MAIN SOUTH ROAD/STURT ROAD...

CIV Consulting

MAIN SOUTH ROAD/STURT ROAD INTERSECTIONBridge Design Detailed Design

6/10/2014University of South Australia

CIV Consulting

CIV Consulting Mawson Lakes Boulevard Mawson Lakes SA 5095 10 June 2014

Attention: Mark Hennessy

University of South Australia School of Natural and Built Environments Mawson Lakes SA 5095

Re: Detailed Design

CIV consulting is pleased to present to you the Bridge Design for the Detailed Design for the

Main South Road & Sturt Road Intersection Upgrade.

The Detailed Design will give a detailed explanation of the Bridge Design through the entire

duration of the Main South Road & Sturt Road Intersection Upgrade with the support of

calculations of all the vital sections of the bridge.

Please feel free to contact us for any further discussion and concerns.

Yours sincerely,

Matthew PilcherProject Manager CIV Consulting

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Proprietary Information StatementThe information contained in this document produced by CIV Consulting is solely for the use of the Client

identified on the cover sheet for the purpose for which it has been prepared and CIV Consulting undertakes

no duty to or accepts any responsibility to any third party who may rely upon this document.

All rights reserved. No section or element of this document may be removed from this document,

reproduced, electronically stored or transmitted in any form without the written permission of CIV

Consulting.

Project Manager Quality Assurance Manager

Revision HistoryRevision Date Prepared Reviewed Approved0 03/6/14 CIV

CONSULTINGBRIDGE DESIGN

D. TET 　

1 05/06/14 CIV CONSULTINGBRIDGE DESIGN

J. ROGERS 　


D.TET 　

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J. ROGERS/D.TET

　


J. ROGERS 　M.PILCHER

Table of Contents

1. Bridge design.............................................................................................................................................. 1

1.1. Scope...........................................................................................................................1

1.2. Proposal.......................................................................................................................1

1.3. Final design Layout.....................................................................................................2

1.4. Design loads................................................................................................................3

1.4.1. Wind loads...........................................................................................................3

1.4.2. Earthquake loads..................................................................................................7

1.4.3. Dead loads..........................................................................................................14

1.4.4. Traffic loads.......................................................................................................15

1.4.5. Combination loads.............................................................................................16

1.5. Structural Analysis....................................................................................................18

1.5.1. Super T analysis.................................................................................................18

1.5.2. Headstock analysis.............................................................................................20

1.5.3. Column/secant pile wall analysis.......................................................................23

1.6. Super T Design..........................................................................................................26

1.7. Decking Design.........................................................................................................35

1.8. Headstock Design......................................................................................................53

1.9. Pier Design................................................................................................................70

1.10. Pile Design.............................................................................................................74

1.11. Pile Cap Design.....................................................................................................81

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1.12. Crash Barrier Design.............................................................................................90

1.12.1. Parapet............................................................................................................90

1.12.2. Pier Crash Barrier...........................................................................................93

1.13. Bearing design.......................................................................................................96

1.14. Summary................................................................................................................97

1.15. References..............................................................................................................98

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List of Figures

Figure 1: Intersection of the project...................................................................................1

Figure 2: Final layout of bridge.........................................................................................2

Figure 3: M1600 moving loads diagram..........................................................................15

Figure 4: Vehicle axles applied to bridge........................................................................18

Figure 5: Maximum bending moment (mid span of critical Super T).............................19

Figure 6: Maximum shear force (end restraints of worst-case Super T)..........................19

Figure 7– Bending moment diagram................................................................................21

Figure 8 – Shear force diagram........................................................................................21

Figure 9 - Abutment bending and shear diagram.............................................................22

Figure 10 – Short term deflection of headstock at mid-span...........................................23

Figure 11: Circular column chart (f’c =40, g= 0.8)..........................................................23

Figure 12: Bridge loading................................................................................................24

Figure 13: Loading form soil...........................................................................................25

Figure 14: Super T diagram and dimensions...................................................................26

Figure 15: Deck ultimate design moments.......................................................................36

Figure 16: Deck serviceability design moments..............................................................36

Figure 17: Loading width................................................................................................74

Figure 18: column chart f'c =40 MPa g= 0.8...................................................................76

Figure 19: Typical Elastomeric bearing from Granor......................................................96

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List of Tables

Table 1: Ultimate Design Loads......................................................................................20



Table 4: Development and Splice lengths for deformed bars.....................80

Table 5: Pile Cap Summary.............................................................................................89

Table 6: Parapet specifications.........................................................................................91

Table 7: Barrier specifications.........................................................................................94

Table 8: Specification of chosen bearing.........................................................................97

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1. Bridge design

1.1.Scope

The Bridge Design Team was accountable for the design of the bridge structure, located at

the Main South Road Sturt Road intersection in Bedford Park. As well as the bridge structure,

this team is also accountable for the pile foundations. Included in the design is the Super T

beams, the headstocks and decking for bridge. These three sections of the bridge were

analysed through SPACEGASS in order to determine a final design in terms of dimensions

and reinforcement that is needed. The pile, pier and pile cap reinforcement are also designed

from the analysis of the Super T’s and decking. Parapet’s and crash barriers are also required

for safety, which are design to Australia Standards, which have been provided. Wind and

earthquake loadings are considered and are designed accordingly.

Figure 1: Intersection of the project

1.2.Proposal

The Main South and Sturt Road underpass will be constructed of 30 Super T’s that will be

joined on the top flange using shear studs. A deck will be placed on top of the Super T’s with

parapets along the edge of the road. Due to the total distance across the underpass two spans

are required, a 5-column pier system has been implemented for the centre of the two spans.

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The bridge design section goes through the design loadings that the underpass will be

subjected to which includes wind loads, earthquake loads, dead loads and traffic loads and

well as load combinations that were used. This detailed design contains structural analysis of

the Super T beams, headstocks and pier/column analysis. Components include the design of

the Super T beam, decking, headstock, piers/columns, piles, crash barriers and the bearings

for geometrical alignment.

1.3.Final design Layout

Below is a computer-generated image of what the intersection will look like with an

underpass. It has incorporated all the relevant sections of the bridge such as the span, the 5-

column pier system that supports the centre of the bridge and the secant pile wall on each

side. The image also illustrates the parapets that sit at the top of Sturt Road and run along the

length of the bridge. The piers are protected by a series of crash barriers to prevent damage to

the piers in the event of an accident. Each of these components of the bridge will be covered

in the later sections of the report.

Figure 2: Final layout of bridge

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Project Title: Main South Road / Sturt Road Underpass– Detailed Design

Subject: Wind LoadsJob Number: 01 Contract: Sturt Road Bridge

DesignDate: 05/06/2014 Prepared: D TetSheet: 01 of 04 Checked: J NgoClient: DPTI Approved: XXXX

1.4.Design loads

Wind loads are an important design aspect of the structural design that needs to be considered

when designing for a lateral load. A comparison is to be done with the earthquake loads to

determine which lateral load is greater out of the two. As the structure itself is symmetrical,

the greater of the two loads need designed for as the critical case. The following two sections

go into further depth the loads due to wind and earthquakes.

1.4.1. Wind loads

Winds loads are designed with accordance with AS 5100.2. Cl 16.2.1 states that the design

wind speed shall be derived by the appropriate regional basic design wind speeds, taking into

account average return interval (Vr), geographical location, terrain category, shielding (Ms)

and height above ground (Mz,cat). The average return interval is given in Cl 16.2.2 for ultimate

and serviceability limit states. The other factors are specified in AS 1170.2.

Design wind speed

The average return interval that is to be adopted for ultimate limit sites is 2000 year, which is

given as 44 m/s. For serviceability limit states, what should be adopted is 20 years, which is

37 m/s. These values are given in AS 1170.2, Table 3.1. The geographical location of the

underpass is region A so all the following values are in correspondence with this region. The

terrain category that corresponds with the intersection is 3. From this, the height multiplier

can be determined from table 4.1 in AS 1170.2. The height of the structure is 8 m, which

gives a value of 0.83. The shielding multiplier of the structure is an underpass and Main

South Road is below grade whereas Sturt Road is on grade. The shielding multiplier is given

as 1, as of in table 4.3 in AS 1170.2.

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DesignDate: 05/06/2014 Prepared: D TetSheet: 02 of 04 Checked: J NgoClient: DPTI Approved: J Rogers

V ultimate=V r × M z , cat × M s

¿48 × 0.83× 1

¿40 m / s

V serviceablility=V r× M z , cat × M s

¿37×0.83 × 1

¿32m /s

For serviceability, Cl 16.2.2 in AS 5100.2 states that this value is too low meaning that the

design speed shall be taken as 35 m/s instead.

Transverse wind load

As given in Cl 16.3.1 of AS 5100.2, the transverse wind load is taken as if it were acting

horizontally at the centroids in the areas which are appropriate.

Ultimate design transverse wind load (W ¿tu)=0.0006 V u

2 A t Cd

Serviceability designtransverse wind load (W ¿ts)=0.0006 V s

2 A t Cd

Where Vu is the design wind speed for ultimate limit states, Vs is the design wind speed for

serviceability states, At is the area of the structure for the calculation of wind load and Cd is

the drag coefficient.

In order to work out At, Cl 16.3.2 (a) in AS 5100.2 is followed as the superstructure has solid

parapets. The area that the wind will act on the bridge will the length of the bridge on Main

South Road, 43 m multiplied by the surface to surface level including the parapets. This is

approximately 9.3 m (8 m + 1.3 m).

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At=length ×d epth

¿43× 9.3

¿400 m2

Cl 16.3.3 (a) in AS 5100.2 is followed in order to determine the Cd for the structure as well as

figure 16.3.3. This means working out the b/d ratio of the underpass, where b is the overall

width of the bridge between outer faces of parapets, which is 30.5 m, and depth of the

structure which has been previous stated

bd

ratio=30.59.3

¿3.3

Using figure 16.3.3, the drag coefficient is approximately 1.41.

W ¿tu=0.0006 × 402× 400 ×1.41

¿537 kN

W ¿ts=0.0006 ×352 × 400× 1.41

¿414 kN

Longitudinal wind load

In order to work out the longitudinal wind load, Cl 16.5 is to be followed for ultimate and

serviceability limit states.

Ultimate design vertical wind load (W ¿vu)=0.6 V u

2 A pCL 10−3

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Serviceability design vertical wind load (W ¿vs )=0.6 V s

2 A pCL 10−3

The above equations can be used as the inclination of the bridge is less than 5°. Regardless,

the modelling of the bridge will be done in a flat plan for simplicity which is further

explained in section 1.5.1 under Super T Analysis. Ap is the bridge area in plan and CL is the

lift coefficient, which is provided as 0.75. Ap is the length of the bridge multiplied by how

wide it is. The length is 43 m and the width is 30.5 m.

Ap=43 ×30.5

¿1312 m2

W ¿vu=0.6 V u

2 Ap CL 10−3

¿0.6× 402× 1312× 0.75 ×10−3

¿937 kN

W ¿vs=0.6 V u

2 Ap CL 10−3

¿0.6×352×1312 ×0.75 ×10−3

¿723kN

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1.4.2. Earthquake loads


Subject: Earthquake loadsJob Number: 01 Contract: Sturt Road Bridge

DesignDate: 05/06/2014 Prepared: C LiSheet: 01 of 07 Checked: J NgoClient: DPTI Approved: D Tet

For the design of bridge structures, earthquake effects shall be considered in accordance with

this Australian Standard AS5100.2-2004. Some factors to be used in the calculation of

earthquake effects which are in common with those given in AS1170.4 earthquake actions in

Australia.

