m.
Here, the fall height differs from the one in figure 5.5.7 (2.82 m)
and a correction must be made with the aid of figure 5.5.12.
HAROOX400
WELDOX 700
12 mm"""", - 4.0 mm/m reduced fall height
2_82 - 2_0 = 0.82
i.e_ - 4.0 . 0.82 = - 3.28 mm
10 mm"""", - 3.0 mm/m reduced fall height
- 3.0' 0_82 = - 2.46 mm
But 6. Pm = 10 mm 'is a requirement, which means that if figure
5.5.7 is to be used, then
6. Pm = 10 + 3.28 = 13.28
6. Pm = 10 + 2.46 = 12.46
mm
mm
the plate thicknesses are then
t = 11.2 mm
t = 8.2 mm
i.e. 12 mm
i.e. 8 mm
Figure 5.5.11
Figure 5.5.12
ON:~----------------------+---~
1------"""'""::::--=-=-----t---I-1 HARDOX 400
WElDOX 700
10~------------~~--~~~~~
LlPm carr mm!m(reduced fall height)
-4.0
/
1--------------~--~~~~S355
20~-----------------~~
-3.0
/
30t-------------------+~-~
-2.0
40L-------------------+--~
Llpmm
Parameters:
t= 16 mm
m=500 kg
h=2.82 m
-1.0
/
/
Influence of fall height
This has also been investigated, but on a smaller scale, and it
has been found that at constant energy, the maximum plastic
deformation decreases with decreasing fall height.
This is of no importance when comparing two steels and the
same m and h. However, the fact that the deformation is always
less when the fall height is reduced is important to bear in mind
when sizing the plate_ Figure 5.5.12 can be used to correct for
different plate thicknesses when the fall height deviates from
the one assumed here (2.82 m).
Example 5.5.2
In designing a dumper body for a haulage vehicle, the designer
has two steels to choose from: WELDOX 700 and HARDOX 400.
The body must be able to withstand 155 kg, from a height of
2.82 m, and the customers complain when the dents (~Pm) are
larger than 10 mm.
What plate thicknesses are required for the two steels?
5:40
Plate thickness in mm
o
I
I
6
8
1
I
10
12
Example 5.5.3
A tipper for very heavy duty has to be able to withstand boulders
weighing 800 kg falling from 2.5 m without the maximum plastic
deformation in the bottom exceeding 20 mm. What plate thicknesses are required for grades WELDOX 700 and HARDOX 400?
The body has the following dimensions:
R~re5.5.1~
...----
3000
,. ____ ~
Dynamic strength - Design of panels against impact
Solution:
Solution:
Input energy Einput := 300·9.81· 1500 = 4414· 103 Nmm
WELOOX 700
HARDOX400
Input energy Einput = mgh = 800· 9.81 . 2500
= 19620' 103 Nmm
as typ = 750 N/mm2
700 N/mm2 steel
=WELDOX 700
HARDOX400
as typ = 750 N/mm2
as typ = 1050 N/mm2
as typ = 1200 N/mm2
Check of accuracy of calculation
Check of accuracy of calculation
I
t.Pmperm == (
(Einput ) 1/3
as
I
29.7 mm
26.5 mm
Einput ) l/~
"
as
t.'l.Pmperm = 18 mm
OK, since t.'l. Pm = 20 mm
I
I
I
t= -
5
I
Einput
V :It. as' t::.. Pm
36
t = 15.6 mm
Agrees well with diagram!
16 mm according to
our standard
1920 kg
Plate mass 2400 kg
_ ~ . t.'l. Pm 2
36
t = 7.6 mm
t = 9.7 mm
t= 19 mm
i.e. 20 mm
t.'l.Pmperm = 16.1 mm
I
OK, since t::.. Pm = 15 mm
these thicknesses must still apply after 5000 hours
Thickness allowance due to wear
Difference = 480 kg
(see our price list)
Plate price _ _ /tonne
_ _ /tonne
Plate price _ _ apiece
_ _ apiece
HARDOX400
WELD OX 700
5000
1.5· 2000 = 3.75
Wear is in reverse proportion
to hardness: 250 HB for
WELD OX 700, 360 HB (typical
value 400) for HARDOX 400
t = 13.45 mm
Difference:
1.5. 250. 5000 = 2.6 mm
360 2000
Example 5.5.4
A leading dumptruck manufacturer wishes to introduce a new
body on the market. The body was previously made of 700
N/mm2 steel (e.g. WELDOX 700), but the customers are now
starting to demand more wear-resistant bodies.
The dumptruck manufacturer has carried out measurements of
wear, and in the middle of the body, wear amounts to about 1.5
mm/2000 hours on the 700 NJmm steel.
The dumptrucks are intended for use in road haulage, which
means that their unladen weight is very important.
The dumptruck manufacturer intends to guarantee that the
body will be able to withstand boulders weighing 300 kg falling
from a height of 1.5 m without creating dents larger than 15 mm,
even after 5000 hours. What plate thickness should be chosen
and what is the most economical alternative if saving of 1 kg of
weight is worth GBP 1,00 and if the designer has to choose
between WELDOX 700 and HARDOX 400?
The body has the following dimensions (figure 5.5.14).
t = 14 mm (new rolling)
t = 10.2 mm
t = 10 mm
Plate mass
1075 kg
768 kg
Difference: 307 kg
Plate price (see our price list)
(Prices May '81)
15,90 + 18,20 = 34,10 GBPI 15,90 + 17,40'" 33,30 GBPI
tonne
tonne GBP 25,58
Plate cost GBP 36.66
Difference
GBP n,20/body
Taking into account value of weight reduction
at GBP 1,00 per kg:
Difference GBP 384,20/machine
Figure 5.5.14
counted as bottom during e.g. loading
from side with loader.
5:41
Dynamic strength - Design of panels against impact
Example 5.5.5
A dumptruck owner has sustained a large dent Ll Pm = 30 mm in
his new body made of 10 mm HARD OX 400. The customer
claims that the new bodies are of poorer Quality than the old ones
WELDOX 700 and demands a new body from the dumptruck
supplier.
He has one loader, and this can only lift 2.0 m above the bottom
of the body.
The body is the same as in 5.5.4, Le.
it is supposed to withstand 300 kg from 1.5 m.
There is no doubt that the customer has loaded his new body
very hard.
How hard?
I S 355
WELD OX 700
HARDOX400
typical values
1050
380 N/mm2
750
16mm
14.7 mm
!J.. Pmperm 20 mm
OK, larger than 10 mm max, value according to the
stipulation. In other words, the formula can be used.
Os =
V
t=
5
2
- - ' 6. Pm
36
Einput
:n; • os' I':. Pm
Solution:
I
11112mm
mm
t = 16 mm
i.e. 16 mm
9.3 mm
10mm
according to our standard range
10 mm I':. Pm = 30 mm
Plate weight:
960 kg
Differences:
HARDOX400
After having glanced through 5.5.7-10, we see that 10 mm and
/':, Pm = 30 mm agree with the curve for HARDOX 400, i.e. 800 kg
from 2.82 m.
The boulder that hit the body from max. 2.5 m was of the
following size:
800· 2.82 == 900 kg
2.5
[600 kg
[720 kg
120 kg
240 kg
Example 5.5.7
so the claim is rejected!
Example 5.5.6
A manufacturer of loader buckets is intending to introduce a new
bucket that he calls "Controlling Stones' on the market.
The bucket is particularly suitable for large boulders weighing
about 1 tonne, and is supposed to be able to withstand a collision
with such a boulder at a speed of 10 km/hour without suffering
larger dents in its shell than 10 mm. What plate thickness should
be used for S 355, WELD OX 700 and HARDOX 400?
How great is the difference in the weight (mass) of the bucket
when it has the dimensions shown in figure 5.5.15?
Designing underbody protection plates (to protect the undersides of off-road vehicles) is often difficult owing to the complex
mechanics involved, for example, in collisions with stones. A
forwarder drives up onto a stump with one front wheel while
straddling a stone, slips on the stump and falls onto the stone,
the underbody protection plate absorbing the force of the collision.
Normally, underbody protection plates are made of S 355
(SS 50); sometimes they are made of 700 N/mm2 steel, i,e.
WELD OX 700.
The thicknesses are usually 10 mm for S 355 and 8 mm for
WELDOX 700.
What would the plate thickness be with HARDOX 400?
Solution:
The easiest way is to perform a comparative calculation with a
known case, e.g. 8 mm WELDOX 700, and assume that maximum dents of 15 mm are permitted, considering the same
machine and te same load case.
The known case gives:
-r )
5 !J..p 3
Einput = :n; • as' f (I':. Pm + 36 .
