9 Headed Anchor Design[1]

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PCI 6th

EditionPCI 6th

Edition

Headed Concrete Anchors (HCA)

Presentation OutinePresentation Outine

Research Background

Steel Capacity

Concrete Tension Capacity

Tension Example

Concrete Shear Capacity

Shear Example

Interaction Example

Background for

Headed Concrete Anchor Design

Background for

Anchorage to concrete and the design of

welded headed studs has undergone a

significant transformation since the Fifth

Edition of the Handbook.

³Concrete Capacity Design´ (CCD) approach

has been incorporated into ACI 318-02

Appendix D

Headed Concrete Anchor Design HistoryHeaded Concrete Anchor Design History

The shear capacity equations are based on

PCI sponsored research

The Tension capacity equations are based onthe ACI Appendix D equations only modified

for cracking and common PCI variable names

Background for

PCI sponsored an extensive research project,conducted by Wiss, Janney, Elstner

Associates, Inc., (WJE), to study design

criteria of headed stud groups loaded inshear and the combined effects of shear andtension

Section D.4.2 of ACI 318-02 specifically

permits alternate procedures, providing thetest results met a 5% fractile criteria

Supplemental ReinforcementSupplemental Reinforcement

Appendix D, Commentary

³« supplementary reinforcement in the direction of load, confining reinforcement, or both, can greatly

enhance the strength and ductility of the anchor connection.´

³Reinforcement oriented in the direction of load andproportioned to resist the total load within thebreakout prism, and fully anchored on both side of the breakout planes, may be provided instead of calculating breakout capacity.´

HCA Design PrinciplesHCA Design Principles

Performance based on the location of thestud relative to the member edges

Shear design capacity can be increased withconfinement reinforcement

In tension, ductility can be provided byreinforcement that crosses the potentialfailure surfaces

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 inunconfined concrete

Where feasible, connection failure should bedefined 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 thestud

Thicker plates may be required for bendingresistance or to ensure a more uniform loaddistribution to the attached studs

Stainless Steel StudsStainless Steel Studs

Can be welded to either stainless steel or mild carbon steel

Fully annealed stainless steel studs are

recommended when welding stainless steelstuds to a mild carbon steel base metal

Annealed stud use has been shown to beimperative for stainless steel studs welded to

carbon steel plates subject to repetitive or cyclic loads

Stud DimensionsStud Dimensions

Table 6.5.1.2

Steel CapacitySteel 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 F ut

and not F y

Steel CapacitySteel Capacity

JVs = JNs = J·n·Ase·f ut

J = steel strength reduction factor

= 0.65 (shear)= 0.75 (tension)

Vs = nominal shear strength steel capacity

Ns = nominal tensile strength steel capacity

n = number of headed studs in group Ase = nominal area of the headed stud shank

f ut = ultimate tensile strength of the stud steel

Material PropertiesMaterial Properties

Adapted from AWS D1.1-02

Table 6.5.1.1 page 6-11

Concrete CapacityConcrete Capacity

ACI 318-02, Appendix D, ³ Anchoring toConcrete´

Cover many types of anchors

In general results in more conservativedesigns than those shown in previouseditions of this handbook

Cracked ConcreteCracked 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% fractileThe 5% fractile

ACI 318-02, Section D.4.2 states, in part:

³«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 allowedto fall below

expected

5% Failures

Capacity

Test strength

This allows us to say with 90 percentconfidence that 95 percent of the test actualstrengths exceed the equation thus derived

Determination of the coefficient , associatedwith the 5 percent fractile () ± Based on sample population,n number of tests

± x the sample mean

± is the standard deviation of the sample set

Example values of based on sample sizeare:

n = = 1.645n = 40 = 2.010

n = 10 = 2.568

Strength Reduction FactorStrength Reduction Factor

Function of supplied confinement reinforcement

J= 0.75 with reinforcement

J = 0.70 with out reinforcement

Notation DefinitionsNotation Definitions

± de1, de2, de3, de4

Stud Layout

± x1, x2,«

± y1, y2, «

± X, Y

Critical Dimensions

± BED, SED

Concrete Tension FailureModesConcrete Tension FailureModes

Design tensile strength is the minimum of the

following modes:

