Summary of Equations and Tables - En1992-2 Eurocode 2 - Design of Concrete Structures 2C Part 2 - Concrete Bridges (3) (1)

8/10/2019 Summary of Equations and Tables - En1992-2 Eurocode 2 - Design of Concrete Structures 2C Part 2 - Concrete Bri

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SUMMARY OF EQUATIONS

AND TABLES

EN 1992-2 Eurocode 2:

Design of Concrete Structures,

Part 2: Concrete Bridges


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DURABILITY AND COVER TO REINFORCEMENT

Relaxation loss of low relaxation prestressing tendon.

gfgfdgfddgfdgfdgfdgfdsg

Class 1: gfhgfhgfhgfhgfhgfhgfhgfhgfhClass 2: Class 3:

t = long term relaxation losses period in hours (given, exp 57 yrs take as 500,000 hrs)

Cover for deck slab

Where, = Minimum cover requirement (table 4.4-2) = Bar size (given)

= 0 mm (additional safety element) = 0 mm (minimum cover, in use of stainless steel)

= 0 mm (additional protection, exp coating)

,where = 10 mm (recommended) and = Previous step.


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STRUCTURAL ANALYSIS

Simplified Criteria for Second-Order Effect

Effective flange width for a box girder

a) Considering mid-span firsti) For cantilever portion,

l 0 = 0.7 l 2 ,where beff,i= 0.2b i + 0.1l 0< 0.2l 0

ii) For internal flange,

beff,i= 0.2b i + 0.1l 0< 0.2l 0 where bi for internal flange = bi/2

*Also compare with the available width of internal flange, bi for internal flange = bi/2. If bi/2 is less thanthan the calculated at i and ii, so the least value will be taken.

Make Conclusion. (compare with available width) b) Considering the supports

i) l 0 = 0.15( l 1+ l 2)ii) The cantilever portion has effective width given by :

beff,i= 0.2b i + 0.1l 0< 0.2l 0iii) Similarly, the internal flange associated with the web has effective width :

beff,i= 0.2b i + 0.1l 0< 0.2l 0 ; bi for internal flange = bi/2iv) Finally, the total width of flange acting with an outer web is:

beff,i= b eff,I + bwv) Make Conclusion. (compare with available width)

Effective length of cantilevering pier, l 0

l o = l . max Where,

k = < 0.1 and = rad/kNm , = kNm/rad

Slenderness about the minor axis,

lim ,where lim = and = ; f cd = where = 0.85 , = 1.5

= , where i =


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General Method Second-order Non-linear Analysis Second-order Analysis Based onNormal Stiffness

Final Moment at the base of the pier clause 5.8.7, EN 1992-2 Eurocode 2: Design of ConcreteStructures, Part 2: Concrete Bridges

K c = ,where k 1 = and k 2 = 0.2 K s =1.0 E cd = (in MPa) where = 1.2, EI = K c E cd I c + K s E s I s (in Nmm

2) , N B = (in kN)

l i . i = l . 0 . h where h = and 0 = (unit in mm)

M Ed = , and M 0Ed = lateral load x length +axial load x l i . i

General Method Second-order Non-linear Analysis Second-order Analysis Based onNominal Curvatures

Final Moment at the base of the pier clause 5.8.8, EN 1992-2 Eurocode 2: Design of ConcreteStructures, Part 2: Concrete Bridges

= , where yd = and E = 200GPa , d = , i s = i K = 1 + . ef ; where = 0.35 + - , =

= K r . K . , where K r = 1.0

M 2 = N Ed . e 2 where e2 = and c =

l i . i = l . 0 . h where h = and 0 = (unit in mm)

M Ed = M 0Ed + M 2, M 0Ed = lateral load x length +axial load x l i . i

Immediate Loss of Prestress For Post-Tensioning

Immediate loss of prestress in a concrete box girder

Stressing force per tendon, where Stress in the concrete, where P 0 = no. of tendon x P max ;Where the equation is for

3 conditions :


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i. End supportsii. Midspan

iii. Piers

Average stress, Losses due to the elastic deformation of concrete, P el P el = A p E p [ ] , where elastic loss ,

*Make a conclusion about the loss and convert to percentage. Compare with stress limit = 0.6 f ck .Sketch the diagram of loss of prestress from friction and anchorage draw-in. The value in thediagram can be calculated as ;

P(x) / P max = e - ( + kx) ,

For small value of ( + kx), equation can be written as

P(x) / P max = 1 - kx

Aad = ad E p A p , where E p= 200 x 10 3 where ad = 6mm to 12mm

* Make a conclusion about the shaded region. Construct force-distance diaghram and compare withformulae A ad.

