AAB - Example Problems Handout

LRFD Steel Design

AASHTO LRFD Bridge Design Specifications

Example Problems

Created July 2007

This material is copyrighted by

The University of Cincinnati and

Dr. James A Swanson.

It may not be reproduced, distributed, sold, or stored by any means, electrical or

mechanical, without the expressed written consent of The University of Cincinnati and

Dr. James A Swanson.

July 31, 2007

LRFD Steel Design

AASHTO LRFD Bridge Design Specification

Example Problems Case Study: 2-Span Continuous Bridge.......................................................................................1 Case Study: 1-Span Simply-Supported Bridge .........................................................................63 Case Study: 1-Span Truss Bridge...............................................................................................87 Ad-Hoc Tension Member Examples Tension Member Example #1 ..........................................................................................105 Tension Member Example #2 ..........................................................................................106 Tension Member Example #3 ..........................................................................................108 Tension Member Example #4 ..........................................................................................110 Ad-Hoc Compression Member Examples Compression Member Example #1 .................................................................................111 Compression Member Example #2 .................................................................................112 Compression Member Example #3 .................................................................................114 Compression Member Example #4 .................................................................................116 Compression Member Example #5 .................................................................................119 Compression Member Example #6 .................................................................................121 Compression Member Example #7 .................................................................................123 Ad-Hoc Flexural Member Examples Flexure Example #1 ..........................................................................................................127 Flexure Example #2 ..........................................................................................................129 Flexure Example #3 ..........................................................................................................131 Flexure Example #4 ..........................................................................................................134 Flexure Example #5a ........................................................................................................137 Flexure Example #5b........................................................................................................141 Flexure Example #6a ........................................................................................................147 Flexure Example #6b........................................................................................................152 Ad-Hoc Shear Strength Examples Shear Strength Example #1 .............................................................................................159 Shear Strength Example #2 .............................................................................................161

Ad-Hoc Web Strength and Stiffener Examples Web Strength Example #1 ...............................................................................................165 Web Strength Example #2 ...............................................................................................168 Ad-Hoc Connection and Splice Examples Connection Example #1....................................................................................................175 Connection Example #2....................................................................................................179 Connection Example #3....................................................................................................181 Connection Example #4....................................................................................................182 Connection Example #5....................................................................................................185 Connection Example #6a..................................................................................................187 Connection Example #6b .................................................................................................189 Connection Example #7....................................................................................................190

James A Swanson Associate Professor University of Cincinnati Dept of Civil & Env. Engineering 765 Baldwin Hall Cincinnati, OH 45221-0071

Ph: (513) 556-3774 Fx: (513) 556-2599

[email protected]

2- Span Continuous Bridge Example AASHTO-LRFD 2007 ODOT LRFD Short Course - Steel Page 1 of 62

1. PROBLEM STATEMENT AND ASSUMPTIONS: A two-span continuous composite I-girder bridge has two equal spans of 165 and a 42 deck width. The steel girders have Fy = 50ksi and all concrete has a 28-day compressive strength of fc = 4.5ksi. The concrete slab is 91/2 thick. A typical 2 haunch was used in the section properties. Concrete barriers weighing 640plf and an asphalt wearing surface weighing 60psf have also been applied as a composite dead load. HL-93 loading was used per AASHTO (2004), including dynamic load allowance.

3 spaces @ 12' - 0" = 36' - 0" 3'-0"

42' - 0" Out to Out of Deck

39' - 0" Roadway Width

9 (typ)

23/4" Haunch (typ)

3'-0"

References:

Barth, K.E., Hartnagel, B.A., White, D.W., and Barker, M.G., 2004, Recommended Procedures for Simplified Inelastic Design of Steel I-Girder Bridges, ASCE Journal of Bridge Engineering, May/June Vol. 9, No. 3

Four LRFD Design Examples of Steel Highway Bridges, Vol. II, Chapter 1A Highway Structures Design Handbook, Published by American Iron and Steel Institute in cooperation with HDR Engineering, Inc. Available at http://www.aisc.org/

-- 1 --


Positive Bending Section (Section 1)

Negative Bending Section (Section 2)

2. LOAD CALCULATIONS: DC dead loads (structural components) include:

Steel girder self weight (DC1) Concrete deck self weight (DC1) Haunch self weight (DC1) Barrier walls (DC2)

DW dead loads (structural attachments) include:

Wearing surface (DW) 2.1: Dead Load Calculations

Steel Girder Self-Weight (DC1): (Add 15% for Miscellaneous Steel)

(a) Section 1 (Positive Bending)

A = (15)(3/4) + (69)(9/16) + (21)(1) = 71.06 in2

( ) ( )Lbft

inft

2sec 1 2 1.15

490 pcf71.06 in 278.112

tionW

= = per girder

(b) Section 2 (Negative Bending) A = (21)(1) + (69)(9/16) + (21)(2-1/2) = 112.3 in2

( ) ( )Lbft

inft

2sec 2 2 1.15

490 pcf112.3 in 439.512

tionW

= = per girder

-- 2 --


Deck Self-Weight (DC1):

( )Lbft

inft

2150 pcf(9.5")(144") 1,42512

deckW

= = per girder

Haunch Self-Weight (DC1):

Average width of flange: 21"(66') 15"(264') 16.2"66' 264'

+ =+

Average width of haunch: ( ) ( )12 16.2"16.2" (2)(9") 25.2" + + =

( )( )( )

Lbft2in

ft

2" 25.2"

12(150 pcf ) 52.5haunchW

= = per girder

Barrier Walls (DC2):

( ) Lbft

(2 each) 640 plf320.0

4 girdersbarriersW

= = per girder

Wearing Surface (DW):

Lbft4 girders

(39')(60 psf ) 585fwsW = = per girder The moment effect due to dead loads was found using an FE model composed of four frame elements. This data was input into Excel to be combined with data from moving live load analyses performed in SAP 2000. DC1 dead loads were applied to the non-composite section (bare steel). All live loads were applied to the short-term composite section (1n = 8). DW (barriers) and DC2 (wearing surface) dead loads were applied to the long-term composite section (3n = 24).

-- 3 --


Unfactored Dead Load Moment Diagrams from SAP

-8,000

-7,000

-6,000

-5,000

-4,000

-3,000

-2,000

-1,000

0

1,000

2,000

3,000

4,000

0 30 60 90 120 150 180 210 240 270 300 330

Station (ft)

Mom

ent (

kip-

ft)

DC1

DW

DC2

Unfactored Dead Load Shear Diagrams from SAP

-200

-150

-100

-50

0

50

100

150

200

0 30 60 90 120 150 180 210 240 270 300 330

Station (ft)

Shea

r (k

ip)

DC1

DW

DC2

-- 4 --


The following Dead Load results were obtained from the FE analysis:

The maximum positive live-load moments occur at stations 58.7 and 271.3 The maximum negative live-load moments occur over the center support at station 165.0

Max (+) Moment Stations 58.7 and 271.3

Max (-) Moment Station 165.0

DC1 - Steel: 475k-ft -1,189k-ft DC1 - Deck: 2,415k-ft -5,708k-ft

DC1 - Haunch: 89k-ft -210k-ft DC1 - Total: 2,979k-ft -7,107k-ft

DC2: 553k-ft -1,251k-ft DW 1,011k-ft -2,286k-ft

-- 5 --


2.2: Live Load Calculations The following design vehicular live load cases described in AASHTO-LRFD are considered: 1) The effect of a design tandem combined with the effect of the lane loading. The design tandem consists of two 25kip axles spaced 4.0 apart. The lane loading consists of a 0.64klf uniform load on all spans of the bridge. (HL-93M in SAP) 2) The effect of one design truck with variable axle spacing combined with the effect of the 0.64klf lane loading. (HL-93K in SAP)

3) For negative moment between points of contraflexure only: 90% of the effect of a truck-train combined with 90% of the effect of the lane loading. The truck train consists of two design trucks (shown below) spaced a minimum of 50 between the lead axle of one truck and the rear axle of the other truck. The distance between the two 32kip axles should be taken as 14 for each truck. The points of contraflexure were taken as the field splices at 132 and 198 from the left end of the bridge. (HL-93S in SAP)

4) The effect of one design truck with fixed axle spacing used for fatigue loading.

All live load calculations were performed in SAP 2000 using a beam line analysis. The nominal moment data from SAP was then input into Excel. An Impact Factor of 1.33 was applied to the truck and tandem loads and an impact factor of 1.15 was applied to the fatigue loads within SAP.

-- 6 --


Unfactored Moving Load Moment Envelopes from SAP

-6,000

-4,000

-2,000

0

2,000

4,000

6,000

0 30 60 90 120 150 180 210 240 270 300 330

Station (ft)

Mom

ent (

kip-

ft)

Single Truck

Tandem

Tandem

Two Trucks

Single Truck

Contraflexure PointContraflexure Point

Fatigue

Fatigue

Unfactored Moving Load Shear Envelopes from SAP

-200

-150

-100

-50

0

50

100

150

200

0 30 60 90 120 150 180 210 240 270 300 330

Station (ft)

Shea

r (k

ip)

Single Truck

Tandem

Fatigue

-- 7 --


The following Live Load results were obtained from the SAP analysis:

The maximum positive live-load moments occur at stations 73.3 and 256.7 The maximum negative live-load moments occur over the center support at station 165.0

Max (+) Moment Stations 73.3 and 256

Max (-) Moment Station 165

HL-93M 3,725k-ft -3,737k-ft HL-93K 4,396k-ft -4,261k-ft HL-93S N/A -5,317k-ft Fatigue 2,327k-ft -1,095k-ft

Before proceeding, these live-load moments will be confirmed with an influence line analysis.

-- 8 --


2.2.1: Verify the Maximum Positive Live-Load Moment at Station 73.3:

Tandem:

Lane:

8kip

32kip 32kip

25kip25kip

0.640kip/ft

Single Truck:

-20

-10

0

10

20

30

40

0 15 30 45 60 75 90 105 120 135 150 165 180 195 210 225 240 255 270 285 300 315 330

Station (ft)

Mom

ent (

k-ft

/ kip

)

Tandem: ( )( ) ( )( )+ =kip kip k-ftk-ft k-ftkip kip25 33.00 25 31.11 1,603 Single Truck: ( )( ) ( )( ) ( )( )+ + =kip kip kip k-ftk-ft k-ft k-ftkip kip kip8 26.13 32 33.00 32 26.33 2,108 Lane Load: ( )( ) =2 k-ftkip k-ftft kip0.640 2,491 1,594

(IM)(Tandem) + Lane: ( )( ) + =k-ft k-ft k-ft1.33 1,603 1,594 3,726 (IM)(Single Truck) + Lane: ( )( ) + =k-ft k-ft k-ft1.33 2,108 1,594 4,397 GOVERNS The case of two trucks is not considered here because it is only used when computing negative moments.

