SEAoT State Conference Seminar Topics. Early knowledge needed. 11/5/2009. Wind Versus Seismic Which Controls? The Code

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1 /5/29 SEAoT State Conference 29 Austin, Texas Wind Versus Seismic Which Controls? by Larry Griffis P.E. Walter P. Moore and Associates, Inc. Seminar Topics ASCE 7 Simplified Wind Provisions ASCE 7 Seismic Provisions: ELF Method Equivalent Lateral Force Procedure Controlling Wind and Seismic Torsion Other Design Tips Preliminary design example for wind and seismic loads The Code The Challenge East Coast Engineers 26 IBC invokes seismic design all across US not just western US Seismic design impacts many more designs than in previous codes Engineers need to know early in design: Does wind or seismic control the design Early knowledge needed. Selection of proper structural system Architectural planning for structural system Budgeting for structural costs Fast-track structural delivery

/5/29 Wind versus Seismic Behavior V E Elastic Response Lateral Load V y V s R R d Fully Yielded Strength Design Force Level First Significant Yield Ω ο Seismic Requirements D s C d D E Drift

2 /5/29 Wind versus Seismic Behavior V E Elastic Response Lateral Load V y V s R R d Fully Yielded Strength Design Force Level First Significant Yield Ω ο Seismic Requirements D s C d D E Drift Spectral Accelerations S S, S (S DS, S D ) ASCE 7 Seismic Map USGS web site: Latitude / Longitude Importance Factor I 2

3 /5/29 Site Soil Conditions Strong Influence on Seismic Design Site Coefficients F a, F v Design Spectral Accelerations S DS, S D Fundamental Period T Fundamental Period T Base Shear (V) Seismic Design T C u T a ASCE 7-5 addendum or C s =.44S DS I 3

4 F x /5/29 Response Modification Coefficient R Concrete Systems Seismic Design Category Restrictions B C D E System Category R Ω o C d A F Concrete: Shear Wall Frame Interaction SW/Frame Interaction 4 /2 2 /2 4 NL NL NP NP NP NP (Ord SW + Ord MF) 2 Ordinary Shear Wall Building Frame System 5 2 /2 4 /2 NL NL NL NP NP NP 3 Ordinary Shear Wall + IMF Dual System 5 /2 2 /2 4 /2 NL NL NL NP NP NP 4 Ordinary Shear Wall + SMF Dual System 6 2 /2 5 NL NL NL NP NP NP 5 Special Shear Wall + Int MF Dual System 6 /2 2 /2 5 NL NL NL 6 / 24 / 6 6 Special Shear Wall + SMF Dual System 7 2 /2 /2 NL NL NL NL NL NL 5 7 Special Moment Frame Moment Frame /2 NL NL NL NL NL NL Response Modification Coefficient R Steel Systems Steel: Steel Systems w/o Seismic Steel Sys w/o Seismic NL NL NL NP NP NP 2 Ord Steel Conc Braced Frame Building Frame System 3 /4 2 3 /4 NL NL NL 35 / 6 35 / 6 NP / 6 3 Ord Moment Frame Moment Frame System 3 /2 3 NL NL NL NP / 65 NP / 65 NP / Int Moment Frame Moment Frame System 4 /2 3 4 NL NL NL 35 / 65 NP / 65 NP / 65 5 Sp Steel Conc Braced Frame Building Frame System NL NL NL Sp Steel Conc Br Frame + IMF Dual System 6 2 /2 5 NL NL NL 35 / 65 NP / 65 NP / 65 6 Sp Steel Plate Shear Wall Building Frame System NL NL NL Buckling Restr Braced Frame Building Frame System /2 NL NL NL 6 6 (w/o MC) 6 Steel Ecc Braced Frame Building Frame System NL NL NL 6 6 (w/o MC) 6 Special Truss MF Moment Frame /2 NL NL NL 6 NP 6 Sp Stl Conc Br Frame + SMF Dual System 7 2 /2 5 /2 NL NL NL NL NL NL 7 Buckling Restr Braced Frame Building Frame System 8 2 /2 5 NL NL NL 6 6 (w/ MC) 7 Steel Ecc Braced Frame Building Frame System NL NL NL 6 6 (w/ MC) 7 Special Steel Moment Frame Moment Frame System /2 NL NL NL NL NL NL 7 Buckling Restr Br Frame + SMF Dual System 8 2 /2 5 NL NL NL NL NL NL 7 Steel Ecc Braced Frame + SMF Dual System 8 2 /2 4 NL NL NL NL NL NL 7 Sp Steel Plate Shear Wall + SMF Dual System 8 2 /2 6 /2 NL NL NL NL NL NL Bldg. Density (pcf) Building Weight W vs No. Floors Building Wt. (pcf) No. Floors Stl Conc Building Density Concrete Steel Period T (sec) T=.28h.8 MF Conc. T=.6h.9 MF Steel. T=.232h.5 h > 3 ft. T=h / 75 h < 3 ft. Period Formulas Wind and Seismic T=h/75 h/ h/5 EBF Steel All other No. Floors Building Ht. h (ft.) Distribution of Base Shear F x Seismic Force Diagram Seismic Force F x k= k=.25 k=.5 k=.75 k= F x 7 6 Height h

