design of ridge mohamed.doc

8/10/2019 design of ridge MOHAMED.doc

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Design Example Precast/Prestressed Concrete Girder

1. Problem Definition

Span Data

Overall Girder Length =106 FT

Design Span =105 FT

Girder is simply supported

Skew = 0

Bridge Cross Section Data

Number Lanes = 4

Number Girders = 6

Girder Spacing = 9.00 FTRoadway Width = 48.00 FT

Overall Width = 51.00 FT

Deck Thickness

Actual = 9.00 IN

Structural = 8.00 IN

Girder Type

PCI BT-72 (72 in. deep bulb-tee)

Location: Interior

Dead Load

Future Wearing Surface =0.025 KSF

Barrier Weight =0.418 KLF

Live Load

HL-93 - Design Truck + Design Lane Loads

Girder Concrete

fc = 6.0 KSI

fci= 4.5 KSI

wc= 0.150 KCF

Deck Concrete

fc = 4.0 KSI

wc= 0.150 KCF

Prestressing Steel

Type: 0.5-IN Diameter 270 KSI Low-

Relaxation Seven-Wire Strand

Eps= 28,500 KSI

Pull: 75%

H = 75% (Relative Humidity)

Time to Release = 18 HRS

Profile: 2-pt. Depressed

Depression Point: 0.45L = 0.45 (106 FT)

Reinforcing Steel (Non-Prestressed)

fy=60 KSI

Es = 29,000 KSI

1


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Figure 1: Cross Section of Bridge With Six PCI BT-72 Bulb Tee Girders at 9-0 Spacing

2


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2. Analysis

2.1 Section Properties

2.1.1 Bare Girder:

The LRFDSpecs allow the inclusion of transformed strand in the section propertiesfor a

prestressed member (Article 5.9.1.4). For simplicity, the contribution of the strand to the

section properties is neglected in this example.

Properties of PCI BT-72:

A = 767.0 IN2

I = 545,894 IN4

h = 72.00 IN

yb = 36.60 IN

yt = 35.40 IN

Sb= 545,894 IN4/ 36.60 IN = 14,915 IN

3

St= 545,894 IN4/ 35.40 IN = 15,421 IN

3

2.1.2 Composite Section

Figure 2: Cross Section of Single Girder with

Composite Deck

Note: Any thickening of the slab over the top flange of the girder (i.e., a haunch or build-up)

will be neglected in the computation of the section properties of the composite section.

However, if they are detailed in the plans, they should be included as additional dead load.

Effective deck width: (LRFD 4.6.2.6.1)

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One-quarter span length = (105 FT / 4)(12) = 315 IN

One-half flange width + (12)(deck thickness):

(42 IN) / 2 + (12) (8 IN) = 117 IN

Average spacing of adjacent girders = (9 FT) (12) = 108 IN (Controls)

Transformed deck width = (n) (effective width) = (0.8165) (108 IN) = 88.182 IN

n = = = = 0.8165

Comonent !rea "# ! "# ! ("# $

"#c)2

Io Ic

%irder &6&.00 '6.60 280&2 2&'111 558* 81*005

+ffective ,eck &05.5 &6.00 5'61 2*&'' '&62 '010*6

Total 1&2.50 81686 1120101

ybcg= (A yb) / A = 81,686 IN3/ 1,472.5 IN = 55.47 IN

ytcg= h - ybc= 72.00 IN - 55.47 IN = 16.53 IN

ytcd= hc- ybc= 80.00 IN - 55.47 IN = 24.53 IN

Sbcg= Ic/ ybc= 1,120,101 IN4/ 55.47 IN = 20,193 IN

3

Stcg= Ic/ ytcg= 1,120,101 IN4/ 16.53 IN = 67,762 IN

3

Stcd= (Ic/ ytcd) / n = (1,120,101 IN4/ 24.53 IN) / (0.8165) = 55,925 IN

3

2.2 Moments and Shears

2.2.1 Dead Loads

2.2.1.1Girder Dead Load at Release

The moments for this condition are computed separately from other moments because the full

length of the girder is used in computing these moments, rather than the design span

(distance from center-to-center of bearings).The full length is used because, when the girder

cambers upward in the prestressing bed after release, its only points of contact with the bed

(and therefore its support locations) will be at the ends of the girder.

