University of Melbourne
CVEN90049 Structural Theory and Design 2
Design Project: Assignment 01
A five-storey office has the general layout shown in Figure 1. In addition to the selfweight of the structural members, a superimposed dead load of 1 kPa should be allowed
for. The live load rating of the office floor is 3 kPa. The roof should be designed for a
superimposed dead load of 0.5 kPa and live load rating of 1 kPa. Wind load is assumed to
be resisted by the shear walls located at the edges of the building and need not be
considered.
Design for a typical floor slab, beam and column. You are required to provide two options:
1. Reinforced concrete slab supported by reinforced
concrete beams and columns
2. Reinforced concrete slab supported by steel beams
and columns.
Note:
1. If you decide to change the beam layout for the steel options, the reinforced
concrete slab should normally be re-designed. You are not required to redesign
the slab for this assignment but you should take into account the change of
thickness (and hence self-weight) in your design.
2. The slab cannot be assumed to provide continuous lateral restraints for the steel beams.
The design should be supported by calculations to ensure that the slab, beams and
columns and their connections (when applicable) are adequate in resisting the design
load. The calculations for any horizontal bracings, should they be used in the design
option ii, are not required for this assignment. However, the locations and details of
their connection should be included in the final drawings
University of Melbourne
CVEN90049 Structural Theory and Design 2
Design Project: Assignment 01
Summary
This report is the part 1 of the continuous assignment and contains three individual task. The
main objective of this report is to understand the application of Space Gass for calculating the
maximum bending moment and reaction forces. Furthermore, for task 01 three conceptual
deigns of the building are sketched. For each design its merits and demerits are also discussed
and one particular designed is finalized. For task 2, model is validated by comparing the bending
moment and reaction force values from the computer analysis with results from manual
calculation. For task 3, two load cases are applied to the selected layout and summary of
maximum bending moment and shear force is calculated.
TASK 01:
Conceptual design of the building
Brainstorm with your partner and come up with three sketches of possible columns and
beams layout. Discuss the merits and limitations of each sketch. You are to choose one
layout for further analysis.
Layout 01: Conventional Slab design with flat slab system
Slab area: 7500 mm x 10000 mm
 10000/7500 = 1.33 < 2 (Therefore, Two way slab)
Advantages:
 Economical structure since the number of columns are less.
 Can be constructed in less time, since no complex structure is involved.
University of Melbourne
CVEN90049 Structural Theory and Design 2
Design Project: Assignment 01
 Carpet area is increased due to less number of columns
Disadvantages:
 Design will be difficult since the slabs are two way.
 Extra care while designing since load is distributed on less number of column and beam.
Layout 02: One Way Slab (Conventional Slab design with flat slab system)
Slab area: 5000 mm x 10000 mm
 10000/5000 = 2 = 2 (Therefore, one way slab)
Advantages:
 Designs calculation will be easier, since it’s a one way slab design
 Less complexity Involved, since structure is simple.
Disadvantages:
 For one way slabs, due to their huge difference in lengths, load is not transferred to the
shorter beams
University of Melbourne
CVEN90049 Structural Theory and Design 2
Design Project: Assignment 01
Layout 03 (hollow core slab design)
Slab area: 3000 mm x 4000 mm
4000/3000 = 1.33 < 2 (Therefore, Two way slab)
Advantages:
 reducing the cost and self-weight
 easy construction
 excellent sound and fire resistance
Disadvantages:
 hard to strengthen and repair
 hard to lift and move
 easy to be damage during transport
 expensive for small span
University of Melbourne
CVEN90049 Structural Theory and Design 2
Design Project: Assignment 01
Chosen Layout: Layout 03
Reason for choosing
Since, the design has one way slab, calculations will be much easier than other two. Load
distribution will be also uniform and thickness of the column will not be that much high unlike
Layout 01. Layout 02 will be also more economical than layout 03, since the number of column
are far lesser than layout 03. In layout 03, slab is designed as hollow slab, which will be very
expensive to construct as compared to the other layouts.
Preliminary Dimension Slab
Preliminary Dimension Beam
Preliminary Dimension Column
University of Melbourne
CVEN90049 Structural Theory and Design 2
Design Project: Assignment 01
TASK 02:
University of Melbourne
CVEN90049 Structural Theory and Design 2
Design Project: Assignment 01
Maximum +ve moment locations
18.75 – 10x = 10
x = 1.875 meters and 8.125 meters (symmetrical)
Maximum -ve moment at 5 meters
Maximum +ve moment = 0.5 x 1.875 x 18.75 = 17.58 kN
ii. Validation of model using Space Gas
Fig. 1: Reaction forces for the continues beam
Fig. 2: Shear Force Diagram for the continuous beam
Fig. 3: Bending Moment diagram for the continuous beam
Fig. 4: Position vs bending
moment from 0-5m
University of Melbourne
CVEN90049 Structural Theory and Design 2
Design Project: Assignment 01
TASK 03:
Design Load Combinations:
Density for Reinforced Concrete Grade 32 = 2450kg/m2
Tributary area of 5m acting along the beam and thickness of slab is 180mm = 0.18m
We know that,
Live Load Rating for Office floor is 3 kPa
Superimposed Dead load is 1 kPa
For Roof, Dead load = 0.5 kPa and Live Load = 1 kPa
Assuming floors provides the maximum bending moment and Shear force
Self-weight of slab = 25.5 kN/m3 x 0.18 m x 1 m = 4.41 kNm
Now, WSDL = 1 kPa x 1 m = 1 kNm
o Therefore, total Dead load, G = 4.41 + 1 = 5.51 kNm
WLL = 3 kPa x 1 = 3 kNm
o Therefore, Q = 3 kN/m
As per assignment there are two load cases:
Case 1: 1.2G + 1.5Q = 1.2 x (5.51) + 1.5 x (3) = 10.992 kN/m
Fig. 5: Position vs bending
moment from 5-10m
University of Melbourne
CVEN90049 Structural Theory and Design 2
Design Project: Assignment 01
Case 2: 0.8G = 0.8 x (5.41) = 4.328 kN/m
Assumptions:
 Design calculations for roof are exempted for the preliminary calculations, since the
loads on the roof are smaller as compared to the floors.
 The design layout has one way slab, therefore analysis is done only on the critical
direction i.e. (15m) of the slabs.
 For max. bending moment and shear force calculations, slab is treated as a beam with 1
meter width 15 meter length and 0.18 m depth.
 For design calculations, external support is taken as pinned and internal support as
roller supports.
 For this report, Wind load is exempted since wind load is assumed to be resisted by
the shear walls located at the edges of the building.
SPACE GASS ANALYSIS:
The considered beam has 3 spans therefore, it will have total 8 combination for application of
loads on its different spans. Following table shows the different load combinations:
University of Melbourne
CVEN90049 Structural Theory and Design 2
Design Project: Assignment 01
Figure 6. Maximum & minimum bending moment from all load combinatons
Figure 7. Maximum & minimum shear force from all load combinatons
University of Melbourne
CVEN90049 Structural Theory and Design 2
Design Project: Assignment 01
Maximum bending moment and shear force from Space Gass (For appendix for individual
load case BM and SF diagrams)
Maximum Bending Moment

