PCI 6 th Edition Lateral Component Design

Presentation Outline Architectural Components –Earthquake Loading Shear Wall Systems –Distribution of lateral loads –Load bearing shear wall analysis –Rigid diaphragm analysis

Architectural Components Must resist seismic forces and be attached to the SFRS Exceptions –Seismic Design Category A –Seismic Design Category B with I=1.0 (other than parapets supported by bearing or shear walls).

Seismic Design Force, F p Where: a p = component amplification factor from Figure

Seismic Design Force, F p Where: R p = component response modification factor from Figure

Seismic Design Force, F p Where: h = average roof height of structure S DS = Design, 5% damped, spectral response acceleration at short periods W p = component weight z= height in structure at attachment point < h

Cladding Seismic Load Example Given: –A hospital building in Memphis, TN –Cladding panels are 7 ft tall by 28 ft long. A 6 ft high window is attached to the top of the panel, and an 8 ft high window is attached to the bottom. –Window weight = 10 psf –Site Class C

Cladding Seismic Load Example Problem: –Determine the seismic forces on the panel Assumptions –Connections only resist load in direction assumed –Vertical load resistance at bearing is 7 1 / 2 ” from exterior face of panel –Lateral Load (x-direction) resistance is 4 1 / 2 ” from exterior face of the panel –Element being consider is at top of building, z / h =1.0

Solution Steps Step 1 – Determine Component Factors Step 2 – Calculate Design Spectral Response Acceleration Step 3 – Calculate Seismic Force in terms of panel weight Step 4 – Check limits Step 5 – Calculate panel loading Step 6 – Determine connection forces Step 7 – Summarize connection forces

Step 1 – Determine a p and R p Figure a p R p

Step 2 – Calculate the 5%-Damped Design Spectral Response Acceleration Where: S MS = F a S S S s = 1.5 From maps found in IBC 2003 F a = 1.0 From figure

Step 3 – Calculate F p in Terms of W p Wall Element: Body of Connections: Fasteners:

Step 4 – Check F p Limits Wall Element: Body of Connections: Fasteners:

Step 5 – Panel Loading Gravity Loading Seismic Loading Parallel to Panel Face Seismic or Wind Loading Perpendicular to Panel Face

Step 5 – Panel Loading Panel Weight Area = in2 Wp=485(28)=13,580 lb Seismic Design Force Fp=0.48(13580)=6518 lb

Step 5 – Panel Loading Upper Window Weight Height =6 ft W window =6(28)(10)=1680 lb Seismic Design Force –Inward or Outward –Consider ½ of Window Wp=3.0(10)=30 plf Fp=0.48(30)=14.4 plf 14.4(28)=403 lb –Wp=485(28)=13,580 lb Seismic Design Force –Fp=0.48(13580)=6518 lb

Step 5 – Panel Loading Lower Window Weight –No weight on panel Seismic Design Force –Inward or outward –Consider ½ of window height=8 ft Wp=4.0(10)=40 plf Fp=0.48(30)=19.2 plf 19.2(28)=538 lb

Step 5 Loads to Connections Dead Load Summary W p (lb) z (in) W p z (lb-in) Panel13, ,110 Upper Window 1, ,230 Lower Window Total 15,26064,470

Step 6 Loads to Connections Equivalent Load Eccentricity z=64,470/15,260=4.2 in Dead Load to Connections –Vertical =15,260/2=7630 lb –Horizontal = 7630 ( )/32.5 =774.7/2=387 lb

Step 6 – Loads to Connections Seismic Load Summary F p (lb) y (in) F p y (lb-in) Panel6, ,871 Upper Window ,852 Lower Window Total 7,459258,723

Step 6 – Loads to Connections Seismic Load Summary F p (lb) z (in) F p z (lb-in) Panel6, ,331 Upper Window Lower Window ,836 Total 7,45941,973

Step 6 – Loads to Connections Center of equivalent seismic load from lower left y=258,723/7459 y=34.7 in z=41,973/7459 z=5.6 in

Step 6 – Seismic In-Out Loads Equivalent Seismic Load y=34.7 in Fp=7459 lb Moments about Rb Rt=7459( )/32.5 Rt=1652 lb Force equilibrium Rb= Rb=5807 lb

Step 6 – Wind Outward Loads Outward Wind Load Summary F p (lb) y (in) F p y (lb-in) Panel3, ,060 Upper Window 1, ,480 Lower Window 1, Total 6,860267,540

Step 6 – Wind Outward Loads Center of equivalent wind load from lower left y=267,540/6860 y=39.0 in Outward Wind Load Fp=6,860 lb FpFp

