Anchor Example

S. K. Ghosh Associates Inc.

www.skghoshassociates.com

1

- 1 -

Example Problem

½-in. Headed boltsASTM F 1554 Grade 36fyt = 36,000 psifuta = 58,000 psif’c = 4000 psihef = 12 in.ha = 18 in.SDC CConcrete is assumed

cracked at service load

8" 4"

6"

6"

4"

14"

18"

20"

Steel base plateWide flange column

- 2 -

Example Problem

e Nx = 1 in.

e´Ny = 1 in.

8" 4"

6"

6"

4"

20"

e Nx

e Ny

Nua



2

- 3 -

D.5 – Design Requirements for Tensile Loading

D.5.1 – Steel Strength

D.5.2 – Concrete Breakout Strength

D.5.3 – Pullout Strength

D.5.4 – Concrete Side-Face Blowout Strength of a Headed Anchor

- 4 -

Tension Failure

Check for possible compression in anchor group due to load eccentricity e´Nx :

NuaNua

Mua = Nua x 1

Nua/6Nua/6

-[Mua/8]/3

= Nua/8

+ [Mua/8]/3 = 5Nua/24



3

- 5 -

Tension Failure

Check for possible compression in anchor group due to load eccentricity e´Ny :

NuaNua

Mua = Nua x 1

Nua/6Nua/6

-[Mua/12]/2

= Nua/8

+ [Mua/12]/2 = 5Nua/24Nua/6

- 6 -

Tension Failure

Check for possible compression in anchor group due to load eccentricities:

Maximum tension = Nua/6 + Nua/24 + Nua/24 = Nua/4

Minimum tension = Nua/6 – Nua/24 – Nua/24 = Nua/12 ……..No compression



4

- 7 -

D.5.1 – Steel Failure (Tension)

- 8 -


Calculating for one bolt:

Ase = 0.142 in.2

futa = 58,000 psi < 1.9fya (= 68,400 psi)…………OK< 125,000 psi………………….OK

Nsa = 0.142 x 58,000/1000 = 8.24 kips



5

- 9 -


Strength reduction factor, φ:

φ = 0.75 for ductile steel elements under tension…….D.4.4(a)

For SDC C or above, a factor of 0.75 is applied..……D.3.3.3

This factor is not required in steel failure modes in ACI 318-08

0.75φ = 0.75 × 0.75 = 0.56

- 10 -


Design strength of a single bolt for steel failure in tension:

0.75φNsa = 0.56 × 8.24 = 4.61 kips



6

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Equating with the maximum demand on a single bolt:Nua/4 = 4.61 kips

Thus maximum Nua = 18.44 kips = 0.75φNsa for the group of bolts

This is the maximum tension demand that the whole bolt formation can support for steel failure in tension

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D.5.2 – Concrete Breakout (Tension)



7

- 13 -

D.1 – Definitions

Anchor Group

hef = 12 in.Spacing in X-direction = 8 in. < 3 × 12 in. .......OKSpacing in Y-direction = 6 in. < 3 × 12 in. .......OK

The current bolt formation acts as a group in tension.

- 14 -

D.5.2.3 – Anchor Close to 3 or 4 Edges

In the example problem, number of edge distances that are less than 1.5hef = 2

hef need not be modified

hef = 12 in.



8

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D.5.2.1 – Projected Area ANCO

ANco = 9 × 122 = 1296 in.2

- 16 -

D.5.2.1 – Projected Area ANC

ANc = [18+8+4] ×[4+6+6 +18]

= 1020 in.2

ANc < nANco (= 6 × 1296)

…………….OK

8" 4"

6"6"

4"

1.5x12 = 18"

1.5x

12 =

18"



9

- 17 -

D.5.2.2 – Basic Concrete Breakout

= 63.1 kips

For cast-in headed bolts with 11 in. ≤ hef ≤ 25 in.