Locations of interest at release conditions:

1.Transfer point (LRFD 5.8.2.3)

l t= 60 db= 60 (0.5 IN) = 30.0 IN = 2.5 FT

2.Depression point

x = 0.45 L = 0.45 (106 FT) = 47.7 FT

3.Midspan

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x = 0.5 L = 0.5 (106 FT) = 53.0 FT

Girder Dead Load

wgdl= (767 IN2/ 144) (0.150 KCF) = 0.799 KLF

Mgdli=

L = 106 FT (overall girder length)

Mgdli= 0.799 (x / 2) (106 - x) = 42.35 x - 0.400 x2

2.2.1.2Girder Dead Load - Final

L = 105 FT (bearing to bearing)

Mgdl= 41.95 x - 0.400 x2

Vgdl= = 41.95 - 0.799 x

2.2.1.3 Deck Dead Load (Structural Deck)

Structural Deck Thickness = 8.0 IN

wddl= ((8 IN x 108 IN) /144) (0.150 KCF) = 0.900 KLF

L = 105 FT

Mddl= 47.25 x - 0.450 x2

Vddl= 47.25 - 0.900 x

2.2.1.4 Additional Non-Composite Dead Load (Non-Structural Deck)

Non-Structural Deck Thickness = 1.0 IN

wncdl= ((1 IN x 108 IN) /144) (0.150 KCF) = 0.1125 KLFL = 105 FT

Mncdl= 5.906 x - 0.0563 x2

Vncdl= 5.906 - 0.1125 x

2.2.1.5 Composite Dead Load -Barriers

Barriers: (2) (0.418 KLF / barrier) = 0.836 KLF

wcdl= 0.836 KLF / (6 girders) = 0.1393 KLF / girder (LRFD 4.6.2.2.1)

L = 105 FT

Mcdl= 7.313 x - 0.0697 x2

Vcdl= 7.313 - 0.1393 x

2.2.1.6 Composite Dead Load -Future Wearing Surface

Future Wearing Surface: (48.00 FT) (0.025 KSF) = 1.200 KLF

wfws= 1.200 KLF / (6 girders) = 0.200 KLF

L = 105 FT

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Mcdl= 10.500 x - 0.100 x2

Vcdl= 10.500 - 0.200 x

2.2.2 Live Loads

2.2.2.1Distribution Factors (LRFD 4.6.2.2.1)

To use thesimplified live load distribution factor formulae, the following

conditions must be met:

Width of deck is constant O.K.

Number of girders, Nb, 4 O.K. (Nb= 6)

Girders parallel and same stiffness O.K.

Roadway part of overhang, de, 3.0 FT O.K. (de= 1.25 FT)

Curvature < 4o

O.K. (Curvature = 0o)

Bridge Type: k (LRFD Table 4.6.2.2.1-1)

Distribution Factorfor Moment(2 or More Lanes loaded):

(LRFD 4.2.2.2b-1)

Provided that: 3.5S16 S = 9.00 FT O.K.

4.5ts12.0 ts= 8.00 IN O.K.

20L < 240 L = 105 FT O.K.

Nb4 Nb= 6 O.K.

Kg= n (I + A eg2

) (LRFD 4.6.2.2.1-1)

n = = = = 1.2247(Note that this is the reciprocal of the value for

n computed earlier)

eg= yt+ (ts/ 2) = 35.40 IN + (8 IN / 2) = 39.40 IN

Kg= 1.2247 [545,894 IN4+ (767 IN

2) (39.40 IN)

2] = 2,126,758 IN

4

=0.7423 lanes/ girder

Distribution Factorfor Shear(2 or More Lanes Loaded):

(LRFD 4.6.2.2.3a-1)

Provided that the following condition is met in addition to the conditions specified above:

10,000Kg(=2,126,758)7,000,000 O.K.

=0.8839lanes / girder

2.2.2.2 Live Load Effects

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Figure 3: LRFD Design Truck and Design Lane Load

At Midspan:

Design Truck will govern over Design Tandem for this span.

Mtruck= 18L - 280 (Maximum moment at midspan)

Mtruck= (18) (105 FT) - 280 = 1,610.0 K-FT

Mlane= = = 882.0 K-FT

Dynamic Load Allowance (Impact Factor) (LRFD Table 3.6.2.1-1)

1 + IM = 1 + 0.33 = 1.33 (applied only to truck portion of live load)

MLL+I= DF [Mlane+ 1.33 (Mtruck)] = (0.7423) [882.0 + (1.33) (1,610.0)] =2,244.2 K-FT

!t 6.06-T from Centerine of /earin (Critica ection for hear$ see ection '.2.1.1 #eow)

3tr4ck=

3tr4ck= ('2 I7) (*8.*& -T) ('2 I7) (8.*& -T) (8 I7) (&0.*& -T)9 (6.05' -T) : 105 -T

3tr4ck= '&2.0 $-T

3ane= = (0.6 ;-) (6.05' -T) (105 -T $ 6.05' -T) : 2 = 1*1.& $-T

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MSLnc= 1,101.1 + 1,240.3 + 155.0 = 2,496.4 K-FT

Acting on the composite girder, MSLc:

MSLc= 192.0 + 275.6 + 2,244.2 = 2,711.8 K-FT

2.3.3Service III

This load combination is a special combination for Service Limit State stress checks thatapplies only

to tension in prestressed concretestructures with the objective of crack control.

