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CIV3222 Assignment 2B – Group 38 Andrew Cameron 23394684, Kyle Mitchell 24231207, Jasim Qureshi
24210587 & Xuan Kai Zhao 25385461
i
Executive Summary From project 2A:
Limit State Maximum Bending Moment (KNm)
Maximum Shear Force (KN)
ULS 6644.04 1255.38 SLS 4237.58 787
The 42 prestress strands chosen for our design were 15.2mm diameter super stands with a
tensile strength of 1750MPa, with the lower row a distance of 130mm from the soffit.
Moment capacity 𝜙𝑀𝑢 > 𝑀∗ = 6.64 𝑀𝑁 To ensure shear capacity, use 2 legs of R20 stirrups at a spacing of 150 mm. Total losses = 26.76% Deflection at transfer without slab = -14.07 Deflection at transfer with slab = -13.24 Short term deflection at service = 8.459 Long term deflection = 16.919 To ensure safe slab design use 14mm diameter with a bar spacing of 250mm in both the longitudinal and transfer direction.
Stress Checks
Stage 1
Top 8.77930381 < 20 ok
Bottom 19.6999737 < 20 ok
Stage 2
Top 12.0182898 < 25 ok
Bottom 17.490181 < 25 ok
Stage 3
Top 5.96970827 < 25 ok
Mid 17.0900537 < 25 ok
Bottom 2.78459611 < 25 ok
ii
Table of Contents
Executive Summary i
Summary of Part A 1
Introduction 2 Beam Section Design 3
Section properties 3
Verifying the minimum section modulus 5
Prestressing Force Checks 6
Cable limits 6
Stress Checks 7
Magnel diagram 8
De-bonding of prestressed cables 8
Flexural Strength 11
Shear Design 14 Section Capacity 15
Ultimate Strength of the Concrete 15
Web Shear Cracking 15
Shear Reinforcement addition 16
Pre-tensioning Losses 17 Elastic Loss (I) 17
Creep (TD) 18
Shrinkage Loss (TD) 20
Relaxation Loss (TD) 21
Total loss 21
Deflection 22 Deflection at Transfer Before Slab 22
Deflection at Transfer With Slab 23
Short-Term Deflection 23
Long Term Deflection 24
Slab Design 25 Loading 25
Flexure Design 26
Shear Design 28
Shrinkage and Temperature Design 28
iii
Conclusion 30
1
Summary of Part A
Part A of this report the following tasks were carried out.
Determined the loadings on the bridge through the use of the M1600 design standard.
These include not only traffic loads, but also superimposed dead loads (ie. the load from
the traffic barrier, etc.)
We then assessed the grillage system of the Super-T Beams and Deck for ULS and SLS
conditions. For the grillage analysis two computer programs were used, SPACEGASS and
LUSAS, to check the validity of each load case.
Different loads cases were utilized to simulate different road conditions, designing ultimately for the
worst-case scenario whether that were ULS or SLS.
The dimensions of the bridge were 10.35m width and 25m long. The dimensions of this bridge segment
were put into the application SPACEGASS and LUSAS to create a model of the design using T3 super-T
beams.
Table 1 below contains results obtained from Part A, showing the maximum bending moments and
shear forces from both ULS and SLS conditions.
Table 1 – Part A results
Limit State Maximum Bending Moment (KNm)
Maximum Shear Force (KN)
ULS 6644.04 1255.38 SLS 4237.58 787
2
Introduction
As in the first part of this project, a bridge is to be designed to span 25m over a small body of water.
In the first part of this project, grillage analysis was performed on the bridge, determining the maximum
forces and bending moments that would be experienced by the bridge, as well as the deflections the
members will experience. In this report, emphasis is placed on the design of individual members of the
bridge, more specifically the Super T-beam and slab.
The Super T-beams will comprise of a concrete cross-section, pre-stressed steel tendons, and the
concrete slab to be used as the base for the road. Design of these Super T-beams will focus on
calculating the number and location of pre-stressed steel strands, as the cross-section of the concrete
beam is given. During the design of these beams, checks will be made to ensure the beam satisfies
strength criteria at transfer and under applied service loads - as well as checking the short-term, long-
term, and overall deflection. The requirement and spacing of stirrups used for web reinforcement will
also be determined in order to design the beam for shear.
ULS conditions will be used for strength and shear design, whilst SLS conditions will be used for
calculating the Magnel’s diagram, stresses and deflections.
The cast-in-situ concrete slabs that will be placed on top of the Super T-beams, will be designed, along
with the amount of steel reinforcement needed will also be determined.
Finally, the cross-sections of the Super T-beams, along with the cast-in-situ concrete slabs, can be drawn
at mid-span and the end of the girder.
These calculated and optimized stirrups and pre-stressing tendons will reduce upon the deflection
obtain in Part A.
3
Beam Section Design
To ensure adequate strength and serviceability in the T3 Super T-Beams that will span the 25m bridge
prestressing tendons are required. Pre-stressing tendons are used to compress the concrete into
compression and reduce and tensile stresses that may occur during bending. The tendons initially give
these Super T-beams a hogging shape but when in service the increased moment cause the beams to
sag but due to the initial hogging this is much less than a deflection without pre-stressed tendons.
For this reason the design of pre-stressed tendons is a critical part of a beam design and as such is
considered first in the design of the bridge. In order to determine the amount of pre-stressing strands
required, the initial pre-stressing force required for the beam must be determined. The pre-stressed
strands will be located in the bottom flange of the beam at a distance of 130 mm from the bottom fibre.
Section properties
In order to ease the calculation process, the Super T-beam dimensions were simplified when they were
at transfer. The dimensional properties when the beam was a service with the overlaying slab was taken
directly from space gass as calculated in part A of this project. The simplified beam at transfer is shown
in the figure below.
538.75 538.75
100100
29
2
920.5
83
3
Figure 1 – Simplified Cross-Section
From this simplified design the required beam properties without a slab were calculated.
Table 2 - Geometric properties before slab
Area 516198.5 mm^2
Iz 1.54851E+11 mm^4
yt 713.3310882 mm
yb -486.6689118 mm
ztb 217081825.8 mm^3
zbb -318185960.3 mm^3
4
e -306.6689118 mm
The geometric beam properties with the overlaying concrete slab are shown in the table below.
The prestressing tendons chosen for this design were 15.2mm super strands with a straight horizontal
profile. The number of strands originally chosen for this design was 36. It was found that more strands
were needed for this design with 42 strands being chosen for the final beam design. Originally prestress
losses were taken as 20% but with further calculates it was found to be approximately 26.67%, this is
shown later in the report. The characteristic strength of concrete was taken as 50Mpa and at transfer it
was taken as 40Mpa as the concrete is at a younger age. For this design the strength of the slab was
taken to be the same as the beam to simplify the design process. The beam properties are shown in the
table below.
