Download - 9 Headed Anchor Design
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 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 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
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 = ·n·Ase·fut
Where = steel strength reduction factor
= 0.65 (shear)= 0.75 (tension)Vs = nominal shear strength steel capacityNs = nominal tensile strength steel capacityn = number of headed studs in groupAse = nominal area of the headed stud shankfut = ultimate tensile strength of the stud steel
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
5% FailuresCapacity
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
= 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
– 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 Breakout Strength
Where:Ccrb = Cracked concrete factor, 1 uncracked, 0.8 CrackedAN = Projected surface area for a stud or grouped,N =Modification for edge distanceCbs = Breakout strength coefficient
Ncb Ncbg Cbs AN Ccrb ed,N
Cbs 3.33
f 'chef
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
2No ef ef efA 2 1.5 h 2 1.5 h 9 h
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 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 1.01.5 h
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.
Where:e′N = eccentricity of the tensile force relative to the center of the stud groupe′N ≤ s/2
ec,NN
ef
1 1.02 e'1 3 h
Pullout Strength
• Nominal pullout strength
WhereAbrg = bearing area of the stud head = area of the head – area of the shankCcrp = cracking coefficient (pullout)
= 1.0 uncracked = 0.7 cracked
Npn 11.2Abrg f 'cCcrp
Side-Face Blowout Strength
• For a single headed stud located close to an edge (de1 < 0.4hef)
WhereNsb = Nominal side-face blowout strengthde1 = Distance to closest edgeAbrg = Bearing area of head
Nsb 160de1 Abrg f 'c
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:
Where:
e2
e1
d1 d4
1
de2de1
3
Side-Face Blowout
• For multiple headed anchors located close to an edge (de1 < 0.4hef)
Whereso = spacing of the outer anchors along the edge in the groupNsb = nominal side-face blowout strength for a single anchor previously defined
osbg sb
e1
sN 1 N6 d
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 plateNominal stud length = 8 inf′c = 4000 psi (normal weight concrete)
fy = 60,000 psi
Solution Steps
Step 1 – Determine effective depthStep 2 – Check for edge effectStep 3 – Check concrete strength of stud groupStep 4 – Check steel strength of stud group
Step 5 – Determine tension capacityStep 6 – Check confinement steel
Step 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 applyde,min = 4 in.
Step 3 – Breakout Strength
cbs
ef
cbg bs e1 ef ef ed,n crb
f ' 4000C 3.33 3.33 74.5lbsh 8
From design aid, case 4N C d X 1.5h de3 Y 1.5h C
0.8 0.75 74.5 4 16 12 6 8 12 1.01000 37.2kips
Step 3 – Pullout Strength
Abrg 0.79in2 4studs
Npn (11.2)Abrg f 'cCcrp
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 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
2
1000 32 1.4100037,000
1.20 3.437.2
0.75 60 1.2 0.68
cre
u
uvf
y e
AV
VAf
in
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
WhereVco3 = 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 Strength
Where:λ = lightweight concrete factorBED = distance from back row of studs to
front edge
Vco3 16.5 f 'c BED 1.33
de3 y de3 Y
X Spacing factor
Where:X = Overall, out-to-out dimension of outermoststuds in back row of anchoragenstuds-back= Number of studs in back row
Cx3 0.85
X3BED nstuds back
Eccentricity Factor
Wheree′v = Eccentricity of shear force on a group of
anchors
Cev3 1
1 0.67e'vBED
1.0 when e'v X2
Cracked Concrete Factor
Uncracked concreteCvcr = 1.0
For cracked concrete,Cvcr = 0.70 no reinforcement
orreinforcement < 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 shouldbe considered when:
where the Side Edge distance (SED) as shown
0.2
SEDBED 3.0
Corner Shear Strength
Where:Ch3 = Member thickness coefficient Cev3 = Eccentric shear coefficient Cvcr = Member cracking coefficientCc3 = Corner influence coefficient
Vc3 Vco3 Cc3 Ch3 Cev3 Cvcr
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 de1 ≤ 2.5BED
Cc3 0.7
SEDBED
3 1.0
Side 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
SEDBED 3.0
Side 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 Strength
Where:de1 = Distance from side stud to side edge (in.)
do = Stud diameter (in.)
