Design of Welded Steel Plate Girders

7/30/2019 Design of Welded Steel Plate Girders

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4 Design ofWelded Steel Plate GirdersIn accordance with the 1992 Fifteenth Edition AASHTO Specifications andRevisionsby Ca ltrans Bridge Design Specifications

ContentsNotations and Abbreviations ................... .......................................... 4-1

4.0 Introduction ......... ............................................... ............................. .. 4-9 4.1 General Design Considerations ..................................................... .. 4-10 4.2 Design Loads .................................... .................................. .... .. ....... 4-11 4.3 Design for Maximum Loads ........................ .. ... ... ........................... 4-11

4.3.1 Braced Sections ... .................................................................................. 4-11 4.3.2 Unbraced Sections ........ .... ............................................................ ......... 4-13 4.3.3 Shear Capacity and Design .................................. .................................. 4-14

4.4 Composite Girders ................ ....... ............. .. .. ..... .............................. 4-16 4.5 Fatigue Design .......................................... .............. ........................ 4-18

4.5.1 Factors Affecting Fatigue Performance ................................................. 4-18 4.5.2 Applied Stress Range............................................................................. 4-19 4.5.3 Allowable Stress Range ........................................................... .............. 4-20 4.5.4 Type of Loading ................................ .................................................... 4-20 4.5.5 Stress Category ...................................................................................... 4-20


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4.8 Design Example Problem .......... ............................ .. ....................... .4-23 Design E ~ a m p l e Solution

4.9 Loading .. .. ................................................. .......... .. ......... ...... ........... 4-25 4.9.1 Dead Load ................. .. ..... ............................... .. ..................... ................ 4-25 4.9.2 LiveLoad .................................... .......... ............. .. ..... .. .. .. ....................... 4-26

4.10 Composite Section Design ........................................... ...... ............. 4-29 4.10.1 Design Loads .................... ............................................ ....................... 4-29 4.10.2 Fatigue Loads .................................................. .............. .. .. .................. 4-30 4.10.3 Girder Section ...................................................................................... 4-31 4.10.4 Width to Thickness Ratios ................................................ .. ..... ............ 4-37 4.10.5 Bracing Requirements .................................................. ....................... 4-37 4.10.6 Fatigue Requirements .......................................................................... 4-38

4.10.6.1 Applied and Allowable Stress Ranges ........................................ 4-48 4.10.7 Shear Design ........................................................................................ 4-49

4.10.7.1 Moment and Shear Interaction ................................................... 4-50 4.10.7.2 Transverse Stiffener Design .................................................... .. .. 4-51

4.11 Non-Composite SectionDesign ............................................ ... .. .....4-52 4.11 .1 Design Loads ................................................ ....................................... 4-52 4.11.2 Girder Section ...................................................................................... 4-53 4.1 1.3 Width to Thickness Ratios ................................................................... 4-55 4.11.4 Bracing Requirements ......................................................................... 4-56 4 .11 .5 Fatigue Requirements .......................................................................... 4-56


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4.12 Flange-to-Web Weld ... ... ........ ......................... .. ....... ........................ 4-63 4.12.1 Weld Design .......... .. ........................... .......... ... .... ................................. 4-63 4. 12.2 Fatigue Check .................................. .................................................... 4-64

4.13 Shear Connectors ........ ..... ..... .. ............. .... ........................................4-65 4.13.1 Fatigue Design ..................................................................................... 4-65 4.13.2 Ultimate Strength .................................................................. ......... ...... 4-68 4. 13.3 Shear Connectors at Points of Contraflexure ...................................... 4-69

4.14 Bearing Stiffener at Pier 2 ... ..... .......... ............................................4-70 4.15 Splice Plate Connection TO BEADDED ATA FUTURE DATE .... 4-75 4.16 Bridge Design System Computer Output. ....................... .. .... 4-A to 4-1


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Notations

Design ofWelded Steel Plate Girders

References within parentheses are to Bridge Design Specifications, Section 10. A = areaofcrosssection (Articles 10.37.1.1, 10.34.4, 10.48.1.1 , 10.48.2.1, 10.48.4.2,

10.48.5.3 and 10.55.1 )= bending moment coefficient (Article 10.50.1.1.2)= amplification factor (Articles 10.37.1.1 and 10.55.1)= product of area and yield point for bottom flange of steel section (Article10.50.1.1.1)= product of area and yield point of that part of reinforcing which lies in the

compression zone of the slab (Article 10.50.1.1.1)(AFy)tf = productofarea and yield pointfortop flange of steel section (Article 10.50.1.1.1)(AFy)w = product of area and yield point for web of steel section (Article 10.50.1.1.1 )AJ = area of flange (Articles 10.39 4.4.2, t"0.48.2.1 , 10.53.1.2, and 10.56.3)Ajc = area of compression flange (Article 10.48.4.1)= total area of longitudinal reinforcing steel at the interior support within theeffective flange width (Article 10.38.5.1.2)= total area of longitudinal slab reinforcement steel for each beam over interiorsupport (Article 10.38.5.1.3)= area of steel section (Articles 10.38.5.1.2, 10.54.1.1 , and 10.54.2.1 )

= area of web of beam (Article 10.53.1.2)= distance from center of bolt under consideration to edge of plate in inches(Articles 10.32.3.3.2 and 10.56.2)

a = spacing of transverse stiffeners (Article 10.39.4.4.2)a =depth of stress block (Figure 10.50A)a =ratio of numerically smaller to the larger end moment (Article 10.54.2.2)


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b

bbb'b'

ccc

DD

Dsddd

= distance from edge of plate or edge of perforation to the point of support(Article 10.35.2.3)= unsupported distance between points of support (Article 10.35.2.7)= flange width between webs (Articles 10.37.3.1, 10.39.4.2.10.51.5.1, and 10.55 3)= widthofstiffeners(Articles 10.34.5.2, 10.34.6, 10.37.2.4, 10.39.4.5.1,and 10.55.2= width of a projecting flange element, angle, or stiffener (Articles 10.34.2.2,

10.37.3.2, 10.39.4.5.1 , 10.48.1, 10.48.2, 10 .48.5.3, 10.50, 10.51.5.5, and 10.55.3)= web buckling coefficient (Articles 10.34.4, 10.48.5.3. 10.48.8. and 10.50(e))= compressive force in the slab (Article 10.50.1.1.1)= equivalent moment factor (Anicle 10.54.2.1 )= compressive force in top portion of steel section (Article 10.50.1.1.1 )= bending coefficient (Table 10.32.1A, Article 10.48.4.1)= columnslenderness ratiodividing elasticand inelastic buckling(Table 10.32.1A)= coefficient about X-axis (Article 10.36)= coefficient about the Y-axis (Article 10.36)= buckling stress coefficient (Article 10.51 .5.2)= clear distance between flanges, inches (Article 10.15.2)= clear unsupported distance between flange components (Articles 10.34.3,10.34.4, 10.34.5, 10.37.2, 10.48.1 , 10.48.2, 10.48.5, 10.48.6, 10.48.8, 10.49.2,10.49.3.2, 10.50(d), 10.50.1.1.2, 10.50.2.1, and 10.55.2)= clear distance between the neutral axis and the compression flange (Articles10.48.2.1(b), 10.48.4.1, 10.49.2, 10.49.3 and, 10.50(d))= moments caused by dead load acting on composite girder (Article 10.50.1.2.2)= distance to the compression flange from the neutral axis for plastic bending,

inches (Articles 10.50.1.1.2 and 10.50.2.1)= moments caused by dead load acting on steel girder (Article 10.50.1.2.2)= bolt diameter (Table 10.32.3B)= diameter of stud, inches (Article 10.38.5.1)= depth of beam or girder, inches (Article 10 .13, Table 10.32.1A, Articles

10.48.2, 10.48.4.1, and 10.50.1.1.2)
http:///reader/full/10.32.1Ahttp:///reader/full/10.32.1Ahttp:///reader/full/10.32.1Ahttp:///reader/full/10.32.3Bhttp:///reader/full/10.32.3Bhttp:///reader/full/10.32.3Bhttp:///reader/full/10.32.1Ahttp:///reader/full/10.32.1Ahttp:///reader/full/10.32.3Bhttp:///reader/full/10.32.1A


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= modulus of elasticity of concrete. psi (Anicle 10.38.5.1.2)= maximum induced stress in the bottom flange r t i e ! ~ 10.20.2.1)= maximum compressive stress, psi (Article 10.41.4.6)= allowable axial unit stress (Table 10.32.1 A and Articles 10.36, 10.37 1.2, and

10.55.1)= allowable bending unit stress (Table 10.32.1A and Articles 10.37.1.2 and 10.55.1 )= buckling stress of the compression flange plate or column (Articles 10.51.1,

10.515 10.54.1.1, and 10.54.2.1)Fbx = compressive bending stress penniued abom the X-axis (Anicle 10.36)Fb). = compressive bending stress pennitted about theY-axis (A rticle 10.36)Fo = maximum horizontal force (Article 10.20.2.2)Fe = Euler buckling stress (Articles 10.37 1 , 10.54.2.1 , and 10.55.1)F/ = Euler stress divided by a factor of safety (Anicle 10.36)Fp = computed bearing stress due to design load (Table 10.32.3B)Fs = limiting bending stress (Article 10.34.4)Fsr = allowable range of stress (Table 10.3.1A)F = specified minimum yield point of the reinforcing steel (Articles 10.38.5.1.2)F.S. = factor of safety (Table 10.32.1A and Articles 10.32.1 and 10.36)

= specified minimum tensile strength (Tables 10.32.1A and 10.32.3B, Article10.18.4)= tensile strength of electrode classification (Table 10.56A and Anicle 10.3?.2)= allowable shear stress (Tables 10.32. 1A, 10.32.3B and Articles 10.32.2.10.32.3, 10.34.4, 10.40.2.2)= shear strength of a fastener (Article 10.56.1.3)= combined tension and shear in bearing-type connections (Article 10.56.1.3)= specified minimum yield point of steel (Articles 10.15 .2.1, 10.15.3, 10.16.11,

10.32.1, 10.32.4, 10.34, 10.35, 10.37.1.3, 10.38.5, 10.39.4,10.40.2.2, 10.41.4.6,10.46, 10.48, 10.49, 10.50, 10.51.5, and 10.54)= specified minimum yield strength of the flange (Anicle 10.48.1.1 ,and 10.53.1)


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BRIDGE DESIGN PRACTICE DECEMBER 1995tizluans= range of stress due to live load plus impact, in the slab reinforcement over thesuppon (Anicle 10.38.5.1.3)

Is = maximum longitudinal bending stress in the flange of the panels on either sideof the transverse stiffener (Article 10.39.4.4)fr = tensile stress due to applied loads (Anicles 10.32.3.3.3 and 10.56.1.3.2)!v = unit shear stress (Anicles 10.32.3.2.3 and 10.34.4.4)fbx = computed compressive bending stress about the x axis (Anicle 10.36)fb) = computed compressive bending stress about they axis (Article 10.36)g = gage between fasteners , inches (Articles 10.16.14 and 10.24.5)H = height of stud, inches (Article 10.38.5.1.1 )h = average flange thickness of the channel flange, inches (Article 10.38.5.1.2)I = moment of inertia, in.4 (Articles 10.34.4, 10.34.5, 10.38.5.1.1 , 10.48.5.3, and10.48.6.3)