Table 4 - Beam Design Properties
prestress losses 26.67%
Beam length 25 m
Ec 34800 Mpa
Strand type 15.2 mm
Area 143 mm^2
Min Breaking load 250 KN
Min Tensile strength 1750 Mpa
No. of strands 42
Ap 6006 mm
Pi 7882875 N
Initial prestress 75%
σ pi 1312.5 Mpa
f'ci 40 Mpa
f'c 50 Mpa
Ep 195000 Mpa
Table 3 - Geometric properties with slab
Area 915644.1 mm^2
Iz 2.17733E+11 mm^4
yt 514.5171 mm
yb -865.4829 mm
ztcomp 423178938.1 mm^3
zbcomp -251573774.6 mm^3
e -719.4829 mm
5
The maximum allowable stresses in the concrete at services and transfer can then be determined based
on α1 = β1 = 0.5 and α2 = β2 = 0.25. Using the below formulas the allowable stresses can be calculated.
𝜎𝑐𝑖 = 0.5 ∗ 𝑓𝑐𝑖′
𝜎𝑡𝑖 = −0.25 ∗ √𝑓𝑐𝑖′
𝜎𝑐𝑠 = 0.5 ∗ 𝑓𝑐𝑖′
𝜎𝑡𝑠 = −0.25 ∗ √𝑓𝑐𝑖′
To preform stress checks on the beam it necessary to know the maximum moments acting on the beam
at transfer and at service.
Table 6 – Maximum Moments
Mi 1008.2 KNm
MSLS 4237.6 KNm
Mslab 703.1 KNm
Masl 2526.2 KNm
Verifying the minimum section modulus The beam must first be checked to verify the section modulus meets the minimum modulus based on
the maximum allowed stresses. The minimum section modulus for a composite beam was calculated
using the formulas below.
Table 5 - Max allowable stresses
σ ti -1.58113883 Mpa
σ ci 20 Mpa
σ ts -1.767766953 Mpa
σ cs 25 Mpa
6
|𝑍𝑏, 𝑐𝑜𝑚𝑝| ≥𝑀𝑎𝑠𝑙
(𝛼 ∗ 𝜎𝑐𝑖 − 𝜎𝑡𝑠) +(1 − 𝛼) ∗ 𝑀𝑖 + 𝑀𝑠𝑙𝑎𝑏
𝑍𝑏, 𝑏
|𝑍𝑏, 𝑏| ≥(1 − 𝛼) ∗ 𝑀𝑖 + 𝑀𝑠𝑙𝑎𝑏
(𝛼 ∗ 𝜎𝑐𝑖 − 𝜎𝑡𝑠) +𝑀𝑎𝑠𝑙
𝑍𝑏, 𝑐𝑜𝑚𝑝
The following table shows that beam sections do meet this criteria.
Table 7 - Verify the minimum section modulus
Min Modulus (mm^3) Actual Modulus (mm^3) Check Zbb 142013152.2 318185960.3 OK Zbcomp 183516380.7 251573774.6 OK
Prestressing Force Checks To check that the pre-stressing in the beam is at an appropriate amount the following checks are made.
𝑃𝑖 ≥𝑍𝑡, 𝑏 ∗ 𝜎𝑡𝑖 − 𝑀𝑖
𝑘𝑡 + 𝑒𝑏
In this design the eccentricity is below the neutral axis and causes the denominator of the above
formula to be negative, as this is the case the above inequality is reversed.
𝑃𝑖 ≥𝑍𝑏, 𝑏 ∗ 𝜎𝑡𝑠 − 𝑀𝑎𝑙𝑠 ∗
𝑍𝑏, 𝑏𝑍𝑏, 𝑐𝑜𝑚𝑝
− 𝑀𝐼𝐼
𝛼(𝑘𝑏 + 𝑒𝑏)
The table below shows that prestresses applied to this beam are acceptable.
Table 8 - Verify the minimum section modulus
Checking Pi (KN) Actual Pi (KN) Check Pi ≤ -11868.2 7882.9 OK Pi ≥ 6274.7 7882.9 OK
Cable limits Checks must also be made to determine that the location of pre-stressing cable are adequate at both
service and a transfer of the beam. Due to the addition of the concrete slab a transfer the neutral axis of
the beam changes and so too does the eccentricities of the pre-stressed cables. To determine if the
eccentricities of the cables are the values the following checks must be performed.
At transfer:
𝑒𝑏 ≥1
𝑝𝑖∗ (𝑍𝑡, 𝑏 ∗ 𝜎𝑡𝑖 − 𝑀𝑖) − 𝑘𝑡
7
At service:
𝑒𝑏 ≤1
𝛼𝑝𝑖∗ (𝑍𝑏, 𝑏 ∗ 𝜎𝑡𝑠 − 𝑀𝑎𝑠𝑙 ∗ (
𝑍𝑏, 𝑏
𝑍𝑏, 𝑐𝑜𝑚𝑝) − 𝑀𝐼𝐼) − 𝑘𝑏
The table below shows that the eccentricities caused by the pre-stressed cables in this beam are
acceptable.
Table 9 - Verify that the eccentricities are adequate
eb extremities (mm) Actual eb (mm) Check eb ≥ -591.979 -306.669 OK eb ≤ 118.356 -719.483 OK
Stress Checks Stress checks must be performed at each stage of the bridge construction process to determine if the
estimated stresses will be greater than the allowable stress. The stress checks were determined at the
mid span of the beam. Stage 1 of the stress check is determining whether the stresses in the beam at
transfer without a slab are acceptable. Stage 2 is determine if the stresses are acceptable when the wet
slab of concrete is placed over the beam and no forces other than self-weight are applied to the beam.
Stage 3 checks the beam when it is at service and is acting compositely with the slab to resisting the
combined live and dead loads of the bridge. The following table summarises the result of the stress
checks during the three stages.
Table 10 - Stress Checks
Stage 1
Top 8.77930381 < 20 ok
Bottom 19.6999737 < 20 ok
Stage 2
Top 12.0182898 < 25 ok
Bottom 17.490181 < 25 ok
Stage 3
Top 5.96970827 < 25 ok
Mid 17.0900537 < 25 ok
Bottom 2.78459611 < 25 ok
During stage 1 both the top and bottom sections of the beam are in compression and both are less than
maximum allowable compressive stress, so the beam is acceptable. In stage 2 both the top and bottom
sections of the beam are once again in compression and are less than the maximum allowable stresses.
