STR4801 Assignment 2026 |Year Module Advanced Structural Steel| Design 2026

UNIVERSITY OF SOUTH AFRICA
College of Science, Engineering and Technology


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STR4801: Advanced Structural Steel Design

Assignment 1 | Year Module

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STR4801
Module Code:
Advanced Structural Steel Design
Module Name:
2026
Due Date:
100
Total Marks:




Submitted in partial fulfilment of the require-
ments for Advanced Structural Steel Design, UNISA

,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


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


For the two horizontal welds:



L2 = 2(250) = 500 mm



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,UNISA | STR4801 Advanced Structural Steel Design


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


3. Determine the eccentricity of the load


The distance from the column centreline to the load is:



250 + 350 = 600 mm


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,UNISA | STR4801 Advanced Structural Steel Design


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


4. Determine the applied moment on the weld group


Mu = Vu e


Using e = 540 mm:



Mu = (450)(540)

Mu = 243 000 kNmm


Using the unrounded centroid:



Mu = (450)(540.48)

Mu ≈ 243 216 kNmm


Therefore, use:



Mu ≈ 243 000 kNmm




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, UNISA | STR4801 Advanced Structural Steel Design



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


6. Determine Iwy


For the vertical weld:



Iwy1 = 550(59)2

Iwy1 = 1.91455 × 106 mm4




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