LRFD Steel Design. AASHTO LRFD Bridge Design Specifications. Example Problems

Transcription

1 LRFD Steel Design AASHTO LRFD Bridge Design Specifications Example Problems Created July 007

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3 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, 007

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5 LRFD Steel Design AASHTO LRFD Bridge Design Specification Example Problems Case Study: -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 # Tension Member Example # Tension Member Example # Tension Member Example # Ad-Hoc Compression Member Examples Compression Member Example # Compression Member Example #...11 Compression Member Example # Compression Member Example # Compression Member Example # Compression Member Example # Compression Member Example # Ad-Hoc Flexural Member Examples Flexure Example # Flexure Example #...19 Flexure Example # Flexure Example # Flexure Example #5a Flexure Example #5b Flexure Example #6a Flexure Example #6b...15 Ad-Hoc Shear Strength Examples Shear Strength Example # Shear Strength Example #...161

6 Ad-Hoc Web Strength and Stiffener Examples Web Strength Example # Web Strength Example # Ad-Hoc Connection and Splice Examples Connection Example # Connection Example # Connection Example # Connection Example # Connection Example # Connection Example #6a Connection Example #6b Connection Example #

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

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9 1. PROBLEM STATEMENT AND ASSUMPTIONS: A two-span continuous composite I-girder bridge has two equal spans of 165 and a 4 deck width. The steel girders have F y = 50 and all concrete has a 8-day compressive strength of f c = 4.5. The concrete slab is 9 1 / thick. A typical ¾ haunch was used in the section properties. Concrete barriers weighing 640 plf and an asphalt wearing surface weighing 60 psf have also been applied as a composite dead load. HL-93 loading was used per AASHTO (004), including dynamic load allowance. 4' - 0" Out to Out of Deck 39' - 0" Roadway Width 9½ (typ) 3 / 4 " Haunch (typ) 3'-0" 3 1' - 0" = 36' - 0" 3'-0" References: Barth, K.E., Hartnagel, B.A., White, D.W., and Barker, M.G., 004, 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 - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 1 of

10 Positive Bending Section (Section 1) Negative Bending Section (Section ). LOAD CALCULATIONS: DC dead loads (structural components) include: Steel girder self weight (DC1) Concrete deck self weight (DC1) Haunch self weight (DC1) Barrier walls (DC) DW dead loads (structural attachments) include: Wearing surface (DW).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 ) + (1 )(1 ) = in W 490 pcf Lb = in ( 1.15) = 78.1 per girder ft in ( 1 ft ) section1 (b) Section (Negative Bending) A = (1 )(1 ) + (69 )(9/16 ) + (1 )(-1/ ) = 11.3 in W 490 pcf Lb = 11.3 in ( 1.15) = per girder ft in ( 1 ft ) section - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page of

11 Deck Self-Weight (DC1): W deck 150 pcf Lb = (9.5")(144") 1,45 = per girder ft in ( 1 ft ) Haunch Self-Weight (DC1): Average width of flange: 1"(66') + 15"(64') = 16." 66' + 64' Average width of haunch: ( ) ( ) 1 16." ()(9") + 16." 5." + = W haunch ( " )( 5." ) Lb = (150 pcf ) 5.5 = per girder ft in ( 1 ft ) Barrier Walls (DC): W barriers ( ) 30.0 Lb ft ( each) 640 plf = = per girder 4 girders Wearing Surface (DW): (39')(60 psf ) Lb W fws = = 585 per girder ft 4 girders 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 000. 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 DC (wearing surface) dead loads were applied to the long-term composite section (3n = 4). - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 3 of

12 Unfactored Dead Load Moment Diagrams from SAP 4,000 3,000 DC1,000 1,000 DW Moment (-ft) 0-1,000 -,000-3,000 DC -4,000-5,000-6,000-7,000-8, Station (ft) Unfactored Dead Load Shear Diagrams from SAP DC DW Shear () 0-50 DC Station (ft) - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 4 of

13 The following Dead Load results were obtained from the FE analysis: The maximum positive live-load moments occur at stations 58.7 and 71.3 The maximum negative live-load moments occur over the center support at station Max (+) Moment Stations 58.7 and 71.3 Max (-) Moment Station DC1 - Steel: 475 k-ft -1,189 k-ft DC1 - Deck:,415 k-ft -5,708 k-ft DC1 - Haunch: 89 k-ft -10 k-ft DC1 - Total:,979 k-ft -7,107 k-ft DC: 553 k-ft -1,51 k-ft DW 1,011 k-ft -,86 k-ft - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 5 of

14 .: 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 5 axles spaced 4.0 apart. The lane loading consists of a 0.64 klf uniform load on all spans of the bridge. (HL-93M in SAP) ) The effect of one design truck with variable axle spacing combined with the effect of the 0.64 klf 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 3 axles should be taken as 14 for each truck. The points of contraflexure were taken as the field splices at 13 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 000 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. - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 6 of

15 Unfactored Moving Load Moment Envelopes from SAP 6,000 4,000 Single Truck Tandem,000 Fatigue Moment (-ft) 0 -,000 Tandem Fatigue Contraflexure Point Contraflexure Point -4,000 Single Truck Two Trucks -6, Station (ft) Unfactored Moving Load Shear Envelopes from SAP Tandem Fatigue Single Truck Shear () Station (ft) - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 7 of

16 The following Live Load results were obtained from the SAP analysis: The maximum positive live-load moments occur at stations 73.3 and 56.7 The maximum negative live-load moments occur over the center support at station Max (+) Moment Stations 73.3 and 56 Max (-) Moment Station 165 HL-93M 3,75 k-ft -3,737 k-ft HL-93K 4,396 k-ft -4,61 k-ft HL-93S N/A -5,317 k-ft Fatigue,37 k-ft -1,095 k-ft Before proceeding, these live-load moments will be confirmed with an influence line analysis. - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 8 of

17 ..1: Verify the Maximum Positive Live-Load Moment at Station 73.3 : 5 5 Tandem: Single Truck: Lane: /ft Moment (k-ft / ) Station (ft) k-ft k-ft Tandem: ( )( ) ( )( ) = 1,603 k-ft k-ft k-ft k-ft Single Truck: ( )( ) ( )( ) ( )( ) =,108 k-ft Lane Load: ( )( ) = 0.640,491 1,594 k-ft k-ft ft k-ft k-ft k-ft (IM)(Tandem) + Lane: ( 1.33)( 1,603 ) + 1,594 = 3,76 k-ft k-ft k-ft (IM)(Single Truck) + Lane: ( 1.33)(,108 ) + 1,594 = 4,397 GOVERNS The case of two trucks is not considered here because it is only used when computing negative moments. - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 9 of

18 ..: Verify the Maximum Negative Live-Load Moment at Station : 5 5 Tandem: Single Truck: Two Trucks: Lane: /ft Moment (k-ft / ) Station (ft) k-ft k-ft Tandem: ( )( ) ( )( ) = 94.0 k-ft k-ft k-ft k-ft Single Truck: ( )( ) ( )( ) ( )( ) Two Trucks: = 1,318 k-ft ( 8 )( ) ( )( ) ( )( ) k-ft k-ft k-ft +... k-ft k-ft k-ft ( )( ) ( )( ) ( )( ) =,630 Lane Load: ( )( ) = k-ft ,918,508 k-ft k-ft ft k-ft k-ft k-ft (IM)(Tandem) + Lane: ( 1.33)( 94.0 ) +,508 = 3,737 k-ft k-ft k-ft (IM)(Single Truck) + Lane: ( 1.33)( 1,318 ) +,508 = 4,61 k-ft k-ft k-ft ,630,508 5,405 GOVERNS (0.90){(IM)(Two Trucks) + Lane}: ( ) ( )( ) + = - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 10 of

19 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 K g, the longitudinal stiffness parameter, we must first compute the sections properties of the girders..3: Braking Force The Breaking Force, BR, is taken as the maximum of: A) 5% of the Design Truck ( )( ) BR = = Single Lane B) 5% of the Design Tandem ( )( ) BR = = 1.50 Single Lane C) 5% of the Design Truck with the Lane Load. ( ) ( ) ( )( )( ft ) BRSingle Lane = ' = D) 5% of the Design Tandem with the Lane Load. ( ) ( ) ( )( )( ft ) BRSingle Lane = ' = Case (A) Governs: Net ( Single Lane )(# )( ) ( )( 3)( 0.85) BR = BR Lanes MPF = = This load has not been factored - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 11 of

20 .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. v C = f ( ) gr where: ft v - Highway design speed ( sec ) f - 4 / 3 for all load combinations except for Fatigue, in which case it is 1.0 ft g - The acceleration of gravity ( ) R - The radius of curvature for the traffic lane (ft). sec Suppose that we have a radius of R = 600 and a design speed of v = 65 mph = ft / sec. ft ( sec ) ft ( )( ) 4 C = = ' sec ( )( )(# )( ) ( 7 )( 0.67)( 3)( 0.85) 115. CE = Axle Loads C Lanes MPF = = This force has not been factored The centrifugal force acts horizontally in the direction pointing away from the center of curvature and at a height of 6 above the deck. Design the cross frames at the supports to carry this horizontal force into the bearings and design the bearings to resist the horizontal force and the resulting overturning moment. - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 1 of

21 .5: Wind Loads For the calculation of wind loads, assume that the bridge is located in the open country at an elevation of 40 above the ground. Take Z = 40 Open Country V o = 8.0 mph Z o = 0.3 ft Horizontal Wind Load on Structure: (WS) Design Pressure: P V = P = P DZ D B B VB V 10, 000 DZ mph ( ) P B - Base Pressure - For beams, P B = 50 psf when V B = 100 mph. (Table ) V B - Base Wind Velocity, typically taken as 100 mph. V 30 - Wind Velocity at an elevation of Z = 30 (mph) V DZ - Design Wind Velocity (mph) Design Wind Velocity: V Z = 30 VDZ.5Vo ln V B Z o ft mph = (.5)( 8.0 ) Ln = ft mph ( ) ( ) ( mph ) psf P = 50 = 55.9 D mph ( 10, 000 ) psf P D The height of exposure, h exp, for the finished bridge is computed as h exp h = 71.5" " + 4" = 15.3" = 10.44' exp The wind load per unit length of the bridge, W, is then computed as: psf ( )( ) W = ' = lbs ft WS = ' = 19.6 lbs Total Wind Load: ( )( )( ) H, Total ft lbs 1 H, Abt ft lbs 1 H, Pier ft WS = ' = For End Abutments: ( )( )( ) WS = ' = For Center Pier: ( )( )( )( ) - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 13 of

22 Vertical Wind Load on Structure: (WS) When no traffic is on the bridge, a vertical uplift (a line load) with a magnitude equal to 0 psf times the overall width of the structure, w, acts at the windward quarter point of the deck. PV ( )( w) ( )( ) = 0 = 0 4' = 840 psf psf lbs ft ' = 77. lbs Total Uplift: ( )( )( ) lbs 1 For End Abutments: ( )( )( ) ft ' = ft ' = lbs 1 For Center Pier: ( )( )( )( ) ft Wind Load on Live Load: (WL) The wind acting on live load is applied as a line load of 100 lbs/ft acting at a distance of 6 above the deck, as is shown below. This is applied along with the horizontal wind load on the structure but in the absence of the vertical wind load on the structure. WL P D - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 14 of

23 3. SECTION PROPERTIES AND CALCULATIONS: 3.1: Effective Flange Width, b eff : For an interior beam, b eff is the lesser of: Leff 13' = = 33' = 396" 4 4 bf 15" 1 t + = s (1)(8.5") + = 109.5" S = (1')(1 in ft ) = 144" For an exterior beam, b eff is the lesser of: Leff 13' = = 33' = 198.0" 4 4 bf 15" 1 t + = s (1)(8.5") + = 109.5" S 1' in + de = + 3' ( 1 ft ) = 108.0" Note that L eff was taken as 13.0 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 #, A s = 13.0 in located 4.5 from the top of the slab was taken from an underlying example problem first presented by Barth (004). - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 15 of

24 3.: Section 1 Flexural Properties Bare Steel t b A y Ay I x d Ad I X Top Flange ,78 17,79 Web , , ,301 Bot Flange ,15 19, , I Total = 53,157 Y = S BS1,top = 1,37 S BS1,bot = 1,733 Short-Term Composite (n = 8) t b A y Ay I x d Ad I X Slab , ,86 33,56 Haunch Top Flange ,669 1,670 Web , , ,988 35,387 Bot Flange ,900 69, , I Total = 140,51 n : 8.00 Y = S ST1,top = 11,191 S ST1,bot =,415 Long-Term Composite (n = 4) t b A y Ay I x d Ad I X Slab , ,885 3,119 Haunch Top Flange ,506 6,507 Web , , ,549 19,948 Bot Flange ,101 44, , I Total = 10,676 n : 4.00 Y = S LT1,top = 4,04 S LT1,bot =,16 Cracked Section t b A y Ay I x d Ad I X Rebar ,77 73,77 Top Flange ,717 55,718 Web , , ,913 64,31 Bot Flange ,159.8 I Total = 193,764 Y = S CR1,top = 5,84 S CR1,bot = 5,156 These section properties do NOT include the haunch or sacrificial wearing surface. - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 16 of

25 3.3: Section Flexural Properties Bare Steel t b A y Ay I x d Ad I X Top Flange , ,841 4,843 Web , , ,01 19,411 Bot Flange ,361 34, , I Total = 96,64 Y = 6.83 S BS,top =,116 S BS,bot = 3,60 Short Term Composite (n = 8) t b A y Ay I x d Ad I X Slab , ,941 70,641 Haunch Top Flange , ,07 8,08 Web , , ,005 4,403 Bot Flange , , , I Total = 39,734 n : 8.00 Y = 5.3 S ST,top = 11,88 S ST,bot = 4,590 Long-Term Composite (n = 4) t b A y Ay I x d Ad I X Slab , ,393 53,66 Haunch Top Flange , ,983 1,985 Web , , ,670 Bot Flange ,395 77, , I Total = 168,704 n : 4.00 Y = S LT,top = 5,135 S LT,bot = 4,55 Cracked Section t b A y Ay I x d Ad I X Rebar , ,313 6,313 Top Flange , ,55 33,57 Web , , ,35 Bot Flange ,786 49, ,016.3 I Total = 16,006 Y = 3.04 S CR,top = 3,115 S CR,bot = 3,93 These section properties do NOT include the haunch or sacrificial wearing surface. - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 17 of

26 4. DISTRIBUTION FACTOR FOR MOMENT 4.1: Positive Moment Region (Section 1): Interior Girder One Lane Loaded: DF DF S S K g M1, Int+ = L 1Lts Kg = n( I + Aeg) K K g g = 8(53,157 in + (71.06 in )(46.8") ) = 1,67, 000 in ' 1 ' 1, 67, 000 in = ' (1)(165')(8.5") M1, Int+ 3 DF 1, + = M Int 0.1 In these calculations, the terms e g and K g include the haunch and sacrificial wearing surface since doing so increases the resulting factor. Note that t s in the denominator of the final term excludes the sacrificial wearing surface since excluding it increases the resulting factor. Two or More Lanes Loaded: DF DF DF S S K g M, Int+ = L 1Lts M, Int+ 3 M, Int ' 1 ' 1,67, 000 in = ' 1(165')(8.5") = 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. 8.5 DF = M1, Ext+ 1 = Multiple Presence: DF M1,Ext+ = (1.) (0.7083) = Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 18 of

27 Two or More Lanes Loaded: DF M,Ext+ = e DF M,Int+ d e e = = = DF M,Ext+ = (0.9348) (0.7781) = : Negative Moment Region (Section ): 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 and = Interior Girder One Lane Loaded: DF DF S S K g M1, Int = L 1Lts Kg = n( I + Aeg) K K g g = 8(96, 64 in + (11.3 in )(5.17") ) = 3, 18, 000 in ' 1 ' 3, 18,000 in = ' (1)(165 ')(8.5") M1, Int 3 DF 1, = M Int 0.1 Two or More Lanes Loaded: DF DF DF S S K g M, Int = L 1Lts M, Int 3 M, Int ' 1 ' 3, 18, 000 in = ' (1)(165')(8.5") = Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 19 of

28 Exterior Girder One Lane Loaded: Same as for the positive moment section: DF M1,Ext- = Two or More Lanes Loaded: DF M,Ext- = e DF M,Int- d e = e = = DF M,Ext- = (0.9348) (0.857) = : Minimum Exterior Girder Distribution Factor: DF Ext, Min N L = + N b X Ext N b N L x e One Lane Loaded: 1 (18.0')(14.5') DF M1, Ext, Min= + = () (18 ') + (6 ') Multiple Presence: DF M1,Ext,Min = (1.) (0.615) = Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 0 of

29 Two Lanes Loaded: 14.5' ' 3'.5' 3' ' 3' 3' P 1 P DF M, Ext, Min (18.0')(14.5' +.5') = + = () (18') + (6 ') Lane 1 (1') Lane (1') Multiple Presence: DF M,Ext,Min = (1.0) (0.950) = ' 1' 6' Three Lanes Loaded: The case of three lanes loaded is not considered for the minimum exterior distribution factor since the third truck will be placed to the right of the center of gravity of the girders, which will stabilize the rigid body rotation effect resulting in a lower factor. 4.4: Moment Distribution Factor Summary Strength and Service Moment Distribution: Positive Moment Negative Moment Interior Exterior Interior Exterior 1 Lane Loaded: Lanes Loaded: For Simplicity, take the Moment Distribution Factor as 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 Since the multiple presence factor for 1-lane loaded is 1., these factors can be obtained by divided the first row of the table above by 1.. Positive Moment Negative Moment Interior Exterior Interior Exterior 1 Lane Loaded: For Simplicity, take the Moment Distribution Factor as everywhere for the Fatigue load combination Multiplying the live load moments by this distribution factor of yields the table of nominal girder moments shown on the following page. - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 1 of

30 Nominal Girder Moments for Design Nominal Moments Station (LL+IM)+ (LL+IM)- Fat+ Fat- DC1 DC DW (ft) (k-ft) (k-ft) (k-ft) (k-ft) (k-ft) (k-ft) (k-ft) Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page of

31 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 5.1: Interior Girder One Lane Loaded: S DF V 1,Int= ' = = Two or More Lanes Loaded: DF V,Int S S = ' 1' = 0. + = : Exterior Girder One Lane Loaded: Lever Rule, which is the same as for moment: DF V1,Ext = Two or More Lanes Loaded: DF V,Ext = e DF V,Int e = d e ' = = DF V,Ext = (0.7500) (1.08) = : Minimum Exterior Girder Distribution Factor - The minimum exterior girder distribution factor applies to shear as well as moment. DF V1,Ext,Min = DF V,Ext,Min = Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 3 of

32 5.4: Shear Distribution Factor Summary Strength and Service Shear Distribution: Shear Distribution Interior Exterior 1 Lane Loaded: Lanes Loaded: For Simplicity, take the Shear Distribution Factor as 1.08 everywhere for Strength and Service load combinations. Fatigue Shear Distribution: For Fatigue, the distribution factor is based on the one-lane-loaded situations with a multiple presence factor of Since the multiple presence factor for 1-lane loaded is 1., these factors can be obtained by divided the first row of the table above by 1.. Shear Distribution Interior Exterior 1 Lane Loaded: For Simplicity, take the Shear Distribution Factor as 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. - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 4 of

33 Nominal Girder Shears for Design Nominal Shears Station (LL+IM)+ (LL+IM)- Fat+ Fat- DC1 DC DW (ft) () () () () () () () Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 5 of