= 64.2 kips

Nb = 64.2 kips

- 18 -

D.5.2.4 – Eccentricity Effect

In the example problem:

hef = 12 in., e Nx = 1 in., e Ny = 1in.ψec,Nx = 1 / [ 1 + 2x1/3x12 ] = 0.95 ≤ 1 ……OKψec,Ny = 1 / [ 1 + 2x1/3x12 ] = 0.95 ≤ 1 ……OKψec,N = ψec,Nx ψec,Ny = 0.90



10

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D.5.2.5 – Edge Effect


hef = 12 in., ca,min = 4 in. ≤ 1.5hef

ψed,N = 0.7 + 0.3x4/(1.5x12) = 0.77

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D.5.2.6 – Cracking Effect


Concrete is cracked at service load

ψc,N = 1.0

Provide crack control reinforcement per 10.6.4



11

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D.5.2.7 – Post-installed Anchors in Uncracked Concrete without Supplementary Reinforcement


Anchors are cast-in place

ψcp,N = 1.0

- 22 -

D.5.2.1 – Concrete Breakout Strength of Anchor Group (Tension)

…………(D-5)


Ncbg = 1020 1296

0.90 x 0.77 x 1.0 x 1.0 x 64.2

= 35.0 kips



12

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φ = 0.75 for cast-in headed bolts (assuming Condition A) …….D.4.4(c)


0.75φ = 0.75 x 0.75 = 0.56

- 24 -


Design strength of the bolt group for concrete breakout in tension:

0.75φNcbg= 0.56 x 35.0 = 19.6 kips



13

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D.5.3 – Pullout Strength

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D.5.3.6 – Pullout Strength


Anchors used - 1/2" cast-in headed bolts with Hex head

Abrg = 0.291 in.2

NP = 8 x 0.291 x 4000 / 1000 = 9.31 kips

For cracked concrete under service load, ψc,P = 1.0

Npn = 1.0 x 9.31 = 9.31 kips

This is the pullout strength of one anchor



14

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φ = 0.70 for cast-in headed studs (Condition B applies for pullout failure) …….D.4.4(c)


0.75φ = 0.75 x 0.70 = 0.53

- 28 -


Design strength of a single bolt for pullout in tension:

0.75φNpng= 0.53 x 9.31 = 4.89 kips



15

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Equating with the maximum demand on a single bolt:

Nua/4 = 4.89 kips

Thus max. Nua = 19.7 kips = 0.75φNpng for the group of bolts

This is the maximum tension demand that the whole bolt formation can support for pullout failure in tension

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D.5.4 – Side-face Blowout



16

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D.5.4.2 – Side-face Blowout


Anchors used - Headed boltsMinimum edge distance in X-direction = 4 in. < 0.4hef (= 4.8 in.)Minimum edge distance in Y-direction = 4 in. < 0.4hef (= 4.8 in.)

Side-face blowout possible in both X- and Y-directions.Nsbg needs to be calculated in both directions and the smaller

strength would govern.

- 32 -


Failure in X-direction

6"

6"

4"

8" 4"

Bolts under consideration

ca1 = 4 in.

Anchor spacing = 6 in. < 6ca1

…......OK

s = 6 + 6 = 12 in.

Abrg = 0.291 in.2

f c = 4000 psi



17

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Nsbg = 32,752/1000 = 32.8 kips

- 34 -


Strength reduction factor, Φ:

φ = 0.75 for cast-in headed bolts (assuming Condition A) …….D.4.4(c)


0.75φ = 0.75 x 0.75 = 0.56



18

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Design strength of the bolt group in side-face blowout under tension:

0.75φNsbg = 0.56 x 32.8 = 18.4 kips

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Nua

Mua = Nua x 1

Nua/2 – Nua/8 = 3Nua/8

Nua/2 + Nua/8 = 5Nua/8

Proportion of tension demand carried by considered column of bolts = 5/8

The maximum tension demand that the whole bolt formation can support in side-face blowout

= 18.4x8/5 = 29.5 kips




19

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6"

6"

4"

8" 4"

Bolts under consideration

ca1 = 4 in.

Anchor spacing = 8 in. < 6ca1

…......OK

s = 8 in.

Abrg = 0.291 in.2

f c = 4000 psi

Failure in Y-direction

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Nsbg = 29,106/1000 = 29.1 kips




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Design strength of the bolt group in side-face blowout under tension:

0.75φNsbg = 0.56 x 29.1 = 16.4 kips

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Nua

Mua = Nua x 1

Nua/2 + Nua/8 = 5Nua/8

Proportion of tension demand carried by considered row of bolts = 5/12

The maximum tension demand that the whole bolt formation can support in side-face blowout

= 16.4x12/5 = 39.3 kips

Nua

Mua = Nua x 1

Nua/3 + Nua/12 = 5Nua/12Nua/3

Nua/3 - Nua/12 = Nua/4




21

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Governing side-face blowout strength of the whole bolt formation is the smaller of those computed by considering failure in X-direction (29.5 kips) and in Y-direction (39.3 kips)

= 29.5 kips

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D.5 – Design for Tensile Loading

