AAB - Example Problems Handout

LRFD Steel Design

AASHTO LRFD Bridge Design Specifications

Example Problems

Created July 2007

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The University of Cincinnati and

Dr. James A Swanson.

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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 f’c = 4.5ksi. The concrete slab is 91/2” thick. A typical 2¾” 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

9½” (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

( )( ) Lb

ftinft

2sec 1 2 1.15490 pcf71.06 in 278.1

12tionW

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

= = per girder

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

( )( ) Lb

ftinft

2sec 2 2 1.15490 pcf112.3 in 439.5

12tionW

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

= = 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: ( ) ( )1

2 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-ft

kip kip25 33.00 25 31.11 1,603

Single Truck: ( )( ) ( )( ) ( )( )+ + =kip kip kip k-ftk-ft k-ft k-ftkip kip kip

8 26.13 32 33.00 32 26.33 2,108

Lane Load: ( )( ) =2 k-ftkip k-ftft kip

0.640 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-ft

kip kip25 18.51 25 18.45 924.0

Single Truck: ( )( ) ( )( ) ( )( )+ + =kip kip kip k-ftk-ft k-ft k-ftkip kip kip

8 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 kip

0.640 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 doesn’t 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 ( )ft

sec f - 4/3 for all load combinations except for Fatigue, in which case it is 1.0 g - The acceleration of gravity ( )2

ftsec

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)

( ) ( )( )2

2mphpsf 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: ( )( )( ) kiplbs

ft840 2 165' 277.2= For End Abutments: ( )( )( ) kiplbs 1

ft 2840 165' 69.30= For Center Pier: ( )( )( )( ) kiplbs 1

ft 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 IXTop 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)

t b A y Ay Ix d Ad2 IXSlab 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)

t b A y Ay Ix d Ad2 IXSlab 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

t b A y Ay Ix d Ad2 IXRebar 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

t b A y Ay Ix d Ad2 IXTop 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)

t b A y Ay Ix d Ad2 IXSlab 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)

t b A y Ay Ix d Ad2 IXSlab 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

t b A y Ay Ix d Ad2 IXRebar 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

+= + =

+⎡ ⎤⎣ ⎦


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 0.3625.012 '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, η, was applied as is shown below.

i i iQ Qη γ= ∑ η is taken as the product of ηD, ηR, and ηI, and is taken as not less than 0.95. For this example, ηD and ηI are taken as 1.00 while ηR 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 η 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 ksi

1 4

2,779 12 70" 30.68"24.67

53,157 inDCf−

= =

k-ft2 515.8DCM =


2 4

515.8 12 70" 46.33"1.427

102,676 inDCf−

= =

ksi ksi ksi

, 24.67 1.427 26.09comp DLf = + =

k-ft, 258.6Fat NegM =


4

258.6 12 70" 58.19"0.261

140,521 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. ( ) ( )n

f FΔ ≤ Δγ

( ) ( )13

2TH

n

FAFN

Δ⎛ ⎞Δ = ≥⎜ ⎟⎝ ⎠

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

( ) ( )( )( )k-ft inft ksi

4

1, 236 12 58.19" 1.00"6.036

140,521 infγ

−Δ = =

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

( ) ksiksi12.0 6.00

2 2TH

FΔ= = 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

Δ⎛ ⎞ = =⎜ ⎟⎝ ⎠

, the infinite life governs.

( ) ksi6.00

nFΔ =

Since ( ) ( )ksi ksi6.036 6.00

nf Fγ Δ = > = , the detail is not satisfactory.

-- 33 --


Calculate the design life of the part under consideration:

Since ( ) ( ) is greater than

2TH

Ffγ

ΔΔ , solve for N in the following equation.

( )13Af

Nγ ⎛ ⎞Δ ≤ ⎜ ⎟

⎝ ⎠

( ) ( )8 3

63 3ksi

44 10 ksi 20.01 10 cycles6.036

ANfγ

×≤ = = ×

Δ

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 ( )( )in

ft42 ' 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:

34

8.5"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 ( )( )in

ft42 ' 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:

12

8.5"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: ΔTruck = 2.442” Deflection due to the Design Lane Load: ΔLane = 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 lane’s 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

⎛ ⎞Δ = = ←⎜ ⎟⎜ ⎟

⎝ ⎠⎛ ⎞

Δ = + =⎡ ⎤⎜ ⎟ ⎣ ⎦⎜ ⎟⎝ ⎠

The limiting deflection for this bridge is:

( )( )in

ft165' 122.475" OK

800 800LimitL

Δ = = = ←

-- 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 IMc

BS 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 1.301,327 in 4,204 in 4,204 in 11,191 in

ff = + ++

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

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 1.301,327 in 4,204 in 4,204 in 11,191 in

ff = + ++

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

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 1.301,733 in 2,216 in 2,216 in 2,415 in

ff = + ++

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

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 1.301,733 in 2,216 in 2,216 in 2,415 in

ff = + ++

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

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 1.302,116 in 5,135 in 5,135 in 11,828 in

ff = + ++

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

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 1.303,602 in 4,255 in 4,255 in 4,590 in

ff = + ++

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

ksi50.39ff =

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

ksi

ksi ksi ksi00.95 50.39 (0.95)(1.00)(50 ) 47.502 2

lf h yf

ff 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 ( )2

9

/c

kD 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 =

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

ft 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

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

f = + ++⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

ksi55.08tf =

( )ksi

ksi ksi

0

50.39 72.5" 2.5" 050.39 55.08

32.14

cc cf

c t

fD d tf f

⎛ ⎞−= − ≥⎜ ⎟⎜ ⎟+⎝ ⎠⎛ ⎞

= − ≥⎜ ⎟+⎝ ⎠′′=

( )2 2

9 941.49

/ 32.14"69"

c

kD D

= = =⎛ ⎞⎜ ⎟⎝ ⎠

ksi

ksi2

916

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

69""

crwF = =⎛ ⎞⎜ ⎟⎝ ⎠

This is larger than fc…O.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

3.76cp

w yc

D Et F

≤

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

( )( )ksi kip

ksi kip916

ksi kip34

' 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.

( ) ( )kip

kip

3,5548.5"3,560

8.486 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 fM f S Mφ+ ≤ 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 p

t

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 Mφ+ ≤ + ≤ 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 doesn’t 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 2hRβ ρ ρ

β

+ −=

+

( )( )( )( )

916(2) 57.19" "2 3.064

1" 21"n w

fn

D tA

= = =β

ksi

ksi

501.0 0.714370

yw

n

Ff

ρ ρ= ≤ → = =

( )

( )( )

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, 1

2, 2 " 69" 52.23" 19.27"n TopD = + − =

12, 52.23" 2 " 49.73"n BottomD = − = Governs

( )312 3

12 2hRβ ρ ρ

β

+ −=

+

( )( )( )( )

916

12

(2) 49.73" "2 1.0662 " 21"

n w

fn

D tA

= = =β

ksi

ksi

501.0 0.714370

yw

n

Ff

ρ ρ= ≤ → = =

( )

( )( )

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 ) 65.39l

f h yff

f 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 ( )2

9

/c

kD 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 =

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

ft 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

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

f = + ++⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

ksi55.08tf =

( )ksi

ksi ksi

0

50.39 72.5" 2.5" 050.39 55.08

32.14

cc cf

c t

fD d tf f

⎛ ⎞−= − ≥⎜ ⎟⎜ ⎟+⎝ ⎠⎛ ⎞

= − ≥⎜ ⎟+⎝ ⎠′′=

( )2 2

9 941.49

/ 32.14"69"

c

kD D

= = =⎛ ⎞⎜ ⎟⎝ ⎠

ksi

ksi2

916

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

69""

crwF = =⎛ ⎞⎜ ⎟⎝ ⎠

This is larger than fc…O.K.

-- 48 --


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


Check 2

3.76cp

w yc

D Et F

≤

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

( )( )ksi kip

ksi kip916

ksi kip34

' 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 fM f S Mφ+ ≤ 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 p

t

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-ft13 (13,568 ) (0) (1.00)(15,320 )u xt nl fM f S Mφ+ ≤ + ≤ 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 doesn’t 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 5.70c

w 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 ff f Fφ+ ≤

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 LL

buBS 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, let’s look at the shear capacity.

-- 52 --


12.3 Vertical Shear Capacity At the strength Limit, the following must be satisfied u nV Vφ≤ 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 kip1.00 340.6 340.6nVφ = = 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 2

5 55 5 7.58396"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 CVφ = φ = = 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 2

5 55 5 6.148144"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 ( ) ( )

916

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 kip1.00 684.8 684.8nVφ = = 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)

Shea

r (k

ip)

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 7/8”diameter 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

r

sr

nZpV

≤ fsr

V QV

I= (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 dα= ≥ (6.10.10.2-1)

34.5 4.28Log( )Nα = − (6.10.10.2-2) 6 ksi34.5 4.28Log(55.84 10 ) 1.343α = − × =

( )( ) ( )2 2ksi 7 7

8 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 2SectionQ

⎡ ⎤ ⎛ ⎞= + + + − =⎢ ⎥ ⎜ ⎟⎝ ⎠⎣ ⎦

( )( ) 3

2

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

⎡ ⎤ ⎛ ⎞= + + + − =⎢ ⎥ ⎜ ⎟⎝ ⎠⎣ ⎦

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 Qφ φ= =

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

( )2 278 " 0.6013 in

4scA π⎛ ⎞= =⎜ ⎟⎝ ⎠

' 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 kip

stud stud0.85 36.08 30.67sc nQφ = =

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) (in3) (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 f’c = 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

( )( )

pcf2 lbs

ft2inft

49095.5 in 1.15 373.712

section1W⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟= =

⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

per girder

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

( )( )

pcf2 lbs

ft2inft

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

( )( )

pcf2 lbs

ft2inft

490118.5 in 1.15 463.712

section1W⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟= =

⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

per girder

(d) Average Girder Self Weight

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

ft

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 12DeckW

⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

per girder

-- 66 --


Haunch Self-Weight (DC1):

Average width of haunch: 14’’

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

pcflbs1

2 ft2inft

15014 4 2 9 '' 4 '' 94.3312

haunchW⎛ ⎞⎜ ⎟= + =⎜ ⎟⎝ ⎠

per girder

Barrier Walls (DC2):

( )( )plflbsft

2 each 640213.3

6 girdersbarriersW⎛ ⎞⎜ ⎟= =⎜ ⎟⎝ ⎠

per girder

Wearing Surface (DW):

( )( )psflbsft

61.0' 60610.0

6 Girderswearing_surfaceW = = per girder

The moment effect due to dead loads was found using an FE model composed of six frame elements to model the bridge (a node was placed at mid-span). 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). The maximum moments at mid-span are easily computed since the bridge is statically determinate.

( )( )2lbs2

ft k-ft1,

413.3 166.3'1, 429

8 8DC SteelwLM

⎡ ⎤⎛ ⎞⎢ ⎥= = =⎜ ⎟⎢ ⎥⎝ ⎠ ⎣ ⎦

( )( )2lbs2ft k-ft

1,1, 267 166.3'

4,3798 8DC Deck

wLM⎡ ⎤⎛ ⎞⎢ ⎥= = =⎜ ⎟⎢ ⎥⎝ ⎠ ⎣ ⎦

( )( )2lbs2ft k-ft

2,213.3 166.3'

737.48 8DC Barriers

wLM⎡ ⎤⎛ ⎞⎢ ⎥= = =⎜ ⎟⎢ ⎥⎝ ⎠ ⎣ ⎦

( )( )2lbs2ft k-ft610.0 166.3'

2,1098 8DW

wLM⎡ ⎤⎛ ⎞⎢ ⎥= = =⎜ ⎟⎢ ⎥⎝ ⎠ ⎣ ⎦

-- 67 --


The maximum shear forces at the ends of the girder are also easily computed.

( )( )lbsft kip

1,413.3 166.3'

34.372 2DC Steel

wLV⎡ ⎤⎛ ⎞= = =⎢ ⎥⎜ ⎟

⎝ ⎠ ⎢ ⎥⎣ ⎦

( )( )lbsft kipt

1,1, 267 166.3'

105.42 2DC Deck

wLV⎡ ⎤⎛ ⎞= = =⎢ ⎥⎜ ⎟

⎝ ⎠ ⎢ ⎥⎣ ⎦

( )( )lbsft kip

2,213.3 166.3'

17.742 2DC Barriers

wLV⎡ ⎤⎛ ⎞= = =⎢ ⎥⎜ ⎟

⎝ ⎠ ⎢ ⎥⎣ ⎦

( )( )lbsft kip610.0 166.3'

50.722 2DW

wLV⎡ ⎤⎛ ⎞= = =⎢ ⎥⎜ ⎟

⎝ ⎠ ⎢ ⎥⎣ ⎦

-- 68 --


2b. 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)

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 within SAP.

-- 69 --


Unfactored HL-93 Moment Envelopes from SAP

-6,000

-4,000

-2,000

0

2,000

4,000

6,000

0 30 60 90 120 150

Station (ft)

Mom

ent (

kip-

ft)

Single Truck

Tandem

The following results were obtained from the SAP analysis:

• The maximum positive live-load moments occur at stations 83.15’

Station 40.16’- Section 1 Station 58.15’- Section 2 Station 83.15’- Section 3 HL-93M 3,614k-ft 4,481k-ft 4,911k-ft HL-93K 4,322k-ft 5,238k-ft 5,821k-ft HL-93S N/A N/A N/A

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

-- 70 --


2c. Verify the Maximum Positive Live-Load Moment at Station 83.15’:

Tandem: ( )( ) ( )( )kip kip k-ftk-ft k-ft

kip kip25 41.58 25 39.58 2,029+ =

Single Truck: ( )( ) ( )( ) ( )( )kip kip kip k-ftk-ft k-ft k-ftkip kip kip8 34.57 32 41.58 32 34.57 2,713+ + =

Lane Load: ( )( ) k-ftk-ft k-ftkip kip0.640 3, 457 2, 212=

(IM)(Tandem) + Lane: ( )( ) k-ft k-ftk-ftkip1.33 2, 029 2, 212 4,911+ =

(IM)(Single Truck) + Lane: ( )( ) k-ft k-ftk-ftkip1.33 2,713 2, 212 5,821+ = GOVERNS

The case of two trucks is not considered here because it is only used when computing negative moments.

Single Truck

Tandem :

Lane

8kip

32kip 32kip

25kip25kip

0.640kip/ft

:

05

1015202530354045

0 15 30 45 60 75 90 105 120 135 150 165

Station (ft)

Mom

ent (

k-ft

/ kip

)

-- 71 --


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. 3. SECTION PROPERTIES AND CALCULATIONS: 3a. Effective Flange Width, bs:

For an interior beam, bs is the lesser of:

( )( )

( )( )inft

14"12 12 8.5" 109 ''2 2

11.33' 12 135.96 ''

166.3' 41.58' 498.9 ''4 4

fs

eff

bt

SL

⎧• + = + =⎪⎪⎪• = =⎨⎪⎪• = = =⎪⎩

Therefore, bs = 109”

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. Note: At this point one should also check the effective of the outside girders as well. For this example, however, I will proceed sing the effective width for the interior girders.

-- 72 --


3b. Section 1 Flexural Properties

Bare Steel


95.50 2,787.86 ITotal = 72,886

Y = 29.19 SBS1,top = 1,759SBS1,bot = 2,497

Short-Term Composite (N=8)

t b A y Ay Ix d Ad2 IX

Slab 8.5000 109.00 115.81 74.88 8,671.46 697.29 -20.65 49,365 50,062Haunch 0.0000 14.0000 0.00 70.63 0.00 0.00 -16.40 0 0Top Flange 1.1250 14.0000 15.75 70.06 1,103.48 1.66 -15.83 3,948 3,950Web 0.6875 68.0000 46.75 35.50 1,659.63 18,014.33 18.73 16,399 34,414Bot Flange 1.5000 22.0000 33.00 0.75 24.75 6.19 53.48 94,381 94,387

211.31 11,459.32 ITotal = 182,813

n = 8.00 Y = 54.23 SST1,top = 11,150SST1,bot = 3,371

Long-Term Composite (N=24)


134.10 5,678.35 ITotal = 130,491

n = 24.00 Y = 42.34 SLT1,top = 4,614SLT1,bot = 3,082

Single Span Bridge Example - Section 1

-- 73 --


3c. Section 2 Flexural Properties

Bare Steel

t b A y Ay Ix d Ad2 IXTop Flange 2.0000 14.00 28.00 71.00 1,988.00 9.33 -40.08 44,978 44,987Web 0.5625 68.00 38.25 36.00 1,377.00 14,739.00 -5.08 987 15,726Bot Flange 2.0000 22.00 44.00 1.00 44.00 14.67 29.92 39,391 39,405

110.25 3,409.00 ITotal = 100,119

Y = 30.92 SBS1,top = 2,437SBS1,bot = 3,238



226.06 12,239.70 ITotal = 216,871

n = 8.00 Y = 54.14 SST1,top = 12,145SST1,bot = 4,006



148.85 6,352.57 ITotal = 159,101

n = 24.00 Y = 42.68 SLT1,top = 5,426SLT1,bot = 3,728


-- 74 --


3d. Section 3 Flexural Properties

Bare Steel


118.50 3,451.89 ITotal = 107,546

Y = 29.13 SBS1,top = 2,487SBS1,bot = 3,692



234.31 12,326.02 ITotal = 240,366

n = 8.00 Y = 52.61 SST1,top = 12,158SST1,bot = 4,569


t b A y Ay Ix d Ad2 IXSlab 8.5000 109.00 38.60 76.63 2,958.04 232.43 -35.82 49,544 49,777Haunch 0.0000 14.00 0.00 72.38 0.00 0.00 -31.57 0 0Top Flange 2.0000 14.0000 28.00 71.38 1,998.50 9.33 -30.57 26,174 26,184Web 0.5625 68.0000 38.25 36.38 1,391.34 14,739.00 4.43 749 15,488Bot Flange 2.3750 22.0000 52.25 1.19 62.05 24.56 39.61 81,990 82,015

157.10 6,409.93 ITotal = 173,463

n = 24.00 Y = 40.80 SLT1,top = 5,494SLT1,bot = 4,251


-- 75 --


4. DISTRIBUTION FACTOR FOR MOMENT 4a. Section 1:

Interior Girder - One Lane Loaded:

( )( )( )( )

( )( )( )

0.10.4 0.3

3

2

24 2

4

0.4 0.3 4

3

0.0614 12

(8) 72,890 in 95.5 in 49.06"

2,422,000 in

11.33' 11.33' 2, 422,000 in0.0614 166.3' 12 166.3' 8.5"

gM1,Int,Sec1

S

g g

g

g

M1,Int,Sec1

KS SDFL Lt

K n I Ae

K

K

DF

⎛ ⎞⎛ ⎞ ⎛ ⎞= + ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= +

= +

=

⎛ ⎞ ⎛ ⎞= + ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

0.1

0.4994M1,Int,Sec1DF

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

=

Interior Girder - Two or More Lanes Loaded:

( )( )( )

0.10.4 0.3

3

0.10.6 0.2 4

3

0.0759.5 12

11.33' 11.33' 2, 422,000 in0.0759.5 166.3' 12 166.3' 8.5"

0.7703

gM2,Int,Sec1

S

M2,Int,Sec1

M2,Int,Sec1

KS SDFL Lt

DF

DF

⎛ ⎞⎛ ⎞ ⎛ ⎞= + ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞⎛ ⎞ ⎛ ⎞= + ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠=

-- 76 --



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 18 5' 0 7500

11 33'M ,Ext,Sec.DF ..

