1
INTRODUCTION
This document contains a summary of class notes made during lectures at the University of Pretoria.
Please note that, because these notes were produced during lectures, that there may be typing and
calculation errors in the document. We tried our best to limit this to the minimum.
If you detect errors in the document, please notify the lecturer as soon as possible.
2
OUTLAY OF OUR CONTACT SESSIONS
The figure on the next slide shows a layout of topics that will be dicussed during contact time with
Michiel Heyns.
Design of welds
𝑎
Static design
Variable amplitude
loading
𝑎𝑐𝑟𝑖
Maximum principal stress, 𝜎1 or 𝜎3
Von-Mises:
𝜎𝑣𝑚 =
1
𝜎1 − 𝜎2 2 + 𝜎1 − 𝜎3 2 + 𝜎2 − 𝜎3 2
2
Tresca (Maximum shear stress): 𝜏 =
Maximum principal strain: 𝜀1
Buckling
Creep
𝜎1 −𝜎3
2
Fracture
Mechanics
Fatigue analysis
𝑎𝑑
𝑎𝑖
Fatigue analysis
Cycles
Miner’s
Δ𝜎1
damage
𝑛𝑖
𝐷=
𝑁𝑖
rule:
Fracture Mechanics
Crack propagation
𝑑𝑎
𝑑𝑁
Δ𝜎2
𝑁1
𝑁2
Endurance,
𝑁, (cycles)
𝑑𝑎
= 𝐶 Δ𝐾 +
𝑑𝑁
𝑚
Stress intensity: Δ𝐾 = 𝛽Δ𝜎 𝜋𝑎[𝑀𝑃𝑎. 𝑚]
Critical crack size
 Plastic collapse
𝜎𝑟𝑒𝑚 ≥ 𝑓𝑦
 Fracture
𝐾 = 𝜎𝛽 𝜋𝑎 ≥ 𝐾𝐼𝑐
3
Equations to calculate static stress
Consider the geometry shown below. The I-beam of length 𝐿 is welded with weld of size 𝑑 all around
as shown. The dimensions of the geometry is indicated in the figure. What is the stress at Points A
and B?
𝐹𝐻
𝑦
𝑊
𝐹𝑣
A
𝑡𝑓
𝑡𝑤
𝐻
𝑥
B
8
Solution:
The stress equation on a cross-section is given as follows:
𝑀𝑥 𝑦 𝑀𝑦 𝑥 𝐹𝑧
𝜎=
−
+
𝐼𝑥𝑥
𝐼𝑦𝑦
𝐴
The average shear stress is:
𝑉
𝜏=
𝐴
In the example above, say: 𝐻 = 400 𝑚𝑚, 𝐷 = 300 𝑚𝑚, 𝑡𝑓 = 10 𝑚𝑚, 𝑡𝑤 = 8 𝑚𝑚.
