Design of Tall Bldg
Design of Tall Bldg EGN 5439
Popular in Course
verified elite notetaker
Popular in General Engineering
This 136 page Class Notes was uploaded by Theodora Gleichner on Monday October 12, 2015. The Class Notes belongs to EGN 5439 at Florida International University taught by Luis Prieto-Portar in Fall. Since its upload, it has received 35 views. For similar materials see /class/221783/egn-5439-florida-international-university in General Engineering at Florida International University.
Reviews for Design of Tall Bldg
Report this Material
What is Karma?
Karma is the currency of StudySoup.
Date Created: 10/12/15
EGN5439 The Design of Tall Buildings Lecture 26 Analysis of a Shear Wall Find the shears and moments in a 20st0ry building s shear walls L A PrietoPortar 2008 1quot 4 i This lecture s problem is like this shearwall at a building in Coral Gables LeJeune and Valencia Find the shears and moments in the shear walls of a nonproportionate nontwisting building The typical floor plan of a 20story of ce building is shown on the next slide and the elevation of the building on the slide after that one Notice that the shear walls are oriented in the yaxis only to provide simplicity to the problem A real structure would have shear walls in both the X and yaxes Walls 3 enclose the elevator and stair cores Notice that all the shear wall thicknesses are a function of their location along the vertical portion of the building The wind pressure is assumed to be uniform for simplicity at a value of 15 kNm2 throughout the building s height Each floor is 35 m high and all floors have the same height Shear walls 1 8 m long all the way up and with the fallowing widths 0 mm 0 t0 2 m 300 mm 21 t0 455 m 200 mm 455 t0 70 m hear walls 2 7m long from 0 t0 2 m 5 m from 21 t0 70 m 500 mm 0 t0 2 m 300 mm 21 t0 455 m 200 mm 455 t0 70 m Shear walls 3 6 m by 6 m long all the way up 400 mm 0 t0 2 m 200 mm 21 to 70 m g N E 8 m long all the way up All floors are 35 m high 200 mm 455 to 70 mf0r floors 14 to 20 g2 P Shear wall thtekness 300 mm 21 to 455 mf0r floors 7 to 13 z Shear wall thickness 450 mm 0 to 21 mf0r floors 1 t0 6 4 ngld mat foundatwn Step 1 Determine the inertias In of each wall or each of the three vertical zones There are three walls in three different vertical zones 9 different values 1 Wall 1 bottom 6 oors ILL 2 Wall 1 middle 7 oors ILM 3 Wall 1 upper 7 oors 4 Wall 2 lower 6 oors 5 Wall 2 middle 7 oors I 6 Wall 2 upper 7 oors 7 Wall 3 lower 6 oors 8 Wall 3 middle 7 oors I3M 9 Wall 3 upper 7 oors 3045n0ampnf 192nf 12 12 3 3 030m8m 2128 12 12 3 020 8 3 AUELL JQLEL85m4 12 12 3 3 2L W 143714 12 12 3 3 2MO3Om5m 2 31m 12 12 3 2 3 2UO Om5m 211114 12 12 3 3 2 4 3 2 2 43 ALQh4qu OM6 5 0 A28fb247h 12 12 12 12 3 3 2 2 3 2 23 J1 zhz y22A O 0 36 229211226Om4 12 12 12 12 3 3 2 2 3 2 23 QU L kht0 m n 56MO A29Yu1m260m4 12 12 12 12 Step 2 Determine the ratios k and Q of the inertias Iii of each wall Recall that the total external shear Q and the total external moment M at the story level i will be distributed between all the walls at that level in the ratio of their flexural rigidities E1 The resulting shear and moment in any wall j at that level i is given by jS Qi and M l M l where E1 E1 0 Q EIl 1 EIl E1 E1 0r EIl 1 E1 EIji is the flexural rigidity of the wall j at the