= =

Multiple Presence: DFM1,Ext,Sec1 = (1.2) (0.7500) = 0.9000

Exterior Girder - Two or More Lanes Loaded:

DFM2,Ext,Sec1 = e DFM2,Int,Sec1

0.77

9.12.167 '0.77 1.008

9.1

ede

e

= +

= + =

DFM2,Ext+ = (1.008) (0.7703) = 0.7765

4b. Section 2:

Interior Girder – One Lane Loaded:

( )( )( )( )

( )( )

0.10.4 0.3

3

2

24 2

4

0.4 0.3 4

0.0614 12

(8) 100,100 in 110.3 in 47.83"

2,819,000 in

11.33' 11.33' 2,819,000 in0.0614 166.3' 12 166.3' 8.5

gM1,Int,Sec2

S

g g

g

g

M1,Int,Sec2

KS SDFL Lt

K n I Ae

K

K

DF

⎛ ⎞⎛ ⎞ ⎛ ⎞= + ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= +

= +

=

⎛ ⎞ ⎛ ⎞= + ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ( )

0.1

3"

0.5061M1,Int,Sec2DF

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

=

-- 77 --


Interior Girder – Two or More Lanes Loaded:

( )( )( )

0.10.4 0.3

3

0.10.6 0.2 4

3

0.0759.5 12

11.33' 11.33' 2,819,000 in0.0759.5 166.3' 12 166.3' 8.5"

0.7809

gM2,Int,Sec2

S

M2,Int,Sec2

M2,Int,Sec2

KS SDFL Lt

DF

DF

⎛ ⎞⎛ ⎞ ⎛ ⎞= + ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞⎛ ⎞ ⎛ ⎞= + ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠=

Exterior Girder - One Lane Loaded:

Same as for the positive moment section: DFM1,Ext,Sec2 = 0.9000



e = 1.008 (same as before)

DFM2,Ext,Sec2 =(1.008) (0.7809) = 0.7871

4c. Section 3:

Interior Girder – One Lane Loaded:

( )( ) ( )( )( )

( ) ( )( )

0.10.4 0.3

1, , 3 3

2

24 2

4

0.4 0.3 4

1, , 3 3

0.0614 12

8 107,500 in 118.5 in 50.00"

3, 230,000 in

11.33' 11.33' 3, 230,000 in0.0614 166.3' 12 166.3' 8.5"

gM Int Sec

S

g g

g

g

M Int Sec

KS SDFL Lt

K n I Ae

K

K

DF

⎛ ⎞⎛ ⎞ ⎛ ⎞= + ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= +

= +

=

⎛ ⎞ ⎛ ⎞= + ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

0.1

1, , 3 0.5122M Int SecDF

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

=

-- 78 --


Interior Girder - Two or More Lanes Loaded:

( )( ) ( )

0.10.4 0.3

2, , 3 3

0.10.6 0.2 4

2, , 3 3

2, , 3

0.0759.5 12

11.33' 11.33' 3,230,000 in0.0759.5 166.3' 12 166.3' 8.5"

0.7906

gM Int Sec

S

M Int Sec

M Int Sec

KS SDFL Lt

DF

DF

⎛ ⎞⎛ ⎞ ⎛ ⎞= + ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎜ ⎟= + ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠=


Same as for the positive moment section: DFM1,Ext,Sec3 = 0.9000



e = 1.008 (same as before)

DFM2,Ext,Sec3 =(1.008) (0.7906) = 0.7969 4d. Minimum Exterior Girder Distribution Factor:

,

2

L

Ext Minb

N

ExtL

Nb

X eN

DFN x

= +∑

∑

-- 79 --


One Lane Loaded:

( ) ( )( ) ( ) ( )

1, , 2 2 2

28.33' 25.5 '1 0.48816 2 28.33' 17.0 ' 5.667 '

M Ext MinDF⎡ ⎤⎣ ⎦= + =

⎡ ⎤+ +⎣ ⎦


Two Lanes Loaded:

( ) ( ) ( )

( ) ( ) ( )2, , 2 2 2

28.33' 25.5 ' 13.5 '2 0.82506 2 28.33' 17.0 ' 5.667 '

M Ext MinDF+⎡ ⎤⎣ ⎦= + =

⎡ ⎤+ +⎣ ⎦


Three Lanes Loaded:

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

3, , 2 2 2

28.33' 25.5 ' 13.5 ' 1.5 '3 1.0116 2 28.33' 17.0 ' 5.667 '

M Ext MinDF+ +⎡ ⎤⎣ ⎦= + =

⎡ ⎤+ +⎣ ⎦


Four Lanes Loaded:

( ) ( ) ( ) ( ) ( )

( ) ( ) ( )4, , 2 2 2

28.33' 25.5 ' 13.5 ' 1.5 ' 10.5 '4 1.0456 2 28.33' 17.0 ' 5.667 '

M Ext MinDF+ + + −⎡ ⎤⎣ ⎦= + =

⎡ ⎤+ +⎣ ⎦


Five Lanes Loaded:

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

( ) ( ) ( )5, , 2 2 2

28.33' 25.5 ' 13.5 ' 1.5 ' 10.5 ' 22.5 '5 0.83676 2 28.33' 17.0 ' 5.667 '

M Ext MinDF+ + + − + −⎡ ⎤⎣ ⎦= + =

⎡ ⎤+ +⎣ ⎦


-- 80 --


4d. Moment Distribution Factor Summary

Positive Moment Section # Lanes Loaded Interior Exterior

1 0.4994 0.9000 ≥ 0.5857 2 0.7703 0.7765 ≥ 0.8250 1 3 0.7703 0.7765 ≥ 0.8589 1 0.5061 0.9000 ≥ 0.5857 2 0.7809 0.7871 ≥ 0.8250 2 3 0.7809 0.7871 ≥ 0.8589 1 0.5122 0.9000 ≥ 0.5857 2 0.7906 0.7969 ≥ 0.8250 3 3 0.7906 0.7969 ≥ 0.8589

For Simplicity, take the Moment Distribution Factor as 0.9000 everywhere. Multiplying the live load moments by this distribution factor of 0.9000 yields the table of “nominal” girder moments shown below.

Nominal Girder Moments from Visual Analysis and SAP

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

0.0 0.0 0.0 0.0 0.0 0.08.0 986.5 0.0 1086.2 135.6 457.7

16.1 1868.0 0.0 2064.8 257.4 869.024.1 2645.5 0.0 2935.8 365.5 1233.732.1 3319.1 0.0 3699.2 459.8 1552.040.2 3889.3 0.0 4355.1 540.3 1823.949.2 4404.3 0.0 4959.7 614.2 2073.358.2 4795.1 0.0 5424.7 670.8 2264.366.5 5054.9 0.0 5729.3 707.8 2389.174.8 5202.6 0.0 5912.1 730.0 2464.083.1 5238.6 0.0 5973.0 737.4 2489.083.1 5238.5 0.0 5973.0 737.4 2489.091.5 5202.6 0.0 5911.6 729.9 2463.899.8 5055.0 0.0 5728.3 707.7 2388.7

108.1 4795.7 0.0 5423.3 670.6 2263.7108.1 4795.2 0.0 5423.3 670.6 2263.7117.1 4404.5 0.0 4957.8 614.0 2072.4126.1 3890.2 0.0 4352.6 540.0 1822.9126.1 3889.5 0.0 4352.6 540.0 1822.9134.2 3319.4 0.0 3696.9 459.5 1551.1142.2 2645.7 0.0 2933.8 365.2 1232.9150.2 1868.1 0.0 2063.3 257.2 868.3158.3 986.6 0.0 1085.3 135.5 457.3166.3 1.1 0.0 0.0 0.0 0.0

Nominal Moments

-- 81 --


5. FACTORED 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 for Fatigue = 15%) 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 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, η, was applied as is shown below.

i i iQ Qη γ= ∑ η is taken as the product of ηD, ηR, and ηI, and is taken as not less than 0.95. For this example, ηD, ηR, and ηI are taken as 1.00. Using these load combinations, the shear and moment envelopes shown on the following pages were developed.

-- 82 --


Station LL+ LL- DC1 DC2 DW Total + Total -(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 0.08.0 1726.3 0.0 1357.7 169.5 686.6 3940.1 0.0

16.1 3268.9 0.0 2581.0 321.8 1303.4 7475.1 0.024.1 4629.6 0.0 3669.7 456.9 1850.6 10606.8 0.032.1 5808.5 0.0 4624.0 574.7 2328.1 13335.3 0.040.2 6806.2 0.0 5443.9 675.4 2735.8 15661.3 0.049.2 7707.5 0.0 6199.7 767.8 3109.9 17784.8 0.058.2 8391.4 0.0 6780.8 838.5 3396.5 19407.2 0.066.5 8846.1 0.0 7161.6 884.7 3583.7 20476.2 0.074.8 9104.5 0.0 7390.1 912.5 3696.0 21103.1 0.083.1 9167.5 0.0 7466.3 921.7 3733.5 21289.0 0.083.1 9167.4 0.0 7466.3 921.7 3733.5 21288.9 0.091.5 9104.6 0.0 7389.5 912.4 3695.7 21102.2 0.099.8 8846.3 0.0 7160.4 884.6 3583.1 20474.4 0.0

108.1 8392.4 0.0 6779.1 838.3 3395.6 19405.3 0.0108.1 8391.6 0.0 6779.1 838.3 3395.6 19404.5 0.0117.1 7707.8 0.0 6197.3 767.5 3108.7 17781.2 0.0126.1 6807.8 0.0 5440.8 675.0 2734.3 15657.9 0.0126.1 6806.6 0.0 5440.8 675.0 2734.3 15656.7 0.0134.2 5808.9 0.0 4621.1 574.4 2326.6 13331.0 0.0142.2 4630.0 0.0 3667.2 456.5 1849.3 10603.1 0.0150.2 3269.2 0.0 2579.1 321.5 1302.5 7472.2 0.0158.3 1726.5 0.0 1356.7 169.4 686.0 3938.5 0.0166.3 1.9 0.0 0.0 0.0 0.0 1.9 0.0

Strength I Moments

Station DC1 DC2 DW Total + Total -(ft) (k-ft) (k-ft) (k-ft) (k-ft) (k-ft)

0.0 0.0 0.0 0.0 0.0 0.07.2 1629.3 203.4 686.6 2519.2 0.0

14.5 3097.2 386.1 1303.4 4786.7 0.021.7 4403.7 548.2 1850.6 6802.5 0.028.9 5548.8 689.7 2328.1 8566.6 0.036.2 6532.6 810.5 2735.8 10078.9 0.043.4 7439.6 921.3 3109.9 11470.8 0.050.6 8137.0 1006.2 3396.5 12539.7 0.057.8 8593.9 1061.7 3583.7 13239.3 0.065.1 8868.1 1095.0 3696.0 13659.1 0.072.3 8959.5 1106.1 3733.5 13799.1 0.079.5 8959.5 1106.1 3733.5 13799.1 0.086.8 8867.4 1094.9 3695.7 13658.0 0.094.0 8592.5 1061.5 3583.1 13237.1 0.0

101.2 8134.9 1005.9 3395.6 12536.4 0.0108.5 8134.9 1005.9 3395.6 12536.4 0.0115.7 7436.7 920.9 3108.7 11466.3 0.0122.9 6529.0 810.0 2734.3 10073.3 0.0130.1 6529.0 810.0 2734.3 10073.3 0.0137.4 5545.4 689.3 2326.6 8561.2 0.0144.6 4400.7 547.9 1849.3 6797.8 0.0151.8 3094.9 385.9 1302.5 4783.2 0.0159.1 1628.0 203.2 686.0 2517.2 0.0166.3 0.0 0.0 0.0 0.0 0.0

Strength IV Moments

-- 83 --


Station LL+ LL- DC1 DC2 DW Total + Total -(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 0.07.2 1282.4 0.0 1086.2 135.6 457.7 2961.9 0.0

14.5 2428.3 0.0 2064.8 257.4 869.0 5619.5 0.021.7 3439.2 0.0 2935.8 365.5 1233.7 7974.2 0.028.9 4314.9 0.0 3699.2 459.8 1552.0 10026.0 0.036.2 5056.0 0.0 4355.1 540.3 1823.9 11775.3 0.043.4 5725.6 0.0 4959.7 614.2 2073.3 13372.8 0.050.6 6233.6 0.0 5424.7 670.8 2264.3 14593.4 0.057.8 6571.4 0.0 5729.3 707.8 2389.1 15397.6 0.065.1 6763.4 0.0 5912.1 730.0 2464.0 15869.4 0.072.3 6810.2 0.0 5973.0 737.4 2489.0 16009.6 0.079.5 6810.1 0.0 5973.0 737.4 2489.0 16009.5 0.086.8 6763.4 0.0 5911.6 729.9 2463.8 15868.7 0.094.0 6571.5 0.0 5728.3 707.7 2388.7 15396.3 0.0

101.2 6234.4 0.0 5423.3 670.6 2263.7 14591.9 0.0108.5 6233.8 0.0 5423.3 670.6 2263.7 14591.3 0.0115.7 5725.8 0.0 4957.8 614.0 2072.4 13370.0 0.0122.9 5057.2 0.0 4352.6 540.0 1822.9 11772.7 0.0130.1 5056.3 0.0 4352.6 540.0 1822.9 11771.8 0.0137.4 4315.2 0.0 3696.9 459.5 1551.1 10022.6 0.0144.6 3439.4 0.0 2933.8 365.2 1232.9 7971.3 0.0151.8 2428.5 0.0 2063.3 257.2 868.3 5617.3 0.0159.1 1282.5 0.0 1085.3 135.5 457.3 2960.7 0.0166.3 1.4 0.0 0.0 0.0 0.0 1.4 0.0

Service II Moments

-- 84 --


Strength Limit Moment Envelopes

0

5,000

10,000

15,000

20,000

25,000

0 30 60 90 120 150 180

Station (ft)

Mom

ent (

kip-

ft)

Strength I

Strength IV

Max (@ 83.14') = 21289k-ft

Service II Moment Envelope

0

2,500

5,000

7,500

10,000

12,500

15,000

17,500

0 30 60 90 120 150 180

Station (ft)

Mom

ent (

kip-

ft)

Max (@ 83.14') = 16,010k-ft

-- 85 --

-- 86 --

Single-Span Truss Bridge Example AASHTO-LRFD 2007 ODOT LRFD Short Course - Steel Created July 2007: Page 1 of 17

SINGLE-SPAN TRUSS BRIDGE DESIGN EXAMPLE 1. PROBLEM STATEMENT AND ASSUMPTIONS: Consider the truss bridge shown in Figure 1 below. The truss is simply supported with a span length of 112’–0” and a width (c-c of the trusses) of 19’–6”. The truss is made up of 7 panels that are each 16’-0” in length. Floor beams span between the truss panel points perpendicular to traffic and support stringers that span 16’-0” in the direction of traffic. Finally, the noncomposite W10 x 88 stringers support a 6” thick reinforced concrete deck. The simply supported stringers (6 across in each panel) are spaced at 3’ - 6” laterally. 1) Determine maximum and minimum axial forces in members 1-2, 1-4, 9-11, 9-10, and 10-13

due to an HL-93 Loading. 2) Determine the maximum moment in the stringer members due to the HL-93 Loading The entire truss superstructure is made up of W14 x 109 members except for the bottom chord, which is made up of MC 12 x 35 members. You may assume that the trucks drive down the center of the bridge (they really do, by the way) and as a result, the truck loads are approximately equally distributed between the trusses. To be on the safe side, however, assume that each truss carries 75% of the single lane. Model the truss as a determinate structure with pinned joints even though the actual truss has very few joints that are truly pinned. You may use a computer program for your truss analysis if you wish. I would suggest that you use SAP2000, Visual Analysis, or another similar FE package to model the truss. Disregard the lower limit of L = 20’ on the span length for computing distribution factors for the stringer members. Think about what is appropriate for the multiple presence factor.

Figure 1 - Tyler Road Bridge, Delaware County, OH

-- 87 --


Figure 2 - Truss Layout

18' - 0" Clear Roadway

19' - 6" cc Trusses

6, W10 x 88 Stringers @ 3'-6" cc

6" Thick Reinforced Concrete Deck

Figure 3 - Truss Cross Section

-- 88 --


Compute the Maximum and Minimum Forces in Critical Members of the Truss: The following Influence Lines were obtained from SAP 2000:

-- 89 --


-- 90 --


-- 91 --


Consider Member 1-2 of the Truss:

Tandem: ( )kip kip kipkipkip

96' 4 '(25 ) (25 ) 1.415 69.2896'

⎡ ⎤−⎛ ⎞+ − = −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

Truck: ( )kip kip kip kipkipkip

16' 14' 96' 14'(8 ) (32 ) (32 ) 1.415 85.3816' 96'

⎡ ⎤− −⎛ ⎞ ⎛ ⎞+ + − = −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

Lane: ( )( )( ) kipkip kip1

ft kip20.640 1.415 (112') 50.71− = − Combining the HL-93 Components with impact applied appropriately: (IM)(Tandem) + Lane: ( )( ) ( )kip kip kip1.33 69.28 50.71 142.9− + − = −

(IM)(Truck) + Lane: ( )( ) ( )kip kip kip1.33 85.38 50.71 164.3− + − = − GOVERNS

Apply the Truss Distribution Factor: Assume that each truss carries 75% of the HL-93 load effect P1-2 = -123.2kip

-- 92 --




96' 4 '(25 ) (25 ) 1.127 55.1896'

⎡ ⎤−⎛ ⎞+ =⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦


16 ' 14' 96' 14'(8 ) (32 ) (32 ) 1.127 68.0016' 96'

⎡ ⎤− −⎛ ⎞ ⎛ ⎞+ + =⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦


ft kip20.640 1.127 (112') 40.39= Combining the HL-93 Components with impact applied appropriately: (IM)(Tandem) + Lane: ( )( ) ( )kip kip kip1.33 55.18 40.39 113.8+ =

(IM)(Truck) + Lane: ( )( ) ( )kip kip kip1.33 68.00 40.39 130.8+ = GOVERNS

Apply the Truss Distribution Factor: Assume that each truss carries 75% of the HL-93 load effect P1-4 = 98.12kip

-- 93 --




64' 4 '(25 ) (25 ) 2.254 109.264'

⎡ ⎤−⎛ ⎞+ − = −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦


48' 14' 64' 14'(8 ) (32 ) (32 ) 2.254 141.348' 64'

⎡ ⎤− −⎛ ⎞ ⎛ ⎞+ + − = −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦




Apply the Truss Distribution Factor: Assume that each truss carries 75% of the HL-93 load effect P9-11 = -201.5kip

-- 94 --


Consider Member 9-10 of the Truss: Member 9-10 of the truss is a zero force member. It may see some bending moment due to its rigid connection to the floor beam but it will not experience a net axial force due to live load. P9-10 = 0.000kip

-- 95 --



Tandem: ( )kip kip kipkip

kip(25 ) (25 ) 1.972 98.60⎡ ⎤+ =⎣ ⎦


48' 12'(8 ) (32 ) (32 ) 1.972 138.048'

⎡ ⎤−⎛ ⎞ + + =⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

Lane: ( )( ) ( ) kipkip kip 1

ft kip 20.640 1.972 (96') (16') 80.77⎡ ⎤+ =⎣ ⎦ Combining the HL-93 Components with impact applied appropriately: (IM)(Tandem) + Lane: ( )( ) ( )kip kip kip1.33 98.60 80.77 211.9+ =


Apply the Truss Distribution Factor: Assume that each truss carries 75% of the HL-93 load effect P10-13 = 198.2kip

-- 96 --


Consider Member 10-11 of the Truss – Tensile Force:

8kip 32kip 32kip

25kip 25kip

0.640kip/ft

0.5128kip/kip

Tandem:

Truck:

Lane:

IL Mem 10-11:

0.5124kip/kip

1 4 7 10 13 16 19 21


48' 4 '(25 ) (25 ) 0.5128 24.5748'

⎡ ⎤−⎛ ⎞+ =⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦


48' 28' 48' 14'(8 ) (32 ) (32 ) 0.5128 29.7448' 48'

⎡ ⎤− −⎛ ⎞ ⎛ ⎞+ + =⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦


ft kip20.640 0.5128 (56') 9.189= Combining the HL-93 Components with impact applied appropriately: (IM)(Tandem) + Lane: ( )( ) ( )kip kip kip1.33 24.57 9.189 41.87+ =


Apply the Truss Distribution Factor: Assume that each truss carries 75% of the HL-93 load effect P+

10-11 = 36.56kip

-- 97 --


Consider Member 10-11 of the Truss – Compressive Force:

8kip32kip32kip

25kip 25kip

0.640kip/ft

0.5128kip/kip

Tandem:

Truck:

Lane:

IL Mem 10-11:

0.5124kip/kip

1 4 7 10 13 16 19 21


48' 4 '(25 ) (25 ) 0.5124 24.5548'

⎡ ⎤−⎛ ⎞+ − = −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦


48' 28' 48' 14'(8 ) (32 ) (32 ) 0.5124 29.7248' 48'

⎡ ⎤− −⎛ ⎞ ⎛ ⎞+ + − = −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦




Apply the Truss Distribution Factor: Assume that each truss carries 75% of the HL-93 load effect P-

10-11 = -36.53kip

-- 98 --


Member Force Summary:

Member Max Tension Max Compression 1-2 0.000kip 123.2kip 1-4 98.12kip 0.000kip

9-11 0.000kip 201.5kip 9-10 0.000kip 0.000kip

10-13 198.2kip 0.000kip 10-11 36.56kip 36.53kip

-- 99 --


Compute the Moment Distribution Factor for the Stringers in the Floor System: Interior Girder –

One Lane Loaded:

0.10.4 0.3

1, 3

2

4 2 2

4

0.10.4 0.3 4

1, 3

1,

0.0614 12

( )

(8)(534 in (25.9 in )(8.40") )

18,890 in

3.5 ' 3.5 ' 18,890 in0.06

14 16 ' 12(16 ')(6.0")

gM Int

s

g g

g

g

M Int

M Int

KS SDF

L Lt

K n I Ae

K

K

DF

DF

= +

= +

= +

=

= +

=

⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

0.3965


The bridge is designed for a single traffic lane. Exterior Girder –

One Lane Loaded:

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

( )

1,

1.75')0.2500

(3.50 ') 0.2500

/ 2 (

M Ext

P

DF

PR = =

=

The Multiple Presence Factor would generally be applied but in this case, there is only a single design lane so it is not used.


The bridge is designed for a single traffic lane.

Minimum Exterior Girder Distribution Factor:

-- 100 --


,

2

L

Ext Minb

N

ExtL

Nb

X eN

DFN

x= +

∑

∑

One Lane Loaded:

8'-9"

P/2 P/2

3'-0"2'-0"

1'-9"

5'-3"

3'-0"

4'-0"

1, , 2 2 2

1 (4.00 ')(8.75')0.3299

6 (2) (8.75') (5.25') (1.75')M Ext MinDF = + =

+⎡ ⎤+⎣ ⎦

The Multiple Presence Factor would generally be applied but in this case, there is only a single design lane so it is not used.

Moment Distribution Factor Summary: Interior Stringer: DFM1,Int = 0.3965 Exterior Stringer (Lever Rule): DFM1, Ext = 0.2500 Exterior Stringer (Minimum): DFM1, Ext = 0.3299

For simplicity, take the moment distribution factor as 0.3965 for all stringers.