All load factors are equal to 1.0 for this problem, except thatthe live load is reduced by a factor of 0.8.

-or moment at midsan

!ctin on the non$comosite irder 3;nc(same as for ervice I).

!ctin on the comosite irder 3;c

3;c= 1*2.0 2&5.6 (0.8)(22.2) = 226'.0 $-T

2.'. -ati4e(not required)

According to LRFD 5.5.3.1, Fatigue need not be checked for concrete deck slabs in multigirder

applications. Fatigue of the reinforcement need not be checked for fully prestressed components

designed to have extreme fiber tensile stress due to Service III Limit State within the tensile stress

limit specified in Article 5.9.4.2.2b. Fatigue of concrete is checked indirectly by satisfying the

compression stress limit of 0.4 '

cf for the load combination specified in LRFD 5.9.4.2.1.

2.'.5 trenth I

This oad com#ination is the enera com#ination for trenth ;imit tate desin. ince the

str4ct4re is sim" s4orted the maim4m va4es for the oad factors are 4sed #eca4sethe" rod4ce the reatest effect (see ;>-, Ta#e '..1$2).

No distinction is made #etween moments and shears aied to the non$comosite orcomosite sections for strenth com4tations. The factored oads are aied to thecomosite section.

The foowin oad factors a"

,ead ;oad $ Comonent and !ttachments 1.25 ,C

,ead ;oad $ ?earin 4rface and @tiities 1.50 ,?


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34= (1.25) 2'*.' 26*.5 ''.& 1.&9 (1.50) (5*.*) (1.&5) (50*.6)

34= &'0.' 8*.* 8*1.8 = 1&12.0 $-T

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Summary of Dead and Live Load Effects

Moments at Release (K-FT)

Component Transfer Pt. (2.5 FT) 0.45L (47.7 FT) Midspan (53.0 FT)

Girder 103.4 1,111 1,122

Moments and Shears

!t Critica ection for hear(6.05' -T from Center of 4ort)

3idsan (52.50 -T)

Comonent < (I7) 3 ($-T) 3 ($-T)

Acting on Non-Composite Girder:

%irder '&.1 2'*.' 1101.1

,eck (tr4ct4ra) 1.8 26*.5 120.'

!dditiona Non$Comosite 5.2 ''.& 155.0

SUBTOTAL 84.1 542.5 2,496.4

Acting on Composite Girder:

/arriers 6.5 1.& 1*2.0

-4t4re ?earin 4rface *.' 5*.* 2&5.6

;ive ;oad Imact *8.5 50*.6 22.2

SUBTOTAL - Seri!e " 114.3 611.2 2,711.8

SUBTOTAL - Seri!e """ --- --- 2,263

TOTAL - Stren#t$ " 299.7 1,712 7,7%1

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3. Design

3.1 Flexural Design

3.1.1 Strand Patterns at End of Girder and at Midspan

A trial-and-error procedure is used to determine the strand pattern. For simplicity, the trial

designs that were performed to arrive at the strand patterns shown below are not included as

part of this design example.

The section at midspan is considered first. Generally, strands are added to the section in

pairs, filling the available strand locations from the bottom, until the stress limits and strength

requirements at midspan are satisfied.

Theend pattern is then determined by draping strandsas required to satisfy the stress limits

at the end of the girder at release. Other methods, which are not considered in this example,

could also be used to control stresses at the end of the girder.

Draped Strands

No. ofStrands

Dist. fromBottom

2 68 IN

2 66 IN

2 64 IN

2 62 IN

Straight Strands

No. of

Strands

Dist. from

Bottom2 8 IN

6 6 IN

10 4 IN

10 2 IN

Figure 4: Strand Pattern at End and Midspan

3.1.2 Properties of Assumed Strand Patterns

For the following computations, all c. g. dimensions are measured from the bottom of the girder.

