Dear Customer: American Association o State Highwa and Transportation Oicials ERRATA John R. Njord, President Executie Director Utah Department o Transportation John Horsle Executie Director Due to errors ound ater the publication had been completed, AASHTO has reprinted the pages and added two point pages listed below in order to make the ollowing errata changes to the AASHTO Guide Speciications or Horizontall Cured Steel Girder Highwa Bridges with Design Examples or I-Girder and Box-Girder Bridges: Page No(s). Aected Section Errata Change Horizontall Cured Steel Box Girder Example pp. 333 336 Appendix E.6 Add the ollowing paragraph immediatel beore the last paragraph on p. 334: In certain situations, the Engineer ma wish to consider the eect o the access hole in the diaphragm on the section properties o the eectie diaphragm section. From separate but similar calculations to the aboe, considering the eect o an assumed 36-in deep access hole centered in the middle o the example diaphragm, the maximum shear stress in the eectie bottom lange due to the diaphragm shear is calculated to be 6.13 ksi. Thereore, in this particular case, the eect o the access hole on the bottom lange shear stress is not signiicant. pp. 339 340 Appendix E.9 Change H55 to HS5 on p. 339 pp. 341 34 Appendix E.10 Change H55 to HS5 on p. 34 pp. 355 356 Appendix E.1 Add the ollowing to the end o the text on p. 355: Eq. (A10) gies the normal distortional warping stress at an point in the cross section. The alue o C w is obtained rom either Figure A, A3, or A9. Cw m ( or T ) σ dw = (A10) I β a where: = distance along the transerse ertical axis o the box rom the neutral axis to the point under consideration Eq. (A11) gies the axial (brace) orce due to distortional orces applied to the box. The alue o C b is obtained rom either Figure A5 or A10. Executie Oice: 444 N. Capitol St., N.W., Suite 49, Washington, D.C. 0001 Telephone (0) 64-5800 Teleax (0) 64-5806 Telex 49000095800 HTO
Page No(s). Aected Section Errata Change Horizontall Cured Steel Box Girder Example F C a+ b 1+ h a 1 + b b = b ( m or T) a (A11) Since onl two loading positions or concentrated loads are considered in the charts, it is oten necessar to interpolate between igures. The principle o superposition applies or more than one torque. Figure A11 shows the eect o β on the inluence line or diaphragm orces when the diaphragm is rigid. pp. 355 356 Appendix E.1 Add Figures A and A3 beore Figure A4 on p. 356 pp. 359 360 Appendix E.13 In the equation or tran on p. 359, which is the third equation rom the top o the page, 50 ksi should be 0 ksi. pp. 361 36 Appendix E.14 In the middle o p. 36, in the equation or tran, 50 ksi should be 0 ksi. pp. 363 364 Appendix E.14 Change H55 to HS5 on p. 363 pp. 365 366 Appendix E.15 Change H55 to HS5 on p. 366 The ollowing new pages hae been added to accommodate omitted igures: pp. 357 357.1 Appendix E.1 Moe down Figures A5 through A9 Add Figure A10 on p. 357.1 pp. 357. 358 Appendix E.1 Add Figure A11 on p. 357. Please substitute the original pages o text with the enclosed pages. We apologize or an inconenience this ma hae caused. AASHTO Publications Sta October 003
HORIZONTALLY CURVED STEEL BOX-GIRDER DESIGN EXAMPLE 333 R 1 shall be taken as: where: R = 1 97 k 1 k + + 4 F k s Eq. (10-5) k = plate buckling coeicient k s = shear buckling coeicient Tr: k = 4.0 and k s = 5.34 Since the denominator o Eq (10-5) is approximatel equal to 1.0 in this case, R 1 = 97 k = 97 x 4.0 = 194. Since 191 < 194, use Eq. (10-4) to compute the critical lange stress. F cr = 50 x 1.0 = 50 ksi A er rigid longitudinal lange stiener is required to proide k = 4.0 to ensure that a node orms at the stiener. The Engineer should tr to match the stiener size to the required critical lange stress in order to tr and obtain the most economical solution. Alwas assuming k equal to 4.0, which requires the largest size longitudinal stiener(s), ma not be the optimum solution or cases where the resulting higher critical lange stress is not necessar. Since a critical lange stress less than 50 ksi would be satisactor or this case, tr k =.0 instead o 4.0, which will result in a lower critical lange buckling stress, but which will also result in a signiicantl smaller longitudinal lange stiener. Since the denominator o Eq. (10-5) is again approximatel equal to 1.0 with k taken equal to.0: Thereore, compute R. R 1 = 97 x.0 = 137 < 191 R = 10 k 1 k -0.4 + ( -0.4) + 4 1. F k s Eq. (10-7) R = 10.0 1 1.18.0 1.0-0.4 + ( 1.0-0.4) + 4 1. 50 5.34 97 R = = 97 1.18.0 0.833 0.6 + (1.0-0.4 ) + 4 50 5.34 Check b R < F R. Since 137 < 191 < 97, use Eq. (10-6) to compute the critical lange stress. s 1 t b R F π - F cr = F - 0.4 1 - sin R - R1 s t Eq. (10-6)
