AAB%20-%20Example%20Problems%20Handout by Mojtaba Asghari

LRFD Steel Design AASHTO LRFD Bridge Design Specifications Example Problems

Created July 2007

This material is copyrighted by The University of Cincinnati and Dr. James A Swanson. It may not be reproduced, distributed, sold, or stored by any means, electrical or mechanical, without the expressed written consent of The University of Cincinnati and Dr. James A Swanson. July 31, 2007

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

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

James A Swanson Associate Professor University of Cincinnati Dept of Civil & Env. Engineering 765 Baldwin Hall Cincinnati, OH 45221-0071 Ph: (513) 556-3774 Fx: (513) 556-2599 James.Swanson@uc.edu

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.

42' - 0" Out to Out of Deck 39' - 0" Roadway Width

9½” (typ)

23/4" Haunch (typ)

3'-0"

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

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/ 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 1 of 62 -- 1 --

Negative Bending Section (Section 2)

Positive Bending Section (Section 1)

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 ⎛

2 ⎜ 490

Wsec tion1 = 71.06 in ⎜ ⎜ ⎝

⎞

pcf ⎟

(

12 inft

)

⎟ (1.15 ) = 278.1 ft ⎟ ⎠

per girder

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

2⎜

Wsec tion 2 = 112.3 in ⎜ ⎜ ⎝

⎞

490 pcf ⎟

(

12 inft

)

⎟ (1.15 ) = 439.5 ⎟ ⎠

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

Lb ft

per girder

AASHTO-LRFD 2007 Page 2 of 62 -- 2 --

Deck Self-Weight (DC1):

Wdeck

⎛ ⎜ 150 pcf = (9.5")(144") ⎜ 2 ⎜ 12 inft ⎝

(

)

⎞ ⎟ Lb ⎟ = 1, 425 ft per girder ⎟ ⎠

Haunch Self-Weight (DC1): ⎛ 21"(66') + 15"(264') ⎞ ⎟ = 16.2" 66'+ 264' ⎝ ⎠

Average width of flange: ⎜

Average width of haunch:

Whaunch

( 1 2 ) ⎡⎣(16.2"+ (2)(9") ) + 16.2"⎤⎦ = 25.2"

⎛ ⎞ ( 2")( 25.2") ⎟ ⎜ Lb =⎜ 2 ⎟ (150 pcf ) = 52.5 ft per girder in ⎜ 12 ft ⎟ ⎝ ⎠

(

)

Barrier Walls (DC2): ⎛ (2 each) ( 640 plf ) ⎞ ⎟ = 320.0 Lb ft per girder ⎜ ⎟ 4 girders ⎝ ⎠

Wbarriers = ⎜

Wearing Surface (DW):

W fws =

(39')(60 psf ) 4 girders

= 585 Lb per girder ft

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).

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 3 of 62 -- 3 --

Unfactored Dead Load Moment Diagrams from SAP 4,000 DC1

3,000 2,000 DW

1,000

Moment (kip-ft)

0 DC2

-1,000 -2,000 -3,000 -4,000 -5,000 -6,000 -7,000 -8,000 0

120

150

180

210

240

270

300

330

Station (ft)

Unfactored Dead Load Shear Diagrams from SAP 200 DC1

150

100 DW

Shear (kip)

50 DC2

-50

-100

-150

-200 0

120

150

180

210

240

270

300

330

Station (ft)

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 4 of 62 -- 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’

DC1 - Steel: DC1 - Deck: DC1 - Haunch: DC1 - Total: DC2: DW

Max (+) Moment Stations 58.7’ and 271.3’ 475k-ft 2,415k-ft 89k-ft 2,979k-ft 553k-ft 1,011k-ft

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

Max (-) Moment Station 165.0’ -1,189k-ft -5,708k-ft -210k-ft -7,107k-ft -1,251k-ft -2,286k-ft

AASHTO-LRFD 2007 Page 5 of 62 -- 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.

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 6 of 62 -- 6 --

Unfactored Moving Load Moment Envelopes from SAP 6,000

Single Truck

4,000 Tandem

Moment (kip-ft)

2,000

Fatigue

0 Fatigue Tandem

-2,000 Contraflexure Point

Contraflexure Point

-4,000

Single Truck Two Trucks

-6,000 0

120

150

180

210

240

270

300

330

270

300

330

Station (ft)

Unfactored Moving Load Shear Envelopes from SAP 200 Single Truck

150 Tandem

100 Fatigue

Shear (kip)

-50

-100

-150

-200 0

120

150

180

210

240

Station (ft)

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 7 of 62 -- 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’

HL-93M HL-93K HL-93S Fatigue

Max (+) Moment Stations 73.3’ and 256’ 3,725k-ft 4,396k-ft N/A 2,327k-ft

Max (-) Moment Station 165’ -3,737k-ft -4,261k-ft -5,317k-ft -1,095k-ft

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

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 8 of 62 -- 8 --

2.2.1: Verify the Maximum Positive Live-Load Moment at Station 73.3â&#x20AC;&#x2122;: 25kip

25kip

Tandem: 32kip 32kip 8kip

Single Truck:

0.640kip/ft

Lane:

Moment (k-ft / kip)

40 30 20 10 0 -10 -20 0

105

120

135

150

165

180

195

210

225

240

255

270

285

300

315

330

Station (ft)

Tandem: Single Truck: Lane Load:

( 25 ) ( 33.00 ) + ( 25 ) ( 31.11 ) = 1, 603 ( 8 ) ( 26.13 ) + ( 32 ) ( 33.00 ) + ( 32 ) ( 26.33 ) = 2,108 kip

kip

k-ft

kip

k-ft

kip

( 0.640 ) ( 2, 491 ) = 1,594 ft

kip

k-ft

kip

k-ft

kip

k-ft

kip

k-ft

kip

(IM)(Tandem) + Lane:

(1.33 ) (1, 603

(IM)(Single Truck) + Lane:

(1.33 ) ( 2,108

k-ft

) + 1,594

k-ft

) + 1,594

k-ft

= 3,726

k-ft

= 4,397

k-ft

GOVERNS

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

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 9 of 62 -- 9 --

2.2.2: Verify the Maximum Negative Live-Load Moment at Station 165.0’: 25kip

25kip

Tandem: 32kip 32kip 8kip

Single Truck: 32kip 32kip

32kip 32kip

8kip

Two Trucks:

0.640kip/ft

Lane:

Station (ft) 0

105

120

135

150

165

180

195

210

225

240

255

270

285

300

315

330

Moment (k-ft / kip)

0 -5 -10 -15 -20

Tandem: Single Truck:

( 25 ) (18.51 ) + ( 25 ) (18.45 ) = 924.0 ( 8 ) (17.47 ) + ( 32 ) (18.51 ) + ( 32 ) (18.31 ) = 1,318 kip

kip

k-ft

kip

k-ft

kip

k-ft

kip

k-ft

kip

k-ft

kip

k-ft

kip

k-ft

kip

k-ft

kip

k-ft

kip

(IM)(Tandem) + Lane:

(1.33 ) ( 924.0

(IM)(Single Truck) + Lane:

(1.33 ) (1,318

(0.90){(IM)(Two Trucks) + Lane}:

kip

k-ft

( 0.640 ) ( 3,918 ) = 2,508 kip

k-ft

kip

Lane Load:

k-ft

kip

( 8 ) (17.47 ) + ( 32 ) (18.51 ) + ( 32 ) (18.31 ) + ... ... + ( 8 ) (16.72 ) + ( 32 ) (18.31 ) + ( 32 ) (18.51 ) = 2, 630 kip

Two Trucks:

kip

k-ft

) + 2,508

( 0.90 ) ⎡⎣(1.33 ) ( 2,630

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

k-ft

= 3,737

k-ft

= 4, 261

) + 2,508

k-ft

⎤⎦ = 5, 405k-ft GOVERNS

AASHTO-LRFD 2007 Page 10 of 62 -- 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

BRSingle Lane = ( 0.25 ) ( 8kip + 32kip + 32kip ) = 18.00kip B) 25% of the Design Tandem

BRSingle Lane = ( 0.25 ) ( 25kip + 25kip ) = 12.50kip C) 5% of the Design Truck with the Lane Load.

(

)

kip ⎤ BRSingle Lane = ( 0.05 ) ⎡⎣( 8kip + 32kip + 32kip ) + ( 2 )(165') 0.640 kip ft ⎦ = 14.16

D) 5% of the Design Tandem with the Lane Load.

(

)

kip ⎤ BRSingle Lane = ( 0.05 ) ⎡⎣( 25kip + 25kip ) + ( 2 )(165' ) 0.640 kip ft ⎦ = 13.06

Case (A) Governs:

BRNet = ( BRSingle Lane ) ( # Lanes )( MPF ) = (18.00kip ) ( 3)( 0.85 ) = 45.90kip

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

This load has not been factored…

AASHTO-LRFD 2007 Page 11 of 62 -- 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.

C= f

v2 gR

(3.6.3-1)

where:

( secft )

- Highway design speed

- 4/3 for all load combinations except for Fatigue, in which case it is 1.0

- The acceleration of gravity

- The radius of curvature for the traffic lane (ft).

( ) ft sec 2

Suppose that we have a radius of R = 600’ and a design speed of v = 65mph = 95.33ft/sec. ft 2 ⎤ ⎛ 4 ⎞ ⎡⎢ ( 95.33 sec ) ⎥ = 0.6272 C =⎜ ⎟ ⎝ 3 ⎠ ⎢⎣ 32.2 secft 2 ( 600 ') ⎥⎦

(

)

CE = ( Axle Loads )( C )( # Lanes )( MPF )

= ( 72kip ) ( 0.6272 )( 3)( 0.85 ) = 115.2kip

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.

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 12 of 62 -- 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’

V o = 8.20mph Z o = 0.23ft

Open Country

Horizontal Wind Load on Structure: (WS) Design Pressure: 2

⎛V ⎞ VDZ 2 PD = PB ⎜ D Z ⎟ = PB 2 10, 000mph ⎝ VB ⎠

(3.8.1.2.1-1)

- Base Pressure - For beams, PB = 50psf when VB = 100mph. - Base Wind Velocity, typically taken as 100mph. - Wind Velocity at an elevation of Z = 30’ (mph) - Design Wind Velocity (mph)

PB VB V30 VDZ

(Table 3.8.1.2.1-1)

Design Wind Velocity:

⎛V ⎞ ⎛ Z ⎞ VDZ = 2.5Vo ⎜ 30 ⎟ ln ⎜ ⎟ ⎝ VB ⎠ ⎝ Z o ⎠ = ( 2.5 ) ( 8.20

PD = ( 50

psf

)

(105.8 )

mph

(3.8.1.1-1)

ft ⎛ 100 ⎞ ⎛ 40 ⎞ ) ⎜⎝ 100 ⎟⎠ Ln ⎜ 0.23ft ⎟ = 105.8mph ⎝ ⎠

mph 2

(10, 000 ) mph 2

= 55.92psf

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

hexp

hexp = 71.5"+ 11.75"+ 42" = 125.3" = 10.44 ' The wind load per unit length of the bridge, W, is then computed as:

W = ( 55.92psf ) (10.44 ' ) = 583.7 lbs ft kip = ( 583.7 lbs ft ) ( 2 )(165' ) = 192.6

Total Wind Load:

WS H ,Total

For End Abutments:

kip 1 WS H , Abt = ( 583.7 lbs ft ) ( 2 ) (165' ) = 48.16

For Center Pier:

WS H , Pier

kip 1 = ( 583.7 lbs ft ) ( 2 ) ( 2 ) (165' ) = 96.31

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 13 of 62 -- 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.

PV = ( 20psf ) ( w ) = ( 20psf ) ( 42 ' ) = 840 lbs ft

Total Uplift:

(840 lbsft ) ( 2 )(165') = 277.2kip

For End Abutments:

(840 lbsft ) ( 12 ) (165') = 69.30kip

kip 1 For Center Pier: ( 840 lbs ft ) ( 2 ) ( 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 6â&#x20AC;&#x2122; 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

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 14 of 62 -- 14 --

3. SECTION PROPERTIES AND CALCULATIONS: 3.1: Effective Flange Width, beff:

For an interior beam, beff is the lesser of: ⎧ Leff 132' = = 33' = 396" ⎪• 4 ⎪ 4 bf ⎪ 15" = (12)(8.5") + = 109.5" ⎨• 12ts + 2 2 ⎪ ⎪• S = (12')(12 in ft ) = 144" ⎪ ⎩ For an exterior beam, beff is the lesser of: ⎧ Leff 132' = = 33' = 198.0" ⎪• 4 ⎪ 4 bf ⎪ 15" = (12)(8.5") + = 109.5" ⎨• 12ts + 2 2 ⎪ ⎪ S ⎛ 12' ⎞ + 3' ⎟ (12 inft ) = 108.0" ⎪• + d e = ⎜ ⎝ 2 ⎠ ⎩ 2

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).

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 15 of 62 -- 15 --

3.2: Section 1 Flexural Properties Bare Steel

Top Flange Web Bot Flange

t 0.7500 0.5625 1.0000

b 15.00 69.00 21.00

A 11.25 38.81 21.00

Ix Ay 791.72 0.53 1,377.84 15,398.86 10.50 1.75

y 70.38 35.50 0.50

71.06 Y=

Ad2 17,728 902 19,125

IX 17,729 16,301 19,127

2,180.06

ITotal =

53,157

30.68

SBS1,top =

1,327

SBS1,bot =

1,733

d -39.70 -4.82 30.18

Short-Term Composite (n = 8)

Slab Haunch Top Flange Web Bot Flange

t 8.5000 0.0000 0.7500 0.5625 1.0000

b 109.50 15.00 15.0000 69.0000 21.0000

8.00

A 116.34 0.00 11.25 38.81 21.00 187.41

y 75.00 70.75 70.38 35.50 0.50

Ix Ay 8,725.78 700.49 0.00 0.00 791.72 0.53 1,377.84 15,398.86 10.50 1.75 10,905.84

Ad2

d -16.81 -12.56 -12.18 22.69 57.69

58.19

32,862 0 1,669 19,988 69,900 ITotal =

33,562 0 1,670 35,387 69,901 140,521

SST1,top = SST1,bot =

11,191 2,415

Long-Term Composite (n = 24)

Slab Haunch Top Flange Web Bot Flange

t 8.5000 0.0000 0.7500 0.5625 1.0000

b 109.50 15.00 15.0000 69.0000 21.0000

24.00

A 38.78 0.00 11.25 38.81 21.00 109.84

y 75.00 70.75 70.38 35.50 0.50

Ix Ay 2,908.59 233.50 0.00 0.00 791.72 0.53 1,377.84 15,398.86 10.50 1.75 5,088.66

Ad2 31,885 0 6,506 4,549 44,101 ITotal =

IX 32,119 0 6,507 19,948 44,103 102,676

SLT1,top = SLT1,bot =

4,204 2,216

d -28.67 -24.42 -24.05 10.83 45.83

46.33

Cracked Section

Rebar Top Flange Web Bot Flange

t 4.5000 0.7500 0.5625 1.0000

b 15.0000 69.0000 21.0000

A 13.02 11.25 38.81 21.00 84.08

y 75.25 70.38 35.50 0.50

Ix Ay 979.76 791.72 0.53 1,377.84 15,398.86 10.50 1.75 3,159.82 37.58

Ad2 73,727 55,717 48,913 5 ITotal =

IX 73,727 55,718 64,312 7 193,764

SCR1,top = SCR1,bot =

5,842 5,156

d -75.25 -70.38 -35.50 -0.50

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

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 16 of 62 -- 16 --

3.3: Section 2 Flexural Properties Bare Steel

Top Flange Web Bot Flange

t 1.0000 0.5625 2.5000

b 21.00 69.00 21.00

A 21.00 38.81 52.50

Ix Ay 1,512.00 1.75 1,436.06 15,398.86 65.63 27.34

y 72.00 37.00 1.25

112.31 Y=

Ad2 42,841 4,012 34,361

IX 42,843 19,411 34,388

3,013.69

ITotal =

96,642

26.83

SBS2,top =

2,116

SBS2,bot =

3,602

d -45.17 -10.17 25.58

Short Term Composite (n = 8)

Slab Haunch Top Flange Web Bot Flange

t 8.5000 0.0000 1.0000 0.5625 2.5000

b 109.50 21.00 21.0000 69.0000 21.0000

8.00

A 116.34 0.00 21.00 38.81 52.50 228.66

y 76.75 72.50 72.00 37.00 1.25

Ix Ay 8,929.38 700.49 0.00 0.00 1,512.00 1.75 1,436.06 15,398.86 65.63 27.34 11,943.07

d -24.52 -20.27 -19.77 15.23 50.98

52.23

Ad2

69,941 0 8,207 9,005 136,454 ITotal =

70,641 0 8,208 24,403 136,481 239,734

SST2,top = SST2,bot =

11,828 4,590

Long-Term Composite (n = 24)

Slab Haunch Top Flange Web Bot Flange

t 8.5000 0.0000 1.0000 0.5625 2.5000

b 109.50 15.00 21.0000 69.0000 21.0000

24.00

A 38.78 0.00 21.00 38.81 52.50 151.09

y 76.75 72.50 72.00 37.00 1.25

Ix Ay 2,976.46 233.50 0.00 0.00 1,512.00 1.75 1,436.06 15,398.86 65.63 27.34 5,990.15

Ad2 53,393 0 21,983 272 77,395 ITotal =

IX 53,626 0 21,985 15,670 77,423 168,704

SLT2,top = SLT2,bot =

5,135 4,255

d -37.10 -32.85 -32.35 2.65 38.40

39.65

Cracked Section

Rebar Top Flange Web Bot Flange

t 4.5000 1.0000 0.5625 2.5000

b 21.0000 69.0000 21.0000

A 13.02 21.00 38.81 52.50 125.33

y 77.00 72.00 37.00 1.25

Ix Ay 1,002.54 1,512.00 1.75 1,436.06 15,398.86 65.63 27.34 4,016.23 32.04

Ad2 26,313 33,525 953 49,786 ITotal =

IX 26,313 33,527 16,352 49,813 126,006

SCR2,top = SCR2,bot =

3,115 3,932

d -44.96 -39.96 -4.96 30.79

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

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 17 of 62 -- 17 --

4. DISTRIBUTION FACTOR FOR MOMENT 4.1: Positive Moment Region (Section 1): Interior Girder – One Lane Loaded: 0.4 0.3 ⎛ S ⎞ ⎛ S ⎞ ⎛ Kg ⎟ ⎜ ⎟ ⎜ ⎝ 14 ⎠ ⎝ L ⎠ ⎜⎝ 12 Lt s3

DFM 1, Int + = 0.06 + ⎜

⎞ ⎟⎟ ⎠

0.1

2 K g = n( I + Aeg ) 4

K g = 8(53,157 in + (71.06 in )(46.82") ) K g = 1, 672, 000 in

0.4 0.3 4 ⎛ 12 ' ⎞ ⎛ 12 ' ⎞ ⎛ 1, 672, 000 in ⎞ ⎟ ⎟ ⎜ ⎟ ⎜ ⎝ 14 ⎠ ⎝ 165 ' ⎠ ⎝ (12)(165 ')(8.5")3 ⎠

0.1

DFM 1, Int + = 0.06 + ⎜ DFM 1, Int + = 0.5021

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: ⎞ ⎟⎟ ⎠

0.1

DFM 2, Int +

0.6 0.2 ⎛ S ⎞ ⎛ S ⎞ ⎛ Kg = 0.075 + ⎜ ⎟ ⎜ ⎟ ⎜⎜ ⎝ 9.5 ⎠ ⎝ L ⎠ ⎝ 12 Lt s3

DFM 2, Int +

0.6 0.2 4 ⎛ 12 ' ⎞ ⎛ 12 ' ⎞ ⎛ 1, 672, 000 in ⎞ = 0.075 + ⎜ ⎟ ⎟ ⎜ ⎟ ⎜ ⎝ 9.5 ⎠ ⎝ 165 ' ⎠ ⎝ 12(165 ')(8.5")3 ⎠

0.1

DFM 2, Int + = 0.7781

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.

DFM 1, Ext + =

8.5 12

= 0.7083

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

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 18 of 62 -- 18 --

Two or More Lanes Loaded:

DFM2,Ext+ = e DFM2,Int+

e = 0.77 + = 0.77 +

de 9.1 1.5 9.1

= 0.9348

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:

DFM 1, Int −

0.4 0.3 ⎛ S ⎞ ⎛ S ⎞ ⎛ Kg = 0.06 + ⎜ ⎟ ⎜ ⎟ ⎜ ⎝ 14 ⎠ ⎝ L ⎠ ⎜⎝ 12 Lts3

⎞ ⎟⎟ ⎠

0.1

2 K g = n( I + Aeg )

K g = 8(96, 642 in 4 + (112.3 in 2 )(52.17 ") 2 ) K g = 3, 218, 000 in 4 0.4 0.3 4 ⎛ 12 ' ⎞ ⎛ 12 ' ⎞ ⎛ 3, 218, 000 in ⎞ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 14 ⎠ ⎝ 165 ' ⎠ ⎝ (12)(165 ')(8.5")3 ⎠

0.1

DFM 1, Int − = 0.06 + ⎜ DFM 1, Int − = 0.5321

Two or More Lanes Loaded: ⎞ ⎟⎟ ⎠

0.1

DFM 2, Int −

0.6 0.2 ⎛ S ⎞ ⎛ S ⎞ ⎛ Kg = 0.075 + ⎜ ⎟ ⎜ ⎟ ⎜⎜ ⎝ 9.5 ⎠ ⎝ L ⎠ ⎝ 12 Lt s3

DFM 2, Int −

0.6 0.2 4 ⎛ 12 ' ⎞ ⎛ 12 ' ⎞ ⎛ 3, 218, 000 in ⎞ = 0.075 + ⎜ ⎟ ⎟ ⎜ ⎟ ⎜ ⎝ 9.5 ⎠ ⎝ 165 ' ⎠ ⎝ (12)(165 ')(8.5")3 ⎠

0.1

DFM 2, Int − = 0.8257

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 19 of 62 -- 19 --

Exterior Girder – One Lane Loaded:

Same as for the positive moment section: DFM1,Ext- = 0.8500 Two or More Lanes Loaded:

DFM2,Ext- = e DFM2,Intd e = 0.77 + e 9.1 = 0.77 +

1.5 9.1

= 0.9348

DFM2,Ext- = (0.9348) (0.8257) = 0.7719 4.3: Minimum Exterior Girder Distribution Factor: NL

Ext , Min

NL Nb

X Ext ∑ e Nb

∑x

One Lane Loaded:

M 1, Ext , Min

1 4

(18.0 ')(14.5 ') (2) ⎡⎣(18 ') 2 + (6 ') 2 ⎤⎦

= 0.6125

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

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 20 of 62 -- 20 --

Two Lanes Loaded: 14.5' 2.5' 2'

M 2 , Ext , Min

2 4

(18.0 ')(14.5 '+ 2.5 ') (2) ⎡⎣(18 ') 2 + (6 ') 2 ⎤⎦

= 0.9250

Multiple Presence: DFM2,Ext,Min = (1.0) (0.9250) = 0.9250 Lane 1 (12')

Lane 2 (12')

12'

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:

1 Lane Loaded: 2 Lanes Loaded:

Positive Moment Interior Exterior 0.5021 0.8500 ≥ 0.7350 0.7781 0.7274 ≥ 0.9250

Negative Moment Interior Exterior 0.5321 0.8500 ≥ 0.7350 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.

1 Lane Loaded:

Positive Moment Interior Exterior 0.4184 0.7083 ≥ 0.6125

Negative Moment Interior Exterior 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.

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 21 of 62 -- 21 --

Nominal Girder Moments for Design Station (ft) 0.0 14.7 29.3 44.0 58.7 73.3 88.0 102.7 117.3 132.0 135.7 139.3 143.0 146.7 150.3 154.0 157.7 161.3 165.0 168.7 172.3 176.0 179.7 183.3 187.0 190.7 194.3 198.0 212.7 227.3 242.0 256.7 271.3 286.0 300.7 315.3 330.0

(LL+IM)+ (k-ft) 0.0 1605.1 2791.4 3572.6 3999.4 4066.7 3842.5 3310.8 2509.4 1508.6 1274.6 1048.4 828.6 615.8 463.3 320.5 185.5 76.4 0.0 76.4 185.5 320.5 463.3 615.8 828.6 1048.4 1274.6 1508.6 2509.4 3310.8 3842.5 4066.7 3999.4 3572.6 2791.4 1605.1 0.0

(LL+IM)(k-ft) 0.0 -280.7 -561.3 -842.0 -1122.7 -1403.4 -1684.0 -1964.7 -2245.4 -2547.5 -2660.0 -2793.3 -2945.6 -3115.6 -3371.3 -3728.6 -4105.0 -4496.9 -4918.1 -4496.9 -4105.0 -3728.6 -3371.3 -3115.6 -2945.6 -2793.3 -2660.0 -2547.5 -2245.4 -1964.7 -1684.0 -1403.4 -1122.7 -842.0 -561.3 -280.7 0.0

Nominal Moments Fat+ Fat(k-ft) (k-ft) 0.2 0.0 645.6 -68.9 1127.9 -137.9 1449.4 -206.8 1626.1 -275.8 1647.9 -344.7 1599.4 -413.7 1439.3 -482.6 1148.6 -551.6 763.6 -620.5 651.3 -637.8 539.1 -655.0 425.3 -672.2 310.8 -689.5 221.9 -706.7 158.6 -724.0 98.8 -741.2 49.4 -758.4 0.1 -775.6 49.4 -758.4 98.8 -741.2 158.6 -724.0 221.9 -706.7 310.8 -689.5 425.3 -672.2 539.1 -655.0 651.3 -637.8 763.2 -620.6 1148.6 -551.6 1439.3 -482.6 1599.4 -413.7 1647.9 -344.7 1626.1 -275.8 1449.4 -206.8 1127.9 -137.9 645.6 -68.9 0.2 0.0

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

DC1 (k-ft) 0.0 1309.9 2244.5 2799.9 2978.6 2779.3 2202.1 1248.4 -84.8 -1793.1 -2280.8 -2794.0 -3333.2 -3898.1 -4488.6 -5105.1 -5747.2 -6415.3 -7108.8 -6415.3 -5747.2 -5105.1 -4488.6 -3898.1 -3333.2 -2794.0 -2280.8 -1793.1 -84.8 1248.4 2202.1 2779.3 2978.6 2799.9 2244.5 1309.9 0.0

DC2 (k-ft) 0.0 240.0 412.0 515.0 549.7 515.8 413.2 242.3 2.5 -305.4 -393.2 -485.2 -581.5 -682.1 -787.0 -896.2 -1009.7 -1127.5 -1249.5 -1127.5 -1009.7 -896.2 -787.0 -682.1 -581.5 -485.2 -393.2 -305.4 2.5 242.3 413.2 515.8 549.7 515.0 412.0 240.0 0.0

DW (k-ft) 0.0 440.3 755.6 944.7 1008.3 946.1 757.9 444.4 4.6 -560.2 -721.2 -890.0 -1066.7 -1251.3 -1443.7 -1643.9 -1852.1 -2068.1 -2291.9 -2068.1 -1852.1 -1643.9 -1443.7 -1251.3 -1066.7 -890.0 -721.2 -560.2 4.6 444.4 757.9 946.1 1008.3 944.7 755.6 440.3 0.0

AASHTO-LRFD 2007 Page 22 of 62 -- 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:

S 25.0 12 ' = 0.36 + = 0.8400 25.0

DFV 1,Int = 0.36 +

Two or More Lanes Loaded:

DFV 2 ,Int

S ⎛ S ⎞ = 0.2 + − ⎜ ⎟ 12 ⎝ 35 ⎠

= 0.2 +

12 ' ⎛ 12 ' ⎞ −⎜ ⎟ = 1.082 12 ⎝ 35 ⎠

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

de 10 1.5' = 0.60 + = 0.7500 10

e = 0.60 +

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 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 23 of 62 -- 23 --

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

1 Lane Loaded: 2 Lanes Loaded:

Shear Distribution Interior Exterior 0.8400 0.8500 ≥ 0.7350 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:

1 Lane Loaded:

Shear Distribution Interior Exterior 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.

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 24 of 62 -- 24 --

Nominal Girder Shears for Design Station (ft) 0.0 14.7 29.3 44.0 58.7 73.3 88.0 102.7 117.3 132.0 135.7 139.3 143.0 146.7 150.3 154.0 157.7 161.3 165.0 168.7 172.3 176.0 179.7 183.3 187.0 190.7 194.3 198.0 212.7 227.3 242.0 256.7 271.3 286.0 300.7 315.3 330.0

(LL+IM)+ (kip) 144.9 123.5 103.5 85.0 68.1 52.8 39.4 27.8 18.0 10.0 8.3 6.7 5.5 4.3 3.2 2.2 1.3 0.0 0.0 170.1 166.2 162.3 158.4 154.5 150.5 146.5 142.5 138.6 122.3 105.7 89.1 72.7 56.7 41.4 26.8 20.3 19.7

(LL+IM)(kip) -19.7 -20.3 -26.8 -41.4 -56.7 -72.7 -89.1 -105.7 -122.3 -138.6 -142.5 -146.5 -150.5 -154.5 -158.4 -162.3 -166.2 -170.1 -173.9 -0.5 -1.3 -2.2 -3.2 -4.3 -5.5 -6.7 -8.3 -10.0 -18.0 -27.8 -39.4 -52.8 -68.1 -85.0 -103.5 -123.5 -144.9

Nominal Shears Fat+ Fat(kip) (kip) 50.8 -4.7 44.6 -4.7 38.5 -6.4 32.6 -11.1 26.9 -17.2 21.4 -23.2 16.3 -29.0 11.5 -34.6 7.3 -39.9 3.9 -44.9 3.4 -46.0 2.8 -47.2 2.3 -48.3 1.8 -49.4 1.4 -50.4 1.0 -51.5 0.6 -52.4 0.3 -53.4 54.3 -54.3 53.4 -0.3 52.4 -0.6 51.5 -1.0 50.4 -1.4 49.4 -1.8 48.3 -2.3 47.2 -2.8 46.0 -3.4 44.9 -3.9 39.9 -7.3 34.6 -11.5 29.0 -16.3 23.2 -21.4 17.2 -26.9 11.1 -32.6 6.4 -38.5 4.7 -44.6 4.7 -50.8

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

DC1 (kip) 115.0 88.8 62.5 36.3 10.1 -16.1 -42.3 -68.6 -94.8 -121.0 -127.6 -134.1 -140.7 -147.2 -153.8 -160.3 -166.9 -173.4 -180.0 173.4 166.9 160.3 153.8 147.2 140.7 134.1 127.6 121.0 94.8 68.6 42.3 16.1 -10.1 -36.3 -62.5 -88.8 -115.0

DC2 (kip) 20.6 15.9 11.2 6.5 1.8 -2.9 -7.6 -12.3 -17.0 -21.7 -22.8 -24.0 -25.2 -26.4 -27.5 -28.7 -29.9 -31.0 -32.2 31.0 29.9 28.7 27.5 26.4 25.2 24.0 22.8 21.7 17.0 12.3 7.6 2.9 -1.8 -6.5 -11.2 -15.9 -20.6

DW (kip) 37.6 29.0 20.5 11.9 3.3 -5.3 -13.9 -22.4 -31.0 -39.6 -41.7 -43.9 -46.0 -48.2 -50.3 -52.5 -54.6 -56.8 -58.9 56.8 54.6 52.5 50.3 48.2 46.0 43.9 41.7 39.6 31.0 22.4 13.9 5.3 -3.3 -11.9 -20.5 -29.0 -37.6

AASHTO-LRFD 2007 Page 25 of 62 -- 25 --

6. FACTORED SHEAR AND MOMENT ENVELOPES

The following load combinations were considered in this example: Strength I: Strength IV:

1.75(LL + IM) + 1.25DC1 + 1.25DC2 + 1.50DW 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. Q = ∑ηiγ i Qi

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).