At Midspan (and between Depression Points):

Depression Point Location:

0.45 L = 0.45 (105 FT) = 47.25 FT from CL bearing = 47.75 FT from end of girder

c.g. @ midspan = [(12 strands)(2 IN) + (12)(4 IN) +(8)(6 IN) + (4)(8 IN)] / 36 strands = 4.22 IN

eccentricity @ midspan = eCL= yb- c.g. @ midspan = 36.60 - 4.22 =32.38 IN

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At End of Girder:

c.g. @ end = [(10 strands)(2 IN) + (10)(4 IN) +(6)(6 IN) + (2)(8 IN) + (2)(62 IN) + (2)(64

IN) + (2)(66 IN) +(2)(68 IN)] / 36 strands = 17.56 IN

eccentricity @ end = eend= yb- c.g. @ end = 36.60 - 17.56 =19.04 IN

At Transfer Length from End of Member (2.5 FT):

l t= 60 db= 60 (0.5 IN) = 30 IN = 2.5 FT (LRFD 5.8.2.3)

c.g. of strand pattern @ end = 17.56 IN

c.g. of strand pattern @ depression point = 4.22 IN

c.g. @ transfer point = 17.56 IN - (2.50 FT / 47.25 FT) (17.56 IN - 4.22 IN) = 16.85 IN

eccentricity @ transfer point = etr= yb- c.g. @ t.p. = 36.60 - 16.85 = 19.75 IN

At Critical Location for Shear (6.06 FT from CL bearing; 6.56 FT from end of member):

c.g. @ 6.06 FT = 17.56 IN - (6.56 FT / 47.75 FT) (17.56 IN - 4.22 IN) = 15.73 IN

eccentricity @ 6.06 FT = ecv= 36.60 - 15.73 = 20.87 IN

Total Area of Prestressing Strands:

Aps= (36 strands) (0.153 IN2) =5.508 IN

2

C. G. of Straight Strands:

c.g. straight= [(10 strands)(2 IN) + (10)(4 IN) +(6)(6 IN) + (2)(8 IN)] / 28 strands = 4.00 IN

eccentricity of straight strands = estr= yb- c.g. straight = 36.60 - 4.00 = 32.60 IN

Area of Straight Prestressing Strands:

Apss= (28 strands) (0.153 IN2) =4.284 IN

2

3.1.3 Prestress Losses

3.1.3.1 Components of Prestress Loss

Elastic Shortening

(LRFD 5.9.5.2.3a-1)

fcgp= Stress at c.g. of strands at release (at midspan)

Scgp= Section modulus of bare girder at c.g. of strand pattern

Scgp= I / eCL= (545,894 IN

4

) / (32.38 IN) = 16,860 IN

3

fi= Stress in strands after release = 0.70 fpu= 189.0 KSI (assumed) (LRFD 5.9.5.2.3a)

Pi= Apsfi= (5.508 IN2)(189.0 KSI) = 1,041.0 KIP

fcgp=

fcgp= = 2.558 KSI

Eps= 28,500 KSI (LRFD 5.4.4.2)

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Pi= (Aps) (fpi) = (5.508 IN2) (182.77 KSI) =1,006.7 KIP

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3.1.3.3Prestress Loss and Effective Prestress after All Losses (Final)

Com4te fina restress oss

fT= f+ f> fC> f>2 (;>-, 5.*.5.1$1)

fT= 1&.*' 5.&5 21.*5 2.1* = &.82 I (23.62 &o' initi *re+tre++, '*)

Com4te effective stress and force after ossesfe= 202.5 $ &.82 = 15.68 I

7e= (!s) (fe) = (5.508 IN2) (15.68 I) = 852.0 I7

Check effective stress (;>-, Ta#e 5.*.'$1)

!dd effect of ive oad to effective restress fe. This is the aroach 4sed in the tdecifications #4t the ;>-, is not cear on this.

Chane in stress in #ottom row of strands

fe fs= 15.68 6.2 = 160.*2 I A 0.8 f"= (0.8) (2' I) = 1*. I B..

