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PCI 6 th Edition Headed Concrete Anchors (HCA)

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Presentation Outine Research Background Steel Capacity Concrete Tension Capacity Tension Example Concrete Shear Capacity Shear Example Interaction Example

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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 Appendix D

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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

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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 specifically permits alternate procedures, providing the test results met a 5% fractile criteria

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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.

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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

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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

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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

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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

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Stud Dimensions Table Page 6-12

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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 F ut and not F y

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Steel Capacity V s = N s = ·n·A se ·f ut Where = steel strength reduction factor = 0.65 (shear) = 0.75 (tension) V s = nominal shear strength steel capacity N s = nominal tensile strength steel capacity n = number of headed studs in group A se = nominal area of the headed stud shank f ut = ultimate tensile strength of the stud steel

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Material Properties Adapted from AWS D Table page 6-11

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Concrete Capacity ACI , 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

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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

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The 5% fractile ACI , 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% Failures Capacity Test strength

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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

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The 5% fractile Example values of κ based on sample size are: n = κ = n = 40 κ = n = 10 κ = 2.568

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Strength Reduction Factor Function of supplied confinement reinforcement = 0.75 with reinforcement = 0.70 with out reinforcement

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Notation Definitions Edges –d e1, d e2, d e3, d e4 Stud Layout –x1, x2, … –y1, y2, … –X, Y Critical Dimensions –BED, SED

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Concrete Tension Failure Modes Design tensile strength is the minimum of the following modes: –Breakout N cb : usually the most critical failure mode –Pullout N ph : function of bearing on the head of the stud –Side-Face blowout N sb : studs cannot be closer to an edge than 40% the effective height of the studs

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Concrete Tension Strength N cb : Breakout N ph : Pullout N sb : Side-Face blowout T n = Minimum of

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Concrete Breakout Strength Where: C crb = Cracked concrete factor, 1 uncracked, 0.8 Cracked A N = Projected surface area for a stud or group ed,N =Modification for edge distance C bs = Breakout strength coefficient

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Effective Embedment Depth h ef = 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.

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Projected Surface Area, A n Based on 35 o A N - calculated, or empirical equations are provided in the PCI handbook Critical edge distance is 1.5h ef

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No Edge Distance Restrictions For a single stud, with d e,min > 1.5h ef

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Side Edge Distance, Single Stud d e1 < 1.5h ef

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Side Edge Distance, Two Studs d e1 < 1.5h ef

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Side and Bottom Edge Distance, Multi Row and Columns d e1 < 1.5h ef d e2 < 1.5h ef

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Edge Distance Modification ed,N = modification for edge distance d e,min = minimum edge distance, top, bottom, and sides PCI also provides tables to directly calculate N cb, but C bs, C crb, and ed,N must still be determined for the in situ condition

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Determine Breakout Strength, N cb The PCI handbook provides a design guide to determine the breakout area

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Determine Breakout Strength, N cb First find the edge condition that corresponds to the design condition

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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 group e N s/2

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Pullout Strength Nominal pullout strength Where A brg = bearing area of the stud head = area of the head – area of the shank C crp = cracking coefficient (pullout) = 1.0 uncracked = 0.7 cracked

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Side-Face Blowout Strength For a single headed stud located close to an edge (d e1 < 0.4h ef ) Where N sb = Nominal side-face blowout strength d e1 = Distance to closest edge A brg = Bearing area of head

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Side-Face Blowout Strength If the single headed stud is located at a perpendicular distance, d e2, less then 3d e1 from an edge, N sb, is multiplied by: Where:

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Side-Face Blowout For multiple headed anchors located close to an edge (d e1 < 0.4h ef ) Where s o = spacing of the outer anchors along the edge in the group N sb = nominal side-face blowout strength for a single anchor previously defined

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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 plate Nominal stud length = 8 in f c = 4000 psi (normal weight concrete) f y = 60,000 psi

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Example: Stud Group Tension Problem: Determine the design tension strength of the stud group

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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

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Step 1 – Effective Depth

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Step 2 – Check for Edge Effect Design aid, Case 4 X = 16 in. Y = 8 in. d e1 = 4 in. d e3 = 6 in. d e1 and d e3 > 1.5h ef = 12 in. Edge effects apply d e,min = 4 in.

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Step 2 – Edge Factor

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Step 3 – Breakout Strength

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Step 3 – Pullout Strength

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Step 3 – Side-Face Blowout Strength d e,min = 4 in. > 0.4h ef = 4 in. > 0.4(8) = 3.2 in. Therefore, it is not critical

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Step 4 – Steel Strength

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Step 5 – Tension Capacity The controlling tension capacity for the stud group is Breakout Strength

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Step 6 – Check Confinement Steel Crack plane area = 4 in. x 8 in. = 32 in. 2

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Step 6 – Confinement Steel Use 2 - #6 L-bar around stud group. These bars should extend l d past the breakout surface.

