PCI 6th Edition

Headed Concrete Anchors (HCA)

Presentation Outine

• Research Background• Steel Capacity• Concrete Tension Capacity• Tension Example• Concrete Shear Capacity• Shear Example• Interaction Example

Background for Headed Concrete Anchor Design

• Anchorage to concrete and the design of welded headed studs has undergone a significant transformation since the Fifth Edition of the Handbook.

• “Concrete Capacity Design” (CCD) approach has been incorporated into ACI 318-02 Appendix D

Headed Concrete Anchor Design History

• The shear capacity equations are based on PCI sponsored research

• The Tension capacity equations are based on the ACI Appendix D equations only modified for cracking and common PCI variable names

Background forHeaded Concrete Anchor Design

• PCI sponsored an extensive research project, conducted by Wiss, Janney, Elstner Associates, Inc., (WJE), to study design criteria of headed stud groups loaded in shear and the combined effects of shear and tension

• Section D.4.2 of ACI 318-02 specifically permits alternate procedures, providing the test results met a 5% fractile criteria

Supplemental Reinforcement

Appendix D, Commentary

“… supplementary reinforcement in the direction of load, confining reinforcement, or both, can greatly enhance the strength and ductility of the anchor connection.”

“Reinforcement oriented in the direction of load and proportioned to resist the total load within the breakout prism, and fully anchored on both side of the breakout planes, may be provided instead of calculating breakout capacity.”

HCA Design Principles

• Performance based on the location of the stud relative to the member edges

• Shear design capacity can be increased with confinement reinforcement

• In tension, ductility can be provided by reinforcement that crosses the potential failure surfaces

HCA Design Principles

• Designed to resist – Tension– Shear– Interaction of the two

• The design equations are applicable to studs which are welded to steel plates or other structural members and embedded in unconfined concrete

HCA Design Principles

• Where feasible, connection failure should be defined as yielding of the stud material

• The groups strength is taken as the smaller of either the concrete or steel capacity

• The minimum plate thickness to which studs are attached should be ½ the diameter of the stud

• Thicker plates may be required for bending resistance or to ensure a more uniform load distribution to the attached studs

Stainless Steel Studs

• Can be welded to either stainless steel or mild carbon steel

• Fully annealed stainless steel studs are recommended when welding stainless steel studs to a mild carbon steel base metal

• Annealed stud use has been shown to be imperative for stainless steel studs welded to carbon steel plates subject to repetitive or cyclic loads

Stud Dimensions

• Table 6.5.1.2• Page 6-12

Steel Capacity

• Both Shear and Tension governed by same basic equation

• Strength reduction factor is a function of shear or tension

• The ultimate strength is based on Fut and not Fy

Steel Capacity

Vs = Ns = ·n·Ase·fut

Where = steel strength reduction factor

= 0.65 (shear)= 0.75 (tension)Vs = nominal shear strength steel capacityNs = nominal tensile strength steel capacityn = number of headed studs in groupAse = nominal area of the headed stud shankfut = ultimate tensile strength of the stud steel

Material Properties

• Adapted from AWS D1.1-02• Table 6.5.1.1 page 6-11

Concrete Capacity

• ACI 318-02, Appendix D, “Anchoring to Concrete”

• Cover many types of anchors• In general results in more conservative

designs than those shown in previous editions of this handbook

Cracked Concrete

• ACI assumes concrete is cracked• PCI assumes concrete is cracked• All equations contain adjustment factors for cracked

and un-cracked concrete• Typical un-cracked regions of members

– Flexural compression zone– Column or other compression members– Typical precast concrete

• Typical cracked regions of members– Flexural tension zones– Potential of cracks during handling

The 5% fractile

• ACI 318-02, Section D.4.2 states, in part: “…The nominal strength shall be based on the 5

percent fractile of the basic individual anchor strength…”

• Statistical concept that, simply stated,– if a design equation

is based on tests, 5 percent of the tests are allowed to fall below expected

5% FailuresCapacity

Test strength

The 5% fractile

• This allows us to say with 90 percent confidence that 95 percent of the test actual strengths exceed the equation thus derived

• Determination of the coefficient κ, associated with the 5 percent fractile (κσ) – Based on sample population,n number of tests– x the sample mean– σ is the standard deviation of the sample set