= moment of inertia of stiffener (Articles 10.37.2, 10.39.4.4.1 , and 10.51.5.4)=moment of inertia of transverse stiffeners (Anicle 10.39.4.4.2)= moment of inertiaofmember about the vertical axis in the plane of the web, in.4(Article 10.48.4.1)= moment of inertiaof compression flange about the vertical axis in the plane of

the web, in.4 (Table 10.32.1A, Anicle 10.48.4.1)J = required ratio of rigidity of one transverse stiffener to that of the web plate(Articles 10.34.4.7 and 10.48.5.3)J = in.4 (Table 10.32.1A, Article 10.48.4.1) St. Venant torsional constantK = effective length factor in planeof buckling(Table 10.32.1A andArticles 10.37,10.54.1 and 10.54.2)= effective length factor in the plane of bending (Article 10.36)

= constant: 0.75 for rivets; 0.6 for high-strength bolts with thread excluded fromshear plane (Article 10.32.3.3.4)k = buckling coefficient (Articles 10.34.4, 10.39.4.3, 10.48.8, and 10.51.5.4)k = distance from outer face of flange to toe ofweb fillet ofmember to be stiffened(Article 10.56.3)


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= length of member berween points of suppon, inches (Anicle 10.54.1 .1)= limiting unbraced length (A nicle 10.48.4.1)= limiting unbraced length (A nicle I0.48.4.1)= member length (Table 10.32.1 A and Article 10.35.1)= maximum bending moment (Articles 10.48.8, and 10.54.2.1)

M 1 = moments at the ends of a memberM1 & M2 = moments at two adjacent braced points (Table 10.32.1A. Anicles 10.36A and10.48.4.1)Me = co lumn moment (Anicle 10.56.3.2)Mp = full plastic moment of the section (Articles 10.50.1.1.2 and 10.54.2.1)M, = lateral torsional buckling moment or yield moment (Articles 10.48.4.1 and

10.53.1.3)Ms = elastic pier moment for loading producing maximum positive moment inadjacent span (Article l 0.50.1.1.2)Mu = maximum bending strength (Articles 10.48, 10.51.1, 10.53.1, and 10.54.2.1)N1 & N2 = number of shear connectors (Article 10.38.5.1.2)Nc = number of additional connectors for each beam at point of contrafiexure

(Article 10.38.5.1.3)Ns = number of slip planes in a slip critical connection (Articles 10.32.3.2.1 and10.57.3.1)N... = number of roadway design lanes (Article 10.39 2)n = ratio of modulus of elasticity of steel to that of concrete (Article 10.38.1)n = numberoflongitudinalstiffeners (Articles 10.39.4.3, 10.39.4.4 ,and 10.51.5.4)P = allowable compressive axial load on members (Article 10.35.1)P = axialcompressiononthemember(Articles 10.48.1.1, 10.48.2.1,and 10.54.2.1)P,P.,P2&P3 = force in the slab (Article 10.38.5.1.2) Ps = allowable slip resistance (Article 10.32 2.2.1 ) Pu = maximum axial compression capacity (Article 10.54.1.1) p = allowable bearing (Article 10.32.4.2)
http:///reader/full/10.32.1Ahttp:///reader/full/10.32.1A


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Rev = a range of stress involving both tension and compressipn during a stress cycle(Table 10.3.1B)= vertical force at connections of vertical stiffeners to longitudinal stiffeners(Article 10.39.4.4.8)= vertical web force (Article 10.39.4.4.7)

r = radius of gyration, inches (Articles 10.35.1 , 10.37.1 , 10.41.4.6, 10.48.6.3,10.54.1.1 , 10.54.2.1 , and 10.55.1)= radius of gyration in plane of bending (Article 10.36)

ry = radius of gyration with respect to theY- Y axis (Article 10.48.1.1)r' = radius ofgyration in inchesof the compression flange about the axis in the plane

oftheweb(Table 10.32.1A, Article 10.48.4.1)s = allowable rivet or bolt unit stress in shear (Article 10.32.3.3.4)s = section modulus, in.3 (Articles 10.48.2, 10.51.1, 10.53.1.2, and 10.53.1.3)s = pitch of any two successive holes in the chain (Article 10.16.14.2)= range of horizontal shear (Article 10.38.5.1.1)= section modulus of transverse stiffener, in3 (Articles 10.39.4.4 and 10.48.6.3)= section modulus oflongirudinal or transverse stiffener, in.3 (Article 10.48.6.3)= ultimate strength of the shear connector (Article 10.38.5.1.2)= section modulus with respect to the compression flange. in.3 (Table 10.32.1A,

Article 10.48.4.1)s = computed rivet or bolt unit stress in shear (Article 10.32.3.3.4)T = range in tensile stress (Table 10.3.1B)T = direct tension per bolt due to external load (Articles 10.32.3 and 10.56.2)T = arch rib thrust at the quarter point from dead + live + impact loading

(Articles 10.37.1 and 10.55.1)= thickness of the thinner outside plate or shape (Article 10.35.2)r = thickness of members in compression (Article 10.35.2)

= thickness of thinnest part connected, inches (Articles 10.32.3.3.2 and 10.56.2)= computed rivet or bolt unit stress in tension, including any stress due to prying


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= thickness of flange delivering concentrated force (Ankle 10.56.3.2)= thickness of flange of member to be stiffened (Article 10.56.3.2)= thickness of the flange (Articles 10.37.3. 10.55.3 and 10.39.4.3)= thickness of stiffener (Anicle 10.37.2 and 10.55.2)= slab thickness (Articles 10.38.5. 1.2 , 10.50.1.1.1, 10.50.1.1.2)= web thickness, inches (Articles 10.15.2.1, 10.34.3 , 10.34.4, 10.34.5, 10.37.2.

10.48. 10.49.2, 10.49.3, 10.55.2, and 10.56.3)l rj = thickness of top flange (Article 10.50.1.1.1 )t ' = thickness of o u t ~ t a n d i n stiffener element (Articles 10.39.4.5.1 and 10.51.5.5)

= shearing force (Articles 10.35.1, 10.48.5.3, 10.48.8, and 10.51.3)= shear yielding strength of the web (Ankles 10.48.8 and 10.53.1.4)= range of shear due to live loads and impact, kips (Article 10.38.5.1.1)= maximum shear force (Articles 10.34.4, 10.48.5.3, 10.48 .8, and 10.53.1.4)= vertical shear (Article 10.39.3 .1)= design shear for a web (Articles 10.39.3.1 and 10.51.3)w = length of a channel shear connector, inches (Article 10.38.5. 1.2)= roadway width between curbs in feet or barriers if curbs are not used (Article

10.39.2.1)= fraction of a wheel load (Article 10.39.2)

w = lengthof a channel shearconnector in inches measured in a transverse directionon the flange of a girder (Article 10.38.5.1.1)

w = unit weight of concrete, lb. per cu. ft. (Article 10.38.5. 1.2)w = width of flange between longitudinal stiffeners (Articles 10.39.4.3, 10.39.4.4.and 10.51.5.4)y = ratio of web plate yield strength to stiffener plate yield strength (Anicles

10.34.4 and 10.48.5.3)= distance from the neutral axis to the extreme outer fiber, inches (Article 10.15.3)

y = location of steel sections from neutral axis (Article 10.50.1.1.1 )z = plastic section modulus (Articles 10.48.1 , 10.53.1.1. and 10.54.2.1)Z, = allowable range of horizontal shear, in pounds on an individual connector


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e = angle of inclinationof the web plate to the vertical (Articles 10.39.3.1 and 10.51.3)= ratio of total cross sectional area to the cross sectioqal area of both flangesIll (Article 10.15.2)= distance from the outer edge of the tension flange to the neutral axis divided by

the depth of the stee l section (Articles 10.40.2 and 10.53.1.2)= amount of camber, inches (Article 10.15.3)= dead load camber in inches at any point (Article 10.15.3)= maximum value of ..DL inches (Article 10.15.3)= reduction factor (Articles 10.38.5.1.2, 10.56.1.1, and 10.56.1.3)= longitudinal sti ffener coefficient (Articles 10.39.4.3 and 10.51.5.4)= slip coefficient in a slip-critical joint (Article 10.57.3)

AbbreviationsBDS = Bridge Design Specifications manual


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4.0 IntroductionThis section illustrates Load Factor Design (LFD) for a continuous. welded. structural steelgirder highway bridge. composite for positive live load moments according to Section I0 ofthe Bridge Design Specifications (BDS).In addition to being classified as symmetrical or unsymmetrical as shown in Figure 4-1 ,steel girders can be further categorized as follows: Compact Composite Non-compact Non-composite Braced Hybrid Unbraced Transversely stiffened Longitudinally stiffened

Symmetrical Unsymmetrical

Figure 4-1 Types of Steel GirdersThe steel girders designed by Caltrans are usually welded plate girders. Typically these arenon-compact and transversely stiffened; they can be either braced or unbraced. The use oflongitudinal stiffeners should be avoided if possible as they lead to complicated details and,


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BRIDGE DESIGN PRAcnCE DECEMBER 1995til/transc s : : ; ~ ~ - Flange Pia e

Bea ring Stiffen ers -......., II Web Plate " '." lTransverse Stiffener

Sole Plate

Figure 4-2 Details of Welded Steel Plate Girder

4.1 General Design ConsiderationsMembers designed by the load factor design (LFD) method are proportioned for a numberof design loads. They are required to meet three main theoretical load levels:1. Maximum Design Load2. Overload3. Service LoadThe maximum design load and overload requirements are based on multiples of the serviceloads with certain other coefficients necessary to insure the required capabilities of thestrucrure. The maximum design load criteria insures the strucrures capability of withstand-ing a few passages of exceptionally heavy vehicles.The overload criteria insures control of permanent deformation in a member caused byoccasional overweight vehicles as specified in BDS, Article 10.57.Service loads are utilized for the serviceability criteria to limit the live load deflection and


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4.2 Design LoadsThe moments and shears are determined by subjecting the girder to the design loads. Elasticanalysis is used to calculate the various straining actions.The design loads are given by For HS20: 1.3 [ D + 513(L + /)]

For permit loading: 1. widely spaced 1.3 [D + (L + / )HS2o+ 1.15 (L + / ) P13l2. closely spaced 1.3 [D + (L + /)Pl3]Where D =dead load, L =live load (HS20, P13), / =impact. The factor 1.3 is included tocompensate for uncenainties in strength, theory, loading, analysis and material properties.Also, the factors 5/3 and 1.15 are incorporated to allow for variability in overloads.

