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PCI 6th Edition PCI 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 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 HistoryHeaded 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
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 ReinforcementSupplemental 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 PrinciplesHCA 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 PrinciplesHCA 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 PrinciplesHCA 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 StudsStainless 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 DimensionsStud Dimensions
• Table 6.5.1.2• Page 6-12
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 Fut and not Fy
Steel CapacitySteel Capacity
Vs = Ns = ·n·Ase·fut
Where = steel strength reduction factor
= 0.65 (shear)= 0.75 (tension)
Vs = nominal shear strength steel capacity
Ns = nominal tensile strength steel capacityn = number of headed studs in group
Ase = nominal area of the headed stud shank
fut = 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, “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 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: “…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
The 5% fractileThe 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% fractileThe 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 FactorStrength Reduction Factor
Function of supplied confinement reinforcement
= 0.75 with reinforcement
= 0.70 with out reinforcement
Notation DefinitionsNotation Definitions
• Edges– de1, de2, de3, de4
• Stud Layout– x1, x2, …– y1, y2, …– X, Y
• Critical Dimensions– BED, SED
Concrete Tension Failure ModesConcrete 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 StrengthConcrete Tension Strength
Ncb: Breakout
Nph: Pullout
Nsb: Side-Face blowout
Tn = 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
N
cbN
cbgC
bsA
NC
crb
ed,N
C
bs3.33
f 'c
hef
Effective Embedment DepthEffective 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, AnProjected 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 RestrictionsNo 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 StudSide Edge Distance, Single Stud
de1 < 1.5hef
N e1 ef efA d 1.5 h 2 1.5 h
Side Edge Distance, Two StudsSide 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
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 ModificationEdge 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 1.0
1.5 h
Determine Breakout Strength, NcbDetermine Breakout Strength, Ncb
• The PCI handbook provides a design guide to determine the breakout area
Determine Breakout Strength, NcbDetermine Breakout Strength, Ncb
• First find the edge condition that corresponds to the design condition
Eccentrically LoadedEccentrically 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
ec,N
N
ef
11.0
2 e'1
3 h
Pullout StrengthPullout Strength
• Nominal pullout strength
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
N
pn11.2A
brgf '
cC
crp
Side-Face Blowout StrengthSide-Face Blowout Strength
• For a single headed stud located close to an edge (de1 < 0.4hef)
Where
Nsb = Nominal side-face blowout strength
de1 = Distance to closest edge
Abrg = Bearing area of head
N
sb160d
e1 A
brg 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, is multiplied by:
Where:
e2
e1
d1
d
4
1
de2
de1
3
Side-Face BlowoutSide-Face Blowout
• For multiple headed anchors located close to an edge (de1 < 0.4hef)
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
osbg sb
e1
sN 1 N
6 d
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. headed studs welded to ½ in thick plate
Nominal stud length = 8 in
f′c = 4000 psi (normal weight concrete)
fy = 60,000 psi
Example: Stud Group TensionExample: Stud Group Tension
Problem:
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 capacityStep 6 – Check confinement steel
Step 1 – Effective DepthStep 1 – Effective Depth
ef pl hs1h L t t "8
31 1 8" " " "2 8 8 8"
hef
L tpl
tns
18
8 12
38
18
8in
Step 2 – Check for Edge EffectStep 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 apply
de,min = 4 in.
Step 2 – Edge FactorStep 2 – Edge Factor
e,mined,N
ef
d0.7 0.3 1.0
1.5 h
4in. .7 0.3
1.5 8in
0.8
Step 3 – Breakout StrengthStep 3 – Breakout Strength
cbs
ef
cbg bs e1 ef ef ed,n crb
f ' 4000C 3.33 3.33 74.5lbs
h 8
From design aid, case 4
N C d X 1.5h de3 Y 1.5h C
0.8 0.75 74.5 4 16 12 6 8 12 1.0
1000
37.2kips
Step 3 – Pullout StrengthStep 3 – Pullout Strength
Abrg
0.79in2 4studs
Npn
(11.2)Abrg
f 'cC
crp
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
Ns nA
sef
ut
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
T
nN
cbg37.2kips
Step 6 – Check Confinement SteelStep 6 – Check Confinement Steel
• Crack plane area = 4 in. x 8 in. = 32 in.2
2
1000 32 1.4100037,000
1.20 3.4
37.20.75 60 1.2
0.68
cre
u
uvf
y e
AV
VA
f
in
Step 6 – Confinement SteelStep 6 – Confinement Steel
Use 2 - #6 L-bar around stud group.
