UNIVERSITY OF SOUTH AFRICA (UNISA)
College of Science, Engineering and Technology
⋄
Advanced Structural Steel Design
STR4801 Year Module
⋄
Module Code: STR4801
Module Name: Advanced Structural Steel Design
Assignment No.: Assignment 1
Due Date: 2026
Semester: Year Module
Submitted in partial fulfilment of the requirements for Advanced Structural Steel Design
at the University of South Africa.
,QUESTION 1 [25 MARKS]
A steel bracket to support an ultimate load of 450 kN is welded to the flange of a 305 x 305 x 198
kg/m H-column as shown in Figure Q1 below. Determine a suitable size for the fillet weld shown
if the bracket is Grade 350W and E80XX electrodes are used.
FIGURE Q1
,UNISA | STR4801 Advanced Structural Steel Design
Solution
1. Given data
Vu = 450 kN
Lv = 550 mm
Lh = 250 mm
There are two horizontal welds:
2Lh = 2(250) = 500 mm
Total weld length:
Lw = 550 + 500
Lw = 1050 mm
The weld group is therefore made up of:
• one vertical weld of 550 mm
• two horizontal welds of 250 mm each
The calculation method agrees with the available worked solution, which determines the
weld-group centroid using 1050 mm total weld length.
2. Determine the centroid of the weld group
Take the vertical weld as the reference axis.
For the vertical weld:
L1 = 550 mm
x1 = 0
Page 1 of 9
,UNISA | STR4801 Advanced Structural Steel Design
For the two horizontal welds:
L2 = 2(250) = 500 mm
The centroid of each horizontal weld is:
250
x2 = = 125 mm
2
Therefore,
P
Li xi
x̄ = P
Li
(550)(0) + (500)(125)
x̄ =
1050
62500
x̄ =
1050
x̄ = 59.52 mm
Approximately,
a = 59 mm
The vertical centroid is at the centre of the weld group:
ȳ = 0
Hence,
a + b = 250
b = 250 − 59
b = 191 mm
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,UNISA | STR4801 Advanced Structural Steel Design
3. Determine the eccentricity of the load
The distance from the column centreline to the load is:
250 + 350 = 600 mm
The load eccentricity measured from the centroid of the weld group is:
e = 600 − a
Using a = 59 mm:
e = 600 − 59
e = 541 mm
Using the rounded value e = 540 mm:
e ≈ 540 mm
This is also the eccentricity obtained in the available worked solution.
4. Determine the applied moment on the weld group
Mu = Vu e
Using e = 540 mm:
Mu = (450)(540)
Mu = 243000 kNmm
Using the unrounded centroid:
Page 3 of 9
, UNISA | STR4801 Advanced Structural Steel Design
Mu = (450)(540.48)
Mu ≈ 243216 kNmm
Therefore, use:
Mu ≈ 243000 kNmm
5. Determine the second moment of area Iwx
For the vertical weld:
L3
Iwx1 =
12
5503
Iwx1 =
12
Iwx1 = 13.8646 × 106 mm4
For the two horizontal welds:
Iwx2 = 2(250)(275)2
Iwx2 = 37.8125 × 106 mm4
Therefore,
Iwx = Iwx1 + Iwx2
Iwx = 13.8646 × 106 + 37.8125 × 106
Iwx = 51.677 × 106 mm4
Approximately,
Iwx = 51 × 106 mm4
Page 4 of 9
College of Science, Engineering and Technology
⋄
Advanced Structural Steel Design
STR4801 Year Module
⋄
Module Code: STR4801
Module Name: Advanced Structural Steel Design
Assignment No.: Assignment 1
Due Date: 2026
Semester: Year Module
Submitted in partial fulfilment of the requirements for Advanced Structural Steel Design
at the University of South Africa.
,QUESTION 1 [25 MARKS]
A steel bracket to support an ultimate load of 450 kN is welded to the flange of a 305 x 305 x 198
kg/m H-column as shown in Figure Q1 below. Determine a suitable size for the fillet weld shown
if the bracket is Grade 350W and E80XX electrodes are used.
FIGURE Q1
,UNISA | STR4801 Advanced Structural Steel Design
Solution
1. Given data
Vu = 450 kN
Lv = 550 mm
Lh = 250 mm
There are two horizontal welds:
2Lh = 2(250) = 500 mm
Total weld length:
Lw = 550 + 500
Lw = 1050 mm
The weld group is therefore made up of:
• one vertical weld of 550 mm
• two horizontal welds of 250 mm each
The calculation method agrees with the available worked solution, which determines the
weld-group centroid using 1050 mm total weld length.
2. Determine the centroid of the weld group
Take the vertical weld as the reference axis.
For the vertical weld:
L1 = 550 mm
x1 = 0
Page 1 of 9
,UNISA | STR4801 Advanced Structural Steel Design
For the two horizontal welds:
L2 = 2(250) = 500 mm
The centroid of each horizontal weld is:
250
x2 = = 125 mm
2
Therefore,
P
Li xi
x̄ = P
Li
(550)(0) + (500)(125)
x̄ =
1050
62500
x̄ =
1050
x̄ = 59.52 mm
Approximately,
a = 59 mm
The vertical centroid is at the centre of the weld group:
ȳ = 0
Hence,
a + b = 250
b = 250 − 59
b = 191 mm
Page 2 of 9
,UNISA | STR4801 Advanced Structural Steel Design
3. Determine the eccentricity of the load
The distance from the column centreline to the load is:
250 + 350 = 600 mm
The load eccentricity measured from the centroid of the weld group is:
e = 600 − a
Using a = 59 mm:
e = 600 − 59
e = 541 mm
Using the rounded value e = 540 mm:
e ≈ 540 mm
This is also the eccentricity obtained in the available worked solution.
4. Determine the applied moment on the weld group
Mu = Vu e
Using e = 540 mm:
Mu = (450)(540)
Mu = 243000 kNmm
Using the unrounded centroid:
Page 3 of 9
, UNISA | STR4801 Advanced Structural Steel Design
Mu = (450)(540.48)
Mu ≈ 243216 kNmm
Therefore, use:
Mu ≈ 243000 kNmm
5. Determine the second moment of area Iwx
For the vertical weld:
L3
Iwx1 =
12
5503
Iwx1 =
12
Iwx1 = 13.8646 × 106 mm4
For the two horizontal welds:
Iwx2 = 2(250)(275)2
Iwx2 = 37.8125 × 106 mm4
Therefore,
Iwx = Iwx1 + Iwx2
Iwx = 13.8646 × 106 + 37.8125 × 106
Iwx = 51.677 × 106 mm4
Approximately,
Iwx = 51 × 106 mm4
Page 4 of 9