8
In stage 2 there are no stresses in the slab as it is cannot resist strains in the bridge. In the final stage
when the beam is acting compositely with the slab the stress in both the slab and top and bottom of the
beam is less than the maximum allowable stresses. From this stress check it can be concluded that the
beam is designed appropriately.
Magnel diagram A Magnel diagram was produced to show that the section is designed accordingly. An eccentricity with a
maximum value was chosen to allow as small a prestressing force as possible to satisfy the design
criteria.
Figure 2 – Mangel Dieagram
De-bonding of prestressed cables The limit of the prestressing cable in the beam are shown in the figure below.
-5E-07
-4E-07
-3E-07
-2E-07
-1E-07
0
0.0000001
0.0000002
0.0000003
0.0000004
0.0000005
-1500 -1000 -500 0 500 1000 1500
1/P
i (N
-1)
eccentrcity (mm)
Euqation 1 Equation 2 equation 3 Equation 4
euqation 5 e (tranfer) e (services)
9
Figure 3 – Cable Limits
The limits between the upper and lower cable limits increase when the prestressing force is increased.
When the initial 36 strands was chosen for this the design the amount of prestressing force in the beam
was too low and the limits were too restrictive. By increasing the number of strands to 42 the
prestressing forced was increased and the limits became larger.
To allow straight section of prestressed sections de-bonding is required along the beam. By de-bonding
sections of the prestressed tendons the amount of force that they transfer at certain lengths of the
beam can be controlled. The figure below shows the about of prestressing forces required along the
length of the beam.
Figure 4 – Prestress vs Distance
0
200
400
600
800
1000
1200
1400
1600
0 5 10 15 20 25 30
Dis
tan
ce f
rom
so
ffit
?? T
OP
? (m
m)
Distance along beam (mm)
Cable Limits
Centroid
Lower
Upper
0
1000
2000
3000
4000
5000
6000
7000
8000
9000
0 5 10 15 20 25
Forc
e (
KN
)
Distance along beam
Prestress vs. Dist
10
The stresses as a result of the de-bonding along the beam can be seen in the table below.
Table 11 - Applied stresses
Length (m) σti σci σcs σts
0 2.06748562 11.4342808 1.55061421 8.57571062
2.5 3.73944534 10.2935892 8.567876 3.7881958
5 5.72902034 13.2178115 14.0980644 3.2264391
7.5 6.65788685 12.5840939 17.5365374 0.88054678
10 8.59353051 19.8267172 20.9168147 4.99676804
12.5 8.77930381 19.6999737 21.9099425 4.31920828
15 8.59353051 19.8267172 21.1922408 4.80885905
17.5 6.65788685 12.5840939 17.9002251 0.63242144
20 5.72902034 13.2178115 14.1965526 3.15924569
22.5 3.73944534 10.2935892 8.35007794 3.93678815
25 2.06748562 11.4342808 1.56365077 8.56681645 At 2.5m increments along the beam it can be seen that no stresses are great than the maximum
allowable stresses. The figure below also illustrates that the eccentricity in the beam is at a safe level at
transfer as well as service.
Figure 5 – Eccentricity Limits
-1500
-1000
-500
0
500
1000
1500
2000
0 5 10 15 20 25
e lower ti e lower ci e upper ts e upper cs
Beam top Beam Bottom Tendon
11
To meet the required prestressing forces the tendons were de-bonded according the table below. The
arrangement of the tendons within the beam can be seen in the appendix.
Table 12 - Tendons and de-bonding length Strands per row
Strand Number
1 2 3 4 5 6 7 8 9 10 11 12 13 14
Row C 7.5 0 7.5 0 2.5 0 2.5 2.5 0 7.5 0 7.5 0 2.5 14
Row B 0 7.5 0 7.5 0 7.5 0 0 7.5 0 7.5 0 7.5 0 14
Row A 2.5 0 7.5 0 7.5 0 2.5 2.5 0 2.5 0 7.5 0 7.5 14
Flexural Strength With the amount of prestressing steel required calculated, the moment capacity of the Super-T beam
was checked in order to ensure failure does not take place. The strength was checked at transfer and
service
The first step was to propose a possible depth of the neutral axis at failure for transfer and service. To
do this, and conduct the flexural strength check, it was assumed that the beam was under reinforced to
ensure ductile failure and the compression block was inside the flange.
The following table displays necessary parameters for calculation:
Table 13 - Flexural Strength Parameters
Ultimate Moment Capacity
Parameter Unit
ϕ - 0.8
ϒ - .766(
transfer) 0.696(service)
Ap mm2 6006
fp MPa 1750
(tb 6.3.1)
Ec MPa 34800
( tb3.1.2 AS3600)
Zt and Zb mm^3 217.08*10^6
and -318*10^6
With these necessary parameters determined, calculation was conducted to calculate the depth
of the neutral axis at failure for transfer, wet concrete and service.
𝑃𝑖 = 0.75 × 𝑓𝑝 × 𝐴𝑝 = .75 × 1750 × 6006 = 7882875𝑁
𝑃𝑒 = .85 × 𝑓𝑝 = 6306300𝑁
12
𝑓𝑝𝑦 = .85 × 𝑓𝑝 = 1487.5
Assuming that the strands have yielded:
𝐴𝑝𝑓𝑝𝑦 = .85𝑓′𝑐 × 𝛾𝑏𝑑𝑛
The difference between the service and transfer for these calculations comes solely from the
difference in the strength of concrete (40mpa vs 50mpa)
Solving for 𝑑𝑛 gives 171.52 𝑚𝑚2 at transfer and 151.013 𝑚𝑚2 at service. These values are
within the flange of the beam.
The next step is to determine strains for each component. This is to ensure that the steel yields
at failure, as we do not want failure, which is sudden and unpredictable.