± Breakout

JNcb: usually the most critical failure mode

± Pullout

JNph: function of bearing on the head of the stud

± Side-Face blowout

JNsb: studs cannot be closer to an edge than 40% theeffective height of the studs

Concrete Tension StrengthConcrete Tension Strength

JNcb: Breakout

: Pullout

JNsb: Side-Face blowout

JTn = Minimum of

Concrete Breakout StrengthConcrete Breakout Strength

Where:

Ccrb = Cracked concrete factor, 1 uncracked, 0.8 Cracked

AN = Projected surface area for a stud or group

=ed,N =Modification for edge distance

Cbs = Breakout strength coefficient

cbg! C

! 3.33 P f 'c

Effective Embedment DepthEffective Embedment Depth

hef = effective embedment depth

For headed studs welded to a plate

flush with the surface, it is the nominallength 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 thePCI handbook

Critical edgedistance is 1.5hef

No Edge Distance RestrictionsNo Edge Distance Restrictions

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

No ef ef ef A 2 1.5 h 2 1.5 h 9 h« » « »! ! - ½ - ½

Side Edge Distance, Single StudSide Edge Distance, Single Stud

de1 < 1.5hef

N e1 ef ef A d 1.5 h 2 1.5 h!

Side Edge Distance, Two StudsSide Edge Distance, Two Studs

de1 < 1.5hef

N e1 ef ef A d X 1.5 h 2 1.5 h!

Side and Bottom Edge Distance,

Multi Row and Columns

Side and Bottom Edge Distance,

Multi Row and Columns

de1 < 1.5hef

de2< 1.5hef

N e1 ef e2 ef A d X 1.5 h d Y 1.5 h!

Edge DistanceModificationEdge DistanceModification

=ed,N = modification for edge distance

de,min = minimum edge distance, top, bottom, andsides

PCI also provides tables to directly calculate JNcb, butCbs , Ccrb, and =ed,N must still be determined for the in

situ condition

d0.7 0.3 1.0

¨ ¸] ! e© ¹ª º

Determine Breakout Strength, J N cb

The PCI handbook

provides a design

guide to determine

the breakout area

Determine Breakout Strength, J N cb

First find the edge

condition that

corresponds to the

design condition

Eccentrically LoadedEccentrically Loaded

When the load application cannot be logicallyassumed concentric.

Where:

e N = eccentricity of the tensile force relative

to the center of the stud groupe N s/2

2e '1 3 h

] ! e¨ ¸

Pullout StrengthPullout Strength

Nominal pullout strength

Abrg = bearing area of the stud head

= area of the head ± area of the shank

Ccrp = cracking coefficient (pullout)

= 1.0 uncracked= 0.7 cracked

pn! 11.2 A

brg f '

Side-Face Blowout StrengthSide-Face Blowout Strength

For a single headed stud located close to anedge (de1 < 0.4hef )

Nsb = Nominal side-face blowout strength

de1 = Distance to closest edge Abrg = Bearing area of head

Nsb ! 160 de1 Abrg f 'c

Side-Face Blowout StrengthSide-Face Blowout Strength

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

Where:

¨ ¸© ¹

Side-Face BlowoutSide-Face Blowout

For multiple headed anchors located close to anedge (de1 < 0.4hef )

so = spacing of the outer anchors along the

edge in the groupNsb = nominal side-face blowout strength for

a single anchor previously defined

sbg sbe1

sN 1 N

¨ ¸!

Example: Stud Group TensionExample: 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. J headed studs welded to ½ in thick plate

Nominal stud length = 8 in

f c = 4000 psi (normal weight concrete)

f y = 60,000 psi

Example: Stud Group TensionExample: Stud Group Tension

Probl em:

Determine the design

tension strength of the

stud group

Solution StepsSolution Steps

Step 1 ± Determine effective depth

Step 2 ± Check for edge effect

Step 3 ± Check concrete strength of stud group

Step 4 ± Check steel strength of stud group

Step 5 ± Determine tension capacity

Step 6 ± Check confinement steel

Step 1 ± Effective DepthStep 1 ± Effective Depth

ef pl hs1

h L t t "8

31 18" " " "2 8 8

! L t pl

! 8 12

Step 2 ± Check for Edge EffectStep 2 ± Check for Edge Effect

Design aid, Case 4

X = 16 in.