5 /5/29 Seismic Shear Diagram Seismic Moment Diagram Story V Overturning Moment k = 2. 7 Building Height h k =. Building Height h k =. k = Story Shear Overturning Moment C vx versus k C vx vs k Calculation Base M from V k vs M arm -jh k= k=.25 k=.5 k=.75 k= V % Building Height h M = V x jh jh Eqv Moment Arm M = V x jh V = Base shear C vx k The Wind Pressure Equation p = I ½ ρv 2 K d Kz K zt C p G f Wind Requirements I ½ρV 2 K z K d K zt C p G f Importance of Building Velocity Pressure Terrain Exposure Directionality Factor Topographic Effect (Wind speed up) Shape Coefficient Gust Effect Factor 5

/5/29 ASCE 7-5 Wind Speed Map ASCE 7-5 Wind Pressure Equation Involves 48 different variables Requires solution to 24 equations Wind Variable Categories (6 categories, 48 variables) Building Geometry

6 /5/29 ASCE 7-5 Wind Speed Map ASCE 7-5 Wind Pressure Equation Involves 48 different variables Requires solution to 24 equations Wind Variable Categories (6 categories, 48 variables) Building Geometry (5 variables) : B, L, h, Cpw, Cpw, z Building Properties (3 variables) I, T, β Wind Speed (4 variables) V, V z, qz, qh Wind Climate (8 variables) Kz, Gf, I z, z, gq, gr, gv, Q, R, L z, RN, Rh, RB, RL, N, ηh, ηl, ηb Terrain Exposure (8 variables) α, zg, α, b, c, l, ε, zmin Site Topographic Features (9 variables) Kzt, K, K2, K3, H, LB, x, μ, γ Metal Building Type Residential Low Rise Simple Diaphragm Wind Load Method Building Types -Use Method 2: Low Rise (Figure 6-) Envelope Approach from ASCE 7-5 Commentary C6.5. -Use Method : Low Rise (Figure 6-2) Simplified Envelope Approach from ASCE 7-5 Commentary C6.4 Low Rise Building Mean roof height h 6 feet Mean roof height h does not exceed least horizontal dimension Simple Diaphragm Building Building in which both windward and leeward wind loads are transmitted through floor and roof diaphragms to the same MWFRS 6

/5/29 Simple Diaphragm Building Main Wind Force Resisting System ( MWFRS) floor diaphragms The Basis of Simplification Simplified Wind Design Simple diaphragm buildings h 6 feet Generally flat roofs

7 /5/29 Simple Diaphragm Building Main Wind Force Resisting System ( MWFRS) floor diaphragms The Basis of Simplification Simplified Wind Design Simple diaphragm buildings h 6 feet Generally flat roofs Based on ASCE 7-5 Figure 6-6 Method 2 Traditional Directional Approach from ASCE 7-5 Commentary C6.5. Assumptions Rigid diaphragm buildings h = 5 6 ft. Period T = h/75 seconds (upper bound) Damping = % (lower bound) L/B =.5.., 2. (interpolate between) I =. No topographic effects (k zt = ) Simplified Method Figure 6-6. The building shall be a simple diaphragm building as defined in Section The building shall have a mean roof height h 6 ft. 3. The ratio of L/B shall not be less than.5 nor more than The fundamental frequency (hertz) of the building used to determine the Gust Effect Factor Gf defined in Section shall not be less 75/h where h is in feet. 5. The structural damping ratio β of the building used to determine the Gust Effect Factor Gf defined in Section shall not be less than one percent (%) of critical. 6. The arrangement of elements of the MWFRS (walls, braced frames, moment frames) is symmetric about each principal building axis direction. 7