4.3 Design for Maximum LoadsWelded plate girders of normal proportions are not likely to satisfy the requirements for acompact section, which is capable of developing full plastic stress distribution. Usuallywelded plate girders are non-compact braced or unbraced sections.Thenon-compact bracedsection is a section that can develop yield strength in thecompressionflange before the onsetof local buckling, but it cannot resist inelastic deformation required for full plastic stressdistribution.

4.3.1 Braced Sections For non-compact braced section;

Mu = FyS .................................................................... ......................................... (10-97) Where Fy= yield stress and S = elastic section modulus. The section modulus consequently must be proportioned so that


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For the relationship to be permitted. the following criteria must be satisfied:1. Width-thickness ratio of the compression flange:

~ ' : : ; 2-; ......................................................................................................... (10-98) where b' = width of projecting flange element = !!._

t = flange thickness 2 2. Depth - thickness ratio of the web:

De < 15,400~ - T , (10-99)

Where Deis the depth of the web in compression and t.., is the web thickness. However,for a symmetrical section this ratio can be exceeded by providing transverse stiffenersand meetingE..::; 3/t;OO ............................................................................ (10-103) and (10.50(d)) t,.. F1or for an unsymmetrical section

~ ::; 1F ,O ......................................................................................... ........... (10-119) 3. Spacing of lateral bracing of the compression flange:

Lt,::; 20,000,000Af (10- 1()())F>,dwhere A1= cross sectional area of compression flange d = total depth of gi rder


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4.3.2 Unbraced SectionsWhen a girder does not meet the lateral bracing requirement, the section is considered anunbraced section and its ultimate moment capacity is given by:

Mu = M,Rb ..................................................................................... .............. ... (10-102a) WhereMu is themaximum bending strength,M ,i s the lateral torsional buckling moment,andRb is a bending capacity reduction factor.When the compression region of a bending member does not have adequate lateral suppon,the membermay deflect laterally in a torsional mode before thecompressive bending stressreaches the yield stress. This mode of failure is known as "lateral torsional buckling" orsimply "lateral buckling".The tendency of the compression flange to twist is resisted by a combination ofSt Venant andwarping torsion. In resisting lateral buckling by warping torsion, thecompression flange acts asa column susceptible to buckling in the lateral direction. In closed sections, such as box girdersor rubes, torsional stiffness is generally very largeand lateral bucklingis nota concern. However,for open sections, such as plate girders, lateral buckling must be considered. Because of thecomplexity of the theoretical expressions for lateral buckling stress that take into account thesimultaneous resistance to lateral buckling afforded by St Venant and warping torsion,conservative simplified expressions have been developed for design use that consider theeffectsseparately. Plate girders, usually deep girders, are controlled by warping torsion since the effectof St Venant torsion is small. The ultimate momentcapacity for unbraced section, as used inAASHTO Specifications prior to the fifteen edition, is:

2M =F..S[l- .3: ._(4) ]" Y 4n2 b'

This equation treats the compression flange as a column, provided that the compressionflange is not smaller in width than the tension flange. When using the equation, the momentcapacity may be increased20% when the ratio of theend moments is less than 0.7,but cannotexceed FyS. The specifications also limit the stress in the top flange of a composite girder to0.6Fy under dead load. However, if the width of the compression flange is smaller than thetension flange, then the above equation is unconservative and the moment capacity should


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4.3.3 Shear Capacity and DesignThe shear capacity of girder webs with transverse stiffeners is given by:

0.87(1-C)Vu=Vp C+ 0 ; t y ......................................................... ...................... (10-113) ~ l + ( ~ r

This equation combines the "beam action" and the "tension field action." The first term ofthe equation represents web buckling under shear and the second term represents theadditional post-buckling strength.

VP =plastic shear capacity= 0.58FyD t.., ......................................................... (10-114) and

C = web buckling shear stress web shear yield stress

Depending on the value ofDlt,.. the web can be one of three cases which is given in Article10.48:D < 6,000-Jk . C=l.O. Yielding: lw ~ .

. b klin 6, YJO.Jk < D< 7.soo.Jk . C 6,000.Jke asuc uc. In 1 g: = D ) ~ .................. (10-115)..{F; - lw - jF; .t,.,

D 7,soo.Jk 4.5xl07k3. Elastic buckling: -> . c= 2 ( 10-116)lw JF; . ( ~ ) Fy
http:///reader/full/C6,000.Jkhttp:///reader/full/C6,000.Jkhttp:///reader/full/C6,000.Jkhttp:///reader/full/C6,000.Jkhttp:///reader/full/C6,000.Jkhttp:///reader/full/C6,000.Jkhttp:///reader/full/C6,000.Jkhttp:///reader/full/7,soo.Jkhttp:///reader/full/7,soo.Jkhttp:///reader/full/7,soo.Jkhttp:///reader/full/C6,000.Jkhttp:///reader/full/7,soo.Jk


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BRIDGE DESIGN PRAcriCE DECEMBER 1995

Generally, the effect of bending on the shear strength of a girder can be ignored. However,if the bendingMexceeds 0.75M11 and the shear capacity is calculated from Equation 10-113,then the shear at that section should be limited to:

V=Vu[2.2-l.6 ; ] ................................... .................................................... (10-117)

Spacing of transverse stiffeners along a girder should not exceed d0 determined from V"formula nor 3D. However, for transversely stiffened plate girders with Dlt.,., >150, thestiffener spacing shall not exceed

v[ 260 2] to ensure efficient handling, fabrication and erection of the girder.Dlt...,At simply supported ends of girders, the first stiffener space shall be such that the appliedshear will not exceed the plastic or buckling shear force:

V = CVP ............................. ............................................................................... (10-1 12) and the maximum spacing is limited to 1.5D.Transverse stiffeners should be proportioned so that the width-thickness ratio shall be

b' 2,600- ~ rr;:- (10-104)t "\\FYAlso, the gross cross-sectional areaof each one-sided stiffeneror pairof two-sided stiffenersshall be at least

A=[0.15BDt.... ( 1 - C ) ~ -18tl J r ~ O ........................................................ (10-105) where Y=ratio of web yield strength to stiffener yield strength; B = 1.0 for stiffener pairs;B = 2.4 for single plates, and; C is the value used in computation of V,..In addition, the required moment of inertia of stiffeners with respect to midplane ofweb is


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BRIDGE D ESIGN PRACTICE D ECEMBER 1995tizltrans4.4 Composite Girders

ln the non-composite type of steel girder bridge, the entire dead load and live load of thesuperstructure is supponed by the steel girders alone, with the deck only transmitting loadsto the girders. However, in composite construction, the concrete deck is keyed to the steelgirders by mechanical means and may thus be considered a component pan of the girder.

"b 8 .. ( Shear connectors.~ h ., I. . i j I .y( ConcreteDeck

Web

Figure 4-3 Details of Composite Steel GirderFigure 4-3 shows a section and elevation view of a typical composite girder. The concretedeck is keyed to the steel girder by shearconnectors, therefore, the deck serves as additionalupper flange area for the steel girder.In accordancewith BDS, Article 10.38.3.1, the assumed effective width of theconcrete deckshall not exceed the following:(a) one-fourth of the span length of the girder.(b) the distance center to center of girders.(c) twelve times the least thickness of slabSince the modulus of elasticity of the concrete deck is different from thatof the steel girders,


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BRIDGE DESIGN PRACTICE DECEMBER 1995tbluansconverted to an equivalent area of steel, the section may be considered to be a steel girdercomposed of (1) the original steel girder and (2) an additional rectangular flange ofwidthb composite bridgesteel girder is usually designedas acompositefor live load and noncomposite for dead load. Since intermediate temporary suppons are not normally usedduring deck placement, the steel girder alone bas to carry its own weight in addition to theweight of the deck. Once the concrete hardens the girder and deck will act as a compositesection. Usually three types of loading act on the girder:1. Dead load (weight of girder and slab)2. Additional dead load (rail, AC overlay)3. Live loadFor design purposes the girder is considered a non-composite section for dead load and a fullcomposite section for live load. However, for additional dead load (AC overlay + rail) thegirderwill act as a partiallycomposite section. This is because the additional dead load willcause sustained stress on the concrete section. Due to this sustained stress, the concretewillundergo plastic flow, and its effectiveness in resisting stress will be reduced. The mainreason of this plastic flow is the creep of concrete. One conservative way to account for thecreep of concrete under sustained loading is to reduce the elastic modulus Ec to 1/3Ec whichmeans increasing n to 3n as in the BDS Article 10.38.1.4.

SustainedLoading

VJ VJ Q)....Ci5

Creep or plastic flowStrain


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BRIDGE DESIGN PRACTICE. DECEMBER 1995tblt:rans4.5 Fatigue Design

The fatigue provisions of the bridge design specifications were developed from research andstudies of failures in the field with respect to in-plane bending: out-of-plane bending is notaddressed. Details for main load carrying members, such as butt weld at tension flanges andstiffener welds, are familiar to designers. However, the effects of connections to the mainmembers are not as familiar and have beena sourceofan increasing numberof fatigue problems.Fatigue may be defined as the initiation and/or propagation of cracks due to repeatedvariation of normal stresses which include a tensile component. Therefore, fatigue is theprocess of cumulative damage that is caused by repeated fluctuating loads.Fatigue damage for a component that is subjected to normally elastic stress fluctuationsoccurs at regions of stress raisers. After a certain number of load fluctuations, theaccumulated damage causes the initiation and subsequent propagation of a crack or cracksin the plastically damaged regions. This process can and in many cases does cause fractureof components. The more severe the stress concentration, the shorter the time to initiate afatigue crack for the same stress cycle.

4.5.1 Factors Affecting Fatigue PerformanceMany parameters affect the fatigue performance of structural components. They includeparameters related to stress, geometry and properties of. the component, and externalenvironment.The stress parameters include stress range, constant or variable loading and frequency. Thegeometry and properties of the component include stress raisers, size, stress gradient andmechanical properties of the base metal and weldment. The external environment parameters include temperature and aggressiveness of the environment. The major factors thatgovern fatigue are: applied stress range number of load cycles applied type of detail


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IIi' BRIDGE D ESIGN PRACTICE DECEMBER 1995lb!UCIIIS

4.5.2 Applied Stress RangeThe appliedstress range may be defined as the algebraic difference between extreme stressesresulting from the passage of load across the structu re. If. as in a compression member, thestress range remains within compressive values there is no fatigue considerations.

1 cyclernrnQ)....U5

StressRangeTime 'Figure 4-5 Constant Amplitude Cycles

The above figure represents the simplest stress history which is the constant-amplitudecyclic-stressfluctuation .The stress range is the algebraic difference between the maximumstress,fmaz, and the minimum stress,fmin, in the cycle.

fs,=Jmaz + !fminlThe other type of stress history is the variable-amplitude random-sequence stress history asshown in the Figure 4 -6. This is a very complex history and cannot be represented by ananalytical function. The truck loading on bridges is a panicularexampleofthis stress history.