3.1.4 Midspan

3.1.4.1 Concrete Stresses Due to Loads

Sign convention for stresses: (+)= Compression

(-)= Tension

%irder ,ead ;oad $ !t >eease with ; = 106 -T

ft= = 12 = 0.8&' I f #= 12 = $ 0.*0' I

%irder ,ead ;oad $ -ina with ; = 105 -T

ft= = 12 = 0.85& I f #= 12 = $ 0.886 I

,eck (tr4ct4ra) ,ead ;oadft= 12 = 0.*65 I f #= 12 = $ 0.**8 I

!dditiona Non$Comosite ,ead ;oad (Non$tr4ct4ra ,eck)

ft= 12 = 0.121 I f #= 12 = $ 0.125 I

Comosite ,ead ;oad $ /arriers -4t4re ?earin 4rface

ft= 12 = 0.08' I f #= 12 = $ 0.2&8 I

ftd= 12 = 0.100 I

;ive ;oad I

ft= 12 = 0.'*& I

ftd= 12 = 0.82 I

;ive ;oad IIIf#= 12 = $ 1.06& I

'.1..2 Concrete tresses ,4e to 7restress

!t >eease

/ottomf#= = = '.*8 I

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Toft= = = $ 0.801 I

!fter a ;osses (-ina)

/ottomf#= = = 2.*60 I

To

ft= Pe1

AeCL

St

= = $ 0.6&8 I

'.1..' Concrete tresses at ervice ;imit tate $ /efore ;osses (!t >eease)

Note: Stresses at depression point will be more critical at release.

Service I:

Bottom of Girder (Compressive Stress):

fb= 3.498 - 0.903 = 2.595 KSICheck limiting stress: (LRFD 5.9.4.1.1)

2.595 KSI $= -= -0.201 KSI -0.200 KSI O.K.

3.1.4.4 Concrete Stresses at Service Limit States - After All Losses (Final)

Service III (Tensile Stress in Bottom of Girder):

fb= 2.960 -0.886 - 0.998 - 0.125 - 0.278 - 1.067 = - 0.394 KSI

Check limiting stress: (LRFD 5.9.4.2.2b)

- 0.394 KSI > = = - 0.465 KSI O.K.

ervice I (Comressive tress in To of %irder) (;>-, 5.*..2.1)

Comressive stress d4e to the s4m of effective restress and ermanent oads

ft= $ 0.6&8 0.85& 0.*65 0.121 0.08' = 1.'8 I

Check imitin stress

1.'8 I A = 0.456.00KSI( )= 2.&0 I B..

Comressive stress d4e to the s4m of effective restress ermanent oads and transientoads (f4 service oad)

ft= $ 0.6&8 0.85& 0.*65 0.121 0.08' 0.'*& = 1.&5 I

Check imitin stress

1.&5 I A = 0.60(1.0)(6.00) = '.60 I B.

where w= 1.0 for to fane of irder with comosite deck #eca4se san:thickness ratio offane m4st #e ess than 15. ee ;>-, 5.&..&.2.

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17/26

Comressive stress d4e to ive oad and one$haf the s4m of effective restress andermanent oads

ft= 0.'*& 0.5(1.'8) = 1.0&1 I

Check imitin stress

1.0&1 I A = = 2.0 I B.

ervice I (Comressive tress in To of ,eck) (;>-, 5.*..2.1)

Comressive stress d4e to the s4m of effective restress and ermanent oads

ft= 0.100 I

Check imitin stress

0.100 I A = = 1.80 I B..

Comressive stress d4e to the s4m of effective restress ermanent oads and transient

oads (f4 service oad)

ft= 0.100 0.82 = 0.582 I

Check imitin stress

0.582 I A = 0.604.00KSI( )= 2.0 I B.

where w= 1.0 for deck #etween tis of irder fanes #eca4se

san:thickness = (*.00 -T '.5 -T) : (8 IN : 12 IN:-T) = 8.2 A 15. ee ;>-, 5.&..&.2.

Comressive stress d4e to ive oad and one$haf the s4m of effective restress andermanent oads

ft= 0.82 0.5(0.100) = 0.5'2 I

Check imitin stress

0.5'2 I A = = 1.60 I B.

'.1..5 trenth ;imit tate (trenth I)

Com4te nomina moment caacit" 3n

Check whether section #ehaves as a rectan4ar #eam or a T$#eam

= deth of ne4tra ais (;>-, 5.&.'.1.1$)

(;>-, 5.&.'.1.1$2)

k = 2 1.0 $ (2' I : 2&0 I)9 = 0.280

d= h hf$ c.. D midsan = &2.00 8.00 $ .22 = &5.&8 IN d= 6.2& In at critica sectionfor shear9

ince no mid tension or comression reinforcement is #ein considered terms areeiminated. Note that the f4 effective (not transformed) deck width is 4sed in thiscom4tation since the deck concrete strenth is 4sed.

= .682 IN

ince c = .682 IN A hf= 8.00 IN the ne4tra ais is in the deck confirmin rectan4ar

#eam #ehavior.