level i and HE i represents the summation of the exural rigidities of all the walls at level 139 Since the modulus of elasticity E is the same for all the walls it is only required to find the ratios of the moments of inertia I s The ratio of the moments of inertia between two levels of a same shear wall is the parameter k Thus k is the ratio of a particular wall s stiffness on a speci c floor to the value of the stiffness of that entire floor Therefore the parameter k for each wall j at floor x is dependent of the inertias of the walls above and below oor x This is expressed as above below above Xj below Xj kxj n kxj n above below Zia Zia where lax and bej are the inertias of the wall j just above and just below the floor X For example Wall 1 at the 14th floor j 1 andx 14 above 8 5 kabove X above 36 x 20 am 391 85 2105260 21x j1 below kbelow Ixj klbfllow L x 1283105260 20 below 2 i1 All the other values for all the other two shear walls 2 and 3 at this transition floor 14 are also obtained in the same manner The change in the parameter k for Wall 1 at the 13 to 14th floor transition is below above Akxj Ak141 kxj kxj 044036 008 All the other k parameters are found for the this Wall 1 and then for Walls 2 and 3 All the values of k are tabulated on slide 12 along With other parameters we still need to calculate Finally a check can be made of k at each floor level through this relation 2kg mm 1 The next ratio is p which expands the value of k to the two floors at a transition of stiffness like the slab between the 13th and 14th floors It is the stiffness of the floor above or below to the combined stiffness of both oors The ratio p is defined by these two relations above below above Xj below Xj px above below and pxj above below 96quot 96quot X X For example the value of p for Wall 1 at 14th floor slab is above above Ixj above xj Iabove Ibelow p14 1 8 8 39 xj xj below below Ixj below 060 x1 gown p1 85128 These values are checked at each floor by below pxj in this case 0400601 GOOD above lpxj Step 3 Determine the parameter a The value of the parameter aat each change of floor is defined by the relation n above ax Z pxj xj i1 For example for Wall 1 at 14th oor slab 14 1 3 0514 ZpabmAk141 040008 040002 050 010 0010 11 Step 4 Determine the parameter The parameter 6 is used to find the secondary moments that enter the shear wall At a transition level of the stiffness the value of 6 is given by the relations 1 gbove pgboveAkxj axkgbove and 1 ax 1 b l b l b l x e 0W pxje OWAkxj axkxje 0W 1 ax For example for Wall 1 at 14 oor slab Cl 0V8 1 Cl 0V8 Cl 0V8 1 14 1 am 1011 Ak141 1141143911 1 001 040 008 001 036 0036 1 1 if pffijWAk141 amkffiiw 060 008 001 044 0045 1 0514 1 001 Table of lnem39al Pammelers caml p and Coe cients 1 and 0 Transition 1an Wall Tm k above k below Ak p above p below 0 above 0 below 14th oor 1 044 034 011 04 06 003 008 2 011 025 014 012 082 003 011 3 045 041 004 036 064 0 004 a 0030 7th oor 1 036 044 008 040 060 004 005 2 009 011 002 040 060 001 001 3 006 045 010 050 050 005 006 0L 0010 Step 5 Find the External Moments M on the Building at the Slab Level ofi Consider the Wind load on each oor level The Wind is assumed to be uniform over the entire height of the building With a pressure of 15 kNmz T this 1 to a quot39 load W along the Width of the building 