/5/29 Wind Pressure Equation p= qgf Cp qi ( GC ) p - General Equation (6-23) p = qg f C p - For simple diaphragm buildings ( q C q C ) p z = Gf z pw + h pl - windward, leeward walls Height (ft) 6 4 2

8 /5/29 Wind Pressure Equation p= qgf Cp qi ( GC ) p - General Equation (6-23) p = qg f C p - For simple diaphragm buildings ( q C q C ) p z = Gf z pw + h pl - windward, leeward walls Height (ft) h=6 ft. V=2 MPH Exposure C T=h/75 Damping=% Wind Pressure Vs Height ASCE 7-5 ASCE 7-5 Exact Simplified p =.4p 6 2 p Pressure (psf) Height (ft) Story Shear Vs Height Exact vs Simplified Story Shear (pounds) Story Moment Vs Height Exact vs Simplified Wind Load Equations Height (ft) Moment (foot-pounds) Pressure (psf): p z = p ( - z / h) + (z / h) p h Story Shear (pounds): v z =.5(h - z) [(p ( - z / h) + p h ( + z / h)] Overturning Moment (ft.-pounds): m z = /3 (h - z) 2 [.5p ( z / h) + p h ( +.5 z / h)] z p z p p h z z v z m z 8

9 /5/29 Why is Building Period Important? Related to mass and stiffness of building Stiffness affects drift and motion perception Mass affects wind forces Mass affects seismic forces Mass affects motion perception Period affects Gust Effect Factor, thus p Buildings with high periods interact more with the wind Note: Higher Period T is conservative (opposite from seismic!) Higher T is softer more flexible building Period Building Period T T = f { ( mass / stiffness ) } n T = 4 i= m i k i m i = w i / g floor i k i = story stiffness floor i w i = weight floor i Period Formulas Wind Load Proposed Period Formulas Period T (sec). h/75 h/ h/ h Recommended 4. h/75 Period Equations: 3. T = h/.75 h < 3 ft 2. T =.232h.5 h > 3 ft. 3 ft Bldg Ht h (ft) Period T (sec) Period Formulas Wind and Seismic T=.28h.8 9 MF Conc. T=h/75 h/ h/5 8 T=.6h.9 7 MF Steel. 6 T=.232h.5 h > 3 ft. EBF Steel 5 T=h / 75 4 h < 3 3 All other Building Ht. h (ft.) T Service or Ultimate? Steel Buildings: T proportional (/.8 stiffness).5 =.2 Concrete Buildings: T proportional (/.7 stiffness).5 =.2 9

10 /5/29 Anatomy of Wind p = I ½ ρv 2 K d * K z K zt C p G I Importance of Building ½ρV 2 Velocity Pressure K z Terrain Exposure K d Directionality Factor K zt Topographic Effect (Wind speed up) C p Shape Coefficient Gust Effect Factor G f h (ft) Velocity Pressure Exposure Cofficient vs Height Exp B: Kz = 2.(h/2) 2/7 Exp C: Kz = 2.(h/9) 2/9.5 Exp D: Kz = 2.(h/7) 2/ Kz Normalized Velocity Pressure Exposure Coefficients Normalized Gust Effect Factor x Vel. Pressure Exp. Coef. Gust Effect Factor Height h (ft) V = 9 mph β = % T =.232 h.5 h > 3 ft. T = h/75 h 3 ft. Kz: Exp C/B Gf x Kz: Exp C/B Kz: Exp D/B Gf x Kz: Exp D/B ft. 5 ft. Rato (Kz Exp C,D/ Kz Exp B), Ratio [(Gf x Kz) Exp C,D/ (Gf x Kz) Exp B] Wind Amplification Factor - Gustiness of the wind Accounts for the loading effects in the along-wind direction (parallel to the direction of the wind) due to wind turbulence-structure interaction. Accounts for along-wind loading effects due to dynamic amplification for flexible structures. Complex equation: ( gq Q + g R R ) +. 7I zbar G = f gv I zbar 5 Gust Effect Factor ( V, terrainexposure, B, L,h,T ) G f = f,β Seven key parameters Wind velocity (V) Building geometry (B, L, h) Building properties (T,β) Terrain exposure parameters (Table 6-2 ASCE 7-5) Exposure α zg ε zmin α b c l ft. ft. ft. B 7. 2 / /3 3 C / /5 5 D.5 7 / /8 7 Gust Factor Simplification Use L/B ratio instead B, L separately (reduces one variable) Relate T to building height (reduces one variable) Use lower bound damping ratio (β = %) β = % for typical steel buildings β = 2% for typical concrete buildings Thus, G f = f (V, exposure, h)