-- 101 --


Compute the Maximum Bending Moment in the Stringers of the Floor System:

32kip

25kip 25kip

0.640kip/ft

Tandem:

Truck:

Lane:

IL Moment@ CL Stringer

4.00k-ft/kip

4 spaces @ 4'-0" = 16'-0"

Tandem: ( )kip kip k-ftk-ftkip

8' 4 '(25 ) (25 ) 4.00 150.08'

⎡ ⎤−⎛ ⎞+ =⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

Truck: ( )kip k-ftk-ft

kip(32 ) 4.00 128.0= Lane: ( )( )( ) k-ftkip 1 k-ft

ft kip20.640 4.00 (16') 20.48= In this case, since the axle spacing is substantial relative to the beam length, we should consider the more general approach for computing maximum moment. For two equal point loads, P, separated by a distance, a, the maximum bending moment in a simply supported span is: when a < 0.5858L,

2

2 2MaxP aM LL⎛ ⎞= −⎜ ⎟⎝ ⎠

when a ≥ 0.5858L

4Max

PLM =

-- 102 --


Tandem: 2kip

k-ft(25 ) 4'16' 153.1(2)(16') 2MaxM ⎛ ⎞= − =⎜ ⎟

⎝ ⎠ (an increase of 2.067%)

Truck: kip

k-ft(32 )(16') 128.0(4)MaxM = = (no change)

Combining the HL-93 Components with impact applied appropriately: (IM)(Tandem) + Lane: ( )( ) ( )k-ft k-ft k-ft1.33 153.1 20.48 224.1+ = GOVERNS

(IM)(Truck) + Lane: ( )( ) ( )k-ft k-ft k-ft1.33 128.0 20.48 190.7+ =

Apply the Stringer Distribution Factor: Each stringer carries 0.3965 lanes of the HL-93 loading MStringer = 88.86k-ft

-- 103 --

-- 104 --

ODOT-LRFD Short Course - Steel AASHTO Tension Member Example #1

AASHTO-LRFD 2007 Created July 2007: Page 1 of 1

AASHTO Tension Member Example #1: Problem: A tension member is made up from a bar of M270-50 material that is 6” wide and 1” thick. It is bolted at its ends by six, 7/8” diameter bolts arranged in two staggered rows as is shown below. If the governing factored load, Pu, is 200kip, determine whether or not the member is adequate. The member is 4’-0” long. Solution: Check Minimum Slenderness Ratio:

3 2

0.2887"12 12

minmin

I bt trA bt

= = = =

inft(4 ' 0")(12 ) 166.2

0.2887"min

Lr

−= =

Since 140 < L / rmin <200, the slenderness is ok so long as the member is not subjected to stress reversals. Compute the Design Strength: Gross Section Yielding: Pn = Fy Ag = (50ksi)(6”)(1”) = 300.0kip φPn = (0.95)(300.0kip) = 285.0kip Net Section Fracture: Pn = Fu Ae = Fu U An

( ) ( )2

7 18 8

2

(1.5")6" (2) " " 1"4(3.0")

4.188 in

nA⎡ ⎤⎛ ⎞

= − + +⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

=

Pn = (65ksi)(1.00)(4.188 in2) = 272.2kip

φPn = (0.80)(272.2kip) = 217.8kip

3" 1"

1.5"

3"3"

3"3"

3"

1.5" 1.5"

NSF Governs - φPn = 218kip

Since Pu < φPn (200kip < 218kip) the member is adequate.

U is 1.00 here because the section is composed of a single element that is connected. Therefore the load is “transmitted directly to each of the elements within the cross section.”

-- 105 --



AASHTO Tension Member Example #2: Problem: A C12x30 is used as a tension member (L = 8’-6”) as is shown in the sketch below. The channel is made of M270-36 material and is attached to the gusset plate with 7/8” diameter bolts. Calculate the design tensile capacity, φPn, of the member considering the failure modes of gross section yielding and net section fracture. Solution: Check Minimum Slenderness Ratio: rmin = 0.762” (from the AISC Manual)

inft(8.5 ')(12 ) 133.9

0.762"min

Lr

= =

Since L/rmin < 140, the slenderness is ok. Compute the Design Strength: Gross Section Yielding: Pn = Fy Ag = (36ksi)(8.81 in2) = 317.2kip φPn = (0.95)(317.2kip) = 301.3kip Net Section Fracture: Pn = Fu Ae = Fu U An

A

3"3" 6"

Pu

A

Section A-A

C12 x 30

( ) ( )( )2 27 1

8 88.81 in (2) " " 0.510" 7.790 innA = − + = U = 0.85 since there are ≥ 3 fasteners in the direction of stress Pn = (58ksi)(0.85)(7.790 in2) = 384.0kip

φPn = (0.80)(384.0kip) = 307.2kip

Gross Section Yielding Governs - φPn = 301kip

-- 106 --



Side Note:

Note that if the AISC shear lag provisions were used that Case 2 from AISC Table D3.1 would apply:

0.674"1 1 0.92519.00"

xUL

= − = − = …. for net section fracture, φPn = 334.3kip

In this case, however, the design strength is unaffected since gross yielding governs.

-- 107 --



AASHTO Tension Member Example #3: Problem: Determine the design strength of the W10x60 member of M270-50 steel. As is shown, the member is connected to two gusset plates – one on each flange. The end connection has two lines of 3/4” diameter bolts in each flange - five in each line.

A

ASection A-A

5 spaces @ 3”

W10 x 60

Gusset Plates

Solution: Check Minimum Slenderness Ratio:

rmin = 2.57” (from the AISC Manual)

min

1402.57"

L Lr

= ≤ this is satisfied so long as L ≤ 359.8” = 29’-113/4”

Compute the Design Strength: Gross Section Yielding: Pn = Fy Ag = (50ksi)(17.6 in2) = 880.0kip φPn = (0.95)(880.0kip) = 836.0kip

-- 108 --



Net Section Fracture: Pn = Fu Ae = Fu U An ( ) ( )( )2 23 1

4 817.6 in (4) " " 0.680" 15.22 innA = − + =

Check ?

23fb d≥ … ( ) ( )( )

?2

310.1" 10.2"≥ OK U = 0.90 since bf > 2/3d and there are ≥ 3 fasteners in the direction of stress.

Pn = (65ksi)(0.90)(15.22 in2) = 890.4kip

φPn = (0.80)(890.4kip) = 712.3kip

Net Section Fracture Governs - φPn =712kip

Side Note:

Note that if the AISC shear lag provisions were used that Case 7a from AISC Table D3.1 would apply:

Check ?

23fb d≥

( ) ( )( )?

2310.1" 10.2"≥ OK

∴ U = 0.90 Alternatively, Table D3.1 Case 2 can be applied:

0.884"1 1 0.926312.0"

xUL

= − = − =

The value of U = 0.9263 can be used. Pn = (65ksi)(0.9263)(15.22 in2) = 916.4kip

φPn = (0.80)(916.4kip) = 733.1kip

x x

The connection eccentricity x is taken as the distance from the faying surface to the CG of a WT5x30.

Since Net Section Fracture governs the capacity of this member, the overall design strength of the member would be increased to 733kip.

-- 109 --



AASHTO Tension Member Example #4: Problem: An L6x4x1/2, M270-36, is welded to a gusset plate. The long leg of the angle is attached using two, 8” long fillet welds. Compute the strength of the angle in tension. Solution: Check Minimum Slenderness Ratio:

rmin = rz = 0.864” (from the AISC Manual)

2400.864"min

L Lr

= ≤

this is satisfied so long as L ≤ 207.4” = 17’-33/8”

Compute the Design Strength: Gross Section Yielding: Pn = Fy Ag = (36ksi)(4.75 in2) = 171.0kip φPn = (0.95)(171.0kip) = 162.5kip Net Section Fracture: Pn = Fu Ae = Fu U An Lacking other guidance, AISC Table D3.1 Case 2 will be applied:

0.981"1 1 0.87748.0"

xUL

= − = − =

Pn = (58ksi) (0.8774)(4.75 in2) = 241.7kip φPn = (0.80)(241.7kip) = 193.4kip

Gross Section Yielding Governs - φPn =163kip

-- 110 --

ODOT-LRFD Short Course - Steel AASHTO Compression Member Example #1


AASHTO Compression Member Example #1: Problem: Compute the design compressive strength of a W14x74 made of M270-50 steel. The column has a length of 20 ft and can be treated as pinned-pinned. Solution: Check Local Buckling:

Flange: ?

2f

f y

b Ekt F≤ λf = 6.41 (Tabulated)

? 29,0006.41 0.56 13.550

≤ = OK

Web: ?

w y

h Ekt F≤ λw = 25.4 (Tabulated)

? 29,00025.4 1.49 35.950

≤ = OK

Compute Flexural Buckling Capacity: Slenderness Ratios:

inft

inft

(1.00)(20 ')(12 ) 39.74 120 OK6.04"

(1.00)(20 ')(12 ) 96.77 120 OK2.48"

x

y

KLr

KLr

⎛ ⎞ = = <⎜ ⎟⎝ ⎠

⎛ ⎞ = = <⎜ ⎟⎝ ⎠

Since the effective slenderness ratio is larger for the y axis than the x axis, y-axis buckling will govern. 2 2 ksi

ksi

96.77 50 1.63629,000

y

y

FKLr E

⎛ ⎞⎛ ⎞ ⎛ ⎞λ = = =⎜ ⎟⎜ ⎟ ⎜ ⎟π π⎝ ⎠ ⎝ ⎠ ⎝ ⎠

(6.9.4.1-3)

Since λ ≤ 2.25, Inelastic Buckling Governs

( )( )( )( )1.636 ksi 2 kip0.66 0.66 50 21.8 in 549.6n y sP F Aλ= = = (6.9.4.1-1)

φPn = (0.90)(549.6kip) φPn = 495kip

-- 111 --



AASHTO Compression Member Example #2 Problem: Compute the axial compressive design strength based on flexural buckling (no torsional or flexural-torsional buckling). Assume that the cross-sectional elements are connected such that the built-up shape is fully effective. All plates are 4” thick. Solution: Compute Section Properties:

Ir =A

32

12xbhI Ad= +∑

( )( ) ( ) ( ) ( )( )33 2 4" 30"- 2 4"36" 4" 30" 4"2 36" 4" 212 2 2 12

⎡ ⎤⎡ ⎤ ×⎛ ⎞ ⎢ ⎥= + × − +⎢ ⎥⎜ ⎟ ⎢ ⎥⎝ ⎠⎢ ⎥⎣ ⎦ ⎣ ⎦

456,150 in=

3

2

12yhbI Ad= +∑

( )( ) ( )( )( ) ( )( )( )33 230"- 2 4" 4"4" 36" 36" 4"2 2 30"- 2 4" 4"

12 12 2 2

⎡ ⎤⎡ ⎤ × ⎛ ⎞⎢ ⎥= + + × × −⎢ ⎥ ⎜ ⎟⎢ ⎥⎝ ⎠⎢ ⎥⎣ ⎦ ⎣ ⎦

476,390 in=

( ) ( )( )( ) 22 36" 4" + 2 30"- 2 4" 4" 464.0 insA = × × × =

Since 4 456,154.67 in 76,394.67 inx yI I= < = , x-axis buckling controls

4

256,150 in 11.0 in464.0 in

xx

s

Ir =A

= =

-- 112 --



Check Local Buckling (Section 6.9.4.2):

( )36" 2 4"7.00

4"bt

−= =

?

y

b Ekt F≤ (6.9.4.2-1)

ksi

ksi29,0007.00 1.40 33.72

50≤ = OK

Calculate the Nominal Compressive Strength (Section E3 page 16.1-33):

Slenderness Ratios: KLr

where:

0.8K = (Section 4.6.2.5)

inft40 ft 12 480"x yL L= = × =

( )( )( )

0.8 480 in34.91

11.0 inx

x

KLr

= =

2 2 ksi

ksi34.91 50 0.2129

29,000y

s

FKLr E

⎛ ⎞ ⎛ ⎞λ = = =⎜ ⎟ ⎜ ⎟π π⎝ ⎠⎝ ⎠ (6.9.4.1-3)

Since 2.25λ ≤ , Inelastic Flexural Buckling Governs

( )( )0.2129 ksi 2 kip0.66 0.66 50 464.0 in 21,240n y sP F Aλ= = = (6.9.4.1-1)

( )( )kip kip

c 0.90 21,240 19,110nPφ = =

-- 113 --

ODOT-LRFD Short Course - Steel AASHTO Compression Member Example #3 v2


AASHTO Compression Member Example #3: Problem: Determine the effective length factor, K, for column AB in the frame shown below. Column AB is a W10x88 made of A992 steel. W16x36 beams frame into joint A and W16x77 beams frame into joint B. The frame is unbraced and all connections are rigid. Consider only buckling in the plane of the page about the sections’ strong axes.

W16 x 36

L=24'

W16 x 77

L=24'

W10 x 88

L=14'

A

B

4 @ 24'

8 @

14'

A

B

Solution:

-- 114 --

ODOT-LRFD Short Course - Steel AASHTO Compression Member Example #3 v2


Determine the Effective Length Factor:

4

4

(2)(534 in )(14 ')

3.0652 (2)(448 in )3 (24 ')

CA

G

ILGIL

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠= = =⎛ ⎞ ⎛ ⎞⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

∑

∑

4

4

(2)(534 in )(14 ')

1.2372 (2)(1,110 in )3 (24 ')

CB

G

ILGIL

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠= = =⎛ ⎞ ⎛ ⎞⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

∑

∑

For unbraced frames:

5.75.7)(0.46.1

+++++

=BA

BABA

GGGGGGK

(1.6)(3.065)(1.237) (4.0)(3.065 1.237) 7.5 1.615(3.065 1.237 7.5)

K + + += =

+ +

The factor of 2/3 appears in the denominator to reflect the fact that the far ends of the girders are “fixed” connections.

GB K GA

-- 115 --



AASHTO Compression Member Example #4: Problem: Check to see if a built-up section will work to resist a factored load of Pu = 209kip. The column is to be fabricated from two C10x15.3as is shown in the figure to the right. The steel is M270-36 and the effective length is 20’ with respect to all axes.

If the column is adequate, determine the thickness of the battens.The battens are 8” long and 6” deep and are also made of M270-36 steel.

Solution: Check Local Buckling:

Flange: 2.60" 5.960.436"

f

f

bbt t= = =

ksi?

ksi

29,0000.56 0.56 15.8936y

b Et F≤ = = OK

Web: 2 10" (2)(0.436") 38.03

0.240"f

w

d tbt t

− −= = =

ksi?

ksi

29,0001.49 1.49 42.2936y

b Et F≤ = = OK

9"

a

x

y

10"

Compute Section Properties: As = (2) (4.48 in2) = 8.96 in2 IX = (2) (Ix) = (2)(67.3in4) = 134.6 in4

2

4 2 49"(2) 2.27 in (4.48 in ) 0.634" 138.5 in2YI

⎡ ⎤⎛ ⎞= + − =⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

4

2

134.6 in 3.88"8.96 in

XX

IrA

= = = 4

2

138.5 in 3.93"8.96 in

YY

IrA

= = =

-- 116 --



Slenderness Ratios:

inft(20 ')(12 ) 61.86

3.88"X

KLr

⎛ ⎞⎛ ⎞ = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

inft(20 ')(12 ) 61.07

3.93"Y

KLr

⎛ ⎞⎛ ⎞ = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

It appears as though X axis buckling will govern but since the battens will be subjected to shear if the section buckles about its Y axis, this slenderness ratio must be modified.

Batten Spacing:

34i

max

KLa rr

⎛ ⎞⎛ ⎞≤ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

ri = ry = 0.711” (for one channel) max X

KL KLr r

⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

( )( )(0.711") 0.75 61.86a ≤

32.98"a ≤ use 9 battens @ a = 30”

Modified Slenderness Ratio – Y-axis Buckling:

The modified slenderness ratio is calculated as,

22 2

20.82(1 )m o ib

KL KL ar r r

αα

⎛ ⎞⎛ ⎞ ⎛ ⎞= + ⎜ ⎟⎜ ⎟ ⎜ ⎟ +⎝ ⎠ ⎝ ⎠ ⎝ ⎠ (6.9.4.3.1-1)

rib = 0.711” h = 9” – (2)(0.634”) = 7.73”

7.732" 5.442 (2)(0.711")ib

hr

α = = =

( ) ( )22

2

2

(5.44) 30"61.07 0.82 71.700.711"1 (5.44)m

KLr

⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟= + =⎜ ⎟ ⎜ ⎟⎜ ⎟+⎝ ⎠ ⎝ ⎠⎝ ⎠

Now we can see that after the Y axis slenderness ratio is modified, Y axis buckling actually governs over X axis buckling.

-- 117 --



Column Design Capacity:

2 2 ksi

ksi

71.70 36 0.646629,000

y

y

FKLr E

⎛ ⎞⎛ ⎞ ⎛ ⎞λ = = =⎜ ⎟⎜ ⎟ ⎜ ⎟π π⎝ ⎠ ⎝ ⎠ ⎝ ⎠

(6.9.4.1-3)

Since λ ≤ 2.25, Inelastic Buckling Governs ( )( )( )( )0.6466 ksi 2 kip0.66 0.658 36 8.96 in 246.1n y sP F Aλ= = = (6.9.4.1-1)

φPn = (0.90)(246.1kip) φPn = 221kip Since φPn > ΣγQ, the column is adequate.

Batten Design: Assume that there are inflection points half way between the battens and design for a shear equal to 2% of the compressive design strength (AISC Section E6. Pg 16.1-39)

2Mu,Batten

2.21 kip

2.21 kip

Vu = (0.02)(221kip) = 4.42kip

kipkip

channel4.42 2.21

2=

ΣM Mu,Batten = 33.15k-in

3(6") 1812Batten

tI t= =

18 63Batten

tS t= =

for first yield, My = Fy S Let φFy S ≥ Mu,Batten

k-in

ksi 3

33.15(1.00)(36 )(6 in )

t ≥

t ≥ 0.153” use t = 5/16” (Min Thickness) use PL6 x 8 x 5/16 Battens

-- 118 --



AASHTO Compression Member Example #5: Problem: Find the design strength of a WT15x146 made of M270-50 steel. KL = 24’ for buckling in all directions. Use the provisions in the AISC Specification to determine the Flexural-Torsional Buckling strength of the column. Solution:

Check Local Buckling:

Flange: 15.3" 4.142 (2)(1.85")

f

f

bbt t= = =

ksi?

ksi

29,0000.56 13.550y

b Ekt F≤ = = OK

Web: 15.7w

b ht t= = (Tabulated)

ksi?

ksi

29,0000.75 18.150y

b Ekt F≤ = = OK

Calculate the buckling load for Flexural Buckling about the X-Axis:

( )( ) 22 ksiinft

ksi

24 ' 12 50 0.7219(4.48")( ) 29,000

yX

X

FKLr E

⎛ ⎞ ⎛ ⎞⎛ ⎞λ = = =⎜ ⎟ ⎜ ⎟⎜ ⎟π π⎝ ⎠ ⎝ ⎠⎝ ⎠

Since λX < 2.25, Inelastic Buckling Governs

( )( )( )( )0.7219 ksi 2 kip0.66 50 42.9 in 1,586nP = = (6.9.4.1-1)

-- 119 --



Calculate the Critical Stress for Flexural-Torsional Buckling about the Y-axis:

( ), , , ,

2, ,

41 1

2cr Y cr Z cr Y cr Z

crft

cr Y cr Z

F F F F HF

H F F

⎛ ⎞+ ⎜ ⎟= − −⎜ ⎟+⎝ ⎠

(AISC E4-2)

( )( ) 22 ksiinft

ksi

24 ' 12 50 1.131(3.58")( ) 29,000

yY

Y

FKLr E

⎛ ⎞ ⎛ ⎞⎛ ⎞λ = = =⎜ ⎟ ⎜ ⎟⎜ ⎟π π⎝ ⎠ ⎝ ⎠⎝ ⎠

Since λY < 2.25, Inelastic Buckling Governs

( )( )( )1.131 ksi ksi

, 0.66 50 31.15ncr Y

s

PFA

= = = (6.9.4.1-1)

2 2 2 x y

o o og

I Ir x y

A+

= + + …… 1.85"3.62" 2.695"2oy = − = (AISC E4-7)

( )4 4

22 2 22

(861 in 549 in )(0.00) 2.695 40.13 in42.9 inor +

= + + =

ksi 4

ksi, 2 2 2

(11,200 )(37.5 in ) 244.0(42.9 in )(40.13 in )cr Z

o

GJFAr

= = = (AISC E4-3)

2 2 2 2

2 2

(0.000") (2.695")1 1 0.819040.13 in

o o

o

x yHr+ +

= − = − = (AISC E4-8)

( )ksi ksi ksi ksi

ksi2ksi ksi

31.15 244.0 (4)(31.15 )(244.0 )(0.819)1 1 30.37(2)(0.819) 31.15 244.1

crftF⎛ ⎞⎛ ⎞+ ⎜ ⎟= − − =⎜ ⎟⎜ ⎟⎝ ⎠ +⎝ ⎠

(AISC E4-2)

Pn = AsFcrft = (42.9 in2)(30.37ksi) = 1,303kip Since 1,303kip < 1,586kip, Flexural-Torsional Buckling Governs φPn = (0.90)(1,303kip) = 1,170kip

-- 120 --



AASHTO Compression Member Example #6: Problem: Find the design strength of a C12x30 made of A36 steel. KLy = 7’ and KLx = KLz = 14’. x

y

Solution: Check Local Buckling:

Flange: 3.17" 6.3270.501"

f

f

bbt t= = =

ksi?

ksi

29,0000.56 15.8936y

b Ekt F≤ = = OK

Web: 2 12.0" (2)(0.501") 21.56

0.510"f

w w

d tb ht t t

− −= = = =

ksi?

ksi

29,0001.49 42.2936y

b Ekt F≤ = = OK

Since both the flange and the web are non-slender, local buckling is OK.

Buckling Strength: Note that the axes of the channel are not arranged properly for the equations in the AISC Specification. These axes need to be rearranged so that the y axis is the axis of symmetry.

Using this modified set of axes, note that KLx = 7’ and KLy = KLz = 14’.

