PCI 6th Edition Headed Concrete Anchors (HCA) Presentation

PCI 6th Edition Headed Concrete Anchors (HCA) Presentation

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 for Headed 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 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 = nAsefut Where

= 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 fut = 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

Capacity 5% Failures 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 , associated with the 5 percent fractile ()

Based on sample population,n number of tests x the sample mean is the standard deviation of the sample set The 5% fractile Example values of based on sample size are: n= = 1.645 n = 40 = 2.010

n = 10 = 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 Pullout Nph: function of bearing on the head of the stud Side-Face blowout Nsb: studs cannot be closer to an edge than 40% the effective height of the studs Concrete Tension Strength

Ncb: Breakout Tn = Minimum of Nph: Pullout Nsb: Side-Face blowout Concrete Breakout Strength Ncb Ncbg Cbs AN Ccrb ed,N 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 Cbs 3.33 f 'c hef 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 ANo 2 1.5 hef 2 1.5 hef 9 hef 2 Side Edge Distance, Single Stud de1 < 1.5hef AN de1 1.5 hef 2 1.5 hef Side Edge Distance, Two Studs

de1 < 1.5hef AN de1 X 1.5 hef 2 1.5 hef Side and Bottom Edge Distance, Multi Row and Columns de1 < 1.5hef de2< 1.5hef AN de1 X 1.5 hef de2 Y 1.5 hef

Edge Distance Modification ed,N = modification for edge distance de,min ed,N 0.7 0.3 1.0 1.5 hef 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 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. ec,N 1 1.0

2 e'N 1 3 h ef Where: eN = eccentricity of the tensile force relative

the center of the stud group eN s/2 to Pullout Strength Nominal pullout strength Npn 11.2Abrg f 'c Ccrp Where 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 Side-Face Blowout Strength For a single headed stud located close to an edge (de1 < 0.4hef) Nsb 160de1 Abrg f 'c Where Nsb = Nominal side-face blowout strength de1 = Distance to closest edge

Abrg = Bearing area of head 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: de2 1 d

e1 4 Where: 1 de2 de1 3

Side-Face Blowout For multiple headed anchors located close to an edge (de1 < 0.4hef) Nsbg so 1 Nsb 6 de1

Where so = spacing of the outer anchors along the edge in the group Nsb = nominal side-face blowout strength for a single anchor previously defined 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) in. headed studs welded to in thick plate Nominal stud length = 8 in fc = 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 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 Depth hef L tpl tns 1

8 8 1 3 1 2 8 8 8in hef L tpl ths 1 " 8 3 1 8"

" " 1 " 2 8 8 8" Step 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 Factor de,min ed,N 0.7 0.3 1.0 1.5 hef

4in. .7 0.3 1.5 8in 0.8 Step 3 Breakout Strength f 'c 4000 Cbs 3.33 3.33 74.5lbs

hef 8 From design aid, case 4 Ncbg Cbs de1 X 1.5hef de3 Y 1.5hef ed,n Ccrb 0.8 0.75 74.5 4 16 12 6 8 12 1.0 1000 37.2kips

Step 3 Pullout Strength Abrg 0.79in2 4studs Npn (11.2) Abrg f 'c Ccrp 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 1000 Acr 1000 32 1.4 e Vu 37,000 1.20 3.4 Vu 37.2

Avf f y e 0.75 60 1.2 0.68in 2 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 SED

3.0 BED Corner Edge Shear Strength, Modified Vc3 SED 0.2 3.0 BED Side Edge Shear Strength, Vc1

SED 0.2 BED Front Edge Shear Strength Vc3 Vco3 Cx3 Ch3 Cev3 Cvcr Where Vco3 = Concrete breakout strength, single anchor Cx3 =X spacing coefficient Ch3 = Member thickness coefficient Cev3 = Eccentric shear force coefficient Cvcr = Member cracking coefficient

Single Anchor Strength 1.33 Vco3 16.5 f 'c BED Where: = lightweight concrete factor BED = distance from back row of studs to front edge

de3 y de3 Y X Spacing factor Cx3 X 0.85 nstuds back 3BED Where: X = Overall, out-to-out dimension of outermost

studs in back row of anchorage nstuds-back= Number of studs in back row Thickness Factor h for h 1.75BED BED Ch3 1 for h > 1.75BED Ch3 0.75 Where: h = Member thickness

Eccentricity Factor Cev3 1 X 1.0 when e'v e' 2

v 1 0.67 BED Where ev = Eccentricity of shear force on a group of anchors Cracked Concrete Factor Uncracked concrete Cvcr = 1.0

For cracked concrete, Cvcr = 0.70 no reinforcement or 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. Corner Shear Strength

A corner condition should be considered when: SED 0.2 3.0 BED where the Side Edge distance (SED) as shown Corner Shear Strength

Vc3 Vco3 Cc3 Ch3 Cev3 Cvcr Where: Ch3 = Member thickness coefficient Cev3 = Eccentric shear coefficient Cvcr = Member cracking coefficient Cc3 = Corner influence coefficient Corner factor SED Cc3 0.7

1.0 BED 3 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 Side Edge Shear Strength In this case, the shear force is applied parallel to the side edge, de1

SED 0.2 3.0 BED 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 Side Edge Shear Strength

Vc1 Vco1 CX1 CY1 Cev1 Cvcr Where: Vco1 = nominal concrete breakout strength for a single stud CX1 = X spacing coefficient CY1 = Y spacing coefficient Cev1 = Eccentric shear coefficient Single Anchor Strength 1.33

0.75 d Vco 87 f 'c de1 o Where: de1 = Distance from side stud to side edge (in.) do = Stud diameter (in.)

