Designer Supplement GSD-1-19 Intermediate Crossframes and Intermediate Diaphragms January 18, 2019

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1 Designer Supplement GSD-1-19 Intermediate Crossframes and Intermediate Diaphragms January 18, Overview The purpose of Standard Bridge Drawing, GSD-1-19, is to address updated design requirements for intermediate crossframes not previously addressed in the GSD-1-96 Standard Bridge Drawing. An intermediate diaphragm was added for shallow, rolled beam members. End crossframes and scupper details remain unchanged from GSD This drawing includes four separate plan sheets, which separately address the following steel bridge details: Sheet 1/4: Intermediate Crossframes This drawing addresses standard intermediate crossframes, for use with tangent steel beams or girders designed with line-girder methodologies per the ODOT Bridge Design Manual. Sheet 2/4: Intermediate Diaphragms - This drawing addresses standard intermediate diaphragms, for use with tangent steel rolled beams ( 36 deep) designed with line-girder methodologies per the ODOT Bridge Design Manual. Sheet 3/4: End Crossframe Details This drawing remains unchanged from Standard Bridge Drawing GSD Sheet 4/4: GSD Scupper Details - This drawing remains unchanged from Standard Bridge Drawing 2 Plan Preparation Requirements: 2.1 Sheet 1/4 Intermediate Crossframes Limitations for Intermediate Crossframes This drawing provides standard intermediate crossframe details for use with tangent steel beams or girders designed with line-girder methodologies per the ODOT Bridge Design Manual. The intermediate crossframes are applicable to rolled beam depths or girder web depths between 36 and 72 as well as the spacing and overhang limits shown on the sheet. Crossframe Spacing: The intermediate crossframe designs are applicable for intermediate crossframe spacings up to 25 feet for positive moment regions and up to 15 feet for negative moment regions for bridge widths up to 70 feet wide. For bridge decks wider than 70 feet, the intermediate crossframe designs are applicable for spacings up to 21 feet for positive moment regions and up to 15 feet for negative moment regions. Page 1

2 Designer Supplement GSD-1-19 Intermediate Crossframes and Intermediate Diaphragms January 18, Intermediate Crossframe Types There are three separate types of intermediate crossframes, considered equivalent for design purposes and each with different fabrication and constructability pros and cons. Unless the contract plans specify otherwise, the contractor may choose either the Type A, B or C intermediate crossframes in accordance with the limitations specified in GSD The three types of intermediate crossframes include: Type A: The National Steel Bridge Alliance (NSBA) preferred crossframe, includes crossframe angles welded to gusset plates, which are field bolted to beam/girder stiffener connection plates. This is NSBA s preferred type of crossframe for fabrication and erection efficiencies. Depending on beam/girder depth and deck overhang width, crossframe angle size and number of bolts required vary. Type B: The field-welded intermediate crossframe, similar to the Type 3 intermediate crossframe from GSD-1-96, includes an erection bolt for steel beam/girder erection and field welded connection prior to the deck pour. Limitations include angle size and minimum weld length, depending on beam/girder depth and deck overhang width. Type C: The field-bolted, knock-down type crossframe, similar to the Type 4 intermediate crossframe from GSD-1-96, is fully bolted prior to the deck pour. Use of this type of intermediate crossframe is not permitted for shallow beams/girders with larger overhangs and/or wider beam/girder spacings. The number of bolts required varies depending on deck overhang width and beam/girder depth. 2.2 Sheet 2/4 - Intermediate Diaphragms Sheet 2/4: Intermediate Diaphragms - This drawing addresses standard intermediate diaphragms, for use with tangent steel rolled beams up to 36 deep that are designed with linegirder methodologies per the ODOT Bridge Design Manual. Beam Spacing: The standard intermediate diaphragms are applicable to beam spacings up to 10.5 feet. Diaphragm Spacing: The intermediate diaphragm designs are applicable for diaphragm spacings up to 25 feet in positive moment regions and up to 15 feet in negative moment regions for bridge decks up to 70 feet wide. For bridge decks wider than 70 feet, the intermediate diaphragm designs are applicable for spacings up to 21 feet for positive moment regions and up to 15 feet for negative moment regions due to the influence from the increasing ODOT finishing machine loading. Page 2

3 Designer Supplement GSD-1-19 Intermediate Crossframes and Intermediate Diaphragms January 18, Estimated Quantities Assume the use of the Type A Crossframes to determine the total weight of structural steel for the Estimated Quantities. If the pay unit is Lump Sum, provide the estimated weight on the Estimated Quantities Form as defined in Section 1504 of the L&D Manual, Vol. 3. If the pay unit is Pound, provide the following note on the Estimated Quantities plan sheet: Item 513 Structural Steel Members, Level : This Total Weight is based on the use of Type A Crossframes. Provide the unit cost for Structural Steel using the total weight provided, regardless of any change to the total weight resulting from the selection of Type B or Type C crossframes in lieu of Type A. 3 Detail Information 3.1 Design Criteria The intermediate crossframes and intermediate diaphragms were designed in accordance with the following references: AASHTO Guide Design Specifications for Bridge Temporary Works, 2nd Ed. [GSBTW] AASHTO LRFD Bridge Design Specifications, 7 th Edition, 2014 [AASHTO] Ohio Bridge Design Manual, 2007, including the 2018 updates [ODOT BDM] ASCE 7-10 Minimum Design Loads and Associated Criteria for Buildings and Other Structures [ASCE] FHWA Steel Bridge Design Handbook Volume 13, Bracing System Design, 2015 [FHWA] 3.2 Design Methodology Intermediate crossframes and diaphragms were designed for the following: Externally applied forces o Wind loads ASCE (Construction), AASHTO (Final) o Cantilever bracket loading during deck construction GSBTW Bracing forces o Based on in-plane bending moment of the beam/girder. FHWA o Can be simplified to 0.02 M br (M br = in-plane bending moment of the beam/girder) based on AISC Steel Construction Manual, 15 th Edition, App a. Bracing Stiffness FHWA The cantilever bracket used for construction of the deck overhang creates a torsional force in the beam/girder that is carried by the intermediate bracing. This cantilever bracket loading governs the bracing design for large deck overhangs. Loads for the cantilever bracket during the deck pour were developed as follows: Page 3

4 Designer Supplement GSD-1-19 Intermediate Crossframes and Intermediate Diaphragms January 18, 2019 Dead loads plus allowance for form loads GSBTW Horizontal Construction Load GSBTW Area live loads GSBTW Deck finishing machine loads per ODOT BDM 3.3 Sample Design Calculations Refer to the Appendix for a sample bracing calculation. The sample calculation provided is for a Type B Intermediate Crossframe. The Type B Crossframe is temporarily connected with an erection bolt, then field-welded prior to deck construction. The design of Type A and Type C crossframes are similar, but with the design of the bolted connections. Page 4