Step 6 – Wind Outward Loads Moments about Rb Rt=7459( )/32.5 Rt=2427 lb Force equilibrium Rb= Rb=4433 lb

Step 6 – Wind Inward Loads Outward Wind Reactions Rt=2427 lb Rb=4433 lb Inward Wind Loads –Proportional to pressure Rt=(11.3/12.9)2427 lb Rt=2126 lb Rb=(11.3/12.9)4433 lb Rb=3883 lb

Step 6 – Seismic Loads Normal to Surface Load distribution (Based on Continuous Beam Model) –Center connections =.58 (Load) –End connections = 0.21 (Load)

Step 6 – Seismic Loads Parallel to Face Parallel load = lb

Step 6 – Seismic Loads Parallel to Face Up-down load

Step 6 – Seismic Loads Parallel to Face In-out load

Step 7 – Summary of Factored Loads 1.Load Factor of 1.2 Applied 2.Load Factor of 1.0 Applied 3.Load Factor of 1.6 Applied

Distribution of Lateral Loads Shear Wall Systems For Rigid diaphragms –Lateral Load Distributed based on total rigidity, r Where: r=1/D D=sum of flexural and shear deflections

Distribution of Lateral Loads Shear Wall Systems Neglect Flexural Stiffness Provided: –Rectangular walls –Consistent materials –Height to length ratio < 0.3 Distribution based on Cross-Sectional Area

Distribution of Lateral Loads Shear Wall Systems Neglect Shear Stiffness Provided: –Rectangular walls –Consistent materials –Height to length ratio > 3.0 Distribution based on Moment of Inertia

Distribution of Lateral Loads Shear Wall Systems Symmetrical Shear Walls Where: F i = Force Resisted by individual shear wall k i =rigidity of wall i  r=sum of all wall rigidities V x =total lateral load

Distribution of Lateral Loads “Polar Moment of Stiffness Method” Unsymmetrical Shear Walls Force in the y-direction is distributed to a given wall at a given level due to an applied force in the y- direction at that level

Unsymmetrical Shear Walls Where: V y = lateral force at level being considered K x,K y = rigidity in x and y directions of wall  K x,  K y = summation of rigidities of all walls T = Torsional Moment x = wall x-distance from the center of stiffness y = wall y-distance from the center of stiffness Distribution of Lateral Loads “Polar Moment of Stiffness Method”

Unsymmetrical Shear Walls Force in the x-direction is distributed to a given wall at a given level due to an applied force in the y- direction at that level. Distribution of Lateral Loads “Polar Moment of Stiffness Method”

Unsymmetrical Shear Walls Where: V y =lateral force at level being considered K x,K y =rigidity in x and y directions of wall  K x,  K y =summation of rigidities of all walls T=Torsional Moment x=wall x-distance from the center of stiffness y=wall y-distance from the center of stiffness Distribution of Lateral Loads “Polar Moment of Stiffness Method”

Unsymmetrical Shear Wall Example Given: –Walls are 8 ft high and 8 in thick

Unsymmetrical Shear Wall Example Problem: –Determine the shear in each wall due to the wind load, w Assumptions: –Floors and roofs are rigid diaphragms –Walls D and E are not connected to Wall B Solution Method: –Neglect flexural stiffness h/L < 0.3 –Distribute load in proportion to wall length

Solution Steps Step 1 – Determine lateral diaphragm torsion Step 2 – Determine shear wall stiffness Step 3 – Determine wall forces

Step 1 – Determine Lateral Diaphragm Torsion Total Lateral Load V x =0.20 x 200 = 40 kips

Step 1 – Determine Lateral Diaphragm Torsion Center of Rigidity from left

Step 1 – Determine Lateral Diaphragm Torsion Center of Rigidity y=center of building

Step 1 – Determine Lateral Diaphragm Torsion Center of Lateral Load from left x load =200/2=100 ft Torsional Moment M T =40( )=1236 kip-ft

Step 2 – Determine Shear Wall Stiffness Polar Moment of Stiffness

Step 3 – Determine Wall Forces Shear in North-South Walls

Step 3 – Determine Wall Forces Shear in North-South Walls

Step 3 – Determine Wall Forces Shear in North-South Walls

Step 3 – Determine Wall Forces Shear in East-West Walls

Load Bearing Shear Wall Example Given:

Load Bearing Shear Wall Example Given Continued: –Three level parking structure –Seismic Design Controls –Symmetrically placed shear walls –Corner Stairwells are not part of the SFRS Seismic Lateral Force Distribution LevelC vx FxFx Total941