11 /5/29 Height (ft) Exp B; V = 9 mph Exp B; V = 2 mph Exp B; V = 5 mph Exp C; V = 9 mph Exp C; V = 2 mph Exp C; V = 5 mph Exp D; V = 9 mph Exp D; V = 2 mph Exp D; V = 5 mph Gust Effect Factor Vs Height L/B= T=h/75 % Damping β = % T = h/75 sec. L/B = Gust Effect Factor Height h (ft.) Gust Effect Factor vs Height V = 9 mph V = 2 mph V = 5 mph L = 5 ft. Exp C Wind Exp D B =5 ft. Exp B L/B = T = h/75 h < 75 ft. T =.232h.5 h 75 ft. β = 2% Gust Effect Factor G f Height h (ft.) L = ft. Exp C Wind Exp D B = 2 ft. Exp B L/B =.5 T = h/75 ft. h 75 ft. T =.236 h.5 h > 75 ft. β = 2% Gust Effect Factor vs Height V = 9 mph V = 2 mph V = 5 mph Gust Effect Factor Gf Height (ft.) Wind L = 2 ft. Gust Effect Factor vs Height B = ft. Exp C Exp D Exp B L/B = 2. T = h/75 h 75 ft T =.232h.5 h > 75 ft. β = 2% Gust Effect Factor Gf Solve Wind Pressure Equation at Height h and z = 5 ft. ( q C q C ) p z = Gf z pw + h pl p top = Cxp h h Pressure (psf): p = p 5 p z = p ( - z / h) + (z / h) p top C varies:. at h=5 ft to.4 at h=6 ft p z Wind Pressure MWFRS (psf) Exposure C 2 V(mph) h(ft.), L/B Notes: V=basic wind speed (mph), Figure 6- L=horizontal building dimension measured parallel to direction of wind (ft) B=horizontal building dimension measured normal to direction of wind (ft) Wind L B h ph p5

12 /5/29 Wind Load Equations Pressure (psf): p z = p ( - z / h) + (z / h) p h z p z p h p Story Shear (pounds): v z =.5(h - z) [(p ( - z / h) + p h ( + z / h)] z v z Dealing with Torsion Overturning Moment (ft.-pounds): m z = /3 (h - z) 2 [.5p ( z / h) + p h ( +.5 z / h)] z m z Sources of Wind Torsion Inherent torsion Center of pressure not at center of rigidity (e ip ) Center of mass not at center of rigidity (e im ) Accidental torsion Variation in center of pressure caused by turbulence (e a ) Causes - Torsional Wind Loading Center of pressure not at center of rigidity Strive for e ip = e ip.5b Center of mass not at center of rigidity Strive for e im = -.5B e im.5b Accidental wind torsion e a =.5B at.75w Minimizing the Effects Torsional Wind Loads Wind Torsion Align the center of pressure and center of rigidity as close as possible (Goal is zero inherent torsion). Maximum eccentricity e ip =.5B Avoid putting too much lateral load resistance at or near the center of the building. Spread some resistance at building perimeter if possible. Avoid having the torsional period as the first period (should be third period for normal - 2 ft buildings in plan) Study mode shapes Conform to minimum period recommendations Principal axis y x Principal axis 2 ki = k c.r. c.r. c.p. k2 k2j = k2 k22 k23 k3 d2j = d2 d23 d22.5b e2 B W.75W di = di d3 d2 e L Control location and stiffness of outer MWFRS s 2