1. The bridge classification: this bridge is classed as a Type 2 bridge: (that is designed to

carry large volumes of traffic or bridges over other roadways, railways or buildings)

2. The product of acceleration coefficient.

3. The site factors.

Acceleration coefficient:

a=0.1 in the Adelaide area: from table 2.3 (AS1170.4:1993)

Site factor: S=1.25 comprised of mainly stiff or hard clays or controlled fill from table 2.4 (a)

(AS1170.4:1993)

Hence:

a × S=0.1×1.25=0.125

And the bridge earthquake design category from table 14.3.1(AS 5100.2:2004) is BEDC-2

Requirements for category BEDC-2:

Static bridge analysis

Consider both horizontal and vertical earthquake forces

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Subject: Earthquake loadingJob Number: 01 Contract: Sturt Road Bridge


Static analysis: Horizontal

The formula for horizontal forces is as described in 14.5.2 (AS5100.2:2004) and below

HU¿ =I (CS

R f )Gg

With the limits-

H U¿ ≥0.02 Gg

HU¿ ≤ I ( 2.5 a

Rf)Gg

Where:

- I= Importance factor=1 because it is a type 2 bridge from table 14.5.3 (AS5100.2)

- S= Site factor =1.25 (calculated above)

- R f = Structural response factor=3 from table 14.5.5 (AS5100.4:2004 Bridges with

simply supported spans)

- Gg= total un-factored dead load (calculated below)

- C= Earthquake design coefficient as described in 14.5.4 (AS5100.2 :2004) and below

For the earthquake design coefficient C, the equation is given as below:

C=1.25 a

T23

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Where:

a=0.1 (From AS1170.4:1993)

T=0.063√δ

Where:

δ= 5WL4

384 EI

Where:W =Serviceability load combinations=1.3 Q+1.3 G=336.38 kN /m

Where: L=Span=24m (one simply supported span in total 24 meters)

E=32800 (40 Mpa concrete)

I decking=bd3

12=160∗305003 /12=3.78× 1014 mm4

I super T=1.97 × 1011mm4 (Table H 2 ( B ) (2 ) AS 5100.5−2004)

I total=I decking+ I superT=3.78 ×1014+1.97 × 1011=3.78 ×1014m m4

δ= 5 wl4

384 EI= 5× 336.38× 240004

384 × 32800× 3.78× 1014 =0.12 mm (L

250=176 mm)

Gg= total unfactored dead load=Gdeck+G superTee+Gpier+Gheadstock (calculated below)

Gdeck=0.16 ×24× 30.5× 24=2810.88 kN

Gsuper T=19.6 kN /m ×15×24=7056 kN

G pier=24 ×1 m×1m× 6m× 5=720 kN

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Gheadstock=Gmiddleheadstock+Gside headstock

Gmiddle headstock=24×3.5× 1× 22=1848 kN

Gside headstock=24 ×0.4 × 1× 22× 1=211.2kN

Gheadstock=1848+211.2=2059.2kN

∴Gg=2810.88+7056+720+2059.2=12646.08 kN

T=0.063√δ=0.063√0.12=0.022 seconds

Hence, C=1.25 a

T23

=1.25 × 0.1

0.0223

=1.6

Hence

H Hor¿ =I (CS

R f)G g

HU¿ =1 ×( 1.6× 1.25

3 )×12646.08=8430.72kN

Because

HU¿ =8430.72kN ≤ I( 2.5 a

R f )G g=1 × 2.5 ×0.13 × 8430.72=702.56 kN

∴Hhor¿ =702.56 kN=29.27 kN /m

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Static analysis: Vertical:

The formula for horizontal forces is as described in 14.5.2. (AS5100.2:2004) and below

H vert¿ =I ( CS

R f )Gg

Where:

I= Importance factor=1 because it is a type 2 bridge from table 14.5.3 (AS5100.2)

S=site factor=1.25 (calculated above)

R f=Structural response factor=3 (calculated above) from table 14.5.5 (AS5100.4:2004)

Gg= total un-factored dead load=Gdeck+G superTee+G pier+Gheadstock (calculated below)

C= Earthquake design coefficient as described in 14.5.4 (AS 5100.2:2004) and below

C=1.25 a

T23

For the earthquake design coefficient C, the equation is given as below:

C=1.25 a

T23

Where:

a=0.1 (From AS1170.4:1993)

T=0.063√δ

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DesignDate: 05/06/2014 Prepared: C LiSheet: 06 of 07 Checked: J NgoClient: DPTI Approved: D TetWhere:

δ= 5 WL4

384 EI

W =Serviceability load combinations=1.3 Q+1.3 G=336.38 kN /m

Where: L=Span=24m (simple supported span in total 24 meters)

E=32800 (40 Mpa concrete)

I decking=bd3

12=30500× 1603/12=1.04 ×1010mm4

I super T=1.97× 1011mm4 (Table H 2 ( B ) (2 ) AS 5100.5−2004)

I total=I decking+ I superT=1.04 × 1010+1.97 × 1011=2.07× 1012m m4

δ= 5 wl4

384 EI= 5× 336.38× 240004

384 × 32800× 2.07 ×1012 =22.15 mm (L

250=96 mm)

Gg= total unfactored dead load=Gdeck+G superTee+Gpier+Gheadstock (calculated below)

Gdeck=0.16 ×24× 30.5× 24=2810.88 kN

Gsuper T=19.6 kN /m ×15×24=7056 kN

G pier=24 ×1 m×1m× 6m× 5=720 kN

Gheadstock=Gmiddleheadstock+Gside headstock

Gmiddle headstock=24×3.5× 1× 22=1848 kN

Gside headstock=24 ×0.4 × 1× 22× 1=211.2kN

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Gheadstock=1848+211.2=2059.2 kN

∴Gg=2810.88+7056+720+2059.2=12646.08 kN

T=0.063√δ=0.063√22.15=0.3 seconds

Hence, C=1.25 a

T23

=1.25 × 0.1

1.9723

=0.28

H vert¿ =I ( CS

R f )Gg

H❑¿ =1 ×( 0.28× 1.25

3 )×12646.08=1481.3 kN=61.72 kN /m

However, according to section 14.5.6 of (AS5100.2:2004) the vertical earthquake force shall

not be less than 50% of the maximum horizontal earthquake force in either direction. And the

vertical earthquake force could be considered independently of the horizontal earthquake

forces.

∴H vert¿ =61.72 kN /m

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Subject: Dead LoadsJob Number: 01 Contract: Sturt Road Bridge

DesignDate: 29/05/2014 Prepared: A GreigSheet: 01 of 01 Checked: J NgoClient: DPTI Approved: D Tet

1.4.3. Dead loads

Dead Load of Structure

AS 5100.2 Cl 5 defines dead loads as two separate quantities. The first being the dead load of

the structure, this takes into account the self-weight of all the structural elements and any

non-structural elements that will unlikely change during construction and use of the structure.

In this case, the dead loads of the structure contain:

- Super T self-weight (Calculated by software, SPACE GASS)

- Decking self-weight: 7.68 kN/m, load width used was 2m which is the width of a

super T (defined in Decking Design, section 1.7)

- Parapet self-weight: 13kN/m (defined in Parapet, section 1.11.1)

The load factor for the dead load of the structure can be taken for ultimate design and

serviceability can be taken from AS5100.2 T 5.2. In this case:

- Ultimate Design: 1.2G

- Serviceability: 1G

Super imposed Dead Load

The second dead load to consider is the super imposed dead load which is the self-weight of

all the non-structural elements that can be removed. In this case the superimposed dead loads

are:

- Bitumen (24.5 kN/m3, given from Earthworks/ Road works Department)

- Services (1 kPa)

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Subject: Traffic LoadsJob Number: 01 Contract: Sturt Road Bridge


1.4.4. Traffic loads

AS5100.2 Cl 6.2.3 defines the moving traffic loads that will be the most critical case for the

Sturt Road Bridge. The vehicle length has been taken as 25m.

Figure 3: M1600 moving loads diagram

Figure 2 was extracted from AS5100.2 Fig. 6.2.4. It shows the loads that are applied onto the

bridge including the moving point loads, UDLs and spacing between axle loads.

In this case, all the modelling was completed on SPACE GASS where the software already

had this traffic loads readily available to use.

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Subject: Combination LoadsJob Number: 01 Contract: Sturt Road Bridge


1.4.5. Combination loads

The load factor for the dead load of the structure can be taken for ultimate design and

serviceability can be taken from AS5100.2 T 5.3. In this case:

- Ultimate Design: 2G

- Serviceability: 1.3G

The load factor for the traffic loads will need to be calculated. This can be found in AS1500.2

Cl. 6.7.2, “Dynamic Load Allowance, Magnitude.”

(1+α )× theload factor× the actionunder consideration

The Dynamic Load Allowance (α ) can be taken from AS5100.2 T6.7.2. For an M1600 load, it

is given as 0.3.

The load factor can be given from AS5100.2 T6.10 (A). For an M1600 moving traffic load,

the values can be taken as:

- Ultimate: 1.8

- Serviceability: 1

Hence the ultimate limit state can be given as:

Ultimate Traffic Load Factor= (1+0.3 )× 1.8

Ultimate Traffic Load Factor=2.34

The serviceability state can be given as:

Serviceability Traffic Load Factor=(1+0.3 )× 1

Serviceability Traffic Load Factor=1.3

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Subject: Combination LoadsJob Number: 01 Contract: Sturt Road Bridge


All load factors and loads have been applied into SPACE GASS to analyse the moving loads

under different positions on the bridge to account for the worst possible case which will be

detailed in section 1.5.1, Super T Analysis.

A UDL was given for the traffic loads and this was applied to each of the super T’s.

Pedestrian loads were not considered due to the traffic load giving the worst case. Hence all

the Super T’s have the same reinforcement and the worst case of the vehicle would govern

the worst case of pedestrian.

The load factor can be given from AS5100.2 T6.10 (A). For an M1600 moving traffic load,

the values can be taken as:

- Ultimate: 1.8

- Serviceability: 1

Hence the ultimate limit state can be given as:

Ultimate Traffic Load Factor= (1+0.3 )× 1.8

Ultimate Traffic Load Factor=2.34

The serviceability state can be given as:

Serviceability Traffic Load Factor=(1+0.3 )× 1

Servi ceability Traffic Load Factor=1.3

All load factors and loads have been applied into SPACE GASS to analyse the moving loads

under different positions on the bridge to account for the worst possible case, which will be

detailed in section 1.5.1, Super T Analysis.

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1.5.Structural Analysis

1.5.1. Super T analysis

In order to do the structural analysis of the Super T’s SPACE GASS was used as the

modelling software. Figure 3 shows the rendered version of the super T’s with the vehicle

loadings being applied to the bridge at an arbitrary point in time (may not be worst case). The

load cases defined in Dead Loads (section 1.4.3) and Traffic Loads (Section 1.4.4) have all

been inputted into SPACE GASS to give an ultimate result to begin design calculations.