Solution:
10 km/h
Figure 5.5.15
2
Einput =:n;' 750· 8
5
(15 + 36"
15
(4)
(Nmml
3
Einput = 3.366' 106 Nmm
With HARDOX 400, the plate thickness is:
t=
,V
.
Einput
:n;·1050·15
t = 6.06 Le. 6 mm
The velocity of the boulder plus loader after the collision is:
v = ]2 . 10 = 9.3 km/h = 2.58 m/s
16
i.e. the boulder was imparted an energy of
m; = 1000· 2.582
2
2
= 3328 Nm
Or, conversely, the boulder "hit the bucket" with
3328 Nm = 3328· 1()3 Nmm
Check of accuracy of calculation:
!J..
5:42
Pmperm
= (Einput ) 1/3
Os
Check of the formula:
!J..
Pmperm
= (Einput ) 113 = (3.366' 106 )1/3 = 14.7 mm
1050
Os
OK in this case, since it is only a comparative calculation.
If the surface area of the underbody protection plate is
2.8 m2 , the weight difference is 45 kg!
Same example but with S 355.
t = 10 mm and the difference in weight between HARDOX 400
and S 355 is 90 kg!
6 Toughness - brittleness
What is brittle fracture? ...................... 6:1
Conditions for brittle fracture ............... 6: 1
Design philosophies ............................ 6: 1
Different measures of toughness ......... 6:2
Fracture mechanics ............................ 6:2
Comparison between ordinary
steels - WELDOX and HARDOX steels ... 6:4
Toughness requirements ..................... 6:5
Examples ........................................... 6:5
Fracture mechanics data for parent
material and welded joints ................... 6:7
6
6 Toughness - brittleness
Many people cherish the belief that steels of very high strength
must be more brittle than ordinary steels. We now know that this
is not the case; on the contrary, WELDOX and HARD OX steels
exhibit very high toughness in relation to their high strength and
hardness.
This section contains a comparison between ordinary steels on
the one hand and WELDOX and HARDOX steels on the other with
respect to toughness.
As a tool for this comparison, we will use fracture mechanics a relatively new branch of Materials Science, with the aid of which
it is possible to determine whether defects in structures are
critical from the viewpoint of brittle fracture.
Introduction
Nothing (unfortunately) advances engineering science as much
as failures and catastrophes. An example of this is provided by
brittle fracture research, which received new impetus on Monday, March 14, 1938, when the all-welded Hasselt bridge in
Belgium collapsed due to brittle fracture. Over the years, many
famous brittle fractures have occurred with very tragic consequences, such as the Liberty ships that snapped in the middle,
ammonia tanks, steam domes etc.
Nowadays, happily, the steel industry can offer extremely
tough steels down to -196°C, and the fact that brittle failures
occur at all today - fortunately very rarely - is due more to
economic than technical factors: Tough steels are more difficult
to manufacture and therefore more expensive.
It is therefore of great importance for the designer to select a
steel with exactly the toughness level that provides the safety
against brittle fracture that he needs. Toughness costs money!
Brittle fracture is absolutely not unique to WELDOX and
HARDOX steels, but since these steels are harder and possess
higher yield strengths than ordinary steels, it is only appropriate!
that we take a closer look at how tough the WELDOX and
HARDOX steels actually are.
What is brittle fracture?
A brittle fracture is characterized by the fact that the failure is
preceded by negligible plastic deformation immediately adjacent to the fracture surface and that the fracture propagates at
high velocity (750-2000 m/s)*. Crystal cleavages can be seen
under the microscope, which means that the fracture is transcrystalline (cuts through the crystals).
Macroscopically, a chevron pattern (something like a herring
bone pattern) can be seen, and the chevrons point towards the
point of initiation.
*The fracture proceeds unstably, and brittle fracture is therefore a type of unstable fracture propagation.
Three fundamental conditions must be satisfied in order for a
brittle fracture to be initiated (45):
l. Sufficiently high nominal stress in the material.
2. Sufficiently low temperature.
3. Sufficiently high degree of triaxial state of stress.
Beyond these conditions, the strain rate (i) is of great importance in that an increase of the strain rate can be equivalent to
an increase of the nominal stress, a decrease of the temperature or an increase of the degree of triaxiality in the state of stress.
Plate thickness is also of importance for the initia'.':Jn of brittle
fracture, since the degree of triaxiality is greater in a thick material (plane strain) than in a thin one (plane stress).
Some of the circumstances in which these three conditions
may be critical for a welded structure are as follows:
Condition 1. High stresses due to high permissible stress, overload or welding residual stresses of the same order
of magnitude as the yield stress.
Condition 2. Low service temperature.
Condition 3. High degree of triaxial state of stress at notches
and around defects.
It is unrealistic to imagine a welded structure
completely free of defects.
Figure 6.1
a
/
/ /a, = f (Temp)
/
/
/
a, = f (Temp)
Temp
It is easy to satisfy the three conditions given above. This is
illustrated by figure 6.1, which shows two curves: the yield
stress (as) and the initiating stress (a ,) as a function of the
temperature. The welding residual stresses are of the same
order of magnitude as the yield stress. At the same time, the
stress required for initiation decreases with declining temperature. This means that below the critical temperature (T c), the
nominal stresses are greater than those required to initiate the
fracture. aj is also a function of the degree of triaxiality.
It is the 0j curve that is dependent upon the toughness of the
steel (the weld).
The defects do not have to derive solely from fabrication.
They can propagate owing to fatigue or they can be "pure"
fatigue cracks. The cracks can then initiate brittle fracture.
Design philosophies
Two philosophies can be applied in order to avoid brittle fracture.
a. Accept the fact that cracks can form in certain zones, but
that the surrounding (parent) material is tough enough to
stop propagation. This results in the propagation testing of,
chiefly, parent material.
b. Make sure that the material in all zones is tough enough to
prevent the initiation of a brittle fracture from a defect.
Defect size is determined by the fabrication procedure and
by what can be detected by non-destructive testing. This
leads to the initiation testing of all zones.
6:1
Toughness - brittleness
Often, both philosophies are applied in practice, with the
emphasis on b.
In order for the condition in b. to be satisfied in practice, the
parent material must also, in most cases, be relatively tough.
Different measures of toughness
Over the years, a large number of test methods have seen the
light of day, but it would take us too far afield to describe them
here. We shall concentrate on only a few.
The Charpy V notch test
This is by far the most widespread method of testing impact
toughness (also called notch toughness in reference to this test).
The test cannot be counted as a pure initiation test or propagation test; rather, the energy absorbed in initiating and propagating the crack through the specimen is measured.
As a result of the use of this method, the number of brittle
fractures in ships during the 40s and 50s was considerably
reduced. It was found that when the energy absorbed in the
impact test at low service temperatures was under 20J, problems were experienced with brittle fracture in soft steels (such
as S 235). When energy absorption was 27J or more, the
frequency of brittle fracture was very low.
Note that this body of experience is limited to ships (plate
thickness = 15-20 mm, loading conditions, shipyard practice
etc.). To convert these data to other thicknesses, the curve in
figure 6.2 (which is taken from SS 4741) can be used. The
Charpy V test is the simplest and cheapest type of pre-delivery
test, and has proved highly useful for ships in particular. There
is a risk involved in extrapolating these experiences to other
types of steel, plate thicknesses and loadings. The Charpy V test
tells us nothing about how dangerous a defect is. Nor is it
advisable to allow large numbers of catastrophic in-service failures to occur in order to accumulate more experience that could
eventually provide Charpy V values for new structures and
steels. On the other hand, it is quite unrealistic economically to
employ supertough steels.
A more reliable and refined method is needed in order to be
able to determine what defect size should be specified, and
what toughness should be demanded, of the structure in a
given load case and at a given temperature.
Such a method is called fracture mechanics, and is not at all
as complicated as it sounds.
Fracture mechanics
Brittle fracture often start from defects, especially sharp ones.
The state of stress in front of a stress-raising defect (notch)
can be written as follows according to elastic theory:
o max=o nominal (1 + 2
V+)
where a = notch depth
r = notch radius
The radius of e.g. a fatigue crack is less than 0.01 mm, which
means that 0 max > > 0 s' This will mean plasticization of the
crack tip at very low nominal stresses. If r approaches 0,0 max
approaches ! The state in front of the crack tip must therefore
be described in some other way. This is done by means of the
stress intensity factor K, which can be expressed as follows in
front of a through crack in a large plate (se figure 6.3):
(X)
Figure 6.3
o
--I..
Figure 6.2
2a
o
+20~-----------------------r------~
K=o'~'f
f= 1
Note that the unit is stress' length 1l2 = force' length-3!2 e.g.
N/mm3!2 or MN/m3!2.