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, 6 BRIDGE DESIGN I'RACTJCE DECEMBER 1995tizltrtlll6

4.5.3 Allowable Stress RangeThe following items control the allowable stress range.1. type of loading2. stress category (connection detail)3. redundancy

4.5.4 Type of LoadingThe number of cycles has a significant affect on tlle fatigue design. Generally, by increasingthe number of cycles, the allowable stress range would decrease.The number of cycles used for fatigue design depends on the type of road and live load. Forexample, "Case r,which is the mostusedcase for freeways (an average daily truck traffic whichexceeds 2,500), bas the following live load cycles to consider for longitudinal members:1. HS20 (multi-truck) ............................................................................. 2,000,000 cycles 2. HS20 (multi-lane) .................................................................................. 500,000 cycles 3. Single HS20 (truck) .................................................................... over 2,000,000 cycles 4. P Loading (P13 with HS20) .................................................................. 100,000 cycles

4.5.5 Stress CategoryThe main stress categories A, B, C, D, Eand Fare described in Table 10.3.1B and illustratedin Figure 10.3.1 C ofthe Bridge Design Specifications. Thesecategories correspond to platesand rolled beams; we lds and welded beams and plate girders; stiffener and short (less than2") attachments; intermediate (over 2" but less than 4") attachments; long (over 4")attachments and cover plates; and fillet welds in shear, respectively.The most severe connection details are in category E and E'. These should be avoided asmuch as possible because they are regarded as poor details.


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BRIDGE DESIGN PRACTICE D ECEMBER 1995tbltransInsumm ary, the fatigue allowable stress ranges and number ofcycles represent a confidencelimit for 95-percent survival of aU details in a given category. Also. the stress ranges aregoverned by details that have the most severe geometrical discontinuities and/or imperfections. It is important to note that the fa tigue crack/propagation is independent of the strengthof steel. Therefore, the allowable stress ranges are independent of steel strength.

4.6 Charpy V-Notch Impact RequirementsMain load carrying member components subjected to tensile stress are required to provideimpact properties as shown in the table below.These impact requirements vary depending on the type of steel used and the averageminimum service temperature to which the structure may be subjected.The basis and philosophy used to develop these requirements are given in a paper entitled"TheDevelopment of AASHTO Fracrure-Toughness Requirements for Bridge Steels" by John M.Barsom, February 1975, available from the American Iron and Steel Institute,Washington,D.C.Charpy V-notch (CVN) impact values shall conform to the following minimum values:

Table 4-1 Fracture Toughness RequirementsWelded or Fracture-Critical Non-Fracture-CriticalMechanically Grade Thickness Zone l Zone2 Zooe3 Zone 1 Zooe2 Zone3Fastened (Y.P./Y.S.) (Inches) Ft-Lbs@F Ft-Lbs@F Ft-Lbs@F Ft-Lbs@F FtLbs@Of Ft-Lbs@ F

36 t s Jlh1Yl


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BRlDGE DESIGN PRACTICE D ECEMBER 1995

The CVN-impact testing shall be "P" plate frequency testing in accordance with AASHTOT-243 (ASTM A673). For Zone 3 requirements only. Charpy impact tests are required oneach plate at each end. Th e Charpy test pieces shall be coded with respect to heat/platenumber and that code shall be recorded on the mill-test report of the steel supplier with thetest result. I f requested by the Engineer, the broken pieces from each test (three specimens,six halves) shall be packaged and forwarded to the Quality Assurance organization of theState. Use the average of three (3) tests. I f the energy value for more than one of three testspecimens is below the minimum average requirements , or if the energy value for one of thethree specimens is less than two-thirds (213) of the specified minimum average requirements ,a retest shall be made and the energy value obtained from each of the three retest specimensshall equal or exceed the specified minimum average requirements.Zone 1: Minimum Service Temperature 0F and above.Zone 2: Minimum Service Temperature from -1 F to -30F.Zone 3: Minimum Service Temperature from -31F to -60F

4.7 Fracture Control Plan (FCP)The FCP is a plan devised to prevent collapse of steel bridges. Much of the FCP relales todesign, welding,and material properties. Thedesigner has the responsibility for designatingany member or strucrural component as a Fracrure Critical Member (FCM) when failure ofthatmember wouldcause the structure to collapse. The FCP requires the FCM be fabricatedin a qualified shopand inspected by qualified inspectors; requires Nondestructive Inspection(NDI) by qualified testers; supplements the current AWS and AASIITO welding specifications; and specifies material toughness.It is a comprehensive plan whose adoption should improve the overall quality of steelstructures from design through fabrication.For more detailed information see AASIITO's Guide Specification for Fracture CriticalNon-Redwuiant Steel Bridge Members.


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BRIDGE DESIGN PRACTICE DECEMBER 1995

4.8 Design Example ProblemTo illustrate load factor design, portionsof an interior girderof a three-span bridge as shownin Figure 4-7 will be designed. The section in the positive-moment region consists of awelded steel girder acting compositely with the concrete slab. In the negative momentregion, the section is designed as a non.:composite section.Roadway Section: Figure 4-8 Typical SectionSpecification: 1992 Fifteenth Edition AASHTO with Interims and Revisions by CaltransLoading: 1. Dead Load

2. Live Load; HS20-44 and alternative and permit design loadStructural Steel: A709 Grade 50 - assume for web and flanges

A709 Grade 36 - assume for stiffeners, etc.Concrete: fc' = 3,250 psi, modular ratio n =9

BB 503'-0" EB= s ~ o ~ - ~ o - - ~ = 6" ~ - - o ~ - - - - ~ - - - - - - = 2 o o = - - o ~ - - - - - - ~ - - - - ~ 1 s = o ~ ... ~ - 6 ~ r -r

-- ---...N ELEVATION0R T H

- - - + - - - - - - - - + - - - - - - ~ - - - - - + - - - - - - - - + - - 21+00 22+00 23+00 24+00 25+00


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BRIDGE DESIGN PRAcncE DECEMBER 1995lif;liz/trans

44'-0" 40'-6"

2 @ 16' -0 " =32'-0 "

Figure 4-8 Typical Section


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4.9 LoadingSince the spacing between girders exceeds 14 feet (BDS Table 323.1), this is a widelyspaced girder and should be checked for load combinations IH and IPW

IH Group = l .3[D + 513(L + I)HS2olIpwGroup = 1.3[D + (L + I )HS20 + 1.15(L + I)Pl3]

4.9.1 Dead LoadTributo to Interior Girder8'-0 "

~

Figure 4-9 Interior Girder Cross SectionConcrete Slab: Assume transverse deck design has been completed and a 107h" thick dec khas been selected.

Area= (lOVs/12)(16) = 14.50 ft2w = 14 .50 (0.150) = 2.18 k/ft

Steel Girder:w = 0.30 klft (including bracing and fillet welds) (estimated weight)


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lr BRIDGE D ESIGN PRACTICE D ECEMBER 1995tiz/frQJ't!;

4.9.2 Live LoadFor widely spaced girders, the load on each girder will be the reaction of the wheel loadsassuming the deck between the girders acts as a simple beam.Number of traffic lanes:

width of deck between rails= 44- 2(1.75) = 40.5 ftFrom BDS 3.6, a traffic lane is 12 feet wide

40 5number of traffic lanes= = 3.38 :. number of design traffic lanes= 3.12

40 5' between face of rails

7' 13',..3' -1Ie.g. WL e.g. WL cgfl I4' 6' 2' 4' 6' 2' 2' 6' 4' I ! IA tB c16' between girders 16' between girders

12' lane 12' lane 12' lanelFigure 4-10 Location of3 Traffic Lanes for Maximum Load at B (HS20)
http:///reader/full/44-2(1.75http:///reader/full/44-2(1.75http:///reader/full/44-2(1.75


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BRJDGE DESIGN PRACTICE DECEMBER 1995liz/transLive Load contributory reaction at Interior Girder B (3 lanes)

Rs =(l3 +3) +}__=1.4416 16number of live load lanes = 1.44

However, according to BDS 3.12.1, for 3lanes use 90% of maximum.number of design live load lanes= 0.90(1.44) = 1.30 lanes HS20

40.5' between face of rails

9' 0 13'I Ie.g. WL e.g. WLI I

lAI.J 4' 6' 2' 12- ~ - ~J

J !s16' between girders J.

6' 4' I-'16' between girders

l cJ12' lane .1. 12' 1ane .I

Figure 4-11 Location of2 Traffic Lanes for Maximum Load at B (HS20)Live Load contributory reaction at Interior Girder B (2 lanes)

13 9Rs=-+-= 1.3816 16
http:///reader/full/0.90(1.44http:///reader/full/0.90(1.44http:///reader/full/Rs=-+-=1.38http:///reader/full/Rs=-+-=1.38http:///reader/full/Rs=-+-=1.38http:///reader/full/Rs=-+-=1.38http:///reader/full/Rs=-+-=1.38http:///reader/full/0.90(1.44http:///reader/full/Rs=-+-=1.38


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40.5' between face of railsI. .. . 9' 13'... . ...... ie.g. WL e.g. WLI (HS20) (P13)I I I I I 4' 6' 2' I 2' 6' 4'.i. .. I.T TI II I ! IA ! ' is

16' between girders 16' between girdersAI. ..I. Jl c

12' lane 12' laneI. .I. JFigure 4-12 Location of2 Traffic Lanes for Maximum Load at B (P13 with HS20)

Live Load contributory reaction at Interior Girder B13Pl3: R8 ::::-:::: 0.8llanes16

9HS20: R8 ::::- :::: 0.56 lanes 16 For lpwGroupIPW :::: 1.3 [D + 0.56 (L + /)HS20 + 1.15 (0.81 )(L + /)pl)]

IPW :::: 1.3D + 0.73 (L + /)HS2o+ 1.22(L + /)Pl3 (See foomote below)


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BRIDGE DESIGN PRACTICE DECEMBER 1995tblt:rans4.10 Composite Section Design

This section illustrates the design ofan interior girder ofa composite section at0.4 poinl ofSpan 1

4.10.1 Design Loads (See Section 4-16, Bridge Design System Computer Output)

Load on Steel Girder Only (Non-composite)Dead Load girder and slab Moment= 1.3(3,590) = 4,667 k-ft Shear = 1.3(-15.2) = -19.8 k

Load on Partially Composite Section (n =3 x 9 =27)Dead Load rail and AC overlay Moment= 1.3(1,221) = 1,587 k-ft Shear = 1.3(-5.2) = -6.8 k

Load on Composite Section (n = 9)Live Load Group IH Maximum Moment = 3.0(2,424) = 7,272 k-ft Associated Shear = 3.0(17.7) = 53.1 k Maximum+ Shear = 3.0(38.7) = 116 k Associated Moment= 3.0(2,319) = 6,957 k-ft Maximum - Shear = 3.0 ( 36.7) = -110 k Associated Moment= 3.0(1,367) =4,101 k-ft


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BRIDGE D ES IGN PRAcnCE DECEMBER 1995

4.1 0.2 Fatigue Loads (Case I Road)

HS20 (Single Truck) Over 2,000,000 Cycles +LLM = 0.81(2,360) = 1,912 k-ft -LLM=0.81(-590)=-478k-ft

HS20 (Multiple Lanes) 2,000,000 Cycles (Truck Load) +LLM = 1.38 (2,360) = 3,257 k-ft - LLM= 1.38(-590) =-814k-ft

HS20 (Multiple La.nes) 500,000 Cycles (Lane Load) +LLM = 1.38 (2.424) = 3,345 k-ft - LLM= 1.38(-805)=-l ,l l lk-ft

P13 with HS20 100,000 Cycles +LLM = 0.56(2,424) + 0.81 (1.15)6,648 = 7,550 k-ft - LLM = 0.56(-805) + 0.81(1.15)(-2,219) = -2,518 k-ft


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6 BRIDGE DESIGN PRACTICE D ECEMBER 1995liz/trans

4.1 0.3 Girder Section

Top FlangeTypically, the maximum transponed length of a steel plate girder is 120 feeL Due toconstruction problems, some erectors limit the length of girder shipping pieces to 85 timesthe flange width. Based on that, for 120 foot length. the width of the compression flange willbe about 18 inches, and thls dimension can be used fo r the first trial size.Try top flange 18" x I"where the thickness of the flange can be obtained from the following equation.