5 Designer Supplement GSD-1-19 Intermediate Crossframes and Intermediate Diaphragms January 18, 2019 Appendix Sample Calculations for Type B Intermediate Crossframe Page 5

6 REFERENCE LIST REFERENCES No. Abbreviation Title 1 GSBTW AASHTO Guide Design Spcifications for Bridge Temporary Works, 2nd Edition, ODOT BDM ODOT Bridge Design Manual, AASHTO AASHTO LRFD Bridge Design Specifications, 7th Edition, ASCE 7 ASCE 7-10 Minimum Design Loads and Associated Criteria for Buildings and Other Structures 5 ODOT SBR-2-13 ODOT Standard Bridge Drawing 57" Single Slope Concrete Median Bridge Railing 6 FHWA FHWA Steel Bridge Design Handbook Volume 13, Bracing System Design, December Battistini Stiffness Behavior of Cross Frames in Steel Bridge Systems 8 AISC AISC Steel Construction Manual, 13th Edition, CalTrans Bridge Design Aids Table of Maximum Moments, Shears, and Reactions Simple Spans 10 AASHTO SS AASHTO Standard Specifications for Highway Bridges, 17th Edition, AISC Mom. AISC Moments Shears and Reactions for Continuous Highway Bridges 12 AISC Torsional Analysis of Structural Steel Members 13 NSBA SS NSBA Short Span Standards, Jan NSBA CS NSBA Continuous Span Standards Field Welded Cross Frame_ xlsm 1 of 55

7 INPUT SUMMARY Beam Depth Angle size P.R. 72 in L5x5x1/2 Check OK Geometry check Beam moment check 0.95 OK Stiffness check 0.29 OK Angle check 0.70 OK Connection check 0.86 OK MATERIALS Steel modulus of elasticity Beam/Girder yield strength Angle yield strength Angle tension strength Stiffener yield strength Stiffener tension strength E = 29, ksi F y = ksi F yl = ksi F ul = ksi F ys = ksi F us = ksi Concrete strength f' c = 4.50 ksi Steel unit weight γ s = pcf Concrete unit weight γ c = pcf GSBTW Field Welded Cross Frame_ xlsm 2 of 55

8 INPUT SUPERSTRUCTURE Span length L = ft Girder spacing S = in Number of girders n g = 7.00 Overhang OVR = in Cross-frame spacing S cf = ft Skew 0.00 deg Simply supported/continuous SS Total number of cross-frames 8 Drop Drop = 2.63 in Strut angle θ strut = atan(drop/s) = 0.02 rad Out-to-out bridge width W deck = 2*OVR+(n g -1)*S = in -Perpendicular to CL of bridge Live load distribution limits AASHTO Table b-1 Spacing 3.50 ft S ft OK Span length ft L ft OK Number of beams 4.00 n g OK ODOT Specified limits Maximum frame spacing S cf ft OK Maximum overhang OVR in OK Girder spacing S ft OK Field Welded Cross Frame_ xlsm 3 of 55

9 INPUT GIRDER Girder/rolled shape Data source Girder name Girder NSBA CS CS Dimensions Depth d = in Top flange width b f1 = 18 in Bottom flange width b f2 = 20 in Top flange thickness t f1 = 1.56 in Bottom flange thickness t f2 = 2 in Web thickness t w = 0.5 in Web depth D = 68 in Beam/Girder Properties Weight w gird = lb/ft Area A = in2 Moment of Inertia I = 94,402 in4 Top section modulus S top = 2,378 in3 Bottom section modulus S bot = 2,962 in3 Top section yield moment M y - = 137,944 kip-in Bottom section yield moment M y + = 171,802 kip-in Used yield moment M f = min(m y -, M y +) = 137,944 kip-in Positive moment effective moment of inertia I eff = 1,830 in4 Negative moment effective moment of inertia I eff = 2,279 in4 Used effective moment of inertia I eff = 1,830 in4 Flange centroid distance h o = d - t f1 /2- t f1 /2 = 70 in Field Welded Cross Frame_ xlsm 4 of 55

10 INPUT DECK Barrier assumed BR-1-13 Deck effective span length S deck = S - b f1 /2- t w /2 = 9.73 ft Thickness t deck =max(8.5", (L+17)*12/36) = 9.00 in ODOT BDM Supported deck width b deck = OVR + S/2 = in Overhang thickness t OVR = t deck + 4" = in ODOT BDM Figure Clear overhang width OVR CL = OVR - b f1 /2 = in Total Deck Area A deck = b deck t deck + OVR CL (t OVR -t deck ) = 1, in2 Bottom of beam to deck centroid in Total deck weight w deck = A deck *γ c = 1, lb/ft Combined girder + deck weight w comb = w + w deck = 1,631 lb/ft % Weight of girder w / w com = 21% Overhang area A OVR = OVR CL *t OVR + b f1 /2*t deck = 588 in2 Overhang centroid eccentricity in Overhang weight w OVR = A OVR γ c = lb/ft AASHTO Live Load distribution limits Slab thickness 4.5 in t s 12 in OK Stiffness constant back-calculated from simplified value in AASHTO Table &. % $ 1.02 & 12 $ % 1,706,202 10,000 in t s 7,000,000 in OK Field Welded Cross Frame_ xlsm 5 of 55

11 INPUT STIFFENER Width b s = 5.00 in Thickness t s = in Vertical web corner clip clip V = 2.00 in Horizontal web corner clip clip H = 1.00 in Weld leg w stiff = 5/16 in Weld thickness t w-stiff = * w stiff = 0.22 in Web stiffener width for stiffness b s-stiff = b s = 5.00 in Beam/Girder Field Welded Cross Frame_ xlsm 6 of 55