-- 121 --



Calculate the buckling load for Flexural Buckling about the X-Axis:

( )( ) 22 ksiinft

ksi

7 ' 12 36 1.528(0.762")( ) 29,000

yx

x

FKLr E

⎛ ⎞ ⎛ ⎞⎛ ⎞λ = = =⎜ ⎟ ⎜ ⎟⎜ ⎟π π⎝ ⎠ ⎝ ⎠⎝ ⎠

Since λx < 2.25, Inelastic Buckling Governs

( )( )( )( )1.528 ksi 2 kip0.66 36 8.81 in 167.3nP = = (6.9.4.1-1) Calculate the Critical Stress for Flexural-Torsional Buckling about the Y-axis: For Singly symmetric Sections:

( ) ⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

+−−

+= 2

411

2 ezey

ezeyezeye FF

HFFH

FFF

(AISC E4-5)

2 ksiksi

2inft

( )(29,000 ) 186.6(14 ')(12 )

4.29"

eyF π= =⎛ ⎞⎜ ⎟⎝ ⎠

(AISC E4-10)

( )

2 ksi 6ksi 4

2 2 2inft

ksi

( )(29,000 )(151 in ) 1(11,200 )(0.861 in )(8.81 in )(4.54")(14 ')(12 )

61.54

ez

ez

F

F

⎡ ⎤π= +⎢ ⎥⎢ ⎥⎣ ⎦

=

(AISC E4-11)

ksi ksi ksi ksi

ksi ksi 2

ksi

(186.6 61.54 ) (4)(186.6 )(61.54 )(0.919")1 1(2)(0.919") (186.6 61.54 )

59.30

e

e

F

F

⎛ ⎞⎛ ⎞+= − −⎜ ⎟⎜ ⎟⎜ ⎟+⎝ ⎠⎝ ⎠=

(AISC E4-5)

ksi

ksi

36 0.607159.30

y

e

FF

λ = = =

( )( )( )(0.6071) ksi 2 kip0.66 36 8.81 in 246.0nP = = (6.9.4.1-1)

since 167.3kip < 246.0kip, Flexural Buckling about the x Axis Governs φPn = (0.90)( 167.3kip) = 151kip

-- 122 --

ODOT-LRFD Short Course – Steel AASHTO Compression Member Example #7


AASHTO Compression Members Example #7: Problem: A pair of L4x4x1/2 angles are used as a compression member. The angles are made of M270-36 steel and have an effective length of 12’. The angles are separated by 3/8” thick connectors. Solution: Check Local Buckling:

12

4.0" 8.0"

bt= =

ksi?

ksi

29,0000.45 12.7736y

b Ekt F≤ = =

Local Buckling is OK

Check the Connector Spacing:

( )( )in

ft12 ' 12119.0

(1.21")X

KLr

⎛ ⎞⎛ ⎞ = =⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

( )( )in

ft12 ' 1278.69

(1.83")Y

KLr

⎛ ⎞⎛ ⎞ = =⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

34i

max

KLa rr

⎛ ⎞⎛ ⎞≤ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

( )3(0.776") 119.0 69.26"4

a ⎛ ⎞≤ =⎜ ⎟⎝ ⎠

Use 5 connectors….. a = 36”

ry = 1.83” from AISC 2L Table 1-15, Pg 1-104.

FullyTensioned

4" 3/8" 4"Y

X

-- 123 --



Check Flexural Buckling about the X-Axis: (Y axis is the axis of symmetry)

2 2 ksi

ksi

119.0 36 1.78129,000

yX

X

FKLr E

⎛ ⎞⎛ ⎞ ⎛ ⎞λ = = =⎜ ⎟⎜ ⎟ ⎜ ⎟π π⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Since λX < 2.25, Inelastic Buckling Governs ( )( )( )( )1.781 ksi 2 kip0.66 36 7.49 in 127.9nP = = (6.9.4.1-1)

Check Flexural-Torsional Buckling about the Y-Axis: For Tees and Double Angles where the Y axis is the Axis of Symmetry:

( ), , , ,

2, ,

41 1

2cr Y cr Z cr Y cr Z

crft

cr Y cr Z

F F F F HF

H F F

⎛ ⎞+ ⎜ ⎟= − −⎜ ⎟+⎝ ⎠

(AISC E4-2)

Since the section is built-up and the connectors will be in shear for Y-axis buckling, we must consider a modified slenderness ratio…

Calculate Modified Slenderness and Y-axis Flexural Buckling Stress:

22 2

20.82(1 )m o ib

KL KL ar r r

⎛ ⎞α⎛ ⎞ ⎛ ⎞= + ⎜ ⎟⎜ ⎟ ⎜ ⎟ + α⎝ ⎠ ⎝ ⎠ ⎝ ⎠ (6.9.4.3.1-1)

2 ib

hr

α = ( )( ) ( )382 1.18" " 2.735"h = + =

rib = ry for a single angle = 1.21”

2.735" 1.130(2)(1.21")

α = =

( )22

22

(0.82)(1.130) 36"78.69 81.24(1 (1.130) ) 1.21"m

KLr

⎛ ⎞ ⎛ ⎞= + =⎜ ⎟ ⎜ ⎟+⎝ ⎠ ⎝ ⎠

-- 124 --



Compute the Y-axis Flexural Buckling Stress, Fcry:

22 ksi

ksi

(81.24) 36 0.8301( ) 29,000

yY

Y

FKLr E

⎛ ⎞⎛ ⎞⎛ ⎞λ = = =⎜ ⎟⎜ ⎟⎜ ⎟π π⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Since λY < 2.25, Inelastic Buckling Governs

( )( )( )0.8301 ksi ksi, 0.66 36 25.43n

cr Ys

PFA

= = = (6.9.4.1-1)

Calculate Torsional Buckling Stress, Fcr,Z:

2.38"or = (AISC Table 1-15, Pg 1-104) ∴ 2 25.664 inor =

J = 0.322 in4 for a single angle (AISC Table 1-7, Pg 1-42)

∴ Jtotal = (2)(0.322 in4) = 0.644 in4

ksi 4

ksi, 2 2

(11,200 )(0.644 in ) 170.0(7.49 in )(5.664 in )cr Z

o

GJFAr 2= = = (AISC E4-3)

H = 0.848 (AISC Table 1-15, Pg 1-104)

( )ksi ksi ksi ksi

ksi2ksi ksi

25.43 170.0 (4)(25.43 )(170.0 )(0.848)1 1 24.79(2)(0.848) 25.43 170.0

crftF⎛ ⎞⎛ ⎞+ ⎜ ⎟= − − =⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠ +⎝ ⎠

Pn = AsFcrft = (7.49 in2) (24.79ksi) = 185.7kip Since 127.9kip < 185.7kip, Flexural Buckling Governs φPn = (0.90)(127.9kip ) = 115kip

-- 125 --

-- 126 --

ODOT-LRFD Short Course - Steel AASHTO Flexure Example #1


AASHTO Flexure Example #1: Problem: Determine the plastic moment of the steel section shown below. Solution Since the section is made up of components of different materials, the location of the PNA must be determined by equating the force above the PNA to the force below the PNA.

( )( )( )

( )( )( )

ksi kip

ksi kip34

ksi kip

(16")(1") 50 800.0

(22") " 36 594.0

8" 2" 70 1,120

c

w

t

P

P

P

= =

= =

= =

Since ( )kip kip kip kip 800.0 594.0 1,394 1,120c w tP P P+ > + = > , the PNA must lie in the web. Define q as the fraction of the web that lies above the PNA.

( )

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

1

800.0 594.0 1 594.0 1,120

0.7694

compression tension

c w w t

P P

P qP q P P

q q

q

=

+ = − +

+ = − +

=

I.e., 76.94% of the web lies above the PNA (acts in compression assuming a positive moment).

( )( )1" 0.7694 22" 17.93"Y = + = from the top of steel

Find the moment arms from the resultant forces to the PNA.

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

1 12 2

1 12 2

1"17.93" 17.43"2 2

0.7694 22" 8.463"

1 1 0.7694 22" 2.537"2"25" 17.93" 6.074"

2 2

cc

wc

wt

tt

td Y

d qh

d q htd d Y

= − = − =

= = =

= − = − =

= − − = − − =

-- 127 --



PL16 x 1, 50ksi

PL22 x 3/4, 36ksi

PL8 x 2, 70ksi

Pc

Pwt

dwc

PNA

50ksi

dcPwc

Pt

dwt

dt

36ksi

36ksi

70ksi

Y

Compute the plastic moment by summing the moments about the PNA.

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

k-in k-ft

800 17.43" 457 8.463" 137 2.537" 1,120 6.074"

24,960 2,080

p i i c c wc wc wt wt t tM Pd P d P d P d Pd= = + + +

= + + +

= =

∑

-- 128 --



AASHTO Flexure Example #2: Problem: Determine the plastic moment capacity for the composite beam shown below. The section is a W30x99 and supports an 8” concrete slab. The dimensions are as shown. Use Fy = 50ksi and f’c = 4ksi. Assume full composite action.

Solution:

Determine the Controlling Compression Force:

( )( )( )( )' ksi kipc0.85 0.85 4 8" 100" 2720s e sP f b t= = =

( )( )2 ksi kip29.1 in 50 1455Steel st yP A F= = =

Assuming full composite action, the shear connectors must carry the smallest of Ps and Psteel. Since Ps > Psteel, the PNA must lie in the slab.

Determine the Location of the PNA:

The PNA location is determined by equating the compressive force in the slab, acting over a depth ac, with the tensile force in the steel section.

=Conc SteelP P '

c0.85 e c st yf b a A F=

( )( )( )( )( )

2 ksi

' ksic

29.1 in 504.279"

0.85 0.85 4 100"st y

ce

A Fa

f b= = = (measured from the top of slab)

100"

8"

W30 x 99:A = 29.1 in2

d = 29.7" bf = 10.5" tf = 0.670" tw = 0.520" Zx = 312 in3

Ix = 3,990 in4

Iy = 128 in4

rx = 11.7" ry = 2.10"

-- 129 --



Determine the Plastic Moment:

The plastic moment is calculated by summing the tension and compression forces about any point. In general, the moments are summed about the PNA. In this case (where the PNA is in the slab) it is simplest to sum moments about either force PSteel or the force Pconc. Note that the tension force in the concrete is ignored.

100"

8"Pconc

Psteel

a1

PNA

0.85f’c

Fy

ac

( ) ( )1 1( ) ( )p Conc SteelM P a P a= =

129.7" 4.279"8" 20.71"

2 2 2 2st c

sd aa t= + − = + − =

( )kip k-in k-ft1455 (20.71") 30,130 2,511pM = = =

-- 130 --



AASHTO Flexure Example #3: Problem: Determine the plastic moment capacity for the composite beam shown below. The section is a W30x99 and supports a 6” thick concrete slab. The dimensions are as shown. Use Fy = 50ksi and f’c = 4ksi. Assume full composite action. Solution: Determine the Controlling Compression Force:

( )( )( )( )' ksi kip

c0.85 0.85 4 6" 50" 1020s s sP f b t= = =

( )( )2 ksi kip29.1 in 50 1455Steel st yP A F= = =

Assuming full composite action, the shear connectors must carry the smallest of Ps and Psteel. Since Ps < Psteel, the PNA must lie in the steel. When this occurs, it is simplest to use the aids in Appendix D of the AASHTO Specification to determine the location of the PNA and plastic moment.

Referring to Table D6.1-1 in Appendix D6.1, Page 6-290:

Determine the forces in the components of the cross section. The forces in the rebar will be conservatively taken as zero (we don’t know what size the rebar is any ways…)

( )( )( )( )

( )( )

' ksi kipc

ksi kip

2 ksi kip

kip

0.85 0.85 4 6" 50" 1020

(0.670")(10.5") 50 351.8

29.1 in (2)(0.670")(10.5") 50 751.5

351.8

s s s

c

w

t c

P f b t

P

P

P P

= = =

= =

⎡ ⎤= − =⎣ ⎦= =

Check Case I ?

t w c sP P P P+ ≥ +

?

kip kip kip kip351.8 751.5 351.8 1020+ ≥ + NO

In this case, I took Aw = Asteel - 2Af. Otherwise, Pc+Pt+Pw ≠ Psteel. If you take Aw = D tw where D = d - 2tf, the plastic moment changes by ~2%

-- 131 --



Check Case II ?

t w c sP P P P+ + ≥

?

kip kip kip kip351.8 751.5 351.8 1020+ + ≥ YES - PNA in Top Flange

50"

6"Ps

Pw

ds PNA

0.85f’c

FyPt

Fy

dw

dt

Pc2

Pc1

First, the location of the PNA within the top flange is determined.

kip kip kip

kip

12

0.670" 751.5 351.8 1,020 1 0.2368"2 351.8

c w t s

c

t P P PYP

⎛ ⎞+ −⎛ ⎞= +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

⎛ ⎞+ −⎛ ⎞= + =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

Next, the distances from the component forces to the PNA are calculated.

6" 0.2368" 3.237"229.7" 0.2368" 14.61"

20.670"29.7" 0.2368" 29.13"

2

s

w

t

d

d

d

= + =

= − =

= − − =

-- 132 --



Finally, the plastic moment is computed.

( ) [ ]

( ) ( )

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

( )

22

kip2 2

kip kip kip

kip 2 k-inin

k-

2

351.8 0.2368" 0.670" 0.2368" ...(2)(0.670")

... 1,020 3.237" 751.5 14.61" 351.8 29.13"

262.5 0.2437 in 24,530

24,590

cp c s s w w t t

c

PM Y t Y P d P d Pdt

⎛ ⎞ ⎡ ⎤= + − + + +⎜ ⎟ ⎣ ⎦⎝ ⎠⎛ ⎞ ⎡ ⎤= + − +⎜ ⎟ ⎣ ⎦⎝ ⎠

⎡ ⎤+ + +⎣ ⎦

⎡ ⎤ ⎡ ⎤= +⎣ ⎦ ⎣ ⎦

= in k-ft2,049=

-- 133 --



AASHTO Flexure Example #4: Problem: Determine the plastic moment capacity for the composite beam shown below for negative flexure. The section is a W30x99 and supports an 8” concrete slab. The dimensions are as shown. Use Fy = 50ksi and f’c = 4ksi. Assume full composite action. The grade 60 reinforcement in the slab is made up of #4 bars, with a clear cover of 17/8”.

Solution: The concrete slab will be in tension, therefore none of the concrete is assumed to be effective. Referring to Table D6.1-2 in Appendix D6.1, Page 6-291:

( )( )2

ksi kip( )(0.5")60 8 94.254rt yrt rtP F A

⎛ ⎞π= = =⎜ ⎟

⎝ ⎠

( )( )2

ksi kip( )(0.5")60 4 47.124rb yrb rbP F A

⎛ ⎞π= = =⎜ ⎟

⎝ ⎠

( )( )

ksi kip

2 ksi kip

kip

(0.670")(10.5") 50 351.8

29.1 in (2)(0.670")(10.5") 50 751.5

351.8

t

w

c t

P

P

P P

= =

⎡ ⎤= − =⎣ ⎦= =

Check Case I: ?

c w t rb rtP P P P P+ ≥ + +

?

kip kip kip kip kip351.8 751.5 351.8 47.12 94.25+ ≥ + + YES - PNA is in Web

12

c t rt rb

w

P P P PDYP

⎡ ⎤− − −⎛ ⎞= +⎢ ⎥⎜ ⎟⎝ ⎠ ⎣ ⎦

Take D as ( )( )2 29.7" 2 0.670" 28.36"fd t− = − =

kip kip kip kip

kip

28.36" 351.8 351.8 94.25 47.12 12 751.5

Y⎡ ⎤− − −⎛ ⎞= +⎜ ⎟ ⎢ ⎥

⎝ ⎠ ⎣ ⎦

11.51"Y = (measured from the bottom of the top flange)

100"

8"

W30 x 99:A = 29.1 in2

d = 29.7" bf = 10.5" tf = 0.670" tw = 0.520" Sx = 269 in3

Zx = 312 in3

Ix = 3,990 in4

Iy = 128 in4

rx = 11.7" ry = 2.10"

Slab Reinforcement:Top Layer: #4 bars @ 6" ccBottom Layer: #4 bars @ 12" cc

-- 134 --



( )( )7 1 1

8 2 211.51" 0.670" 8" 1 " " 18.06"rtd = + + − + =⎡ ⎤⎣ ⎦

( )( )7 1 18 2 211.51" 0.670" 1 " " 14.31"rbd = + + + =⎡ ⎤⎣ ⎦

( )( )1211.51" 0.670" 11.85"td = + =

( )( )12 11.51" 5.755"wtd = =

( )( )12 28.36" 11.51" 8.425"wcd = − =

( ) ( )( )1228.36" 11.51" 0.670" 17.19"cd = − + =

[ ]2 2( )2

wp rt rt rb rb t t c c

PM y D y P d P d Pd P dD

⎛ ⎞ ⎡ ⎤= + − + + + +⎜ ⎟ ⎣ ⎦⎝ ⎠

kip2 2 kip

kip kip kip

751.5 (11.51") (28.36" 11.51") [(94.25 )(18.06") ...(2)(28.36")

... (47.12 )(14.31") (351.8 )(11.85") (351.8 )(17.19")]

⎡ ⎤= + − + +⎣ ⎦

+ + +

k-in k-ft18,110 1,509= =

Not needed when using Table D6.1-2

-- 135 --



The plastic moment can also be computed from “first principles” as well, though it is a bit more involved. What follows is an example of how this would be completed. Determine the Location of the PNA:

Since c w t rb rtP P P P P+ ≥ + + , the PNA is in the web of the section. The location of the PNA within the web is determined by equating the tensile force acting above the PNA with the compressive force acting below it. Assume the PNA lies at a depth Y below the bottom of the top flange.

c wc wt t rb rtP P P P P P+ = + + +

( )( )( ) ( )( )( )kip ksi ksi kip kip kip351.8 50 0.520" 28.36" 50 0.520" 351.8 47.12 94.25Y Y+ − = + + +

11.46 ''Y =

( )ksi kip(50 )(11.46") 0.520" 298.0wtP = =

( )ksi kip(50 )(28.36" 11.46") 0.520" 439.4wcP = − = Determine the Plastic Moment:

The plastic moment is calculated by summing the moments of the tensile and compressive forces about any point. In general, the moments are summed about the PNA. In this case (where the PNA is in the web) note that the tension force in the concrete is ignored.

( )( )7 1 18 2 211.46" 0.670" 8" 1 " " 17.88"rtd = + + − + =⎡ ⎤⎣ ⎦

( )( )7 1 18 2 211.46" 0.670" 1 " " 14.38"rbd = + + + =⎡ ⎤⎣ ⎦

( )( )1211.46" 0.670" 11.80"td = + =

( )( )12 11.46" 5.730"wtd = =

( )( )12 28.36" 11.46" 8.450"wcd = − =

( ) ( )( )1228.36" 11.46" 0.670" 17.24"cd = − + =

( ) ( )( )

kip kip kip

kip kip kip

(94.25 )(17.88'') (47.12 )(14.38") (351.8 )(11.80 '') ...

... (298.0 )(5.730") (439.4 ) 8.450" 351.8 17.24"pM = + + +

+ + +

k-in k-ft18,000 1,500= = The minor difference in between the two answers can be attributed to the fillet area.

-- 136 --

ODOT-LRFD Short Course - Steel AASHTO Flexure Example #5a


AASHTO Flexural Example #5a: Problem: A non-composite W30x99 made of M270-50 steel is used to span 48’. The beam is braced laterally at 12’-0” intervals and is subjected to a factored load of w = 3.75kip/ft, which includes the self weight of the beam. Check to see if the section is adequate considering flexural failure modes at the Strength Limit States. If appropriate, use the provisions in AASHTO Section 6.10.8 to determine capacity. Solution:

Determine Classification of the Section:

Check ?2 5.7c

w yc

D Et F

≤ (6.10.6.2.3-1)

Take D = d - 2tf = 29.7” - (2)(0.670”) = 28.36”

28.36" 14.18"2 2cDD = = =

ksi?

ksi

(2)(14.18") 29,00054.54 5.7 137.3(0.520") 50

= ≤ = OK, ∴ web is non-slender

Check ?

0.3yc

yt

II

≥ (6.10.6.2.3-2)

Since Section is doubly symmetric, Iyc = Iyt OK

-- 137 --



Since the web is non-slender and Eq 6.10.6.2.3-2 is satisfied, we have the option of using either AASHTO Section 6.10.8 or Appendix A6 to determine the flexural capacity of this member. For this example, the provisions of 6.10.8 will be used and we will work with stresses. The following failure modes must be investigated:

• Flange Local Buckling of the Compression Flange Fnc(FLB) • Compression Flange Lateral Buckling Fnc(LTB) • Yielding of Tension Flange Fnt

Investigate Compression Flange Local Buckling:

10.5" 7.8362 (2)(0.670")

fcf

fc

bt

λ = = = (6.10.8.2.2-3)

ksi

ksi

29,0000.38 0.38 9.15250pf

yc

EF

λ = = = (6.10.8.2.2-4)

Since λf < λp, the flange is compact and, ( )nc FLB b h ycF R R F= (6.10.8.2.2-1) Rb = 1.00 (since the web is non-slender) Rh = 1.00 (since the section is rolled and is ∴ non-hybrid) ( )ksi ksi

( ) (1.00)(1.00) 50 50nc FLBF = = Investigate Compression Flange Lateral-Torsional Buckling: The unbraced length of the beam is Lb = 12’-0” = 144.0”.

(10.5") 2.609"1 (14.18")(0.520")1 12 112 1 3 (10.5")(0.670")3

fct

c w

fc fc

br

D tb t

= = =⎛ ⎞ ⎛ ⎞⎛ ⎞++ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠

(6.10.8.2.3-9)

ksi

ksi

29,0001.0 (1.0)(2.609") 62.84"50p t

yc

EL rF

= = = (6.10.8.2.3-4)

Fnc

-- 138 --



( )min 0.7 , 0.5yr yc yw ycF F F F= ≥ ( )ksi ksi(0.7) 50 35yrF = = (Pg 6-110)

ksi

ksi

29,000 ( )(2.609") 235.9"35r t

yr

EL rF

= π = π = (6.10.8.2.3-5)

Since 62.84" 144" 235.9"p b rL L L= < = < = , Inelastic LTB must be investigated.