334 AASHTO GUIDE SPECIFICATIONS FOR HORIZONTALLY CURVED STEEL GIRDER HIGHWAY BRIDGES 40.5 π 97 50 F cr = 50 1.0 0.4 1 sin - - - 1.5 = 47.6 ksi 97 137 - -46.47 = 0.98 < 1.00 OK 47.6 The actual alue o k s will be determined in the next section on the design o the longitudinal lange stiener. The critical stress will then be checked using the actual alue o k s to determine i there is a signiicant change in the stress. The bottom lange at interior supports acting in combination with the internal diaphragm is subject to bending in two directions plus the torsional and diaphragm shear (ignoring through-thickness bending o the lange plate under its own sel-weight). Thereore, Article 10.4..1 also requires that the combined stress in non-composite box langes at supports, due to ertical bending in the box girder and diaphragm and due to the diaphragm plus torsional shear, caused b the actored loads, not exceed the speciied minimum ield stress o the lange. As speciied in Article C10.4..1, or such cases, the combined stress in the lange due to the actored loads can be checked using the general orm o the Huberon Mises-Henck ield criterion as ollows: σ σ σ +σ + 3 F Eq. (C10-3) x x where: σ x = aerage longitudinal stress in the bottom lange due to ertical bending o the girder (ksi) σ = stress in the bottom lange due to ertical bending o the diaphragm oer the bearing sole plate (ksi) = shear stress in the bottom lange due to the sum o the diaphragm and torsional shears (ksi) F = speciied minimum ield stress o the bottom lange (ksi) For a box supported on two bearings (the case in this example), the bottom-lange bending stress due to ertical bending o the diaphragm oer the bearing sole plate (i.e. σ ) is tpicall relatiel small and will be neglected or simplicit in this example. In this case, σ x can simpl be checked instead against the critical stress gien b Eq. (10-), which results rom Eq. (C10-3) when σ is taken equal to zero. From preious calculations on pages 331 and 33, the total actored St. Venant torsional shear stress in the bottom lange is equal to 1.18 ksi. To estimate the shear stress in the bottom lange due to the diaphragm shear, assume a 1 in x 1 in top lange or the diaphragm. As permitted in Article 10.4..1, assume that 18 times the thickness o the bottom (box) lange (18 x 1.5 = 7 in) is eectie with the diaphragm. The diaphragm is assumed to be 78 inches deep and 1 in thick. From separate calculations, the moment o inertia o the eectie section is 11,375 in 4 and the neutral axis is located 31.05 in aboe the mid-thickness o the bottom lange. Subsequent calculations on page 339 indicate that the total actored ertical component o the diaphragm shear is 1,384 kips. The maximum shear stress in the eectie bottom lange due to the diaphragm shear is thereore approximated as: VQ 1,384(7/)(1.5)(31.05) = = = 5.16 ksi It 11, 375(1.5) tot = 1.18 + 5.16 = 6.34 ksi In certain situations, the Engineer ma wish to consider the eect o the access hole in the diaphragm on the section properties o the eectie diaphragm section. From separate but similar calculations to the aboe, considering the eect o an assumed 36-in deep access hole centered in the middle o the example diaphragm, the maximum shear stress in the eectie bottom lange due to the diaphragm shear is calculated to be 6.13 ksi. Thereore, in this particular case, the eect o the access hole on the bottom lange shear stress is not signiicant. The aerage actored longitudinal stress in the bottom lange due to ertical bending, 331) to be 46.47 ksi. Since σ is assumed to equal zero in this case, b Eq. (10-), as discussed preiousl. σ x, was computed earlier (page σ ma be checked against the critical stress gien x
HORIZONTALLY CURVED STEEL BOX-GIRDER DESIGN EXAMPLE 335 F = F Eq. (10-) cr where: = 1 3 F - Eq. (10-3) For this check, in Eq (10-3) is taken as the total shear stress in the lange. 6.34 = 1-3 = 0.976 50 F cr = 50(0.976) = 48.80 ksi -46.47 = 0.95 < 1.0 OK 48.80 Note that the direct use o Equation (C10-3) would eectiel ield the same answer or this case: σ σ σ +σ + 3 F Eq. (C10-3) x x ( 46.47) + 3(6.34) = 47.75 ksi 47.75 50 = 0.95 OK Another equall alid approach, which would inole some additional calculation, would be to instead check the combined principal stresses in the lange according to the ollowing orm o the Huber-on Mises-Henck ield criterion: σ σσ +σ F 1 1 Eq. (C10-1) where: σ,σ 1 = maximum and minimum principal stresses in the bottom lange (ksi) = x σx σ ± + σ +σ (Re: Ugural, A. C. and Fenster, S. K. (1975). Adanced Strength and Applied Elasticit, Elseier North Holland Publishing Co., Inc., New York, NY, pp. 105 107) For this case: 46.47 + 0 46.47 0 σ 1, = ± + ( 6.34) σ 1, = 47.3 ksi, 0.85 ksi ( ) ( )( ) ( ) 47.3 47.3 0.85 + 0.85 = 47.75 ksi 47.75 50 = 0.95 OK
336 AASHTO GUIDE SPECIFICATIONS FOR HORIZONTALLY CURVED STEEL GIRDER HIGHWAY BRIDGES For a box supported on a single bearing, the eect o bending in the plane o the diaphragm is likel to be more signiicant and should be considered. The eectie section speciied in Article 10.4..1 ma be used to compute σ due to ertical bending o the internal diaphragm oer the sole plate. In this case, either Equation (C10-3) or (C10-1) should then be used to make the check. Although not illustrated here, bend buckling o the web must also be checked at the Strength limit state according to the proisions o Section 6. The critical compressie lange stresses, computed aboe, should not exceed the critical compressie web stress (adjusted or the thickness o the lange).
