1. Anchor
Channel Systems
Code
2. HAC
Portfolio
3. HAC
Applications
4. Design
Introduction
Discussion
5. Base material
6. Loading
Calculations
9. Special Anchor
Channel Design
10. Design
Software
11. Best
Practices
12. Instructions
for Use
Discussion
14. Design
Example
13. Field Fixes
Calculations
Step 7: Concrete strength
Concrete pryout strength - perpendicular shear (Anchor a3)
The ICC-ES Acceptance Criteria AC232 includes amendments to the ACI 318
anchoring to concrete provisions. These amendments are given in Section 3.1
Strength Design – Amendments to ACI 318. Part D.6.3.2 (ACI 318-11) and Section
17.5.3.2 (ACI 318-14) of these amendments requires the factor ψ s,N to be modified
when calculating concrete pryout strength in shear. All of the parameters used to
calculate ψ s,N in tension are used except the parameter (N aua,i / N aua,3 ). The shear loads
acting on the anchor elements are substituted for the tension loads such that (V aua,i /
V aua,3 ) is used instead of (N aua,i / N aua,3 ).
These provisions for calculating concrete pryout strength are also given in ESR-3520
Section 4.1.3.3.4.
V cp,y,3 = k cp N cb,3 ESR-3520 Equation (41)
K cp3 = 2.0 ESR-3520 Table 8-6
N cb3 = N b3 · ψ s,N3 · ψ ed,N3 · ψ co,N3 · ψ c,N3 · ψ cp,N3 ESR-3520 Equation (6)
s i
s xx,1
s 1, 3
s 2,3
s cr,N
8. Reinforcing
Bar Anchorage
Code
Step 7: Concrete strength
ESR-3520 Section
4.1.3.3.4.
ACI 318-14 Chapter 17
7. Anchor Channel
Design Code
= spacing between each anchor element = 5.91 in
= distance of each influencing anchor element from anchor element #3
= distance from anchor element #1 to anchor element #3 = 11.812 in
= distance from anchor element #2 to anchor element #3 = 5.906 in
= critical anchor spacing for tension loading (h ef =4.173 in)
1 . 3 h ef
æ
s cr , N = 2 ç ç 2 . 8 -
7 . 1
è
ö
÷ ÷ h ef ³ 3 h ef
ø
ESR-3520 Equation (10)
The parameter ψ s,N is a modification factor that is used to account for the influence of
adjacent anchor elements on the anchor element being considered.
V a ua,1 = tension load on anchor element #1=421 lb
V a ua,2 = tension load on anchor element #2 1461 lb
V a ua,3 = tension load on anchor element #3=1618 lb
The calculated value for V cp,y,3 will be multiplied by a strength reduction factor
(φ-factor) to give a design strength (φV cp,y,3 ).
The calculated φV cp,y,3 value for anchor element #3 will be checked against the factored
load acting on anchor element #3 (V ua3 ) to obtain the % utilization (V ua3 / φV cp,y,3 ).
The anchor element with the highest % utilization will control the design with respect
to concrete pryout failure in shear.
ESR-3520 section
4.1.3.3.3
ACI 318-14 Chapter 17
Concrete breakout strength in perpendicular shear for anchor element #3
ФV cb ≥ V aua
V cb,3 = V b,3 · ψ s,V,3 · ψ co1,V,3 · ψ co2,V,3 · ψ h,V,3 · ψ c,V,3
V b =
ψ s,V =
ψ co,V =
ψ c,V =
ψ h,V =
Pryout: ФN cp,yb
s cr,N = 16.98 in
refer to concrete breakout tension influence of
anchor element #1 on anchor element #3:
1.5
æ 11.812 in ö 421 lbs
ç 1 -
÷
è 16.980 in ø 1618 lbs
= 0.0437
influence of anchor element #2 on anchor
element #3:
1.5
5.906 in ö 1461 lbs
æ
ç 1 -
÷
è 16.980 in ø 1618 lbs
= 0.4756
y s , N , 3 =
1
= 0.658
1 + (0.0437 + 0.4756)
ESR-3520 Equation (30)
Basic concrete breakout strength in shear
Modification factor for anchor spacing
Modification factor for corner effects
Modification factor cracked/uncracked concrete
Modification factor for concrete thickness
Concrete edge breakout: ФV cb,y
V b = ( 1 . 0 ) (10.50) 6,000psi × ( 5.0in ) 3
4
Calculate the basic concrete breakout strength in shear (V b,3 ).
V b = l × α ch, V × f c ' × ( c a1 ) 3
\ V b = 6 , 954 lbs
4
ESR-3520 Equation (31)
λ… Modification for lightweight concrete
Lightweight concrete = 0.75
Sand-Lightweight concrete = 0.85
α ch,V … Influence factor for channel size (10.50, max.)
f´ c … Concrete compressive strength (psi) (8,500 psi, max)
c a1 … Perpendicular edge distance (in.) (edge to center line of channel)
c cr , N = 0.5 s cr , N = ( 0.50 ) 16.98 in
c cr , N = 8.49 in
æ 5.0in ö
÷
è 8.49in ø
= 0 . 767
y ed, N = ç
\ y ed, N
0.5
< 1.0
c a2( + x) = 6.00 in + 0.984 in
c a2( + x) = 6.984 in
æ c
ö
c a 2 ( - x ) = ¥ ® ç ç a2( - x) ÷ ÷
è c cr, N ø
c cr , N = 8 . 49 in
æ 6.984in ö
÷
è 8.49in ø
\ y co, N,3 = 0 . 907
y co, N,3 = ç
0.5
= 1
0.5
< 1.0
ψ s,N = 0.658
ψ ed,N = 0.767
ψ co,N = 0.907
ψ cp,N = 1.0
ψ c,N = 1.0
N b 3 = 14634 lbs
N cb , 3 = 6699 lbs
k cp = 2
Figure 14.1.21 — Design example – reduction factors of V cb
V cp , y , 3 = k cp xN cb , 3
V cp , y , 3 = 2 x 6699 lbs = 13397 lbs
Condition B
j = 0.7
j V cp,y,3 = 9378lbs
V a ua,3 = 1618 lbs
æ 1618 ö
÷ x 100% = 1 8%
è 9378 ø
b cp , v , 3 = ç
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Cast-In Anchor Channel Product Guide, Edition 1 • 02/2019
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