H-1 CHAPTER H DESIGN OF MEMBERS FOR COMBINED FORCES AND TORSION For all interaction equations in Specification Section H, the required forces and moments must include the results of a second-order analysis, as required by Section C of the Specification.This represents a significant change for ASDusers,whoareaccustomedtousinganinteractionequationthatincludesapartialsecond-order amplification. H-2 Example H.1a W-shape Subjected to Combined Compression and Bending About Both Axes (braced frame). Given: Verify if an ASTM A992 W1499 has sufficient available strength to support the axial forces and moments listed below, obtained from a second order analysis that includes P- effects.The unbraced length is 14 ft and the member has pinned ends. KLx =KLy =Lb =14.0 ft LRFDASD Pu =400 kips Mux =250 kip-ft Muy =80.0 kip-ft Pa =267 kips Max =167 kip-ft May =53.3 kip-ft Solution: Material Properties: ASTM A992Fy =50 ksiFu =65 ksi Manual Table 2-3 Try a W1499 Take combined strength parameters from Manual Table 6-1 LRFDASD p= 30.886 10kips at 14.0 ft bx =31.38 10kip-ftat 14.0 ft by = 32.85 10kip-ft

p= 31.3310kipsat 14.0 ft bx =32.0810kip-ftat 14.0 ft by =34.2910kip-ft ManualTable 6-1 Check limit for Equation H1-1aCheck limit for Equation H1-1a uc nPP =400 kips1,130 kips =0.354 Since uc nPP >0.2,pPu +bxMux +byMuy M 1.0 ( )( )( ) = + + 3330.866400kips10 kips1.38250kip-ft10 kip-ft2.8580.0kip-ft10 kip-ft =0.346 +0.345 +0.228 =0.927 M 1.0o.k. /an cPP =267 kips751 kips =0.356 Since /an cPP >0.2,pPa +bxMax +byMay M 1.0 ( )( )( ) = + + 3331.33267kips10 kips2.08167kip-ft10 kip-ft4.2953.3kip-ft10 kip-ft =0.356 +0.347 +0.229 =0.931 M 1.0o.k. ManualTable 4-1 Section H1.1 Manual Table 6-1 simplifies the calculation of Specification Equations H1-1a and H1-1b.A direct application of these equations is shown in Example H.2. H-3 Example H.1bW-shape Column Subjected to Combined Compression and Bending Moment About Both Axes (braced frame) Verify if an ASTM A992 W1499 has sufficient available strength to support the axial forces and moments listed below, obtained from a second order analysis that includes second-order effects.The unbraced length is 14 ft and the member has pinned ends. KLx =KLy =Lb =14.0 ft LRFDASD Pu =400 kips Mux =250 kip-ft Muy =80.0 kip-ft Pa =267 kips Max =167 kip-ft May =53.3 kip-ft Solution: Material Properties: ASTM A992Fy =50 ksiFu =65 ksi Manual Table 2-3 Take the available axial and flexural strengths from the Manual Tables LRFDASD at KLy =14.0 ft,Pc =cPn =1130 kips at Lb =14.0 ft,Mcx =Mnx =642 kip-ft Mcy =Mny =311 kip-ft uc nPP =400 kips1,130 kips =0.354 Since uc nPP >0.2, use Eqn. H1.1a 8 ++9ryrx rc cx cyMM PP M M M 1.0 400 kips 8 250 kip-ft 80.0 kip-ft+ +1130 kips 9 642 kip-ft 311 kip-ft = ( )80.354 + 0.389 +0.2579 =0.929 < 1.0 o.k. at KLy =14.0 ft,Pc =Pn/c=751 kips at Lb =14.0 ft,Mcx =Mnx / =428 kip-ft Mcy =Mny / =207 kip-ft /an cPP =267 kips751 kips =0.356 Since /an cPP >0.2, use Eqn. H1.1a 8 ++9ryrx rc cx cyMM PP M M M 1.0 267 kips 8 167 kip-ft 53.3 kip-ft+ +751 kips 9 428 kip-ft 207 kip-ft = ( )80.356 + 0.390 +0.2579 =0.931 < 1.0 o.k. ManualTable 4-1 Manual Table 3-10 Manual Table 3-2 Eq. H1.1a H-4 Example H.2W-Shape Column Subjected to Combined Compression and Bending Moment About Both Axes (by Specification Section H2) Given: VerifyifanASTMA992W1499showninExampleH.1hassufficientavailablestrength,using Specification Section H2.1.This example is included primarily to illustrate the use of Specification Section H2.KLx =KLy =Lb =14.0 ft LRFDASD Pu =400 kips Mux =250 kip-ft Muy =80.0 kip-ft Pa =267 kips Max =167 kip-ft May =53.3 kip-ft Solution: Material Properties: ASTM A992Fy =50 ksiFu =65 ksi Geometric Properties: W1499A =29.1 in.2 Sx =157 in.3Sy =55.2 in.3 Manual Table 2-3 Manual Table 1-1 Calculate the required flexural and axial stresses LRFDASD 2400kips13.7ksi29.1in.uaPfA= = =3250kip-ft 12in.19.1ksift 157in.uxbwxMfS = = = 380.0kip-ft 12in.17.4ksift 55.2in.uybzyMfS = = = 2267kips9.18ksi29.1in.aaPfA= = =3167kip-ft 12in.12.8ksift 157in.axbwxMfS = = = 353.3kip-ft 12in.11.6ksift 55.2in.aybzyMfS = = = Calculate the available flexural and axial stresses from the available strengths in Example H.1b LRFDASD 21,130kips38.8ksi29.1in.c na c crPF FA= = = =3645kip-ft 12in.49.3ksift 157in.b nxbwxMFS = = = 3311kip-ft 12in.67.6ksift 55.2in.b nybzyMFS = = = 2751kips25.8ksi29.1in.cr nac cF PFA= = = = 3431kip-ft 12in.32.9ksift 157in.nxbwb xMFS = = = 3207kip-ft 12in.45.0ksift 55.2in.nybzb yMFS = = = As shown in the LRFD calculation of Fbz above, the available flexural stresses can exceed the yield stress in cases where the available strength is governed by yielding and the yielding strength is calculated using the plastic section modulus. H-5 Calculate the combined stress ratio LRFDASD ++a bw bza bw bzf f fF F F M 1.0 13.7 ksi 19.1 ksi 17.4 ksi+ +38.8 ksi 49.3 ksi 67.6 ksi=0.998