( ) 1 1 yr b pnc LTB b b h yc b h yc

h yc r p

F L LF C R R F R R F

R F L L

⎡ ⎤⎛ ⎞⎛ ⎞−= − − ≤⎢ ⎥⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟−⎢ ⎥⎝ ⎠⎝ ⎠⎣ ⎦

(6.10.8.2.3-2)

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

ksiksi ksi

( ) ksi

35 144" 62.84"1 1 1.0 1.0 50 1.0 1.0 50235.9" 62.84"1.0 50nc LTB bF C

⎡ ⎤⎛ ⎞ −⎛ ⎞⎢ ⎥⎜ ⎟= − − ≤⎜ ⎟⎜ ⎟ −⎢ ⎥⎝ ⎠⎝ ⎠⎣ ⎦

( )( )( ) ( )( )ksi ksi ksi

( ) 0.8593 50 42.97 50nc LTB b bF C C= = ≤ Compute the Moment Gradient Factor, Cb, for segment BC of the beam, which will be critical.

2

1 1

2 2

1.75 1.05 0.3 2.3bf fCf f

⎛ ⎞ ⎛ ⎞= − + ≤⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ (6.10.8.2.3-7)

k-ft ksi2 2

k-ft ksi

k-ft ksi,

1,080 48.18

810.0 36.13

1,013 45.19

c

o B o

BC mid mid

M M f

M M f

M f

= = → =

= = → =

= → =

Since the BMD is not concave, ( ) ( )ksi ksi ksi ksi

1 22 (2) 45.19 48.18 42.20 36.13mid of f f f= − ≥ = − = ≥

242.20 42.201.75 1.05 0.3 1.061 2.348.18 48.18

1.061

bC ⎛ ⎞ ⎛ ⎞= − + = ≤⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

=

( )( )ksi ksi ksi( )

ksi

1.061 42.97 45.57 50

45.57nc LTBF = = ≤

=

B

C

f2fmidf1

-- 139 --



The governing strength for the compression flange is the smaller of Fnc(FLB) and Fnc(LTB):

Since ksi ksi( ) ( )45.57 50.00nc LTB nc FLBF F= < = , LTB governs the strength of the

compression flange.

ksi( ) 45.57nc nc LTBF F= =

( )ksi ksi(1.00) 45.57 45.57ncFφ = =

Check 13bu l f ncf f F+ ≤ φ (6.10.8.1.1-1)

ksi48.18bu Cf f= = Assume that Strength I Load Combination Governs, ∴ γWS = 0.0 and fl = 0

Since ksi ksi1 48.18 =45.573bu l f ncf f F+ = > φ , the compression flange is not adequate.

Investigate the Strength of the Tension Flange: Since the tension flange is discretely braced, ( )ksi ksi(1.0) 50 50nt h ytF R F= = = (6.10.8.3-1) ( )ksi ksi(1.00) 50 50ntFφ = =

Check 13bu l f ntf f F+ ≤ φ (6.10.8.1.2-1)

ksi48.18bu Cf f= =

Since ksi ksi1 48.18 =50.003bu l f ntf f F+ = < φ , the tension flange is adequate.

Since the compression flange is not adequate, the section is not adequate for flexure.

-- 140 --

ODOT-LRFD Short Course - Steel AASHTO Flexure Example #5b


AASHTO Flexural Example #5b: Problem: A non-composite W30x99 made of M270-50 steel is used to span 48’. The beam is braced laterally at 12’-0” intervals and is subjected to a factored load of w = 3.75kip/ft, which includes the self weight of the beam. Check to see if the section is adequate considering flexural failure modes at the Strength Limit States. If appropriate, use the provisions in AASHTO Appendix A6 to determine capacity. Solution:


Check ?2 5.7c

w yc

D Et F

≤ (6.10.6.2.3-1)

Take D = d - 2tf = 29.7” - (2)(0.670”) = 28.36”

28.36" 14.18"2 2cDD = = =

ksi?

ksi

(2)(14.18") 29,00054.54 5.7 137.3(0.520") 50


Check ?

0.3yc

yt

II

≥ (6.10.6.2.3-2)


-- 141 --



Since the web is non-slender and Eq 6.10.6.2.3-2 is satisfied, we have the option of using either AASHTO Section 6.10.8 or Appendix A6 to determine the flexural capacity of this member. For this example, the provisions of A6 will be used and we will work with moments. The following failure modes must be investigated:

• Flange Local Buckling of the Compression Flange Mnc(FLB) • Lateral-Torsional Buckling Mnc(LTB) • Yielding of Tension Flange Mnt

Compute Web Plasticity Factors, Rpc and Rpt (Section A6.2): Investigate the classification of the web.

Check ?

( )

2cp

cppw D

w

Dt

≤λ (A6.2.1-1)

( ) 20.54

0.09cp

yc cppw D rw

cp

h y

EF D

DMR M

⎛ ⎞λ = ≤ λ ⎜ ⎟

⎛ ⎞ ⎝ ⎠−⎜ ⎟⎜ ⎟

⎝ ⎠

(A6.2.1-2)

5.7 137.3rwyc

EF

λ = = (A6.2.1-3)

Rh = 1.00 (since the section is rolled and is ∴ non-hybrid)

( )( )( )( )

3 ksi k-in k-ft

3 ksi k-in k-ft

269 in 50 13,450 1,121

312 in 50 15,600 1,300

y x y

p x y

M S F

M Z F

= = = =

= = = =

( )( )( )( )

ksi

ksi

( ) 2k-in

k-in

29,00014.18"50 83.76 137.3 137.314.18"0.54 15,600

0.091.0 13,450

cppw D⎛ ⎞λ = = ≤ =⎜ ⎟⎝ ⎠⎛ ⎞

⎜ ⎟−⎜ ⎟⎝ ⎠

( ) 83.76

cppw Dλ =

Mnc

-- 142 --



?

( )

2 (2)(14.18") 54.54 83.760.520" cp

cppw D

w

Dt

= = ≤λ = OK, ∴ web is compact

Since the web is compact,

k-in

k-in

15,600 1.16013,450

ppc

yc

MR

M= = = (A6.2.1-4)

k-in

k-in

15,600 1.16013,450

ppt

yt

MR

M= = = (A6.2.1-5)

Investigate Compression Flange Local Buckling: Investigate the compactness of the compression flange.

10.5" 7.8362 (2)(0.670")

fcf

fc

bt

λ = = = (A6.3.2-3)

ksi

ksi

29,0000.38 0.38 9.15250pf

yc

EF

λ = = = (A6.3.2-4)

Since λf < λpf, the flange is compact and, ( )( )k-in k-in

( ) 1.160 13,450 15,600nc FLB pc ycM R M= = = (A6.3.2-1) Investigate Lateral-Torsional Buckling: The unbraced length of the beam is Lb = 12’-0” = 144.0”.

(10.5") 2.609"1 (14.18")(0.520")1 12 112 1 3 (10.5")(0.670")3

fct

c w

fc fc

br

D tb t

= = =⎛ ⎞ ⎛ ⎞⎛ ⎞++ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠

(A6.3.3-10)

ksi

ksi

29,0001.0 (1.0)(2.609") 62.84"50p t

yc

EL rF

= = = (A6.3.3-4)

-- 143 --



2

1.95 1 1 6.76 yr xcr t

yr xc

F S hE JL rF S h E J

⎛ ⎞= + + ⎜ ⎟

⎝ ⎠ (6.10.8.2.3-5)

min 0.7 , , 0.5xtyr yc h yt yw yc

xc

SF F R F F FS

⎛ ⎞= ≥⎜ ⎟

⎝ ⎠ ( )ksi ksi(0.7) 50 35yrF = = (Pg 6-222)

Sxc = 269 in3 ( )( )1

2 29.7" 0.670" 29.03"fc fth d t t= − + = − =

3 33 1 0.63 1 0.63

3 3 3fc fc fc ft ft ftw

fc ft

b t t b t tD tJb b

⎛ ⎞ ⎛ ⎞= + − + −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ (A6.3.3-9)

( )3 3

4(28.36")(0.520") (10.5")(0.670") 0.670"(2) 1 0.63 3.350 in3 3 10.5"

J⎡ ⎤⎛ ⎞⎛ ⎞= + − =⎢ ⎥⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎣ ⎦

J = 3.350 in4 J = 3.77 in4 from AISC Manual….use J = 3.77 in4

ksi

ksi

29,000 828.635yr

EF

= =

( )( )3

4

269 in 29.03"2,071

3.77 inxcS hJ

= =

( )( )( )21 2,0711.95 2.609" 828.6 1 1 6.76 254.9" 21.24 '

2,071 828.6rL ⎛ ⎞= + + = =⎜ ⎟⎝ ⎠

This value of Lr = 21.24’ agrees well with the value published in AISC on Page 3-15 Since 62.84" 144" 254.9"p b rL L L= < = < = , Inelastic LTB must be investigated.

( ) 1 1 yr xc b pnc LTB b pc yc pc yc

pc yc r p

F S L LM C R F R F

R F L L

⎡ ⎤⎛ ⎞⎛ ⎞−= − − ≤⎢ ⎥⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟−⎢ ⎥⎝ ⎠⎝ ⎠⎣ ⎦

(A6.3.3-2)

-- 144 --



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

( )( )( ) ( )( )

ksi 3k-in k-in

( ) k-in

k-in k-in k-in

35 269 in 144" 62.84"1 1 1.160 13,450 1.160 13,450254.9" 62.84"1.16 13,450

0.9656 15,600 12,990 15,600

nc LTB b

b b

M C

C C

⎡ ⎤⎛ ⎞ −⎛ ⎞⎢ ⎥⎜ ⎟= − − ≤⎜ ⎟⎜ ⎟ −⎢ ⎥⎝ ⎠⎝ ⎠⎣ ⎦

= = ≤

Compute the Moment Gradient Factor, Cb, for segment BC of the beam, which will be critical.

2

1 1

2 2

1.75 1.05 0.3 2.3bM MCM M

⎛ ⎞ ⎛ ⎞= − + ≤⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ (A6.3.3-7)

k-ft2

k-ft

k-ft,

1,080

810.0

1,013

c

o B

BC mid

M M

M M

M

= =

= =

=

Since the BMD is not concave, ( ) ( )k-ft k-ft k-ft k-ft

1 22 (2) 1,013 1,080 946 810mid oM M M M= − ≥ = − = ≥

2946 9461.75 1.05 0.3 1.061 2.31,080 1,080

1.061

bC ⎛ ⎞ ⎛ ⎞= − + = ≤⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

=

( )( )k-in k-in k-in

( ) 1.061 12,990 13,780 15,600nc LTBM = = ≤ k-in

( ) 13,780nc LTBM =

-- 145 --



The governing strength for the compression flange is the smaller of Mnc(FLB) and Mnc(LTB):

Since k-in k-in( ) ( )13,780 15,600nc LTB nc FLBM M= < = , LTB governs the strength of the

compression flange.

k-in k-ft( ) 13,780 1,148nc nc LTBM M= = =

( )k-ft k-ft(1.00) 1,148 1,148ncMφ = =

Check 13u l xc f ncM f S M+ ≤ φ (A6.1.1-1)

k-ft1,080u CM M= = Assume that Strength I Load Combination Governs, ∴ γWS = 0.0 and fl = 0

Since k-ft k-ft1 1,080 =1,1483u l xc f ncM f S M+ = < φ , the compression flange is adequate.

Investigate the Strength of the Tension Flange: Since the tension flange is discretely braced, ( )k-in k-in k-ft(1.160) 13,450 15,600 1,300nt pt ytM R M= = = = (A6.4-1) ( )k-ft k-ft(1.00) 1,300 1,300ntMφ = =

Check 13u l xt f ntM f S M+ ≤ φ (6.10.8.1.2-1)

k-ft1,080u CM M= =

Since k-ft k-ft1 1,080 =1,3003u l xt f ntM f S M+ = < φ , the tension flange is adequate.

Since both flanges are adequate, the section is adequate for flexure.

Note that the benefits of using Appendix A6 are illustrated here since the section was found to be not adequate when the provisions in Section 6.10.8 were used to compute capacity.

-- 146 --



AASHTO Flexural Example #6a: Problem: A non-composite built-up girder made of M270-50 steel is used to span 48’. The beam is braced laterally at 12’-0” intervals and is subjected to a factored load of w = 3.75kip/ft, which includes the self weight of the beam. Check to see if the section is adequate considering flexural failure modes at the Strength Limit States. If appropriate, use the provisions in AASHTO Section 6.10.8 to determine capacity. Solution:


Check ?2 5.7c

w yc

D Et F

≤ (6.10.6.2.3-1)

Take D = 38”

38" 19"2 2cDD = = =

ksi?

ksi38

(2)(19") 29,000101.3 5.7 137.3( ") 50


Check ?

0.3yc

yt

II

≥ (6.10.6.2.3-2)


PL16 x 3/4

PL38 x 3/8

PL16 x 3/4

Ix = 10,730 in4

Iy = 513.2 in4

Sx = 543.1 in3

Sy = 64.15 in3

-- 147 --



Since the web is non-slender and Eq 6.10.6.2.3-2 is satisfied, we have the option of using either AASHTO Section 6.10.8 or Appendix A6 to determine the flexural capacity of this member. For this example, the provisions of 6.10.8 will be used and we will work with stresses. The following failure modes must be investigated:

• Flange Local Buckling of the Compression Flange Fnc(FLB) • Compression Flange Lateral Buckling Fnc(LTB) • Yielding of Tension Flange Fnt

Investigate Compression Flange Local Buckling:

3

4

16" 10.672 (2)( ")

fcf

fc

bt

λ = = = (6.10.8.2.2-3)

ksi

ksi

29,0000.38 0.38 9.15250pf

yc

EF

λ = = = (6.10.8.2.2-4)

Since λf < λp, the flange is non compact and,

( ) 1 1 yr f pfnc FLB b h yc

h yc rf pf

FF R R F

R F

⎡ ⎤⎛ ⎞⎛ ⎞λ −λ= − −⎢ ⎥⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟λ −λ⎢ ⎥⎝ ⎠⎝ ⎠⎣ ⎦

(6.10.8.2.2-2)

0.56rfyr

EF

λ = (6.10.8.2.2-5)

( )min 0.7 , 0.5yr yc yw ycF F F F= ≥ (Pg 6-109)

( )ksi ksi(0.7) 50 35yrF = =

ksi

ksi

29,0000.56 16.1235rf = =λ

Rb = 1.00 (since the web is non-slender) Rh = 1.00 (since the section is non-hybrid)

( )ksi

ksi ksi( ) ksi

35 10.67 9.1521 1 (1.00)(1.00) 50 46.74(1.00)(50 ) 16.12 9.152nc FLBF

⎡ ⎤⎛ ⎞ −⎛ ⎞= − − =⎢ ⎥⎜ ⎟⎜ ⎟−⎝ ⎠⎝ ⎠⎣ ⎦

Fnc

-- 148 --



Investigate Compression Flange Lateral-Torsional Buckling: The unbraced length of the beam is Lb = 12’-0” = 144.0”.

3

83

4

(16") 4.220"1 (19")( ")1 12 112 1 3 (16")( ")3

fct

c w

fc fc

br

D tb t

= = =⎛ ⎞ ⎛ ⎞⎛ ⎞++ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠

(6.10.8.2.3-9)

ksi

ksi

29,0001.0 (1.0)(4.220") 101.6"50p t

yc

EL rF

= = = (6.10.8.2.3-4)

( )min 0.7 , 0.5yr yc yw ycF F F F= ≥ ( )ksi ksi(0.7) 50 35yrF = = (Pg 6-110)

ksi

ksi

29,000 ( )(4.220") 381.6"35r t

yr

EL rF

= π = π = (6.10.8.2.3-5)


( ) 1 1 yr b pnc LTB b b h yc b h yc

h yc r p

F L LF C R R F R R F

R F L L

⎡ ⎤⎛ ⎞⎛ ⎞−= − − ≤⎢ ⎥⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟−⎢ ⎥⎝ ⎠⎝ ⎠⎣ ⎦

(6.10.8.2.3-2)

( )( ) ( )( )( ) ( )( )( )ksi

ksi ksi( ) ksi

35 144" 101.6"1 1 1.0 1.0 50 1.0 1.0 50381.6" 101.6"1.00 50nc LTB bF C

⎡ ⎤⎛ ⎞ −⎛ ⎞⎢ ⎥⎜ ⎟= − − ≤⎜ ⎟⎜ ⎟ −⎢ ⎥⎝ ⎠⎝ ⎠⎣ ⎦

( )( )( ) ( )( )ksi ksi ksi

( ) 0.9546 50 47.73 50nc LTB b bF C C= = ≤

-- 149 --




2

1 1

2 2

1.75 1.05 0.3 2.3bf fCf f

⎛ ⎞ ⎛ ⎞= − + ≤⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ (6.10.8.2.3-7)

k-ft ksi2 2

k-ft ksi

k-ft ksi,

1,080 23.86

810.0 17.90

1,013 22.38

c

o B o

BC mid mid

M M f

M M f

M f

= = → =

= = → =

= → =

Since the BMD is not concave, ( ) ( )ksi ksi ksi ksi

1 22 (2) 22.38 23.86 20.90 17.90mid of f f f= − ≥ = − = ≥

220.90 20.901.75 1.05 0.3 1.061 2.3 1.061

23.86 23.86b bC C⎛ ⎞ ⎛ ⎞= − + = ≤ → =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

( )( )ksi ksi ksi ksi

( ) ( )1.061 47.73 50.64 50 50nc LTB nc LTBF F= = ≤ → = The governing strength for the compression flange is the smaller of Fnc(FLB) and Fnc(LTB):

Since ksi ksi( ) ( )50 > 46.70nc LTB nc FLBF F= = , FLB governs the strength of the compression

flange.

ksi( ) 46.70nc nc FLBF F= =

( )ksi ksi(1.00) 46.70 46.70ncFφ = =

Check 13bu l f ncf f F+ ≤ φ (6.10.8.1.1-1)

ksi23.86bu Cf f= = Assume that Strength I Load Combination Governs, ∴ γWS = 0.0 and fl = 0

Since ksi ksi1 23.86 =46.703bu l f ncf f F+ = < φ , the compression flange is adequate.

B

C

f2fmidf1

-- 150 --



Investigate the Strength of the Tension Flange: Since the tension flange is discretely braced, ( )ksi ksi(1.0) 50 50nt h ytF R F= = = (6.10.8.3-1) ( )ksi ksi(1.00) 50 50ntFφ = =

Check 13bu l f ntf f F+ ≤ φ (6.10.8.1.2-1)

ksi23.86bu Cf f= =

Since ksi ksi1 23.86 =50.003bu l f ntf f F+ = < φ , the tension flange is adequate.

Since both the compression flange and tension flange are adequate, the section is adequate for flexure.

-- 151 --



AASHTO Flexural Example #6b: Problem: A non-composite built-up girder made of M270-50 steel is used to span 48’. The beam is braced laterally at 12’-0” intervals and is subjected to a factored load of w = 3.75kip/ft, which includes the self weight of the beam. Check to see if the section is adequate considering flexural failure modes at the Strength Limit States. If appropriate, use the provisions in AASHTO Appendix A6 to determine capacity. Solution:


Check ?2 5.7c

w yc

D Et F

≤ (6.10.6.2.3-1)

Take D = 38”

38" 19"2 2cDD = = =

ksi?

ksi38

(2)(19") 29,000101.3 5.7 137.3( ") 50


Check ?

0.3yc

yt

II

≥ (6.10.6.2.3-2)


-- 152 --



Since the web is non-slender and Eq 6.10.6.2.3-2 is satisfied, we have the option of using either AASHTO Section 6.10.8 or Appendix A6 to determine the flexural capacity of this member. For this example, the provisions of A6 will be used and we will work with moments. The following failure modes must be investigated:

• Flange Local Buckling of the Compression Flange Mnc(FLB) • Lateral-Torsional Buckling Mnc(LTB) • Yielding of Tension Flange Mnt

Compute Web Plasticity Factors, Rpc and Rpt (Section A6.2): Investigate the classification of the web.

Check ?

( )

2cp

cppw D

w

Dt

≤λ (A6.2.1-1)

( ) 20.54

0.09cp

yc cppw D rw

cp

h y

EF D

DMR M

⎛ ⎞λ = ≤ λ ⎜ ⎟

⎛ ⎞ ⎝ ⎠−⎜ ⎟⎜ ⎟

⎝ ⎠

(A6.2.1-2)

5.7 137.3rwyc

EF

λ = = (A6.2.1-3)

Rh = 1.00 (since the section is rolled and is ∴ non-hybrid)

( )( )( )( )

3 ksi k-in k-ft

3 ksi k-in k-ft

543.1 in 50 27,160 2,263

600.4 in 50 30,020 2,502

y x y

p x y

M S F

M Z F

= = = =

= = = =

( )( )( )( )

ksi

ksi

( ) 2k-in

k-in

29,00019"50 93.68 137.3 137.319"0.54 30,020

0.091.0 27,160

cppw D⎛ ⎞λ = = ≤ =⎜ ⎟⎝ ⎠⎛ ⎞

⎜ ⎟−⎜ ⎟⎝ ⎠

( ) 93.68

cppw Dλ =

Mnc

-- 153 --



?