We need to calculate the second moments of AREA as follows:
2
𝐻 2
𝑊 − 𝑡𝑤
𝐻
1
𝐻 𝑡𝑓 2
1 3
3
𝐼𝑥𝑥 = 2 × 𝑊𝑑 ( ) + 4 ×
𝑑 ( − 𝑡𝑓 ) + 2 ×
(𝐻 − 2𝑡𝑓 ) 𝑑 + 4 × (𝑡𝑓 𝑑 ( − ) +
𝑡 𝑑)
2
2
2
12
2 2
12 𝑓
𝐼𝑦𝑦 = 2 ×
1 3
𝑊 − 𝑡𝑤
𝑊 𝑡𝑤 2
1
𝑊 − 𝑡𝑤 3
𝑡𝑤 2
𝑊 𝑑 + 4 × ((
𝑑) ( − ) +
((
) 𝑑)) + 2 × (𝐻 − 2𝑡𝑓 )𝑑 ( ) + 4
12
2
2
2
12
2
2
× (𝑡𝑓 𝑑 (
𝑊 2
) )
2
H
0.4
W
0.3
t_f
0.01
t_w
0.008
d
0.008
Calculations:
I_xx
0.000446
I_yy
A
0.015872
I_zz
m
m
m
m
m
Mx
My
Fz
Fx
Fy
m4
0.000151 m4
m2
0.000597 m4
500 Nm
0 Nm
0N
N
N
A
x_A
y_A
sigma=
Shear
0.15
0.2
224 219
0
B
m
m
Pa
MPa
0.15
-0.2
-224 219
0
Average shear stress calculation on a round section according to the factored resistance formulae:
𝑉𝑟 = 0.67𝜙𝐴𝑓𝑢
= 0.67𝜙 0.707𝑑𝐿 𝑓𝑢
= 0.67 × 0,67 0.707𝑑𝐿 𝑓𝑢
= 0.45 0.707𝑑𝐿 𝑓𝑢
𝑓𝑜𝑟 𝜙 = 0.67
Average stress on a round object subject to shear forces and torque:
X
𝑦
𝑅
X
X
𝑇
𝑥
𝜏𝐹𝑦 =
𝐹𝑦
𝐴
𝜏𝑇 =
𝑇𝑅
𝐼𝑧𝑧
𝜏𝐹𝑥 =
𝐹𝑥
𝐴
X
This is for average shear stress distribution. Normally we would solve the shear flow
equation that will result in large shear force dependent shear stress on the neutral axis. However,
it seems as if you were only introduced to average shear stress calculation. Do not hesitate to
contact me if you need more information.
4
Variable loads in Excel
The following table and graph was used to demonstrate calcuation of maximum, minimum, mean and
standard deviation of a signal in Excel.
Chart Title
0
100
200
-100
50
150
30
100
-20
80
50
150
0
-200
1
2
3
4
5
6
7
8
9
0
-50
Max
150
-100
Minimum
-200
Mean
9
-150
St. Dev.
101.0445
-200
-250
10
5
Explaining Rainflow counting
The rainflow counting process is, in a nutshell, as follows:
1. Calculate extrema:
a. This is the turning points at the top and bottom sides of the signal.
b. Also called Peak-Valley reduction
c. In most cases the turning points are grouped together to simplify.
i. This is done by specifying the stress intervals in which turning points must be
calculated.
2. Rainflow counting:
a. Output is: stress amplitude (𝜎𝑎𝑖 ), mean stress (𝜎𝑚𝑖 and No. of cycles (𝑛𝑖
3. For weld fatigue:
a. Calculate the stress ranges (Δ𝜎𝑖 = 2𝜎𝑎𝑖 ) and number of cycles, 𝑛𝑖
From the count shown below, we have the following rainflow counted detail:
From
[MPa]
1
2
3
4
5
6
7
8
9
50
100
150 -50
30
-100
100
20
-50
50
0
1
3
-100
-150
4
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
Range Ranges
No. of
[MPa] [MPa]
cycles
50
50
100
80
150
100
200
120
80
150
80
200
120
70
50
5
2
1
ni
Chart Title
0
-50
0
50
-50
100
-50
30
-100
20
-50
To
[MPa]
50
-50
100
-100
30
-50
20
-50
0
2
3
4
5
6
7
76
8
9
9 8
10
11
1
1
0.5
0.5
0.5
0.5
6
STRESS LIFE S-N CURVE
The stress amplitude at 1000 cycles is estimated as 𝑆103 = 0.9𝑆𝑢𝑡 for steel, for 50% probability of
survival (50% probability of failure) in 𝑁.
The stress amplitude at 106 cycles is: 𝑆106 = 0.5𝑆𝑢𝑡 for the same probability of survival mentioned above.
How do we model the S-N curve then?
𝑆103 = 0.9𝑆𝑢𝑡
𝐶 = 0.9𝑆𝑢𝑡 𝑚 103
𝑆106 = 0.5𝑆𝑢𝑡
𝐶 = 0.5𝑆𝑢𝑡 𝑚 106
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒:
0.9𝑆𝑢𝑡 𝑚 103 = 0.5𝑆𝑢𝑡 𝑚 106
0.9𝑆𝑢𝑡 𝑚
(
) = 103
0.5𝑆𝑢𝑡
𝑆103
𝑚 log
=3
𝑆106
3
𝑚=
𝑆 3
log 106
𝑆10
For fatigue calculations, we normally have the stress ranges with their associated number of cycles
over the period which the stress was measured. So, we need to have a formula to give the material
endurance, 𝑁𝑖 , for every stress amplitude, 𝑆𝑖 .