The 5% fractile

• Example values of κ based on sample size are:

n = ∞ κ = 1.645 n = 40 κ = 2.010 n = 10 κ = 2.568

Strength Reduction Factor

Function of supplied confinement reinforcement

= 0.75 with reinforcement = 0.70 with out reinforcement

Notation Definitions

• Edges– de1, de2, de3, de4

• Stud Layout– x1, x2, …– y1, y2, …– X, Y

• Critical Dimensions– BED, SED

Concrete Tension Failure Modes

• Design tensile strength is the minimum of the following modes:– Breakout

Ncb: usually the most critical failure mode

– PulloutNph: function of bearing on the head of the stud

– Side-Face blowoutNsb: studs cannot be closer to an edge than 40% the

effective height of the studs

Concrete Tension Strength

Ncb: Breakout

Nph: Pullout

Nsb: Side-Face blowout

Tn = Minimum of

Concrete Breakout Strength

Where:Ccrb = Cracked concrete factor, 1 uncracked, 0.8 CrackedAN = Projected surface area for a stud or grouped,N =Modification for edge distanceCbs = Breakout strength coefficient

Ncb Ncbg Cbs AN Ccrb ed,N

Cbs 3.33

f 'chef

Effective Embedment Depth

• hef = effective embedment depth• For headed studs welded to a plate flush

with the surface, it is the nominal length less the head thickness, plus the plate thickness (if fully recessed), deducting the stud burnoff lost during the welding process about 1/8 in.

Projected Surface Area, An

• Based on 35o

• AN - calculated, or empirical equations are provided in the PCI handbook

• Critical edge distance is 1.5hef

No Edge Distance Restrictions

• For a single stud, with de,min > 1.5hef

2No ef ef efA 2 1.5 h 2 1.5 h 9 h

Side Edge Distance, Single Stud

de1 < 1.5hef

N e1 ef efA d 1.5 h 2 1.5 h

Side Edge Distance, Two Studs

de1 < 1.5hef

N e1 ef efA d X 1.5 h 2 1.5 h

Side and Bottom Edge Distance, Multi Row and Columns

de1 < 1.5hef

de2< 1.5hef

N e1 ef e2 efA d X 1.5 h d Y 1.5 h

Edge Distance Modification

• ed,N = modification for edge distance

• de,min = minimum edge distance, top, bottom, and sides

• PCI also provides tables to directly calculate Ncb, but Cbs , Ccrb, and ed,N must still be determined for the in situ condition

e,mined,N

ef

d0.7 0.3 1.01.5 h

Determine Breakout Strength, Ncb

• The PCI handbook provides a design guide to determine the breakout area

Determine Breakout Strength, Ncb

• First find the edge condition that corresponds to the design condition

Eccentrically Loaded

• When the load application cannot be logically assumed concentric.

Where:e′N = eccentricity of the tensile force relative to the center of the stud groupe′N ≤ s/2

ec,NN

ef

1 1.02 e'1 3 h

Pullout Strength

• Nominal pullout strength

WhereAbrg = bearing area of the stud head = area of the head – area of the shankCcrp = cracking coefficient (pullout)

= 1.0 uncracked = 0.7 cracked

Npn 11.2Abrg f 'cCcrp

Side-Face Blowout Strength

• For a single headed stud located close to an edge (de1 < 0.4hef)

WhereNsb = Nominal side-face blowout strengthde1 = Distance to closest edgeAbrg = Bearing area of head

Nsb 160de1 Abrg f 'c

Side-Face Blowout Strength

• If the single headed stud is located at a perpendicular distance, de2, less then 3de1 from an edge, Nsb, is multiplied by:

Where:

e2

e1

d1 d4

1

de2de1

3

Side-Face Blowout

• For multiple headed anchors located close to an edge (de1 < 0.4hef)

Whereso = spacing of the outer anchors along the edge in the groupNsb = nominal side-face blowout strength for a single anchor previously defined

osbg sb

e1

sN 1 N6 d

Example: Stud Group Tension

Given:A flush-mounted base plate with four headed studs embedded in a corner of a 24 in. thick foundation slab(4) ¾ in. headed studs welded to ½ in thick plateNominal stud length = 8 inf′c = 4000 psi (normal weight concrete)

fy = 60,000 psi

Example: Stud Group Tension

Problem:Determine the design tension strength of the stud group

Solution Steps

Step 1 – Determine effective depthStep 2 – Check for edge effectStep 3 – Check concrete strength of stud groupStep 4 – Check steel strength of stud group