-- 9 --


2.2.2: Verify the Maximum Negative Live-Load Moment at Station 165.0:

Tandem:

Single Truck:

Lane:

25kip25kip

0.640kip/ft

Two Trucks:

8kip

32kip 32kip

8kip

32kip 32kip

8kip

32kip 32kip

-20

-15

-10

-5

00 15 30 45 60 75 90 105 120 135 150 165 180 195 210 225 240 255 270 285 300 315 330

Station (ft)

Mom

ent (

k-ft

/ kip

)

Tandem: ( )( ) ( )( )+ =kip kip k-ftk-ft k-ftkip kip25 18.51 25 18.45 924.0 Single Truck: ( )( ) ( )( ) ( )( )+ + =kip kip kip k-ftk-ft k-ft k-ftkip kip kip8 17.47 32 18.51 32 18.31 1,318 Two Trucks:

( )( ) ( )( ) ( )( )( )( ) ( )( ) ( )( )

+ + ++ + + =

kip kip kipk-ft k-ft k-ftkip kip kip

kip kip kip k-ftk-ft k-ft k-ftkip kip kip

8 17.47 32 18.51 32 18.31 ...

... 8 16.72 32 18.31 32 18.51 2,630

Lane Load: ( )( ) =2 k-ftkip k-ftft kip0.640 3,918 2,508

(IM)(Tandem) + Lane: ( )( ) + =k-ft k-ft k-ft1.33 924.0 2,508 3,737 (IM)(Single Truck) + Lane: ( )( ) + =k-ft k-ft k-ft1.33 1,318 2,508 4,261 (0.90){(IM)(Two Trucks) + Lane}: ( ) ( )( ) + = k-ft k-ft k-ft0.90 1.33 2,630 2,508 5,405 GOVERNS

-- 10 --


Based on the influence line analysis, we can say that the moments obtained from SAP appear to be reasonable and will be used for design. Before these Service moments can be factored and combined, we must compute the distribution factors. Since the distribution factors are a function of Kg, the longitudinal stiffness parameter, we must first compute the sections properties of the girders. 2.3: Braking Force The Breaking Force, BR, is taken as the maximum of:

A) 25% of the Design Truck ( )( )kip kip kip kip 0.25 8 32 32 18.00Single LaneBR = + + = B) 25% of the Design Tandem

( )( )kip kip kip 0.25 25 25 12.50Single LaneBR = + =

C) 5% of the Design Truck with the Lane Load. ( ) ( ) ( )( )( )kipkip kip kip kip ft0.05 8 32 32 2 165' 0.640 14.16Single LaneBR = + + + = D) 5% of the Design Tandem with the Lane Load. ( ) ( ) ( )( )( )kipkip kip kip ft0.05 25 25 2 165' 0.640 13.06Single LaneBR = + + =

Case (A) Governs: ( )( )( )

( )( )( )

kip kip

#

18.00 3 0.85 45.90

Net Single LaneBR BR Lanes MPF== = This load has not been factored

-- 11 --


2.4: Centrifugal Force A centrifugal force results when a vehicle turns on a structure. Although a centrifugal force doesnt apply to this bridge since it is straight, the centrifugal load that would result from a hypothetical horizontal curve will be computed to illustrate the procedure. The centrifugal force is computed as the product of the axle loads and the factor, C.

2vC f

gR= (3.6.3-1)

where: v - Highway design speed ( )ftsec f - 4/3 for all load combinations except for Fatigue, in which case it is 1.0 g - The acceleration of gravity ( )2ftsec R - The radius of curvature for the traffic lane (ft). Suppose that we have a radius of R = 600 and a design speed of v = 65mph = 95.33ft/sec.

( )( )( )2

2ftsec

ftsec

95.334 0.62723 32.2 600 '

C = =

( )( )( )( )( )( )( )( )kip kip

#

72 0.6272 3 0.85 115.2

CE Axle Loads C Lanes MPF== =

This force has not been factored The centrifugal force acts horizontally in the direction pointing away from the center of curvature and at a height of 6 above the deck. Design the cross frames at the supports to carry this horizontal force into the bearings and design the bearings to resist the horizontal force and the resulting overturning moment.

-- 12 --


2.5: Wind Loads For the calculation of wind loads, assume that the bridge is located in the open country at an elevation of 40 above the ground.

Take Z = 40 Open Country oV = 8.20mph oZ = 0.23ft

Horizontal Wind Load on Structure: (WS) Design Pressure:

2

2 2

mph10,000DZ DZ

D B BB

V VP P PV

= = (3.8.1.2.1-1)

PB - Base Pressure - For beams, PB = 50psf when VB = 100mph. (Table 3.8.1.2.1-1)

VB - Base Wind Velocity, typically taken as 100mph. V30 - Wind Velocity at an elevation of Z = 30 (mph)

VDZ - Design Wind Velocity (mph)

Design Wind Velocity:

( )( )30

ftmph mph

ft

2.5 ln

100 402.5 8.20 Ln 105.8100 0.23

DZ oB o

V ZV VV Z

= = =

(3.8.1.1-1)

( ) ( )( )22mph

psf psf

mph

105.850 55.92

10,000DP = =

The height of exposure, hexp, for the finished bridge is computed as

71.5" 11.75" 42" 125.3" 10.44 'exph = + + = = The wind load per unit length of the bridge, W, is then computed as: ( )( )psf lbsft55.92 10.44 ' 583.7W = =

Total Wind Load: ( )( )( ) kiplbs, ft583.7 2 165' 192.6H TotalWS = = For End Abutments: ( )( )( ) kiplbs 1, ft 2583.7 165' 48.16H AbtWS = = For Center Pier: ( )( )( )( ) kiplbs 1, ft 2583.7 2 165' 96.31H PierWS = =

PD

hexp

-- 13 --


Vertical Wind Load on Structure: (WS) When no traffic is on the bridge, a vertical uplift (a line load) with a magnitude equal to 20psf times the overall width of the structure, w, acts at the windward quarter point of the deck. ( )( ) ( )( )psf psf lbsft20 20 42 ' 840VP w= = =

Total Uplift: ( )( )( ) kiplbsft840 2 165' 277.2= For End Abutments: ( )( )( ) kiplbs 1ft 2840 165' 69.30= For Center Pier: ( )( )( )( ) kiplbs 1ft 2840 2 165' 138.6=

Wind Load on Live Load: (WL) The wind acting on live load is applied as a line load of 100 lbs/ft acting at a distance of 6 above the deck, as is shown below. This is applied along with the horizontal wind load on the structure but in the absence of the vertical wind load on the structure.

WL

PD

-- 14 --


3. SECTION PROPERTIES AND CALCULATIONS: 3.1: Effective Flange Width, beff: For an interior beam, beff is the lesser of:

inft

132' 33' 396"4 4

15"12 (12)(8.5") 109.5"2 2

(12')(12 ) 144"

eff

fs

L

bt

S

= = = + = + = = =

For an exterior beam, beff is the lesser of:

( )inft

132' 33' 198.0"4 4

15"12 (12)(8.5") 109.5"2 2

12' 3' 12 108.0"2 2

eff

fs

e

L

bt

S d

= = = + = + = + = + =

Note that Leff was taken as 132.0 in the above calculations since for the case of effective width in continuous bridges, the span length is taken as the distance from the support to the point of dead load contra flexure. For computing the section properties shown on the two pages that follow, reinforcing steel in the deck was ignored for short-term and long-term composite calculations but was included for the cracked section. The properties for the cracked Section #1 are not used in this example, thus the amount of rebar included is moot. For the properties of cracked Section #2, As = 13.02 in2 located 4.5 from the top of the slab was taken from an underlying example problem first presented by Barth (2004).

-- 15 --


3.2: Section 1 Flexural Properties Bare Steel

t b A y Ay Ix d Ad2 IX

Top Flange 0.7500 15.00 11.25 70.38 791.72 0.53 -39.70 17,728 17,729Web 0.5625 69.00 38.81 35.50 1,377.84 15,398.86 -4.82 902 16,301Bot Flange 1.0000 21.00 21.00 0.50 10.50 1.75 30.18 19,125 19,127

71.06 2,180.06 ITotal = 53,157

Y = 30.68 SBS1,top = 1,327SBS1,bot = 1,733

Short-Term Composite (n = 8)


Slab 8.5000 109.50 116.34 75.00 8,725.78 700.49 -16.81 32,862 33,562Haunch 0.0000 15.00 0.00 70.75 0.00 0.00 -12.56 0 0Top Flange 0.7500 15.0000 11.25 70.38 791.72 0.53 -12.18 1,669 1,670Web 0.5625 69.0000 38.81 35.50 1,377.84 15,398.86 22.69 19,988 35,387Bot Flange 1.0000 21.0000 21.00 0.50 10.50 1.75 57.69 69,900 69,901

187.41 10,905.84 ITotal = 140,521n : 8.00

Y = 58.19 SST1,top = 11,191SST1,bot = 2,415

Long-Term Composite (n = 24)


Slab 8.5000 109.50 38.78 75.00 2,908.59 233.50 -28.67 31,885 32,119Haunch 0.0000 15.00 0.00 70.75 0.00 0.00 -24.42 0 0Top Flange 0.7500 15.0000 11.25 70.38 791.72 0.53 -24.05 6,506 6,507Web 0.5625 69.0000 38.81 35.50 1,377.84 15,398.86 10.83 4,549 19,948Bot Flange 1.0000 21.0000 21.00 0.50 10.50 1.75 45.83 44,101 44,103

109.84 5,088.66 ITotal = 102,676n : 24.00

Y = 46.33 SLT1,top = 4,204SLT1,bot = 2,216

Cracked Section


Rebar 4.5000 13.02 75.25 979.76 -75.25 73,727 73,727Top Flange 0.7500 15.0000 11.25 70.38 791.72 0.53 -70.38 55,717 55,718Web 0.5625 69.0000 38.81 35.50 1,377.84 15,398.86 -35.50 48,913 64,312Bot Flange 1.0000 21.0000 21.00 0.50 10.50 1.75 -0.50 5 7

84.08 3,159.82 ITotal = 193,764

Y = 37.58 SCR1,top = 5,842SCR1,bot = 5,156

These section properties do NOT include the haunch or sacrificial wearing surface.

-- 16 --


3.3: Section 2 Flexural Properties Bare Steel


Top Flange 1.0000 21.00 21.00 72.00 1,512.00 1.75 -45.17 42,841 42,843Web 0.5625 69.00 38.81 37.00 1,436.06 15,398.86 -10.17 4,012 19,411Bot Flange 2.5000 21.00 52.50 1.25 65.63 27.34 25.58 34,361 34,388

112.31 3,013.69 ITotal = 96,642

Y = 26.83 SBS2,top = 2,116SBS2,bot = 3,602

Short Term Composite (n = 8)


Slab 8.5000 109.50 116.34 76.75 8,929.38 700.49 -24.52 69,941 70,641Haunch 0.0000 21.00 0.00 72.50 0.00 0.00 -20.27 0 0Top Flange 1.0000 21.0000 21.00 72.00 1,512.00 1.75 -19.77 8,207 8,208Web 0.5625 69.0000 38.81 37.00 1,436.06 15,398.86 15.23 9,005 24,403Bot Flange 2.5000 21.0000 52.50 1.25 65.63 27.34 50.98 136,454 136,481

228.66 11,943.07 ITotal = 239,734n : 8.00

Y = 52.23 SST2,top = 11,828SST2,bot = 4,590

Long-Term Composite (n = 24)


Slab 8.5000 109.50 38.78 76.75 2,976.46 233.50 -37.10 53,393 53,626Haunch 0.0000 15.00 0.00 72.50 0.00 0.00 -32.85 0 0Top Flange 1.0000 21.0000 21.00 72.00 1,512.00 1.75 -32.35 21,983 21,985Web 0.5625 69.0000 38.81 37.00 1,436.06 15,398.86 2.65 272 15,670Bot Flange 2.5000 21.0000 52.50 1.25 65.63 27.34 38.40 77,395 77,423

151.09 5,990.15 ITotal = 168,704n : 24.00

Y = 39.65 SLT2,top = 5,135SLT2,bot = 4,255

Cracked Section


Rebar 4.5000 13.02 77.00 1,002.54 -44.96 26,313 26,313Top Flange 1.0000 21.0000 21.00 72.00 1,512.00 1.75 -39.96 33,525 33,527Web 0.5625 69.0000 38.81 37.00 1,436.06 15,398.86 -4.96 953 16,352Bot Flange 2.5000 21.0000 52.50 1.25 65.63 27.34 30.79 49,786 49,813

125.33 4,016.23 ITotal = 126,006

Y = 32.04 SCR2,top = 3,115SCR2,bot = 3,932

These section properties do NOT include the haunch or sacrificial wearing surface.

