200-foot simple span bridge girder design using nu2000 v1

8/22/2019 200-Foot Simple Span Bridge Girder Design Using NU2000 V1

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NU2000-200 ft Single Span Example

Given:

Span Data

Overall Girder Length =200 FTDesign Span = 199 FT

Girder is simply supported

Bridge Cross Section Data

Number of girder lines = 6Girder Spacing = 9.00 FT

Roadway Width= 51.00 FT

Overall Width = 53.50 FT

Deck Thickness

Actual = 7.5 INStructural = 7.0 IN

Haunch thickness = 1 IN

Girder Type

NU 2000, with 150 mm (5.91 in.) web

Location: Interior

Dead Load

Future Wearing Surface = 0.025 KSFBarrier Weight/Barrier Line = 0.27

KLF

Live Load

HL-93 - Design Truck + Design Lane

Loads

Girder Concrete

fc (fci/0.8) KSIfci = to be determined

wc = normal weight aggregate concrete

Deck Concrete

fc = 5.0 KSI

Prestressing Steel

Type: 0.6-IN Diameter 270 KSI Low-Relaxation Seven-Wire Strand

Pull: 75%

H = 70 % (Relative Humidity)

Use as many depression points as needed

for design. Max. 12 strands depressed at any

one point.

Reinforcing Steel (Non-Prestressed)

fy = 60 KSIEs = 29,000 KSI

Welded Wire Reinforcement

Use standard reinforcement sizes and

spacing. Use 75 ksi steel for shearreinforcement


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Proposed design steps:

(1)Calculate gross section properties, assuming a girder concrete strength of 10 ksi.(2)Calculate live load distribution factors.(3)Calculate moments and shears at 10th points due to various limit state load combinations.

(4)Approximate number and arrangement of strands based on Service III stress and NCHRPapproximate method. Then use detailed NCHRP 1807 (LRFD 2005) loss method to refine

the number of strands (use spreadsheet and check by hand).

(5)Check Strength I with strands calculated in (4), and determine concrete strength at final(use spreadsheet and check by hand).

(6)Check if the precast section satisfies Strength IV (use spreadsheet and check by hand).(7)Determine concrete strength at release; revise concrete strength at final if needed (use

spreadsheet and check by hand).(8)Determine the design moment diagram at 10th points and check that it is greater than the

Mu diagram at the same points.

(9) Check min. reinforcement limits.

(10) Design shear reinforcement; at least the critical section must be designed by handcalculation, including longitudinal reinforcement anchorage check.

(11) Check interface (horizontal) shear.(12) Design and detail the end zone reinforcement (splitting and confining steel).(13) Estimate midspan camber at release and at erection.(14) Estimate live load deflection.


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1.1 Section Properties

1.1.1 Precast Girder

The LRFD Specs allow the inclusion of transformed strand in the section propertiesfor a prestressed member (Article 5.9.1.4). For initial calculations, the reinforcementamount is not available and contribution of the strand to the section properties is

neglected.

Properties of NU2000:

A = 903.8 IN2

I = 790,592 IN4

h = 78.7 IN

bW = 5.9 IN

yb = 35.7 IN

yt = 43.0 IN

Sb = 790,592 IN4 / 35.7 IN = 22,145 IN3

St = 790,592 IN4 / 43.0 IN = 18,385 IN3

1.1.2 Composite Section [spelling error on this figure Transformed

Girder C.G.35.70"

Deck C.G.

83.20"

108"Effective Width

72.58"Transformed Width

Figure 1.1.2-1: Cross Section of Single Girder with Composite Deck


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Effective deck width: (LRFD 4.6.2.6.1)

One-quarter span length = (199 FT / 4)(12) = 597 IN

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

(48.2 IN) / 2 + (12) (7.0 IN) = 108.1 IN

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

1000

)KSI(f140.0)KCF(W

'c+=

KCF145.01000

5140.0Wcd =+=

KCF150.01000

10140.0Wcg =+=

n =cg

cd

E

E=

5.1

cg

cd

'

cg

'

cd

W

W

f

f

=

5.1

150.0

145.0

10

5

= 0.6720

Transformed deck width = (n) (effective width) = (0.672) (108 IN) = 72.58 IN

Transformed haunch width = (n) (haunch width) = (0.672) (48.2 IN) = 32.39 IN

Table 1.1.2-1: Composite section properties.

Component Transformed

Areayb A yb A(yb- ybc)

2 Io Ic

Girder 903.80 35.70 32,265.66 282,684.28 790,592.00 1,073,276.28Haunch 32.39 79.20 2,565.50 21,586.30 2.70 21,589.00

Effective

Deck508.07 83.20 42,271.26 451,627.27 2,074.61 453,701.88

Total 1,444.26 77,102.43 1,548,567.16

Table Section Properties Used in Design:

ybcg = ( A yb) / A =77,102.43/ 1,444.26 = 53.39 IN

ytcg = h ybc = 78.70 IN - 53.39 IN = 25.31IN

ytcd = hc ybc = 78.7 IN +1 IN+ 7.0 IN - 53.39 IN = 33.31 IN

Sbcg = Ic / ybc = 1,548,567.16 IN4 /53.39 IN = 29,004.82IN3

Stcg = Ic / ytcg = 1,548,567.16 IN4 / 25.31 IN = 61,184.00IN3


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Stcd = (Ic / ytcd) / n = (1,548,567.16 IN4 / 33.31 IN) / (0.6720) = 31,240.98 IN3

Component ybcg ytcg ytcd Sbcg Stcg Stcd

Value 53.39 25.31 33.31 29,004.82 61,184.00 31,240.98


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2. Live Load Distribution Factor for a Typical Interior Beam

Distribution Factors: (LRFD 4.6.2.2.1)

To use the simplified live load distribution factor formula, 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 FTO.K.

Curvature < 4o O.K. (Curvature = 0o)

Bridge Type: k (LRFD Table 4.6.2.2.1-1)

2.1 Distribution Factor for Moment(2 or more lanes loaded)

1.0

3s

g2.06.0

tL0.12

K

L

S

5.9

S075.0DF

+= (LRFD 4.2.2.2b-1)

Provided that: 3.5 S 16 S = 9.00 FT O.K.

4.5 ts 12.0 ts = 7.50 IN O.K.

20 L < 240 L = 200 FT O.K.

Nb 4 Nb = 6 O.K.

Kg = n (I + A eg2) (LRFD 4.6.2.2.1-1)

n = (1/n for composite section calcs = 1/0.6720) = 1.488

eg = yt + (ts / 2) = 43 IN + 1 IN +(7 IN / 2) = 47.50 IN

Kg = 1.488 [790,592 IN4 + (903.8 IN2) (47.50 IN) 2 ] = 4,210,699 IN4

( )( )

1.0

3

42.06.0

IN7FT1990.12

IN699,210,4

FT199

FT0.9

5.9

FT0.9075.0DF

+= = 0.689 lanes / girder

2.2 Distribution Factor for Shear (2 or more lanes loaded)0.2

35

S

12

S2.0DF

+= (LRFD 4.6.2.2.3a-1)

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

above:

10,000 Kg ( = 4,210,699) 7,000,000 O.K.


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0.2

35

FT0.9

12

FT0.92.0DF

+= = 0.884 lanes / girder

Distribution Factor for Bending Moment (DFM):

For all limit states except Fatigue Limit State, for two or more lanes

loaded, DFM = 0.689 lanes/beam.

For one design lane loaded, DFM = 0.450 lanes/beam.

Therefore, in the case of the two design lanes loaded controls, DFM = 0.689

lanes/beam.

Distribution Factor for Shear Force (DFV):

For two or more lanes loaded, DFV = 0.884 lanes/beam[LRFD Table 4.6.2.2.3a-1].

For one design lane loaded, DFV = 0.72 lanes/beam.

Therefore, in the case of two design lanes loaded controls, DFV = 0.884.


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3. Moments and Shears

3.1 Dead Loads

3.1.1 Girder Dead Load at Release

The moments for this condition are computed separately from other moments becausethe full length of the girder is used in computing these moments, rather than thedesign 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 theends of the girder.

Girder Dead Load:

w = (903.8 IN2/ 144) (0.140+10/1000 KCF) = 0.941 KLF

Moment at x distance from end of beam due to self weight at relaese time:

Mgdli = ( )xL2xw

L = 200 FT (overall girder length)

Mgdli = 0.941 (x / 2) (200 - x) = 94.1 x - 0.471 x2

Moment due to self weight during lifting. The lifting points are at 0.1L and 0.9 L.

w = 0.941 KLF

188.2 K-FT 188.2 K-FT

2823.0 K-FT

0.1L 0.8L 0.1L

Locations of interest at release conditions:

1. Transfer point t = 60 db = 60 (0.6 IN) = 36.0 IN = 3.0 FT (LRFD 5.8.2.3)

2. Depression points, yet to be determined

3. Midspan


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See Spreadsheet 3.3 for calculation of the moment at 10 th points. See Table 3.1 for a summary.