34 6. FACTORED SHEAR AND MOMENT ENVELOPES The following load combinations were considered in this example: Strength I: Strength IV: Service II: 1.75(LL + IM) + 1.5DC DC DW 1.50DC DC DW 1.3(LL + IM) + 1.0DC DC + 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 iqi η is taken as the product of η D, η R, and η I, and is taken as not less than For this example, η D and η I are taken as 1.00 while η R is taken as 1.05 since the bridge has 4 girders with a spacing greater than or equal to 1. 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 , Page 6-9 and b, Page 3-5). - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 6 of

35 Strength Limit Moment Envelopes 0,000 15,000 10,000 5,000 Strength IV Strength I Moment (-ft) 0-5,000-10,000 Strength IV -15,000 Strength I -0,000-5, Station (ft) Strength Limit Shear Force Envelope Strength I 400 Strength IV 00 Shear () Station (ft) - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 7 of

36 Service II Moment Envelope 1,500 10,000 7,500 5,000,500 Moment (-ft) 0 -,500-5,000-7,500-10,000-1,500-15,000-17,500-0, Station (ft) Service II Shear Envelope Shear () Station (ft) - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 8 of

37 Factored Fatigue Moment Envelope 1,500 1, Moment (-ft) ,000-1, Station (ft) Factored Fatigue Shear Envelope Shear () Station (ft) - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 9 of

38 Factored Girder Moments for Design Strength I Strength IV Service II Fatigue Station Total + Total - Total + Total - Total + Total - Total + Total - (ft) (k-ft) (k-ft) (k-ft) (k-ft) (k-ft) (k-ft) (k-ft) (k-ft) Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 30 of

39 Factored Girder Shears for Design Strength I Strength IV Service II Fatigue Station Total + Total - Total + Total - Total + Total - Total + Total - (ft) () () () () () () () () Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 31 of

40 7. FATIGUE CHECKS 7.1: Check transverse stiffener to flange weld at Station 73.3: Traffic information: ADTT given as 400. Three lanes are available to trucks. (ADTT) SL = (0.80) (,400) = 1,90 N = (ADTT) SL (365) (75) n = (1,90) (365) (75) (1) = 5.56M 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 f Fat Distance from bottom of section to the detail under investigation y = t f,bottom + D = = 70 k-ft M DC1 =, 779 k-ft M DC = (Pg 4) (Pg 16) k-ft in (,779 )( 1 ft )( 70" 30.68" ) f DC 1 4 = = ,157 in (Pg 16) (Pg 4) (Pg 16) k-ft in ( )( 1 ft )( 70" 46.33" ) = = ,676 in f DC 4 (Pg 16) f comp, DL = = 6.09 M Fat, Neg = 58.6 k-ft f Fat (Pg 30) (Pg 16) k-ft in ( 58.6 )( 1 ft )( 70" 58.19" ) = = ,51 in (Pg 16) Check?? fcomp, DL ffat 6.09 ( )( 0.61 ) 0.51 =, No. Fatigue need not be checked on the top flange at Station Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 3 of

41 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 ( F ) 1 3 A Δ = n N ( ΔF ) TH γ is a load factor of 0.75, which is already included in the fatigue moments. ( f ) k-ft in ( 1, 36 )( 1 ft )( 58.19" 1.00" ) γ Δ = = ,51 in The detail under consideration is a Category C detail. A = 44.0 x and (ΔF) TH = 1.0 ( ΔF ) TH = = The stress in the detail is almost less than the infinite life threshold A = = N Since 1 3 ( ΔF ) A = = N TH is less than 6.00, the infinite life governs. ( Δ F ) = 6.00 n Since ( f ) ( F) γ Δ = > = 6.00, the detail is not satisfactory. n - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 33 of

42 Calculate the design life of the part under consideration: Since γ ( Δ f ) is greater than ( ΔF ) TH, solve for N in the following equation. A γ ( Δf ) N A N = = 3 3 γ ( Δf ) ( ) cycles 1, cycles = 10,40 days 10, 40 days = 8.55 years ( = 8y, 6m, 19d, h, 38min... ) Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 34 of

43 8. CHECK CROSS_SECTION PROPORTION LIMITS Web Proportions D 69" 150 = O.K. t " w 9 16 Flange Proportions bf 15" 1 = O.K. t ()( ") f 3 4 bf 1" 1 = O.K. t ()(1") f bf 1" 1 = O.K. t ()( ") f 1 Check ODOT Criteria for Flange Width b f? D 69" +.5 1" +.5 = 14" 6 6 O.K. D 69" bf,min = = =11.50" O.K. 6 6 t = 1.1 t =(1.1)( ") " O.K. 9 5 f,min w 16 8 I yc ( 3 4 ")(15") = O.K. I 33 (1")(1") yt I 3 yc (.5")(1") = O.K. I 3 (1")(1") yt - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 35 of

44 9. CHECK SERVICE LIMIT STATE 9.1: Check Absolute Deflection of the Bridge: Section Section 1 The cross section of Section 1 that is used for computing deflections is shown above. The entire deck width is used (as opposed to just the effective width that was used earlier) and the haunch and sacrificial wearing surface have been neglected. AASHTO permits the use of the stiffness of parapets and structurally continuous railing but ODOT does not. The transformed width of the bridge deck is in ( 4' )( 1 ) w = = 8 ft ' 63.00" Using the bottom of the steel as a datum, the location of the CG of the deck can be found as: 8.5" y 1" 69" 3 c = " + = 75.00" The CG of this composite cross section is found as: Y ( 63" )( 8.5" ) i( 75.00" ) + ( 4)( in ) i( 30.68" ) ( 63" )( 8.5" ) + ( 4)( in ) = = 59.63" Now the moment of inertia of the section can be found as: ( 63" )( 8.5" ) 3 ( )( )[ ] ( )( ) ( )( )[ ] 4 Concrete + 63" 8.5" 75.00" 59.63" = 19,700 in Steel 4 53,160 in in 30.68" 59.63" = 450,900 in I 1, total = 580,600 in 4 580,600 in I = = 4 Girders 4 4 in 1 145,100 Girder - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 36 of

45 Section The cross section of Section that is used for computing deflections is shown above. The transformed width of the bridge deck is in ( )( ) 4' 1 w ' = ft = 63.00" 8 Using the bottom of the steel as a datum, the location of the CG of the deck can be found as: 8.5" y 1 c = " + 69" + 1" + = 76.75" The CG of this composite cross section is found as: Y ( 63" )( 8.5" ) i( 76.75" ) + ( 4)( 11.3 in ) i( 6.83" ) ( 63" )( 8.5" ) + ( 4)( 11.3 in ) = = 53.98" Now the moment of inertia of the section can be found as: ( 63" )( 8.5" ) 3 ( )( )[ ] ( )( ) ( )( )[ ] 4 Concrete + 63" 8.5" 76.75" 53.98" = 80,900 in Steel 4 96,640 in in 6.83" 53.98" = 717,700 in I total = 998,600 in 4 998,600 in I = = 4 Girders 4 4 in 49, 700 Girder - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 37 of

46 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 , the loading includes (1) the design truck alone and () the lane load with 5% 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 in 4 I = 49,700 in 4 I = 145,100 in 4 From the analysis: Deflection due to the Design Truck with Impact: Δ Truck =.44 Deflection due to the Design Lane Load: Δ Lane = These deflections are taken at Stations 79. and The model was broken into segments roughly 5 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) ( 4) Δ 1 = (.44" ) = 1.558" Governs ( 0.85)( 3) ( 4) Δ = ( 0.844" ) + ( 0.5)(.44" ) = 0.974" The limiting deflection for this bridge is: in ( )( ) L 165' 1 Δ ft Limit = = =.475" OK Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 38 of

47 9.: Check the Maximum Span-to-Depth Ratio: Section From Table , (1) the overall depth of a composite I-beam in a continuous span must not be less than 0.03L and () the depth of the steel in a composite I-beam in a continuous span must not be less than 0.07L. in () L ( )( )( ft ) in ( ) L ( )( )( ) = ' 1 = 63.36" OK 0.07 = ' 1 = 53.46" OK ft - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 39 of

48 9.3: Permanent Deformations - Section 1 At the Service Limit State, the following shall be satisfied for composite sections Top Flange: f 0.95RF f h yf fl Bottom Flange f RF f h yf Per a, 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. f c + M M M M = + + DC1 DC DW LL+ IM S S S S BS LT LT ST Top Flange, Positive Moment It is not immediately evident to me whether the factored stress at 58.7 or 73.3 will govern. k-ft in k-ft in k-ft in k-ft in (, 979 )(1 ) (549.7 )(1 ) (1, 008 )(1 ) (3, 999 )(1 ) f = f 1,37 in 4,04 in 4,04 in 11,191 in ft ft ft ft, f = f, k-ft in k-ft in k-ft in k-ft in (, 779 )(1 ) (515.8 )(1 ) (946.1 )(1 ) (4, 067 )(1 ) f = f 1,37 in 4,04 in 4,04 in 11,191 in ft ft ft ft, f = f, The stress at 58.7 governs. f f = f R F f 0.95 h yf (0.95)(1.00)(50 ) = O.K. Note: The bending moments in the above calculations come from page while the moments of inertia are found on page Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 40 of

49 Bottom Flange, Positive Moment k-ft in k-ft in k-ft in k-ft in (, 979 )(1 ) (549.7 )(1 ) (1, 008 )(1 ) (3, 999 )(1 ) ft ft ft ft + f = f 1,733 in,16 in,16 in,415 in, f = f, k-ft in k-ft in k-ft in k-ft in (, 779 )(1 ) (515.8 )(1 ) (946.1 )(1 ) (4, 067 )(1 ) ft ft ft ft + f = f 1,733 in,16 in,16 in,415 in, f = f, The stress at 58.7 governs. f f = The load factor for wind under Service II is 0.00, f l = 0 fl 0 f f RhFyf (0.95)(1.00)(50 ) = No Good. Note: The bending moments in the above calculations come from page while the moments of inertia are found on page Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 41 of

50 9.4: Permanent Deformations - Section Top Flange, Negative Moment k-ft in k-ft in k-ft in k-ft in (7,109 )(1 ) (1, 50 )(1 ) (, 9 )(1 ) (4, 918 )(1 ) ft ft ft ft + f = f,116 in 5,135 in 5,135 in 11,88 in, f = f,165 f? 0.95 R F (0.95)(1.00)(50 ) = No Good. f h yf Bottom Flange, Negative Moment k-ft in k-ft in k-ft in k-ft in (7,108 )(1 ft ) (1, 50 )(1 ft ) (, 9 )(1 ft ) (4, 918 )(1 ft ) f = f 3,60 in 4,55 in 4,55 in 4,590 in f = f The load factor for wind under Service II is 0.00, f l = 0 f fl R F (0.95)(1.00)(50 ) = No Good. f h yf - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 4 of

51 9.5: Bend Buckling Checks At the Service Limit State, all sections except composite sections in positive flexure shall satisfy: f c F crw where: F crw 0.9Ek = D tw and k = 9 ( D / D) c Section 1 Not Applicable Section k-ft in k-ft in k-ft in k-ft (7,108 )(1 ) in ft (1, 50 )(1 ) ft (, 9 )(1 ) (4, 918 )(1 ) ft ft f = c 3,60 in 4,55 in 4,55 in 4,590 in f = c k-ft in k-ft in k-ft in k-ft (7,108 )(1 ) in ft (1, 50 )(1 ) ft (, 9 )(1 ) (4, 918 )(1 ) ft ft f = t,116 in 5,135 in 5,135 in 11,88 in f = t f c Dc = 0 d tcf fc f + t = ( 7.5" ).5" = k = = = ( D / D) 3.14" c 69" (0.90)(9,000 )(41.49) F = = crw 69" 9 16 " This is larger than f c O.K. - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 43 of

52 10. CHECK STRENGTH LIMIT STATE 10.1: Section 1 Positive Flexure Section Classification ( , Pg ) Check D t w cp 3.76 E F yc Find D cp, the depth of the web in compression at M p (compression rebar in the slab is ignored). ( )( ) P = F bt = (50 ) 1" 1" = 1,050 t yt t t P = F Dt = (50 )(69")( ") = 1,941 w yw w P = F b t = (50 )(15")( ") = 56.5 c s yc c c ' c s s P = 0.85 f b t = (0.85)(4.5 )(109.5")(8.5") = 3,560 Since P t + P w +P c < P s 3,554 < 3,560, the PNA lies in the slab. Y Y P + P + P 3,554 Ps 3,560 c w t ( t ) ( 8.5" ) = s = = 8.486" from top of slab D = Y = 8.486" p Since none of the web is in compression, D cp = 0 and the web is compact. For Composite Sections in Positive Flexure, ( , Pg ) 1 M u + fs l xt φ fmn M u = 13,568 k-ft from Page 30; take f l = 0 3 D t = / = D t = 7.95 (The haunch is not included in D t, as per ODOT Exceptions) Dp Since D p =8.486 > 0.1D t = 7.95, Mn = M p Dt M p 8.486" = (3,554 ) 79.5" 30.68" k-in = 157,500 = 13,130 k-ft M n k-ft 8.486" = ( 13,130 ) 1.07 ( 0.7) = 13, " k-ft - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 44 of

53 ?? 1 k-ft k-ft u 3 l xt φ f n (13,568 ) (0) (1.00)(13,060 ) M + f S M + No Good. Note that the check of M n 1.3RM h y has not been made in the above calculations. This section would satisfy the Article B6. so this check doesn t need to be made. Check the ductility requirement to prevent crushing of the slab: p?? t ( )( ) D 0.4 D 8.486" " = 33.9" O.K. The Section is NOT Adequate for Positive Flexure at Stations 58.7 and 71.3 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 70 steel in the top and bottom flanges. - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 45 of

54 Hybrid Girder Factors Will Now be Required: Compute the Hybrid Girder Factor, R h, for Section 1: Per AASHTO Commentary Pg 6-95, D n shall be taken for the bottom flange since this is a composite section in positive flexure. D n, Bottom = 58.19" 1" = 57.19" R h 3 ( ) 1 + β 3ρ ρ = 1 + β 9 ( )( 16 ) ( 1" )( 1" ) Dt () 57.19" " n w β = = = A fn R h, Section 1 ( ) Fyw 50 ρ = 1.0 ρ = = f (3)(0.7143) (0.7143) = = ( )( ) n Compute the Hybrid Girder Factor, R h, for Section : For the short-term composite section, D = + = 1 ntop, " 69" 5.3" 19.7" D = = Governs 1 nbottom, 5.3" " 49.73" R h 3 ( ) 1 + β 3ρ ρ = 1 + β 9 ( )( 16 ) ( 1 ")( 1" ) Dt () 49.73" " n w β = = = A fn R h, Section ( ) Fyw 50 ρ = 1.0 ρ = = f (3)(0.7143) (0.7143) = = ( )( ) n - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 46 of

55 11. RECHECK SERVICE LIMIT STATE WITH 70 KSI FLANGES 11.1: Permanent Deformations - Section 1 At the Service Limit State, the following shall be satisfied for composite sections Top Flange: f 0.95RF f h yf fl Bottom Flange f RF f h yf Top Flange, Positive Moment From before: f = f, ? f 0.95R F (0.95)(0.966)(70 ) = O.K. f h yf Bottom Flange, Positive Moment f = The load factor for wind under Service II is 0.00, f l = 0 f, f? l 0 f f RhFyf (0.95)(0.966)(70 ) = O.K. 11.: Permanent Deformations - Section Top Flange, Negative Moment From before: f = f,165 f? 0.95 R F (0.95)(0.9833)(70 ) = O.K. f h yf Bottom Flange, Negative Moment f = The load factor for wind under Service II is 0.00, f f l = 0 f f? l R F (0.95)(0.9833)(70 ) = O.K. f h yf - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 47 of

56 11.3: Bend Buckling Checks At the Service Limit State, all sections except composite sections in positive flexure shall satisfy: f c F crw where: F crw 0.9Ek = D tw and k = 9 ( D / D) c Section 1 - Not Applicable Section k-ft in k-ft in k-ft in k-ft (7,108 )(1 ) in ft (1, 50 )(1 ) ft (, 9 )(1 ) (4, 918 )(1 ) ft ft f = c 3,60 in 4,55 in 4,55 in 4,590 in f = c k-ft in k-ft in k-ft in k-ft (7,108 )(1 ) in ft (1, 50 )(1 ) ft (, 9 )(1 ) (4, 918 )(1 ) ft ft f = t,116 in 5,135 in 5,135 in 11,88 in f = t f c Dc = d tcf 0 fc f + t = ( 7.5" ).5" = k = = = ( D / D) 3.14" c 69" (0.90)(9,000 )(41.49) F = = crw 69" 9 16 " This is larger than f c O.K. - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 48 of

57 1. RECHECK STRENGTH LIMIT STATE WITH 70 KSI FLANGES 1.1: Section 1 - Positive Flexure Section Classification ( , Pg ) Check D t w cp 3.76 E F yc Find D cp, the depth of the web in compression at M p (compression rebar in the slab is ignored). ( )( ) P = F bt = (70 ) 1" 1" = 1, 470 t yt t t P = F Dt = (50 )(69")( ") = 1,941 w yw w P = F b t = (70 )(15")( ") = c s yc c c ' c s s P = 0.85 f b t = (0.85)(4.5 )(109.5")(8.5") = 3,560 Since P t + P w +P c > P s 4,199 > 3,560, the PNA is NOT in the slab. Check Case I? P + P P + P t w c s? 1, , ,560 NO Check Case II P + P + P P t w c s?? 1, , ,560 YES - PNA in Top Flange Y tc Pw + Pt P s = 1 + Pc 0.750" 1, , 470 3,560 = + 1 = " (from the top of steel) D p = = Since none of the web is in compression, D cp = 0 and the web is compact. For Composite Sections in Positive Flexure, ( , Pg ) 1 M u + fs l xt φ fmn M u = 13,568 k-ft from Page 30; take f l = 0 3 D t = / = D t = Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 49 of

58 (The haunch is not included in D t, as per ODOT Exceptions) Dp Since D p = > 0.1D t = 7.95, Mn = M p Dt Determine M p : The distances from the component forces to the PNA are calculated. d d d s w t 8.5" = " = 4.554" 69" = ( 0.75" " ) = 34.05" 1" = 70.75" " = 69.95" The plastic moment is computed. M P = Y + t Y + Pd + P d + Pd ( in ) ( ) [ ] c p c s s w w t t tc ( ) ( ) = " " " +... ()(0.750") ( )( ) ( )( ) ( )( ) , " + 1, " + 1, " k-in = in + 185,100 = 185,300 k-in k-ft =15,440 M n k-ft 8.804" = ( 15,440 ) 1.07 ( 0.7) = 15, " k-ft?? 1 k-ft k-ft u 3 l xt φ f n (13,568 ) (0) (1.00)(15,30 ) M + f S M + O.K. Note that the check of M n 1.3RM h y has not been made in the above calculations. This section would satisfy the Article B6. so this check doesn t need to be made. Check the ductility requirement to prevent crushing of the slab: p?? t ( )( ) D 0.4 D 8.804" " = 33.9" O.K. The Section is Adequate for Positive Flexure at Stations 58.7 and 71.3 with 70 Flanges - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 50 of