Y = 8 in.

de1 = 4 in.

de3 = 6 in.

de1 and de3 > 1.5hef = 12 in.

Edge effects apply

de,min = 4 in.

Step 2 ± Edge FactorStep 2 ± Edge Factor

d0.7 0.3 1.0

4in..7 0.3

1.5 8in

¨ ¸] ! e© ¹ª º

¨ ¸! © ¹ª º!

Step 3 ± Breakout StrengthStep 3 ± Breakout Strength

cbg bs e1 ef ef ed,n crb

f ' 4000C 3.33 3.33 74.5lbs

From design aid, case 4

N C d X 1.5h de3 Y 1.5h C

0.80.75 74.5 4 16 12 6 8 12 1.0

37.2kips

¨ ¸! © ¹

ª º!

Step 3 ± Pullout StrengthStep 3 ± Pullout Strength

! 0.79in2 4studs

! J (11.2) Abrg

f 'c C

! 0.7(11.2)(3.16)(4)(1.0)

! 99.1kips

Step 3 ± Side-Face Blowout StrengthStep 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 StrengthStep 4 ± Steel Strength

! J n Ase

! 0.75(4)(0.44)(65)

! 85.8kips

Step 5 ± Tension CapacityStep 5 ± Tension Capacity

The controlling tension capacity for the stud

group is Breakout Strength

cbg! 37.2kips

Step 6 ± Check Confinement SteelStep 6 ± Check Confinement Steel

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

1000 32 1.41000

37,000 1.20 3.4

0.75 60 1.2

Step 6 ± Confinement SteelStep 6 ± Confinement Steel

Use 2 - #6 L-bar

around stud group.

These bars should

extend ld past thebreakout surface.

Concrete Shear StrengthConcrete Shear Strength

The design shear strength governed byconcrete failure is based on the testing

The in-place strength should be taken as theminimum value based on computing both theconcrete and steel

J Vc(failure mode)

! J Vco(failure mode)

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

cracking influence

Front Edge Shear Strength, Vc3Front Edge Shear Strength, Vc3

BED" 3.0

Corner Edge Shear Strength, Modified Vc3Corner Edge Shear Strength, Modified Vc3

0.2 e SED

BEDe 3.0

Side Edge Shear Strength, Vc1Side Edge Shear Strength, Vc1

SEDBED

Front Edge Shear StrengthFront Edge Shear Strength

Vco3 = Concrete breakout strength, single anchor Cx3 =X spacing coefficient

Ch3 = Member thickness coefficient

Cev3 = Eccentric shear force coefficient

Cvcr = Member cracking coefficient

c3! J V

Single Anchor StrengthSingle Anchor Strength

Where:

= lightweight concrete factor BED = distance from back row of studs to

front edge

co3! 16.5 P f '

c BED 1.33

! de3 y§ ! de3 Y

Thickness FactorThickness Factor

Where:

h = Member thickness

! 0.75h

BEDfor h e 1.75 BED

! 1 for h > 1.75 BED

i ii i

Eccentricity FactorEccentricity Factor

e v = Eccentricity of shear force on a group of

anchors

1 0.67 e '

e 1.0 when e 'v

C k d C FC k d C F

Cracked Concrete FactorCracked Concrete Factor

Uncracked concreteCvcr = 1.0

For cracked concrete,

Cvcr = 0.70 no reinforcement

reinforcement < No. 4 bar

= 0.85 reinforcement No. 4 bar

= 1.0 reinforcement. No. 4 bar and

confined within stirrups with a

spacing 4 in.

C Sh S hC Sh S h

Corner Shear StrengthCorner Shear Strength

A corner condition should

be considered when:

where the Side Edge

distance (SED) as

BED e 3.0

C Sh St thC Sh St th

Corner Shear StrengthCorner Shear Strength

Where:

Ch3 = Member thickness coefficient

Cev3 = Eccentric shear coefficient

Cvcr = Member cracking coefficient

Cc3 = Corner influence coefficient

c3! J V

C f tC f t

Corner factorCorner factor

For the special case of a large X-spacing studanchorage located near a corner, such thatSED/BED > 3, a corner failure may still result,

if de1 2.5BED

c3! 0.7

BED3 e 1.0

Sid Ed Sh St thSid Ed Sh St th

Side Edge Shear StrengthSide Edge Shear Strength

In this case, the shear force is applied parallelto the side edge, de1

Research determined that the corner influencecan be quite large, especially in thin panels