Figure 4: Vehicle axles applied to bridge

From this analysis, the maximum bending moment on each Super T can be found as well as

all the shear forces and reaction forces that can be taken into account for the load transfer to

the headstock. The connection between the headstock and Super T is taken as a pinned

connection as transferring moment is not a good option on bridges.

When the analysis was run, a total of 186 different load cases due to the moving axle

locations have been defined and enveloped to find the worst-case scenario. The analysis

showed that the fourth Super T from the edge is the most critical which produces the worst-

case scenario for bending moment. Figure 5 shows the maximum bending moment at mid

span of the fourth super T in from the edge. The grey lines to the left and right of the bending

moment diagram are the other super T’s which have been filtered out to clearly show the

ultimate bending moment.

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Figure 5: Maximum bending moment (mid span of critical Super T)

Figure 6 displays the shear force of the worst-case Super T from the edge that has the

maximum negative shear force.

Figure 6: Maximum shear force (end restraints of worst-case Super T)

Table 1 displays all the critical cases that need to be considered for design. It is important to

note that the load cases have been enveloped to give the worst case on every situation so table

1’s results may not be at equilibrium:

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Table 1: Ultimate Design Loads

Ultimate Case Load/MomentMoment 4775 kNmPositive Shear Force at Abutment 1542 kNNegative Shear Force at Abutment -1589 kNReaction at Abutment 1543 kNPositive Shear Force at Columns 1305 kNNegative Shear Force at Columns -1444 kNReaction at Column 2320 kN

Since all the Super T’s have been tied together properly (similar during construction from the

use of connectors) in the analysis model, the load distribution between every super T was

even so all the ultimate moments and shear forces were very similar. This means there was an

even load distribution due to the decking, so the design for all components can be taken from

the ultimate values from table 1.

Since there was such great load distribution between the Super T’s the shear force at one end

of the super T may not equal the shear force at the end of the restraint as displayed in table 1.

1.5.2. Headstock analysis

Modeling’s of the headstocks at each abutment and at mid-span were undertaken using

SPACE GASS structural analysis software. The loads acquired from structural analysis of the

Super T Bridge (section 1.5) were applied directly to simulate the loading conditions. These

factored loads were applied as concentrated loads at the location of each super T rather than

applying a conservative UDL to the headstocks. Structural analysis provides the critical

bending moments, shear forces, support reactions and displacements required for design

purposes.

Headstock at mid-span

The mid-span headstock was modeled as a continuous beam supported by pinned columns to

ensure no moment was transferred to the pile cap beneath. As a result traditional bending

moment and shear force diagrams were produced as shown in figures 7 and 8. The design

actions can be seen in Table 2.

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Figure 7– Bending moment diagram

Figure 8 – Shear force diagram


Ultimate Case Load / MomentPositive moment at support 7194 kNmNegative moment at mid-span 10955 kNmShear at support 6546 kNReaction at column/pier (Axial load) 10778 kN

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Headstocks at abutments

Both abutment headstocks share the same design; this was modeled as a continually

supported beam, however the secant pile wall support had spring restraints to simulate the

effects of bearing on soil. Due to the concentrated loads from the Super T Bridge this created

a series of small bending moments and shear surrounding each Super T. Figure 9 shows a

typical pile sitting beneath the transferred load, which governed the design. The design

actions can be seen in Table 3.

Figure 9 - Abutment bending and shear diagram


Ultimate Case Load / MomentPositive moment 261 kNmNegative moment 110 kNmShear force 784 kN

Deflection

Short-term deflection was analyzed for each headstock. Figure 9 shows the maximum short-

term deflection incurred by the headstock at mid-span. This value is well below the

recommended maximum of 25mm. The headstocks at each abutment experience no

deflection due to the nature of their supports.

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Figure 10 – Short term deflection of headstock at mid-span

1.5.3. Column/secant pile wall analysis

To design the reinforcement of the pile wall supporting the bridge, column charts are selected

as it is a more effective and simpler way to design it. Several column charts have been

considered as shown in figure 11 below.

Figure 11: Circular column chart (f’c =40, g= 0.8)

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According to different specifications of the column that for example, concrete strength f’c

and the ratio of the reinforcement depth, a different column is applied. X axial locates the

moment capacity, and the Y axial locates the axial force capacity. The reinforcement ratio

lines define the area that is within the safety of design the loading. Each of the columns have

been designed with a concrete strength (f’c) of 40 MPa.

Loadings

Figure 12: Bridge loading

Figure 12 above shows the indicated axial load from the bridge. The first and last loadings

are given as a UDL (universally distributed load) due to it being a pile wall. The middle

supports are five piers that to be designed as column structure. The load has been taken 6500

kN as this is the worse design situation of the five piers.

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Figure 13: Loading form soil

The loading from the soil is only applied to the pile wall as they are soil pressure of the soils

adjacent to the underpass. The soil pressure has been converted as single shear force with

height aapproximately 2.9 m for design purpose.

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1.6.Super T Design


Subject: Super T DesignJob Number: 01 Contract: Sturt Road Bridge

DesignDate: 05/06/2014 Prepared: D TetSheet: 01 of 09 Checked: A GreigClient: DPTI Approved: J RogersThrough the analysis of the Super T beams in the program “SPACEGASS” in section 1.5.1,

the values obtained can be used in order to design the reinforcement as well as the

prestressing that will be required for the design. This table of values have been given in Table

1 of section 1.5.1 (Super T Analysis). The Super T will be designed as a fully prestressed

section. Below, Figure 14 is the dimension of the Super T, which is a national template. The

dimensions correspond with properties of a T4 Super T. The diameter of the strands has been

chosen as 15.2 mm.

Figure 14: Super T diagram and dimensions

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DesignDate: 05/06/2014 Prepared: D TetSheet: 02 of 09 Checked: A GreigClient: DPTI Approved: J Rogers

An initial ductility value (ku) is taken, which is generally less than the limit, 0.36. The general

range is normally between 0.15-0.25. 0.2 will be used for ku. Using the follow process below

will determine if this ductility chosen is sufficient.

b d2 ≥ M ¿

∅ α2 f 'c γ ku(1−

γ ku

2 )

Where:

b = 2000 mm, D = 1660(1500mm super T depth plus 160mm of decking) mm, cover = 50 mm, d = D – dc –(15.2 mm diameter strand/2)= 1552.4 mm, ∅=0.8. F’c is given as 40 MPa. α 2=1−0.003 f '

c

¿1−0.003× 40

¿0.88

As this is larger than 0.85, 0.85 is used instead from Cl 8.1.3 (b) in AS3600.

γ=1.05−0.007 f 'c

¿1.05−0.007 ×40

¿0.77

∴2000 ×1552.42≥ 4775 ×103

0.8× 0.85 × 40× 0.77×0.2 ×(1−0.77×0.252 )

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4.8 × 109≥ 1.2 ×109 ∴ sufficent




In order to determine the amount of prestressing that will be required, the stresses at the top

and bottom of the Super T can be used in order to work out the initial prestressing force, p i.

Generally, the top tensile stress at transfer shouldn’t exceed 0.6√40¿3.79 MPa) and the

bottom tensile stress under full load shouldn’t exceed 0.6√40¿3.79 MPa). The section

modulus of the Super T is given below. These values are from AS 5100.5 appendix H table

H2 (B) (2). The only additional part is for the top which includes the decking section

modulus.

Ztop=Z top for Super T+Z for Decking

¿268 ×106+2000 × (160 )2

6=2.77 ×108mm4

Zbottom=257.8 × 106 mm4

σtop=nPiAg

−n PieZtop

+ MZtop

σbottom=nPiAg

+ n Pi eZ bottom

+ MZbottom

Where n=1, Ag has been given in AS 5100.5 Table H2 (B) (2) which is 627.1 x 103 mm2 plus

the area of the deck which is 320 x103 mm4 and M is M*. This means Ag is calculated to be

947.1x103 mm4. The neutral axis of the beam is calculated from the equation below. It is the

summation of the areas multiplied by the centroid of each one divided by the total area of the

section

NA=∑ A × yA total

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DesignDate: 05/06/2014 Prepared: D TetSheet: 04 of 09 Checked: A GreigClient: DPTI Approved: J RogersFrom this equation, NA is found to be 566.66 mm.

e=NA−cover=566.66−50=516.66 mm

Calculating Pi using Ztop.

−3.79= 1 × Pi947.1 ×103 −

Pi×516.662.77 ×108 + 4775 ×106

2.77 ×108

From simply rearranging the equation, Pi can be solved. Pi is calculated to be 16627.22 kN.

For the bottom stress, n is given as 0.8 due to losses.

−3.79= 0.8 × Pi947.1 ×103 + 0.8× Pi×516.66

257 ×106 + 4665 ×106

257 ×106

Once again through rearranging, Pi is calculated to be 1548.21 kN. The higher of the two

values are selected in order to choose the prestressing to be used.

Jacking force , Pe=0.8 × Pi=13299.38 kN

Taking into account friction in the ducts, the jacking force is given as

P jj=Pe

0.943=13299.38

0.943=14103.27 kN

The strands are limited to 80% of their breaking loading meaning,

Ppu=14103.27

0.8=16629.08 kN

Total strands=Ppu

Jacking forceof 15.2 mmdiameter strand

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¿ 16629.08212

=83.18 strands

Adopt 84 strands at 15.2mm diameter

A check is done to see if any reinforcement is required to be added for Mu.

M u=Apt σ pu z p

σ pu=f pb(1−k1k 2

γ)

K1 is given as AS 3600 Cl 8.1.7 for bonded tendons. Each strand has a nominal area of 143

mm2. Fpb is given as 1750 MPa.

k 2=1

b d p f 'c

( Apt f pb)

¿ 12000× 1552.4 f '

c

(84 × 143 ×1750 )

¿0.17

σ pu=1750(1−0.4 × 0.170.77 )

¿1596.88 kN

Taking account 20% of losses, fpy is given as 1357.34 kN.

z p=1552.4 (1−0.77 × 0.2× 12)

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¿1432.87 mm




M u=Apt σ pu z p

¿(84 ×143)× 1596.88 ×1432.87

¿22551 kNm

As it is greater than 5965.75 kN (M*/0.8), it is sufficient. The amount of strands can be

decreased by half in order to have a reasonable amount (Mu=13742.45 kNm)

Adopt 42 strands at 15.2mm diameter as it is sufficient

Web shearing

As this is a Super T, web shearing will need to be looked at as opposed to flexural shear as

this will create the most critical case.

V uc=V t+PV

Pv=Pe × 4 hL

¿6006 ×( 4× 516.6624000 )

¿613.38 kN

In order to work out Vt, the process of using Mohr’s circle is used

The second moment of inertia of the Super T is to be calculated, as shown below. I is given

for the Super T in AS 5100.5 Table H2 (B) (2) as 197320 x 106 mm4). I for the deck is to be

calculated

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I deck=2000 ×1602

12




¿6.82 ×108 mm4

I g=197320 ×106+6.82× 108

¿1.98 ×1011mm4

σ cx=MyI

−Pe

Ag−

Pe eyI

¿ 4775 ×106

1.98× 1011−7123 ×103

947100−7123× 103× 516.66× 566.66

1.98× 1011

¿−18.03 MPa

τ xy=V t ×Q

I bv

Where Q is shown below

Q= (2000× 200 )×(566.66−( 2352 ))+(757 ×(566.66−( 235

2 )2 ))

¿2.11×108m m3

The following two equations are used to work out Vt

τ xy=V t ×Q

I bv

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Where bv is equal to 577 mm.