Com4te deth of comression #ock a

= 0.85 for = .0 I (deck concrete) (;>-, 5.&.2.2)

a = (0.85) (.682 IN) = '.*80 IN a = '.*& IN at critica section for shear9

Compute average stress in prestressing strand at strength limit state:

(LRFD 5.7.3.1.1-1)

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18/26

=265.3 KSI [fps= 264.6 KSI at critical section for shear]

Compute nominal moment capacity, Mn:

(LRFD 5.7.3.2.2-1)

= 107,827 K-IN =8,986 K-FT

Compute factored moment resistance, Mr:

= 1.0 for flexure (LRFD 5.5.4.2)

Mr=Mn= 8,986 K-FT

Compare factored moment resistance, Mrto required moment, Mu:

Mr= 8,986 K-FT > Mu= 7,701 K-FT O.K.

3.1.4.6 Reinforcement Limits

Check Maximum Reinforcement:

(LRFD 5.7.3.3.1-1)

Since there is no mild steel, de= dp= 75.78 IN

= 0.0620.42 O.K.

Check 3inim4m >einforcement

3rthe esser of 1.2 3cror 1.'' 34 (;>-, 5.&.'.'.2)

Com4te 3cr4sin td ecs !rtice *.18.2.1

(td ecs *.18.2.1)

where

= = 0.588 I (;>-, 5..2.6)

= comressive stress in concrete d4e to effective restress forces on" (after

E osses) at etreme fi#er of section where tensie stress is ca4sed #"eterna" aied oads

= f#after osses (see ection '.1..2) = 2.*26 I

= the non$comosite dead oad moment

= 1101.1 120.' 155.0 = 2*6. $-T

= = comosite section mod44s for the tension face

= non$comosite section mod44s for the tension face

Mcr= 5*1' $-T $ 88' $-T = 50'0 $-T

1.2 3cr= 1.2 (50'0) = 60'6 $-T O/0S +in!e 1.2 !r 1.33 u

1.'' 34= 1.'' (&&01) = 1022 $-T

Mr= 8*86 $-T F 1.2Mcr= 60'6 $-T 0..

'.1.5 +nd and Transfer 7oint at >eease

Stre++e+ on need to e !$e!ed t ree+e t t$i+ o!tion +in!e o++e+ it$ tie iredu!e t$e !on!rete +tre++e+ in# t$e e++ !riti!.

7-18


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'.1.5.1 Com4te Concrete tresses ,4e to ;oads (%irder Bn")

ft= = 12 = 0.080 I f #= 12 = $ 0.08' I

'.1.5.2 Com4te Concrete tresses ,4e to 7restress

tresses d4e to restress are eG4a to Hero at the ends.

!t Transfer 7oint

/ottomf#= = = 2.66 I

Toft= = = 0.02' I

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'.1.5.' Check Concrete tresses at ervice ;imit tate $ /efore ;osses (!t >eease)

ervice I

/ottom of %irder (Comressive tress)

f#= 2.66 $ 0.08' = 2.56' I

Check stress imit (;>-, 5.*..1.1)2.56' I A = = 2.&0 I B..

To of %irder (Tensie tress)

ft= 0.02' 0.080 = 0.10' I

Check stress imit witho4t #onded a4iiar" reinforcement (;>-, 5.*..1.2)

0.10' I F $ = $ = $ 0.201 I $ 0.200 I B..

'.1.6 ,eression 7oint (0.5;) at >eease

Stre++e+ on need to e !$e!ed t ree+e t t$i+ o!tion +in!e id+*n i #oern 'or'in +tre++ !ondition+.

'.1.6.1 Com4te Concrete tresses ,4e to ;oads (%irder Bn")

ft= 12 = 0.865 I f #= 12 = $ 0.8* I

'.1.6.2 Com4te Concrete tresses ,4e to 7restress

/ottom

f#= '.*8 I (+e + t id+*n)

To

ft= $ 0.801 I (+e + t id+*n)

'.1.6.' Check Concrete tresses at ervice ;imit tate $ /efore ;osses (!t >eease)

ervice I/ottom of %irder (Comressive tress)

f#= '.*8 $ 0.8* = 2.60 I

Check stress imit in concrete (;>-, 5.*..1.1)

2.60 I A = = 2.&0 I B..

To of %irder (Tensie tress)

ft= $ 0.801 0.865 = 0.06 I

Check stress imit witho4t #onded a4iiar" reinforcement (;>-, 5.*..1.2)

0.06 I F $

= $= $ 0.201 I $ 0.200 I B..