Vco 87 f 'c de1 1.33 do 0.75
X 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 nx x
2.5de1
2 nsides
Cx1 1.0 when x = 0
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
Where:ny = Number of Y-rows
Y = Out-to-out Y-row spacing (in) = y (in)
Y1 y
0.25y
Y1 y ye1
C 1.0 for n 1 (one Y - row)n Y
C 0.15 n for n 10.6 d
Eccentricity Factor
Where:ev1 = Eccentricity form shear load to
anchorage centroid
v1ev1
e1
eC 1.0 1.04 d
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)
Where:Vcp = nominal pry-out shear strength (lbs)
Vcp 215y n f 'c (do)1.5 (hef )0.5
y
y4do
for yd
20
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 of1 ½ in from the centerline. The
panel has #5 confinement bars.
Solution Steps
Step 1 – Check corner conditionStep 2 – Calculate steel capacityStep 3 – Front Edge Shear StrengthStep 4 – Calculate shear capacity coefficientsStep 5 – Calculate shear capacity
Step 4 – Shear Capacity Coefficient
1.33co3 c
1.33
V 16.5 f ' BED16.5 1 5000 12 4
1000 47.0kips
• Concrete Breakout Strength, Vco3
Step 4 – Shear Capacity Coefficient
Cx3 0.85X
3BED nstuds back
0.854
316 0.93
0.93
• X Spacing Coefficient, Cx3
Step 4 – Shear Capacity Coefficient
Check if h 1.75BED 8 1.7516 OK
Ch3 0.75 hBED
0.75 816
0.53
• Member Thickness Coefficient, Ch3
Step 4 – Shear Capacity Coefficient
Check if e'v X2 1.5
42 OK
Cev3 1
1 0.67e'vBED
1.0
1
1 0.671.516
0.94
•Eccentric Shear Force Coefficient, Cev3
Step 4 – Shear Capacity Coefficient
• Member Cracking Coefficient, Cvcr– Assume uncracked region of member
• #5 Perimeter Steel
Cvcr 1.0
0.75
Step 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
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 concreteλ = 1.0(8) – 1/2 in diameter studsAse = 0.20 in.2 Nominal stud length = 6 in.fut = 65,000 psi (Table 6.5.1.1)Vu = 25 kipsNu = 4 kipsColumn size: 18 in. x 18 in.
• Provide ties around vertical bars in the column to ensure confinement: = 0.75
• Determine effective depthhef = L + tpl – ths – 1/8 in
= 6 + 0.5 – 0.3125 – 0.125 = 6.06 in
Solution Steps
Step 1 – Determine applied loadsStep 2 – Determine tension design
strengthStep 3 – Determine shear design strengthStep 4 – Interaction Equation
Step 1 – Determine applied loads
• Determine net Tension on Tension Stud Group
• Determine net Shear on Shear Stud Group
Nhu Vu edc
Nu
25 6
10 4
19.0kips
Vu Vu
2
252
12.5kips
Step 2 – Concrete Tension Capacity
cb bs N crb ed,N
cbs
ef
N e1 e2 ef
e,mined,N
ef
cb
N C A Cf ' 6000C 3.33 3.33 1 104.8h 6.06
A d X d Y 3h 6 6 6 3 3 6.06 381.24d 60.7 0.3 0.7 0.3 0.8981.5h 1.5 6.06
0.75 381.24 104.8 0.898N 26.9kips1000
Step 3 – Concrete Shear Capacity
c1 co1 X1 Y1 ev1 vcr1.33 0.75
co c e1 o1.33 0.75
x10.25 0.25
yY1
e1
ev1
vcr
c1
V V C C C CV 87 f ' d d
87 1 6000 6 0.5 43.7kipsC 2
n Y 2 3C 0.15 0.15 0.580.6 d 0.6 6C 1.0C 1.0
V 0.75 43.7 2 0.58 1 1 38.0kips
Step 4 – Interaction
• Check if Interaction is required
If Vu 0.2 Vn Interaction is not Required12.5 0.2 33.8
12.5 6.76 - Interaction Required
If Nhu 0.2 Nn Interaction is not Required19 0.2 26.9
19 5.38 - Interaction Required
Step 4 – Interaction
Nhu
Nn
Vu
Vn
19.026.9
12.533.8 0.71 0.37 1.08 1.2
ORNhu
Nn
53
vu
Vn
53
0.71 53 0.37 5
3 0.75 1.0