SEAU 5 th Annual Education Conference 1. ASCE Concrete Provisions. Concrete Provisions. Concrete Strengths. Robert Pekelnicky, PE, SE

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ASCE 41-13 Concrete Provisions Robert Pekelnicky, PE, SE Principal, Degenkolb Engineers Chair, ASCE 41 Committee* *The view expressed represent those of the

Concrete Provisions Cast-in-place Beam-Column Moment Frames Slab-Column Frames Concrete Frames with Infill Precast Frames Cast-in-place Shear Walls Precast

1 ASCE Concrete Provisions Robert Pekelnicky, PE, SE Principal, Degenkolb Engineers Chair, ASCE 41 Committee* *The view expressed represent those of the author, not the standard s committee as a whole. Concrete Provisions Cast-in-place Beam-Column Moment Frames Slab-Column Frames Concrete Frames with Infill Precast Frames Cast-in-place Shear Walls Precast Shear Walls Concrete Braced Frames Diaphragms Cast-in-place & Precast Concrete Strengths Concrete strengths based on value specified on the drawings, value tested, or default based on age. 1.5 factor to translate to expected strength SEAU 5 th Annual Education Conference 1

2 Reinforcing Steel Strengths Rebar strengths based on value specified on the drawings, value tested, or default based on age/astm factor to translate to expected strength Reinforcing Steel Strengths 1.25 factor to translate to expected strength Columns Axial force is always force-controlled (tension will be deformation-controlled in ASCE 41-17). Flexure ductility based on shear reinforcement and axial force. Condition i Conforming ties, flexure governed = V@Mp/V 0 < 0.6 Condition ii Nonconforming ties, flexure governed = V@Mp/V 0 < 0.6 Condition iii - Shear governed = V@Mp/V 0 > 1.0 Condition iv Nonconforming lap splices V@Mp is the shear force in the column when Mp is applied at both ends. V 0 is the shear capacity in the column based on the amount of anticipated ductility. SEAU 5 th Annual Education Conference 2

3 Beams Moment frame beams are deformation controlled for both flexure and shear. M-factor for moment frame beams based on shear reinforcement and reinforcement ratio. Slab must be included in beam capacity. Lap splice and development length may affect capacity. Joint Regions Beam-column joint regions may be considered deformation-controlled with m = 1 for all primary components. For secondary components, m-factor varies based on shear reinforcing in joint and ratio of joint capacity to joint shear at beam yield. Slabs Slab-column frames may be primary or secondary components. Flexure in slabs is deformation controlled if the reinforcement is mild steel. Flexure is force controlled if capacity is only based on crushing of concrete in post tensioned concrete. Shear and punching shear are force controlled. Ductility is dependent on amount of bottom reinforcement passing over the column core and the ratio of gravity force to punching shear capacity. SEAU 5 th Annual Education Conference 3

4 Moment Frame Joint region def-cont. Foundation elements almost always force-contr. Columns can be defcont. or force-cont. MF beam def-cont. Soil actions can be def-cont. or force-cont. Shear Walls & Wall Piers Shear and flexure are deformation-controlled actions Ductility affected by presence of boundary elements Ductility affected by lap splice lengths Ductility affected by ratio of shear to flexure Ductility affected by axial load Ductility reductions for precast walls Coupling Beams Should be considered primary components Shear and flexure are deformation-controlled actions Ductility affected by presence of diagonal reinforcement Ductility affected by the presence of conforming shear reinforcement Ductility affected by ratio of shear to flexure SEAU 5 th Annual Education Conference 4

5 Diaphragms Shear and flexure are deformation-controlled actions. Ductility based on shear wall m-factors Connections to walls, collectors and frames are force-controlled Collector Elements / Drag Elements May be treated as deformation-controlled using column provisions Collector connections to walls and frames are force-controlled Shear Wall Joint region def-cont. Wall piers & spandrels def-cont. Collector beam defcont. Columns can be defcont. or force-cont. Collector to wall connection force-cont. Foundation elements almost always force-contr. Soil actions can be def-cont. or force-cont. SEAU 5 th Annual Education Conference 5

Concrete Material Testing Provisions Usual or comprehensive data collection. Performance levels greater than Life Safety require comprehensive data collection.