12 INPUT CROSS-FRAME Type Type A Description Field Welded Angle Size L5x5x1/2 Angle width b = 5.00 in Centroid from face y b = 1.42 in Thickness t = 0.50 in Gross area A g = 4.79 in2 Radius of gyration r x = 1.53 in Vertical offset OFF V = 2.00 in Horizonal offset OFF H = 1.00 in Strut Geometry Strut Angle θ strut = atan(drop/s) = 0.02 rad Bottom of strut left end Y 1 = t f2 + OFF V = 4.00 in Top of strut left end Y 2 = Y 1 + b f *cos(θ strut ) + (b s -OFF H )*tan(θ) = 9.08 in Centroid line at center of girder Y left = Y 2 - (OFF H + t w /2)tan(θ) - (b-y b )cos(θ) = 5.48 Bottom of strut at right end Y 3 = t f2 + OFF V + Drop = Top of strut at right end Y 4 = Y 3 + b*cos(θ strut ) + (b s - OFF H )*tan(θ strut ) = Centroid line at center of girder Y right = Y 3 + (OFF H + t w /2)*tan(θ) + y b *cos(θ) = 6.63 in in 8.07 in AD Diagonal Geometry Bottom of Angle Y 1 = Y 2 (strut) + OFF V = in Top of angle Y 2 = d + Drop - t f1 - OFF V = in Left edge of angle X 1 = t w /2 + OFF H = 1.25 in Right edge of angle X 2 = S - X 1 = in Field Welded Cross Frame_ xlsm 7 of 55

13 INPUT Angle θ = atan((y 2 -Y 1 )/(X 2 -X 1 )) = 0.45 rad Left End Bottom Corner X 3 = X 1 + b*sin(θ) = 3.42 in Centroid at end of Angle X 4 = X 3 - y b *sin(θ) = 2.80 in Y 4 = Y 1 + y b *cos(θ) = Angle centroid line intersection with web Y 5 = Y 4 - X 4 tan(θ) = in in Right end right corner Y 6 = Y 2 - b*cos(θ) = in Centroid at end of angle X 7 = X 2 - y b sin(θ) = in Y 7 = Y 6 + y b cos(θ) = Angle centroid line intersection with web Y 8 = Y 7 + (S- X 7 )tan(θ) = in in Field Welded Cross Frame_ xlsm 8 of 55

14 INPUT BC Diagonal Geometry Top of angle Y 1 = d - t f1 - OFF V = in Bottom of angle Y 2 = Y 2 (strut) + Drop + OFF V = in Left edge of angle X 1 = t w /2 + OFF H = 1.25 in Right edge of angle X 2 = S - X 1 = in Angle θ = atan((y 2 -Y 1 )/(X 2 -X 1 )) = rad Left End Top Corner X 3 = X 1 - b*sin(θ) = 3.26 in Centroid at end of Angle X 4 = X 3 + (b-y b )*sin(θ) = 1.82 in Y 4 = Y 1 -(b-y b )*cos(θ) = Angle centroid line intersection with web Y 5 = Y 4 - X 4 tan(θ) = in in Right end right corner Y 6 = Y 2 + b*cos(θ) = in Centroid at end of angle X 7 = X 2 + (b-y b )sin(θ) = in Y 7 = Y 6 + y b cos(θ) = Angle centroid line intersection with web Y 8 = Y 7 + (S- X 7 )tan(θ) = in in Overlap Check for applicable Girder Y 1 (BC)- Y 1 (AD) = in 2b + OFFv = OK -Fails if angles would overlap. Gap between working lines 2.00 in Bottom of girder to bottom diagonal working line in Top of girder to top diagonal working line 6.64 in Field Welded Cross Frame_ xlsm 9 of 55

15 INPUT Cross-Frame Schematic ANGLE GEOMETRY Left Right X (in) Y (in) X (in) Y (in) L (in) θ (rad) θ (deg) AC Strut AD Diagonal BC Diagonal Truss model height h b = Y BC(Left) - Y AC(Left) = in Minimum Angle θ = min( θ AD - θ AC, θ BC - θ AC ) = deg deg OK -Arbitrary minimum angle to ensure angles act like a truss. Maximum angle length L max = in Angles must pass the limiting slenderness ratios for both tension and compression Radius of gyration r x = 1.53 in Tension slenderness ratio L max /r x = OK -Tension slenderness ratio for secondary member AASHTO Effective length factor K = Assumes angles act like pinned-pinned Compression slenderness ratio L max K/r x = OK AASHTO Field Welded Cross Frame_ xlsm 10 of 55

16 INPUT CONNECTIONS FIELD CONNECTION BOLT Grade Diameter Hole type Hole diameter A325 d b = in Oversized d h = in A b = πd 2 / in2 Tensile strength F ub = ksi AASHTO Hole reduction factor R p = 0.90 AASHTO Allow holes punched full size Minimum bolt tension P t = 19 kip AASHTO Table Bolt Edge Distance e h = 1.50 Minimum edge distance e hmin = in OK AASHTO Table End Distances AASHTO Minimum clear end distance e hbrg = in Angle edge clear distance L cl = e hmin - d h /2 = in OK Field Welded Cross Frame_ xlsm 11 of 55

17 INPUT Bolt Stiffener Clear Distances Angle Side b s - OFF (in) H cs (in) L cs (in) AC Left AD Left BC Left Angle clear distance geometry: Where: b s = Stiffener width OFF = Horizontal offset between web and angle e = Edge distance θ = Diagonal angle esin(θ) = Horizontal distance from corner of angle to bolt line at end of angle ecos(θ) = Horizontal from bolt line at end of angle to center of bolt. H cs = Horizontal distance from center of bolt to edge of stiffener H cs = b s - OFF - esin(θ) - ecos(θ) L cs = H cs /cos(θ) - d h /2 Change e to (b-e) for BC Diagonal Minimum stiffener clear distance L cs = 1.81 in Minimum clear distance check Min. edge distance e hmin = 0.88 in OK Field Welded Cross Frame_ xlsm 12 of 55

18 INPUT WELD Weld leg w = 5/16 in Weld thickness t = 0.707w = 0.22 in Classification strength of weld metal F e70 = ksi Reduced length for compression capacity Lcomp AC AD BC Stiffener compression length -Stiffeners are checked for compression buckling. Length taken along the centroid line of the angle from the face of the web to th end of the angle. $ cos/ tan/ Lstiff (in) AC 1.07 AD 2.73 BC $ ()* $ cos/ COMPILED GEOMETRY CHECKS Geometry checks on this page Check of weld lengths on analysis tab Overall geometry check OK OK OK Field Welded Cross Frame_ xlsm 13 of 55

19 LOADS DESCRIPTION: Calculate applied loads on cross frames per GSBTW (based on ASCE 7) and AASHTO LRFD -Load convention: = PosiQve Load, = NegaQve Load. WIND LOADING (ERECTION): GSBTW/ASCE Girder/Beam Depth D = in Diaphragm/Cross Frame Spacing Spa = ft Design Height Above Ground H = ft Exposure Category C -Includes flat open country and grasslands (ASCE 7 C.3.(a)) Velocity Exposure Pressure Coefficient K z = 1.09 Table C.1 Field Welded Cross Frame_ xlsm 14 of 55