Time-dependent Loss

Time-dependent loss of prestress in a concrete box girder

*Shrinkage from clause 3.1.4Remaining autogeneous shrinkage = (1- as ) x ca( where , ( ) and after 30days, Remaining drying shrinkage = (1- ds ) x (k h . cd,0 ) ,where where t s = 3Total Shrinkage, cs = Remaining autogeneous shrinkage + Remaining drying shrinkage


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*Creep Coefficient from clause 3.1.4 or Annex B- can be referred to creep coefficient calculation above.

*Relaxation from clause 3.3.2

i) Concrete stress at the tendon centroid,

*

+ ,where

ii) Concrete stress at the extreme fibre, * + ,where where A = 0.945 and B = 0.956* Compare with clause 5.10.2.2(5) and similar to clause 7.2.If c , Q P and bot 0.6f ck and 0.45f ck ,the creep factor would have need adjustment according to clause 3.1.4, basic the calculation on the

stress at the extreme fibre.

For Midspan,

, where

E p = 200GPa , (t,t s ) = 1.5 and A p = no. of tendon x area

*convert to percentage, p,c + s + r / n . P max , where n = no. of tendon. Finally, find total loss fromelastic loss, friction and draw-in and creep, shrinkage and relaxation.

ULTIMATE LIMIT STATES

Effect of Prestressing at Ultimate, Serviceability and Fatigue Limit State

Simply supported pre-tensioned beam with straight, fully bonded tendon

(clause 3.1.3(3)) where and refer to table.(i) Stresses at transfer at beam end (critical location)

Elastic loss of prestress force, , where ,where steel relaxation is assuming 1% to take place before transfer , and P 0 =n . %stress . CTS where n = no of strands


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or, more accurately, (take the lowest) P mo = 0.99( P 0 ) - P el

*Bottom fibre stress = compressive limit of where f ck (t) = f cm(t) 8 [Clause 5.10.2.2(5)] *Top fibre stress =

(ii) Final serviceability stresses at midspan

*short term lossescalculate on short-term losses

*Long term lossesi) The total steel Relaxation : ii) Concrete shrinkage remaining after transfer is found to be 300 x 10 -6 [ clause 3.1.4]iii) Stresses due to beam self-weight alone are :

*Bottom fibre stress = - M / w p,1 *Top fibre stress = M / w p,2

(iii) Sketch stress diagram combine of Stresses at transfer at beam end (critical location) +Final serviceability stresses at midspan

(iv) Compare the maximum concrete stress with 0.45 f ck (t). If the maximum stress > 0.45 f ck (t), redesign by reduce the creep factor or replaced f cm(t) with f ck (t).

Ultimate Limit State Singly Reinforced Concrete Deck Slab

Reinforced concrete deck slab

,where K av = M/bd 2 f av

*To ensure reinforcement is yielding, check limit with this formulae;

*Check again the x/d using this equation ;


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with z = d x *Decide to use the size of bar. The A s of bar need to be larger than the calculated.

*The average stress, f avand centroid ratio, :

i ) Parabolic rectangularf av = f cd ( 1 - )

= 1 - i i) Bilinear

f av = f cd ( 1 0.5 )

= 1 - i i i ) Simpli fi ed r ectangular

f av = f cd

= / 2

* Calculate back K av and x/d ,

K av = ( 1 ) = ( )2

*Reinforcement will yield by inspection with x = x 200 ( in unit mm) and z = d x.

* Recalculate steel area, A s ,

with z = d x * Check on the moment resistance with the new steel area, A s M = A s f yd d( 1 - ) , unit in kNm/m where

Voided reinforced concrete deck slab

*Determine the slab depth above hole = slab depth below hole and effective depth, d


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D = h cover link bar /2

A s = x (bar /2) 2 x ( b / spacing) ; spacing = in diagram

The average stress, f av and centroid ratio, :

i ) Parabolic rectangularf av = f cd ( 1 - )

= 1 - i i) Bilinear

f av = f cd ( 1 0.5 )

= 1 - i i i ) Simpli fi ed r ectangular

f av = f cd

= / 2

= where =

*Check against limit to ensure reinforcement is yielding ;

and make conclusion

X = x D , z = D x ,

MRd = x A s x z for m width

*To find required increase in As

to resist MM rq = 1.4 x M

Kav = , = Less limit for reinforcement yield,


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X = * D < (height diameter of void)/2 mm , z = D- x , A st , A s(add) = *Provide at mm centres

Doubly Reinforced Concrete - Clause 6.1, EN 1992-2 Eurocode 2: Design of Concrete Structures, Part 2: Concrete Bridges

;

If the value less than x/d ,

For compression reinforcement to yield :

;

; ; Taking moments about top fibre to find moment of resistance :

Flanged Beams

Flanged reinforced concrete beam

; ; ; ;


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; { } ;

* + ; Reinforcement strain = }0.0035

Pretressed Concrete Beam (Additional Sub-section)

Prestressed C oncrete M beam. - Clause 6.1(2) P, EN 1992-2 Eurocode 2: Design of ConcreteStructures, Part 2: Concrete Bridges

f pd = , ,

*consider horizontal top branch and a neutral axis depth, obtained by trial and error.