휀𝑝𝑖 =𝑃𝑒
𝐸𝑝𝐴𝑝=
6306300
195 × 109 × 6006 × 10−6= 5384.6 𝑚𝑖𝑐𝑟𝑜𝑠𝑡𝑟𝑎𝑖𝑛
휀𝑒 =1
𝐸𝑐(
𝛼𝑃
𝐴+
𝛼𝑃 × 𝑒2
𝐼) 𝑤ℎ𝑒𝑟𝑒 𝛼𝑃 𝑖𝑠 𝑃𝑒 𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒
휀𝑒 =1
34800(
6306300
516198.5+
6306300 × (−306.7)2
1.55 × 1011) = 461.03 𝑚𝑖𝑐𝑟𝑜𝑠𝑟𝑎𝑖𝑛 𝑻𝑹𝑨𝑵𝑺𝑭𝑬𝑹
휀𝑒 =1
34800(
6306300
915644.1+
6306300 × (−719.5)2
2.18 × 1011) = 628.2 𝑚𝑖𝑐𝑟𝑜𝑠𝑟𝑎𝑖𝑛 𝑺𝑬𝑹𝑽𝑰𝑪𝑬
* Decompression strain differs between service and transfer due to the fact that the slab’s
composite action changes section properties including the distance of the eccentricity to the
centroid
휀𝑢 = .003 ×𝑑𝑝 − 𝑑𝑛
𝑑𝑛= .003 ×
1020 − 171.52
171.52= 14,840.9 𝑚𝑖𝑐𝑟𝑜𝑠𝑡𝑟𝑎𝑖𝑛 𝑻𝑹𝑨𝑵𝑺𝑭𝑬𝑹
휀𝑢 = .003 ×𝑑𝑝 − 𝑑𝑛
𝑑𝑛= .003 ×
1234 − 151.01
171.52= 21,514.46 𝑚𝑖𝑐𝑟𝑜𝑠𝑡𝑟𝑎𝑖𝑛 𝑺𝑬𝑹𝑽𝑰𝑪𝑬
13
* dp varies between service and transfer due to the fact that it is the distance from the top to
the eccentricity and composite action increases this distance, as a result effecting ultimate
strain for the section.
𝑒𝑝𝑠 = 휀𝑒 + 휀𝑢 = 461.03 + 14,840.9 = 15,301.93 𝑚𝑖𝑐𝑟𝑜𝑠𝑡𝑟𝑎𝑖𝑛 𝑻𝑹𝑨𝑵𝑺𝑭𝑬𝑹
𝑒𝑝𝑠 = 휀𝑒 + 휀𝑢 = 628.2 + 21,514.46 = 22142.66 𝑚𝑖𝑐𝑟𝑜𝑠𝑡𝑟𝑎𝑖𝑛 𝑺𝑬𝑹𝑽𝑰𝑪𝑬
휀𝑃𝑈 = 휀𝑝𝑖 + 휀𝑝𝑠 = 5384.6 + 15,301.93 = 20686.53 𝑚𝑖𝑐𝑟𝑜𝑠𝑡𝑟𝑎𝑖𝑛 𝑻𝑹𝑨𝑵𝑺𝑭𝑬𝑹
휀𝑃𝑈 = 휀𝑝𝑖 + 휀𝑝𝑠 = 5384.6 + 22,142.66 = 27527.26 𝑚𝑖𝑐𝑟𝑜𝑠𝑡𝑟𝑎𝑖𝑛 𝑺𝑬𝑹𝑽𝑰𝑪𝑬
휀𝑝𝑦 =𝑓𝑝𝑦
𝐸𝑝=
1487.5
195 × 10^3= 7630𝑚𝑖𝑐𝑟𝑜𝑠𝑡𝑟𝑎𝑖𝑛
Since 휀𝑝𝑢 ≫ 휀𝑝𝑦 , our initial assumption that fpy =sigmapy is okay.
Now we check the neutral axis depth ratio to ensure that it is below 0.4 as we require under-
reinforced type failure.
𝑘𝑢 =𝑑𝑛
𝑑𝑝=
171.52
1020= .168 < 0.4 𝑻𝑹𝑨𝑵𝑺𝑭𝑬𝑹
𝑘𝑢 =𝑑𝑛
𝑑𝑝=
151.013
1234= .122 < 0.4 𝑺𝑬𝑹𝑽𝑰𝑪𝑬
OK
We can now calculate the moment capacity at transfer and service as such:
𝜙𝑀𝑢 = 𝜙𝐴𝑝𝐹𝑃𝑌 (𝑑𝑝 −𝛾𝑑𝑛
2)
𝜙𝑀𝑢 = 0.8 × 6006 × 1487.5 × (1020 −. 766 × 171.52
2) = 6.82𝑀𝑁 𝑎𝑡 𝑻𝑹𝑨𝑵𝑺𝑭𝑬𝑹
𝜙𝑀𝑢 > 𝑀∗ = 1𝑀𝑁 𝑂𝐾
𝜙𝑀𝑢 = 0.8 × 6006 × 1487.5 × (1234 −. 696 × 151.013
2) = 8.44 𝑀𝑁 𝑎𝑡 𝑺𝑬𝑹𝑽𝑰𝑪𝑬
14
𝜙𝑀𝑢 > 𝑀∗ = 6.64 𝑀𝑁 𝑂𝐾
As expected, the moment capacity at transfer is lower than at service and this is due to the lack
of composite action at transfer. Additionally, a wet slab would provide the same results as
transfer due to the lack of composite action.
The calculations above display the fact that the flexural strength at service and transfer is
sufficient to carry the loadings determined from the LUSAS and SPACEGASS analysis from part
a. In this case, the number and type of strands provided sufficient flexural strength. It is
important to note however that in the event that the moment was not satisfied, extra
reinforcements would need to be provided based on the amount of moment capacity that is
exceeded. We would then calculate the additional number of required strands and the new
neutral axis depth to ensure that it is within the slab. With these parameters calculated, the
new moment capacity would be calculated and compared with M*.
Shear Design The complete shear capacity consists of the shear strength provided by concrete and the shear strength
provided by the stirrups.
Table 14 - Important parameters
Parameter Unit Value
Φ - 0.7
bv mm 200 Do=effective depth (depth-
cover) mm 1550
V*max kN
1125.38 ( obtained
from part a)
𝜷𝟏 - 1.1
𝜷𝟐 - 1
𝜷𝟑 - 1 Fcp(transfer strength of concrete)
Mpa 40
Pv 0
Apt Mm^2 6006
Do Mm 1380
15
With these parameters, listed, firstly, it was checked wether the section dimensions were
adequate for shear strength.
Section Capacity The beam is first required to have and adequate section to guard against crushing failure of the
concrete in the web.
𝜙𝑉𝑢𝑚𝑎𝑥 = 𝜙 × 0.2 × 𝑓′𝑐 × 𝑏𝑣𝑑0 = 0.7 × 0.2 × 50 × 200 × 1550 = 1750𝑘𝑛
𝜙𝑉𝑢𝑚𝑎𝑥 > 𝑉∗ = 1125.38𝑘𝑛
OK: Section dimensions are adequate for shear
Ultimate Strength of the Concrete In order to guard against flexural shear cracking, shear reinforcement may be required. The
shear strength of the concrete must first be determined by the following equation.