~
For stress intensity:
-20
~
1 MN/m3!2 = 31.623 N/mm3!2 = 0.9101 ksi viii
.a
IV
iii
Co
E
.! -40
For stress:
...u
1 MN/m2 = 1 N/mm2 = 0.1449 ksi
'1...:
The stress intenSity of all cracks can be described as follows:
VI
C
~
K=o·\f'3ta·f
-60
C and CMn steels in the welded state
27J for steels withoB < 450 N/mm2
40J for steels with oB > 450 N/mm2
-80
9
//
6mm
-1001-1------'-:....-------,----+------1
-60
-40
-20
0
+20
Testing temperature·C
6:2
It is only f which expresses the crack's relative size, position,
interaction with other stress raisers etc, that is a little tricky.
There are handbooks olK values for different geometries, see
ref. (29, 30), and charts of the most common ones are shown in
figures 5.4.10-17.
If K increases, it eventually reaches a critical value Kc at
which unstable crack growth (brittle fracture) occurs. Kc is called fracture toughness. Compare different fracture criteria:
o = OB plain specimen
K = Kc sharply notched specimen
Toughness - brittleness
In other words, Kc is a material characteristic that depends on
temperature, microstructure, location in plate, environment etc.
The equation K = (J •
length a
vn:a. f provides a measure of the crack
Figure 6.5
Toughness
I load
load
~l~
I
displacement
displacement
Critical crack size ac occurs when K = Kc for a given stress (J:
I
I
a- )2. (-f-1 )2
. _ 1 (K c
ac - 1C
displacement
a yield,f max
f is usually"'" 1, and for a through crack (2 ac ) in a large plate,
the following formula applies:
a =~(~)2
c
1C
(J
ductile
brittle behaviour
Here, then, is a method of calculating the size of cracks which
can be permitted when the values of Kc and a and the location
of the crack f are known.
Fracture toughness exhibits a temperature dependence similar to that of Charpy V toughness, and its dependence on thickness is illustrated by figure 6.4. Above a given thickness (te l, Kc
is independent of thickness and is called K1C if the crack is
loaded perpendicular to its plane, for example as shown in
figure 6.3.
Figure 6.4
Temp.
plastic collapse
I n the brittle region
~/ (1 _v 2 )
2'(Js'
E
KI2 (l _v 2 )
E
Jlc = 2 'a s ·oc
v = 0.3 for steel
E = modulus of elasticity
In the ductile region:
Temp = constant
I
Oc ""
I
I
K2
c
2a s ' E
K/
J c "'" -E-
-----------------K~
Plate thickness
Table 6.6 presents some fracture toughness data for parent
material and welded joints in ordinary steels and in WELDOX and
HARDOX steels. Figure 6.6 provides explanations for table 6.1
regarding the location of the specimens.
The scatter of the
values is about ± 10%.
Kc
The consequence of this is that at plate thicknesses less than
tc (and this is usually the case), the value of Kc must be that
appropriate to the plate thickness concerned.
Fracture toughness works best for thick plates and materials
that behave in a relatively brittle fashion, e.g. at low temperature,
but it can also be used for materials with a ductile fracture
behaviour, as long as the appropriate value of Kc is used.
There are other fracture mechanics methods that are more
suitable for ductile fracture modes, for example
Figure 6.6
COD (Crack Opening Displacement), which is a semi-empirical
method and describes the critical valueo c (mm) that the crack
tip is capable of opening before unstable crack growth occurs.
COD can be used for non-linear relationships.
Jet the J integral, which is an energy method and can describe
non-linear relationships. Unstable growth takes place when
J "" J c .
These methods complement each other depending upon the
fracture behaviour of the material, see figure 6.5. COD and J c
are also thickness-dependent! The following approximate relationships apply between Kc, COD (o c) and J c:
WM = Weld Metal
HAZ = Heat-Affected Zone (junction + 1 mm)
SA = Submerged-Arc
MSA= Manual Shielded-Arc, covered stick
6:3
Toughness - brittleness
Comparison of the toughnesses of ordinary steels
- WElDOX and HARDOX steels.
We will compare the Charpy Vvalues and critical crack size ac
when the steels are loaded up to their respective yield strengths
(see figure 6.7, whose values are taken from table 6.6).
The table shows that the WELDOX materials are very tough and
can withstand defects in the parent material as well as the HS
steels (e.g. Domex 390 or S 355), despite the fact that the steels
are loaded up to their respectiv'e yield strengths!
Figure 6.7
a c mm
70
60
Parent material
t = 20 mm along the specimen (l-T)
WELDOX 500 Q
50
a, = l.. (~)2
40
o.,......---D
as
;r
/; ---
,,'"
,
//
,-
"
Domex 390/ "
, /WELDOX 700
20
10
X
./ /
30
WELDOX 700 < 45 mm
,//1
/"
"
/,,-/
,,-
Table 6.4
. .
Toughness in:
Parent material
r;f/
-120
-100
The steels are welded with roughly the same heat input.
Comparisons show that roughly the same defect sizes can be
withstood!
This is very importantinformation when WELDOX and HARDOX
steels are to be welded with the same welding methods, personnel, equipment etc. We know that we will obtain the same types
of welding defects and problems as with ordinary steels, since
WELDOX and HARDOX steels are fundamentally equivalent to
ordinary steels with regard to weldability.
Furthermore, we have access to the same equipment for
nondestructive testing. Since defect size is largely the same
when the steels are loaded to their respective yield strengths and
is of an order that can be detected, it must be evident that
WELDOX and HARDOX steels are no more prone to brittle
fracture than ordinary steels.
The most important information is provided by experience. It
tells us that in more than one million tonnes of WELDOX steel
plate delivered and used in advanced structures, there have been
no brittle fractures to our knowledge.
The toughness of the weld metal can vary within very wide
limits, as is evident from ref. (38) and table 6.6.
At -20°C, for example, Kc can vary between 46 and 180
MN/m3!2, the corresponding ac values for S 355 J2 being 5 and
84 mm, respectively, depending upon filler material, heat input
and number of passes! It seems as if the weakest link at the
present time is the weld metal, which is, of course, where the
most welding defects occur.
-80
-60
-40
-20
±O
'c
Table 6.1
WELDOX 500 P WELDOX 700
-120'
25
- 80'
- 60'
155
Steel manufacturer
Welding shop
Heat treatment
Weld metal
Manufacturer of filler material,
powder flux, gas etc.
Welding shop
Heat treatment
Steel manufacturer to some extent
(melting)
- 40'
- 20'
+ 0'
+ 20'
WELDOX 700
t < 45 mm
20
60
80
128
170
180
60
157
90
170
182
210
122
130
190
Steel manufacturer
Heat treatment
HAZ
Charpy V Longitudinal specimen (L-T) Typical values, Joules
Domex 390'
Influenced by:
This is illustrated in figure 6.8, which shows the impact toughness of different zones in WELDOX 700 welded by means of the
submerged-arc method with two different heat inputs.
Figure 6.8
221
WELDOX 700
217
20mm
Submerged-arc NiCrMo 2.5, 0 4 - OK 10.61
*) Y.P. 390 N/mm2
1Ox 10 Charpy-V
2.5 kJ/mm
Table 62
Yield strength (guaranteed value)
Steel
N/mm2
390
500
700
900
Domex 390
WELDOX 500
WELDOX 700
WELDOX 900
The following comparison can be made as in figure 6.7 for the
heat-affected zone (HAZ). See table 6.3.
Table 6.3
20
HAZ
ac mm
S 355 J2
WELDOX 500
8
19
-60'C
-30
-20
6:4
I
WELDOX 700
20
16
-40
±O
+40
I
I
I
-40
±O
+40 Co
Toughness - brittleness
Toughness requirements
It is almost impossible to provide general rules for the selection
of toughness to avoid brittle fracture. It is, of course, possible to
design with a large margin of safety, but toughness costs money. On the other hand, so does brittle fracture. In the case of
simpler structures, what is mainly required is experience and
judgement. When it comes to more complicated structures (e.g.
offshore drilling platforms) and structures where human life is at
stake, extensive tests are sometimes necessary.
The following are some of the faetors which influence the choice
of toughness:
Design
Temperature
Plate thickness
Steel type
Load, stress, state of stress, strain rate etc.
Welds (heat input, filler metal, electrode care, welding environment etc.)
Heat treatment
Scope of inspection etc.
For structures regulated by standards and specifications, the
toughness level is usually specified. See table 6.5, which is
taken from ref. (73).
Table 6.5
taken from the most recent edition of the Swedish Regulations
for Cranes (73).
Quality class requirements for steel in welded structures.
lowest quality class at respective material thickness, mm
t, mm
Operating
temperature
T,GC
Steels except
EHS steels
5.;;T
-40 ... T <5
EHS steels
t ... 20
20 <t ... 40 40 <t ... 100
B
B
C
B
B
0
5 ... T
0
0
0
-40 ... T <5
E
E
E
A value of 39 Joules applies for EHS steels
There is no guaranteed value of K" for S 355 J2 in table 6.6.
