5. ADVANTAGES OF PRESTRESSED
CONCRETE
Section remains un-cracked under service loads
Reduction of steel corrosion
Increase in durability.
Full section is utilized
Higher moment of inertia (higher stiffness)
Less deformations (improved serviceability).
Increase in shear capacity
Suitable for use in pressure vessels, liquid retaining structures.
Improved performance (resilience) under dynamic and fatigue loading.
B) High span-to-depth ratios
Larger spans possible with prestressing (bridges, buildings with large column-
free spaces)
For the same span, less depth compared to RC member.
Reduction in self weight
More aesthetic appeal due to slender sections
More economical sections.
6. DISADVANTAGES OF PRESTRESS
CONCRETE
If wires/strands are stressed individually
inside the Prestressing needs skilled
technology. Hence, it is not as common as
reinforced concrete.
The use of high strength materials is costly.
There is additional cost in auxiliary
equipments.
There is need for quality control and
inspection.
8. REFERENCE PROBLEM:-
DESIGN OF POST –TENSIONED PRESTRESSED
CONCRETE TEE BEAM AND SLAB BRIDGE DECK
FOR A NATIONAL HIGHWAY CROSSING TO
SUIT THE FOLLOWING –
9. GIVEN DATA :
EFFECTIVE SPAN= 30m
WIDTH OF ROAD= 7.5m
KERBS= 600mm on each side
FOOTPATH= 1.5m wide on each side
THICKNESS OF WEARING COAT= 80mm
LIVE LOAD= I.R.C class AA tracked
vehicle
10. For deck slab , adopt M20 grade concrete
for prestressed concrete girders , adopt
M50 grade concrete with cube strength at
transfer as 40N/mm2.
Loss ratio= 0.85
Spacing of cross girder= 5m
Adopt Fe-415 grade HYSD bars
Strands of 15.2mm – 7ply conforming to
IS:6006-1983 are available for use.
11. DESIGN THE GIRDERS AS CLASS I
TYPE MEMBERS
Permissible stresses:
For M20 grade concrete and Fe-415 grade Hysd bars (IRC:
21-2000)
σ cb= 6.7N/mm2
σ st= 200N/mm2
m= 10
n=[1/(1+((σ st)/(m×σ cb)) )]=0.25 ; j=(1-n/3)=0.916
Q= 0.5 σ cb nj = 0.767
For M50 grade concrete (I.R.C:18-2000)
fck=50N/mm2
fci=40N/mm2
fct=0.45fci=(0.45 *40) =18N/mm2
fw=0.33fck =(0.33*50) = 16N/mm2
ftt=ftw =0 (clss 1 type member)
12. Ec= 5700 𝑓𝑐𝑘 =5700 √50 = 40305 N/mm2= 40KN/mm2
CROSS-SECTION OF DECK :
4 main girdes are provided at 2.5 m intervals.
Thickness of deck slab =250mm
Wearing coat= 80mm
Kerbs 60mm wide by 300mm deep are provided.
The main girders are precast & the slab connecting the girders is CAST IN SITU.
Spacing of cross girders= 5m
Spacing of main girders= 2.5m
MODULUS OF ELASTICITY OF CONCRETE IN GIRDERS
13. DESIGN OF INTERIOR SLAB PANEL
1. BENDING MOMENT
Dead wt. of slab =(1x1x0.25x24)= 64KN/m2
Dead wt. of W.C.= (0.08x22x1x1)= 1.76KN/m2
Total dead load= 7.76KN/m2
Live load is IRC class AA tracked vehicle , one wheel is placed at the
center of the panel.