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 26 of 62 -- 26 --

Strength Limit Moment Envelopes 20,000 15,000 Strength I

10,000 Strength IV

Moment (kip-ft)

5,000 0 -5,000 Strength IV

-10,000 -15,000 Strength I

-20,000 -25,000 0

120

150

180

210

240

270

300

330

270

300

330

Station (ft)

Strength Limit Shear Force Envelope 800 Strength I

600

400 Strength IV

Shear (kip)

200

-200

-400

-600

-800 0

120

150

180

210

240

Station (ft)

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 27 of 62 -- 27 --

Service II Moment Envelope 12,500 10,000 7,500 5,000

Moment (kip-ft)

2,500 0 -2,500 -5,000 -7,500 -10,000 -12,500 -15,000 -17,500 -20,000 0

120

150

180

210

240

270

300

330

270

300

330

Station (ft)

Service II Shear Envelope 600

400

Shear (kip)

200

-200

-400

-600 0

120

150

180

210

240

Station (ft)

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 28 of 62 -- 28 --

Factored Fatigue Moment Envelope 1,500

1,000

Moment (kip-ft)

500

-500

-1,000

-1,500 0

120

150

180

210

240

270

300

330

270

300

330

Station (ft)

Factored Fatigue Shear Envelope 50 40 30 20

Shear (kip)

10 0 -10 -20 -30 -40 -50 0

120

150

180

210

240

Station (ft)

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 29 of 62 -- 29 --

Factored Girder Moments for Design Station (ft) 0.0 14.7 29.3 44.0 58.7 73.3 88.0 102.7 117.3 132.0 135.7 139.3 143.0 146.7 150.3 154.0 157.7 161.3 165.0 168.7 172.3 176.0 179.7 183.3 187.0 190.7 194.3 198.0 212.7 227.3 242.0 256.7 271.3 286.0 300.7 315.3 330.0

Strength I Total + Total (k-ft) (k-ft) 0.0 0.0 5677.1 -515.7 9806.0 -1031.5 12403.3 -1547.2 13567.8 -2062.9 13287.4 -2578.7 11687.1 -3094.4 8740.0 -3610.2 4621.6 -4237.1 2772.1 -8317.5 2342.0 -9533.2 1926.4 -10838.2 1522.6 -12230.6 1131.6 -13707.1 851.2 -15392.8 588.9 -17317.3 340.9 -19328.3 140.4 -21420.1 0.0 -23617.1 140.4 -21420.1 340.9 -19328.3 588.9 -17317.3 851.2 -15392.8 1131.6 -13707.1 1522.6 -12230.6 1926.4 -10838.2 2342.0 -9533.2 2772.1 -8317.5 4621.6 -4237.1 8740.0 -3610.2 11687.1 -3094.4 13287.4 -2578.7 13567.8 -2062.9 12403.3 -1547.2 9806.0 -1031.5 5677.1 -515.7 0.0 0.0

Strength IV Total + Total (k-ft) (k-ft) 0.0 0.0 3134.6 0.0 5374.1 0.0 6708.8 0.0 7145.1 0.0 6679.8 0.0 5312.9 0.0 3047.7 0.0 11.2 -133.5 0.0 -4187.3 0.0 -5347.3 0.0 -6566.4 0.0 -7845.7 0.0 -9184.5 0.0 -10582.9 0.0 -12041.3 0.0 -13559.1 0.0 -15137.1 0.0 -16774.1 0.0 -15137.1 0.0 -13559.1 0.0 -12041.3 0.0 -10582.9 0.0 -9184.5 0.0 -7845.7 0.0 -6566.4 0.0 -5347.3 0.0 -4187.3 11.2 -133.5 3047.7 0.0 5312.9 0.0 6679.8 0.0 7145.1 0.0 6708.8 0.0 5374.1 0.0 3134.6 0.0 0.0 0.0

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

Service II Total + Total (k-ft) (k-ft) 0.0 0.0 4280.7 -383.1 7393.0 -766.2 9349.1 -1149.4 10222.6 -1532.5 10004.2 -1915.6 8787.0 -2298.7 6551.1 -2681.8 3432.8 -3153.9 2059.3 -6268.9 1739.8 -7195.8 1431.1 -8190.4 1131.1 -9251.2 840.6 -10375.8 632.3 -11657.1 437.4 -13117.1 253.3 -14642.7 104.3 -16229.6 0.0 -17895.9 104.3 -16229.6 253.3 -14642.7 437.4 -13117.1 632.3 -11657.1 840.6 -10375.8 1131.1 -9251.2 1431.1 -8190.4 1739.8 -7195.8 2059.3 -6268.9 3432.8 -3153.9 6551.1 -2681.8 8787.0 -2298.7 10004.2 -1915.6 10222.6 -1532.5 9349.1 -1149.4 7393.0 -766.2 4280.7 -383.1 0.0 0.0

Fatigue Total + Total (k-ft) (k-ft) 0.2 0.0 484.2 -51.7 845.9 -103.4 1087.1 -155.1 1219.6 -206.8 1235.9 -258.6 1199.5 -310.3 1079.5 -362.0 861.5 -413.7 572.7 -465.4 488.5 -478.3 404.3 -491.3 318.9 -504.2 233.1 -517.1 166.5 -530.0 119.0 -543.0 74.1 -555.9 37.1 -568.8 0.1 -581.7 37.1 -568.8 74.1 -555.9 119.0 -543.0 166.5 -530.0 233.1 -517.1 318.9 -504.2 404.3 -491.3 488.5 -478.3 572.4 -465.4 861.5 -413.7 1079.5 -362.0 1199.5 -310.3 1235.9 -258.6 1219.6 -206.8 1087.1 -155.1 845.9 -103.4 484.2 -51.7 0.2 0.0

AASHTO-LRFD 2007 Page 30 of 62 -- 30 --

Factored Girder Shears for Design Station (ft) 0.0 14.7 29.3 44.0 58.7 73.3 88.0 102.7 117.3 132.0 135.7 139.3 143.0 146.7 150.3 154.0 157.7 161.3 165.0 168.7 172.3 176.0 179.7 183.3 187.0 190.7 194.3 198.0 212.7 227.3 242.0 256.7 271.3 286.0 300.7 315.3 330.0

Strength I Total + Total (kip) (kip) 479.5 -34.5 390.5 -35.5 304.0 -46.9 220.1 -72.4 138.9 -99.3 92.5 -158.9 68.9 -239.1 48.6 -319.7 31.5 -400.1 17.5 -480.2 14.5 -500.0 11.7 -519.8 9.6 -539.7 7.6 -559.6 5.7 -579.3 3.9 -599.0 2.2 -618.7 0.0 -638.3 0.0 -657.9 638.3 -0.9 618.7 -2.2 599.0 -3.9 579.3 -5.7 559.6 -7.6 539.7 -9.6 519.8 -11.7 500.0 -14.5 480.2 -17.5 400.1 -31.5 319.7 -48.6 239.1 -68.9 158.9 -92.5 99.3 -138.9 72.4 -220.1 46.9 -304.0 35.5 -390.5 34.5 -479.5

Strength IV Total + Total (kip) (kip) 272.8 0.0 210.6 0.0 148.4 0.0 86.2 0.0 24.0 0.0 0.0 -38.2 0.0 -100.4 0.0 -162.6 0.0 -224.8 0.0 -287.0 0.0 -302.6 0.0 -318.1 0.0 -333.7 0.0 -349.2 0.0 -364.8 0.0 -380.3 0.0 -395.9 0.0 -411.4 0.0 -427.0 411.4 0.0 395.9 0.0 380.3 0.0 364.8 0.0 349.2 0.0 333.7 0.0 318.1 0.0 302.6 0.0 287.0 0.0 224.8 0.0 162.6 0.0 100.4 0.0 38.2 0.0 0.0 -24.0 0.0 -86.2 0.0 -148.4 0.0 -210.6 0.0 -272.8

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

Service II Total + Total (kip) (kip) 379.7 -26.9 309.0 -27.7 240.2 -36.6 173.4 -56.5 108.9 -77.5 72.1 -124.8 53.8 -188.6 37.9 -252.7 24.6 -316.8 13.7 -380.5 11.3 -396.3 9.2 -412.1 7.5 -427.9 5.9 -443.7 4.4 -459.4 3.0 -475.1 1.7 -490.8 0.0 -506.4 0.0 -522.0 506.4 -0.7 490.8 -1.7 475.1 -3.0 459.4 -4.4 443.7 -5.9 427.9 -7.5 412.1 -9.2 396.3 -11.3 380.5 -13.7 316.8 -24.6 252.7 -37.9 188.6 -53.8 124.8 -72.1 77.5 -108.9 56.5 -173.4 36.6 -240.2 27.7 -309.0 26.9 -379.7

Fatigue Total + Total (kip) (kip) 38.1 -3.5 33.5 -3.5 28.9 -4.8 24.5 -8.3 20.2 -12.9 16.1 -17.4 12.2 -21.8 8.6 -26.0 5.5 -29.9 3.0 -33.7 2.5 -34.5 2.1 -35.4 1.7 -36.2 1.4 -37.0 1.0 -37.8 0.8 -38.6 0.5 -39.3 0.2 -40.0 40.7 -40.7 40.0 -0.2 39.3 -0.5 38.6 -0.8 37.8 -1.0 37.0 -1.4 36.2 -1.7 35.4 -2.1 34.5 -2.5 33.7 -2.9 29.9 -5.5 26.0 -8.6 21.8 -12.2 17.4 -16.1 12.9 -20.2 8.3 -24.5 4.8 -28.9 3.5 -33.5 3.5 -38.1

AASHTO-LRFD 2007 Page 31 of 62 -- 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 f comp , DL ≤ 2 f Fat Distance from bottom of section to the detail under investigation y = tf,bottom + D = 1.00” + 69.00” = 70” (Pg 16)

(Pg 24)

( 2, 779 ) (12 ) ( 70"− 30.68") = 24.67 k-ft

M DC1 = 2, 779k-ft

f DC1 =

in ft

ksi

53,157 in 4 (Pg 16)

(Pg 24)

(Pg 16)

( 515.8 ) (12 ) ( 70"− 46.33") = 1.427 k-ft

M DC 2 = 515.8k-ft

f DC 2 =

in ft

102, 676 in

ksi

(Pg 16)

f comp , DL = 24.67 ksi + 1.427 ksi = 26.09ksi (Pg 16)

(Pg 30)

( 258.6 ) (12 ) ( 70"− 58.19") = 0.261 k-ft

M Fat , Neg = 258.6k-ft

f Fat =

in ft

140,521 in

ksi

(Pg 16)

Check f comp , DL ≤ 2 f Fat

26.09ksi ≤ ( 2 ) ( 0.261ksi ) = 0.521ksi , No. ?

Fatigue need not be checked on the top flange at Station 73.3.

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 32 of 62 -- 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.

γ ( Δf ) ≤ ( ΔF )n 1

( ΔF ) n

⎛ A ⎞ 3 ( ΔF )TH =⎜ ⎟ ≥ 2 ⎝N⎠

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

γ ( Δf ) =

(1, 236 ) (12 ) ( 58.19"− 1.00") = 6.036 k-ft

in ft

140,521 in

ksi

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

( ΔF )TH 2

12.0ksi = 6.00ksi 2

The stress in the detail is almost less than the infinite life threshold

8 3 ⎛ A ⎞ 3 ⎛ 44 × 10 ksi ⎞ 3 = = 4.375ksi ⎜ ⎟ ⎜ 6 ⎟ ⎝ N ⎠ ⎝ 52.56 × 10 ⎠

( ΔF )TH ⎛ A ⎞3 = 6.00ksi , the infinite life governs. Since ⎜ ⎟ = 4.375ksi is less than 2 ⎝N⎠ ( ΔF )n = 6.00ksi Since γ ( Δf ) = 6.036ksi > ( F )n = 6.00ksi , the detail is not satisfactory.

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 33 of 62 -- 33 --

Calculate the design life of the part under consideration: Since γ ( Δf ) is greater than

( ΔF )TH 2

, solve for N in the following equation.

⎛ A ⎞3 γ ( Δf ) ≤ ⎜ ⎟ ⎝N⎠

N≤

γ ( Δf )

44 × 108 ksi3

( 6.036 )

ksi 3

= 20.01×106 cycles

20.01×106 cycles = 10, 420 days 1,920

10, 420 days = 28.55 years 365

( = 28y, 6m, 19d, 2h, 38min...)

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

☺

AASHTO-LRFD 2007 Page 34 of 62 -- 34 --

8. CHECK CROSS_SECTION PROPORTION LIMITS Web Proportions

•

69"

D ≤ 150 tw

9 " 16

= 122.7 ≤ 150

O.K.

Flange Proportions

•

bf ≤ 12 2t f

15" = 10.00 ≤12 (2)( 3 4 ")

O.K.

•

bf ≤ 12 2t f

21" = 10.50 ≤12 (2)(1")

O.K.

•

bf ≤ 12 2t f

21" = 4.200 ≤12 (2)(2 12 ")

O.K.

Check ODOT Criteria for Flange Width ? ⎛D ⎞ b f ≥ ⎜ + 2.5 ⎟ ≥ 12" ⎝6 ⎠

• b f ,min =

→

⎛ 69" ⎞ + 2.5 ⎟ = 14" ⎜ ⎝ 6 ⎠

D 69" = =11.50" 6 6

• t f ,min = 1.1tw =(1.1)( 916 ")

O.K.

O.K. 5

O.K.

• 0.1 ≤

I yc ≤ 10 I yt

0.1 ≤

( 3 4 ")(15")3 = 0.2733 ≤ 10 (1")(21")3

O.K.

• 0.1 ≤

I yc ≤ 10 I yt

0.1 ≤

(2.5")(21")3 = 2.500 ≤ 10 (1")(21")3

O.K.

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 35 of 62 -- 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 w ' =

( 42 ') (12 inft ) 8

= 63.00"

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

yc = 1"+ 69"+ 3 4 "+

8.5" = 75.00" 2

The CG of this composite cross section is found as:

( 63")(8.5")i( 75.00") + ( 4 ) ( 71.06 in 2 )i( 30.68") = = 59.63" ( 63")( 8.5") + ( 4 ) ( 71.06 in 2 )

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

Concrete

→

Steel

→

( 63")( 8.5") 12

+ ( 63")( 8.5") [ 75.00"− 59.63"] = 2

129, 700 in 4

( 4 ) ( 53,160 in 4 ) + ( 4 ) ( 71.06 in 2 ) [30.68"− 59.63"]

450,900 in 4

I1,total = 580, 600 in 4

I1 = 2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

580, 600 in 4 in 4 = 145,100 Girder 4 Girders AASHTO-LRFD 2007 Page 36 of 62

-- 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 w ' =

( 42 ') (12 inft ) = 63.00" 8

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

yc = 2 1 2 "+ 69"+ 1"+

8.5" = 76.75" 2

The CG of this composite cross section is found as:

( 63")(8.5")i( 76.75") + ( 4 ) (112.3 in 2 )i( 26.83") = = 53.98" ( 63")( 8.5") + ( 4 ) (112.3 in 2 )

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

Concrete

→

Steel

→

( 63")( 8.5") 12

+ ( 63")( 8.5") [ 76.75"− 53.98"] = 2

280,900 in 4

( 4 ) ( 96, 640 in 4 ) + ( 4 ) (112.3 in 2 ) [ 26.83"− 53.98"]

717, 700 in 4

I total = 998, 600 in 4

I2 =

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

998, 600 in 4 in 4 = 249, 700 Girder 4 Girders

AASHTO-LRFD 2007 Page 37 of 62 -- 37 --

The following model, which represents the stiffness of a single girder, was used to compute absolute liveload 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 = 249,700 in4

I = 145,100 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:

⎛ ( 0.85 )( 3) ⎞ Δ1 = ⎜⎜ ⎟⎟ ( 2.442") = 1.558" 4 ( ) ⎝ ⎠

← Governs

⎛ ( 0.85 )( 3) ⎞ Δ 2 = ⎜⎜ ⎟⎟ ⎡⎣( 0.8442") + ( 0.25 )( 2.442") ⎤⎦ = 0.9274" ⎝ ( 4) ⎠ The limiting deflection for this bridge is:

Δ Limit =

(165') (12 inft ) L = = 2.475" 800 800

← OK

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 38 of 62 -- 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.

(1) ( 2)

→

0.032 L = ( 0.032 )(165') (12 inft ) = 63.36"

→

0.027 L = ( 0.027 )(165') (12 inft ) = 53.46"

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 39 of 62 -- 39 --

9.3: Permanent Deformations - Section 1 At the Service Limit State, the following shall be satisfied for composite sections

Top Flange:

f f ≤ 0.95Rh Fyf

Bottom Flange

ff +

fl ≤ 0.95Rh Fyf 2

⎛ M DC 1 ⎞ ⎛M ⎞ ⎛M + 1.00 ⎜ DC 2 ⎟ + 1.00 ⎜ DW ⎟ ⎝ S LT ⎠ ⎝ S LT ⎝ S BS ⎠

fc = 1.00 ⎜

⎛ M LL + IM ⎞ ⎞ ⎟ + 1.30 ⎜ S ⎟ ⎠ ⎝ ST ⎠

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

⎛ (2, 979

f f ,58.7 = 1.00 ⎜

⎝

k-ft

)(12 in ft ) ⎞

1,327 in

⎛ (549.7

⎟ + 1.00 ⎜ ⎠

⎝

k-ft

)(12 in ft ) ⎞

4,204 in

⎛ (1, 008

⎟ + 1.00 ⎜ ⎠

⎝

k-ft

)(12 in ft ) ⎞

4,204 in

⎛ (3, 999

⎟ + 1.30 ⎜ ⎠

⎝

k-ft

)(12 in ft ) ⎞

11,191 in

⎟ ⎠

f f ,58.7 = 36.96ksi

⎛ (2, 779

f f , 73.3 = 1.00 ⎜

⎝

k-ft

)(12 in ft ) ⎞

1,327 in

⎛ (515.8

⎟ + 1.00 ⎜ ⎠

⎝

k-ft

)(12 in ft ) ⎞

4,204 in

⎛ (946.1

k-ft

⎟ + 1.00 ⎜ ⎠

⎝

)(12 in ft ) ⎞

4,204 in

⎛ (4, 067

⎟ + 1.30 ⎜ ⎠

⎝

k-ft

)(12 in ft ) ⎞

11,191 in

⎟ ⎠

f f ,73.3 = 34.97 ksi The stress at 58.7’ governs. ff = 36.96ksi. f f ≤ 0.95Rh Fyf

→

36.96ksi ≤ (0.95)(1.00)(50ksi ) = 47.50ksi

O.K.

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

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 40 of 62 -- 40 --

Bottom Flange, Positive Moment

⎛ (2, 979

f f ,58.7 = 1.00 ⎜

⎝

k-ft

)(12 in ft ) ⎞

1,733 in

⎛ (549.7

⎟ + 1.00 ⎜ ⎠

⎝

k-ft

)(12 in ft ) ⎞

2,216 in

⎛ (1, 008

⎟ + 1.00 ⎜ ⎠

⎝

k-ft

)(12 in ft ) ⎞

2,216 in

⎛ (3, 999

⎟ + 1.30 ⎜ ⎠

⎝

k-ft

)(12 in ft ) ⎞

2,415 in

⎟ ⎠

f f ,58.7 = 54.90ksi

⎛ (2, 779 )(12 inft ) ⎞ ⎛ (515.8 )(12 inft ) ⎞ ⎛ (946.1 )(12 inft ) ⎞ ⎛ (4, 067 )(12 inft ) ⎞ 1.00 1.00 1.30 = 1.00 ⎜ + + + ⎟ ⎜ 2,216 in ⎟ ⎜ 2,216 in ⎟ ⎜ 2,415 in ⎟ ⎝ 1,733 in ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ k-ft

f f , 73.3

k-ft

f f ,73.3 = 53.43ksi The stress at 58.7’ governs. ff = 54.90ksi. The load factor for wind under Service II is 0.00, ∴ fl = 0ksi ff +

fl ≤ 0.95Rh Fyf 2

→

54.90ksi +

0ksi ≤ (0.95)(1.00)(50ksi ) = 47.50ksi 2

No Good.

Note: The bending moments in the above calculations come from page 22 while the moments of inertia are found on page 17.

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 41 of 62 -- 41 --

9.4: Permanent Deformations - Section 2 Top Flange, Negative Moment

⎛ (7,109 )(12 inft ) ⎞ ⎛ (1, 250 )(12 inft ) ⎞ ⎛ (2, 292 )(12 inft ) ⎞ ⎛ (4, 918 )(12 inft ) ⎞ 1.00 1.00 1.30 = 1.00 ⎜ + + + ⎟ ⎜ 5,135 in ⎟ ⎜ 5,135 in ⎟ ⎜ 11,828 in ⎟ ⎝ 2,116 in ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ k-ft

f f ,165

k-ft

f f ,165 = 55.08ksi f f ≤ 0.95 Rh Fyf

→

55.08ksi ≤(0.95)(1.00)(50ksi ) = 47.50ksi

No Good.

Bottom Flange, Negative Moment

⎛ (7,108

f f = 1.00 ⎜

⎝

k-ft

)(12 in ft ) ⎞

3,602 in

⎛ (1, 250

⎟ + 1.00 ⎜ ⎠

⎝

k-ft

)(12 in ft ) ⎞

4,255 in

⎛ (2, 292

⎟ + 1.00 ⎜ ⎠

⎝

k-ft

)(12 in ft ) ⎞

4,255 in

⎛ (4, 918

⎟ + 1.30 ⎜ ⎠

⎝

k-ft

)(12 in ft ) ⎞

4,590 in

⎟ ⎠

f f = 50.39 ksi The load factor for wind under Service II is 0.00, ∴ fl = 0ksi ff +

fl ≤ 0.95 Rh Fyf 2

→

50.39 ksi +

0ksi ≤ (0.95)(1.00)(50ksi ) = 47.50ksi 2

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

No Good.

AASHTO-LRFD 2007 Page 42 of 62 -- 42 --

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

f c ≤ Fcrw where: Fcrw =

0.9 Ek

⎛D⎞ ⎜ ⎟ ⎝ tw ⎠

and

( Dc / D )

Section 1

Not Applicable

Section 2

⎛ (7,108 )(12 ⎝ 3,602 in k-ft

fc = 1.00 ⎜

)⎞

⎛ (1, 250 )(12 ⎝ 4,255 in k-ft

⎟ + 1.00 ⎜

⎠

in ft

)⎞

⎠

⎛ (2, 292 )(12 ⎝ 4,255 in k-ft

⎟ + 1.00 ⎜

in ft

)⎞

⎠

⎛ (4, 918 )(12 ⎝ 4,590 in k-ft

⎟ + 1.30 ⎜

)⎞

⎟ ⎠

f c = 50.39 ksi

⎛ (7,108 )(12 ⎝ 2,116 in k-ft

ft = 1.00 ⎜

in ft

)⎞

⎛ (1, 250 )(12 ⎝ 5,135 in k-ft

⎟ + 1.00 ⎜ ⎠

in ft

)⎞

⎛ (2, 292 )(12 ⎝ 5,135 in

⎟ + 1.00 ⎜ ⎠

k-ft

in ft

)⎞

⎛ (4, 918 )(12 ⎝ 11,828 in k-ft

⎟ + 1.30 ⎜ ⎠

⎟ ⎠

ft = 55.08ksi

⎛ − fc ⎞ Dc = ⎜⎜ ⎟⎟ d − tcf ≥ 0 ⎝ f c + ft ⎠ ⎛ ⎞ 50.39ksi 72.5") − 2.5" ≥ 0 =⎜ ksi ksi ⎟ ( ⎝ 50.39 + 55.08 ⎠ = 32.14′′ k=

( Dc / D )

Fcrw =

⎛ 32.14" ⎞ ⎜ ⎟ ⎝ 69" ⎠

= 41.49

(0.90)(29, 000ksi )(41.49)

⎛ 69" ⎞ ⎜ 9 "⎟ ⎝ 16 ⎠

= 71.96 ksi

This is larger than fc…O.K.

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 43 of 62 -- 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 Dcp E ≤ 3.76 tw Fyc

Find Dcp, the depth of the web in compression at Mp (compression rebar in the slab is ignored). Pt = Fyt bt tt = (50ksi ) ( 21")(1") = 1,050kip Pw = Fyw Dtw = (50ksi )(69")(916 ") = 1,941kip Pc = Fyc bc tc = (50ksi )(15")( 3 4 ") = 562.5kip Ps = 0.85 f c'bs ts = (0.85)(4.5ksi )(109.5")(8.5") = 3,560kip 3,554kip < 3,560kip, the PNA lies in the slab.

Since Pt + Pw +Pc < Ps

⎡ P + Pw + Pt ⎤ ⎡ 3,554kip ⎤ Y = ( ts ) ⎢ c = 8.5" ( ) ⎥ ⎢ kip ⎥ Ps ⎣ 3,560 ⎦ ⎣ ⎦ Y = 8.486 " ↓ from top of slab ∴ D p = Y = 8.486 "

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) Mu +

1 f S ≤ φf Mn 3 l xt

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) Dp ⎞ ⎛ Since Dp =8.486 > 0.1Dt = 7.925, M n = M p ⎜ 1.07 − 0.7 ⎟ Dt ⎠ ⎝ ⎛

M p = (3,554kip ) ⎜ 79.25"− ⎝

= 157,500

k-in

8.486" ⎞ − 30.68" ⎟ 2 ⎠

= 13,130k-ft

⎡ ⎛ 8.486" ⎞ ⎤ M n = 13,130k-ft ⎢1.07 − ( 0.7 ) ⎜ = 13,060k-ft ⎟ ⎥ ⎝ 79.25" ⎠ ⎦ ⎣

(

)

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 44 of 62 -- 44 --

M u + 13 f l S xt ≤ φ f M n

(13,568k-ft ) + (0) ≤(1.00)(13,060k-ft )

No Good.

Note that the check of M n ≤ 1.3Rh M y 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: ?

D p ≤ 0.42 Dt

→ 8.486" ≤ ( 0.42 )( 79.25") = 33.29"

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.

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 45 of 62 -- 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. Dn , Bottom = 58.19"− 1" = 57.19" Rh =

(

12 + β 3ρ − ρ 3

)

β=

12 + 2 β

ρ=

Rh , Section 1 =

2 Dn tw (2) ( 57.19")( 916 ") = = 3.064 Afn (1")( 21") Fyw fn

≤ 1.0

→ ρ=

12 + ( 3.064 ) ⎡⎣(3)(0.7143) − (0.7143)3 ⎤⎦ 12 + ( 2 )( 3.064 )

50ksi = 0.7143 70ksi

= 0.9626

Compute the Hybrid Girder Factor, Rh, for Section 2: For the short-term composite section, Dn ,Top = 2 1 2 "+ 69"− 52.23" = 19.27" Dn , Bottom = 52.23"− 2 1 2 " = 49.73"

Rh =

(

12 + β 3ρ − ρ 3

)

Governs

β=

12 + 2 β

ρ=

Rh , Section 2 =

2 Dn tw (2) ( 49.73")( 916 ") = = 1.066 Afn ( 2 12 ")( 21") Fyw fn

≤ 1.0

→ ρ=

12 + (1.066 ) ⎡⎣(3)(0.7143) − (0.7143)3 ⎤⎦ 12 + ( 2 )(1.066 )

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

50ksi = 0.7143 70ksi

= 0.9833

AASHTO-LRFD 2007 Page 46 of 62 -- 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:

f f ≤ 0.95Rh Fyf

Bottom Flange

ff +

fl ≤ 0.95Rh Fyf 2

Top Flange, Positive Moment

From before: f f ,58.7 = 36.96ksi f f ≤ 0.95Rh Fyf

36.96ksi ≤(0.95)(0.9626)(70ksi ) = 64.01ksi

→

O.K.

Bottom Flange, Positive Moment f f ,58.7 = 54.90ksi

ff +

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

fl ≤ 0.95Rh Fyf 2

→

54.90ksi +

0ksi ? ≤(0.95)(0.9626)(70ksi ) = 64.01ksi 2

O.K.

11.2: Permanent Deformations - Section 2 Top Flange, Negative Moment

From before: f f ,165 = 55.08ksi f f ≤ 0.95Rh Fyf

→

55.08ksi ≤(0.95)(0.9833)(70ksi ) = 65.39ksi

O.K.

Bottom Flange, Negative Moment f f = 50.39 ksi The load factor for wind under Service II is 0.00, ∴ fl = 0ksi ff +

fl ≤ 0.95 Rh Fyf 2

→

50.39ksi +

0ksi ? ≤(0.95)(0.9833)(70ksi ) = 65.39ksi 2

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

O.K.

AASHTO-LRFD 2007 Page 47 of 62 -- 47 --

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

f c ≤ Fcrw where: Fcrw =

0.9 Ek

⎛D⎞ ⎜ ⎟ ⎝ tw ⎠

and

( Dc / D )

Not Applicable

Section 1 -

Section 2

⎛ (7,108 )(12 ⎝ 3,602 in k-ft

fc = 1.00 ⎜

)⎞

⎛ (1, 250 )(12 ⎝ 4,255 in k-ft

⎟ + 1.00 ⎜

⎠

in ft

)⎞

⎠

⎛ (2, 292 )(12 ⎝ 4,255 in k-ft

⎟ + 1.00 ⎜

in ft

)⎞

⎠

⎛ (4, 918 )(12 ⎝ 4,590 in k-ft

⎟ + 1.30 ⎜

)⎞

⎟ ⎠

f c = 50.39ksi

⎛ (7,108 )(12 ⎝ 2,116 in k-ft

ft = 1.00 ⎜

in ft

)⎞

⎛ (1, 250 )(12 ⎝ 5,135 in k-ft

⎟ + 1.00 ⎜ ⎠

in ft

)⎞

⎛ (2, 292 )(12 ⎝ 5,135 in

⎟ + 1.00 ⎜ ⎠

k-ft

in ft

)⎞

⎛ (4, 918 )(12 ⎝ 11,828 in k-ft

⎟ + 1.30 ⎜ ⎠

⎟ ⎠

ft = 55.08ksi

⎛ − fc ⎞ Dc = ⎜⎜ ⎟⎟ d − tcf ≥ 0 ⎝ f c + ft ⎠ ⎛ ⎞ 50.39ksi 72.5") − 2.5" ≥ 0 =⎜ ksi ksi ⎟ ( ⎝ 50.39 + 55.08 ⎠ = 32.14′′ k=

( Dc / D )

Fcrw =

⎛ 32.14" ⎞ ⎜ ⎟ ⎝ 69" ⎠

= 41.49

(0.90)(29, 000ksi )(41.49)

⎛ 69" ⎞ ⎜ 9 "⎟ ⎝ 16 ⎠

= 71.96 ksi

This is larger than fc…O.K.