Step 5 – Determine tension capacityStep 6 – Check confinement steel

Step 1 – Effective Depth

ef pl hs1h L t t "8

31 1 8" " " "2 8 8 8"

hef L tpl tns 18

8 12 3

8 18

8in

Step 2 – Check for Edge Effect

Design aid, Case 4X = 16 in.Y = 8 in.de1 = 4 in.de3 = 6 in.de1 and de3 > 1.5hef = 12 in.

Edge effects applyde,min = 4 in.

Step 2 – Edge Factor

e,mined,N

ef

d0.7 0.3 1.01.5 h4in. .7 0.3 1.5 8in

0.8

Step 3 – Breakout Strength

cbs

ef

cbg bs e1 ef ef ed,n crb

f ' 4000C 3.33 3.33 74.5lbsh 8

From design aid, case 4N C d X 1.5h de3 Y 1.5h C

0.8 0.75 74.5 4 16 12 6 8 12 1.01000 37.2kips

Step 3 – Pullout Strength

Abrg 0.79in2 4studs

Npn (11.2)Abrg f 'cCcrp

0.7(11.2)(3.16)(4)(1.0) 99.1kips

Step 3 – Side-Face Blowout Strength

de,min = 4 in. > 0.4hef

= 4 in. > 0.4(8) = 3.2 in.

Therefore, it is not critical

Step 4 – Steel Strength

Ns nAse fut

0.75(4)(0.44)(65) 85.8kips

Step 5 – Tension Capacity

The controlling tension capacity for the stud group is Breakout Strength

Tn Ncbg 37.2kips

Step 6 – Check Confinement Steel

• Crack plane area = 4 in. x 8 in. = 32 in.2

2

1000 32 1.4100037,000

1.20 3.437.2

0.75 60 1.2 0.68

cre

u

uvf

y e

AV

VAf

in

Step 6 – Confinement Steel

Use 2 - #6 L-bar around stud group.

These bars should extend ld past the breakout surface.

Concrete Shear Strength

• The design shear strength governed by concrete failure is based on the testing

• The in-place strength should be taken as the minimum value based on computing both the concrete and steel

Vc(failure mode) Vco(failure mode) C

Vco(failure mode) anchor strength Cx(failure mode) x spacing influence Cy(failure mode) y spacing influence Ch(failure mode) thickness influence Cev(failure mode) eccentricity influence Cc(failure mode) corner influence Cvcr cracking influence

Front Edge Shear Strength, Vc3

SEDBED

3.0

Corner Edge Shear Strength, Modified Vc3

0.2

SEDBED 3.0

Side Edge Shear Strength, Vc1

SEDBED 0.2

Front Edge Shear Strength

WhereVco3 = Concrete breakout strength, single anchor

Cx3 =X spacing coefficient

Ch3 = Member thickness coefficient

Cev3 = Eccentric shear force coefficient

Cvcr = Member cracking coefficient

Vc3 Vco3 Cx3 Ch3 Cev3 Cvcr

Single Anchor Strength

Where:λ = lightweight concrete factorBED = distance from back row of studs to

front edge

Vco3 16.5 f 'c BED 1.33

de3 y de3 Y

X Spacing factor

Where:X = Overall, out-to-out dimension of outermoststuds in back row of anchoragenstuds-back= Number of studs in back row

Cx3 0.85

X3BED nstuds back

Thickness Factor

Where:h = Member thickness

Ch3 0.75 hBED for h 1.75BED

Ch3 1 for h > 1.75BED

Eccentricity Factor

Wheree′v = Eccentricity of shear force on a group of

anchors

Cev3 1

1 0.67e'vBED

1.0 when e'v X2

Cracked Concrete Factor

Uncracked concreteCvcr = 1.0

For cracked concrete,Cvcr = 0.70 no reinforcement

orreinforcement < No. 4 bar

= 0.85 reinforcement ≥ No. 4 bar= 1.0 reinforcement. ≥ No. 4 bar and confined within stirrups

with a spacing ≤ 4 in.