20 LOADS Topographic Factor K zt = 1.00 Conservative -No hills, ridges, or escarpments are assumed. Wind Directionality Factor K d = 1.00 Table C.2 -Conservatively ignored because GSBTW is limited to tower structures and are < Basic Wind Speed V = mph Figure C.1 (a) Field Welded Cross Frame_ xlsm 15 of 55

21 LOADS Gust-Effect Factor G = 0.85 ASCE 7 C.6 Length/exposed height ratio B/s = 45 For ASCE Table C.5 Clearance ratio s/h = D/H = 0.12 ERECTION STAGE Velocity Exposure Pressure Coefficient K z = 1.09 Table C.1 Force Coefficient C f = 1.95 Table C.5, Case A Velocity pressure q h = *K z K zt K d V 2 = psf Gross area of solid surface A g = D * Spa = ft2 Force applied over cross frame spacing F tot = q h GCfA g = 9,119 lb GSBTW (C-2) Top & bottom lateral load F = F tot /2 = 4.56 kip DECK STAGE -Include an additional 2' of wind height above top of girder for deck formwork, machinary. Height allowance for deck stage H deck = 2.00 ft Deck height H +2' = ft Velocity Exposure Pressure Coefficient K z = 1.10 Table C.1 Clearance ratio s/h = (D + 2')/(H + 2') = 0.15 Force Coefficient C f = 1.95 Table C.5, Case A Velocity pressure q h = *K z K zt K d V 2 = psf Top area Agt = (D/2+2')* Spa = ft2 Force applied to top of cross-frame F T = q h GC f A gt = 7,674 lb Bottom area A gb = D/2 * Spa = ft2 Force applied to top of cross-frame FT = q h GC f A gt = 4,593 lb Top lateral load F WS-T = 7.67 kip Bottom lateral load F WS-B = 4.59 kip Field Welded Cross Frame_ xlsm 16 of 55

22 LOADS WIND LOADING (FINAL) (AASHTO LRFD): Wind Speed, STR III = mph AASHTO Figure Velocity Exposure Pressure Coefficient K z = 1.09 Table C Gust-Effect Factor G = 1.00 Table Drag Coefficient C d = 1.30 Table P z = psf Final force applied to bottom half of girder/beam F = P z * D/2 * Spa = 3, lb 3.58 kip Field Welded Cross Frame_ xlsm 17 of 55

23 LOADS DEAD LOAD Impact % I = 0.00 GSBTW Impact not included because the entire 25' of deck is not placed at once. The horizontal live load and higher concrete weight accounts for any additional loading from the concrete placement. ERECTION STAGE -No direct loading on the cross-frame from dead load during erection stage. DECK STAGE -Torsion is applied to the cross-frame as a result of the overhang deck concrete. Only the part of the overhang supported by the bracket is applied to the cross frame. From ODOT BDM Overhang OVR = in Platform total overhang OVR plat = OVR + 5" + 12" + 5" = in Bracket support to centerline of girder b brack = OVR plat - 16" = in Width of area load applied b deck = OVR - b brack /2 = in Overhang thickness t OVR = in 10 psf D Included for Walkway weight Resisted by = OVRplat Resisted Directly by Beam Overhang weight w OVR = b deck * t OVR * γ c = lb/ft Eccentricity x OVR = OVR - b deck /2 = in Cross frame spacing S cf = 300 in Girder spacing S = in Overhang moment M OVR = w OVR x OVR S cf = kip-in Platform weight p D = psf bbrack -Platform weight is included with C D (Construction dead load) in the GSBTW, but is included with the other dead loads for -ODOT BDM c.E weight of the formwork used for the platform. Platform width w plat = 5" + 12" + 5" = in ODOT BDM Assumptions for TAEG Bracket Design Input Total load V plat = p d *w plat *S cf = 0.46 kip Moment arm x plat -2 = w plat /2 + OVR = in Platform moment M plat = V plat * x plat = kip-in Field Welded Cross Frame_ xlsm 18 of 55

24 LOADS Overhang bracket weight p brack = lb Overhang bracket spacing s brack = in Overhang bracket distributed weight w brack = p brack / s brack = 1.04 lb/in Overhang assumed eccentricity e brack = OVR/2 = in Bracket moment M brack = w brack * S cf * e brack = 7.5 kip-in ODOT BDM c.F ODOT BDM c.F Dead load moment M DC = M OVR + M plat + M brack = kip-in Truss height h b = in Top of frame lateral load F DT = -M DC /h b = kip Bottom of frame lateral load F DB = M DC /h b = 5.31 kip Top of frame impact load F IT = F DT = 0.00 kip Bottom of frame impact load F IB = F DB I = 0.00 kip Field Welded Cross Frame_ xlsm 19 of 55

25 LOADS CONSTRUCTION LIVE LOAD ODOT BDM c & GSBTW Weight of equipment is assumed to be the machine wheel loads and area load are specified in ODOT BDM c.E.6. The linear additional load from the GSBTW has been removed because the machine wheel loads are based on the larges screed machi Used for typical bridges. The area load on the walkway is for people walking along the deck. The area load on the deck is for loading from the finishing machine on the wet concrete. All live loads are applied simultaneously. WHEEL LOAD Skew Skew = 0.00 deg Nearest 5 degrees of skew 0.00 deg Bridge out-to-out width measured perpendicular to centerline of bridge W deck = in Centerline to centerline of screed rails W = W deck + 5 in = ODOT BDM Assumptions for TAEG Bracket Data Input Rail-to-Rail length multiplier k W = 1.00 W ODOT BDM c.E Table Skew extra end length L extra = 0.00 ft Total machine length L machine = min(120ft, W * k W + L extra ) = 71 ft ODOT BDM c.E Cross Frame Spacing S cf = ft Distribution % Dist% = 96% -Distribution% calculated by distributing the Bidwell wheel loads through the brackets as concentrated torsion moments using the AISC Torsional Analysis of Structural Steel Members Design Guide. Wheel load P wheel =4x(2.2 kip + max(0, L machine - 36)*0.012 kip)xdist% = kip -All 4 wheels combined Wheel load application length L wheel = 36" + 31" + 36" = 103 in Distributed load P wheel / L wheel = 0.10 kip/in Cross-frame spacing S fc = 300 in Applied load P wheel-app = P whee l/l wheel * min(s fc, L wheel ) = kip Moment arm x wheel = OVR + 2.5in = in Vertical load V wheel = P wheel-app = kip Cross frame moment M wheel = P wheel-app * x wheel = kip-in Wheel Load Distribution % RESISTED BY SINGLE CROSS FRAME 120% 100% 80% 60% 40% 20% 0% CROSS-FRAME SPACING (FT) Field Welded Cross Frame_ xlsm 20 of 55