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 48 of 62 -- 48 --

12. RECHECK STRENGTH LIMIT STATE WITH 70KSI FLANGES 12.1: Section 1 - Positive Flexure Section Classification (§6.10.6.2, Pg. 6.98 – 6.99)

Check

2 Dcp E ≤ 3.76 tw Fyc

Find Dcp, the depth of the web in compression at Mp (compression rebar in the slab is ignored). Pt = Fyt bt tt = (70ksi ) ( 21")(1") = 1, 470kip Pw = Fyw Dtw = (50ksi )(69")(916 ") = 1,941kip Pc = Fyc bc tc = (70ksi )(15")( 3 4 ") = 787.5kip Ps = 0.85 f c'bs ts = (0.85)(4.5ksi )(109.5")(8.5") = 3,560kip 4,199kip > 3,560kip, the PNA is NOT in the slab.

Since Pt + Pw +Pc > Ps ?

Check Case I Pt + Pw ≥ Pc + Ps ?

1, 470kip + 1,941kip ≥ 787.5kip + 3,560kip

Check Case II Pt + Pw + Pc ≥ Ps ?

1, 470kip + 1,941kip + 787.5kip ≥ 3,560kip

YES - PNA in Top Flange

⎞ ⎛ t ⎞ ⎛ P + P − Ps Y = ⎜ c ⎟⎜ w t + 1⎟ Pc ⎝ 2 ⎠⎝ ⎠ kip kip kip ⎞ ⎛ 0.750" ⎞ ⎛ 1,941 + 1, 470 − 3,560 =⎜ + 1⎟ = 0.3040" (from the top of steel) ⎟⎜ kip 787.5 ⎝ 2 ⎠⎝ ⎠ 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) Mu +

1 f S ≤ φf Mn 3 l xt

Mu = 13,568k-ft from Page 30; take fl = 0

Dt = 1” + 69” + 3/4” + 8.5” = 79.25”

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

0.1Dt = 7.925”

AASHTO-LRFD 2007 Page 49 of 62 -- 49 --

(The haunch is not included in Dt, as per ODOT Exceptions) Dp ⎞ ⎛ Since Dp = 8.804” > 0.1Dt = 7.925”, M n = M p ⎜1.07 − 0.7 ⎟ Dt ⎠ ⎝ 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" dw = 2 1" dt = 70.75"− − 0.3040" = 69.95" 2 ds =

The plastic moment is computed. ⎛P ⎞ 2 M p = ⎜ c ⎟ ⎡Y 2 + ( tc − Y ) ⎤ + [ Ps d s + Pw d w + Pd t t] ⎦ ⎝ 2tc ⎠ ⎣ ⎛ 787.5kip ⎞ ⎡ 2 2 =⎜ ⎟ ⎣( 0.3040") + ( 0.750"− 0.3040") ⎤⎦ + ... ⎝ (2)(0.750") ⎠

... + ⎡⎣( 3,560kip ) ( 4.554") + (1,941kip ) ( 34.05") + (1, 470kip ) ( 69.95") ⎤⎦

= ( 525 kip in ) ⎡⎣0.2913 in 2 ⎤⎦ + ⎡⎣185,100k-in ⎤⎦ = 185,300k-in = 15, 440k-ft ⎡ ⎛ 8.804" ⎞ ⎤ k-ft M n = 15, 440k-ft ⎢1.07 − ( 0.7 ) ⎜ ⎟ ⎥ = 15,320 79.25" ⎝ ⎠ ⎣ ⎦

(

)

M u + 13 f l S xt ≤ φ f M n

(13,568k-ft ) + (0) ≤(1.00)(15,320 k-ft )

O.K.

D p ≤ 0.42 Dt

→ 8.804" ≤ ( 0.42 )( 79.25") = 33.29"

O.K.

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

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 50 of 62 -- 50 --

12.2: Section 2 - Negative Flexure Section Classification (§6.10.6.2, Pg. 6.98 – 6.99)

Check

2 Dc E ≤ 5.70 tw Fyc

Dc is the depth of the web in compression for the cracked section. Dc = 32.04” – 21/2” = 29.54” 2 Dc (2)(29.54") E 29,000ksi = = 105.0 < 5.70 = 5.70 = 137.3 tw (916 ") Fyc 50ksi

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: fbu +

1 f ≤ φ f Fnc 3 l

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.

⎛ M DC1 ⎞ ⎛ 1.25M DC 2 + 1.50 M DW + 1.75M LL − ⎞ ⎟+⎜ ⎟ SCR ⎝ S BS ⎠ ⎝ ⎠

fbu = 1.25 ⎜

k-ft k-ft k-ft ⎛ (7,109 k-ft )(12 inft ) ⎞ ⎛ ⎡⎣ (1.25)(1, 250 ) + (1.50)(2, 292 ) + (1.75)(4, 918 ) ⎤⎦ (12 inft ) ⎞ fbu = 1.25 ⎜ ⎟ ⎟+⎜ 3 3 3,932 in ⎝ 3,602 in ⎠ ⎝ ⎠

fbu = 71.13ksi 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.

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 51 of 62 -- 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.

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 52 of 62 -- 52 --

12.3 Vertical Shear Capacity At the strength Limit, the following must be satisfied

Vu ≤ φVn For an unstiffened web,

Vn = Vcr = CV p Check,

D Ek , ≤ 1.12 tw Fyw

D 69" = = 122.7 tw 916 " Since there are no transverse stiffeners, k = 5

(29, 000ksi )(5) 1.12 = 60.31 (50ksi ) Since

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

D Ek , elastic shear buckling of the web controls. > 1.40 tw Fyw

1.57 ⎛ kE ⎞ 1.57 ⎛ (5)(29, 000ksi ) ⎞ = ⎜ ⎟ ⎟ = 0.3026 2 2 ⎜ (50ksi ) ⎛ D ⎞ ⎜⎝ Fyw ⎟⎠ ⎛ 69" ⎞ ⎝ ⎠ ⎜ ⎟ ⎜ 9 "⎟ ⎝ 16 ⎠ ⎝ tw ⎠

V p = 0.58 Fyw Dt w = (0.58)(50 ksi )(69")( 9 16 ") = 1,126 kip Vn = CV p = (0.3026)(1,126 kip ) = 340.6 kip

(

)

φVn = (1.00 ) 340.6 kip = 340.6 kip

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.

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 53 of 62 -- 53 --

Try adding transverse stiffeners spaced at do = 8’ = 96”

k =5+

5 ⎛ do ⎞ ⎜D⎟ ⎝ ⎠

=5+

5 ⎛ 96" ⎞ ⎜ 69" ⎟ ⎝ ⎠

= 7.583

D (29, 000ksi )(7.583) (29, 000ksi )(7.583) = 122.7 , 1.12 , = 74.28 1.40 = 92.85 (50ksi ) (50ksi ) tw Since

D Ek > 1.40 , elastic shear buckling of the web controls. tw Fyw

1.57 ⎛ Ek ⎞ 1.57 ⎛ (29, 000ksi )(7.583) ⎞ C= ⎟= ⎟ = 0.4589 2 ⎜ 2 ⎜ (50ksi ) ⎛ D ⎞ ⎝⎜ Fyw ⎠⎟ ⎛ 69" ⎞ ⎝ ⎠ ⎜ ⎟ ⎜ 9 "⎟ ⎝ 16 ⎠ ⎝ tw ⎠ φVn = φCV p = (1.00)(0.4589)(1,126 kip ) = 516.5kip

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”

k =5+

5 ⎛ do ⎞ ⎜D⎟ ⎝ ⎠

=5+

5 ⎛ 144" ⎞ ⎜ 69" ⎟ ⎝ ⎠

= 6.148

D (29, 000ksi )(6.148) (29, 000ksi )(6.148) , = 122.7 , 1.12 = 66.88 1.40 = 83.60 (50ksi ) (50ksi ) tw Since

D Ek , elastic shear buckling of the web controls. > 1.40 tw Fyw

1.57 ⎛ Ek ⎞ 1.57 ⎛ (29, 000ksi )(6.148) ⎞ = ⎟ ⎟ = 0.3721 2 ⎜ 2 ⎜ (50ksi ) ⎛ D ⎞ ⎝⎜ Fyw ⎠⎟ ⎛ 69" ⎞ ⎝ ⎠ ⎜ ⎟ ⎜ 9 "⎟ ⎝ 16 ⎠ ⎝ tw ⎠

Without TFA: Vn = CV p = (0.3721)(1,126 kip ) = 418.9 kip

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 54 of 62 -- 54 --

With TFA: Since

2 Dtw (2)(69")( 916 ") = = 1.056 ≤ 2.5 , ( b fct fc + b ft t ft ) ( (21")(2 12 ") + (21")(1") )

⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ 0.87(1 − C ) ⎥ (0.87)(1 0.3721) − ⎢ ⎢ ⎥ = (1,126kip ) ⎢0.3721 + Vn = V p ⎢C + ⎥ ⎥ 2 2 ⎛ do ⎞ ⎥ ⎛ 144" ⎞ ⎥ ⎢ ⎢ 1+ ⎜ 1+ ⎜ ⎟ ⎟ ⎢ ⎢⎣ ⎝ 69" ⎠ ⎥⎦ ⎝ D ⎠ ⎦⎥ ⎣ Vn = (1,126kip )(0.6082) = 684.8kip

(

)

φVn = (1.00 ) 684.8kip = 684.8kip

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.

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 55 of 62 -- 55 --

Strength Limit Shear Capacity 800 Tension Field Action

600

Strength I

400

Strength IV

Shear (kip)

200

-200

-400

-600

-800 0

120

150

180

210

240

270

300

330

Station (ft)

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 56 of 62 -- 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. be

ts + thaunch − t flange 2 9.5" = + 2.75"− 0.75" = 6.75" 2

ts thaunch

tc bc

Use 7/8” x 61/2” shear studs D tw h

AASHTO requires that the ratio of /d be greater than or equal to 4.0. tt bt

h ≥ 4.0 d 6 12 " = 7.429 ≥ 4.0 7 " 8

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

p≤

nZ r Vsr

Vsr =

Vf Q I

(6.10.10.1.2-1 & 3)

where: n Zr Vsr Vf Q I

Number of studs per row Fatigue resistance of a single stud Horizontal fatigue shear range per unit length Vertical shear force under fatigue load combination 1st moment of inertia of the transformed slab about the short-term NA 2nd moment of inertia of the short-term composite section

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 57 of 62 -- 57 --

Zr = α d 2 ≥

5.5d 2 2

(6.10.10.2-1)

α = 34.5 − 4.28Log( N )

(6.10.10.2-2)

α = 34.5 − 4.28Log(55.84 ×106 ) = 1.343ksi 2 ⎛ 5.5 ⎞ 7 2 Z r = (1.343ksi ) ( 7 8 ") ≥ ⎜ ⎟ ( 8 ") ⎝ 2 ⎠ = 1.028kip ≥ 2.105kip

→

Z r = 2.105kip

Q = Atc d c ⎡ (109.5")( 9.5") ⎤ ⎛ 9.5" ⎞ − 58.19" ⎟ = 2,511 in 3 QSection 1 = ⎢ ⎥ ⎜ 1"+ 69"+ 2.75"+ 8 2 ⎠ ⎣ ⎦⎝ ⎡ (109.5")( 9.5") ⎤ ⎛ 9.5" ⎞ − 52.23" ⎟ = 3, 481 in 3 QSection 2 = ⎢ ⎥ ⎜ 2.5"+ 69"+ 2.75"+ 8 2 ⎠ ⎣ ⎦⎝

I Section 1 = 140,500 in 4

I Section 2 = 239, 700 in 4

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, V f = 38.13kip − −3.53kip = 41.66kip

( 41.66 )( 2,511 in ) = 0.7445 = (140,500 in ) kip

Vsr

p≤

( 0.7445 ) kip inch

= 8.482

kip inch

For 4 studs in each row:

For 3 studs in each row:

( 3) ( 2.105ksi )

p≤

in row

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

( 4 ) ( 2.105ksi )

( 0.7453 ) kip inch

in = 11.31 row

AASHTO-LRFD 2007 Page 58 of 62 -- 58 --

Strength Limit:

Qr = φsc Qn Qn = 0.5 Asc

φsc = 0.85 f c' Ec ≤ Asc Fu

(6.10.10.4.3-1)

2 ⎛π ⎞ Asc = ⎜ ⎟ ( 7 8 ") = 0.6013 in 2 ⎝4⎠

f c' = 4.5ksi Since n = 8, Ec =

Es 29, 000ksi = = 3, 625ksi 8 n

Fu = 60ksi Qn = ( 0.5 ) ( 0.6013 in 2 )

( 4.5 )( 3, 625 ) ≤ ( 0.6013 in )( 60 ) ksi

ksi

kip kip = 38.40 stud ≤ 36.08 stud

kip kip φscQn = ( 0.85 ) ( 36.08 stud ) = 30.67 stud

n+ =

n− =

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

Pp + Pn Qr

AASHTO-LRFD 2007 Page 59 of 62 -- 59 --

Positive moment - Section 1: Station 0.0’ - 73.3’

Pp = Min ( PConcrete , Psteel )

PConcrete = 0.85 f c'bets

= ( 0.85 ) ( 4.5ksi ) (109.5")( 9.5") = 3,979kip

PSteel = Fyw Dtw + Fft b ft t ft + Ffc b fc t fc = ( 70ksi ) ⎡⎣(15")( 0.75") + ( 21")(1") ⎤⎦ + ( 50ksi ) ( 69")( 0.5625") = 4,198kip Pp = 3,979kip

n+ =

Pp Qr

3,979kip = 129.7studs kip 30.67 stud

3 Studs per Row:

129.7studs = 44rows studs 3 row

→

( 73.3'− 0 ') (12 inft ) = 20.46 inch row ( 44 − 1)

→

Say 20"

→

( 73.3'− 0 ') (12 inft ) = 27.49 inch row ( 33 − 1)

→

Say 24"

4 Studs per Row:

129.7studs = 33rows studs 4 row

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 60 of 62 -- 60 --

Negative Moment - Section 2: Station 73.3’ - 165.0’

Pn = Min ( Psteel , PCrack )

PCrack = 0.45 f c'bets

= ( 0.45 ) ( 4.5ksi ) (109.5")( 9.5") = 2,107 kip

PSteel = Fyw Dtw + Fft b ft t ft + Ffc b fc t fc = ( 70ksi ) ⎡⎣( 21")( 2.5") + ( 21")(1") ⎤⎦ + ( 50ksi ) ( 69")( 0.5625") = 7, 086kip Pn = 2,107 kip

n− =

Pp + Pn Qr

3,979kip + 2,107 kip = 198.4studs kip 30.67 stud

3 Studs per Row:

198.4studs = 67 rows studs 3 row

→

(165.0 '− 73.3') (12 inft ) = 16.67 inch row ( 67 − 1)

→ Say 16"

→

(165.0 '− 73.3') (12 inft ) = 22.48 inch row ( 50 − 1)

→ Say 20"

4 Studs per Row:

198.4studs = 50rows studs 4 row

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 61 of 62 -- 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. 3 Studs Per Row Station (ft) 0.0 14.7 29.3 44.0 58.7 73.3 88.0 102.7 117.3 132.0 135.7 139.3 143.0 146.7 150.3 154.0 157.7 161.3 165.0 168.7 172.3 176.0 179.7 183.3 187.0 190.7 194.3 198.0 212.7 227.3 242.0 256.7 271.3 286.0 300.7 315.3 330.0

Vf (kip) 41.66 37.01 33.68 32.79 33.04 33.46 33.98 34.59 35.38 36.62 37.07 37.53 37.98 38.42 38.88 39.34 39.81 40.26 81.44 40.26 39.81 39.34 38.88 38.42 37.98 37.53 37.07 36.62 35.38 34.59 33.98 33.46 33.04 32.79 33.68 37.01 41.66

V sr

(in )

2,511 2,511 2,511 2,511 2,511 2,511 2,511 2,511 2,511 2,511 3,481 3,481 3,481 3,481 3,481 3,481 3,481 3,481 3,481 3,481 3,481 3,481 3,481 3,481 3,481 3,481 3,481 2,511 2,511 2,511 2,511 2,511 2,511 2,511 2,511 2,511 2,511

140,521 140,521 140,521 140,521 140,521 140,521 140,521 140,521 140,521 140,521 239,734 239,734 239,734 239,734 239,734 239,734 239,734 239,734 239,734 239,734 239,734 239,734 239,734 239,734 239,734 239,734 239,734 140,521 140,521 140,521 140,521 140,521 140,521 140,521 140,521 140,521 140,521

(in )

kip

( /in) 0.7444 0.6613 0.6018 0.5859 0.5904 0.5979 0.6071 0.6181 0.6323 0.6543 0.5383 0.5449 0.5514 0.5579 0.5645 0.5713 0.5780 0.5847 1.1826 0.5847 0.5780 0.5713 0.5645 0.5579 0.5514 0.5449 0.5383 0.6543 0.6323 0.6181 0.6071 0.5979 0.5904 0.5859 0.6018 0.6613 0.7444

p Fat

p Str

(in) 8.48 9.55 10.49 10.78 10.70 10.56 10.40 10.22 9.99 9.65 11.73 11.59 11.45 11.32 11.19 11.05 10.93 10.80 5.34 10.80 10.93 11.05 11.19 11.32 11.45 11.59 11.73 9.65 9.99 10.22 10.40 10.56 10.70 10.78 10.49 9.55 8.48

(in) 20.00 20.00 20.00 20.00 20.00 20.00 16.00 16.00 16.00 16.00 16.00 16.00 16.00 16.00 16.00 16.00 16.00 16.00 16.00 16.00 16.00 16.00 16.00 16.00 16.00 16.00 16.00 16.00 16.00 16.00 16.00 20.00 20.00 20.00 20.00 20.00 20.00

4 Studs Per Row

p max (in) 8.48 9.55 10.49 10.78 10.70 10.56 10.40 10.22 9.99 9.65 11.73 11.59 11.45 11.32 11.19 11.05 10.93 10.80 5.34 10.80 10.93 11.05 11.19 11.32 11.45 11.59 11.73 9.65 9.99 10.22 10.40 10.56 10.70 10.78 10.49 9.55 8.48

p Fat

p Str

(in) 11.31 12.73 13.99 14.37 14.26 14.08 13.87 13.62 13.32 12.87 15.64 15.45 15.27 15.09 14.92 14.74 14.57 14.40 7.12 14.40 14.57 14.74 14.92 15.09 15.27 15.45 15.64 12.87 13.32 13.62 13.87 14.08 14.26 14.37 13.99 12.73 11.31

(in) 24.00 24.00 24.00 24.00 24.00 24.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 20.00 24.00 24.00 24.00 24.00 24.00 24.00

p max (in) 11.31 12.73 13.99 14.37 14.26 14.08 13.87 13.62 13.32 12.87 15.64 15.45 15.27 15.09 14.92 14.74 14.57 14.40 7.12 14.40 14.57 14.74 14.92 15.09 15.27 15.45 15.64 12.87 13.32 13.62 13.87 14.08 14.26 14.37 13.99 12.73 11.31

The arrangement of shear studs is shown below.

2- Span Continuous Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Page 62 of 62 -- 62 --

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 ksi f’c = 4.5 . 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.

Single-Span Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Created July 2007: Page 1 of 21 -- 63 --

172' - 4" Total Girder Length

G 2

G 3

G 4

G 5

G 6

166' - 4" cc Bearings

Cross Frames Spaced @ 22' - 0" cc G 1

Single-Span Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Created July 2007: Page 2 of 21 -- 64 --

Positive Bending Section (Section 2)

Positive Bending Section (Section 1)

Positive Bending Section (Section 3)

Single-Span Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 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 ⎛ ⎛ 490pcf ⎞ ⎞ per girder ⎟ ⎟ (1.15 ) = 373.7 lbs Wsection1 = ⎜ 95.5 in 2 ⎜ ft ⎜ (12 in )2 ⎟ ⎟ ⎜ ft ⎝ ⎠⎠ ⎝

(b) Section 2 A = (14”)(2”) + (68”)(0.5625”) + (22”)(2”) = 110.25 in2 ⎛ ⎛ 490pcf ⎞ ⎞ per girder ⎟ ⎟ (1.15 ) = 431.4 lbs Wsection1 = ⎜110.3 in 2 ⎜ ft ⎜ (12 in )2 ⎟ ⎟ ⎜ ft ⎝ ⎠⎠ ⎝ (c) Section 3

A = (14”)(2”) + (68”)(0.5625”) + (22”)(2.375”) = 118.5 in2 ⎛ ⎛ 490pcf ⎞ ⎞ per girder ⎟ ⎟ (1.15 ) = 463.7 lbs Wsection1 = ⎜118.5 in 2 ⎜ ft ⎜ (12 in )2 ⎟ ⎟ ⎜ ft ⎝ ⎠⎠ ⎝

(d) Average Girder Self Weight Wave =

( 2 )( 40.17 ') ( 373.7 lbsft ) + ( 2 )(18.0 ') ( 431.4 lbsft ) + ( 50.0 ') ( 463.7 lbsft ) 166.3'

= 413.3 lbs ft

Deck Self-Weight (DC1):

⎛ ( 9.5'')( 64.0 ') ⎞ ⎛ 150pcf ⎞ per girder WDeck = ⎜ ⎟ = 1,267 lbs ⎟ ⎜⎜ ft in ⎟ 6 Girders 12 ⎝ ⎠ ⎝ ( ft ) ⎠

Single-Span Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Created July2007: Page 4 of 23 -- 66 --

Haunch Self-Weight (DC1): Average width of haunch: 14’’

⎛ 150pcf ⎞ per girder ⎟ = 94.33 lbs Whaunch = (14 )( 4 ) + 2 ( ( 12 ) ( 9 '')( 4 '') ) ⎜ ft ⎜ (12 in )2 ⎟ ft ⎝ ⎠

(

)

Barrier Walls (DC2):

Wbarriers

⎛ ( 2 each ) ( 640plf ) ⎞ per girder ⎟ = 213.3 lbs =⎜ ft ⎜ ⎟ 6 girders ⎝ ⎠

Wearing Surface (DW):

Wwearing_surface =

( 61.0') ( 60psf ) 6 Girders

per girder = 610.0 lbs ft

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 noncomposite 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.

M DC1,Steel

2 ⎤ ⎛ wL2 ⎞ ⎡ ( 413.3 lbs ft ) (166.3' ) ⎢ ⎥ = 1, 429k-ft =⎜ = ⎟ 8 8 ⎥⎦ ⎝ ⎠ ⎢⎣

M DC1, Deck

2 166.3' ) ⎤ ⎛ wL2 ⎞ ⎡ (1, 267 lbs ft ) ( ⎥ = 4,379k-ft =⎜ ⎟=⎢ 8 ⎥⎦ ⎝ 8 ⎠ ⎢⎣

M DC 2, Barriers

M DW

2 ⎤ ⎛ wL2 ⎞ ⎡ ( 213.3 lbs ft ) (166.3' ) ⎢ ⎥ = 737.4k-ft =⎜ = ⎟ 8 ⎥⎦ ⎝ 8 ⎠ ⎢⎣

2 166.3' ) ⎤ ⎛ wL2 ⎞ ⎡ ( 610.0 lbs ft ) ( ⎥ = 2,109k-ft =⎜ ⎟=⎢ 8 8 ⎥⎦ ⎝ ⎠ ⎢⎣

Single-Span Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Created July2007: Page 5 of 23 -- 67 --

The maximum shear forces at the ends of the girder are also easily computed. lbs ⎛ wL ⎞ ⎡ ( 413.3 ft ) (166.3' ) ⎤ kip VDC1,Steel = ⎜ = ⎢ ⎥ = 34.37 ⎟ 2 2 ⎝ ⎠ ⎢⎣ ⎥⎦

VDC1, Deck

lbs ⎛ wL ⎞ ⎡ (1, 267 ft ) (166.3' ) ⎤ kipt =⎜ ⎥ = 105.4 ⎟=⎢ 2 ⎝ 2 ⎠ ⎣⎢ ⎦⎥

lbs ⎛ wL ⎞ ⎡ ( 213.3 ft ) (166.3' ) ⎤ kip VDC 2, Barriers = ⎜ = ⎥ = 17.74 ⎟ ⎢ 2 ⎝ 2 ⎠ ⎣⎢ ⎦⎥ lbs ⎛ wL ⎞ ⎡ ( 610.0 ft ) (166.3' ) ⎤ kip VDW = ⎜ = ⎢ ⎥ = 50.72 ⎟ 2 2 ⎝ ⎠ ⎢⎣ ⎥⎦

Single-Span Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Created July2007: Page 6 of 23 -- 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)

Single-Span Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Created July2007: Page 7 of 23 -- 69 --

Unfactored HL-93 Moment Envelopes from SAP 6,000 Single Truck

4,000 Tandem

Moment (kip-ft)

2,000

-2,000

-4,000

-6,000 0

120

150

Station (ft)

The following results were obtained from the SAP analysis: •

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

HL-93M HL-93K HL-93S

Station 40.16’- Section 1 3,614k-ft 4,322k-ft N/A

Station 58.15’- Section 2 4,481k-ft 5,238k-ft N/A

Station 83.15’- Section 3 4,911k-ft 5,821k-ft N/A

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

Single-Span Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Created July2007: Page 8 of 23 -- 70 --

2c. Verify the Maximum Positive Live-Load Moment at Station 83.15â&#x20AC;&#x2122;: kip

kip

Tandem: kip

32 8

kip

Single Truck:

0.640kip/ft

Moment (k-ft / kip)

Lane

45 40 35 30 25 20 15 10 5 0 0

105

120

135

150

165

Station (ft)

Tandem: Single Truck: Lane Load:

( 25 ) ( 41.58 ) + ( 25 ) ( 39.58 ) = 2, 029 (8 ) ( 34.57 ) + ( 32 ) ( 41.58 ) + ( 32 ) ( 34.57 ) = 2, 713 kip

kip

k-ft kip

kip

k-ft kip

( 0.640 )(3, 457 ) = 2, 212 k-ft kip

k-ft kip

k-ft

k-ft kip

kip

k-ft kip

k-ft

(IM)(Tandem) + Lane:

+ 2, 212 k-ft = 4, 911k-ft (1.33 ) ( 2, 029 k-ft kip )

(IM)(Single Truck) + Lane:

+ 2, 212 k-ft = 5,821k-ft (1.33 ) ( 2, 713 k-ft kip )

GOVERNS

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

Single-Span Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Created July2007: Page 9 of 23 -- 71 --

3. SECTION PROPERTIES AND CALCULATIONS: 3a. Effective Flange Width, bs: For an interior beam, bs is the lesser of: bf ⎧ 14" = (12 )( 8.5") + = 109 '' ⎪•12ts + 2 2 ⎪⎪ ⎨•S = (11.33')(12 in ft ) = 135.96 '' ⎪ L ⎪• eff = 166.3' = 41.58' = 498.9 '' ⎪⎩ 4 4

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.

Single-Span Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Created July2007: Page 10 of 23 -- 72 --

3b. Section 1 Flexural Properties

Single Span Bridge Example - Section 1 Bare Steel t

Ad2

Top Flange

1.1250

14.00

15.75

70.06

1,103.48

1.66

-40.87

26,308

26,310

Web Bot Flange

0.6875 1.5000

68.00 22.00

46.75 33.00

35.50 0.75

1,659.63 18,014.33 24.75 6.19

-6.31 28.44

1,860 26,696

19,874 26,702

2,787.86

ITotal =

72,886

29.19

SBS1,top =

1,759

SBS1,bot =

2,497

95.50 Y=

Short-Term Composite (N=8) t

Ad2

Slab

8.5000

109.00

115.81

74.88

8,671.46

697.29

-20.65

49,365

50,062

Haunch Top Flange

0.0000 1.1250

14.0000 14.0000

0.00 15.75

70.63 70.06

0.00 1,103.48

0.00 1.66

-16.40 -15.83

0 3,948

0 3,950

Web Bot Flange

0.6875 1.5000

68.0000 22.0000

46.75 33.00

35.50 0.75

1,659.63 18,014.33 24.75 6.19

18.73 53.48

16,399 94,381

34,414 94,387

11,459.32

ITotal =

182,813

54.23

SST1,top =

11,150

SST1,bot =

3,371

211.31 n=

8.00

Long-Term Composite (N=24) t

Ad2

Slab

8.5000

109.00

38.60

74.88

2,890.49

232.43

-32.53

40,856

41,089

Haunch

0.0000

14.00

0.00

70.63

0.00

-28.28

Top Flange

1.1250

14.0000

15.75

70.06

1,103.48

1.66

-27.72

12,102

12,104

Web Bot Flange

0.6875 1.5000

68.0000 22.0000

46.75 33.00

35.50 0.75

1,659.63 18,014.33 24.75 6.19

6.84 41.59

2,189 57,089

20,203 57,095

5,678.35

ITotal =

130,491

42.34

SLT1,top =

4,614

SLT1,bot =

3,082

134.10 n=

24.00

Single-Span Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Created July2007: Page 11 of 23 -- 73 --

3c. Section 2 Flexural Properties

Single Span Bridge Example - Section 2 Bare Steel t

Ad2

Top Flange

2.0000

14.00

28.00

71.00

1,988.00

9.33

-40.08

44,978

44,987

Web Bot Flange

0.5625 2.0000

68.00 22.00

38.25 44.00

36.00 1.00

1,377.00 14,739.00 44.00 14.67

-5.08 29.92

987 39,391

15,726 39,405

3,409.00

ITotal =

100,119

30.92

SBS1,top =

2,437

SBS1,bot =

3,238

110.25 Y=

Short-Term Composite (N=8) t

Ad2

Slab

8.5000

109.00

115.81

76.25

8,830.70

697.29

-22.11

56,600

57,297

Haunch Top Flange

0.0000 2.0000

14.0000 14.0000

0.00 28.00

72.00 71.00

0.00 1,988.00

0.00 9.33

-17.86 -16.86

0 7,956

0 7,966

Web Bot Flange

0.5625 2.0000

68.0000 22.0000

38.25 44.00

36.00 1.00

1,377.00 14,739.00 44.00 14.67

18.14 53.14

12,591 124,264

27,330 124,279

12,239.70

ITotal =

216,871

54.14

SST1,top =

12,145

SST1,bot =

4,006

226.06 n=

8.00

Long-Term Composite (N=24) t

Ad2

Slab

8.5000

109.00

38.60

76.25

2,943.57

232.43

-33.57

43,514

43,746

Haunch

0.0000

14.00

0.00

72.00

0.00

-29.32

Top Flange

2.0000

14.0000

28.00

71.00

1,988.00

9.33

-28.32

22,462

22,472

Web Bot Flange

0.5625 2.0000

68.0000 22.0000

38.25 44.00

36.00 1.00

1,377.00 14,739.00 44.00 14.67

6.68 41.68

1,705 76,425

16,444 76,439

6,352.57

ITotal =

159,101

42.68

SLT1,top =

5,426

SLT1,bot =

3,728

148.85 n=

24.00

Single-Span Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Created July2007: Page 12 of 23 -- 74 --

3d. Section 3 Flexural Properties

Single Span Bridge Example - Section 3 Bare Steel t

Ad2

Top Flange

2.0000

14.00

28.00

71.38

1,998.50

9.33

-42.25

49,970

49,980

Web Bot Flange

0.5625 2.3750

68.00 22.00

38.25 52.25

36.38 1.19

1,391.34 14,739.00 62.05 24.56

-7.25 27.94

2,008 40,796

16,747 40,820

3,451.89

ITotal =

107,546

29.13

SBS1,top =

2,487

SBS1,bot =

3,692

118.50 Y=

Short-Term Composite (N=8) t

Ad2

Slab

8.5000

109.00

115.81

76.63

8,874.13

697.29

-24.02

66,819

67,516

Haunch Top Flange

0.0000 2.0000

14.0000 14.0000

0.00 28.00

72.38 71.38

0.00 1,998.50

0.00 9.33

-19.77 -18.77

0 9,865

0 9,874

Web Bot Flange

0.5625 2.3750

68.0000 22.0000

38.25 52.25

36.38 1.19

1,391.34 14,739.00 62.05 24.56

16.23 51.42

10,076 138,137

24,815 138,161

12,326.02

ITotal =

240,366

52.61

SST1,top =

12,158

SST1,bot =

4,569

234.31 n=

8.00

Long-Term Composite (N=24) t

Ad2

Slab

8.5000

109.00

38.60

76.63

2,958.04

232.43

-35.82

49,544

49,777

Haunch

0.0000

14.00

0.00

72.38

0.00

-31.57

Top Flange

2.0000

14.0000

28.00

71.38

1,998.50

9.33

-30.57

26,174

26,184

Web Bot Flange

0.5625 2.3750

68.0000 22.0000

38.25 52.25

36.38 1.19

1,391.34 14,739.00 62.05 24.56

4.43 39.61

749 81,990

15,488 82,015

6,409.93

ITotal =

173,463

40.80

SLT1,top =

5,494

SLT1,bot =

4,251

157.10 n=

24.00

Single-Span Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Created July2007: Page 13 of 23 -- 75 --

4. DISTRIBUTION FACTOR FOR MOMENT 4a. Section 1: Interior Girder - One Lane Loaded:

DFM1,Int,Sec1

0.4 0.3 ⎛ S ⎞ ⎛ S ⎞ ⎛ Kg ⎞ = 0.06 + ⎜ ⎟ ⎜ ⎟ ⎜ 3 ⎟ ⎝ 14 ⎠ ⎝ L ⎠ ⎝ 12 LtS ⎠

0.1

K g = n ( I + Aeg2 )

(

K g = (8) 72,890 in 4 + ( 95.5 in 2 ) ( 49.06")

)

K g = 2, 422, 000 in 4 0.4

DFM1,Int,Sec1

⎛ 11.33' ⎞ ⎛ 11.33' ⎞ = 0.06 + ⎜ ⎟ ⎜ ⎟ ⎝ 14 ⎠ ⎝ 166.3' ⎠

0.3

⎛ 2, 422, 000 in 4 ⎞ ⎜ ⎟ ⎜ (12 )(166.3' )( 8.5")3 ⎟ ⎝ ⎠

0.1

DFM1,Int,Sec1 = 0.4994 Interior Girder - Two or More Lanes Loaded:

DFM2,Int,Sec1

0.4 0.3 ⎛ S ⎞ ⎛ S ⎞ ⎛ Kg ⎞ = 0.075 + ⎜ ⎟ ⎜ ⎟ ⎜ 3 ⎟ ⎝ 9.5 ⎠ ⎝ L ⎠ ⎝ 12 LtS ⎠

DFM2,Int,Sec1

⎛ 11.33' ⎞ ⎛ 11.33' ⎞ = 0.075 + ⎜ ⎟ ⎜ ⎟ ⎝ 9.5 ⎠ ⎝ 166.3' ⎠

0.6

0.2

0.1

⎛ 2, 422, 000 in 4 ⎞ ⎜ ⎟ ⎜ (12 )(166.3')( 8.5")3 ⎟ ⎝ ⎠

0.1

DFM2,Int,Sec1 = 0.7703

Single-Span Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Created July2007: Page 14 of 23 -- 76 --

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.