59 1.: Section - Negative Flexure Section Classification ( , Pg ) Check D t w c 5.70 E F yc D c is the depth of the web in compression for the cracked section. D c = / = 9.54 Dc ()(9.54") E 9,000 = = < 5.70 = 5.70 = t ( ") F 50 w 9 16 yc 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 , which will provide a somewhat conservative determination of the flexural resistance. For Composite Sections in Negative Flexure, ( , Pg ) The Compression Flange must satisfy: 1 fbu + fl φ f F 3 nc Per a, 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 c, 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. f bu M 1.5M M M DC1 DC DW LL = S S BS CR f bu k-ft k-ft k-ft k-ft in in (7,109 )(1 (1.5)(1, 50 ) (1.50)(, 9 ) (1.75)(4, 918 ) (1 ft ) + + ft ) = ,60 in 3,93 in f = bu Since f bu is greater than F yc, it is obvious that a strength computed based on the provisions in will not be adequate. - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 51 of

60 As it stands here, this girder is clearly not adequate over the pier. The compression flange is overstressed as per the provisions in There are still other options to explore, though, before increasing the plate dimensions. 1. Since the web is non-slender for Section 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 M p instead of M y as was determined in 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 ~,000 k-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 50, 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. - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 5 of

61 1.3 Vertical Shear Capacity At the strength Limit, the following must be satisfied V u φv n For an unstiffened web, Vn = Vcr = CVp D Ek Check, 1.1, t F w yw D 69" = = 1.7 t " w 9 16 Since there are no transverse stiffeners, k = 5 (9, 000 )(5) (50 ) 1.1 = (9,000 )(5) (50 ) 1.40 = D Ek Since > 1.40, elastic shear buckling of the web controls. t F w yw 1.57 ke 1.57 (5)(9, 000 ) C = = = D F yw 69" (50 ) 9 t 16 " w V 9 p = 0.58 FywDtw = (0.58)(50 )(69")( 16 ") = 1,16 V n = CV = = p (0.306)(1,16 ) ( )( ) φ V n = = No Good. This strength is adequate from and This strength is not adequate near the end supports or near the pier, however. - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 53 of

62 Try adding transverse stiffeners spaced at d o = 8 = k = 5+ = 5+ = do 96" D 69" D = 1.7, t w (9,000 )(7.583) (50 ) 1.1 = 74.8, (9,000 )(7.583) (50 ) 1.40 = 9.85 D Ek Since > 1.40, elastic shear buckling of the web controls. t F w yw 1.57 Ek 1.57 (9, 000 )(7.583) C = = = D F yw 69" (50 ) 9 t 16 " w φ Vn =φ CVp = (1.00)(0.4589)(1,16 ) = 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 d o = 1 = k = 5+ = 5+ = do 144" D 69" D = 1.7, t w (9,000 )(6.148) (50 ) 1.1 = 66.88, (9,000 )(6.148) (50 ) 1.40 = D Ek Since > 1.40, elastic shear buckling of the web controls. t F w yw 1.57 Ek 1.57 (9, 000 )(6.148) C = = = D F yw 69" (50 ) 9 t 16 " w Without TFA: V n = CV = = p (0.371)(1,16 ) Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 54 of

63 With TFA: Since Dt ()(69")( 9 w 16 ") = = , 1 ( b ) ((1")( ") (1")(1") fctfc + bftt + ft ) 0.87(1 C) (0.87)( ) Vn = Vp C+ (1,16 ) = + do 144" D 69" V = = n (1,16 )(0.608) ( )( ) φ V n = = 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. - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 55 of

64 Strength Limit Shear Capacity Tension Field Action Strength I 400 Strength IV 00 Shear () Station (ft) - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 56 of

65 1.4: 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. ts h= + thaunch tflange 9.5" = +.75" 0.75" = 6.75" t s t c b c b e t haunch Use 7 / 8 x 6 1 / shear studs AASHTO requires that the ratio of h / d be greater than or equal to 4.0. h? d 6 1 " = 7 8 " OK AASHTO requires a center-to-center transverse spacing of 4d and a clear edge distance of 1. t t D b t t w 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 nz p V sr r V sr VQ f = ( & 3) I where: n - Number of studs per row Z r - Fatigue resistance of a single stud V sr - Horizontal fatigue shear range per unit length V f - Vertical shear force under fatigue load combination Q - 1 st moment of inertia of the transformed slab about the short-term NA I - nd moment of inertia of the short-term composite section - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 57 of

66 Z r 5.5d = αd ( ) α = Log( N) ( ) 6 α = Log( ) = Z r 5.5 = ( )( 8" ) ( 8" ) = Z =.105 r Q= A d tc c Q Section 1 Q Section ( 109.5" )( 9.5" ) 9.5" 3 = 1" + 69" +.75" " =,511 in 8 ( 109.5" )( 9.5" ) 9.5" 3 =.5" + 69" +.75" + 5.3" = 3, 481 in 8 I Section 1 = 140,500 in 4 I Section = 39,700 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. V = = At the abutment, f ( ) 3 ( )(,511 in ) 4 ( 140,500 in ) V = = sr inch For 3 studs in each row: ( 3)(.105 ) ( inch ) p = 8.48 in row For 4 studs in each row: ( 4)(.105 ) ( inch ) p = in row - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 58 of

67 Strength Limit: Q = φ Q φ = 0.85 r sc n sc Q = 0.5A f E A F ( ) ' n sc c c sc u A sc π = = 4 7 ( ) 8 " in f = ' c 4.5 Since n = 8, E c Es 9, 000 = = = 3,65 n 8 F u = 60 Q = n ( 0.5)( in ) ( 4.5 )( 3,65 ) ( in )( 60 ) = stud stud ( )( stud ) φ Q = = sc n stud n + = P Q p r n P p + = Q r P n - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 59 of

68 Positive moment - Section 1: Station (, ) P = Min P P p Concrete steel P P = 3,979 p = 0.85 f bt = " 9.5" = 3,979 ' Concrete c e s ( )( )( )( ) PSteel = FywDtw + Fftbfttft + Ffcbfctfc ( ) ( )( ) ( )( ) ( )( )( ) = 70 15" 0.75" + 1" 1" " 0.565" = 4,198 Pp 3,979 n + = = = 19.7 Q r 3 Studs per Row: studs row stud studs in ( 73.3' 0' )( 1 ft ) ( ) studs 19.7 = = = rows inch 44 p 0.46 row Say 0" 4 Studs per Row: studs row in ( 73.3' 0' )( 1 ft ) ( ) studs 19.7 = = = rows inch 33 p 7.49 row Say 4" - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 60 of

69 Negative Moment - Section : Station (, ) P = Min P P n steel Crack P P n =,107 = 0.45 f bt = " 9.5" =,107 ' Crack c e s ( )( )( )( ) PSteel = FywDtw + Fftbfttft + Ffcbfctfc ( ) ( )( ) ( )( ) ( )( )( ) = 70 1".5" + 1" 1" " 0.565" = 7, 086 Pp + P n 3,979 +,107 n = = = Q Studs per Row: studs row r stud studs in ( 165.0' 73.3' )( 1 ft ) ( ) studs = = = rows inch 67 p row Say 16" 4 Studs per Row: studs row in ( 165.0' 73.3' )( 1 ft ) ( ) studs = = = rows inch 50 p.48 row Say 0" - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 61 of

70 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 4 Studs Per Row Station V f Q I V sr p Fat p Str p max p Fat p Str p max (ft) () (in 3 ) (in 4 ) ( / in ) (in) (in) (in) (in) (in) (in) , , , , , , , , , , , , , , , , , , , , ,481 39, ,481 39, ,481 39, ,481 39, ,481 39, ,481 39, ,481 39, ,481 39, ,481 39, ,481 39, ,481 39, ,481 39, ,481 39, ,481 39, ,481 39, ,481 39, ,481 39, , , , , , , , , , , , , , , , , , , , , The arrangement of shear studs is shown below. - Span Continuous Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Page 6 of

71 ONE-SPAN INELASTIC I-GIRDER BRIDGE DESIGN EXAMPLE 1. PROBLEM STATEMENT AND ASSUMPTIONS: A single span composite I-girder bridge has span length of and a 64 deck width. The steel girders have F y = 50 and all concrete has a 8-day compressive strength of f c = 4.5. The concrete slab is 9.5 thick. A typical 4 haunch was used in the section properties. Concrete barriers weighing 640 plf and an asphalt wearing surface weighing 60 psf have also been applied as a composite dead load. HL-93 loading was used per AASHTO (004), including dynamic load allowance. Single-Span Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Created July 007: Page 1 of

72 G 1 G G 3 G 4 G 5 G 6 Cross Frames ' - 0" cc 166' - 4" cc Bearings 17' - 4" Total Girder Length Single-Span Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Created July 007: Page of

73 Positive Bending Section (Section 1) Positive Bending Section (Section ) Positive Bending Section (Section 3) Single-Span Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Created July007: Page 3 of

74 . LOAD CALCULATIONS: DC dead loads (structural components) include: Steel girder self weight (DC1) Concrete deck self weight (DC1) Haunch self weight (DC1) Barrier (DC) DW dead loads (structural attachments) include: Wearing surface (DW), Including FWS a. Dead Load Calculations Steel Girder Self-Weight (DC1): (a) Section 1 A = (14 )(1.15 ) + (68 )( ) + ( )(1.5 ) = 95.5 in W section1 (b) Section 490 = 95.5 in 1.15 = pcf lbs ( ) ft in ( 1 ft ) per girder A = (14 )( ) + (68 )(0.565 ) + ( )( ) = in 490 W section1 = in 1.15 = in ( 1 ft ) (c) Section 3 pcf lbs ( ) ft per girder A = (14 )( ) + (68 )(0.565 ) + ( )(.375 ) = in W section1 490 = in 1.15 = in ( 1 ft ) pcf lbs ( ) ft per girder (d) Average Girder Self Weight W ave ( )( ')( lbs ) + ( )( 18.0' )( lbs ) + ( 50.0' )( lbs ) ft ft ft lbs ft = = 166.3' Deck Self-Weight (DC1): W Deck ( )( ) 9.5'' 64.0' 150 = = 1,67 6 Girders 1 pcf in ( ) ft lbs ft per girder Single-Span Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Created July007: Page 4 of

75 Haunch Self-Weight (DC1): Average width of haunch: W haunch = (( 14)( 4) + ( ( )( 9'' )( 4'' ))) = in ( 1 ft ) pcf lbs ft per girder Barrier Walls (DC): W barriers plf ( each)( 640 ) = = girders lbs ft per girder Wearing Surface (DW): psf ( 61.0' )( 60 ) lbs W wearing_surface = = per girder ft 6 Girders 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 000. 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 DC (wearing surface) dead loads were applied to the long-term composite section (3n = 4). The maximum moments at mid-span are easily computed since the bridge is statically determinate. M M DC1, Steel DC1, Deck lbs ( )( 166.3' ) wl ft = = = 1, lbs ( 1, 67 )( 166.3' ) wl ft = = = 4, k-ft k-ft M DC, Barriers lbs ( 13.3 )( 166.3' ) wl ft = = = k-ft M DW lbs ( )( 166.3' ) wl ft = = =, k-ft Single-Span Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Created July007: Page 5 of

76 The maximum shear forces at the ends of the girder are also easily computed. V V DC1, Steel DC1, Deck ( lbs )( 166.3' ) wl ft = = = ( 1, 67 lbs )( 166.3' ) wl ft = = = t V DC, Barriers ( 13.3 lbs )( 166.3' ) wl ft = = = V DW ( lbs )( 166.3' ) wl ft = = = 50.7 Single-Span Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Created July007: Page 6 of

77 b. 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 5 axles spaced 4.0 apart. The lane loading consists of a 0.64 klf uniform load on all spans of the bridge. (HL-93M in SAP) ) The effect of one design truck with variable axle spacing combined with the effect of the 0.64 klf 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 3 axles should be taken as 14 for each truck. The points of contraflexure were taken as the field splices at 13 and 198 from the left end of the bridge. (HL-93S in SAP) All live load calculations were performed in SAP 000 using a beam line analysis. The nominal moment data from SAP was then input into Excel. An Impact Factor of 1.33 was applied to the truck and tandem loads within SAP. Single-Span Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Created July007: Page 7 of

78 Unfactored HL-93 Moment Envelopes from SAP 6,000 Single Truck 4,000 Tandem,000 Moment (-ft) 0 -,000-4,000-6, Station (ft) The following results were obtained from the SAP analysis: The maximum positive live-load moments occur at stations Station Section 1 Station Section Station Section 3 HL-93M 3,614 k-ft 4,481 k-ft 4,911 k-ft HL-93K 4,3 k-ft 5,38 k-ft 5,81 k-ft HL-93S N/A N/A N/A Before proceeding, these live-load moments will be confirmed with an influence line analysis. Single-Span Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Created July007: Page 8 of

79 c. Verify the Maximum Positive Live-Load Moment at Station : 5 5 Tandem : Single Truck : Lane /ft Moment (k-ft / ) Station (ft) Tandem: k-ft k-ft ( )( ) ( )( ) k-ft =,09 Single Truck: k-ft k-ft k-ft ( )( ) ( )( ) ( )( ) k-ft =,713 Lane Load: k-ft k-ft k-ft ( )( ) ,457 =,1 (IM)(Tandem) + Lane: k-ft k-ft k-ft ( )( ) 1.33,09 +, 1 = 4,911 (IM)(Single Truck) + Lane: k-ft k-ft k-ft ( )( ) 1.33,713 +,1 = 5,81 GOVERNS The case of two trucks is not considered here because it is only used when computing negative moments. Single-Span Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Created July007: Page 9 of

80 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 K g, the longitudinal stiffness parameter, we must first compute the sections properties of the girders. 3. SECTION PROPERTIES AND CALCULATIONS: 3a. Effective Flange Width, b s : For an interior beam, b s is the lesser of: bf 14" 1ts + = ( 1)( 8.5" ) + = 109'' in S = ( 11.33' )( 1 ft ) = '' Leff 166.3' = = 41.58' = 498.9'' 4 4 Therefore, b s = 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 AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Created July007: Page 10 of

81 3b. Section 1 Flexural Properties Bare Steel Single Span Bridge Example - Section 1 t b A y Ay I x d Ad I X Top Flange , ,308 6,310 Web , , ,860 19,874 Bot Flange ,696 6, , I Total = 7,886 Y = 9.19 S BS1,top = 1,759 S BS1,bot =,497 Short-Term Composite (N=8) t b A y Ay I x d Ad I X Slab , ,365 50,06 Haunch Top Flange , ,948 3,950 Web , , ,399 34,414 Bot Flange ,381 94, ,459.3 I Total = 18,813 n = 8.00 Y = 54.3 S ST1,top = 11,150 S ST1,bot = 3,371 Long-Term Composite (N=4) t b A y Ay I x d Ad I X Slab , ,856 41,089 Haunch Top Flange , ,10 1,104 Web , , ,189 0,03 Bot Flange ,089 57, , I Total = 130,491 n = 4.00 Y = 4.34 S LT1,top = 4,614 S LT1,bot = 3,08 Single-Span Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Created July007: Page 11 of

82 3c. Section Flexural Properties Bare Steel Single Span Bridge Example - Section t b A y Ay I x d Ad I X Top Flange , ,978 44,987 Web , , ,76 Bot Flange ,391 39, , I Total = 100,119 Y = 30.9 S BS1,top =,437 S BS1,bot = 3,38 Short-Term Composite (N=8) t b A y Ay I x d Ad I X Slab , ,600 57,97 Haunch Top Flange , ,956 7,966 Web , , ,591 7,330 Bot Flange ,64 14, ,39.70 I Total = 16,871 n = 8.00 Y = S ST1,top = 1,145 S ST1,bot = 4,006 Long-Term Composite (N=4) t b A y Ay I x d Ad I X Slab , ,514 43,746 Haunch Top Flange , ,46,47 Web , , ,705 16,444 Bot Flange ,45 76, ,35.57 I Total = 159,101 n = 4.00 Y = 4.68 S LT1,top = 5,46 S LT1,bot = 3,78 Single-Span Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Created July007: Page 1 of

83 3d. Section 3 Flexural Properties Bare Steel Single Span Bridge Example - Section 3 t b A y Ay I x d Ad I X Top Flange , ,970 49,980 Web , , ,008 16,747 Bot Flange ,796 40, , I Total = 107,546 Y = 9.13 S BS1,top =,487 S BS1,bot = 3,69 Short-Term Composite (N=8) t b A y Ay I x d Ad I X Slab , ,819 67,516 Haunch Top Flange , ,865 9,874 Web , , ,076 4,815 Bot Flange , , ,36.0 I Total = 40,366 n = 8.00 Y = 5.61 S ST1,top = 1,158 S ST1,bot = 4,569 Long-Term Composite (N=4) t b A y Ay I x d Ad I X Slab , ,544 49,777 Haunch Top Flange , ,174 6,184 Web , , ,488 Bot Flange ,990 8, , I Total = 173,463 n = 4.00 Y = S LT1,top = 5,494 S LT1,bot = 4,51 Single-Span Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Created July007: Page 13 of

84 4. DISTRIBUTION FACTOR FOR MOMENT 4a. Section 1: Interior Girder - One Lane Loaded: DF g g g M1,Int,Sec S S K g = L 1LtS ( g) K = n I + Ae K K DF 4 ( ( )( ) ) = (8) 7,890 in in 49.06" =,4,000 in M1,Int,Sec1 DF M1,Int,Sec ' 11.33', 4,000 in = ' ' 8.5" = ( )( )( ) Interior Girder - Two or More Lanes Loaded: DF DF DF M,Int,Sec1 M,Int,Sec1 M,Int,Sec S S K g = L 1LtS ' 11.33', 4, 000 in = ' ( 1)( 166.3' )( 8.5" ) = Single-Span Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Created July007: Page 14 of

85 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. 85'. DF M 1,Ext,Sec1 = = ' Multiple Presence: DF M1,Ext,Sec1 = (1.) (0.7500) = Exterior Girder - Two or More Lanes Loaded: DF M,Ext,Sec1 = e DF M,Int,Sec1 d e e = ' e = = DF M,Ext+ = (1.008) (0.7703) = b. Section : Interior Girder One Lane Loaded: DF g g g M1,Int,Sec S S K g = L 1LtS ( g) K = n I + Ae K K DF 4 ( ( )( ) ) = (8) 100,100 in in 47.83" =,819, 000 in M1,Int,Sec DF M1,Int,Sec ' 11.33',819,000 in = = ' ( 1)( 166.3' )( 8.5" ) Single-Span Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Created July007: Page 15 of

86 Interior Girder Two or More Lanes Loaded: DF DF DF M,Int,Sec M,Int,Sec M,Int,Sec S S K g = L 1LtS ' 11.33',819,000 in = ' ( 1)( 166.3' )( 8.5" ) = Exterior Girder - One Lane Loaded: Same as for the positive moment section: DF M1,Ext,Sec = Exterior Girder - Two or More Lanes Loaded: DF M,Ext,Sec = e DF M,Int,Sec e = (same as before) DF M,Ext,Sec =(1.008) (0.7809) = c. Section 3: Interior Girder One Lane Loaded: DF g g g S S Kg M1, Int, Sec L 1LtS ( g) K = n I + Ae K K ( ) 4 ( 8) 107,500 in ( in )( 50.00" ) = 3,30,000 in = + = ' 11.33' 3, 30, 000 in DF = ' ( 1)( 166.3' )( 8.5" ) 1,, 3 = 0.51 M1, Int, Sec3 3 DF M Int Sec Single-Span Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Created July007: Page 16 of

