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ASCE 41‐13 Hands‐On Approach Concrete & Masonry Provisions
SEAU 5th Annual Education Conference 1
ASCE 41-13 Concrete ProvisionsRobert Pekelnicky, PE, SE
Principal, Degenkolb EngineersChair, ASCE 41 Committee*
*The view expressed represent those of the author, not the standard’s committee as a whole.
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 Provisions
Concrete strengths based on value specified on the drawings, value tested, or default based on age.
Concrete Strengths
1.5 factor to translate to expected strength
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Rebar strengths based on value specified on the drawings, value tested, or default based on age/ASTM.
Reinforcing Steel Strengths
1.25 factor to translate to expected strength
Reinforcing Steel Strengths
1.25 factor to translate to expected strength
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/V0 < 0.6
Condition ii – Nonconforming ties, flexure governed = V@Mp/V0 < 0.6
Condition iii - Shear governed = V@Mp/V0 > 1.0
Condition iv – Nonconforming lap splices
V@Mp is the shear force in the column when Mp is applied at both ends.
V0 is the shear capacity in the column based on the amount of anticipated ductility.
Columns
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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.
Beams
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
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.
Slabs
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Moment Frame
Columns can be def-cont. or force-cont.
Foundation elements almost always force-contr.
Soil actions can be def-cont. or force-cont.
Joint region def-cont.
MF beam def-cont.
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
Shear Walls & Wall Piers
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
Coupling Beams
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Shear and flexure are deformation-controlled actions.
Ductility based on shear wall m-factors
Connections to walls, collectors and frames are force-controlled
Diaphragms
May be treated as deformation-controlled using column provisions
Collector connections to walls and frames are force-controlled
Collector Elements / Drag Elements
Shear Wall
Columns can be def-cont. or force-cont.
Foundation elements almost always force-contr.
Soil actions can be def-cont. or force-cont.
Joint region def-cont.
Collector beam def-cont.
Wall piers & spandrels def-cont.
Collector to wall connection force-cont.
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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 41-17.
Concrete Material Testing Provisions
Building Overview
180’ x 150’
W1 = 4,400 kips
WTyp = 4,100k
W4 = 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
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Wall Details
Column Details
Compare m-factors against BSE-2E to BSE-1E ratio.
BSE-2E: SXS = 1.86
BSE-1E: SXS = 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: mLS = 2 & mCP = 3 → mCP/ mLS = 1.5
Nonconforming walls in flexure, low axial & shear: mLS = 2.5 & mCP = 4 → mLS / mCP = 1.6
Collapse Prevention @ BSE-2E will govern the evaluation.
What Earthquake Hazard Should be Used for BPOE?
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Even if LDP is used, it is still advisable to calculation LSP base shear.
W = 16,500 k
Ta = Cthn = 0.020*52.7^0.75 = 0.39 sec
Sa = SXS = 1.86 since Ta < TS
C1C2 = 1.1 for 0.3 < T < 1.0 sec and max m-factor is 4 for Collapse Prevention @ BSE-2E
Cm = 0.8 for concrete shear wall 3 stories or taller
V =0.8*1.1*1.86*16,500
V = 27,000 kips
LSP Base Shear Calculation @ BSE-2E
Use expected concrete strength, f’ce, to determine Ec.
Model concrete members with cracked sections.
Table 10-5
Beams & Sabs = 0.3Ig Columns varies between 0.3Ig and 0.7Ig depending on axial force.
Shear walls 0.5Ig
Apply mass as uniform over diaphragm.
Apply gravity load at columns for P-Delta.
Include accidental torsion.
Model interior flat plate slab-column frames.
Modeling Assumptions
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 f11 and f22 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.
Shear Wall Modeling
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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.
Diaphragm Modeling – Rigid vs. Stiff
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.
Slab-Column Interior Frame Modeling
ALWAYS CHECK MASSES!