DesignDate: 05/06/2014 Prepared: D TetSheet: 08 of 09 Checked: A GreigClient: DPTI Approved: J RogersThe following two equations are used to work out Vt

τ xy=V t ×Q

I bv

Where bv is equal to 577 mm.

σ c1=0.36√ f 'c=√ (σcx 0.5 )2+ τxy

2+0.5 σ cx

σ c1=0.36√40=√(−18.03 ×0.5 )2+( V t ×2.11×108

1.98× 1011×577 )2

+0.5× (−18.03 )

Through rearranging the equation, Vt can be solved which calculated to be 3667 kN.

∴V uc=3667+613.38=4280.4 kN

∅ 0.5V uc=0.7× 0.5 ×4280.4=1498.14 kN

As this value is larger less the worst case V* as shown in section 1.5.1, this means that Vu.min

needs to be calculated.

V u , min=V uc +0.10√ f 'c bv do≥ V uc+0.6 bv do

¿4465.24+0.10√40 × .577 × .1425≥ 4465.24+0.6 × .577 × .1425

¿Φ 4800 kN ≥Φ 4280.9 kN

¿3360.3 kN

As ΦVu, min is greater than V*, minimum area of shear steel used to determine it.

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A sv , min=0.06√ f 'c bv(

sf sv ,min

)≥+0.35 bv(s

f sv , min)




A sv ,min

s=0.06√40 ×(577

500 )≥+0.35×( 577500 )

¿1.52 mm2/mm

Using either 0.75D or 500 as the spacing, try 2N24 @ 500 cts

A sv ,min

s=2 × 450

500

¿1.8mm2/mmok but can go smaller

Try 6N12 @ 400 cts

A sv ,min

s6× 110

400

¿1.65 mm2/mm

Adpot 6N12 ligs @400 cts

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1.7.Decking Design


Subject: Decking DesignJob Number: 01 Contract: Sturt Road Bridge

DesignDate: 05/06/2014 Prepared: M EaSheet: 01 of 18 Checked: A GreigClient: DPTI Approved: D TetLoad combinations considered are as followed:

Ultimate limit state combination

PE + ultimate traffic load

PE + ultimate pedestrian traffic loads

PE + ultimate wind loads

PE + earthquakes

Serviceability limit state combination

PE + ultimate traffic load + 0.7*ultimate pedestrian traffic loads

PE + ultimate traffic load + 0.7*ultimate pedestrian traffic loads + 0.5*ultimate wind load

PE + ultimate traffic load + 0.7*ultimate pedestrian traffic loads + 0.5*earthquakes

SPACEGASS Analysis

The deck is designed as a one-way slab transversely across the width of the bridge by having

the Super T beam web underneath as supports. The distance between the supports is 1m. A

variety of positions for truck wheels are placed as eccentrically as possible to give adverse

effects. The loadings applied here is the same as the loadings used for Super T analysis

which is complied to AS5100.2.

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DesignDate: 05/6/2014 Prepared: M EaSheet: 02 of 18 Checked: A GreigClient: DPTI Approved: D Tet

Figure 15: Deck ultimate design moments

Figure 16: Deck serviceability design moments

Bending Strength

Ultimate: 𝑀∗=26.505𝑘𝑁𝑚 𝑜𝑟 𝑀∗=−14.696𝑘𝑁𝑚xlii

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Serviceability: 𝑀∗=14.818𝑘𝑁𝑚 𝑜𝑟 𝑀∗=−12.480𝑘𝑁𝑚𝑓′c = 40𝑀𝑃𝑎 𝑓sy = 500𝑀𝑃𝑎𝐸c=32.8𝐺𝑃𝑎 𝐸s=200𝐺𝑃𝑎Project Title: Main South Road / Sturt Road Underpass– Detailed Design


DesignDate: 05/6/2014 Prepared: M EaSheet: 03 of 18 Checked: A. GreigClient: DPTI Approved: D Tet

Exposure Classification: B1, Minimum cover: 40mm𝐷=160𝑚𝑚, Try N12 bars top and bottom𝑑 = 160−40−12/2 = 114𝑚𝑚Design for 1m wide strip 𝛼2 = 0.85

γ=0.85−0.007 (f c '−28)

γ=0.85−0.007 ×(40−28)

γ=0.766

f cf' =0.6√ f c

'

f cf' =0.6 ×√40

f cf' =3.79

Minimum design bending moment:

According to AS5100.2, clause 8.1.4.1, the ultimate strength in bending (Muo) shall not be less

than (Muo)min cacculated below.

M uo ,min=1.2[Z ( f cf ’+ PA s )+P e ]

Where Z = bD2

6 = 1000∗1602

6 = 4.27 × 106 mm3

M uo ,min=1.2 ×[4.27× 106 ×(3.79+0)+0]

M uo ,min=19.4 kNm

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A s tmin=0.0025 bd

A s tmin=0.0025 ×1000 ×114

A s tmin=285 mm2




Top Reinforcement (Negative)

M uo=14.70.8

=18.37 kNm< M uo , min

M uo ,min=19.4 kNm

Z ≈ 0.925 d=0.925× 114

Z=105.45 mm

M u o=A st f sy Z

A st=M uof sy Z

A st=19.4 ×10 6

500× 105.45=368 mm 2

Try N 12 bars at 275 cts(400 mm2/m)

T=A st f sy

T=440× 500

T=200 kN

C=T=0.85 f c ’ γk udb

ku=200× 103

0.85× 40×0.766 × 1000× 114

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ku=0.067<0.36, ductility is OK




M uo=200 ×(0.114−0.5 × 0.766 ×0.067 ×0.114 )

M uo=22.2 kNm

ϕ M uo=0.8 ×22.2=17.8 kNm

ϕ M uo>M∗¿14.7 kNm¿)

Adopt N12 bars at 275cts

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DesignDate: 05/06/2014 Prepared: M EaSheet: 06 of 18 Checked: A GreigClient: DPTI Approved: D TetBottom Reinforcement (Positive)

M u o=26.5050.8

=33.1 kNm

Z ≈ 0.925 d

Z=0.925×114

Z=105.45 mm

M uo=A st f sy Z

A st=M uof sy Z

¿ 33.1 ×106

500× 105.45=628 mm2

Try N 12 bars at 175 cts(629 mm2m )

T=A st f sy

T=629 ×500

T=315 kN

C=T=0.85 f c ’ γk udb

ku=315× 103

0.85× 40×0.766 × 1000× 114

ku=0.106<0.36, ductility is OK

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M uo=315 × (0.114−0.5 ×0.766 × 0.106× 0.114 )

Muo =34.5𝑘𝑁𝑚ϕ M uo=0.8∗34.5=27.6 kNm

ϕM uo>M∗¿26.5 kNm (Ok)Use N12 at 275cts top and N12 at 175cts bottom in primary direction

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DesignDate: 05/06/2014 Prepared: M EaSheet: 08 of 18 Checked: A. GreigClient: DPTI Approved: D TetCrack Control

Crack Control for Flexure (AS5100.5 clause 9.4.1)

The centre-to-centre spacing of bars in each direction shall be not greater than the

lesser of 2.0Ds or 300mm. Maximum spacing = min(2*160, 300) = 300mm.

The steel stress (fscr) calculated shall not exceed the maximum steel stress given in

Table 9.4.1(A). From this table, maximum Steel Stress, 𝑓scr = 295𝑀𝑃𝑎.

Calculate stress in steel assuming section is cracked, assuming top steel is in tension

Positive

n=Es

E c

n= 20032.8

n=6.1 ≈6.5

A st 1=550 mm2

A st 2=400 m m2

b×d❑n

2

2+ (n−1 ) A st 2(d❑n−d2)=n A st 1(d1−dn)

1000 ×d❑n

2

2+(6.5−1)∗400∗(dn−46)=6.5∗550∗(114−dn)

500 dn2+2200 dn – 101200=407550 – 3575 dn

500 dn2+5775 dn−306350=0

dn = 19.6mm above top steel

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I cr=b dn

3

3+(n−1 ) A st 2 ( dn−d2 )2+n A st 1 (d1−dn )2

= 1000*(19.6)3/3 + (6.5-1)*400*(19.6-46)2 + 6.5*550*(114-19.6)2

I cr=1000× (19.6 )3

3+(6.5−1 )× 400 × (19.6−46 )2+6.5× 550× (114−19.6 )2

I cr=33× 106 mm4

σ st 1=M yI cr

σ st 1=14.818× (46−19.6 )

33×106

σ st 1=11.9 MPa<295 MPa Ok

σ st 2=MyIcr

σ st 2=14.818× (114−19.6 )

33 ×106

σ st 2=42.4 MPa<295 MPaOk

The calculated steel stress (fscr) shall not exceed 0.8fsy.

0.8fsy = 0.8*500 = 400MPa

As above, 𝜎st1=11.9𝑀𝑃𝑎<400𝑀𝑃𝑎 (Ok)𝜎st2=42.4𝑀𝑃𝑎<400𝑀𝑃𝑎 (Ok)

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Negative

n=Es

E c

n= 20032.8

n=6.1≈6.5

A st 1=400 m m2

A st 2=550 mm2

b ×d❑n

2

2+ (n−1 ) A st 2(d❑n−d2)=n A st 1 (d1−dn )

1000 ×dn2

2+(6.5−1 )× 550×(d❑n−46)=6.5∗400∗(114−dn)

500 dn2+3025 dn – 126500=296400 – 2600 dn

500 dn2+5625 dn−422900=0

dn = 24mm above top steel

I cr=bd n3

3+ (n−1 ) A st 2 (dn−d2 )2+n A st 1 (d1−dn )2

= 1000*(24)3/3 + (6.5-1)*550*(24-46)2 + 6.5*400*(114-24)2 = 23*106 mm4

I cr=1000 × (24 )3

3+ (6.5−1 ) ×550 × (24−46 )2+6.5 ×400 × (114−24 )2

I cr=23 ×106 mm4

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σ st 1=M yI cr

σ st 1=12.480× (46−24 )

23× 106

σ st 1=11.9 MPa<295 MPa(Ok)

σ st 2=MyIcr

σ st 2=12.480× (114−24 )

23 ×106

σ st 2=48.8 MPa<295 MPa (Ok)

The calculated steel stress (fscr) shall not exceed 0.8fsy.

0.8fsy = 0.8*500 = 400MPa

As above, 𝜎st1=11.9𝑀𝑃𝑎<400𝑀𝑃𝑎 (Ok)𝜎st2=48.8𝑀𝑃𝑎<400𝑀𝑃𝑎 (Ok)

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Crack Control for Shrinkage and Temperature (AS5100.5 Clause 9.4.3)

Minimum steel met in primary direction.

Steel required in secondary direction:

(6.0−2.5𝜎cp )𝑏𝐷×10-3=(6−0)×1000×160×10-3 =960𝑚𝑚2

(6.0−2.5 σcp ) bD× 10−3

¿ (6−0 ) ×1000 ×160 ×10−3

¿960 m m2

Use N12 at 225cts top and bottom (978mm2/m) in secondary direction

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DesignDate: 5/6/2014 Prepared: M EaSheet: 13 of 18 Checked: A GreigClient: DPTI Approved: D TetEffects of concentrated load

Punching Shear

According to AS5100.2, clause 6.2.1, the W80 wheel load consists of an 80kN load

uniformly distributed over a contact area of 400mm x 250mm.