Estimate Kc! S 355, 27J at - 20°C.
Kc 5355 "" 50 + 0.9 . 27 = 75 MN/mm3/2
ac = _1_ (~
n 390
)2 = 0.011 m = 11 mm
Required Kc for WELDOX 700
Kowx 700= ;700, ~
Kcwx 700= 133 MN/mm 3l2
this means that the required Charpy V value is
Cvwx 700=
133 -50
0.9
= 92 J
According to Table 6.1, the typical value for WELD OX 700 at
-20°C = 210 J, and there doesn't appear to be any problem.
The thickness reduction entails a reduced degree of triaxiality,
and according to figure 6.2, the testing temperature for WELDOX
700", O°C when the service temperature is -30°C.
92 J at ± O°C is a less stringent requirement than 92 J at
-20°C.
It would be completely impractical for the steel mills if they
were all to specify individual Charpy V values. Therefore, 27 J has
been established for S 235 JR - S 355 JR and 40 J for EHS.
We are trying to find a testing temperature that has 40 J and
corresponds approximately to 92 J at ± O°C.
For This material and this thickness, it has been calculated that
a reduction of the temperature by 20°C is equivalent to a
reduction of the toughness by 50 J.
.. 92 J - 50 J = 42 J at -20°C.
40 J at -20°C can be accepted WELDOX 700 D. If the structure is such that human lives are at risk in the event of a brittle
fracture, then 40 J at -40°Cshould be chosen (WELDOX 700 E).
.. WELDOX 700 0 and if a greater margin of safety is
required, WELDOX 700 E.
Example 6.1
A manufacturer has used S 355 JO (27 J at ± 0°) in his welded
structures. The company has been buying from a very good steel
mill for 10 years and has in actual fact been getting much
tougher steels than requested, often 27J at - 40°C! The company is therefore building up its experience with tougher steels than
they realize and may be in trouble if they switch steel suppliers,
for example.
Example 6.3
Which defect is the most dangemus?
a) Surface crack of elliptical shape as shown in figure 6.9
Figure 6.9
Example 6.2
A designer has found. reason to switch material in a welded
structure exposed to -30°C from S 355 JO (as = 390 N/mm2) to
WELDOX 700, and he wonders what toughness he should choose
in his structure that will enable him to reduce plate thickness from
19 mm to 390/100·19= 11 mm, i.e. 12 mm. He knows the
relationship between service temperature and testing temperature as illustrated in figure 6.2 and an empirical relationship between Charpy V and Kc.
Kc= 50 + 0.9' Cv
Cv in J, Kc in MN/M3I2
Cv;;o27J
Solution:
Proceed via the same critical crack size
ac=_l (~)
n
Os
Os because the designer exploits the full yield strength of the
steel, and there are welding residual stresses here!
Toughness - brittleness
b) Internal crack of elliptical shape as shown in figure 6.10
Solution:
Figure 6.10
Figure 5.4.17 gives
15
5
As the plates are not stress-relieved, a = as is reached in the
weld.
alc = 0.5-+flO
(+)= 0.82
Compare the stress intensities
a) Ka is obtained from figure 5.4.17
a=5
alc =
2c = 15
~ = 0.67 = -+110 (-ca )= 0.76
ac = 0.377'
7.5
(~y
as
Ka 1.12·a o v;r:5·0.76= 3.37ao
b) 2a = 5
2c = 15
~ = 2.5 = 0.33 = -+flO (~c )= 0.90
c
-2CY'C
7.5
Kb = ao' ~. 0.90 = 2.52ao
Ka/Kb = 1.33
'.' Surface cracks are the most dangerous and visual inspection
should be taken seriously.
If we have bending stress in the plate, the surface defects are
even more dangerous and the internal defects less dangerous.
Example 6.4
What defect sizes in the form of surface cracks of elliptical
shape 2 . c = 4 . a . a, Le alc = 0.5, can a welded joint (welded
by means of the submerged-arc method) resist at -20°C with
WELDOX 700 and S 355, respectively, when the structures are
not stress-relieved? Plate thickness = 20 mm.
6:6
Kc from table 6.6
S 355 <1s = 350 N/mm'
HAZ
Weld
Parent
material
WElDOX 700 ':;5=700 N/mm'
Parent
material
HAZ
metal
Weld
metal
167*
22
113**
10
Kc MN/m 3l2
190
88
46
200
acmm
111
24
7
40
* -30° for WELDOX 700
** No data available. Must be estimated e.g. via Charpy V. The
Charpy V value for 1.7 kJ/mm weld metal at -20"C is taken
from fig. 6.8, which gives 70J. Kc is estimated from
Kc=50+0.9·Cy
Cy ";i!:27J
Kc =50 + 0.90' 70 = 113 J.
The example shows that we can tolerate the same defect size in
WELDOX 700 as in S 355 at -20°C and 20 mm plate thickness
when the welded joint is loaded to the yield strength of the
respective steel!
Toughness ~ brittleness
Table 6.6 Cl
Some fracture mechanics data. Typical values for parent meterial. Explanations of locations, see fig. 6.6
Steel
BS 43 D, S 275 J2
BS 50 D, S 355 J2 G3
Domex 390D'
WELDOX 500 Q
T
mm
Loca·
Temp.
lion
·C
MN/m3/2
kN/m
20
20
20
L-T
T-l
- 20
- 20
- 20
210
141
190
215
54
96
30
112
-120
- 40
- 20
+ 20
+ 20
-120
- 80
- 40
- 20
-120
- 80
- 40
-120
- 80
- 40
- 20
- 40
- 20
± 0
- 40
- 40
- 40
- 40
- 40
60
140
140
140
60
130
130
130
87
20
20
20
20
50
20
40
L-T
L-T
L-T
l-T
l-T
L-T
L-T
L-T
WELDOX 700 > 40 mm
20
L-T
WELDOX 700
20
L-T
HARDOX 400
20-40
50
60
20
20
L-T
L-T
L-T
L-T
T -L
WELDOX 900
Kc
80
180
230
. 195
45
70
200
85
160
230
225
155
200
210
110
75
60
160
147
Je
COD
dcmm
175
Cv
J
60
90
132
Weld·
iog
method
Filler
metal
Number
of
passes
1\
Heat Rei.
input
KJ/mm
/
128
170
190
Parent material
80
130
270
260
25
155
180
0.10
0.26
0.29
100
210
110
• Y.P. 390 N/mm2
/
32
32
32
33
33
33
33
34
35
35
35
36
35
35
35
33
33
33
33
37
\
Table 6.6 b
Some fracture mechanics data. Typical values for welded joints.
Steel
T
mm
Loca·
lion
Temp.
·C
Kc
MN/m3/2
Je
kN/m
COO
dcmm
Cv
J
BS 43 D, S 275 J2
25
L - TI
HAZ
-20
105
55
0.75
20
BS 50 0, S 355 J2 G3
25
L - TI
HAZ
-20
88
39
0.79
30
SS 50 0, S 355
25
L - TI
WM
-40
44
0.05
10
-20
± 0
46
66
0.08
0.07
22
38
-40
90
0.22
18
-20
± 0
III
0.18
0.30
42
200
-40
115
0.13
60
-20
± 0
180
124
0.23
0.23
91
143
0.67
70
SS 50 D, S 355
L - TI
Weld·
ing
method
Filler
metal
OK 12.24
25
L - TI
WM
De·
scending
vertical
Ref.
Heat
input
KJ/mm
2
5
36
2
5
36
2
6.2
38
OK 1061
OK 21.82
MSA
77
MSA
38
38
PhC 6H
5
5
38
PhC 6H
PhC 6H
5
5
5
5
38
38
Ph 27
21
0.8
38
Ph 27
Ph 27
21
21
0.8
0.8
38
38
2
5
36
WM
BS SOD, S 355
of
passes
SA
OK 12.24
On both OK 10.61
sides
SA
OK 12.24
On both OK 10.61
sides
SA
Number
SA
OK 12.24
On both OK 10.61
sides
WELDOX 500
25
L- TI
HAZ
-20
llO
WELDOX 700 > 40 mm
20
T
HAZ
-u
-60
108
0.03
SA
S3NiMoCr
6
2.7
39
-30
167
SA
SA
OK 10.61
OK 10.61
OK 10.61
6
?
2.7
7.0
39
39
20
7.0
2.7
39
39
62
20
T
HAZ
-u
-60
88
0.18
0.02
-30
50
T -U
HAZ
-60
98
109
0.03
0.03
SA
SA
OK 10.61
OK 10.61
143
140
0.04
0.06
39
T -SI
HAZ
-30
-80
2.7
40
SA
OK 10.61
4.5
39
-40
198
0.24
SA
OK 10.61
4.5
39
?