3.1.1.2 Girder Dead Load - Final

For all final stress analysis, bearing-to-bearing span should be used.

L = 199 FT (bearing to bearing)

Mgdl = 93.63 x - 0.471 x2

Vgdl =

x

2

Lw = 93.63 - 0.941 x

3.1.1.3 Deck Slab Loading

Deck Thickness = 7.5 IN

wddl = ((7.5 IN x 108 IN) /144) (0.140+5/1000 KCF) = 0.816 KLF

L = 199 FT

Mddl

= 81.19 x - 0.408 x2

Vddl = 81.19 - 0.816 x

3.1.1.4 Haunch Weight

wncdl = ((1.0 IN x 48.2 IN) /144) (0.145 KCF) = 0.049 KLF

L = 199 FT

Mncdl = 4.876 x 0.024 x2

Vncdl = 4.876 0.049 x

3.1.1.5 Composite Dead Load - Barriers

Barriers: (2) (0.27 KLF / barrier) = 0.54 KLF

wcdl = 0.54 KLF / (6 girders) = 0.09 KLF / girder (LRFD 4.6.2.2.1)

Mcdl = 8.955 x - 0.045 x2

Vcdl = 8.955 - 0.09x

3.1.1.6 Composite Dead Load - Future Wearing Surface

Future Wearing Surface: (51.00 FT) (0.025 KSF) = 1.275 KLF

wfws = 1.275 KLF / (6 girders) = 0.213 KLF

L = 199 FTMcdl = 21.194 x - 0.117 x

2

Vcdl = 21.194 - 0.106 x


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3.2 Live load

Bending Moment at Mid-Span

Design truck (design tandem should be checked, but this does not govern for relatively longspans).

Design Truck will govern over Design Tandem for this span.

8 kips

14 ft 14 ft

32 kips 32 kips

For maximum moment due to truck load at mid-span, position the rear axles 14 ft apart. Themaximum moment occurs when the section being considered is halfway between the resultant ofall three loads and the next point load. Also, the case where the 32 kip center axle load is directly

over the midspan section should be checked.

AR

32 kyyx

R32 k 8 k

AR

FT33.972

2881432x =

+=

FT33.22

33.914y =

=

KIP84.36199

)33.22/199(72R A =

+=

FTK76.292,3)33.233.9(322/19984.36MTruck =+=

Lane load (distributed load):


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q = 0.64 klf

RRAA

KIP68.632

19964.0R A =

=

FTK08.31682/1992/)2/19964.0(2/19968.63MLane ==

Moment distribution factor for live load 0.689 and impact factor 1.33 for truck load,

MLL+I = DF [Mlane + 1.33 (Mtruck)] = 0.689(3283.3961.33+3168.081 ) = 5191.6 K-FT

The shear critical section is 5.70 ft away from the support (see section 10.1.1).1. Truck load:

The influence line of the mid-span shear force is as follows:

5.7/199=0.029

5.7 ft

0.97

The layout of the truck load is as follows:

32k

R

32k 8k

5.7 ft

( ) KIP56.66)287.5199(8)147.5199(32)7.5199(32199

1R =++=


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66.56 K

34.56K

2.56 K

5.44 K

5.7 ft

Vtruck= 66.56 KIPS

3.2.3.2. Lane load

The layout of the lane load is as follows:

R

0.64k/ft

KIP08.602

)7.5199(64.0

199

1RV

2

left =

==

VLL+I = DF [Vlane + 1.33 Vtruck] = 0.896 (1.3366.56+60.08) = 133.3 KIP

3.3 Applicable Limit States (LRFD Table 3.4.1-1)

DCQStrengthIVu 5.1: =

3.3.1 Service I

This load combination is the general combination for Service Limit State stress

checks and applies to all conditions other than Service III.

DC5.1Q:IVStrength u =

)(8.0)(0.1: IMLLDWDCQIIIService u +++=

)(0.1)(0.1: IMLLDWDCQIService u +++=

)(75.15.125.1: IMLLDWDCQIStrength u +++=


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All load factors are equal to 1.0 for this problem.

For moment at mid-span:

Acting on the non-composite girder, MSLnc:

MSLnc = 4,658.1+ 3,768.3 + 269.2 + 240.3 = 8,935.8 K-FT

Acting on the composite girder, MSLc:MSLc = 534.6 + 1,158.3 + 5,191.5 = 6,884.4 K-FT

3.3.2 Service III

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

applies only to tension in prestressed concrete structures with the objective of crack

control.

All load factors are equal to 1.0 for this problem, except that the live load is reduced by afactor of 0.8.

Moments at mid-span:

Acting on the non-composite girder, MSLnc (same as for Service I).

Acting on the composite girder, MSLc:

MSLc = 534.6 + 1158.3 + (0.8)(5,191.5) = 5846.1 K-FT

3.3.3 Fatigue

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

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

prestressed components designed to have extreme fiber tensile stress due to Service III LimitState 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.

3.3.4 Strength I

This load combination is the general combination for Strength Limit State design.Since the structure is simply supported, the maximum values for the load factors are

used because they produce the greatest effect (see LRFD Table 3.4.1-2).

No distinction is made between moments and shears applied to the non-composite orcomposite sections for strength computations. The factored loads are applied to the

composite section.

The following load factors apply:

Dead Load - Component and Attachments 1.25 DC

Dead Load - Wearing Surface and Utilities 1.50 DW

Vehicular Live Load and Impact 1.75 LL and IM


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Mu or Vu = 1.25DC + 1.50DW + 1.75(LL + IM)

For moment at midspan, Mu:

Mu = (1.25) [4,658.1 + 3,768.3 + 269.2 + 240.3 + 534.6] + (1.50) (1,158.3) + (1.75)

(5,191.5)

Mu = 22,660.6 K-FT

For shear at the critical section for shear, Vu :

Vu = (1.25) [88.3 + 71.4 + 5.1 + 4.6 + 10.1] +(1.50) (21.9) + (1.75) (133.3)

Vu = 490.5 KIP

For moment at the critical section for shear, Mu:

Mu = (1.25) [518.4 + 419.4 + 30.0 + 26.7 + 59.5] + (1.50) (128.9) + (1.75) (577.8)

Mu = 2,521.9 K-FT

3.3.5 Strength IV

This design limit state checks a precast members strength under its own weight plus the

weight of all loads applied before composite action takes effect. If this condition is satisfied,

there is no need to check the compressive stress limit of 0.45 'cf due to effective prestress plus

dead load.

The following load factors apply:

Dead Load - Component and Attachments: DC 1.50

Moment at mid-span should be determined for self weight, deck slab weight and haunch

weight, Mu:

Mu = (1.5) [4658.1 + 4037.4+ 240.3] =13,403.6 K-FT

The bending moments and shear forces for all required limit states at 10th

points of the span

are shown in Tables 3.1 and 3.2:


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Table 3.1 Bending Moments for a Typical Interior Girder

DF 0.689 moment

DF 0.884 shear Table of Moments, K-FT

Location 0 critical 0.1 L 0.2 L 0.3 L 0.4 L 0.5 L 0.6 L 0.7 L

ft 0 5.70 19.9 39.8 59.7 79.6 99.5 119.4 139.3

Acting on Non-comp. Section

Girder, klf 0.941 0 518.4 1676.9 2608.5 3912.8 4471.7 4658.1 4471.7 3912.8

Deck (Structural) , klf 0.761 0 419.4 1356.6 2411.7 3165.4 3617.6 3768.3 3617.6 3165.4

Additional Non-composite 0.054 0 30.0 96.9 172.3 226.1 258.4 269.2 258.4 226.1

haunch, klf 0.049 0 26.7 86.5 153.8 201.8 230.6 240.3 230.6 201.8

Subtotal: 0 994.5 3216.9 5346.2 7506.0 8578.3 8935.8 8578.3 7506.0

Acting on Compostie section

Barrier, KLF 0.108 0 59.5 192.5 342.2 449.1 513.2 534.6 513.2 449.1

Future wearing Surface 0.234 0 128.9 417.0 741.3 973.0 1112.0 1158.3 1112.0 973.0

Live Load + Impact 0 577.8 1878.2 3367.8 4414.9 5040.8 5191.5 5040.8 4414.9

SUBTOTAL-Service I 0 766.2 2487.7 4451.3 5837.0 6666.0 6884.4 6666.0 5837.0

SUBTOTAL-Service III 0 650.6 2112.0 3777.7 4954.0 5657.9 5846.1 5657.9 4954.0

Total-Strength I 0 2521.9 8174.0 14116.1 19129.5 21853.8 22660.6 21853.8 19129.5

Total-Strength IV 0 1491.7 4825.3 8019.4 11259.1 12867.5 13403.6 12867.5 11259.1

Moments at release, K-FT

Location 0 transfer 0.1 L 0.2 L 0.3 L 0.4 L 0.5 L 0.6 L 0.7 L

ft 0 1.50 20 40 60 80 100 120 140

Girder, KLF 0.941 0 140.1 1693.8 3011.2 3952.2 4516.8 4705.0 4516.8 3952.2


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Table 3.2 Shear Forces for a Typical Interior Girder