-- 17 --


4. DISTRIBUTION FACTOR FOR MOMENT 4.1: Positive Moment Region (Section 1): Interior Girder

One Lane Loaded:

0.10.4 0.3

1, 3

2

4 2 2

4

0.4 0.3 4

1, 3

0.0614 12

( )

8(53,157 in (71.06 in )(46.82") )

1,672, 000 in

12 ' 12 ' 1, 672, 000 in0.06

14 165 ' (12)(165 ')(8.5")

gM Int

s

g g

g

g

M Int

KS SDF

L Lt

K n I Ae

K

K

DF

+

+

= +

= += +=

= +

0.1

1, 0.5021M IntDF + =

In these calculations, the terms eg and Kg include the haunch and sacrificial wearing surface since doing so increases the resulting factor. Note that ts in the denominator of the final term excludes the sacrificial wearing surface since excluding it increases the resulting factor.

Two or More Lanes Loaded:

0.10.6 0.2

2, 3

0.10.6 0.2 4

2, 3

2,

0.0759.5 12

12 ' 12 ' 1,672, 000 in0.075

9.5 165 ' 12(165 ')(8.5")

0.7781

gM Int

s

M Int

M Int

KS SDF

L Lt

DF

DF

+

+

+

= +

= +

=

Exterior Girder

One Lane Loaded:

The lever rule is applied by assuming that a hinge forms over the first interior girder as a truck load is applied near the parapet. The resulting reaction in the exterior girder is the distribution factor.

1,

8.50.7083

12M ExtDF + = = Multiple Presence: DFM1,Ext+ = (1.2) (0.7083) = 0.8500

-- 18 --



DFM2,Ext+ = e DFM2,Int+

0.779.11.5

0.77 0.93489.1

ede = +

= + =

DFM2,Ext+ = (0.9348) (0.7781) = 0.7274

4.2: Negative Moment Region (Section 2): The span length used for negative moment near the pier is the average of the lengths of the adjacent spans. In this case, it is the average of 165.0 and 165.0 = 165.0. Interior Girder

One Lane Loaded:

0.10.4 0.3

1, 3

2

4 2 2

4

0.4 0.3 4

1, 3

0.0614 12

( )

8(96, 642 in (112.3 in )(52.17") )

3, 218, 000 in

12 ' 12 ' 3, 218,000 in0.06

14 165 ' (12)(165 ')(8.5")

gM Int

s

g g

g

g

M Int

KS SDF

L Lt

K n I Ae

K

K

DF

= +

= += +=

= +

0.1

1, 0.5321M IntDF =


0.10.6 0.2

2, 3

0.10.6 0.2 4

2, 3

2,

0.0759.5 12

12 ' 12 ' 3, 218, 000 in0.075

9.5 165 ' (12)(165 ')(8.5")

0.8257

gM Int

s

M Int

M Int

KS SDF

L Lt

DF

DF

= +

= +

=

-- 19 --


Exterior Girder One Lane Loaded:

Same as for the positive moment section: DFM1,Ext- = 0.8500


DFM2,Ext- = e DFM2,Int-

0.779.11.5

0.77 0.93489.1

dee = +

= + =

DFM2,Ext- = (0.9348) (0.8257) = 0.7719

4.3: Minimum Exterior Girder Distribution Factor:

,

2

L

Ext Minb

N

ExtL

Nb

X eN

DFN x

= +

One Lane Loaded:

1, , 2 2

1 (18.0 ')(14.5 ')0.6125

4 (2) (18 ') (6 ')M Ext MinDF = + =+

Multiple Presence: DFM1,Ext,Min = (1.2) (0.6125) = 0.7350

-- 20 --


Two Lanes Loaded:

12'3'

2' 3'

14.5'

6'

P1 P2

Lane 1 (12')

3' 2' 3' 3'

2.5'

Lane 2 (12')

2 , , 2 2

2 (18.0 ')(14.5 ' 2.5 ')0.9250

4 (2) (18 ') (6 ')M Ext MinDF

+= + =+

Multiple Presence: DFM2,Ext,Min = (1.0) (0.9250) = 0.9250

Three Lanes Loaded:

The case of three lanes loaded is not considered for the minimum exterior distribution factor since the third truck will be placed to the right of the center of gravity of the girders, which will stabilize the rigid body rotation effect resulting in a lower factor.

4.4: Moment Distribution Factor Summary Strength and Service Moment Distribution: Positive Moment Negative Moment Interior Exterior Interior Exterior

1 Lane Loaded: 0.5021 0.8500 0.7350 0.5321 0.8500 0.7350 2 Lanes Loaded: 0.7781 0.7274 0.9250 0.8257 0.7719 0.9250

For Simplicity, take the Moment Distribution Factor as 0.9250 everywhere for the Strength and Service load combinations. Fatigue Moment Distribution: For Fatigue, the distribution factor is based on the one-lane-loaded situations with a multiple presence factor of 1.00. Since the multiple presence factor for 1-lane loaded is 1.2, these factors can be obtained by divided the first row of the table above by 1.2. Positive Moment Negative Moment Interior Exterior Interior Exterior

1 Lane Loaded: 0.4184 0.7083 0.6125 0.4434 0.7083 0.6125 For Simplicity, take the Moment Distribution Factor as 0.7083 everywhere for the Fatigue load combination Multiplying the live load moments by this distribution factor of 0.9250 yields the table of nominal girder moments shown on the following page.

-- 21 --


Nominal Girder Moments for Design

Station (LL+IM)+ (LL+IM)- Fat+ Fat- DC1 DC2 DW(ft) (k-ft) (k-ft) (k-ft) (k-ft) (k-ft) (k-ft) (k-ft)

0.0 0.0 0.0 0.2 0.0 0.0 0.0 0.014.7 1605.1 -280.7 645.6 -68.9 1309.9 240.0 440.329.3 2791.4 -561.3 1127.9 -137.9 2244.5 412.0 755.644.0 3572.6 -842.0 1449.4 -206.8 2799.9 515.0 944.758.7 3999.4 -1122.7 1626.1 -275.8 2978.6 549.7 1008.373.3 4066.7 -1403.4 1647.9 -344.7 2779.3 515.8 946.188.0 3842.5 -1684.0 1599.4 -413.7 2202.1 413.2 757.9

102.7 3310.8 -1964.7 1439.3 -482.6 1248.4 242.3 444.4117.3 2509.4 -2245.4 1148.6 -551.6 -84.8 2.5 4.6132.0 1508.6 -2547.5 763.6 -620.5 -1793.1 -305.4 -560.2135.7 1274.6 -2660.0 651.3 -637.8 -2280.8 -393.2 -721.2139.3 1048.4 -2793.3 539.1 -655.0 -2794.0 -485.2 -890.0143.0 828.6 -2945.6 425.3 -672.2 -3333.2 -581.5 -1066.7146.7 615.8 -3115.6 310.8 -689.5 -3898.1 -682.1 -1251.3150.3 463.3 -3371.3 221.9 -706.7 -4488.6 -787.0 -1443.7154.0 320.5 -3728.6 158.6 -724.0 -5105.1 -896.2 -1643.9157.7 185.5 -4105.0 98.8 -741.2 -5747.2 -1009.7 -1852.1161.3 76.4 -4496.9 49.4 -758.4 -6415.3 -1127.5 -2068.1165.0 0.0 -4918.1 0.1 -775.6 -7108.8 -1249.5 -2291.9168.7 76.4 -4496.9 49.4 -758.4 -6415.3 -1127.5 -2068.1172.3 185.5 -4105.0 98.8 -741.2 -5747.2 -1009.7 -1852.1176.0 320.5 -3728.6 158.6 -724.0 -5105.1 -896.2 -1643.9179.7 463.3 -3371.3 221.9 -706.7 -4488.6 -787.0 -1443.7183.3 615.8 -3115.6 310.8 -689.5 -3898.1 -682.1 -1251.3187.0 828.6 -2945.6 425.3 -672.2 -3333.2 -581.5 -1066.7190.7 1048.4 -2793.3 539.1 -655.0 -2794.0 -485.2 -890.0194.3 1274.6 -2660.0 651.3 -637.8 -2280.8 -393.2 -721.2198.0 1508.6 -2547.5 763.2 -620.6 -1793.1 -305.4 -560.2212.7 2509.4 -2245.4 1148.6 -551.6 -84.8 2.5 4.6227.3 3310.8 -1964.7 1439.3 -482.6 1248.4 242.3 444.4242.0 3842.5 -1684.0 1599.4 -413.7 2202.1 413.2 757.9256.7 4066.7 -1403.4 1647.9 -344.7 2779.3 515.8 946.1271.3 3999.4 -1122.7 1626.1 -275.8 2978.6 549.7 1008.3286.0 3572.6 -842.0 1449.4 -206.8 2799.9 515.0 944.7300.7 2791.4 -561.3 1127.9 -137.9 2244.5 412.0 755.6315.3 1605.1 -280.7 645.6 -68.9 1309.9 240.0 440.3330.0 0.0 0.0 0.2 0.0 0.0 0.0 0.0

Nominal Moments

-- 22 --


5. DISTRIBUTION FACTOR FOR SHEAR The distribution factors for shear are independent of the section properties and span length. Thus, the only one set of calculations are need - they apply to both the section 1 and section 2 5.1: Interior Girder

One Lane Loaded:

1 0.36 25.012 '0.36 0.840025.0

V ,IntSDF = +

= + =


2

2

2

0.212 35

12 ' 12 '0.2 1.08212 35

V ,IntS SDF = +

= + =

5.2: Exterior Girder

One Lane Loaded: Lever Rule, which is the same as for moment: DFV1,Ext = 0.8500 Two or More Lanes Loaded:

DFV2,Ext = e DFV2,Int

0.60101.5 '0.60 0.750010

ede = +

= + =

DFV2,Ext = (0.7500) (1.082) = 0.8115

5.3: Minimum Exterior Girder Distribution Factor - The minimum exterior girder distribution factor applies to shear as well as moment. DFV1,Ext,Min = 0.7350 DFV2,Ext,Min = 0.9250

-- 23 --


5.4: Shear Distribution Factor Summary Strength and Service Shear Distribution:

Shear Distribution Interior Exterior

1 Lane Loaded: 0.8400 0.8500 0.7350 2 Lanes Loaded: 1.082 0.6300 0.9250

For Simplicity, take the Shear Distribution Factor as 1.082 everywhere for Strength and Service load combinations. Fatigue Shear Distribution: For Fatigue, the distribution factor is based on the one-lane-loaded situations with a multiple presence factor of 1.00. Since the multiple presence factor for 1-lane loaded is 1.2, these factors can be obtained by divided the first row of the table above by 1.2.