40 m Consider only one half of the building to simplify therefore the w 12 15 kNm240 m 30 kNm For th A4 M14 straddling the transition oor 14 that is the moments at oors 13 14 and 15 wL2 30 kN m 70 4902 M15 2 2 26615 kN m 2 30 kN 70 455 2 M14WL m 9004 kN m 2 2 2 30 kN 70 420 2 M13W2L m 11760 kN m Step 6 Determine the Prima Moments M2 on each wall i Using the external moments Mj obtained in Step 5 the primary moments Mp above and below each level x can be determined from above above below below MIN kxj Mx and MIN kxj Mx For example the primary moments for wall 1 at level 14 are M532 2 kffffeMM O369 004 kN m 3240 kN m M 5 51 kfjfjwMM 0449 004 kN m 3 960 kN m At other oor levels 139 the primary moment M p is found from M p xj kijx For example for Wall 1 at the 15th and the 13th oors the primary moments are Mp151 k151M15 O366 615 kN m 2380 kN m Mp131 k131M13 O441 1 760 kN m 5170 kN m Step 7 Determine the Secondary Moments M2 on each wall 139 The secondary moments MS can be determined from above above below below Ms xj I6xj Mx and Ms xj I6xj Mx For example for Wall 1 at the 14th floor slab level M533 2 l ijeMM 0049 004 kN m 324 kN m 1330mm 005x9 004 kN m 405 kN m s 141 The secondary moments two levels above and below a transition slab level are given by M above above s x2j s xj above above MW 0268MH M below below s x l j s xj M below below s x 2j s xj For example the secondary moments in Wall 1 at the slab level of floors 12 13 15 and 16 are Mfffgj 02682Mfbf6vf1 02682324 kN m 23 kN m Mfg st 0268Mfbfsvf1 0268324 kN m 87 kN m Mffig 0268Mf 1031f1 0268 405 kN m 109 kN m Mffiaz 02682Mffi f1 02682 405 kN m 29 kN m Floors further above and below these can be analyzed using this progression Notice the rapidly diminishing values of the moments as we move away from the transition 14th oor slab level Step 8 Determine the Total Moments M on each wall i At Wall j at a transition level x the total moment is the sum of the primary and secondary moments above above Mtotal xj Mp xj Ms xj A check to see if the results are correct ensue from below below M M Hf M total xj s xj For example for Wall 1 at the 14th oor slab level M Mm Mji fzj 3 250 kN m 324 kN m 3574 kN m and as a check M M3133 Mffiifj 3980 kN m 405 kN m 3575 kN m total l4l total xj At any other oor level i for Wall j can be found by Mtotal ij Mp ij Ms ij For example Wall 1 at the slab level of the 15th oor MWHi1 Mp15 1Ms15 1 2388 kN m O268324 kN m 2301 kN m Step 9 Determine the Shear 0 within a storyhigh region of a shear wall The shear Q is found as the difference in moments between the top and bottom of the story being calculated divided by the story height For example for Wall 1 at the 14th floor slab level above above 141 m 3574kN m 2301kN m Q141 2364 kN h 35 m The values for the remaining 19 shears are left as an exercise 1111110 ofMomm in the Simon Wells kNm 74911 1 Wail 2 111211 3 External Pn39mmy Secondary Total Primary Seoomity Totoi Primary Secondiny Total Critical Moment Moment Moment Moment Moment Moment Moment Moment Moment Moment Floor Level M Ma Ma MW M p33 M 31 Mmg Mai M553 M3333 15111 oor 6615 2388 87 2301 582 22 560 3645 109 3754 14111110032600 9004 3250 324 3574 792 81 873 4961 405 4556 14111150333153 9004 3980 405 3575 972 99 873 4052 504 4556 13110232 11050 5193 109 5300 1200 20 1290 5292 135 