N z¿=80 kN × Ultimate Load Factor × (1+α )

N z¿=80×1.8 × (1+0.4 )

N z¿=202 kN

M v¿=wa (1−a )

Ln

M v¿=202× 0.6 ×0.4

1=48.5 kNm

a2=D¿+dom

2+

dom

2

a2=0.25+ 0.1082

+ 0.1082

=0.358m

a=0.4+ 0.1082

+ 0.1082

=0.508 m

u=2 ( a+a2 )

u=2× (0.358+0.508 )

u=1.732m

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f cv=0.17 (1+2/ βh)√ f c' ≤0.34 √ f c

'

β h= D¿

b¿

βh=0.4

0.25=1.6

f cv=0.17 (1+ 21.6 )√40 ≤0.34 √40

f cv=2.42 MPa ≤2.15 MPa

f cv=2.15 MPa

V uo=1732 ×108 ×2.15

V uo=402 kN

ϕ V uo=402× 0.7

ϕ V uo=281 kN>N Z¿ =202 kN (Ok )

Moments caused by concentrated load (AS3600, Clause 9.6)

bef=load width+2.4 a[1.0−( aLn )]

bef=0.25+2.4 × 0.6×[1−(0.61 )]

bef=0.826m

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M ¿=wa (1−a )

Ln

M ¿=202 ×0.6 × 0.41

=48.5 kNm

According to AS5100.2, clause 8.1.4.1, the ultimate strength in bending (Muo) shall not be less than (Muo)min calculated below.

M uo ,min=1.2[Z ( f cf ’+ PA s )+P e ]

Where Z = bD2

6 = 826∗1602

6 = 3.52 ×106 mm3

M uo ,min=1.2 ×[3.52 ×106 ×(3.79+0)+0]M uo ,min=16 kNm

A s tmin=0.0025 bdA s tmin=0.0025 × 0826× 114

A s tmin=235 mm2

M uo=48.50.8

M uo=60.6 kNm>M uo, min

M uo ,min=60.6 kNm

Z ≈ 0.925d

Z=0.925 ×114

Z=105.45 mm

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M uo=A st f sy Z

A st=M uo

f sy Z

A st=60.6 ×106

500× 105.45

A st=1150m m2

Try N12 bars at 75cts (1467mm2/m)

T=A st f sy

T=1467 ×500

T=734 kN

C=T=0.85 f c' γ ku db

ku=734 ×103

0.85× 40×0.766 × 826 ×114

ku=0.247<0.36, ductility OKϕ M uo=0.8 ×734 × (0.114−0.5 × 0.766× 0.247 ×0.114 )

ϕ M uo=60.6 kNm>M∗¿48.5 kNm (Ok)

Hence, bottom bars would need to change to N12 bars at 75cts.

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Development Length

Lsy .t=k7 k8 f sy Ab

( 2 a+db )√ f c'

Lsy .t=1 ×1.7× 500× 110

(70+12 ) √40

Lsy .t=180 mm<25 k 7 d b

25 k 7d b=25∗1∗12=300 mm

Hence, Lsy.t = 300mm

Lsy .c=20d b

Lsy .c=20∗12

Lsy .c=240 mm

Hence, Lsy.c = 240mm

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Splicing

The lap length for bars in tension shall be not less than Lsy.t (AS5100.2, clause 13.1.3).

According to AS5100.2, clause 13.2.5 (a), for fsy = 500MPa, lap length for compression

member shall be not less than (0.125fsy -22)db = (0.125*500 – 22)*12 = 486mm ≈ 500mm

Therefore, lap length for tension bar = 300mm

lap length for compression bar =500mm.

Top reinforcement will be comprised of 12m, 12m and 7.4m N12 bars with 500mm lap

length while bottom reinforcement will be comprised of 11.6m, 12m and 7.4m N12 bars with

300mm lap length. Secondary direction steel will be comprised of three 11m long bars with

500mm laps, staggered between adjacent bars. See drawings for more details.

For the detailed drawings of the deck see drawing BD01 (cross section of deck detail) and

BD02 (plan view of the deck detail).

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1.8.Headstock Design


Subject: Headstock designJob Number: 01 Contract: Sturt Road Bridge

DesignDate: 01/06/2014 Prepared: B HopkinsSheet: 01 of 17 Checked: A GreigClient: DPTI Approved: D Tet

In total there are three headstocks required for the Sturt Road bridge; two at each abutment

sharing the same design and one at mid-span. Headstocks will be designed as reinforced

concrete beams.

Concrete strength

Exposure classification – B1 (Surface of member in above-ground exterior environment

within 50km from coastline)

Surface and exposure environment – 3(b) AS3600 T 4.3

Minimum f’c = 32 MPa AS3600 T 4.4

Adopt 40 MPa for construction purposes

Concrete cover

Required cover = 40 mm (standard formwork) AS3600 T 4.10.3.2

Headstock at abutments

Headstocks continually supported by secant pile wall abutment

Beam dimensions

Trial dimensions; b = width of beam (mm), D = depth of beam (mm)

b = 1000, D = 400

Length = 45m

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DesignDate: 01/06/2014 Prepared: B HopkinsSheet: 02 of 17 Checked: A GreigClient: DPTI Approved: D TetDesign loads

Super T transfer load (per abutment) = 1543kN x 15 super T beams = 23145kN

Self weight = 1.0m x 0.4m x 24kN/m3 x 44m = 422.4kN

Total loadings from bridge = 23567 kN

Bending design at critical sections

Top reinforcement

Critical section: negative moment, top reinforcement

M* = -110 kNm

Determine area of tensile steel required, assuming N12 bars

M* = φ Mu = 110 kNm

Mu = 110/0.8 = 138 kNm

d = D – cover – ligature – ½ bar diameter

d = 400 – 40 – 12 – 6 = 342mm

Zu = 0.85d = 0.85 x 342 = 291mm

Mu = T.Zu

T = (138 x 106) / 291 = 474 x 103 N

T = Ast Fsy

Ast = (474 x 103) / 500 = 948mm2

Try 10 N12 bars, Ast = 1100mm2

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Actual T = 1100 x 500 = 550 kN

Check the beam is ductile, ku < 0.36

∑ Forces = 0, Hence T = C = α2 f’c b γ kud

α2 = 1.0 – 0.003f’c = 1 – 0.003(40) = 0.88 (> 0.85, hence adopt 0.85)

γ = 1.05 – 0.007f’c = 1.05 – 0.007(40) = 0.77 (< 0.85, hence adopt 0.77)

ku = (550 x 103) / (0.85 x 40 x 1000 x 0.77 x 342) = 0.06 (< 0.36, OK ductile)

Re-calculate accurate capacity

Accurate lever arm, zu = d - ½ γ kud = 342 – 0.5 x 0.77 x 0.06 x 342 = 334mm

Mu = 550 x 0.334 = 184 kNm

Design strength, φ Mu = 0.8 x 184 = 147 kNm > 110 kNm (OK)

Check fit

1000mm > 2(40) cover + 2(12) ligs + 10(12) bars + 9(40) min spacing = 624mm (OK)

Minimum steel requirement

A st ≥ [∝b ×( Dd )

2

×( f 'ct ,f

f sy)]bw d AS3600 8.1.6.1(2)

αb = 0.20 (rectangular section)

f 'ct , f =0.6√ f '

c = 0.6 x √40 = 3.79 MPa

[∝b ×( Dd )

2

×( f 'ct , f

f sy)]bw d = [0.20 ×( 400

342 )2

×( 3.79500 )] x 1000 x 342 = 709mm2 < Ast,top OK

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Hence, adopt 10 N12 bars top reinforcement



DesignDate: 01/06/2014 Prepared: B HopkinsSheet: 04 of 17 Checked: A GriegClient: DPTI Approved: D Tet

Bottom reinforcement

Critical section: positive moment, bottom reinforcement

M* = 261 kNm


M* = φ Mu = 261 kNm

Mu = 261/0.8 = 326 kNm


d = 400 – 40 – 12 – 8 = 340mm

Zu = 0.85d = 0.85 x 340 = 289mm

Mu = T.Zu

T = (326 x 106) / 289 = 1128 x 103 N

T = Ast Fsy

Ast = (1128 x 103) / 500 = 2256mm2


Actual T = 2200 x 500 = 1100 kN



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α2 = 1.0 – 0.003f’c = 1 – 0.003(40) = 0.88 (> 0.85, hence adopt 0.85)




γ = 1.05 – 0.007f’c = 1.05 – 0.007(40) = 0.77 (< 0.85, hence adopt 0.77)




Mu = 1100 x 0.324 = 356 kNm


Check fit



A st ≥ [∝b ×( Dd )

2

×( f 'ct ,f

f sy)]bw d AS3600 8.1.6.1(2)


f 'ct , f =0.6√ f '

c = 0.6 x √40 = 3.79 MPa

[∝b ×( Dd )

2

×( f 'ct , f

f sy)]bw d = [0.20 ×( 400

340 )2

×( 3.79500 )] x 1000 x 340 = 713mm2 < Ast,top OK

Hence, adopt 11 N16 bars bottom reinforcement

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Shear design

Ultimate shear force

V* = 784 kN

Check if concrete has sufficient shear capacity without shear reinforcement

V uc=β1 β2 β3 bv do f cv( A st

bv do)

13 AS3600 8.2.7.1

β1=1.1(1.6−do

1000 )≥ 1.1

d0=¿ 400 – 40 – 12 – ½(12) = 342mm

β1=1.1(1.6− 3421000 ) = 1.38 (> 1.1, OK)

β2=β3=1

bv=¿ 1000mm

f cv=f 'c

13 = 401/3 = 3.42 (< 4 Mpa, OK)

A st=¿ 1100mm2

V uc=1.38 ×1 ×1× 1000 ×342× 3.42( 11001000× 342 )

13

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V uc=¿ 238 kN

Check ductile behaviour

V ¿≤ 0.5∅V uc




0.5 x 0.7 x 238 = 96 kN (< V*, hence does not satisfy and shear reinforcement required)

Check shear capacity with minimum shear steel

V ¿≤∅V u , min

V u , min=V uc+0.10√ f 'c bv do ≥V uc+0.6 × bv ×do

V uc+0.10√ f ' c bv do = 238 ×103+0.10√40 × 1000× 342 = 454 kN

V uc+0.6×bv × do=238 ×103+0.6× 1000 ×342=¿443 kN

∴V u , min = 454kN

∅V u ,min=0.7× 454 = 318 kN

∴V ¿>∅V u , min

Hence, shear reinforcement shall be provided in accordance with Clause 8.2.10 AS3600

V us=(A sv f sy , f do/s )cot θv

V* ≤ φ (Vuc + Vus)

∴ φ Vus = 784 – 0.7 x 238 = 617 kN

Vus = 617 / 0.7 =881 kN

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s = 0.5D = 0.5(400) = 200mm

Taking cotΦv = 1(conservative)