Model mass = 17,100 kips (104% of hand calc)
Story masses match up
Modal Properties
T1,EW = 0.51s w/ 75% EW mass
T2,EW = 0.14s w/ 10% EW mass
T1,NS = 0.33s w/ 80% NS mass
T2,NS = 0.09s w/ 14% NS mass
T1,Torsion = 0.55s w/ 5% EW mass
Modal Analysis Results
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V = 25,200 kips (without C1C2)
V = 1.1*25,200 = 27,900 kips
V = 27,900/16,500 = 1.7*W
East-West Direction Results
Story Fx Displacement Drift
4th 10,500 kips 9.1 in 1.3%
3rd 8,100 kips 7.2 in 1.5%
2nd 5,700 kips 4.8 in 1.5%
1st 3,600 kips 4.6 in 1.4%
Shear Walls – Flexure
Shear Walls – Shear
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Reinf. Ratio – 0.62in2/(12”x12”) = 0.0043 > 0.0015
Pg = 1,540k → = ,
.0.08
Mud = 204,000 k-ft & Vud = 6,500 k
Mce = 37,400 k-ft
DCRM = 204,000/34,600 = 5.5
12 384 2 1.0 5,500 0.0043 50,000
1,700
DCRv = 6,500/1,700 = 3.9 < DCRM → Flexure Governed
Shear Wall Example – Line 2
DCRM = 204,000/34,600 = 5.5
37,400. 1,400
1,400,000
12 384 5,5004.1
No confined boundary
mCP = 4 < DCRM → Wall Overstressed
DCRM/mCP = 1.4
For shear mCP = 3 < DCRV → Wall overstressed in shear if flexure capacity retrofit
Shear Wall Example – Line 2
Pe = 4,400 k
Puf = 1,540/2 – 4,400 = -3,630 k (tension)
Puf = 1,540/2 + 4,400 = 5,200 k (compression)
Tce = 494 k << Tuf
What to do when tension has DCR > 1?
Treat as deformation controlled with flexure m-factor
1.0
Shear Wall Example – Line C
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Coupling Beams
Shear
Flexure
Mud = 5,800 k-ft & Vud = 1,200 k
Mce = 400 k-ft
DCRM = 5,800/400 = 14.5
V = 400x2/6 = 133k133,000
12 48 5,5003.1
m = 5 < DCRM = 5,800/400 = 14.5
12 48 3 1.0 5,500 0.0043 50,000
250
DCRv = 1,200/250 = 4.8 < m = 3
Shear Wall Example – Line C
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.
Nonconforming Coupling Beams
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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.
Diaphragms
Vud = 1,100k @ roof
qud = 700/180 = 3.9 k/ft (force between 1-2)
Lowest reinforcement ratio
12-#9 Top & 12 - #5 Bottom = 0.0049
9 3 1.0 5,500 0.0049 50,000
4.2 /
DCR = 3.9/4.2 = 0.93 < DCRwall
Check direct transfer to wall 2-#5 @ 12”
qcl = 1.4x40x2x.31 = 25 k/ft (Shear Friction)
quf = 1,100/32 = 35 k/ft > qcl → Collector req’d
Diaphragm Example – Line 2
May be treated as deformation controlled using column provisions.
Tension will be deformation controlled in ASCE 41-17.
Collector connections to walls and frames are force-controlled.
Collector Elements / Drag Elements
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6 - #7 with 2 #7 dropping off every 30’
Vud = 1,100k @ roof
qud = 1,100/180 = 6.1 k/ft
Qud = 6.1x90 = 550 k
Qce = 6x0.6x50 = 180 k
DCR = 550/180 = 3.0 < DCRwall
Collector Example – Line 2
Concrete shear wall buildings experience major collapses due to failure of the gravity framing.
Always perform an explicit evaluation of some representative secondary frames.
Deformation Compatibility
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 Ieff = 0.33Ig Determine Ieqiv of flat slab using equivalent frame method.
Start with Ig = bt3/12 using b = half the bay width in each direction.
Calculate Ieqiv, then calculate beqiv = 12(Ieqiv/0.33)/t3
Input beam element in model with dimensions of t, beqiv and 0.3 cracked section modifier.
Slab-Column Interior Frame Modeling
1∑
1∑
10.33
4 4 4 4
9
ℓ 1 ℓReference reinforced concrete text book on equivalent frame for more information on Kt.
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Column strip width is 180”
Column strip reinf. = 17-#9 Top & 5-#6 Bottom
No explicit direction to pass bottom bars through column
Me = 100 k-ft
Capacity is based on column strip moment minus gravity moment.
M-ce = 460 – 274 = 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
vcl = 4√4,200 = 0.26 ksi
Vg = 160 kips
Vuf = 160
Muf = 2x100= 200 k-ft
. .
,0.25ksi < vcl
Punching shear is ok.
Slab-Column Frame Slab Example
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/V0 < 0.6
Condition ii – Conforming ties, flexure-shear transition 0.6 < V@Mp/V0 < 1.0
Condition ii – Nonconforming ties, flexure governed = V@Mp/V0 < 0.6
Condition iii - Shear governed = V@Mp/V0 > 1.0
Condition iv – Nonconforming lap splices
V@Mp is the shear force in the column when Mp is applied at both ends.
V0 is the shear capacity in the column based on the amount of anticipated ductility
Concrete Frame Columns
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Concrete Frame Columns
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.
Column Shear Capacity
Section 10.3.4 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.
Read the Standard!