𝑆106 𝑚 6
(
) 10 𝑓𝑜𝑟 𝑆103 ≥ 𝑆𝑖 ≥ 𝑆106
𝑁 = { 𝑆𝑖
∞
𝑓𝑜𝑟 𝑆𝑖 < 𝑆106
Where you make provision for the stress concentration notch fatigue and other reductions factors,
remember that:
0.9𝑆𝑢𝑡
𝑆103 =
𝐶𝑠𝑖𝑧𝑒 𝐶𝑙𝑜𝑎𝑑 𝐶𝑇𝑒𝑚𝑝 𝐶𝑟𝑒𝑙
𝐾𝐹′
0.5𝑆𝑢𝑡
𝑆106 =
𝐶𝑠𝑖𝑧𝑒 𝐶𝑙𝑜𝑎𝑑 𝐶𝑠𝑢𝑟𝑓 𝐶𝑇𝑒𝑚𝑝 𝐶𝑟𝑒𝑙
𝐾𝐹
Affecting factors:
1. Mean stress (Affects 𝑺𝟏𝟎𝟑 and 𝑺𝟏𝟎𝟔
a. Use Goodman mean stress correction.
b. Calculate equivalent completely reversed stress amplitude, 𝑆𝑎 for signal with amplitude
𝜎𝑎 and mean 𝜎𝑚 on a material with ultimate tensile strength 𝑆𝑢𝑡 .
𝜎𝑎 𝜎𝑚
+
=1
𝑆𝑎 𝑆𝑢𝑡
𝜎𝑎
𝑆𝑎 =
𝜎
1− 𝑚
𝑆𝑢𝑡
2. Size (Affects 𝑺𝟏𝟎𝟑 and 𝑺𝟏𝟎𝟔 .
𝑆
3. Loading effect (Affects 𝑺𝟏𝟎𝟑 and 𝑺𝟏𝟎𝟔 : 𝑆𝑒,𝑏𝑒𝑛𝑑𝑖𝑛𝑔 = 𝑒
0.7
4. Surface finish (Affects 𝑺𝟏𝟎𝟔 :
Figure 1: Surface finish factors
5. Surface treatment.
a. Effect depends on residual stresses produced.
b. If residual stresses can be calculated from the treatment done, add this residual stress
to the mean stress and do Goodman mean stress correction.
6. Temperature (Affects 𝑺𝟏𝟎𝟑 and 𝑺𝟏𝟎𝟔 :
𝐸𝑇
6 , 𝐼𝐼𝑊 𝐵𝑢𝑙𝑙𝑒𝑡𝑖𝑛 520
𝑆 𝑇,2×106 =
𝑆
𝐸𝑇𝑎 2×10
1.0 𝑓𝑜𝑟 𝑇 ≤ 450 ℃
𝐶𝑡 =
−3
1 − 5.8 𝑇 − 450 , 𝑓𝑜𝑟 450 < 𝑇 ≤ 550 ℃
7. Reliability (Affects 𝑺𝟏𝟎𝟑 and 𝑺𝟏𝟎𝟔 :
Table 1: Modification factors for different probability of survival
8. Corrosion (Affects 𝑺𝟏𝟎𝟔 and removes the endurance limit)
Figure 2: Surface finish and effect of corrosion
The S-N curve the becomes:
𝑆103
𝐶 = 𝜎𝑎𝑚 𝑁
𝑆106
103
106
Endurance 𝑁
Figure 3: Typical S-N curve for steels
6.1 Class problem from the slides
For the class problem we have:
𝑆𝑒 = 𝑆106 = 132.4 𝑀𝑃𝑎
𝑆103 = 0.9 × 450
= 405 𝑀𝑃𝑎
From this:
3
405
log
132.4
= 6.18
The number of cycles to failure at 200 and 300 MPa.