DFM 1,Ext,Sec1 =

8.5' = 0.7500 11.33'

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

de 9.1 2.167 ' e = 0.77 + = 1.008 9.1 e = 0.77 +

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

4b. Section 2: Interior Girder – One Lane Loaded: 0.4 0.3 ⎛ S ⎞ ⎛ S ⎞ ⎛ Kg ⎞ DFM1,Int,Sec2 = 0.06 + ⎜ ⎟ ⎜ ⎟ ⎜ 3 ⎟ ⎝ 14 ⎠ ⎝ L ⎠ ⎝ 12 LtS ⎠

0.1

K g = n ( I + Aeg2 )

(

K g = (8) 100,100 in 4 + (110.3 in 2 ) ( 47.83")

)

K g = 2,819, 000 in 4 0.4

DFM1,Int,Sec2

⎛ 11.33' ⎞ ⎛ 11.33' ⎞ = 0.06 + ⎜ ⎟ ⎜ ⎟ ⎝ 14 ⎠ ⎝ 166.3' ⎠

0.3

⎛ 2,819, 000 in 4 ⎞ ⎜ ⎟ ⎜ (12 )(166.3' ) ( 8.5 ")3 ⎟ ⎝ ⎠

0.1

DFM1,Int,Sec2 = 0.5061 Single-Span Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Created July2007: Page 15 of 23 -- 77 --

Interior Girder – Two or More Lanes Loaded: 0.4 0.3 ⎛ S ⎞ ⎛ S ⎞ ⎛ Kg ⎞ DFM2,Int,Sec2 = 0.075 + ⎜ ⎟ ⎜ ⎟ ⎜ 3 ⎟ ⎝ 9.5 ⎠ ⎝ L ⎠ ⎝ 12 LtS ⎠

DFM2,Int,Sec2 DFM2,Int,Sec2

0.1

0.6 0.2 4 ⎞ ⎛ 11.33' ⎞ ⎛ 11.33' ⎞ ⎛ 2,819, 000 in = 0.075 + ⎜ ⎜ ⎟ ⎟ ⎜ ⎟ ⎜ 3 ⎝ 9.5 ⎠ ⎝ 166.3' ⎠ ⎝ (12 )(166.3')( 8.5") ⎟⎠ = 0.7809

0.1

Exterior Girder - One Lane Loaded:

Same as for the positive moment section: DFM1,Ext,Sec2 = 0.9000 Exterior Girder - Two or More Lanes Loaded:

DFM2,Ext,Sec2 = e DFM2,Int,Sec2 e = 1.008 (same as before) DFM2,Ext,Sec2 =(1.008) (0.7809) = 0.7871

4c. Section 3: Interior Girder – One Lane Loaded:

DFM 1, Int ,Sec 3

(

⎛S ⎞ = 0.06 + ⎜ ⎟ ⎝ 14 ⎠

K g = n I + Aeg2

(

0.4

⎛S⎞ ⎜ ⎟ ⎝L⎠

0.3

) (

⎛ Kg ⎞ ⎜ 3 ⎟ ⎝ 12 LtS ⎠

)

0.1

K g = ( 8 ) 107,500 in 4 + 118.5 in 2 ( 50.00")

)

K g = 3, 230, 000 in 4 DFM 1, Int , Sec 3

⎛ 11.33' ⎞ = 0.06 + ⎜ ⎟ ⎝ 14 ⎠

0.4

⎛ 11.33' ⎞ ⎜ ⎟ ⎝ 166.3' ⎠

0.3 ⎛

⎞ ⎜ ⎟ 3 ⎜ (12 )(166.3')( 8.5") ⎟ ⎝ ⎠ 3, 230, 000 in 4

0.1

DFM 1, Int ,Sec 3 = 0.5122

Single-Span Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Created July2007: Page 16 of 23 -- 78 --

Interior Girder - Two or More Lanes Loaded: 0.4

DFM 2, Int ,Sec 3

⎛ S ⎞ = 0.075 + ⎜ ⎟ ⎝ 9.5 ⎠

⎛S⎞ ⎜ ⎟ ⎝L⎠

DFM 2, Int ,Sec 3

⎛ 11.33' ⎞ = 0.075 + ⎜ ⎟ ⎝ 9.5 ⎠

0.6

0.3

⎛ Kg ⎞ ⎜ 3 ⎟ ⎝ 12 LtS ⎠

⎛ 11.33' ⎞ ⎜ ⎟ ⎝ 166.3' ⎠

0.2

0.1

⎛ 3, 230, 000 in 4 ⎞ ⎜ ⎟ ⎜ (12 )(166.3')( 8.5")3 ⎟ ⎝ ⎠

0.1

DFM 2, Int ,Sec 3 = 0.7906

Exterior Girder – One Lane Loaded:

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

Exterior Girder - Two or More Lanes Loaded:

DFM2,Ext,Sec3 = e DFM2,Int,Sec3 e = 1.008 (same as before) DFM2,Ext,Sec3 =(1.008) (0.7906) = 0.7969

Ext , Min

4d. Minimum Exterior Girder Distribution Factor:

Single-Span Bridge Example ODOT LRFD Short Course - Steel

NL Nb

X Ext ∑ e Nb

∑x

AASHTO-LRFD 2007 Created July2007: Page 17 of 23 -- 79 --

One Lane Loaded:

DFM 1, Ext , Min =

( 28.33') ⎡⎣( 25.5') ⎤⎦ 1 + = 0.4881 6 2 ⎡( 28.33')2 + (17.0 ')2 + ( 5.667 ' )2 ⎤ ⎣ ⎦

Multiple Presence: DFM1,Ext,Min = (1.2) (0.4881) = 0.5857

Two Lanes Loaded:

DFM 2, Ext , Min =

( 28.33') ⎡⎣( 25.5') + (13.5') ⎤⎦ 2 + = 0.8250 6 2 ⎡( 28.33' )2 + (17.0 ') 2 + ( 5.667 ' ) 2 ⎤ ⎣ ⎦

Multiple Presence: DFM1,Ext,Min = (1.0) (0.8250) = 0.8250

Three Lanes Loaded:

DFM 3, Ext , Min =

3 ( 28.33') ⎡⎣( 25.5') + (13.5' ) + (1.5') ⎤⎦ + = 1.011 6 2 ⎡( 28.33' )2 + (17.0 ')2 + ( 5.667 ' ) 2 ⎤ ⎣ ⎦

Multiple Presence: DFM1,Ext,Min = (0.85) (1.011) = 0.8589 Four Lanes Loaded:

DFM 4, Ext , Min =

4 ( 28.33') ⎡⎣( 25.5' ) + (13.5') + (1.5') + ( −10.5') ⎤⎦ + = 1.045 2 2 2 6 2 ⎡( 28.33') + (17.0 ') + ( 5.667 ') ⎤ ⎣ ⎦

Multiple Presence: DFM1,Ext,Min = (0.65) (1.045) = 0.6791 Five Lanes Loaded:

DFM 5, Ext , Min =

5 ( 28.33' ) ⎡⎣( 25.5') + (13.5') + (1.5') + ( −10.5' ) + ( −22.5') ⎤⎦ + = 0.8367 2 2 2 6 2 ⎡( 28.33') + (17.0 ') + ( 5.667 ') ⎤ ⎣ ⎦

Multiple Presence: DFM1,Ext,Min = (0.65) (0.8367) = 0.5438

Single-Span Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Created July2007: Page 18 of 23 -- 80 --

4d. Moment Distribution Factor Summary Section

# Lanes Loaded 1 2 3 1 2 3 1 2 3

Positive Moment Interior Exterior 0.4994 0.9000 ≥ 0.5857 0.7703 0.7765 ≥ 0.8250 0.7703 0.7765 ≥ 0.8589 0.5061 0.9000 ≥ 0.5857 0.7809 0.7871 ≥ 0.8250 0.7809 0.7871 ≥ 0.8589 0.5122 0.9000 ≥ 0.5857 0.7906 0.7969 ≥ 0.8250 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 (ft) 0.0 8.0 16.1 24.1 32.1 40.2 49.2 58.2 66.5 74.8 83.1 83.1 91.5 99.8 108.1 108.1 117.1 126.1 126.1 134.2 142.2 150.2 158.3 166.3

LL+ (k-ft) 0.0 986.5 1868.0 2645.5 3319.1 3889.3 4404.3 4795.1 5054.9 5202.6 5238.6 5238.5 5202.6 5055.0 4795.7 4795.2 4404.5 3890.2 3889.5 3319.4 2645.7 1868.1 986.6 1.1

Nominal Moments LLDC1 (k-ft) (k-ft) 0.0 0.0 0.0 1086.2 0.0 2064.8 0.0 2935.8 0.0 3699.2 0.0 4355.1 0.0 4959.7 0.0 5424.7 0.0 5729.3 0.0 5912.1 0.0 5973.0 0.0 5973.0 0.0 5911.6 0.0 5728.3 0.0 5423.3 0.0 5423.3 0.0 4957.8 0.0 4352.6 0.0 4352.6 0.0 3696.9 0.0 2933.8 0.0 2063.3 0.0 1085.3 0.0 0.0

Single-Span Bridge Example ODOT LRFD Short Course - Steel

DC2 (k-ft) 0.0 135.6 257.4 365.5 459.8 540.3 614.2 670.8 707.8 730.0 737.4 737.4 729.9 707.7 670.6 670.6 614.0 540.0 540.0 459.5 365.2 257.2 135.5 0.0

DW (k-ft) 0.0 457.7 869.0 1233.7 1552.0 1823.9 2073.3 2264.3 2389.1 2464.0 2489.0 2489.0 2463.8 2388.7 2263.7 2263.7 2072.4 1822.9 1822.9 1551.1 1232.9 868.3 457.3 0.0

AASHTO-LRFD 2007 Created July2007: Page 19 of 23 -- 81 --

5. FACTORED MOMENT ENVELOPES

The following load combinations were considered in this example: Strength I: Strength IV:

1.75(LL + IM) + 1.25DC1 + 1.25DC2 + 1.50DW 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. Q = ∑ηiγ i Qi

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.

Single-Span Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Created July2007: Page 20 of 23 -- 82 --

Station (ft) 0.0 8.0 16.1 24.1 32.1 40.2 49.2 58.2 66.5 74.8 83.1 83.1 91.5 99.8 108.1 108.1 117.1 126.1 126.1 134.2 142.2 150.2 158.3 166.3

LL+ (k-ft) 0.0 1726.3 3268.9 4629.6 5808.5 6806.2 7707.5 8391.4 8846.1 9104.5 9167.5 9167.4 9104.6 8846.3 8392.4 8391.6 7707.8 6807.8 6806.6 5808.9 4630.0 3269.2 1726.5 1.9

LL(k-ft) 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0

Strength I Moments DC1 DC2 (k-ft) (k-ft) 0.0 0.0 1357.7 169.5 2581.0 321.8 3669.7 456.9 4624.0 574.7 5443.9 675.4 6199.7 767.8 6780.8 838.5 7161.6 884.7 7390.1 912.5 7466.3 921.7 7466.3 921.7 7389.5 912.4 7160.4 884.6 6779.1 838.3 6779.1 838.3 6197.3 767.5 5440.8 675.0 5440.8 675.0 4621.1 574.4 3667.2 456.5 2579.1 321.5 1356.7 169.4 0.0 0.0

DW (k-ft) 0.0 686.6 1303.4 1850.6 2328.1 2735.8 3109.9 3396.5 3583.7 3696.0 3733.5 3733.5 3695.7 3583.1 3395.6 3395.6 3108.7 2734.3 2734.3 2326.6 1849.3 1302.5 686.0 0.0

Total + (k-ft) 0.0 3940.1 7475.1 10606.8 13335.3 15661.3 17784.8 19407.2 20476.2 21103.1 21289.0 21288.9 21102.2 20474.4 19405.3 19404.5 17781.2 15657.9 15656.7 13331.0 10603.1 7472.2 3938.5 1.9

Station (ft) 0.0 7.2 14.5 21.7 28.9 36.2 43.4 50.6 57.8 65.1 72.3 79.5 86.8 94.0 101.2 108.5 115.7 122.9 130.1 137.4 144.6 151.8 159.1 166.3

DC1 (k-ft) 0.0 1629.3 3097.2 4403.7 5548.8 6532.6 7439.6 8137.0 8593.9 8868.1 8959.5 8959.5 8867.4 8592.5 8134.9 8134.9 7436.7 6529.0 6529.0 5545.4 4400.7 3094.9 1628.0 0.0

Strength IV Moments DC2 DW (k-ft) (k-ft) 0.0 0.0 203.4 686.6 386.1 1303.4 548.2 1850.6 689.7 2328.1 810.5 2735.8 921.3 3109.9 1006.2 3396.5 1061.7 3583.7 1095.0 3696.0 1106.1 3733.5 1106.1 3733.5 1094.9 3695.7 1061.5 3583.1 1005.9 3395.6 1005.9 3395.6 920.9 3108.7 810.0 2734.3 810.0 2734.3 689.3 2326.6 547.9 1849.3 385.9 1302.5 203.2 686.0 0.0 0.0

Total + (k-ft) 0.0 2519.2 4786.7 6802.5 8566.6 10078.9 11470.8 12539.7 13239.3 13659.1 13799.1 13799.1 13658.0 13237.1 12536.4 12536.4 11466.3 10073.3 10073.3 8561.2 6797.8 4783.2 2517.2 0.0

Total (k-ft) 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0

Single-Span Bridge Example ODOT LRFD Short Course - Steel

Total (k-ft) 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0

AASHTO-LRFD 2007 Created July2007: Page 21 of 23 -- 83 --

Station (ft) 0.0 7.2 14.5 21.7 28.9 36.2 43.4 50.6 57.8 65.1 72.3 79.5 86.8 94.0 101.2 108.5 115.7 122.9 130.1 137.4 144.6 151.8 159.1 166.3

LL+ (k-ft) 0.0 1282.4 2428.3 3439.2 4314.9 5056.0 5725.6 6233.6 6571.4 6763.4 6810.2 6810.1 6763.4 6571.5 6234.4 6233.8 5725.8 5057.2 5056.3 4315.2 3439.4 2428.5 1282.5 1.4

LL(k-ft) 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0

Service II Moments DC1 DC2 (k-ft) (k-ft) 0.0 0.0 1086.2 135.6 2064.8 257.4 2935.8 365.5 3699.2 459.8 4355.1 540.3 4959.7 614.2 5424.7 670.8 5729.3 707.8 5912.1 730.0 5973.0 737.4 5973.0 737.4 5911.6 729.9 5728.3 707.7 5423.3 670.6 5423.3 670.6 4957.8 614.0 4352.6 540.0 4352.6 540.0 3696.9 459.5 2933.8 365.2 2063.3 257.2 1085.3 135.5 0.0 0.0

Single-Span Bridge Example ODOT LRFD Short Course - Steel

DW (k-ft) 0.0 457.7 869.0 1233.7 1552.0 1823.9 2073.3 2264.3 2389.1 2464.0 2489.0 2489.0 2463.8 2388.7 2263.7 2263.7 2072.4 1822.9 1822.9 1551.1 1232.9 868.3 457.3 0.0

Total + (k-ft) 0.0 2961.9 5619.5 7974.2 10026.0 11775.3 13372.8 14593.4 15397.6 15869.4 16009.6 16009.5 15868.7 15396.3 14591.9 14591.3 13370.0 11772.7 11771.8 10022.6 7971.3 5617.3 2960.7 1.4

Total (k-ft) 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0

AASHTO-LRFD 2007 Created July2007: Page 22 of 23 -- 84 --

Strength Limit Moment Envelopes 25,000 Max (@ 83.14') = 21289k-ft Strength I

Moment (kip-ft)

20,000

15,000

Strength IV

10,000

5,000

0 0

120

150

180

Station (ft)

Service II Moment Envelope 17,500 Max (@ 83.14') = 16,010

k-ft

15,000

Moment (kip-ft)

12,500

10,000

7,500

5,000

2,500

0 0

120

150

180

Station (ft)

Single-Span Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Created July2007: Page 23 of 23 -- 85 --

-- 86 --

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 Single-Span Truss Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Created July 2007: Page 1 of 17 -- 87 --

Figure 2 - Truss Layout

6" Thick Reinforced Concrete Deck

6, W10 x 88 Stringers @ 3'-6" cc 18' - 0" Clear Roadway 19' - 6" cc Trusses

Figure 3 - Truss Cross Section

Single-Span Truss Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Created July 2007: Page 2 of 17 -- 88 --

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

Single-Span Truss Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Created July 2007: Page 3 of 17 -- 89 --

Single-Span Truss Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Created July 2007: Page 4 of 17 -- 90 --

Single-Span Truss Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Created July 2007: Page 5 of 17 -- 91 --

Consider Member 1-2 of the Truss:

Tandem:

⎡ kip kip ⎛ 96'− 4' ⎞ ⎤ kip kip ⎢ (25 ) + (25 ) ⎜ 96' ⎟ ⎥ ( −1.415 kip ) = −69.28 ⎝ ⎠ ⎣ ⎦

⎡ ⎛ 16'− 14' ⎞ kip kip ⎛ 96'− 14' ⎞ ⎤ kip kip Truck: ⎢ (8kip ) ⎜ ⎟ + (32 ) + (32 ) ⎜ ⎟ ⎥ ( −1.415 kip ) = −85.38 ⎝ 16' ⎠ ⎝ 96' ⎠ ⎦ ⎣ Lane:

( 0.640 kip ft ) ( 12 ) ( −1.415 kip kip ) (112') = −50.71kip

Combining the HL-93 Components with impact applied appropriately: (IM)(Tandem) + Lane:

(1.33) ( −69.28kip ) + ( −50.71kip ) = −142.9kip

(IM)(Truck) + Lane:

(1.33) ( −85.38kip ) + ( −50.71kip ) = −164.3kip

GOVERNS

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

Single-Span Truss Bridge Example ODOT LRFD Short Course - Steel

P1-2 = -123.2kip

AASHTO-LRFD 2007 Created July 2007: Page 6 of 17 -- 92 --

Consider Member 1-4 of the Truss:

Tandem:

⎡ kip kip ⎛ 96'− 4' ⎞ ⎤ kip kip ⎢ (25 ) + (25 ) ⎜ 96' ⎟ ⎥ (1.127 kip ) = 55.18 ⎝ ⎠⎦ ⎣

⎡ ⎛ 16'− 14' ⎞ kip kip ⎛ 96'− 14' ⎞ ⎤ kip kip Truck: ⎢ (8kip ) ⎜ ⎟ + (32 ) + (32 ) ⎜ ⎟ ⎥ (1.127 kip ) = 68.00 ⎝ 16' ⎠ ⎝ 96' ⎠ ⎦ ⎣ Lane:

( 0.640 kip ft ) ( 12 ) (1.127 kip kip ) (112') = 40.39kip

Combining the HL-93 Components with impact applied appropriately: (IM)(Tandem) + Lane:

(1.33) ( 55.18kip ) + ( 40.39kip ) = 113.8kip

(IM)(Truck) + Lane:

(1.33) ( 68.00kip ) + ( 40.39kip ) = 130.8kip

GOVERNS

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

Single-Span Truss Bridge Example ODOT LRFD Short Course - Steel

P1-4 = 98.12kip

AASHTO-LRFD 2007 Created July 2007: Page 7 of 17 -- 93 --

Consider Member 9-11 of the Truss:

Tandem:

⎡ kip kip ⎛ 64'− 4' ⎞ ⎤ kip kip ⎢(25 ) + (25 ) ⎜ 64' ⎟ ⎥ ( −2.254 kip ) = −109.2 ⎝ ⎠⎦ ⎣

⎡ ⎛ 48'− 14' ⎞ kip kip ⎛ 64'− 14' ⎞ ⎤ kip kip Truck: ⎢ (8kip ) ⎜ ⎟ + (32 ) + (32 ) ⎜ ⎟ ⎥ ( −2.254 kip ) = −141.3 48' 64' ⎝ ⎠ ⎝ ⎠⎦ ⎣ Lane:

( 0.640 kip ft ) ( 12 ) ( −2.254 kip kip ) (112') = −80.78kip

Combining the HL-93 Components with impact applied appropriately: (IM)(Tandem) + Lane:

(1.33) ( −109.2kip ) + ( −80.78kip ) = −226.0kip

(IM)(Truck) + Lane:

(1.33) ( −141.3kip ) + ( −80.78kip ) = −268.6kip

GOVERNS

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

Single-Span Truss Bridge Example ODOT LRFD Short Course - Steel

P9-11 = -201.5kip

AASHTO-LRFD 2007 Created July 2007: Page 8 of 17 -- 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

Single-Span Truss Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Created July 2007: Page 9 of 17 -- 95 --

Consider Member 10-13 of the Truss:

Tandem:

⎡⎣ (25kip ) + (25kip ) ⎤⎦ (1.972 kip kip ) = 98.60kip

⎡ ⎛ 48'− 12' ⎞ kip kip ⎤ kip kip Truck: ⎢ (8kip ) ⎜ ⎟ + (32 ) + (32 ) ⎥ (1.972 kip ) = 138.0 ⎝ 48' ⎠ ⎣ ⎦ Lane:

( 0.640 kip ft )(1.972 kip kip ) ⎡⎣( 12 ) (96') + (16') ⎤⎦ = 80.77kip

Combining the HL-93 Components with impact applied appropriately: (IM)(Tandem) + Lane:

(1.33) ( 98.60kip ) + (80.77kip ) = 211.9kip

(IM)(Truck) + Lane:

(1.33) (138.0kip ) + (80.77kip ) = 264.3kip

GOVERNS

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

Single-Span Truss Bridge Example ODOT LRFD Short Course - Steel

P10-13 = 198.2kip

AASHTO-LRFD 2007 Created July 2007: Page 10 of 17 -- 96 --

Consider Member 10-11 of the Truss – Tensile Force: 25kip

25kip

Tandem:

32kip

8kip

32kip

Truck:

0.640kip/ft

Lane: 0.5128kip/kip

IL Mem 10-11:

0.5124kip/kip

Tandem:

⎡ kip kip ⎛ 48'− 4' ⎞ ⎤ kip kip ⎢ (25 ) + (25 ) ⎜ 48' ⎟ ⎥ ( 0.5128 kip ) = 24.57 ⎝ ⎠⎦ ⎣

⎡ ⎛ 48'− 28' ⎞ kip ⎛ 48'− 14' ⎞ kip ⎤ kip kip Truck: ⎢ (8kip ) ⎜ ⎟ + (32 ) ⎜ ⎟ + (32 ) ⎥ ( 0.5128 kip ) = 29.74 48' 48' ⎝ ⎠ ⎝ ⎠ ⎣ ⎦ Lane:

( 0.640 kip ft ) ( 12 ) ( 0.5128 kip kip ) (56') = 9.189kip

Combining the HL-93 Components with impact applied appropriately: (IM)(Tandem) + Lane:

(1.33) ( 24.57kip ) + ( 9.189kip ) = 41.87kip

(IM)(Truck) + Lane:

(1.33) ( 29.74kip ) + ( 9.189kip ) = 48.74kip

GOVERNS

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

Single-Span Truss Bridge Example ODOT LRFD Short Course - Steel

P+10-11 = 36.56kip

AASHTO-LRFD 2007 Created July 2007: Page 11 of 17 -- 97 --

Consider Member 10-11 of the Truss – Compressive Force: 25kip

25kip

Tandem:

32kip

8kip

Truck:

0.640kip/ft

Lane: 0.5128kip/kip

IL Mem 10-11:

0.5124kip/kip

Tandem:

⎡ kip kip ⎛ 48'− 4' ⎞ ⎤ kip kip ⎢ (25 ) + (25 ) ⎜ 48' ⎟ ⎥ ( −0.5124 kip ) = −24.55 ⎝ ⎠⎦ ⎣

⎡ ⎛ 48'− 28' ⎞ kip ⎛ 48'− 14' ⎞ kip ⎤ kip kip Truck: ⎢ (8kip ) ⎜ ⎟ + (32 ) ⎜ ⎟ + (32 ) ⎥ ( −0.5124 kip ) = −29.72 48' 48' ⎝ ⎠ ⎝ ⎠ ⎣ ⎦ Lane:

( 0.640 kip ft ) ( 12 ) ( −0.5124 kip kip ) (56') = −9.182kip

Combining the HL-93 Components with impact applied appropriately: (IM)(Tandem) + Lane:

(1.33) ( −24.55kip ) + ( −9.182kip ) = −41.83kip

(IM)(Truck) + Lane:

(1.33) ( −29.72kip ) + ( −9.182kip ) = −48.71kip

GOVERNS

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

Single-Span Truss Bridge Example ODOT LRFD Short Course - Steel

P-10-11 = -36.53kip

AASHTO-LRFD 2007 Created July 2007: Page 12 of 17 -- 98 --

Member Force Summary: Member 1-2 1-4 9-11 9-10 10-13 10-11

Max Tension 0.000kip 98.12kip 0.000kip 0.000kip 198.2kip 36.56kip

Single-Span Truss Bridge Example ODOT LRFD Short Course - Steel

Max Compression 123.2kip 0.000kip 201.5kip 0.000kip 0.000kip 36.53kip

AASHTO-LRFD 2007 Created July 2007: Page 13 of 17 -- 99 --

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

⎛ S ⎞ ⎛ S ⎞ ⎛ Kg ⎞ = 0.06 + ⎜ ⎟ ⎜ ⎟ ⎜ 3 ⎟ ⎝ 14 ⎠ ⎝ L ⎠ ⎝ 12 Lt s ⎠ 0.4

DFM 1, Int

0.3

0.1

K g = n( I + Aeg2 ) K g = (8)(534 in 4 + (25.9 in 2 )(8.40") 2 ) K g =18,890 in 4 4 ⎛ 3.5' ⎞ ⎛ 3.5' ⎞ ⎛ 18,890 in ⎞ = 0.06 + ⎜ ⎟ ⎜ ⎟ ⎜ 3 ⎟ ⎝ 14 ⎠ ⎝ 16 ' ⎠ ⎝ 12(16 ')(6.0") ⎠ 0.4

DFM 1, Int

0.3

0.1

DFM 1, Int = 0.3965

Two or More Lanes Loaded:

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. R=

( P / 2 ) (1.75') = 0.2500 P (3.50 ')

DFM 1, Ext = 0.2500

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

The bridge is designed for a single traffic lane.

Minimum Exterior Girder Distribution Factor:

Single-Span Truss Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Created July 2007: Page 14 of 17 -- 100 --

Ext , Min

NL Nb

X Ext ∑ e Nb

∑x

One Lane Loaded: 4'-0" 2'-0"

3'-0"

P/2

1'-9" 5'-3" 8'-9"

M 1, Ext , Min

1 6

(4.00 ')(8.75') (2) ⎡⎣ (8.75') 2 + (5.25') 2 + (1.75') 2 ⎤⎦

= 0.3299

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: Exterior Stringer (Lever Rule): Exterior Stringer (Minimum):

DFM1,Int = 0.3965 DFM1, Ext = 0.2500 DFM1, Ext = 0.3299

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

Single-Span Truss Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Created July 2007: Page 15 of 17 -- 101 --

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

25kip

Tandem:

32kip

Truck:

0.640kip/ft

Lane: 4.00k-ft/kip

IL Moment @ CL Stringer 4 spaces @ 4'-0" = 16'-0"

Tandem:

⎡ kip kip ⎛ 8'− 4' ⎞ ⎤ k-ft k-ft ⎢ (25 ) + (25 ) ⎜ 8' ⎟ ⎥ ( 4.00 kip ) = 150.0 ⎝ ⎠⎦ ⎣

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

( 0.640 kip ft ) ( 12 ) ( 4.00 k-ft kip ) (16') = 20.48k-ft

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 P ⎛ a⎞ M Max = L − ⎜ ⎟ 2L ⎝ 2⎠ when a ≥ 0.5858L M Max =

PL 4

Single-Span Truss Bridge Example ODOT LRFD Short Course - Steel

AASHTO-LRFD 2007 Created July 2007: Page 16 of 17 -- 102 --

Tandem:

M Max

(25kip ) ⎛ 4' ⎞ k-ft = (an increase of 2.067%) ⎜ 16'− ⎟ = 153.1 (2)(16') ⎝ 2⎠

Truck: M Max =

(32kip )(16') = 128.0k-ft (4)

(no change)

Combining the HL-93 Components with impact applied appropriately: (IM)(Tandem) + Lane:

(1.33) (153.1k-ft ) + ( 20.48k-ft ) = 224.1k-ft

(IM)(Truck) + Lane:

(1.33) (128.0k-ft ) + ( 20.48k-ft ) = 190.7 k-ft

GOVERNS

Apply the Stringer Distribution Factor: Each stringer carries 0.3965 lanes of the HL-93 loading Æ

Single-Span Truss Bridge Example ODOT LRFD Short Course - Steel

MStringer = 88.86k-ft

AASHTO-LRFD 2007 Created July 2007: Page 17 of 17 -- 103 --

-- 104 --

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:

1.5"

Check Minimum Slenderness Ratio:

I min bt 3 t2 = = = 0.2887" 12bt 12 A

rmin =

(4 '− 0")(12 inft ) L = = 166.2 0.2887" rmin

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:

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.”