F, = ~ ~ 6 . ~ =9.84 .......................................................................... (10-98) b' 9- = -=9


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BRIDGE DESIGN PR.ACI1CE D ECEMBER 1995

Composite Concrete SlabFrom Aniclel0.38.3 , the effective flange width of the slab shall not exceed:l . ~ s length= lA (150) = 37.5 ft2. Spacing between girders= 16ft3. Twelve times the slab thickness= 12( 107/s) = 131 in. r control

131"

Figure 4-13 Deck Effective Width

Effective concrete area= 131 (107/s) = 1,425 in.2

For a full composite section witbfc' = 2,900- 3,500 psi, n = Es =9 ......... (10.38.1.3). ~For a partially composite section, n = 3(9) = 27 .......................................... (10.38.1.4)

Calculation ofmoment of inertia can be done using the composite girder worksheet or with"COMP" on mM mainframe. Both methods are illustrated on the following pages.


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-- --

,. BRIDGE D ESIGN PRACriCE DECEMBER 1995tblt:rans

'STATE OF CALIFORNIA OEPARTI.IENTOF TRANSPORTATION liCOMPOSITE WELDED GIRDER WORKSHEET05-00124 ( REV. 1191) lli'Job: Shee1 of

1!1 Interior .~

O . Compostle Welded Gtrder By:EJc1enorSpan: Spaang: Date:

Section for Slab and GirderLoads Size Area y Ay Ayl.-. lCD I 0 Top Range =...H._ X 1.0 = __1L _H_ 1,7611 172,872

@ Web = ...i_ x o/'1 = __fl._ 2/l,70 147,015~, l Q) Bo l Flange = _JL x 1'/:1 .,_E _ 20.3 15.2~.... IA = 105 -4,7511 319,902.. J , e .G. 116,080'6' 100 Web " ItJ-1--- ICI>- = 365,982v.l I. ~ = Y o = ~ -'J7 >< IA = - 215.4:66 "y, = 53.2.:1J_L I II ~

Ie.G. = I 150,717 ln. I I;Section fo r Curb Loads, Railing, Utilities (n =27) 11IA= 105 11,754 319.902'


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~~ f l~ I fJ )z - (' )~ * * ~PF3=QUIT PF5=PR I NT --* COMPOSI TE GIRDER CROSS SECTION ANALYSIS/FLANGE DESIGNpl =SLAB - - - : BAR : STEEL : - - - - - -- WED - - - - - - - FLANGE - - - - - - - - *

* DIM. EFF. : DIM. : BAR : DIM. TOP BOTTOM .. ~Q A AREA : 8 : AREA : Y DEPTH T WIDTH T WIDTH T '05.4 1425.0 8.9 0.0 15.3 96 . 0 10 . 0 18 . 0 8 .0 1 8 . 0 12 . 0 * APPLIED MOMENTS - - - - : l~ * DEAD CURB & LIVE : N ALLOW * fj'~

.. LOAD RAIL LOAD : FS * d* 4667.0 1587.0 9880 . 0 9.0 50.0 * C/)... * * * * * * * * * OUTPUT * * * * * * Cl0 LOADS C/ I C/I C/I Y I Q/ I (JC* ..* (BARS) (TOP FL) (BOT FL) BAR ..DEAD 4.237 3.605 45.3 150723


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,. BRIDGE DESIGN PRACTICE DECEMBER 1995tiz/trans

Load Factor CheckCalculate CheckSteel Girder Design X Section b' and _Q FatigueSymm - Unsymm t lw

Reduce Non Check ShearAllowable Compact V, LbStress

YesCalculate CheckVU> d0 \1, Lb

'-----------------1 Revise:1. Flange Size2. Web, twYes 3. do. or4. Fy

NoDesign' - - - - - - - - - - - - - t ~ TransverseStiffener

Revise Most economicaltw. do between do. tw. andstiffener sizes?Checlc1. Range-WebWelds2. Shear


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BRIDGE D ESIGN PRACTICE DECEMBER 1995

l~ Xg I.L....It)cDM

II..)(

cS' t-tw

b' - 2.200 _)t - -:J!!';compression flange

Transverse Stiffener BDS 10.48.5.3b' < 2.600 t ..p;A = [ 0 . 1 5 B D t w ( 1 - C ~ - 1 8 t ~ ]yI=

Transverse StiffenerWeb {transversely stiffened) BDS 1 0.48.8

Vp=0.85FyDtwVu.=Vp C+ 0.87(1-C)F(M if M > 0.75MuV= Vu [2.2 - 1.6:J

Bending1. BDS 10.48.2


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lr BRIDGE DESIGN PRAcnCE DECEMBER 1995tbltrans

4.10.4 Width to Thickness RatiosRequirements for braced non-compact sectionsa) Outstanding leg of compression flange, non-compact

b' ~ = 9 . 8 4 .... .................................... (10.48.2.1)t "'Fy Ia" lb ' = ~ = 9 (ol2 CJ)

b' 9- = -


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lbltransSince the spacing between the cross-frames will be 20 feet, the section is unbraced for noncomposite dead load, and therefore, a reduction in allowable stresses is required due to buckling.Allowable stress

Fer= 0.6 Fy= 30 ksi ................. ...................................................................... (10.50(h))or

F: ]F. [l-3_ (Lb2cr y 47r2 b'Lo = 20 (12) = 240 in.b' = 18/2 = 9Fy =50 ksiE = 29 X 103 ksi

23 240f;;, = 50[1- (SO) ( ) ] =45.3 ksi :. Fer= 30 ksi controls47r 2(29xl03) 9Applied DL stress in top flange

!D L =19.8 ksi .......................................................................................... from page 4-33since Fer= 30 ksi >!D L = 19.8 ksi, top flange bracing of 20 feet is okay.

4.10.6 Fatigue RequirementsThe allowable fatigue stress ranges are dependent on (BDS Table 10.3.1A):1. redundant or non-redundant load path structures2. stress cycles3. stress categories


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BRIDGE DESIGN PRACTICE DECEMBER 1995lbltransType of Connection

Find: Illustration and Number Figure 4-16 (10.3.1C) Illustrative Examples

Find: Illustrated Example Number. Read Description and Stress Category Table 4-5 (1 0.3.1 B)

Find: Stress Cycle Table 4.2 (10.3.2A)*

N o ~ Yes~ - - - - - - - - - - - < > - - - - - - - - - - - - .?Find: Find:Stress Range, Fsr. Stress Range, Fsr.Non-redundant RedundantTable 4.4 (1 0.3.1 A)* Table 4.4 (10.3.1A)

BRIDGE DESIGN PRAcncE D ECEMBER 1995


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Table 4-2 Number of Cycles for Case ILoading Cycles

Pl3 with HS20 100,000HS20 Lane Loading 500,000HS20 Truck Loading 2.000,000HS20 Single Truck Loading over 2,000,000

Due to the uncenainty involved in predicting future traffic levels, it is specified that "CaseI" be used forall d.esigns. This also insures that permit vehicles are considered since P13 withHS20 (a t 100,000 cycles) has a strong influence on the fatigue behavior.The most common types of connections found in plate girders are:1. Transverse stiffeners2. Butt weld of flange plates3. Gusset plates for lateral bracing4. Flange-to-web weldTheseconnectionsand othersare illustrated inFigure 4-16(lllustrative Examples) and describedinTable4-5.Table4-5 isused to select the category which maiChes the detail being consideredTh e four connec tions listed above have been marked on Figure 4-16 and Table 4-5 and theresul ts s ummarized be low:

Table 4-3 Common Types of Bridge ConnectionsType ofConnection Stress Category l1lustratioo

I Toe of transverse stiffeners Tor Rev. c 62 Butt weld at flanges Tor Rev. B 8, lO3 Gusset for lateral bracing (bolt gusset to flange) Tor Rev. B 214 Aange-to-web weld Sbear F 9



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lif;Iii/trans

The applicable stress ranges are now read from BDS Table 10.3.1A and shown below :Table 4-4 Allowable Fatigue Stress Range

CyclesType ofConnection Category 100,000(Permit) 500.000(HS20 Lane) 2,000.000(HS20 Truck) Over 2.000,000(HS20Single Truck)

I Toe of transverse stiffeners c 35.5 ksi 21 ksi 13 ksi 12 ksi2 Butt weld at flanges B 49 ksi 29 ksi 18 ksi 16 ksi3 Gusset for lateral bracing B 49 ksi 29 ksi 18 ksi 16 ksi4 Aange-to-web weld F 15 ksi 12ksi 9 ksi 8 ksi

BRIDGE DESIGN PRAcnCE D ECEMBER 1995


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J .AI_Of_9 -no11n91>

210I 10Q) 18""'":0~ 3 -c;--z 7:7-a> 11Q)(/ )c0:;::C'CS:::J 4(i5

2' Rad. - ~ -20c ) - ~ -5 13

WELD CONDITION Cat~ ~ - - - - . P i o t : : e Ec ) ~ , _ _ c0___ .__...E 4 * ~ R e r i k > P i o t : : eE 4 * ~ - ReinL ,__.., . B

1 ; BRIDGE D ESIGN PRACTICE D ECEMBER 1995


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liz/trans

Table 4-5 (BDS Table 10.3.1B)Stress Illustrative Category Example Kind of (See Table (See Figure General Condition Situation Stress 10.3.1A) 10.3.1C)

Plain Member Base metal with rolled or cleaned surface. Flamecut Tor Rev a A l , 2edges with ANSI smoo thness of 1.000 or Jess .Built-Up Members Base metal and weld metal io members of built-up T or Rev B 3, 4,5 . 7plates or shapes. (without attachment


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Table 4-5 (continued)

General Condition SituationBase metal and weld metal in or adjacent to fullpenetration groove weld splices at transitions in widthor thickness, with welds ground to provide slopes nosteeper than 1 to 2V2, with grinding in the direction ofthe applied stress. and weld soundness established bynondestructive inspection:(a) AASIITO M270 Grades 100/IOOW(ASTM A709)base metal(b) Other base metalsBase metal and weld metal in or adjacent to fullpenetration groove weld splices, with or withouttransitions having slopes no greater than 1to 2112, whenthe reinforcementis not removed aod weld soundnessis established by nondestructive inspection.