Shear Distribution Interior Exterior

1 Lane Loaded: 0.7000 0.7083 0.6125 For Simplicity, take the Shear Distribution Factor as 0.7083 everywhere for the Fatigue load combination. Multiplying the live load shears by these distribution factors yields the table of nominal girder shears shown on the following page.

-- 24 --


Nominal Girder Shears for Design

Station (LL+IM)+ (LL+IM)- Fat+ Fat- DC1 DC2 DW(ft) (kip) (kip) (kip) (kip) (kip) (kip) (kip)

0.0 144.9 -19.7 50.8 -4.7 115.0 20.6 37.614.7 123.5 -20.3 44.6 -4.7 88.8 15.9 29.029.3 103.5 -26.8 38.5 -6.4 62.5 11.2 20.544.0 85.0 -41.4 32.6 -11.1 36.3 6.5 11.958.7 68.1 -56.7 26.9 -17.2 10.1 1.8 3.373.3 52.8 -72.7 21.4 -23.2 -16.1 -2.9 -5.388.0 39.4 -89.1 16.3 -29.0 -42.3 -7.6 -13.9

102.7 27.8 -105.7 11.5 -34.6 -68.6 -12.3 -22.4117.3 18.0 -122.3 7.3 -39.9 -94.8 -17.0 -31.0132.0 10.0 -138.6 3.9 -44.9 -121.0 -21.7 -39.6135.7 8.3 -142.5 3.4 -46.0 -127.6 -22.8 -41.7139.3 6.7 -146.5 2.8 -47.2 -134.1 -24.0 -43.9143.0 5.5 -150.5 2.3 -48.3 -140.7 -25.2 -46.0146.7 4.3 -154.5 1.8 -49.4 -147.2 -26.4 -48.2150.3 3.2 -158.4 1.4 -50.4 -153.8 -27.5 -50.3154.0 2.2 -162.3 1.0 -51.5 -160.3 -28.7 -52.5157.7 1.3 -166.2 0.6 -52.4 -166.9 -29.9 -54.6161.3 0.0 -170.1 0.3 -53.4 -173.4 -31.0 -56.8165.0 0.0 -173.9 54.3 -54.3 -180.0 -32.2 -58.9168.7 170.1 -0.5 53.4 -0.3 173.4 31.0 56.8172.3 166.2 -1.3 52.4 -0.6 166.9 29.9 54.6176.0 162.3 -2.2 51.5 -1.0 160.3 28.7 52.5179.7 158.4 -3.2 50.4 -1.4 153.8 27.5 50.3183.3 154.5 -4.3 49.4 -1.8 147.2 26.4 48.2187.0 150.5 -5.5 48.3 -2.3 140.7 25.2 46.0190.7 146.5 -6.7 47.2 -2.8 134.1 24.0 43.9194.3 142.5 -8.3 46.0 -3.4 127.6 22.8 41.7198.0 138.6 -10.0 44.9 -3.9 121.0 21.7 39.6212.7 122.3 -18.0 39.9 -7.3 94.8 17.0 31.0227.3 105.7 -27.8 34.6 -11.5 68.6 12.3 22.4242.0 89.1 -39.4 29.0 -16.3 42.3 7.6 13.9256.7 72.7 -52.8 23.2 -21.4 16.1 2.9 5.3271.3 56.7 -68.1 17.2 -26.9 -10.1 -1.8 -3.3286.0 41.4 -85.0 11.1 -32.6 -36.3 -6.5 -11.9300.7 26.8 -103.5 6.4 -38.5 -62.5 -11.2 -20.5315.3 20.3 -123.5 4.7 -44.6 -88.8 -15.9 -29.0330.0 19.7 -144.9 4.7 -50.8 -115.0 -20.6 -37.6

Nominal Shears

-- 25 --


6. FACTORED SHEAR AND MOMENT ENVELOPES

The following load combinations were considered in this example: Strength I: 1.75(LL + IM) + 1.25DC1 + 1.25DC2 + 1.50DW Strength IV: 1.50DC1 + 1.50DC2 + 1.50DW Service II: 1.3(LL + IM) + 1.0DC1 + 1.0DC2 + 1.0DW

Fatigue: 0.75(LL + IM) (IM = 15% for Fatigue; IM = 33% otherwise) Strength II is not considered since this deals with special permit loads. Strength III and V are not considered as they include wind effects, which will be handled separately as needed. Strength IV is considered but is not expected to govern since it addresses situations with high dead load that come into play for longer spans. Extreme Event load combinations are not included as they are also beyond the scope of this example. Service I again applies to wind loads and is not considered (except for deflection) and Service III and Service IV correspond to tension in prestressed concrete elements and are therefore not included in this example. In addition to the factors shown above, a load modifier, , was applied as is shown below.

i i iQ Q = is taken as the product of D, R, and I, and is taken as not less than 0.95. For this example, D and I are taken as 1.00 while R is taken as 1.05 since the bridge has 4 girders with a spacing greater than or equal to 12. Using these load combinations, the shear and moment envelopes shown on the following pages were developed. Note that for the calculation of the Fatigue moments and shears that is taken as 1.00 and the distribution factor is based on the one-lane-loaded situations with a multiple presence factor of 1.00 (AASHTO Sections 6.6.1.2.2, Page 6-29 and 3.6.1.4.3b, Page 3-25).

-- 26 --


Strength Limit Moment Envelopes

-25,000

-20,000

-15,000

-10,000

-5,000

0

5,000

10,000

15,000

20,000

0 30 60 90 120 150 180 210 240 270 300 330

Station (ft)

Mom

ent (

kip-

ft)

Strength I

Strength IV

Strength IV

Strength I

Strength Limit Shear Force Envelope

-800

-600

-400

-200

0

200

400

600

800

0 30 60 90 120 150 180 210 240 270 300 330

Station (ft)

Shea

r (k

ip)

Strength IV

Strength I

-- 27 --


Service II Moment Envelope

-20,000

-17,500

-15,000

-12,500

-10,000

-7,500

-5,000

-2,500

0

2,500

5,000

7,500

10,000

12,500

0 30 60 90 120 150 180 210 240 270 300 330

Station (ft)

Mom

ent (

kip-

ft)

Service II Shear Envelope

-600

-400

-200

0

200

400

600

0 30 60 90 120 150 180 210 240 270 300 330

Station (ft)

Shea

r (k

ip)

-- 28 --


Factored Fatigue Moment Envelope

-1,500

-1,000

-500

0

500

1,000

1,500

0 30 60 90 120 150 180 210 240 270 300 330

Station (ft)

Mom

ent (

kip-

ft)

Factored Fatigue Shear Envelope

-50

-40

-30

-20

-10

0

10

20

30

40

50

0 30 60 90 120 150 180 210 240 270 300 330

Station (ft)

Shea

r (k

ip)

-- 29 --


Factored Girder Moments for Design

Station Total + Total - Total + Total - Total + Total - Total + Total -(ft) (k-ft) (k-ft) (k-ft) (k-ft) (k-ft) (k-ft) (k-ft) (k-ft)

0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.2 0.014.7 5677.1 -515.7 3134.6 0.0 4280.7 -383.1 484.2 -51.729.3 9806.0 -1031.5 5374.1 0.0 7393.0 -766.2 845.9 -103.444.0 12403.3 -1547.2 6708.8 0.0 9349.1 -1149.4 1087.1 -155.158.7 13567.8 -2062.9 7145.1 0.0 10222.6 -1532.5 1219.6 -206.873.3 13287.4 -2578.7 6679.8 0.0 10004.2 -1915.6 1235.9 -258.688.0 11687.1 -3094.4 5312.9 0.0 8787.0 -2298.7 1199.5 -310.3

102.7 8740.0 -3610.2 3047.7 0.0 6551.1 -2681.8 1079.5 -362.0117.3 4621.6 -4237.1 11.2 -133.5 3432.8 -3153.9 861.5 -413.7132.0 2772.1 -8317.5 0.0 -4187.3 2059.3 -6268.9 572.7 -465.4135.7 2342.0 -9533.2 0.0 -5347.3 1739.8 -7195.8 488.5 -478.3139.3 1926.4 -10838.2 0.0 -6566.4 1431.1 -8190.4 404.3 -491.3143.0 1522.6 -12230.6 0.0 -7845.7 1131.1 -9251.2 318.9 -504.2146.7 1131.6 -13707.1 0.0 -9184.5 840.6 -10375.8 233.1 -517.1150.3 851.2 -15392.8 0.0 -10582.9 632.3 -11657.1 166.5 -530.0154.0 588.9 -17317.3 0.0 -12041.3 437.4 -13117.1 119.0 -543.0157.7 340.9 -19328.3 0.0 -13559.1 253.3 -14642.7 74.1 -555.9161.3 140.4 -21420.1 0.0 -15137.1 104.3 -16229.6 37.1 -568.8165.0 0.0 -23617.1 0.0 -16774.1 0.0 -17895.9 0.1 -581.7168.7 140.4 -21420.1 0.0 -15137.1 104.3 -16229.6 37.1 -568.8172.3 340.9 -19328.3 0.0 -13559.1 253.3 -14642.7 74.1 -555.9176.0 588.9 -17317.3 0.0 -12041.3 437.4 -13117.1 119.0 -543.0179.7 851.2 -15392.8 0.0 -10582.9 632.3 -11657.1 166.5 -530.0183.3 1131.6 -13707.1 0.0 -9184.5 840.6 -10375.8 233.1 -517.1187.0 1522.6 -12230.6 0.0 -7845.7 1131.1 -9251.2 318.9 -504.2190.7 1926.4 -10838.2 0.0 -6566.4 1431.1 -8190.4 404.3 -491.3194.3 2342.0 -9533.2 0.0 -5347.3 1739.8 -7195.8 488.5 -478.3198.0 2772.1 -8317.5 0.0 -4187.3 2059.3 -6268.9 572.4 -465.4212.7 4621.6 -4237.1 11.2 -133.5 3432.8 -3153.9 861.5 -413.7227.3 8740.0 -3610.2 3047.7 0.0 6551.1 -2681.8 1079.5 -362.0242.0 11687.1 -3094.4 5312.9 0.0 8787.0 -2298.7 1199.5 -310.3256.7 13287.4 -2578.7 6679.8 0.0 10004.2 -1915.6 1235.9 -258.6271.3 13567.8 -2062.9 7145.1 0.0 10222.6 -1532.5 1219.6 -206.8286.0 12403.3 -1547.2 6708.8 0.0 9349.1 -1149.4 1087.1 -155.1300.7 9806.0 -1031.5 5374.1 0.0 7393.0 -766.2 845.9 -103.4315.3 5677.1 -515.7 3134.6 0.0 4280.7 -383.1 484.2 -51.7330.0 0.0 0.0 0.0 0.0 0.0 0.0 0.2 0.0

FatigueStrength I Strength IV Service II

-- 30 --


Factored Girder Shears for Design

Station Total + Total - Total + Total - Total + Total - Total + Total -(ft) (kip) (kip) (kip) (kip) (kip) (kip) (kip) (kip)