( )38

2 (2)(19") 101.3 93.68" cp

cppw D

w

Dt

= = >λ = ∴ web is non compact

Since the web is non compact,

( )

( )

1 1 c

c

w pw Dh yc p ppc

p rw pw D yc yc

R M M MR

M M M

⎡ ⎤⎛ ⎞⎛ ⎞ λ −λ= − − ≤⎢ ⎥⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟λ −λ⎢ ⎥⎝ ⎠⎝ ⎠⎣ ⎦

(A6.2.2-4)

Where,

( ) ( )c cp

cppw D pw D rw

c

DD

⎛ ⎞λ = λ ≤ λ⎜ ⎟

⎝ ⎠ (A6.2.2-6)

19 ''93.68 137.319 ''

93.68 137.3

⎛ ⎞= ≤⎜ ⎟⎝ ⎠

= ≤

k-in k-in k-in

k-in k-in k-in

(1.00)(27,160 ) 101.3 93.68 (30,020 ) (30,020 )1 1(30,020 ) 137.3 93.68 (27,160 ) (27,160 )pcR

⎡ ⎤⎛ ⎞ −⎛ ⎞= − − ≤⎢ ⎥⎜ ⎟⎜ ⎟−⎝ ⎠⎝ ⎠⎣ ⎦

1.087 1.1051.087

= ≤=

( )

( )

1 1 c

c

w pw Dh yt p ppt

p rw pw D yt yt

R M M MR

M M M

⎡ ⎤⎛ ⎞⎛ ⎞ λ −λ= − − ≤⎢ ⎥⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟λ −λ⎢ ⎥⎝ ⎠⎝ ⎠⎣ ⎦

(A6.2.2-5)

k-in k-in k-in

k-in k-in k-in

(1.00)(27,160 ) 101.3 93.68 (30,020 ) (30,020 )1 1(30,020 ) 137.3 93.68 (27,160 ) (27,160 )ptR

⎡ ⎤⎛ ⎞ −⎛ ⎞= − − ≤⎢ ⎥⎜ ⎟⎜ ⎟−⎝ ⎠⎝ ⎠⎣ ⎦

1.087 1.1051.087

= ≤=

Investigate Compression Flange Local Buckling: Investigate the compactness of the compression flange.

3

4

16" 10.672 (2)( ")

fcf

fc

bt

λ = = = (A6.3.2-3)

ksi

ksi

29,0000.38 0.38 9.15250pf

yc

EF

λ = = = (A6.3.2-4)

-- 154 --



Since λf >λpf, the flange is non compact and,

( ) 1 1 yr xc f pfnc FLB pc yc

pc yc rf pf

F SM R M

R M

⎡ ⎤⎛ ⎞⎛ ⎞λ −λ= − −⎢ ⎥⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟λ −λ⎢ ⎥⎝ ⎠⎝ ⎠⎣ ⎦

(A6.3.2-2)


xc

SF F R F F FS

⎛ ⎞= ≥⎜ ⎟

⎝ ⎠ ; ( )ksi ksi(0.7) 50 35yrF = = (Pg 6-222)

0.95 crf

yr

EkF

λ = (A.6.3.2-5)

''

''38

4 4 0.397438

c

w

kDt

= = = (A6.3.2-6)

( )( )

( )ksi

ksi

29,000 0.39740.95 17.24

35rf = =λ

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

ksi 3k-in

( ) k-in

35 543.1 in 10.67 9.1521 1 1.087 27,16017.24 9.1521.087 27,160nc FLBM

⎡ ⎤⎛ ⎞ −⎛ ⎞⎢ ⎥⎜ ⎟= − − ⎜ ⎟⎜ ⎟ −⎢ ⎥⎝ ⎠⎝ ⎠⎣ ⎦

k-in k-ft27,550 2,296= = Investigate Compression Flange Lateral-Torsional Buckling: The unbraced length of the beam is Lb = 12’-0” = 144.0”.

3

83

4

(16") 4.220"1 (19")( ")1 12 112 1 3 (16")( ")3

fct

c w

fc fc

br

D tb t

= = =⎛ ⎞ ⎛ ⎞⎛ ⎞++ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠

(A6.3.3-10)

ksi

ksi

29,0001.0 (1.0)(4.220") 101.6"50p t

yc

EL rF

= = = (A6.3.3-4)

2

1.95 1 1 6.76 yr xcr t

yr xc

F S hE JL rF S h E J

⎛ ⎞= + + ⎜ ⎟

⎝ ⎠ (6.10.8.2.3-5)


xc

SF F R F F FS

⎛ ⎞= ≥⎜ ⎟

⎝ ⎠ ( )( )ksi ksi0.7 50 35yrF = = (Pg 6-222)

-- 155 --



( )( ) 312 438" " 38.75"fc fth D t t= + + = + =

3 33 1 0.63 1 0.63

3 3 3fc fc fc ft ft ftw

fc ft

b t t b t tD tJb b

⎛ ⎞ ⎛ ⎞= + − + −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ (A6.3.3-9)

( )3 33 3 3

48 4 4(38")( ") (16")( ") "(2) 1 0.63 5.035 in3 3 16"

J⎡ ⎤⎛ ⎞⎛ ⎞= + − =⎢ ⎥⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎣ ⎦

ksi

ksi

29,000 828.635yr

EF

= =

( )( )3

4

543.1 in 38.75"4,180

5.035 inxcS hJ

= =

( )( )( )21 4,1801.95 4.220" 828.6 1 1 6.76 396.8" 33.06 '

4,180 828.6rL ⎛ ⎞= + + = =⎜ ⎟⎝ ⎠


( ) 1 1 yr xc b pnc LTB b pc yc pc yc

pc yc r p

F S L LM C R M R M

R M L L

⎡ ⎤⎛ ⎞⎛ ⎞−= − − ≤⎢ ⎥⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟−⎢ ⎥⎝ ⎠⎝ ⎠⎣ ⎦

(A6.3.3-2)

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

( )( )( ) ( )( )

ksi 3k-in k-in

( ) k-in

k-in k-in k-in

35 543.1 in 144" 101.6"1 1 1.087 27,160 1.087 27,160396.8" 101.6"1.087 27,160

0.9488 29,520 28,010 29,520

nc LTB b

b b

M C

C C

⎡ ⎤⎛ ⎞ −⎛ ⎞⎢ ⎥⎜ ⎟= − − ≤⎜ ⎟⎜ ⎟ −⎢ ⎥⎝ ⎠⎝ ⎠⎣ ⎦

= = ≤

-- 156 --




2

1 1

2 2

1.75 1.05 0.3 2.3bM MCM M

⎛ ⎞ ⎛ ⎞= − + ≤⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ (A6.3.3-7)

k-ft2

k-ft

k-ft,

1,080

810.0

1,013

c

o B

BC mid

M M

M M

M

= =

= =

=

Since the BMD is not concave, ( ) ( )k-ft k-ft k-ft k-ft

1 22 (2) 1,013 1,080 946 810mid oM M M M= − ≥ = − = ≥

2946 9461.75 1.05 0.3 1.061 2.31,080 1,080

1.061

bC ⎛ ⎞ ⎛ ⎞= − + = ≤⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

=

( )( )k-in k-in k-in

( ) 1.061 28,010 29,720 29,520nc LTBM = = ≤ k-in k-ft

( ) 29,520 2,460nc LTBM = = The governing strength for the compression flange is the smaller of Mnc(FLB) and Mnc(LTB):

Since k-ft k-ft( ) ( )2, 460 > 2, 296nc LTB nc FLBM M= = , ∴ FLB governs the strength of the

compression flange.

k-ft( ) 2, 296nc nc FLBM M= =

( )k-ft k-ft(1.00) 2,296 2,296ncMφ = =

Check 13u l xc f ncM f S M+ ≤ φ (A6.1.1-1)

k-ft1,080u CM M= = Assume that Strength I Load Combination Governs, ∴ γWS = 0.0 and fl = 0

Since k-ft k-ft1 1,080 =2,2963u l xc f ncM f S M+ = < φ , the compression flange is adequate.

-- 157 --



Investigate the Strength of the Tension Flange: Since the tension flange is discretely braced, ( )k-in k-in k-ft(1.087) 27,160 29,520 2,460nt pt ytM R M= = = = (A6.4-1) ( )k-ft k-ft(1.00) 2,460 2,460ntMφ = =

Check 13u l xt f ntM f S M+ ≤ φ (6.10.8.1.2-1)

k-ft1,080u CM M= =

Since k-ft k-ft1 1,080 =2,4603u l xt f ntM f S M+ = < φ , the tension flange is adequate.

Since both flanges are adequate, the section is adequate for flexure.

Note that the benefits of using Appendix A6 are illustrated here. Even though the capacity was found to be adequate in both Examples #6a and #6b, using Appendix A6, the capacity was found to be 16% greater than the capacity found using the provisions in Section 6.10.8.

-- 158 --

ODOT-LRFD Short Course - Steel AASHTO Shear Strength Example #1


AASHTO Shear Strength Example #1: Problem: Check the beam shown below to see if it has adequate shear strength and web strength to resist the factored loads shown. The beam is a W27x94 made of M270-50 steel.

Solution: Draw the shear force diagram.

From the diagram, Vu = 60kip.

Referring to Section 6.10.9.2 of the Specification, Check Design Shear Strength:

Vn = CVp (6.10.9.2-1)

0.58p y wV F Dt= (6.10.9.2-2)

2 26.9" (2)(0.745") 25.41"fD d t= − = − =

[ ]ksi kip(0.58)(50 ) (25.41")(0.490") 361.1pV = =

75kip

12' 12'12'

30kip

SFD(kip)

60kip

-15kip

-45kip

-- 159 --



Since the web is unstiffened, k = 5.00.

( )( )( )

ksi

ksi

29,000 5.0053.85

50y

EkF

= =

( )( )1.12 1.12 53.85 60.31y

EkF

= = 25.41" 51.860.490"w

Dt= =

Since 51.86 1.12 60.31w y

D Ekt F

= < = , shear yielding governs and,

C = 1.00 (6.10.9.3.2-4) Vn = CVp = (1.00)(361.1kip) = 361.1kip φVn = (1.00)(361.1kip) = 361.1kip > Vu = 60kip O.K.

-- 160 --



AASHTO Shear Strength Example #2: Problem: A built-up section made of M270-50 steel, is used as a beam. Determine the design shear capacity of the beam and determine if the beam can sustain a factored shear force of 242kip. Solution: Referring to Section 6.10.9.2 of the Specification:

Vn = CVp (6.10.9.2-1)

0.58p y wV F Dt= (6.10.9.2-2)

[ ]ksi kip38(0.58)(50 ) (38")( ") 413.3pV = =

Since the web is unstiffened, k = 5.00.

( )( )( )

ksi

ksi

29,000 5.0053.85

50y

EkF

= =

( )( )1.12 1.12 53.85 60.31y

EkF

= =

( )( )1.40 1.40 53.85 75.39y

EkF

= = 3

8

38" 101.3"w

Dt= =

Since 101.3 1.40 75.39w y

D Ekt F

= > = , Elastic shear buckling governs and,

( )( )( )

( )( )( )

ksi

2 2 ksi

29,000 5.001.571.57 0.443750/ 101.3yw

EkCFD t

⎛ ⎞⎛ ⎞⎜ ⎟= = =⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

(6.10.9.3.2-6)

Vn = CVp = (0.4437) (413.3kip) = 183.4kip φVn = (1.00)(183.4kip) = 183.4kip < Vu = 242kip

-- 161 --



Try adding transverse stiffeners to the web to increase the shear strength. A panel aspect ratio of 1.25 to 1.50 looks good…

o48"1.25 d 1.25 (1.25)(38") 47.5" say 48" 1.26338"

o od dDD D

→ = = ∴ = =

For stiffened webs,

2 2

5 55 5 8.134( / ) (1.263)o

kd D

= + = + = (6.10.9.3.2-7)

( )( )( )

ksi

ksi

29,000 8.13468.68

50y

EkF

= =

( )( )1.12 68.68 1.12 76.92y

EkF

= =

( )( )1.40 68.68 1.40 96.15y

EkF

= =

Since 101.3 1.40 96.15w y

D Ekt F

= > = , Elastic shear buckling governs and,

( )( )( )

( )( )( )

ksi

2 2 ksi

29,000 8.1341.571.57 0.721850/ 101.3yw

EkCFD t

⎛ ⎞⎛ ⎞⎜ ⎟= = =⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

(6.10.9.3.2-6)

Vn = CVp = (0.7218) (413.3kip) = 298.3kip φVn = (1.00)(298.3kip) = 298.3kip > Vu = 242kip

-- 162 --



The previous calculations were based on the buckling strength of the web. For interior panels where:

( )2 2.5w

fc fc ft ft

Dtb t b t

≤+

(6.10.9.3.2-1)

[ ]3

8

3 34 4

(2)(38")( ") 28.5 1.188 2.5(16")( ") (16")( ") 24.0

= = ≤+

OK

Tension Field Action can be developed:

2

0.87(1 )

1n p

o

CV V CdD

⎡ ⎤⎢ ⎥

−⎢ ⎥= +⎢ ⎥⎛ ⎞⎢ ⎥+ ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

(6.10.9.3.2-2)

( )kip kip

2

(0.87)(1 0.7218)(413.3 ) (0.7218) 360.41 1.263

nV⎡ ⎤

−⎢ ⎥= + =⎢ ⎥+⎣ ⎦

φVn = (1.00)(360.4kip) = 360.4kip

-- 163 --

-- 164 --

ODOT-LRFD Short Course - Steel AASHTO Web Strength Example #1


AASHTO Web Strength Example #1: Problem: Check the beam shown below to see if it has adequate shear strength and web strength to resist the factored loads shown. The beam is a W27x94 made of M270-50 steel.

Solution: Draw the shear force diagram.

Referring to Section D6.5 of the Specification, Check the End Reactions for Web Yielding and Web Crippling: (Assume that the bearing length, N, is 3-1/4”.) Check Web Yielding

Since the supports are likely to be at a distance less than or equal to d from the end of the member:

(2.5 )n yw wR k N F t= + (D6.5.2-3)

ksi kip((2.5)(1.34") 3.25")(50 )(0.490") 161.7nR = + =

kip kip(1.0)(161.7 ) 161.7nRφ = = > 60kip O.K.

SFD(kip)

60kip

-15kip

-45kip

75kip

12' 12'12'

30kip

-- 165 --



Check Web Crippling

Since the supports are likely to be at a distance less than or equal to d/2 from the end of the member.

Check Nd

:

3.25" 0.1208 0.2026.9"

= ≤

Therefore, (D6.5.3-3) controls: 1.5

20.40 1 3 yw fwn w

f w

EF ttNR td t t

⎡ ⎤⎛ ⎞⎛ ⎞⎢ ⎥= + ⎜ ⎟⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎝ ⎠⎣ ⎦

1.5 ksi ksi

2 3.25" 0.490" (29,000 )(50 )(0.745")0.40(0.490") 1 326.9" 0.745" 0.490"nR

⎡ ⎤⎛ ⎞⎛ ⎞= +⎢ ⎥⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎢ ⎥⎣ ⎦

2 ksi kip(0.09604 in )(1.193)(1485 ) 170.2nR = =

kip kip(0.80)(170.2 ) 136.2nRφ = = > 60kip O.K.

Check the Interior Concentrated Loads for Web Yielding and Web Crippling: (Assume that the bearing length, N, is 3.25”) Check Web Yielding

Since the applied load is located at a distance greater than d from the end of the member.

(5 )n yw wR k N F t= + (D6.5.2-2)

ksi kip((5)(1.34") 3.25")(50 )(0.490") 243.8nR = + =

kip kip(1.0)(243.8 ) 243.8nRφ = = > 75kip O.K.

-- 166 --



Check Web Crippling

Since the applied load is located at a distance greater than d/2 from the end of the member.


20.80 1 3 yw fwn w

f w

EF ttNR td t t

⎡ ⎤⎛ ⎞⎛ ⎞⎢ ⎥= + ⎜ ⎟⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎝ ⎠⎣ ⎦

1.5 ksi ksi

2 3.25" 0.490" (29,000 )(50 )(0.745")0.80(0.490") 1 326.9" 0.745" 0.490"nR

⎡ ⎤⎛ ⎞⎛ ⎞= +⎢ ⎥⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎢ ⎥⎣ ⎦

2 ksi kip(0.1921 in )(1.193)(1485 ) 340.3nR = =

kip kip(0.80)(340.3 ) 272.2nRφ = = > 75kip O.K.

The Web Yielding and Web Crippling Strengths are Satisfactory

-- 167 --



AASHTO Web Strength Example #2: Problem: A built-up section made of M270-50 steel, is used as a beam. It was determined in AASHTO Shear Strength Example #2 that intermediate stiffeners were required to develop adequate shear strength in the web. Determine the required size of these intermediate stiffeners. And check the web to see if an end reaction of 128kip can be supported. Solution:

Design the intermediate stiffeners that were added to increase the shear strength: The moment of inertia of the intermediate stiffeners should satisfy the smaller of: 3

t wI bt J≥ (6.10.11.1.3-1) and

1.54 1.3

40ywt

t

FDIE

ρ ⎛ ⎞≥ ⎜ ⎟

⎝ ⎠ (6.10.11.1.3-2)

where: It - Moment of inertia of the stiffener pair about the mid-thickness of the web.

2

2.5 2.0 0.5/o

DJd D

⎛ ⎞= − ≥⎜ ⎟

⎝ ⎠ use…

( )22.5 2.0 0.5/o

Jd D

= − ≥ (6.10.11.1.3-3)

-- 168 --



( )

22

2.5 2.0 0.4332 in 0.548"/ 38"

J = − = − ≥ take J = 0.50

b = smaller of do and D, b = 38” ρ = larger of Fyw / Fcrs and 1.00

ksi

ksi ksi2 2

38

0.31 (0.31)(29,000 ) 35.11 50 6"

"

crs ys crs

t

p

EF F Fbt

= ≤ = = ≤⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

50 / 35.11 1.00 =1.424ρ ρ= ≥ 3 3 43

8(38")( ") (0.5) 1.002 inwbt j = =

1.5 1.54 1.3 4 1.3 ksi

4ksi

(38") (1.424) 50 5.909 in40 40 29,000

ywt FDE

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

ρ

3( )(2 )

12s s w

tt b tI +

= take bs = 6”

[ ]338 3( ) (2)(6") "

( )157.9 in12

st s

tI t

+= =

3 4

3 3

1.002 in 0.006345"157.9 in 157.9 in

wp

bt Jt ≥ = = …say 38 "st =

Using a 6” wide stiffener is based on the assumption that 6” bar stock can be used, which should be readily available. Base the width of 3/8” on engineering judgment of minimum thickness of stiffener.

-- 169 --



Referring to SectionD6.5 of the Manual, Check the End Reactions for Web Yielding and Web Crippling: The bearing length, N, is 9” and we’ll assume that 3/8” fillet welds connect the flanges and web. This gives an effective “k” distance of 3/4” + 3/8” = 1.125” Check Web Yielding

Since the supports are likely to be at a distance less than or equal to d from the end of the member.

(2.5 )n yw wR k N F t= + (D6.5.2-3)

ksi kip38((2.5)(1.125") 9")(50 )( ") 221.5nR = + =

kip kip(1.0)(221.5 ) 221.5nRφ = = > 128kip O.K.

Check Web Crippling

Since the supports are likely to be at a distance less than or equal to d/2 from the end of the member.

Check Nd

:

34

9" 0.2278 0.2038" (2)( ")

= >+


2 40.40 1 0.2 yw fwn w

f w

EF ttNR td t t

⎡ ⎤⎛ ⎞⎛ ⎞⎢ ⎥= + − ⎜ ⎟⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎝ ⎠⎣ ⎦

1.5 ksi ksi3 3

2 8 438 3 3 3

4 4 8

(4)(9") " (29,000 )(50 )( ")(0.40)( ") 1 0.2(38" (2)( ") " ( ")nR

⎡ ⎤⎛ ⎞⎛ ⎞= + −⎢ ⎥⎜ ⎟⎜ ⎟+ ⎝ ⎠⎝ ⎠⎢ ⎥⎣ ⎦

2 ksi kip(0.05625 in )(1.252)(1,703 ) 119.9nR = =

kip kip(0.80)(119.9 ) 95.92nRφ = = <128kip No Good.

The web strength is satisfactory with regard to web yielding but not for web crippling. Bearing stiffeners will need to be added. (Technically speaking, stiffeners are required by AASHTO at all bearing locations on built-up sections anyways…)

-- 170 --



Design the bearing stiffeners that need to be added to increase the web crippling strength: Check local buckling of the bearing stiffener:

0.48t py

Eb tF

≤ take Fy = 50ksi (6.10.11.2.2-1)

ksi

ksi

29,0000.48 11.5650t p pb t t≤ = take bs = 7”

7" 0.6055"11.56 11.56

tp

bt ≥ = = take 58 "st = (7” x 5/8” bar stock may be used)

Check the bearing stiffeners as an effective column section:

( ) ( )( ) ( )( )3 35 3 5 3112 8 8 8 8

4

" 7" " 7" 6.75" " "

154.7 in

I ⎡ ⎤= + + + −⎣ ⎦=

( )( )( ) ( )( ) 25 3

8 8" 2 7" 6.75" " 11.28 inA = + =⎡ ⎤⎣ ⎦

4

2

154.7 in 3.704"11.28 in

IrA

= = =

(0.75)(38") 7.695

3.704"KLr

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2 2 ksi

ksi

7.695 36 0.00744829,000

yFKLr E

⎛ ⎞⎛ ⎞ ⎛ ⎞λ = = =⎜ ⎟⎜ ⎟ ⎜ ⎟π π⎝ ⎠ ⎝ ⎠ ⎝ ⎠ Inelastic Buckling (6.9.4.1-3)

0.66n y sP F Aλ=

( )( ) ( )( )0.007448 ksi 2 kip0.66 36 11.28 in 404.8= = (6.9.4.1-1)

( )( )kip kip0.90 404.8 364.3nPφ = =

In this solution, it is assumed that the bearing stiffener is located at the middle of the 9” wide plate. Thus, there is at least 4.5” of web between the stiffener and the end of the girder, which is greater than 9tw.