20
6:7
7 High surface pressures
Introduction, where the problems
occur anr how they can be solved
with WELDOX and HARDOX steels ......... 7:1
Design principles ................................. 7:1
Guidelines and examples ..................... 7:2
Lugs and how to design them
against: play, fatigue failure
and crack propagation ........................ 7:2
7
7 High surface pressures
Introduction
Figure 7.1
High surface pressures are frequently encountered, are often
critical design criteria, and can cause problems in some cases.
Steel against steel
P N/mm (width)
Roll against flat surface
The problems can take the form of:
Surface fatigue
Pitting
Mangling
Surface deformation
Cylinder against cylinder
Surface fatigue
occurs in e.g. roller races in rolling-contact bearings, trunnions
on cement kilns and drum barkers. The failure often starts at
the location of a maximum shear stress (a little below the surface) and near a defect (often slag). The result is spalling with
subsequent pitting.
oH=0.591· -JP.E'
Pitting
occurs in gearing in the presence of lubricants and high surface
pressures, according to Niemann when the Hertz surface pressureoH =0.27' HB. (HB = Brinell hardness.)
oH=0.591· - J P ' E '
Mangling
occurs when the yield stress of the steel has been exceeded
and the steel is "kneaded out", e.g. rollers, roller races.
Surface deformation
involves deformation of the surface on both a microscopic scale
i.e. the peaks of the surface profile are flattened (smoothing) and a macroscopic scale, i.e. the entire hole, for example, is
deformed plastically.
Some rules of thumb for avoiding spalling or
pitting
A factor kH that has been found empirically to work for permissible Hertz surface pressureoH in N/mm 2 is often given
0Hperm"" kH ' HB . 10
where HB is the Brinell hardness of the softest rolling body.
- Stationary and slow movement
(static 0.5 - 0.6)
EHS and AR steels have been used very successfully to
combat all these problems.
EHS and AR steels have
high yield strength - counteracts mangling and surface deformation
High hardness - counteracts surface damage such as pitting
high hardness through their entire thickness - counteracts
spalling
High purity - counteracts fatigue failures in the surface zone
Telescopic jibs with rollers kH = 0.10 - 0.15
Pure fatigue loading must be avoided when the pOint of contact is completely fixed over a long period of time.
- Railway rails
(OH"" 1100 N/mm2)
kH =0.5
(V max -'=30 m/s)
- Bridge bearings
kH = O. 5 static
- Cranes (wheel contact) kH = 0.2 - 0.3
- The lower the travelling speed, the higher the value.
- Gears, precision-made and well lubricated
10 < V <20 m/s ref. (41)
Some design principles
Rolling contact
occurs between rollers and roller races/ways (e.g. cement kilns
and drum barkers, railway wheels-rails, bridge bearings, rolling
contact bearings).
If adequate lubrication is not provided between the surfaces,
the Hertz surface pressure is a relatively good parameter for
design.
Some formulas for the Hertz surface pressureoH (N/mm 2 )
are given in figure 7.1.
- Roller races for steel furnace, unlubricated in very dusty environment (Kaldo in Oxeliisund) kH = 0.14 HB = 320 V =
11 m/so
- Roller races for cement
kilns and drum barkers
- For rolling bearingsoH =
v = 1.5 - 3 m/s
0.5 HB W = total number of
6VW
revolutions, accroding to ref. (41).
7:1
High surface pressures
Surface deformation
The hardness of the bolts must not be lower than that of the
plate.
is brought about by plastic flow at the peaks of the surface
profile and, if the surface pressure is very high, a deformation of
underlying material.
This phenomenon can be related to the yield strength of the
material, and a hardness measurement is, after all, nothing but
a quantification of the resistance of the surface to plastic deformation.
Problems of this kind can arise in connection with:
8.8
10.9
12.9
14.9
225
280
330
390
of e.g. roller races occurs frequently when the Hertz surface
pressure exceeds the yield stress of the steel (surface).
This occurred in S 355 (HB '" 150 cr styp = 380 N/mm2) at crH
= 410 N/mm2.
and in this case, kH was'" 0.3.
EHS and AR steels, of course, possess high yield strength
and great hardness (see table 7.1).
In other words, the designer must also check that the Hertz
surface pressure does not exceed the yield strength of the
steel.
WELDOX and HARDOX steels can,beused with success here.
In this case, a warning concerning the soft zone created by the
gas cutting of WELDOX and HARDOX steel is in order ..
Since gas cutting involves heating above the tempenng temperature (overtempering), a heat-affected soft zone will ~e
created in the plate. Figure 7.2 shows how hardness vanes at
different distances into the plate from the gas-cut edge. A hard
zone is created immediately adjacent to the edge. This is caused
by enrichment with carbon and alloying elements, which 10c~11y
increase the hardenability of the steel and harden upon coohng,
thereby giving rise to the hardness.
.
2-5 mm into the plate, we encounter the soft zone. Since e.g.
WELDOX and HARDOX steels have the same basic analysis as
S 355 (BS 50), they will have the same hardness in this region
after overtempering as S 355 (BS 50). In order to be able to
exploit the high hardness of the WELDOX and HARDOX steels in
edges that have been gas-cut, we must machine these edges to
a depth of 3 - 6 mm, depending upon the degree of heating
involved in the gas cutting process. Such an edge has to be
machined anyway in order to provide good roller contact, for
example.
Some guidelines and examples
Bearing pressure
can sometimes be completely decisive in determining which
plate thickness should be used. To permit full exploitation of the
strength of e.g. bolts, the plate and the bolt should have roughly
the same hardness, which is not the case with 8.8 bolts and
ordinary steels, for example, where the bolt is much harder than
the steel.
The Swedish Regulations for Bolted Connections StBK-N3
(42) give the following permissible bearing stresses for ordinary
and exceptional combinations of loads for two different bolted
connection classes SI and S2, see table 7.2.
SI is intended for structures under static loads.
8.8 bolts can be used for both SI and S2.
Table 7.2
Permissible bearing pressure p according to StBK-N3 in N/mm 2 .
Class of
bolted
connection
ord.
exc.
ord.
exc.
ord.
exc.
SI, S2
260
300
380
440
390
450
V.P. 390 NI mm'
S 355
HBmin according to SMS 2265
Mangling
Excessively high bearing pressure in e.g. bolted joints.
Mangling of roller races.
Fatigue load or extreme loads in lugs of various types give rise
to play, which can be very troublesome.
S 235
Bolt
Lugs
Recommended bearing pressure for EHS and AR steel in N/mm 2 •
Class of
bolted
connection
WElDOX 500
WElDOX 700
HARDOX 400
ord.
exc.
ord.
exc.
ord.
exc.
SI, S2
490
570
690
800
900
1040
Lugs for hydraulic cylinders, articulation jOints etc. are structural
elements that are subjected to very heavy loads and to which
special attention must be devoted.
.. . ..
Problems associated with lugs can be very serious and Irntating.
Play
leads to poor precision, high impact stresses, noise, insecurity
on the pert of the operator, gives a poor impression of quality
etc.
Table 7.1
:
(J sguaranteed
(J styp
(J Bguaranteed
HBtyp
HBguaranteed
N/mm2
Steel t - 6 60 mm
350
min 510
380
150
-
WELDOX 500
500
590- 750
540
210
-
WELDOX 700
700
780-930
750
260
-
WELDOX 900
900
940-1100
950
310
-
WELDOX 960
960
980-1150
1000
320
-
HARDOX 400
900
-
1000
400
360-440
-
1300
480
min 450
S 355 (BS 50)
HARDOX 500
Choosing the right steel is a simple way of solving many problemsl
7:2
High surface pressures
Figure 702
Hardness after gas cutting WELDOX 700
Hardness after gas cutting S 355 (SS 50. St 52-3)
HVlO
450
HVlO
450
,
40 0
35 0
''."
\'\
30 0
400
...
- - - Just below the plate surface
- - - Middle of the plate thickness
I
~ WELD OX 760 > 40 mm
~
\\
"\ \
\
35 0
\
\
\
\ \
30 0
I
\
I \
\ \
25 0
20 0
\
/
..::;.=
25 0
i/'-I'
\
V\
\ I'-- "
\
\
\.
I
)
\
:..::
20 0
~
NWELDOX 700 < -40 mm
150
T
- - - Just below the plate surface
- - - Middle of the plate thickness
""
~
t' __
=
15 0
2
4
6
Distance from the cut edge, mm
T
2
4
6
Distance from the cut edge, mm
Locations for hardness measurements
Hardness measured in the middle of the
Fatigue
causes the structure to fail in service, which can somtimes be
catastrophic.
Figure 7.3.
Failure under extreme load
is at least as serious as fatigue failure.