26 LOADS AREA LOAD -Live load is applied on the slab and walkway. Live load on slab p slab = psf GSBTW Platform total overhang OVR plat = OVR + 5" + 12" + 5" = in ODOT BDM Assumptions for Bracket support to centerline of girder b brack = OVR plat - 16" = in TAEG Bracket Data Input Width of area load applied b slab = OVR/2 = in -includes area from edge of deck to halfway between bracket and centerline of girder. Load applied on overhang w slab = p slab *b slab = 3.33 lb/in Area load moment arm x slab = 3 b slab /4 = in Live load on walkway p walk = psf GSBTW Width of area load applied b walk = 16" = in -includes area of platform Load applied on overhang w walk = p walk *b walk = 1.11 lb/in Area load moment arm x walk = OVR plat - b walk /2 = in Vertical load V area = (w slab + w walk )*S fc = 1.33 kip Cross frame moment M area = (w slab x slab +w walk x walk )*S fc = kip-in LINEAR LOAD (REMOVED, SEE DISCUSSION ABOVE) Live load w lin = 0.00 lb/ft GSBTW Maximum applied length L lin = 240 in GSBTW C Applied length L lin-app = min(l lin, S fc ) = 240 in Moment arm x lin = OVR = 48 in Vertical load V lin = w lin L lin = 0 kip Cross frame moment M lin = w lin L lin-app x lin = 0 kip-in COMBINED LIVE LOAD Total vertical load V LL = V wheel + V area + V lin = kip Total live load moment M LL = M wheel + M area + M lin = kip-in Truss height h b = in Top of frame lateral load F DT = M DC /h b = kip Bottom of frame lateral load F DB = -M DC /h b = 9.11 kip Field Welded Cross Frame_ xlsm 21 of 55

27 LOADS HORIZONTAL LOADING GSBTW % of Vertical load applied horizontally. Load is distributed spatially in proportion to the mass. Can be applied in either direction, but not simultaneous with wind loading. ERECTION STAGE -Only the weight of the girder is included Girder weight w = lb/ft Diaphragm spacing S cf = ft Total load 2F Ch = 0.02wS cf = 0.17 kip Top and bottom applied load F Ch = 0.09 kip DECK STAGE -Girder weight, deck weight, and construction live load are included in the horizontal loading. -Half of the girder weight is applied to the bottom of the cross-frame. -All other loads are applied at the top of the cross-frame. Deck weight w deck = 1, lb/ft Load applied at top of cross-frame. F ChT = 0.02*((w + w deck /2)S cf + V LL ) = 0.73 kip Load applied a bottom of cross-frame F ChB = 0.02*w*S cf /2 = 0.09 kip Field Welded Cross Frame_ xlsm 22 of 55

28 LOADS GIRDER MOMENT The strength requirements in NSBA are based on the moment applied to the girder. For a given uniform load, the maximum positive moment occurs at the center of a simple span. Similarly, the maximum negative moment comes over the center of a two-span bridge with equal spans. -Both of these moments = 0.125wL 2 (AISC Table 3-22c & 3-23 Case 1). -As discussed in the AISC Moments Shear and Reactions for Continuous Highway Bridges, negative moments can be increased by up t 5% due to an increase in the moment of inertia over a pier up to 50%. -This is applied to plate girders in continuous bridges, but not wide flange or simply supported bridges. Rolled Shape/Girder Girder Simple span/continuous girder SS Girder multiplier m gird = for Continuous span, plate girders only Span length L = ft ERECTION STAGE Loading consists of self weight of girder and cross-frames. Assume the weight of the cross-frames & accessories = 10% of girders. GSBTW Load combination 1 is used for the erection stage because it maximizes dead load. Girder weight w girder = 0.35 kip/ft Moment Mf-erect = 1.4 * 1.1 * mgird * w girder * L 2 *0.125 = 1, kip-ft Field Welded Cross Frame_ xlsm 23 of 55

29 LOADS DECK STAGE Loading includes the girders, cross-frames, deck, and construction live load. Girder weight w girder = 0.35 kip/ft Deck weight w deck = 1.28 kip/ft Dead load moment M D-Deck = (1.1w girder + w deck )L 2 * = 5, kip-ft Deck moment M D-Deck-Only = (1.1w girder + w deck )L 2 * = 4, kip-ft -Used for impact calculation which is based on the dead load being placed during this stage only. Construction live load Wheel load P wheel = kip -Calculated above for construction live load From AISC Table 3-23, Case 31 Continuous span moment M wheel = P wheel *L/(6 (3)) = kip-ft -Two equal spans, concentrated load at any point Simple span moment M wheel = P wheel *L/4 = kip-ft -Find maximum moment over support from Used wheel moment M wheel = kip-ft concentrated load. Area load p area = psf Applied width w Area = OVR plat + S/2 = ft Distributed load w area = w area *p area = 0.22 kip/ft Area load moment M area = w area *L 2 * kip-ft Linear load -Conservatively collapsed to concentrated load. Linear load total vertical force V lin = 0.00 kip $ :$8; $ $8 $8 4$ 4$ :$ 88 % ; $ :$ 38 ; Continuous span moment Distributed load moment M LL-lin = V lin L/(6 (3)) = 0.00 kip-ft Simple span moment M LL-lin = V lin L/4 = 0.00 kip-ft Used linear moment M LL-lin = 0.00 kip-ft 0$ 38 8 $ $ $ % 3 $% 3 3 7$ Total live load moment M LL = M wheel + M dist = 1, kip-ft 6 Impact 7$ 6 3 % of the Dead load M I = Impact%*M D-Deck = 0.00 kip-ft GSBTW GSBTW LRFD 1 GSBTW LRFD 2 Deck stage moment M 1 = 1.4M D-Deck = 7, kip-ft M 2 = 1.2M D-Deck + 1.6*(M LL + M I ) = 8, kip-ft M deck = m grid * max(m 1, M 2 ) = 8, kip-ft Check that the deck stage moment doesn't exceed the 90% of yield moment during deck placement moment for simple spans. 110% M y = 12, kip-ft P.R. Check AASHTO Minimum Depth requirement (AASHTO Table ) 0.65 OK Composite depth d + t deck = in 0.04L 76.8 in 0.95 OK Steel depth d = in 0.033L in 0.89 OK Field Welded Cross Frame_ xlsm 24 of 55