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Evaluate first floor interior column
Puf = 630 kips
Puf/Agf’c = 630/(24x24x4.2) = 0.26
Mud = 220 k-ft
Mce = 890 k-ft → V=2*890/(12.5-0.75) = 151k
Vud = 22 k
Mud/(Vudd) = 220x12/(22x22) = 5.4, use 4
0.44 50,000 2212
6 5,5004
1630,000
6 5,500 24 240.8 24 24 135
0.40
24 120.0014
Frame Column Example
V@Mp/V0 = 151/135 = 1.1
Condition iii
Puf/Agf’c = 0.26 & 0.0014
Virtually no ductility, even as a secondary component!
Frame Column Example
Evaluate first floor interior column @ Top
Puf = 630kips
Mce = 890 k-ft
DCR = 220 / 890 = 0.25
@ deficient lap splice, fs = 35 ksi
Mce = 780 k-ft
DCR = 220 / 780 = 0.28
Frame Column Example
P (k ip)
Mx (k-ft)
3500
-500
10000
fs=0.5fy
fs=0
(Pmax)
(Pmin)
fs=0.5fy
fs=0
1
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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.
Evaluation Summary
Beams & Moment Frame Connection m-factors
Evaluate second floor beam.
2-#9 top bars & 2-#8 bottom bars @ joint
Mud = 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
Moment Frame Beam Example
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Check shear ratio and whether shear controlled
V@MCE = (750+170)/(18.75-1.17) = 52 kips
Vce = 2x16x28x√6,000 +0.22x50,000x28/10
Vce = 100 kips > V@MCE → Flexure controlled
V@MCE/bd√f’c = 52,000/(16x28x√6,000) = 1.5
Moment Frame Beam Example
Check reinforcement ratios
- = 4.6in2/(16”x28”) = 0.010
+= 1.6in2/(16”x28”) = 0.0036.
0.85 0.75 6,00075,000
87,00087,000 75,000
b= 0.027
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.
Moment Frame Beam Example
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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 = 16”+14” = 30”
2x beam cl to column edge = 2x8” = 16”
1.0 10 6,000 14 16 65 kips
,,
2 1,400 1228
205 995
DCR = 995/65 = 15 > m = 1.0 → No good
Even if joint were secondary component, m = 3 → No good
Joint Regions
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)
Reinforced Masonry
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Reinforced Masonry Example
1 story Building90’ x 135’25’ tall8” CMU fully grouted walls
15 psf roof load 80 psf wall load
SDS = 1.0T = 0.6 s (calc’d using diaph)C1C2 = 1.0 from m < 2Cm = 1.0
Reinforced Masonry Example
Check Diaphragm
V = 1.0*1.0*(15*90*135 + 80*135*2*25/2)V = 450 kips
Qud = 450/90 = 5.0 k/ft
Qce = 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
#4 @ 48 vertical and horizontal
#8 in each end cell and T&B of spandrel450 k
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Wall In-Plane Evaluation
Build model of frame to determine forces
f‘m = 1,300 psi => f’me = 1.3*1,500 = 1,950 psify = 40 ksi => f’me = 1.25*40 = 50 ksi
Wall shear: Vud = 150 k / pier & Wall moment: Mud = 3,000 k-ft / pierM/Vd = 1.5
Vce = 104 k => DCR = 150/104 = 1.4 < m = 2 ok for shear
Mce = 900 k-ft => DCR = 3,000/900 = 3.3
fae/fme = 0.01 L/heff = 0.75gfy/fme = 0.001*50/2 = 0.025g = v + h = 2*0.2/(48*8) = 0.001
m = 4.8 from interpolation
Force-controlled actions
Revised design force equations
Equivalency to ASCE 7-10
ka factor used to account for diaphragm flexibility
1.0 ≤ ka ≤ 2.0
kh factor used to account for force distribution over height ofbuildings with rigid diaphragms
Out‐of‐Plane Strength
Diaphragm Anchorage
Out-of-Plane Wall Strength & Anchorage
Force-controlled actions
Revised design force equations
Equivalency to ASCE 7-10
ka factor used to account for diaphragm flexibility
1.0 ≤ ka ≤ 2.0
kh factor used to account for force distribution over height ofbuildings with rigid diaphragms
Out‐of‐Plane Strength
Diaphragm Anchorage
Out-of-Plane Wall Strength & Anchorage
Currently out-of-plane in ASCE 41 is more conservative than ASCE 7 for CP in the BSE-2N. will be changing in ASCE 41-17 to address this.
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Wall Out-of-Plane Evaluation
Check wall for out-of-plane forces
Fp = 0.4*1.0*1.65*80 = 53 psf
Muf = 53*252/8 = 4.1 k-ft/ft (excluding p-delta)
Muf = 5.0 k-ft/ft (including p-delta)
Mcl = 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
Fp = 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
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)
Unreinforced Masonry
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ASCE 41-13 Concrete ProvisionsMasonry ProvisionRobert Pekelnicky, PE, SE
Principal, Degenkolb EngineersChair, ASCE 41 Committee*
*The view expressed represent those of the author, not the standard’s committee as a whole.