Shear Force, K

Location 0 critical 0.1 L 0.2 L 0.3 L 0.4 L 0.5 L 0.6 L 0.7 L

ft 0 5.70 19.9 39.8 59.7 79.6 99.5 119.4 139.3

Acting on Non-comp. Section

Girder 0.941 93.6 88.3 74.9 56.2 37.5 18.7 0.0 -18.7 -37.5

Deck (Structural) 0.761 75.7 71.4 60.6 45.4 30.3 15.1 0.0 -15.1 -30.3

Additional Non-composite 0.054 5.4 5.1 4.3 3.2 2.2 1.1 0.0 -1.1 -2.2

haunch 0.049 4.8 4.6 3.9 2.9 1.9 1.0 0.0 -1.0 -1.9

Subtotal: 179.6 169.3 143.7 107.8 71.8 35.9 0.0 -35.9 -71.8

Acting on Compostie section

Barrier 0.108 10.7 10.1 8.6 6.4 4.3 2.1 0.0 -2.1 -4.3

Future wearing Surface 0.234 23.3 21.9 18.6 14.0 9.3 4.7 0.0 -4.7 -9.3

Live Load + Impact 138.9 133.3 119.4 101.2 84 68 53.2 -68 -84

SUBTOTAL-Service I 172.9 165.4 146.6 121.6 97.6 74.8 53.2 -74.8 -97.6

SUBTOTAL-Service III 145.1 138.7 122.7 101.4 80.8 61.2 42.6 -61.2 -80.8 Total-Strength I 515.9 490.5 427.2 340.8 256.1 173.6 93.1 -173.6 -256.1 Total-Strength IV 320.5 302.1 256.4 192.3 128.2 64.1 0.0 -64.1 -128.2


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4. Total number and arrangement of strands

4.1 Estimated Number of Strands

The total number and arrangement of strands was estimated using service III and NCHRP 18-07

approximate loss method. Refer to attached Excel sheet 4.1 and Tables 4.1 and 4.2. The total

number of bottom strands can be found = 85-0.6 in. strands: 46 straight strands at the bottomflange, which is the maximum number of strands that the bottom flange can accommodate. The

rest of the strands will be harped in three groups at 0.2L, 0.3L, and 0.4 L as shown in Figures 4-1

and 4-2. The strands can be harped using a hold down device or steel pipes attached to theprestressing bed.

Table 4.1 Prestress loss using NCHRP 18-07 approximate method.Sub-

Total

Change Net Change Net Loss

202.5

(1) Elastic shortening due to Pi 7.020 -38.1 164.4

(2) Elastic shortening due to self weight -1.776 9.6 174.0Elastic, Prestress transfer, fpES 5.244 28.5

(3) Shrinkage 12(1.7-0.01H)(5/(1+f'ci) -7.1 167.0

(4) Creep 10(fpiApsAg)(1.7-0.01H)(5/(1+f'ci)) -24.3 142.7

(5) Relaxation, fpR 2.5 -2.5 140.2

Total long-term, fpLT 33.9

(6) Elastic due to deck weight Mdecketr-fin/Ibm-tr-fin *n -1.673 7.9 148.0

(7) Elastic due to superimposed DL (on composite section) MADLecomp-tr/Icomp-tr *n -0.487 2.3 150.3

Elastic, Deck + SIDL -10.2

Total Prestress Loss prior to LL, fpT 52.2

(8) Elastic due to LL MLLecomp-net/Icomp-net *n -7.9

Total loss including gain due to LL 44.3

fpsfcgs

]I

e

A

1[Af

reltrgdr

2

reltr

reltrgdr

pspi

+ cgips fnf =

reltrgdr

2 reltrgdr

I

eM

Table 4.2 Bottom fiber stresses (using transformed/net section properties and approximate

NCHRP 18-07 method).Cause Initial Final

Pi (transf. section, release) 7.735 7.735

Mg (transf., release) -2.169 -2.169

Loss (net section, precast) -1.594

deck weight (transf., service) -2.039

SIDL (transf. composite) -0.552

LL (transf., composite) -1.910

Net 5.566 -0.528

Code Limit 4.500 -0.601

'

cf19.0

4.2 Number of Strands Using Detailed NCHRP 18-07 Method

Use service III and NCHRP 18-07 (LRFD 2005) detailed loss method. For this iterative process

to be done quickly, the Loss spreadsheet is used. Refer to Spreadsheet 4.2 and Tables 4.3 and

4.4. The total number of bottom strands can be refined to 82-0.6 in. strands: 46 straight strands at

the bottom flange and 36 harped strands. To avoid excessive uplift force on the prestressing bed,a maximum of 12 strands will be harped at any point. Thus, the 36 strands will be harped in three


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groups. The drape point locations will be assumed at 0.2L, 0.3L, and 0.4 L as shown in figures 5-1 and 5-2. Further design checks may require altering this arrangement. In addition, 4 straight top

strands will be used for mild reinforcement support and for control of top cracking at prestress

release. For the top strands, the tension will be specified at 10 ksi. The effect of the top strandson the total prestress force will be ignored.

Table 4.3: Prestress loss using NCHRP 18-07 detailed method.Prestress Loss Using NCHRP 18-07 Detailed Method

Project Name: PCI Handbook Page 2-19 12DT32 member with 208-D1 Strand Pattern

Designer: MKT Date: 40407

Prestress Loss (ksi)

Loading Loss Components

Combinations

Change Net Change Net

0 202.5

(1) Elastic shortening due to Pi =fpiA ps 6.809 6.809 -37.0 165.5

(2) Elastic shortening due to self weight -1.786 5.023 9.7 175.2

Elastic, Prestress transfer, fcgp and fpES 5.023 5.023 -27.3

(3) Shrinkage between release and deck place fpSR=bid*Ep*Kid -0.149 4.874 -3.6 171.6

(4) Creep between release and deck place fpCR=ni*fcgp*ybid* -0.565 4.309 -13.7 157.9

(5) Relaxation between release and deck place (assumption) fpR2 -0.049 4.260 -1.2 156.7

Total long-term (initial to deck placemnt)id -0.763 4.260 -18.6 -18.6

(6) Elastic due to deck weight Mdecketr-fin/Ibm-tr-fin*n -1.681 2.579 7.9 164.6

(7) Elastic due to superimposed DL (on composite sectioMADLecomp-fin/Icomp-fin*n -0.489 2.090 2.3 166.9

Elastic, Deck + SIDL: Dfcd and DfpED -2.170 2.090 10.2 -8.4

(8) Shrinkage of beam bet.deck place and final fpSD=bdf*Ep*Kdf -0.406 1.684 -1.9 165.0

(9) Creep of beam bet.deck place and final, initial loads fpCD1 =fcgp*(ybif-ybid)*Kdf*ni -1.334 0.350 -7.2 157.7

(10) Creep of beam due to deck and SIDL fpCD2=fcd*ybdf*Kdf*n 1.399 1.749 6.6 164.3

(11) Relaxation between deck place and final (assumption) fpR3 -0.255 1.494 -1.2 163.1

(12) Shrinkage of deck 0.134 1.628 0.6 163.7

Total long-term (deck placemnt to final)df -0.462 1.628 -3.1 -11.5

(13) Elastic due to LL n*MLLecomp_tr/Icomp_tr -1.693 -0.065 8.0 -3.5

Total loss including gain due to LL -30.8

fcgp fps

]1

[2

reltrgdr

reltr

reltrgdr

pspiI

e

AAf

+cgips fnf =

reltrgdr

reltrgdr

I

eM

2

( )( )

bdfdf

c

dpc

cdf

cddddf

c

p

pSS KI

ee

A

EA

E

Ef

7.01

1

7.01+

+

+=

Table 4.4: Bottom fiber stresses (using transformed/net section properties and detailed

NCHRP 18-07 method).

Cause Final

Pi (transf. section, release) 7.503

Mg (transf., release) -2.180

Loss (net section, precast) -0.841

deck weight (transf., service) -2.047SIDL (transf. composite) -0.554

Loss (net, composite) -0.136

LL (transf., composite) -1.918

Net -0.174

70

Code Limit -0.601

'

cf19.0


19/50

19

Figure 4-1: Harped strands profile

Section at (0.5L)End Section (0.0L) Figure 4-2: Pretensioning scheme

All strands eccentricities are shown in Table 4.5 below:

Table 4-5: Pretensioning eccentricity at different sectionsShear critical Develop.Leng.