The major industries, and especially the aviation industry,
have studied lugs thoroughly, and this has resulted in design
standards.
Small companies may have difficulty finding suitable rules.
Here are some:
Prevention of play
Tlie holes in lugs that are subjected to fatigue load have a
tendency to become loose, i.e. develop play. This is due either
to deformation of the whole lug under extreme load or to plastic
deformation of the peaks of the surface profile (and possibly the
underlying material).
Abrasive corrosion is very common in these contexts and
greatly accelerates the process.
Lugs are often fabricated from heavy plate by means of gas
cutting, with a subsequent drilling operation to complete the
hole.
Reaming is less common.
The roughness of the surface after drilling can be very large
"" 10 - 100,u m.
Machine reaming gives about 6-16,u m.
A higher tolerance grade than ITl2 (i.e. 250,u m on 040 mm)
is seldom achieved by twist drilling.
This means that tlie "surface" is very readily deformed on
steels with low yield stresses.
A link bearing is often built into the lug, see figure 7.3. SKF
specifies M7 in the lug for link bearings in hydraulic cylinders,
which requires reaming. This is not done for cost reasons,
however; instead, the link bearing is simply pressed into a
drilled hole and often develops play after a while.
Increasing the interference of the fit makes the bearing
very difficult to fit.
7:3
High surface pressures
The use of WELDOX 700 steels in lugs to reduce play has been
highly successful!
In most cases, the problem has been completely eliminated!
The lugs are usuallYimade of WELDOX 700, gas-cut with Cl
radial machining allowance of about 5 mm (due to the
overtempered zone) and then simply drilled.
Since WELDOX 700 has a much higher yield stress (twice as
high) than S 355 which is normally used, the risk of plastic
deformation of the whole lug is also drastically reduced.
This is evident from figure 7.4, which shows the plastic deformation of the hole as a function of the stress in the hole crosssection crnet' from ref. (47), (48), (49), (50) and (51).
Note that the holes used in the tests for figure 7.4 complied
with a tolerance of H7 (Le. reamed), and if the holes are only
drilled with e.g. H12, the deformation will be about 10011 m
greater.
Most fatigue failures occur at section C-C, where it is common to have changes of section with high stress concentrations
(Kt) and, not too uncommonly, poor welds! The design of this
section must therefore be given special attention. It can be
designed in accordance with the principles of chapter 5.
At sections A and B, the fatigue crack often starts from the
edge of the hole. For this reason, a chamfering of the hole to
about 1.5· 45 is very beneficial.
This is also verified in figure 7.6, which shows the dimensions
and principal stress distributions of a lug.
The figure shows that stress concentrations are caused by the
hole and the pin.
The maximum stress a max is of interest and can be calculated
according to (46):
0
a max = a net • Kt
°net =
Figure 7.4
N/mm2
Plug
t· (w-d)
as per fig. 7.6
A = d/w
d/w"" 0.4-0.6
I'l did", 0.3-0.4 "10 (diametral play)
1000~------~~~----------r----------.~
w
WELDOX 700
500·4f------~~~----------r_--------~~
-P/2
S 355 (BS 50, St 52·3)
There are also calculation methods that take into account uneven
pressure distribution in the hole (along the pin) according to
(46), but further studies are required in order to verify these
methods.
I'J d H7
p
Figure 7.7
O+-----------+---------~r_--------~~
o
50
100
150
I'l pl.,um
G rmax
a rmax = anot . Kt
). = dfw
N/mm2
1000
Kt =). + 1/).
R = 0, p = 1
700
Fatigue
in lugs is, naturally, of frequent occurrence and the consequences are often serious, since it leads to a failure of function of
e.g. a hydraulic cylinder.
Figure 7.5 shows different pOints of initiation for the fatigue
failure of a longitudinally loaded lug (more interesting and often
more critical than a transversely loaded lug).
500
t--..
400
-
WELDOX 700
'""i.",
I III
S 355(BS 50)
-
300
200
Figure 7.5
100
104
- r - - - - - - -+--- P
10 5
106
108
10 7
Figure 7.7 can be used in design against fatigue.
The graph in figure 7.7 has been plotted from ref. (44), (46),
(47), (48), (49), (50), (51), (52) and on the basis of experience.
Note that the graph has been transformed too rma in the hole
according to the above formula and applies for p =
i.e. a full
load spectrum.
t
Observations in connection with lugs subjected to longitudinal
tension and compression.
7:4
High surface pressures
Lugs subjected to tensile and compressive loads (see figure 7.5)
are usually said to be suhjected to alternating load since the
nominal stress a nom changes sign, see figure 7.8. The stress
distribution when + a nom acts towards the right and gives
a max 1 is shown at the top of the figure.
Figure 7.8
3. Calculate a r ma. = a r net' Kt
4. Compare with figure 7.7 and put
a maXdiagram
2
a maxperm
This gives a probability of failure"" 10-3
if we have spectrum (variable amplitude) loading, use the Palmgren-Miner cumulative damage rule and put
.r ~ = 1 as a criterion
Ni
Example 7.1
w
Check a lug as per figure 7.9 for fatigue. The lug is to sustain
2 . 106 load cycles. It is incorporated in an ejector mechanism for
flour sacks, load spectrum p = I, and according to the designer,
the bearing pressure is 60 N/mm2.
Unoml
Solution: Checking the bearing pressure
Pi
°nom::::;:
w:t
a
p max
= _P_ = 4.8' 104 = 60 N/mm2
d ·t
40 . 20
Critical paint A
4.B· 104
20 (BO -40)
0rnet = -:--:-c-~:-:- = 60 N/mm2
o nominal
°max
The compressive force does not contribute to a r net at point A
Kt = 40lBO + _~l:-::- = 2.5
40/80
a r max = 2.5' 60 = 150 N/mm
2
0rmaxdiagram = 150· 2 = 300 N/mm2
according to figure 7.7, we can readily permit 2· 106 load
cycles.
Will the weld hold?
Assume QB = 10-3
Weld class Sv2. Kx = 5.0 case 28
Nd = 2· 106
According to table 5.2.5a, this givesa perm = 44 N/mm2
When a nom changes direction, most of the stresses will not
have to go past the hole, but will rather be taken up by the pin
directly. The stresses that go past the hole will return and, upon
passing the edge of the hole, will give rise to tensile stresses
0max2! See the lower figure.
This means that the edge of the hole on longitudinally loaded
lugs will not be subjected to compressive stress (in the tangential direction)!
(4.8 + 0.3)' 104
a rnorm {weld } =
BD. 20
= 31 N/mm2
OK'
Example 7.2
How much can the radius R of the lug in example 7.1 be reduced
if WELDOX 700 is used instead of S 355, and Nd = 10'?
'.' The stress ratio ~ O.
If the lug is connected by a welded jOint, the joint will naturally
be subjected to alternating stress if a nom changes sign.
Other rules apply to transversely loaded lugs.
Solution:
Rwx 700 = 40·
Calculation procedure for checking
longitudinally loaded lugs:
W
Plug
I
1. Caculatearnet=arnom'--=
-----'=-w- d
t· (w - d)
Note: onlya~O is included!
2. Calculate Kt = l + III
wherel = d/w
a rmaxdiagram
S 355
OrmaXdiagrarnWELDOX 700
300
= 40 - - - - 27 mm
450
i.e. 40 - 27 = 13 mm
Example 7.3
The lug in figure 7.9 is also used in a street sweeper, which
has the following estimated load spectrum during its expected
life. Fatigue failure occurs. What should be done?
7:5
High surtace pressures
(J min
N/mm2
I
All three points must be checked according to re!. (53).
°max
ni
180
144
128
110
0.4 . 105
10 · 105
20 · 105
25 · 105
N/mm2
0
0
0
-110
1. Pperm = 0.9' (w-d), i'a perm
Failure load a perm = as
"Yield load" a perm = a s' Note that this load gives plastic
deformations as per figure 7.4
2. Pperm = k2 ' C . a perm
k2 as per figure 7.10
Figure 7.9
3. Pperm = d· t· Pperm
Pperm as per section "Bearing pressure".
Figure 7.10
o
o
4.8 tonnes
0.3 tonnes
1.5-t------,:------'----,:-----7'--1
R = 40
t = 20
1.4+-----t----->~-t-_¥-----i
Solution:
Use the Palmgren-Miner cumulative damage rule and only include a rnom whose
l.3~+-----+-----1-~-__\_---1
a min >-: O.
Kt = 2.5
a rmax = Kt 'a rnom
1. 2 -t--.---r--.,----.--i-,.-.-......-.--t--.--.--.-,--t--o
0.5
1.0
1.5
cid
ni
a rmax2
N/mm
arnom2
N/mm
0.4· 105
10 · 105
20 · 105
25 · 105
450
360
320
275
180
144
128
llO
Ni (as per graph
in fig 7.7 for
1:~
S 355
Ni
1.7. 105
16 . 105
45 .105
200 . 105 *
0.23
0.625
0.444
0.125
--
Cracks in lugs
Fracture mechanics can also be applied to lugs, and ref. (54)
gives the following expressions for the stress intensity factor for
a through crack.
