ASCE 41‐13 Hands‐On Approach Concrete & Masonry … Part 04-ASCE_41_Concrete... · ASCE 41‐13 Hands‐On Approach Concrete & Masonry Provisions SEAU 5thAnnual Education Conference

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



Rebar strengths based on value specified on the drawings, value tested, or default based on age/ASTM.

Reinforcing Steel Strengths


Reinforcing Steel Strengths


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



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



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


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



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.



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



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?



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



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



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



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



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



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



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.



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



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!



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!


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


P (k ip)

Mx (k-ft)

3500

-500

10000

fs=0.5fy

fs=0

(Pmax)

(Pmin)

fs=0.5fy

fs=0

1



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



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


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


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.




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


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



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



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.



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



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.

Documents

ASCE 41‐13 Hands‐On Approach Concrete & Masonry … Part 04-ASCE_41_Concrete... · ASCE 41‐13 Hands‐On Approach Concrete & Masonry Provisions SEAU 5thAnnual Education Conference