D.5.1 – Steel Strength = 18.44 kipsD.5.2 – Concrete Breakout Strength = 19.6 kipsD.5.3 – Pullout Strength = 19.7 kipsD.5.4 – Concrete Side-Face Blowout Strength = 29.5 kips

Group strength in tension = 18.44 kips, governed by steel strength

Suitable for application in buildings assigned to SDC C or higher



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D.6 – Design for Shear Loading

D.6.1 – Steel StrengthD.6.2 – Concrete Breakout StrengthD.6.3 – Concrete Pryout Strength

- 44 -

8" 4"

6"

6"

4"

e Vx

20"

e Vx = 1 in.

e´Vy = 0



23

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Shear FailureCheck for possible reversal in direction of shear in anchor group due to load eccentricity e´Vx :

VuaVua

Vua/6

Vua/6

Vua/6

Vua/6

Vua/6

Vua/6

+ [Tua/8]/3

+ [Tua/8]/3

+ [Tua/8]/3

- [Tua/8]/3

- [Tua/8]/3

- [Tua/8]/3

Tua = Vua x 1

- 46 -

Shear Failure

Check for possible reversal in direction of shear in anchor group due to load eccentricity e´Vx :

Maximum shear = Vua/6 + Vua/24= 5Vua/24

Minimum shear = Vua/6 – Vua/24= Vua/8 ……..No shear reversal



24

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D.6.1 – Steel Failure (Shear)

- 48 -

D.6.1.2 – Steel Strength (Shear)

In the example problem, cast-in headed bolts are usedCalculating for one bolt:

Ase = 0.142 in.2

futa = 58,000 psi < 1.9fya (= 68,400 psi)…………OK< 125,000 psi………………….OK

Vsa = 0.142 x (0.6x58,000) / 1000 = 4.94 kips



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φ = 0.65 for ductile steel elements under shear…….D.4.4(a)


This factor is not required in steel failure modes in ACI 318-08

0.75φ = 0.75 x 0.65 = 0.49

- 50 -


Design strength of a single bolt for steel failure in shear:

0.75φVsa= 0.49 x 4.94 = 2.42 kips



26

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Equating with the maximum demand on a single bolt:5Vua/24 = 2.42 kips

Thus max. Vua = 0.75φVsa = 11.62 kips for the group of bolts

This is the maximum shear demand that the whole bolt formation can support for steel failure in shear in Y-direction

Shear in Y-direction

- 52 -


Equating with the maximum demand on a single bolt:Vua/6 = 2.42 kips

Thus max. Vua = 0.75φVsa = 14.52 kips for the group of bolts

This is the maximum shear demand that the whole bolt formation can support for steel failure in shear in X-direction

Shear in X-direction



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D.6.2 – Concrete breakout (Shear)

- 54 -

ACI 318-08 Section D.1 - Definitions

Anchor GroupIn the example problem:

In X-direction, ca1 = 4 in.Row spacing = 6 in. < 3 x 4 in. .......OK

In Y-direction, ca1 = 4 in.Column spacing = 8 in. < 3 x 4 in. .......OK

The example bolt formation acts as a group for shear acting in X- and in Y- directions.



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Shear in X-Direction

8" 4"

6"

6"

4"

vuax

- 56 -


Failure Mode 1: Concrete breakout occurs from the anchors closest to the edge

8" 4"

4"

1.5ca1

ca1 = 4 in.ha = 18 in. > 1.5ca1 (= 6 in.) ….OKca2,top > 1.5ca1 ……………OKca2,bottom < 1.5ca1

Only one edge distance < 1.5ca1

ca1 need not be revised

Vuax/2

6"

6"



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AVc = [1.5ca1 +6+6+4][Min(ha, 1.5ca1)]

= [1.5x4 + 6 + 6 + 4][1.5x4] = 132 in.2

AVco = 4.5ca12 = 4.5x42 = 72 in.2

Number of bolts, n = 3AVc < nAVco ……………OK

AVc/AVco = 1.83

8" 4"

4"

1.5ca1

6"

6"

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le = hef = 12 in. > 8do = 4 in. le = 4 in.

= 3796/ 1000 = 3.8 kips

Basic concrete breakout strength



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Eccentricity factor

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Edge factorca2, bottom = 4 in. < 1.5ca1 (= 6 in.)

= 0.90



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Cracking factor

Concrete is cracked at service load. It is assumed that supplementary reinforcement comprising of No. 4 bars is present between the anchors and the edges. No stirrup is used to enclose the reinforcement.