K = a nom' V If a' [5.38 - 12.3
1.43
++ (+ y19.1
* extrapolate the curve for safety's sake.
for wld = 2.40
' .. 1: ~i > 1 and we quite definitely get fatigue failure
I
The function
(+) in tabular form
Change the material to WElDOX 700
aIr
a rmax
450
360
320
275
ni
0.4· 105
10 · 105
20 · 105
25 · 105
Ni
WELDOX 700
100· 105
1000· 105
2800· 105
-
ni
Ni
0.004
0.010
0.007
--0.021
This gives a much better result. since 1: ~i. < 1.
I
o
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
(5.38 - .... )
5.380
4.327
3.574
3.052
2.698
2.464
2.312
2.214
2.152
2.123
2.103
For corner cracks, the following approximation can be used:
Extreme loads
Three points have to be checked when it comes to the extreme
K = a nom' fu· f
load that a lug can bear.
(+ )
1. Width - diameter ratio d/w
2. Edge distance C according to figure 7.6
3. Bearing pressure
7:6
(+)
a = 2 mm
a = 3 mm
2.58
2.64
8
8 Wear
Wear, general ..................................... 8: 1
Choice of material ............................... 8:2
Design principles rules of thumb .................................... 8:2
Some hints concerning
constructional details .......................... 8:3
8 Wear
Wear, general
Figure 8.1
Knowledge of the physics of wear scarcely extends beyond a
narrow circle of specialists.
In practice, many designers have proceeded by trial and
error, with more or less success, in their attemrts to reduce
wear, without really knowing what they have been dOing in
terms of the physics of wear.
Even though we are still in the early stages of research into
the phenomenon of wear, it may be of interest to examine
briefly what has been found so far and to show how many wear
problems can already be tackled today through a suitable
choice of material.
Some basic concepts
When solid bodies come into mechanical surface contact,
phenomena occur that can be grouped under the heading
tribology - the science of friction, lubrication and wear.
The phenomenon of friction has been known since ancient
times, and we are all familiar with the law of friction F = fI . N
from mechanics.
What is less well known is that the actual area of contact
between two bodies is several orders of magnitude smaller than
the nominal contact area, and when movement occurs between
the bodies, projecting micro-irregularities (asperities) collide
with each other and give rise to considerable plastic deformations and very high temperatures in the surface layer.
The object of lubrication is to separate the surfaces and to
prevent contact between the asperities. In practice, this is impossible owing to the fact that the asperities break up the lubricant film and owing to the imperfect stability of the film itself.
Wear can be defined as the loss of material from the surface
layer of a body under mechanical·contact.
Mechanisms of wear
The fundamental phenomenon for all wear is fracture due to
a. shearing-off of adhesive contact bridges
b. chip cutting action
c. impact
d. fatigue
These fundamental mechanisms of wear seldom occur in isolation, but interact with each other. In addition, other phenomena
such as heating, plastic deformation or corrosion interact with
the fracture mechanisms.
In addition, a practical tribological system is controlled by
system parameters such as forces, movements, design etc.
In other words, practical cases of wear are extremely complex, so it is perhaps not so strange that the literature contains
references to an abundance of "wear mechanisms" with imaginative names, many of which try to describe the entire system
rather than verifiable wear mechanisms.
A good rule of thumb in analysing wear problems is to consider the actual wear process in terms of one of the types of
fracture listed above.
Abrasive wear
Abrasive wear is the type of wear that is of most interest to us.
If a hard particle is dragged across a softer surface, it can act
as a cutting tool- a so-called 'abrasive element'.
Material removal takes place by the formation of chips, leaving
a scratch in the surface, figure 8.1. However, most of the fric-
tional energy is always converted to heat, and the temperatures
in the surface layer become so high that phase transformations
take place. Figure 8.2 illustrates schematically what happens
during abrasion.
Corrosion increases abrasive wear, since the corrosion products are often brittle and therefore sensitive to abrasion.
Corrosion must be included in the analyses, and if the corrosive medium is very aggressive or the period of service is very
long, corrosion and abrasion can play a decisive role.
Figure 8.2
Direction of
abrasion
Severe deformatIOn and
phase transformatIOn
"Friction martensite"m9!J!IJ;:mH!l!l
700 Hv max
Deformation and
temperature effects
Strain-hardened
600 Hv
Deformation
Strain-hardened to
a lesser degree '
400 Hv
200,,01
As-supptied
Depth
Erosion
Erosion occurs when a surface is bombarded by a stream of
particles. Two types of fracture dominate: chip cutting and impact. Cutting predominates at small angles of incidence, impact
at large angles. In the extreme case of glanCing incidence, erosion grades over into abrasion.
8:1
Wear
Material aspects of the wear of wear parts
Figure 8.4
The predominant wear mechanism in e.g. construction, forestry
and transport equipment is the abrasive-erosive and, to some
extent, corrosive wear of wear parts,
The most common material that is exposed to the wear in
such applications is AR steel.
AR (Abrasion Resistant) steels are steel alloys based on Mn,
Cr, Mo, W, B and heat-treated to great hardness.
Modern AR steels have found many successful areas of
application because they:
- possess their great hardness from the start
- exhibit considerable toughness, despite their great hardness
- are readily weldable
- are bendable
- resist abrasion-erosion very well
- can take hard impacts without cracking
- function simultaneously as structural steels.
Tonnes loaded
50000
40000
30000
20000
10000
None of the competing materials - such as 13% Mn steel,
Nihard, cemented carbide and wear rubber - can boast such a
wide range of usage as AR steels.
On the other hand, these materials can perform better than
AR steels in special cases of wear.
100
200
300
400
500
Brinell hardness
Choice of material
Faced with this complex picture of wear, it might be useful to
review some si!l1ple guidelines for material selection.
For many applications, it has been found that hardness is a
very good material parameter for determining the wear resistance of the material.
Figure 8.3 provides a good overview and guidance.
The fact that hardness is a good measure of wear resistance
is evident from figure 8.4, which is taken from a study conducted by the Swedish Mine-Owners' Association on loader buckets
for loading iron ore. The curve shows almost exact proportionality between life and hardness.
Design principles
We all understand now that wear is a complex problem, making
it difficult to give any general rules for deSign.
The following guidelines must therefore be applied with
discretion and some caution.
In doubtful cases, consult the steel manufacturer.
- Try to ascertain which wear mechanism dominates
- If corrosion is a serious problem, allow for a reduction of plate
thickness (one-sided material loss, uniform corrosion)
0.5 mm in 10 years
0.75 mm per year
Figure 8.3 The importance of different factors in the abrasive
wear of steel.
~
STRUCTURE·
DEPENDENT
I
--
small
soft
IMPULSE
PIECE SIZE
MATERIAL
HARDNESS
~
large.
-
hard
I
WEAR NOT
CORRELATED
WITH
HARDNESS
Table 8.1
NOT
STRUCTURE·
DEPENDENT
WEAR
CORRELATED
WITH
HARDNESS
Industrial air, coast, urban area
Structure operating alternately
in water and air
- Learn from experience of previous installations (actual plate
thickness at start, n;Jmber of tonnes loaded, hours, trips,
hardness of material, reduction of plate thickness).
- If experience is available from previous installations, extrapo·
late to the hardness of the new material if abrasive-erosive
wear is involved.
- If experience is not available, use some of the following
typical values and extrapolate to the hardness of the new
mater a1.
1. Truck body (tippers, dumptrJcks, hcfulage vehicles):
WELDOX 600 (240 HB)
3 mm/lOOO hours
HARDOX 400 (400 HB)
0.65 mm/lOOO hours
The hardness of different steels
S 355 (SS 50)
WELDOX 500
WELDOX 700
WELDOX900
WELDOX960
HARDOX400
HARDOX500
OSguaranteed
N/mm'
OSguaranteed
N/mm'
350
500
700
900
960
900
min 510
570-720
780-930
940-1100
980-1150
aStyp
HBtyp
HBguaranteed
N/mmz
380
540
750
950
1000
1050
1300
150
210
260
310
320
400
480
360-440
450-560
Table 8.1 shows that the AR steels have very high yield stresses and are thus better equipped to withstand shocks and impacts (see
section entitled "Design of panels against impact).
8:2
Wear
2. Loader buckets (shelll...,truck body multiplied by 5.
Figure 8.6
3. Bucket cutting edges, 480 HB, working in iron ore, 50 mm,
Wear plate + (e.g.) Philips C6
300 - 600 hours life, depending on design of cutting edge.