Remember to first compare the stress amplitudes with the modified
0.5𝑆𝑢𝑡
𝑆106 =
𝐶𝑠𝑖𝑧𝑒 𝐶𝑙𝑜𝑎𝑑 𝐶𝑠𝑢𝑟𝑓 𝐶𝑇𝑒𝑚𝑝 𝐶𝑟𝑒𝑙
𝐾𝐹
To prevent unnecessary calculations. You will be caught out by this in the tests.
𝑚=
S_ut
K_F
K_F'
S_10^3
S_10^6
m
450 MPa
1.7
1
405 MPa
132.3529 MPa
6.17638
Stress
Endurance
amplitude
[MPa]
200
78 090
300
6 383
endurance
limit:
7
STRESS LIFE FATIGUE EXAMPLE PROBLEM
You have a steel part that is subjected to the stress spectrum shown in the table below, which was
measured over a period of 2 years.
Stress
amplitude
[MPa]
200
100
200
100
Stress
No.
mean
of cycles
[MPa]
ni
100
10000
-200
50000
0
100000
0
10000
The material is steel with ultimate tensile strength 700 MPa. The specimen contains a notch with
theoretical stress concentration factor 2. Assume a notch radius of 0.01 inch (0.254 mm).
The machined specimen will be used at a temperature of 300 °C.
What is the fatigue life of the component for a 95% probability of survival (5% probability of crack
initiation)?
Answer
Estimate points on the S-N curve
The stress amplitude on the S-N curve at 1000 cycles is given as:
0.9𝑆𝑢𝑡
𝑆103 =
𝐶𝑠𝑖𝑧𝑒 𝐶𝑙𝑜𝑎𝑑 𝐶𝑇𝑒𝑚𝑝 𝐶𝑟𝑒𝑙
𝐾𝐹′
The stress amplitude on the S-N curve at 106 cycles is given as:
0.5𝑆𝑢𝑡
𝑆106 =
𝐶𝑠𝑖𝑧𝑒 𝐶𝑙𝑜𝑎𝑑 𝐶𝑠𝑢𝑟𝑓 𝐶𝑇𝑒𝑚𝑝 𝐶𝑟𝑒𝑙
𝐾𝐹
Mean stress correction
The following equation will be used:
𝜎𝑎
𝑆𝑎 =
𝜎
1− 𝑚
𝑆𝑢𝑡
Notch fatigue factors
From the lecture notes:
300 1.8
𝑎=(
) × 10−3 𝑖𝑛𝑐ℎ
𝑆𝑢𝑡
300
=(
) × 10−3
100
= 0.0072
Then:
2−1
𝑎
1+
𝑟
= 1.58
𝐾𝑓 = 1 +
Assume 𝑟 = 0.01 𝑖𝑛𝑐ℎ.
At 1000 cycles:
Determine the factor q from the graph below and calculate:
𝐾𝐹′ − 1
𝑞 = 0.2 =
𝐾𝐹 − 1
𝐾𝐹′ = 𝑞 𝐾𝐹 − 1 + 1
= 0.2 1.58 − 1 + 1
= 1.12
Affecting factors
Surface finish (Affects 𝑺𝟏𝟎𝟔 :
From the curve it is estimated at 𝐶𝑠𝑢𝑟𝑓 = 0.75.
Temperature (Affects 𝑺𝟏𝟎𝟑 and 𝑺𝟏𝟎𝟔 :
AT 300 °C, 𝐶𝑇 = 1.
𝐶𝑡 =
1.0 𝑓𝑜𝑟 𝑇 ≤ 450 ℃
−3
1 − 5.8 𝑇 − 450 , 𝑓𝑜𝑟 450 < 𝑇 ≤ 550 ℃
Reliability (Affects 𝑺𝟏𝟎𝟑 and 𝑺𝟏𝟎𝟔 :
For a probability of survival of 95%, 𝐶𝑟𝑒𝑙 = 0.868.