𝑉𝑢𝑐 = 𝛽1𝛽2𝛽3𝑏𝑣𝑑𝑜 ((𝐴𝑝 + 𝐴𝑠𝑡)𝑓𝑐
′
𝑏𝑣𝑑𝑜)
13
+ 𝑉𝑜 + 𝑃𝑣
𝑀𝑜 = ‖𝑍𝑏‖ × (𝛼𝑃
𝐴+
𝛼𝑃 × 𝑒
𝑍𝑏) = 251.6 × 106 × (
6306300
915644.1+
6306300 × −719.5
−251.6 × 106) = 4598𝑘𝑛𝑚
𝑉𝑜 =𝑀𝑜
𝑀𝑜∗ × 𝑉𝑜∗=
4598
1125.38 × 6644.04 = 778.83
We can now solve for Vuc
𝜙𝑉𝑢𝑐 = 0.7 × 1.1 × 200 × 1550 ((6006) × 50
200 × 1550)
13
+ 778.83 + 0 = 205.175𝑘𝑛
NOT OK
Web Shear Cracking We are now required to check the cracking in the web of the section (taken at the section
centroid).
The following equation allows us to find the shear strength of concrete, Vt.
𝑉𝑡 =𝑣 ∗ 𝐼𝑔 ∗ 𝑏𝑣
𝑄 𝑤ℎ𝑒𝑟𝑒 𝑄 𝑖𝑠 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑜𝑝 ℎ𝑎𝑙𝑓 𝑜𝑓 𝑡ℎ𝑒 𝑇 𝑏𝑒𝑎𝑚
𝑉𝑡 = 1078.11𝑘𝑛
16
Where v is the shear stress at the section centroid. This if found by knowing the principle tensile stress.
𝜎1 = 0.33 ∗ √𝑓𝑐′ = 2.33
𝜎 =𝑃
𝐴𝑔= 8.61
And solving for v in the following equation;
𝑣 → 𝜎1 = √(𝜎
2)
2
+ 𝑣2 −𝜎
2→ 𝑣 = 5.05
Where the normal stress is equal: Our capacity is then determined by the following equation:
𝜙𝑉𝑢𝑐 = 𝜙(𝑉𝑡 + 𝑃𝑣) = .7 × 1078.11 = 754.7𝑘𝑛
NOT OK Flexural and web shear cracking checks have failed but smallest capacity governs.
Shear Reinforcement addition
Since our previous checks have failed, we must input shear reinforcements. We can first check if minimum shear reinforcement is sufficient.
𝜙𝑉𝑢𝑚𝑖𝑛 = 0.7(𝑉𝑢𝑐 + 0.6 ∗ 𝑏𝑣 ∗ 𝑑𝑜) = 0.7 × (293.11 + .6 × 200 × 1550) = 310.175𝑘𝑛
𝜙𝑉𝑢𝑚𝑖𝑛 < 𝑉∗, 𝑀𝑜𝑟𝑒 𝑡ℎ𝑎𝑛 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑚𝑒𝑛𝑡 𝑖𝑠 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑.
𝑉∗ = 𝜙𝑉𝑢 = 𝜙(𝑉𝑢𝑐 + 𝑉𝑢𝑠) → 𝑉𝑢𝑠 =𝑉∗
𝜙−
𝑉𝑢𝑐
𝜙=
1125.38
0.7+
293.11
. 7= 1314.58𝑘𝑛
𝜃𝑣 = 𝜃𝑣 = 30 + 15 [𝑉∗ − 𝜙𝑉𝑢𝑚𝑖𝑛
𝜙𝑉𝑢𝑚𝑎𝑥 − 𝜙𝑉𝑢𝑚𝑖𝑛] = 45.085
Use 2 legs of R20 stirrups, resulting in an area of
𝐴𝑠𝑣 = 628.32
17
𝑆 = 628.32 ×250 × 1250 × 1
1314.58= 148.9𝑚𝑚
𝑆𝑚𝑎𝑥 = min (𝐷
2, 300) = min (
1380
2, 300) 𝑢𝑠𝑒 150𝑚𝑚 𝑠𝑝𝑎𝑐𝑖𝑛𝑔.
Finally: to ensure sufficient shear capacity, use 2 legs of R20 stirrups at a spacing of 150 mm
Pre-tensioning Losses The process of prestress loss applied in beams can mainly be divided into 2 stages. The first stage losses
occurs immediately when jacking force is applied in tendons, for pre-tensioned PSC, the elastic loss is
mainly occupied which caused by deformation of PSC. The second stage loss is time-depended loss, in
this stage, some loss like creep, shrinkage and relaxation should be taken into consideration, this stage
last from the moment of concrete was poured to 10000days. The data for following calculations is
presented as below:
parameters value
A 516198.5
I1 1.549E+11
I2 2.177E+11
Ec 34800
Ap 6006
Pi 7882875
Ep 195000
e -306.6689
Mi 1.008E+09
Msus 2.217E+09
Elastic Loss (I) Losses due to Elastic Shortening:
∆𝑃𝜀 =𝐴𝑝𝐸𝑝
𝐸𝑐(
𝑃𝑗
𝐴+
𝑃𝑗𝑒2
𝐼+
𝑀𝑖𝑒
𝐼)
=6006𝑚𝑚 × 195000𝑀𝑝𝑎
348000𝑀𝑝𝑎(
7882875𝑁
516198.5𝑚𝑚2+
7882875𝑁 × (−306.6689𝑚𝑚)2
1.549 × 1011𝑚𝑚4
+1008200195𝑁𝑚𝑚 × (−306.6689𝑚𝑚)
1.549 × 1011𝑚𝑚4 ) = 607859.8869𝑁
18
This is:
%∆𝑃𝜀 =∆𝑃𝜀
𝑃𝑖=
607859.8869𝑁
7882875N= 7.71%
Creep (TD) From the table 6.1.8 (A) we know that the basic creep factor is 2.0
The drying perimeter is the perimeter of half the exposed or drying perimeter (including half the
perimeter of any internal voids):
𝑢𝑒 = 𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟𝑒𝑥𝑡𝑒𝑟𝑛𝑎𝑙 +1
2𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙
= 2 × (2000 + 1380) + 0.5 × 2 × (1027 + 814 + 2 × (1380 − 255 − 292))
= 10267𝑚𝑚
The hypothetical thickness, th, is the ratio of the concrete area A to ue:
𝑡ℎ =2𝐴