V=3.7 m
B=2.5m
14. 𝑢 = 𝑥 + 2 × 𝐷 = 0.85 + 2 × 0.08 = 1.01𝑚
X=wheel load contact area along the
span
D=Depth of the wearing coatU=effective length of load dispersion
v=(x'+2 × D)=(3.6+2×0.08)=3.76m
X=1.01m
X=0.85m
350KN
80mm
80mm
360KN
X=3.76m
X=3.6m
15. (U/B)=(1.01/2.5)=0.404
(V/L)=(3.76/5.0)=0.752
K=B/L=2.5/5.0=0.5
Referring to Pigeaud’s curves
m1=0.098 & m2=0.02
MB=W(m1+0.15m1)
=350(0.098+0.15x0.02)=35.35KNm
Similarily, ML=W(m2+.15m1)
=350(0.02+0.15x0.098)=12.14KNm
As the slab is continuous, design
BM=0.8M. Design B.M. , including Impact & continuity factor is given by:
MB(short span)=(1.25x0.8x35.35)=35.35KNm
ML(long span)=(1.25x0.8x12.14)=12.14KNm
16. SHEAR FORCE
Dispersion in the direction of the
span=[0.85+2(0.08+0.25)]=1.51m
For maxm shear , load is kept such that the whole dispersion is in
the span. So , the load is kept at (1.51/2)=0.755m from the edge of
the beam.
17. Effective width of slab =𝐾𝑥 1 −
𝑥
2
+ 𝑏𝑤 on which the load acts
Where,
L=effective span
X=Distance of center of gravity of load from nearer support
bw=track contact area cover the road surface of the slab in a direction at
right angles to the span plus twice the thickness of the wearing coat or
the surface finish above the structural slab.
K=A constant depending upon (B/L) ratio.
Breadth of cross girder= 20mm
Clear length of panel=(5-0.2)=4.8m
(B/L)=(4.8/2.3)=2.08
K for continuous slab obtained as 2.6
Effective width of slab=2.6 × 0.755 1 −
0.755
2.3
+ 3.6 + 2 × 0.08 = 5.079𝑚
Load per metre width=(350/5.079)=70KN
Shear force/meter width= 70 2.3 − 0.755 ÷ 2.3 = 47𝐾𝑁
Shear force with impact=(1.25x47)=58.75Kn
18. Design moments & Shear forces
Total MB=(35.35+3.76)=39.11KNm
Total ML=(12.14+1.32)=13.46KNm
Design of slab section & reinforcement
Effective depth d =
𝑀
𝑄×𝑏
= 39.11 × 106
0.767 × 1000
Adopt effective depth d=230mm
Ast =
𝑀
𝜎𝑠𝑡×𝑗×𝑑
= 39.11 × 106
÷ 200 × 0.916 × 230 = 928𝑚𝑚2
But min. reinforcement using HYSD bars according to IRC:18-2000 is 0.15% of
cross section .
Hence,
Ast=(0.0015x1000x250)=375mm2
Use 10mm dia. bars at 150mm centres (Ast=524mm2) for crack control(IRC21-
2000)
19. Check for shear stess(As per IRC:21-2003)
Nominal shear stress= ɽv=(v/bd)= 58.75 × 103
÷ 1000 × 230 = 0.255𝑁/𝑚𝑚2
At support section Ast=942mm2
Hence the ratio 100𝐴 𝑠𝑡/𝑏𝑑 =(100x942)/(1000x230)=0.40
From Table 12B of IRC:21-2000 & Table 12C of IRC:21-2000
The permissible shear stress in concrete slab = Kɽv=(1.1x0.25)=0.275 N/mm2
Kɽv>ɽv=0.255N/mm2
Design of longitudinal girders
(a)Reaction Factors:
Using Courbon’s Theory , the I.R.C. class AA loads are arranged for max. eccentricity.
Courbon’s method is popular due to smilicity of computations as detailed below:-
When live loads are positione nearer to the kerb the centre of gravity of live load acts
eccentrically with the centre of gravity of girder system. Due to this eccentricity, the loads
shared by each girder is increased or decresed depending upon the position of girders. This is
calculated by Courbon’s theory by a reaction factor given by :
Rx
∑𝑊
𝑛
[1 +
∑𝐼
∑𝑑 𝑥
2.𝐼
𝑑 𝑥. 𝑒
20. The live load bending moments & shear forces are computed for each of the girders. The
maximum design moments & shear forces are obtained by adding the live load & dead load
bending moments . The reinforcements in the main longitudinal girders are designed for
the maximum moments and shears developed in the girders
e
1.2m
kerb
dx
0.85
W
W
21. kerb
kerb
W1 W1
W
e=1.1m
A B C D
2.5m
2.5m
2.5m
2.05m1.625 m
Reaction factor for outer girder A is
RA=
RA=0.764W1
Reaction factor for inner girder B is
RB=
RB=0.588W1
If W= A x Le load =700KN
W1=0.5W = 350KN
RA=0.764 x W1=0.764 x 350 = 267.4=0.382 W
RB=(0.588 x 0.5 W) = 0.294 W = 102.9
22. Dead load from slab
per girder
1.5m
7.5m
1000
300
250
180 mm wc
250mmRC slab
Footpath
Weight of :
Parapet railing (lumpsum) = 0.92KN/m
Footh path & kerb =(0.3 x 1.5 x 24) = 10.08KN/m
Deck slab = (0.25 x 1.5 x 24) = 9.00KN/m
20KN/m
Total dead load of deck =[(2 x 20)+(7.76 x 7.5)]=98.2 KN/m
It is assumed that the deck load is shared equally by all the 4
girders .