Taking Fy = 50ksi here is a conservative assumption since I am not sure what material will actually be used.

Taking Fy = 36ksi here is a conservative assumption since I am not sure what material will be used.

-- 171 --



Check bearing stress between the end of the bearing stiffeners and the loaded flange: 1.4n pn ysR A F= (6.10.11.2.3-1)

The corners of the stiffeners are clipped 1” horizontal and 2 1/2” vertical to provide clearance for the flange-to-web welds

( ) 25

8(2)(7" 1") " 7.50 inpnA = − = 2 ksi kip(1.4)(7.50 in )(36 ) 378.0nR = = kip kip(1.00)(378.0 ) 378.0nRφ = =

The capacity of the bearing stiffeners is governed by the “equivalent column” capacity. φRn = 364kip, which is greater than the reaction of 128kip.

-- 172 --



Just for fun ☺, lets check the capacity of 2 pairs of 7” x 5/8” interior bearing stiffeners: The local buckling check will be the same as for the single pair of bearing stiffeners. Check the bearing stiffeners as an effective column section:

( ) ( )( ) ( )( )( )3 3 45 3 5 31

12 8 8 8 8(2) " 7" " 7" (2)(3.375") 7" (2) " " 309.5 inI ⎡ ⎤= + + + + − =⎣ ⎦

( )( )( ) ( ) ( ) 25 3

8 8" 4 7" (2) 3.375" 7" " 22.66 inA ⎡ ⎤= + + =⎡ ⎤⎣ ⎦⎣ ⎦

4

2

309.5 in 3.696"22.66 in

IrA

= = = (0.75)(38") 7.7113.696"

KLr

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2 2 ksi

ksi

7.711 36 0.00747929,000

yFKLr E

⎛ ⎞⎛ ⎞ ⎛ ⎞λ = = =⎜ ⎟⎜ ⎟ ⎜ ⎟π π⎝ ⎠ ⎝ ⎠ ⎝ ⎠ Inelastic Buckling (6.9.4.1-3)

( )( )( )( )0.007479 ksi 2 kip0.66 0.66 36 22.66 in 813.2n y sP F Aλ= = = (6.9.4.1-1)

( )( )kip kip0.90 813.2 731.9nPφ = = The bearing stress between the end of the bearing stiffeners and the loaded flange would be twice that calculated for a single pair of stiffeners: φRn = (2)(378.0kip) = 756.0kip. The strength is governed again by the “equivalent column” capacity, φRn = 732kip.

-- 173 --

-- 174 --

ODOT-LRFD Short Course - Steel AASHTO Connection Example #1


AASHTO Connection Example #1: Problem: A C12x30 is used as a tension member as is shown in the sketch below. The channel is made of M270-36 material and is attached to the gusset plate with 8, 7/8” diameter M164 (A325) bolts. The gusset is 5/8” thick and made of M270-36 steel. Calculate the design capacity, φPn, of the connection considering the failure modes of bolt shear, bolt bearing, and block shear. Also compute the load which will cause slip of the connection. Solution: Shear Strength of the Bolts:

Assume that the threads are included in the shear plane of the connection.

0.38n b ub sR A F N= (6.13.2.7-2)

( )2 278 " 0.6013 in

4bA π⎛ ⎞= =⎜ ⎟⎝ ⎠

For A325 bolts, Fub = 120ksi

Bolts are in single shear so Ns = 1

( )( ) kip2 ksibolt(0.38) 0.6013 in 120 (1) 27.42nR = =

( )kip kipbolt bolt(0.80) 27.42 21.94nRφ = =

For all 8 bolts, ( )kip kip

bolt(8 bolts) 21.94 175.5nRφ = =

A

3"

3"3"

1.5"

3" 6"

Pu

A

Section A-A

C12 x 30

3"

-- 175 --



Check Bearing Strength: Interior Bolts Bearing on the Channel Web: ( )( )( )71 1

2 8 163" 2 " " 2.063"cL = − + = Since Lc = 2.063” > 2d = 1.75”, 2.4n uR dtF= ( ) ( ) kipksi7

8 bolt(2.4) " (0.510") 58 62.11nR = = (6.13.2.9-1)

Interior Bolts Bearing on the Gusset Plate: ( )( )( )71 1

2 8 163" 2 " " 2.063"cL = − + = Since Lc = 2.063” > 2d = 1.75”, 2.4n uR dtF= ( )( )( ) kipksi7 5

8 8 bolt(2.4) " " 58 76.13nR = = (6.13.2.9-1)

End Bolts Bearing on the Channel Web: ( )( )71 1

2 8 161.5" " " 1.031"cL = − + = Since Lc = 1.031” < 2d = 1.75”, 1.2n c uR L tF= ( ) ( ) kipksi

bolt(1.2) 1.031" (0.510") 58 36.60nR = = (6.13.2.9-2)

End Bolts Bearing on the Gusset Plate: (Assume that the end distance on the gusset is 11/2”) ( )( )71 1

2 8 161.5" " " 1.031"cL = − + = Since Lc = 1.031” < 2d = 1.75”, 1.2n c uR L tF= ( )( )( ) kipksi5

8 bolt(1.2) 1.031" " 58 44.85nR = = (6.13.2.9-2)

For all 8 Bolts: ( ) ( ) ( )kip kip kip kip

bolt bolt bolt(2 bolts) 44.85 (4 bolts) 62.11 (2 bolts) 36.60 411.3nR = + + =

( )kip kip(0.80) 411.3 329.1nRφ = =

-- 176 --



Since the channel web is thinner than the gusset plate and they’re made of the same material, block shear of the channel will govern over block shear of the gusset plate. Check Block Shear in the Channel Web: 2(6")(0.510") 3.060 intgA = = ( )( ) 271 1

2 8 8(6") (2) " " (0.510") 2.550 intnA = − + =⎡ ⎤⎣ ⎦ [ ] 2(2) (1.5") (3)(3") (0.510") 10.71 invgA = + = ( ) 27 1

8 8(2) (1.5") (3)(3") (3.5) " " (0.510") 7.140 invnA = + − + =⎡ ⎤⎣ ⎦

?

?2 2 2

0.58

2.550 in (0.58)(7.140 in ) 4.141 in NO!

tn vnA A≥

≥ =

0.58n u vn y tgR F A F A∴ = + (6.13.4-2) ( )( ) ( )( )ksi 2 ksi 2 kip (0.58) 58 7.140 in 36 3.060 in 350.3nR = + =

( )kip kip(0.80) 350.3 280.3nRφ = =

The Shear Strength of the Bolts Governs, φRn = 176kip

-- 177 --



Check the Slip Capacity of the Connection: n h s s tR K K N P= (6.13.2.8-1)

Kh = 1.00 (for standard holes) Ks = 0.33 (assume Class A surface) Ns = 1 Pt = 39kip (from Table 6.13.2.8-1 for M164 Bolts)

( ) kipkip

bolt(1.00)(0.33)(1) 39 12.87nR = =

For All 8 Bolts: ( )kip kip

bolt(8 bolts) 12.87 103.0nR = =

-- 178 --



AASHTO Connection Example #2: Problem: An 8” long WT 10.5 x 66 is attached to the bottom flange of a beam as is shown below. This hanger must support a factored load of 120kip. Given that 4, 1” diameter M164 (A325) bolts are used to attach the hanger to the beam, investigate the adequacy of the bolts and tee flange. Solution: Prying must be investigated in this situation:

33

8 20ub tQ Pa

⎛ ⎞= −⎜ ⎟⎝ ⎠

(6.13.2.10.4-1)

181

7" 1 " 2.375"2 2

tgb k= − = − = (k1 = 11/8” for W21 x 132)

( )( ) ( )( )1 1

2 2 12.4" 7" 2.700"f ta b g= − = − =

3(3)(2.375") (1.04") 0.2736

(8)(2.700") 20u u uQ P P⎛ ⎞

= − =⎜ ⎟⎝ ⎠

( )kip kip1.274 (1.274) 120 152.8u u u uT Q P P= + = = = Tensile Resistance of the Bolts: 0.76n b ubT A F= (6.13.2.10.2-1)

( )2 21" 0.7854 in4bA π⎛ ⎞= =⎜ ⎟

⎝ ⎠

Fub = 120ksi

( )( ) kip2 ksi

bolt(0.76) 0.7854 in 120 71.63nT = = ( )kip kipbolt bolt(0.80) 71.63 57.30nTφ = =

For All 4 Bolts: ( )kip kip

bolt(4 bolts) 57.30 229.2nTφ = = OK

-- 179 --



Check the Strength of the Flange of the WT: Assume that a plastic mechanism forms between the bolt lines and stem. Moment Equilibrium about at the fillet:

22u

uP bM M ⎛ ⎞→ = ⎜ ⎟

⎝ ⎠∑

( )kip

k-in

120 (2.375")4

71.25

uM⎛ ⎞⎜ ⎟=⎜ ⎟⎝ ⎠

=

For Safety, p uM Mφ ≥ . ( )2

ksi k-in(8")(1.04") 50 108.24 4p y

LtM F ⎛ ⎞= = =⎜ ⎟⎝ ⎠

( )k-in k-in(1.00) 108.2 108.2pMφ = = OK

Mu

Pu/2

Mu

Tu

QQ T T

Pu

b

-- 180 --



AASHTO Connection Example #3: Problem: Assuming an unfactored fatigue load of 60kip, determine the fatigue life of the tension bolts in the previous example. Solution: For Safety, ( ) ( )n

f Fγ Δ ≤ Δ (6.6.1.2.2-1)

( ) ( ) ( )( )( )

kipksi

2

(0.75) 1.274 6018.24

(4) 0.7854 inbolts

Pf

A

⎡ ⎤γ Δ ⎣ ⎦γ Δ = = =

( ) ( )13

2TH

n

FAFN

Δ⎛ ⎞Δ = ≥⎜ ⎟⎝ ⎠

For infinite life, ( ) ( ) ksiksi31.0 15.5

2 2TH

n

FF

ΔΔ = = =

Since ( ) ( )ksi ksi18.24 15.5

nf Fγ Δ = > Δ = , the bolts will have a finite life

For finite life, ( ) ( )13

n

Af FN

⎛ ⎞γ Δ ≤ Δ = ⎜ ⎟⎝ ⎠

( )( ) ( )

8 3

3 3ksi

17.1 10 ksi 281,800 cycles18.24

ANf

×≤ = =

γ Δ

Note that if prying is not included, ( ) ksi14.32fγ Δ = and the calculations would incorrectly show that the bolts have an infinite fatigue life.

-- 181 --



AASHTO Connection Example #4: Problem: Suppose that the hanger depicted in Examples #2 and #3 is subjected to a force that is applied at an angle as is shown below. Determine if the connection is adequate in this configuration. Solution:

kip2 0.8944 107.35u

u uPV P= = =

kip

kipbolt

107.3 26.834 boltsuV = =

kip1.274 0.5694 68.335

uu u

PT P= = =

kip

kipbolt

68.33 17.084 boltsuT = =

Assume that the threads are included in the shear plane of the connection. 0.38n n b ub sV R A F N= = (6.13.2.7-2)

( )( ) kip2 ksibolt(0.38) 0.7854 in 120 (1) 35.81nV = =

kipboltkipbolt

26.83 0.749135.81

u

n

VV

= = , 2

0.76 1 un b ub

n

VT A FV

⎛ ⎞∴ = − ⎜ ⎟φ⎝ ⎠

(6.13.2.11-2)

( )( ) ( )( )

2kipbolt kip2 ksi

boltkipbolt

26.83(0.76) 0.7854 in 120 1 25.11

(0.80) 35.81nT⎛ ⎞

= − =⎜ ⎟⎜ ⎟⎝ ⎠

( )kip kip

bolt bolt(0.80) 25.11 20.09nTφ = = Since kip kip

bolt bolt20.09 17.08n uT Tφ = > = , the bolts are OK for the loading shown.

-- 182 --



Check Bearing of the Flange of the WT:

It is given that the WT is 8” long. Since the minimum spacing is 3”, we’ll assume that an end distance of 2” is provided resulting in a spacing of 4” bolt-to-bolt.

Interior Bolts: ( )( )( )1 1

2 164" 2 1" " 2.938"cL = − + = Since Lc = 2.938” > 2d = 2”, 2.4n uR dtF= (6.13.2.9-1) ( ) ( ) kipksi

bolt(2.4) 1" (1.04") 65 162.2nR = =

End Bolts: ( )( )1 1

2 162" 1" " 1.469"cL = − + = Since Lc = 1.469” < 2d = 2”, 1.2n c uR L tF= (6.13.2.9-2) ( ) ( ) kipksi

bolt(1.2) 1.469" (1.04") 65 119.1nR = =

For all 4 Bolts: ( ) ( )kip kip kip

bolt bolt(2 bolts) 162.2 (2 bolts) 119.1 562.8nR = + =

( )kip kip(0.80) 562.8 450.1nRφ = =

Since kip kip450 107n uR Vφ = > = , the flange of the WT will be OK in bearing.

Note that since the flange thickness of the W24x176 is greater than that of the WT10.5x66 and they are made of the same material, bearing of the WT will govern over bearing of the W24x176.

-- 183 --



Check Shear in the Stem of the WT: 0.58n g yR A F= (6.13.5.3-2) [ ]( )ksi kip(0.58) (8")(0.650") 50 150.8nR = = ( )kip kip(1.00) 150.8 150.8nRφ = = Since kip kip150.8 107.3n uR Vφ = > = , the stem will be satisfactory in shear.

-- 184 --



AASHTO Connection Example #5: Problem: An L6 x 4 x 1/2, M270-36, is welded to a 3/8” thick gusset plate made of M270-50 steel. The long leg of the angle is attached using two, 8” long fillet welds. The capacity of the angle was previously computed as φPn = 163kip based on Gross Yielding. Determine the weld size required to develop the full capacity of the member. Solution: Design the Welds: Use an E70 Electrode since the gusset has a strength of Fu = 65ksi. The maximum weld size is 1/2” - 1/16” = 7/16”. Since the gusset and angle are both less than 3/4” thick, the minimum weld size that can be used is 1/4”. , 2 ,0.6n weld e exx w n memberR F A Pφ = φ ≥ φ

( )( )( ) ( ) ( )( )ksi kip, 0.6 0.80 70 (0.7071) 2 8" 163n weldR wφ = × ≥⎡ ⎤⎣ ⎦

kip

716kip

inch

163 0.4288" Say "380.1

w ≥ = →

-- 185 --



Check Tension for the Gusset: Check tension on the Whitmore section: Since the overall width of the gusset is not given, I’ll check the Whitmore width assuming that it governs. If the overall width is less than the Whitmore width, these calculations will be unconservative. Compute the width of the Whitmore section: ( )( ) ( )6" 2 8" 30 15.24"wL Tan= + ° = Gross Section Yielding: ( ) ( )( )ksi kip3

8(0.95) 50 15.24" " 271.4n y gP F Aφ = φ = =⎡ ⎤⎣ ⎦

Net Section Fracture:

( ) ( )( ) ( )ksi kip38(0.80) 65 15.24" " 1.00 297.1n u eP F Aφ = φ = =⎡ ⎤⎣ ⎦ (Taking U = 1.00)

Check Block Shear in the Gusset Plate: ( ) 23

8(6") " 2.250 intg tnA A= = = ( )( ) 238(2) 8" " 6.000 invg vnA A= = =

? ?

2 2 20.58 2.250 in (0.58)(6.000 in ) 3.480 in NO!tn vnA A≥ → ≥ = 0.58n u vn y tgR F A F A∴ = + (6.13.4-2) ( )( ) ( )( )ksi 2 ksi 2 kip (0.58) 65 6.000 in 50 2.250 in 338.7nR = + =

( )kip kip(0.80) 338.7 271.0nRφ = =

Use 7/16” x 8” Fillet Welds made with an E70 Electrode

-- 186 --

ODOT-LRFD Short Course - Steel AASHTO Connection Example #6a


AASHTO Connection Example #6a: Problem: Use the elastic vector method to compute the maximum force on any bolt in the

eccentrically loaded bolt group shown in the figure below. The bolts are all the same size. (Example 4.12.1 from Salmon & Johnson)

B

C D

E F

P3"4"

3"3"

A

Solution:

( )( )

( )( ) ( )( ) ( )

( )( )

( ) ( )

kip k-in

2 2

2 2 22 2

4

k-in 2 2

4

ksi

2ksi kip

24 3" 2" 120

4 (2") (3") 2 2" 1"4

47.12 in

120 (2") (3")Corner Bolts:

47.12 in9.182

9.182 1" 7.2114

Tr TJ

J Ad A d

J

J

TrJ

V

τ = → = + =

= =

π⎛ ⎞⎡ ⎤= + + ⎜ ⎟⎢ ⎥⎣ ⎦ ⎝ ⎠=

+τ = =

τ =

π⎛ ⎞= =⎜ ⎟⎝ ⎠

∑ ∑

-- 187 --

ODOT-LRFD Short Course - Steel AASHTO Connection Example #6a


Force acts perpendicular to line drawn from bolt to C.G. Breaking force into horizontal and vertical components…

( ) ( )

( ) ( )

kip kip

kip kip

2 2 7.211 4.0003.606 3.606

3 3 7.211 6.0003.606 3.606

y

x

V V

V V

⎛ ⎞ ⎛ ⎞= = = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Add evenly distributed vertical force to the vertical, torsional force for Bolt B…

kipkip kip

kip

2 2 kip 2 kip 2 kip

244.000 8.0006

6.000

( 8.000 ) (6.000 ) 10.00

y

x

total x y

V

V

V V V

−= − + = −

=

= + = − + =

Check These Results with a Spreadsheet Solution:

Px: 0 xCG: 0Py: -24 yCG: 0ex: 5ey: 0 Σ d2: 60.00

T = -120.0 Vmax = 10.0

Bolt x y d2 Vx Vy Vtotal

A -2.00 3.00 13.00 -6.0 4.0 6.0B 2.00 3.00 13.00 -6.0 -4.0 10.0C -2.00 0.00 4.00 0.0 4.0 0.0D 2.00 0.00 4.00 0.0 -4.0 8.0E -2.00 -3.00 13.00 6.0 4.0 6.0F 2.00 -3.00 13.00 6.0 -4.0 10.0

Everything checks out OK.

-- 188 --

ODOT-LRFD Short Course - Steel AASHTO Connection Example #6b


AASHTO Connection Example #6b: Problem: Use the simplified equations to solve the previous example problem. (Example

4.12.1 from Salmon & Johnson)

B

C D

E F

P3"4"

A

Solution:

( )( )

( ) ( )( )

kip k-in

22 2 2

2 2

24 3" 2" 120

4 (2") (3") 2 2"

60.00 in

T

d

d

= − − = −

⎡ ⎤= + +⎣ ⎦=

∑∑

Looking at Bolt B:

( )( )

( )( )

k-inkip

, 2

k-inkip

, 2

120 3"6.000

60.00 in

120 2"4.000

60.00 in

B x

B y

V

V

⎛ ⎞−⎜ ⎟= − =⎜ ⎟⎝ ⎠

⎛ ⎞−⎜ ⎟= = −⎜ ⎟⎝ ⎠

( ) ( )2kip2kip kip

,

kip,

246.000 4.0006

10.00

B total

B total

V

V

⎡ ⎤⎛ ⎞−= + − +⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦

=

-- 189 --



AASHTO Connection Example #7: Problem: Detail a splice between two non-composite W30 x 99 M270 Gr 50. Take Mu at

the location of the splice as 810k-ft and take φMn as 1,300k-ft.