Corner Shear Strength

A corner condition shouldbe considered when:

where the Side Edge distance (SED) as shown

0.2

SEDBED 3.0

Corner Shear Strength

Where:Ch3 = Member thickness coefficient Cev3 = Eccentric shear coefficient Cvcr = Member cracking coefficientCc3 = Corner influence coefficient

Vc3 Vco3 Cc3 Ch3 Cev3 Cvcr

Corner factor

• For the special case of a large X-spacing stud anchorage located near a corner, such that SED/BED > 3, a corner failure may still result, if de1 ≤ 2.5BED

Cc3 0.7

SEDBED

3 1.0

Side Edge Shear Strength

• In this case, the shear force is applied parallel to the side edge, de1

• Research determined that the corner influence can be quite large, especially in thin panels

• If the above ratio is close to the 0.2 value, it is recommended that a corner breakout condition be investigated, as it may still control for large BED values

0.2

SEDBED 3.0

Side Edge Shear Strength

Vc1 Vco1 CX1 CY1 Cev1 Cvcr

Where:Vco1 = nominal concrete breakout strength for a

single studCX1 = X spacing coefficient CY1 = Y spacing coefficientCev1 = Eccentric shear coefficient

Single Anchor Strength

Where:de1 = Distance from side stud to side edge (in.)

do = Stud diameter (in.)

Vco 87 f 'c de1 1.33 do 0.75

X Spacing Factor

Where:nx = Number of X-rowsx = Individual X-row spacing (in.)nsides =Number of edges or sides that influence the X direction

Cx1 nx x

2.5de1

2 nsides

Cx1 1.0 when x = 0

X Spacing Factor

• For all multiple Y-row anchorages located adjacent to two parallel edges, such as a column corbel connection, the X-spacing for two or more studs in the row:

Cx1 = nx

Y Spacing Factor

Where:ny = Number of Y-rows

Y = Out-to-out Y-row spacing (in) = y (in)

Y1 y

0.25y

Y1 y ye1

C 1.0 for n 1 (one Y - row)n Y

C 0.15 n for n 10.6 d

Eccentricity Factor

Where:ev1 = Eccentricity form shear load to

anchorage centroid

v1ev1

e1

eC 1.0 1.04 d

Back Edge Shear Strength

• Under a condition of pure shear the back edge has been found through testing to have no influence on the group capacity

• Proper concrete clear cover from the studs to the edge must be maintained

“In the Field” Shear Strength

• When a headed stud anchorage is sufficiently away from all edges, termed “in-the-field” of the member, the anchorage strength will normally be governed by the steel strength

• Pry-out failure is a concrete breakout failure that may occur when short, stocky studs are used

“In the Field” Shear Strength

• For hef/de ≤ 4.5 (in normal weight concrete)

Where:Vcp = nominal pry-out shear strength (lbs)

Vcp 215y n f 'c (do)1.5 (hef )0.5

y

y4do

for yd

20

Front Edge Failure Example

Given:Plate with headed studs as shown, placed in a position where cracking is unlikely. The 8 in. thick panel has a 28-day concrete strength of 5000 psi. The plate is loaded with an eccentricity of1 ½ in from the centerline. The

panel has #5 confinement bars.

Example

Problem:Determine the design shear strength of the stud group.

Solution Steps

Step 1 – Check corner conditionStep 2 – Calculate steel capacityStep 3 – Front Edge Shear StrengthStep 4 – Calculate shear capacity coefficientsStep 5 – Calculate shear capacity