30 FINAL STAGE Deck moment M D-Deck = 5, kip-ft LOADS Barrier area A bar = 4.75'*1.75' - 0.5*4.75'*10.875" = 6.16 ft^2 ODOT SBR Conservatively assume the weight of a SBR " Single Slope Concrete Bridge Railing. Barriers N bar = 2 Girders n g = 7 Barrier weigh/girder w bar = γ c *N bar *n g = 0.28 kip/ft Barrier moment M D-Bar = kip-ft Dead load moment Wearing surface M D-Final = M D-Deck + M D-Bar = 6, kip-ft p DW = psf Out-to-out of deck w deck = ft Distributed load w DW = p DW * w deck / n g = 0.61 kip/ft Wearing course moment M DW = w DW L 2 * , kip-ft Live Load Live load tables used to calculate beam bending moments. Distribution factor taken from AASHTO. Calculate distribution factor: &. AASHTO Stiffness parameter % 1.02 AASHTO $ Girder spacing S = 10.5 ft Span length L = ft Overhang OVR = 4.00 ft Barrier BR-1-13 Center of girder to face of barrier d e = OVR = 2.5 ft -0' for TST-1-99, 1.5' for Other barriers. Interior girder Single lane distribution factor Multi-lane distribution factor AASHTO b-1 Exterior girder Single lane distribution factor -Lever rule. Assume TST-1-99 barrier with face at edge of overhang. First wheel is placed 2' from edge of barrier, second wheel 8' from barrier is the multiple presence factor from AASHTO Table Multiple lane distribution factor e = d e /9.1 = 1.04 D Fm-ext = eg m-int = Maximum exterior girder distribution factor DF = Moment on Simple Span Moment with impact taken from CalTrans Bridge Design Aids Table of Maximum Moments, Shears, and Reactions Simple Spans. Simply supported moment M SS = 5, kip-ft Distributed moment? ?+ ) &.@ $ &.D $ &.% &. 12.0$ % 12.0$ %?+ EF G 0.5 : 8 G ; &. &. M LL+I = M SS * DF = 5, kip-ft Field Welded Cross Frame_ xlsm 25 of 55

31 LOADS Simply Supported HL-93 With Impact Moment TRUCK MOMENT (KIP-FT) Simply Supported Design Span SPAN LENGTH (FT) Moment on Continuous Span Curve fit to AISC Mom. two equal span negative moments. Two-span truck bending moment M HS20 = xL xL = 2, kip-ft 3000 Two-Span HS20-44 Truck Moment TRUCK MOMENT (KIP-FT) y = x x R² = 1 Two Spans Design Span Poly. (Two Spans) SPAN LENGTH (FT) Lane load bending moment Distributed moment Impact Live load + impact moment Simply Supported/Simple span Used live load + Impact moment AASHTO LRFD Load Combinations Strength I Final stage moment M lane = 0.64kip/ft * L 2 * = 2, kip-ft M LL = DFx(M HS20 + M lane ) = 4, kip-ft M I = 33%xM LL = 1, kip-ft M LL+I = M LL + M I = 5, kip-ft SS M LL+I = 5, kip-ft M 1 = 1.25M D + 1.5M DW M LL+I = 19, kip-ft M final = m gird * M 1 = 19, kip-ft Check that the final stage moment doesn't exceed the 110% of yield moment during final loading. P.R. Check moment for simple spans. M y = 11, kip-ft 0.00 OK Field Welded Cross Frame_ xlsm 26 of 55

32 LOADS BRACING STRENGTH MOMENT FHWA Unbraced length L b = 300 in Span length L = 1,920 in Number of span braces n = 8 Steel modulus of elasticity E = 29,000 ksi Effective moment of inertia I eff = 1,830 in4 Moment modification factor C bb = 1.00 Flange centroid distance h o = in Erection stage Girder moment M f-erect = 20,550 kip-in Strength moment: 6 I 0.005$ $ kip-in FHWA (14) JKL M N ( Deck Stage Erection design moment M f-deck = 98,096 kip-in Final stage kip-in FHWA (14) -Bracing only needed for continuous spans because the deck braces the compression flange for simply supported spans. Simply Supported/Simple span 6 IE1O 0.005$ $6 JKL M N ( Erection design moment M f-final = 0 kip-in 6 IE1O 0.005$ $6 JKL M N ( SS 0.00 kip-in FHWA (14) Field Welded Cross Frame_ xlsm 27 of 55

33 LOADS LOAD FACTORS AND LOAD COMBINATIONS TABULATED LOADING Top of Cross-Frame Force Bottom of Cross-Frame Force Stage DL I LL CH WS DL I LL CH WS Erection Deck Final Load Factor Comb. Stage Spec Load Comb. DL I LL C H WS Mbr 1 Erection GSBTW LRFD Erection GSBTW LRFD Erection GSBTW LRFD Erection LRFD Service II Deck GSBTW LRFD Deck GSBTW LRFD Deck GSBTW LRFD Deck LRFD Service II Final LRFD Strength III Final LRFD Service II Comb. FT (kip) FB (kip) Mbr Field Welded Cross Frame_ xlsm 28 of 55

34 CROSS-FRAME STIFFNESS REQUIRED STIFFNESS FHWA Reduction factor φ = 0.75 FHWA Span length L = 1, in Number of span braces n = 8.00 braces Steel modulus of elasticity E = 29, ksi Effective moment of inertia I eff = 1, in^4 Moment modification factor C bb = 1.00 Skew skew = 0.00 deg BEAM MOMENTS Erection stage M f1 = 20,550 kip-in Deck Stage M f2 = 98,096 kip-in Final stage M f3 = 0 kip-in REQUIRED STIFFNESS P 0 2.4$6 1 QJKL M cos / Erection stage β t = 6,112 kip-in/rad Deck Stage β t = 139,262 kip-in/rad Final stage β t = 0 kip-in/rad FHWA (13) & (19) Field Welded Cross Frame_ xlsm 29 of 55

