AASHTO GHC-2003 AASHTO Guide Specification for Horizontally Curved Steel Girder Highway Bridges (Revision 4 Errata 10 2003)《平面曲线钢箱梁公路桥的AASHTO指南规范.修改件4 勘误表 10 2003[替代 AASHTO LRFD I-.pdf
《AASHTO GHC-2003 AASHTO Guide Specification for Horizontally Curved Steel Girder Highway Bridges (Revision 4 Errata 10 2003)《平面曲线钢箱梁公路桥的AASHTO指南规范.修改件4 勘误表 10 2003[替代 AASHTO LRFD I-.pdf》由会员分享,可在线阅读,更多相关《AASHTO GHC-2003 AASHTO Guide Specification for Horizontally Curved Steel Girder Highway Bridges (Revision 4 Errata 10 2003)《平面曲线钢箱梁公路桥的AASHTO指南规范.修改件4 勘误表 10 2003[替代 AASHTO LRFD I-.pdf(436页珍藏版)》请在麦多课文档分享上搜索。
1、 John R. Njord, President Executive Director Utah Department of Transportation American Association of State Highway and John Horsley Transportation Officials Executive Director Executive Office: 444 N. Capitol St., N.W., Suite 249, Washington, D.C. 20001 Telephone (202) 624-5800 Telefax (202) 624-5
2、806 Telex 49000095800 HTO ERRATA Dear Customer: Due to errors found after the publication had been completed, AASHTO has reprinted the pages and added two point pages listed below in order to make the following errata changes to the AASHTO Guide Specifications for Horizontally Curved Steel Girder Hi
3、ghway Bridges with Design Examples for I-Girder and Box-Girder Bridges: Page No(s). Affected Section Errata Change Horizontally Curved Steel Box Girder Example pp. 333336 Appendix E.6 Add the following paragraph immediately before the last paragraph on p. 34: In certain situations, the Engineer may
4、wish to consider the effect of the access hole in the diaphragm on the section properties of the effective diaphragm section. From separate but similar calculations to the above, considering the effect of an assumed 36-in deep access hole centered in the middle of the example diaphragm, the maximum
5、shear stress in the effective bottom flange due to the diaphragm shear is calculated to be 6.13 ksi. Therefore, in this particular case, the effect of the access hole on the bottom flange shear stress is not significant. pp. 339340 Appendix E.9 Change “H525” to “HS25” on p. 339 pp. 341342 Appendix E
6、.10 Change “H525” to “HS25” on p. 342 pp. 355356 Appendix E.12 Add the following to the end of the text on p. 355: Eq. (A10) gives the normal distortional warping stress at any point in the cross section. The value of Cwis obtained from either Figure A2, A3, or A9. ()wdwCymor TIa=A (A10) where: y =
7、distance along the transverse vertical axis of the box from the neutral axis to the point under consideration Eq. (A11) gives the axial (brace) force due to distortional forces applied to the box. The value of Cbis obtained from either Figure A5 or A10. 2 Page No(s). Affected Section Errata Change H
8、orizontally Curved Steel Box Girder Example ()2bbab1morT2hFCa a21b+=+A(A11) Since only two loading positions for concentrated loads are considered in the charts, it is often necessary to interpolate between figures. The principle of superposition applies for more than one torque. Figure A11 shows th
9、e effect of on the influence line for diaphragm forces when the diaphragm is rigid.” pp. 355356 Appendix E.12 Add Figures A2 and A3 before Figure A4 on p. 356 pp. 359360 Appendix E.13 In the equation for “ftran” on p. 359, which is the third equation from the top of the page, “50 ksi” should be “20
10、ksi.” pp. 361362 Appendix E.14 In the middle of p. 362, in the equation for “ftran,” “50 ksi” should be “20 ksi.” pp. 363364 Appendix E.14 Change “H525” to “HS25” on p. 363 pp. 365366 Appendix E.15 Change “H525” to “HS25” on p. 366 The following new pages have been added to accommodate omitted figur