5150 35352 31054 13025 239 14015 3354 239 3055 13904 0 13904 031133534 35015 15919 1030 14339 3390 1030 4900 15200 0 15200 05115325234 35015 12101 2030 14333 9040 4000 4900 14304 1333 15200 3311132 41344 13392 034 13153 10300 1091 11453 10005 350 15013 31310311011535 73500 24696 0 24696 18449 0 18449 30356 0 30356 This table provides all the moments for the 6 oors straddling the critical levels floors 6 7 and 8 plus floors 13 14 and 15 The moments for the other 14 floors are left as an exercise Floor level 0 I Wall type 1 wall type 2 Wall type 3 Tonal external shear Change level Total external Change level 4480 mi lue of shear kN 2047 This shear force diagram for the at the critical oors 67 and 1314 Wall ype 1 Wall type 2 Wall type 3 Tozal external momen 15 H a Total external 3 moment L 10 i a a 5 7 O 3 c g xo C v ox m c Q Q m m co c m a 1 of moment kNm lt OI C D Notice that the moment diagram is not as seriously affected by the dramatic exchange of force couples at the critical levels oors 6 to 7 and 13 to 1 References 1 Tall Building Structures Smith BS and Coull A John Wiley amp Sons Inc New York 1991 2 ETABS 3D Analysis of Building Systems Computer and Structures Inc Berkeley CA 2001 EGN5439 The Design of Tall Buildings Lecture 24 What is a shear wall L PrietoPortar 2008 of the Wall are s ffened by using shear Walls Many pans onhe m 1 Some Consider the simple analogy of a typical residential interior stud wall The wall is 12 long by 8 high sitting on a concrete slab l2 e 8 file 4 Floor Slob A stud wall is built from 2 4 wooden or steel studs placed every 16 on center with 2 4 pieces at the top and the bottom 2x4quot Top PIale 2x4 S39ruds 16quot 00 2x4quot S ill Plate 80 HF The studs are secured with toenailing 10d nails 30 1120 EH 7H i L r Toemqimmmg 1 d Na s USU my 30 deg KEN Consider now the application of a large horizontal load to the stud wall coming from wind or seismic loads 1000 lbs The shear wall will rack and some nails are pulled out 1000 lbs Nails are pulled out The dotted lines represents the original structure As the load is continually applied the stud frame racks more and more until it collapses aazggl mm mm m mm m mm m Mm mm mm mm mm mm mm m mm mm mm m mm mm mm m mm mm mm mm m mm mm mm mm H mm M H mm mm mm m m mm m m m m m w m m w m m w m m m m m m m m m m m n S m m m m m m m m Now reinforce the structure with 4 X 8 plywood or drywall sheets The dotted lines represent the stud walls 80 EH When the shear wall is now pushed it does not deform but rather slides to the right Finally the panel is anchored to the oor slab I I I I I I I I I I I I I I I I I I I I I I I I I I I I I II I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I II I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I II II II I39IIII39 II II II 39 II 39 VV II39lII II II II II II II II II39 II39 I I I I I I I I y I I I I I II I I I When the load is reapplied shear appears in the middle of the panel tension on the left and compression on the right 1000 lbs IIOSS3IduIO3 tension The shears at the center of the panel cancel out However the tension and compression forces create a couple which is equal to a moment V uogssonquo tension MOMENT U H H l U H 1 l I I l l l his is identical to a tall building that bends under Wind and seismic In I in Ts I I lI39L39 u39f Illr ll In