HORIZONTALLY CURVED STEEL BOX-GIRDER DESIGN EXAMPLE 339 E.9 GIRDER STRESS CHECK SECTION 5-5 G NODE 36 DESIGN OF INTERNAL DIAPHRAGM Tr a 1-in thick A36 diaphragm plate. Compute the maximum actored ertical shear in the diaphragm. The dead load shears will be computed b conseratiel summing the ertical shears in the critical web o G acting on each side o the interior pier section. The shears include the St. Venant torsional shear. Load Shear (kips) Source Steel 46 + 47 = 93 3D Finite Element Analsis Deck 185 + 185 = 370 (in critical web) SupImp 96 + 10 = 198 Total DL 661 The lie load shear in the critical web at the interior pier is goerned b two lanes o lane loading plus impact. The lie load shear in the diaphragm will be computed b summing the ertical shears in the critical web acting on each side o the interior-pier section and subtracting two times the concentrated load portion o the lane loading plus impact (i.e., one concentrated load per lane). For HS5 loading, the concentrated load portion o the lane loading is equal to 6.0 x 1.5 = 3.5 kips. The impact actor is conseratiel taken equal to 1.4 (Table 3.5.6.). Thereore: Load Shear (kips) Lie Load 163 + 170 = 333 (3.5)(1.4) = 4 HS5 + Impact Determine the total actored ertical shear in the diaphragm at the Strength limit state. V = 1.3(93 + 370 + 198) + 1.3(5/3)(4) = 1,384 kips Compute the shear capacit according to AASHTO Eq. (10-113). Separate calculations indicate that C = 1.0. V = CV AASHTO Eq. (10-113) u p V p = 0.58FDt w = 0.58(36)(78)(1.0) = 1, 69 kips V u = 1.0(1,69) = 1,69 kips 1,384 = 0.85 < 1.0 OK 1,69 The internal diaphragm is subject to ertical bending oer the bearing sole plates in addition to shear. Thereore, Article 10... requires that the principal stresses in internal support diaphragms be computed and that the combined principal stresses not exceed the speciied minimum ield stress o the diaphragm. Compute the maximum actored shear stress in the diaphragm web. First, separate out the shears due to ertical bending (V b ) and due to St. Venant torsion (V T ). The sum o the total Steel plus Deck shears is equal to 93 + 370 = 463 kips. Reerring to the calculations on page 331, the shear low in the non-composite box is computed as: T 6 SF = = = 0.0193 kips/in A (56.0)(1) o V T = 0.0193(80.4) = 15.5 kips
340 AASHTO GUIDE SPECIFICATIONS FOR HORIZONTALLY CURVED STEEL GIRDER HIGHWAY BRIDGES The ertical component o V T is computed as: 78 (V T ) = 15.5 = 15.06 kips 80.4 V b = 463-15.06 = 447.9 kips The sum o the total Superimposed Dead Load and Lie Load plus Impact shears is equal to 198 + 5/3(4) = 601 kips. Reerring to the calculations on page 33, the shear low in the composite box is computed as: T -1,956 SF = = = 1.33 kips/in A (61.0)(1) o The ertical component o V T is computed as: V T = 1.33(80.4) = 106.9 kips 78 (V T) = 106.9 = 103.7 kips 80.4 V b = 601-103.7 = 497.3 kips The actored shear stress due to torsion in the diaphragm web is thereore equal to: ( ) = 1.3(0.0193/1.0 + 1.33/1.0) = 1.75 ksi T The aerage actored shear stress due to ertical bending in the diaphragm web is equal to: 1.3(447.9 + 497.3) ( ) b = = 15.75 ksi 78(1.0) Thereore, the total actored shear stress in the diaphragm web is equal to: ( ) ( ) = + = 1.75 + 15.75 = 17.50 ksi T b As speciied in Article C10..., the combined principal stresses in the diaphragm due to the actored loads can be checked using the general orm o the Huber-on Mises-Henck ield criterion as ollows: σ σσ +σ F 1 1 Eq. (C10-1) where: σ1, σ = critical maximum and minimum principal stresses in the diaphragm (ksi) = z σ σz ± + σ +σ Eq. (C10-) σ = stress in the diaphragm due to ertical bending o the diaphragm oer the bearing sole plate (ksi) σ z = stress in the diaphragm due to bending o the diaphragm about its longitudinal axis (ksi) = shear stress in the diaphragm (ksi) F = speciied minimum ield stress o the diaphragm (ksi) For a box supported on two bearings (the case in this example), the stress in the diaphragm due to ertical bending o the diaphragm oer the bearing sole plate (i.e., σ ) is tpicall relatiel small and will be neglected or simplicit in this example. σ z is also tpicall neglected. From Equation (C10-), i no bending is assumed, the two principal stresses are simpl equal tensile and compressie stresses with a magnitude equal to the shear stress as ollows:
HORIZONTALLY CURVED STEEL BOX-GIRDER DESIGN EXAMPLE 341 ( ) ( ) σ = 0 ± 0 + 17.50 =± 17.50 ksi 1, Checking the combined principal stresses according to Equation (C10-1): ( ) ( )( ) ( ) 17.50 17.50 17.50 + 17.50 = 30.31 ksi 30.31 = 0.84 OK 36 It should be noted that or this case with no bending assumed where σ = σ 1, the preceding calculation is equialent to simpl limiting the principal stress, σ1, in the diaphragm, which is equal to the shear stress in the diaphragm due to the actored loads, to the shear ield stress, τ = F/ 3, as ollows: F 36 τ = = = 0.78 ksi 3 3 17.50 = 0.84 OK 0.78 In this case, this check is essentiall equialent to the shear check illustrated on page 339. Howeer, the Engineer is asked to compute the principal stresses in the diaphragm because when the magnitude o the principal tensile stress due to the actored loads is deemed signiicant, the Engineer ma wish to also consider the magnitude o this stress under the actored atigue lie load plus impact. A large principal tensile stress under this loading condition ma preclude the use o certain atigue-sensitie details on the diaphragm. Thus, een