PROJECT PRESENTATION

Submitted by –

ABHISHEK GOEL MAMTA YADAV

PRASHANT PATEL AMBER KATIYAR

OBJECTIVE:Design of Post- Tensioned Prestressed Concrete Tee Beam And Slab Bridge Deck

GUIDE: Dr. J. GIRISH

TYPES OF BRIDGES

Arch Bridges

Reinforced Slab Bridges

CANTILEVER BRIDGE

PRE-TENSIONING AND POST-TENSIONINGThe prestress in a structure is influenced by

either of the two processes: Pre-tensioning

Post-tensioning

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.

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.

DESIGN OF BRIDGETYPE OF LOADING

CLASS AA LOADING

REFERENCE PROBLEM:-

DESIGN OF POST –TENSIONED PRESTRESSED CONCRETE TEE BEAM AND SLAB BRIDGE DECK FOR A NATIONAL HIGHWAY CROSSING TO SUIT THE FOLLOWING –

GIVEN DATA :

EFFECTIVE SPAN= 30mWIDTH OF ROAD= 7.5mKERBS= 600mm on each side FOOTPATH= 1.5m wide on each side THICKNESS OF WEARING COAT= 80mmLIVE LOAD= I.R.C class AA tracked

vehicle 

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= 0.85Spacing of cross girder= 5mAdopt Fe-415 grade HYSD barsStrands of 15.2mm – 7ply conforming to

IS:6006-1983 are available for use.

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)

Ec= 5700 =5700 = 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

DESIGN OF INTERIOR SLAB PANEL

1. BENDING MOMENTDead 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

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

(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)=35.35KNmML(long span)=(1.25x0.8x12.14)=12.14KNm

SHEAR FORCE

Dispersion in the direction of the span=[0.85+2(0.08+0.25)]=1.51mFor 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.

Effective width of slab = on which the load acts Where,L=effective spanX=Distance of center of gravity of load from nearer supportbw=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= 20mmClear 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=Load per metre width=(350/5.079)=70KN

Shear force/meter width=

Shear force with impact=(1.25x47)=58.75Kn

Design moments & Shear forcesTotal MB=(35.35+3.76)=39.11KNmTotal ML=(12.14+1.32)=13.46KNm

Design of slab section & reinforcement

Effective depth d = Adopt effective depth d=230mmAst =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)

Check for shear stess(As per IRC:21-2003)

Nominal shear stress=v=(v/bd)=

At support section Ast=942mm2

Hence the ratio =(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

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

WW

kerb kerb

W1 W1

W

e=1.1m

A B C D

2.5m2.5m

2.5m

2.05m1.625 m

Reaction factor for outer girder A is RA=RA=0.764W1Reaction factor for inner girder B is RB=RB=0.588W1If W= A x Le load =700KNW1=0.5W = 350KNRA=0.764 x W1=0.764 x 350 = 267.4=0.382 WRB=(0.588 x 0.5 W) = 0.294 W = 102.9

Dead load from slab per girder

1.5m7.5m

1000

300

250

180 mm wc

250mmRC slab

Footpath

Weight of :Parapet railing (lumpsum) = 0.92KN/mFooth path & kerb =(0.3 x 1.5 x 24) = 10.08KN/mDeck 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

Dead load of main girder

1800

500

300

200

200

200

1200

200

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)=1800mmThe 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/mDead weight of bottom flange= (0.5 x 0.4 x 24)=4.80= 10.32 KN/mWeight of cross girder = (0.2 x 1.25 x 24)= 6KN/m (assuming rectangular section)

Cross-section of the main girder is assumed for calculations.

1800

1200

1150

400

200

500

Dead load moments & shear in main girders Reaction from deck slab on each girder = 24.55 KN/mWeight of cross girder = 6KN/m Reaction on main girder =(6 x 2.5) = 15KN/mSelf weight of main girder = 10.32KN/mTotal 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:

5m

30m

15KN

Mmax=[(0.126 ×34.87 × 302)+(0.25×15 ×30)+(15×10)+(15×5) ]=4261KNmDead load shear at support  Vmax= [(0.5 ×34.87×30)+(0.5×75) ]=561KN

Live load bending moment in girderSpan of girder = 30mImpact factor (class AA) =10%The live load is placed centrally on the span

Bending Moment at centre of span = 0.5(6.6+7.5) *700 =4935 kN.mB.M including impact and reaction factors for outer girders is,Live load B.M.=(4935*1.18*0.382) = 2074 kN.mFor 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 6.11Reaction of W2 on girder B = (350*0.45)/2.5 =63 kNReaction of W2 on girder A=(350+63) =413 kNTotal load on girder B =(350+63)= 413 kNMaximum reaction in girder B =(413*28.2)/30 =388 kNMaximum reaction on girder A = (287*28.2)/30 =270 kNMaximum live load shear with impact factor in inner girder = (388*1.1) =427 kNOur girder = (270 *1.1) =297 kN

Design Bending Moments and shear forces  Table - Abstract of design moments and shear Forces in main Girders

Bending Moment

D.L.B.M L.L.B.M TOTAL B.M.

UNITS

OUTER GIRDERINNER GIRDER

42614261

20741596

63355857

KN.mKN.m

SHEAR FORCE

D.L.S.F. L.L.S.F. TOTAL S.F.

UNITS

OUTER GIRDERINNER GIRDER

561

561

297

427

858

988

KN

KN

Properties Of Main Girder SectionThe 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 ModulusFc k = 50 N/mm2 ɳ = 18 N/mm2

Fc t = 18 N/mm2 Mg =4261N/mm2

Fc i = 40 N/mm2 Mq = 2074kN.mFt t = Ft w = 0 Md = (Mg + Mq) = 6335 KN.mFcw = 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.

Pre stressing Forces

Allowing for two rows of cable, cover required =200 mmMaximum possible eccentricity e = (1050 – 200) = 850 mmPre 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 KNUsing Freyssinet system, anchorage type 7K-15 (7strands of 15.2 mm diameter) in 65 mmCables ducts, (IS: 6006-1983)Force in each cable = (7 ×.8 ×260.7) = 1459 KNNumber of cables = (6053/1459) =5Area 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

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

Check for StressesFor the centre of span section, we haveP = 6053 KNe = 850 mmA = 0.73 ×106 mm2

Zb = 2.78 ×108 mm3

Zt = 3.89 ×108 mm3

H = 0.85Mg = 4261 KN.mMq = 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

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.

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