Prestrain = s1,2,3. = s + Prestrain

s

Fs = no. of strands As f pd

Fc = refer to formulae flanged beam

uk =

M = F c1 ( x a 1 ) + F c2 ( x a2 ) + Fc 3 ( x a 3 ) + [As x 2f pd ] x (3 x [high of bar])

+ [15A s x fpd] x [high of bar], [high of bar]=refer to figure below


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Reinforced Concrete Column (Additional Sub-section)

Reinforced concrete pier

cu2

, cu2 = 0.0035

Mpa (compression) (find As with given As)

Brittle Failure of Members with Prestress (Additional Sub-section)

Post-tensioned concrete box girder

(MNm)


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Members Not Requiring Design Shear Reinforcement

Reinforced concrete deck slab.

d = h cover link bar /2

kN/m

Reinforced concrete column

( ) (kN)Prestressed Sections Uncracked In Flexure Shear Tension

Post-tensioned concrete box girder, un-cracked in flexure

where tendons A s f pk 0.70 0.75 10 -3


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( )

Members Requiring Design Shear Reinforcement

Voided reinforced concrete deck slab

If V Ed >V Rd,c therefore design shear resistance is required.

(a) Try vertical links =90 =45


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(b) Consider =45 =45

(c) =90 =21.8

Shear At Points of Contraflexure Section Cracked in Flexure

Post-tensioned concrete box girder without tendon drape

(a) Consider =90 =45

(b) By using =90 =41.5

If so the arrangements is adequate

(c) By using =45 =29.5


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Take the lowest V Rd


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APPENDIX

Summary of Tables and Graphs


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Table 1 Sectional areas per meter width for various bar spacing (mm)

Bar size Spacing of bars

(mm) 50 75 100 125 150 175 200 250 300

6 566 377 283 226 189 162 142 113 94

8 1010 671 503 402 335 287 252 201 168

10 1570 1050 785 628 523 449 393 314 262

12 2260 1510 1130 905 754 646 566 452 377

16 4020 2680 2010 1610 1340 1150 1010 804 670

20 6280 4190 3140 2510 2090 1800 1570 1260 1050

25 9820 6550 4910 3930 3270 2810 2450 1960 1640

32 16100 10700 8040 6430 5360 4600 4020 3220 2680

40 25100 16800 12600 10100 8380 7180 6280 5030 4190

Table 2 Asv / s v for varying stirrup diameter and

spacing

Stirrupdiameter

(mm)

Stirrup spacing (mm)

85 90 100 125 150 175 200 225 250 275 300

8 1.183 1.118 1.006 0.805 0.671 0.575 0.503 0.447 0.402 0.366 0.335

10 1.847 1.744 1.57 1.256 1.047 0.897 0.785 0.698 0.628 0.571 0.523

12 2.659 2.511 2.26 1.808 1.507 1.291 1.13 1.004 0.904 0.822 0.753

16 4.729 4.467 4.02 3.216 2.68 2.297 2.01 1.787 1.608 1.462 1.34


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Table 3 Sectional areas of groups of bars (mm)

Barsize

Number of bars

(mm) 1 2 3 4 5 6 7 8 9 10

6 28.3 56.6 84.9 113.2 141.5 169.8 198.1 226.4 254.7 283

8 50.3 100.6 150.9 201.2 251.5 301.8 352.1 402.4 452.7 503

10 78.5 157 235.5 314 392.5 471 549.5 628 706.5 785

12 113 226 339 452 565 678 791 904 1017 1130

16 201 402 603 804 1005 1206 1407 1608 1809 2010

20 314 628 942 1256 1570 1884 2198 2512 2826 3140

25 491 982 1473 1964 2455 2946 3437 3928 4419 4910

32 804 1608 2412 3216 4020 4824 5628 6432 7236 8040

40 1260 2520 3780 5040 6300 7560 8820 10080 11340 12600

Documents

Summary of Equations and Tables - En1992-2 Eurocode 2 - Design of Concrete Structures 2C Part 2 - Concrete Bridges (3) (1)