ψc,V = 1.2

- 62 -


………(D-22)

Vcbg = 132 72

1.0 x 0.90 x 1.2 x 3.8

= 7.52 kips




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φ = 0.75 for Condition A …….D.4.4(c) For SDC C or above, a factor of 0.75 is applied

..……D.3.3.30.75φ = 0.75 x 0.75 = 0.56


Strength reduction factor, φ

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Design strength of the first row in concrete breakout in shear:

0.75φVcbg= 0.56 x 7.52 = 4.21 kips

This needs to be compared against the fraction of the total shear demand causing this failure




33

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Vuax/2 = 4.21 kips Vuax = 8.42 kips

This is the total shear demand that the whole group can support before Mode 1 failure takes place

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Failure Mode 2: Concrete breakout occurs from the anchors in the 2nd column from the edge

4"

1.5ca1

ca1 = 8 + 4 = 12 in.ha = 18 in. = 1.5ca1 ……….OKca2,top > 1.5ca1 ……………OKca2,bottom < 1.5ca1



Vuax

6"

6"

8" 4"



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AVc = [1.5ca1 +6+6+4][Min(ha, 1.5ca1)]= [1.5x12 + 6 + 6 + 4][1.5x12] = 612 in.2

AVco = 4.5ca12 = 4.5x122 = 648 in.2


AVc/AVco = 0.94


4"

1.5ca1

Vuax

6"

6"

8" 4"

- 68 -



= 19,724/ 1000 = 19.7 kips





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Eccentricity factor


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Edge factorca2, bottom = 4 in. < 1.5ca1 (= 18 in.)

= 0.77




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Cracking factorConcrete is cracked at service load. It is assumed that supplementary reinforcement comprising of No. 4 bars is present between the anchors and the edges. No stirrup is used to enclose the reinforcement.

ψc,V = 1.2


- 72 -


………(D-22)

Vcbg = 612 648 1.0 x 0.77 x 1.2 x 19.7

= 17.19 kips




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..……D.3.3.30.75φ = 0.75 x 0.75 = 0.56



- 74 -


Design strength of the first row in concrete breakout in shear:

0.75φVcbg= 0.56 x 17.19 = 9.62 kips





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Vuax = 9.62 kips



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Design concrete breakout strength of the anchor group based on two failure modes:

1st Mode: 8.42 kips2nd Mode: 9.62 kips

0.75φVcbg,x = 8.42 kips



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Shear in Y-Direction

8" 4"

6"

6"

4"

e Vx = 1"

vuay

20"

- 78 -



8" 4"

4"

1.5ca1

ca1 = 4 in.ha = 18 in. > 1.5ca1 …….OKca2,right = 4 in. < 1.5ca1

ca2,left = 20 in. > 1.5ca1 ……OK



Vuay/3

20"



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8" 4"

4"

1.5ca1

AVc = [1.5ca1 + 8 + 4][Min(ha, 1.5ca1)]= [1.5x4 + 8 + 4][1.5x4] = 108 in.2

AVco = 4.5ca12 = 4.5x42 = 72 in.2


AVc/AVco = 1.5

20"

- 80 -




= 3796/ 1000 = 3.8 kips




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Eccentricity factor

- 82 -



Edge factorca2, right = 4 in. < 1.5ca1 (= 6 in.)

= 0.90



42

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ψc,V = 1.2

- 84 -



………(D-22)

Vcbg = 10872 0.86 x 0.90 x 1.2 x 3.8

= 5.29 kips



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φ = 0.75 for Condition A…….D.4.4(c)


0.75φ = 0.75 x 0.75 = 0.56


- 86 -



Design strength of the first row for concrete breakout in shear:

0.75φVcbg= 0.56 x 5.29 = 2.96 kips This needs to be compared against the fraction of the

total shear demand causing this failure



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Vuay/3 = 2.96 kips Vuay = 8.88 kips


- 88 -


Failure Mode 2: Concrete breakout occurs from the anchors in the 2nd row from the edge

ca1 = 4 + 6 = 10 in.ha = 18 in. > 1.5ca1 …….OKca2,right = 4 in. < 1.5ca1

ca2,left = 20 in. > 1.5ca1 ……OK



8" 4"

1.5ca1

4"

6"

2Vuay/3

20"



45

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AVc = [1.5ca1 + 8 + 4][Min(ha, 1.5ca1)]= [1.5x10 + 8 + 4][1.5x10] = 405 in.2

AVco = 4.5ca12 = 4.5x102 = 450 in.2


AVc/AVco = 0.9

8" 4"

1.5ca1

4"

6"


20"

- 90 -



= 15,005 / 1000 = 15 kips





46

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Eccentricity factor


- 92 -


Edge factor

ca2, right = 4 in. < 1.5ca1 (= 15 in.)