87 Interior Girder - Two or More Lanes Loaded: DF DF DF S S Kg M, Int, Sec L 1LtS M, Int, Sec3 3 M, Int, Sec3 = ' 11.33' 3,30,000 in = ' ( 1)( 166.3' )( 8.5" ) = Exterior Girder One Lane Loaded: Same as for the positive moment section: DF M1,Ext,Sec3 = Exterior Girder - Two or More Lanes Loaded: DF M,Ext,Sec3 = e DF M,Int,Sec3 e = (same as before) DF M,Ext,Sec3 =(1.008) (0.7906) = d. Minimum Exterior Girder Distribution Factor: DF Ext, Min N L = + N b X Ext N b N L x e Single-Span Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Created July007: Page 17 of

88 One Lane Loaded: ( 8.33' ) ( 5.5' ) ( ) + ( ) + ( ) 1 = + = ' 17.0' 5.667' DF M 1, Ext, Min Multiple Presence: DF M1,Ext,Min = (1.) (0.4881) = Two Lanes Loaded: ( 8.33' ) ( 5.5' ) + ( 13.5' ) ( ) + ( ) + ( ) = + = ' 17.0' 5.667' DF M, Ext, Min Multiple Presence: DF M1,Ext,Min = (1.0) (0.850) = Three Lanes Loaded: ( 8.33' ) ( 5.5' ) + ( 13.5' ) + ( 1.5' ) ( ) + ( ) + ( ) 3 = + = ' 17.0' 5.667' DF M 3, Ext, Min Multiple Presence: DF M1,Ext,Min = (0.85) (1.011) = Four Lanes Loaded: ( 8.33' ) ( 5.5' ) + ( 13.5' ) + ( 1.5' ) + ( 10.5' ) ( ) + ( ) + ( ) 4 = + = ' 17.0' 5.667' DF M 4, Ext, Min Multiple Presence: DF M1,Ext,Min = (0.65) (1.045) = Five Lanes Loaded: ( 8.33' ) ( 5.5' ) + ( 13.5' ) + ( 1.5' ) + ( 10.5' ) + (.5' ) ( ) + ( ) + ( ) 5 = + = ' 17.0' ' DF M 5, Ext, Min Multiple Presence: DF M1,Ext,Min = (0.65) (0.8367) = Single-Span Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Created July007: Page 18 of

89 4d. Moment Distribution Factor Summary Section 1 3 # Lanes Positive Moment Loaded Interior Exterior For Simplicity, take the Moment Distribution Factor as everywhere. Multiplying the live load moments by this distribution factor of yields the table of nominal girder moments shown below. Nominal Girder Moments from Visual Analysis and SAP Nominal Moments Station LL+ LL- DC1 DC DW (ft) (k-ft) (k-ft) (k-ft) (k-ft) (k-ft) Single-Span Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Created July007: Page 19 of

90 5. FACTORED MOMENT ENVELOPES The following load combinations were considered in this example: Strength I: Strength IV: Service II: 1.75(LL + IM) + 1.5DC DC DW 1.50DC DC DW 1.3(LL + IM) + 1.0DC DC + 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 iqi η is taken as the product of η D, η R, and η I, and is taken as not less than For this example, η D, η R, and η I are taken as Using these load combinations, the shear and moment envelopes shown on the following pages were developed. Single-Span Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Created July007: Page 0 of

91 Strength I Moments Station LL+ LL- DC1 DC DW Total + Total - (ft) (k-ft) (k-ft) (k-ft) (k-ft) (k-ft) (k-ft) (k-ft) Strength IV Moments Station DC1 DC DW Total + Total - (ft) (k-ft) (k-ft) (k-ft) (k-ft) (k-ft) Single-Span Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Created July007: Page 1 of

92 Service II Moments Station LL+ LL- DC1 DC DW Total + Total - (ft) (k-ft) (k-ft) (k-ft) (k-ft) (k-ft) (k-ft) (k-ft) Single-Span Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Created July007: Page of

93 Strength Limit Moment Envelopes 5,000 Strength I Max (@ 83.14') = 189 k-ft 0,000 Moment (-ft) 15,000 10,000 Strength IV 5, Station (ft) Service II Moment Envelope 17,500 Max (@ 83.14') = 16,010 k-ft 15,000 1,500 Moment (-ft) 10,000 7,500 5,000, Station (ft) Single-Span Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Created July007: Page 3 of

94

95 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 11 0 and a width (c-c of the trusses) of 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-, 1-4, 9-11, 9-10, and due to an HL-93 Loading. ) 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 1 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 SAP000, Visual Analysis, or another similar FE package to model the truss. Disregard the lower limit of L = 0 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 AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Created July 007: Page 1 of

96 Figure - Truss Layout 6" Thick Reinforced Concrete Deck 6, W10 x 88 3'-6" cc 18' - 0" Clear Roadway 19' - 6" cc Trusses Figure 3 - Truss Cross Section Single-Span Truss Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Created July 007: Page of

97 Compute the Maximum and Minimum Forces in Critical Members of the Truss: The following Influence Lines were obtained from SAP 000: Single-Span Truss Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Created July 007: Page 3 of

98 Single-Span Truss Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Created July 007: Page 4 of

99 Single-Span Truss Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Created July 007: Page 5 of

100 Consider Member 1- of the Truss: 96' 4' (5 ) + (5 ) = ' Tandem: ( ) 16' 14' 96' 14' (8 ) + (3 ) + (3 ) = ' 96' Truck: ( ) (11') = Lane: ( )( )( ) ft Combining the HL-93 Components with impact applied appropriately: (IM)(Tandem) + Lane: ( )( ) ( ) = (IM)(Truck) + Lane: ( )( ) ( ) + = GOVERNS Apply the Truss Distribution Factor: Assume that each truss carries 75% of the HL-93 load effect P 1- = -13. Single-Span Truss Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Created July 007: Page 6 of

101 Consider Member 1-4 of the Truss: 96' 4' (5 ) + (5 ) 1.17 = ' Tandem: ( ) 16' 14' 96' 14' (8 ) + (3 ) + (3 ) 1.17 = ' 96' Truck: ( ) (11') = Lane: ( )( )( ) ft Combining the HL-93 Components with impact applied appropriately: (IM)(Tandem) + Lane: ( )( ) ( ) = (IM)(Truck) + Lane: ( )( ) ( ) = GOVERNS Apply the Truss Distribution Factor: Assume that each truss carries 75% of the HL-93 load effect P 1-4 = 98.1 Single-Span Truss Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Created July 007: Page 7 of

102 Consider Member 9-11 of the Truss: 64' 4' (5 ) + (5 ).54 = ' Tandem: ( ) 48' 14' 64' 14' (8 ) + (3 ) + (3 ).54 = ' 64' Truck: ( ) (11') = Lane: ( )( )( ) ft Combining the HL-93 Components with impact applied appropriately: (IM)(Tandem) + Lane: ( )( ) ( ) = 6.0 (IM)(Truck) + Lane: ( )( ) ( ) = GOVERNS Apply the Truss Distribution Factor: Assume that each truss carries 75% of the HL-93 load effect P 9-11 = Single-Span Truss Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Created July 007: Page 8 of

103 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. P 9-10 = Single-Span Truss Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Created July 007: Page 9 of

104 Consider Member of the Truss: Tandem: ( ) (5 ) + (5 ) 1.97 = ' 1' (8 ) + (3 ) + (3 ) 1.97 = ' Truck: ( ) ft 1.97 (96') + (16') = Lane: ( )( ) ( ) Combining the HL-93 Components with impact applied appropriately: = 11.9 (IM)(Tandem) + Lane: ( )( ) ( ) (IM)(Truck) + Lane: ( )( ) ( ) = 64.3 GOVERNS Apply the Truss Distribution Factor: Assume that each truss carries 75% of the HL-93 load effect P = 198. Single-Span Truss Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Created July 007: Page 10 of

105 Consider Member of the Truss Tensile Force: 5 5 Tandem: Truck: /ft Lane: / IL Mem 10-11: / 48' 4' (5 ) + (5 ) = ' Tandem: ( ) 48' 8' 48' 14' (8 ) + (3 ) + (3 ) = ' 48' Truck: ( ) 1 Lane: ( )( )( ) (56') = ft Combining the HL-93 Components with impact applied appropriately: (IM)(Tandem) + Lane: ( )( ) ( ) = (IM)(Truck) + Lane: ( )( ) ( ) = GOVERNS Apply the Truss Distribution Factor: Assume that each truss carries 75% of the HL-93 load effect P = Single-Span Truss Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Created July 007: Page 11 of

106 Consider Member of the Truss Compressive Force: 5 5 Tandem: Truck: /ft Lane: / IL Mem 10-11: / 48' 4' (5 ) + (5 ) = ' Tandem: ( ) 48' 8' 48' 14' (8 ) + (3 ) + (3 ) = ' 48' Truck: ( ) 1 Lane: ( )( )( ) (56') = 9.18 ft Combining the HL-93 Components with impact applied appropriately: (IM)(Tandem) + Lane: ( )( ) ( ) = (IM)(Truck) + Lane: ( )( ) ( ) = GOVERNS Apply the Truss Distribution Factor: Assume that each truss carries 75% of the HL-93 load effect P = Single-Span Truss Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Created July 007: Page 1 of

107 Member Force Summary: Member Max Tension Max Compression Single-Span Truss Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Created July 007: Page 13 of

108 Compute the Moment Distribution Factor for the Stringers in the Floor System: Interior Girder One Lane Loaded: DF DF DF M1, Int 3 14 L 1Lts g g g ( ) g 4 (8)(534 in (5.9 in )(8.40") ) 4 M1, Int 3 M1, Int S S K g = K = n I + Ae K K = + = 18,890 in ' 3.5' 18,890 in = ' 1(16')(6.0") = 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. ( P ) / ( 1.75') R = = 0.500P (3.50') DF = M1, Ext 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 AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Created July 007: Page 14 of

109 DF Ext, Min N L = + N b X Ext Nb NL x e One Lane Loaded: 4'-0" '-0" 3'-0" 3'-0" P/ P/ 8'-9" 5'-3" 1'-9" 1 (4.00')(8.75') DF M1, Ext, Min= + = 6 () (8.75') + (5.5') + (1.75') 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: DF M1,Int = Exterior Stringer (Lever Rule): DF M1, Ext = Exterior Stringer (Minimum): DF M1, Ext = For simplicity, take the moment distribution factor as for all stringers. Single-Span Truss Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Created July 007: Page 15 of

110 Compute the Maximum Bending Moment in the Stringers of the Floor System: 5 5 Tandem: 3 Truck: /ft Lane: 4.00 k-ft / IL CL Stringer 4 4'-0" = 16'-0" 8' 4' (5 ) + (5 ) 4.00 = ' Tandem: k-ft ( ) Truck: k-ft ( ) k-ft (3 ) 4.00 = Lane: k-ft k-ft ( )( )( ) = ft k-ft (16') 0.48 In this case, since the axle spacing is substantial relative to the beam length, we should consider the more general approach for computing maximum moment. For two equal point loads, P, separated by a distance, a, the maximum bending moment in a simply supported span is: when a < L, P a MMax = L L when a L PL M Max = 4 Single-Span Truss Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Created July 007: Page 16 of

111 Tandem: M Max (5 ) 4' = 16' = ()(16') k-ft (an increase of.067%) Truck: (3 )(16') k-ft M Max = = 18.0 (no change) (4) Combining the HL-93 Components with impact applied appropriately: k-ft k-ft k-ft = 4.1 GOVERNS (IM)(Tandem) + Lane: ( )( ) ( ) = (IM)(Truck) + Lane: ( )( ) ( ) k-ft k-ft k-ft Apply the Stringer Distribution Factor: Each stringer carries lanes of the HL-93 loading M Stringer = k-ft Single-Span Truss Bridge Example AASHTO-LRFD 007 ODOT LRFD Short Course - Steel Created July 007: Page 17 of

112

113 AASHTO Tension Member Example #1: Problem: A tension member is made up from a bar of M70-50 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, P u, is 00, determine whether or not the member is adequate. The member is 4-0 long. Solution: 1.5" 3" 1.5" 1" Check Minimum Slenderness Ratio: 1.5" r min 3 = Imin bt t 0.887" A = 1bt = 1 = 3" 3" L r min in (4' 0")(1 ft ) = = " 3" 3" Since 140 < L / r min <00, the slenderness is ok so long as the member is not subjected to stress reversals. 3" Compute the Design Strength: Gross Section Yielding: P n = F y A g = (50 )(6 )(1 ) = φp n = (0.95)(300.0 ) = 85.0 Net Section Fracture: A n P n = F u A e = F u U A n (1.5") = 6" () " + " + 1" 4(3.0") = in 7 1 ( 8 8 ) ( ) P n = (65 )(1.00)(4.188 in ) = 7. φp n = (0.80)(7. ) = 17.8 ODOT-LRFD Short Course - Steel AASHTO Tension Member Example #1 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. NSF Governs - φp n = 18 Since P u < φp n (00 < 18 ) the member is adequate. AASHTO-LRFD 007 Created July 007: Page 1 of

114 AASHTO Tension Member Example #: Problem: A C1x30 is used as a tension member (L = 8-6 ) as is shown in the sketch below. The channel is made of M70-36 material and is attached to the gusset plate with 7/8 diameter bolts. Calculate the design tensile capacity, φp n, of the member considering the failure modes of gross section yielding and net section fracture. Solution: Check Minimum Slenderness Ratio: Section A-A r min = 0.76 (from the AISC Manual) L r min in (8.5')(1 ft ) = = " 3" 6" 3" Since L/r min < 140, the slenderness is ok. Compute the Design Strength: Gross Section Yielding: P n = F y A g = (36 )(8.81 in ) = 317. A A φp n = (0.95)(317. ) = C1 x 30 Net Section Fracture: P u P n = F u A e = F u U A n 7 1 ( ) ( 8 8 )( ) A = 8.81 in () " + " 0.510" = in n U = 0.85 since there are 3 fasteners in the direction of stress P n = (58 )(0.85)(7.790 in ) = φp n = (0.80)(384.0 ) = 307. Gross Section Yielding Governs - φp n = 301 ODOT-LRFD Short Course - Steel AASHTO Tension Member Example # AASHTO-LRFD 007 Created July 007: Page 1 of

115 Side Note: Note that if the AISC shear lag provisions were used that Case from AISC Table D3.1 would apply: U x 0.674" = 1 = 1 = for net section fracture, φp n = L 9.00" In this case, however, the design strength is unaffected since gross yielding governs. ODOT-LRFD Short Course - Steel AASHTO Tension Member Example # AASHTO-LRFD 007 Created July 007: Page of

116 AASHTO Tension Member Example #3: Problem: Determine the design strength of the W10x60 member of M70-50 steel. As is shown, the member is connected to two gusset plates one on each flange. The end connection has two lines of 3/4 diameter bolts in each flange - five in each line. A Gusset Plates W10 x A Section A-A Solution: Check Minimum Slenderness Ratio: r min =.57 (from the AISC Manual) L r min L = 140 this is satisfied so long as L = / 4.57" Compute the Design Strength: Gross Section Yielding: P n = F y A g = (50 )(17.6 in ) = φp n = (0.95)(880.0 ) = ODOT-LRFD Short Course - Steel AASHTO Tension Member Example #3 AASHTO-LRFD 007 Created July 007: Page 1 of

117 Net Section Fracture: P n = F u A e = F u U A n 3 1 ( ) ( 4 8 )( ) A = 17.6 in (4) " + " 0.680" = 15. in n Check b f? 3 d ( 10.1" ) ( 3 )( 10." ) OK? U = 0.90 since b f > /3d and there are 3 fasteners in the direction of stress. P n = (65 )(0.90)(15. in ) = φp n = (0.80)(890.4 ) = 71.3 Net Section Fracture Governs - φp n =71 Side Note: Note that if the AISC shear lag provisions were used that Case 7a from AISC Table D3.1 would apply: Check b f? 3 d? ( 10.1" ) ( 3 )( 10." ) U = 0.90 OK Alternatively, Table D3.1 Case can be applied: x x x 0.884" U = 1 = 1 = L 1.0" The value of U = can be used. P n = (65 )(0.963)(15. in ) = The connection eccentricity x is taken as the distance from the faying surface to the CG of a WT5x30. φp n = (0.80)(916.4 ) = Since Net Section Fracture governs the capacity of this member, the overall design strength of the member would be increased to 733. ODOT-LRFD Short Course - Steel AASHTO Tension Member Example #3 AASHTO-LRFD 007 Created July 007: Page of

118 AASHTO Tension Member Example #4: Problem: An L6x4x1/, M70-36, 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: r min = r z = (from the AISC Manual) L r min L = " this is satisfied so long as L 07.4 = / 8 Compute the Design Strength: Gross Section Yielding: P n = F y A g = (36 )(4.75 in ) = φp n = (0.95)(171.0 ) = 16.5 Net Section Fracture: P n = F u A e = F u U A n Lacking other guidance, AISC Table D3.1 Case will be applied: U x 0.981" = 1 = 1 = L 8.0" P n = (58 ) (0.8774)(4.75 in ) = 41.7 φp n = (0.80)(41.7 ) = Gross Section Yielding Governs - φp n =163 ODOT-LRFD Short Course - Steel AASHTO Tension Member Example #4 AASHTO-LRFD 007 Created July 007: Page 1 of

119 AASHTO Compression Member Example #1: Problem: Compute the design compressive strength of a W14x74 made of M70-50 steel. The column has a length of 0 ft and can be treated as pinned-pinned. Solution: Check Local Buckling: Flange: Web: b f t w f? E k λ f = 6.41 (Tabulated) F y y h? E k λ w = 5.4 (Tabulated) t F? 9, = 13.5 OK 50? 9, = 35.9 OK 50 Compute Flexural Buckling Capacity: Slenderness Ratios: KL r (1.00)(0')(1 ) 6.04" in = ft = < x OK KL r (1.00)(0')(1 ).48" in = ft = < y OK Since the effective slenderness ratio is larger for the y axis than the x axis, y-axis buckling will govern. KL Fy λ= = = rπ y E π 9, 000 ( ) P ( ) ( )( )( ) Since λ.5, Inelastic Buckling Governs λ n = 0.66 FyAs = in = ( ) φp n = (0.90)(549.6 ) φp n = 495 ODOT-LRFD Short Course - Steel AASHTO Compression Member Example #1 AASHTO-LRFD 007 Created July 007: Page 1 of

120 AASHTO Compression Member Example # 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: r= I A 3 I x bh = + Ad 1 ( ) 3 3 ( 36" )( 4" ) 30" 4" ( 4" ) 30"-( 4" ) ( 36" 4" ) 4 = + + = 56,150 in hb I y = + Ad ( 4" )( 36" ) ( 30"-( 4" ))( 4" ) 36" 4" = + + (( 30"-( 4" )) 4" ) = 76,390 in (( ) ) ( ) ( ) A = 36" 4" + 30"- 4" 4" = in s Since I x 4 4 Iy = 56, in < = 76, in, x-axis buckling controls I x 56,150 in r= x = = 11.0 in A in s 4 ODOT-LRFD Short Course - Steel AASHTO Compression Member Example # AASHTO-LRFD 007 Created July 007: Page 1 of

121 Check Local Buckling (Section ): b t ( ) 36" 4" = = " b? E k ( ) t F y 9, = 33.7 OK 50 Calculate the Nominal Compressive Strength (Section E3 page ): Slenderness Ratios: KL r where: K = 0.8 (Section ) L = L = 40 ft 1 = 480" x y in ft KL r x x ( 0.8)( 480 in) ( 11.0 in) = = F y KL λ= = = 0.19 rs π E π 9,000 ( ) Since λ.5, Inelastic Flexural Buckling Governs λ n y s ( )( ) 0.19 P = 0.66 F A = in = 1, 40 ( ) ( )( ) φ c = , 40 = 19,110 P n ODOT-LRFD Short Course - Steel AASHTO Compression Member Example # AASHTO-LRFD 007 Created July 007: Page of