6 Concrete Material Testing Provisions Usual or comprehensive data collection. Performance levels greater than Life Safety require comprehensive data collection. Concrete core samples based on: Strengths specified on drawings, Number of different strengths specified, Types of elements, Number of stories, Building area/volume of concrete, and Coefficient of variation of tested Reinforcing steel: No tests for usual and coupon samples for comprehensive. Post-installed anchors: No testing in ASCE 41-13, but testing requirements in ASCE Building Overview 180 x 150 W 1 = 4,400 kips W Typ = 4,100k W 4 = 4,000k W = 16,500k Solid walls along Lines 2 and 5 Walls with openings along Lines B, C and E Coupled C shaped core wall Total height is 52-8 BPOE Perfomrance Wall Elevations SEAU 5 th Annual Education Conference 6

Wall Details Column Details What Earthquake Hazard Should be Used for BPOE?

6, Collapse Prevention in the BSE-2E will be the more severe performance objective.

7 Wall Details Column Details What Earthquake Hazard Should be Used for BPOE? Compare m-factors against BSE-2E to BSE-1E ratio. BSE-2E: S XS = 1.86 BSE-1E: S XS = 1.19 BSE-2E/BSE-1E = 1.89/1.19 = 1.6 If ratio of Collapse Prevention m-factor to Life Safety m-factor is less than 1.6, Collapse Prevention in the BSE-2E will be the more severe performance objective. Walls controlled by shear w/ axial: m LS = 2 & m CP = 3 m CP / m LS = 1.5 Nonconforming walls in flexure, low axial & shear: m LS = 2.5 & m CP = 4 m LS / m CP = 1.6 Collapse BSE-2E will govern the evaluation. SEAU 5 th Annual Education Conference 7

LSP Base Shear Calculation @ BSE-2E Even if LDP is used, it is still advisable to calculation LSP base shear. W = 16,500 k T a = C t h n = 0.020*52.7^0.75 = 0.39 sec S a = S XS = 1.

86*16,500 V = 27,000 kips Modeling Assumptions Use expected concrete strength, f ce, to determine E c. Model concrete members with cracked sections. Table 10-5 Beams & Sabs = 0.

8 LSP Base Shear BSE-2E Even if LDP is used, it is still advisable to calculation LSP base shear. W = 16,500 k T a = C t h n = 0.020*52.7^0.75 = 0.39 sec S a = S XS = 1.86 since T a < T S C 1 C 2 = 1.1 for 0.3 < T < 1.0 sec and max m-factor is 4 for Collapse BSE-2E C m = 0.8 for concrete shear wall 3 stories or taller V =0.8*1.1*1.86*16,500 V = 27,000 kips Modeling Assumptions Use expected concrete strength, f ce, to determine E c. Model concrete members with cracked sections. Table 10-5 Beams & Sabs = 0.3I g Columns varies between 0.3I g and 0.7I g depending on axial force. Shear walls 0.5I g Apply mass as uniform over diaphragm. Apply gravity load at columns for P-Delta. Include accidental torsion. Model interior flat plate slab-column frames. Shear Wall Modeling Model using shell elements Mesh is critical. Too course (i.e. one shell per floor) produces artificially high stiffness. Proper section cuts to get shear and moments. Modify f 11 and f 22 stiffness for cracking, not thickness. Model using frame elements Acceptable alternate. Confirm frame elements include shear deformations. Use frame provisions to approximate joint stiffness. Remember cracked section modifiers. In this example use shells for vertical walls and wall piers and frame elements for coupling beams. SEAU 5 th Annual Education Conference 8

Diaphragm Modeling Rigid vs. Stiff Diaphragm modeling can affect horizontal distribution of forces.

Explicit diaphragm very important to capture basement back-stay effect.

Use LSP to determine if slab-column frames are primary. Note: Can use LDP story forces in a static analysis instead of LSP forces.