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3.2 Shear Design

'.2.1 Transverse hear >einforcement

In this example, the girder will be designed for vertical shear at the critical

section for shear. In a full design, other sections along the length of the girderwould have to be designed as well.

'.2.1.1 Critica ection for hear

Shear design using theSectional Design Modelis an iterative process that begins by

assuming a value for.

!ss4me an initia va4e for the incination of the comression fied of 25J.

Critica section for shear is reater of (;>-, 5.8.'.2)

K 0.5 dvcot or

K dv

Com4te dv

dv = +ffective shear deth

= ,istance #etween res4tants of tensie and comressive forces

The deth of the comression #ock a was com4ted in determinin the momentcaacit" of the section (see ection '.1..5).

= (&2.0 8.0 $15.&') $ ('.*&:2) = 62.28 IN

/4t dvneed not #e taken ess than the reater of (;>-, 5.8.2.&)

0.* de= (0.*) (80$15.&') = (0.*) (6.2&) = 5&.8 IN

0.&2 h = (0.&2) (80) = 5&.60 IN

Therefore 4se dv= 62.28 IN

Critica section for shear is reater of

K 0.5 dvcot= 0.5 (62.28) cot(25J) = 66.&8 IN or O/0S (;>-, 5.8.'.2)

K dv= 62.28 IN

Note dvwi overn for F 26.6J.

Therefore the critica section for shear is

0.50 -T 66.&8 IN : 12 = 6.06 -T from centerine of s4ort.

!t the critica section for shear


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'.2.1.' %overnin +G4ations for hear


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This is the same va4e as was ass4med so converence has #een achieved.

With these values, the concrete contribution,Vc, can now be computed.

=78.7 KIP

3.2.1.5 Required Shear Reinforcement, Vs

Required Vs= Vu/- Vc Vp= 299.7 / 0.9 78.7 19.6 = 234.7 KIP

Assuming vertical stirrups,

(LRFD C5.8.3.3-1)

Compute Avon an IN2/FT basis (s = 12 IN):

= 0.351 IN

2/FT

Check minimum transverse reinforcement:

(LRFD 5.8.2.5)

= 0.093 IN2/FT < 0.351 IN

2/FT O.K.

Check maximum stirrup spacing: (LRFD 5.8.2.7-2)

Vu= 299.7 KIP > 0.1 fcbvdv= (0.1) (6.00) (6.00) (62.28) = 224.2 KIP

Therefore, maximum stirrup spacing is 12 IN.

Use #4 stirrups @ 12 IN (Av,provd= 0.40 IN2/FT)

3.2.2 Interface Shear Reinforcement

In this example, the girder will be designed for interface shear at the initial

critical section for shear. In a full design, other sections along the length of

the girder would have to be designed as well.

The width of the shear interfaceis equal to the width of the top flange of the

girder, which is42.00 IN. Therefore, bv= 42.00 IN.

Assume that the top surface of the girder was intentionally roughened to an

amplitude of 0.25 IN and cleaned prior to placement of the deck concrete.

The requirement for intentional roughening of the top of the girder should be

indicated on the plans.Compute the factored horizontal shear, Vh:

Vh= Vu/ de (LRFD C5.8.4.1-1)

The definition for degiven for this equation is the same as dv. Therefore use dv

as computed above.

Vh= Vu/ dv= 299.7/ 62.28 = 4.82 KIPS/IN

Since VhVnand= 0.9,

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Vn reqd= Vh/= 4.82 / 0.9 =5.36 KIPS/IN

Check limits on Vn:

Vn0.2 fcAcvor 0.8 Acv

Using the deck concrete strength of 4.0 KSI for fc, the two limits are equal.

Acv = area of concrete engaged in shear transfer

= = (42.0 IN) (1.0 IN) = 42.0 IN2

Use = 1.0 IN to compute Vhon a per inch basis.

Vn reqd= 5.36 KIPS/IN0.2 fcAcv= 0.2 (4.0) (42.0) = 33.6 KIPS / IN O.K.

Compute the nominal interface shear resistance, Vn:

(LRFD 5.8.4.1-1)

where:

c = 0.100 KSI and= 1.000 for an intentionally roughened surface (LRFD 5.8.4.2)

Avf = area of shear reinforcement crossing the shear plane

Pc = permanent net compressive force normal to the shear plane

=0

Solve for the required Avf:

= 0.0193 IN

2/INor 0.232 IN

2/FT

Minimum steel requirement:

(LRFD 5.8.4.1-4)

Avf= (0.05) (42 IN) (12 IN) / 60 KSI =0.42 IN2/ FT (Controls)

Use 2 # 4 @ 12 IN (Av provd= 0.40 IN2/FTSay OK Note that this limit depends directly on the width

of the interface more steel required for a wider interface)

3.3 Longitudinal Reinforcement Requirement

In this example, the longitudinal reinforcement requirement will bechecked at

the critical section for shear. The Specifications require that this requirement

must be satisfied at each section of the girder.Therefore, in a full design,

other sections along the length of the girder would also have to be checked.