Asv = (881 x 103 x 200) / (500 x 342) = 1030mm2




5 N16 stirrups, Asv = 1000mm2 ∴ need closer spacing

s = (1000 x 500 x 342) / (881 x 103) = 195mm

Adopt 5 N16 stirrups @ 195 cts

Reinforcement detailing

Flexural Steel

Continue bars full length of beam

Lapped splices for bars in tension

Lsy .t .lap=k7 Lsy . t ≥ 29 k1db AS3600 13.2.2

k7 = 1.25

Lsy .tb=0.5 k1 k3 f sy db

k2√ f ' c

k1 = 1.3 (horizontal bar with more than 300mm concrete below bar)

k2 = (132 – db)/100 = (132 – 12)/100 = 1.2

k3 = 1.0 – 0.15(cd – db) / db = 1.0 – 0.15(40 – 12) / 12 = 0.65

cd = 40mm AS3600 13.1.2.3(A)

Lsy .tb=0.5 ×1.3 ×0.65 ×500 ×12

1.2×√40 = 334mm

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Lsy .t .lap = 1.25 x 334 = 418mm

29k1 db = 29 x 1.3 x 12 = 452mm

Hence, adopt 452mm lapped splice for tension bars




Lapped splices for bars in compression

Lsy .cb=0.22 f sy

√ f ' c

db ≥ 0.0435 f sy db or 200mm, whichever is greater AS3600 13.1.5.2

Lsy .cb=0.22×500

√40×12 = 209mm

0.0435 x 500 x 12 = 261mm

Hence, adopt 261mm lapped splice for compression bars

Shear ligatures

Keep N16 @ 175 centres full length of beam

Headstock at mid-span

Headstocks supported by 5 piers at 11m centres

Beam dimensions

Trial dimensions; b = width of beam (mm), D = depth of beam (mm)

b = 3500, D = 1000

Length = 45m

Design loads

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Super T transfer load (mid-span) = 2320kN x 15 super T beams = 34800 kN

Self weight = 1.0m x 3.5m x 24kN/m3 x 44m = 3696 kN

Total loading from bridge = 38496 kN




Bending design at critical sections

Top reinforcement

Critical section: internal support, negative moment, top reinforcement

M* = -10955 kNm


M* = φ Mu = 10955 kNm

Mu = 10955/0.8 = 13694 kNm

d = D – cover –ligature – bar diameter (as there are two layers of reo bar)

d = 1000 – 40 – 12 – 50 = 898mm

Zu = 0.85d = 0.85 x 898 = 763mm

Mu = T.Zu

T = (13694 x 106) / 763 = 17.94 x 106 N

T = Ast Fsy

Ast = (17.94 x 106) / 500 = 35880mm2


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Actual T = 35280 x 500 = 17640 kN






α2 = 1.0 – 0.003f’c = 1 – 0.003(40) = 0.88 (> 0.85, hence adopt 0.85)

γ = 1.05 – 0.007f’c = 1.05 – 0.007(40) = 0.77 (< 0.85, hence adopt 0.77)




Mu = 17640 x 0.824 = 14533 kNm


Check fit



A st ≥ [∝b ×( Dd )

2

×( f 'ct ,f

f sy)]bw d AS3600 8.1.6.1(2)


f 'ct , f =0.6√ f '

c = 0.6 x √40 = 3.79 MPa

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[∝b ×( Dd )

2

×( f 'ct , f

f sy)]bwd = [0.20 ×( 1000

898 )2

×( 3.79500 )] x 3500 x 898 = 5909mm2 < Ast,top OK

Hence, adopt 18 N50 bars top reinforcement (acting as two rows of 9 bars)

Bottom reinforcement

Critical section: end span, positive moment, and bottom reinforcement




M* = 7194 kNm


M* = φ Mu = 7194 kNm

Mu = 7194/0.8 = 8993 kNm


d = 1000 – 40 – 12 – 18 = 930mm

Zu = 0.85d = 0.85 x 930 = 791mm

Mu = T.Zu

T = (8446 x 106) / 791 = 11.38 x 106 N

T = Ast Fsy

Ast = (11.35 x 106) / 500 = 22751mm2


Actual T = 20400 x 500 = 1020 kN

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α2 = 1.0 – 0.003f’c = 1 – 0.003(40) = 0.88 (> 0.85, hence adopt 0.85)

γ = 1.05 – 0.007f’c = 1.05 – 0.007(40) = 0.77 (< 0.85, hence adopt 0.77)







Mu = 1020 x 0.887 = 9049 kNm


Check fit



A st ≥ [∝b ×( Dd )

2

×( f 'ct ,f

f sy)]bw d AS3600 8.1.6.1(2)


f 'ct , f =0.6√ f '

c = 0.6 x √40 = 3.79 MPa

[∝b ×( Dd )

2

×( f 'ct , f

f sy)]bwd = [0.20 ×( 1000

930 )2

×( 3.79500 )] x 3500 x 930 = 5705mm2 < Ast,top OK

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Hence, adopt 20 N36 bars bottom reinforcement

Shear design

Ultimate shear force

V* = 6546 kN

Check if concrete has sufficient shear capacity without shear reinforcement




V uc=β1 β2 β3 bv do f cv( A st

bv do)

13 AS3600 8.2.7.1

β1=1.1(1.6−do

1000 )≥ 1.1

d0=¿ 1000 – 40 – 12 – ½(50) = 923mm

β1=1.1(1.6− 9231000 ) = 0.74 (adopt 1.1)

β2=β3=1

bv=¿ 3500mm

f cv=f 'c

13 = 401/3 = 3.42 (< 4 Mpa, OK)

A st=¿ 39200mm2

V uc=1.1× 1× 1× 3500× 923 ×3.42( 392003500 ×923 )

13

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V uc=¿ 2793 kN

Check ductile behaviour

V ¿≤ 0.5∅V uc

0.5 x 0.7 x 2793 = 978 kN (< V*, hence does not satisfy and shear reinforcement required)

Check shear capacity with minimum shear steel

V ¿≤∅V u , min

V u , min=V uc+0.10√ f 'c bv do ≥V uc+0.6 × bv ×do




V uc+0.10√ f ' c bv do = 2793 ×103+0.10√40× 3500× 923 = 4836 kN

V uc+0.6 ×bv × do=2793 ×103+0.6× 3500 ×923=¿4731 kN

∴V u , min = 4836kN

∅V u ,min=0.7× 4836 = 3385 kN

∴V ¿>∅V u , min

Hence, shear reinforcement shall be provided in accordance with Clause 8.2.10 AS3600

V us=(A sv f sy , f do/s )cot θv

V* ≤ φ (Vuc + Vus)

∴ φ Vus = 6546 – 0.7 x 2793 = 4591 kN

Vus = 4591 / 0.7 = 6558 kN

s = 300mm

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Taking cotΦv = 1(conservative)

Asv = (6558 x 103 x 300) / (500 x 923) = 4263mm2

5 N24 stirrups, Asv = 4500mm2

Adopt 5 N24 stirrups @ 300 cts

Reinforcement detailing

Flexural steel

Continue bars full length of beam




Where bars are to be lapped the following lapped splices apply

Lapped splices for bars in tension

Lsy .t .lap=k7 Lsy . t ≥ 29 k1db AS3600 13.2.2

k7 = 1.25

Lsy .tb=0.5 k1 k3 f sy db

k2√ f ' c

k1 = 1.3 (horizontal bar with more than 300mm concrete below bar)

k2 = (132 – db)/100 = (132 – 36)/100 = 0.96

k3 = 1.0 – 0.15(cd – db) / db = 1.0 – 0.15(40 – 36) / 36 = 0.98

cd = 40mm AS3600 13.1.2.3(A)

Lsy .tb=0.5 ×1.3 ×0.98 ×500 × 36

0.96 ×√40 = 1888mm

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Lsy .t .lap = 1.25 x 1888 = 2360mm

29k1 db = 29 x 1.3 x 36 = 1357mm

Hence, adopt 2360mm lapped splice for tension bars

Lapped splices for bars in compression

Lsy .cb=0.22 f sy

√ f ' c

db ≥ 0.0435 f sy db or 200mm, whichever is greater AS3600 13.1.5.2

Lsy .c b=0.22 ×500

√40× 36 = 626mm




0.0435 x 500 x 36 = 783mm

Hence, adopt 783mm lapped splice for compression bars

NOTE: Lapped splices shall not be used for bars in compression or tension with diameter

larger than 40mm; hence N50 bars to be welded in accordance with AS3600 standards

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1.9.Pier Design


Subject: Pier DesignJob Number: 01 Contract: Sturt Road Bridge

DesignDate: 05/06/2014 Prepared: J FangSheet: 01 of 03 Checked: H T LiClient: DPTI Approved: D Tet

The diameter of single pile is 900mm in the design, and f’c=40MPa

The width is 1000mm, depth is 1000mm, and length is 5500mm

Loading profile:

The axial loading is 10778kN

Alse, eccecc=width× 0.05=1000 ×0.05=50mm

Therefore moment=axiel loading × ecc1000

=10778 ×501000

=538.9 kNm

Check capacity:

M ¿=538.9+3295=3833.9 kNm , N ¿=10778 kN

So, M ¿

b D2 =325× 1000

1000 ×10002 =0 MPa and N ¿

bD=10778 ×1000

1000 ×1000=10.8 MPa

cover=D ×(1−g)/2, whichD=1000 and g=0.8

Thus, cover=1000 × 1−0.82

=100 mm

A st=width ×depth × p=1000 ×1000 × 0.01=10000 mm2

Try 10N36 with Ast=10200m2

d=D× g=1000 ×0.7=700 mm

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And reinforcin gW =width × g×2=1000 × 0.7 ×2=1400 mm

spacing=W−10 ×3610

=1400−10 ×3610

=104 mm

Therefore, adopt 10N36 with Ast= 10200m2, spacing 104mm

Buckle load:

Check if it is short column, assuming the overall height of the pile is 8m and k = 0.85 for top

pinned, bottom fixed.

Le=k ×length , which k=0.85 , length=5500, thus Le=0.85 ×5500=4675 mm

And r=0.3× D=300

So, Le

r=4675

300=16 ≤ 25 this is a short column

The moment should be increased by the moment magnifier δ; hence the strength of a short

column depends on its cross-section properties and length.

For buckling moment, assuming 10N36, Ast = 8160 mm2

M c=f sy × A st× zu

106 , whichf sy=500,A st=10200 mm2,

zu=d−0.5 ×ku ×W × γ=700−0.5× 0.545×1400 ×0.77=406 mm

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Thus M c=f sy × A st × zu

106 =500×10200 × 406106 =2071.8 kNm




Buckling load Nc,

N c=π2

( Le /1000 )2×

182× d1000

× M c × Φ

1+βd

βd=G

G+Q ,which G=axiel loading1.5

=107781.5

=7185 kN and

Q=axiel loading1.3

=107781.3

=8291 kN

Thus βd=7185

7185+8291=0.46So, N c=

π2

( Le /1000 )2×

182× d1000

× M c × Φ

1+βd=48841.6 kN

To find δ,

δ=km

1− N ¿

N c

k m=0.6−0.4 ×M 1

¿

M 2¿ =0.6−0.4=0.2<0.4, so take k m=0.4

N ¿

N c= 10778

48841.6=0.22

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Then,

δ=km

1− N ¿

N c

= 0.41−0.22

=0.51∨≥1




The result of the moment magnifier δ=1, buckle does not affect the moment

A st=0.01 × width×depth=0.01×1000 × 1000=10000 mm2

Therefore, adopt 10N36 with Ast= 10200m2, spacing 124mm

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1.10. Pile Design


Subject: Pile designJob Number: 01 Contract: Sturt Road Bridge

DesignDate: 05/06/2014 Prepared: H T LiSheet: 01 of 05 Checked: A GreigClient: DPTI Approved: D Tet

For the pile wall on both end of the bridge, there will be an overlap of the piles which will be

1/8 of the diameter as a result of it being a wall (refer to the Earth/Roadworks team). The

piles will only be reinforced from phase to phase for convenience.