Corrosion (Affects 𝑺𝟏𝟎𝟔 and removes the endurance limit)
The S-N curve for endurance
𝑚=
𝑆106 𝑚 6
(
) 10
𝑁 = { 𝑆𝑖
∞
Sut
KF'
KF
S1000
Se
m=
700 MPa
1.12
1.58
488.25 MPa
144.2089
5.66
3
𝑆 3
log 106
𝑆10
𝑓𝑜𝑟 𝑆103 ≥ 𝑆𝑖 ≥ 𝑆106
Csize
Csurf
CTemp
Crel
Cload
𝑓𝑜𝑟 𝑆𝑖 < 𝑆106
1
0.75
1
0.868
1
Stress
Stress
No.
CompletelyEndurance Damage
amplitude mean
of cycles rev stress
[MPa]
[MPa]
ni
[MPa]
Ni
Di
200
100
10000
233
65 507
0.15
100
-200
50000
78 INF
0.00
200
0
100000
200 156 848
0.64
100
0
10000
100 INF
0.00
Total damage =
0.79
Period =
2 years
Fatigue life =
2.5 years
8
WELD FATIGUE
8.1 Problem Statement
A flat section of thickness 28.95 mm is joined to a flat section of thickness 20 mm using a manual
shielded metal arc welding process to produce the double V-groove weld of the butt joint shown in
Figure 4. Both sections are from 300W structural steel and a normal-match electrode was used. The
2 mm root face has a root opening of 1 mm and the sections chamfered to produce a groove angle 𝛼 =
60°. The weld preparation was done in compliance with AWS D1.1:2008 as summarised in Table 3.
The thick section is ground in the direction of the principal stress to produce a taper of 1:5. The height
of the weld convexity is measured at 8% of the weld width with smooth transition to the plate surface.
The welding procedure specification required the use of weld run-on and run-off pieces that were
removed afterwards. The edges are ground flush with the surface with grinding marks in the direction
of the axial stress, which is also the direction of the principal stress in this case. Welding was done
from both sides and ultrasonic testing done to confirm the absence of sub-surface defects in the weld.
The stress spectrum of the joint over a period of 2 years is as summarised in Table 2. For your
information, the following has been included:
 EN 1993-1-9:2005 Table 8.3.
The design requires a safe life assessment method with high consequence of failure. The operating
temperature is 250 °C and the surfaces corrosion protected using International Paint’s HT-10 epoxy
paint. No post weld heat treatment was done. No post weld improvement (peening, grinding, dressing,
etc.) were done to the weld detail.
Table 2: Stress spectrum on the joint over a period of two years
Number of cycles
𝝈𝒎𝒂𝒙
𝝈𝒎𝒊𝒏
[MPa]
[MPa]
200
100
100 000
50
-75
50 000
40
0
1 000 000
Figure 4: Tapered joint
Table 3: AWS D1.1 double V-groove weld for butt joints (2008:96)
Please answer the following (Please use EN 1993-1-9:2005 as reference):
1. What is the partial factor for fatigue for this problem? [5% of marks]
2. What other factors need to be taken into account in this problem? [5% of marks]
3. What is the detail category for this joint? [5% of marks]
4. What value must you take into account for the thickness effect in this case? [5% of marks]
5. Where do you expect the crack to initiate first in this joint? [5% of marks]
6. If the stress spectrum refers to a block loading applied to the joint over a period of 8 years, what
estimate of fatigue life would you make for the welded joint for a probability of failure of 5% that is, for a probability of survival of 95%? [70% of marks]
7. What improvement in detail category is possible in this case with hammer peening? [5% of
marks]
Table 4: EN 1993-1-9 Table 8.3 for transverse butt welds (2005:22)
8.2 Solution
Question 1: Partial factor for fatigue
For a design philosophy of safe life assessment method with high consequence of failure, the partial
factor for fatigue is 𝛾𝑀𝑓 = 1.35.