0.0 479.5 -34.5 272.8 0.0 379.7 -26.9 38.1 -3.514.7 390.5 -35.5 210.6 0.0 309.0 -27.7 33.5 -3.529.3 304.0 -46.9 148.4 0.0 240.2 -36.6 28.9 -4.844.0 220.1 -72.4 86.2 0.0 173.4 -56.5 24.5 -8.358.7 138.9 -99.3 24.0 0.0 108.9 -77.5 20.2 -12.973.3 92.5 -158.9 0.0 -38.2 72.1 -124.8 16.1 -17.488.0 68.9 -239.1 0.0 -100.4 53.8 -188.6 12.2 -21.8

102.7 48.6 -319.7 0.0 -162.6 37.9 -252.7 8.6 -26.0117.3 31.5 -400.1 0.0 -224.8 24.6 -316.8 5.5 -29.9132.0 17.5 -480.2 0.0 -287.0 13.7 -380.5 3.0 -33.7135.7 14.5 -500.0 0.0 -302.6 11.3 -396.3 2.5 -34.5139.3 11.7 -519.8 0.0 -318.1 9.2 -412.1 2.1 -35.4143.0 9.6 -539.7 0.0 -333.7 7.5 -427.9 1.7 -36.2146.7 7.6 -559.6 0.0 -349.2 5.9 -443.7 1.4 -37.0150.3 5.7 -579.3 0.0 -364.8 4.4 -459.4 1.0 -37.8154.0 3.9 -599.0 0.0 -380.3 3.0 -475.1 0.8 -38.6157.7 2.2 -618.7 0.0 -395.9 1.7 -490.8 0.5 -39.3161.3 0.0 -638.3 0.0 -411.4 0.0 -506.4 0.2 -40.0165.0 0.0 -657.9 0.0 -427.0 0.0 -522.0 40.7 -40.7168.7 638.3 -0.9 411.4 0.0 506.4 -0.7 40.0 -0.2172.3 618.7 -2.2 395.9 0.0 490.8 -1.7 39.3 -0.5176.0 599.0 -3.9 380.3 0.0 475.1 -3.0 38.6 -0.8179.7 579.3 -5.7 364.8 0.0 459.4 -4.4 37.8 -1.0183.3 559.6 -7.6 349.2 0.0 443.7 -5.9 37.0 -1.4187.0 539.7 -9.6 333.7 0.0 427.9 -7.5 36.2 -1.7190.7 519.8 -11.7 318.1 0.0 412.1 -9.2 35.4 -2.1194.3 500.0 -14.5 302.6 0.0 396.3 -11.3 34.5 -2.5198.0 480.2 -17.5 287.0 0.0 380.5 -13.7 33.7 -2.9212.7 400.1 -31.5 224.8 0.0 316.8 -24.6 29.9 -5.5227.3 319.7 -48.6 162.6 0.0 252.7 -37.9 26.0 -8.6242.0 239.1 -68.9 100.4 0.0 188.6 -53.8 21.8 -12.2256.7 158.9 -92.5 38.2 0.0 124.8 -72.1 17.4 -16.1271.3 99.3 -138.9 0.0 -24.0 77.5 -108.9 12.9 -20.2286.0 72.4 -220.1 0.0 -86.2 56.5 -173.4 8.3 -24.5300.7 46.9 -304.0 0.0 -148.4 36.6 -240.2 4.8 -28.9315.3 35.5 -390.5 0.0 -210.6 27.7 -309.0 3.5 -33.5330.0 34.5 -479.5 0.0 -272.8 26.9 -379.7 3.5 -38.1

Strength I Strength IV Service II Fatigue

-- 31 --


7. FATIGUE CHECKS 7.1: Check transverse stiffener to flange weld at Station 73.3: Traffic information: ADTT given as 2400. Three lanes are available to trucks. (ADTT)SL = (0.80) (2,400) = 1,920 N = (ADTT)SL (365) (75) n = (1,920) (365) (75) (1) = 52.56M Cycles Check Top Flange Weld: Fatigue need only be checked when the compressive stress due to unfactored permanent loads is less than twice the maximum tensile stress due to factored fatigue loads.

Check ?

, 2comp DL Fatf f Distance from bottom of section to the detail under investigation

y = tf,bottom + D = 1.00 + 69.00 = 70

k-ft1 2,779DCM = ( )( )( )k-ft inft ksi1 42,779 12 70" 30.68" 24.6753,157 inDCf= =

k-ft2 515.8DCM = ( )( )( )k-ft inft ksi2 4515.8 12 70" 46.33" 1.427102,676 inDCf

= = ksi ksi ksi, 24.67 1.427 26.09comp DLf = + =

k-ft, 258.6Fat NegM = ( )( )( )k-ft inft ksi4258.6 12 70" 58.19" 0.261140,521 inFatf

= =

Check ?

, 2comp DL Fatf f ( )( )?ksi ksi ksi26.09 2 0.261 0.521 = , No. Fatigue need not be checked on the top flange at Station 73.3.

(Pg 24) (Pg 16)

(Pg 16)

(Pg 24) (Pg 16)

(Pg 16)

(Pg 30) (Pg 16)

(Pg 16)

-- 32 --


Check Bottom Flange Weld: The permanent loads at Station 73.3 cause tension in the bottom flange, thus by inspection fatigue needs to be checked. ( ) ( )nf F

( ) ( )13

2TH

n

FAFN

= is a load factor of 0.75, which is already included in the fatigue moments.

( ) ( )( )( )k-ft inft ksi41, 236 12 58.19" 1.00" 6.036140,521 inf = =

The detail under consideration is a Category C detail. A = 44.0 x 108 ksi3 and (F)TH = 12.0 ksi

( ) ksi ksi12.0 6.002 2

THF = = The stress in the detail is almost less than the

infinite life threshold

118 3 33 ksi

6

44 10 ksi 4.37552.56 10

AN

= =

Since ( )13 ksi ksi4.375 is less than 6.00

2TH

FAN

= = , the infinite life governs. ( ) ksi6.00nF =

Since ( ) ( )ksi ksi6.036 6.00nf F = > = , the detail is not satisfactory.

-- 33 --


Calculate the design life of the part under consideration:

Since ( ) ( ) is greater than 2

THF

f , solve for N in the following equation.

( )13Af

N ( ) ( )

8 36

3 3ksi

44 10 ksi 20.01 10 cycles6.036

ANf

= =

620.01 10 cycles 10,420 days

1,920 =

( )10, 420 days 28.55 years = 28y, 6m, 19d, 2h, 38min...365

=

-- 34 --


8. CHECK CROSS_SECTION PROPORTION LIMITS

Web Proportions

916

69"150 122.7 150"w

Dt

= O.K.

Flange Proportions

34

15"12 10.00 122 (2)( ")

f

f

bt

= O.K.

21"12 10.50 122 (2)(1")

f

f

bt

= O.K.

1

2

21"12 4.200 122 (2)(2 ")

f

f

bt

= O.K. Check ODOT Criteria for Flange Width

? 69"2.5 12" 2.5 14"

6 6fDb + + = O.K.

,min69"= =11.50"

6 6fDb = O.K.

9 5

,min 16 81.1 =(1.1)( ") "wft t = O.K.

33

43

( ")(15")0.1 10 0.1 0.2733 10(1")(21")

yc

yt

II

= O.K.

3

3(2.5")(21")0.1 10 0.1 2.500 10(1")(21")

yc

yt

II

= O.K.

-- 35 --


9. CHECK SERVICE LIMIT STATE 9.1: Check Absolute Deflection of the Bridge: Section 6.10.4.1

Section 1

The cross section of Section 1 that is used for computing deflections is shown above. The entire deck width is used (as opposed to just the effective width that was used earlier) and the haunch and sacrificial wearing surface have been neglected. AASHTO permits the use of the stiffness of parapets and structurally continuous railing but ODOT does not.

The transformed width of the bridge deck is ( )( )inft42 ' 12' 63.00"

8w = =

Using the bottom of the steel as a datum, the location of the CG of the deck can be found as:

3 48.5"1" 69" " 75.00"

2cy = + + + =

The CG of this composite cross section is found as:

( )( ) ( ) ( )( ) ( )

( )( ) ( )( )2

2

63" 8.5" 75.00" 4 71.06 in 30.68"59.63"

63" 8.5" 4 71.06 inY

+= =+i i

Now the moment of inertia of the section can be found as:

( )( ) ( )( )[ ]( )( ) ( )( )[ ]

32 4

24 2 4

41,

63" 8.5"Concrete 63" 8.5" 75.00" 59.63" 129,700 in

12Steel 4 53,160 in 4 71.06 in 30.68" 59.63" 450,900 in

580,600 intotalI

+ = + =

=

44

in1 Girder

580,600 in 145,1004 Girders

I = =

-- 36 --


Section 2

The cross section of Section 2 that is used for computing deflections is shown above.

The transformed width of the bridge deck is ( )( )inft42 ' 12' 63.00"

8w = =

Using the bottom of the steel as a datum, the location of the CG of the deck can be found as:

128.5"2 " 69" 1" 76.75"

2cy = + + + =

The CG of this composite cross section is found as:

( )( ) ( ) ( )( ) ( )

( )( ) ( )( )2

2

63" 8.5" 76.75" 4 112.3 in 26.83"53.98"

63" 8.5" 4 112.3 inY

+= =+i i

Now the moment of inertia of the section can be found as:

( )( ) ( )( )[ ]( )( ) ( )( )[ ]

32 4

24 2 4

4

63" 8.5"Concrete 63" 8.5" 76.75" 53.98" 280,900 in

12Steel 4 96,640 in 4 112.3 in 26.83" 53.98" 717,700 in

998,600 intotalI

+ = + =

=

44

in2 Girder

998,600 in 249,7004 Girders

I = =

-- 37 --


The following model, which represents the stiffness of a single girder, was used to compute absolute live-load deflections assuming the entire width of the deck to be effective in both compression and tension. The live load component of the Service I load combination is applied. Based on AASHTO Section 3.6.1.3.2, the loading includes (1) the design truck alone and (2) the lane load with 25% of the design truck. The design truck and design lane load were applied separately in the model and will be combined below. The design truck included 33% impact.

I = 145,100 in4 I = 145,100 in4I = 249,700 in4

From the analysis: Deflection due to the Design Truck with Impact: Truck = 2.442 Deflection due to the Design Lane Load: Lane = 0.8442 These deflections are taken at Stations 79.2 and 250.8. The model was broken into segments roughly 25 long in the positive moment region and 7 long in the negative moment region. A higher level of discretization may result in slightly different deflections but it is felt that this level of accuracy was acceptable for deflection calculations. Since the above results are from a single-girder model subjected to one lanes worth of loading, distribution factors must be applied to obtain actual bridge deflections. Since it is the absolute deflection that is being investigated, all lanes are loaded (multiple presence factor apply) and it is assumed that all girders deflect equally. Given these assumptions, the distribution factor for deflection is simply the number of lanes times the multiple presence factor divided by the number of girders. Looking at the two loading criteria described above:

( )( )( ) ( )

( )( )( ) ( ) ( )( )

1

2

0.85 32.442" 1.558" Governs

4

0.85 30.8442" 0.25 2.442" 0.9274"

4

= = = + =

The limiting deflection for this bridge is:

( )( )inft165' 12 2.475" OK

800 800LimitL = = =

-- 38 --


9.2: Check the Maximum Span-to-Depth Ratio: Section 6.10.4.1 From Table 2.5.2.6.3-1, (1) the overall depth of a composite I-beam in a continuous span must not be less than 0.032L and (2) the depth of the steel in a composite I-beam in a continuous span must not be less than 0.027L.