If the above ratio is close to the 0.2 value, it isrecommended that a corner breakout conditionbe investigated, as it may still control for largeBED values

0.2 eSED

BEDe 3.0

Sid Ed Sh St thSid Ed Sh St th

Side Edge Shear StrengthSide Edge Shear Strength

c1! J V

Where:Vco1 = nominal concrete breakout strength for a

single studCX1 = X spacing coefficientCY1 = Y spacing coefficient

Cev1 = Eccentric shear coefficient

Si l A h St thSi l A h St th

Single Anchor StrengthSingle Anchor Strength

Where:

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

do = Stud diameter (in.)

co! 87 P f '

e1 1.33

do 0.75

X S i F tX S i F t

X Spacing FactorX Spacing Factor

Where:

nx = Number of X-rows

x = Individual X-row spacing (in.)

nsides =Number of edges or sides that influencethe X direction

! nx x

2.5 de1

2 nsides

! 1.0 when x = 0

X S i F tX S i F t

X Spacing FactorX Spacing Factor

For all multiple Y-row anchorages locatedadjacent to two parallel edges, such as acolumn corbel connection, the X-spacing for

two or more studs in the row:

Cx1 = nx

Y Spacing FactorY Spacing Factor

Where:

ny = Number of Y-rows

Y = Out-to-out Y-row spacing (in) = 7y (in)

Y 1 y y

C 1.0 for n 1 (one Y - row)

n Y C 0.15 n for n 1

Where:

ev1 = Eccentricity form shear load to

anchorage centroid

eC 1.0 1.0

¨ ¸! e© ¹ª º

Back Edge Shear StrengthBack Edge Shear Strength

Under a condition of pure shear theback edge has been found throughtesting to have no influence on the

group capacity Proper concrete clear cover from the

studs to the edge must be maintained

³In the Field´ Shear Strength³In the Field´ Shear Strength

When a headed stud anchorage is sufficiently

away from all edges, termed ³in-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

For hef /de 4.5 (in normal weight concrete)

Where:

Vcp = nominal pry-out shear strength (lbs)

cp! J 215 ]

y n f '

] y ! y

for yd

Front Edge Failure ExampleFront Edge Failure Example

Given:

Plate with headed studs as shown, placed in a positionwhere cracking is unlikely. The 8 in. thick panel has a28-day concrete strength of 5000 psi. The plate isloaded with an

eccentricity of

1 ½ in from the

centerline. The

panel has #5

confinement bars.

ExampleExample

Probl em:

Determine the design shear strength of

the stud group.

Step 1 ± Check corner condition

Step 2 ± Calculate steel capacity

Step 3 ± Front Edge Shear StrengthStep 4 ± Calculate shear capacity coefficients

Step 5 ± Calculate shear capacity

Step 1 Check Corner ConditionStep 1 Check Corner Condition

Step 1 ± Check Corner ConditionStep 1 ± Check Corner Condition

Not a Corner Condition

BEDu 3

48 412 4

! 3.25

Step 2 Calculate Steel CapacityStep 2 Calculate Steel Capacity

Step 2 ± Calculate Steel CapacityStep 2 ± Calculate Steel Capacity

JVns = J·ns·An·f ut

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

Step 3 ± Front Edge Shear StrengthStep 3 ± Front Edge Shear Strength

Front Edge Shear Strength

c3! J V

Step 4 ± Shear Capacity CoefficientStep 4 ± Shear Capacity Coefficient

1.33 V 16.5 f ' BED

16.5 1 5000 12 4

47.0kips

Concrete Breakout Strength, Vco3

Step 4 Shear Capacity CoefficientStep 4 Shear Capacity Coefficient

Cx3 ! 0.85 X

3 BED e nstudsback

! 0.85 4

3 16! 0.93 e

! 0.93

X Spacing Coefficient, Cx3

Check if h e 1.75 BED

8 e 1.75 16 OK

! 0.75h

! 0.75

! 0.53

Member Thickness Coefficient, Ch3

Check if e 'v

21.5 e

Cev3 ! 1

1 0.67 e '

