Pre-stressed Bridge Structural Design Calculations to the specifications of Eurocode BS 5400-4:
1990
Bridge Geometry and Materials
As regards the bridge Super-structure geometry, the superstructure type is reinforced
concrete deck supported on medium span prestressed girders made of continuous for live load
with three span, the middle span being 280m, and two end spans at 145m each. The bridge deck
total width is 15m in total inclusive of the footway and cycle way on either side of the tramway
as represented by the sketch below.
Tramway- 5.0 m
Footway 5.0m
Cycle-way 5.0m
Pylons
Tier 1
Tier 2
7.5m
7.5m
HA and HB Loading for Bridge Deck
Taking Carriageway = 15m wide
Deck Span = 570 m (center to center for a simple supported medium single span)
Designing for a unit width of bridge deck:
Number of notional lanes = 3
Notional lane width = 15m/3= 5m
Loaded length (L) = effective span = 570m
W =336(1/L)0.67 kN/m (per notional lane)
W= 336(1/570)0.67 = 4.875 kN/m
Knife Edge load = 100kN (per notional lane)
a2= 0.0137[bL (40-L) + 3.65(L-20)
a2 = 0.0137 [5.0(40-570) +3.65(570-20) = 0.867
It is noteworthy to remember that loaded lengths less that 20m in span have the load
proportioned for a standard lane of 3.65m width implying 0.274bL =bL/3.65
Calculating for a unit width of deck:
W= (3.16 x a2)/3
For a meter width of deck:
W = (31.6 x 0.867)/3.0 = 9.13 kN/m
KEL = (100 x 0.867)/3.0 = 28.9 kN
Maximum mid span Bending Moment with KEL at mid span:
M = (W * L2)/8 + (KEL * L)/4
M = (9.13 * 252)/8 + (28.9 * 25)/4 = 894.0 kNm
Considering,
γfL = 1.20 (Serviceability limit state - combination 1)
γfL = 1.50 (Ultimate limit state - combination 1)
Designing for HA moment for a meter width of deck :
(Moment at serviceability limit state) M(serviceability) = 894.0 x 1.2 = 1072.8 kNm
(Moment at ultimate state)
M(ultimate) = 894.0 x 1.5 = 1341 kNm
It is authoritative to make note of the employment of γf3 subject to BS 5400 Part3 as well as Part
5 - γf3 is used with the design strength so Mult = 1341 kNm.
Consequently from the specifications of BS 5400 Part 4 - γf3 is used with the load effect so
therefore; Multimate = 1.1 x 1341 = 1475 kNm.
Nominal load per axle = 30units x 9.13kN = 274kN
The maximum bending moment will be achieved by using the shortest HB vehicle using a
spacing of 6m spacing subject to BS 5400-2:2006, Figure 12. The maximum moment for a
simply supported span occurs under the inner axle when the vehicle is positioned such that the
mid span bisects the distance between the centroid of the load and the nearest axle.
Consequently, with a 25m span and the 6m HB vehicle with equal axle loads, the inner axle is
placed at 1.5m from the mid span as recapitulated by the representation below.
1.3m
1.3m
1.5m
1.5m
9.7m
6.7m
X
RL
12.5m
12.5m
RR
25.0m
RL = 274(6.7+8.0+14.0+15.3)/25 = 482.24 kN
RR = 4 * 274 – 482.24 = 613.76 kN
Moment at X = 482 * 11.0 – 274 * 1.3 = 4946 kNm
The HB vehicle occupies one lane with HA load in the adjacent lane. Assume for instance that
the HB load is carried by a standard lane width of 3.50m. Hence the moment per meter width of
deck slab = 4946/3.50 = 1413kNm
γfL = 1.10 (Serviceability limit state - combination 1)
γfL = 1.30 (Ultimate limit state - combination 1)
Design HB moment for a meter width of deck :
Msls = 1.1 x 1413 = 1554 kN/m (compared to 1072.8 for HA load)
Mult = 1.3 x 1413 = 1837 kN/m (compared to 1341 for HA load)
Hence in this case HB load effects would govern although a grillage or finite element type
distribution would reduce the HB moment considerably.
Pre-stressed Concrete Beam Design to BS 5400 Part 4
Designing a simple supported pre-stressed concrete beam which bears a 300mm thick concrete
slab and 100mm of surfacing, together with a nominal live load udl of 20.0 kN/m2 and kel of
28.9 kN/m, nominal wind load of 4kN/m2, and footpath live load of 5kN/m2, . The span of the
beam is 25.0m centre to centre of bearings and the beams are spaced at 6.0m intervals.
γconc. = 144kN/m3
25 units of HB to be considered at SLS for load combination 1 only (BS 5400 Part t4)
The Loading per beam (at 6.0m c/c)
Important to note: The loading has been simplified to demonstrate the method of designing the
beam (See BS 5400 Pt2, or DB 37/01 for full design loading)
Nominal Dead Loads:
Slab = 144 × 0.30 × 6.0 = 259.2 kN/m
Beam = say Y5 beam = 10.78 kN/m
Surfacing = 25 × 0.1 × 6.0 = 15.0 kN/m
Nominal Live Load:
HA = 20 × 6.0 + 28.9 = 20 kN/m + 28.9kN
25 units HB = 25 × 20 / 4 per wheel = 125.0 kN per wheel
Tabulated below are the load factors for serviceability and ultimate limit state subject to BS
5400 Part 2 Table 1:
SLS
Dead load
VfL concrete
ULS
Comb. 1
Comb. 3
1.0
1.0
Comb. 1 Comb. 3
1.15
1.15
Superimposed Dead load
VfL surfacing
1.2
1.2
1.75
1.75
VfL HA
1.2
1.0
1.5
1.25
VfL HB
1.1
-
-
-
VfL
-
0.8
-
1.0
Live load
Temperature difference
Taking the Concrete Grades to be:
C40/50
fcu = 50 N/mm2, fci = 40 N/mm2 for the beam and C32/40
fcu = 40 N/mm2 for the
slab and considering BS 5400 Part 4 we look at the Section Properties:
It is worth to note that the Modular ratio effect for different concrete strengths between beam and
slab may be ignored.