9 Diaphragm Modeling Rigid vs. Stiff Diaphragm modeling can affect horizontal distribution of forces. Stiff diaphragms (modeled with shell or membrane elements) will represent the effective stiffness of the diaphragm. Explicit diaphragm very important to capture basement back-stay effect. Slab-Column Interior Frame Modeling Use equations in the commentary to determine the effective slab width. Use LSP to determine if slab-column frames are primary. Note: Can use LDP story forces in a static analysis instead of LSP forces. Over 95% of seismic force resisted by the shear walls. Keep slab-column frames in model for deformation compatibility. Modal Analysis Results ALWAYS CHECK MASSES! Model mass = 17,100 kips (104% of hand calc) Story masses match up Modal Properties T 1,EW = 0.51s w/ 75% EW mass T 2,EW = 0.14s w/ 10% EW mass T 1,NS = 0.33s w/ 80% NS mass T 2,NS = 0.09s w/ 14% NS mass T 1,Torsion = 0.55s w/ 5% EW mass SEAU 5 th Annual Education Conference 9

10 East-West Direction Results V = 25,200 kips (without C 1 C 2 ) V = 1.1*25,200 = 27,900 kips V = 27,900/16,500 = 1.7*W Story Fx Displacement Drift 4 th 10,500 kips 9.1 in 1.3% 3 rd 8,100 kips 7.2 in 1.5% 2 nd 5,700 kips 4.8 in 1.5% 1 st 3,600 kips 4.6 in 1.4% Shear Walls Flexure Shear Walls Shear SEAU 5 th Annual Education Conference 10

11 Shear Wall Example Line 2 Reinf. Ratio 0.62in 2 /(12 x12 ) = > P g = 1,540k, = M ud = 204,000 k-ft & V ud = 6,500 k M ce = 37,400 k-ft DCR M = 204,000/34,600 = , ,000 1,700 DCR v = 6,500/1,700 = 3.9 < DCR M Flexure Governed Shear Wall Example Line 2 DCR M = 204,000/34,600 = ,400. 1,400 1,400, , No confined boundary m CP = 4 < DCR M Wall Overstressed DCR M /m CP = 1.4 For shear m CP = 3 < DCR V Wall overstressed in shear if flexure capacity retrofit Shear Wall Example Line C P e = 4,400 k P uf = 1,540/2 4,400 = -3,630 k (tension) P uf = 1,540/2 + 4,400 = 5,200 k (compression) T ce = 494 k << T uf What to do when tension has DCR > 1? Treat as deformation controlled with flexure m-factor 1.0 SEAU 5 th Annual Education Conference 11

12 Coupling Beams Flexure Shear Shear Wall Example Line C M ud = 5,800 k-ft & V ud = 1,200 k M ce = 400 k-ft DCR M = 5,800/400 = 14.5 V = 400x2/6 = 133k 133, , m = 5 < DCR M = 5,800/400 = , , DCR v = 1,200/250 = 4.8 < m = 3 Nonconforming Coupling Beams What to do when coupling beams are significantly overstressed? Use different effective stiffness modifier? Treat as secondary? Delete them from Model? Will change behavior of building. Period increases from 0.51s to 0.65s (62% increase in stiffness) Confirm coupling beams will not present a falling hazard. Nonlinear analysis can be appropriate to allow coupling beams to yield and drop load. SEAU 5 th Annual Education Conference 12

13 Diaphragms Shear and flexure are deformation-controlled actions. Ductility based on shear wall m-factors for diaphragms. Ductility based on frame m-factors for chords and collectors. Connections to walls, collectors and frames are force-controlled. Diaphragm Example Line 2 V ud = roof q ud = 700/180 = 3.9 k/ft (force between 1-2) Lowest reinforcement ratio 12-#9 Top & 12 - #5 Bottom = / , ,000 DCR = 3.9/4.2 = 0.93 < DCR wall Check direct transfer to wall 12 q cl = 1.4x40x2x.31 = 25 k/ft (Shear Friction) q uf = 1,100/32 = 35 k/ft > q cl Collector req d Collector Elements / Drag Elements May be treated as deformation controlled using column provisions. Tension will be deformation controlled in ASCE Collector connections to walls and frames are force-controlled. SEAU 5 th Annual Education Conference 13

Collector Example Line 2 6 - #7 with 2 #7 dropping off every 30 V ud = 1,100k @ roof q ud = 1,100/180 = 6.1 k/ft Q ud = 6.