or cases where no bending o the diaphragm is assumed, the Engineer ma still need to consider the critical principal stress range under the moing atigue lie load when highl atigue-sensitie details are present on the diaphragm. It should be noted that the direction o the principal tensile stress ma change or dierent positions o the lie load. For a box supported on a single bearing, the eect o the bending stress in the plane o the diaphragm is likel to be more signiicant and should be considered. As speciied in Article 10.4..1, a width o the bottom (box) lange equal to 18 times its thickness ma be considered eectie with the diaphragm in resisting bending. For cases where bending o the diaphragm is considered, more than one loading condition ma need to be inestigated in order to determine the critical principal stresses in the diaphragm. For the two-bearing arrangement selected in this example, the section through the access hole is not critical. Howeer, the designer should ensure that suicient section is proided around the access hole to carr the torsional shear low without reinorcement o the hole. For a box supported on a single bearing, the section through the access hole is critical and additional stiening and/or reinorcement around the hole ma be necessar.
34 AASHTO GUIDE SPECIFICATIONS FOR HORIZONTALLY CURVED STEEL GIRDER HIGHWAY BRIDGES E.10 GIRDER STRESS CHECK SECTION 5-5 G NODE 36 DESIGN OF BEARING STIFFENERS Compute the actored reactions. Figure E.10.1 Internal Diaphragm and Bearing Stieners at Pier o Girder Looking Upstation Reaction Location Load Let Right Source Steel 79 93 3D Finite Element Analsis Deck 38 370 (Not tabulated) SupImp 198 6 Total DL 515 k 489 k Lie Load HS5 + Impact 5 k 88 k 78 k 17 k uplit loads. Note that the reported lie load plus impact reactions at the let and right bearings are not caused b coincident 5 R let = 1.3 515 + (5 ) = 1,16 kips 3 5 R right = 1.3 489 + (88 ) = 1,60 kips 3 (controls) Ignore uplit. Assume that the bearings are ixed at the piers. Thus, there will be no expansion causing eccentric loading on the bearing stieners. Design the bearing stieners at this location according to the proisions o Article 6.7. Use bars with F = 50 ksi. Compute the maximum permissible width-to-thickness ratio o the stiener plates according to Eq. (6-13). where: bs E 9, 000 0.48 = 0.48 = 11.6 t F 50 s F = speciied minimum ield stress o the stieners Compute the eectie area o the diaphragm to which the stieners are attached (t w = 1.0 in) according to the proisions o Article 6.7. A = 18t = 18 x 1.0 = 18.0 in d w
HORIZONTALLY CURVED STEEL BOX-GIRDER DESIGN EXAMPLE 355 Eq. (A6) tacitl assumes that cross bracing is eectie in both compression and tension. I the bracing slenderness is large, the bracing is onl eectie in tension, and A b in Equation A6 should be one-hal the area o one brace. The stresses deried rom distortion o the box can be determined analogousl b soling the BEF problem. Moment in the BEF is analogous to normal distortional stress, σ, dw and delection in the BEF is analogous to distortional transerse bending stress, σ t. The reactions in the BEF are analogous to the orces in cross bracing, F b. Solutions or these three components are presented in graphical orm in Figures A through A10. [Ed. note: onl the pages containing Figures A4 through A9 are reproduced here. Figure A6 is used in this example.] These igures gie a C alue which is used in appropriate equations A8, A10, A11. These graphs show relationships or uniorm torque, m, or concentrated torque, T, at midpanel or diaphragms. The igures gie the appropriate C alues or a gien box geometr, β, loading, diaphragm stiness, q, and spacing, l. The designer is able to determine the distortion-related stresses, and estimate how diaphragm spacing and stiness ma be best modiied i necessar. Eq. (A8) gies transerse bending stresses at the top or bottom corners o the box section, depending on the determination o F d in Equations A9a or A9b. The critical stress ma be in either the web or lange. The AASHTO Speciication limits the range o the transerse bending stresses to 0,000 psi. [Ed. note: the preceding limit is speciied onl in AASHTO Article 10.39.3.. or straight boxes and not in the 1993 Guide Spec or the Guide Speciications.] Thereore, the torsion in both directions oten must be determined. The stress range is the sum o absolute alues o stresses due to opposite torques. 1 σ t = C t F d β ( m l or T) (A8) a where: m = uniorm torque per unit length T = concentrated torque F d = b or bottom corner o box S (A9a) F d = a b or top corner o box S a + b (A9b) where: S = section modulus o transerse member (see Figure A1c) Eq. (A10) gies the normal distortional warping stress at an point in the cross section. The alue o C w is obtained rom either Figure A, A3, or A9. Cw m ( or T ) σ dw = (A10) I β a where: = distance along the transerse ertical axis o the box rom the neutral axis to the point under consideration Eq. (A11) gies the axial (brace) orce due to distortional orces applied to the box. The alue o C b is obtained rom either Figure A5 or A10. F C a+ b 1+ h a 1 + b b = b ( m or T) a (A11) Since onl two loading positions or concentrated loads are considered in the charts, it is oten necessar to interpolate between igures. The principle o superposition applies or more than one torque. Figure A11 shows the eect o β on the inluence line or diaphragm orces when the diaphragm is rigid.