Pn = Fu Ae = Fu U An ⎡ ⎛ (1.5") 2 ⎞ ⎤ An = ⎢6"− (2) ( 7 8 "+ 1 8 ") + ⎜ ⎟ ⎥ (1") ⎝ 4(3.0") ⎠ ⎦ ⎣ = 4.188 in 2

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

NSF Governs - φPn = 218kip

φPn = (0.80)(272.2kip) = 217.8kip

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

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

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

Check Minimum Slenderness Ratio: rmin = 0.762” (from the AISC Manual) (8.5')(12 inft ) L = = 133.9 0.762" rmin

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

C12 x 30

Net Section Fracture: Pu

Pn = Fu Ae = Fu U An

An = ( 8.81 in 2 ) − (2) ( 7 8 "+ 18 ")( 0.510") = 7.790 in 2 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

ODOT-LRFD Short Course - Steel AASHTO Tension Member Example #2

AASHTO-LRFD 2007 Created July 2007: Page 1 of 2 -- 106 --

Side Note:

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

x 0.674" = 1− = 0.9251 9.00" L

…. for net section fracture, φPn = 334.3kip

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

ODOT-LRFD Short Course - Steel AASHTO Tension Member Example #2

AASHTO-LRFD 2007 Created July 2007: Page 2 of 2 -- 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.

Gusset Plates

W10 x 60

Section A-A

A 5 spaces @ 3”

Solution:

Check Minimum Slenderness Ratio: rmin = 2.57” (from the AISC Manual)

L rmin

L ≤ 140 2.57"

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

ODOT-LRFD Short Course - Steel AASHTO Tension Member Example #3

AASHTO-LRFD 2007 Created July 2007: Page 1 of 2 -- 108 --

Net Section Fracture: Pn = Fu Ae = Fu U An An = (17.6 in 2 ) − (4) ( 3 4 "+ 1 8 ")( 0.680") = 15.22 in 2 ?

Check b f ≥ 2 3 d

(10.1") ≥ ( 2 3 )(10.2")

…

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: x

Check

bf ≥ 2 3 d ?

(10.1") ≥ ( 2 3 )(10.2")

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

U = 1−

x 0.884" = 1− = 0.9263 L 12.0"

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

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

φPn = (0.80)(916.4kip) = 733.1kip Since Net Section Fracture governs the capacity of this member, the overall design strength of the member would be increased to 733kip. ODOT-LRFD Short Course - Steel AASHTO Tension Member Example #3

AASHTO-LRFD 2007 Created July 2007: Page 2 of 2 -- 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) L L = ≤ 240 rmin 0.864" 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: U = 1−

x 0.981" = 1− = 0.8774 L 8.0" φPn = (0.80)(241.7kip) = 193.4kip

Pn = (58ksi) (0.8774)(4.75 in2) = 241.7kip

Gross Section Yielding Governs - φPn =163kip

ODOT-LRFD Short Course - Steel AASHTO Tension Member Example #4

AASHTO-LRFD 2007 Created July 2007: Page 1 of 1 -- 110 --

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: Web:

bf 2t f

≤k

E Fy

h ? E ≤k tw Fy

29, 000 = 13.5 OK 50

29, 000 = 35.9 50

λf = 6.41 (Tabulated)

6.41 ≤ 0.56

λw = 25.4 (Tabulated)

25.4 ≤ 1.49

Compute Flexural Buckling Capacity: Slenderness Ratios:

(1.00)(20 ')(12 inft ) ⎛ KL ⎞ = = 39.74 < 120 ⎜ ⎟ 6.04" ⎝ r ⎠x

(1.00)(20 ')(12 inft ) ⎛ KL ⎞ = = 96.77 < 120 ⎜ ⎟ 2.48" ⎝ r ⎠y

Since the effective slenderness ratio is larger for the y axis than the x axis, y-axis buckling will govern. 2 2 ksi ⎞ ⎛ KL ⎞ Fy ⎛ 96.77 ⎞ ⎛ 50 λ=⎜ = = 1.636 ⎟ ⎜ ⎟ ⎜ ksi ⎟ ⎝ r π ⎠ y E ⎝ π ⎠ ⎝ 29, 000 ⎠

(6.9.4.1-3)

Since λ ≤ 2.25, Inelastic Buckling Governs

(

)

Pn = 0.66λ Fy As = 0.66(1.636) ( 50ksi )( 21.8 in 2 ) = 549.6kip

(6.9.4.1-1)

φPn = (0.90)(549.6kip) φPn = 495kip ODOT-LRFD Short Course - Steel AASHTO Compression Member Example #1

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

Ix = ∑

bh3 + Ad 2 12

2 ⎡ ( 4") ( 30"- ( 2 × 4") )3 ⎤ ⎡ ( 36")( 4")3 ⎛ 30" 4" ⎞ ⎤ ⎥ = 56,150 in 4 = 2⎢ + ( 36" × 4") ⎜ − ⎟ ⎥ + 2⎢ 12 2⎠ ⎥ 12 ⎢ ⎥ ⎝ 2 ⎢⎣ ⎦ ⎣ ⎦

hb3 + Ad 2 12 2 ⎡ ( 30"- ( 2 × 4") ) ( 4")3 ⎡ ( 4")( 36")3 ⎤ 36" 4" ⎞ ⎤ ⎛ = 2⎢ + ( 30"- ( 2 × 4") ) × 4" ⎜ − ⎟ ⎥ ⎥ + 2⎢ 12 12 2⎠ ⎥ ⎢⎣ ⎝ 2 ⎢⎣ ⎥⎦ ⎦ 4 = 76,390 in Iy = ∑

(

)

As = 2 ( 36" × 4") + 2 ( 30"- ( 2 × 4") ) × 4" = 464.0 in 2

Since I x = 56,154.67 in 4 < I y = 76,394.67 in 4 , x-axis buckling controls

rx =

Ix 56,150 in 4 = = 11.0 in As 464.0 in 2

ODOT-LRFD Short Course - Steel AASHTO Compression Member Example #2

AASHTO-LRFD 2007 Created July 2007: Page 1 of 2 -- 112 --

Check Local Buckling (Section 6.9.4.2): b 36" − 2 ( 4") = = 7.00 t 4" b? E ≤k t Fy

(6.9.4.2-1)

7.00 ≤1.40

29, 000ksi = 33.72 50ksi

Calculate the Nominal Compressive Strength (Section E3 page 16.1-33): Slenderness Ratios: KL r

where: K = 0.8 (Section 4.6.2.5) in Lx = Ly = 40 ft × 12 ft = 480"

KLx ( 0.8 )( 480 in ) = = 34.91 rx (11.0 in )

2 ⎛ KL ⎞ Fy ⎛ 34.91 ⎞ 50ksi λ=⎜ = = 0.2129 ⎟ ⎜ ⎟ ksi ⎝ rs π ⎠ E ⎝ π ⎠ 29, 000

(6.9.4.1-3)

Since λ ≤ 2.25 , Inelastic Flexural Buckling Governs

(

Pn = 0.66λ Fy As = 0.660.2129 50ksi

(

)( 464.0 in ) = 21, 240 2

kip

(6.9.4.1-1)

)

φc Pn = ( 0.90 ) 21, 240kip = 19,110kip

ODOT-LRFD Short Course - Steel AASHTO Compression Member Example #2

AASHTO-LRFD 2007 Created July 2007: Page 2 of 2 -- 113 --

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 sectionsâ&#x20AC;&#x2122; strong axes.

W16 x 36 L=24'

B W10 x 88 L=14'

8 @ 14'

W16 x 77 L=24'

4 @ 24'

Solution:

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

AASHTO-LRFD 2007 Created July 2007: Page 1 of 2 -- 114 --

Determine the Effective Length Factor:

⎛ (2)(534 in 4 ) ⎞ ⎛I⎞ ⎜ ⎟ ∑ ⎜⎝ L ⎟⎠ (14 ') ⎝ ⎠ = 3.065 C GA = = 4 ⎛I⎞ ⎛ in ) ⎞ ∑ ⎜⎝ L ⎟⎠ ⎛⎜ 23 ⎞⎟ ⎜ (2)(448 ⎟ (24 ') G ⎝ ⎠⎝ ⎠ ⎛ (2)(534 in 4 ) ⎞ ⎛I⎞ ⎜ ⎟ ∑ ⎜⎝ L ⎟⎠ (14 ') ⎝ ⎠ = 1.237 C GB = = 4 ⎛I⎞ ⎛ ⎛ 2 ⎞ (2)(1,110 in ) ⎞ ∑ ⎜⎝ L ⎟⎠ ⎜ 3 ⎟ ⎜ (24 ') ⎟ G ⎝ ⎠⎝ ⎠ For unbraced frames: K=

1.6G AG B + 4.0(G A + G B ) + 7.5 G A + G B + 7.5

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

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

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

AASHTO-LRFD 2007 Created July 2007: Page 2 of 2 -- 115 --

AASHTO Compression Member Example #4: Problem: y

10"

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 C10x15.3 as 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: 2.60" b bf = = = 5.96 t t f 0.436" a

Flange:

b? E 29, 000ksi ≤ 0.56 = 0.56 = 15.89 OK t Fy 36ksi Web:

b d − 2t f 10"− (2)(0.436") = = = 38.03 t tw 0.240" b? E 29, 000ksi ≤ 1.49 = 1.49 = 42.29 OK t Fy 36ksi

Compute Section Properties:

As = (2) (4.48 in2) = 8.96 in2 IX = (2) (Ix) = (2)(67.3in4) = 134.6 in4 2 ⎡ ⎞ ⎤ 4 2 ⎛ 9" IY = (2) ⎢ 2.27 in + (4.48 in ) ⎜ − 0.634" ⎟ ⎥ = 138.5 in 4 ⎝ 2 ⎠ ⎦⎥ ⎣⎢

rX =

IX 134.6 in 4 = = 3.88" 8.96 in 2 A

rY =

ODOT-LRFD Short Course - Steel AASHTO Compression Member Example #4

IY 138.5 in 4 = = 3.93" 8.96 in 2 A

AASHTO-LRFD 2007 Created July 2007: Page 1 of 3 -- 116 --

Slenderness Ratios: ⎛ (20 ')(12 inft ) ⎞ ⎛ KL ⎞ = ⎜ ⎟ ⎜ ⎟ = 61.86 ⎝ r ⎠ X ⎝ 3.88" ⎠

⎛ KL ⎞ ⎛ (20 ')(12 inft ) ⎞ ⎜ ⎟ =⎜ ⎟ = 61.07 ⎝ r ⎠Y ⎝ 3.93" ⎠

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: ⎛ 3 ⎞ ⎛ KL ⎞ a ≤ ri ⎜ ⎟ ⎜ ⎟ ⎝ 4 ⎠ ⎝ r ⎠ max ⎛ KL ⎞ ⎛ KL ⎞ ⎜ ⎟ =⎜ ⎟ ⎝ r ⎠ max ⎝ r ⎠ X

ri = ry = 0.711” (for one channel) a ≤ (0.711") ( 0.75 )( 61.86 ) a ≤ 32.98"

use 9 battens @ a = 30”

Modified Slenderness Ratio – Y-axis Buckling: The modified slenderness ratio is calculated as, 2

α2 ⎛ a ⎞ ⎛ KL ⎞ ⎛ KL ⎞ ⎜ ⎟ ⎜ ⎟ = ⎜ ⎟ + 0.82 (1 + α 2 ) ⎝ rib ⎠ ⎝ r ⎠m ⎝ r ⎠o rib = 0.711”

α=

(6.9.4.3.1-1)

h = 9” – (2)(0.634”) = 7.73”

h 7.732" = = 5.44 2rib (2)(0.711")

⎛ KL ⎞ ⎜ ⎟ = ⎝ r ⎠m

⎛ (5.44) 2 ⎞ ⎛ 30" ⎞ 2 ⎟⎜ = 71.70 ( 61.07 ) + 0.82 ⎜⎜ 2 ⎟ ⎝ 0.711" ⎟⎠ 1 (5.44) + ( ) ⎝ ⎠ 2

Now we can see that after the Y axis slenderness ratio is modified, Y axis buckling actually governs over X axis buckling. ODOT-LRFD Short Course - Steel AASHTO Compression Member Example #4

AASHTO-LRFD 2007 Created July 2007: Page 2 of 3 -- 117 --

Column Design Capacity: 2 2 ksi (6.9.4.1-3) ⎞ ⎛ KL ⎞ Fy ⎛ 71.70 ⎞ ⎛ 36 0.6466 λ=⎜ = = ⎟ ⎜ ⎟ ⎜ ksi ⎟ ⎝ r π ⎠ y E ⎝ π ⎠ ⎝ 29, 000 ⎠ Since λ ≤ 2.25, Inelastic Buckling Governs

(

)

Pn = 0.66λ Fy As = 0.658( 0.6466) ( 36ksi )( 8.96 in 2 ) = 246.1kip

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

2.21 kip

4.42kip kip = 2.21 channel 2 ΣM Î Mu,Batten = 33.15k-in 2Mu,Batten

I Batten

t (6")3 = = 18t 12

S Batten = 2.21 kip

18t = 6t 3

for first yield, My = Fy S Let φFy S ≥ Mu,Batten 33.15k-in t≥ (1.00)(36ksi )(6 in 3 )

t ≥ 0.153”

use t = 5/16”

(Min Thickness)

use PL6 x 8 x 5/16 Battens

ODOT-LRFD Short Course - Steel AASHTO Compression Member Example #4

AASHTO-LRFD 2007 Created July 2007: Page 3 of 3 -- 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" b bf = = = 4.14 t 2t f (2)(1.85") b? E 29, 000ksi ≤k = 0.56 = 13.5 OK t Fy 50ksi

Web:

b h = = 15.7 (Tabulated) t tw b? E 29, 000ksi ≤k = 0.75 = 18.1 t Fy 50ksi

Calculate the buckling load for Flexural Buckling about the X-Axis: 2 ksi in ⎛ KL ⎞ Fy ⎛ ( 24 ') (12 ft ) ⎞ ⎛ 50 λX = ⎜ = ⎜ ⎟ ⎜ ⎟ ksi ⎝ r π ⎠ X E ⎝ (4.48")(π) ⎠ ⎝ 29, 000 2

⎞ ⎟ = 0.7219 ⎠

Since λX < 2.25, Inelastic Buckling Governs

(

)

Pn = 0.66( 0.7219) ( 50ksi )( 42.9 in 2 ) = 1,586kip

ODOT-LRFD Short Course - Steel AASHTO Compression Member Example #5

(6.9.4.1-1)

AASHTO-LRFD 2007 Created July 2007: Page 1 of 2 -- 119 --

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

Fcr ,Y + Fcr , Z ⎛⎜ 4 Fcr ,Y Fcr , Z H 1− 1− 2 ⎜ 2H Fcr ,Y + Fcr , Z ) ( ⎝

⎞ ⎟ ⎟ ⎠

(AISC E4-2)

2 ksi in ⎞ ⎛ KL ⎞ Fy ⎛ ( 24 ') (12 ft ) ⎞ ⎛ 50 λY = ⎜ =⎜ = 1.131 ⎟ ⎜ ⎟ ksi ⎟ ⎝ r π ⎠Y E ⎝ (3.58")(π) ⎠ ⎝ 29, 000 ⎠ 2

Since λY < 2.25, Inelastic Buckling Governs

Fcr ,Y =

(

)

Pn = 0.66(1.131) ( 50ksi ) = 31.15ksi As

r o2 = xo2 + yo2 +

Ix + I y

…… yo = 3.62"−

(6.9.4.1-1) 1.85" = 2.695" 2

(AISC E4-7)

(861 in 4 + 549 in 4 ) r = (0.00) + ( 2.695 ) + = 40.13 in 2 2 42.9 in 2 o

GJ (11, 200ksi )(37.5 in 4 ) = = 244.0ksi 2 2 2 (42.9 in )(40.13 in ) Aro

(AISC E4-3)

xo2 + yo2 (0.000") 2 + (2.695") 2 H = 1− = 1− = 0.8190 40.13 in 2 r o2

(AISC E4-8)

Fcr , Z =

Fcrft

⎛ 31.15ksi + 244.0ksi =⎜ (2)(0.819) ⎝

⎛ ⎞ ⎞⎜ (4)(31.15ksi )(244.0ksi )(0.819) ⎟ = 30.37 ksi (AISC E4-2) ⎟ ⎜1 − 1 − ksi ksi 2 ⎟ ⎠ ( 31.15 + 244.1 ) ⎝ ⎠

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

ODOT-LRFD Short Course - Steel AASHTO Compression Member Example #5

AASHTO-LRFD 2007 Created July 2007: Page 2 of 2 -- 120 --

AASHTO Compression Member Example #6: Problem: y

Find the design strength of a C12x30 made of A36 steel. KLy = 7’ and KLx = KLz = 14’. x

Solution:

Check Local Buckling:

b? E 29, 000ksi ≤k = 0.56 = 15.89 OK 36ksi t Fy

Flange:

3.17" b bf = = = 6.327 t t f 0.501"

Web:

b h d − 2t f 12.0"− (2)(0.501") = = = = 21.56 0.510" t tw tw b? E 29, 000ksi ≤k = 1.49 = 42.29 36ksi t Fy

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’.

ODOT-LRFD Short Course - Steel AASHTO Compression Member Example #6

AASHTO-LRFD 2007 Created July 2007: Page 1 of 2 -- 121 --

Calculate the buckling load for Flexural Buckling about the X-Axis: 2 ksi in ⎞ ⎛ KL ⎞ Fy ⎛ ( 7 ' ) (12 ft ) ⎞ ⎛ 36 λx = ⎜ =⎜ = 1.528 ⎟ ⎜ ⎟ ksi ⎟ ⎝ r π ⎠ x E ⎝ (0.762")(π) ⎠ ⎝ 29, 000 ⎠ 2

Since λx < 2.25, Inelastic Buckling Governs

(

)

Pn = 0.66(1.528) ( 36ksi )( 8.81 in 2 ) = 167.3kip

(6.9.4.1-1)

Calculate the Critical Stress for Flexural-Torsional Buckling about the Y-axis: For Singly symmetric Sections: Fe =

Fey =

Fey + Fez ⎛⎜ 4 Fey Fez H 1− 1− 2H ⎜ (Fey + Fez )2 ⎝

(π2 )(29, 000ksi ) ⎛ (14 ')(12 ) ⎞ ⎜ 4.29" ⎟ ⎝ ⎠ in ft

⎞ ⎟ ⎟ ⎠

(AISC E4-5)

= 186.6ksi (AISC E4-10)

⎡ (π2 )(29, 000ksi )(151 in 6 ) ⎤ 1 ksi 4 (11, 200 )(0.861 in ) Fez = ⎢ + ⎥ 2 2 in (8.81 in )(4.54") 2 (AISC E4-11) ( (14 ')(12 ft ) ) ⎣⎢ ⎦⎥ Fez = 61.54ksi ⎛ (186.6ksi + 61.54ksi ) ⎞ ⎛ (4)(186.6ksi )(61.54ksi )(0.919") ⎞ Fe = ⎜ − − 1 1 ⎜ ⎟ ⎟⎜ ⎟ (2)(0.919") (186.6ksi + 61.54ksi ) 2 ⎝ ⎠⎝ ⎠

(AISC E4-5)

Fe = 59.30ksi

λ=

Fy Fe

36ksi = 0.6071 59.30ksi

Pn = ( 0.66(0.6071) )( 36ksi )( 8.81 in 2 ) = 246.0kip

(6.9.4.1-1)

since 167.3kip < 246.0kip, Flexural Buckling about the x Axis Governs φPn = (0.90)( 167.3kip) = 151kip ODOT-LRFD Short Course - Steel AASHTO Compression Member Example #6

AASHTO-LRFD 2007 Created July 2007: Page 2 of 2 -- 122 --

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.

/8"

Solution:

Check Local Buckling:

b 4.0" = = 8.0 1 " t 2

Fully Tensioned

b? E 29, 000ksi ≤k = 0.45 = 12.77 t Fy 36ksi Local Buckling is OK

Check the Connector Spacing: ⎛ (12 ') (12 inft ) ⎞ ⎛ KL ⎞ = ⎜ ⎟ = 119.0 ⎜ ⎟ ⎝ r ⎠ X ⎝ (1.21") ⎠ ry = 1.83” from AISC 2L Table 1-15, Pg 1-104.

in ⎛ KL ⎞ ⎛ (12 ') (12 ft ) ⎞ = ⎟ = 78.69 ⎜ ⎟ ⎜ ⎝ r ⎠Y ⎝ (1.83") ⎠

⎛ 3 ⎞ ⎛ KL ⎞ a ≤ ri ⎜ ⎟ ⎜ ⎟ ⎝ 4 ⎠ ⎝ r ⎠ max

⎛3⎞ a ≤ (0.776") ⎜ ⎟ (119.0 ) = 69.26" ⎝4⎠

Use 5 connectors….. a = 36”

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

AASHTO-LRFD 2007 Created July 2007: Page 1 of 3 -- 123 --

Check Flexural Buckling about the X-Axis: (Y axis is the axis of symmetry) 2 2 ksi ⎛ KL ⎞ Fy ⎛ 119.0 ⎞ ⎛ 36 λX = ⎜ = ⎟ ⎜ ⎟ ⎜ ksi ⎝ r π ⎠ X E ⎝ π ⎠ ⎝ 29, 000

(

⎞ ⎟ = 1.781 ⎠ Since λX < 2.25, Inelastic Buckling Governs

)

Pn = 0.66(1.781) ( 36ksi )( 7.49 in 2 ) = 127.9kip

(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:

Fcrft

Fcr ,Y + Fcr , Z ⎛⎜ 4 Fcr ,Y Fcr , Z H = 1− 1− 2 ⎜ 2H ( Fcr ,Y + Fcr ,Z ) ⎝

⎞ ⎟ ⎟ ⎠

(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:

α2 ⎛ a ⎞ ⎛ KL ⎞ ⎛ KL ⎞ 0.82 = + ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ (1 + α 2 ) ⎝ rib ⎠ ⎝ r ⎠m ⎝ r ⎠o

α=

h 2rib

(6.9.4.3.1-1)

h = ( 2 )(1.18") + ( 3 8 ") = 2.735"

rib = ry for a single angle = 1.21” α=

2.735" = 1.130 (2)(1.21")

⎛ KL ⎞ ⎜ ⎟ = ⎝ r ⎠m

(0.82)(1.130) 2 ⎛ 36" ⎞ ( 78.69 ) + ⎜ ⎟ = 81.24 (1 + (1.130) 2 ) ⎝ 1.21" ⎠ 2

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

AASHTO-LRFD 2007 Created July 2007: Page 2 of 3 -- 124 --

Compute the Y-axis Flexural Buckling Stress, Fcry: 2

2 ksi ⎞ ⎛ KL ⎞ Fy ⎛ (81.24) ⎞ ⎛ 36 λY = ⎜ = = 0.8301 ⎜ ⎟ ⎜ ⎟ ksi ⎟ ⎝ r π ⎠Y E ⎝ (π) ⎠ ⎝ 29, 000 ⎠ Since λY < 2.25, Inelastic Buckling Governs

Fcr ,Y =

(

)

Pn = 0.66( 0.8301) ( 36ksi ) = 25.43ksi As

(6.9.4.1-1)

Calculate Torsional Buckling Stress, Fcr,Z: ro = 2.38" (AISC Table 1-15, Pg 1-104) J = 0.322 in4 for a single angle

∴ r o2 = 5.664 in 2

(AISC Table 1-7, Pg 1-42)

∴ Jtotal = (2)(0.322 in4) = 0.644 in4 Fcr ,Z =

GJ (11, 200ksi )(0.644 in 4 ) = = 170.0ksi 2 2 2 Ar o (7.49 in )(5.664 in )

(AISC E4-3)

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

Fcrft

⎛ 25.43ksi + 170.0ksi =⎜ (2)(0.848) ⎝

⎛ ⎞ ⎞⎜ (4)(25.43ksi )(170.0ksi )(0.848) ⎟ = 24.79ksi ⎟ ⎜1 − 1 − ksi ksi 2 ⎟ ⎠⎜ ( 25.43 + 170.0 ) ⎟⎠ ⎝

Pn = AsFcrft = (7.49 in2) (24.79ksi) = 185.7kip Since 127.9kip < 185.7kip,

Flexural Buckling Governs

φPn = (0.90)(127.9kip ) = 115kip

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

AASHTO-LRFD 2007 Created July 2007: Page 3 of 3 -- 125 --

-- 126 --

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.

Pc = (16")(1") ( 50ksi ) = 800.0kip Pw = (22") ( 3 4 ") ( 36ksi ) = 594.0kip Pt = ( 8")( 2") ( 70ksi ) = 1,120kip Since Pc + Pw > Pt

(800.0

kip

+ 594.0kip = 1,394kip > 1,120kip ) , the

PNA must lie in the web. Define q as the fraction of the web that lies above the PNA. Pcompression = Ptension Pc + qPw = (1 − q ) Pw + Pt

(800.0 ) + q ( 594.0 ) = (1 − q ) ( 594.0 ) + (1,120 ) kip

kip

q = 0.7694

I.e., 76.94% of the web lies above the PNA (acts in compression assuming a positive moment). Y = 1"+ ( 0.7694 )( 22") = 17.93" from the top of steel

Find the moment arms from the resultant forces to the PNA. tc 1" = 17.93"− = 17.43" 2 2 1 1 = ( 2 ) qh = ( 2 )( 0.7694 )( 22") = 8.463"

dc = Y − d wc

d wt = ( 1 2 )(1 − q ) h = ( 1 2 )(1 − 0.7694 )( 22") = 2.537" dt = d − Y −

tt 2" = 25"− 17.93"− = 6.074" 2 2

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

AASHTO-LRFD 2007 Created July 2007: Page 1 of 2 -- 127 --

PL16 x 1, 50ksi

50ksi

36ksi

Pwc

PL22 x 3/4, 36ksi

dwc

36ksi

PNA

Pwt

dwt dt

PL8 x 2, 70ksi

70ksi

Compute the plastic moment by summing the moments about the PNA. M p = â&#x2C6;&#x2018; Pd i i = Pc d c + Pwc d wc + Pwt d wt + Pd t t

= ( 800kip ) (17.43") + ( 457 kip ) ( 8.463") + (137 kip ) ( 2.537") + (1,120kip ) ( 6.074") = 24,960k-in = 2, 080k-ft

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

AASHTO-LRFD 2007 Created July 2007: Page 2 of 2 -- 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. 100"

Solution:

Determine the Controlling Compression Force: Ps = 0.85 f c'bets = ( 0.85 ) ( 4ksi ) ( 8")(100") = 2720kip PSteel = Ast Fy = ( 29.1 in 2 )( 50ksi ) = 1455kip Assuming full composite action, the shear connectors must carry the smallest of Ps and Psteel.

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"

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. PConc = PSteel ac =

Ast Fy 0.85 f c'be

0.85 f c'be ac = Ast Fy

( 29.1 in )( 50 ) = 4.279" (measured from the top of slab) = ( 0.85) ( 4 ) (100") 2

ksi

ODOT-LRFD Short Course - Steel AASHTO Flexure Example #2

AASHTO-LRFD 2007 Created July 2007: Page 1 of 2 -- 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" 0.85f’c

Pconc

PNA

Psteel

M p = ( PConc ) (a1 ) = ( PSteel ) (a1 )

a1 =

d st a 29.7" 4.279" + ts − c = + 8"− = 20.71" 2 2 2 2

M p = (1455kip ) (20.71") = 30,130k-in = 2,511k-ft

ODOT-LRFD Short Course - Steel AASHTO Flexure Example #2

AASHTO-LRFD 2007 Created July 2007: Page 2 of 2 -- 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: Ps = 0.85 f c'bs ts = ( 0.85 ) ( 4ksi ) ( 6")( 50") = 1020kip PSteel = Ast Fy = ( 29.1 in 2 )( 50ksi ) = 1455kip 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…) Ps = 0.85 f c'bs ts = ( 0.85 ) ( 4ksi ) ( 6")( 50") = 1020kip Pc = (0.670")(10.5") ( 50ksi ) = 351.8kip

Pw = ⎡⎣ 29.1 in 2 − (2)(0.670")(10.5") ⎤⎦ ( 50ksi ) = 751.5kip Pt = Pc = 351.8kip 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%

Check Case I Pt + Pw ≥ Pc + Ps ?

351.8kip + 751.5kip ≥ 351.8kip + 1020kip ODOT-LRFD Short Course - Steel AASHTO Flexure Example #3

NO AASHTO-LRFD 2007 Created July 2007: Page 1 of 3

-- 131 --

Check Case II Pt + Pw + Pc ≥ Ps ?

351.8kip + 751.5kip + 351.8kip ≥ 1020kip

YES - PNA in Top Flange

50" 0.85f’c

Ps Pc1

6" Fy

PNA

Pc2

First, the location of the PNA within the top flange is determined. ⎞ ⎛ t ⎞ ⎛ P + P − Ps Y = ⎜ c ⎟⎜ w t + 1⎟ Pc ⎝ 2 ⎠⎝ ⎠ kip kip kip ⎞ ⎛ 0.670" ⎞ ⎛ 751.5 + 351.8 − 1, 020 =⎜ + 1⎟ = 0.2368" ⎟⎜ kip 351.8 ⎝ 2 ⎠⎝ ⎠

Next, the distances from the component forces to the PNA are calculated. 6" + 0.2368" = 3.237" 2 29.7" dw = − 0.2368" = 14.61" 2 0.670" dt = 29.7"− − 0.2368" = 29.13" 2 ds =

ODOT-LRFD Short Course - Steel AASHTO Flexure Example #3

AASHTO-LRFD 2007 Created July 2007: Page 2 of 3 -- 132 --

Finally, the plastic moment is computed. ⎛P ⎞ 2 M p = ⎜ c ⎟ ⎡Y 2 + ( tc − Y ) ⎤ + [ Ps d s + Pw d w + Pd t t] ⎦ ⎝ 2tc ⎠ ⎣ ⎛ 351.8kip ⎞ ⎡ 2 2 =⎜ ⎟ ⎣( 0.2368") + ( 0.670"− 0.2368") ⎤⎦ + ... ⎝ (2)(0.670") ⎠

(

... + ⎡⎣(1, 020kip ) ( 3.237") + ( 751.5kip ) (14.61") + ( 351.8kip ) ( 29.13") ⎤⎦

)

2 k-in ⎡ ⎤ ⎡ ⎤⎦ = 262.5 kip in ⎣ 0.2437 in ⎦ + ⎣ 24,530

= 24,590k-in = 2, 049k-ft

ODOT-LRFD Short Course - Steel AASHTO Flexure Example #3

AASHTO-LRFD 2007 Created July 2007: Page 3 of 3 -- 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”. 100"

Solution:

The concrete slab will be in tension, therefore none of the concrete is assumed to be effective.