0 < DCR wall Deformation Compatibility Concrete shear wall buildings experience major collapses due to failure of the gravity

Use coefficient in the commentary to reduce the slab width. 2. Use modified equivalent frame assumption.

33I g Determine I eqiv of flat slab using equivalent frame method.

14 Collector Example Line #7 with 2 #7 dropping off every 30 V ud = roof q ud = 1,100/180 = 6.1 k/ft Q ud = 6.1x90 = 550 k Q ce = 6x0.6x50 = 180 k DCR = 550/180 = 3.0 < DCR wall Deformation Compatibility Concrete shear wall buildings experience major collapses due to failure of the gravity framing. Always perform an explicit evaluation of some representative secondary frames. Slab-Column Interior Frame Modeling 1. Use coefficient in the commentary to reduce the slab width. 2. Use modified equivalent frame assumption. Use effective moments of inertia including cracking. For slab I eff = 0.33I g Determine I eqiv of flat slab using equivalent frame method. Start with I g = bt 3 /12 using b = half the bay width in each direction. Calculate I eqiv, then calculate b eqiv = 12(I eqiv/0.33)/t 3 Input beam element in model with dimensions of t, b eqiv and 0.3 cracked section modifier l 1 l 4 4 Reference reinforced concrete text book on equivalent frame for more information on K t. SEAU 5 th Annual Education Conference 14

15 Slab-Column Frame Slab Example Column strip width is 180 Column strip reinf. = 17-#9 Top & 5-#6 Bottom No explicit direction to pass bottom bars through column M e = 100 k-ft Capacity is based on column strip moment minus gravity moment. M - ce = = 186 k-ft DCR = 0.5 M + ce = 65 (-274) = 339 k-ft DCR = 0.3 Slab will remain essentially elastic. Slab-Column Frame Slab Example Check punching shear as force-controlled v cl = 4 4,200 = 0.26 ksi V g = 160 kips V uf = 160 M uf = 2x100= 200 k-ft ksi < v, cl Punching shear is ok. Concrete Frame Columns Axial force is always force-controlled (tension will be deformation-controlled in ASCE 41-17) Flexure ductility based on shear reinforcement and axial force Condition i Conforming ties, flexure governed = /V 0 < 0.6 Condition ii Conforming ties, flexure-shear transition 0.6 < /V 0 < 1.0 Condition ii Nonconforming ties, flexure governed = /V 0 < 0.6 Condition iii - Shear governed = /V 0 > 1.0 Condition iv Nonconforming lap splices is the shear force in the column when M p is applied at both ends. V 0 is the shear capacity in the column based on the amount of anticipated ductility SEAU 5 th Annual Education Conference 15

16 Concrete Frame Columns Column Shear Capacity k = coefficient that varies from 1.0 when ductility demand (DCR) is less than 2 to 0.7 when ductility demand is greater than 6. M/(Vd) shall not be less than 2 or greater than 4. d can be approximated as 0.8h. Read the Standard! Section requires steel contribution to shear strength to be reduced by 50% if tie spacing greater than d/2 and to not contribute if spacing greater than d. SEAU 5 th Annual Education Conference 16

P (kip) 3500-500 0 (Pmax) 1 (Pmin) fs=0 fs=0.5fy 1000 Mx (k-ft) ASCE 41 13 Hands On Approach Frame Column Example Evaluate first floor interior column P uf = 630 kips P uf /A g f c = 630/(24x24x4.

0014 24 12 630,000 1 6 5,500 24 24 0.8 24 24 135 Frame Column Example V @Mp /V 0 = 151/135 = 1.1 Condition iii P uf /A g f c = 0.26 & 0.0014 Virtually no ductility, even as a secondary component!

