Transcript Slide 1

PCI 6th Edition
Headed Concrete Anchors (HCA)
Presentation Outine
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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 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 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
fVs = fNs = f·n·Ase·fut
Where
f = 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 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% 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 Factor
Function of supplied confinement reinforcement
f = 0.75 with reinforcement
f = 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
fNcb: usually the most critical failure mode
– Pullout
fNph: function of bearing on the head of the stud
– Side-Face blowout
fNsb: studs cannot be closer to an edge than 40% the
effective height of the studs
Concrete Tension Strength
fTn = Minimum of
fNcb: Breakout
fNph: Pullout
fNsb: Side-Face blowout
Concrete Breakout Strength
Ncb  Ncbg  Cbs  AN  C crb   ed,N
Where:
Ccrb = Cracked concrete factor, 1 uncracked, 0.8 Cracked
AN = Projected surface area for a stud or group
Yed,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
•
Yed,N = modification for edge distance
 ed,N
 de,min
 0.7  0.3 
 1.5  hef

  1.0

• de,min = minimum edge distance, top, bottom, and
sides
• PCI also provides tables to directly calculate fNcb, but
Cbs , Ccrb, and Yed,N must still be determined for the in
situ condition
Determine Breakout Strength, fNcb
• The PCI handbook
provides a design
guide to determine
the breakout area
Determine Breakout Strength, fNcb
• 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:
e′N = eccentricity of the tensile force relative
to the center of the stud group
e′N ≤ s/2
Pullout Strength
• Nominal pullout strength
Npn  11.2  A brg  f 'c  C crp
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  160  de1  A brg  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) ¾ in. f 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 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  t pl  t ns  1
8 1
 8in
2
 3
8
8
 1
8
hef  L  t pl  t hs  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
 ed,N
 de,min 
 0.7  0.3 
  1.0
 1.5  hef 
 4in. 
 .7  0.3 

 1.5  8in 
 0.8
Step 3 – Breakout Strength
Cbs
f 'c
4000
 3.33 
 3.33 
 74.5lbs
hef
8
From design aid, case 4
Ncbg  f  Cbs   de1  X  1.5hef    de3  Y  1.5hef    ed,n  C crb
 0.8 
  0.75  74.5  4  16  12  6  8  12  
1.0 

 1000 
 37.2kips
Step 3 – Pullout Strength
A brg  0.79in2  4studs
fNpn  f  (11.2)  A brg  f 'c  C crp
 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
fNs  f  n  A se  fut
 0.75(4)(0.44)(65)
 85.8kips
Step 5 – Tension Capacity
The controlling tension capacity for the stud
group is Breakout Strength
fTn  Ncbg  37.2kips
Step 6 – Check Confinement Steel
• Crack plane area = 4 in. x 8 in. = 32 in.2
e 
1000  Acr  
Vu

1000  32 1.4 
37, 000
 1.20  3.4
Vu
37.2
Avf 


ff 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
fVc(failure mode)  f  Vco(failure mode)   C
Vco(failure mode)  anchor strength
C x(failure mode)  x spacing influence
C y(failure mode)  y spacing influence
Ch(failure mode)  thickness influence
C ev(failure mode)  eccentricity influence
C c(failure mode)  corner influence
C vcr
 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
fVc3  fVco3  C x3  Ch3  C ev3  C vcr
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
 
