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PCI 6th
EditionPCI 6th
Edition
Headed Concrete Anchors (HCA)
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Presentation OutinePresentation 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
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.
³Concrete Capacity Design´ (CCD) approach
has been incorporated into ACI 318-02
Appendix D
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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
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Background for
Headed Concrete Anchor Design
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 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
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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.´
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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
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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 inunconfined concrete
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HCA Design PrinciplesHCA Design Principles
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
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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
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Stud DimensionsStud Dimensions
Table 6.5.1.2
Page 6-12
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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
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Steel CapacitySteel Capacity
JVs = JNs = J·n·Ase·f ut
Where
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
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Material PropertiesMaterial Properties
Adapted from AWS D1.1-02
Table 6.5.1.1 page 6-11
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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
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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
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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
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The 5% fractileThe 5% fractile
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
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The 5% fractileThe 5% fractile
Example values of based on sample sizeare:
n = = 1.645n = 40 = 2.010
n = 10 = 2.568
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Strength Reduction FactorStrength Reduction Factor
Function of supplied confinement reinforcement
J= 0.75 with reinforcement
J = 0.70 with out reinforcement
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Notation DefinitionsNotation Definitions
Edges
± de1, de2, de3, de4
Stud Layout
± x1, x2,«
± y1, y2, «
± X, Y
Critical Dimensions
± BED, SED
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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
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Concrete Tension StrengthConcrete Tension Strength
JNcb: Breakout
JNph
: Pullout
JNsb: Side-Face blowout
JTn = Minimum of
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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
cb! N
cbg! C
bs A
N C
crb]
ed,N
Cbs
! 3.33 P f 'c
hef
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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.
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Projected Surface Area, An
Projected Surface Area, An
Based on 35o
AN - calculated, or empirical equations
are provided in thePCI handbook
Critical edgedistance is 1.5hef
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No Edge Distance RestrictionsNo Edge Distance Restrictions
For a single stud, with de,min > 1.5hef
2
No ef ef ef A 2 1.5 h 2 1.5 h 9 h« » « »! ! - ½ - ½
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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!
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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!
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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!
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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
e,min
ed,N
ef
d0.7 0.3 1.0
1.5 h
¨ ¸] ! e© ¹ª º
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Determine Breakout Strength, J N cb
Determine Breakout Strength, J N cb
The PCI handbook
provides a design
guide to determine
the breakout area
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Determine Breakout Strength, J N cb
Determine Breakout Strength, J N cb
First find the edge
condition that
corresponds to the
design condition
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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
ec,N
N
ef
11.0
2e '1 3 h
] ! e¨ ¸
© ¹ª º
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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
pn! 11.2 A
brg f '
c C
crp
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Side-Face Blowout StrengthSide-Face Blowout Strength
For a single headed stud located close to anedge (de1 < 0.4hef )
Where
Nsb = Nominal side-face blowout strength
de1 = Distance to closest edge Abrg = Bearing area of head
Nsb ! 160 de1 Abrg f 'c
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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:
e2
e1
d1d
4
¨ ¸© ¹
ª º
1 ed
e2
de1
e 3
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Side-Face BlowoutSide-Face Blowout
For multiple headed anchors located close to anedge (de1 < 0.4hef )
Where
so = spacing of the outer anchors along the
edge in the groupNsb = nominal side-face blowout strength for
a single anchor previously defined
o
sbg sbe1
sN 1 N
6 d
¨ ¸!
© ¹ª º
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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
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Example: Stud Group TensionExample: Stud Group Tension
Probl em:
Determine the design
tension strength of the
stud group
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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
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Step 1 ± Effective DepthStep 1 ± Effective Depth
ef pl hs1
h L t t "8
31 18" " " "2 8 8
8"
!
!
!
hef
! L t pl
t ns
18
! 8 12
38
18
! 8in
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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.
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Step 2 ± Edge FactorStep 2 ± Edge Factor
e,min
ed,N
ef
d0.7 0.3 1.0
1.5 h
4in..7 0.3
1.5 8in
0.8
¨ ¸] ! e© ¹ª º
¨ ¸! © ¹ª º!
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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
0.80.75 74.5 4 16 12 6 8 12 1.0
1000
37.2kips
! ! !
! J ]
¨ ¸! © ¹
ª º!