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

Referring to Table D6.1-2 in Appendix D6.1, Page 6-291: ⎛ (π)(0.5") 2 ⎞ kip Prt = Fyrt Art = ( 60 ) ( 8 ) ⎜ ⎟ = 94.25 4 ⎝ ⎠ 2 ⎛ (π)(0.5") ⎞ kip Prb = Fyrb Arb = ( 60ksi ) ( 4 ) ⎜ ⎟ = 47.12 4 ⎝ ⎠ ksi

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"

Pt = (0.670")(10.5") ( 50ksi ) = 351.8kip

Pw = ⎡⎣ 29.1 in 2 − (2)(0.670")(10.5") ⎤⎦ ( 50ksi ) = 751.5kip Pc = Pt = 351.8kip ?

Check Case I: Pc + Pw ≥ Pt + Prb + Prt ?

351.8kip + 751.5kip ≥ 351.8kip + 47.12kip + 94.25kip ⎛ D ⎞ ⎡ P − P − Prt − Prb ⎤ Y = ⎜ ⎟⎢ c t + 1⎥ Pw ⎝ 2 ⎠⎣ ⎦ ⎛ 28.36" ⎞ ⎡ 351.8 Y =⎜ ⎟⎢ ⎝ 2 ⎠⎣

kip

YES - PNA is in Web

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

− 351.8kip − 94.25kip − 47.12kip ⎤ + 1⎥ 751.5kip ⎦

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

ODOT-LRFD Short Course - Steel AASHTO Flexure Example #4

AASHTO-LRFD 2007 Created July 2007: Page 1 of 3 -- 134 --

d rt = 11.51"+ 0.670"+ 8"− ⎡⎣1 7 8 "+ ( 1 2 )( 1 2 ") ⎤⎦ = 18.06" d rb = 11.51"+ 0.670"+ ⎡⎣1 7 8 "+ ( 1 2 )( 1 2 ") ⎤⎦ = 14.31" dt = 11.51"+ ( 1 2 )( 0.670") = 11.85" d wt = ( 1 2 )(11.51") = 5.755" d wc = ( 1 2 )( 28.36"− 11.51") = 8.425"

Not needed when using Table D6.1-2

d c = ( 28.36"− 11.51") + ( 1 2 )( 0.670") = 17.19"

⎛ P ⎞ M p = ⎜ w ⎟ ⎡⎣ y 2 + ( D − y ) 2 ⎤⎦ + [ Prt d rt + Prb d rb + Pd t t + Pc d c ] ⎝ 2D ⎠ 751.5kip ⎡⎣ (11.51") 2 + (28.36"− 11.51") 2 ⎤⎦ + [(94.25kip )(18.06") + ... = (2)(28.36") ... + (47.12kip )(14.31") + (351.8kip )(11.85") + (351.8kip )(17.19")] = 18,110k-in = 1,509k-ft

ODOT-LRFD Short Course - Steel AASHTO Flexure Example #4

AASHTO-LRFD 2007 Created July 2007: Page 2 of 3 -- 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 Pc + Pw ≥ Pt + Prb + Prt , 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. Pc + Pwc = Pwt + Pt + Prb + Prt 351.8kip + ( 50ksi ) ( 0.520") ( 28.36"− Y ) = ( 50ksi ) ( 0.520") (Y ) + 351.8kip + 47.12kip + 94.25kip

Y = 11.46 '' Pwt = (50ksi )(11.46") ( 0.520") = 298.0kip Pwc = (50ksi )(28.36"− 11.46") ( 0.520") = 439.4kip

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. d rt = 11.46"+ 0.670"+ 8"− ⎡⎣1 7 8 "+ ( 1 2 )( 1 2 ") ⎤⎦ = 17.88" d rb = 11.46"+ 0.670"+ ⎡⎣1 7 8 "+ ( 1 2 )( 1 2 ") ⎤⎦ = 14.38" dt = 11.46"+ ( 1 2 )( 0.670") = 11.80" d wt = ( 1 2 )(11.46") = 5.730" d wc = ( 1 2 )( 28.36"− 11.46") = 8.450" d c = ( 28.36"− 11.46") + ( 1 2 )( 0.670") = 17.24"

M p = (94.25kip )(17.88'') + (47.12kip )(14.38") + (351.8kip )(11.80 '') + ... ... + (298.0kip )(5.730") + (439.4kip ) ( 8.450") + ( 351.8kip ) (17.24")

= 18, 000k-in = 1,500k-ft The minor difference in between the two answers can be attributed to the fillet area. ODOT-LRFD Short Course - Steel AASHTO Flexure Example #4

AASHTO-LRFD 2007 Created July 2007: Page 3 of 3 -- 136 --

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 Dc ? E ≤ 5.7 tw Fyc

(6.10.6.2.3-1)

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

D 28.36" = = 14.18" 2 2

? (2)(14.18") 29, 000ksi = 54.54 ≤ 5.7 = 137.3 (0.520") 50ksi

Check

I yc I yt

OK, ∴ web is non-slender

≥ 0.3

(6.10.6.2.3-2)

Since Section is doubly symmetric, Iyc = Iyt ODOT-LRFD Short Course - Steel AASHTO Flexure Example #5a

OK AASHTO-LRFD 2007 Created July 2007: Page 1 of 4

-- 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

Fnc

Investigate Compression Flange Local Buckling: λf =

λ pf

b fc 2t fc

10.5" = 7.836 (2)(0.670")

(6.10.8.2.2-3)

E 29, 000ksi = 0.38 = 0.38 = 9.152 Fyc 50ksi

(6.10.8.2.2-4)

Since λf < λp, the flange is compact and,

Fnc ( FLB ) = Rb Rh Fyc

(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) Fnc ( FLB ) = (1.00)(1.00) ( 50ksi ) = 50ksi

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

rt =

b fc ⎛ 1 Dc tw ⎞ 12 ⎜1 + ⎟⎟ ⎜ ⎝ 3 b fc t fc ⎠

L p = 1.0rt

(10.5") ⎛ ⎛ 1 ⎞ (14.18")(0.520") ⎞ 12 ⎜ 1 + ⎜ ⎟ ⎟ ⎝ ⎝ 3 ⎠ (10.5")(0.670") ⎠

E 29, 000ksi = (1.0)(2.609") = 62.84" Fyc 50ksi

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

= 2.609"

(6.10.8.2.3-9)

(6.10.8.2.3-4)

AASHTO-LRFD 2007 Created July 2007: Page 2 of 4 -- 138 --

Fyr = (0.7) ( 50ksi ) = 35ksi

Fyr = min ( 0.7 Fyc , Fyw ) ≥ 0.5Fyc Lr = π rt

(Pg 6-110)

E 29, 000ksi = (π)(2.609") = 235.9" Fyr 35ksi

(6.10.8.2.3-5)

Since L p = 62.84" < Lb = 144" < Lr = 235.9" , Inelastic LTB must be investigated. ⎡ ⎛ F Fnc ( LTB ) = Cb ⎢1 − ⎜1 − yr ⎜ ⎢⎣ ⎝ Rh Fyc

⎞⎛ Lb − Lp ⎟⎜ ⎟⎜ Lr − Lp ⎠⎝

⎞⎤ ⎟⎟ ⎥ Rb Rh Fyc ≤ Rb Rh Fyc ⎠ ⎥⎦

(6.10.8.2.3-2)

⎡ ⎛ ⎞ ⎛ 144"− 62.84" ⎞ ⎤ 35ksi ksi ksi ⎟⎜ Fnc ( LTB ) = Cb ⎢1 − ⎜1 − ⎟ ⎥ (1.0 )(1.0 ) ( 50 ) ≤ (1.0 )(1.0 ) ( 50 ) ksi ⎜ ⎟ ⎢ ⎝ (1.0 ) ( 50 ) ⎠ ⎝ 235.9"− 62.84" ⎠ ⎥ ⎣ ⎦ Fnc ( LTB ) = ( Cb )( 0.8593) ( 50ksi ) = ( Cb ) ( 42.97 ksi ) ≤ 50ksi

Compute the Moment Gradient Factor, Cb, for segment BC of the beam, which will be critical. 2

⎛ f ⎞ ⎛ f ⎞ Cb = 1.75 − 1.05 ⎜ 1 ⎟ + 0.3 ⎜ 1 ⎟ ≤ 2.3 ⎝ f2 ⎠ ⎝ f2 ⎠

(6.10.8.2.3-7) C

M 2 = M c = 1, 080k-ft →

f 2 = 48.18ksi

M o = M B = 810.0k-ft →

f o = 36.13ksi

M BC ,mid = 1, 013k-ft

f mid = 45.19ksi

→

fmid

Since the BMD is not concave, f1 = 2 f mid − f 2 ≥ f o = (2) ( 45.19ksi ) − ( 48.18ksi ) = 42.20ksi ≥ 36.13ksi 2

⎛ 42.20 ⎞ ⎛ 42.20 ⎞ Cb = 1.75 − 1.05 ⎜ ⎟ + 0.3 ⎜ ⎟ = 1.061 ≤ 2.3 ⎝ 48.18 ⎠ ⎝ 48.18 ⎠ = 1.061 Fnc ( LTB ) = (1.061) ( 42.97 ksi ) = 45.57 ksi ≤ 50ksi = 45.57 ksi

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

AASHTO-LRFD 2007 Created July 2007: Page 3 of 4 -- 139 --

The governing strength for the compression flange is the smaller of Fnc(FLB) and Fnc(LTB): Since Fnc ( LTB ) = 45.57 ksi < Fnc ( FLB ) = 50.00ksi , LTB governs the strength of the compression flange. Fnc = Fnc ( LTB ) = 45.57 ksi

φFnc = (1.00) ( 45.57 ksi ) = 45.57 ksi Check fbu +

1 f l ≤ φ f Fnc 3

(6.10.8.1.1-1)

fbu = f C = 48.18ksi Assume that Strength I Load Combination Governs, ∴ γWS = 0.0 and fl = 0 Since fbu +

1 f l = 48.18ksi > φ f Fnc =45.57 ksi , the compression flange is not adequate. 3

Investigate the Strength of the Tension Flange: Since the tension flange is discretely braced, Fnt = Rh Fyt = (1.0) ( 50ksi ) = 50ksi

(6.10.8.3-1)

φFnt = (1.00) ( 50ksi ) = 50ksi

Check fbu +

1 f l ≤ φ f Fnt 3

(6.10.8.1.2-1)

fbu = f C = 48.18ksi Since fbu +

1 f l = 48.18ksi < φ f Fnt =50.00ksi , the tension flange is adequate. 3

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

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

AASHTO-LRFD 2007 Created July 2007: Page 4 of 4 -- 140 --

AASHTO Flexural Example #5b: Problem:

Determine Classification of the Section: Check

2 Dc ? E ≤ 5.7 tw Fyc

(6.10.6.2.3-1)

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

D 28.36" = = 14.18" 2 2

? (2)(14.18") 29, 000ksi = 54.54 ≤ 5.7 = 137.3 (0.520") 50ksi

Check

I yc I yt

OK, ∴ web is non-slender

≥ 0.3

(6.10.6.2.3-2)

Since Section is doubly symmetric, Iyc = Iyt ODOT-LRFD Short Course - Steel AASHTO Flexure Example #5b

OK AASHTO-LRFD 2007 Created July 2007: Page 1 of 6

-- 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) Mnc(LTB) • Lateral-Torsional Buckling Mnt • Yielding of Tension Flange

Mnc

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

2 Dcp tw

≤ λ pw( Dcp )

λ pw( Dcp ) =

(A6.2.1-1)

E Fyc

⎛ Dcp ⎞ ≤ λ rw ⎜ ⎟ ⎛ 0.54M p ⎞ ⎝ Dc ⎠ − 0.09 ⎟ ⎜⎜ ⎟ R M h y ⎝ ⎠

(A6.2.1-2)

λ rw = 5.7

E = 137.3 Fyc

(A6.2.1-3)

Rh = 1.00 (since the section is rolled and is ∴ non-hybrid) M y = S x Fy = ( 269 in 3 )( 50ksi ) = 13, 450k-in = 1,121k-ft

M p = Z x Fy = ( 312 in 3 )( 50ksi ) = 15, 600k-in = 1,300k-ft

λ pw( Dcp ) =

29, 000ksi 50ksi

⎛ 14.18" ⎞ = 83.76 ≤ 137.3 ⎜ ⎟ = 137.3 ⎝ 14.18" ⎠ ⎛ ( 0.54 ) (15, 600k-in ) ⎞ ⎜ − 0.09 ⎟ ⎜ (1.0 ) (13, 450k-in ) ⎟ ⎝ ⎠ 2

λ pw( Dcp ) = 83.76

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

AASHTO-LRFD 2007 Created July 2007: Page 2 of 6 -- 142 --

2 Dcp tw

? (2)(14.18") = 54.54 ≤ λ pw( Dcp ) = 83.76 0.520"

OK, ∴ web is compact

Since the web is compact, R pc =

R pt =

Mp M yc Mp M yt

15, 600k-in = 1.160 13, 450k-in

(A6.2.1-4)

15, 600k-in = 1.160 13, 450k-in

(A6.2.1-5)

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

b fc

2t fc

λ pf = 0.38

10.5" = 7.836 (2)(0.670")

(A6.3.2-3)

E 29, 000ksi = 0.38 = 9.152 50ksi Fyc

(A6.3.2-4)

Since λf < λpf, the flange is compact and, M nc ( FLB ) = R pc M yc = (1.160 ) (13, 450k-in ) = 15, 600k-in

(A6.3.2-1)

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

b fc

⎛ 1 Dc tw ⎞ 12 ⎜1 + ⎜ 3 b t ⎟⎟ fc fc ⎠ ⎝

L p = 1.0rt

(10.5") ⎛ ⎛ 1 ⎞ (14.18")(0.520") ⎞ 12 ⎜ 1 + ⎜ ⎟ ⎟ ⎝ ⎝ 3 ⎠ (10.5")(0.670") ⎠

E 29, 000ksi = (1.0)(2.609") = 62.84" 50ksi Fyc

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

= 2.609"

(A6.3.3-10)

(A6.3.3-4)

AASHTO-LRFD 2007 Created July 2007: Page 3 of 6 -- 143 --

E Lr = 1.95 rt Fyr

⎛ Fyr S xc h ⎞ 1 + 1 + 6.76 ⎜ ⎟ S xc h ⎝ E J ⎠ J

(6.10.8.2.3-5)

⎛ ⎞ S Fyr = min ⎜ 0.7 Fyc , Rh Fyt xt , Fyw ⎟ ≥ 0.5Fyc Fyr = (0.7) ( 50ksi ) = 35ksi S xc ⎝ ⎠

(Pg 6-222)

Sxc = 269 in3 h = d − ( 1 2 ) ( t fc + t ft ) = 29.7"− 0.670" = 29.03" J=

3 D tw3 b fc t fc + 3 3

⎛ t fc ⎜⎜1 − 0.63 b fc ⎝

⎞ b ft t 3ft ⎟⎟ + 3 ⎠

⎛ t ft ⎜⎜ 1 − 0.63 b ft ⎝

⎞ ⎟⎟ ⎠

(A6.3.3-9)

⎡ (10.5")(0.670")3 ⎛ (28.36")(0.520")3 ⎛ 0.670" ⎞ ⎞ ⎤ 4 J= + (2) ⎢ 1 − ( 0.63) ⎜ ⎟ ⎟ ⎥ = 3.350 in ⎜ 3 3 ⎝ 10.5" ⎠ ⎠ ⎦ ⎝ ⎣

J = 3.350 in4 Æ

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

E 29, 000ksi = = 828.6 Fyr 35ksi 3 S xc h ( 269 in ) ( 29.03") = = 2, 071 3.77 in 4 J

Lr = (1.95 )( 2.609")( 828.6 )

1 ⎛ 2, 071 ⎞ 1 + 1 + 6.76 ⎜ ⎟ = 254.9" = 21.24 ' 2, 071 ⎝ 828.6 ⎠

This value of Lr = 21.24’ agrees well with the value published in AISC on Page 3-15 Since L p = 62.84" < Lb = 144" < Lr = 254.9" , Inelastic LTB must be investigated. ⎡ ⎛ F S M nc ( LTB ) = Cb ⎢1 − ⎜1 − yr xc ⎢⎣ ⎜⎝ R pc Fyc

⎞⎛ Lb − Lp ⎟⎜ ⎟⎜ Lr − Lp ⎠⎝

⎞⎤ ⎟⎟ ⎥ R pc Fyc ≤ R pc Fyc ⎠ ⎥⎦

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

(A6.3.3-2)

AASHTO-LRFD 2007 Created July 2007: Page 4 of 6 -- 144 --

⎡ ⎛ ( 35ksi )( 269 in3 ) ⎞⎟ ⎛ 144"− 62.84" ⎞⎤⎥ (1.160 ) 13, 450k-in ≤ (1.160 ) 13, 450k-in M nc ( LTB ) = Cb ⎢1 − ⎜1 − ( ) ( ) ⎜ ⎟ ⎢ ⎜⎝ (1.16 ) (13, 450k-in ) ⎟⎠ ⎝ 254.9"− 62.84" ⎠ ⎥ ⎣ ⎦ = ( Cb )( 0.9656 ) (15, 600k-in ) = ( Cb ) (12,990k-in ) ≤ 15, 600k-in

Compute the Moment Gradient Factor, Cb, for segment BC of the beam, which will be critical. 2

⎛M ⎞ ⎛M ⎞ Cb = 1.75 − 1.05 ⎜ 1 ⎟ + 0.3 ⎜ 1 ⎟ ≤ 2.3 ⎝ M2 ⎠ ⎝ M2 ⎠

(A6.3.3-7)

M 2 = M c = 1, 080k-ft M o = M B = 810.0k-ft M BC ,mid = 1, 013k-ft Since the BMD is not concave, M 1 = 2M mid − M 2 ≥ M o = (2) (1, 013k-ft ) − (1, 080k-ft ) = 946k-ft ≥ 810k-ft 2

⎛ 946 ⎞ ⎛ 946 ⎞ Cb = 1.75 − 1.05 ⎜ + 0.3 ⎜ ⎟ ⎟ = 1.061 ≤ 2.3 ⎝ 1, 080 ⎠ ⎝ 1, 080 ⎠ = 1.061

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

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

AASHTO-LRFD 2007 Created July 2007: Page 5 of 6 -- 145 --

The governing strength for the compression flange is the smaller of Mnc(FLB) and Mnc(LTB): Since M nc ( LTB ) = 13, 780k-in < M nc ( FLB ) = 15, 600k-in , LTB governs the strength of the compression flange. M nc = M nc ( LTB ) = 13, 780k-in = 1,148k-ft

φM nc = (1.00) (1,148k-ft ) = 1,148k-ft Check M u +

1 f l S xc ≤ φ f M nc 3

(A6.1.1-1)

M u = M C = 1, 080k-ft Assume that Strength I Load Combination Governs, ∴ γWS = 0.0 and fl = 0 Since M u +

1 fl S xc = 1, 080k-ft < φ f M nc =1,148k-ft , the compression flange is adequate. 3

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

(A6.4-1)

φM nt = (1.00) (1,300k-ft ) = 1,300k-ft Check M u +

1 fl S xt ≤ φ f M nt 3

(6.10.8.1.2-1)

M u = M C = 1, 080k-ft Since M u +

1 f l S xt = 1, 080k-ft < φ f M nt =1,300k-ft , the tension flange is adequate. 3

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.

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

AASHTO-LRFD 2007 Created July 2007: Page 6 of 6 -- 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. PL16 x 3/4

Solution:

PL38 x 3/8

Ix = 10,730 in4 Iy = 513.2 in4 Sx = 543.1 in3 Sy = 64.15 in3

PL16 x 3/4

Determine Classification of the Section: Check

2 Dc ? E ≤ 5.7 tw Fyc

(6.10.6.2.3-1)

Take D = 38” Dc =

D 38" = = 19" 2 2

? (2)(19") 29, 000ksi = 101.3 ≤ 5.7 = 137.3 ( 38 ") 50ksi

Check

I yc

OK, ∴ web is non-slender

≥ 0.3

I yt Since Section is doubly symmetric, Iyc = Iyt

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

(6.10.6.2.3-2) OK AASHTO-LRFD 2007 Created July 2007: Page 1 of 5

-- 147 --

Fnc

Investigate Compression Flange Local Buckling: λf =

λ pf

b fc 2t fc

16" = 10.67 (2)( 3 4 ")

(6.10.8.2.2-3)

E 29, 000ksi = 0.38 = 0.38 = 9.152 Fyc 50ksi

(6.10.8.2.2-4)

Since λf < λp, the flange is non compact and, ⎡ ⎛ F ⎞⎛ λ − λ pf Fnc ( FLB ) = ⎢1 − ⎜ 1 − yr ⎟⎜ f ⎜ ⎟⎜ ⎣⎢ ⎝ Rh Fyc ⎠⎝ λ rf − λ pf

λrf = 0.56

⎞⎤ ⎟⎟ ⎥ Rb Rh Fyc ⎠ ⎦⎥

E Fyr

(6.10.8.2.2-2) (6.10.8.2.2-5)

Fyr = min ( 0.7 Fyc , Fyw ) ≥ 0.5Fyc

(Pg 6-109)

Fyr = (0.7) ( 50ksi ) = 35ksi

λrf = 0.56

29, 000ksi = 16.12 35ksi

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

⎡ ⎛ ⎞ ⎛ 10.67 − 9.152 ⎞ ⎤ 35ksi ksi ksi = ⎢1 − ⎜1 − ⎟ ⎥ (1.00)(1.00) ( 50 ) = 46.74 ksi ⎟ ⎜ ⎣ ⎝ (1.00)(50 ) ⎠ ⎝ 16.12 − 9.152 ⎠ ⎦

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

AASHTO-LRFD 2007 Created July 2007: Page 2 of 5 -- 148 --

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

b fc ⎛ 1 Dc tw 12 ⎜1 + ⎜ 3b t fc fc ⎝

L p = 1.0rt

⎞ ⎟⎟ ⎠

(16") ⎛ ⎛ 1 ⎞ (19")( 38 ") ⎞ 12 ⎜1 + ⎜ ⎟ ⎟ ⎝ ⎝ 3 ⎠ (16")( 3 4 ") ⎠

E 29, 000ksi = (1.0)(4.220") = 101.6" Fyc 50ksi

Fyr = min ( 0.7 Fyc , Fyw ) ≥ 0.5Fyc Lr = π rt

= 4.220"

(6.10.8.2.3-9)

(6.10.8.2.3-4)

Fyr = (0.7) ( 50ksi ) = 35ksi

E 29, 000ksi = (π)(4.220") = 381.6" Fyr 35ksi

(Pg 6-110)

(6.10.8.2.3-5)

Since L p = 101.6" < Lb = 144" < Lr = 381.6" , Inelastic LTB must be investigated. ⎡ ⎛ F Fnc ( LTB ) = Cb ⎢1 − ⎜1 − yr ⎜ ⎢⎣ ⎝ Rh Fyc

⎞⎛ Lb − Lp ⎟⎜ ⎟⎜ Lr − Lp ⎠⎝

⎞⎤ ⎟⎟ ⎥ Rb Rh Fyc ≤ Rb Rh Fyc ⎠ ⎥⎦

(6.10.8.2.3-2)

⎡ ⎛ ⎞ ⎛ 144"− 101.6" ⎞ ⎤ 35ksi ksi ksi ⎟⎜ Fnc ( LTB ) = Cb ⎢1 − ⎜1 − ⎟ ⎥ (1.0 )(1.0 ) ( 50 ) ≤ (1.0 )(1.0 ) ( 50 ) ksi ⎜ ⎟ 1.00 ) ( 50 ) ⎠ ⎝ 381.6"− 101.6" ⎠ ⎥ ⎢ ⎣ ⎝ ( ⎦ Fnc ( LTB ) = ( Cb )( 0.9546 ) ( 50ksi ) = ( Cb ) ( 47.73ksi ) ≤ 50ksi

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

AASHTO-LRFD 2007 Created July 2007: Page 3 of 5 -- 149 --

Compute the Moment Gradient Factor, Cb, for segment BC of the beam, which will be critical. 2

⎛ f ⎞ ⎛ f ⎞ Cb = 1.75 − 1.05 ⎜ 1 ⎟ + 0.3 ⎜ 1 ⎟ ≤ 2.3 ⎝ f2 ⎠ ⎝ f2 ⎠

(6.10.8.2.3-7) C

M 2 = M c = 1, 080k-ft →

f 2 = 23.86ksi

M o = M B = 810.0k-ft →

f o = 17.90ksi

M BC ,mid = 1, 013k-ft

f mid = 22.38ksi

→

fmid

Since the BMD is not concave, f1 = 2 f mid − f 2 ≥ f o = (2) ( 22.38ksi ) − ( 23.86ksi ) = 20.90ksi ≥ 17.90ksi 2

⎛ 20.90 ⎞ ⎛ 20.90 ⎞ Cb = 1.75 − 1.05 ⎜ ⎟ + 0.3 ⎜ ⎟ = 1.061 ≤ 2.3 ⎝ 23.86 ⎠ ⎝ 23.86 ⎠

→ Cb = 1.061

Fnc ( LTB ) = (1.061) ( 47.73ksi ) = 50.64ksi ≤ 50ksi

→ Fnc ( LTB ) = 50ksi

The governing strength for the compression flange is the smaller of Fnc(FLB) and Fnc(LTB): Since Fnc ( LTB ) = 50ksi > Fnc ( FLB ) = 46.70ksi , FLB governs the strength of the compression flange. Fnc = Fnc ( FLB ) = 46.70ksi

φFnc = (1.00) ( 46.70ksi ) = 46.70ksi Check fbu +

1 f l ≤ φ f Fnc 3

(6.10.8.1.1-1)

fbu = fC = 23.86ksi Assume that Strength I Load Combination Governs, ∴ γWS = 0.0 and fl = 0 Since fbu +

1 f l = 23.86ksi < φ f Fnc =46.70ksi , the compression flange is adequate. 3

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

AASHTO-LRFD 2007 Created July 2007: Page 4 of 5 -- 150 --

Investigate the Strength of the Tension Flange: Since the tension flange is discretely braced, Fnt = Rh Fyt = (1.0) ( 50ksi ) = 50ksi

(6.10.8.3-1)

φFnt = (1.00) ( 50ksi ) = 50ksi

Check fbu +

1 f l ≤ φ f Fnt 3

(6.10.8.1.2-1)

fbu = fC = 23.86ksi Since fbu +

1 f l = 23.86ksi < φ f Fnt =50.00ksi , the tension flange is adequate. 3

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

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

AASHTO-LRFD 2007 Created July 2007: Page 5 of 5 -- 151 --

AASHTO Flexural Example #6b: Problem:

Determine Classification of the Section: Check

2 Dc ? E ≤ 5.7 tw Fyc

(6.10.6.2.3-1)

Take D = 38” Dc =

D 38" = = 19" 2 2

? (2)(19") 29, 000ksi = 101.3 ≤ 5.7 = 137.3 ( 38 ") 50ksi

Check

I yc I yt

OK, ∴ web is non-slender

≥ 0.3

(6.10.6.2.3-2)

Since Section is doubly symmetric, Iyc = Iyt ODOT-LRFD Short Course - Steel AASHTO Flexure Example #6b

OK AASHTO-LRFD 2007 Created July 2007: Page 1 of 7

-- 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

Mnc

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

2 Dcp tw

≤ λ pw( Dcp )

λ pw( Dcp ) =

(A6.2.1-1)

E Fyc

⎛ Dcp ⎞ ≤ λ rw ⎜ ⎟ ⎛ 0.54M p ⎞ ⎝ Dc ⎠ − 0.09 ⎟ ⎜⎜ ⎟ R M h y ⎝ ⎠

(A6.2.1-2)

λ rw = 5.7

E = 137.3 Fyc

(A6.2.1-3)

Rh = 1.00 (since the section is rolled and is ∴ non-hybrid) M y = S x Fy = ( 543.1 in 3 )( 50ksi ) = 27,160k-in = 2, 263k-ft M p = Z x Fy = ( 600.4 in 3 )( 50ksi ) = 30, 020k-in = 2,502k-ft

λ pw( Dcp ) =

29, 000ksi 50ksi

⎛ 19" ⎞ = 93.68 ≤ 137.3 ⎜ ⎟ = 137.3 ⎝ 19" ⎠ ⎛ ( 0.54 ) ( 30, 020k-in ) ⎞ ⎜ − 0.09 ⎟ ⎜ (1.0 ) ( 27,160k-in ) ⎟ ⎝ ⎠ 2

λ pw( Dcp ) = 93.68

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

AASHTO-LRFD 2007 Created July 2007: Page 2 of 7 -- 153 --

2 Dcp tw

? (2)(19") = 101.3 > λ pw( Dcp ) = 93.68 3 " 8

∴ web is non compact

Since the web is non compact, ⎡ ⎛ Rh M yc R pc = ⎢1 − ⎜ 1 − Mp ⎢⎣ ⎜⎝

⎞ ⎛ λ w − λ pw( Dc ) ⎞ ⎤ M p M p ≤ ⎟⎥ ⎟⎟ ⎜⎜ ⎟ ⎥ M yc M yc λ − λ rw pw D ( ) ⎠⎝ c ⎠⎦

(A6.2.2-4)

Where, ⎛D ⎞ λ pw( Dc ) = λ pw( Dcp ) ⎜ cp ⎟ ≤ λ rw ⎝ Dc ⎠ ⎛ 19 '' ⎞ = 93.68 ⎜ ⎟ ≤ 137.3 ⎝ 19 '' ⎠ = 93.68 ≤ 137.3

(A6.2.2-6)

⎡ ⎛ (1.00)(27,160k-in ) ⎞ ⎛ 101.3 − 93.68 ⎞ ⎤ (30, 020k-in ) (30, 020k-in ) R pc = ⎢1 − ⎜ 1 − ≤ ⎟⎜ ⎟⎥ (30, 020k-in ) ⎠ ⎝ 137.3 − 93.68 ⎠ ⎦ (27,160k-in ) (27,160k-in ) ⎣ ⎝ = 1.087 ≤ 1.105 = 1.087 ⎡ ⎛ Rh M yt R pt = ⎢1 − ⎜1 − Mp ⎢⎣ ⎜⎝

⎞ ⎛ λ w − λ pw( Dc ) ⎞ ⎤ M p M p ≤ ⎟⎥ ⎟⎟ ⎜⎜ ⎟ ⎠ ⎝ λ rw − λ pw( Dc ) ⎠ ⎦⎥ M yt M yt

(A6.2.2-5)