Groove WeldedAttachments -LongitudinallyLoadedb

Base metal adjacent to details attached by full or partialpenetration groove welds when the delaillength,L, inthe direction of stress, is less than 2 inches.Base metal adjacent to details attached by full or partialpenetration groove welds when the detail length,L, inthe direction of stress, is between 2 inches and 12 timesthe plate thickness but less than 4 inches.Base metal adjacent to delails attached by full or partialpenetration groove welds wben the delaillength, L. inthe direction of stress, is grearer than 12 times the platethickness or greater than 4 inches:(a)(b)

Detail thickness


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Table 4-5 (continued)Stress Illustrative

Category Example Kind of (See Table (See Figure

General Conclition Situation Stress 10.3.1A) 10.3.1C) (c) 6 inches > Transition raclius 2 inches. D(d) 2 inches > Transition raclius ~ 0 inches. E- F o r all transition raclii without end welds ground Tor Rev E 16smooth.

Groove Welded Detail base metal attaChed by full penetration grooveAnactunents welds with a transition raclius. R. regardless of theTransversely detail length and with weld soundness transverse toLoadedb.c the direction of stress established by nondestructiveinspection:

-With equal plate thickness and reinforcement Tor Rev 16removed.

(a) Transition raclius 24 inches. B(b) 24 inches >Transition r a d i u s ~ 6 inches. c(c) 6 inches > Transition radius 2 inches . D(d) 2 inches > Transition radius 0 inches. E- With equal plate thickness and reinforcement not Tor Rev 16removed.(a) Transition radius 6 inches. c(b) 6 inches> Transition r a d i u s ~ 2 inches.(c) 2 inches >Transition radius 0 inches. E-Wi th unequal plate thickness and reinforcement Tor Rev 16removed.(a) Transition radius ~ 2 inches. D(b) 2 inches> Transition a d i u s ~ 0 inches. E- F o r all transition radii with unequal plate thickness Tor Rev E 16and reinforcement not removed.

Fillet Welded Base metal at details connected with transversely loaded

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Table 4-5 (continued)

General Condition SituationFillet Welded Base metal adjacent to details attached by fillet weldsAtt achments- length. L. in the direction of sttess, is Jess than 2 inchesLongitudinally and stud-type shear connectors.Loaded b.c.e

Base metal adjacent to details attached by fillet weldswith length, L. in the direction of stress, between2 inches and 12 times the plate thickness but lessthan 4 inches.Base metal adjacent to details attached by fillet weldswith length. L. in the direction of stress greater than12 times the plate thickness or greater than 4 inches:(a) Detail thickness < 1.0 inch.(b) Detail t h i c k n e s s ~ 1.0 inch.Base metal adjacent to details attached by fillet weldswith a transition radius. R. regardless of the detail length:-Wi th the end welds ground smooth(a) Transition radius 2 inches.(b) 2 inches> Transition a d i u s ~ 0 inch.- For all transition radii without the end weldsground smooth.

Fillet Welded Detail base metal attached by fillet welds with aAttachments transition radius. R. regardless of the detail lengthTransversely Loaded (shear stress on the throat of fillet welds governedwith the weld in by Category F):the direction ofprincipal sttess b.c - With the end welds ground smooth

(a) Transition radius ~ 2 inches.

Kind of Sttess

Tor Rev

T or Rev

T or Rev T or Rev

T or Rev

Tor Rev

Tor Rev

Sttess IllusttativeCategory Example(See Table (See Figure10.3.1A) 10.3.1C)

c 15. 17. 18. 20

D 15. 17

E 7. 9. 15. 17E' 7.9. 15

16DEE 16

16D

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Table 4-5 (continued)Stress lllustrative

Category ExampleKind of (See Table (See Figure

General Condition Situation Stress 10.3. 1A) 10.3.1C)Mechanically Base metal at gross section of high strength bolted slip Tor Rev B 2 1 Fastened resistant connections. except axially loaded joints which Connections induce out-of-plane bending in connecting materials.

- ,...........w '4..- ~ ................ . . _ . . . . . . . . ~ " " " " " ' - .. ..... .... " ..Base metal at ne t section of high strengtb bolted Tor Rev B 21bearing-type connections.Base metal at ne t section of riveted connections. Tor Rev D 2 1

a ' 'T' signifies range in tensile stress only. "Rev"' signifies a range ofstress involving both tensionand compression during a stress cycle.

b "Longitudinally Loaded" signifies direction of applied stress is parallel to the longitudinal ax.isof the weld. "Transversely Loaded" signifies direction of applied stress is perpendicular to thelongitudinal axis of the weld.

cTransversely loaded panial penetration groove welds are prohibited.d Allowable fatigue stress range on throat of fillet welds transversely loaded is a function of theeffective throat and plate thickness. (SeeFrankandFisher, Journal of the Structural Division, ASCE.

Vol. 105. No. ST9. Sept 1979.)

0.06+0.79 )S, =Sf Pl.Itpi[where Sf is equal to the allowable stress range for Category C given in Table 10.3.1A. This assumesno penetration at the weld root

e Gusset plates attached to girder flange surfaces with only transverse fillet welds are prohibited.

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4.10.6.1 Applied and Allowable Stress Ranges1. HS20 (Multiple Lanes) 2,000,000 cycles (Truck)

+LLM = 3,257 k-ft-LLM = -814 k-ftStress ran e = 3,257(12) (82.6)+ 814(12) (45.3) g 393,606 150,717

= 8.20 + 2.94 = 11.1 ksi < 13 k.si < 18 ksi Okay for Category Band C2. HS20 (Multiple Lanes) 500.000 (Lane Load)

+LLM = 3,345 k-ft-LLM = -1,11 1 k-ftStress ran e = 3,345(12) (82.6) + 1,111(12) (45.3) g 393,606 150,7 17

= 8.42 + 4.01 = 12.4 k.si < 21 ksi < 29 ksi Okay for Category Band C3. Pl3 with HS20 100,000 cycles

+LLM = 7,550 k-ft- LLM = -2,518 k-ftStress ran e = 7,550(12) (82.6) + 2,518(12) (45.3)

g 393,606 150,717 = 19.01 + 9.08 =28.1 ksi


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4.10.7 Shear DesignThe shear capacity, Vu ,of the section is dependent on the yield strength and thickness of theweb and the spacing of the transverse stiffener as

0.87(1- C)Vu =Vp C + ~ ............................ .... ....................................... ........ (10-113) + ( ~ r

where:vp = plastic shear capacity= 0.58FyDt.., .. ........................................ ................................................. (10-1 14)=0.58(50) 96 (5/g) = 1.740 kc = ratio of buckling shear stress to shear yield stress

The stiffeners are usually spaced equally between cross frames up to a maximum of 3D asspecified in BDS Article 10.48.8.3.Maximum d0 = 3(96) = 288 inchesTry do= 20 feet= 240 inches= spacing between cross-frames

5k = buckling coefficient= 5 + =5 + 2 = 5.8~ (2:) 6,CXJO.Jk = 6,ooo-J5i =64_6 and 7,500-Jk =7,500.J5.8 =80_.,JF; .J50,000 .,JF; .J50,OCJO 8

7 500D = ~ =154 > -Jk =80.8t,., Sjg JF:. _ 4.5x 107 k _ 4.5xl07(5.8) _. .C - - - 0.22....................................................... (10-116) 2 2

BRIDGE DESIGN PRACilCE DECEMBER 1995
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0Vu = 1740 0.22+ 87(l - 022) = 821 k1 + ( 2 9 ~ r

Applied Shear= Vmax = -19.8-6.8 - 110 = - 137 kVmax = 137 k < Vu = 821 k Okay

Check requirements for handlingD 96 .- =- = 153.6 >150 ................................................................................. (10.48.8.3) tw Sh

2 2260 ) ( 260 ). d0 S D -- = 96 -- = 275 inches( Dlt,.. 153.6d0 = 240 inches S 275 inches Okay

Spacing of transverse stiffeners and cross-frames is 20 feet.As might be expected due to low sheardemands at the0.4 point, on ly minimal stiffeners arerequired. However, as the design check moves closer to the supports, where the shear ishigher, the spacing of the stiffeners may become much closer.

4.10.7.1 Moment and Shear InteractionMoment- shear interaction ................................................................................. ( 10.48.8.2) I f M > 0.75 Mu then a reduction in the allowable shear, V, must be made. LetM=Mu

V M

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4.1 0. 7.2 Transverse Stiffener DesignMoment of inertia required:

I =dof..,3J ................................................................................ ......................... . (10-106) where:

1= ~ r 2 ~ 0 . 5 .................................................................................... ( 10-107) 296 )= 2.5 - -2 =-1.60 use J =0.5( 240

Area required:

A= [0.15BDtw(l- C ) ~ -l8t:,Jrwhere:

Y = Ratio of web plaxe yield to stiffener yield= 50 =1.3936 B =1.0 for stiffener pairs

A = I 5 ( L 0 ) 9 6 ( ~ 1 - 0 . 2 2 ) ~ ~ ~ - 1 8 ( ~ ) ' } 3 9=-8.15


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The width of stiffener is preferred to be at least 6 inches to aJlow adequate space for crossframe connection.

b' 2.600 = 13.7 !--ct.Webr .j36,(X)()6Let b' = 6 tmm =- - =0.44 inch13.7 ~-,

Try 6" x W' stiffenersb' 6 !Va" 6"r = f = l 2 < 13 .7 Okay

'- 6.31"1= t(6.31)3 {2) I3Figure 4-17

= 83.7 in.4 > ~ =29 .3 in.4 Okay Web and Stiffener Cross SectionUse 6" x 1h" stiffeners

4.11 Non-Composite Section DesignThis section iUustrates the design ofnon-composite section at Pier 2.

4.11.1 DesignLoads (SeeSection4-16, BridgeDesignSystemComputer0utput)

Dead Load Girder and Slab

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liz/transLive Loads1. Live Load Group IH

Maximum moment = 3.0(-3,292) = - 9,876 k-ftAssociated shear = 3.0(94.9) = 285 kMaximum shear = 3.0(110) =330 kAssociated moment = 3.0(-2,634) = -7,902 k-ft

2. Live Load Group l pwMaximum moment = 0.73(-3,292) + 1.22(-5,548) = -9,172 k-ftAssociated shear =0.73(94.9) + 1.22(217) = 334 kMaximum shear =0.73(110) + 1.22(279) = 421 kAssociated moment =0.73(-2,634) + 1.22(4,569) = -7 ,497 k-ft

Fatigue Loads (Case I Road)1. HS20 (single truck) over 2,000,000 cycles

+LIM =0.81(322) = 261 k-ft-UM =0.81(-1,476) =-1,196k-ft

2. HS20 (Multiple Lanes) 2,000,000 cycles (truck)+LIM = 1.38(322) = 444 k-ft-LIM =1.38(-1,476) = -2,037 k-ft

3. HS20 (Multiple Lanes) 500,000 cycles (Lane)+LIM = 1.38(365) = 504 k-ft-LIM = 1.38(-3,292) = - 4,543 k-ft

4. Pl3 with HS20 100,000 cycles+LIM = 0.56(365) + 1.15 (0.81) 1,110 = 1,238 k-ft-UM = 0.56(-3,292) + (1.15) 0.81 (-5,548) =-7 ,011 k-ft

4.11.2 Girder Section

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tizltl'an.sAnother method would be to use shear connectors at the maximum spacing of 24 inchesthrough the negative moment area.A symmetrical steel section will be used.