a tall building the columns on the windward side may go into tension and the columns on the leeward side have an increase of compression This is the same model we just saw for the stud wall Because the tall building is subject to this force couple it de ects primarily in exure bending Therefore a shear wall is perhaps a misnomer since or model stud wall panel is primarily in uenced by the tension on one side and the compression on the other side also creating a moment Hence our shear wal is really a short and stubby exural wall All shear walls behave like this including the large and heavy reinforced concrete shear walls we use for our tall buildings in our cities EGN5439 The Design of Tall Buildings Lecture 18 Posttensianed Cancrete Slabs Basic Concepts Conventionally Reinforced Structures Versus PostTensioned Concrete Structures l i i i ONVENTIQNAQY REINFORCED CONCRETE STRUCTURES Regular Strength Concrete Initial Crack occurs soon Reinforcement gt Steel Rebar Fy 60 ksi Typical Passive Reinforcement After Concrete Cracks Stresses and Deformations have to occur before rebar starts working POSTTENSIONED CONCRETE IN FLOOR SYSTEMS Span Length is generally subject to deflection issues Typical spanslab thickness Ratio 28 i i Post Tensioned Concrete Structures High Strength Concrete High Strength Strands Tendons Fy 250 ksi Active Reinforcement Induces Balanced stresses in concrete EffiCi39ent Design POSTTENSIONED CONCRETE In Floor Systems Much Larger SpanDepth Ratio 45 average stress psi PLAIN CONCRETE MEMBER D 2 900 f6 75 N 7 7 N 3974 JJ Q v 2 i1 40 001 002 003 average Wain average stress psi REINFORCED CONCRETE MEMBER yieldmg W I quotl E F I I 1quot t I a w 3 0 average strain average stress Elsi I PRESTRESSED MEMBER 7 39 i h yielding g I I 2 900 e 6 I 5 60C 4 3 300 2 l g 1 x t 7 I F 339 D O 001 002 1003 average strain POSTTENSIONED CONCRETE USES OF PRESTRESSING IN THE USA IN THE LAST DECADE 19902000 65 PRETENSION 35 POSTTENSION PRECAST PRETENSIONING STIFF STEADY ELEMENTS ABUTMENT PRESTRESSING STEEL STEP 1 TENSIONING OF PRESTRESSING STRANDS IN STRESSING BED BEFORE CONCRETE IS POURED STIFF STEADY 1 ELEMENT39S 39 ELEMENTS ABUTMENT quotCENTROIDALquot AXIS STEP 2 CASTING OF CONCRETE AROUND TENSIONED STRANDS Fc0 MEMBER SHORTENING STEP 3 CUT STRAND FROM STRESSING BED CAUSING STRESSES AND SHORTENING OF THE MEMBER POSTTENSIONING PVC DUCT CORRUGATED METAL GROUTED OR BONDED CAST CONCRETE Fc0T0 mO m220 01 lm Emame Mad Nquot MZMOZZO 0 quot m m vanqz mmm ZO magtzu gtOgtZm gtNumzmu OOZOWm4m STEP 3 ANCHORING OF STRAND 7 r W LEUE HI El Fquot P RESTRESSIN 31TH E U SA 60 O 30 BUILDINGS STRUCTURAL MEMBERS LEDNEME STTENSIONING HMS a39APPLICATIONS WW Ghenm tvm beam Flat Plate Slab with wide shallow beams Slab With C0lumn capitals and drop panels Waffle Slab Two way slab on beams l TYPES OF POSTTENSIONED SLABS POSTTENSIONED CONCRETE Design Concerns oServiceability Cracks Deflections Minimum Rebar Requirements oSafety Collapse Strength QFIexural OShear 3 quotI l39 139 39n 3 W lquot l39 3 o 7 u 75 quot39 a r V j l L I l H 1 x r l l l l 2 k ti 39 t r l lll x 397 l l l Basic Load Combinations Transfer 10 DL 0 LL 108 PreyStressing Service 10 DL 10 LL 10 PreStressing Safety 14 DL 17 LL 10 PreStressing Overload Material Deficiencies Long Term 3 D L 3 LL 10 PreStressing Losses due to materia