⎡ ⎛ (1.00)(27,160k-in ) ⎞ ⎛ 101.3 − 93.68 ⎞ ⎤ (30, 020k-in ) (30, 020k-in ) R pt = ⎢1 − ⎜1 − ≤ ⎟⎜ ⎟⎥ (30, 020k-in ) ⎠ ⎝ 137.3 − 93.68 ⎠ ⎦ (27,160k-in ) (27,160k-in ) ⎣ ⎝ = 1.087 ≤ 1.105 = 1.087 Investigate Compression Flange Local Buckling: Investigate the compactness of the compression flange. λf =

b fc 2t fc

λ pf = 0.38

16" = 10.67 (2)( 3 4 ")

(A6.3.2-3)

E 29, 000ksi = 0.38 = 9.152 Fyc 50ksi

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

(A6.3.2-4)

AASHTO-LRFD 2007 Created July 2007: Page 3 of 7 -- 154 --

Since λf >λpf, the flange is non compact and, ⎡ ⎛ F S ⎞⎛ λ − λ pf ⎞ ⎤ M nc ( FLB ) = ⎢1 − ⎜1 − yr xc ⎟⎜ f R M (A6.3.2-2) ⎟⎜ λ rf − λ pf ⎟⎟ ⎥⎥ pc yc ⎢⎣ ⎝⎜ R pc M yc ⎠⎝ ⎠⎦ ⎛ ⎞ S Fyr = min ⎜ 0.7 Fyc , Rh Fyt xt , Fyw ⎟ ≥ 0.5 Fyc ; Fyr = (0.7) ( 50ksi ) = 35ksi (Pg 6-222) S xc ⎝ ⎠ Ekc λrf = 0.95 (A.6.3.2-5) Fyr kc =

4 = D tw

4 38''

= 0.3974

(A6.3.2-6)

3 '' 8

( 29, 000 ) ( 0.3974 ) = 17.24 λ = 0.95 ( 35 ) ⎡ ⎛ ( 35 )( 543.1 in ) ⎟⎞ ⎛ 10.67 − 9.152 ⎞⎥⎤ (1.087 ) 27,160 = ⎢1 − ⎜1 − ( ) ⎜ ⎟ ⎢ ⎝⎜ (1.087 ) ( 27,160 ) ⎠⎟ ⎝ 17.24 − 9.152 ⎠ ⎥ ⎣ ⎦ ksi

ksi

k-in

M nc ( FLB )

k-in

= 27,550k-in = 2, 296k-ft Investigate Compression Flange Lateral-Torsional Buckling: The unbraced length of the beam is Lb = 12’-0” = 144.0”. rt =

b fc ⎛ 1 Dc tw 12 ⎜1 + ⎜ 3b t fc fc ⎝

L p = 1.0rt

⎞ ⎟⎟ ⎠

(16") ⎛ ⎛ 1 ⎞ (19")( 38 ") ⎞ 12 ⎜ 1 + ⎜ ⎟ ⎟ ⎝ ⎝ 3 ⎠ (16")( 3 4 ") ⎠

= 4.220"

E 29, 000ksi = (1.0)(4.220") = 101.6" Fyc 50ksi

E Lr = 1.95 rt Fyr

⎛ Fyr S xc h ⎞ 1 + 1 + 6.76 ⎜ ⎟ S xc h ⎝ E J ⎠ J

(A6.3.3-10)

(A6.3.3-4)

(6.10.8.2.3-5)

⎛ ⎞ S Fyr = min ⎜ 0.7 Fyc , Rh Fyt xt , Fyw ⎟ ≥ 0.5 Fyc Fyr = ( 0.7 ) ( 50ksi ) = 35ksi (Pg 6-222) S xc ⎝ ⎠

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

AASHTO-LRFD 2007 Created July 2007: Page 4 of 7 -- 155 --

h = D + ( 1 2 ) ( t fc + t ft ) = 38"+ 3 4 " = 38.75" J=

3 t fc D tw3 b fc t fc ⎛ + ⎜⎜1 − 0.63 3 3 ⎝ b fc

⎞ b ft t 3ft ⎟⎟ + 3 ⎠

⎛ t ft ⎜⎜1 − 0.63 b ft ⎝

⎞ ⎟⎟ ⎠

(A6.3.3-9)

⎡ (16")( 3 4 ")3 ⎛ (38")( 3 8 ")3 ⎛ 3 " ⎞ ⎞⎤ + (2) ⎢ J= 1 − ( 0.63) ⎜ 4 ⎟ ⎟ ⎥ = 5.035 in 4 ⎜ 3 3 ⎝ 16" ⎠ ⎠ ⎦ ⎝ ⎣

E 29, 000ksi = = 828.6 Fyr 35ksi 3 S xc h ( 543.1 in ) ( 38.75") = = 4,180 J 5.035 in 4

Lr = (1.95 )( 4.220")( 828.6 )

1 ⎛ 4,180 ⎞ 1 + 1 + 6.76 ⎜ ⎟ = 396.8" = 33.06 ' 4,180 ⎝ 828.6 ⎠

Since L p = 101.6" < Lb = 144" < Lr = 396.8" , Inelastic LTB must be investigated. ⎡ ⎛ F S M nc ( LTB ) = Cb ⎢1 − ⎜1 − yr xc ⎢⎣ ⎜⎝ R pc M yc

M nc ( LTB )

⎞⎛ Lb − Lp ⎟⎜ ⎟⎜ Lr − Lp ⎠⎝

⎞⎤ ⎟⎟ ⎥ R pc M yc ≤ R pc M yc ⎠ ⎥⎦

(A6.3.3-2)

⎡ ⎛ 35ksi )( 543.1 in 3 ) ⎞ ⎛ 144"− 101.6" ⎞ ⎤ ( k-in k-in ⎢ ⎜ ⎟ = Cb 1 − 1 − ⎟ ⎥ (1.087 ) ( 27,160 ) ≤ (1.087 ) ( 27,160 ) k-in ⎟ ⎜ ⎜ 1.087 ) ( 27,160 ) ⎠ ⎝ 396.8"− 101.6" ⎠ ⎥ ⎢ ⎣ ⎝ ( ⎦ = ( Cb )( 0.9488 ) ( 29,520k-in ) = ( Cb ) ( 28, 010k-in ) ≤ 29,520k-in

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

AASHTO-LRFD 2007 Created July 2007: Page 5 of 7 -- 156 --

Compute the Moment Gradient Factor, Cb, for segment BC of the beam, which will be critical. 2

⎛M ⎞ ⎛M ⎞ Cb = 1.75 − 1.05 ⎜ 1 ⎟ + 0.3 ⎜ 1 ⎟ ≤ 2.3 ⎝ M2 ⎠ ⎝ M2 ⎠

(A6.3.3-7)

M 2 = M c = 1, 080k-ft M o = M B = 810.0k-ft M BC ,mid = 1, 013k-ft Since the BMD is not concave, M 1 = 2M mid − M 2 ≥ M o = (2) (1, 013k-ft ) − (1, 080k-ft ) = 946k-ft ≥ 810k-ft 2

⎛ 946 ⎞ ⎛ 946 ⎞ Cb = 1.75 − 1.05 ⎜ + 0.3 ⎜ ⎟ ⎟ = 1.061 ≤ 2.3 ⎝ 1, 080 ⎠ ⎝ 1, 080 ⎠ = 1.061

M nc ( LTB ) = (1.061) ( 28, 010 k-in ) = 29, 720k-in ≤ 29,520k-in M nc ( LTB ) = 29,520k-in = 2, 460k-ft The governing strength for the compression flange is the smaller of Mnc(FLB) and Mnc(LTB): Since M nc ( LTB ) = 2, 460k-ft > M nc ( FLB ) = 2, 296k-ft , ∴ FLB governs the strength of the compression flange. M nc = M nc ( FLB ) = 2, 296k-ft

φM nc = (1.00) ( 2, 296k-ft ) = 2, 296k-ft Check M u +

1 f l S xc ≤ φ f M nc 3

(A6.1.1-1)

M u = M C = 1, 080k-ft Assume that Strength I Load Combination Governs, ∴ γWS = 0.0 and fl = 0 Since M u +

1 fl S xc = 1, 080k-ft < φ f M nc =2,296k-ft , the compression flange is adequate. 3

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

AASHTO-LRFD 2007 Created July 2007: Page 6 of 7 -- 157 --

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

(A6.4-1)

φM nt = (1.00) ( 2, 460k-ft ) = 2, 460k-ft Check M u +

1 fl S xt ≤ φ f M nt 3

(6.10.8.1.2-1)

M u = M C = 1, 080k-ft Since M u +

1 f l S xt = 1, 080k-ft < φ f M nt =2,460k-ft , the tension flange is adequate. 3

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.

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

AASHTO-LRFD 2007 Created July 2007: Page 7 of 7 -- 158 --

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. 75kip

12'

30kip

12'

Solution: Draw the shear force diagram. 60kip

SFD(kip) -15kip -45kip kip

From the diagram, Vu = 60 . Referring to Section 6.10.9.2 of the Specification, Check Design Shear Strength: Vn = CVp

(6.10.9.2-1)

V p = 0.58Fy Dtw

(6.10.9.2-2)

D = d − 2t f = 26.9"− (2)(0.745") = 25.41" V p = (0.58)(50ksi ) [ (25.41")(0.490") ] = 361.1kip

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

AASHTO-LRFD 2007 Created July 2007: Page 1 of 2 -- 159 --

Since the web is unstiffened, k = 5.00. Ek = Fy

( 29, 000 ) ( 5.00 ) = 53.85 ( 50 ) ksi

ksi

D 25.41" = = 51.86 tw 0.490"

1.12

Ek = (1.12 )( 53.85 ) = 60.31 Fy

Since

D Ek = 51.86 < 1.12 = 60.31 , shear yielding governs and, tw Fy

C = 1.00

(6.10.9.3.2-4)

Vn = CVp = (1.00)(361.1kip) = 361.1kip Ď&#x2020;Vn = (1.00)(361.1kip) = 361.1kip > Vu = 60kip

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

O.K.

AASHTO-LRFD 2007 Created July 2007: Page 2 of 2 -- 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)

V p = 0.58Fy Dtw

(6.10.9.2-2)

V p = (0.58)(50ksi ) [ (38")( 3 8 ") ] = 413.3kip

Since the web is unstiffened, k = 5.00. Ek = Fy

( 29, 000 ) ( 5.00 ) = 53.85 ( 50 ) ksi

ksi

1.12

Ek = (1.12 )( 53.85 ) = 60.31 Fy

1.40

Ek = (1.40 )( 53.85 ) = 75.39 Fy

Since

D Ek = 101.3 > 1.40 = 75.39 , Elastic shear buckling governs and, tw Fy

D 38" = = 101.3 tw 38 "

⎛ Ek ⎞ (1.57 ) ⎛ ( 29, 000ksi ) ( 5.00 ) ⎞ ⎜ ⎟ = 0.4437 C= ⎟= 2 ⎜ 2 ⎟ ( 50ksi ) ( D / tw ) ⎜⎝ Fy ⎟⎠ (101.3) ⎜⎝ ⎠ 1.57

(6.10.9.3.2-6)

Vn = CVp = (0.4437) (413.3kip) = 183.4kip φVn = (1.00)(183.4kip) = 183.4kip < Vu = 242kip ODOT-LRFD Short Course - Steel AASHTO Shear Strength Example #2

AASHTO-LRFD 2007 Created July 2007: Page 1 of 3 -- 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… do 1.25 D

→ d o 1.25 D = (1.25)(38") = 47.5" say 48" ∴

d o 48" = = 1.263 D 38"

For stiffened webs, k = 5+

Ek = Fy

Since

5 5 = 5+ = 8.134 2 (d o / D) (1.263) 2

(6.10.9.3.2-7)

( 29, 000 ) (8.134 ) = 68.68 ( 50 ) ksi

ksi

1.12

Ek = ( 68.68 )(1.12 ) = 76.92 Fy

1.40

Ek = ( 68.68 )(1.40 ) = 96.15 Fy

D Ek = 101.3 > 1.40 = 96.15 , Elastic shear buckling governs and, tw Fy ⎛ Ek ⎞ (1.57 ) ⎛ ( 29, 000ksi ) ( 8.134 ) ⎞ ⎜ ⎟ = 0.7218 C= ⎟= 2 ⎜ 2 ⎟ ( 50ksi ) ( D / tw ) ⎜⎝ Fy ⎟⎠ (101.3) ⎜⎝ ⎠ 1.57

(6.10.9.3.2-6)

Vn = CVp = (0.7218) (413.3kip) = 298.3kip φVn = (1.00)(298.3kip) = 298.3kip > Vu = 242kip ODOT-LRFD Short Course - Steel AASHTO Shear Strength Example #2

AASHTO-LRFD 2007 Created July 2007: Page 2 of 3 -- 162 --

The previous calculations were based on the buckling strength of the web. For interior panels where: 2 Dtw ≤ 2.5 ( b fct fc + b ft t ft )

(6.10.9.3.2-1)

(2)(38")( 38 ") 28.5 = = 1.188 ≤ 2.5 [(16")( 3 4 ") + (16")( 3 4 ")] 24.0

Tension Field Action can be developed:

⎡ ⎤ ⎢ ⎥ 0.87(1 − C ) ⎥ ⎢ Vn = V p ⎢C + 2 ⎥ ⎛ do ⎞ ⎥ ⎢ 1+ ⎜ ⎟ ⎢ ⎝ D ⎠ ⎥⎦ ⎣

(6.10.9.3.2-2)

⎡ ⎤ (0.87)(1 − 0.7218) ⎥ ⎢ = 360.4kip Vn = (413.3 ) (0.7218) + 2 ⎢ ⎥ 1 + (1.263) ⎣ ⎦ kip

φVn = (1.00)(360.4kip) = 360.4kip

ODOT-LRFD Short Course - Steel AASHTO Shear Strength Example #2

AASHTO-LRFD 2007 Created July 2007: Page 3 of 3 -- 163 --

-- 164 --

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. 75kip

12'

30kip

12'

Solution: Draw the shear force diagram. 60kip

SFD(kip) -15kip -45kip

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:

Rn = (2.5k + N ) Fywtw

(D6.5.2-3)

Rn = ((2.5)(1.34") + 3.25")(50ksi )(0.490") = 161.7 kip φRn = (1.0)(161.7 kip ) =161.7 kip > 60kip

O.K.

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

AASHTO-LRFD 2007 Created July 2007: Page 1 of 3 -- 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

N : d

3.25" = 0.1208 ≤ 0.20 26.9" Therefore, (D6.5.3-3) controls: Rn = 0.40t

2 w

1.5 ⎡ ⎛ tw ⎞ ⎤ EFywt f N ⎛ ⎞ ⎢1 + 3 ⎜ ⎟ ⎜ ⎟ ⎥ tw ⎢ ⎝ d ⎠ ⎜⎝ t f ⎟⎠ ⎥ ⎣ ⎦

1.5 ksi ksi ⎡ ⎛ 3.25" ⎞⎛ 0.490" ⎞ ⎤ (29, 000 )(50 )(0.745") Rn = 0.40(0.490") 2 ⎢1 + 3 ⎜ ⎟⎜ ⎟ ⎥ 0.490" ⎝ 26.9" ⎠⎝ 0.745" ⎠ ⎥⎦ ⎢⎣

Rn = (0.09604 in 2 )(1.193)(1485ksi ) =170.2kip φRn = (0.80)(170.2kip ) =136.2kip > 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. Rn = (5k + N ) Fywtw

(D6.5.2-2)

Rn = ((5)(1.34") + 3.25")(50ksi )(0.490") = 243.8kip φRn = (1.0)(243.8kip ) = 243.8kip > 75kip

O.K.

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

AASHTO-LRFD 2007 Created July 2007: Page 2 of 3 -- 166 --

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

Therefore, (D6.5.3-2) controls: Rn = 0.80tw2

1.5 ⎡ ⎛ tw ⎞ ⎤ EFywt f N ⎛ ⎞ ⎢1 + 3 ⎜ ⎟ ⎜ ⎟ ⎥ tw ⎢ ⎝ d ⎠ ⎜⎝ t f ⎟⎠ ⎥ ⎣ ⎦

1.5 ksi ksi ⎡ ⎛ 3.25" ⎞⎛ 0.490" ⎞ ⎤ (29, 000 )(50 )(0.745") Rn = 0.80(0.490") ⎢1 + 3 ⎜ ⎥ ⎟⎜ ⎟ 0.490" ⎝ 26.9" ⎠⎝ 0.745" ⎠ ⎦⎥ ⎣⎢ 2

Rn = (0.1921 in 2 )(1.193)(1485ksi ) = 340.3kip φRn = (0.80)( 340.3kip ) = 272.2kip > 75kip

O.K.

The Web Yielding and Web Crippling Strengths are Satisfactory

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

AASHTO-LRFD 2007 Created July 2007: Page 3 of 3 -- 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: I t ≥ bt w3 J

(6.10.11.1.3-1)

and 1.5

D 4 ρt1.3 ⎛ Fyw ⎞ It ≥ ⎜ ⎟ 40 ⎝ E ⎠

(6.10.11.1.3-2)

where: It - Moment of inertia of the stiffener pair about the mid-thickness of the web. 2

⎛ D ⎞ 2.5 − 2.0 ≥ 0.5 J = 2.5 ⎜ ⎟ − 2.0 ≥ 0.5 use… J = 2 ( do / D ) ⎝ do / D ⎠

ODOT-LRFD Short Course - Steel AASHTO Web Strength Example #2

(6.10.11.1.3-3)

AASHTO-LRFD 2007 Created July 2007: Page 1 of 6 -- 168 --

2.5

( 48"/ 38")

− 2.0 = −0.4332 in 2 ≥ 0.5

b = smaller of do and D,

take J = 0.50

b = 38”

ρ = larger of Fyw / Fcrs and 1.00 0.31E

Fcrs =

⎛ bt ⎜⎜ ⎝ tp

⎞ ⎟⎟ ⎠

≤ Fys

Fcrs =

(0.31)(29, 000ksi ) ⎛ 6" ⎞ ⎜ 3 "⎟ ⎝ 8 ⎠

ρ = 50 / 35.11 ≥ 1.00 ρ =1.424 btw3 j = (38")( 3 8 ")3 (0.5) = 1.002 in 4 1.5

D 4 ρt1.3 ⎛ Fyw ⎞ ⎜ ⎟ 40 ⎝ E ⎠

It =

= 35.11ksi ≤ 50ksi

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.

1.5

(38") 4 (1.424)1.3 ⎛ 50 ksi ⎞ = ⎜ ksi ⎟ 40 ⎝ 29, 000 ⎠

(ts )(2bs + tw )3 12

= 5.909 in 4

take bs = 6”

(t ) [ (2)(6") + 38 "] = (ts )157.9 in 3 It = s 12 3

tp ≥

btw3 J 1.002 in 4 = = 0.006345" 157.9 in 3 157.9 in 3

ODOT-LRFD Short Course - Steel AASHTO Web Strength Example #2

…say ts = 3 8 "

AASHTO-LRFD 2007 Created July 2007: Page 2 of 6 -- 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. Rn = (2.5k + N ) Fywtw

(D6.5.2-3)

Rn = ((2.5)(1.125") + 9")(50ksi )( 3 8 ") = 221.5kip φRn = (1.0)(221.5kip ) = 221.5kip > 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. N : d 9" = 0.2278 > 0.20 38"+ (2)( 3 4 ")

Check

Therefore, (D6.5.3-4) controls: Rn = 0.40t

2 w

1.5 ⎡ ⎛ tw ⎞ ⎤ EFywt f 4 N ⎛ ⎞ ⎢1 + ⎜ − 0.2 ⎟ ⎜ ⎟ ⎥ tw ⎢ ⎝ d ⎠ ⎜⎝ t f ⎟⎠ ⎥ ⎣ ⎦

1.5 ⎡ ⎛ ⎞ ⎛ 3 8 " ⎞ ⎤ (29, 000ksi )(50ksi )( 3 4 ") (4)(9") Rn = (0.40)( 8 ") ⎢1 + ⎜ − 0.2 ⎟ ⎜ ⎟ ⎥ ( 3 8 ") ⎠ ⎝ 3 4 " ⎠ ⎥⎦ ⎢⎣ ⎝ (38"+ (2)( 3 4 ") 3

Rn = (0.05625 in 2 )(1.252)(1, 703ksi ) =119.9kip φRn = (0.80)(119.9kip ) = 95.92kip <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…) ODOT-LRFD Short Course - Steel AASHTO Web Strength Example #2

AASHTO-LRFD 2007 Created July 2007: Page 3 of 6 -- 170 --

Design the bearing stiffeners that need to be added to increase the web crippling strength: Check local buckling of the bearing stiffener: bt ≤ 0.48t p

bt ≤ 0.48t p tp ≥

E Fy

take Fy = 50ksi

29, 000ksi = 11.56t p 50ksi

bt 7" = = 0.6055" 11.56 11.56

(6.10.11.2.2-1)

take bs = 7”

take ts = 5 8 "

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

(7” x 5/8” bar stock may be used)

Check the bearing stiffeners as an effective column section: 3 3 I = ( 112 ) ⎡( 5 8 ")( 7"+ 3 8 "+ 7") + ( 6.75"− 5 8 ")( 3 8 ") ⎤ ⎣ ⎦ 4 = 154.7 in

A = ⎡⎣( 5 8 ")( 2 )( 7") + ( 6.75")( 3 8 ") ⎤⎦ = 11.28 in 2 r=

I 154.7 in 4 = = 3.704" A 11.28 in 2

⎛ KL ⎞ ⎛ (0.75)(38") ⎞ ⎜ ⎟=⎜ ⎟ = 7.695 ⎝ r ⎠ ⎝ 3.704" ⎠ 2 2 ksi ⎞ ⎛ KL ⎞ Fy ⎛ 7.695 ⎞ ⎛ 36 λ=⎜ = = 0.007448 ⎜ ⎟ ⎜ ⎟ ksi ⎟ ⎝ r π ⎠ E ⎝ π ⎠ ⎝ 29, 000 ⎠

Pn = 0.66λ Fy As

(

Inelastic Buckling (6.9.4.1-3) Taking Fy = 36ksi here is a conservative assumption since I am not sure what material will be used.

)

= 0.66( 0.007448) ( 36ksi )(11.28 in 2 ) = 404.8kip

(6.9.4.1-1)

φPn = ( 0.90 ) ( 404.8kip ) = 364.3kip 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. ODOT-LRFD Short Course - Steel AASHTO Web Strength Example #2

AASHTO-LRFD 2007 Created July 2007: Page 4 of 6 -- 171 --

Check bearing stress between the end of the bearing stiffeners and the loaded flange: Rn = 1.4 Apn Fys

(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 Apn = (2)(7"− 1") ( 5 8 ") = 7.50 in 2

Rn = (1.4)(7.50 in 2 )(36ksi ) = 378.0kip φRn = (1.00)(378.0kip ) = 378.0kip

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

ODOT-LRFD Short Course - Steel AASHTO Web Strength Example #2

AASHTO-LRFD 2007 Created July 2007: Page 5 of 6 -- 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 I = ( 112 ) ⎡ (2) ( 5 8 ")( 7"+ 3 8 "+ 7") + ( (2)(3.375") + 7"− (2) ( 5 8 ") ) ( 3 8 ") ⎤ = 309.5 in 4 ⎣ ⎦

A = ⎡⎣( 5 8 ")( 4 )( 7") + ⎡⎣(2) ( 3.375") + 7"⎤⎦ ( 3 8 ") ⎤⎦ = 22.66 in 2 r=

I 309.5 in 4 = = 3.696" A 22.66 in 2

2 2 ksi ⎛ KL ⎞ Fy ⎛ 7.711 ⎞ ⎛ 36 λ=⎜ = ⎟ ⎜ ⎟ ⎜ ksi ⎝ r π ⎠ E ⎝ π ⎠ ⎝ 29, 000

(

⎛ KL ⎞ ⎛ (0.75)(38") ⎞ ⎜ ⎟=⎜ ⎟ = 7.711 ⎝ r ⎠ ⎝ 3.696" ⎠ ⎞ ⎟ = 0.007479 ⎠

)

Pn = 0.66λ Fy As = 0.66( 0.007479) ( 36ksi )( 22.66 in 2 ) = 813.2kip

Inelastic Buckling (6.9.4.1-3)

(6.9.4.1-1)

φPn = ( 0.90 ) ( 813.2kip ) = 731.9kip

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. ODOT-LRFD Short Course - Steel AASHTO Web Strength Example #2

AASHTO-LRFD 2007 Created July 2007: Page 6 of 6 -- 173 --

-- 174 --

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: Section A-A

Shear Strength of the Bolts: Assume that the threads are included in the shear plane of the connection. 3"

Rn = 0.38 Ab Fub N s

1.5"

(6.13.2.7-2)

2 ⎛π⎞ Ab = ⎜ ⎟ ( 7 8 ") = 0.6013 in 2 ⎝4⎠

For A325 bolts, Fub = 120ksi Bolts are in single shear so Ns = 1

(

)(

)

kip Rn = (0.38) 0.6013 in 2 120ksi (1) = 27.42 bolt

(

C12 x 30

)

kip kip φRn = (0.80) 27.42 bolt = 21.94 bolt

(

For all 8 bolts, φRn = (8 bolts) 21.94

kip bolt

) = 175.5

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

Pu kip

AASHTO-LRFD 2007 Created July 2007: Page 1 of 4 -- 175 --

Check Bearing Strength: Interior Bolts Bearing on the Channel Web: Lc = 3"− ( 2 )( 1 2 )( 7 8 "+ 116 ") = 2.063"

Since Lc = 2.063” > 2d = 1.75”,

(

)

kip Rn = (2.4) ( 7 8 ") (0.510") 58ksi = 62.11 bolt

Rn = 2.4dtFu

(6.13.2.9-1)

Interior Bolts Bearing on the Gusset Plate: Lc = 3"− ( 2 )( 1 2 )( 7 8 "+ 116 ") = 2.063"

Since Lc = 2.063” > 2d = 1.75”,

(

)

kip Rn = (2.4) ( 7 8 ")( 5 8 ") 58ksi = 76.13 bolt

Rn = 2.4dtFu

(6.13.2.9-1)

End Bolts Bearing on the Channel Web: Lc = 1.5"− ( 1 2 )( 7 8 "+ 116 ") = 1.031"

Since Lc = 1.031” < 2d = 1.75”,

(

)

kip Rn = (1.2) (1.031") (0.510") 58ksi = 36.60 bolt

Rn = 1.2 Lc tFu

(6.13.2.9-2)

End Bolts Bearing on the Gusset Plate: (Assume that the end distance on the gusset is 11/2”) Lc = 1.5"− ( 1 2 )( 7 8 "+ 116 ") = 1.031"

Since Lc = 1.031” < 2d = 1.75”,

(

)

kip Rn = (1.2) (1.031")( 5 8 ") 58ksi = 44.85 bolt

Rn = 1.2 Lc tFu

(6.13.2.9-2)

For all 8 Bolts:

(

)

(

)

(

)

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

(

)

φRn = (0.80) 411.3kip = 329.1kip

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

AASHTO-LRFD 2007 Created July 2007: Page 2 of 4 -- 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: Atg = (6")(0.510") = 3.060 in 2

Atn = ⎡⎣(6") − (2) ( 1 2 )( 7 8 "+ 18 ") ⎤⎦ (0.510") = 2.550 in 2 Avg = (2) [ (1.5") + (3)(3") ] (0.510") = 10.71 in 2

Avn = (2) ⎡⎣ (1.5") + (3)(3") − (3.5) ( 7 8 "+ 18 ") ⎤⎦ (0.510") = 7.140 in 2

Atn ≥ 0.58 Avn ?

2.550 in 2 ≥(0.58)(7.140 in 2 ) = 4.141 in 2

NO!

∴ Rn = 0.58Fu Avn + Fy Atg

(

)(

(6.13.4-2)

) (

)(

)

Rn = (0.58) 58ksi 7.140 in 2 + 36ksi 3.060 in 2 = 350.3kip

(

)

φRn = (0.80) 350.3kip = 280.3kip

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

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

AASHTO-LRFD 2007 Created July 2007: Page 3 of 4 -- 177 --

Check the Slip Capacity of the Connection: Rn = K h K s N s Pt

(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)

(

)

kip Rn = (1.00)(0.33)(1) 39kip = 12.87 bolt

For All 8 Bolts:

(

)

kip Rn = (8 bolts) 12.87 bolt = 103.0kip

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

AASHTO-LRFD 2007 Created July 2007: Page 4 of 4 -- 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: ⎛ 3b t 3 ⎞ Qu = ⎜ − ⎟ P ⎝ 8a 20 ⎠

(6.13.2.10.4-1)

gt 7" − k1 = − 1 1 8 " = 2.375" 2 2

(k1 = 11/8” for W21 x 132)

a = ( 1 2 ) ( b f − gt ) = ( 1 2 )(12.4"− 7") = 2.700" ⎛ (3)(2.375") (1.04")3 ⎞ Qu = ⎜ − ⎟ Pu = 0.2736 Pu 20 ⎠ ⎝ (8)(2.700")

Tu = Qu + Pu = 1.274 Pu = (1.274) (120kip ) = 152.8kip

Tensile Resistance of the Bolts: Tn = 0.76 Ab Fub

(6.13.2.10.2-1)

2 ⎛π⎞ Ab = ⎜ ⎟ (1") = 0.7854 in 2 ⎝4⎠

Fub = 120ksi kip Tn = (0.76) ( 0.7854 in 2 )(120ksi ) = 71.63 bolt

For All 4 Bolts:

(

)

kip φTn = (4 bolts) 57.30 bolt = 229.2kip

ODOT-LRFD Short Course - Steel AASHTO Connection Example #2

(

)

kip kip φTn = (0.80) 71.63 bolt = 57.30 bolt

AASHTO-LRFD 2007 Created July 2007: Page 1 of 2 -- 179 --

Check the Strength of the Flange of the WT: Assume that a plastic mechanism forms between the bolt lines and stem. Q

Moment Equilibrium about at the fillet:

∑ M → 2M

⎛ Pb⎞ =⎜ u ⎟ ⎝ 2 ⎠

⎛ (120kip ) (2.375") ⎞ ⎟ Mu = ⎜ ⎜ ⎟ 4 ⎝ ⎠ k-in = 71.25

For Safety, φM p ≥ M u .

Tu Mu

Pu/2 b

Lt 2 ⎛ (8")(1.04") ⎞ ksi k-in Fy = ⎜ ⎟ ( 50 ) = 108.2 4 4 ⎝ ⎠ k-in φM p = (1.00) (108.2 ) = 108.2k-in OK Mp =

ODOT-LRFD Short Course - Steel AASHTO Connection Example #2

AASHTO-LRFD 2007 Created July 2007: Page 2 of 2 -- 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, γ ( Δf ) ≤ ( ΔF )n

(6.6.1.2.2-1)

kip γ ( ΔP ) (0.75) ⎡⎣(1.274 ) ( 60 ) ⎤⎦ γ ( Δf ) = = = 18.24ksi Abolts (4) ( 0.7854 in 2 )

⎛ A ⎞ 3 ( ΔF )TH =⎜ ⎟ ≥ 2 ⎝N⎠ 1

( ΔF )n

For infinite life,

( ΔF )n =

( ΔF )TH 2

31.0ksi = = 15.5ksi 2

Since γ ( Δf ) = 18.24ksi > ( ΔF )n = 15.5ksi , the bolts will have a finite life 1

For finite life, γ ( Δf ) ≤ ( ΔF )n

N≤

( γ ( Δf ) )

⎛ A ⎞3 =⎜ ⎟ ⎝N⎠

17.1× 108 ksi3

(18.24 )

ksi 3

= 281,800 cycles

Note that if prying is not included, γ ( Δf ) = 14.32ksi and the calculations would incorrectly show that the bolts have an infinite fatigue life.