X Spacing Factor Cx1 nx x 2.5de1 2 nsides Cx1 1.0 when x = 0 Where: nx = Number of X-rows

x = Individual X-row spacing (in.) nsides =Number of edges or sides that influence the X direction 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 CY1 1.0 for ny 1 (one Y - row) CY1 n y Y

0.25 0.6 de1 0.15 ny for ny 1 Where: ny = Number of Y-rows Y = Out-to-out Y-row spacing (in) = y (in)

Eccentricity Factor Cev1 ev1 1.0 1.0 4 de1 Where: ev1 = Eccentricity form shear load to anchorage centroid

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) Vcp 215 y n f 'c (do )1.5 (hef )0.5 Where: Vcp = nominal pry-out shear strength (lbs)

y y y for 20 4 do d 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 of 1 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 condition Step 2 Calculate steel capacity Step 3 Front Edge Shear Strength Step 4 Calculate shear capacity coefficients Step 5 Calculate shear capacity Step 1 Check Corner Condition

SED 3 BED 48 4 3.25 12 4 Not a Corner Condition Step 2 Calculate Steel Capacity Vns = nsAnfut

= 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 Concrete Breakout Strength, Vco3 Vco3 16.5 f 'c BED

16.5 1 47.0kips 1.33 5000 12 4

1000 1.33 Step 4 Shear Capacity Coefficient X Spacing Coefficient, Cx3 X Cx3 0.85 nstuds back 3BED 4

0.85 0.93 316 0.93 Step 4 Shear Capacity Coefficient Member Thickness Coefficient, Ch3 Check if h 1.75BED 8 1.7516 OK h Ch3 0.75

BED 8 0.75 16 0.53 Step 4 Shear Capacity Coefficient Eccentric Shear Force Coefficient, Cev3 X 4 1.5 OK

2 2 1 1.0 e' 1 0.67 v BED Check if e'v Cev3

1 1.5 1 0.67 16 0.94 Step 4 Shear Capacity Coefficient Member Cracking Coefficient, Cvcr Assume uncracked region of member

Cvcr 1.0 #5 Perimeter Steel 0.75 Step 5 Shear Design Strength Vcs = Vco3Cx3Ch3Cev3Cvcr = 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 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 fc = 6000 psi normal weight concrete = 1.0 (8) 1/2 in diameter studs Ase = 0.20 in.2

Nominal stud length = 6 in. fut = 65,000 psi (Table 6.5.1.1) Vu = 25 kips Nu = 4 kips Column size: 18 in. x 18 in. Provide ties around vertical bars in the column to ensure confinement: = 0.75 Determine effective depth hef = L + tpl ths 1/8 in = 6 + 0.5 0.3125 0.125 = 6.06 in

Solution Steps Step 1 Determine applied loads Step 2 Determine tension design strength Step 3 Determine shear design strength Step 4 Interaction Equation Step 1 Determine applied loads Determine net Tension on Tension Stud Group

Determine net Shear on Shear Stud Group Nhu Vu e dc Nu 4

25 6 10 19.0kips Vu Vu 2 25

2 12.5kips Step 2 Concrete Tension Capacity Ncb Cbs AN Ccrb ed,N f 'c 6000 Cbs 3.33 3.33 1 104.8 hef

6.06 AN de1 X de2 Y 3hef 6 6 6 3 3 6.06 381.24 ed,N de,min 6 0.7 0.3 0.7 0.3 0.898 1.5hef 1.5 6.06

Ncb 0.75 381.24 104.8 0.898 1000 26.9kips Step 2 Steel Tension Capacity

Ns n Ase fut Ns 0.75 4 0.2 65 1000 39.0kips Step 2 Governing Tension

Ncb 26.9kips Ns 39.0kips Nn 26.9kips Step 3 Concrete Shear Capacity Vc1 Vco1 CX1 CY1 Cev1 Cvcr Vco 87 f 'c de1 87 1 6000 6 1.33

1.33 do 0.5 0.75 0.75 43.7kips

Cx1 2 CY1 n y Y

0.25 0.6 de1 0.25 2 3 0.15 0.6 6 0.15 0.58

Cev1 1.0 Cvcr 1.0 Vc1 0.75 43.7 2 0.58 1 1 38.0kips Step 3 Steel Shear Capacity Vs n Ase fut Vs 0.65 4 0.2 65

1000 33.8kips Step 3 Governing Shear Vc 38.0kips Vs 33.8kips Vn 33.8kips Step 4 Interaction Check if Interaction is required

If Vu 0.2 Vn Interaction is not Required 12.5 0.2 33.8 12.5 6.76 - Interaction Required

If Nhu 0.2 Nn Interaction is not Required 19 0.2 26.9 19 5.38 - Interaction Required Step 4 Interaction Nhu

Vu 19.0 12.5 0.71 0.37 1.08 1.2 Nn Vn 26.9 33.8 OR N hu

Nn 5 3 v u Vn 5

3 5 5 0.37 0.71 3

3 0.75 1.0 Questions?

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