4. Bark pipelines
S 235, BS 40 A, St 37-2 (140 HB)
WELDOX 600 (240 HB)
HARDOX 400 (400 HB)
0.16 mm/month
0.10 mm/month
0.05 mm/month
5. Rotating drum barkers
shell
lifters
HARDOX 400 0.75 mm/year
HARDOX 400 2.25 mm/year
~\~
6. Loading bin for iron ore, opening 1000 x 500 (mm 2 )
HARDOX 400 (400 HB) in bottom (450 slope) 20.5 mm worn
off after 107 000 tonnes (piece size 0-50 mm).
- The wear must not be allowed to reduce the plate thickness
beyond the minimum required for strength reasons (stress,
stiffness, impact stress, fatigue etc. ).
- Carry out an economic comparison of different alternatives,
and don't forget repairs and downtime costs for the replacement of wear parts.
Note: It is unquestionably economical to switch from e.g. S 355
(BS 500) to HARDOX 400 in cases involving abrasive wear.
This is clearly shown by our experience.
Figure 8.7
Some hints concerning constructional details
Shape
Shape is very important when it comes to sliding materials. A
change of direction of the sliding material causes local wear.
Softly rounded corners are therefore important in cyclones
and pipes.
Pipe joints must not cause misalignment of the pipes.
Flatness is therefore an important requirement for wear
plates!
Increase the thickness where wear is greatest, for example if
the wear looks like that shown in figure 8.5. Service life can
then be extended considerably by, for example, choosing a
more appropriate shape for a new part, for example as shown in
figure 8.5.
Gas-cut edges subjected to abrasive wear suffer very little
wear at first owing to the very hard surface layer. After this layer
has been worn through down to the overtempered soft zone
(3-5 mm wide), wear proceeds more rapidly, eventually slowing
down to the wear rate that is specific for the steel.
Figure 8.5
,
\
I
\
e.g. OK 48.30
HARDOX 400
Stud welding can often be a good alternative for fastening
wear plates that have to be replaced quickly.
Naturally, bolted joints are also acceptable.
Drilling recommendations are provided in our metal-working
booklets.
Fan and pump impellers, as well as their housings, intended
for the transport of dust or slurry can be made of AR steel for
best results.
When fabricating the irnpellers, it is important to locate the
welds (jnci. heat-affected zones) in such a way that they do not
wear out faster than the unaffected plate.
One way of tackling this problem is shown in figure 8.8.
I
,
\
Figure 8.8
I
,
Proposed shape of
\~new part
\
\
\
\
\
\
-~--~
A
right!
B
wrong!
Worn part
The importance of this phenomenon must be judged from
case to case. The antidote is milling of the gas-cut edge to
remove the soft lone, see figure 7.2.
Fastening of wear parts
The easiest way to fasten wear parts is by welding, since EHS
and AR steels offer very good weldability.
Filler material is to be chosen on the basis of the real need for
hardness. It is usually possible to use filler metal intended for
much softer steels (undermatchingl.
Therefore, make sure that the welds are protected against
wear, for example as shown in figure 8.6. If the welds are
exposed to wear, sealing runs should be made with hardfacing
metal, see figure 8.7.
Soft zone -
Rapid wear
9 References
9
9 List of references
1. Jarfall, Dimensionering mot utmattning dell och 2 Mekanresultat 77004.
2. Schnittiger, TillfOrlitlighetsanalys fOr Mekanister Inst. for
Maskinelement KTH 1972.
3. Roark, Formulas for Stress and Strain 4th edition, McGrawHill.
4. Johnson, Impact Strength of Materials, Arnold.
5. Schijve J, Proceedings Rimforsa, Sweden Aug 1977.
6. Gurney T R, Fatigue of welded structures Cambridge Univ.
Press 1968.
7. Olivier R. und Ritter W. Wohlerlinienkatalog fUr Schweissverbindungen au~ Baustahlen. Teil 1 und 2, DVS Bereichte 56/1 und 56/2.'
8. Swedish Regulations for Welded Steel Structures 74 StBKN2. National Swedish Committee on Regulations for Steel
Structures, 1974. (Distr. by Swedish Institute of Steel Construction. )
9. Richards K G, Fatigue Strength of welded structures The
Welding Institute May 1969.
10. Petersson, Stress concentration design factors.
11. Bergqvist L, Sperle J 0, Utmattningsprov av svetsad balk i
HT -stal STU-rapport 73-3983, 75-4487.
12. Friis L E, Utmattningsprov yid hog spanningsniva Laboratorierapport LM 39174 Svenskt Stal, Oxelosund.
13. Sperle J 0, Utmattning yid laga lastcykeltal JK-rapport D
22. Sperle J 0, Medelspanningars inverkan pa utmattningshallfastheten fOr svetsforband, Examensarbete, Inst for Svetsteknologi KTH.
23. Hansen B, Svejsespaendinger og svejsedeformationer. Formelsamling, K 72001115. Svejsecentralen, Copenhagen.
24. Stalbyggnad - Detaljutformning Stalbyggnadsinstitutet,
Stockholm.
25. Stalbyggnadshandboken, NJA Svensk Byggtjanst Box 1403
Stockholm.
26. Kloth Willi H C, Atlas der Spannungsfelder in technischen
Bauteilen, Verlag Stahleisen 1961 Dusseldorf.
27. Falck J, Teknisk Tidskrift 1940 sid 93 -.
28. Timoshenko, Theory of plates and Shells McGraw-Hill,
2nd. ed.
29. Sih, Handbook of stress intensity factors. Lehigh Univ,
Bethlehem Pennsylvania 1973.
30. Formelsamling i Hallfasthetslara publikation 104. Institutionen for Hallfasthetslara, Tekniska Hogskolan, Stockholm.
31. Gurney T R, An analysis of some fatigue crack propagation
data for steels subjected to pulsating tension loading.
Welding inst res rep E/59178.
32. Bergqvist L, Sperie J 0, Fracture toughness across the roiling direction. SSF 152 High Tensile Steel in Shipbuilding
V. The Swedish Ship Research Foundation.
33. Sonander C, Trogen H, Bestamning av brottseghet hos
Domex 400 och OX 802 med icke linjara metoder. Laboratorierapport LM 150174 Svenskt Stal, Oxelosund.
145.
14. Sperle J 0, Utmattning hos svetsforband yid spektrumbelastning, JK-rapport D 266.
34. Markstrom, Experimentell undersokning av provstavstjocklekens inverkan pa he, Rapport 20, Hallfasthetslara Tekniska Hogskolan Stockholm 1977.
15. Blomberg F, Kompendium i maskinkonstruktion AK. Inst.
for maskinkonstruktion KTH.
35. Sonander C, Specialrapport OX 602, Svenskt Stal, Oxelosund 1973.
16. Stuber M and Rolfe, Effective Utili2ation of High-Yield
Strength Steels in Fatigue. WRC Bulletin 243.
36. Westerberg, Markstrom, Sproda zoners betydelse for seghetsegenskaperna hos svetsforband. Jernkontorets Forskningsrapport D 205 1977.
17. Abathi, Albrecht and Irwin, Fatigue of Periodically Overloaded Stiffner Detail, Journal of the Structural Division
Nov 1976.
18. Bergqvist L, Sperle J 0, Utmattningshallfastheten hos hoghallfasta stal med termiskt skurna snittytor, STU -rapport
75-5680.
19. Wylde J G, The effect of axial misalignment on the fatigue
strength of transverse butt welded joints. Welding Institute
Research Reports 99/1979.
20. Johansson B G, Hallsten K E, Hur bra skall man bygga
fartygsskrovet, Nordiska Skeppstekniska motet i Oslo 197609-30, Kockums AB Malmo.
21. Gurney T R, Some recent work relating to the influence of
residual stresses on fatigue strength, Proceeding Residual
stresses in welded construction and their effects, London
Nov 1977.
37. Bergqvist L,. Brottmekanisk undersokning av OX 812
Grundmaterial, Laboratorierapport LR 194176 Svenskt Stal,
Oxelosund.
38. Brottseghet hos svetsgods, Mekanresultat 78001, Sveriges
Mekanforbund 1978, Box 5506, 11485 Stockholm.
39. Brottseghet i svetsfOrband med OX 802, Preliminara resultat. TUS 400173.
40. Cruciform jOints and their optimisation for fatigue, IIW-doc
XIII-750-74.
41. Boestad G, Kompendium i Maskinelement 3A, Tekniska
Hogskolan 1966.
42. StBK-N3 - Regulations for Bolted Connections 76. National
Swedish Committee on Regulations for Steel Structures,
Swedish Institute of Steel Construction.
9:1
9:2
SSAB 5vE'ns<t S;~': AB
:3
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Price SEK 250:-
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