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Step 3 ± Pullout StrengthStep 3 ± Pullout Strength
Abrg
! 0.79in2 4studs
JNpn
! J (11.2) Abrg
f 'c C
crp
! 0.7(11.2)(3.16)(4)(1.0)
! 99.1kips
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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
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Step 4 ± Steel StrengthStep 4 ± Steel Strength
JNs
! J n Ase
f ut
! 0.75(4)(0.44)(65)
! 85.8kips
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Step 5 ± Tension CapacityStep 5 ± Tension Capacity
The controlling tension capacity for the stud
group is Breakout Strength
JT
n! N
cbg! 37.2kips
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Step 6 ± Check Confinement SteelStep 6 ± Check Confinement Steel
Crack plane area = 4 in. x 8 in. = 32 in.2
2
1000 32 1.41000
37,000 1.20 3.4
37.2
0.75 60 1.2
0.68
! !
!
! ! !
!
J Q
cre
u
uvf
y e
A
V
V A
f
in
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Step 6 ± Confinement SteelStep 6 ± Confinement Steel
Use 2 - #6 L-bar
around stud group.
These bars should
extend ld past thebreakout surface.
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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
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J Vc(failure mode)
! J 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
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Front Edge Shear Strength, Vc3Front Edge Shear Strength, Vc3
SED
BED" 3.0
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Corner Edge Shear Strength, Modified Vc3Corner Edge Shear Strength, Modified Vc3
0.2 e SED
BEDe 3.0
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Side Edge Shear Strength, Vc1Side Edge Shear Strength, Vc1
SEDBED
0.2
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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
J V
c3! J V
co3 C
x3 C
h3 C
ev3 C
vcr
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Single Anchor StrengthSingle Anchor Strength
Where:
= lightweight concrete factor BED = distance from back row of studs to
front edge
V
co3! 16.5 P f '
c BED 1.33
! de3 y§ ! de3 Y
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Thickness FactorThickness Factor
Where:
h = Member thickness
Ch3
! 0.75h
BEDfor h e 1.75 BED
Ch3
! 1 for h > 1.75 BED
i ii i
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Eccentricity FactorEccentricity Factor
Where
e v = Eccentricity of shear force on a group of
anchors
Cev3
!1
1 0.67 e '
v
BED
¨
ª©
¸
º¹
e 1.0 when e 'v
eX
2
C k d C FC k d C F
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Cracked Concrete FactorCracked Concrete Factor
Uncracked concreteCvcr = 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.
C Sh S hC Sh S h
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Corner Shear StrengthCorner Shear Strength
A corner condition should
be considered when:
where the Side Edge
distance (SED) as
shown
0.2 e
SED
BED e 3.0
C Sh St thC Sh St th
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Corner Shear StrengthCorner Shear Strength
Where:
Ch3 = Member thickness coefficient
Cev3 = Eccentric shear coefficient
Cvcr = Member cracking coefficient
Cc3 = Corner influence coefficient
J V
c3! J V
co3 C
c3 C
h3 C
ev3 C
vcr
C f tC f t
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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
C
c3! 0.7
SED
BED3 e 1.0
Sid Ed Sh St thSid Ed Sh St th
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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
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Side Edge Shear StrengthSide Edge Shear Strength
J V
c1! J V
co1 C
X 1 C
Y 1 C
ev1 C
vcr
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
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Single Anchor StrengthSingle Anchor Strength
Where:
de1 = Distance from side stud to side edge (in.)
do = Stud diameter (in.)
V
co! 87 P f '
c d
e1 1.33
do 0.75
X S i F tX S i F t
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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
Cx1
! nx x
2.5 de1
2 nsides
Cx1
! 1.0 when x = 0
X S i F tX S i F t
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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
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Y Spacing FactorY Spacing Factor
Where:
ny = Number of Y-rows
Y = Out-to-out Y-row spacing (in) = 7y (in)
Y 1 y
0.25
y
Y 1 y y
e1
C 1.0 for n 1 (one Y - row)
n Y C 0.15 n for n 1
0.6 d
! !
! e "
Eccentricity FactorEccentricity Factor
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Eccentricity FactorEccentricity Factor
Where:
ev1 = Eccentricity form shear load to
anchorage centroid
v1ev1
e1
eC 1.0 1.0
4 d
¨ ¸! e© ¹ª º
Back Edge Shear StrengthBack Edge Shear Strength
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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
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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
used
³In the Field´ Shear Strength³In the Field´ Shear Strength
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³In the Field´ Shear Strength³In the Field´ Shear Strength
For hef /de 4.5 (in normal weight concrete)
Where:
Vcp = nominal pry-out shear strength (lbs)
J V
cp! J 215 ]
y n f '
c (d
o)1.5
(hef
)0.5
] y ! y
4 do
for yd
e 20
Front Edge Failure ExampleFront Edge Failure Example
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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
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ExampleExample
Probl em:
Determine the design shear strength of
the stud group.