𝑢𝑒= 178.36644𝑚𝑚
19
From table below we have:
𝑘2 = 0.76
For the maturity coefficient, since the age at first loading is 28 days, the strength ratio is 1.0, giving 𝑘3 =
1.1
Hence the design creep is:
∅𝑐𝑐 = 𝑘2𝑘3∅𝑐𝑐,𝑏 = 0.76 × 1.1 × 2 = 1.672
The sustained moment is:
𝑀𝑠𝑢𝑠 = 𝑀𝑖 + 𝑀𝑆𝐼𝐷𝐿 + 𝑀𝐿𝐿𝜓
= 1008200195𝑁𝑚𝑚 + 703125000𝑁𝑚𝑚 + 2526254805𝑁𝑚𝑚 × 0.2
= 2.21 × 109𝑁𝑚𝑚
Therefore, we can calculate the creep loss:
∆𝑃𝑐𝑟𝑒𝑒𝑝 =𝐴𝑝𝐸𝑝∅𝑐𝑐
𝐸𝑐(
𝑃𝑖
𝐴+
𝑃𝑖𝑒2
𝐼+
𝑀𝑠𝑢𝑠𝑒
𝐼)
=6006𝑚𝑚 × 195000𝑀𝑝𝑎
348000𝑀𝑝𝑎(
7882875𝑁
915644.1𝑚𝑚2+
7882875𝑁 × (−306.6689𝑚𝑚)2
1.177 × 1011𝑚𝑚4
+2216576156Nmm × (−306.6689𝑚𝑚)
1.177 × 1011𝑚𝑚4 ) = 500353.07𝑁
This is:
20
%∆𝑃𝑐𝑟𝑒𝑒𝑝 =∆𝑃𝑐𝑟𝑒𝑒𝑝
𝑃𝑖=
500353.07𝑁
7882875N= 6.35%
Shrinkage Loss (TD) The basic shrinkage strain is set as 휀𝑐𝑠,𝑏 = 850 × 10−6
As mentioned before
𝑡ℎ =2𝐴
𝑢𝑒= 178.36644𝑚𝑚
From the table below
𝑘1 = 0.72
The design shrinkage strain is:
휀𝑐𝑠 = 𝑘1휀𝑐𝑠,𝑏 = 0.72 × 850 × 10−6 = 6.12 × 10−4
Since there is no reinforcement in the section:
∆𝑃𝜀 = 𝐴𝑝𝐸𝑝휀𝑐𝑠 = 0.72 × 0.00085 × 195000𝑀𝑝𝑎 × 6006𝑚𝑚 = 716756.04N
This is:
%∆𝑃𝑠ℎ𝑟𝑖𝑛𝑘𝑎𝑔𝑒 =∆𝑃𝑠ℎ𝑟𝑖𝑛𝑘𝑎𝑔𝑒
𝑃𝑖=
716756.04N
7882875N= 9.09%
21
Relaxation Loss (TD) For typical strand:
𝑅𝑏 = 2%
𝑘4 = 𝑙𝑜𝑔 [5.4𝑗1
6⁄ ] = 𝑙𝑜𝑔 [5.4 × 30 × 3651
6⁄ ] = 1.4056
Since 𝑓𝑝 = 0.75
𝑘5 = 1.25
𝑘6 = 1
Hence:
𝑅 = 𝑘4𝑘5𝑘6𝑅𝑏 = 1.4056 × 1.25 × 1 × 2% = 3.51%
%∆𝑃𝑅 = 𝑅 (1 −∆𝜎𝑐+𝑠ℎ
𝜎𝑝𝑖) = 𝑅𝑃𝑖 = 3.51% × 8772875𝑁 = 277010.031N
Total loss Total losses are thus:
%∆𝑃𝑡𝑜𝑡𝑎𝑙 = %∆𝑃𝜀 + 𝑅 + %∆𝑃𝑠ℎ𝑟𝑖𝑛𝑘𝑎𝑔𝑒 + %∆𝑃𝑐𝑟𝑒𝑒𝑝
= 7.71% + 6.35% + 9.09% + 3.51% = 26.76%
And the total loss is:
22
𝑃𝑡𝑜𝑡𝑎𝑙 = ∆𝑃𝜀 + ∆𝑃𝑐𝑟𝑒𝑒𝑝 + ∆𝑃𝑠ℎ𝑟𝑖𝑛𝑘𝑎𝑔𝑒 + %∆𝑃𝑅 = 2101979𝑁
Deflection In order for the bridge design to be satisfactory it must have a small suitable deflection, ensuring users
feel comfortable using the bridge. In order to satisfy these conditions, deflection must be determined
for both short-term and long-term.
Deflection at Transfer Before Slab Initially, the Super T-beams must satisfy deflection conditions under transfer loads only. The
values shown below in Table xx were found using the following equations. In order to carry out
these equations we must covert our forces acting on the bridge into a UDL this is done using
the bottom equation below, which is appropriate to work out deflection under a UDL.
𝛿𝑇𝑜𝑡𝑎𝑙 = 𝛿𝐿𝑜𝑎𝑑𝑠 + 𝛿𝑃𝑟𝑒𝑠𝑡𝑟𝑒𝑠𝑠
𝛿𝑇𝑜𝑡𝑎𝑙 = 5𝑤𝐿4
384𝐸𝐼+
𝑃𝑒𝑒𝐿2
8𝐸𝐼
𝑤𝑒𝑞 =8𝑀∗
𝐿2
Table 15 - Deflection at Transfer
Parameter Unit Value
L m 25
E Mpa 32000
I 𝑚𝑚4 1.54851E+11
𝑷𝒆 KN 5780.895973
e mm -306.66891
weq KN/m 12.907
Deflection mm -14.07
As shown above, the deflection due to transfer loads is equal to 14.07 mm upwards. To satisfy
deflection standards, the deflection must be less than L/600, being 41.67 mm
Deflection = 14.07 mm < Dl = 41.67 mm
Therefore, the deflection at transfer is adequate.
23
Deflection at Transfer With Slab Initially, the Super T-beams must satisfy deflection conditions under transfer loads only. In this
section the geometric properties of the beam includes the 180mm slab. The values shown
below in Table xx were found using the following equations.
Table 16 - Deflection at Transfer
Parameter Unit Value
L m 25
E Mpa 34800
I 𝑚𝑚4 2.18+11
𝑷𝒆 KN 5780.895973
e mm -719.48
weq KN/m 21.904
Deflection mm -13.24
As shown above, the deflection due to transfer loads is equal to 13.25 mm upwards. To satisfy
deflection standards, the deflection must be less than L/600, being 41.67 mm
Deflection = 13.25 mm < Dl = 41.67 mm
Therefore the deflection at transfer is adequate.