( ) ( )( )( )( ) ( )( )( )

inft

inft

1 0.032 0.032 165' 12 63.36" OK

2 0.027 0.027 165' 12 53.46" OK

L

L

= = = =

-- 39 --


9.3: Permanent Deformations - Section 1 At the Service Limit State, the following shall be satisfied for composite sections Top Flange: 0.95f h yff R F

Bottom Flange 0.952

lf h yf

ff R F+ Per 6.10.1.1.1a, elastic stresses at any location in a composite section shall consist of the sum of stresses caused by loads applied separately to the bare steel, short-term composite section, and long-term composite section.

1 21.00 1.00 1.00 1.30DC DC DW LL IMcBS LT LT ST

M M M Mf

S S S S+= + + +

Top Flange, Positive Moment It is not immediately evident to me whether the factored stress at 58.7 or 73.3 will govern.

k-ft k-ft k-ft k-ft

,58.7 3 3 3 3

in in in inft ft ft ft(2, 979 )(12 ) (549.7 )(12 ) (1, 008 )(12 ) (3, 999 )(12 )1.00 1.00 1.00 1.30

1,327 in 4,204 in 4,204 in 11,191 inf

f = + ++

ksi,58.7 36.96ff =

k-ft k-ft k-ft k-ft

,73.3 3 3 3 3

in in in inft ft ft ft(2, 779 )(12 ) (515.8 )(12 ) (946.1 )(12 ) (4, 067 )(12 )1.00 1.00 1.00 1.30

1,327 in 4,204 in 4,204 in 11,191 inf

f = + ++

ksi,73.3 34.97ff =

The stress at 58.7 governs. ff = 36.96ksi.

ksi ksi ksi 0.95 36.96 (0.95)(1.00)(50 ) 47.50f h yff R F = O.K.

Note: The bending moments in the above calculations come from page 22 while the moments of inertia are found on page 16.

-- 40 --


Bottom Flange, Positive Moment

k-ft k-ft k-ft k-ft

,58.7 3 3 3 3

in in in inft ft ft ft(2, 979 )(12 ) (549.7 )(12 ) (1, 008 )(12 ) (3, 999 )(12 )1.00 1.00 1.00 1.30

1,733 in 2,216 in 2,216 in 2,415 inf

f = + ++

ksi,58.7 54.90ff =

k-ft k-ft k-ft k-ft

,73.3 3 3 3 3

in in in inft ft ft ft(2, 779 )(12 ) (515.8 )(12 ) (946.1 )(12 ) (4, 067 )(12 )1.00 1.00 1.00 1.30

1,733 in 2,216 in 2,216 in 2,415 inf

f = + ++

ksi,73.3 53.43ff =

The stress at 58.7 governs. ff = 54.90ksi. The load factor for wind under Service II is 0.00, fl = 0ksi

ksi

ksi ksi ksi 54.90 00.95 (0.95)(1.00)(50 ) 47.502 2

lf h yf

ff R F + + = No Good. Note: The bending moments in the above calculations come from page 22 while the moments of inertia are found on page 17.

-- 41 --


9.4: Permanent Deformations - Section 2

Top Flange, Negative Moment

k-ft k-ft k-ft k-ft

,165 3 3 3 3

in in in inft ft ft ft(7,109 )(12 ) (1, 250 )(12 ) (2, 292 )(12 ) (4, 918 )(12 )1.00 1.00 1.00 1.30

2,116 in 5,135 in 5,135 in 11,828 inf

f = + ++

ksi,165 55.08ff =

?

ksi ksi ksi0.95 55.08 (0.95)(1.00)(50 ) 47.50f h yff R F = No Good.

Bottom Flange, Negative Moment

k-ft k-ft k-ft k-ft

3 3 3 3

in in in inft ft ft ft(7,108 )(12 ) (1, 250 )(12 ) (2, 292 )(12 ) (4, 918 )(12 )1.00 1.00 1.00 1.30

3,602 in 4,255 in 4,255 in 4,590 inf

f = + ++

ksi50.39ff =

The load factor for wind under Service II is 0.00, fl = 0ksi

ksiksi ksi ksi00.95 50.39 (0.95)(1.00)(50 ) 47.50

2 2l

f h yff

f R F+ + = No Good.

-- 42 --


9.5: Bend Buckling Checks At the Service Limit State, all sections except composite sections in positive flexure shall satisfy: c crwf F where:

2

0.9crw

w

EkF

Dt

=

and ( )29

/ck

D D=

Section 1 Not Applicable Section 2

k-ft k-ft k-ft k-ftin in in inft ft ft ft

3 3 3 3

(7,108 )(12 ) (1, 250 )(12 ) (2, 292 )(12 ) (4, 918 )(12 )1.00 1.00 1.00 1.30

3,602 in 4,255 in 4,255 in 4,590 inc

f = + ++

ksi50.39cf =


3 3 3 3

(7,108 )(12 ) (1, 250 )(12 ) (2, 292 )(12 ) (4, 918 )(12 )1.00 1.00 1.00 1.30

2,116 in 5,135 in 5,135 in 11,828 int

f = + ++

ksi55.08tf =

( )ksiksi ksi

0

50.39 72.5" 2.5" 050.39 55.08

32.14

cc cf

c t

fD d tf f

= + = +

=

( )2 29 9

41.49/ 32.14"

69"c

kD D

= = =

ksi

ksi2

916

(0.90)(29, 000 )(41.49)71.96

69""

crwF = =

This is larger than fcO.K.

-- 43 --


10. CHECK STRENGTH LIMIT STATE 10.1: Section 1 Positive Flexure

Section Classification (6.10.6.2, Pg. 6.98 6.99)

Check 2

3.76cpw yc

D Et F

Find Dcp, the depth of the web in compression at Mp (compression rebar in the slab is ignored).

( )( )ksi kipksi kip9

16ksi kip3

4' ksi kip

(50 ) 21" 1" 1,050

(50 )(69")( ") 1,941

(50 )(15")( ") 562.5

0.85 (0.85)(4.5 )(109.5")(8.5") 3,560

t yt t t

w yw w

c yc c c

s c s s

P F b t

P F Dt

P F b t

P f b t

= = == = == = == = =

Since Pt + Pw +Pc < Ps 3,554kip < 3,560kip, the PNA lies in the slab.

( ) ( ) kipkip3,5548.5" 3,5608.486 8.486" from top of slab "

c w ts

s

p

P P PY tP

Y YD

+ + =

= == =

Since none of the web is in compression, Dcp = 0 and the web is compact. For Composite Sections in Positive Flexure, (6.10.7.1, Pg. 6.101 6.102)

13u xt nl f

M f S M+ Mu = 13,568k-ft from Page 30; take fl = 0

Dt = 1 + 69 + 3/4 + 8.5 = 79.25 0.1Dt = 7.925 (The haunch is not included in Dt, as per ODOT Exceptions)

Since Dp =8.486 > 0.1Dt = 7.925, 1.07 0.7 pn pt

DM M

D =

kip

k-in k-ft

8.486"(3,554 ) 79.25" 30.68"2

157,500 13,130

pM =

= =

( ) ( )k-ft k-ft8.486"13,130 1.07 0.7 13,06079.25"nM = =

-- 44 --


? ?

k-ft k-ft13 (13,568 ) (0) (1.00)(13,060 )u xt nl fM f S M+ + No Good.

Note that the check of 1.3 h yn R MM has not been made in the above calculations. This section would satisfy the Article B6.2 so this check doesnt need to be made.

Check the ductility requirement to prevent crushing of the slab:

( )( )? ?0.42 8.486" 0.42 79.25" 33.29"p tD D = O.K.

The Section is NOT Adequate for Positive Flexure at Stations 58.7 and 271.3

The Girder failed the checks for service limits and has failed the first of several checks at the strength limit state. At this point I will investigate the strength of a section with 70ksi steel in the top and bottom flanges.

-- 45 --


Hybrid Girder Factors Will Now be Required: Compute the Hybrid Girder Factor, Rh, for Section 1:

Per AASHTO Commentary Pg 6-95, Dn shall be taken for the bottom flange since this is a composite section in positive flexure.

, 58.19" 1" 57.19"n BottomD = = ( )312 3

12 2hR

+ = + ( )( )( )( )

916(2) 57.19" "2 3.064

1" 21"n w

fn

D tA

= = =

ksi

ksi

501.0 0.714370

yw

n

Ff

= = =

( )( )( )

3

, 1

12 3.064 (3)(0.7143) (0.7143)0.9626

12 2 3.064h SectionR

+ = =+ Compute the Hybrid Girder Factor, Rh, for Section 2: For the short-term composite section, 12, 2 " 69" 52.23" 19.27"n TopD = + =

1

2, 52.23" 2 " 49.73"n BottomD = = Governs

( )312 312 2h

R

+ = +

( )( )( )( )

916

12

(2) 49.73" "2 1.0662 " 21"

n w

fn

D tA

= = =

ksi

ksi

501.0 0.714370

yw

n

Ff

= = =

( )( )( )

3

, 2

12 1.066 (3)(0.7143) (0.7143)0.9833

12 2 1.066h SectionR

+ = =+

-- 46 --


11. RECHECK SERVICE LIMIT STATE WITH 70KSI FLANGES 11.1: Permanent Deformations - Section 1 At the Service Limit State, the following shall be satisfied for composite sections Top Flange: 0.95f h yff R F

Bottom Flange 0.952

lf h yf

ff R F+

Top Flange, Positive Moment

From before: ksi,58.7 36.96ff =

?

ksi ksi ksi 0.95 36.96 (0.95)(0.9626)(70 ) 64.01f h yff R F = O.K.

Bottom Flange, Positive Moment

ksi,58.7 54.90ff = The load factor for wind under Service II is 0.00, fl = 0ksi

?ksi

ksi ksi ksi 54.90 00.95 (0.95)(0.9626)(70 ) 64.012 2

lf h yf

ff R F + + = O.K. 11.2: Permanent Deformations - Section 2

Top Flange, Negative Moment

From before: ksi,165 55.08ff =

?ksi ksi ksi0.95 55.08 (0.95)(0.9833)(70 ) 65.39f h yff R F = O.K.

Bottom Flange, Negative Moment

ksi50.39ff = The load factor for wind under Service II is 0.00, fl = 0ksi

ksi ?ksi ksi ksi00.95 50.39

2 2(0.95)(0.9833)(70 ) 65.39lf h yf

ff R F+ + = O.K.