35 CROSS-FRAME STIFFNESS BRACING STIFFNESS FHWA FHWA Figure 9: Field Welded Cross Frame_ xlsm 30 of 55

36 CROSS-FRAME STIFFNESS Steel modulus of elasticity E = 29, ksi Girder spacing S = in Span length L = 1, in Number of girders n g = 7.00 girders Cross frame height Diagonal length h b = in L c = in Horizontal angle area A h = 4.79 in^2 Horizontal angle eccentricity ӯ h = 1.42 in Horizontal angle thickness t h = in Diagonal angle area A c = 4.79 in^2 Horizontal angle eccentricity ӯ c = 1.42 in Horizontal angle thickness t c = 0.50 in Girder moment of inertia I x = 94,402 in^4 Stiffener thickness t s = 0.38 in ANGLE STIFFNESS REDUCTION Battistini (10) -Redduced for eccentrically connected angles. Horizontal reduction R Xh = 0.45 Horizontal effective area A h = 2.17 Diagonal reduction R Xc = 0.45 Diagonal effective area A c = 2.17 CROSS FRAME STIFFNESS FHWA Initial state P K N % 1.32E+06 kip-in/rad -Compression system $ -Erection& Deck State Final state P K N % 4.83E+05 kip-in/rad -Tension system 2$ % WEB DISTORTIONAL STIFFNESS R FHWA Not included for full depth web stiffener with a cross frame. P IN-PLANE STIFFNESS OF GIRDERS FHWA P S 24 J 1 KL F 7.58E+05 kip-in/rad FHWA (18) J $ % COMBINED STIFFNESS Initial state U WI) N 0.159X P T P P P β T = 4.82E+05 kip-in/rad FHWA (15) Final state β T = 2.95E+05 kip-in/rad Required Provided P.R. Erection stage 6.11E E+05 kip-in/rad 0.01 OK Deck Stage 1.39E E+05 kip-in/rad 0.29 OK Final stage 0.00E E+05 kip-in/rad 0.00 OK Combined stiffness check 0.29 OK Field Welded Cross Frame_ xlsm 31 of 55

37 ANALYSIS TRUSS ANALYSIS GEOMETRY Span length Truss height S = in h b = in Drop h D = 2.63 in AC Length L AC = (S 2 + h D 2 ) = in Angle θ AC = ATAN(h D /S) = 0.02 rad 1.19 degrees AD Length L AD = (S 2 + (h b + h D ) 2 ) = in Angle θ AD = ATAN((h b + h D )/S) = 0.46 rad degrees BC Length L BC = (S 2 + (h b - h D ) 2 ) = in Angle θ BC = ATAN((h b - h D )/S) = 0.43 rad degrees Field Welded Cross Frame_ xlsm 32 of 55

38 ANALYSIS APPLIED LOADINGS -From Loads calculatin -FT & FB are the loads from each load combination applied to the top and bottom of the truss respectively. Diagonal Diagonal Strut Total Force/Moment Comb. Stage L.C. FT (kip) FB (kip) BC AD AC F M 1 Erection LRFD Erection LRFD Erection LRFD Erection Service II Deck LRFD Deck LRFD Deck LRFD Deck Service II Final Strength III Final Service II Maximum load on erection bolt from applied loads P bolt-f = 5.01 kip Field Welded Cross Frame_ xlsm 33 of 55

39 ANALYSIS MAX MEMBER FORCES Strength Erection Deck Final Member Tension Comp. Tension Comp. Tension Comp. AD BC AC Service II Member Erection Deck Final AD BC AC During the erection stage, the cross frame behaves as the Compression system shown in FHWA Figure 9, and there is no force on the strut. -During the deck stage, the formwork is assumed to act as a top member, and loads are distributed as shown in that system. This is conservative when calculating loads on the angles. -Because moments shown in FHWA Figure 9 can be applied in the other direction, the loads are calculated for one direction, but the opposite is added to the applied loads when checking the angle capacity. -The length of each individual diagonal is substituted for Lc in FHWA Figure 9 to account for the effect of the drop. Erection strength moment M br = kip-in Applied force F = M br /h b = 0.68 kip Spacing S = in AD Length L AD = in BC Length L BC = in AD Force F AD = 0.76 kip BC Force F BC = 0.75 kip Maximum force on erection bolt Pbolt = Pbolt-f + max(fad, FBC) = 5.77 kip Deck strength moment M br = kip-in Applied force F = M br /h b = in AD Length L AD = in BC Length L BC = in AC Length L AC = in AD Force F AD = 2FL AD /S = kip BC Force F BC = 2FL BC /S = kip AC Force F AC = -F = kip Field Welded Cross Frame_ xlsm 34 of 55

40 ANALYSIS Final strength moment M br = 0.00 kip-in Applied force F = M br /h b = 0.00 in AD Length L AD = in BC Length L BC = in AC Length L AC = in AD Force F AD = 2FL AD /S = 0.00 kip BC Force F BC = 2FL BC /S = 0.00 kip AC Force F AC = -F = 0.00 kip FINAL MEMBER FORCES Erection Deck Final Member Tension Comp. Tension Comp. Tension Comp. AD BC AC Final condition does not govern and is no considered further. Maximum top and bottom stiffener force -Used for checking buckling in the stiffener plate. Top stiffener plate P stiff-top = min(ad, BC) = kip -Maximum compression in either the AD or BC max member forces including bracing forces. Bottom stiffener plate P stiff-bot = min(ac) = kip -Maximum compression in the AC max member forces including bracing forces. Field Welded Cross Frame_ xlsm 35 of 55

41 ANALYSIS STIFFENER PLATE WELD Stiffener width b s = 5.00 in Top flange width b f1 = in Bottom flange width b f2 = in Web depth D = in Web thickness t w = 0.50 in Vertical stiffener clip O 1 = 2.00 in Horizontal stiffener clip O 2 = 1.00 in Top horizontal weld length L 1 = min(b f1 /2 - t w /2 - O 2, b s -O 2 ) = in Vertical weld length L 2 = D - 2O 1 = in Bottom horizontal weld length L 3 = min(b f2 /2 - t w /2 - O 1, b s -O 2 )= in Total weld length L = L 1 + L 2 + L 3 = in Centroid from face of web x 1 = 0.33 in Centroid from center of web y 1 = 0.00 in Moment of inertia about centroid I x = 31, in^3 I y = in^3 I p = 31, in^3 Top leg eccentricity c x1 = L 1 + O 1 - x 1 = 5.67 in c y1 = L 2 /2 + O 2 - y 1 = in Bottom leg eccentricity c x3 = L 3 + O 1 - x 1 = 5.67 in Weld group equations: $ $ $ 2 % $ % 2 $ $ $ % $ 2 $ L F $ $ 2 $ % c y3 = L 2 /2 + O 2 + y 1 = 12 $ $ % $ 2 L Z $ % 12 $ $ $ $ % % 12 $ % % $ % Total force F taken from truss analysis above Total moment M taken from truss analysis above + Mbr in Field Welded Cross Frame_ xlsm 36 of 55

42 ANALYSIS Top leg weld force F M r px r py r mx r my r u Comb. kip kip-in kip/in kip/in kip/in kip/in kip/in , , , , Bottom leg weld force F M rpx rpy rmx rmy ru Comb. kip kip-in kip/in kip/in kip/in kip/in kip/in , , , , Maximum Weld Force w w = 2.35 kip/in Weld thickness t = 0.22 in Maximum weld stress w w /2*t = 5.33 ksi Field Welded Cross Frame_ xlsm 37 of 55

