HCUB60008: Structural Analysis and Design Project - Civil Engineering

This assignment presents a comprehensive structural engineering project. It begins with the application of the unit load method to determine reactions in a continuous beam, followed by the analysis of a rigid-jointed frame, including sketching the bending moment diagram and deflected shape. The core of the assignment involves the analysis and design of a steel portal frame, incorporating calculations for vertical and horizontal forces, bending moments, and shear forces. The solution includes the determination of suitable steel sections for the columns and rafters, along with a trust diagram, shear force diagram, and bending moment diagram. The assignment draws upon multiple references to support the analysis and design methodologies. This project aims to demonstrate a strong understanding of structural principles and practical application in civil engineering.

Part 1
We have a 10 m continuous beam with three supports 5 m apart. We are required to obtain the
reactions using unit load method.
W W
P
1 1 5m 2 5m 3
In this solution we apply unit load as described in diagram above (Williams, 2013). The loads are
represented as unit load and the equation for the reactions formulated as below.
R1= 1
L [ ( W L2
2 )+ ( P ( L−2 ) )−( R2∗L
2 ) ]
R3= 1
L [ ( W L2
2 )+ ( P ( 2 ) ) −( R2∗L
2 ) ]
R1 +R2 + R3= ( P+ ( WL ) )
Substituting the values and formulating we find linear equations
R2 +2 R3=24
R1 +R2 + R3=30
2 R1+ R2=36
Solving we obtain

R1=6 kN , R2=24 kN , R3 =0 kN
Part 2
we are given a rigid-jointed frame with the corresponding E and I for different members. We
need to use the information to sketch the bending moment diagram and the deflected shape.
We start by taking the moment at A
∑ M A =0
( V c∗6 )− ( 5∗6∗3 ) =0
V C= 5∗6∗3
6 =15 kN
V A = ( 5∗6 )−15=15 kN
Calculating the moments at crucial points (Hibbeler, 2013)
M B =0 kNm
M centre= ( 15∗3 )− (5∗3∗1.5 )=22.5 kNm
M C= ( 15∗6 ) − ( 5∗6∗3 )=0 kNm
B C
22.5 kNm

A MA
Deflection in members is found using the calculated moments and the given values of Young’s
modulus and I.
Deflected shape
Deflection δ= M
EI CITATION Wil13 \l 2057 (McKenzie, 2013)
For section B – C
δB−C = M
E∗2 I
For section A – B
δ A− B= M
EI
The deflected shape will be

Part 3
We have size portal frames. We need to analyze and design the frame.
C
B D
0.4kN/m2 0.125kN/m2
HA A E HE
There are both vertical and horizontal forces and reactions acting on the portal frame. Vertical
forces are from the weight of the structure, imposed loads and all other forces acting on the
structure in the y direction. Horizontal forces results from wind forces acting on the sides of the
structure.
Vertical forces and reactions
Ultimate load Q=1.4 Gk+1.6Qk
Q= [ 1.4∗ [ ( 0.2∗3∗8 ) + ( 3∗10.02 ) ]∗2 ] +[2∗1.6∗( 10.2∗3∗8 ) ]
Q=151.368 kN
The ultimate load will be supported by the end vertical reactions A and E. Downward forces
must balance the forces acting upwards for the structure to be stable. Since the structure is
VA VE

uniformly loaded, the ultimate load will be equally shared between the two end reactions
(Philpot, 2012). Thus;
V A =V E=75.684 kN
Horizontal reactions
∑ horizontal forces=0
( 0.4∗8∗3 ) +H A = ( 0.125∗8∗3 ) +H E
9.6+ H A=3+ HE
H A −HE =−6.6 kN
The forces caused by wind are acting on the same area size thus the reactions will be in the
proportion of H A : H E=0.125 :0.4
H A = 0.125
0.125+ 0.4∗−6.6=−1.571 kN
H E= 0.4
0.125+0.4 ∗−6.6=−5.029 kN
We now calculate the bending moments in both horizontal and vertical directions.
Moments along vertical axis; Moments at the supports are equal to zero.
M AV
=0(support )
M BV
=0
M CV
= ( 75.684∗5 )− [ ( 0.2+0.1+0.6 ) ( 10.02∗3 )∗2.5 ]
¿ 310.785 kNm

M D V
=0
M E V
=0
Moments on the horizontal axis;
M A H
=0
M B H
= ( −1.572∗8 ) +(0.4∗8∗3∗4 )
¿ 25.824 kNm
M CH
= (−1.572∗8.61 ) + ( 0.4∗8∗3∗( 0.61+4 ) )
¿ 30.720 kNm
M E H
=0 kNm
M D H
= ( −5.029∗8 ) +(0.125∗8∗3∗4 )
¿−28.232 kNm
M CH
= (−5.029∗8.61 ) +(0.125∗8∗3∗4.61)
¿−29.470 kNm
Trust diagram
0.4 kN/m2

1.571 kN 5.029 kN
75.684 kN 75.684 kN
Shear force diagram (McKenzie, 2013)
54.084 kN
75.684 kN 75.684 kN
8.029 kN 2.029 kN
1.571 kN 5.029 kN
Bending moment diagram 310.785 kNm
30.720 kNm
0 kNm B 25.824 kNm 28.232 kNm D
A E

Suitable section
The best section for the column is WB 600 @ 1.450 kN/m this is in relation to the unit
weightcarried by the column being equal to ( 0.2+0.1+0.6+ 0.4 )=1.3 kN /m
For rafters use HB 350 @0.724 kN/m which is greater than imposed loads plus the service loads
= 0.3 kN/m (Sha, 2013).

References
Hibbeler, R. C., 2013. Engineering Mechanics: Dynamics. In: Engineering mechanics. s.l.:Pearson, pp. 76-
109.
McKenzie, W., 2013. Design of Structural Elements. In: s.l.:Macmillan International Higher Education, pp.
46-65.
McKenzie, W. M., 2013. Examples in Structural Analysis, Second Edition. In: s.l.:CRC Press, pp. 453-463.
Philpot, T. A., 2012. Mechanics of Materials: An Integrated Learning System. In: s.l.:John Wiley & Sons,
pp. 234-310.
Sha, W., 2013. Steels: From Materials Science to Structural Engineering. In: s.l.:Springer Science &
Business Media, pp. 13-76.
Williams, C. D., 2013. Analysis of Statically Indeterminate Structures. In: s.l.:Literary Licensing, LLC, pp.
101-132.