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

Combinations | Maximum Positive (kNm) | Maximum Negative(kNm) |

1 | 21.95 | -27.48 |

2 | 20.86 | –30.26 |

3 | 25.42 | -19.15 |

4 | 20.86 | -30.26 |

5 | 24.26 | -21.93 |

6 | 15.2 | -19.15 |

7 | 24.26 | -21.93 |

8 | 8.64 | -10.82 |

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

Combinations | Maximum Positive(kN) | Maximum Negative(kN) |

1 | 32.98 | -32.98 |

2 | 33.53 | -30.26 |

3 | 31.31 | -31.31 |

4 | 30.26 | -33.53 |

5 | 23.09 | -31.87 |

6 | 27.48 | -27.48 |

7 | 31.87 | -23.09 |

University of Melbourne

CVEN90049 Structural Theory and Design 2

Design Project: Assignment 01

8 | 12.98 | -12.98 |

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:

Category | Positon | Calculated Values |
Maximum values in Space Gass (kNm) |
Scenario | Critcal locaton |
Diﬀerence | Percentage |

Negatve design moment |
5 m and 10 m from lef (interior support) |
-25.01 kNm | -30.26 kNm | Combi. 2 and Combi. 04 |
5 m and 10 m from lef |
-5.25 | 17.3% |

Positve design moment |
In both end spans, mid span 1 and mid span 2 |
20.68 kNm | 25.42 kNm | Combi. 3 | 2.08 m and 12.92 m from lef |
4.74 | 18.6% |

Shear force |
Va=Vd | 25.01 kN | 23.65 kN | Combi. 3 | 0 m from lef | -1.36 | 5.4% |

Actng Shear Force diagram

University of Melbourne

CVEN90049 Structural Theory and Design 2

Design Project: Assignment 01

Vb=Vc | 25.83 kN | 30.26 kN | Combi. 2 & Combi. 4 |
5 m and 10 m from lef |
4.43 | 14.6% |

Vb’=Vc’ | 28.76 kN | 33.53 kN | Combi. 2 & Combi. 4 |
5 m and 10 m from lef |
4.77 | 14.2% |

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