1 0.67 1.5

ª© ¸

! 0.94

Eccentric Shear Force Coefficient, Cev3

Member Cracking Coefficient, Cvcr

± Assume uncracked region of member

#5 Perimeter Steel

vcr! 1.0

J ! 0.75

Step 5 ± Shear Design StrengthStep 5 ± Shear Design Strength

Step 5 Shear Design StrengthStep 5 Shear Design Strength

JVcs = J·Vco3·Cx3·Ch3·Cev3·Cvcr

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

= 16.3 kips

InteractionInteraction

Trilinear Solution

Unity curve with a 5/3 exponent

Interaction CurvesInteraction Curves

Combined Loading ExampleCombined Loading Example

Given:

A ½ in thick plate withheaded studs for attachment of a steel

bracket to a column asshown at the right

Probl em:

Determine if the studsare adequate for theconnection

Example ParametersExample Parameters

f c = 6000 psi normal weight concrete = 1.0

(8) ± 1/2 in diameter studs

Ase = 0.20 in.

Nominal stud length = 6 in.

f ut = 65,000 psi (Table 6.5.1.1)

Vu = 25 kips

Nu = 4 kipsColumn size: 18 in. x 18 in.

Provide ties around vertical bars in thecolumn to ensure confinement: J = 0.75

Determine effective depth

hef = L + tpl ± ths ±1/8 in

= 6 + 0.5 ± 0.3125 ± 0.125 = 6.06 in

Step 1 ± Determine applied loads

Step 2 ± Determine tension design

strengthStep 3 ± Determine shear design strength

Step 4 ± Interaction Equation

Step 1 ± Determine applied loadsStep 1 ± Determine applied loads

p ppp pp

Determine netTension on Tension

Stud Group

Determine net Shear

on Shear StudGroup

25 6 10

! 19.0kips

! 12.5kips

Step 2 ± Concrete Tension CapacityStep 2 ± Concrete Tension Capacity

p p yp p y

cb bs N crb ed,N

N e1 e2 ef

N C A C

f ' 6000C 3.33 3.33 1 104.8

h 6.06

A d X d Y 3h 6 6 6 3 3 6.06 381.24

d 60.7 0.3 0.7 0.3 0.898

1.5h 1.5 6.06

0.75 381.24 104.8 0.898N 26.9kips1000

J ! J ]

! P ! !

] ! ! !

Step 2 ± Steel Tension CapacityStep 2 ± Steel Tension Capacity

p p yp p y

s se ut

N n A f

0.75 4 0.2 65N 39.0kips

Step 2 ± Governing TensionStep 2 ± Governing Tension

p gp g

N 26.9kips N 39.0kips

N 26.9kips

J ! J !

Step 3 ± Concrete Shear CapacityStep 3 ± Concrete Shear Capacity

p p yp p y

c1 co1 X 1 Y 1 ev1 vcr

1.33 0.75

co c e1 o

1.33 0.75

0.25 0.25

V V C C C C

V 87 f ' d d

87 1 6000 6 0.5 43.7kips

n Y 2 3C 0.15 0.15 0.58

0.6 d 0.6 6

V 0.75 43.7 2 0.58 1 1 38.0kips

« »- ½! ! !

Step 3 ± Steel Shear CapacityStep 3 ± Steel Shear Capacity

p p yp p y

s se ut

V n A f

0.65 4 0.2 65 V 33.8kips1000

Step 3 ± Governing ShearStep 3 ± Governing Shear

p gp g

V 38.0kips V 33.8kips

V 33.8kips

J ! J !

Step 4 ± InteractionStep 4 ± Interaction

Check if Interaction is required

0.2 J Vn Interaction is not Required

12.5 0.2 33.8

12.5 " 6.76 - Interaction Required

If Nhu

0.2 JNn Interaction is not Required

19 0.2 26.9 19 " 5.38 - Interaction Required

Step 4 ± InteractionStep 4 ± Interaction

33.8! 0.71 0.37 ! 1.08 e 1.2

! 0.71 53 0.37

53 ! 0.75 e 1.0

Questions?Questions?

9 Headed Anchor Design[1]

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