Question 2: Other influencing factors to take into account
Temperature
The new characteristic strength for the temperature compensated application is:
Δ𝜎𝐶
Δ𝜎𝐶,𝐻𝑇 =
× 0.9
𝛾𝑀𝑓
Size effect
The factor for size is:
1
0.2
𝑘𝑠 = { 25
( )
𝑡
𝑡 ≤ 25 𝑚𝑚
𝑡 > 25 𝑚𝑚
Question 3: Detail category
Based on the information supplied, the detail category for this joint is 90.
Question 4: Thickness effect
No thickness effect need to be included, because, the crack is expected to initiate in the thinner section,
which is 20 mm in thickness.
Question 5: Point of crack initiation
The crack is expected to initiate at the weld toe in the thinner 20 mm plate.
Question 6: Damage and life calculation
S-N curve
The modified characteristic strength of the weld detail is:
Δ𝜎𝐶
Δ𝜎𝐶,𝑚𝑜𝑑 =
𝑘𝑘
𝛾𝑀𝑓 𝑠 𝑇
This is the fatigue strength at 𝑁𝐶 = 2 × 106 .
The constant amplitude fatigue limit is then:
1
𝑁𝐶 𝑚1
Δ𝜎𝐷 = Δ𝜎𝐶,𝑚𝑜𝑑 ( ) , 𝑚1 = 3
𝑁𝐷
The cut-off limit is:
1
𝑁𝐷 𝑚2
Δ𝜎𝐿 = Δ𝜎𝐷 ( ) , 𝑤ℎ𝑒𝑟𝑒 𝑚2 = 5
𝑁𝐿
The endurance, 𝑁𝑅 , for any stress range, Δ𝜎𝑅 , is then:
Δ𝜎𝐶,𝑚𝑜𝑑 𝑚1
(
) 𝑁𝐶 𝑓𝑜𝑟 1.5𝜎𝑦 ≥ Δ𝜎𝑅 ≥ Δ𝜎𝐷 , 𝑎𝑛𝑑 𝑚1 = 3
Δ𝜎𝑅
𝑚2
𝑁𝑅 =
Δ𝜎𝐷
(
) 𝑁𝐷
𝑓𝑜𝑟 Δ𝜎𝐷 ≥ Δ𝜎𝑅 ≥ Δ𝜎𝐿 , 𝑚2 = 5
Δ𝜎𝑅
∞
𝑓𝑜𝑟 Δ𝜎𝑅 < Δ𝜎𝐿
{
Damage and life calculation
Based on the information provided to us, the fatigue life is 13.8 years.
90
ΔσD
44.2 MPa
1.35
ΔσL
24.3 MPa
Detail category
Partial factor for fatigue
Thickness modification
Temperature modification
ΔσC,mod
1
m1
3
0.9
m2
5
60.0 MPa
Endurance: NC
2.00E+06
Endurance: ND
5.00E+06
Endurance: NL
1.00E+08
σmax
[MPa]
σmin
[MPa]
200
50
40
100
-75
0
ΔσR
[MPa]
100
125
40
ni
NR
Di
100 000
432 000
50 000
221 184
1 000 000
8 245 044
Total damage =
Period =
Fatigue life =
0.231481
0.226056
0.121285
0.578823
8 years
13.8 years
Question 7: Peening effect
Based on the information provided to us, peening will result in a fatigue life of 28.8 years for 5%
probability of crack initiation.
112
ΔσD
55.0 MPa
Partial factor for fatigue
1.35
ΔσL
30.2 MPa
Thickness modification
1
m1
3
0.9
m2
5
Detail category
Temperature modification
ΔσC,mod
74.7 MPa
Endurance: NC
2.00E+06
Endurance: ND
5.00E+06
Endurance: NL
1.00E+08
σmax
[MPa]
σmin
[MPa]
200
50
40
100
-75
0
ΔσR
[MPa]
100
125
40
ni
NR
100 000
832 550
50 000
426 266
1 000 000
24 607 671
Total damage =
Period =
Fatigue life =
Di
0.120113
0.117298
0.040638
0.278048
8 years
28.8 years