122 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 A99 steel. W16x36 beams frame into joint A and W16x77 beams frame into joint B. The frame is unbraced and all connections are rigid. Consider only buckling in the plane of the page about the sections strong axes. A W16 x 36 L=4' A W10 x 88 L=14' B 14' B W16 x 77 L=4' 4' Solution: ODOT-LRFD Short Course - Steel AASHTO Compression Member Example #3 v AASHTO-LRFD 007 Created July 007: Page 1 of

123 Determine the Effective Length Factor: G B K G A G A 4 I ()(534 in ) L (14 ') C = = = I ()(448 in ) L G 3 (4') G B 4 I ()(534 in ) L (14 ') C = = = I ()(1,110 in ) L G 3 (4') For unbraced frames: K = 1.6G A GB + 4.0( GA + G G + G A B B ) (1.6)(3.065)(1.37) + (4.0)( ) K = = ( ) The factor of /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 v AASHTO-LRFD 007 Created July 007: Page of

124 AASHTO Compression Member Example #4: Problem: Check to see if a built-up section will work to resist a factored load of P u = 09. The column is to be fabricated from two C10x15.3 as is shown in the figure to the right. The steel is M70-36 and the effective length is 0 with respect to all axes. 10" y x If the column is adequate, determine the thickness of the battens. The battens are 8 long and 6 deep and are also made of M70-36 steel. 9" Solution: Check Local Buckling: Flange: b t bf.60" = = = 5.96 t 0.436" f a? b E 9, = 0.56 = OK t F 36 y Web: b t d tf 10" ()(0.436") = = = t 0.40" w? b E 9, = 1.49 = 4.9 OK t F 36 y Compute Section Properties: A s = () (4.48 in ) = 8.96 in I X = () (I x ) = ()(67.3in 4 ) = in 4 I Y = ().7 in + (4.48 in ) 0.634" = in 4 9" 4 r X 4 I X in = = = 3.88" A 8.96 in r Y 4 IY in = = = 3.93" A 8.96 in ODOT-LRFD Short Course - Steel AASHTO Compression Member Example #4 AASHTO-LRFD 007 Created July 007: Page 1 of

125 Slenderness Ratios: in KL (0')(1 ft ) = = r 3.88" X KL (0')(1 ) r 3.93" in = ft = Y 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 r i = r y = (for one channel) KL KL = r r max X a (0.711") ( 0.75)( 61.86) a 3.98" use 9 a = 30 Modified Slenderness Ratio Y-axis Buckling: The modified slenderness ratio is calculated as, α 0.8 (1 m o + α ) ib KL KL a = + r r r ( ) r ib = h = 9 ()(0.634 ) = 7.73 h 7.73" α = = = 5.44 r ()(0.711") ib = ( 61.07) = r m ( 1 + (5.44) ) 0.711" KL (5.44) 30" 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 007 Created July 007: Page of

126 Column Design Capacity: KL Fy ( ) λ= = = rπ y E π 9, 000 Since λ.5, Inelastic Buckling Governs P ( ) ( )( )( ) λ n = 0.66 FyAs = in = 46.1 ( ) φp n = (0.90)(46.1 ) φp n = 1 Since φp n > 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 % of the compressive design strength (AISC Section E6. Pg ).1 V u = (0.0)(1 ) = =.1 channel ΣM M u,batten = k-in M u,batten I Batten 3 t(6") = = 18t 1 S Batten 18t = = 6t 3.1 for first yield, M y = F y S Let φf y S M u,batten t k-in 3 (1.00)(36 )(6 in ) t 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 007 Created July 007: Page 3 of

127 AASHTO Compression Member Example #5: Problem: Find the design strength of a WT15x146 made of M70-50 steel. KL = 4 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: b t bf 15.3" = = = 4.14 t ()(1.85") f? b E 9, 000 k = 0.56 = 13.5 OK t F 50 y b h Web: = = 15.7 (Tabulated) t t w? b E 9, 000 k = 0.75 = 18.1 OK t F 50 y Calculate the buckling load for Flexural Buckling about the X-Axis: in ( )( ) KL Fy 4' 1 ft 50 λ X = = = rπ X E (4.48")( π) 9,000 ( ) ( )( )( ) in 1,586 Since λ X <.5, Inelastic Buckling Governs P n = = ( ) ODOT-LRFD Short Course - Steel AASHTO Compression Member Example #5 AASHTO-LRFD 007 Created July 007: Page 1 of

128 Calculate the Critical Stress for Flexural-Torsional Buckling about the Y-axis: F crft FcrY, + F crz, 4FcrY, FcrZ, H = 1 1 H F F ( cr, Y + cr, Z ) (AISC E4-) in ( )( ) KL Fy 4' 1 ft 50 λ Y = = = rπ Y E (3.58")( π) 9,000 Since λ Y <.5, Inelastic Buckling Governs ( ) ( )( ) P F, A n cr Y = = = ( ) s I + I r = x + y + x y o o o Ag 1.85" y o = 3.6" =.695" (AISC E4-7) (861 in in ) 4.9 in 4 4 r o = (0.00) + (.695) + = in GJ 4 (11, 00 )(37.5 in ) cr, Z = = = 44.0 (AISC E4-3) Aro (4.9 in )(40.13 in ) F H x + y (0.000") + (.695") = 1 = 1 = (AISC E4-8) o o r o in F crft (4)(31.15 )(44.0 )(0.819) = 1 1 = ()(0.819) ( ) (AISC E4-) P n = A s F crft = (4.9 in )(30.37 ) = 1,303 Since 1,303 < 1,586, Flexural-Torsional Buckling Governs φp n = (0.90)(1,303 ) = 1,170 ODOT-LRFD Short Course - Steel AASHTO Compression Member Example #5 AASHTO-LRFD 007 Created July 007: Page of

129 AASHTO Compression Member Example #6: Problem: Find the design strength of a C1x30 made of A36 steel. KL y = 7 and KL x = KL z = 14. y x Solution: Check Local Buckling: Flange: Web: b bf 3.17" b? E 9, 000 = = = 6.37 k = 0.56 = OK t tf 0.501" t Fy 36 b h d tf 1.0" ()(0.501") = = = = 1.56 t t t 0.510" w w? b E 9, 000 k = 1.49 = 4.9 OK t F 36 y 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 KL x = 7 and KL y = KL z = 14. ODOT-LRFD Short Course - Steel AASHTO Compression Member Example #6 AASHTO-LRFD 007 Created July 007: Page 1 of

130 Calculate the buckling load for Flexural Buckling about the X-Axis: in ( )( ) Fy 7' 1 ft 36 x E (0.76")( π) 9,000 KL λ x = = = 1.58 rπ ( ) ( )( )( ) Since λ x <.5, Inelastic Buckling Governs 1.58 P n = in = ( ) Calculate the Critical Stress for Flexural-Torsional Buckling about the Y-axis: For Singly symmetric Sections: F e F ( ) ey + Fez 4FeyFezH = 1 1 H Fey + Fez (AISC E4-5) F ey ( π )(9,000 ) = = in (14 ')(1 ft ) 4.9" (AISC E4-10) F F ez ez = + (11,00 )(0.861 in ) in ((14 ')(1 ft )) (8.81 in )(4.54") 6 ( π )(9,000 )(151 in ) 4 1 = (AISC E4-11) F F e e ( ) (4)(186.6 )(61.54 )(0.919") = 1 1 ()(0.919") ( ) = (AISC E4-5) Fy 36 λ= = = F e ( )( )( ) (0.6071) P n = in = 46.0 ( ) since < 46.0, Flexural Buckling about the x Axis Governs φp n = (0.90)( ) = 151 ODOT-LRFD Short Course - Steel AASHTO Compression Member Example #6 AASHTO-LRFD 007 Created July 007: Page of

131 AASHTO Compression Members Example #7: Problem: A pair of L4x4x1/ angles are used as a compression member. The angles are made of M70-36 steel and have an effective length of 1. The angles are separated by 3 / 8 thick connectors. 4" Y 3 / 8 " 4" Solution: Check Local Buckling: X b 4.0" 8.0 t = " = b? E 9, 000 k = 0.45 = 1.77 t F 36 1 y Fully Tensioned Local Buckling is OK Check the Connector Spacing: in ( )( ) KL 1' 1 ft = = r X (1.1") in ( )( ) KL 1' 1 ft = = r Y (1.83") r y = 1.83 from AISC L Table 1-15, Pg KL a ri 4 r max 3 a (0.776") ( 119.0) = 69.6" 4 Use 5 connectors.. a = 36 ODOT-LRFD Short Course Steel AASHTO Compression Member Example #7 AASHTO-LRFD 007 Created July 007: Page 1 of

132 Check Flexural Buckling about the X-Axis: (Y axis is the axis of symmetry) KL Fy λ X = = = rπ X E π 9, 000 Since λ X <.5, Inelastic Buckling Governs ( ) ( )( )( ) P n = in = 17.9 ( ) Check Flexural-Torsional Buckling about the Y-Axis: For Tees and Double Angles where the Y axis is the Axis of Symmetry: F crft FcrY, + F crz, 4FcrY, FcrZ, H = 1 1 H F F ( cr, Y + cr, Z ) (AISC E4-) 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: α 0.8 (1 m o +α ) ib KL KL a = + r r r ( ) h α= 3 ( )( ) ( 8 ) r ib h = 1.18" + " =.735" r ib = r y for a single angle = " α= = ()(1.1") KL (0.8)(1.130) 36" ( ) r m (1 + (1.130) ) 1.1" = = 81.4 ODOT-LRFD Short Course Steel AASHTO Compression Member Example #7 AASHTO-LRFD 007 Created July 007: Page of

133 Compute the Y-axis Flexural Buckling Stress, F cry : Fy (81.4) 36 Y E ( π) 9,000 KL λ Y = = = rπ Since λ Y <.5, Inelastic Buckling Governs ( ) ( )( ) P F, A n cr Y = = = ( ) s Calculate Torsional Buckling Stress, F cr,z : r =.38" (AISC Table 1-15, Pg 1-104) o r = in o J = 0.3 in 4 for a single angle (AISC Table 1-7, Pg 1-4) J total = ()(0.3 in 4 ) = in 4 4 GJ (11, 00 )(0.644 in ) cr, Z = = = (AISC E4-3) Ar o (7.49 in )(5.664 in ) F H = (AISC Table 1-15, Pg 1-104) F crft (4)(5.43 )(170.0 )(0.848) = 1 1 = 4.79 ()(0.848) ( ) P n = A s F crft = (7.49 in ) (4.79 ) = Since 17.9 < 185.7, Flexural Buckling Governs φp n = (0.90)(17.9 ) = 115 ODOT-LRFD Short Course Steel AASHTO Compression Member Example #7 AASHTO-LRFD 007 Created July 007: Page 3 of

134

135 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. c w ( ) 3 ( 4 )( ) ( )( )( ) P = (16")(1") 50 = P = (") " 36 = P = 8" " 70 = 1,10 t Since Pc Pw Pt ( ,394 1,10 ) + > + = >, the PNA must lie in the web. Define q as the fraction of the web that lies above the PNA. Pcompression = Ptension P + qp = q P + P ( 1 ) c w w t ( ) + q( ) = ( 1 q)( ) + ( 1,10 ) q = I.e., 76.94% of the web lies above the PNA (acts in compression assuming a positive moment). Y = 1" + ( )( " ) = 17.93" from the top of steel Find the moment arms from the resultant forces to the PNA. tc 1" dc = Y = 17.93" = 17.43" d 1 1 wc = ( ) qh= ( )( )( " ) = 8.463" d 1 1 wt = ( )( 1 q) h= ( )( )( " ) =.537" tt " dt = d Y = 5" 17.93" = 6.074" ODOT-LRFD Short Course - Steel AASHTO Flexure Example #1 AASHTO-LRFD 007 Created July 007: Page 1 of

136 PL16 x 1, P c 36 PL x 3/4, 36 Y P wc d c d wc 36 PNA P wt d wt PL8 x, P t d t Compute the plastic moment by summing the moments about the PNA. ( 800 )( 17.43" ) ( 457 )( 8.463" ) ( 137 )(.537" ) ( 1,10 )( 6.074" ) M = Pd = Pd + P d + P d + Pd p i i c c wc wc wt wt t t = k-in k-ft 4,960,080 = = ODOT-LRFD Short Course - Steel AASHTO Flexure Example #1 AASHTO-LRFD 007 Created July 007: Page of

137 AASHTO Flexure Example #: 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 F y = 50 and f c = 4. Assume full composite action. 100" 8" Solution: Determine the Controlling Compression Force: s ( )( )( )( ) P = 0.85 f bt = " 100" = 70 P ' c e s ( )( ) = A F = 9.1 in 50 = 1455 Steel st y Assuming full composite action, the shear connectors must carry the smallest of P s and P steel. W30 x 99: A = 9.1 in d = 9.7" b f = 10.5" t f = 0.670" t w = 0.50" Z x = 31 in 3 I x = 3,990 in 4 I y = 18 in 4 r x = 11.7" r y =.10" Since P s > P steel, 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 a c, with the tensile force in the steel section. P = P Conc Steel ' 0.85 fcba e c = AstFy a c ( 9.1 in )( 50 ) AF st = = = 4.79" (measured from the top of slab) 0.85 fb " y ' c e ( )( )( ) ODOT-LRFD Short Course - Steel AASHTO Flexure Example # AASHTO-LRFD 007 Created July 007: Page 1 of

138 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 P Steel or the force P conc. Note that the tension force in the concrete is ignored. 100" 8" a c 0.85f c P conc PNA a 1 P steel F y ( ) ( ) M = P ( a ) = P ( a ) p Conc 1 Steel 1 dst ac 9.7" 4.79" a1 = + ts = + 8" = 0.71" ( ) M = 1455 (0.71") = 30,130 =,511 p k-in k-ft ODOT-LRFD Short Course - Steel AASHTO Flexure Example # AASHTO-LRFD 007 Created July 007: Page of

139 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 F y = 50 and f c = 4. Assume full composite action. Solution: Determine the Controlling Compression Force: P s ( )( )( )( ) = 0.85 f bt = " 50" = 100 ' c s s P ( )( ) = A F = 9.1 in 50 = 1455 Steel st y Assuming full composite action, the shear connectors must carry the smallest of P s and P steel. Since P s < P steel, 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 6-90: 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 ) s c ( )( )( )( ) P = 0.85 f bt = " 50" = 100 ' c s s ( ) P = (0.670")(10.5") 50 = ( ) Pw = 9.1 in ()(0.670")(10.5") 50 = Pt = Pc = In this case, I took A w = A steel - A f. Otherwise, P c +P t +P w P steel. If you take A w = D t w where Check Case I? P + P P + P D = d - t f, the plastic moment changes by ~% t w c s? NO ODOT-LRFD Short Course - Steel AASHTO Flexure Example #3 AASHTO-LRFD 007 Created July 007: Page 1 of

140 Check Case II P + P + P P t w c s?? YES - PNA in Top Flange 50" 0.85f c 6" F y P s P c1 d s PNA P c d w P w d t F y P t First, the location of the PNA within the top flange is determined. Y tc Pw + Pt P s = 1 + Pc 0.670" , 00 = + 1 = 0.368" Next, the distances from the component forces to the PNA are calculated. d d d s w t 6" = " = 3.37" 9.7" = 0.368" = 14.61" 0.670" = 9.7" 0.368" = 9.13" ODOT-LRFD Short Course - Steel AASHTO Flexure Example #3 AASHTO-LRFD 007 Created July 007: Page of

141 Finally, the plastic moment is computed. M P = Y + t Y + Pd + P d + Pd ( ) [ ] c p c s s w w t t tc ( ) ( ) = 0.368" " 0.368" +... ()(0.670")... + ( 1, 00 )( 3.37" ) + ( )( 14.61" ) + ( )( 9.13" ) k-in ( 6.5 in ) in 4,530 = + = 4,590 k-in k-ft =,049 ODOT-LRFD Short Course - Steel AASHTO Flexure Example #3 AASHTO-LRFD 007 Created July 007: Page 3 of

142 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 F y = 50 and f c = 4. Assume full composite action. The grade 60 reinforcement in the slab is made up of #4 bars, with a clear cover of 1 7 / " Solution: 8" The concrete slab will be in tension, therefore none of the concrete is assumed to be effective. Referring to Table D6.1- in Appendix D6.1, Page 6-91: ( π)(0.5") Prt = Fyrt Art = ( 60 )( 8 ) = ( π)(0.5") Prb = Fyrb Arb = ( 60 )( 4 ) = P = (0.670")(10.5") 50 = t c t ( ) ( ) Pw = 9.1 in ()(0.670")(10.5") 50 = P = P = Slab Reinforcement: Top Layer: #4 6" cc Bottom Layer: #4 1" cc W30 x 99: A = 9.1 in d = 9.7" b f = 10.5" t f = 0.670" t w = 0.50" S x = 69 in 3 Z x = 31 in 3 I x = 3,990 in 4 I y = 18 in 4 r x = 11.7" r y =.10" Check Case I:? P + P P + P + P c w t rb rt? YES - PNA is in Web Y D Pc Pt Prt P rb = + 1 Pw Take D as ( )( ) d t f = 9.7" 0.670" = 8.36" Y 8.36" = Y = 11.51" (measured from the bottom of the top flange) ODOT-LRFD Short Course - Steel AASHTO Flexure Example #4 AASHTO-LRFD 007 Created July 007: Page 1 of

143 8 ( )( ) ( )( ) 1 ( )( ) 1 ( )( 11.51" ) 5.755" 1 ( )( 8.36" 11.51" ) 8.45" ( ) 1 ( )( ) d rt = = 11.51" 0.670" 8" 1 7 " 1 1 " 18.06" d rb = 11.51" " + 1 8" + " = 14.31" d = 11.51" " = 11.85" t d = = wt d = = wc d = 8.36" 11.51" " = 17.19" c Not needed when using Table D6.1- Pw M p = y + ( D y) + [ Prtdrt + Prbdrb + Pd t t + Pd c c] D = (11.51") + (8.36" 11.51") + [(94.5 )(18.06") +... ()(8.36")... (47.1 )(14.31") (351.8 )(11.85") (351.8 )(17.19")] = = k-in k-ft 18,110 1,509 ODOT-LRFD Short Course - Steel AASHTO Flexure Example #4 AASHTO-LRFD 007 Created July 007: Page of

144 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 ( )( )( Y) ( )( )( Y) " 8.36" = " Y = 11.46'' wt ( ) ( ) P = = wc (50 )(11.46") 0.50" 98.0 P = = (50 )(8.36" 11.46") 0.50" 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. 8 ( )( ) ( )( ) 1 ( )( ) 1 ( )( 11.46" ) 5.730" 1 ( )( 8.36" 11.46" ) 8.450" ( ) 1 ( )( ) d rt = = 11.46" 0.670" 8" 1 7 " 1 1 " 17.88" d rb = 11.46" " + 1 8" + " = 14.38" d = 11.46" " = 11.80" t d = = wt d = = wc d = 8.36" 11.46" " = 17.4" c M = p (94.5 )(17.88'') (47.1 )(14.38") (351.8 )(11.80'')... ( ) ( )( )... + (98.0 )(5.730") + (439.4 ) 8.450" " = = k-in k-ft 18, 000 1,500 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 007 Created July 007: Page 3 of