Short-Term Deflection In order to determine the short-term deflection due to the live load, the moments acting on the
beam must be turning into an equivalent UDL this is done by converting the maximum moment
acting at SLS conditions into a UDL.
Table 17 - Short Term Deflection
Parameter Unit Value
L m 25
E Mpa 34800
I 𝑚𝑚4 2.18+11
𝑷𝒆 KN 5780.895973
e mm -719.48
weq KN/m 54.24
Deflection mm 8.459
As shown above, the short term deflection is equal to 8.459mm downwards. To satisfy
deflection standards, the deflection must be less than L/600, being 41.67 mm
24
Deflection = 8.459 mm < Dl = 41.67 mm
Long Term Deflection The total deflection experienced by the Super T-beam is simply the short term deflection
multiplied by a factor, 𝑘𝑐𝑠 where 𝑘𝑐𝑠 is determined by the amount of prestressing and
reinforcement in the Super T-beam.
𝛿𝐿𝑜𝑛𝑔𝑇𝑒𝑟𝑚 = 𝑘𝑐𝑠𝛿𝑆ℎ𝑜𝑟𝑡𝑇𝑒𝑟𝑚
𝑘𝑐𝑠 = (2 − 1.2 (𝐴𝑠𝑐
𝐴𝑠𝑡)) > 0.8
Since we did not have reinforcement in the Super T-beams the 𝑘𝑐𝑠 factor equalled to two.
Therefore the long term deflection was twice that of the short term deflection. The deflection
limit of the beam is equal to L/600, being 41.67 mm (AS5100.2).
𝛿𝐿𝑜𝑛𝑔𝑇𝑒𝑟𝑚 = 16.919mm < Dl = 41.67 mm
Therefore, deflection of the Super T-beam is adequate.
25
Slab Design The overlay slab for this bridge provides the support necessary to carry the wearing surface of the road
and to transfer those loads on to the type 3 Super-T beams. The slab has been designed in the
transverse direction with a design width of 1m used for checking strength capacities. The overlaying slab
will need to be reinforced with steel bars to ensure it has adequate strength in flexure. To ensure that
shrinkage is not an issue for the slab an appropriate amount of steel must be provided in both the
longitudinal and transverse direction. As the slab acts compositely with the beams the deflection will be
governed by the beams and will not need to be considered in the slab design.
To design the slab the critical location for the slab in relation to the Super-T beams needs to be
determined. This maximum distance was either the gap between the webs of one super T beam or the
distance between the webs of adjacent super T beams. Given the geometry of the type 3 Super-T beam
this distance was calculated as follows
𝑆𝑝𝑎𝑛 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑤𝑒𝑏𝑠 =1027 + 840
2= 933.5𝑚𝑚
𝑆𝑝𝑎𝑛 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑤𝑒𝑏𝑠 𝑜𝑓 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑏𝑒𝑎𝑚𝑠 = 2000 − 933.5 = 1066.5𝑚𝑚
For the design of slab the span of 1.067m was used as this will provide the largest bending moments and
shear. While the slab acts as a continuous beam across all of the super T beam beams this span length
was modelled as fixed at both ends. The slab was designed using the same procedure as a singular
reinforced concrete beam however in reality the reinforcement will be placed in the top and bottom of
the slab to resist both hogging and sagging of the beam.
The design followed the procedure for that of a singly reinforced concrete slab, however in practice the
determined reinforcement arrangement will be used on the top and bottom of the slab, to resist the
hogging and sagging that occurs. When designing the amount of reinforcement need for the slab a cover
of 35mm was used.
Loading For the loading of the slab three parts were considered, the dead load, the superimposed dead load and
the live load acting on the bridge. The most critical span between super T beams would between the
centre beam and the one next to it or the outside beam of the bridge and the one next to that. This is
due to the locations of where the W80 wheel load could possibly be acting. For the design the loading
will be based on this section of the bridge, where the asphalt was only considered as part of the super
imposed dead load. The dead loads for the slab were calculated as shown below.
𝑆𝑒𝑙𝑓 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑙𝑎𝑏 = (1.8𝑚 ∗ 1𝑚) ∗ 25 𝑘𝑁/𝑚3
𝑆𝑒𝑙𝑓 𝑤𝑒𝑖𝑔ℎ𝑡 = 4.5 𝑘𝑁/𝑚3
𝐴𝑠𝑝ℎ𝑎𝑙𝑡 𝐿𝑜𝑎𝑑 = (0.055𝑚 ∗ 1𝑚) ∗ 22 𝑘𝑁/𝑚3
𝐴𝑠𝑝ℎ𝑎𝑙𝑡 𝐿𝑜𝑎𝑑 = 1.21 𝑘𝑁/𝑚3
26
Applying load factors of 1.2 for the self-weight and 2 for the SIDL, the ultimate limit state dead load can
be found.
𝑈𝐿𝑆 𝐷𝑒𝑎𝑑 𝐿𝑜𝑎𝑑 = (1.2 ∗ 4.5) + (2 ∗ 1.21) = 7.82 𝑘𝑁/𝑚3
For the live loads on the slab the W80 wheel load acts as a UDL 400mm wide. The total load of the UDL
is 80kN and it will be positioned at mid span of the considered slab. As the load is positioned at the top
of the beam, the actual span at the middle height of the cross section of the slab needs to be
determined. This is done based on a 45° stress distribution. That span can be found as follows
𝑙𝑜𝑎𝑑 𝑠𝑝𝑎𝑛 = 0.4 + 2 ∗ (0.180
2+ 0.055) = 0.69𝑚
Where 0.4m is the width of tyre, and 0.18 is the thickness of the slab and 0.055 is the thickness of the
asphalt. From this the wheel load can be described as a UDL as shown below.
𝐿𝑖𝑣𝑒 𝐿𝑜𝑎𝑑 =80
0.69= 115.94 𝑘𝑁/𝑚3
Applying the load factor of 1.8 and also the DLA of 0.4 the ULS loading can be found as shown in the
calculation below.
𝑈𝐿𝑆 𝐿𝑖𝑣𝑒 𝐿𝑜𝑎𝑑 = (1 + 0.4) ∗ (1.8 ∗ 115.94) = 292.17 𝑘𝑁/𝑚3
Flexure Design The design for bending moment was calculated based on the moments developed in the beam by dead
and live loads separately and then added together using superposition. The critical moment was known
to be the hogging moment which occurs at the reactions.