Solution: The splice is designed for the larger of:

k-ft k-ft

, , k-ft810 1,300 1,0552 2

u Beam n BeamM M+ φ += =

( )( )k-ft k-ft

,0.75 0.75 1,300 975.0n BeamMφ = =

Since 1,055k-ft > 975.0k-ft, Mu,Splice = 1,055k-ft A) Flange Splice: In this case, it makes no difference which flange is the “controlling flange” and which one is the “non-controlling flange,” (Since the beam is non-composite and we are assuming that moment could be either positive or negative). For the Controlling flange:

1 0.75 2

cfcf f yf f yf

h

fF F F

R⎛ ⎞⎛ ⎞= + αφ ≥ αφ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

(6.13.6.1.4c-1)

( )( )( )k-ft 29.7" 0.670"in

ft 2 2 ksi4

810 1235.36

3,990 incff−

= =

Rh = 1.00, since the beam is non-hybrid. α = 1.00. Since we are assuming that φMn = φMp, Fn = Fyf

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

ksiksi ksi

ksi ksi ksi

1 35.36 1.00 1.00 50 0.75 1.00 1.00 502 1.00

42.68 37.50 42.68

cf

cf

F

F

⎛ ⎞⎛ ⎞= + ≥⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠= ≥ → =

-- 190 --



For the Non-Controlling Flange:

0.75 ncfncf cf f yf

h

fF R F

R= ≥ αφ (6.13.6.1.4c-3)

ksi35.36ncf cff f= =

ksi

ksi42.68 1.20735.36

cfcf

cf

FR

f= = =

( ) ( )( )( )( )

ksiksi

ksi ksi ksi

35.361.207 0.75 1.00 1.00 501.00

42.68 37.50 42.68

ncf

ncf

F

F

= ≥

= ≥ → =

For the Compression Flange: ,u Comp cf eP F A= In compression, Ae is taken as the gross area of the flange. ( )( ) 210.5" 0.670" 7.035 ine gA A= = = ( )( )ksi 2 kip

, 42.68 7.035 in 300.3u CompP = =

For the Tension Flange: ,u Ten cf eP F A=

In tension, u ue n g

y y

FA A AF

⎛ ⎞φ= ≤⎜ ⎟⎜ ⎟φ⎝ ⎠

(6.13.6.4c-2)

For 1” diameter bolts, ( ) ( )( ) ( ) 21

810.5" 2 1 " 0.670" 5.528 innA = − =⎡ ⎤⎣ ⎦

-- 191 --



( )( )( )( ) ( )

ksi2 2

ksi

2 2 2

0.80 655.528 in 7.035 in

0.95 50

6.052 in 7.035 in 6.052 in

e

e

A

A

⎛ ⎞⎜ ⎟= ≤⎜ ⎟⎝ ⎠

= ≤ → =

( )( )ksi 2 kip

, 42.68 6.052 in 258.3u TenP = =

Proceed assuming that the flange splice will consist of plate on both the outside and inside of the flange. Assume that the flange force will be equally distributed between in the inner and outer plates (we’ll check the validity of this assumption later). Also assume that the outer splice plate will be 10.5” wide (the same width as the flange) with two rows of 1” diameter M164 (A325) bolts.

kip

kip,

258.3 129.12u TenP = =

kipkip

,300.3 150.2

2u CompP = =

For the Outer Flange Splice Plate:

Gross Yielding (Tension): ,n y y g u TenP F A Pφ = φ ≥ (6.8.2.1-1)

( )( )( )( )ksi kip0.95 50 10.5" 129.1n outerP tφ = ≥

( )( )( )

kip

ksi

516

129.10.95 50 10.5"

0.2589" say "

outert ≥

≥ →

Net Section Fracture (Tension): ,n u u n u TenP F A U Pφ = φ ≥ (6.8.2.1-2)

( )( ) ( ) ( )( ) ( )( )ksi kip180.80 65 10.5" 2 1" " 1.00 129.1n outerP tφ = − + ≥⎡ ⎤⎣ ⎦

( )( )( )

kip

ksi

516

129.10.80 65 8.25"

0.3010" say "

outert ≥

≥ →

-- 192 --



Gross Yielding (Compression): ,n c y g u CompP F A Pφ = φ ≥ (6.13.6.1.4c-4)

( )( )( )( )ksi kip0.90 50 10.5" 150.2n outerP tφ = ≥

( )( )( )

kip

ksi

38

150.20.90 50 10.5"

0.3179" say "

outert ≥

≥ →

For the Inner Flange Splice Plates: The widths of the inner splice plates will be roughly equal the flange width of the section minus the thickness of the web and fillets. ( )( )1

1612 10.5" 2 1 " 8.375"inner fW b k= − = − = Take the width of each of the two inner plates as 4”.

Gross Yielding (Tension): ,n y g u TenP F A Pφ = φ ≥ (6.8.2.1-1)

( )( )( )( )( )ksi kip0.95 50 2 4.00" 129.1n InnerP tφ = ≥

( )( )( )( )

kip

ksi

38

129.10.95 50 2 4.00"

0.3297" say "

innert ≥

≥ →

Net Section Fracture (Tension): ,n u n u TenP F A U Pφ = φ ≥ (6.8.2.1-2)

( )( )( ) ( ) ( ) ( )( )ksi kip180.80 65 2 4.00" 1" " 1.00 129.1n InnerP tφ = − + ≥⎡ ⎤⎣ ⎦

-- 193 --



( )( )( )

kip

ksi

716

129.10.80 65 5.75"

0.4318" say "

innert ≥

≥ →

Gross Yielding (Compression): ,n c y g u CompP F A Pφ = φ ≥ (6.13.6.1.4c-4)

( )( )( )( )( )ksi kip0.90 50 2 4.00" 150.2n InnerP tφ = ≥

( )( )( )( )

kip

ksi

716

150.20.90 50 2 4.00"

0.4172" say "

Innert ≥

≥ →

For a flange splice with inner and outer splice plates, the flange design force at the strength limit state may be assumed divided equally to the inner and outer plates and their connections when the areas of the inner and outer plates do not differ by more than 10% (Commentary, Page 6-191). ( )( ) 23

810.5" " 3.938 inOuterA = = ( )( )( ) 27162 4.00" " 3.500 inInnerA = =

( ) ( ) ( )( ) ( )

2 2

2 2

2 3.938 in 3.500 in11.76%

3.938 in 3.500 inOuter Inner

Ave

A AA

−−= =

+

Since the difference area is greater than 10%, either (1) the assumption that the flange force is evenly divided between the outer and inner plates must be modified, or (2) the inner plate thickness must be increased to 1/2”, which would result in a difference in area between the outer and inner plates of less than 2%. The second option will be selected for the case of this example. Outer Flange Splice Plate: 101/2” x 3/8” Inner Flange Splice Plates: 4” x 1/8”

-- 194 --



Check Bolt Shear in the Flange Splice: Assume that the threads are included in the shear plane of the connection. Bolts are in double shear since both inside and outside splice plates are used. 0.38n b ub sR A F N= (6.13.2.7-2)

( )2 21" 0.7854 in4bA π⎛ ⎞= =⎜ ⎟

⎝ ⎠ For A325 bolts, Fub = 120ksi

( )( ) kip2 ksibolt(0.38) 0.7854 in 120 (2) 71.63nR = =

( )kip kipbolt bolt(0.80) 71.63 57.30nRφ = =

Determine the number of flange bolts required:

kip

kipbolt

300.3 5.24 bolts say 6 bolts57.30fbn = = →

2 1/2" 3 1/2" 2 1/2" 2 1/2" 2 1/2"3 1/2" 3 1/2" 3 1/2"

10 1 / 2"

2"6

1 / 2"2"

1/2" gap between ends of beams

PL 241/2” x 101/2" x 3/8"

PL 241/2" x 4" x 1/2" Each Side

W30 x 99W30 x 99

1" dia M164 Bolts (12 places)

24 1/2"

-- 195 --



Check Bolt Bearing in the Flange Splice: Interior Bolts Bearing on the Beam Flange: ( )( )( )1 1 1

2 2 163 " 2 1" " 2.438"cL = − + = Since Lc = 2.433” > 2d = 2.0”, 2.4n uR dtF= ( ) ( ) kipksi

bolt(2.4) 1" (0.670") 65 104.5nR = = (6.13.2.9-1)

Interior Bolts Bearing on the Splice Plates: ( )( )( )1 1 1

2 2 163 " 2 1" " 2.438"cL = − + = Since Lc = 2.433” > 2d = 2.0”, 2.4n uR dtF= ( )( )( ) kipksi3 1

8 2 bolt(2.4) 1" " " 65 136.5nR = + = (6.13.2.9-1)

End Bolts Bearing on the Beam Flange: ( )( )1 1 1

2 2 162 " 1" " 1.969"cL = − + = Since Lc = 1.969” < 2d = 2.0”, 1.2n c uR L tF= ( ) ( ) kipksi

bolt(1.2) 1.969" (0.670") 65 102.9nR = = (6.13.2.9-2)

End Bolts Bearing on the Splice Plates: ( )( )1 1 1

2 2 162 " 1" " 1.969"cL = − + = Since Lc = 1.969” < 2d = 2.0”, 1.2n c uR L tF= ( )( )( ) kipksi3 1

8 2 bolt(1.2) 1.969" " " 65 134.4nR = + = (6.13.2.9-2)

For all 6 Bolts: ( ) ( ) ( )kip kip kip kip

bolt bolt bolt(2 bolts) 104.5 (2 bolts) 104.5 (2 bolts) 102.9 623.9nR = + + =

( )kip kip(0.80) 623.9 499.1nRφ = = OK

-- 196 --



Check Slip of the Flange Splice: Bolted connections for flange splices shall be designed as slip-critical connections for the flange design force. As a minimum, for checking slip of the flange splice bolts, the design force for the flange under consideration shall be taken as the Service II design stress, Fs, times the smaller gross flange area on either side of the splice. Take the Service II moment as 548k-ft

slip s gP F A= where ( )( )( )

( )( )k-ft 29.7" 0.670"in

ft 2 2 ksi4

548 1223.92

1.00 3,990 ins

sh

fFR

−= = = (6.13.6.1.4c-5)

( )( )( )ksi kip23.92 10.5" 0.670" 168.3slipP = =

The slip resistance of a single bolt is taken as: n h s s tR K K N P= (6.13.2.8-1)

Kh = 1.00 (for standard holes) Ks = 0.33 (assume Class A surface) Ns = 2 Pt = 51kip (from Table 6.13.2.8-1 for M164 Bolts)

( ) kipkip

bolt(1.00)(0.33)(2) 51 33.66nR = =

Determine the number of flange bolts required:

kip

kipbolt

168.3 5.00 bolts 6 bolts will work33.66fbn = = →

-- 197 --



Check Block Shear of the Beam Flange:

( )( )( ) 22 2" 0.670" 2.680 intgA = =

( ) ( )( ) ( ) 21 1822 2" 1" " 0.670" 1.926 intnA ⎡ ⎤= − + =⎣ ⎦

( )( )( ) 21 1 1

2 2 22 3 " 3 " 2 " 0.670" 12.73 invgA = + + =

( ) ( )( ) ( ) 21 12 82 9 " 2.5 1" " 0.670" 8.961 invnA = − + =⎡ ⎤⎣ ⎦

?

?2 2 2

0.58

1.926 in (0.58)(8.961 in ) 5.198 in NO!

tn vnA A≥

≥ =

0.58n u vn y tgR F A F A∴ = + (6.13.4-2) ( )( ) ( )( )ksi 2 ksi 2 kip (0.58) 65 8.961 in 50 2.680 in 471.8nR = + =

( )kip kip(0.80) 471.8 377.5nRφ = = OK

3 1/2" 2 1/2"3 1/2"

2"6

1 / 2"2"

Shear

Shear

Tens

ion

-- 198 --



B) Web Splice: The web splice is to be designed for the following actions at the Strength Limit:

1. Vuw - The direct shear force. 2. Mvuw - The moment on the web bolts caused by the eccentricity of Vuw 3. Muw - The portion of the bending moment in the beam that is carried by the web. 4. Huw - The horizontal force resulting from the relocation of the beam moment from the

ENA location to the mid-height of the beam.

The shear force in the beam at the location of the splice is Vu = 45kip and the nominal shear capacity of the beam is φVn = 427.7kip.

1. Determine the direct shear force acting on the web splice, Vuw:

( )( )( )

?

?kip kip kip

0.5

45 0.5 1.00 427.7 213.8

u v nV V< φ

< =

Since 0.5u v nV V< φ , ( )( )kip kip1.5 1.5 45 67.5uw uV V= = = (6.13.6.1.4b-1)

2. Determine the moment, Mvuw, that is caused by the eccentricity of the direct shear, Vuw:

Assuming the arrangement of bolts shown on Page 12, the distance from the CG of the bolt group on one side of the splice to the CL of the splice is,

( )( ) ( )( )1 11 1 1

2 2 22 23 " 2 " " 4.50"e = + + =

( )( )kip

k-in k-ft

4.50" 67.5

303.8 25.31

vuw uwM e V= =

= = I used a gap of 1/2” here to be

conservative. I understand that most splices will use much narrower gaps - these calculations will be conservative for smaller gaps.

-- 199 --



3. Determine the portion of the beam moment that is carried by the web splice, Muw:

2

12w

uw h cf cf ncft DM R F R f= − (C6.13.6.1.4b-1)

ksi42.68cfF = (Positive since it’s in tension) 1.207cfR = (from Before) ksi35.36ncff = − (Negative since it’s in compression)

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

( )

2ksi ksi

3 ksi k-in k-ft

0.520" 28.36"1.00 42.08 1.207 35.36

12

34.85 in 85.36 2,975 247.9

uwM⎡ ⎤

= − −⎢ ⎥⎢ ⎥⎣ ⎦

= = =

4. Determine the horizontal force that results from moving the beam moment, Huw:

12w

uw h cf cf ncft DH R F R f= + (C6.13.6.1.4b-2)

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

( )

ksi ksi

2 ksi kip

0.520" 28.36"1.00 42.68 1.207 35.36

12

1.229 in 0.000 0.00

uwH⎡ ⎤

= + −⎢ ⎥⎣ ⎦

= =

In this case, the ENA is at the mid-height of the beam. Since Huw is the horizontal force that results from the eccentricity of the ENA relative to the mid-height of the beam, it makes sense that Huw is zero.

-- 200 --



The total moment acting on the web splice is, k-in k-in k-in303.8 2,975 3, 279Total vuw uwM M M= + = + = The total actions acting on the web splice are as shown below on the left. To determine the forces acting on the bolts using the Elastic Vector Method, tables in the AISC Manual of Steel Construction will be used for preliminary investigations. These tables are set up to account for the shear force, Vuw, but not the moment, MTotal. This can be accommodated by computing a fictitious shear force, P, that when applied over the eccentricity, e, results in the same actions as the actually applied shear and moment.

k-in

kip3, 279 728.74.50"

P = =

P = 728.7kip

A

B

C

D

E

F

G

H

1 2

2 1/2" 3 1/2" 2 3/4"

e = 41/2"

-- 201 --



-- 202 --



From Table 7-8 on Page 7-38 of the 13th Ed. of the AISC Manual of Steel Construction,

kip

min kipbolt

728.7 12.7257.30

u

n

PCr

= = =φ

From the Table for e = 4.00”, S = 3.00”, and for 8 bolts in a row, C = 13.2 From the Table for e = 5.00”, S = 3.00”, and for 8 bolts in a row, C = 12.2 The average of these two values is 12.7. Although this is slightly smaller than 12.72, the proposed configuration will probably still work since our horizontal spacing is 31/2” instead of 3”. Elastic Vector Method for the Web Splice:

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

2 2 2 222, , , ,

2 2 2 2 22 2

4 4

4 4 1.75" 1.5" 4.5" 7.5" 10.5" 805 in

x y D y C y B y Ad d d d d d

d

⎡ ⎤Σ = + + + +⎢ ⎥⎣ ⎦⎡ ⎤Σ = + + + + =⎣ ⎦

Examine Bolt A2:

( )( )k-in

kip, 2 2

3, 279 10.5" 42.77805 inT X

T yVd

= = =Σ

( )( )k-in

kip, 2 2

3, 279 1.75" x 7.128805 inT Y

TVd

= = =Σ

The direct shear force is,

( )kip

kip, bolt

67.54.219

16 boltsD YV = =

( ) ( )2 2kip kip kip kip42.77 7.128 4.219 44.25TotalV = + + =

From this, an actual value of C can be computed, which will be useful when investigating slip resistance.

kip

kipbolt

728.7 16.4744.25

TotalActual

bolt

PCP

= = =

-- 203 --



The calculations shown on the previous page have been validated using the spreadsheet shown here.

M: 2975Px: 0 xCG: 0Py: 67.5 yCG: 0ex: 4.5ey: 0 Σ d2: 805.00

T = 3278.75 Vmax = 44.2459

Bolt x y d2 Vx Vy Vtotal

A1 -1.75 10.50 113.31 42.7663 -7.1277 42.8651B1 -1.75 7.50 59.31 30.5474 -7.1277 30.6856C1 -1.75 4.50 23.31 18.3284 -7.1277 18.5578D1 -1.75 1.50 5.31 6.1095 -7.1277 6.7667E1 -1.75 -1.50 5.31 -6.1095 -7.1277 6.7667F1 -1.75 -4.50 23.31 -18.3284 -7.1277 18.5578G1 -1.75 -7.50 59.31 -30.5474 -7.1277 30.6856H1 -1.75 -10.50 113.31 -42.7663 -7.1277 42.8651A2 1.75 10.50 113.31 42.7663 7.1277 44.2459B2 1.75 7.50 59.31 30.5474 7.1277 32.5866C2 1.75 4.50 23.31 18.3284 7.1277 21.5563D2 1.75 1.50 5.31 6.1095 7.1277 12.8867E2 1.75 -1.50 5.31 -6.1095 7.1277 12.8867F2 1.75 -4.50 23.31 -18.3284 7.1277 21.5563G2 1.75 -7.50 59.31 -30.5474 7.1277 32.5866H2 1.75 -10.50 113.31 -42.7663 7.1277 44.2459

-- 204 --



Check Flexural Yielding of the Web Splice Plates: Take Muw = 3,262k-in

( )( ) 23

12 322

py

p pp p

M dM y M FI d td t

⎡ ⎤ ⎛ ⎞⎢ ⎥σ = = = ≤ φ⎜ ⎟⎢ ⎥ ⎝ ⎠⎣ ⎦

Solve for tp: ( )( )

( ) ( )( )k-in

2 2 ksi

3 3,2623 0.2787"26.5" 1.00 50

pp y

Mtd F

≥ = =φ

Use one PL261/2” x 171/2” x 5/16” each side of web. Check Shear Yielding of the Web Splice Plates: Take Vuw = 67.5kip

( )( )( )( )( )

( )( )( ) ( )( ) ( )ksi kip516

0.58 2

1.00 0.58 26.5" 2 " 50 480.3 OK

uw n p p yV V d t F≤ φ = φ

= =⎡ ⎤⎣ ⎦

Check Shear Rupture of the Web Splice Plates: Take Vuw = 67.5kip

( )( )( )( )( )

( )( ) ( ) ( )( ) ( )( ) ( ),

ksi kip518 16

0.58 2

0.80 0.58 26.5" 8 1 " 2 " 65 329.9 OK

uw n p net p uV V d t F≤ φ = φ

= − =⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

-- 205 --



Check Bearing of the Bolts in the Web Splice: Edge Bolts Bearing on the Beam Web: ( )( )1 1 1

2 2 162 " 1" " 1.969"cL = − + = Since Lc = 1.969” < 2d = 2.0”, 1.2n c uR L tF= ( ) ( ) kipksi

bolt(1.2) 1.969" (0.520") 65 79.85nR = = (6.13.2.9-2)

Edge Bolts Bearing on the Splice Plates: ( )( )1 1 1

2 2 162 " 1" " 1.969"cL = − + = Since Lc = 1.969” < 2d = 2.0”, 1.2n c uR L tF= ( )( )( )( ) kipksi5

16 bolt(1.2) 1.969" 2 " 65 95.99nR = = (6.13.2.9-2)

Summary of Splice Plate Bearing: Bearing on the beam web governs.

( )( )

kipbolt

kip kipbolt bolt

79.85

0.80 79.85 63.88 OK

n

n

R

R

=

φ = =

-- 206 --



Check Slip of the Bolts in the Web Splice: Take the Service II moment as 548k-ft. From the slip check on the flange splice, Fs and fs were determined to be 23.92ksi and Pslip was determined to be 168.3kip. Take the Service II shear force at the location of the splice to be Vsw = 30.4kip. The web splice is to be designed for the following actions at the Service II Limit

1. Vsw - The direct shear force. 2. Mvsw - The moment on the web bolts caused by the eccentricity of Vsw 3. Msw - The portion of the bending moment in the beam that is carried by the web. 4. Hsw - The horizontal force resulting from the relocation of the beam moment from the

ENA location to the mid-height of the beam.

1. The direct shear force acting on the web splice is given as, Vsw = 30.4kip:

2. Determine the moment, Mvsw, that is caused by the eccentricity of the direct shear, Vsw:

Assuming the arrangement of bolts shown on Page 12, the distance from the CG of the bolt group on one side of the splice to the CL of the splice is,

( )( ) ( )( )1 11 1 1

2 2 22 23 " 2 " " 4.50"e = + + =

( )( )kip

k-in k-ft

4.50" 30.4

136.8 11.40

vsw swM e V= =

= =

-- 207 --



3. Determine the portion of the beam moment that is carried by the web splice, Msw:

2

12w

sw s ost DM f f= − (C6.13.6.1.4b-1 mod)

( )( ) ( ) ( )

( )

2ksi ksi

3 ksi k-in k-ft

0.520" 28.36"23.92 23.92

12

34.85 in 47.84 1,667 138.9

swM⎡ ⎤

= − −⎢ ⎥⎢ ⎥⎣ ⎦

= = =

4. Determine the horizontal force that results from moving the beam moment, Hsw:

12w

sw s ost DH f f= + (C6.13.6.1.4b-2 mod)

( )( ) ( ) ( )

( )

ksi ksi

2 ksi kip

0.520" 28.36"23.92 23.92

12

1.229 in 0.000 0.00

swH⎡ ⎤

= + −⎢ ⎥⎣ ⎦

= =

The total moment acting on the web splice is, k-in k-in k-in136.8 1,667 1,804Total vsw swM M M= + = + = The fictitious shear force, P, that when applied over the eccentricity, e, results in the same actions as the actually applied shear and moment is determined as,

k-in

kip1,804 400.84.50"

P = =

The largest bolt force in the web splice due to the Service II combination can be determined as,

kip

kipbolt

400.8 24.3416.47

TotalBolt

PPC

= = =

This force is well below the slip capacity of kip

bolt33.66 that was computed on Page 8. OK

-- 208 --



Final Splice Detail: The final splice configuration is shown below. Technically speaking, fatigue should also be checked for beam flanges and flange splice plates at the location of the splice.

2 1/2" 3 1/2" 2 1/2" 2 1/2" 2 1/2"3 1/2" 3 1/2" 3 1/2"

2"6

1 / 2"2"

2 1/2" 2 1/2"3 1/2" 2 1/2" 2 1/2"3 1/2"

2 3 / 4"

2 3 / 4"

7 Sp

aces

@ 3

"

2"2"

2"2"

Outer Flange Splice Plate: PL 241/2” x 101/2" x 3/8" Each Flange

Inner Flange Splice Plates: PL 241/2" x 4" x 1/2" Each Flange

1" dia M164 Bolts (12 Places Each Flange)

1" dia M164 Bolts (32 Places)

Web Splice Plates: PL 261/2” x 171/2" x 5/16" Each Side of Web

W30 x 99 W30 x 99

-- 209 --

AAB - Example Problems Handout

Documents

elastic vector

special permit

bend buckling

64klf uniform

high dead

span continuous

service limit

influence