13 /5/29 Wind Torsion Wind Torsion B = horizontal plan dimension of the building normal to the wind L = horizontal plan dimension of the building parallel to the wind c.r. = center of rigidity, c.p. = center of wind pressure ki = stiffness of frame I parallel to major axis k2j = stiffness of frame J parallel to major axis 2 di = distance of frame I to c.r. perpendicular to major axis d2j = distance of frame J to c.r. perpendicular to major axis 2 e = distance from c.p. to c.r. perpendicular to major axis e2 = distance from c.p. to c.r. perpendicular to major axis 2 J = polar moment of inertial of all MWFRS wind frames in the building W = wind load as required by standard Vi = wind force in frame i parallel to major axis V2j = wind force in frame j parallel to major axis 2 x, y = coordinates for center of rigidity from the origin of any convenient x,y axes n i= i= n x iki y iki n m i= i= 2 x = y = n J = n xi d i + x2 jd k i k i= j= i V i = (.75W ) k (.75W )( e +.5B) n i= k i i + J k d i i (.75W ) k2 j (.75W )( e2 +.5B) k2 j d V2 j = + m J k2 j j= 2 j 2 2 j Accidental Wind Torsion ASCE 7-5 Lateral Deflection e =.5B W V V 2 e a =.5B δ δ2 δmax X axis k k 2 k = k 2 Elastic shear center k.5r.5r R B Y axis k 2 δavg =.5(δ + δ 2 ) δ max from computer analysis Minimizing Torsional Effects Place MWFRS s to minimize inherent torsion Controlling Wind Torsion d i J n (.45B e ) ki (.45B e2 ) i= and d 2 j J m j= k 2 j Maintain : δmax δ avg =.5(δ + δ 2 ).4 under wind and seismic loading - including req d code eccentricty (see similar equations for using simplified seismic provisions ASCE 7-5 Section 2.4..) 3

For rectangular buildings, as L/B increases from. to

14 /5/29 Simplified Method Maximize distance Wind Torsion 7. For square buildings with L/B=., the combined stiffness of the two most separated lines of the MWFRS in each direction shall be at least two thirds of the total stiffness in each principal axis direction. For rectangular buildings, as L/B increases from. to 2. or decreases from. to.5, the combined stiffness of the two most separated lines of the MWFRS in each direction shall be proportionally increased from two thirds of the total stiffness to at least 8% of the total stiffness in each principal axis direction. Principal axis Principal axis 2 ki = k c.r. c.r. c.p. k2 k2j = k2 k22 k23 di = di d3 d2 e L 8. For square buildings with L/B =., the distance between the two most separated lines of the MWFRS in each major axis direction shall be at least two thirds of the dimension of the building perpendicular to the axis direction under consideration. For rectangular buildings as L/B increases from. to 2. or decreases from. to.5, the distance between the two most separated lines of the MWFRS in each principal axis direction shall be proportionally increased from two thirds of the dimension of the building perpendicular to the axis direction under consideration to % of the dimension. y x k3 d2j = d2 d23 d22.5b e2 B W.75W Control location and stiffness of outer MWFRS s Seismic Detailing Always (even when wind controls) Horizontal Structural Irregularities Table 2.2- System Requirements Avoid horizontal structural irregularities Table 2.3- Avoid vertical structural irregularities Table Vertical Structural Irregularities Obtain wind p from simplified method Obtain seismic base shear from ELF method Compute base shears for each Obtain forces at each level Draw/compare story V, M diagrams for building Minimize torsion Seismic detailing - always 4

15 /5/29 Example Problem story Concrete Building Hotel with large ballroom Downtown St. Louis Missouri Exposure C wind Site Class D - seismic Seismic: Obtain site latitude, longitude Latitude: deg, Longitude: -9.2 deg 2. Obtain S DS, S D Fa =.337 Fv = 2.3 S DS =.56 g S D =.238 g 3. Determine Occupancy Category: (from ASCE 7-5 Table -) Occupancy Category III 4. Determine seismic design category: (from Tables.6-,.6-2) S DS =.56 g 5. Determine Importance Factor (seismic) (Table.5- ASCE 7-5) S D =.238 g Seismic Design Category (SDC) D 6. Determine structural system type (Table 2.2- ASCE 7-5) 5