Load Bearing Shear Wall Example Problem: –Determine the tension steel requirements for the load bearing shear walls in the north-south direction required to resist seismic loading

Load Bearing Shear Wall Example Solution Method: –Accidental torsion must be included in the analysis –The torsion is assumed to be resisted by the walls perpendicular to the direction of the applied lateral force

Solution Steps Step 1 – Calculate force on wall Step 2 – Calculate overturning moment Step 3 – Calculate dead load Step 4 – Calculate net tension force Step 5 – Calculate steel requirements

Step 1 – Calculate Force in Shear Wall Accidental Eccentricity=0.05(264)=13.2 ft Force in two walls Seismic Lateral Force Distribution LevelC vx FxFx Total

Step 1 – Calculate Force in Shear Wall Force at each level Level 3 F 1W =0.500(270)=135 kips Level 2 F 1W =0.333(270)= 90 kips Level 1 F 1W =0.167(270)= 45 kips Seismic Lateral Force Distribution LevelC vx FxFx Total941

Step 2 – Calculate Overturning Moment Force at each level Level 3 F 1W =0.500(270)=135 kips Level 2 F 1W =0.333(270)= 90 kips Level 1 F 1W =0.167(270)= 45 kips Overturning moment, M OT M OT =135(31.5)+90(21)+45(10.5) M OT =6615 kip-ft

Step 3 – Calculate Dead Load Load on each Wall –Dead Load =.110 ksf (all components) –Supported Area = (60)(21)=1260 ft 2 W wall =1260(.110)=138.6 kips Total Load W total =3(138.6)=415.8~416 kips

Step 4 – Calculate Tension Force Governing load Combination U=[ (0.24)]D+1.0EEq a U=0.85D+1.0E Tension Force

Step 5 – Reinforcement Requirements Tension Steel, A s Reinforcement Details –Use 4 - #8 bars = 3.17 in 2 –Locate 2 ft from each end

Rigid Diaphragm Analysis Example Given:

Rigid Diaphragm Analysis Example Given Continued: –Three level parking structure (ramp at middle bay) –Seismic Design Controls –Seismic Design Category C –Corner Stairwells are not part of the SFRS Seismic Lateral Force Distribution LevelC vx FxFx Total941

Rigid Diaphragm Analysis Example Problem: –Part A Determine diaphragm reinforcement required for moment design –Part B Determine the diaphragm reinforcement required for shear design

Solution Steps Step 1 – Determine diaphragm force Step 2 – Determine force distribution Step 3 – Determine statics model Step 4 – Determine design forces Step 5 – Diaphragm moment design Step 6 – Diaphragm shear design

Step 1 – Diaphragm Force, F p F p, Eq F p = 0.2·I E ·S DS ·W p + V px but not less than any force in the lateral force distribution table

Step 1 – Diaphragm Force, F p F p, Eq F p =(1.0)(0.24)(5227)+0.0=251 kips F p =471 kips Seismic Lateral Force Distribution LevelC vx FxFx Total941

Step 2 – Diaphragm Force, F p, Distribution Assume the forces are uniformly distributed –Total Uniform Load, w Distribute the force equally to the three bays

Step 3 – Diaphragm Model Ramp Model

Step 3 – Diaphragm Model Flat Area Model

Step 3 – Diaphragm Model Flat Area Model –Half of the load of the center bay is assumed to be taken by each of the north and south bays w 2 = /2=0.89 kip/ft –Stress reduction due to cantilevers is neglected. –Positive Moment design is based on ramp moment

Step 4 – Design Forces Ultimate Positive Moment, +M u Ultimate Negative Moment Ultimate Shear

Step 5 – Diaphragm Moment Design Assuming a 58 ft moment arm T u =2390/58=41 kips Required Reinforcement, A s –Tensile force may be resisted by: Field placed reinforcing bars Welding erection material to embedded plates

Step 6 – Diaphragm Shear Design Force to be transferred to each wall –Each wall is connected to the diaphragm, 10 ft Shear/ft=V wall /10=66.625/10=6.625 klf –Providing connections at 5 ft centers V connection =6.625(5)= kips/connection

Step 6 – Diaphragm Shear Design Force to be transferred between Tees –For the first interior Tee V transfer =V u -(10)0.59=47.1 kips Shear/ft=V transfer /60=47.1/60=0.79 klf –Providing Connections at 5 ft centers V connection =0.79(5)=4 kips

Questions?