145 AASHTO Flexural Example #5a: Problem: A non-composite W30x99 made of M70-50 steel is used to span 48. The beam is braced laterally at 1-0 intervals and is subjected to a factored load of w = 3.75 / 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 to determine capacity. Solution: Determine Classification of the Section: Check D t w c? E 5.7 ( ) F yc Take D = d - t f = ()(0.670 ) = 8.36 D 8.36" D c = = = 14.18"? ()(14.18") 9, 000 = = OK, web is non-slender (0.50") 50 Check I I yc yt? 0.3 ( ) Since Section is doubly symmetric, I yc = I yt OK ODOT-LRFD Short Course - Steel AASHTO Flexure Example #5a AASHTO-LRFD 007 Created July 007: Page 1 of

146 Since the web is non-slender and Eq is satisfied, we have the option of using either AASHTO Section or Appendix A6 to determine the flexural capacity of this member. For this example, the provisions of will be used and we will work with stresses. The following failure modes must be investigated: Flange Local Buckling of the Compression Flange F nc(flb) Compression Flange Lateral Buckling F nc(ltb) Yielding of Tension Flange F nt F nc Investigate Compression Flange Local Buckling: bfc 10.5" λ f = = = ( ) t ()(0.670") fc E 9, 000 λ pf = 0.38 = 0.38 = 9.15 ( ) F 50 yc Since λ f < λ p, the flange is compact and, F = R R F ( ) nc( FLB) b h yc R b = 1.00 (since the web is non-slender) R h = 1.00 (since the section is rolled and is non-hybrid) ( ) F ( ) = (1.00)(1.00) 50 = 50 nc FLB Investigate Compression Flange Lateral-Torsional Buckling: The unbraced length of the beam is L b = 1-0 = r t bfc (10.5") = = =.609" 1 Dt 1 (14.18")(0.50") c w b 3 (10.5")(0.670") fct fc ( ) L p E 9, 000 = 1.0 rt = (1.0)(.609") = 6.84" ( ) F 50 yc ODOT-LRFD Short Course - Steel AASHTO Flexure Example #5a AASHTO-LRFD 007 Created July 007: Page of

147 = min ( 0.7, ) 0.5 yr ( ) F F F F yr yc yw yc F = (0.7) 50 = 35 (Pg 6-110) L r E 9, 000 =π rt = ( π )(.609") = 35.9" ( ) F 35 yr Since L = 6.84" < L = 144" < L = 35.9", Inelastic LTB must be investigated. p b r F yr Lb L p F ( ) = C 1 1 R R F R R F RF h yc Lr L p nc LTB b b h yc b h yc ( ) F = ( 1.0)( 50 ) 35.9" 6.84" " 6.84" nc( LTB) C b ( )( )( ) ( )( ) F ( ) = C = C nc LTB b b ( )( )( ) ( )( )( ) Compute the Moment Gradient Factor, C b, for segment BC of the beam, which will be critical. C b f f = f f ( ) C M = M = 1, 080 f = k-ft c M M f M k-ft o = B = o = BC, mid k-ft 1, 013 fmid = = Since the BMD is not concave, 1 mid ( ) ( ) f = f f f = () = o B f 1 f mid f C b = = = nc( LTB) ( )( ) F = = = ODOT-LRFD Short Course - Steel AASHTO Flexure Example #5a AASHTO-LRFD 007 Created July 007: Page 3 of

148 The governing strength for the compression flange is the smaller of F nc(flb) and F nc(ltb) : Since F nc( LTB) = < Fnc ( FLB) = 50.00, LTB governs the strength of the compression flange. F nc = F ( ) = nc LTB ( ) φ = (1.00) = F nc Check 1 fbu + fl φ f Fnc ( ) 3 f bu = f = C Assume that Strength I Load Combination Governs, γ WS = 0.0 and f l = 0 Since 1 fbu + fl = > φ f Fnc =45.57, the compression flange is not adequate. 3 Investigate the Strength of the Tension Flange: Since the tension flange is discretely braced, F ( ) nt = RhFyt = (1.0) 50 = 50 ( ) ( ) φ = (1.00) 50 = 50 F nt Check 1 fbu + fl φ f Fnt ( ) 3 f bu = f = C Since 1 fbu + fl = < φ f Fnt =50.00, 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 007 Created July 007: Page 4 of

149 AASHTO Flexural Example #5b: Problem: A non-composite W30x99 made of M70-50 steel is used to span 48. The beam is braced laterally at 1-0 intervals and is subjected to a factored load of w = 3.75 / ft, which includes the self weight of the beam. Check to see if the section is adequate considering flexural failure modes at the Strength Limit States. If appropriate, use the provisions in AASHTO Appendix A6 to determine capacity. Solution: Determine Classification of the Section: Check D t w c? E 5.7 ( ) F yc Take D = d - t f = ()(0.670 ) = 8.36 D 8.36" D c = = = 14.18"? ()(14.18") 9, 000 = = OK, web is non-slender (0.50") 50 Check I I yc yt? 0.3 ( ) Since Section is doubly symmetric, I yc = I yt OK ODOT-LRFD Short Course - Steel AASHTO Flexure Example #5b AASHTO-LRFD 007 Created July 007: Page 1 of

150 Since the web is non-slender and Eq is satisfied, we have the option of using either AASHTO Section 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 M nc(flb) Lateral-Torsional Buckling M nc(ltb) Yielding of Tension Flange M nt M nc Compute Web Plasticity Factors, R pc and R pt (Section A6.): Investigate the classification of the web. Check D t w cp? λ (A6..1-1) pw( D ) cp E Fyc Dcp λ pw( Dcp ) = λ rw 0.54M D p c 0.09 RM h y (A6..1-) E λ rw = 5.7 = (A6..1-3) F yc R h = 1.00 (since the section is rolled and is non-hybrid) M M ( )( ) ( )( ) = S F = 69 in 50 = 13, 450 = 1,11 y x y 3 k-in k-ft = Z F = 31 in 50 = 15, 600 = 1,300 p x y 3 k-in k-ft 9, " λ pw( D cp ) = = = k-in ( 0.54)( 15, 600 ) 14.18" 0.09 k-in ( 1.0)( 13, 450 ) λ pw( D cp ) = ODOT-LRFD Short Course - Steel AASHTO Flexure Example #5b AASHTO-LRFD 007 Created July 007: Page of

151 D? cp ()(14.18") = = λ pw( D ) = OK, web is compact cp t 0.50" w Since the web is compact, R pc k-in M p 15, 600 = = = (A6..1-4) k-in M 13, 450 yc R pt k-in M p 15, 600 = = = (A6..1-5) k-in M 13, 450 yt Investigate Compression Flange Local Buckling: Investigate the compactness of the compression flange. bfc 10.5" λ f = = = (A6.3.-3) t ()(0.670") fc E 9, 000 λ pf = 0.38 = 0.38 = 9.15 (A6.3.-4) F 50 yc Since λ f < λ pf, the flange is compact and, M k-in ( )( ) k-in nc FLB = RpcM yc = = (A6.3.-1) ( ) , , 600 Investigate Lateral-Torsional Buckling: The unbraced length of the beam is L b = 1-0 = r t bfc (10.5") = = =.609" 1 Dt 1 (14.18")(0.50") c w b 3 (10.5")(0.670") fct fc (A ) L p E 9, 000 = 1.0 rt = (1.0)(.609") = 6.84" (A ) F 50 yc ODOT-LRFD Short Course - Steel AASHTO Flexure Example #5b AASHTO-LRFD 007 Created July 007: Page 3 of

152 E J Fyr Sxch Lr = 1.95 rt F yr S xc h E J ( ) S xt F = min 0.7 F, R F, F 0.5F Sxc yr yc h yt yw yc ( ) F yr = (0.7) 50 = 35 (Pg 6-) S xc = 69 in 3 ( )( fc ft ) h= d 1 t + t = 9.7" 0.670" = 9.03" J Dt bfc t fc t fc bft t w ft t ft = b fc 3 b ft (A ) J 3 3 (8.36")(0.50") (10.5")(0.670") 0.670" = + () 1 ( 0.63) = in " 4 J = in 4 J = 3.77 in 4 from AISC Manual.use J = 3.77 in 4 E 9, 000 F = yr 35 = Sxch J 88.6 ( 69 in 3 )( 9.03" ) =, in = L r 1, 071 = ( 1.95)(.609" )( 88.6) = 54.9" = 1.4', This value of L r = 1.4 agrees well with the value published in AISC on Page 3-15 Since L = 6.84" < L = 144" < L = 54.9", Inelastic LTB must be investigated. p b r FyrS xc Lb L p M ( ) = C 1 1 R F R F RpcF yc Lr L p nc LTB b pc yc pc yc (A6.3.3-) ODOT-LRFD Short Course - Steel AASHTO Flexure Example #5b AASHTO-LRFD 007 Created July 007: Page 4 of

153 M 3 ( )( ) C b ( 1.16)( 13, 450 ) ( Cb)( )( ) ( Cb)( ) in 144" 6.84" = , , " 6.84" nc( LTB) k-in = , 600 = 1,990 15, 600 k-in k-in k-in k-in k-in ( )( ) ( )( ) Compute the Moment Gradient Factor, C b, for segment BC of the beam, which will be critical. C b M M = M M (A ) M M M o = M = 1,080 = M = BC, mid c B = 1,013 k-ft k-ft k-ft Since the BMD is not concave, 1 mid ( ) ( ) M = M M M = () 1, 013 1, 080 = o k-ft k-ft k-ft k-ft C b = ,080 + = 1,080 = ( )( ) M ( ) = ,990 = 13,780 15,600 nc LTB k-in k-in k-in M nc( LTB) = 13,780 k-in ODOT-LRFD Short Course - Steel AASHTO Flexure Example #5b AASHTO-LRFD 007 Created July 007: Page 5 of

154 The governing strength for the compression flange is the smaller of M nc(flb) and M nc(ltb) : Since M k-in k-in nc( LTB) = 13,780 < Mnc( FLB) = 15,600, LTB governs the strength of the compression flange. M nc = M = = k-in k-ft nc( LTB) 13,780 1,148 k-ft ( ) φ = (1.00) 1,148 = 1,148 M nc k-ft Check 1 M u + fs l xc φ fmnc (A ) 3 M u = M = 1,080 C k-ft Assume that Strength I Load Combination Governs, γ WS = 0.0 and f l = 0 Since 1 k-ft k-ft Mu + flsxc = 1, 080 < φ fmnc=1,148, the compression flange is adequate. 3 Investigate the Strength of the Tension Flange: Since the tension flange is discretely braced, M ( ) k-in k-in k-ft nt = RptM yt = (1.160) 13, 450 = 15, 600 = 1,300 (A6.4-1) k-ft ( ) φ = (1.00) 1,300 = 1,300 M nt k-ft Check 1 M u + fs l xt φ fmnt ( ) 3 M u = M = 1,080 C k-ft 1 k-ft k-ft Since Mu + flsxt = 1, 080 < φ fmnt=1,300, 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 were used to compute capacity. ODOT-LRFD Short Course - Steel AASHTO Flexure Example #5b AASHTO-LRFD 007 Created July 007: Page 6 of

155 AASHTO Flexural Example #6a: Problem: A non-composite built-up girder made of M70-50 steel is used to span 48. The beam is braced laterally at 1-0 intervals and is subjected to a factored load of w = 3.75 / 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 to determine capacity. Solution: PL16 x 3/4 PL38 x 3/8 I x = 10,730 in 4 I y = 513. in 4 S x = in 3 S y = in 3 PL16 x 3/4 Determine Classification of the Section: Check D t w c? E 5.7 ( ) F yc Take D = 38 D 38" D c = = = 19"? ()(19") 9,000 = = OK, web is non-slender ( ") Check I I yc yt? 0.3 ( ) Since Section is doubly symmetric, I yc = I yt OK ODOT-LRFD Short Course - Steel AASHTO Flexure Example #6a AASHTO-LRFD 007 Created July 007: Page 1 of

156 Since the web is non-slender and Eq is satisfied, we have the option of using either AASHTO Section or Appendix A6 to determine the flexural capacity of this member. For this example, the provisions of will be used and we will work with stresses. The following failure modes must be investigated: Flange Local Buckling of the Compression Flange F nc(flb) Compression Flange Lateral Buckling F nc(ltb) Yielding of Tension Flange F nt F nc Investigate Compression Flange Local Buckling: bfc 16" λ f = = = ( ) t ()( ") fc 3 4 E 9, 000 λ pf = 0.38 = 0.38 = 9.15 ( ) F 50 yc Since λ f < λ p, the flange is non compact and, F yr λ f λ pf F ( ) = 1 1 R R F RF h yc λrf λpf nc FLB b h yc ( ) E λ rf = 0.56 ( ) F yr ( ) ( ) F = min 0.7 F, F 0.5F (Pg 6-109) yr yc yw yc F = (0.7) 50 = 35 yr 9, 000 λ rf = 0.56 = R b = 1.00 (since the web is non-slender) R h = 1.00 (since the section is non-hybrid) F nc ( FLB ) = 1 1 (1.00)(1.00) ( 50 ) = (1.00)(50 ) ODOT-LRFD Short Course - Steel AASHTO Flexure Example #6a AASHTO-LRFD 007 Created July 007: Page of

157 Investigate Compression Flange Lateral-Torsional Buckling: The unbraced length of the beam is L b = 1-0 = r t bfc (16") = = = 4.0" 1 (19")( 3 1 Dt 8 ") c w (16")( 3 3 b 4 ") fct fc ( ) L p E 9, 000 = 1.0 rt = (1.0)(4.0") = 101.6" ( ) F 50 yc = min ( 0.7, ) 0.5 yr ( ) F F F F yr yc yw yc F = (0.7) 50 = 35 (Pg 6-110) L r E 9, 000 =π rt = ( π )(4.0") = 381.6" ( ) F 35 yr Since L = 101.6" < L = 144" < L = 381.6", Inelastic LTB must be investigated. p b r F yr Lb L p F ( ) = C 1 1 R R F R R F RF h yc Lr L p nc LTB b b h yc b h yc ( ) F = ( 1.00)( 50 ) 381.6" 101.6" " 101.6" nc( LTB) C b ( )( )( ) ( )( ) F ( ) = C = C nc LTB b b ( )( )( ) ( )( )( ) ODOT-LRFD Short Course - Steel AASHTO Flexure Example #6a AASHTO-LRFD 007 Created July 007: Page 3 of

158 Compute the Moment Gradient Factor, C b, for segment BC of the beam, which will be critical. C b f f = f f ( ) C M = M = 1, 080 f = 3.86 k-ft c M = M = f = M k-ft o B o BC, mid = = k-ft 1, 013 fmid.38 Since the BMD is not concave, 1 mid ( ) ( ) f = f f f = () = o B f 1 f mid f C b = = Cb = F ( )( ) = = F = 50 nc( LTB) nc( LTB) The governing strength for the compression flange is the smaller of F nc(flb) and F nc(ltb) : Since flange. F nc( LTB) = 50 > Fnc ( FLB) = 46.70, FLB governs the strength of the compression F nc = F ( ) = nc FLB ( ) φ = (1.00) = F nc Check 1 fbu + fl φ f Fnc ( ) 3 f bu = f = C 3.86 Assume that Strength I Load Combination Governs, γ WS = 0.0 and f l = 0 Since 1 fbu + fl = 3.86 < φ f Fnc =46.70, the compression flange is adequate. 3 ODOT-LRFD Short Course - Steel AASHTO Flexure Example #6a AASHTO-LRFD 007 Created July 007: Page 4 of

159 Investigate the Strength of the Tension Flange: Since the tension flange is discretely braced, F ( ) nt = RhFyt = (1.0) 50 = 50 ( ) ( ) φ = (1.00) 50 = 50 F nt Check 1 fbu + fl φ f Fnt ( ) 3 f bu = f = C 3.86 Since 1 fbu + fl = 3.86 < φ f Fnt =50.00, 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 007 Created July 007: Page 5 of

160 AASHTO Flexural Example #6b: Problem: A non-composite built-up girder made of M70-50 steel is used to span 48. The beam is braced laterally at 1-0 intervals and is subjected to a factored load of w = 3.75 / ft, which includes the self weight of the beam. Check to see if the section is adequate considering flexural failure modes at the Strength Limit States. If appropriate, use the provisions in AASHTO Appendix A6 to determine capacity. Solution: Determine Classification of the Section: Check D t w c? E 5.7 ( ) F yc Take D = 38 D 38" D c = = = 19"? ()(19") 9,000 = = OK, web is non-slender ( ") Check I I yc yt? 0.3 ( ) Since Section is doubly symmetric, I yc = I yt OK ODOT-LRFD Short Course - Steel AASHTO Flexure Example #6b AASHTO-LRFD 007 Created July 007: Page 1 of

161 Since the web is non-slender and Eq is satisfied, we have the option of using either AASHTO Section 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 M nc(flb) Lateral-Torsional Buckling M nc(ltb) Yielding of Tension Flange M nt M nc Compute Web Plasticity Factors, R pc and R pt (Section A6.): Investigate the classification of the web. Check D t w cp? λ (A6..1-1) pw( D ) cp E Fyc Dcp λ pw( Dcp ) = λ rw 0.54M D p c 0.09 RM h y (A6..1-) E λ rw = 5.7 = (A6..1-3) F yc R h = 1.00 (since the section is rolled and is non-hybrid) M M ( )( ) ( )( ) = S F = in 50 = 7,160 =, 63 y x y 3 k-in k-ft = Z F = in 50 = 30,00 =,50 p x y 3 k-in k-ft 9, " λ pw( D cp ) = = = k-in ( 0.54)( 30, 00 ) 19" 0.09 k-in ( 1.0)( 7,160 ) λ pw( D cp ) = ODOT-LRFD Short Course - Steel AASHTO Flexure Example #6b AASHTO-LRFD 007 Created July 007: Page of

162 D? cp ()(19") = = 101.3>λ pw( D ) = web is non compact t w 3 8 " cp Since the web is non compact, R M λ λ M M c Rpc = 1 1 M λ λ M M Where, Dcp λ pw( Dc) =λpw( Dcp) λrw Dc 19'' = '' = h yc w pw( D ) p p p rw pw( Dc ) yc yc (A6..-4) (A6..-6) R pc = 1 1 = = k-in k-in k-in (1.00)(7,160 ) (30, 00 ) (30, 00 ) k-in k-in k-in (30, 00 ) (7,160 ) (7,160 ) R pt R pt R M λ λ M M c = 1 1 M λ λ M M h yt w pw( D ) p p p rw pw( Dc ) yt yt = 1 1 = = k-in k-in k-in (1.00)(7,160 ) (30, 00 ) (30, 00 ) k-in k-in k-in (30, 00 ) (7,160 ) (7,160 ) (A6..-5) Investigate Compression Flange Local Buckling: Investigate the compactness of the compression flange. bfc 16" λ f = = = (A6.3.-3) t ()( ") fc 3 4 E 9, 000 λ pf = 0.38 = 0.38 = 9.15 (A6.3.-4) F 50 yc ODOT-LRFD Short Course - Steel AASHTO Flexure Example #6b AASHTO-LRFD 007 Created July 007: Page 3 of