Strand type 0 L 0.031 L 0.06 L 0.1 L 0.2 L 0.3 L 0.4 L 0.5 L

12 draped strands 32.2 27.75 23.59 17.85 3.5 3.5 3.5 3.5

12 draped strands 50.2 45.89 41.86 36.30 22.40 8.5 8.5 8.5

12 draped strands 68.2 63.96 60.00 54.53 40.85 27.18 13.5 13.5

46 straight strands 4.09 4.09 4.09 4.09 4.09 4.09 4.09 4.09

82 Bottom strands 24.33 22.43 20.65 18.20 12.06 8.03 6.03 6.03

0 . 4 L0 . 3 L

0 . 2 L


20/50

20

4.3 Hand Calculation of Prestressed Losses Using Detailed NCHRP 18-07 Method

4.3.1 Shrinkage Strain

4.3.1.1 Precast

First period (from release to Final).

tsthscpbif kkkkk00048.0S = 1kcp =

One day steam curing

038.1735

)2.3(941064ks =

=

0.1)70(0143.000.2kh ==

588.05.71

5=

+=fk

61.0)588.0)(0.1)(038.1)(1(ksh ==

( )( )

998.0120000)5.7(461

120000ktd =+

=

6

bif 10x292S=

Second period (from release to deck placement).

tsthscpbid kkkkk00048.0S = 1kcp =

One day steam curing

038.1735

)2.3(941064ks =

=

0.1)70(0143.000.2kh == 588.0

5.71

5kst =

+=

61.0)588.0)(0.1)(038.1)(1(ksh == ( )

( )656.0

160)5.7(461

160=

+

=tdk

6

bid 10x192S=

bidbifbdf SSS = 6

bdf 10x100S=

4.3.1.2 Deck Slab

tsthscpI-di kkkkk00048.0S = 1kcp =

for 7 days moisture curing

0.1735

)2/0.7(941064ks =

=

0.1)70(0143.000.2kh ==


21/50

21

833.051

5kst =

+=

( )( )

998.06020000)5.7(461

6020000ktd =

+

=

at final

6

ddf 10x399)998.0)(833.0)(0.1)(0.1)(0.1(00048.0S==

4.3.2 Creep Coefficient

4.3.2.1 Precast

First period (from release to final).

tlahsst kkkkk9.1bif =

955.0735

)85.3(941064ks =

=

588.0

5.71

5=

+

=fk

0.1)70(008.056.1 ==hck

( ) 0.1118.0 == lala tk ( )

( )998.0

120000)5.7(461

120000ktd =

+

=

065.1)998.0)(0.1)(0.1)(588.0)(955.0(9.1bif ==

Second period (release to deck placement).

tlahsst kkkkk9.1bid =

955.0735

)85.3(941064ks =

=

588.05.71

5=

+=fk

0.1)70(008.056.1 ==hck ( ) ( ) 0.11 118.0118.0 === lala tk

( )

( )656.0

160)5.7(461

160=

+

=tdk

700.0)656.0)(0.1)(0.1)(588.0)(955.0(9.1bid ==

Third period (deck placement to final).

tlahsst kkkkk9.1bdf =

955.0735

)85.3(941064ks =

=

588.05.71

5=

+=fk

0.1)70(008.056.1 ==hck


22/50

22

( ) ( ) 617.060 118.0118.0 === lala tk ( )

( )998.0

602000)5.7(461

6020000=

+

=tdk

657.0)998.0)(617.0)(0.1)(588.0)(955.0(9.1bdf ==

4.3.2.2 Deck Slab

tlahsfddf kkkkk9.1=

0.1735

)5.3(941064=

=sk

0.1)5(8.01

5=

+=fk

0.1)70(008.056.1 ==hck ( ) ( ) 0.11 118.0118.0 === lala tk

( )( )

998.012000)5(461

120000 =+

=tdk

877.1)998.0)(0.1)(0.1)(0.1)(0.1(9.1bid ==

4.3.3 Transformed Section Properties

Transformed Section Properties at Release ( non-composite ):26.982)1( inAnAA pigti =+=

inA

AYYbi 32.33==

4412,854 inI ti =

Transformed Section Properties at stage II (composite):212.510,1)1( inAnAA pigt =+=

232.51 inA

AYYb ==

4755,666,1 inI t =

4.3.4 Losses at the first period

Strand stress at service from the spreadsheet:

ksi0.17286.02.16.62.79.13.29.72.17.136.33.275.202fs +++++=

Bottom fiber stress at Service III:


23/50

23

=cf ti

bg

ti

bp

ti

iI

y*M

I

ye

A

1P

+

+

i_net

i_net_bi_net_p

i_net

II

ye

A

1P

[Due to time dependent losses from release to deck placement]

tr

trd

I

y*M+

TrCom

Tr_ComSIDL

I

y*M

+

+

Tr_net

Tr_net_bTr_net_p

Tr_net

III

ye

A

1P

[Due to time dependent losses from deck placement to final]

Tr_Com

Tr_ComLL

I

y*M

+

= 7.503-2.180-0.841-2.047-0.554-0.136-1.918= -0.174 KSI

KSI601.0f19.0KSI174.0 'c ==


24/50

24

5. Strength I at Mid-Span

The reinforcement bars in the deck slab and the top flange strands are ignored in thesecalculations. The I-beam top flange is assumed to be a rectangle of equal area to the actual flange

area. The depth can found to be 3.93 in for the actual width of 48.2 in. of the NU I girder top

flange shape. The strain-compatibility approach was used to calculate the mid- span section

strength. See Table 5.1 for the results. The last cycle of the iterative analysis will be redone bylong hand below to explain the spreadsheet analysis and to check its results (See attached Excel

Spreadsheet 5.1 for Strength I).

Straight strands As1=460.217=9.982 IN2 y1 = 81.60 IN from top of slab

Harped Strands As2=360.217=7.812 IN2

y2 = 78.20 IN from top of slab

From the spreadsheet, the value of the compression block depth, is a = 11.06 IN This indicates

that the compression block is within the top flange of the beam: 11.07-7 (slab) 1 (haunch) =

3.07 IN.

According to the average beta formula source PCI BDM and LRFD,

Deck 1 = 0.85-0.05(fc-4) = 0.85-0.05(5-4) = 0.8Girder 1 = 0.85-0.05(fc-4) = 0.85-0.05(10-4) = 0.65

Average 1 = (Ac1fc11+ Ac2fc22)/ (Ac1fc1 + Ac2fc2) = 0.76

c = a/ 1 = 14.55 IN

The compression force in the deck slab and moment due to force at the top of the deck slab is

Fc1 = 0.855(9127) = 3213.00 KIPMc1 = 3213.00 7/2 = 11245.50 K-IN

Similarly, the compression force in the haunch and moment due to this force at the top fiber ofthe deck slab are 204.85 kips and 1536.38 kip-in; the compression force in the beam top flangeand moment due to this force at the top of the deck slab are 1259.17 kips, and 12008.27 kip- in.

The spreadsheet is able to treat each strand layer individually. For convenience in the handcalculation check, we will assume two groups of strands, the straight strands and the harped

strands, clustered at two points.

Straight strands strain: 020.028500

1721

c

82.61003.01 =+

=

Using the Power formula in PCI Bridge Design Manual Section 8.2.2.5,

( ){ }270

4.1121

27613887f36.7/136.7

ps

psps

+

+=

fps1 = 263.21 KSI

Fs1 = 263.219.982=2627.36 kipsMoment at top face Ms1 = 2627.36 82.61 = 217045.84 kip-IN

Tension force in the harped strands and moment due to this force at the top of the deck slab is


25/50

25

Straight strands strain: 019.028500

1721

c

2.78003.01 =+

=

Using Power formula fps2 = 262.37 KSI

Fs1 = 262.37 7.182=2,049.66 kips

Ms1 = 2,049.66 78.2 = 160,283.4 K-IN

Sum of forces = 2,049.66 +2627.36 -204.85 -1259.17-3213.00 = 0.0

Since strain in the extreme bottom stand layer is larger than 0.005, the resistance factor =1.

Sum of M =1 (160,283.4 +217,045.84 -12,008.27 -1536.38 -

11245.50)/12=29378.3 K-FT > Mu = 22393.3 K-FTThus the provided strength is about 30% larger than the required strength. This check will not

need to be revisited unless the final number of strands has been extremely reduced. This also

tells us that the assumed final concrete strength of 10 KSI may be too high. The Strength IV

check should confirm this. We also have an option to try to use a lower deck strength at thistime.