𝑀∗(𝑑𝑒𝑎𝑑 𝑙𝑜𝑎𝑑) =𝑈𝐿𝑆 𝑑𝑒𝑎𝑑 𝑙𝑜𝑎𝑑 ∗ 𝑙2
12
𝑀∗(𝑑𝑒𝑎𝑑 𝑙𝑜𝑎𝑑) =7.82 ∗ 1.0672
12= 0.7419 𝑘𝑁𝑚
To determine the moment due to the live load the following formula was used.
𝑀𝑎 = −𝑀𝑏 =𝑤 ∗ 𝑙2
12∗ (1 −
6𝑎2
𝑙2+
4𝑎3
𝑙3)
Where a is the distance either side on the live load UDL acting on the considered slab span.
27
𝑀∗(𝑙𝑖𝑣𝑒 𝑙𝑜𝑎𝑑) =292.17 ∗ 1.0672
12(1 − 6 (
0.18852
1.0672 ) + 4 (0.18853
1.0673 )) = 23.14 𝑘𝑁𝑚
The total moment could then be found by adding both the moment caused by the dead loads to the
moment caused by the live loads. As shown below.
𝑀∗ = 0.7419 + 23.14 = 23.88 𝑘𝑁𝑚
To determine the moment capacity of the slab the design principles were taken as the same for
designing a reinforced concrete beam. The design procedure for the capacity is shown below.
Ø𝑀𝑢 = 0.8𝐴𝑠𝑡𝑓𝑠𝑦𝑑 (1 −𝛾𝑘𝑢
2)
𝑘𝑢 =𝐴𝑠𝑡𝑓𝑠𝑦
𝛼2𝑓′𝑐𝛾𝑏𝑑
𝛾 = 0.85 − 0.007(𝑓′𝑐 − 28)
𝛼2 = 1 − 0.003𝑓′𝑐
𝐴𝑠𝑡 – The area of steel in tension
𝑓𝑠𝑦 - The yield stress of steel (400 MPa steel was used)
𝑑 - The distance from the extreme compressive fibre to the centroid of the tensile reinforcement
𝑏 - The width of the slab
𝑓′𝑐 - The characteristic compressive strength of concrete
𝛾 = 0.85 − 0.007(50 − 28) = 0.696
𝛼2 = 1 − 0.003 × 50 = 0.85
To determine the minimum about of steel required to provide adequate capacity again bending a trial
and error approach was done via excel. This led to an arrangement of using 14mm diameter bars at a
spacing of 250mm. The area of steel can be calculated as followed.
𝐴𝑠𝑡 = 4 ∗𝜋 ∗ 142
4= 615.75 𝑚𝑚2
𝑘𝑢 =615.75 ∗ 400
0.85 ∗ 50 ∗ 0.696 ∗ 1000 ∗ 138= 0.0603
28
Ø𝑀𝑢 = 0.8 ∗ 615.75 ∗ 400 ∗ 138 (1 −0.696 ∗ 0.0603
2) = 26.62 𝑘𝑁𝑚
Ø𝑀𝑢 > 𝑀∗, 𝑆𝑜 𝑡ℎ𝑒 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑖𝑠 𝑔𝑜𝑜𝑑
Shear Design To design the beam to resist shear the maximum shear forces experienced needed to calculated. The
edge of the slab near the web of the Super-T beam is where the maximum shear forces are experienced.
Due to the symmetry of the load and the slab, the maximum shear force at the reactions will be half that
of the total loads. The calculation for determining the maximum shear acting on the slab is shown
below.
𝑉∗ =1
2(
𝑈𝐿𝑆 𝑑𝑒𝑎𝑑 𝑙𝑜𝑎𝑑 ∗ 𝑙
2+
𝑈𝐿𝑆 𝑙𝑖𝑣𝑒 𝑙𝑜𝑎𝑑 ∗ 𝑙
2)
𝑉∗ =7.82 ∗ 1.067
2+
292.17 ∗ 0.69
2= 104.97 𝑘𝑁
The design shear capacity of a concrete slab without shear reinforcement can be found using the
following formula
Ø𝑉𝑢 = 0.7 ∗ 0.17√𝑓′𝑐𝑏𝑑𝑜
𝑑𝑜 - The distance from extreme compressive fibre to the bottom of the tensile reinforcement
Ø𝑉𝑢 = 0.7 × 0.17√50 × 1000 × 145 = 122.01 𝑘𝑁
Ø𝑉𝑢 > 𝑉∗, 𝑆𝑜 𝑡ℎ𝑒 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑖𝑠 𝑔𝑜𝑜𝑑
Therefore the slab does not need to be designed for shear reinforcement.
Shrinkage and Temperature Design The effects of shrinkage due to temperature change can be resisted by providing a minimum amount of
steel in both the longitudinal and transverse direction. The minimum amount of steel required to resist
shrinkage is calculated below.
𝐴𝑠𝑡𝑚𝑖𝑛 = 0.0025𝑏𝑑
When considering the slab in the longitudinal direction the minimum amount of steel needed is as
followed.
𝐴𝑠𝑡𝑚𝑖𝑛 = 0.0025 ∗ 1000 ∗ 180 = 450 𝑚𝑚2
This is less than the amount of steel already provided to resiting bending moments so no more steel
reinforcement is needed.
29
When considering the slab in the transverse direction the minimum amount of steel needed is as
followed.
𝐴𝑠𝑡𝑚𝑖𝑛 = 0.0025 ∗ 1067 ∗ 180 = 480.15 𝑚𝑚2
To provide this minimum amount of steel in the transfer direction 14mm diameter bar with a spacing of
250mm should also be used. By using the same bars and arrangements in the longitudinal direction the
construction of the slab is simplified. The final slab design will have a symmetrical reinforcement
arrangement with 35mm of cover in both the top and bottom section of the slab. There will also be
reinforcement in the longitudinal direction positioned closest to the surface of the overlaying slab.
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Conclusion
In this report, we determined the necessary reinforcement required for the five Super T-beams used in
our bridge, as well as the amount of 15.2mm diameter strands required for appropriate prestressing.
We decided to use 42 strands in a combination using three rows with 14 strands in each row. Using
appropriate calculations we were able to determine the prestress losses at 26.67% of the original
prestress force. Our beams also passed appropriate stress checks and strength calculations. Graphs such
as the Magnel Diagram were created to visually show lower and upper limits for eccentricities possible
for our beam cross section. Two legs of R20 stirrups are required to satisfy shear design for the beams.
Lastly our bridge underwent deflection checks to ensure it was within the limits of span/600, as it only
deflected 16.17mm over the long term, deflections were well within the allowable limits and our slab
was reinforced with 14mm diameter bars with 250mm spacing, while keeping a symmetrical
reinforcement arrangement.