These bars should extend ld past the breakout surface.
Concrete Shear StrengthConcrete 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, Vc3Front Edge Shear Strength, Vc3
SED
BED 3.0
Corner Edge Shear Strength, Modified Vc3Corner Edge Shear Strength, Modified Vc3
0.2
SED
BED3.0
Side Edge Shear Strength, Vc1Side Edge Shear Strength, Vc1
SED
BED 0.2
Front Edge Shear StrengthFront Edge Shear Strength
Where
Vco3 = 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 StrengthSingle Anchor Strength
Where:λ = lightweight concrete factorBED = distance from back row of studs to
front edge
V
co316.5 f '
cBED 1.33
d
e3 y d
e3 Y
X Spacing factorX Spacing factor
Where:X = Overall, out-to-out dimension of outermost
studs in back row of anchorage
nstuds-back= Number of studs in back row
C
x30.85
X
3BEDn
studs back
Thickness FactorThickness Factor
Where:h = Member thickness
Ch3
0.75h
BED for h 1.75BED
Ch3
1 for h > 1.75BED
Eccentricity FactorEccentricity Factor
Where
e′v = Eccentricity of shear force on a group of anchors
Cev3
1
1 0.67e'
v
BED
1.0 when e'v
X
2
Cracked Concrete FactorCracked 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 StrengthCorner Shear Strength
A corner condition shouldbe considered when:
where the Side Edge distance (SED) as shown
0.2
SED
BED3.0
Corner Shear StrengthCorner Shear Strength
Where:
Ch3 = Member thickness coefficient
Cev3 = Eccentric shear coefficient
Cvcr = Member cracking coefficient
Cc3 = Corner influence coefficient
Vc3
Vco3
Cc3
Ch3
Cev3
Cvcr
Corner factorCorner 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
C
c30.7
SED
BED3 1.0
Side Edge Shear StrengthSide 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
SED
BED3.0
Side Edge Shear StrengthSide 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 StrengthSingle Anchor Strength
Where:
de1 = Distance from side stud to side edge (in.)
do = Stud diameter (in.)
V
co87 f '
c d
e1 1.33 d
o 0.75
X Spacing FactorX 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
n
xx
2.5de1
2 nsides
Cx1
1.0 when x = 0
X Spacing FactorX 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 FactorY Spacing Factor
Where:
ny = Number of Y-rows
Y = Out-to-out Y-row spacing (in) = y (in)
Y1 y
0.25
y
Y1 y ye1
C 1.0 for n 1 (one Y - row)
n YC 0.15 n for n 1
0.6 d
Eccentricity FactorEccentricity Factor
Where:
ev1 = Eccentricity form shear load to anchorage centroid
v1ev1
e1
eC 1.0 1.0
4 d
Back Edge Shear StrengthBack 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“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“In the Field” Shear Strength
• For hef/de ≤ 4.5 (in normal weight concrete)
Where:Vcp = nominal pry-out shear strength (lbs)
V
cp 215
yn f '
c(d
o)1.5 (h
ef)0.5
y
y
4do
for y
d20
Front Edge Failure ExampleFront 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 Example
Problem:
Determine the design shear strength of the stud group.