43 ANALYSIS ANGLE WELDS Minimum weld size L min = max(1.5", 4*w) = 1.5 in AASHTO Angle Properties L5x5x1/2 Angle width b = 5.00 in Centroid from face y b = 1.42 in Center of angle to centroid of angle e L = b/2 - y b = 1.08 in A-D DIAGONAL WELD Angle θ = deg 0.43 rad Stiffener width b s = 5.00 in Web-angle gap gap = 1.00 in Horizontal overlap L h = 4.00 Maximum L3 Minimum weld length Bottom Weld Length L 1 = 2.12 in OK Top Weld Length L 3 = 4.39 in OK Front Weld Length L 2 = 5.49 in OK Front weld centroid from back of angle x 2 = 3.26 in Total length L = in Centroid from end of angle x 1 = 2.48 in Centroid from middle of angle width y 1 = 0.47 in Moment of inertia about X Axis I xx = in^3 I yy = in^3 I p = I xx + I yy = Field Welded Cross Frame_ xlsm 38 of 55

44 ANALYSIS Weld group moment off inertia equations: L FF $ 2 $ 12 $ $ % 2 L ZZ $ % 12 $ $ 2 $ $ % $ $ 12 [ [ $ % % 12 $ % $ % 2 Force in weld: Determined using a unit load along the centroid of the angle. Force/in from direct loading r px = -1/L = kip/in/kip r py = 0 = kip/in/kip Load eccentricity: e = max(e L - y 1, e L + y 1 ) = 1.55 in e is taken as the maximum possible eccentricity because Cross-frame weld is not specific to orientation of angle. \ )F 1]^Z L *, \ )Z 1]^F L * Location cx cy rmx rmy ru in in kip/in/in kip/in/in kip/in/in 1 Bot.-left Bot.-right Top-right Top-left Maximum load / axial load on angle kip/in/in Field Welded Cross Frame_ xlsm 39 of 55

45 B-C DIAGONAL WELD Angle θ = deg rad ANALYSIS Stiffener width b f = 5.00 in Web-angle gap gap = 1.00 in Horizontal overlap L h = 4.00 Maximum L1 Minimum weld length Bottom Weld Length L 1 = 4.32 OK Top Weld Length L 3 = 2.27 OK Front Weld Length L 2 = 5.40 in OK Front weld centroid from back of angle x 2 = 3.30 in Total length L = in Centroid from end of angle x 1 = 2.48 in Centroid from middle of angle width y 1 = in Moment of inertia about X Axis I xx = in^3 I yy = in^3 I p = I xx + I yy = Force in weld: Determined using a unit load along the centroid of the angle. Force/in from direct loading rpx = -1/L = kip/in/kip rpy = 0 = kip/in/kip Load eccentricity: e = max(el - y1, el + y1) = 1.51 in e is taken as the maximum possible eccentricity because Cross-frame weld is not specific to orientation of angle. \ )F 1]^Z L *, \ )Z 1]^F L * Location cx cy rmx rmy ru in in kip/in/in kip/in/in kip/in/in 1 Bot.-left Bot.-right Top-right Top-left Maximum load / axial load on angle kip/in/in Field Welded Cross Frame_ xlsm 40 of 55

46 ANALYSIS A-C STRUT WELD Angle θ = 1.18 deg 0.02 rad Stiffener width bf = 5.00 in Web-angle gap gap = 1.00 in Horizontal overlap Lh = 4.00 Maximum L3 Minimum weld length Bottom Weld Length L 1 = 3.90 OK Top Weld Length L 3 = 4.00 OK Front Weld Length L 2 = 5.00 in OK Front weld centroid from back of angle x 2 = 3.95 in Total length L = in Centroid from end of angle x 1 = 2.74 in Centroid from middle of angle width y 1 = 0.02 in Moment of inertia about X Axis I xx = in^3 I yy = in^3 I p = I xx + I yy = Force in weld: Determined using a unit load along the centroid of the angle. Force/in from direct loading r px = -1/L = kip/in/kip r py = 0 = kip/in/kip Load eccentricity: e = max(e L - y 1, e L + y 1 ) = 1.10 in e is taken as the maximum possible eccentricity because Cross-frame weld is not specific to orientation of angle. \ )F 1]^Z L *, \ )Z 1]^F L * Location cx cy rmx rmy ru in in kip/in/in kip/in/in kip/in/in 1 Bot.-left Bot.-right Top-right Top-left Maximum load / axial load on angle kip/in/in Field Welded Cross Frame_ xlsm 41 of 55

47 ANALYSIS WELD FORCE Member forces from Truss Analysis above. A-D Weld force B-C Weld force A-C Weld Force kip/in/in kip/in/in kip/in/in Diagonal Diagonal Chord Weld Force Stage Spec. L.C. BC AD AC BC AD AC Erection GSBTW LRFD GSBTW LRFD GSBTW LRFD LRFD Service II Deck GSBTW LRFD GSBTW LRFD GSBTW LRFD LRFD Service II Maximum Weld Force w w = 3.47 kip/in Weld thickness t = 0.22 in Maximum weld stress w w /t = ksi All weld lengths are adequate OK Field Welded Cross Frame_ xlsm 42 of 55

48 ANGLE DESIGN CHECKS DESCRIPTION: Check angle capacity Field Welded Cross Frame_ xlsm 43 of 55

49 ANGLE DESIGN CHECKS AD DIAGONAL Angle Size Modulus of Elasticity Yield Strength L5x5x1/2 E = 29, ksi F yl = ksi Ultimate Strength F ul = ksi Angle leg width b = 5.00 in Angle Thickness t = in Angle Area (Gross) A g = 4.79 in Radius of gyration, x r x = 1.53 in Angle Length L = in NONSLENDER MEMBER ELEMENTS CHECK AASHTO Slenderness ratio b/t = Plate buckling coefficient k = 0.45 AASHTO Table K Slenderness Limit ` AASHTO Z Nonslender b/t k (E/F y ) OK $ a2 $ b80 $ \ Angle effective length F \ F AASHTO \ $ a2 $ c80 \ F \ F COMPRESSION CAPACITY Slender Element Reduction Factor, Q Q = 1.00 AASHTO Table Equivalent nominal yield resistance P o = QF y A g = kip AASHTO kip AASHTO kip AASHTO Resistance factor, Compression φ c = 0.95 AASHTO Compressive Resistance 7 O 7 i K $ \ d e 0.658d f 7 ( a2 7 g ( a2 7 7 ( h0.44 P r = φ c P n = kip Field Welded Cross Frame_ xlsm 44 of 55