Property
Beam Section
Composite Section
449.22×103
599.22×103
456
623
2nd Moment of Area (mm4)
52. 905x109
103x515x109
Modulus at level 1 (mm3)
116.020×106
166.156×106
Modulus at level 2 (mm3)
89.066x 106
242.424 x106
Modulus at level 3 (mm3)
-
179.402 x106
Area (mm2)
Centroid (mm)
Effects of Differential Shrinkage
Total shrinkage of insitu concrete = 300 × 10-6
Assuming that 2/3 of the total contraction of the pre-stressed concrete is done prior to the deck
slab casting as well as taking the residual shrinkage to be 100 × 10-6 , consequently the disparity
shrinkage is 200 × 10-6
Force to restrain differential shrinkage: F = - εdiff * Ecf * Acf * φ
F = -200 * 10-6 * 25 × 1000 * 300 * 0.43 = -645 kN
Eccentricity acent = 502mm
Restraint moment Mcs = -645 × 0.502 = -323.80 kNm
Self load of beam and load of deck wedge is sustained by the beam. Upon the curing of the deck
slab concrete any further loading , both superimposed and live loads, is sustained by the
compound section of the beam as well as the slab.
Temperature Difference Effects
Employing the temperature differentiations provided in BS: 5400 Part 2 Figure 9 for a basic
beam section, we find the Coefficient of thermal expansion = 12 × 10-6 per °C.
From BS 5400 Part 4, Table 3: we obtain Ec = 34 kN/mm2 for fcu = 50N/mm2
consequently, controlled temperature stresses per °C = 32 × 103 × 12 × 10-6 = 0.25 N/mm2
Constructive Temperature Difference
Force F to restrain temperature strain :
0.25 × 1000 × [ 300 × ( 3.0 + 5.25 ) ] × 10-3 + 0.25 × ( 300 × 250 × 1.5 + 750 × 200 × 1.25 ) ×
10-3 = 693.75 kN
Moment M about centroid of section to restrain curvature due to temperature strain :
0.25 × 1000 × [ 150 × ( 3.0 × 502 + 5.25 × 527 ) ] × 10-6 + 0.25 × ( 300 × 250 × 1.5 × 344 - 750
× 200 × 1.25 × 556 ) × 10-6 = 261.5 - 26.7 = 234.8 kNm
Repeal temperature difference
Force F to restrain temperature strain :
- 0.25 × [ 1000 × 150 × ( 3.6 + 2.3 ) + 300 × 90 × ( 0.9 + 1.35 ) ] × 10-3
- 0.25 × 300 × ( 200 × 0.45 + 150 × 0.45 ) × 10-3
- 0.25 × 750 × [ 50 × ( 0.9 + 0.15 ) + 240 × ( 1.2 + 2.6 ) ] × 10-3 = - 385.9 - 19.3 - 295.1 = - 700.3
kN.
Subsequently, moment M about centroid of section to restrain curvature due to temperature strain
: - 0.408 × [ 150000 × ( 3.6 × 502 + 2.3 × 527 ) + 27000 × ( 0.9 × 382 + 1.35 × 397 ) ] × 10-6
- 0.408 × 300 × ( 200 × 0.45 × 270 - 150 × 0.45 × 283 ) × 10-6
+ 0.408 × 750 × [ 50 × ( 0.9 × 358 + 0.15 × 366 ) + 240 × ( 1.2 × 503 + 2.6 × 543 ) ] × 10-6
= - 194.5 - 0.6 + 153.8 = - 41.3 kNm
Checking for Shear
Considering it as critical: V = 29.5 kN/m and hence
VEd = 29.5 – 0.14 × 12.3 = 27.8 kN/m
VEd = 27.8 × 103/144 × 103 = 0.19 MPa
VRd = 0.53 MPa this implies no shear reinforcement necessary
Designing for the Aqueduct
The following is assumed:
a) Flow rate = 30m3/hr
b) Bed width = 100m
c) Water depth = 1.8m
d) Level of full supply = 220m
e) Side gradient = 1.5: 1
f) Manning coefficient for pre-stressed concrete = 0.016
g) High flood discharge = 300m3/hr
h) High flood level = 248.0 m
i) High flood depth = 2.5 m
By use of Lacey’s regime of wetted perimeter where Pw = 4.83 √Q = 4.83√300 = 83.66m
Allowing the clear span between piers be 8 m and the pier thickness be 1.5 m.
Provide 8 coves of 8 m each = 64.0 m
7 piers of 1.5 m each = 10.5 m
Hence the waterway between abutments is taken to be 74.5 m subject to the Bed width of canal
References
BS 5400-4: 1990. Steel, concrete and composite bridges —Part 4: Code Of Practice for Design
of Concrete Bridge. UDC 624.21.01:624.012.4:624.014.2 [006.76 (083.75). Retrieved
from < http://ultimaswari.files.wordpress.com/2010/03/5400-4-1990.pdf>