Dead load /girder = (98.2/4) = 24.55KN/m
23. Dead load of main girder
1800
500
300
200
200
200
1200
200
24. The overall depth of the girder is assumed as
1800mm at the rate of 60mm per meter of span .
Span of the girder =30m.
Overall depth=(60 x 30)=1800mm
The bottom flange is selected so that 4 to 6 cables
are easily accommodated in the flange.
The section of main girder selected is as shown in
the figure.
Dead weight of rib = (0.2 x 1.15 x 24)=5.52KN/m
Dead weight of bottom flange= (0.5 x 0.4 x
24)=4.80= 10.32 KN/m
Weight of cross girder = (0.2 x 1.25 x 24)= 6KN/m
(assuming rectangular section)
25. Cross-section of the main girder is assumed for calculations.
1800
1200
1150
400
200
500
26. Dead load moments & shear in main
girders
Reaction from deck slab on each girder = 24.55 KN/m
Weight of cross girder = 6KN/m
Reaction on main girder =(6 x 2.5) = 15KN/m
Self weight of main girder = 10.32KN/m
Total dead load on girder =(24.55+10.32) = 34.87 KN/m
The max. dead load bending moment & shear force is computed
using the loads as shown in figure:
27. 5m
30m
15KN
Mmax=[(0.126 ×34.87 × 302)+(0.25×15 ×30)+(15×10)+(15×5) ]=4261KNm
Dead load shear at support
Vmax= [(0.5 ×34.87×30)+(0.5×75) ]=561KN
Live load bending moment in girder
Span of girder = 30m
Impact factor (class AA) =10%
The live load is placed centrally on the span
28. Bending Moment at centre of span = 0.5(6.6+7.5) *700 =4935
kN.m
B.M including impact and reaction factors for outer girders is,
Live load B.M.=(4935*1.18*0.382) = 2074 kN.m
For Inner girder, B.M.= (4935*1.1*0.294) =1596 kN.m
Live Load Shear force in Girder
For estimating the live load in the girders, the I.R.C. class AA loads
are placed as
Shown in the fig 6.11
Reaction of W2 on girder B = (350*0.45)/2.5 =63 kN
Reaction of W2 on girder A=(350+63) =413 kN
Total load on girder B =(350+63)= 413 kN
Maximum reaction in girder B =(413*28.2)/30 =388 kN
Maximum reaction on girder A = (287*28.2)/30 =270 kN
Maximum live load shear with impact factor in inner girder =
(388*1.1) =427 kN
Our girder = (270 *1.1) =297 kN
29. Design Bending Moments and shear forces
Table - Abstract of design moments and shear Forces in main Girders
Bendin
g
Moment
D.L.B.M L.L.B.M TOTAL
B.M.
UNITS
OUTER
GIRDER
INNER
GIRDER
4261
4261
2074
1596
6335
5857
KN.m
KN.m
SHEAR
FORCE
D.L.S.F. L.L.S.F. TOTAL
S.F.