-- 47 --


11.3: Bend Buckling Checks At the Service Limit State, all sections except composite sections in positive flexure shall satisfy: c crwf F where:

2

0.9crw

w

EkF

Dt

=

and ( )29

/ck

D D=

Section 1 - Not Applicable Section 2


3 3 3 3

(7,108 )(12 ) (1, 250 )(12 ) (2, 292 )(12 ) (4, 918 )(12 )1.00 1.00 1.00 1.30

3,602 in 4,255 in 4,255 in 4,590 inc

f = + ++

ksi50.39cf =


3 3 3 3

(7,108 )(12 ) (1, 250 )(12 ) (2, 292 )(12 ) (4, 918 )(12 )1.00 1.00 1.00 1.30

2,116 in 5,135 in 5,135 in 11,828 int

f = + ++

ksi55.08tf =

( )ksiksi ksi

0

50.39 72.5" 2.5" 050.39 55.08

32.14

cc cf

c t

fD d tf f

= + = +

=

( )2 29 9

41.49/ 32.14"

69"c

kD D

= = =

ksi

ksi2

916

(0.90)(29, 000 )(41.49)71.96

69""

crwF = =

This is larger than fcO.K.

-- 48 --


12. RECHECK STRENGTH LIMIT STATE WITH 70KSI FLANGES 12.1: Section 1 - Positive Flexure


Check 2

3.76cpw yc

D Et F

Find Dcp, the depth of the web in compression at Mp (compression rebar in the slab is ignored).

( )( )ksi kipksi kip9

16ksi kip3

4' ksi kip

(70 ) 21" 1" 1,470

(50 )(69")( ") 1,941

(70 )(15")( ") 787.5

0.85 (0.85)(4.5 )(109.5")(8.5") 3,560

t yt t t

w yw w

c yc c c

s c s s

P F b t

P F Dt

P F b t

P f b t

= = == = == = == = =

Since Pt + Pw +Pc > Ps 4,199kip > 3,560kip, the PNA is NOT in the slab.

Check Case I ?

t w c sP P P P+ +

?

kip kip kip kip1, 470 1,941 787.5 3,560+ + NO

Check Case II ?

t w c sP P P P+ +

?kip kip kip kip1, 470 1,941 787.5 3,560+ + YES - PNA in Top Flange

kip kip kip

kip

12

0.750" 1,941 1,470 3,560 1 0.3040" (from the top of steel)2 787.5

c w t s

c

t P P PYP

+ = + + = + =

Dp = 8.5 + 0.3040 = 8.804

Since none of the web is in compression, Dcp = 0 and the web is compact. For Composite Sections in Positive Flexure, (6.10.7.1, Pg. 6.101 6.102)

13u xt nl f

M f S M+ Mu = 13,568k-ft from Page 30; take fl = 0

Dt = 1 + 69 + 3/4 + 8.5 = 79.25 0.1Dt = 7.925

-- 49 --


(The haunch is not included in Dt, as per ODOT Exceptions)

Since Dp = 8.804 > 0.1Dt = 7.925, 1.07 0.7 pn pt

DM M

D =

Determine Mp:

The distances from the component forces to the PNA are calculated.

( )

8.5" 0.3040" 4.554"2

69" 0.75" 0.3040" 34.05"2

1"70.75" 0.3040" 69.95"2

s

w

t

d

d

d

= + =

= =

= =

The plastic moment is computed.

( ) [ ]( ) ( )

( )( ) ( )( ) ( )( )( )

22

kip2 2

kip kip kip

2 k-inkipin

k-

2

787.5 0.3040" 0.750" 0.3040" ...(2)(0.750")

... 3,560 4.554" 1,941 34.05" 1,470 69.95"

525 0.2913 in 185,100

185,300

cp c s s w w t t

c

PM Y t Y P d P d Pdt

= + + + + = + +

+ + + = +

= in k-ft15,440=

( ) ( )k-ft k-ft8.804"15,440 1.07 0.7 15,32079.25"nM = =

? ?k-ft k-ft1

3 (13,568 ) (0) (1.00)(15,320 )u xt nl fM f S M+ + O.K. Note that the check of 1.3 h yn R MM has not been made in the above calculations. This section would satisfy the Article B6.2 so this check doesnt need to be made.

Check the ductility requirement to prevent crushing of the slab:

( )( )? ?0.42 8.804" 0.42 79.25" 33.29"p tD D = O.K.

The Section is Adequate for Positive Flexure at Stations 58.7 and 271.3 with 70ksi Flanges

-- 50 --


12.2: Section 2 - Negative Flexure


Check 2 5.70cw yc

D Et F

Dc is the depth of the web in compression for the cracked section. Dc = 32.04 21/2 = 29.54

ksi

ksi916

2 (2)(29.54") 29,000105.0 5.70 5.70 137.3( ") 50

c

w yc

D Et F

= = < = = The web is non-slender. Since the web is non-slender we have the option of using the provisions in Appendix A to determine the moment capacity. I will first determine the capacity using the provisions in 6.10.8, which will provide a somewhat conservative determination of the flexural resistance. For Composite Sections in Negative Flexure, (6.10.8.1, Pg. 6.105 6.114)

The Compression Flange must satisfy:

13 ncbu l f

f f F+

Per 6.10.1.1.1a, elastic stresses at any location in a composite section shall consist of the sum of stresses caused by loads applied separately to the bare steel, short-term composite section, and long-term composite section. In 6.10.1.1.1c, though, it states that for the Strength Limit, the short-term and long-term composite sections shall consist of the bare steel and the longitudinal rebar. In other words, for determining negative moment stresses over the pier, we can use the factored moment above with the properties for the cracked section.

1 21.25 1.50 1.751.25 DC DC DW LLbuBS CR

M M M Mf

S S+ += +

k-ft k-ft k-ftk-ft

3 3

ininftft

(1.25)(1, 250 ) (1.50)(2, 292 ) (1.75)(4, 918 ) (12 )(7,109 )(12 )1.25

3,602 in 3,932 inbuf

+ += +

ksi71.13buf =

Since fbu is greater than Fyc, it is obvious that a strength computed based on the provisions in 6.10.8 will not be adequate.

-- 51 --


As it stands here, this girder is clearly not adequate over the pier. The compression flange is overstressed as per the provisions in 6.10.8. There are still other options to explore, though, before increasing the plate dimensions.

1. Since the web is non-slender for Section 2 in Negative Flexure, we have the option of using the provisions in Appendix A6 to determine moment capacity. This would provide an upper bound strength of Mp instead of My as was determined in 6.10.8.

2. The provisions in Appendix B6 allow for redistribution of negative moment from the region

near the pier to the positive moment region near mid-span for sections that satisfy stringent compactness and stability criteria. If this section qualifies, as much as ~2,000k-ft may be able to be redistributed from the pier to mid-span, which could enable the plastic moment strength from Appendix A6 to be adequate. (This solution may even work with the flange strength at 50ksi, but I doubt it)

Despite the fact that the girder appears to have failed our flexural capacity checks, lets look at the shear capacity.

-- 52 --


12.3 Vertical Shear Capacity At the strength Limit, the following must be satisfied u nV V For an unstiffened web,

n cr pV V CV= =

Check, 1.12w yw

D Ekt F

,

916

69" 122.7"w

Dt

= = Since there are no transverse stiffeners, k = 5

ksi

ksi

(29,000 )(5)1.12 60.31(50 )

= ksi

ksi

(29,000 )(5)1.40 75.39(50 )

=

Since 1.40w yw

D Ekt F

> , elastic shear buckling of the web controls.

ksi

2 2 ksi

916

1.57 1.57 (5)(29,000 ) 0.3026(50 )69"

"yw

w

kECFD

t

= = =

ksi kip9

160.58 (0.58)(50 )(69")( ") 1,126p yw wV F Dt= = =

kip kip(0.3026)(1,126 ) 340.6n pV CV= = =

( )( )kip kip1.00 340.6 340.6nV = = No Good. This strength is adequate from 16 100 and 230 - 314. This strength is not adequate near the end supports or near the pier, however.

-- 53 --


Try adding transverse stiffeners spaced at do = 8 = 96

2 25 55 5 7.583

96"69"

o

kdD

= + = + =

122.7w

Dt

= , ksi

ksi

(29,000 )(7.583)1.12 74.28(50 )

= , ksi

ksi

(29,000 )(7.583)1.40 92.85(50 )

=

Since 1.40w yw

D Ekt F


ksi

2 2 ksi

916

1.57 1.57 (29,000 )(7.583) 0.4589(50 )69"

"yw

w

EkCFD

t

= = =

kip kip(1.00)(0.4589)(1,126 ) 516.5n pV CV = = = O.K.

This capacity is fine but we may be able to do better if we account for tension field action. Try adding transverse stiffeners spaced at do = 12 = 144

2 25 55 5 6.148

144"69"

o

kdD

= + = + =

122.7w

Dt

= , ksi

ksi

(29,000 )(6.148)1.12 66.88(50 )

= , ksi

ksi

(29,000 )(6.148)1.40 83.60(50 )

=

Since 1.40w yw

D Ekt F


ksi

2 2 ksi

916

1.57 1.57 (29,000 )(6.148) 0.3721(50 )69"

"yw

w

EkCFD

t

= = =

Without TFA: kip kip(0.3721)(1,126 ) 418.9n pV CV= = =

-- 54 --


With TFA:

Since ( ) ( )9

16

12

2 (2)(69")( ") 1.056 2.5(21")(2 ") (21")(1")

w

fc fc ft ft

Dtb t b t

= = ++ ,

kip

2 2

0.87(1 ) (0.87)(1 0.3721)(1,126 ) 0.3721144"1169"

n p

o

CV V CdD

= + = + ++

kip kip(1,126 )(0.6082) 684.8nV = =

( )( )kip kip1.00 684.8 684.8nV = = O.K.

This TFA strength is adequate near the pier but TFA is not permitted in the end panels. The following stiffener configuration should provide adequate shear strength.

-- 55 --


Strength Limit Shear Capacity

-800

-600

-400

-200

0

200

400

600

800

0 30 60 90 120 150 180 210 240 270 300 330

Station (ft)

S

h

e

a

r

(

k

i

p

)

Strength IV

Strength I

Tension Field Action

-- 56 --


12.4: Horizontal Shear Strength Per ODOT Standard practice, shear studs will be used to transfer horizontal shear between the concrete deck and top flange of the steel girder. ODOT prefers the use of 7/8diameter studs. Ideally, the studs should extend to the mid-thickness of the deck. Using this criterion, the height of the studs can be determined.

29.5" 2.75" 0.75" 6.75"

2

shaunch flange

th t t= +

= + =

Use 7/8 x 61/2 shear studs AASHTO requires that the ratio of h/d be greater than or equal to 4.0.

?