ODOT-LRFD Short Course - Steel AASHTO Connection Example #3

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

Vu =

2 Pu = 0.8944 Pu = 107.3kip 5

Vu =

107.3kip kip = 26.83 bolt 4 bolts

Tu =

1.274 Pu = 0.5694 Pu = 68.33kip 5

Tu =

68.33kip kip = 17.08 bolt 4 bolts

Assume that the threads are included in the shear plane of the connection. Vn = Rn = 0.38 Ab Fub N s

(6.13.2.7-2)

(

)(

)

kip Vn = (0.38) 0.7854 in 2 120ksi (1) = 35.81 bolt

kip Vu 26.83 bolt = = 0.7491 , kip Vn 35.81 bolt

Tn = (0.76) ( 0.7854 in

(

∴Tn = 0.76 Ab Fub

)(120 ) ksi

⎛ V ⎞ 1− ⎜ u ⎟ ⎝ φVn ⎠

(6.13.2.11-2)

⎛ ( 26.83 kip bolt ) ⎞ kip 1− ⎜ = 25.11 bolt ⎜ (0.80) ( 35.81 kip bolt ) ⎟⎟ ⎝ ⎠ 2

)

kip kip φTn = (0.80) 25.11 bolt = 20.09 bolt

kip kip > Tu = 17.08 bolt , the bolts are OK for the loading shown. Since φTn = 20.09 bolt

ODOT-LRFD Short Course - Steel AASHTO Connection Example #4

AASHTO-LRFD 2007 Created July 2007: Page 1 of 3 -- 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: Lc = 4"− ( 2 )( 1 2 )(1"+ 116 ") = 2.938"

Since Lc = 2.938” > 2d = 2”, Rn = 2.4dtFu

(6.13.2.9-1)

(

)

kip Rn = (2.4) (1") (1.04") 65ksi = 162.2 bolt

End Bolts: Lc = 2"− ( 1 2 )(1"+ 116 ") = 1.469"

Since Lc = 1.469” < 2d = 2”, Rn = 1.2 Lc tFu

(6.13.2.9-2)

(

)

kip Rn = (1.2) (1.469") (1.04") 65ksi = 119.1 bolt

For all 4 Bolts:

(

)

(

)

kip kip Rn = (2 bolts) 162.2 bolt + (2 bolts) 119.1 bolt = 562.8kip

(

)

φRn = (0.80) 562.8kip = 450.1kip

Since φRn = 450kip > Vu = 107 kip , 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.

ODOT-LRFD Short Course - Steel AASHTO Connection Example #4

AASHTO-LRFD 2007 Created July 2007: Page 2 of 3 -- 183 --

Check Shear in the Stem of the WT: Rn = 0.58 Ag Fy

(6.13.5.3-2)

Rn = (0.58) [ (8")(0.650") ] ( 50ksi ) = 150.8kip φRn = (1.00) (150.8kip ) = 150.8kip Since φRn = 150.8kip > Vu = 107.3kip , the stem will be satisfactory in shear.

ODOT-LRFD Short Course - Steel AASHTO Connection Example #4

AASHTO-LRFD 2007 Created July 2007: Page 3 of 3 -- 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”. φRn , weld = 0.6φe 2 Fexx Aw ≥ φPn , member φRn, weld = ( 0.6 )( 0.80 ) ( 70ksi ) × ⎡⎣(0.7071) ( w ) ⎤⎦ ( 2 )( 8") ≥ 163kip w≥

163kip = 0.4288" kip 380.1 inch

→ Say

ODOT-LRFD Short Course - Steel AASHTO Connection Example #5

AASHTO-LRFD 2007 Created July 2007: Page 1 of 2 -- 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: Lw = 6"+ ( 2 )( 8") Tan ( 30° ) = 15.24"

Gross Section Yielding:

(

)

(

)

φPn = φFy Ag = (0.95) 50ksi ⎡⎣(15.24")( 3 8 ") ⎤⎦ = 271.4kip Net Section Fracture: φPn = φFu Ae = (0.80) 65ksi ⎡⎣(15.24")( 3 8 ") ⎤⎦ (1.00 ) = 297.1kip

(Taking U = 1.00)

Check Block Shear in the Gusset Plate: Atg = Atn = (6") ( 3 8 ") = 2.250 in 2 ?

Atn ≥ 0.58 Avn

Avg = Avn = (2) ( 8")( 3 8 ") = 6.000 in 2 ?

→

2.250 in 2 ≥(0.58)(6.000 in 2 ) = 3.480 in 2

∴ Rn = 0.58Fu Avn + Fy Atg

(

)(

(6.13.4-2)

) (

Rn = (0.58) 65ksi 6.000 in 2 + 50ksi

(

NO!

)( 2.250 in ) = 338.7 2

kip

)

φRn = (0.80) 338.7 kip = 271.0kip

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

ODOT-LRFD Short Course - Steel AASHTO Connection Example #5

AASHTO-LRFD 2007 Created July 2007: Page 2 of 2 -- 186 --

AASHTO Connection Example #6a:

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)

Problem:

Solution:

τ=

Tr J

→ T = ( 24kip ) ( 3"+ 2") = 120k-in

J = ∑ Ad 2 = A∑ d 2

(

⎡ J = ⎢( 4 ) (2")2 + (3") 2 ⎣ J = 47.12 in 4

) + ( 2)( 2") ⎤⎥⎦ ⎛⎜⎝ π4 ⎞⎟⎠ (1") 2

Tr (120 Corner Bolts: τ = = J τ = 9.182ksi

k-in

)(

(2") 2 + (3") 2

47.12 in

)

2 ⎛π⎞ V = ( 9.182ksi ) ⎜ ⎟ (1") = 7.211kip ⎝4⎠

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

AASHTO-LRFD 2007 Created July 2007: Page 1 of 2 -- 187 --

Force acts perpendicular to line drawn from bolt to C.G. Breaking force into horizontal and vertical components… ⎛ 2 ⎞ ⎛ 2 ⎞ kip kip Vy = ⎜ ⎟ (V ) = ⎜ ⎟ ( 7.211 ) = −4.000 ⎝ 3.606 ⎠ ⎝ 3.606 ⎠ ⎛ 3 ⎞ ⎛ 3 ⎞ kip kip Vx = ⎜ ⎟ (V ) = ⎜ ⎟ ( 7.211 ) = 6.000 3.606 3.606 ⎝ ⎠ ⎝ ⎠ Add evenly distributed vertical force to the vertical, torsional force for Bolt B… Vy = −4.000

kip

−24kip + = −8.000kip 6

Vx = 6.000kip Vtotal = Vx 2 + Vy 2 = (−8.000kip ) 2 + (6.000kip ) 2 = 10.00kip

Check These Results with a Spreadsheet Solution: Px:

xCG:

Py:

-24

yCG:

ex:

ey:

Σd:

60.00

-120.0

Vmax =

10.0

Bolt A B C D E F

x -2.00 2.00 -2.00 2.00 -2.00 2.00

y 3.00 3.00 0.00 0.00 -3.00 -3.00

13.00 13.00 4.00 4.00 13.00 13.00

Vx -6.0 -6.0 0.0 0.0 6.0 6.0

Vtotal

4.0 -4.0 4.0 -4.0 4.0 -4.0

6.0 10.0 0.0 8.0 6.0 10.0

Everything checks out OK.

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

AASHTO-LRFD 2007 Created July 2007: Page 2 of 2 -- 188 --

AASHTO Connection Example #6b:

Use the simplified equations to solve the previous example problem. (Example 4.12.1 from Salmon & Johnson)

Problem:

Solution:

T = ( 24 kip ) ( −3"− 2") = −120 k-in

∑d ∑d

= ( 4 ) ⎡⎣ (2") 2 + (3") 2 ⎤⎦ + ( 2 )( 2")

= 60.00 in 2

Looking at Bolt B:

VB , x

VB , y

⎛ ( − 120 k-in ) ( 3") ⎞ ⎟ = 6.000 kip = −⎜ 2 ⎜ 60.00 in ⎟ ⎝ ⎠ k-in ⎛ ( − 120 ) ( 2") ⎞ ⎟ = − 4.000 kip =⎜ ⎜ 60.00 in 2 ⎟ ⎝ ⎠

VB ,total =

( 6.000 )

kip 2

⎡ ⎛ − 24 kip + ⎢ ( − 4.000 kip ) + ⎜ ⎝ 6 ⎣

⎞⎤ ⎟⎥ ⎠⎦

VB ,total = 10.00 kip ODOT-LRFD Short Course - Steel AASHTO Connection Example #6b

AASHTO-LRFD 2007 Created July 2007: Page 1 of 1 -- 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: M u , Beam + φM n, Beam 2

810k-ft + 1,300k-ft = 1, 055k-ft 2

(

)

0.75 φM n, Beam = ( 0.75 ) 1,300k-ft = 975.0k-ft 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 ⎞ ⎛ f cf Fcf = ⎜ ⎟ ⎜ + αφ f Fyf ⎝ 2 ⎠ ⎜⎝ Rh

(810 ) (12 )( = k-ft

f cf

⎞ ⎟⎟ ≥ 0.75 αφ f Fyf ⎠ in ft

29.7" 2 4

(6.13.6.1.4c-1)

− 0.670" 2 )

3,990 in

= 35.36ksi

Rh = 1.00, since the beam is non-hybrid. α = 1.00. Since we are assuming that φMn = φMp, Fn = Fyf ksi ⎛ 1 ⎞ ⎛ 35.36 Fcf = ⎜ ⎟ ⎜ + (1.00 )(1.00 ) 50ksi ⎜ 2 1.00 ⎝ ⎠⎝

(

= 42.68ksi ≥ 37.50ksi

⎞

) ⎟⎟ ≥ ( 0.75)(1.00 )(1.00 ) ( 50 ) ksi

⎠

→ Fcf = 42.68ksi

ODOT-LRFD Short Course - Steel AASHTO Connection Example #7

AASHTO-LRFD 2007 Created July 2007: Page 1 of 20 -- 190 --

For the Non-Controlling Flange: Fncf = Rcf

f ncf

≥ 0.75 αφ f Fyf

(6.13.6.1.4c-3)

f ncf = f cf = 35.36ksi Rcf =

Fcf f cf

Fncf = (1.207 )

42.68ksi = 1.207 35.36ksi

(

35.36ksi ≥ ( 0.75 )(1.00 )(1.00 ) 50ksi 1.00

= 42.68ksi ≥ 37.50ksi

)

→ Fncf = 42.68ksi

For the Compression Flange: Pu ,Comp = Fcf Ae In compression, Ae is taken as the gross area of the flange. Ae = Ag = (10.5")( 0.670") = 7.035 in 2

(

)(

)

Pu ,Comp = 42.68ksi 7.035 in 2 = 300.3kip For the Tension Flange: Pu ,Ten = Fcf Ae ⎛φ F In tension, Ae = ⎜ u u ⎜ φ y Fy ⎝

⎞ ⎟⎟ An ≤ Ag ⎠

(6.13.6.4c-2)

For 1” diameter bolts, An = ⎡⎣(10.5") − ( 2 )(1 18 ") ⎤⎦ ( 0.670") = 5.528 in 2

ODOT-LRFD Short Course - Steel AASHTO Connection Example #7

AASHTO-LRFD 2007 Created July 2007: Page 2 of 20 -- 191 --

( (

⎛ ( 0.80 ) 65ksi Ae = ⎜ ⎜ ( 0.95 ) 50ksi ⎝

) ⎞⎟ 5.528 in ≤ 7.035 in ( ) ) ⎟⎠ 2

= 6.052 in 2 ≤ 7.035 in 2

(

)(

→ Ae = 6.052 in 2

)

Pu ,Ten = 42.68ksi 6.052 in 2 = 258.3kip

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. Pu ,Ten =

258.3kip = 129.1kip 2

Pu ,Comp =

300.3kip = 150.2kip 2

For the Outer Flange Splice Plate: Gross Yielding (Tension): φPn = φ y Fy Ag ≥ Pu ,Ten

(

(6.8.2.1-1)

)

φPn = ( 0.95 ) 50ksi (10.5") ( touter ) ≥ 129.1kip touter ≥

129.1kip

( 0.95) ( 50ksi ) (10.5")

≥ 0.2589"

→ say

16 "

Net Section Fracture (Tension): φPn = φu Fu AnU ≥ Pu ,Ten

(

(6.8.2.1-2)

)

φPn = ( 0.80 ) 65ksi ⎡⎣(10.5") − ( 2 )(1"+ 18 ") ⎤⎦ ( touter ) (1.00 ) ≥ 129.1kip touter ≥

129.1kip

( 0.80 ) ( 65ksi ) ( 8.25")

≥ 0.3010"

→ say

16 "

ODOT-LRFD Short Course - Steel AASHTO Connection Example #7

AASHTO-LRFD 2007 Created July 2007: Page 3 of 20 -- 192 --

Gross Yielding (Compression): φPn = φc Fy Ag ≥ Pu ,Comp

(

(6.13.6.1.4c-4)

)

φPn = ( 0.90 ) 50ksi (10.5") ( touter ) ≥ 150.2kip touter ≥

150.2kip

( 0.90 ) ( 50ksi ) (10.5")

≥ 0.3179"

→ say

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. Winner = b f − 2k1 = 10.5"− ( 2 )(1 116 ") = 8.375"

Take the width of each of the two inner plates as 4”.

Gross Yielding (Tension): φPn = φFy Ag ≥ Pu ,Ten

(

φPn = ( 0.95 ) 50ksi tinner ≥

(6.8.2.1-1)

) ( 2)( 4.00") ( t

Inner

) ≥ 129.1kip

129.1kip

( 0.95) ( 50ksi ) ( 2 )( 4.00")

≥ 0.3297"

→ say

Net Section Fracture (Tension): φPn = φFu AnU ≥ Pu ,Ten

(

φPn = ( 0.80 ) 65ksi

(6.8.2.1-2)

) ( 2) ⎡⎣( 4.00") − (1"+

ODOT-LRFD Short Course - Steel AASHTO Connection Example #7

)⎤⎦ ( tInner ) (1.00 ) ≥ 129.1kip

AASHTO-LRFD 2007 Created July 2007: Page 4 of 20 -- 193 --

tinner ≥

129.1kip

( 0.80 ) ( 65ksi ) ( 5.75")

≥ 0.4318"

→ say

16 "

Gross Yielding (Compression): φPn = φc Fy Ag ≥ Pu ,Comp

(

φPn = ( 0.90 ) 50ksi t Inner ≥

(6.13.6.1.4c-4)

) ( 2)( 4.00") ( t

Inner

) ≥ 150.2kip

150.2kip

( 0.90 ) ( 50ksi ) ( 2 )( 4.00")

≥ 0.4172"

→ say

16 "

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 6191). AOuter = (10.5")( 3 8 ") = 3.938 in 2

AOuter − AInner AAve

AInner = ( 2 )( 4.00")( 7 16 ") = 3.500 in 2

( 2 ) ( 3.938 in 2 ) − ( 3.500 in 2 )

( 3.938 in ) + ( 3.500 in ) 2

= 11.76%

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”

ODOT-LRFD Short Course - Steel AASHTO Connection Example #7

Inner Flange Splice Plates: 4” x 1/8”

AASHTO-LRFD 2007 Created July 2007: Page 5 of 20 -- 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. Rn = 0.38 Ab Fub N s

(6.13.2.7-2)

2 ⎛π⎞ Ab = ⎜ ⎟ (1") = 0.7854 in 2 ⎝4⎠

(

)(

For A325 bolts, Fub = 120ksi

)

kip Rn = (0.38) 0.7854 in 2 120ksi (2) = 71.63 bolt

(

)

kip kip φRn = (0.80) 71.63 bolt = 57.30 bolt

Determine the number of flange bolts required: 300.3kip n fb = = 5.24 bolts kip 57.30 bolt

→

say 6 bolts

6 1/2"

10 1/2"

24 1/2"

2 1/2"

3 1/2"

2 1/2"

2 1/2" 1

W30 x 99

3 1/2"

2 1/2"

/2" gap between ends of beams

PL 241/2” x 101/2" x 3/8"

W30 x 99

PL 241/2" x 4" x 1/2" Each Side 1" dia M164 Bolts (12 places)

ODOT-LRFD Short Course - Steel AASHTO Connection Example #7

AASHTO-LRFD 2007 Created July 2007: Page 6 of 20 -- 195 --

Check Bolt Bearing in the Flange Splice: Interior Bolts Bearing on the Beam Flange: Lc = 3 1 2 "− ( 2 )( 1 2 )(1"+ 116 ") = 2.438"

Since Lc = 2.433” > 2d = 2.0”,

(

)

kip Rn = (2.4) (1") (0.670") 65ksi = 104.5 bolt

Rn = 2.4dtFu

(6.13.2.9-1)

Interior Bolts Bearing on the Splice Plates: Lc = 3 1 2 "− ( 2 )( 1 2 )(1"+ 116 ") = 2.438"

Since Lc = 2.433” > 2d = 2.0”,

(

)

kip Rn = (2.4) (1") ( 83 "+ 12 ") 65ksi = 136.5 bolt

Rn = 2.4dtFu

(6.13.2.9-1)

End Bolts Bearing on the Beam Flange: Lc = 2 1 2 "− ( 1 2 )(1"+ 116 ") = 1.969"

Since Lc = 1.969” < 2d = 2.0”,

(

)

kip Rn = (1.2) (1.969") (0.670") 65ksi = 102.9 bolt

Rn = 1.2 Lc tFu

(6.13.2.9-2)

End Bolts Bearing on the Splice Plates: Lc = 2 1 2 "− ( 1 2 )(1"+ 116 ") = 1.969"

Since Lc = 1.969” < 2d = 2.0”,

(

)

kip Rn = (1.2) (1.969") ( 83 "+ 12 ") 65ksi = 134.4 bolt

Rn = 1.2 Lc tFu

(6.13.2.9-2)

For all 6 Bolts:

(

)

(

)

(

)

kip kip kip Rn = (2 bolts) 104.5 bolt + (2 bolts) 104.5 bolt + (2 bolts) 102.9 bolt = 623.9kip

(

)

φRn = (0.80) 623.9kip = 499.1kip

ODOT-LRFD Short Course - Steel AASHTO Connection Example #7

AASHTO-LRFD 2007 Created July 2007: Page 7 of 20 -- 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 Pslip = Fs Ag

(

)

0.670" 548k-ft (12 inft )( 29.7" fs 2 â&#x2C6;&#x2019; 2 ) where Fs = = = 23.92ksi 4 Rh (1.00 ) 3,990 in

(

)

(6.13.6.1.4c-5)

)

Pslip = 23.92ksi (10.5")( 0.670") = 168.3kip

The slip resistance of a single bolt is taken as: Rn = K h K s N s Pt

(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)

(

)

kip Rn = (1.00)(0.33)(2) 51kip = 33.66 bolt

Determine the number of flange bolts required: n fb =

168.3kip = 5.00 bolts kip 33.66 bolt

â&#x2020;&#x2019;

ODOT-LRFD Short Course - Steel AASHTO Connection Example #7

6 bolts will work

AASHTO-LRFD 2007 Created July 2007: Page 8 of 20 -- 197 --

Check Block Shear of the Beam Flange: 3 1/2"

2 1/2"

3 1/2"

Shear

Atg = ( 2 )( 2")( 0.670") = 2.680 in 2

6 1/2"

Tension

Atn = ( 2 ) ⎡⎣ 2"− ( 12 ) (1"+ 18 ") ⎤⎦ ( 0.670") = 1.926 in 2 Avg = ( 2 )( 3 1 2 "+ 3 1 2 "+ 2 1 2 ")( 0.670") = 12.73 in 2

Shear

Avn = ( 2 ) ⎡⎣9 1 2 "− ( 2.5 )(1"+ 18 ") ⎤⎦ ( 0.670") = 8.961 in 2

Atn ≥ 0.58 Avn ?

1.926 in 2 ≥ (0.58)(8.961 in 2 ) = 5.198 in 2

NO!

∴ Rn = 0.58Fu Avn + Fy Atg

(

)(

(6.13.4-2)

) (

Rn = (0.58) 65ksi 8.961 in 2 + 50ksi

(

)( 2.680 in ) = 471.8

)

φRn = (0.80) 471.8kip = 377.5kip

kip

ODOT-LRFD Short Course - Steel AASHTO Connection Example #7

AASHTO-LRFD 2007 Created July 2007: Page 9 of 20 -- 198 --

B) Web Splice: The web splice is to be designed for the following actions at the Strength Limit: 1. 2. 3. 4.

Vuw Mvuw Muw Huw -

The direct shear force. The moment on the web bolts caused by the eccentricity of Vuw The portion of the bending moment in the beam that is carried by the web. 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: ?

Vu < 0.5φvVn ?

(

)

45kip < ( 0.5 )(1.00 ) 427.7 kip = 213.8kip Since Vu < 0.5φvVn ,

(

)

Vuw = 1.5Vu = (1.5 ) 45kip = 67.5kip

(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, e = ( 12 ) ( 3 1 2 ") + 2 1 2 "+ ( 12 ) ( 1 2 ") = 4.50"

(

M vuw = e Vuw = ( 4.50") 67.5kip

)

= 303.8k-in = 25.31k-ft

ODOT-LRFD Short Course - Steel AASHTO Connection Example #7

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.

AASHTO-LRFD 2007 Created July 2007: Page 10 of 20 -- 199 --

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

M uw

tw D 2 Rh Fcf − Rcf f ncf 12

(C6.13.6.1.4b-1)

Fcf = 42.68ksi

(Positive since it’s in tension)

Rcf = 1.207

(from Before)

f ncf = −35.36ksi

(Negative since it’s in compression)

⎡ ( 0.520")( 28.36")2 ⎤ =⎢ ⎥ (1.00 ) 42.08ksi − (1.207 ) −35.36ksi 12 ⎢⎣ ⎥⎦

(

)

(

)

= 34.85 in 3 85.36ksi = 2,975k-in = 247.9k-ft

Determine the horizontal force that results from moving the beam moment, Huw: H uw =

tw D Rh Fcf + Rcf f ncf 12

(C6.13.6.1.4b-2)

⎡ ( 0.520")( 28.36") ⎤ ksi ksi H uw = ⎢ ⎥ (1.00 ) 42.68 + (1.207 ) −35.36 12 ⎣ ⎦

(

)

(

)

= 1.229 in 2 0.000ksi = 0.00kip

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.

ODOT-LRFD Short Course - Steel AASHTO Connection Example #7

AASHTO-LRFD 2007 Created July 2007: Page 11 of 20 -- 200 --

The total moment acting on the web splice is, M Total = M vuw + M uw = 303.8k-in + 2,975k-in = 3, 279k-in

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. 3, 279k-in P= = 728.7 kip 4.50" 2 1/2"

3 1/2"

2 3/4"

P = 728.7kip G

e = 41/2"

ODOT-LRFD Short Course - Steel AASHTO Connection Example #7

AASHTO-LRFD 2007 Created July 2007: Page 12 of 20 -- 201 --

ODOT-LRFD Short Course - Steel AASHTO Connection Example #7

AASHTO-LRFD 2007 Created July 2007: Page 13 of 20 -- 202 --

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

Pu 728.7 kip = = = 12.72 kip φrn 57.30 bolt

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 2 2 Σd 2 = ( 4 ) ⎡ ( 4 ) ( d x ) + d y , D + d y ,C + d y , B + d y , A ⎤ ⎢⎣ ⎥⎦ 2 2 2 2 2 Σd 2 = ( 4 ) ⎡( 4 )(1.75") + (1.5") + ( 4.5") + ( 7.5") + (10.5") ⎤ = 805 in 2 ⎣ ⎦

Examine Bolt A2:

(

)

(

)

VT , X

3, 279k-in (10.5") T y = = = 42.77 kip 2 2 805 in Σd

VT ,Y

3, 279k-in (1.75") T x = = = 7.128kip 2 2 Σd 805 in

The direct shear force is,

( 67.5 ) = 4.219 = kip

VD ,Y

VTotal =

16 bolts

( 42.77 ) + ( 7.128 kip 2

kip

kip bolt

+ 4.219kip

)

= 44.25kip

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

PTotal 728.7 kip = = 16.47 kip Pbolt 44.25 bolt

ODOT-LRFD Short Course - Steel AASHTO Connection Example #7

AASHTO-LRFD 2007 Created July 2007: Page 14 of 20 -- 203 --

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

2975 0

xCG:

Py:

67.5

yCG:

ex:

4.5

ey:

ÎŁd:

805.00

3278.75

Vmax =

44.2459

Bolt A1 B1 C1 D1 E1 F1 G1 H1 A2 B2 C2 D2 E2 F2 G2 H2

x -1.75 -1.75 -1.75 -1.75 -1.75 -1.75 -1.75 -1.75 1.75 1.75 1.75 1.75 1.75 1.75 1.75 1.75

y 10.50 7.50 4.50 1.50 -1.50 -4.50 -7.50 -10.50 10.50 7.50 4.50 1.50 -1.50 -4.50 -7.50 -10.50

113.31 59.31 23.31 5.31 5.31 23.31 59.31 113.31 113.31 59.31 23.31 5.31 5.31 23.31 59.31 113.31

Vtotal

42.7663 30.5474 18.3284 6.1095 -6.1095 -18.3284 -30.5474 -42.7663 42.7663 30.5474 18.3284 6.1095 -6.1095 -18.3284 -30.5474 -42.7663

-7.1277 -7.1277 -7.1277 -7.1277 -7.1277 -7.1277 -7.1277 -7.1277 7.1277 7.1277 7.1277 7.1277 7.1277 7.1277 7.1277 7.1277

42.8651 30.6856 18.5578 6.7667 6.7667 18.5578 30.6856 42.8651 44.2459 32.5866 21.5563 12.8867 12.8867 21.5563 32.5866 44.2459

ODOT-LRFD Short Course - Steel AASHTO Connection Example #7

AASHTO-LRFD 2007 Created July 2007: Page 15 of 20 -- 204 --

Check Flexural Yielding of the Web Splice Plates: Take Muw = 3,262k-in σ=

M y ⎡ 12 =⎢ 3 I ⎢ d p 2t p ⎣

( )( )

Solve for tp:

⎤⎛ M d p ⎥⎜ 2 ⎥⎝ ⎦

⎞ 3M ⎟ = 2 ≤ φFy ⎠ d pt p

(

)

( 3) 3, 262k-in 3M tp ≥ 2 = = 0.2787" d p φFy ( 26.5")2 (1.00 ) 50ksi

(

)

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

( )( 2t p )( Fy )

Vuw ≤ φVn = ( φ )( 0.58 ) d p

(

)

= (1.00 )( 0.58 )( 26.5") ⎡⎣( 2 )( 516 ") ⎤⎦ 50ksi = 480.3kip

Check Shear Rupture of the Web Splice Plates: Take Vuw = 67.5kip

(

Vuw ≤ φVn = ( φ )( 0.58 ) d p ,net

)( 2t p ) ( Fu )

(

)

= ( 0.80 )( 0.58 ) ⎡⎣( 26.5") − ( 8 )(1 1 8 ") ⎤⎦ ⎡⎣( 2 )( 516 ") ⎤⎦ 65ksi = 329.9kip

ODOT-LRFD Short Course - Steel AASHTO Connection Example #7

AASHTO-LRFD 2007 Created July 2007: Page 16 of 20 -- 205 --

Check Bearing of the Bolts in the Web Splice: Edge Bolts Bearing on the Beam Web: Lc = 2 1 2 "− ( 1 2 )(1"+ 116 ") = 1.969"

Rn = 1.2 Lc tFu

Since Lc = 1.969” < 2d = 2.0”,

(

)

kip Rn = (1.2) (1.969") (0.520") 65ksi = 79.85 bolt

(6.13.2.9-2)

Edge Bolts Bearing on the Splice Plates: Lc = 2 1 2 "− ( 1 2 )(1"+ 116 ") = 1.969"

Rn = 1.2 Lc tFu

Since Lc = 1.969” < 2d = 2.0”,

(

)

kip Rn = (1.2) (1.969")( 2 )( 516 ") 65ksi = 95.99 bolt

(6.13.2.9-2)

Summary of Splice Plate Bearing: Bearing on the beam web governs. kip Rn = 79.85 bolt

(

)

kip kip φRn = ( 0.80 ) 79.85 bolt = 63.88 bolt

ODOT-LRFD Short Course - Steel AASHTO Connection Example #7

AASHTO-LRFD 2007 Created July 2007: Page 17 of 20 -- 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. 2. 3. 4.

Vsw - The direct shear force. Mvsw - The moment on the web bolts caused by the eccentricity of Vsw Msw - The portion of the bending moment in the beam that is carried by the web. 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, e = ( 12 ) ( 3 1 2 ") + 2 1 2 "+ ( 12 ) ( 1 2 ") = 4.50"

(

M vsw = e Vsw = ( 4.50") 30.4kip

)

= 136.8k-in = 11.40k-ft

ODOT-LRFD Short Course - Steel AASHTO Connection Example #7

AASHTO-LRFD 2007 Created July 2007: Page 18 of 20 -- 207 --

3. Determine the portion of the beam moment that is carried by the web splice, Msw: M sw =

M sw

tw D 2 f s − f os 12

(C6.13.6.1.4b-1 mod)

⎡ ( 0.520")( 28.36")2 ⎤ =⎢ ⎥ 23.92ksi − −23.92ksi 12 ⎢⎣ ⎥⎦

(

) (

)

= 34.85 in 3 47.84ksi = 1, 667 k-in = 138.9k-ft

4. Determine the horizontal force that results from moving the beam moment, Hsw: H sw =

tw D f s + f os 12

(C6.13.6.1.4b-2 mod)

⎡ ( 0.520")( 28.36") ⎤ ksi ksi H sw = ⎢ ⎥ 23.92 + −23.92 12 ⎣ ⎦

(

) (

)

= 1.229 in 2 0.000ksi = 0.00kip

The total moment acting on the web splice is, M Total = M vsw + M sw = 136.8k-in + 1, 667 k-in = 1,804k-in

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, P=

1,804k-in = 400.8kip 4.50"

The largest bolt force in the web splice due to the Service II combination can be determined as, PBolt

PTotal 400.8kip kip = = = 24.34 bolt 16.47 C

kip that was computed on Page 8. This force is well below the slip capacity of 33.66 bolt

ODOT-LRFD Short Course - Steel AASHTO Connection Example #7

AASHTO-LRFD 2007 Created July 2007: Page 19 of 20 -- 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.

6 1/2"

Outer Flange Splice Plate: PL 241/2” x 101/2" x 3/8" Each Flange

3 1/2"

2 1/2"

3 1/2"

2 1/2"

Inner Flange Splice Plates: PL 241/2" x 4" x 1/2" Each Flange

1" dia M164 Bolts (12 Places Each Flange) W30 x 99

W30 x 99

2 3/4"

7 Spaces @ 3"

1" dia M164 Bolts (32 Places)

2 3/4"

Web Splice Plates: PL 261/2” x 171/2" x 5/16" Each Side of Web

2 1/2"

ODOT-LRFD Short Course - Steel AASHTO Connection Example #7

3 1/2"

2 1/2"

3 1/2"

2 1/2"

AASHTO-LRFD 2007 Created July 2007: Page 20 of 20 -- 209 --