Maximum critical negative moments is -30.26 kNm at 5.0 m & 10 m from the left from
Combinations 2 and combination 4.
Maximum critical positive moments is 25.42 kNm at 2.08 m & 12.92 m from the left from
Combinations 3.
Maximum Shear Force

University of Melbourne
CVEN90049 Structural Theory and Design 2
Design Project: Assignment 01

Maximum critical negative Shear force is -33.53 kN at 5.0 m from the left from Combinations 4.
Maximum critical positive moments is also 33.53 kN at 10 m from the left from Combinations 2.
SIMPLIFIED CALCULATIONS:
Using AS3600 (2018) Cl. 6.10.2.
Let the width of the beam be 300mm,
Fd is the uniformly distributed design load per unit length = 10.992 (From Case 1, 1.2 G. + 0.8 Q)
Therefore,
Ln1 = 5000-300-300/2= 4550 mm
Ln2 = 5000-300/2-300/2= 4700 mm
Ln3 = 5000-300-300/2= 4550 mm
University of Melbourne
CVEN90049 Structural Theory and Design 2
Design Project: Assignment 01
Negative Bending Moments: (Using AS3600 (2018) Cl. 6.10.2.2)
 0 m from left,
o At interior faces of exterior supports for members built integrally with their
supports
 For slabs and beams where the support is a beam
 = FdLn2/24
 =10.992 x (4.55)2/24
 = -9.48 kNm
 5 m from left,
o At the first interior support
 more than two spans
 = FdLn2/10
 =10.992 x (4.77)2/10
 = – 25.01 kNm
 10 m from left,
o At the first interior support
 more than two spans
 = FdLn2/10
 =10.992 x (4.77)2/10
 = – 25.01 kNm
 15 m from left,
o At interior faces of exterior supports for members built integrally with their
supports
 For slabs and beams where the support is a beam
 = FdLn2/24
 =10.992 x (4.55)2/24
 = -9.48 kNm
Positive Bending Moments: (Using AS3600 (2018) Cl. 6.10.2.3)
 In both end spans, mid span 1 and mid span 2
o = FdLn2/11
o =10.992 x (4.55)2/11
o = 20.68 kNm
 In interior spans for Ductility Class N
o = FdLn2/16
o =10.992 x (4.70)2/16
o = 15.07 kNm
Shear Force Calculations: (Using AS3600 (2018) Cl. 6.10.2.4)
University of Melbourne
CVEN90049 Structural Theory and Design 2
Design Project: Assignment 01
 In an end span, VA
o At the face of the end support
 VA = FdLn/2 = 10.992 x (4.55)/2 = 25.01 kN
 In an end span, VB ’
o At the face of the interior support
 VB ’ = 1.15 FdLn/2 = 1.15 x 10.992 x (4.55) /2 = 28.76 kN
 In interior spans, VB
o At the face of supports
 VB = FdLn/2 = 10.992 x (4.70)/2 = 25.83 kN
 In interior spans, VC
o At the face of supports
 VC = FdLn/2 = 10.992 x (4.70)/2 = 25.83 kN
 In interior spans, VC ’
o At the face of supports
 VC’ = = 1.15 FdLn/2 = 1.15 x 10.992 x (4.55) /2 = 28.76 kN
 In an end span, VD
o At the face of the end support
 VD = FdLn/2 = 10.992 x (4.55)/2 = 25.01 kN
 In an end span, VAB mid-span
o At mid-span
 VD = FdLn/7 = 10.992 x (4.55)/7 = 7.14 kN
 In interior spans, VBC mid-span
o At mid-span
 VD = FdLn/8 = 10.992 x (4.70)/8 = 6.45 kN
COMPARISON OF SPACE GASS AND MANUAL CALCULATION:

Actng Shear Force diagram
University of Melbourne
CVEN90049 Structural Theory and Design 2
Design Project: Assignment 01

Above table shows the discrepancies between the values obtained from space gass and
Manual Calculation. It is to be noted that percentage deviation between both is not more
than 20 percentage. Reason for this might be that the Space gass have considered all the
possible load combinations which can be made from case 1 and case 2. While manual
calculation depends upon the formulations given in code and it wold not have taken all
the combination into account.
Another reason might be