Vco3  16.5    f 'c  BED
1.33
Where:
λ = lightweight concrete factor
BED = distance from back row of studs to
front edge
 de3   y  de3  Y
X Spacing factor
C x3
X
 0.85 
 nstudsback
3  BED
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  1.75  BED
BED
 1 for h > 1.75  BED
Ch3  0.75
Ch3
Where:
h = Member thickness
Eccentricity Factor
C ev3 
1
 e' 
1  0.67   v 
 BED 
 1.0 when e 'v 
X
2
Where
e′v = 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
fVc3  fVco3  C c3  Ch3  C ev3  C vcr
Where:
Ch3 = Member thickness coefficient
Cev3 = Eccentric shear coefficient
Cvcr = Member cracking coefficient
Cc3 = Corner influence coefficient
Corner factor
C c3
SED
 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
fVc1  fVco1  C X1  C Y1  C ev1  C vcr
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
   d 
Vco  87    f 'c  de1
1.33
0.75
o
Where:
de1 = Distance from side stud to side edge (in.)
do = Stud diameter (in.)
X Spacing Factor
C x1 
nx  x
2.5  de1
 2  nsides
C x1  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
C Y1  1.0 for ny  1 (one Y - row)
C Y1
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) = Sy (in)
Eccentricity Factor
C ev1
 e v1 
 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)
fVcp  f  215   y  n  f 'c  (do )1.5  (hef )0.5
Where:
Vcp = nominal pry-out shear strength (lbs)
y 
y
4  do
y
for
 20
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 ½ 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
fVns = f·ns·An·fut
= 0.65(4)(0.20)(65) = 33.8 kips
Step 3 – Front Edge Shear Strength
• Front Edge Shear Strength
fVc3  fVco3  C x3  Ch3  C ev3  C vcr
Step 4 – Shear Capacity Coefficient
• Concrete Breakout Strength, Vco3
Vco3  16.5    f 'c  BED 
1.33

16.5 1

 47.0kips

5000 12  4 
1000
1.33
Step 4 – Shear Capacity Coefficient
• X Spacing Coefficient, Cx3
X
 nstudsback
3  BED
4
 0.85 
 0.93 
3 16
 0.93
C x3  0.85 
Step 4 – Shear Capacity Coefficient
• Member Thickness Coefficient, Ch3
Check if h  1.75  BED
8  1.75 16 OK
Ch3
h
 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 
C ev3

1
 1.5 
1  0.67  
 16 
 0.94
Step 4 – Shear Capacity Coefficient
• Member Cracking Coefficient, Cvcr
– Assume uncracked region of member
C vcr  1.0
• #5 Perimeter Steel
f  0.75
Step 5 – Shear Design Strength
fVcs = f·Vco3·Cx3·Ch3·Cev3·Cvcr
= 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 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 Parameters
f′c = 6000 psi normal weight concrete
λ = 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: f = 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
fNcb  f  Cbs  AN  C crb   ed,N
Cbs
f 'c
6000
 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  0.7  0.3
fNcb
de,min
1.5hef
6
 0.7  0.3
 0.898
1.5  6.06 
0.75  381.24 104.8  0.898 


 26.9kips
1000
Step 2 – Steel Tension Capacity
fNs  f  n  A se  fut
fNs
0.75  4  0.2  65 


 39.0kips
1000
Step 2 – Governing Tension
fNcb  26.9kips
fNs  39.0kips
fNn  26.9kips
Step 3 – Concrete Shear Capacity
fVc1  f  Vco1  C X1  C Y1  C ev1  C vcr
Vco  87    f 'c   de1 
1.33
 87 1 6000  6 
1.33
  do 
 0.5 
0.75
0.75
 43.7kips
C x1  2
C Y1
n


y
 Y
0.25
0.6  de1
2  3  
 0.15 
 0.15  0.58
0.6  6 
0.25
C ev1  1.0
C vcr  1.0
fVc1  0.75  43.7  2  0.58 11  38.0kips
Step 3 – Steel Shear Capacity
fVs  f  n  A se  fut
fVs
0.65  4  0.2  65 


 33.8kips
1000
Step 3 – Governing Shear
fVc  38.0kips fVs  33.8kips
fVn  33.8kips
Step 4 – Interaction
• Check if Interaction is required
 
If Vu  0.2 fVn Interaction is not Required
 
12.5  0.2 33.8
12.5  6.76 - Interaction Required
 
If Nhu  0.2 fNn 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
fNn fVn 26.9 33.8
OR
N 
hu


 fNn 
5
3
 v 
 u 
 fVn 
5
3
   0.37
 0.71
5
3
5
3
 0.75  1.0
Questions?