16 /5/29 Structural System Selection 7. Determine Building Period T a Seismic Design Category Restrictions B C D E System Category R Ω o C d A F Concrete: Shear Wall Frame Interaction SW/Frame Interaction 4 /2 2 /2 4 NL NL NP NP NP NP (Ord SW + Ord MF) 2 Ordinary Shear Wall Building Frame System 5 2 /2 4 /2 NL NL NL NP NP NP 3 Ordinary Shear Wall + IMF Dual System 5 /2 2 /2 4 /2 NL NL NL NP NP NP 4 Ordinary Shear Wall + SMF Dual System 6 2 /2 5 NL NL NL NP NP NP 5 Special Shear Wall + Int MF Dual System 6 /2 2 /2 5 NL NL NL 6 / 24 / 6 6 Special Shear Wall + SMF Dual System 7 2 /2 /2 NL NL NL NL NL NL 5 7 Special Moment Frame Moment Frame /2 NL NL NL NL NL NL T a =.2 h.75 =.2 ().75 T a =.632 seconds Special Shear wall + IMF (Dual System) R = 6 Ω = 2.5 C D = 5 8. Obtain Seismic Response Coefficient C s Design Response Spectrum Specytral Acceleration (%g) S a = S D / (T a ) SD.238 Cs = = R 6.5 T.632 I.25 = C S = T a = Period (sec) Bldg. Density (pcf) 9. Determine estimated building weight W Building Density 3 Building Wt. (pcf) No. Floors Stl Conc Concrete Steel 5 3 pcf No. Floors W = weight density x bldg volume = 3 x [ 5x5x] / = 29,25 kips W = 29, 25 kips. Determine base shear V V = C s W =.724 x 29,25 = 2,8 kips V = 2,8 kips 6

17 /5/29. Distribute base shear up the bldg V = 2,8 kips w i = 2,925 kips ea floor T a =.624 sec >.5 seconds k =.7 by interpolation Seismic: height C vx Fx V M ft kips kips k-ft Draw F x, V and M diagram for seismic loading Wind:. Determine wind Importance Factor I (from ASCE 7-5 Table 6-) 2. Determine design wind pressures from pressure table p h =.5 x 27.7 = 3.9 psf p o =.5 x 2.7 = 25. psf I =.5 (wind) p h = 3.9 psf p o = 25. psf Wind Pressure MWFRS (psf) Exposure C 2 V(mph) h(ft.), L/B Notes: V=basic wind speed (mph), Figure 6- L=horizontal building dimension measured parallel to direction of wind (ft) B=horizontal building dimension measured normal to direction of wind (ft) L ph Wind h B p5 3. Determine F x, V, M at each floor Wind: height p Fx V M ft psf kips kips kip-ft Draw F x, V and M diagram for wind loading 7

18 /5/29 Compare loading diagrams Seismic/Wind Force (F x) vs Height 9 Compare Shear Diagrams Story Shear vs Height Wind 8 7 Height (ft) Seismic Height (ft) Wind Seismic Force (kips). 5.,.,5. 2,. 2,5. Story Shear (kips) Compare Moment Diagrams Height (ft) Overturning Moment vs Height Wind Seismic 25, 5, 75,, 25, 5, Moment (kip-ft) Conclusions Seismic loading may control design (even in the central US!) High seismic loads from low site classification (Site Class D) and Importance Factor (I =.25) Check wind and seismic drift Seismic detailing always! Control location of LLRS for wind and seismic torsion control Wrap-Up Simple procedures for wind and seismic load calculations Useful comparisons for preliminary design Control wind and seismic torsion Seismic detailing always! Minimize vertical and horizontal irregularities (trade steel/concrete for granite) 8

CHAPTER 5. T a = 0.03 (180) 0.75 = 1.47 sec 5.12 Steel moment frame. h n = = 260 ft. T a = (260) 0.80 = 2.39 sec. Question No.

CHAPTER 5. T a = 0.03 (180) 0.75 = 1.47 sec 5.12 Steel moment frame. h n = = 260 ft. T a = (260) 0.80 = 2.39 sec. Question No. CHAPTER 5 Question Brief Explanation No. 5.1 From Fig. IBC 1613.5(3) and (4) enlarged region 1 (ASCE 7 Fig. -3 and -4) S S = 1.5g, and S 1 = 0.6g. The g term is already factored in the equations, thus