Solution StepsSolution 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 ConditionStep 1 – Check Corner Condition
Not a Corner Condition
SED
BED3
48 4
12 43.25
Step 2 – Calculate Steel CapacityStep 2 – Calculate Steel Capacity
Vns = ·ns·An·fut
= 0.65(4)(0.20)(65) = 33.8 kips
Step 3 – Front Edge Shear StrengthStep 3 – Front Edge Shear Strength
• Front Edge Shear Strength
Vc3
Vco3
Cx3
Ch3
Cev3
Cvcr
Step 4 – Shear Capacity CoefficientStep 4 – Shear Capacity Coefficient
1.33
co3 c
1.33
V 16.5 f ' BED
16.5 1 5000 12 4
1000 47.0kips
• Concrete Breakout Strength, Vco3
Step 4 – Shear Capacity CoefficientStep 4 – Shear Capacity Coefficient
Cx3
0.85X
3BEDn
studs back
0.854
3160.93
0.93
• X Spacing Coefficient, Cx3
Step 4 – Shear Capacity CoefficientStep 4 – Shear Capacity Coefficient
Check if h 1.75BED
8 1.7516 OK
Ch3
0.75h
BED
0.758
16 0.53
• Member Thickness Coefficient, Ch3
Step 4 – Shear Capacity CoefficientStep 4 – Shear Capacity Coefficient
Check if e'v
X
21.5
4
2 OK
Cev3
1
1 0.67e'
v
BED
1.0
1
1 0.671.5
16
0.94
•Eccentric Shear Force Coefficient, Cev3
Step 4 – Shear Capacity CoefficientStep 4 – Shear Capacity Coefficient
• Member Cracking Coefficient, Cvcr
– Assume uncracked region of member
• #5 Perimeter Steel
Cvcr1.0
0.75
Step 5 – Shear Design StrengthStep 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
InteractionInteraction
• Trilinear Solution
• Unity curve with a 5/3 exponent
Interaction CurvesInteraction Curves
Combined Loading ExampleCombined 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 ParametersExample Parameters
f′c = 6000 psi normal weight concreteλ = 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 kipsColumn 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 StepsSolution Steps
Step 1 – Determine applied loadsStep 2 – Determine tension design
strengthStep 3 – Determine shear design strengthStep 4 – Interaction Equation
Step 1 – Determine applied loadsStep 1 – Determine applied loads
• Determine net Tension on Tension Stud Group
• Determine net Shear on Shear Stud Group
Nhu
V
ue
dc
Nu
25 6
10 4
19.0kips
Vu
Vu
2
25
2 12.5kips
Step 2 – Concrete Tension CapacityStep 2 – Concrete Tension Capacity
cb bs N crb ed,N
cbs
ef
N e1 e2 ef
e,mined,N
ef
cb
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 26.9kips
1000
Step 2 – Steel Tension CapacityStep 2 – Steel Tension Capacity
s se ut
s
N n A f
0.75 4 0.2 65N 39.0kips
1000
Step 2 – Governing TensionStep 2 – Governing Tension
cb s
n
N 26.9kips N 39.0kips
N 26.9kips
Step 3 – Concrete Shear CapacityStep 3 – Concrete Shear Capacity
c1 co1 X1 Y1 ev1 vcr
1.33 0.75
co c e1 o
1.33 0.75
x1
0.25 0.25
y
Y1e1
ev1
vcr
c1
V V C C C C
V 87 f ' d d
87 1 6000 6 0.5 43.7kips
C 2
n Y 2 3C 0.15 0.15 0.58
0.6 d 0.6 6
C 1.0
C 1.0
V 0.75 43.7 2 0.58 1 1 38.0kips
Step 3 – Steel Shear CapacityStep 3 – Steel Shear Capacity
s se ut
s
V n A f
0.65 4 0.2 65V 33.8kips
1000
Step 3 – Governing ShearStep 3 – Governing Shear
c s
n
V 38.0kips V 33.8kips
V 33.8kips
Step 4 – InteractionStep 4 – Interaction
• Check if Interaction is required
If Vu 0.2 V
n 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 – InteractionStep 4 – Interaction
Nhu
Nn
V
u
Vn
19.0
26.9
12.5
33.80.71 0.37 1.08 1.2
OR
Nhu
Nn
53
v
u
Vn
53
0.71 53 0.37 5
3 0.75 1.0
Questions?Questions?