UNITS
OUTER
GIRDER
INNER
GIRDER
561
561
297
427
858
988
KN
KN
30. Properties Of Main Girder Section
The main girder section is as shown in for computational purpose. The
properties of the section are:
A = 73 × 104 mm2
Y1 = 750 mm,
Y b = 1050 mm,
I = 2924×108 mm4
Z1 = (I / Y b ) = ( 2924 × 10 8 ) / 750 = 3.89×108 mm4
Z b = (I / Y b ) = ( 2924 × 108 ) / 1050 = 2.78×108 mm4
(i) Check for Minimum Section Modulus
Fc k = 50 N/mm2 ɳ = 18 N/mm2
Fc t = 18 N/mm2 Mg =4261N/mm2
Fc i = 40 N/mm2 Mq = 2074kN.m
Ft t = Ft w = 0
Md = (Mg + Mq) = 6335 KN.m
Fcw = 16N/mm2
Fbr = (ɳfct – ftw) =n(.85 ×18 – 0) = 15.3 N/mm2
Ftr = (Fcw = ɳftt) = 16N/mm2
Finf = (ftw/ɳ) + (Md/ɳzb) = 0+ (6335 × 106)/(.85 ×2.78 ×108) = 26.80 N/mm2
Zb = Mq + (1- ɳ)Mg = (1.77×108 mm3< 2.78 ×108)
Hence the section provided is adequate.
31. Pre stressing Forces
Allowing for two rows of cable, cover required =200 mm
Maximum possible eccentricity e = (1050 – 200) = 850 mm
Pre stressing force is obtained as
P = (A.finf .Zb)/(Zb + A.e)
= [(.73 × 106 ×26.80 ×2.78 ×108)/(2.78 ×108) +(.73 ×106V850)]
= 6053 ×103 N
= 6053 KN
Using Freyssinet system, anchorage type 7K-15 (7strands of 15.2
mm diameter) in
65 mm
Cables ducts, (IS: 6006-1983)
Force in each cable = (7 ×.8 ×260.7) = 1459 KN
Number of cables = (6053/1459) =5
Area of each strand = 140mm2
Area of 7 strands in each cable (7 ×140) = 980 mm2
Area of strands in 5 cable = Ap = (5 ×980) = 4900 mm2
The cables are arranged at centre of span sections
32. Permissible Tendon Zone
At support section,
e ≤ (Zb.fct/P) – (Zb/A)
≤(2.78 ×108 ×18)/(6053 ×103) – (2.78 ×108)/(.73 ×106)
≤ 445 mm
e ≥ (Zb.ftw/ɳP) – (Zb/A)
≥ 0 – (2.78 ×108)/(.73 ×106)
≥ -380 mm
The 5 cables are arranged to follow a parabolic
profile with the resultant force having an
eccentricity of 180 mm towards the soffit at the
support section. The position of cables at support
section
33. Check for Stresses
For the centre of span section, we have
P = 6053 KN
e = 850 mm
A = 0.73 ×106 mm2
Zb = 2.78 ×108 mm3
Zt = 3.89 ×108 mm3
H = 0.85
Mg = 4261 KN.m
Mq = 2074 kN.m
(P/A) = (6053 ×103)(.73 ×106) = 8.29 N/mm2
(Pe/zt) = (6053 ×103 ×850)/(3.89 ×108) = 13.22N/mm2
(Pe/Zb) = (6053 ×103 ×850)/(2.78 ×108) = 18.50 N/mm2
(Mg/Zt) = (4261 ×106) /(3.89 ×108) = 10.95 N/mm2
(Mg/Zb)= (4261 ×106) /(2.78 ×108) = 15.32 N/mm2
(Mq/Zt) = (2074 ×106) /(3.89 ×108) = 5.33 N/mm2
(Mq/Zb) = (2074 ×106) /(2.78 ×108) = 7.46 N/mm2
34. At transfer stage :
σt = [(P/A) - (Pe/zt) + (Mg/Zt) = (8.29 – 13.22 + 10.95)
= 6.02 N/mm2
σb = [(P/A) +(Pe/Zb) -(Mg/Zb)] = (8.29 + 18.50 – 15.32]
= 11.47 N/mm2
At working load stage:
σt = [ɳ(P/A) - ɳ(Pe/zb) - (Mg/Zt)+ (Mg/Zt)]
= [0.85 (8.29 – 13.22) + 10.95 + 5.33]
= 12.09 N/mm2(compression)
σb = [ɳ(P/A) - (Pe/Zb) - (Mg/Zb)+ (Mq/Zb)]
= [0.85(8.29 + 18.50) – 15.32 – 7.46]
= -0.01 N/mm2(Tension)
All the stresses at top and bottom fibres at transfer and
service loads are well within the safe permissible limits.
