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Page 3 of 4 <br /> Comment#10: Along Grid F between Grid 4 and Grid 10, there appears to be a discontinuous shear wall <br /> - r►1iest. " condition between levels 1 and 2,please provide calculations indicating how overturning <br /> vv./tm.4.., <br /> forces are resolved due to the discontinuity. Please verb the load path for resolving the <br /> i - t overturning moment developed at the level 2 shear wall from the story shear imposed at level <br /> 3. If columns located at the shear wall discontinuity are intended to resolve the force couples, <br /> , ij <br /> please apply the overstrength, (1,factor for the design of these columns and beam elements(if <br /> applicable)per ASCE 7-10 Sec 12.3.3.3. Please provide diaphragm strength checks for <br /> transfer of shear out of the discontinuous shear wall and through the walls located below <br /> gr level 1. <br /> Response: The RAM Structural System and RAM Concrete structural analysis model, tat, `�b <br /> gif ' and current typical wall reinforcing detailing shown on sheet S4.03,takes into account `y ^`�^' <br /> the fact that we are utilizing a continuous concrete wall with the openings designed as <br /> perforations within a continuous wall. Therefore,the wall from level 2 to level 3 on _ 'f'�r`d'r— <br /> gridline F is a continuous wall from gridline 1 to 12 with any overturning boundary P r-J-..d+-° <br /> 4.-4 requirements at the wall ends.The walls from Level l to 2 along gridline F are broken Cn di JiJw( <br /> T { into 2 continuous perforated wall segments.The first extends from gridline lto the P <4 ...-Es <br /> column located North of gridline 4.The second starts approximately 6'-0"North of <br /> F t C.. nu,1L.ed gridline 10 and extends to the end of wall at gridline 12.The wall from level 1 to the <br /> \''/At nf•'' foundation is a solid continuous wall from gridline 1 to gridline 12. MI end wall d <br /> L . nesarly.tttj reinforcing is continuous from the level it starts down to the foundation. <br /> -rr-4'1z. -two Gcs a i-k.., <br /> In response to your question regarding applying an overstrength ffi,factor for the design <br /> of the columns,beam and slab elements(if applicable)per ASCE 7-10 Sec 12.3.3.3, it is <br /> b1D Lieu M.ocet_ not necessary to design the beams and columns for an overstrength sa,factor since the <br /> -1-µ re 'F.�'TxriB.s 7 In-plane discontinuity in vertical lateral force-resisting elements does not place a col rt.3Lec ' <br /> -In cr.,b t <br /> age, p W,.7 'OU'OUT', demand on the supporting beams or column from overturning.We did however,desigq� <br /> a collector in the 2vd floor slab along -0w� ��r' <br /> -'�'E'SFt� 4.OA174 gridline F to collect the loads due to the wall yew to <br /> ae <br /> i C S 1--mrve discontinuity that starts at the column located north of gridline 4 to the edge of door <br /> opening located approximately 6'-0"North of gridline 10. <br /> k no c...0l\.ec.� <br /> bets 3047 collector bars,15 top and 15 bottom,have been added to the 2°d Floor Slab Top <br /> ,P ra�tdc.J and Bottom Reinforcing Plans S103C,S103D,S103E and S103F. The Collector bars <br /> to em s. start 6'-0"South of gridline 4 and extend 12'-0"North of Gridline 10. See added <br /> calculation sheet B24 for additional information. <br /> Comment ill I: Please provide calculations for diaphragm force checks per ASCE 7-10 Sec 12.10. Please <br /> indicate in the design calculations which design force governed in the design of the <br /> diaphragm. <br /> Response: See calculation sheets A47-A5S for of the Addendum 1 calculations.Sheets <br /> A47-A54 are reports of select diaphragm segments.Sheets A55-A5S provide a summary <br /> of the forces in the diaphragm segments,with the maximum force being highlighted. <br /> The controlling load cases are E6-E9 Seismic-Forces from the RSS model. <br /> Lochsa Engineering www.lochsa.com 1-866-606-9784 <br />