26/50

26

Flexural Strength Using Strain Compatibity and Mast's Variable 0.75 to 1.0

cu

c=

a

Sum of

forces

ANSWER:

Mn kip-in

kip*ft

Units in kips and inches

Concrete Layers f'c Width, W Thick., T Depth, dc 1 Tupper Tlower Revised T Beta1calcuation

1 5.000 108.000 7.000 3.500 0.800 0.000 7.000 7.000 3024 3780

2 5.000 48.200 1.000 7.500 0.800 7.000 8.000 1.000 192.8 241

3 10.000 48.200 3.930 9.531 0.650 8.000 11.930 3.061 959.123074 1475.57396

4 10.000 5.910 74.770 11.930 0.650 11.930 86.700 0.000 0 0

5 86.700 0.850 86.700 86.700 0.000 0 06 86.700 0.850 86.700 86.700 0.000 0 0

7 86.700 0.850 86.700 86.700 0.000 0 0

Steel Layers Area Asi 4175.923074 5496.57396

Steel Layers Area Asi Grade Effective Pr Depth dsi Es Q fpy R K so

Gr 270 1 9.962 270 172 82.610 28500 0.031 243 7.36 1.043 0.0060

Gr 270 2 7.812 270 172 78.200 28500 0.031 243 7.36 1.043 0.0060

Sum of M MAXIMUM :

W1

W2

W3

W4

T2T2 Lower

T2Upper 1

2

3

4

dsi

As i

Table 5.1: Strength I Calculation at Mid-Span Section


27/50

27

6. Strength IV Limit State

This limit state is used to check the capacity of the precast section only. The Excel strain-

compatibility program was used to design the section. The capacity of the precast section

without any mild reinforcement in the compression zone was estimated at:

FTKIPMn = 0.12314 against the demand of :

FTKIPMu = 6.13403 (from table 3-1)

The mild reinforcement was provided in the top flange to obtain the adequate moment capacity.8 # 9 was used to satisfy this limit state. This reinforcement is extended to the 0.4 L. The 6 # 9

are extended to the 0.3 L.

The section at point 0.4L with 6 # 9 in the top flange has capacity of

FTKIPMn

= 0.12922 > FTKIPMu = 5.12867 . OK

The section at 0.3L without any top reinforcement has a capacity of

FTKIPMn

= 0.11774 > FTKIPMu = 51.11259 . OK

For brevity, the hand calculation for the last iteration only is shown below (mid-span).

'

cf =10,000 psi 1 =0.65

c = 55.34 IN

a = 1c = 0.65(55.34) = 35.97 IN

+

=

p

pepps

Ef

cd 1003.0

We can combine prestressing strand into the clusters:

1. 46 straight strands

INdp 61.7409.47.78 ==

INxps310187.7

28500

0.1721

34.55

61.74003.0 =

+

=

2. 12 draped strands- First Cluster

INdp 2.755.37.78 ==

INxps310219.7

28500

0.1721

34.55

2.75003.0

=

+

=

3. 12 draped strands- Second Cluster


28/50

28

INdp 2.705.87.78 ==

INxps310948.6

28500

0.1721

34.55

2.70003.0 =

+

=

4. 12 draped strandsThird Cluster

INdp 2.655.137.78 ==

INxps310678.6

28500

0.1721

34.55

2.65003.0 =

+

=

5. 4 partially tensioned straight strands in the top flange

INdp 0.2=

INxps310541.2

28500

10

34.55

234.55003.0 =

=

Strain in compression steel is found from the ratio:

d = 2.0 IN

INxs3' 10892.2

34.55

234.55003.0 =

=

KSIfEf ysss 60==

KSIfKSIxf ys 6087.83)29000(10892.23 ===

Thus, take KSIfs 60= .

Find the stress of the prestressing strands using the Power formula:

( )( )KSIf

ps

psps 270

4.1121

613,27887

36.7

136.7

+

+=

1. 46 Straight strands

( )( )KSI

x

xfps 80.199

)10187.7(4.1121

613,2788710187.7

36.7

136.73

3 =

+

+=

2. 12 Draped strands- First Cluster


29/50

29

( )( )KSI

x

xfps 54.200

)10219.7(4.1121

613,2788710219.7

36.7

136.73

3 =

+

+=

3. 12 Draped strands- Second Cluster

( )( )KSI

x

xfps 14.194

)10948.6(4.1121

613,2788710948.6

36.7

136.73

3 =

+

+=

4. 12 Draped strands Third Cluster

( )( )KSI

x

xfps 48.187

)10678.6(4.1121

613,2788710678.6

36.7

136.73

3 =

+

+=

5. 4 Partially tensioned straight strands

( )( )KSI

x

xfps 41.72

)10541.2(4.1121

613,2788710541.2

36.7

136.73

3 =

+

+=

Assume the compression flange is divided into separate sections.

The rectangular shape was used to substitute the trapezoidal form of the top flange. The width of

the rectangular shape b is as follows:

b = (48.2 + 5.9)/2=27.05 IN

Compression forces:

KIPAfC cc 8.1048)10(56.2)2.48(85.085.0'

1 ===

KIPAfC cc 0.407)10(77.1)05.27(85.085.0'

2 ===

KIPAfC cc 8.1586)10)(77.156.297.35)(9.5(85.085.0'

3 ===

KIPffAC cpsPS 5.55))10(85.041.72)(217.0(4)85.0('

4 ===

KIPffAC csS 2.366))10(85.00.60)(79.0(9)85.0('

5 ===

Tension forces:

KIPfAT psps 4.19948.199)217.0(461 ===

KIPfAT psps 2.52254.200)217.0(122 ===


30/50

30

KIPfAT psps 5.50514.194)217.0(123 ===

KIPfAT psps 2.48848.187)217.0(123 ===

To derive the moment capacity of the section, we can take the sum of the moments around any

point.

Summing the moments around the bottom of the compression block:

jcj

j

cscsspscpspstop

psipsipsi

i

n yAfyffAyffAyfAM ==

+++=3

1

'''4

1

85.0)85.0()85.0( =

=1,994.4(38.64) + 522.2(39.23) + 505.5(34.23) + 488.3(29.23) + 1,048.8(34.69)

+ 407.0(32.53) + 1,586.8(15.82) + (55.5+366.2)33.97 =

=77,063.6 + 20,485.9 +17,303.3 + 14,273.0 + 36,382.9 + 13,239.7 + 25,103.2 + 14,325 =

218,176.7 KIP*IN = 18,181.4 KIP*FT

The resistance factor shall be taken according to LRFD 5.5.4.1.

For flexure and tension of prestressed concrete, it is equal to 1.

For the unified design method, the following formula for calculation will be used:

00.13/)25075.1(75.0 += extreme

Proposed for LRFD by Seguirant et al. (PCI Journal 2004):

INxextreme310158.11

34.55

27.78003.0 =

=

68.03/)10158.1(25075.1( 3 =+= x Thus take 75.0=

0.636,13)4.181,18(75.0 ==nM KIP-FT

From Table 3-1, the factored moment for Strength IV is equal to:

6.403,13=uM KIP-FT OK

The Excel program treats each layer of strands individually and provides more accurate results.

Refer to Figure 6-2 and the attached Excel program for Strength IV, cell M17.

Its output for INKIPMn = 7.505,13 , which is greater than INKIPMu = 6.403,13

The precast section has a sufficient capacity to resist the ultimate load during the construction

stages.

OK


31/50

31

Flexural Strength Using Strain Compatibity and Mast's Variable = = = = 0.75 to 1.0))))

c=

a

Sum of

forces

Units in kips and inches ANSWE

kip*ft

Concrete Layers f'c Width, W Thick., T Depth, dc 1 Tupper Tlower Revised T Beta1calcuation

1 10.000 48.200 2.560 1.280 0.650 0.000 2.560 2.560 802.048 123

2 10.000 27.050 1.770 3.445 0.650 2.560 4.330 1.770 311.21025 4783 10.000 5.900 74.370 20.152 0.650 4.330 78.700 31.643 1213.5186 1866

Total: 78.700 2330.214 3584

Steel Layers Area Asi Grade Effective Pre Depth dsi Es Q fpy R K eso Grade 60 Bars 1 7.11 60 0.0 2.000 29000 0 60 100 1.096

Gr 270 1 3.472 270 172 76.700 28500 0.031 243.00 7.36 1.043 0

Gr 270 2 3.472 270 172 74.700 28500 0.031 243 7.36 1.043 0Gr 270 3 2.17 270 172 72.700 28500 0.031 243 7.36 1.043 0Gr 270 4 0.868 270 172 70.700 28500 0.031 243 7.36 1.043 0

5 0.868 270 10 2.000 28500 0.031 243 7.36 1.043 06 2.604 270 172 75.200 28500 0.031 243 7.36 1.043 07 2.604 270 172 70.200 28500 0.031 243 7.36 1.043 08 2.604 270 172 65.200 28500 0.031 243 7.36 1.043 0

W1

W2

W3

W4

T2T2Lower

T2Upper 1

2

3

4

dsi

Asi

Figure 6.1: Strength IV Calculations for Mid-span


32/50

32

7. Required Concrete Strength at Release

Because of the extreme length of this beam, it is necessary to set the lifting points some distance

away from the beams ends to help resist buckling of the top flange. We will assume that a careful

lifting analysis has resulted in identifying the lifting points at 20 feet from each end.