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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

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Front Edge Shear Strength, V c3

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Corner Edge Shear Strength, Modified V c3

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Side Edge Shear Strength, V c1

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Front Edge Shear Strength Where V co3 = Concrete breakout strength, single anchor C x3 =X spacing coefficient C h3 = Member thickness coefficient C ev3 = Eccentric shear force coefficient C vcr = Member cracking coefficient

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Single Anchor Strength Where: λ = lightweight concrete factor BED = distance from back row of studs to front edge

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X Spacing factor Where: X = Overall, out-to-out dimension of outermost studs in back row of anchorage n studs-back = Number of studs in back row

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Thickness Factor Where: h = Member thickness

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Eccentricity Factor Where e v = Eccentricity of shear force on a group of anchors

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Cracked Concrete Factor Uncracked concrete C vcr = 1.0 For cracked concrete, C vcr = 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.

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

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Corner Shear Strength Where: C h3 = Member thickness coefficient C ev3 = Eccentric shear coefficient C vcr = Member cracking coefficient C c3 = Corner influence coefficient

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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 d e1 2.5BED

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Side Edge Shear Strength In this case, the shear force is applied parallel to the side edge, d e1 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

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Side Edge Shear Strength Where: V co1 = nominal concrete breakout strength for a single stud C X1 = X spacing coefficient C Y1 = Y spacing coefficient C ev1 = Eccentric shear coefficient

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Single Anchor Strength Where: d e1 = Distance from side stud to side edge (in.) d o = Stud diameter (in.)

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X Spacing Factor Where: n x = Number of X-rows x = Individual X-row spacing (in.) n sides =Number of edges or sides that influence the X direction

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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: C x1 = n x

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Y Spacing Factor Where: n y = Number of Y-rows Y = Out-to-out Y-row spacing (in) = y (in)

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Eccentricity Factor Where: e v1 = Eccentricity form shear load to anchorage centroid

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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

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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

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In the Field Shear Strength For h ef /d e 4.5 (in normal weight concrete) Where: V cp = nominal pry-out shear strength (lbs)

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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 ½ in from the centerline. The panel has #5 confinement bars.

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

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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

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Step 1 – Check Corner Condition Not a Corner Condition

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Step 2 – Calculate Steel Capacity V ns = ·n s ·A n ·f ut = 0.65(4)(0.20)(65) = 33.8 kips

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Step 3 – Front Edge Shear Strength Front Edge Shear Strength

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Step 4 – Shear Capacity Coefficient Concrete Breakout Strength, Vco3

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Step 4 – Shear Capacity Coefficient X Spacing Coefficient, Cx3

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Step 4 – Shear Capacity Coefficient Member Thickness Coefficient, Ch3

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Step 4 – Shear Capacity Coefficient Eccentric Shear Force Coefficient, Cev3

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Step 4 – Shear Capacity Coefficient Member Cracking Coefficient, C vcr –Assume uncracked region of member #5 Perimeter Steel

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Step 5 – Shear Design Strength V cs = ·V co3 ·C x3 ·C h3 ·C ev3 ·C vcr = 0.75(47.0)(0.93)(0.53)(0.94)(1.0) = 16.3 kips

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Interaction Trilinear Solution Unity curve with a 5 / 3 exponent

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Interaction Curves

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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

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Example Parameters f c = 6000 psi normal weight concrete λ = 1.0 (8) – 1/2 in diameter studs A se = 0.20 in. 2 Nominal stud length = 6 in. f ut = 65,000 psi (Table ) V u = 25 kips N u = 4 kips Column size: 18 in. x 18 in.

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Provide ties around vertical bars in the column to ensure confinement: = 0.75 Determine effective depth h ef = L + t pl – t hs – 1 / 8 in = – – = 6.06 in

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Solution Steps Step 1 – Determine applied loads Step 2 – Determine tension design strength Step 3 – Determine shear design strength Step 4 – Interaction Equation

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Step 1 – Determine applied loads Determine net Tension on Tension Stud Group Determine net Shear on Shear Stud Group

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Step 2 – Concrete Tension Capacity

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Step 2 – Steel Tension Capacity

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Step 2 – Governing Tension

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Step 3 – Concrete Shear Capacity

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Step 3 – Steel Shear Capacity

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Step 3 – Governing Shear

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Step 4 – Interaction Check if Interaction is required

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Step 4 – Interaction

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Questions?

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