!'....__ _ _..... I .---------.1_l r- - - . --'-------r"'l---' tI

At= bt d M=Td

---"-- C

Figure 4-18 Equilibrium of ForcesBy equilibrium T = C = FyA1Design Moment Td= FyA..flDesign Moment: Girder+ Slab =- 10.269 k-ft

Rail+ AC = - 3,492 k-ftLive Load = - 9,876 k-ft Mappliod = 23,637 k-ft

- 23,637 k- ftDistance between e.g. of the flanges d = 96 + 2 = 98 inches (assuming 2-inch thick flanges)To make the fabrication of the plate girder easier, the web depth should remain constantthroughout the length of girder. The depth of the web (D = 96") is the same as used in thecomposite area.

T = M = 23,637(12) 2,894 kD 98



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Try: 18" x 2W' flanges and 96" x Sfs" webt)DI =2At(2 +2 + ~ r o . -

2

(15)6 27/s) 3 .=2(18X27/s) - + - +-- (96) =252,961+46,080=299.04lm42 2 12 8

96 Y, = ~ = +2 Vs = 50.9 inches2 fi = Me = 23,637(50.9)12 = 48 _3 ksi b I 299, 041 !b = 48.3 ksi < 50 ksi Okay

4.11.3 Width to Thickness Ratiosa) Outstanding leg of compression flang e - n o n ~ o m p a

: 1 , = 9 . ....................................... .................. .............................(10.48.2.1)

b' 9-=-= 3.13


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4.11.4 Bracing Requirements4 20,000,000 At ................................................... .............. .................... .... ( 10-100) FydAt = Area of compression flange

= 18(27.4) = 51.8 d =96 + 2(27h) =101.8 I 20000000(51.8) 203 5 . h 17...., ~ = . me es= 1eel50,000(101.8)

Let spacing between cross-frames be 15 feet, and no moment reduction due to bracing willbe required.

4.11.5 Fatigue Requirements1. H$20 (Multiple Lanes) 2,000,000 cycles (Truck Load)

(444 + 2,037)50.9Stress range= , (12) = 5.07 ksi < 13 ksi < 18 ksi299 041 Okay for Category B and C

2. H$20 (Multiple Lanes) 500,000 cycles (Lane Load)(504 + 4,543)50.9Stress range= , (12)=10.3 ksi < 21 ksi < 29 ksi299 041

Okay for Category B and C

3. Pl3 with H$20 100,000 cycles{1,238+ 7.011)50.9Stress range= (12) = 16.8ksi < 35.5 ksi < 49 ksi299 041



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4.1 1.6 Shear DesignMaximum Shear capacity, V11

0.87(1-C)VII = Vp C+ ~ ................................................................................ (10-113) v l + ( ~ Jwhere VP = 0.58FyDI ,.. ...................................................................................... (l0-1l4)

= 0.58 (50) 96 (5/s) = 1.740 kwhere d0 = spacing between transverse stiffeners 2260maximum d0 =3D= 3(96) = 288 in ., or for handling = 96 ( ) = 275 in.153.6try d0 = 15ft= 180 in.= spacing between cross-frames

5 5 k = 5 + ~ = 5 + 2 =6.42~ c : ~ )

6,ooo..[f _ 6,000../6.42 _ and 7,5oo.J k_ 7,5oo-J6.42_68 85JF: - .J5o,ooo - :p; - .J5o,ooo -D = =154 > 75oo-Jf =85t,.. Sjg OJF:

(4.5)107k (4.5)(107 )6.42:.C = ( = )2 =0.24 ..................................... .................. (10-11 6) D ) F. ( ~ 50,000t y Sjg"'

,. BRIDGE D ESIGN PRAcnCE DECEMBER 1995lbltl'ans
http:///reader/full/7,5oo.Jkhttp:///reader/full/7,5oo.Jkhttp:///reader/full/7,5oo.Jkhttp:///reader/full/7,5oo-J6.42http:///reader/full/7,5oo-J6.42http:///reader/full/7,5oo-J6.42http:///reader/full/7,5oo.Jkhttp:///reader/full/7,5oo-J6.42


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4.11.6.1 Moment and Shear Interaction1. Maximum Shear- Associated Momentvapplied = 857 k

Mapp!Jed = - 10.269-3.492-7,497 = - 21 ,258 k-ft299 041 1= FyS = Fy(!.) =(50) ( - ) = 24,479 k-ftc 50.9 12

0.75 Mu = 0.75 (24,479) = 18 ,360 k-ftSince M = 21 ,258 k-ft > 0.75 Mu= 18.360 k-ft, a reduction in the allowable shear, V. mustbe made.

21 258~ = 2 . 2 - 1 . 6 M =2.2-1.6( )= 0.81Vu Mu 24, 479V = Vu(0.81 ) = 959 (0.81) = 777 k

Since applied V =857 k >allowable V = 777 k N.G. The section must be revised.The section can be revised by one or more of the following:1. Increase flange size.2. Increase web thickness.3. Reduce stiffener spacing, d0

Try reducing stiffener spacing, d0 = 90 in.

1, 5oo.Jk =1,5oo-Jfo.69=:p; .j5o,ooo 1107 5

,. BRIDGE DESIGN PRACTICE DECEMBER 1995lbltrans
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v. =v, C+r( ~ r .............................................................................. ( 1 ~ 1 1 3 ) 0 87 1 0 41= 1, 740 0.41 + ( - " ) = 1,365 k

~ ( : Jallowable V=0.81Vu=0.81 (1 ,365)= 1,105kapplied V = 857 k < allowable V = 1,105 k Okay

I fweb size were increased, tw = % in., and the stiffener spacing remained the same d0 = 180 in.

V,=V, C+ ~ ................... ............................................................ (10-113) VP= 0.58 FyD tw = 0.58 (50) 96 (%) = 2,088 k k = 6.42 as before

BRIDGE DESIGN PRAcrrcE DECEMBER 1995liz/trans
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21( D )1=2Aj 2 +2 + l o-O web96 2 7jg ) 1 ( 3 ) 3 .=2{18){2 7/s) -+ - +- - (96) =308,257 m.4( 2 2 12 4

M =F_vl ={50) 308,257 =25 234 k-ft u c 50.9{12) '

0.75 Mu = 18,925 k-ft < M = 21,258 k-ft 21 258~ = 2 . 2 - 1 . 6 M =2.2-1.6( ) =0.85Vu Mu 25,234

V = 0.85Vu = 0.85( 1 286) = 1,093 kapplied V = 857 k < allowable V= 1,093 k Okay

So either reduce spacing between transverse stiffeners,d0 = 90 in. ,or increase the sizeof theweb. t,.. =o/4in.For this example use:

t,.. = o/4 in. and d0 = 180 in.2. Maximum moment - associated shear a) IH Group

M = -10,269-3,492- 9,876 = -23,637 k-ft v = 325 + 111 + 285 = 721 k Mu = 25,234 k-ft

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b) lpwGroupM =-10,269-3,492-9,172 =-22,933 k-ftv= 325 + 111 + 334 = 770 kMu =25 ,234 k-ft0.75 Mu = 18,925 k-ft < M = 22.933 k-ft

22 933: . ~ = 2 . 2 - 1 . 6 M =2.2-1.6( ) =0.75V.. M" 25, 234V = 0.75Vu = 0.75(1,286) = 964 k

applied V =770 k < allowable V =964 k Okay4.11.6.2 Transverse Stiffener DesignMoment of inertia required:

I= d0 tlJ (10-106)Where:

J = 2 . 5 ( ~ r - 2 ~ 0 . 5296 )=2.5 180 -2=-1.289(

Use:1=0.5



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Where:

Y = Ratio of web plate yield to stiffener yield50= - = 1.3936

B= 1.0 for stiffener pairsA = 0 . 1 5 ( 1 . 0 ) 9 6 ~ ) ( 1 - 0 . 3 5 857 ~ 1 . 3 94 1,286 4

= -7.57


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4.12 Flange-to-Web Weld

4.12.1 Weld Design@ Pier 2 Vapplied = 857 k = Design Shear 18"Applied shear flow at flange-to-web weld:

VQs=-1 :}.14"\ . Flange-toWebweld 1

where:Q =static moment= 18(27fs)49.4 = 2.556 in.3I = 308,257 in.4

857(2,556) .s = 308,257 =7.11 kim.According toBDS Article 10.23.6. the minimum size of N.A. ___ ___!fillet weld for 27/s" plate is 112", but need not exceed thethicknessof the thinner part joined. Use 1h" fillet welds. Figure 4-20Allowable shear on throat of weld Girder Dimensions

Fu = ultimate strength of base metal or weld metal, whichever is smallerFor A709 Grade 50 Fu = 65 ksiFor we ld metal Fu = 70 k.siUse F,. = 65 ksiFv =0.45(65) =29.3 k.si

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4.12.2 Fatigue Check For flange-to-web weld in shear, the allowable ranges of shear, Fs;, are:

Table 4-6Type ofLoad Cycles Category F,,

Pl3 with HS20 100,000 F 15 ksiHS20 Lane Load 500,000 F 12 ksiHS20 Truck Load 2.000,000 F 9 ~HS20 Single Truck over 2,000,000 F 8 ksi

Allowable shear flow for Fsr for 15 ksi = 2(lh) 0.707(15) = 10.6 k/in.For 12 ksi ............... 2(1/i ) 0.707(12) = 8.48 klin.For 9 ksi ................. 2( 1h) 0.707(9) = 6.36 k/in.For 8 ksi ................. 2(1h) 0.707(8) = 5.66 klin.

AppliedShear Range (See Section 4-16, Bridge Design System Computer Output):1. HS20 (Multiple Lanes) 2,000,000 cycles (Truck Load)Shear range= V, = 1.38(87.6) =121 k

- V,Q - 121(2556) -1 0 kl . 6 36 kl. Okas--- - . m. < . 10 . yI 308,2572. HS20 (Multiple Lanes) 500,000 cycles (Lane Load)

Shear range= V, = 1.38(119) =164 k164(2556)= V,Q = 1.36 klin. < 8.48 klin. OkayI 308,257

3. P13 with HS20 100,000 cyclesShear range= V,= 0.56(119) + 1. 15(0.81)304 = 350 k

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4.13 Shear Connectors

4.13.1 Fatigue DesignThe shear connectors are designed for fatigue and checked for ultimate strength. Maximumspacing equals 24 inches.