Based on NDOR Policy and January-February 2001 PCI paper by Tadros et al., Strength Design

of Pretensioned Flexural Concrete Members at Prestress Transfer, the prestressed member canbe treated as a reinforced concreted column subjected to moment combined with axial

compression force equal to the force in the prestressing steel just before prestress transfer.

Therefore, we can solve for the neutral axis location c and 'cif by using the equilibrium

equations.

Asi

ysi

b j

j=2

j=1

j=3j=4

j=5

j=6

j=8

j=9

j=7

d c jTj


33/50

33

Table 7.1 Excel program for calculating'

cif

Directions:

1 Input Output

2 Sign convention: compressive stress

in concrete and tensile stress in steelare considered positive. Prestress force

is always positive. Moment in the same

sense as prestress force moment is negtive.

3 Units in kips and inches

4 Make sure the compression fibre.

dc & ys based on compressed concrete edge

5 Functions are used below

Output f'c Top Steel Stress

8.468 -14.28

MExt

-35.2875 1.15 0.75 1.15 0.003

CONCRETE LAYERS

Width , b Thick ., T Sum of T Compr . T Depth, dc

1 38.4 5.3 5.3 5.3 2.65 203.52 1464.896 3881.975

2 22.15 5.5 10.8 5.5 8.05 121.825 876.872 7058.819

3 5.9 63.59 74.39 49.24668 35.42334 290.5554 2091.36 74082.95

4 27.05 1.75 76.14 0 0 0 0 048.2 2.56 78.7 0 0 0 0 0

78.7 4433.128 85023.74

STEEL LAYERSInitial Prest fpi Depth ys Stress Grade

Gr 270 1 9.982 202.5 4.09 28500 -0.002867 -81.71194 1 2355.611 9634.449 270 243 1.04 0.031 7.36

2 7.812 202.5 36.3 28500 -0.001821 -51.9032 1 2076.387 75372.86 270 243 1.04 0.031 7.363 0.868 10 76.95 28500 -0.000501 -14.28047 0 0.913884 70.32338 270 243 1.04 0.031 7.36

4432.912 85077.63

0.65 92.38 60.05 0.2158243 0.215824

STRENGTH DESIGN PROGRAM FOR PRESTRESS TRANSFER

DATA FOR CONTROLING

cA cMcF

c

m p

a

K Q R

F M

sEsA s sf sF sMiw pyf

cu

The equations that were used in the analysis are as follows:

1. Strain calculation:

=

cy1003.0 sisi

2. Force equilibrium:

==

+

+=

0way

1way0wf85.0ffAAf85.0F

isi

isi

i

icsi

p

sisi

j

cjc

3. Moment equilibrium:

==

+

+=

0way

1way0Mywf85.0ffAyAf85.0M

isi

isim

i

siicsi

p

sisi

j

cjcjc

4. Power formula to get strand stress:

+

+=

R/1R

py

ssi

ssisi

kf

E1

Q1QEf

By solving the two equilibrium equations, the two variables c and 'cif can be obtained. The

analysis can be done by trying c and 'cif and checking the equilibrium equations.


34/50

34

The following table shows the results for several critical sections.

Table 7.2: Prestress Member Strength at Release

Sections Moment (Kips-IN) fci (KSI.) Top fiber stress (KSI.)

Transfer length 60d=0.015L 3336.786 7.318 -25.36

0.1L 20325.6 7.449 -24.08

0.2L 36134.4 7.712 -20.09

0.3L 47426.4 8.159 -13.06

0.4L 54201.6 7.456 -20.85

lifting point 0.1L -50.8 8.469 -14.27

Max. 'cif 8.469

Note that prestress loss calculation needs to be redone using 8.5 KSI for concrete strength atrelease.

Note:

The following is the check by hand at 0.1L span and the last trial of the analysis.c= 92.38 IN., fci = 8.468 ksi

= 0.85-0.05(fc-4) = 0.85-0.05(8.468 -4) = 0.7

a = c = 60.91 IN.Divide NU2000 into 5 layers:

Table 7.3: Concrete layers

Concrete Layer Width Thickness

1 38.4 5.3

2 22.15 5.5

3 5.9 63.59

4 27.05 1.75

5 48.2 2.56

The force at each concrete layer, and the corresponding bending moment at the edge of thebottom flange for the final trial, are as follows:

1. Concrete layer1Fc1 = 0.858.468 ( 38.45.3)= 1464.896 KipsMc1 = 1464.896 5.3/2 = 3881.975 Kips-IN

2. Concrete layer2Fc2 = 0.858.468 ( 22.155.5) = 876.872 Kips

Mc2 = 876.872(5.5/2+5.3) = 7058.819 Kips-IN

3. Concrete layer 3Fc3 = 0.858.468 ( 5.963.59) = 2091.36 Kips

Mc3 = 2091.36 [(60.91-5.3-5.5)/2+5.3+5.5] = 74082.95 Kips-IN


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4. Concrete layer 4Fc4 = 0 Kips

Mc4 = 0 Kips-IN

5. Concrete layer 5

Fc5 = 0 Kips

Mc5 = 0 Kips-IN

Adding them together, F = 1464.896 +876.872 +2091.36 = 4433.128 Kips

M = 3881.975+7058.819+74082.95=85023.74 Kips-IN

Use

=

c

y1003.0 sisi

Top steel:

00287.097.86

09.41003.0 1s =

=

KS-81.71

27004.1

2850000287.01

031.01031.02850000287.0

kf

E1

Q1QEf

36.7/136.7

R/1R

py

ssi

s1s1s =

+

+=

+

+=

If the strands are within the compression block height, the compression force of the concreteshould be 0.85fc (Ac-As)=0.85 fcAc-0.85 fc As.

0.85fcAs is considered when we calculate force of strands in order to avoid the calculation of

centroid of the concrete area.

externaleriorint FF =

pips

p

sc fA

FF =

pips

p

sisscc fA

fA)AA(f85.0 =

Rearranging the items in the equation, we get

==

++=

0way

1way0wf85.0f

fAAfF

isi

isi

iicsi

p

sisij

cjc

Hence, Fs=9.982(-81.71+1.15/0.75202.5+0.858.468) = 2355.611 Kips.

Ms = 2355.611 4.09=9634.45 Kips-IN

The force calculations of the other strands are the same way. Therefore the total force and

moment of strands are:

Kips4432.90.9142076.42355.61Fs =++=


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36

IN-K85077.6370.3275372.869634.45Ms =++= 04432.9-4433.128FF sc =

0)-35.2875(0.75

1.15-85077.63-85023.74M

MM ext

msc =

Thus, the assumptions of fci

and the neutral axis location c are OK.


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8. Design moment diagram nM and factored moment Mu at 10th

Points

The flexural moment capacities were estimated at 10th

points of the span. For the end section, themoment capacity was based on 8 bottom strands bent in the end diaphragm. The embedment of

those strands was assumed to be equal to 36 inches, which allowed us to develop puf8.0 .

The moment capacity of the section located at the distance equal to the development length wasalso estimated. The development length of prestressing strands was calculated using the formula:

bpepsd dffkL )3

2( = ( LRFD 5.11.4.2-1)

where k= 1.6 for precast , prestressed beams

and KSIfps 21.263= , see section 5.

FTINLd 88.116.1426.0)1723

221.263(6.1 ===

The appropriate eccentricities of the strands were used to estimate the capacity of the sections at

different points of the span (see Table 4-5). The strain-compatibility approach was used for thesecalculations. (See attached files)

0

5000

10000

15000

20000

25000

30000

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Girder Sections X/L

BendingMoment(K-Ft)

Mu

Mn

Figure 8.1: Design moment diagram and the factored moment along the girder line


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9. Minimum Reinforcement Calculations

nM the lesser of 1.2 crM or 1.33 uM (LRFD 5.7.3.3.2)

Compute Mcr using Std. Specs Article 9.18.2.1

( ) rcbcnc/dcpercr fS1SSMSffM += (LRFD Eq. 5.7.3.3.2-1)where

cr ff = 37.0 = KSI00.1037.0 = 1.170 KSI (LRFD revision 2004)

pef = compressive stress in concrete due to effective prestress forces only (after

losses) at the extreme fiber of the section where tensile stress is caused by externallyapplied loads.

Prestressing force is KIPP 6.30600.172)217.0(82 ==

pef = 7.35

592,790

)03.67.35(6.3060

8.903

6.3060 +=+

nc

nc

y

I

Pe

A

P= 7.487 KSI

ncdM / = the non-composite dead load moment

ncdM / = 8,935.8 K-FT (Refer to Table 4-1)

cS = bcgS = composite section modulus for the tension face

bS = non-composite section modulus for the tension face

( )( ) ( )

+= 1

145,22

004,29)12(935,8004,29487.7170.1

3

33

IN

INFTKINKSIKSIMcr

= 251,087.6 33,212.4 = 217,875.2 KIP-IN = 18,156.3KIP FT. CONTROLS

rcfS = 29,004(0.759) = 22,014.0 KIP-FT

1.2 Mcr = 1.2 (18,156.3) = 21,787.5 K-FT GOVERNS since 1.2 Mcr< 1.33 Mu

1.33 Mu = 1.33 (22,660.6) = 30,138.6 K-FT

Moment capacity of the composite section at mid-span is:

FTKIPMn = 5.728,29

(See attached Excel program Strength I mid-span, cell M17)

nM = 29,728.5 K-FT > 1.2Mcr = 21,787.5 K-FT 0.K

The requirement of minimum reinforcement has been met.