356 AASHTO GUIDE SPECIFICATIONS FOR HORIZONTALLY CURVED STEEL GIRDER HIGHWAY BRIDGES Figure A. Uniorm Torque on Continuous Beam Normal Distortional Warping Stress at Midpanel Figure A3. Uniorm Torque on Continuous Beam Normal Distortional Warping Stress at Diaphragm Figure A4. Uniorm Torque on Continuous Beam Distortional Transerse Bending Stress at Midpanel
HORIZONTALLY CURVED STEEL BOX-GIRDER DESIGN EXAMPLE 357 Figure A5. Uniorm Torque on Continuous Beam Diaphragm Force Figure A6. Concentrated Torque at Midpanel on Continuous Beam Distortional Transerse Bending Stress at Load Figure A7. Concentrated Torque at Midpanel on Continuous Beam Distortional Transerse Bending Stress at Diaphragm
357.1 AASHTO GUIDE SPECIFICATIONS FOR HORIZONTALLY CURVED STEEL GIRDER HIGHWAY BRIDGES Figure A8. Concentrated Torque at Diaphragm on Continuous Beam Distortional Transerse Bending Stress at Diaphragm Figure A9. Concentrated Torque at Midpanel on Continuous Beam Normal Distortional Warping Stress at Midpanel Figure A10. Load at Diaphragm on Continuous Beam Diaphragm Force
HORIZONTALLY CURVED STEEL BOX-GIRDER DESIGN EXAMPLE 357. Figure A11. Inluence Line or Rigid Diaphragm With Concentrated Torque
358 AASHTO GUIDE SPECIFICATIONS FOR HORIZONTALLY CURVED STEEL GIRDER HIGHWAY BRIDGES E.13 GIRDER STRESS CHECK SECTION 5-5 G NODE 36 COMPOSITE BOTTOM FLANGE OPTION Assume that the bottom box lange in the negatie moment regions will be redesigned using an unstiened plate with 8 inches o composite concrete according to the proisions o Article 10.4.3. The girder moment due to the added weight o the lange concrete is considered. It is assumed that the concrete is placed ater the steel has been erected and beore the deck is cast. Thus, the bottom lange will be composite or the deck load. The superimposed dead load is added ater the deck and the bottom lange concrete hae hardened. Lie load plus impact will be applied to the composite section. Onl the rebars in the deck will be considered to act compositel with the steel section (with the composite box lange) at the Strength limit state. Table E.13.1 Section Properties o G with 8 inches o 6,000 psi Concrete in Bottom Flange Top Flg 18 x 3 Bot Flg 81 x 1.5 (Node 36) Top Flg 18 x 1.5 Bot Flg 81 x 0.75 (Node 3) Section I (in 4 ) NA (in) I (in 4 ) NA (in) 1. Noncomposite 396,155 41.99 38,466 39.59. Composite lange 3n w/o deck rebars 438,765 38.6 75,140 34.69 3. Composite lange n w/o deck rebars 50,477 3.68 34,44 8.11 4. Composite lange 3n w/deck rebars 454,801 39.0 93,094 36.06 5. Composite lange n w/deck rebars 560,391 35. 390,313 31.61 NA is distance rom bottom o section to the neutral axis. Eectie deck rebar area or 3n equals 0.0/3 = 6.67 in 1. Noncomposite. Steel with bottom lange concrete at 3n without deck rebars 3. Steel with bottom lange concrete at n without deck rebars 4. Steel with bottom lange concrete at 3n and deck rebars at 3n 5. Steel with bottom lange concrete at n and deck rebars at n Properties o bottom lange concrete: c = 6,000 psi ; E c = 4,696 ksi ; n = 6. The bottom lange concrete is assumed to be continuous through the interior support diaphragm as required in Article 10... Bottom transerse bracing members (i.e. the bottom struts o the interior cross rames) are assumed to be located aboe the concrete in this region. Tr a 1.5-inch thick unstiened bottom lange plate. Check the transerse bending stress in the bottom lange plate due to the sel-weight o the plate and the wet concrete. Compute the section modulus o a one-oot wide section o the bottom lange plate. 1 1 3 S = bt = x 1 x 1.5 = 3.13 in /t 6 6 Compute the moment applied to the lange plate due to the weight o the steel and lange concrete. Assume a simple span between webs. 1.5 x 1 in x 3.4 Steel weight per oot o width, w= s = 4.3 pounds/in/t 1