Step 1 – Check Corner Condition

Not a Corner Condition

SEDBED 3

48 412 4 3.25

Step 2 – Calculate Steel Capacity

Vns = ·ns·An·fut

= 0.65(4)(0.20)(65) = 33.8 kips

Step 3 – Front Edge Shear Strength

• Front Edge Shear Strength

Vc3 Vco3 Cx3 Ch3 Cev3 Cvcr

Step 4 – Shear Capacity Coefficient

1.33co3 c

1.33

V 16.5 f ' BED16.5 1 5000 12 4

1000 47.0kips

• Concrete Breakout Strength, Vco3

Step 4 – Shear Capacity Coefficient

Cx3 0.85X

3BED nstuds back

0.854

316 0.93

0.93

• X Spacing Coefficient, Cx3

Step 4 – Shear Capacity Coefficient

Check if h 1.75BED 8 1.7516 OK

Ch3 0.75 hBED

0.75 816

0.53

• Member Thickness Coefficient, Ch3

Step 4 – Shear Capacity Coefficient

Check if e'v X2 1.5

42 OK

Cev3 1

1 0.67e'vBED

1.0

1

1 0.671.516

0.94

•Eccentric Shear Force Coefficient, Cev3

Step 4 – Shear Capacity Coefficient

• Member Cracking Coefficient, Cvcr– Assume uncracked region of member

• #5 Perimeter Steel

Cvcr 1.0

0.75

Step 5 – Shear Design Strength

Vcs = ·Vco3·Cx3·Ch3·Cev3·Cvcr

= 0.75(47.0)(0.93)(0.53)(0.94)(1.0) = 16.3 kips

Interaction

• Trilinear Solution• Unity curve with a 5/3 exponent

Interaction Curves

Combined Loading Example

Given:A ½ in thick plate with headed studs for attachment of a steel bracket to a column as shown at the right

Problem:Determine if the studs are adequate for the connection

Example Parameters

f′c = 6000 psi normal weight concreteλ = 1.0(8) – 1/2 in diameter studsAse = 0.20 in.2 Nominal stud length = 6 in.fut = 65,000 psi (Table 6.5.1.1)Vu = 25 kipsNu = 4 kipsColumn size: 18 in. x 18 in.

• Provide ties around vertical bars in the column to ensure confinement: = 0.75

• Determine effective depthhef = L + tpl – ths – 1/8 in

= 6 + 0.5 – 0.3125 – 0.125 = 6.06 in

Solution Steps

Step 1 – Determine applied loadsStep 2 – Determine tension design

strengthStep 3 – Determine shear design strengthStep 4 – Interaction Equation

Step 1 – Determine applied loads

• Determine net Tension on Tension Stud Group

• Determine net Shear on Shear Stud Group

Nhu Vu edc

Nu

25 6

10 4

19.0kips

Vu Vu

2

252

12.5kips

Step 2 – Concrete Tension Capacity

cb bs N crb ed,N

cbs

ef

N e1 e2 ef

e,mined,N

ef

cb

N C A Cf ' 6000C 3.33 3.33 1 104.8h 6.06

A d X d Y 3h 6 6 6 3 3 6.06 381.24d 60.7 0.3 0.7 0.3 0.8981.5h 1.5 6.06

0.75 381.24 104.8 0.898N 26.9kips1000

Step 2 – Steel Tension Capacity

s se ut

s

N n A f0.75 4 0.2 65N 39.0kips1000

Step 2 – Governing Tension

cb s

n

N 26.9kips N 39.0kips

N 26.9kips

Step 3 – Concrete Shear Capacity

c1 co1 X1 Y1 ev1 vcr1.33 0.75

co c e1 o1.33 0.75

x10.25 0.25

yY1

e1

ev1

vcr

c1

V V C C C CV 87 f ' d d

87 1 6000 6 0.5 43.7kipsC 2

n Y 2 3C 0.15 0.15 0.580.6 d 0.6 6C 1.0C 1.0

V 0.75 43.7 2 0.58 1 1 38.0kips

Step 3 – Steel Shear Capacity

s se ut

s

V n A f0.65 4 0.2 65V 33.8kips1000

Step 3 – Governing Shear

c s

n

V 38.0kips V 33.8kips

V 33.8kips

Step 4 – Interaction

• Check if Interaction is required

If Vu 0.2 Vn Interaction is not Required12.5 0.2 33.8

12.5 6.76 - Interaction Required

If Nhu 0.2 Nn Interaction is not Required19 0.2 26.9

19 5.38 - Interaction Required

Step 4 – Interaction

Nhu

Nn

Vu

Vn

19.026.9

12.533.8 0.71 0.37 1.08 1.2

ORNhu

Nn

53

vu

Vn

53

0.71 53 0.37 5

3 0.75 1.0

Questions?