'.'.1 >eG4ired ;onit4dina -orce

Required Longitudinal Force:

Treqd= (LRFD 5.8.3.5-1)

However, at the inside edge of bearing at the simply-supported ends,

Treqd= (LRFD 5.8.3.5)

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25/26

where:

Vs= shear resistance provided by transverse reinforcement, not to exceed Vu/.

= (Use final values from shear design above) (LRFD C5.8.3.3-1)

= = 267.1 KIP

Vu/= 299.7 KIP / 0.9 = 333.0 KIP, so use computed quantity for Vs.

Treqd= = (333.0 133.6 19.6) cot (25)

= (179.8) (2.14) =385.6 KIP

'.'.2 !vaia#e ;onit4dina -orce

The force to resist Treqdmust be supplied by the reinforcement on the flexural tension side of

the member. In this case, the available reinforcement consists of the straight strands. The

available force that can be provided by these strands at the critical section for shear must be

determinedconsidering the lack of full development due to the proximity to the end of the

girder.

The location at which T must be provided is where the failure crack assumed for this analysis,

which radiates from inside face of the support, crosses the centroid of the straight strands.

The angledetermined above during shear design at this location is used here. The inside

face of the support is 12 IN from the end of the girder.

Figure 5: Assumed Failure Crack and Location Where

Crack Crosses Straight Strands

The total effective prestress force for thestraight strandsis:

7es=!sfpe= 4.284 IN2(154.68 KSI) =662.6 KIP

The distance from the bottom of the girder to the centroid of these strands is:

dg= c.g. straight strands = 4.00 IN

Measured from the end of the girder, the crack crosses the centroid of the straight strands at:

x = = 12 IN + 4.00 IN (cot 25) =20.6 IN

This ocation is within the transfer enth l tso the avaia#e stress is ess than the effective

restress force for the straiht strands. The avaia#e restress force Tavai at is thereforecom4ted ass4min a inear variation in stress from the end of the irder to the transfer

enth. The transfer enth l t is 60 d#or '0 IN. (;>-, 5.11..1.)

7-25


26/26

Tavai= 7es = 662.6 I7 (20.6 IN : '0 IN) = 55.0 I7

ince Tavai= 55.0 I7 F TreGd= '85.6 I7 the straiht strands are adeG4ateto resist thereG4ired onit4dina force at this ocation and no additiona reinforcement is reG4ired.

If the strands had not #een adeG4ate to resist the force additiona mid reinforcement wo4dhave #een added to rovide the remainder of the reG4ired force.

3.4 nchorage !one "ein#orcement$

'..1 !nchorae Lone >einforcement

!rtice 5.10.10.1 reG4ires that the factored #4rstin resistance of a retensioned anchoraeHone #e at east .0M of the tota restressin force. This resistance is rovided #" verticareinforcement cose to the ends of retensioned irders.

The factored #4rstin resistance is iven #"

7r= fs!s (;>-, 5.10.10.1$1)

where

7r= (0.0) 7o = (0.0) (0.&5) (2&0 I) (5.508 IN2)9 = .61 I7

ote T$e tot !in# 'or!e *rior to n o++e+ i+ u+ed + t$e tot *re+tre++in# 'or!e :oin t$i+ !!ution

fsis the workin stress in the reinforcement not to eceed 20 I

ovin for the reG4ired area of reinforcement !s

= 2.2' IN

2

Therefore at east 2.2' IN2of vertica reinforcement m4st #e aced within h:5 = &2 IN : 5 =

1. IN from the end of the mem#er. tirr4s aced for vertica or interface shear can aso#e 4sed to satisf" this reG4irement since this reinforcement is on" reG4ired to resist forces atreease.

'..2 Confinement >einforcement

In accordance with !rtice 5.10.10.2 confinement reinforcement not ess than ' #ars at asacin of not more than 6.0 IN sha #e aced within 1.5 d (sa" 1.5 h = *.00 -T) from theend of the irder. These #ars sha #e shaed to encose the strands.

design of ridge mohamed.doc

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