HORIZONTALLY CURVED STEEL BOX-GIRDER DESIGN EXAMPLE 359 M 1 1 steel = w s = x 0.0043 x 81 = 3.53 k-in/t 8 8 8 1 Concrete weight per oot o width, w c = 1 x 150 x = 8.33 pounds/in/t 1 1 1 1 8 8 M conc = w c = x 0.00833 x 81 = 6.83 k-in/t Compute the maximum transerse bending stress in the lange plate at the Constructibilit limit state. Load actor = 1.4 (Article 3.3). M (3.53 + 6.83) tran = = x 1.4 = 4.63 ksi < 0 ksi OK S 3.13 Although not checked here, Article 10.4..1 also limits the maximum ertical delection o the box lange due to sel-weight and the applied permanent loads to 1/360 times the transerse span between webs. The concrete is to be placed on the bottom lange o the ield section oer each pier. The bottom lange concrete causes a longitudinal girder moment o 880 k-t at Section 5-5 in G rom the inite element analsis. The longitudinal girder moment due to steel weight = 3,154 k-t rom Table D.1. Compute the total moment applied to the steel section. M = 3,154 + ( 880) = 4,034 k-t Check the ertical bending stress in the unstiened plate at the Constructibilit limit state. Compute the ertical bending stress in the bottom lange due to steel weight and concrete. Section properties are rom Table E.13.1. As speciied in Article 10.4.3.1, or loads applied prior to hardening o the concrete, composite box langes are to be designed as non-composite box langes according to the proisions o Article 10.4.. -4,034 x 41.99 x 1 x 1.4 bot lg = = -7.18 ksi 396,155 The critical buckling stress or the non-composite bottom lange with no stiening according to Article 10.4..4.1 equals 4.9 ksi (calculations not shown). -7.18 = 0.9 < 1.00 OK 4.9 state. Compute the actored bottom lange ertical bending stress in the non-composite section at the Strength limit 1.3 b = -7.18 x = -6.67 ksi 1.4 Compute the actored bottom lange ertical bending stress in the composite section due to the deck weight. M deck = 1,7 k-t (Table D.1) As speciied in Article 10.4.3.3, concrete creep is to be considered when checking the compressie steel stress. Concrete compressie stress are to be checked without considering creep. Thereore, use 3n section properties to check the steel stress and n section properties to check the concrete stress. Compute the actored ertical bending stress in the extreme iber o the composite bottom lange due to the deck weight. Use creep properties (3n) rom Table E.13.1. -1, 7 x 38.6 bot = x 1 x 1.3 = -16.69 ksi 438, 765
360 AASHTO GUIDE SPECIFICATIONS FOR HORIZONTALLY CURVED STEEL GIRDER HIGHWAY BRIDGES Compute the actored ertical bending stress in the extreme iber o the composite bottom lange concrete due to the deck weight. Use no creep properties (n) rom Table E.13.1. -1, 7 x (3.68-1.5) 1 bot = x 1 x 1.3 x = -1.93 ksi 50, 477 6. Compute the actored ertical bending stress in the composite bottom lange due to the superimposed dead load ater the deck has hardened. Superimposed dead load moment = 4,473 k-t rom Table D.1. Compute the actored ertical bending stress in the extreme iber o the composite bottom lange due to the superimposed dead load. Use the appropriate creep properties (3n) rom Table E.13.1. -4,473 x 39.0 bot = x 1 x 1.3 = -6.01 ksi 454,801 Compute the actored ertical bending stress in the extreme iber o the composite bottom lange concrete due to the superimposed dead load. Use the appropriate no creep properties (n) rom Table E.13.1. -4, 473 x (35.-1.5) 1 bot = x 1 x 1.3 x = -0.68 ksi 560, 391 6. Compute the actored ertical bending stress in the composite bottom lange due to the lie load. Use the appropriate n section properties rom Table E.13.1. Moment due to lie load = 8,566 k-t rom Table D.1. -8, 566(5/3) x 35. bot = x 1 x 1.3 = -14.00 ksi 560, 391 Compute the actored ertical bending stress in the composite bottom lange concrete due to the lie load. Use the appropriate n section properties rom Table E.13.1. -8,566(5/3) x (35.-1.5) x 1 x 1.3 1 bot = x = -.18 ksi 560, 391 6. Check the total actored ertical bending stress in the steel bottom lange at the Strength limit state. tot = -6.67 + (-16.69) + (-6.01) + (-14.00) = -43.37 ksi As speciied in Article 10.4.3.3, the critical stress or the steel bottom lange in compression at the Strength limit state is gien b Eq. (10-4) as ollows: F = F Eq. (10-4) cr where: = 1 3 F - Eq. (10-3) From Table D.3, the torque due to the steel weight is k-t (the torque due to the wet bottom concrete is neglected since the load is smmetrical and the curature eect is relatiel small). Using calculations similar to those shown on page 331, the enclosed area o the non-composite box is computed to be A = 55.9 t. o T - = = x 1.3 = 0.017 ksi A t (55.9)(1.5)1 o