Solution StepsSolution Steps
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Solution StepsSolution Steps
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
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Step 1 ± Check Corner ConditionStep 1 ± Check Corner Condition
Not a Corner Condition
SED
BEDu 3
48 412 4
! 3.25
Step 2 Calculate Steel CapacityStep 2 Calculate Steel Capacity
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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
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Step 3 ± Front Edge Shear StrengthStep 3 ± Front Edge Shear Strength
Front Edge Shear Strength
J V
c3! J V
co3 C
x3 C
h3 C
ev3 C
vcr
Step 4 ± Shear Capacity CoefficientStep 4 ± Shear Capacity Coefficient
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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
! P
!
!
Concrete Breakout Strength, Vco3
Step 4 ± Shear Capacity CoefficientStep 4 ± Shear Capacity Coefficient
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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
Step 4 ± Shear Capacity CoefficientStep 4 ± Shear Capacity Coefficient
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Step 4 Shear Capacity CoefficientStep 4 Shear Capacity Coefficient
Check if h e 1.75 BED
8 e 1.75 16 OK
Ch3
! 0.75h
BED
! 0.75
8
16
! 0.53
Member Thickness Coefficient, Ch3
Step 4 ± Shear Capacity CoefficientStep 4 ± Shear Capacity Coefficient
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Step 4 Shear Capacity CoefficientStep 4 Shear Capacity Coefficient
Check if e 'v
eX
21.5 e
4
2OK
Cev3 ! 1
1 0.67 e '
v
BED
¨
ª©
¸
º¹
e 1.0
!
1
1 0.67 1.5
16
¨
ª© ¸
º¹
! 0.94
Eccentric Shear Force Coefficient, Cev3
Step 4 ± Shear Capacity CoefficientStep 4 ± Shear Capacity Coefficient
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Step 4 Shear Capacity CoefficientStep 4 Shear Capacity Coefficient
Member Cracking Coefficient, Cvcr
± Assume uncracked region of member
#5 Perimeter Steel
C
vcr! 1.0
J ! 0.75
Step 5 ± Shear Design StrengthStep 5 ± Shear Design Strength
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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
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InteractionInteraction
Trilinear Solution
Unity curve with a 5/3 exponent
Interaction CurvesInteraction Curves
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Interaction CurvesInteraction Curves
Combined Loading ExampleCombined Loading Example
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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
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Example ParametersExample Parameters
f c = 6000 psi normal weight concrete = 1.0
(8) ± 1/2 in diameter studs
Ase = 0.20 in.
2
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.
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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
Solution StepsSolution Steps
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Solution StepsSolution Steps
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
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p ppp pp
Determine netTension on Tension
Stud Group
Determine net Shear
on Shear StudGroup
Nhu
!
Vu e
dc
Nu
!
25 6 10
4
! 19.0kips
Vu
! V
u
2
! 252
! 12.5kips
Step 2 ± Concrete Tension CapacityStep 2 ± Concrete Tension Capacity
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p p yp p y
cb bs N crb ed,N
cbs
ef
N e1 e2 ef
e,min
ed,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 26.9kips1000
J ! J ]
! P ! !
! ! !
] ! ! !
J ! !
Step 2 ± Steel Tension CapacityStep 2 ± Steel Tension Capacity
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p p yp p y
s se ut
s
N n A f
0.75 4 0.2 65N 39.0kips
1000
J ! J
J ! !
Step 2 ± Governing TensionStep 2 ± Governing Tension
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p gp g
cb s
n
N 26.9kips N 39.0kips
N 26.9kips
J ! J !
J !
Step 3 ± Concrete Shear CapacityStep 3 ± Concrete Shear Capacity
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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
x1
0.25 0.25
y
Y 1
e1
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
J ! J
! P
! !
!
« »- ½! ! !
!
!
J ! !
Step 3 ± Steel Shear CapacityStep 3 ± Steel Shear Capacity
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p p yp p y
s se ut
s
V n A f
0.65 4 0.2 65 V 33.8kips1000
J ! J
J ! !
Step 3 ± Governing ShearStep 3 ± Governing Shear
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p gp g
c s
n
V 38.0kips V 33.8kips
V 33.8kips
J ! J !
J !
Step 4 ± InteractionStep 4 ± Interaction
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Check if Interaction is required
If Vu
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
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Nhu
JNn
V
u
J Vn
!19.0
26.9
12.5
33.8! 0.71 0.37 ! 1.08 e 1.2
OR
Nhu
JNn
¨
ª©
¸
º¹
53
vu
J Vn
¨
ª©
¸
º¹
53
! 0.71 53 0.37
53 ! 0.75 e 1.0
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Questions?Questions?