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10. Shear Design

10.1 Transverse Shear Reinforcement

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 girder would have to be designed as

well.

10.1.1 Critical Section for Shear

Critical section for shear is located at distance dv from the support:

dv = Effective shear depth (LRFD revision 2004)

= Distance between the resultants of tensile and compressive forces

The depth of the compression block, a, was computed in determining the moment

capacity of the section (see Section 3.1.4.5). You can also use the output of Excelprogram Strength I, Strain-Compatibility Approach AASHTO LRFD @ critical

section (cell M8).

( )2

[email protected]

ationcriticalgchh

add fgev +== = (78.7 + 7.0 + 1 -22.19) - (10.05/2)

= 59.49 IN

But dv need not be taken less than the greater of (LRFD 5.8.2.7)

0.9 de = (0.9) (78.7 + 7 + 1 22.19) = (0.9) (64.51) = 58.06 IN

0.72 h = (0.72) (78.7 + 7 + 1) = 62.42 IN GOVERNS

Therefore, use dv = 62.42 IN.

The critical section for shear is then

0.50 FT + 62.42 IN / 12 = 5.70 FT from centerline of support.

At the critical section for shear, Vu = 490.5 KIPS.

10.1.2 Component of Shear Resistance from Prestress, Vp

Pf= 82 (0.217)172.0 = 3060.6 KIP

The angle of the center of gravity of the strand profile with respect to horizontal is

= tan-1 [(eCL - eend) / (distance to depression point)]

= tan-1 [((24.33 IN 18.20 IN) / 12) / (0.1(200)) FT] = 1.463o

Vp = Pfsin (1.463o) = (3060.6 KIP) [sin (1.463o)] = 78.2 KIPS

10.1.3 Governing Equations for Shear

Vu Vr = Vn (LRFD 5.8.2.1-2)

= 0.90 for shear (LRFD 5.5.4.2.1)

Vn = Vc + Vs + Vp (LRFD 5.8.3.3-1)


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Compute maximum shear capacity of the section:

Vn max = pvvc Vdbf25.0 + (LRFD 5.8.3.3-2)

Vn max = (0.25) (10.00 KSI) (5.9 IN) (62.42 IN) + 78.2 KIP = 998.9 KIPS

Vn max = (0.90) (985.1) = 998.9 KIPS > Vu = 485.0 KIPS O.K.

10.1.4 Concrete Contribution to Shear Resistance, Vc

vvcc dbf.V = 03160 (LRFD 5.8.3.3-3)

To use this equation, the quantity must be determined. This quantity is a factor that

represents the efficiency of shear transfer by concrete. Note that ( )KSIf0316.0 c =

( )PSIf0.1 c , so a value of 2 would provide a concrete contribution similar to the

familiar simplified value of ( )PSIf2V cc = bd.

To obtain , the quantitiescf

v

and X are needed, where

cf

v

is a relative shear stress

and X is the largest longitudinal strain which occurs within the web of the member.

( )

( ) ( )IN42.62IN9.59.0

KIP2.789.0KIP490.5

db

VVv

vv

pu =

= = 1.265 KSI (LRFD 5.8.3.4.2-1)

cf

v

= KSI

KSI

0.10

25.1= 0.127

We can begin iterations using the initially assumed value of in the formula below.

( )pspss

popspuu

v

u

xAEAE

fAVVNd

M

+

++

=2

cot)(5.05.0

0.002 (LRFD 5.8.3.4.2-2)

New Commentary allows us to avoid an iterative process by simplifying the formula

above to the form of

( )pspss

popspuu

v

u

xAEAE

fAVVNd

M

+

++

=2

)(5.0

0.002

Mu = 2,521.9 K-FT = 30,262.8 K-IN (Summary of Dead and Live Load Effects)

Moment in this formula is not supposed to be taken less than Vu dv (LRFD C5.8.3.3)

0.617,30)42.62(5.490 ==vu dV K-IN Controls

Nu = 0 - no applied axial loads

fpo = 0.7 (270) = 189.0 KSI (LRFD C5.8.3.4.2)

Aps = area of prestressing steel on the flexural tension side of the member,

i.e., the straight strands


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41

Aps = 46(0.217) = 9.982 IN2

( ) ( )( )

( ) ( )[ ]982.9500,282

0.189982.92.785.49042.62

617,30+

=x 0.002

( ) 974,5681.984

487,2842

6.886,14125.490 =

+=x = - 0.00173

Because x is negative, use Eq. 5.8.3.4.2-3:

( )pspsscc

popspuu

v

u

xAEAEAE

fAVVNd

M

++

++

=2

cot)(5.05.0

(LRFD 5.8.3.4.2-3)

or simplify this formula to:

( )pspsscc

popspuuv

u

xAEAEAE

fAVVNd

M

++

++

=2

)(5.0

Ac = Area of concrete on flexural tension side

= Area of girder below h/2 = (78.7+7+1)/2 = 43.35 IN

= (38.4)(5.3) + (5.5)(38.4+5.9)/2 + (43.35-5.3-5.9)(5.9) = 517.4 IN2

( )( )[ ] 932,841,61.984

487,2844.51760622

6.18864125.490 =

+

+=x = -1.44x10-4

From Table 5.8.3.4.2-1, with x = -1.44x10-4

& cf

v

= 0.127, find = 3.05 and

22.21=

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

( ) ( ) ( )ININKSIVC 42.629.50.1005.30316.0= = 112.2 KIPS

Another check was added by AASHTO LRFD revision 2004 to ensure that web concrete does

not crack (mostly necessary for box girders):

LDcw VVV +

From the Summary of Dead and Live Load Effects, we can find:

KIPVV LD 3.3343.1339.211.103.169 =+++=+

cwV can be found using ACI formula

( )pwpcccw VdbffV ++= 3.05.3

' Eq.(11-12)


42/50

42

where pcf is the resultant compressive stress at the centroid of the composite section due to both

prestress and moments resisted by the precast member acting alone, and dis a distance from the

extreme compression fiber to the centroid of prestressed reinforcement or 0.8h , whichever isgreater.

d= 78.7 + 7 + 1 - 22.19 = 64.51 IN0.8h = (78.7 + 7 + 1)0.8 = 69.36 IN CONTROLS

( ) ( )nccnc

ncd

ncc

nc

pc yyI

Myy

I

Pe

A

Pf += / =

=ncdM / 994.5 KIP-FT = 11934 KIP-IN from Summary of Dead and Live Load Effects

P = 82(172)0.217 = 3060.6 KIPS

Section properties can be found in Table 1.1.2-1:

( ) ( )7.3539.53592,790

119347.3539.53

592,790

)19.22(6.3060

8.903

6.3060+=pcf =

3.386 -1.520 + 0.267 = 2.133 KSI

( ) =++= 2.78)36.69(9.5)133.2(3.01000/10005.3cwV 307.2 + 78.2 = 385.4 KIPS > 334.3 KIPSO.K.

10.1.5 Required Shear Reinforcement, Vs

Required Vs = Vu / - Vc Vp = 490.5 / 0.9 112.2 78.2 = 354.6 KIPS

Assuming vertical stirrups,

s

cotdfAV

vyvs

= (LRFD C5.8.3.3-1)

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

=

cotdf

V12A

vy

sv

( )( )

( ) ( ) ( )=

22.21cot42.6275

354.612

INKSI

KIPINA

v = 0.353 IN2/FT

Check minimum transverse reinforcement:

y

vcv

f

sbf0316.0A = (LRFD 5.8.2.5)

( ) ( )

KSI

ININKSIAv

75

1290.500.100316.0= = 0.094 IN2/FT < 0.352 IN2/FT O.K.


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43

Check maximum stirrup spacing: (LRFD 5.8.2.7-2)

Vu = 490.5 KIP > 0.1 fc bv dv = (0.1) (5.9) (10.00) (62.42) = 368.3 KIPS

Therefore, maximum WWR spacing is 12 IN.

For D18 WWR, maximum spacing is

s = 0.18*2*12/0.352 = 12.27 INUse D18 WWR @ 12 IN

Av,provd = 0.18(2)12/12 = 0.36 IN2/FT

The design of the other sections is in the attached spreadsheet.