HORIZONTALLY CURVED STEEL BOX-GIRDER DESIGN EXAMPLE 361 Subsequent calculations (page 366) show that the actored torque on the composite section at the Strength limit state is equal to,481 k-t. To account or the possibilit o concrete creep, it is conseratiel assumed that all o this torque is resisted b the steel lange in this computation, as required b Article 10.4.3.4 at the Strength limit state. The enclosed area o the composite box is computed to be A = 61.0 t. o T -,481 = = = 1.36 ksi A t (61.0)(1.5)(1) o ( ) = 0.017 + 1.36 = 1.38 ksi tot The total shear stress satisies Eq (10-1) in Article 10.4.. b inspection. 1.38 = 1-3 = 0.999 50 F cr = 50(0.999) = 49.95 ksi -43.37 = 0.87 < 1.00 OK 49.95 Compute the actored ertical bending stress in the bottom o the concrete at the Strength limit state assuming no creep. F cr conc = 0.85 x 6 = 5.10 ksi (Article 10.4.3.3) tot = -1.93 + (-0.68) + (-.18) = -4.79 ksi -4.79 = 0.94 < 1.00 OK 5.10
36 AASHTO GUIDE SPECIFICATIONS FOR HORIZONTALLY CURVED STEEL GIRDER HIGHWAY BRIDGES E.14 GIRDER STRESS CHECK SECTION 4-4 G NODE 3 COMPOSITE BOTTOM FLANGE OPTION Tr a 0.75-inch thick lange plate at this section with 8 inches o concrete. Check the transerse bending stress in the bottom lange plate due to the sel-weight o the plate and the wet concrete. Compute the section modulus o a one-oot wide section o the bottom lange plate. 1 1 3 S = bt = x 1 x 0.75 = 1.13 in /t 6 6 Compute the moment applied to the lange plate due to the weight o the steel and lange concrete. Assume a simple span between webs. 0.75 x 1 in x 3.4 Steel weight per oot o width, w s = =.6 pounds/in/t 1 1 1 8 8 M steel = w s = x 0.006 x 81 =.13 k-in/t Compute the maximum transerse bending stress in the lange plate at the Constructibilit limit state. Load actor = 1.4. M conc = 6.83 k-in/t due to the weight o the concrete rom page 359. M (.13 + 6.83) tran = = x 1.4 = 11.10 ksi < 0 ksi OK 3 S 1.13 in The longitudinal girder moment due to the additional concrete in the bottom lange = 358 k-t rom the inite element analsis. The longitudinal girder moment due to the steel weight = 1,896 k-t (Table D.1). M = 1,896 + ( 358) =,54 k-t Check the ertical bending stress in the bottom lange due to the total moment applied to the non-composite section at the Constructibilit limit state. Section properties are rom Table E.13.1. -, 54 x 39.59 x 1 x 1.4 bot lg = = -6.9 ksi 38, 466 Determine the critical stress or the unstiened box lange according to Article 10.4..4.1. F cr = 8.8 ksi (calculations not shown) -6.9 = 0.71 < 1.00 OK 8.8 Adjust the stress or the Strength limit state load actor. 1.3-6.9 x = -5.84 ksi 1.4 Compute the actored ertical bending stress in the composite bottom lange due to the deck weight. Use 3n section properties rom Table E.13.1. Moment due to deck weight = 7,599 k-t rom Table D.1. -7,599 x 34.69 bot = x 1 x 1.3 = -14.95 ksi 75,140
HORIZONTALLY CURVED STEEL BOX-GIRDER DESIGN EXAMPLE 363 Compute the actored ertical bending stress in the extreme iber o the composite bottom lange concrete due to the deck weight. Use n section properties rom Table E.13.1. -7, 599 x (8.11-0.75) 1 bot = x 1 x 1.3 x = -1.61 ksi 34, 44 6. Compute the actored ertical bending stress in the composite bottom lange due to the superimposed dead load. Use the appropriate 3n section properties rom Table E.13.1. Superimposed dead load moment =,610 k-t rom Table D.1. -,610 x 36.06 bot = x 1 x 1.3 = -5.01 ksi 93, 094 Compute the actored ertical bending stress in the composite bottom lange concrete due to superimposed dead load. Use the appropriate n section properties rom Table E.13.1. -, 610 x (31.61-0.75) 1 bot = x 1 x 1.3 x = -0.5 ksi 390, 313 6. Compute the actored ertical bending stress in the composite bottom lange due to the lie load. Use the appropriate n section properties rom Table E.13.1. Moment due to lie load = 5,61 k-t rom Table D.1. -5, 61(5/3) x 31.61 bot = x 1 x 1.3 = -11.8 ksi 390, 313 Compute the actored ertical bending stress in the composite bottom lange concrete due to lie load. Use the appropriate n section properties rom Table E.13.1. -5, 61(5/3) x (31.61-0.75) 1 bot = x 1 x 1.3 x = -1.86 ksi 390, 313 6. Compute the total actored ertical bending stress in the steel bottom lange at the Strength limit state. tot = -5.84 + (-14.95) + (-5.01) + (-11.8) = -37.6 ksi The torque due to the steel weight is -10 k-t (Table D.3). The enclosed area o the non-composite box is computed to be A o = 55. t. T -10 = = x 1.3 = 0.013 ksi A t (55.)(0.75)(1) o The actored torque on the composite section at the Strength limit state is computed to be Loading Torque (Table D.3) Deck 63 x 1.3 8 k-t Superimposed DL 73 x 1.3 355 k-t Lie Load HS5 + Impact 688(5/3) x 1.3 1,491 k-t 1,764 k-t The enclosed area o the composite box is computed to be A o = 60.8 t. T -1,764 = = = 1.61 ksi A t (60.8)(0.75)(1) o ( ) = 0.013 + 1.61 = 1.6 ksi tot
