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Channel Systems
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5. Base material
6. Loading
Calculations
ESR-3520 section
4.1.3.2.3
ACI 318-14 Chapter 17
Calculate the modification factor for corner influence (ψ co,N,3 ).
c a2 is measured parallel to the anchor channel longitudinal axis, and is considered
when calculating the modification factor for corner influence (ψ co,N ).
Since the calculated value for concrete breakout in tension (N cb ) is dependent on
the concrete geometry, it is important to note that values for ca1 and ca2 must be
taken with respect to the relevant edge distances from the anchor element being
considered.
0.5
0.5
æ c
ö
æ c
ö
y co, N,3 = ç ç a2( + x) ÷ ÷ £ 1.0 - and - y co, N,3 = ç ç a2( - x) ÷ ÷ £ 1.0
c
c
è cr, N ø
è cr, N ø
ESR-3520 Equation (16)
The parameter c cr,N corresponds to the maximum edge distance that is assumed with
respect to values for c a1 and c a2 . Any c a1 or c a2 value less than c cr,N must be considered
when calculating ψ ed,N and ψ co,N . If more than one c a1 value is less than c cr,N , the
smallest c a1 value will be used to calculate ψ ed,N . If more than one c a2 value is less than
c cr,N , ψ co,N will be calculated for each c a2 value and the product of these ψ co,N values will
be used to calculate the nominal concrete breakout strength in tension (N cb ).
Calculate the modification factor for cracked/uncracked concrete (ψ c,N,3 )
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14. Design
Example
Calculations
Concrete breakout: ФN cb
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) ÷ ÷
c
cr,
N
è
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
Concrete is typically assumed to be cracked under normal service load conditions. If
uncracked concrete conditions are assumed, an increase in N cb is permitted via the
modification factor ψ c,N . Cracked concrete conditions
if uncracked concrete conditions are assumed,ψ c,N,3 =1.25. → ψ cp,N,3 = 1.0
=
=
=
Concrete breakout strength in Tension continued….
Nominal concrete breakout strength in tension for anchor element #3
ФN cb ≥ N aua
N cb = N b · ψ s,N · ψ ed,N · ψ co,N · ψ cp,N · ψ c,N
→ ψ c,N,3 = 1.0
Cracked concrete conditions
Calculate the modification factor for splitting (ψ cp,N
402
9. Special Anchor
Channel Design
Step 7: Concrete strength
Concrete breakout strength in Tension continued…
c ac
c cr,N
c a,min
8. Reinforcing
Bar Anchorage
Code
Step 7: Concrete strength
ESR-3520 section
4.1.3.2.3
ACI 318-14 Chapter 17
7. Anchor Channel
Design Code
critical edge distance for splitting
critical anchor edge distance
minimum edge distance
Uncracked concrete with no supplementary reinforcement ψ cp,N
if c a,min ≥ c ac 1
if c amin < c ac ⎧⎛ c a,min ⎞ ⎛ c cr,N ⎞ ⎫
ψ cp,N = MAX ⎨ ⎜ ______ ⎜ ; ⎜ _____ ⎜ ⎬
⎩⎝ c ac ⎠ ⎝ c ac ⎠ ⎭
Uncracked concrete with supplementary reinforcement 1
Cracked concrete 1
ESR-3520 Equation (6)
The calculated value for N cb,3 will be multiplied by a strength reduction factor (φ-factor)
to give a design strength (φ Ncb,3 ). Design strengths for this example are summarized in
the table on this page titled Summary for Concrete Breakout in Tension. Concrete breakout: ФN cb
The calculated φ Ncb,3 value for anchor element #3 will be checked against the factored
load acting on anchor element #3 (N aua,3 ) to obtain the % utilization (N aua,3 / φ Ncb3 ). ψ s,N = 0.658
The anchor element with the highest % utilization will control the design with respect
to concrete breakout failure in tension. ψ co,N = 0.907
φ-factors for concrete breakout in tension are given in ESR-3520 Table 8-4.
Ф factor:
Condition A (Ф =0.75) is considered when
• Supplementary reinforcement is present
• Reinforcement does not need to be explicitly designed for the anchor channel
• Arrangement should generally conform to anchor reinforcement
• Development is not required
Condition B (Ф =0.70) is considered when
• No Supplementary reinforcement is present
Assume Condition B
Ф =0.70
The concrete breakout calculations (φ Ncb ) were influenced by the anchor location and
the concrete geometry via the modification factors ψ s,N , ψ ed,N and ψ co,N
ψ c,N = 1.0
N cb = N b × ψ s,N × ψ ed,N × ψ co,N × ψ cp,N × ψ c,N
ψ ed,N = 0.767
ψ cp,N = 1.0
N b 3 = 14634 lbs
N cb , 3 = 14634 x 0.658 x 0.767 x 0.907 x 1.0 x 1.0
N cb , 3 = 6699 lbs
Condition B
j = 0.7
j N cb,3 = 4689lbs
N a ua,3 = 786 lbs
æ 78 6 ö
÷ x 100% = 17%
è 4690 ø
b cb , N , 3 = ç
Cast-In Anchor Channel Product Guide, Edition 1 • 02/2019
403