17 P (kip) (Pmax) 1 (Pmin) fs=0 fs=0.5fy 1000 Mx (k-ft) ASCE Hands On Approach Frame Column Example Evaluate first floor interior column P uf = 630 kips P uf /A g f c = 630/(24x24x4.2) = 0.26 M ud = 220 k-ft M ce = 890 k-ft V=2*890/( ) = 151k V ud = 22 k M ud /(V ud d) = 220x12/(22x22) = 5.4, use , , , , Frame Column Example /V 0 = 151/135 = 1.1 Condition iii P uf /A g f c = 0.26 & Virtually no ductility, even as a secondary component! Frame Column Example Evaluate first floor interior Top P uf = 630kips M ce = 890 k-ft DCR = 220 / 890 = deficient lap splice, fs = 35 ksi M ce = 780 k-ft DCR = 220 / 780 = 0.28 SEAU 5 th Annual Education Conference 17

Evaluation Summary The building does not meet CP in the BSE-2E, nor will it meet LS in the BSE-1E: Walls are overstressed in both shear and flexure by a factor of 1.5 or more.

18 Evaluation Summary The building does not meet CP in the BSE-2E, nor will it meet LS in the BSE-1E: Walls are overstressed in both shear and flexure by a factor of 1.5 or more. Coupling beams are significantly overstressed. Collectors are stronger than walls, unless the walls are strengthened. Existing frame is acceptable for deformation compatibility. Beams & Moment Frame Connection m-factors Moment Frame Beam Example Evaluate second floor beam. 2-#9 top bars & 2-#8 bottom joint M ud = 1,400 k-ft M - ce = 750 k-ft (including 6-#6 slab reinforcement) M + ce = 170 k-ft (including reduction due to bottom bar short splice over column ) 1.25 / 1.25 / 60= 47 ksi SEAU 5 th Annual Education Conference 18

19 Moment Frame Beam Example Check shear ratio and whether shear controlled = ( )/( ) = 52 kips V ce = 2x16x28x 6, x50,000x28/10 V ce = 100 kips > Flexure controlled /bd f c = 52,000/(16x28x 6,000) = 1.5 Moment Frame Beam Example Check reinforcement ratios - = 4.6in 2 /(16 x28 ) = = 1.6in 2 /(16 x28 ) = ,000 87,000 75,000 87,000 75,000 b = Negative bending: ( )/ b = 0.24 For Nonconforming shear & V/bd f c < 3 m = 3.5 Positive bending ( )/ b = -1.0 m would be 4, but lap splice governs, so m = 1.75 Moment Frame Beam Example Positive Moment DCR DCR = 1,400 / 220 = 6.4 m = 1.75 for CP due to deficient splices DCR / m = 3.6 No good Negative Moment DCR DCR = 1,400 / 750 = 1.9 m = 3.5 for CP > DCR Negative moment ok. SEAU 5 th Annual Education Conference 19

20 Joint Regions Capacity of joint based on configuration and amount of column ties in joint region. Area of joint region joint depth times smallest of: Column width; Beam width plus joint depth; or Twice the smaller dimension from beam centerline to column side. Beam-column joint regions may be considered deformation-controlled with m = 1 for all primary components. For secondary components, m-factor varies based on shear reinforcing in joint and ratio of joint capacity to joint shear at beam yield. Joint Regions Interior Beam-Column joint has beam offset from column Joint depth is 14 Joint width: Column width = 20 Beam width plus joint depth = = 30 2x beam cl to column edge = 2x8 = , kips, 2 1,400 12, DCR = 995/65 = 15 > m = 1.0 No good Even if joint were secondary component, m = 3 No good Reinforced Masonry Shear and flexure are deformation-controlled actions Ductility affected by axial load Ductility affected by aspect ratio Ductility affected by reinforcement ratio Out-of-plane actions always force-controlled (Same for reinforced concrete walls) SEAU 5 th Annual Education Conference 20