364 AASHTO GUIDE SPECIFICATIONS FOR HORIZONTALLY CURVED STEEL GIRDER HIGHWAY BRIDGES which satisies Eq. (10-1) in Article 10.4.. b inspection. 1.6 = 1-3 = 0.998 50 F cr = 50(0.998) = 49.90 ksi -37.6 = 0.75 < 1.00 OK 49.90 Compute the total actored ertical bending stress in the bottom lange concrete at the Strength limit state. tot = -1.61 + (-0.5) + (-1.86) = -3.99 ksi F cr conc = 5.10 ksi (rom page 361) -3.99 = 0.78 < 1.00 OK 5.10
HORIZONTALLY CURVED STEEL BOX-GIRDER DESIGN EXAMPLE 365 E.15 GIRDER STRESS CHECK SECTION 5-5 G NODE 36 COMPOSITE BOTTOM FLANGE OPTION DESIGN OF SHEAR CONNECTORS STRENGTH Check the ultimate strength o the shear connectors on the composite bottom lange according to the proisions o Article 10.4.3.5, which reer back to the proisions o Article 7..1. As speciied in Article 10.4.3.5, the radial orce due to curature is ignored. The longitudinal orce to be deeloped is gien b Eq. (7-4), with b d taken as the ull width o the bottom lange concrete. Compute the capacit o one shear connector. Required capacit = 0.85 cbdtd Eq. (7-4) P = 0.85 x 6 ksi x 81 in x 8 in = 3,305 kips H 6 = = 6.86 > 4.0 d 0.875 S = 0.4d E 60,000A AASHTO Eq. (10-67) u c c sc (Note: the upper limit o 60,000A sc in the aboe equation is incorporated in the 000 Interims to the Standard Speciications and is included here.) A = π(0.875 ) /4 = 0.6 in sc S u = 0.4 x 0.875 6 x 4,696 = 51.4 kips > 60(0.6) = 36 kips S u = 36 kips Compute the minimum number o shear connectors required on each side o the pier. No. o shear connectors req d. = P 3,305 = = 108.0 φ S 0.85(36) sc u Tr six studs uniorml spaced across the lange (Figure E.15.1) with 18 rows on each side o the pier (or 108 shear connectors per lange on each side o the pier). The studs must be spaced transersel so that the steel plate slenderness limit o R 1 in Eq (10-4) is satisied, where b is taken as the transerse spacing between the shear connectors (Article 10.4.3.5). Check the computed orce on the critical studs at Node 36. Compute the axial orce in the bottom lange concrete due to the ertical moment. Compute the stresses in the top o the composite bottom lange concrete due to the deck weight without creep. Use the ratio o the distances to the neutral axis. 3.68-9.5 top = -1.93 = -1.44 ksi 3.68-1.5 Compute the stress in the top o the composite bottom lange concrete due to superimposed dead load and lie load without creep. 35. -9.5 top = [-0.68 + (-.18)] = -.19 ksi 35. -1.5
366 AASHTO GUIDE SPECIFICATIONS FOR HORIZONTALLY CURVED STEEL GIRDER HIGHWAY BRIDGES Figure E.15.1 Shear Studs on Composite Bottom Flange Compute the longitudinal orce in the composite bottom lange concrete due to ertical bending. Use the aerage bending stress in the concrete times the area o the concrete. Compute the longitudinal orce per stud. 1.93 + 1.44.86 +.19 F = + x 81 x 8 =,78 kips F L =,78/108 = 5.6 kips/stud Compute the St. Venant torsional shear in the concrete at the Strength limit state. Assume that a single row o studs across the lange will resist the torsional shear in the lange concrete. Loading Torque (Table D.3) Deck 48 x 1.3 6 k-t Superimposed DL 346 x 1.3 450 k-t Lie load HS5 + Impact 966(5/3) x 1.3,093 k-t Total,481 k-t Assume all torsion is applied to the uncracked section without creep. Eectie concrete thickness = Thickness/n = 8 in /6. = 1.9 in. Using calculations similar to those shown on page 33, the enclosed area o the composite box is computed to be A o = 61.0 t. T V = A o b -, 481 81 V = x 61.0 1 = 137 kips Compute the portion o the torsional shear resisted b the concrete b taking the ratio o the eectie concrete thickness to the total thickness o the steel lange plus the eectie concrete. 1.9 in V conc = 137 kips x = 70 kips (1.5 in + 1.9 in)