= 0.78




47

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ψc,V = 1.2


- 94 -


………(D-22)

Vcbg = 405 450

0.94 x 0.78 x 1.2 x 15

= 11.88 kips




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..……D.3.3.30.75φ = 0.75 x 0.75 = 0.56



- 96 -


Design strength of the 2nd row for concrete breakout in shear:

0.75φVcbg= 0.56 x 11.88 = 6.65 kips





49

- 97 -


2Vuay/3 = 6.65 kips Vuay = 9.98 kips



- 98 -


Failure Mode 3: Concrete breakout occurs from the anchors in the 3rd row from the edge

ca1 = 4 + 6 + 6 = 16 in.ha = 18 in. < 1.5ca1

ca2,right = 4 in. < 1.5ca1

ca2,left = 20 in. < 1.5ca1

3 edge distances < 1.5ca1

ca1 needs to be revised

8" 4"

Ca2,left = 20 in.

4"

6"

Vuay

6"

20"



50

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Revised ca1 = Greatest of -ca2,right /1.5 = 4 / 1.5 = 2.67 in.ca2,left /1.5 = 20 / 1.5 = 13.33 in.ha /1.5 = 18 / 1.5 = 12 in.(1/3) spacing between anchors = 8 / 3 = 2.67 in.

ca1 = 13.33 in.

- 100 -


AVc = [1.5ca1 + 8 + 4][Min(ha, 1.5ca1)]= [1.5x13.33 + 8 + 4][18] = 576 in.2

AVco = 4.5ca12 = 4.5x13.332 = 800 in.2


AVc/AVco = 0.72


8" 4"

1.5ca1 = 20 in.

4"

6"

6"

20"



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= 23,093 / 1000 = 23.1 kips



- 102 -


Eccentricity factor




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Edge factor

ca2, right = 4 in. < 1.5ca1 (= 20 in.)

= 0.76


- 104 -



ψc,V = 1.2




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………(D-22)

Vcbg = 576800 0.95 x 0.76 x 1.2 x 23.1

= 14.4 kips


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φ = 0.75 for Condition A ..….D.4.4(c) For SDC C or above, a factor of 0.75 is applied

..……D.3.3.30.75φ = 0.75 x 0.75 = 0.56





54

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Design strength of the 3rd row in concrete breakout in shear:

0.75φVcbg= 0.56 x 14.4 = 8.10 kips



- 108 -


Vuay = 8.10 kips





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Design concrete breakout strength of the anchor group based on three failure modes:

1st Mode: 8.88 kips2nd Mode: 9.98 kips3rd Mode: 8.10 kips

0.75φVcbg,y = 8.10 kips

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D.6.3 – Concrete Pryout



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Single anchor Vcp = kcpNcb ………….(D-29)Group of anchors Vcpg = kcpNcbg ………..(D-30)

where• kcp = 1.0 for hef < 2.5 in.• kcp = 2.0 for hef ≥ 2.5 in.• Ncb computed from Eq. (D-4)• Ncbg computed from Eq. (D-5)

- 112 -


In the current problem:

hef = 12 in. > 2.5 in. kcp = 2.0Ncbg = 35.0 kips

Vcpg = 2 x 35.0 = 70.0 kips



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φ = 0.70 (Condition B applies for pryout failure) …….D.4.4(c)


0.75φ = 0.75 x 0.70 = 0.53

- 114 -


Design strength of the anchor group in concrete pryout

0.75φNcpg= 0.53 x 70.0 = 37.1 kips



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D.6 – Design for Shear Loading

D.6.1 – Steel Strength = 14.52 kips for shear in X-direction= 11.62 kips for shear in Y-direction

D.6.2 – Concrete Breakout Strength = 8.42 kips for shear in X-direction= 8.10 kips for shear in Y-direction

D.6.3 – Concrete Pryout Strength = 37.1 kips for shear in both directions

Group strength for shear in X-direction = 8.42 kips, governed by concrete breakout strength

Group strength for shear in Y-direction = 8.10 kips, governed by concrete breakout strength

Not suitable for application in buildings assigned to SDC C or higher

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For more information…www.skghoshassociates.com

Anchor Example

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