Figure 17: Loading width

For a trial pile size of 900mm in diameter, the worst case loading width is 1575mm

So, the loads will be,

Axial Load (refer to pile analysis),

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517 kNm

× 1.575 m=815 kN



DesignDate: 05/06/2014 Prepared: H T LiSheet: 02 of 05 Checked: A GreigClient: DPTI Approved: D TetMoment caused by the axial load with 0.05D eccentricity (AS3600)

815 kN × 0.9 m×0.05=36.7 kNm

Shear force caused by soil,

288 kNm

× 1.575m=454 kN

Moment caused by soil with height of 2.845 m

454 kN ×2.845 m=1290.5 kNm

To get the factor for column chart from figure 6,

N ¿

Ag=815 ×1000

9002× π4

=1.3 MPa

M ¿

AgD= 1290.5+37

9002× π4 × 0.9

÷ 1000=2.3 MPa

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DesignDate: 05/06/2014 Prepared: H T LiSheet: 03 of 05 Checked: A GreigClient: DPTI Approved: D Tet

According to the column chart

Figure 18: column chart f'c =40 MPa g= 0.8

The required reinforcement ratio is 0.02.

Buckling problems should also be of concern when designing a column.

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Check if it is a short column, assuming the overall height of the pile is 8m and k = 0.85 for

top pinned, bottom fixed. “r” is the radius of gyration =0.25D for a circular column

¿=6000 mm ×0.85=5100 mm

r=0.25 ×900=225 mm



DesignDate: 05/06/2014 Prepared: H T LiSheet: 04 of 05 Checked: A GreigClient: DPTI Approved: D TetSo,

5100225

=23<25 which is a short colmun

Since it is a short column, the buckling problem does not occur the affect the design, but

check for safety concern.

The moment should be increased by the moment magnifier δ; hence the strength of a slender

column depends on its cross-section properties and length.

Buckling moment, assuming 16N32, Ast = 12800 mm2

Mu = Mc with ku = 0.545 and Φ=0.6

Mu=Ast × Fsy× d (1− γku2 )

Mc=500 ×12800 × (0.8× 900 ) ×(1−0.77 ×0.5452 )÷ 106=3641 kNm

Buckling load Nc, assuming βd = 0 when Le/r <=40,

N c=( π2

Le2 )×(182× d0×

ϕ M c

1+βd)

N c=( π2

5.12)×(182 ×0.72 × 0.6 ×36411+0 )=108630 kN

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To find δ,

δ=km/(1−N ¿

N c)≥1



DesignDate: 05/06/2014 Prepared: H T LiSheet: 04 of 05 Checked: A. GreigClient: DPTI Approved: D TetFirstly, km has to be defined, assuming M 1

¿=31 kNm and M ❑2¿=1106kNm

k m=0.6−0.4 ×( 31108630 )=0.59

So,

δ= 0.59

1− 81561104.1

=0.6 ≥1

The result of the moment magnifierδ=1, buckling is not critical for the moment.

Ast=0.02× 9002× π4

=12723 mm2

Adopt 16N32 with Ast= 12800m2, spacing 109mm2

Shear capacity check

Shear load from soil V ¿=453.6 kN

V uc=β1 β2 β3 bv do f cv¿¿

β1=1.1(1.6−do

1000 )=0.77>1.1 so , β1=1.1

β2=1+ 815∗10003.5× 636172.5

=1.37

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β3=1

Assuming the shear area is the middle reinforcing plie,

bv d o=636172.5 mm2


Subject: Pile designJob Number: XXXX Contract: Sturt Road Bridge

DesignDate: 05/06/2014 Prepared: H T LiSheet: 05 of 05 Checked: A. GreigClient: DPTI Approved: D Tet

f cv=f c'

13=40

13=3.42 MPa

A st=12 N 32=12800 mm2

V uc=301.7

0.5 ϕV uc=0.5∗0.7∗301.7=105.6 kN<V ¿not satisfied

Check for minimum shear reinforcement,

V u . min=V uc+0.1√ f c ' bv do ≥V uc+0.6 bv do

Where,

V uc+0.6 bv do=402.64 kN

V uc+0.1√ f c ' bv do=381.99 kN

V u . min=402.64 kN

ϕ V u .min=0.7∗402.64=281.85 kN <V ¿

Extra shear reinforcement is required

Need 346.3 kN more shear capacity,

V us≤( A sv f sy. f d0

s )cot∅ v

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Maximum spacing is of either minimum of 300mm and 0.5, s = 300mm

A sv=346.3 ×300

500 × (900−90 )=256.52 mm2

N12 @ 300 ligatures N10, Ast = 367 mm2


Subject: Pile designJob Number: XXXX Contract: Sturt Road Bridge

DesignDate: 05/06/2014 Prepared: H T LiSheet: 05 of 05 Checked: A. GreigClient: DPTI Approved: D Tet

The starter bar for the connection piles to the headstock was taken by the following table,

Table 4: Development and Splice lengths for deformed bars

Source: ARC reinforcement hand book Page 43

Since the bending moment was applied to the pile, the critical value was taken from tensile

development length of N32 bar size which is 1270 mm

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1.11. Pile Cap Design


Subject: Pile CapJob Number: 01 Contract: Sturt Road Bridge

DesignDate: 06/06/2014 Prepared: A GreigSheet: 01 of 09 Checked: D TetClient: DPTI Approved: J Rogers

Given from the piers, the total pier axial load that will be applied to the pile cap is 10778 kN.

The Pile section (1.9) states that 5 x 900 mm piles shall be used under each pier.

But first, the punching shear will be calculated to determine a suitable depth for the 4400 mm

x 4400 mm pile cap.

The exposure classification for this pile cap is A2. (AS3600-2009 T4.3). Hence the minimum

concrete strength shall be 25 MPa (As3600-2009 T4.4). However for this pile cap, use 40

MPa.

According to AS3600-2009 Table 4.10.3.2, the minimum cover for a 40 MPa strength

concrete with an exposure classification of A2 has a minimum cover of 20 mm. However, in

this case a safe cover of 75 mm was chosen.

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DesignDate: 06/06/2014 Prepared: A GreigSheet: 02 of 09 Checked: D TetClient: DPTI Approved: J RogersPunching Shear Due to Pier

In order to determine the punching shear depth, work with equation:

V ¿≤ ϕV uo

V ¿=10778 kN

ϕ V uo=u dom f cv

AS3600-2009: Cl 9.2.3

f cv=0.17 (1+ 2β h)√ f c

' ≤ 0.34√ f c'

AS3600-2009: Cl 9.2.3

Take the lesser of the results fromf cv:

f c' =40 MPa

βh=1 Due to having a square cross section for the pier. AS3600-2009: Cl 9.2.3

0.17 (1+ 21 )√40=3.23 MPa

0.34 √40=2.15 MPa

Hence adopt f cv=2.15 MPa

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u is equal to the critical shear perimeter which can be taken as:

u=(1000+d )× 4

ϕ=0.7

AS3600-2009: T2.2.2

Hence:

10778 ≤0.7 × (1000+d ) ×4 ×d ×2.15

107780.7 × 4×2.15

≤ 1000 d+d2

0≤ d2+1000 d− 107780.7× 4 × 2.15

928 mm≤ d

Adopt d=950 mm

Include 75 mm cover:

1025 mm≤ D

Round up to the next 100mm,

Adopt D=1100 mm

The critical bending shear is located at d away from the face of the support. In this case it is

950mm. According to drawing BD 8 Rev 0 sheet 1or BD 9 Rev 0 sheet 1, this distance is on

top of a pile in all four directions, hence bending shear force will not be critical and punching

shear will govern the design. The thickness designed by the punching shear will be suitable

that no shear ligatures are required.

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Punching Shear Due to Piles

V ¿≤ ϕV uo

The ultimate Axial Load of the pile can be taken by the distribution of the pier load and self-

weight:

V pile¿ =

10778+1.2(24 × 4.4 × 4.4 ×1.1)5

V ¿=2278 kN

ϕ V uo=u dom f cv

AS3600-2009: Cl 9.2.3

f cv=2.15 MPa

u is equal to the critical shear perimeter, which for a pile will be given as:

u=2 a+a2

bc=√π × 4502

bc=797.6 mm

a=bc+do

2

a=797.6+ 9502

a=1272.6 mm

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a2=bc+dom

a2=bc+do

a2=797.6+950

a2=1748 mm

u=2× 1272.6+1748

u=4292.8 mm

do=950 mm

V ¿≤ ϕV uo

ϕ=0.7

AS3600-2009: T2.2.2

2278 kN ≤ 0.7 × 4292.8× 950 ×2.15 ×10−3

2278 kN ≤ 6139 kN

Hence: Piles are satisfied for punching shear

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DesignDate: 06/06/2014 Prepared: A GreigSheet: 06 of 09 Checked: D TetClient: DPTI Approved: J RogersBending Moment Check

To get the ultimate design moment, take two pile reactions and add them together and

multiply it by the distance from the edge of the column face to the centre of the piles. Two

piles are used as the force due to two piles being in the same line, both causing the moment.

The distance from the edge of the column to the centre of the piles is 800mm.

M x¿=M y

¿ =2278× 2× 0.8

M ¿=3645 kNm

M ¿≤ ϕ M u

ϕ=0.8

AS3600-2009: T2.2.2

M u=ϕz f sy A st

z=0.85 d

d=950 mm

f sy=500 MPa

Rearrange to find A st

3645 kNm≤ 0.8 ×500 ×0.85 × 950× A st

3645 × 106

0.8 ×500 ×0.85 × 950≤ A st

11285m m2≤ A st

Now check if this is greater than minimum required Area of Steel:

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Treat it as a beam:

A st ,min≥ [ α b(Dd )

2

f ct .f'

f sy]bwd

AS3600-2009 Cl 8.1.6.1

α b=0.2 bw=4400 mm

D=1100 mm f sy=500 MPa

d=950 mm f ct . f' =0.6√40

A st ,min≥ [ 0.2( 1100950 )

2

0.6√40

500 ]4400 × 950

A st ,min ≥ 8496 m m2

Adopt: A st=11285m m2

Using the A st, the best possible solution is to use 20N28 bars.

The centre to centre spacing shall be:

s= 4400−75 ×219

s=223.7mm

Adopt 220 mm cts spacing

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Using 220 mm spacing, the spacing from the edge shall be:

( Allowable total spacing )−(Total length¿ spacing)2 +cover

sout=( 4400−75 ×2 )−(220 ×19+28 )

2+75

sout=96 mm¿outer edge of ¿̄

sout=96+14

sout=110mm¿centre of ¿̄

Adopt the distance from the edge of the concrete to the centre of the bar as 110 mm

All detailed cross section reinforcing can be found on Drawing BD 8 Rev. 0 Sheet 1.