21 Reinforced Masonry Example 1 story Building 90 x tall 8 CMU fully grouted walls 15 psf roof load 80 psf wall load S DS = 1.0 T = 0.6 s (calc d using diaph) C 1C 2 = 1.0 from m < 2 C m = 1.0 Reinforced Masonry Example Check Diaphragm V = 1.0*1.0*(15*90* *135*2*25/2) V = 450 kips Q ud = 450/90 = 5.0 k/ft Q ce = 1.7 k/ft. DCR = 5/1.7 = 2.9 Metal deck diaphragms are only deformation-controlled if they are governed by panel yielding or plate bucking. Most untopped metal deck diaphragms will be force-controlled. Wall Evaluation Full length walls will be governed by out-of-plane capacity Pierced walls are 15 long x 20 tall Spandrel beam at roof 7 tall (2 above roof and 5 below roof) Spandrel beams use same m-factors as walls 48 vertical and horizontal #8 in each end cell and T&B of spandrel 450 k SEAU 5 th Annual Education Conference 21

22 Wall In-Plane Evaluation Build model of frame to determine forces f m = 1,300 psi => f me = 1.3*1,500 = 1,950 psi f y = 40 ksi => f me = 1.25*40 = 50 ksi Wall shear: V ud = 150 k / pier & Wall moment: M ud = 3,000 k-ft / pier M/Vd = 1.5 V ce = 104 k => DCR = 150/104 = 1.4 < m = 2 ok for shear M ce = 900 k-ft => DCR = 3,000/900 = 3.3 f ae/f me = 0.01 L/h eff = 0.75 gf y/f me = 0.001*50/2 = g = v + h = 2*0.2/(48*8) = m = 4.8 from interpolation Out-of-Plane Wall Strength & Anchorage Diaphragm Anchorage Out of Plane Strength Force-controlled actions Revised design force equations Equivalency to ASCE 7-10 k a factor used to account for diaphragm flexibility 1.0 k a 2.0 k h factor used to account for force distribution over height of buildings with rigid diaphragms Out-of-Plane Wall Strength & Anchorage Diaphragm Anchorage Force-controlled actions Revised design force equations Out of Plane Strength Equivalency to ASCE 7-10 k a factor used to account for diaphragm flexibility 1.0 k a 2.0 k h factor used to account for force distribution over height of buildings with rigid diaphragms Currently out-of-plane in ASCE 41 is more conservative than ASCE 7 for CP in the BSE-2N. will be changing in ASCE to address this. SEAU 5 th Annual Education Conference 22

23 Wall Out-of-Plane Evaluation Check wall for out-of-plane forces F p = 0.4*1.0*1.65*80 = 53 psf M uf = 53*25 2 /8 = 4.1 k-ft/ft (excluding p-delta) M uf = 5.0 k-ft/ft (including p-delta) M cl = 1.3 k-ft/ft DCR = 5.0/1.3 = 3.8 => NG for out-of-plane wall forces Need to add strong backs Wall Out-of-Plane Evaluation Check connection forces F p = 0.4*1.0*2.0*1.0*1.65*80*(2+25/2) = 1.5 k/ft Check anchors or design new anchors for 1.5 k/ft of force Unreinforced Masonry Shear along the bed joint and rocking are deformation-controlled actions Diagonal tension and toe crushing are force-controlled actions Out-of-plane actions always force-controlled (Same for reinforced concrete and mastonry walls) SEAU 5 th Annual Education Conference 23

24 ASCE Concrete Provisions Masonry Provision Robert Pekelnicky, PE, SE Principal, Degenkolb Engineers Chair, ASCE 41 Committee* *The view expressed represent those of the author, not the standard s committee as a whole. SEAU 5 th Annual Education Conference 24

SEAU 5 th Annual Education Conference 1. Read the Standard! ASCE Analysis Provisions. (not just the tables and equations)

SEAU 5 th Annual Education Conference 1. Read the Standard! ASCE Analysis Provisions. (not just the tables and equations) ASCE 41-13 Robert Pekelnicky, PE, SE Principal, Degenkolb Engineers Chair, ASCE 41 Committee* *The view expressed represent those of the author, not the standard s committee as a whole. ASCE 41-13 Analysis