11、es: pp. 357357.1 Appendix E.12 Move down Figures A5 through A9 Add Figure A10 on p. 357.1 pp. 357.2358 Appendix E.12 Add Figure A11 on p. 357.2 Please substitute the original pages of text with the enclosed pages. We apologize for any inconvenience this may have caused. AASHTO Publications Staff Oct
12、ober 2003 HORIZONTALLY CURVED STEEL BOX-GIRDER DESIGN EXAMPLE 333R1shall be taken as:1222vys97 kR = f1k + + 4 2Fk Eq. (10-5)where: k = plate buckling coefficientks = shear buckling coefficientTry: k = 4.0 and ks= 5.34Since the denominator of Eq (10-5) is approximately equal to 1.0 in this case, 1R =
13、 97 k = 97 x 4.0 = 194.Since 191 4.0d 0.8752ucc scS = 0.4d f E 60,000A AASHTO Eq. (10-67)(Note: the upper limit of 60,000Ascin the above equation is incorporated in the 2000 Interims to theStandard Specifications and is included here.)2 2scA = (0.875 /4 = 0.6 in)2uS = 0.4 x 0.875 6 x 4,696 = 51.4 ki
14、ps 60(0.6) = 36 kipsuS = 36 kipsCompute the minimum number of shear connectors required on each side of the pier.No. of shear connectors reqd. = sc uP3,305= = 108.0S 0.85(36)Try six studs uniformly spaced across the flange (Figure E.15.1) with 18 rows on each side of the pier (for108 shear connector
15、s per flange on each side of the pier). The studs must be spaced transversely so that the steel plateslenderness limit of R1in Eq (10-4) is satisfied, where bfis taken as the transverse spacing between the shearconnectors (Article 10.4.3.5).Check the computed force on the critical studs at Node 36.C
16、ompute the axial force in the bottom flange concrete due to the vertical moment. Compute the stresses inthe top of the composite bottom flange concrete due to the deck weight without creep. Use the ratio of the distances tothe neutral axis.top32.68 9.25f = 1.93 = 1.44 ksi32.68 1.25-Compute the stres
17、s in the top of the composite bottom flange concrete due to superimposed dead load andlive load without creep.top35.22 9.25f = 0.68 + ( 2.18) = 2.19 ksi35.22 1.25- -366 AASHTO GUIDE SPECIFICATIONS FOR HORIZONTALLY CURVED STEEL GIRDER HIGHWAY BRIDGESFigure E.15.1 Shear Studs on Composite Bottom Flang
18、eCompute the longitudinal force in the composite bottom flange concrete due to vertical bending. Use theaverage bending stress in the concrete times the area of the concrete.1.93 + 1.44 2.86 + 2.19F = + x 81 x 8 = 2,728 kips22Compute the longitudinal force per stud.FL = 2,728/108 = 25.26 kips/studCo
19、mpute the St. Venant torsional shear in the concrete at the Strength limit state. Assume that a single row ofstuds across the flange will resist the torsional shear in the flange concrete.Loading Torque (Table D.3)Deck 48 x 1.3 62 k-ftSuperimposed DL 346 x 1.3 450 k-ftLive load HS25 + Impact 966(5/3
20、) x 1.3 2,093 k-ftTotal 2,481 k-ftAssume all torsion is applied to the uncracked section without creep.Effective concrete thickness = Thickness/n = 8 in /6.2 = 1.29 in.Using calculations similar to those shown on page 332, the enclosed area of the composite box is computedto be Ao= 61.0 ft2.foTV = b
21、2A2, 48181V = = 137 kips2 x 61.0 12-Compute the portion of the torsional shear resisted by the concrete by taking the ratio of the effectiveconcrete thickness to the total thickness of the steel flange plus the effective concrete.conc1.29 inV = 137 kips x = 70 kips(1.25 in + 1.29 in)AASHTO Guide Spe
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