Advanced Concrete Design: T-Roff Bridge Project - STEN4005

This assignment presents a comprehensive design of a T-Roff bridge, addressing various aspects of civil engineering design. The project encompasses the specification of relevant loads, including self-weight, deck slab weight, and traffic loads, along with the approach for analysis and design based on AS5100 standards. The solution details the determination of longitudinal sections, cable layout, and prestress losses. It further delves into the T-roff section design, checking stresses at transfer and service, and calculating the ultimate moment capacity, shear strength, and torsion strength. The design process includes the calculation of concrete compressive force, neutral axis depth, and development length, ensuring the structural integrity of the bridge. The document also covers prestressing losses and stress checks at transfer and service, providing a complete analysis of the bridge's structural behavior. This detailed analysis is essential for students studying advanced concrete design and construction.

Design of a Tee Roff Bridge
Loads
Self-weight of girder
¿ 10−6∗
[ (1600+1400
2 ∗320 )−( 1
2∗1400∗( 320−260 ) )+2 ( 150∗( 2200−320 ) ) +2 ( 100∗1700 ) ]∗2650∗9.81∗10−3
¿ 34. 887 kN /m
Weight of deck slab ¿ ( 5∗0.2 )∗2650∗9.81∗10−3=25.99 kN /m
Deck wearing surface ¿ 0.05∗5∗2650∗9.81∗10−3=6.499 kN /m
Span length = 37 m, at mid span;
M g =1
8 ∗34.887∗372=5970 kNm
M s= 1
8∗25.977∗372=4448.74 kNm
M dws= 1
8∗6.499∗372=1112.141kNm
Moment at mid-span due to traffic loads from structural analysis.
M p=655 kNm
M sm=2990 kNm
Ultimate moment capacity (Clause 8.1 AS5100.5)
Calculate ultimate moment, M ¿
Load factor:
For dead load = 1.2
For superimposed dead load = 2.0
For live load SM1600 loading = 1.8
Maximum live load moment

For SM1600 loading ¿ 1.8∗2990=3614 kNm
M ¿=1.2 ( 5970+4448.74+655 )+2.0 ( 1112.141 )=15512.77 kNm
Ultimate moment capacity
Centre of gravity of bottom strands CG
CG= 260
2 =130 mm
Effective depth d p
d p=2200+200+50−130=2320 mm
Strand stress σ pu at ultimate stress (Clause 8.1.7 AS5100.5)

σ pu=f p (1− k1 k2
γ )
k1 =0.4 if f py
f pb
<0.9
k 2= Apt f p
bef d p f c
' = 65∗130∗1790
320∗2320∗50 =0.4075 (Assume no Ast and Asc)
Clause 8.1.3 AS5100.5
γ=1.05−0.007 f c
' =1.05−0.007 ( 50 ) =0.70
Strand stress σ pu=1790 (1− 0.4∗0.4075
0.70 )=1373.19 MPa
Total tensile force in bottom strands
Ft=65∗130∗1373.19∗10−3=11603.42 kN
Check the position of neutral axis.

Concrete compressive force in kerb.
FC1
=α 2 f c
' bd=0.85∗40∗700∗320∗10−3=7616 kN
Adopt concrete grade 40.
Since FC1
< Ft, the compression zone extends to slab.
Compressive force on the slab
FC2
=Ft −FC1
=11603.42−7616=3987.42 kN
Depth of compression zone, x= 3987.42∗103
0.85∗40∗2320 =50.55 mm ≈51 mm
Since x >slab thickness, compression zone does not extend to the girder,
Check for neutral axis depth ku (Clause 8.1.5 AS 5100.5)
Depth of compression zone ¿ 50+51=101 mm

Thus, ku d= 101
0.7 =144.29 mm
ku =101+130
2320 =0.10<0.36 OK
Ultimate moment capacity M u
M u=FC1 (d− 50
2 )+ FC2 (d−50− 51
2 )
¿ 7616∗¿
¿ 26428484.19 kNmm
¿ 26428.48 kNm
∅ M u > M ¿(15512.77 kNm) the section is OK at the mid-span.
Development length
Table 13.3.2.1 AS5100.5
Development length for strands;
At transfer, Lpt=60 db
At higher loads, Lp=180 db
For 15.2 mm strands;
Lpt=60∗15.2=912 mm≈ 910 mm
Lp=180∗15.2=2736 mm ≈ 2740 mm
Check stresses at transfer
Stresses at transfer
Prestressing losses before losses = 11603.42 kN
Clause 3.3.4.4 AS5100.5
Prestressing losses due to relaxation during curing = 10%

To determine elastic shortening loss;
Prestressing compressive stress σ ci= Pi
A g
+ Pi e2
Ig
− M DL e
Ig
¿ 10443.078∗103
1342000 + 10443.078∗103∗2602
1.3323∗1011 − 5970∗106∗260
1.3323∗1011
¿ 19.154 MPa
Elastic modulus of concrete at transfer
Eci =0.043∗2400
3
2∗40
1
2 =31975 MPa
Prestress loss due to elastic shortening;
∆ σ p=σci
Ep
Eci
=19.154∗195∗103
31975 =116.81 MPa
Prestress loss due to elastic shortening ¿ 140∗116.81
200000 ∗100 %=8.2 %
Allowable stress limits (Clause 8.1.6.2 AS5100.5)
Maximum stress limit ¿ 0.6 f cp=0.6∗40=24 MPa
Minimum stress limit ¿−0.25∗ √ f cp=−0.25∗40
1
2 =−1.58 MPa
Hence, transfer stresses at mid-span fall within the allowable limits.
Check Stresses at Service (Clause 3.1.8, AS5100.5)
For girder only

Effective perimeter ue=2 ( 1700+90+1600+2080+1760 ) + ( 1400+1500 ) =17360 mm
Gross sectional area Ag=1069287 m m2
Hypothetical thickness th= 2∗1069287
17360 =125 mm
For composite section

ue=5000+1400+ 2 ( 90+ 1600+ 2080 ) + ( 3200+ ( 2∗1760 ) +1500 )
2 =17111 mm
Ag=1069287+ ( 5000∗200 ) =2069287 mm2
th= 2∗2069287
17111 =242 mm
Therefore, for 1 – 100 days, th=125 mm
For 100 days to 3 years, th=242 mm