behaVior creep wedge seating shrinkage long term creep relaxation Maximum stress at tensile fiber 6 V f c Two Way 12 One way If Maximum Stress is Greater than 2 W j Bonded Reinforcement is required Minimum Reinforcement POSTTENSIONED CONCRETE Minimum Bonded Reinforcement Required 0 Top Rebar support5 ASS 000075 ACf 0 One Way Slab gt Min Rebar 0004 A POSTTENSIONED CONCRETE Stren th Desi n Factored M Demand M Resisting EQUILIBRIUM EQUATIONS Results of structural STRAIN COMPATIBILITY analysis MATERIAL STRENGTH VDemand 05 tau 05dz cross concrete actual equivalent concrete sectlon strains concrete uniform forces 6 stresses stresses STRESSES IN CROSSSECTIONAL ANALYSIS CALCULATION OF RESISTING MOMENT Plain Concrete Beam duh E q STRESS CONTROL BY PRESTRESSING I Axially Prestressed Beam STRESS CONTROL BY PRESTRESSING I Ec centrically Prestressed Beam 2Q P i P i r i nigh3 mi O 2fc STRESS CONTROL BY PRESTRESSING Beam Prestressed with variable Eccentricity 2Q LII5 fr 1 are i RI quot 39 h3 T and uniform compressive stresses at the ends STRESS CONTROL BY PRESTRESSING k MIDSPAN BALANCED LOAD STAGE EQUIVALENT LOAD ON MOMENT FROM CONCRETE FROM TENDON PRESTRESSING fiir F35quot 39539 qr Find 3H3 l J 0 J jugr 51 1 NH 7 I u Equot EQUIVALENT LOADS AND MOMENTS PRODUCED BY PRESTRESSING TENDONS IIEIIEII EQUIVALENTLOADMV DMDMENTFROM CONCRETE FROM TENDON PRESTRESSING hum Hnnr EQUIVALENT LOADS AND MOMENTS PRODUCED BY PRESTRESSING TENDONS Panel It 5 39 39 kmmu Q V pal 39139 l quot371quot I Calmrung L 3932 I l L h In F quot BUILDING IDEALIZATION FOR EQUIVALENT FRAME ANALYSIS BANDED TENDON PERSPECTIVE DISTRIBUTED TENDON PERSPECTIVE r H 5779 ibhm C gkvzokf g w 7 if iii quot3 r n 21 g 4 7 1 71 1 If TOP STRESSES AT POSTTENSIONED SLAB CABLE SHOULD FOLLOW MOMENT DIAGRAM SHAPE 1704 33 1 Nln 1quot i 39 391 E I PUNCHING SHEAR FAILURE SURFACE critical section SHEAR STRESS A329 of critical section DISTRIBUTION DUE TO Vu AND Mu FOR TYPICAL INTERIOR COLUMN polar moment of inertia39A J1 of the critical section 39 about its centroidal axis L Sil IOIHH NELSON PSR STUD Mid 20d COMMON NAIL x 1 W L Iquot 39 h x I I w r r 5 m w 121 m x V g l 394 Io V y quot 7 Ilkr l l D j i u WM v Aquot 4 UNLEOPM TENDONS HANDED TENDONS SEE RELNEORCEMENT PLAN 0 TLES 3quot 0 0 2 LEGS 10 TLES 3quot 00 2 LEGS UNLEORM TENDONS BANDED TENDON S 10 TLES 3quot OC 3 LEGS 10 TLES 3quot QC 3 LEGS nut1511114 39 MuratW J V I 1 UV 1 39 I v r i 39I39 quot39I r I W 39 a u M x quot x l 4 l V vlt 52 Kink WW v w u I39 1 IV 39 3 I o r39 A 391 h I I I I V j vf I III I II IIIIIIIIrIImk I 1 I 13quot 39 r IMI 39 sir i I A I 5 I h I I V 7 y I a I 39 V II 7 7 I 39 gt Y 5 i I I mm quot w r 39 r 391 I is l h Ir7 7 J 4 9 F v M 4 a 4 A r I u I I V quot t I I I I r 39 I39 I I l I 739 jI 39 I v I I I It II Vquot quot 391 quot VI run I V I III 39I I I l I I I I Im 39 n I I I V I h I w I l39 I I I I I I 4 7 39I II I I I I I I A I 7 4 39 M l 1 I 39I g 39 4 1 I 391 39 It I v I 39 I Iquot quot F I I If I y I III V l quot n L II LI I I u I II a I p I I I I I 39 I I I I I l I I I II I J39 I W L I I II I H I 39 I I I I I III I 1P s39 I 39 quot 31 I 39 i 5 y quot i u v A v39 H m 7 V A 391 L 39 v I Iquot I F ll g 394 I 39 41quot quot 3 15 T 39 inquot 1 III A quot mquot mm m V i 1 qi ll39a39m HM M 5 i 39 1 3 7 U 7 I 39 n In l y u i 9quotquot m y r 39lquot l 7 39 A 39 7 