10.2 Longitudinal Reinforcement Requirement

In this example, the longitudinal reinforcement requirement will be checked at the

critical section for shear. The Specifications require that this requirement be satisfiedat each section of the girder. Therefore, in a full design, other sections along the

length of the girder would also have to be checked.

10.2.1 Required Longitudinal Force

Required Longitudinal Force:

Treqd =

+

+

cotVV5.0

VN5.0

d

Mps

uu

v

u (LRFD 5.8.3.5-1)

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

Treqd =

cotVV5.0

Vps

u

(LRFD 5.8.3.5)

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

=s

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

=( )( )( ) ( )

IN

INKSIIN

0.12

22.21cot42.627536.0 2 = 361.7 KIPS

Treqd = ( ) ( )

22.21cot2.787.3615.0

9.0

5.490= (545.0 180.9 78.2) cot (21.22)

= (285.9) (2.58) = 736.5 KIPS

10.2.2 Available Longitudinal Force

The force to resist Treqd must 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 determined considering the lack of full development due tothe proximity to the end of the girder.


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The location at which T must be provided is where the failure crack is assumed forthis analysis, which radiates from the inside face of the support and crosses the

centroid of the straight strands. The angle determined during shear design at thislocation is used here. The inside face of the support is 12 IN from the end of the

girder.

Figure 10.1: Assumed failure crack and location where

crack crosses straight strands.

The total effective prestress force for the straight strands is

Pes = Aps fpe = 9.982 IN2 (172.0 KSI) = 1716.9 KIPS.

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

dg = c.g. straight strands = 4.09 IN.

Measured from the end of the girder, the crack crosses the centroid of the straightstrands at

x = + cotdgb = 12 IN + 4.09 IN (cot 21.22) = 22.53 IN.

This location is within the transfer length, t , so the available stress is less than the

effective prestress force for the straight strands. The available prestress force, Tavail,at x is therefore computed assuming a linear variation in stress from the end of the

girder to the transfer length. The transfer length, t , is 60 db or 36 IN. (LRFD 5.11.4.1.)

Tavail = Pes ( )tx = 1716.9 KIPS (22.53 IN / 36 IN) = 1,074.1 KIPS.

Since Tavail = 1074.1 KIP > Treqd = 736.5 KIPS, the straight strands are adequate to

resist the required longitudinal force at this location, and no additional reinforcement

is required.

If the strands had not been adequate to resist the force, additional mild reinforcement

would have been added to provide the remainder of the required force.

It is good practice to bend at least 8 strands from the bottom layer of the strands.


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11. Interface Shear Reinforcement

The girder will be designed for interface shear at the initial critical section for shear.

The width of the shear interface is equal to the width of the top flange of the girder,which is 48.20 IN. Therefore, bv = 48.20 IN.

Assume that the top surface of the girder is 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 de given for this equation is the same as dv.

Therefore use dv as computed above.

Vh = Vu/ dv = 490.5/ 62.42 = 7.858 KIPS/IN.

Since Vh Vn and = 0.9,

Vn reqd = Vh/ = 7.858 / 0.9 = 9.702 KIPS/IN.

Check limits on Vn:

Vn 0.2 fc Acv or 0.8 Acv :0.8 Acv is controlling the design

Acv = area of concrete engaged in shear transfer

= vb = (48.2 IN) (1.0 IN) = 48.2 IN2

Use = 1.0 IN to compute Vh on a per inch basis.

Vn reqd = 7.858 KIPS/IN 0.2 fc Acv = (0.8) (48.2) = 38.56 KIPS / IN. O.K.

Compute the nominal interface shear resistance, Vn:

cyvfcvn PfAAcV ++= (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:

y

cvreqdnvf

f

AcVA

=

( ) ( )

( )( )6000.1

2.48100.0702.9 =vfA = 0.081 IN2/IN or 0.98 IN2/FT. CONTROLS

Minimum steel requirement:


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46

yvvf

f

sb05.0A (LRFD 5.8.4.1-4)

Avf= (0.05) (48.2 IN) (12 IN) / 60 KSI = 0.482 IN2/ FT

Use 2 # 5 @ 7 IN (Av provd = 1.06 IN2/FT Say OK Note that this limit depends directly

on the width of the interface more steel is required for a wider interface)


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12. Design and detail the end zone reinforcement

12.1 Anchorage Zone Reinforcement

LRFD 5.10.10.1 requires that the factored bursting resistance of a pretensioned

anchorage zone be at least 4.0% of the total prestressing force. This resistance isprovided by vertical reinforcement close to the ends of pretensioned girders.

The factored bursting resistance is given by

Pr = fs As , (LRFD 5.10.10.1-1)

where

Pr = (0.04) Po = (0.04) [(0.75) (270 KSI) (17.794IN2)] = 144.13 KIP

Note: The total jacking force prior to any losses is used as the total prestressing

force Po in this calculation:

fs is the working stress in the reinforcement, not to exceed 20 KSI.

Solving for the required area of reinforcement, As:

( )KSI20

KIP13.144

f

PA

s

rs == = 7.20IN

2

Therefore, at least 7.20 IN2 of vertical reinforcement must be placed within h/5 =

78.7 IN / 5 = 15.7 IN from the end of the member. Stirrups placed for vertical or

interface shear can also be used to satisfy this requirement since this reinforcement is

only required to resist forces at release.

Take # 7 stirrups @ 3 inch spacing

2provided

s IN20.76.0)2(6A ==

Note: The first spacing from the end of the girder to the first stirrup is 3 inches as

well.

12.2 Confinement Reinforcement

In accordance with LRFD 5.10.10.2, confinement reinforcement of not less than # 3

bars at a spacing of not more than 6.0 IN shall be placed within 1.5 d (say 1.5 h =

9.00 FT) from the end of the girder. These bars shall be shaped to enclose thestrands.


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13 and 14. Mid-Span Camber at release and erection and live load deflection

Curvature Values at mid-span

Due to Initial prestressinitialci

initialpspi

IE

eAf

= = 2.08E-05

Due to member weightinitialci

gdr

IE

M

= = -1.18E-05

Due to loss (from initial to erection)finalci

finalpsps

IE

eAf

= = -1.97E-06

Due to dead load on precast girderfinalc

deck

IE

M

= = -1.00E-05

Due to dead load on composite sectioncompositefinalc

dl

IEM

= = -1.76E-06

Due to prestress loss (erection to final)compositelfinalc

compositefinalps2ps

IE

eAf

= = -1.91E-07

Due to live loadcompositefinalc

ll

IE

M

= = -6.08E-06


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49

Approximate Curvature Integration

It is valid fMidspan End and two-po

Initial prestress 2.08E-05 8.49E-06 13.79 13.79 23.85339714

Member weight -1.18E-05 0 -6.99 -6.99 -12.08981378 due to pres

Loss (initial to erection) -1.97E-06 -7.81E-07 -1.30 ------ -1.967053642

Dead load on precast -1.00E-05 0 -5.95 ------ ------ Where

Dead load on composite -1.76E-06 0 -1.04 ------ ------

Loss (erection to final) -1.91E-07 -4.75E-07 -0.14 ------ ------ a = distancLive load -6.08E-06 0 ------ ------ ------

Total 6.80 9.80 due to selfw

Accurate Curvature Integration

End 0.1L 0.2L 0.3L 0.4L Midspan

Initial prestress 8.49E-06 1.29E-05 1.70E-05 1.96E-05 2.08E-05 2.08E-05

Member weight 0 -4.44E-06 -6.76E-06 -9.97E-06 -1.13E-05 -1.18E-05

Loss (initial to erection) -7.81E-07 -1.20E-06 -1.65E-06 -1.88E-06 -1.99E-06 -1.97E-06Dead load on precast 0 -3.77E-06 -6.57E-06 -8.49E-06 -9.62E-06 -1.00E-05

Dead load on composite 0 -6.57E-07 -1.15E-06 -1.49E-06 -1.69E-06 -1.76E-06

Loss (erection to final) -4.75E-07 -4.62E-07 -4.02E-07 -2.93E-07 -2.17E-07 -1.91E-07

Live load 0 -2.29E-06 -4.03E-06 -5.21E-06 -5.90E-06 -6.08E-06Total

The formula for deflection at midspan as a function of curvatures at (span/10) points, for simple spans and for interior spans.

It may be used for end spans without much loss of accuracy. Symmetrical moment diagram about midspan is assumed) .

ElasticCurvature

Elastic CurvatureElastic Initial Erection

)3

7432

6

1(L01.0 543210

2+++++=

The live load deflection is 3.64 IN. The detailed calculations are shown in the attached Excel spreadsheet


50/50

Calculating deflection by the elastic weight or moment-area method,

Curvature

Moment

Load M

!"!

Figure 13.1: Deflection Calculation

The equation in the Excel spreadsheet was developed by calculating the deflection at the

mid-span section by taking the moment of the elastic loads at the mid-span. Assume that

the curvature is symmetrical. Refer to Figure 13.1.

200-foot simple span bridge girder design using nu2000 v1

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