163 Since λ f >λ pf, the flange is non compact and, M FyrS xc λ f λ pf ( ) = 1 1 R M RpcM yc λrf λpf S xt F = min 0.7 F, R F, F 0.5F Sxc nc FLB pc yc yr yc h yt yw yc ; yr ( ) (A6.3.-) F = (0.7) 50 = 35 (Pg 6-) Ekc λ rf = 0.95 (A ) F k c yr 4 4 = = = (A6.3.-6) '' D 38 3 '' t 8 w ( 9, 000 )( ) λ rf = 0.95 = 17.4 ( 35 ) ( )( 3 ) ( 1.087)( 7,160 ) in = , k-in k-ft = 7,550 =, 96 M nc ( FLB ) k-in Investigate Compression Flange Lateral-Torsional Buckling: The unbraced length of the beam is L b = 1-0 = k-in ( )( ) r t bfc (16") = = = 4.0" 1 (19")( 3 1 Dt 8 ") c w (16")( 3 3 b 4 ") fct fc (A ) L p E 9, 000 = 1.0 rt = (1.0)(4.0") = 101.6" (A ) F 50 yc E J Fyr Sxch Lr = 1.95 rt F yr S xc h E J ( ) S xt F = min 0.7 F, R F, F 0.5F Sxc yr yc h yt yw yc ( )( ) F yr = = 35 (Pg 6-) ODOT-LRFD Short Course - Steel AASHTO Flexure Example #6b AASHTO-LRFD 007 Created July 007: Page 4 of

164 ( )( ) 3 h= D+ t + t = 38" + " = 38.75" 1 fc ft 4 J Dt bfc t fc t fc bft t w ft t ft = b fc 3 b ft (A ) J 3 3 (38")( ") (16")( 4") 4" = + () 1 ( 0.63) = in " 4 E 9, 000 F = yr 35 = Sxch J 88.6 ( in 3 )( 38.75" ) = = 4, in L r 1 4,180 = ( 1.95)( 4.0" )( 88.6) = 396.8" = 33.06' 4, Since L = 101.6" < L = 144" < L = 396.8", Inelastic LTB must be investigated. p b r FyrS xc Lb L p M ( ) = C 1 1 R M R M RpcM yc Lr L p nc LTB b pc yc pc yc (A6.3.3-) M 3 ( )( ) C b ( 1.087)( 7,160 ) ( Cb)( )( ) ( Cb)( ) in 144" 101.6" = , , " 101.6" nc( LTB) k-in = , 50 = 8, 010 9, 50 k-in k-in k-in k-in k-in ( )( ) ( )( ) ODOT-LRFD Short Course - Steel AASHTO Flexure Example #6b AASHTO-LRFD 007 Created July 007: Page 5 of

165 Compute the Moment Gradient Factor, C b, for segment BC of the beam, which will be critical. C b M M = M M (A ) M M M o = M = 1,080 = M = BC, mid c B = 1,013 k-ft k-ft k-ft Since the BMD is not concave, 1 mid ( ) ( ) M = M M M = () 1, 013 1, 080 = o k-ft k-ft k-ft k-ft C b = ,080 + = 1,080 = ( )( ) M nc( LTB) = ,010 = 9,70 9,50 k-in k-in k-in M = = k-in k-ft nc( LTB) 9,50,460 The governing strength for the compression flange is the smaller of M nc(flb) and M nc(ltb) : Since M k-ft k-ft nc( LTB) =,460 > Mnc( FLB) =,96, FLB governs the strength of the compression flange. M nc = Mnc( FLB) =,96 k-ft k-ft ( ) φ = (1.00), 96 =, 96 M nc k-ft Check 1 M u + fs l xc φ fmnc (A ) 3 M u = M = 1,080 C k-ft Assume that Strength I Load Combination Governs, γ WS = 0.0 and f l = 0 Since 1 k-ft k-ft Mu + flsxc = 1, 080 < φ fmnc=,96, the compression flange is adequate. 3 ODOT-LRFD Short Course - Steel AASHTO Flexure Example #6b AASHTO-LRFD 007 Created July 007: Page 6 of

166 Investigate the Strength of the Tension Flange: Since the tension flange is discretely braced, M ( ) k-in k-in k-ft nt = RptM yt = (1.087) 7,160 = 9,50 =, 460 (A6.4-1) k-ft ( ) φ = (1.00), 460 =, 460 M nt k-ft Check 1 M u + fs l xt φ fmnt ( ) 3 M u = M = 1,080 C k-ft 1 k-ft k-ft Since Mu + flsxt = 1, 080 < φ fmnt=,460, 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 ODOT-LRFD Short Course - Steel AASHTO Flexure Example #6b AASHTO-LRFD 007 Created July 007: Page 7 of

167 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 W7x94 made of M70-50 steel Solution: 1' 1' 1' Draw the shear force diagram. 60 SFD() -15 From the diagram, V u = Referring to Section of the Specification, Check Design Shear Strength: V n = CV p ( ) V = 0.58F Dt ( ) p y w D= d t f = 6.9" ()(0.745") = 5.41" p [ ] V = = (0.58)(50 ) (5.41")(0.490") ODOT-LRFD Short Course - Steel AASHTO Shear Strength Example #1 AASHTO-LRFD 007 Created July 007: Page 1 of

168 Since the web is unstiffened, k = ( 9, 000 )( 5.00) Ek F = = y ( 50 ) Ek ( )( ) F = = D 5.41" y D Ek Since = < 1.1 = 60.31, shear yielding governs and, t F w y t w = = " C = 1.00 ( ) V n = CV p = (1.00)(361.1 ) = φv n = (1.00)(361.1 ) = > V u = 60 O.K. ODOT-LRFD Short Course - Steel AASHTO Shear Strength Example #1 AASHTO-LRFD 007 Created July 007: Page of

169 AASHTO Shear Strength Example #: Problem: A built-up section made of M70-50 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 4. Solution: Referring to Section of the Specification: V n = CV p ( ) V = 0.58F Dt ( ) p y w p 3 [ 8 ] V = = (0.58)(50 ) (38")( ") Since the web is unstiffened, k = ( 9, 000 )( 5.00) Ek F = = y ( 50 ) Ek 1.1 ( 1.1)( 53.85) F = = y Ek 1.40 ( 1.40)( 53.85) F = = D 38" = = t 3 8 " y D Ek Since = > 1.40 = 75.39, Elastic shear buckling governs and, t F w y w ( 1.57) ( ) ( 9,000 )( 5.00) 1.57 Ek C = = = ( D/ t ) F y ( 50 w ) ( ) V n = CV p = (0.4437) (413.3 ) = φv n = (1.00)(183.4 ) = < V u = 4 ODOT-LRFD Short Course - Steel AASHTO Shear Strength Example # AASHTO-LRFD 007 Created July 007: Page 1 of

170 Try adding transverse stiffeners to the web to increase the shear strength. A panel aspect ratio of 1.5 to 1.50 looks good do D do 48" 1.5 do 1.5 D = (1.5)(38") = 47.5" say 48" = = 1.63 D 38" For stiffened webs, 5 5 k = ( d / D) = + (1.63) = ( ) o ( 9, 000 )( 8.134) Ek F = = y ( 50 ) Ek 1.1 ( 68.68)( 1.1) 76.9 F = = y Ek 1.40 ( 68.68)( 1.40) F = = y D Ek Since = > 1.40 = 96.15, Elastic shear buckling governs and, t F w y ( 1.57) ( ) ( 9,000 )( 8.134) 1.57 Ek C = = = ( D/ t ) F y ( 50 w ) ( ) V n = CV p = (0.718) (413.3 ) = 98.3 φv n = (1.00)(98.3 ) = 98.3 > V u = 4 ODOT-LRFD Short Course - Steel AASHTO Shear Strength Example # AASHTO-LRFD 007 Created July 007: Page of

171 The previous calculations were based on the buckling strength of the web. For interior panels where: Dt w ( bfctfc + bfttft ).5 ()(38")( 3 8 ") 8.5 = = [ ] (16")( ") (16")( ") 4.0 ( ) OK Tension Field Action can be developed: 0.87(1 C) Vn = Vp C+ do 1+ D ( ) V n (0.87)( ) = + = 1 ( ) (413.3 ) (0.718) φv n = (1.00)(360.4 ) = ODOT-LRFD Short Course - Steel AASHTO Shear Strength Example # AASHTO-LRFD 007 Created July 007: Page 3 of

172

173 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 W7x94 made of M70-50 steel ' 1' 1' Solution: Draw the shear force diagram. 60 SFD() 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: R = (.5 k+ N) F t (D6.5.-3) n yw w R = + = n ((.5)(1.34") 3.5")(50 )(0.490") φ R n = (1.0)(161.7 ) = > 60 O.K. ODOT-LRFD Short Course - Steel AASHTO Web Strength Example #1 AASHTO-LRFD 007 Created July 007: Page 1 of

174 Check Web Crippling Since the supports are likely to be at a distance less than or equal to d/ from the end of the member. Check N d : 3.5" " = Therefore, (D ) controls: R n 1.5 N t EF t d t tw w = 0.40tw 1+ 3 f yw f R n " 0.490" (9, 000 )(50 )(0.745") = 0.40(0.490") " 0.745" 0.490" R = ( in )(1.193)(1485 ) = 170. n φ R n = (0.80)(170. ) = 136. > 60 O.K. Check the Interior Concentrated Loads for Web Yielding and Web Crippling: (Assume that the bearing length, N, is 3.5 ) Check Web Yielding Since the applied load is located at a distance greater than d from the end of the member. R = (5 k+ N) F t (D6.5.-) n yw w R = + = n ((5)(1.34") 3.5")(50 )(0.490") 43.8 φ R n = (1.0)(43.8 ) = 43.8 > 75 O.K. ODOT-LRFD Short Course - Steel AASHTO Web Strength Example #1 AASHTO-LRFD 007 Created July 007: Page of

175 Check Web Crippling Since the applied load is located at a distance greater than d/ from the end of the member. Therefore, (D6.5.3-) controls: R n 1.5 N t EF t d t tw w = 0.80tw 1+ 3 f yw f R n " 0.490" (9, 000 )(50 )(0.745") = 0.80(0.490") " 0.745" 0.490" R = (0.191 in )(1.193)(1485 ) = n φ R n = (0.80)( ) = 7. > 75 O.K. The Web Yielding and Web Crippling Strengths are Satisfactory ODOT-LRFD Short Course - Steel AASHTO Web Strength Example #1 AASHTO-LRFD 007 Created July 007: Page 3 of

176 AASHTO Web Strength Example #: Problem: A built-up section made of M70-50 steel, is used as a beam. It was determined in AASHTO Shear Strength Example # 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 18 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 J ( ) 3 w and I t D ρ F t yw 40 E 1.5 ( ) where: I t - Moment of inertia of the stiffener pair about the mid-thickness of the web. J D = do / D use J.5 = ( ) ( d / D) o ODOT-LRFD Short Course - Steel AASHTO Web Strength Example # AASHTO-LRFD 007 Created July 007: Page 1 of

177 .5 J =.0 = in 0.5 take J = 0.50 ( 48"/ 38" ) b = smaller of d o and D, b = 38 ρ = larger of F yw / F crs and E (0.31)(9, 000 ) crs ys crs b 6" t 3 t 8 " p F = F F = = ρ = 50 / ρ=1.44 bt j = (38")( ") (0.5) = 1.00 in w 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 ρ F t yw (38") (1.44) 50 D = = in 40 E 40 9,000 4 I t 3 ( ts)( bs + tw) = take b s = 6 1 I t [ + ] 3 ( ts ) ()(6") " = = ( ts)157.9 in btwj 1.00 in t p = = " say t s = 8 " in in ODOT-LRFD Short Course - Steel AASHTO Web Strength Example # AASHTO-LRFD 007 Created July 007: Page of

178 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 = 1.15 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. R = (.5 k+ N) F t (D6.5.-3) n yw w R = + = n 3 ((.5)(1.15") 9")(50 )( 8 ") 1.5 R n (1.0)(1.5 ) 1.5 φ = = > 18 O.K. Check Web Crippling Since the supports are likely to be at a distance less than or equal to d/ from the end of the member. Check N d : 9" " + ()( ") = > 3 4 Therefore, (D ) controls: R n 1.5 4N t EF t w = 0.40t w d tf tw yw f R n (4)(9") 3 8" (9,000 )(50 )( 4") = (0.40)( 8 ") (38" ()( ") 4" + ( 8") R = ( in )(1.5)(1, 703 ) = n φ R n = (0.80)(119.9 ) = 95.9 <18 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 # AASHTO-LRFD 007 Created July 007: Page 3 of

179 Design the bearing stiffeners that need to be added to increase the web crippling strength: Check local buckling of the bearing stiffener: b t E 0.48tp take F y = 50 ( ) F y 9, 000 bt 0.48tp = 11.56t p take b s = 7 50 Taking F y = 50 here is a conservative assumption since I am not sure what material will actually be used. bt 7" t p = = " take t 5 s = 8 " (7 x 5 / 8 bar stock may be used) Check the bearing stiffeners as an effective column section: ( 1 ) ( 8 ")( 7" 8 " 7" ) ( 6.75" 8 ")( 8 ") I = = in 4 ( 5 8" )( )( 7" ) ( 6.75" )( 3 8" ) 11.8 in A = + = 4 I in r = = = 3.704" A 11.8 in KL (0.75)(38") = = r 3.704" F y KL λ= = = rπ E π 9, 000 P = 0.66 F A ( ) ( )( )( ) Inelastic Buckling ( ) λ n y s = in = ( ) ( )( ) φ = = P n Taking F y = 36 here is a conservative assumption since I am not sure what material will be used. 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 9t w. ODOT-LRFD Short Course - Steel AASHTO Web Strength Example # AASHTO-LRFD 007 Created July 007: Page 4 of

180 Check bearing stress between the end of the bearing stiffeners and the loaded flange: R = 1.4A F ( ) n pn ys The corners of the stiffeners are clipped 1 horizontal and 1/ vertical to provide clearance for the flange-to-web welds 5 ( 8 ) A = ()(7" 1") " = 7.50 in pn R = (1.4)(7.50 in )(36 ) = n φ = = R n (1.00)(378.0 ) The capacity of the bearing stiffeners is governed by the equivalent column capacity. φr n = 364, which is greater than the reaction of 18. ODOT-LRFD Short Course - Steel AASHTO Web Strength Example # AASHTO-LRFD 007 Created July 007: Page 5 of

181 Just for fun, lets check the capacity of 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: ( 1 ) ( 8 )( 8 ) ( ( 8 ))( 8 ) I = () " 7" + " + 7" + ()(3.375") + 7" () " " = in ( 5 8" )( 4)( 7" ) () ( 3.375" ) 7" ( 3 8" ).66 in A = + + = 4 I in r = = = 3.696" A.66 in KL (0.75)(38") = = r 3.696" F y KL λ= = = rπ E π 9, 000 Inelastic Buckling ( ) P ( ) ( )( )( ) λ n = 0.66 FyAs = in = 813. ( ) ( )( ) φ = = P n 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: φr n = ()(378.0 ) = The strength is governed again by the equivalent column capacity, φr n = 73. ODOT-LRFD Short Course - Steel AASHTO Web Strength Example # AASHTO-LRFD 007 Created July 007: Page 6 of

182

183 AASHTO Connection Example #1: Problem: A C1x30 is used as a tension member as is shown in the sketch below. The channel is made of M70-36 material and is attached to the gusset plate with 8, 7/8 diameter M164 (A35) bolts. The gusset is 5/8 thick and made of M70-36 steel. Calculate the design capacity, φp n, of the connection considering the failure modes of bolt shear, bolt bearing, and block shear. Also compute the load which will cause slip of the connection. Solution: Shear Strength of the Bolts: Section A-A Assume that the threads are included in the shear plane of the connection. R = 0.38AF N ( ) n b ub s 3" 6" 3" A b π = = 4 7 ( ) 8 " in 3" 3" 1.5" For A35 bolts, F ub = 10 3" Bolts are in single shear so N s = 1 A A ( )( ) R n = (0.38) in 10 (1) = 7.4 bolt ( bolt ) φ = (0.80) 7.4 = 1.94 R n bolt For all 8 bolts, R n ( bolt ) φ = (8 bolts) 1.94 = P u C1 x 30 ODOT-LRFD Short Course - Steel AASHTO Connection Example #1 AASHTO-LRFD 007 Created July 007: Page 1 of

184 Check Bearing Strength: Interior Bolts Bearing on the Channel Web: ( )( 1 7 )( 1 ) L = 3" " + " =.063" Since L c =.063 > d = 1.75, c 8 16 R n =.4dtF R = (.4)( ")(0.510")( 58 ) = 6.11 ( ) u n 7 8 bolt Interior Bolts Bearing on the Gusset Plate: ( )( 1 7 )( 1 ) L = 3" " + " =.063" Since L c =.063 > d = 1.75, c 8 16 R n =.4dtF R = (.4) 7 5 ( ")( ")( 58 ) = ( ) u n 8 8 bolt End Bolts Bearing on the Channel Web: ( 1 7 )( 1 ) L = 1.5" " + " = 1.031" Since L c = < d = 1.75, R c 8 16 = 1.LtF R n = (1.)( 1.031" )(0.510")( 58 ) = bolt ( ) n c u End Bolts Bearing on the Gusset Plate: (Assume that the end distance on the gusset is 1 1 / ) ( 1 7 )( 1 ) L = 1.5" " + " = 1.031" Since L c = < d = 1.75, R c 8 16 = 1.LtF R = (1.)( 1.031" )( ")( 58 ) = ( ) n c u n 5 8 bolt For all 8 Bolts: ( bolt ) ( bolt ) ( bolt ) R = ( bolts) (4 bolts) ( bolts) = n ( ) φ = (0.80) = 39.1 R n ODOT-LRFD Short Course - Steel AASHTO Connection Example #1 AASHTO-LRFD 007 Created July 007: Page of

185 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: A = (6")(0.510") = in tg ( 1 7 )( 1 ) A = (6") () " + " (0.510") =.550 in tn 8 8 [ ] A = () (1.5") + (3)(3") (0.510") = in vg ( 1 ) A = () (1.5") + (3)(3") (3.5) " + " (0.510") = in vn A tn? 0.58A vn?.550 in (0.58)(7.140 in ) = in NO! R = 0.58FA + FA ( ) n u vn y tg ( )( ) ( )( ) R = (0.58) in in = n ( ) φ = (0.80) = 80.3 R n The Shear Strength of the Bolts Governs, φr n = 176 ODOT-LRFD Short Course - Steel AASHTO Connection Example #1 AASHTO-LRFD 007 Created July 007: Page 3 of

186 Check the Slip Capacity of the Connection: R = KKNP ( ) n h s s t K h = 1.00 (for standard holes) K s = 0.33 (assume Class A surface) N s = 1 P t = 39 (from Table for M164 Bolts) ( ) R n = (1.00)(0.33)(1) 39 = 1.87 bolt For All 8 Bolts: ( bolt ) R = (8 bolts) 1.87 = n ODOT-LRFD Short Course - Steel AASHTO Connection Example #1 AASHTO-LRFD 007 Created July 007: Page 4 of

187 AASHTO Connection Example #: 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 10. Given that 4, 1 diameter M164 (A35) 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: Q u 3 3b t = P 8a 0 ( ) g t 7" b= k 1 1 = 1 8" =.375" (k 1 = 1 1 / 8 for W1 x 13) 1 1 ( )( ) ( )( ) a= b g = 1.4" 7" =.700" f t 3 (3)(.375") (1.04") Q u = Pu = 0.736Pu (8)(.700") 0 ( ) T = Q + P = 1.74 P = (1.74) 10 = 15.8 u u u u Tensile Resistance of the Bolts: T = 0.76A F ( ) n b ub A b π = = 4 ( ) 1" in F ub = 10 ( )( T n = (0.76) in 10 ) = bolt T n ( bolt ) For All 4 Bolts: T n ( bolt ) φ = (0.80) = φ = (4 bolts) = 9. OK bolt ODOT-LRFD Short Course - Steel AASHTO Connection Example # AASHTO-LRFD 007 Created July 007: Page 1 of