T5 7 1quot 3397 39 v u l Y mm mm v if 1 I Lb jp r 139 V quot v quot f c x 139 c 39I 39 39 39 I f ittingin V r I v av 7 m t 1 39I l n HHHII 5 x I k If EA l n u w u a v Q H ml I I I a I39 am Mm 39r 7w u H v gt i r I H r a l L j J 4 I r a I V a i p I J39v a I 5 4 v I I 2 1 H quot p t r i x A g r x u gtI I A 2 1 I 4 I a h 39u 39 l 1 1 l 2 L l J y 39 r I 39 x J I n n f J h i V 39 quotVr A A 391 I 0 7 a I r I a gt3 a37723 39 X j uy l1 71 i W x I M x g x V k I39 I quot V 1 39 39 I 455 7 39 i jg zuzI r gm 3 1m 39 m u r quot 39V v quot V O i V x A s 3 5A 2 W quotg M q 3 1 V quot EXTRUSION PROCESS USED TO PRODUCE UNBONDED SINGLE STRAND TENDONS ONE SET LENGTH OE PVC T 34quot MAXTMUM m 14 TD PVC CUT TO ETT OVER TENDON ELONGATION MEASUREMENT STEP 1 SAME P ECE OF PVC 1 34quot MAX MUM 19 f 7 J m 14quot D PVC CUT TO HT OVER TENDON ELONGATION MEASUREMENT STEP 2 x 39 Il tig l ni Ihh JJIM 1411 3in mmgm ivy va 1 If 7 I 39ll llf I h quot f VtVI39wm lydLJh Hutch gig fair 3 1 39Ily39Jquot i 1 ml 1 W a 1 r 39HH PM n 1 A J 39i lquotquot H a h V39u if 39 L If a ETABS ma t lOIlS Load Comb Lecture 41 EGN5439 The Design af Tall Buildings Define the Static Load Cases El TABS Nnnlinear v 54rTall Buildings Tulnrialr rh View g Ede gum Mew exne ran is etl 555an mm msgan Demng Ojtmns dew 51 X E vameS n QQEQWWMWS 2 em I 7 7 7 7 7 7 7 7 7 0 WaWS ahDeckiectmns W W 39 i I N U kPmDevtes 7 r mg Nanhneav mg vaemes gt m 5m gm WmSuemumm M We mm mm L Stan NanhreavEushavev Cases and seguemax anstmctmn Case tam ans ma Defau t Des qn gm Wen amhas m Nanhneay Cases spenax sew Lan Effects a Miss Sump 2 o Vmw Fi mm Create the Wind Load Cases Define Static Load Ease Names Load 5 all Weight Auto Load T ype Multiplier Lateral Load Iwrwox WIND i In User Defined DE ID DE D 1 LIVE LIVE D WINDX WIND D WINDY WIND D WIND MIKE WIND D U oer Defined WIND MY WIND D U oer Defined Eliok To 39139dd New Load l Modify Load I Modify Lateral Load ll Delete Load l Cancel Create the Rigid Diaphragms Select All then assign shell rigid diaphragm Ln TABS Nonlineal 154 7 Tall Buildings Tulnrial r 370 View 5 5e m yew We new elem We new new We get lmntPmnt V ee ymeyme m5 Hhevletaw hellAvea Jami2mm Laads eememnemee em in Lacal Exes Shell surmess Ma l evs Ell Pigvlahel geav DiSDlaY af Assigns E Swan Law WWW 5 messast Assign Diaphragm 25 neemanemeemeee V mes OEett Mesh Optmns y hula me ganstvamt D Iaphragms 39 Click to Add New Diaphragm ModifyShow Diaphragm Delete Diaphragm UK Cancel xasnnnvannnnzamnn mm vj Jemm vy m j 2w Pmnls iazn Lines momeee 8335 Edges selecled O 0875 0655 ooxlmmbwmd wa baled aa a j o wmvmmbwmd 0875 0875 0875 0655 0655 0655 0655 0655 0655 0655 0655 0655 0655 MX 025 025 025 025 019 019 019 019 019 019 019 019 019 019 019 019 019 019 019 MY ASCE7 Code Combinations 1 1 1 1 1 o 75 P W 39 1 v I CASE 2 PL 12DL 16LL 12DL 08W 12DL 16W 05L 12759quot 07517 07511 LLL LL 07511 07511 05613quot 39 07511 056 I L CASE 4 De ne Load tumhi atiu39ns Combinations Click to EUL SDL Mndifw39E how Enmbn D elete Enmtnzl UK Cancel I awm gt10 pauuag asn 3 39 o awn piwxeqwsn W 1 I u W990 an n 51 V N vauuaqlam g amp PPD39W PDW p psm aglaxn D T WE A pm mum nauamnw L WW WV mnv mamas sum wasquot samequot m3 peal mu 9 9 LL 39 6 39 6 9an 1 u H uI eJqdegq em 01 Bugpeoq em 1ddV M E sIgtlt HJMW MLDSV FE 1 H E mica 53 2 mm 5 sum 4 Sum n nnnsn n nnnm n nnnam