The effective depth for one direction will be 950 mm.

The effective depth for the other direction will be 950−28=922mm

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Table 5 summarises all the key points of the pile cap.

The total length of reinforcing can be calculated by finding the length of reinforcing required, multiplying it by the number of bars and then doubling this for both layers of reinforcing.

Lr=4400−75× 2

Lr=4250 mm ×20 Bars

Lr=85000 ×2

Lr=170000 mm

Hence the total length of reinforcement is 170000 mm or 170 m.

Table 5: Pile Cap Summary

Pier Axial Load 10778 kNPile Diameter 900 mmNo. of Piles 5Load on Each Pile 2278 kNPile Cap Length 4400 mmPile Cap Width 4400 mmCover 75 mmPile Cap Effective Depth One Direction 950 mmPile Cap Effective Depth Second Direction 922 mmPile Cap Total Depth 1100 mmPile cap reinforcing (both directions) 20N28Centre to Centre Spacing of Reo Bars 220 mmEdge of Concrete to Centre of Outer Reo 110 mmTotal Length of Reinforcing 170 m

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1.12. Crash Barrier Design

1.12.1. Parapet

The parapets will be pressure grouted to the bridge deck accompanied by steel bolt

anchors beneath it and connected to one another by end connector rebar dowels to

distribute loads. The face of the parapet will be designed following AS 5100.1 figure

10.6.1 and pattern decoration of the parapet is optional. With a bridge span of 43m, 30

parapets of span 3m will be required to cover both sides. In the event of an accident

where the parapet is damaged and unrepairable they are easily replaced by Bianco Precast

with a small lead time. The minimum height required by the Australian standards AS

5100.1 Cl 12.1 (d) for parapets are 1.3m for a case where cyclists could possibly use the

pedestrian walk way. According to AS 5100.1 Table 10.4, the barrier performance level

should be regular for Sturt Road since the regular performance level provides effective

containment for general traffic on all roads.

Following the regular performance level, the design load is taken from AS 5100.2 Table

11.2.2 and the minimum effective height is taken as 800mm from AS 5100.2 Table

11.2.3.

With only a height of 1.3m, to prevent pedestrians from throwing objects off the

underpass and for safety concerns, a protection screen will be required. The protection

screen will be installed through a connection on the parapets and use either replaceable

glass, acrylic or plastic sheets with a height of 1.7m to clear the 3m minimum height

above the walkway surface required by AS 5100.1 Cl 12.3 (c)i.). Any clear openings

between the protection screens cannot exceed 50mm x 50mm so the supports on the

protection screens will be spaced closer together. Since the protection screen would lie

above the parapet, the screen has to be a material that cannot shatter and fall onto the

traffic flowing on the underpass.

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Subject: Parapet DesignJob Number: 01 Contract: Sturt Road Bridge

DesignDate: 01/06/2014 Prepared: J NgoSheet: 01 of 02 Checked: D TetClient: DPTI Approved: J Rogers

Calculations

Height = 1300mm

Width = 3000mm

Thickness = 300mm (top), 530mm (bottom)

Cover = 40mm (AS 3600 table 4.4/4.10.3.3)

Concrete strength = f’c = 40MPa

Live Load: AS 5100.2 Table 11.2.2

Table 6: Parapet specifications

Barrier Performance

Ultimate longitudinal Outward load (Ft) kN

Ultimate longitudinal Inward load (FL) kN

Vehicle contact length for traverse loads (Lt) and longitudinal loads (LL) m

Ultimate vertical downward load (Fv) kN

Vehicle contact length for vertical loads (Lv) m

Minimum effective height (He) mm

Regular 250 80 1.1 80 5.5 800

Try N12 @ 250 (440mm2)

fsy = 500MPa

Design load = 250 / (1.1 X 1) = 227.27 kN/m2

Designed as two way slab -

Ly/Lx = 3000/1300 = 2.307 > 2

One long edge discontinuous, therefore βx=0.066 and βy=0.028 (Table AS3600 Table

6.10.3.2 (A))

Positive moments at mid-span

M*x = βx x Fd x Lx2

= 0.066 x 227.27 x (1300)2

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= 25.35 kNm


Subject: Parapet DesignJob Number: 01 Contract: Sturt Road Bridge


M*y = βy x Fd x Lx2

= 0.028 x 227.27 x (1300)2

= 10.75 kNm

Negative moments at middle

M*x = 1.33 x 25.35 = 33.7155 kNm

Negative moment at outside

M*x = 0.5 x 10.75 kNm = 5.37727273


d = 300 – 40 – 12 – 12/2 = 242mm

T = fsy x Ast

T = 440 x 500

T = 220 kN

C = T = fsy x Ast

C = a2 x f’c x b x y x ku x d

a2 = 0.85

y = 1.05 – 0.007 x 40

y = 0.77

ku = (220 x 10^3)/ (0.85 x 40 x 1300 x 0.77 x 242)

ku = 0.0267 < 0.36 OK ductility

Mu = C x (d – 0.5 x y x ku x d)

Mu = 220 x 103 (242 – ½(0.77 x 0.06 x 242)

Mu = 52.69 kNm > (33.72/0.8)=42.15 OK

Adopt N12 @ 250

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1.12.2. Pier Crash Barrier

Following the design of the parapet, the barriers under the bridge will have the same profile

from AS 5100.1 figure 10.6.1 with altered dimensions and have the same capability to be

replaced. The height required for the barriers is 1.4m according to AS 5100.1 Table A3 where

the minimum effective height for barriers of special performance levels should be 1.4m. The

barrier performance level is special to prevent damage to the piers and contain heavy vehicles

on Main South road, which is an arterial road. Being a special performance design, the load

case is taken from AS 5100.2 Table A2. The barriers need to be placed along the underpass

of South Road and encase the piers along the median strip. The amount of barriers required

will be 177 along the median strip with 29 to surround the piers. The crash barriers can be hit

from both sides and will need doubled reinforcement rather than reinforcement along the

middle. The designs will vary slightly to follow the curving slope of the underpass but still

maintain structural integrity.

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Subject: Crash Barrier DesignJob Number: 01 Contract: Sturt Road Bridge


Design calculation

Height = 1400mm

Width = 3000mm

Thickness = 300mm (top), 790mm (bottom)

Cover = 40mm (AS 3600 table 4.4/4.10.3.3)

Concrete strength = f’c = 40MPa

Live Load: AS 5100.2 Table A2

Table 7: Barrier specifications

Barrier Performance

Ultimate longitudinal Outward load (Ft) kN

Ultimate longitudinal Inward load (FL) kN

Vehicle contact length for traverse loads (Lt) and longitudinal loads (LL) m

Ultimate vertical downward load (Fv) kN

Vehicle contact length for vertical loads (Lv) m

Minimum effective height (He) mm

Special test level 6 (36t articulated tanker)

750 250 2.4 350 12 1400

Try N16 @ 200 (1000mm2)

fsy = 500MPa

Design load = 750 / (1.1 X 1) = 227.27 kN/m2

Designed as two way slab -

Ly/Lx = 3000/1400 = 2.14 > 2

One long edge discontinuous, therefore βx=0.066 and βy=0.028 (Table AS3600 Table

6.10.3.2 (A))

Positive moments at mid-span

M*x = βx x Fd x Lx2

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Subject: Crash Barrier DesignJob Number: 01 Contract: Sturt Road Bridge


= 0.066 x 312.5 x (1400)2

= 40.43 kNm

M*y = βy x Fd x Lx2

= 0.028 x 312.5 x (1400)2

= 17.15 kNm

Negative moments at middle

M*x = 1.33 x 40.43 = 53.77 kNm

Negative moment at outside

M*x = 0.5 x 17.15 kNm = 8.58 kNm


d = 300 – 40 – 16 – 16/2 = 236 mm

T = fsy x Ast

T = 1000 x 500

T = 500 kN

C = T = fsy x Ast

C = a2 x f’c x b x y x ku x d

a2 = 0.85

y = 1.05 – 0.007 x 40

y = 0.77

ku = (500 x 10^3)/ (0.85 x 40 x 1300 x 0.77 x 236)

ku = 0.062251 < 0.36 OK ductility

Mu = C x (d – 0.5 x y x ku x d)

Mu = 500 x 103 (236 – ½(0.77 x 0.06 x 236)

Mu = 115.2 kNm > (33.72/0.8)=42.15 OK

Adopt N16 @ 200

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1.13. Bearing design

A bearing for a bridge is used in order to sit between the Super T and the headstock to

counteract movement of the bridge that may occur. A bearing can absorb any movement of

the bridge structure itself. This is an item that can be sourced from a company and doesn’t

need to be designed. Due to the cross fall of the road which is 4%, there needs to be a design

for compensation of cross fall, as stated by the Department of Planning, Transport and

Infrastructure’s standard design of structures. In order to compensate for this a tapered plate

is to be proved which matches the cross fall of the road, that is then placed between the

Super-T and the bearing

In order to choose a sufficient bearing for the structure, the main criteria would be choosing a

bearing and size that will withstand the loading and meet the ratings criteria. There are two

types of bearings that can be used from the manufacturer Granor, pot bearing or elastomeric

bearing. Each of these types of bearings have been designed in accordance to AS5100.4. An

elastomeric bearing has been chosen due to its familiarity to companies, as it is commonly

used in bridge design.

Figure 19: Typical Elastomeric bearing from Granor

This type of bearing is made out of high quality elastomers with internal layered sheets of

steel reinforcement. J Series rectangular 600x450 elastomeric bearing has been adopted. The

figure below shows the ratings and specifications of the bearing.

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Table 8: Specification of chosen bearing

Elastomeric Bearing

Overall height (mm)

Rated load at max rotation (kN)

Net Weight (kg)

Max deflection (mm)

Part Number

Series J600x400mmRectangular

86 At max shear

At zero shear

67 33 GJE-03

1676 2088

1.14. Summary

Through the use of the analytical program SPACEGASS, our team has been able to get

values in order to design the components of the bridge. Each section of the report has

gone through the analysis that is required in order to made the bridge safety as well as

satisfying the goals of the project. From the analysis or the Australian Standard of Codes,

all the components have been design according to it all, meaning that the underpass will

be safe and has been designed with due care.

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1.15. References

Granor Rubber and Engineering, no date available, Elastomeric Bearings, <

http://www.granor.com.au/uploads//products/elastomericbearings/Granor_Elastomeric_L

aminated_Bearings.pdf>

Loo, Y.C, Chowdhury, S.H 2013, Reinforced and Prestressed Concrete Second Edition,

Cambridge University Press, New York

Merretz W., Smith G., 1997, Standarisation of and Detailing for Super T Bridge Girders,

Structural Concrete Industries, no date available, viewed 30th May 2014, <

http://www.sciaust.com.au/pdfs/supertstandards.pdf>

Warner R., Faulkes K., Foster S., 2013, Prestressed Concrete 3rd Edition, Pearson

Australia, New South Wales

108

http://www.sciaust.com.au/pdfs/supertstandards.pdf

http://www.granor.com.au/uploads//products/elastomericbearings/Granor_Elastomeric_Laminated_Bearings.pdf

http://www.granor.com.au/uploads//products/elastomericbearings/Granor_Elastomeric_Laminated_Bearings.pdf

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