188 Check the Strength of the Flange of the WT: Assume that a plastic mechanism forms between the bolt lines and stem. Moment Equilibrium about at the fillet: Q T T Q Pb u M M u = ( ) 10 (.375") M u = 4 = 71.5 k-in T u M u P u / b M u P u For Safety, M p M u Lt (8")(1.04") k-in = = 50 = k-in k-in φ = (1.00) 108. = 108. OK φ. M p Fy ( ) M p ( ) ODOT-LRFD Short Course - Steel AASHTO Connection Example # AASHTO-LRFD 007 Created July 007: Page of

189 AASHTO Connection Example #3: Problem: Assuming an unfactored fatigue load of 60, determine the fatigue life of the tension bolts in the previous example. Solution: For Safety, ( f ) ( F ) n ( f ) γ Δ Δ ( ) γ Δ γ Δ = = = 18.4 ( P) (0.75) ( 1.74)( 60 ) Abolts (4)( in ) A Δ = n N ( F ) 1 3 ( ΔF ) For infinite life, ( F ) TH ( ΔF ) 31.0 TH Δ = = = 15.5 n Since ( f ) ( F) γ Δ = 18.4 > Δ = 15.5, the bolts will have a finite life n A γ Δ Δ = n N For finite life, ( f ) ( F) ( γ( Δf )) ( 18.4 ) A N = = 81,800 cycles Note that if prying is not included, ( f ) 14.3 γ Δ = 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 007 Created July 007: Page 1 of

190 AASHTO Connection Example #4: Problem: Suppose that the hanger depicted in Examples # 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: V u Pu = = Pu = V u = = bolts bolt T u 1.74Pu = = Pu = T u = = bolts bolt Assume that the threads are included in the shear plane of the connection. V = R = 0.38A F N ( ) n n b ub s ( )( ) V n = (0.38) in 10 (1) = bolt Vu V = =, n bolt bolt V u Tn = 0.76AbFub 1 φvn ( ) T n ( 6.83 bolt ) ( bolt ) = (0.76)( in )( 10 ) 1 = 5.11 (0.80) bolt ( bolt ) φ = (0.80) 5.11 = 0.09 T n bolt Since φ T = 0.09 > T = 17.08, the bolts are OK for the loading shown. n bolt u bolt ODOT-LRFD Short Course - Steel AASHTO Connection Example #4 AASHTO-LRFD 007 Created July 007: Page 1 of

191 Check Bearing of the Flange of the WT: Interior Bolts: It is given that the WT is 8 long. Since the minimum spacing is 3, we ll assume that an end distance of is provided resulting in a spacing of 4 bolt-to-bolt. ( )( 1 )( 1 ) L = 4" 1" + " =.938" c 16 Since L c =.938 > d =, R n =.4dtF ( ) u End Bolts: ( ) ( ) R n = (.4) 1" (1.04") 65 = 16. bolt ( 1 )( 1 ) L = " 1" + " = 1.469" c 16 Since L c = < d =, R = 1.LtF ( ) n c u ( ) ( ) R n = (1.) 1.469" (1.04") 65 = bolt For all 4 Bolts: ( bolt ) ( bolt ) R = ( bolts) ( bolts) = 56.8 n ( ) φ = (0.80) 56.8 = R n Since φ R = > =, the flange of the WT will be OK in bearing. n 450 Vu 107 Note that since the flange thickness of the W4x176 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 W4x176. ODOT-LRFD Short Course - Steel AASHTO Connection Example #4 AASHTO-LRFD 007 Created July 007: Page of

192 Check Shear in the Stem of the WT: R = 0.58AF ( ) n g y [ ]( ) R = (0.58) (8")(0.650") 50 = n ( ) φ = (1.00) = R n Since φ R = > =, the stem will be satisfactory in shear. n Vu ODOT-LRFD Short Course - Steel AASHTO Connection Example #4 AASHTO-LRFD 007 Created July 007: Page 3 of

193 AASHTO Connection Example #5: Problem: An L6 x 4 x 1/, M70-36, is welded to a 3 / 8 thick gusset plate made of M70-50 steel. The long leg of the angle is attached using two, 8 long fillet welds. The capacity of the angle was previously computed as φp n = 163 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 F u = 65. The maximum weld size is 1 / - 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. φ R = 0.6φ F A φ P n, weld e exx w n, member ( )( )( ) ( w) ( )( ) φ Rnweld, = (0.7071) 8" w = 0.488" Say 7 16 " inch ODOT-LRFD Short Course - Steel AASHTO Connection Example #5 AASHTO-LRFD 007 Created July 007: Page 1 of

194 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 ( )( ) Tan( ) = 6" + 8" 30 = 15.4" Gross Section Yielding: ( ) 3 ( )( 8 ) φ Pn =φ FyAg = (0.95) " " = 71.4 Net Section Fracture: ( ) 3 ( )( 8 ) ( ) φ Pn =φ FuAe = (0.80) " " 1.00 = 97.1 (Taking U = 1.00) Check Block Shear in the Gusset Plate: A tg = A = (6") ( ") =.50 in A = A = ()( 8" )( ") = in tn 3 8 vg vn 3 8 A?? tn Avn = in (0.58)(6.000 in ) in NO! R = 0.58FA + FA ( ) n u vn y tg ( )( ) ( )( ) R = (0.58) in in = n ( ) φ = (0.80) = 71.0 R n Use 7/16 x 8 Fillet Welds made with an E70 Electrode ODOT-LRFD Short Course - Steel AASHTO Connection Example #5 AASHTO-LRFD 007 Created July 007: Page of

195 AASHTO Connection Example #6a: Problem: Use the elastic vector method to compute the maximum force on any bolt in the eccentrically loaded bolt group shown in the figure below. The bolts are all the same size. (Example from Salmon & Johnson) 4" 3" P A B 3" C D 3" E F Solution: Tr τ= T = ( 4 )( 3" + " ) = 10 J J = Ad = A d k-in π J = ( 4 )( (") (3") ) ( )( " ) ( 1" ) J = 47.1 in 4 Corner Bolts: Tr τ= = J τ= 9.18 k-in ( 10 )( (") + (3") ) 47.1 in π V = ( 9.18 ) ( 1" ) = ODOT-LRFD Short Course - Steel AASHTO Connection Example #6a AASHTO-LRFD 007 Created July 007: Page 1 of

196 Force acts perpendicular to line drawn from bolt to C.G. Breaking force into horizontal and vertical components V V y x = ( V) = ( 7.11 ) = = ( V) = ( 7.11 ) = Add evenly distributed vertical force to the vertical, torsional force for Bolt B V V y x = = = V = V + V = ( ) + (6.000 ) = total x y Check These Results with a Spreadsheet Solution: P x : 0 x CG : 0 P y : -4 y CG : 0 e x : 5 e y : 0 Σ d : T = V max = 10.0 Bolt x y d V x V y V total A B C D E F Everything checks out OK. ODOT-LRFD Short Course - Steel AASHTO Connection Example #6a AASHTO-LRFD 007 Created July 007: Page of

197 AASHTO Connection Example #6b: Problem: Use the simplified equations to solve the previous example problem. (Example from Salmon & Johnson) 4" 3" P A B C D E F Solution: T ( )( ) = 4 3" " = 10 k-in d d ( 4 ) (") (3") ( )( " ) = + + = in Looking at Bolt B: V V k-in ( 10 )( 3" ) = = in Bx, k-in ( 10 )( " ) = = in B, y V V B, total B, total 4 = ( ) + ( ) + 6 = ODOT-LRFD Short Course - Steel AASHTO Connection Example #6b AASHTO-LRFD 007 Created July 007: Page 1 of

198 AASHTO Connection Example #7: Problem: Detail a splice between two non-composite W30 x 99 M70 Gr 50. Take M u at the location of the splice as 810 k-ft and take φm n as 1,300 k-ft. Solution: The splice is designed for the larger of: M +φ M + = = 1,055 k-ft k-ft ubeam, nbeam, 810 1,300 k-ft M nbeam, k-ft ( )( ) 0.75 φ = ,300 = k-ft Since 1,055 k-ft > k-ft, M u,splice = 1,055 k-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 fcf F = F +αφ 0.75 αφ F R h cf f yf f yf ( c-1) f cf k-ft 9.7" 0.670" ( 810 )( 1 in )( ) ft = = 3,990 in R h = 1.00, since the beam is non-hybrid. α = Since we are assuming that φm n = φm p, F n = F yf F cf ( 1.00)( 1.00)( 50 ) ( 0.75)( 1.00)( 1.00)( 50 ) = Fcf = = 4.68 ODOT-LRFD Short Course - Steel AASHTO Connection Example #7 AASHTO-LRFD 007 Created July 007: Page 1 of

199 For the Non-Controlling Flange: fncf Fncf = Rcf 0.75 αφ f Fyf ( c-3) R h f ncf = f = cf R cf F cf 4.68 = = = 1.07 f cf F ncf ( 1.07) ( 0.75)( 1.00)( 1.00)( 50 ) = 1.00 Fncf = = 4.68 For the Compression Flange: P = F A ucomp, cf e In compression, A e is taken as the gross area of the flange. A e g ( 10.5" )( 0.670" ) in = A = = ( )( ) P ucomp, = in = For the Tension Flange: P = F A u, Ten cf e φuf u In tension, A = A A yf φ y e n g ( c-) For 1 diameter bolts, ( ) ( )( ) ( ) A 10.5" 1 1 n = 8 " 0.670" = 5.58 in ODOT-LRFD Short Course - Steel AASHTO Connection Example #7 AASHTO-LRFD 007 Created July 007: Page of

200 A e ( 0.80)( 65 ) ( 0.95)( 50 ) = ( 5.58 in ) in Ae = 6.05 in in = 6.05 in ( )( ) P, = in = 58.3 uten 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 (A35) bolts P uten, = = P ucomp, = = 150. For the Outer Flange Splice Plate: Gross Yielding (Tension): φ P =φ F A P ( ) n y y g u, Ten ( )( )( )( t ) φ P = " 19.1 n outer touter 19.1 ( 0.95)( 50 )( 10.5" ) 0.589" say " 5 16 Net Section Fracture (Tension): φ P =φ F AU P ( ) n u u n u, Ten ( )( ) ( ) ( )( ) ( t )( ) φ P " 1" 1 n = + 8 " outer touter 19.1 ( 0.80)( 65 )( 8.5" ) " say " 5 16 ODOT-LRFD Short Course - Steel AASHTO Connection Example #7 AASHTO-LRFD 007 Created July 007: Page 3 of

201 Gross Yielding (Compression): φ P =φ F A P ( c-4) n c y g u, Comp ( )( )( )( t ) φ P = " 150. n outer touter 150. ( 0.90)( 50 )( 10.5" ) " say " 3 8 For the Inner Flange Splice Plates: The widths of the inner splice plates will be roughly equal the flange width of the section minus the thickness of the web and fillets. 1 ( )( ) W = b k = 10.5" 1 " = 8.375" inner f Take the width of each of the two inner plates as Gross Yielding (Tension): φ P =φf A P ( ) n y g u, Ten ( )( )( )( )( t ) φ P = " 19.1 n Inner tinner 19.1 ( 0.95)( 50 )( )( 4.00" ) 0.397" say " 3 8 Net Section Fracture (Tension): φ P =φf AU P ( ) n u n u, Ten ( )( )( ) ( ) ( ) ( t )( ) φ P " 1" 1 n = + 8 " Inner ODOT-LRFD Short Course - Steel AASHTO Connection Example #7 AASHTO-LRFD 007 Created July 007: Page 4 of

202 tinner 19.1 ( 0.80)( 65 )( 5.75" ) " say " 7 16 Gross Yielding (Compression): φ P =φ F A P ( c-4) n c y g u, Comp ( )( )( )( )( t ) φ P = " 150. n Inner tinner 150. ( 0.90)( 50 )( )( 4.00" ) 0.417" say " 7 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 6-191). A = ( 10.5" )( ") = in A = ( )( 4.00" )( ") = in Outer 3 8 Inner 7 16 A Outer A A Ave Inner ( ) ( in ) ( in ) ( in ) + ( in ) = = 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 () the inner plate thickness must be increased to 1 /, which would result in a difference in area between the outer and inner plates of less than %. The second option will be selected for the case of this example. Outer Flange Splice Plate: 10 1 / x 3 / 8 Inner Flange Splice Plates: 4 x 1 / 8 ODOT-LRFD Short Course - Steel AASHTO Connection Example #7 AASHTO-LRFD 007 Created July 007: Page 5 of

203 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. R = 0.38AF N ( ) n b ub s A b π = = 4 ( ) 1" in For A35 bolts, F ub = 10 ( )( ) R n = (0.38) in 10 () = bolt ( bolt ) φ = (0.80) = R n bolt Determine the number of flange bolts required: n fb = = 5.4 bolts say 6 bolts bolt 4 1 / " " 10 1 /" 6 1 /" " 1 / " 3 1 / " 3 1 / 1 / " 1 / " 1 " 3 1 / " 3 1 / " / " 1 / " gap between ends of beams W30 x 99 PL 4 1 / x 10 1 / " x 3 / 8 " W30 x 99 PL 4 1 / " x 4" x 1 / " Each Side 1" dia M164 Bolts (1 places) ODOT-LRFD Short Course - Steel AASHTO Connection Example #7 AASHTO-LRFD 007 Created July 007: Page 6 of

204 Check Bolt Bearing in the Flange Splice: Interior Bolts Bearing on the Beam Flange: ( )( )( ) L = 3 " 1" + " =.438" Since L c =.433 > d =.0, c R n =.4dtFu R n = (.4)( 1" )(0.670") ( 65 ) = bolt ( ) Interior Bolts Bearing on the Splice Plates: ( )( )( ) L = 3 " 1" + " =.438" Since L c =.433 > d =.0, c R n 3 1 =.4dtF R = (.4)( 1" )( " + ")( 65 ) = ( ) u n 8 bolt End Bolts Bearing on the Beam Flange: ( )( ) L = " 1" + " = 1.969" Since L c = < d =.0, R c = 1.LtF R n = (1.)( 1.969" )(0.670")( 65 ) = 10.9 bolt ( ) n c u End Bolts Bearing on the Splice Plates: ( )( ) L = " 1" + " = 1.969" Since L c = < d =.0, R c = 1.LtF R = (1.)( 1.969" )( " + ")( 65 ) = ( ) n c u n 8 bolt For all 6 Bolts: ( bolt ) ( bolt ) ( bolt ) R = ( bolts) ( bolts) ( bolts) 10.9 = 63.9 n ( ) φ = (0.80) 63.9 = OK R n ODOT-LRFD Short Course - Steel AASHTO Connection Example #7 AASHTO-LRFD 007 Created July 007: Page 7 of

205 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, F s, times the smaller gross flange area on either side of the splice. Take the Service II moment as 548 k-ft P = F A where slip s g F s k-ft 9.7" 0.670" ( 548 )( 1 in )( ) fs ft = = = 3.9 ( c-5) R h 4 ( 1.00)( 3,990 in ) ( )( )( ) P = " 0.670" = slip The slip resistance of a single bolt is taken as: R = KKNP ( ) n h s s t K h = 1.00 (for standard holes) K s = 0.33 (assume Class A surface) N s = P t = 51 (from Table for M164 Bolts) ( ) R n = (1.00)(0.33)() 51 = bolt Determine the number of flange bolts required: n fb = = 5.00 bolts 6 bolts will work bolt ODOT-LRFD Short Course - Steel AASHTO Connection Example #7 AASHTO-LRFD 007 Created July 007: Page 8 of

206 Check Block Shear of the Beam Flange: 3 1 / " 3 1 / " 1 / " " tg ( )( " )( 0.670" ).680 in A = = Tension Shear 6 1 /" ( ) " ( )( 1" ") ( 0.670" ) 1.96 in Atn = = 8 vg ( )( 3 1 " 3 1 " 1 ")( 0.670" ) 1.73 in A = + + = Shear " ( ) 1 ( )( 1 ) ( ) A = 9 ".5 1" + " 0.670" = in vn 8 A tn? 0.58A vn? 1.96 in (0.58)(8.961 in ) = in NO! R = 0.58FA + FA ( ) n u vn y tg ( )( ) ( )( ) R = (0.58) in in = n ( ) φ = (0.80) = OK R n ODOT-LRFD Short Course - Steel AASHTO Connection Example #7 AASHTO-LRFD 007 Created July 007: Page 9 of

207 B) Web Splice: The web splice is to be designed for the following actions at the Strength Limit: 1. V uw - The direct shear force.. M vuw - The moment on the web bolts caused by the eccentricity of V uw 3. M uw - The portion of the bending moment in the beam that is carried by the web. 4. H uw - 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 V u = 45 and the nominal shear capacity of the beam is φv n = Determine the direct shear force acting on the web splice, V uw :? u< 0.5φvVn? V ( )( )( ) 45 < = 13.8 Since V < 0.5φ V, u v n V uw ( )( ) = 1.5V = = 67.5 ( b-1) u. Determine the moment, M vuw, that is caused by the eccentricity of the direct shear, V uw : Assuming the arrangement of bolts shown on Page 1, the distance from the CG of the bolt group on one side of the splice to the CL of the splice is, 1 ( )( 1 1 ) 1 ( )( 1 ) e = 3 " + " + " = 4.50" M vuw = e V = uw k-in ( 4.50" )( 67.5 ) = = 5.31 k-ft I used a gap of 1/ here to be conservative. I understand that most splices will use much narrower gaps - these calculations will be conservative for smaller gaps. ODOT-LRFD Short Course - Steel AASHTO Connection Example #7 AASHTO-LRFD 007 Created July 007: Page 10 of

208 3. Determine the portion of the beam moment that is carried by the web splice, M uw : twd M uw = RF h cf Rcf fncf (C b-1) 1 F = 4.68 (Positive since it s in tension) cf R = 1.07 (from Before) cf f = (Negative since it s in compression) ncf M uw ( 0.50" )( 8.36" ) = 1 ( ) ( 1.00)( 4.08 ) ( 1.07)( ) 3 k-in k-ft = in =,975 = Determine the horizontal force that results from moving the beam moment, H uw : twd Huw = Rh Fcf + Rcf fncf (C b-) 1 H uw ( 0.50" )( 8.36" ) = + 1 ( ) = 1.9 in = 0.00 ( 1.00)( 4.68 ) ( 1.07)( ) In this case, the ENA is at the mid-height of the beam. Since H uw is the horizontal force that results from the eccentricity of the ENA relative to the mid-height of the beam, it makes sense that H uw is zero. ODOT-LRFD Short Course - Steel AASHTO Connection Example #7 AASHTO-LRFD 007 Created July 007: Page 11 of

209 The total moment acting on the web splice is, k-in k-in k-in MTotal = Mvuw + Muw = ,975 = 3, 79 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, V uw, but not the moment, M Total. This can be accommodated by computing a fictitious shear force, P, that when applied over the eccentricity, e, results in the same actions as the actually applied shear and moment. k-in 3,79 P = = " 1 /" 3 1 /" 3 /4" A 1 B C D E F G P = 78.7 H e = 4 1 /" ODOT-LRFD Short Course - Steel AASHTO Connection Example #7 AASHTO-LRFD 007 Created July 007: Page 1 of

210 ODOT-LRFD Short Course - Steel AASHTO Connection Example #7 AASHTO-LRFD 007 Created July 007: Page 13 of