VrQSr = = - .................................................................................. ....... ... ................ (10-57) Iwhere:

Sr == range of horizontal shear flowVr ==range of vertical shear due to (LL+l) (Service Load)Q = static moment of the transformed concrete areaI = moment of inertia of the composite section

. l:Zrs =spacmg= - Srwhere:

Zr = allowable range for horizontal shear for welded studs with i d ~ 4, Zr = ad2, where d = diameter of stud and

a == 13,000 for 100,000 cycles = 10,600 for 500,000 cycles = 7,850 for 2,000,000 cycles = 5,.500 for over 2,000,000 cycles

Q = 0.010 from "COMP" program on IBM mainframeI



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1. HS20 (Multiple Lanes) 2,000,000 cycles (Truck Load)Allowable range of horizontal shearAssume 7/s" diameter studs, 3 per rowa= 7,850

7,850( 7/8 )23 . 18U = = 18 k/3 studs ~ spacmo=-T 1,000 ::> s,

Table 4-7Span 1 v, Q/1 s, SpacingAbut 1 122 0.010 1.22 14.80.4Ll 101 0.010 1.01 17.80.7Ll 105 0.010 1.05 17.1

2. HS20 (Multiple Lanes) 500,000 cycles (Lane Load)10,600( 1jg )23 24 3 k/3 ds . 24.3L -Lr = = . stu ~ spacmg = -1,000 s,

Table 4-8Span 1 v, Q/1 S, SpacingAbut 1 134 . 0.010 1.34 18.10.4Ll 97.0 0.010 0.97 25.1 > 24use 240.7L l 110 0.010 1.10 22.1

3. P13 with HS20 100,000 cycles13.000( Vs f3 d . 29.9U , = = 29 9 k /3 stu s ~ spacmg =- 1,000 s,

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4. HS20 (Single Truck) over 2,000.000 cycles~ 5.500( 7/s )23 . 12.6"- 'L r = - =12 6 k/ 3 sruds ~ spacmg =1,000 s,

Table 4-10Span l v, Q/1 s, SpacingAbut 1 7 1.6 0.010 0.716 17.6OALI 59.4 O.OlO 0.594 21.20.7Ll 61.3 0.0 10 0.613 20.6

Spacing for Fatigue

40' 70' 40'Rows@ 10 Rows@ 15 No Studs orcould use max.spacing =24"

ct. ct.Abut 1 Pier 2

Figure 4-21 Spacing ofShear StudsNumber of sruds provided for Span 1:

=(48 +56+ 1) x 3 =315 studsThese calculation are also applicable to Span 3 because the bridge is symmetrical. Span 2calculations are similar.

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4.13.2 Ultimate StrengthThe number of studs provided for fatigue must be checked for the ultimate strength of thestructure.

pNl = Su (10-60)

where:N1 =number of studs between point of maximum positive moment and adjacentend suppon or point of inflection.

= 0.85, a reduction factor= ultimate strength of connector=ultimate force capacity, smaller ofP1= AsFy ............................................................................................. (10-61)P2=0.85J: bts ..................... .............................. .................................. (10-62)

where:As = area of steel sectionFy = yield point of steelf/ = compressive strength of concreteb = effective flange width of concretets =thickness of concrete

The ultimate strength, s...of the stud connector with Hid> 4 is:Su =0.4d2 ..jj/Ec ~ A s c F y ..................................... ............ .................................. (10-66)

where:= w3/ 2 x33.fj(= 3,250 psi, w= 145 pcf, Ec= (145)312 33 3 , 2 5 = 3.3 x 106 psi, d = 118 in.

1=0.4( 7/s )2 250(3.3 x 106) ( - - ) =31.7 klstud1,000 = area of stud section

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therefore P = 3,936 k controlsp 3,936number of studs N1=- = =147S,. 0.85{31.7)

number of studs required in compression flange length of Span 1 = 2 x 147 = 294 studs315 studs provided for fatigue> 294 studs required for strengthFatigue design governs the number of studs in Span 1

4.13.3 Shear Connectors at Points of ContraflexureI fno studsare used over the negative moment area. additional studs are required at the deadload points ofcontraflexure to anchorthe additional deck reinforcement placed over the pier.The minimum amount of reinforcement is 1%of the concrete area. of which two-thirds mustbe placed in the top layer within the effective width.Area of concrete= 14.50 ft2

A/ = total area of longitudinal slab reinforcement over pier= 0.01(14.50) = 0.145 ft2= 20.9 in.2Number of connectors:

M ~ .Nc=- .......................................................................................................... (10-68)z,where:

Nc = number of additional connectors at points of contraflexuref, =range of stress due to live load plus impact in the slab reinforcemenL m a ybe taken as 10 ksi.z, = allowable range of horizontal shear on an individual shear connector= 1263 = 4.2 for 7fg" diameter stud at over 2,000,000 cycles

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4.14 Bearing Stiffener at Pier 2Reaction at Pier 2 (See Section 4-16, Bridge Design System Computer Output):1. DL (Girder+ Slab) = 1.3(240 + 250) = 637 k2. DL (rail +overlay) = 1.3(82 + 85) = 217 k3. Live load- greater of either: a) IH Group = 3.0 (185) = 555k

b) lpw Group= 0.73 (185) + 1.22 (344) =555k R = 637 + 217 + 555 = 1,409 k

According to BDS Anicle 10.34.6. bearing stiffeners are designed as concentrically loadedcolumns.

P 11 =0.85AsFcr (10-150)where:

As= gross effective area of column and

Fa= F, [1- ~ A ; )'] .............................................................. (10-151) when

KI..c < -r-- --p:;- (10-152)or

n2E'er= 2 {10-153)

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where:K = effective length factor

= 0.75 for welded end connectionsr =radius of gyration, Fy =yield of steel. E = 29 x 106 psi, Fer= critical buckling

stress

The stiffeners are A709 Grade 36 steel, Fy= 36 ksifor short columns assumeFer= Fy= 36 ksi

p 1.409 . 2Areqd= -=--=39.1 m.Fe, 36Try:

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39 1 2 0 81 . Try 8" x 3f4" PLstiffener =- = . m.6(8)lmin = ~ ~ ............................ ............... .. ................................ ................ (10-34)

8 ~ 3 6 , 0 0 0= = 0.70 in.< 0.75 in. Okay12 33 000 Area:

2stiffeners= (6)8(3/4) = 36 in .web:

between stiffeners= 12(3/4) = 9 in.2outside stiffeners (18t,..) = 18 (3/4)3/4 = 10.1 in .2

total area= 55.1 in.2 3I = (3)3.4 (16.75) 881 in.4x-x 12

r = (T = ~ 881 =4.00 in.f i 55.1KLc = 0.75(96) = 18.0

r 4.00

~ = 2n2(29xl06) = 126> K4 = 18.0~ T 36,000 r



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Check bearing on end of stiffenersbearing strength =1.5Fy=1.5(36) =54 ksiapplied bearing (assuming 1V2 in. cope on bearing stiffeners)

1409= =48.2 ksi Okay6(8 -1.5)0.75

:. use 6 PL 8" x 3/4" @Pier 2Note: Spacing of bearing stiffeners is normally controlled by the size of the bearing pad.Access for welding should also be considered: the 6 inches shown in Figure 4-22, whileadequate for design purposes, will make welding difficult

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4.15 Splice Plate ConnectionExample and details to be completed at a later date and distributed at that time.

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4.16 Bridge Design System Computer OutputThe following pages areselected parts of "Bridge Design System" for the example problem.

~

INPUT FILE: STRUCTURAL STEEL DESIGN EXAMPLE PAGE 1=


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I I RAME DESCRIPTION~ END SUPPORT CARRY OVER5 i MEM JT . COND OR DEAD LOAD K FACTORS RECALLz i NO LT RT LT RT DIR SPAN I HINGE E UNI SEC LT RT LT RT ~ ! E M., I 1-1 1-1 1-1 1-1 1 -1 1-1 1 -1 1-1 1--1 1--1 1--1 1 -1~ ' 1 1 2 R H 15 0 .0 90.00 0.0 3600. 2.500 .000 0.00 0.00 0 .00 0.00 02 2 3 H 200.0 90.00 0.0 3600. 2.500 .000 0.00 0.00 0 .00 0.00 0{!l I 3 3 4 R H 15 0 .0 90.00 0.0 3600. 2 . 500 .000 0.00 0.00 0 .00 0 .00 0m l p 5 2 20.0 5.00 0 .0 3250. 0.000 .000 0.00 0.00 0 .00 0.00 0~ 5 6 3 p 20 .0 5.00 0.0 3250. 0.000 .000 0.00 0.00 0.00 0.00 0V>

, FRAME PROPERTIESl END SUPPORT CARRY OVER DISTRIBUTION~ MEM JT COND OR FACTORS FACTORS;;I ! NO LT RT LT RT DIR SPAN MIN E"I HINGE B LT RTI LT RT0 1-1 1-1 1 -1 1-1 1 -1 1-1 1-1 1-1 I I I I& 1 1 2 R H 150.0 0.3240E+06 0.0 3600. 0.500 0.000 0.000 0.500m 2 2 3 H 200.0 0.32408+06 0.0 3600. 0.500 0.500 0.500 0.5003 3 4 R H 150.0 0.3240E+06 0.0 3600. 0.000 0.500 0.500 0.000 t::C4 5 2 p 20 .0 0.1625E+05 0.0 3250. 0 . 00 0 0.500 0.000 0.000 225 6 3 p 20.0 0. 1625E+05 0.0 3250. 0.000 0.500 0.000 0.000

~"" FRAME DOES NOT SWAY WITH THIS LOADING ' * *" 0HORIZONTAL MEMBER MOMENTS TRIAL 0 rnMEM 0NO LEFT . 1 PT .2 PT .3 PT . 4 PT .5 PT .6 PT . 7 PT .8 PT .9 PT RIGHT z1 0. 17 41. 2920. 3536. 3590. 3082. 2010. 377. -1819. -4578. -7899 . 2 -789 9. -3399. 101. 2601. 41 01. 4601. 4101. 2601. 101. -3399. -789 9. 3 -7899. -4578. -1819. 377. 2010. 3082. 3590. 3536. 2920. 1741. 0. iORIZONTAL MEMBER SHEARS TRIAL 01 134 .8 97.3 59.8 22.3 -15.2 -52.7 -90.2 - 127.7 -165.2 -202.7 - 240.2 2 250.0 200.0 150.0 100.0 50.0 0.0 -50 .0 -100.0 -150.0 -200.0 -250.0 03 240.2 202.7 165 . 2 127.7 90.2 52.7 15.2 -22.3 -59.8 -97.3 -134.8 ~ ~

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O . B ~ O F I ~ ~ ~ T E N O

Design of Welded Steel Plate Girders

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