12

78

4.0

6 " 7.429 4.0 OK"

hd

=

AASHTO requires a center-to-center transverse spacing of 4d and a clear edge distance of 1. With 7/8 diameter studs, there is room enough transversely to use up to 4 studs in each row. With this in mind, I will investigate the option of either 3 or 4 studs per row. Fatigue Limit State: The longitudinal pitch of the shear studs based on the Fatigue Limit is determined as

rsr

nZpV

fsrV Q

VI

= (6.10.10.1.2-1 & 3) where: n - Number of studs per row Zr - Fatigue resistance of a single stud Vsr - Horizontal fatigue shear range per unit length Vf - Vertical shear force under fatigue load combination Q - 1st moment of inertia of the transformed slab about the short-term NA I - 2nd moment of inertia of the short-term composite section

be

ts

bc

bt

tc

D

tt

tw

thaunch

-- 57 --


2

2 5.52rdZ d= (6.10.10.2-1)

34.5 4.28Log( )N = (6.10.10.2-2) 6 ksi34.5 4.28Log(55.84 10 ) 1.343 = =

( )( ) ( )2 2ksi 7 78 8

kip kip kip

5.51.343 " "2

1.028 2.105 2.105

r

r

Z

Z

= = =

tc cQ A d=

( )( ) 3

1

109.5" 9.5" 9.5"1" 69" 2.75" 58.19" 2,511 in8 2Section

Q = + + + =

( )( ) 3

2

109.5" 9.5" 9.5"2.5" 69" 2.75" 52.23" 3, 481 in8 2Section

Q = + + + =

4 1 140,500 inSectionI = 4 2 239,700 inSectionI = Since the fatigue shear varies along the length of the bridge, the longitudinal distribution of shear studs based on the Fatigue Limit also varies. These results are presented in a tabular format on a subsequent page. To illustrate the computations, I have chosen the shear at the abutment as an example. At the abutment, ( )kip kip kip38.13 3.53 41.66fV = =

( )( )

( )kip 3

kipinch4

41.66 2,511 in0.7445

140,500 insrV = =

For 3 studs in each row: For 4 studs in each row:

( )( )( )

ksiin

rowkipinch

3 2.1058.482

0.7445p = ( )( )( )

ksiin

rowkipinch

4 2.10511.31

0.7453p =

-- 58 --


Strength Limit: 0.85r sc n scQ Q = =

'0.5n sc c c sc uQ A f E A F= (6.10.10.4.3-1)

( )2 27 8 " 0.6013 in4scA = =

' ksi4.5cf =

Since n = 8, ksi

ksi29,000 3,6258

sc

EEn

= = = ksi60uF =

( )( ) ( )( ) ( )( )2 ksi ksi 2 ksi

kip kipstud stud

0.5 0.6013 in 4.5 3,625 0.6013 in 60

38.40 36.08

nQ = =

( )( )kip kipstud stud0.85 36.08 30.67sc nQ = =

p

r

Pn

Q+ = p n

r

P Pn

Q +=

-- 59 --


Positive moment - Section 1: Station 0.0 - 73.3 ( ),p Concrete steelP Min P P=

( )( )( )( )'

ksi kip

0.85

0.85 4.5 109.5" 9.5" 3,979Concrete c e sP f b t=

= =

( ) ( )( ) ( )( ) ( )( )( )ksi ksi kip70 15" 0.75" 21" 1" 50 69" 0.5625" 4,198Steel yw w ft ft ft fc fc fcP F Dt F b t F b t= + +

= + + =

kip3,979pP =

kip

studskipstud

3,979 129.730.67

p

r

Pn

Q+ = = =

3 Studs per Row:

( )( )

( )studs in

ftrows inchrowstuds

row

73.3' 0 ' 12129.7 44 20.46 Say 20"3 44 1

p= = =

4 Studs per Row:

( )( )

( )studs in

ftrows inchrowstuds

row

73.3' 0 ' 12129.7 33 27.49 Say 24"4 33 1

p= = =

-- 60 --


Negative Moment - Section 2: Station 73.3 - 165.0 ( ),n steel CrackP Min P P=

( )( )( )( )'

ksi kip

0.45

0.45 4.5 109.5" 9.5" 2,107Crack c e sP f b t=

= =

( ) ( )( ) ( )( ) ( )( )( )ksi ksi kip70 21" 2.5" 21" 1" 50 69" 0.5625" 7,086Steel yw w ft ft ft fc fc fcP F Dt F b t F b t= + +

= + + =

kip2,107nP =

kip kip

studskipstud

3,979 2,107 198.430.67

p n

r

P Pn

Q + += = =

3 Studs per Row:

( )( )

( )studs in

ftrows inchrowstuds

row

165.0 ' 73.3' 12198.4 67 16.67 Say 16"3 67 1

p= = =

4 Studs per Row:

( )( )

( )studs in

ftrows inchrowstuds

row

165.0 ' 73.3' 12198.4 50 22.48 Say 20"4 50 1

p= = =

-- 61 --


Shear Stud Summary: This table represents that pitch of shear studs required for either 3 or 4 studs per row based on location in the bridge.

Station V f Q I V sr p Fat p Str p max p Fat p Str p max(ft) (kip) (in

3) (in4) (kip/in) (in) (in) (in) (in) (in) (in)0.0 41.66 2,511 140,521 0.7444 8.48 20.00 8.48 11.31 24.00 11.31

14.7 37.01 2,511 140,521 0.6613 9.55 20.00 9.55 12.73 24.00 12.7329.3 33.68 2,511 140,521 0.6018 10.49 20.00 10.49 13.99 24.00 13.9944.0 32.79 2,511 140,521 0.5859 10.78 20.00 10.78 14.37 24.00 14.3758.7 33.04 2,511 140,521 0.5904 10.70 20.00 10.70 14.26 24.00 14.2673.3 33.46 2,511 140,521 0.5979 10.56 20.00 10.56 14.08 24.00 14.0888.0 33.98 2,511 140,521 0.6071 10.40 16.00 10.40 13.87 20.00 13.87

102.7 34.59 2,511 140,521 0.6181 10.22 16.00 10.22 13.62 20.00 13.62117.3 35.38 2,511 140,521 0.6323 9.99 16.00 9.99 13.32 20.00 13.32132.0 36.62 2,511 140,521 0.6543 9.65 16.00 9.65 12.87 20.00 12.87135.7 37.07 3,481 239,734 0.5383 11.73 16.00 11.73 15.64 20.00 15.64139.3 37.53 3,481 239,734 0.5449 11.59 16.00 11.59 15.45 20.00 15.45143.0 37.98 3,481 239,734 0.5514 11.45 16.00 11.45 15.27 20.00 15.27146.7 38.42 3,481 239,734 0.5579 11.32 16.00 11.32 15.09 20.00 15.09150.3 38.88 3,481 239,734 0.5645 11.19 16.00 11.19 14.92 20.00 14.92154.0 39.34 3,481 239,734 0.5713 11.05 16.00 11.05 14.74 20.00 14.74157.7 39.81 3,481 239,734 0.5780 10.93 16.00 10.93 14.57 20.00 14.57161.3 40.26 3,481 239,734 0.5847 10.80 16.00 10.80 14.40 20.00 14.40165.0 81.44 3,481 239,734 1.1826 5.34 16.00 5.34 7.12 20.00 7.12168.7 40.26 3,481 239,734 0.5847 10.80 16.00 10.80 14.40 20.00 14.40172.3 39.81 3,481 239,734 0.5780 10.93 16.00 10.93 14.57 20.00 14.57176.0 39.34 3,481 239,734 0.5713 11.05 16.00 11.05 14.74 20.00 14.74179.7 38.88 3,481 239,734 0.5645 11.19 16.00 11.19 14.92 20.00 14.92183.3 38.42 3,481 239,734 0.5579 11.32 16.00 11.32 15.09 20.00 15.09187.0 37.98 3,481 239,734 0.5514 11.45 16.00 11.45 15.27 20.00 15.27190.7 37.53 3,481 239,734 0.5449 11.59 16.00 11.59 15.45 20.00 15.45194.3 37.07 3,481 239,734 0.5383 11.73 16.00 11.73 15.64 20.00 15.64198.0 36.62 2,511 140,521 0.6543 9.65 16.00 9.65 12.87 20.00 12.87212.7 35.38 2,511 140,521 0.6323 9.99 16.00 9.99 13.32 20.00 13.32227.3 34.59 2,511 140,521 0.6181 10.22 16.00 10.22 13.62 20.00 13.62242.0 33.98 2,511 140,521 0.6071 10.40 16.00 10.40 13.87 20.00 13.87256.7 33.46 2,511 140,521 0.5979 10.56 20.00 10.56 14.08 24.00 14.08271.3 33.04 2,511 140,521 0.5904 10.70 20.00 10.70 14.26 24.00 14.26286.0 32.79 2,511 140,521 0.5859 10.78 20.00 10.78 14.37 24.00 14.37300.7 33.68 2,511 140,521 0.6018 10.49 20.00 10.49 13.99 24.00 13.99315.3 37.01 2,511 140,521 0.6613 9.55 20.00 9.55 12.73 24.00 12.73330.0 41.66 2,511 140,521 0.7444 8.48 20.00 8.48 11.31 24.00 11.31

3 Studs Per Row 4 Studs Per Row

The arrangement of shear studs is shown below.

-- 62 --

Single-Span Bridge Example AASHTO-LRFD 2007 ODOT LRFD Short Course - Steel Created July 2007: Page 1 of 21

ONE-SPAN INELASTIC I-GIRDER BRIDGE DESIGN EXAMPLE 1. PROBLEM STATEMENT AND ASSUMPTIONS: A single span composite I-girder bridge has span length of 166.3 and a 64 deck width. The steel girders have Fy = 50ksi and all concrete has a 28-day compressive strength of fc = 4.5ksi. The concrete slab is 9.5 thick. A typical 4 haunch was used in the section properties. Concrete barriers weighing 640plf and an asphalt wearing surface weighing 60psf have also been applied as a composite dead load. HL-93 loading was used per AASHTO (2004), including dynamic load allowance.

-- 63 --

Single-Span Bridge Example AASHTO-LRFD 2007 ODOT LRFD Short Course - Steel Created July 2007: Page 2 of 21

166'

-4"

cc

Bea

rings

172'

-4"

Tot

al G

irder

Len

gth

G 1

G 2

G 3

G 4

G 5

G 6

Cro

ss F

ram

es S

pace

d @

22'

-0"

cc

-- 64 --

Single-Span Bridge Example AASHTO-LRFD 2007 ODOT LRFD Short Course - Steel Created July2007: Page 3 of 23




-- 65 --


2. LOAD CALCULATIONS: DC dead loads (structural components) include:

Steel girder self weight (DC1) Concrete deck self weight (DC1) Haunch self weight (DC1) Barrier (DC2)

DW dead loads (structural attachments) include:

Wearing surface (DW), Including FWS 2a. Dead Load Calculations

Steel Girder Self-Weight (DC1):

(a) Section 1

A = (14)(1.125) + (68)(0.6875) + (22)(1.5) = 95.5 in2

( ) ( )pcf

2 lbsft2in

ft

49095.5 in 1.15 373.712

section1W = =

per girder

(b) Section 2 A = (14)(2) + (68)(0.5625) + (22)(2) = 110.25 in2

( ) ( )pcf

2 lbsft2in

ft

490110.3 in 1.15 431.412

section1W = =

per girder

(c) Section 3 A = (14)(2) + (68)(0.5625) + (22)(2.375) = 118.5 in2

( ) ( )pcf

2 lbsft2in

ft

490118.5 in 1.15 463.712

section1W = =

per girder

(d) Average Girder Self Weight

( )( )( ) ( )( )( ) ( )( )lbs lbs lbsft ft ft lbsft

2 40.17 ' 373.7 2 18.0 ' 431.4 50.0 ' 463.7413.3

166.3'aveW

+ += =

Deck Self-Weight (DC1):

( )( )( )

pcflbsftin

ft

9.5'' 64.0 ' 150 1,2676 Girders 12Deck

W = =

per girder

-- 66 --


Haunch Self-Weight (DC1):

Average width of haunch: 14

( )( ) ( )( )( )( )( ) ( )pcf

lbs12 ft2in

ft

15014 4 2 9 '' 4 '' 94.3312

haunchW = + =

per girder

Barrier Walls (DC2):

( )( )plf lbsft

2 each 640213.3

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