Headed Bolt Anchor Design using ACI 318-14 Chapter 17 and Quick Anchor

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Headed bolt Design with three rows of anchors By Bodhi Rudra M.S.

1 Given Parameters          

Concrete Depth; ℎ𝑎 = 18 𝑖𝑛 Seismic Design Category; 𝑆𝐷𝐶 𝐶 Concrete is assumed to be cracked Anchor Material : ASTM F1554 Grade 36 𝑓𝑢𝑡𝑎 = 58000 𝑝𝑠𝑖 𝑓𝑦𝑡 = 105000 𝑝𝑠𝑖 𝐴𝑠𝑒,𝑁 = 0.142 𝑖𝑛2 𝐴𝑏𝑟𝑔 = 0.291 𝑖𝑛2 𝑓𝑐′ = 4000 𝑝𝑠𝑖 Loading eccentricities 𝑒𝑥 = 1 𝑖𝑛 𝑒𝑦 = −1 𝑖𝑛  Axial Tension; 𝑁𝑢 = 8 𝑘𝑖𝑝𝑠  Axial Shear Force along X axis 𝑉𝑢𝑥 = 0 𝑘𝑖𝑝𝑠  Axial Shear Force along Y axis; 𝑉𝑢𝑦 = −6 𝑘𝑖𝑝𝑠

Y

X

2 Tension in Anchors 𝑁𝑢 × 1 = 𝑁𝑢

𝑁𝑢 × 1 = 𝑁𝑢

(

𝑁𝑢 𝑁𝑢 − ) 3 12 = 𝑁𝑢 2 8

(

𝑁𝑢 ) 3 = 𝑁𝑢 2 6

𝑁𝑢 𝑁𝑢 5𝑁𝑢 ( 3 + 12) = 24 2

For each column of anchors (there are 2 columns)

(

𝑁𝑢 ) 8 = 𝑁𝑢 3 24

(

𝑁𝑢 ) 8 = 𝑁𝑢 3 24

For each row of anchors (there are 3 rows) 1

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Net forces on each anchor (positive is Tension)

𝑁𝑢 𝑁𝑢 𝑁𝑢 − = 8 24 12

𝑁𝑢 𝑁𝑢 𝑁𝑢 + = 8 24 6

𝑁𝑢 𝑁𝑢 𝑁𝑢 − = 6 24 8

𝑁𝑢 𝑁𝑢 5𝑁𝑢 + = 6 24 24

5𝑁𝑢 𝑁𝑢 𝑁𝑢 − = 24 24 6

5𝑁𝑢 𝑁𝑢 𝑁𝑢 + = 24 24 4

Anchor number 2 carries maximum tension of 𝑁𝑢 8 = = 2 𝑘𝑖𝑝𝑠 4 4

3 Requirements for Tension Loading Our Seismic Design Category is ‘C’, so we need to satisfy additional requirements for seismic design. …17.2.3.1 Tensile component of the strength level earthquake forces applied to anchors exceeds 20% of the total factored anchor tensile force. Hence, we have to design in accordance with 17.2.3.4.3 and corresponding strengths will be determined in accordance with 17.2.3.4.4 …17.2.3.4.2

4 Failure Modes 4.1. Steel strength of anchor in tension (17.4.1) 𝑁𝑠𝑎 = 𝐴𝑠𝑒,𝑁 𝑓𝑢𝑡𝑎

…Eq. 17.4.1.2

𝐴𝑠𝑒,𝑁 = 0.142 𝑖𝑛2 𝑓𝑢𝑡𝑎 = 58000 𝑝𝑠𝑖 2

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𝑓𝑢𝑡𝑎 ≯ min(1.9 𝑓𝑦𝑎 & 125000 𝑝𝑠𝑖) = 125000

...17.4.1.2

Since 𝑓𝑢𝑡𝑎 < 125000; 𝑓𝑢𝑡𝑎 = 58000 𝑝𝑠𝑖 𝑁𝑠𝑎 = 0.142 ×

58000 = 8.236 𝑘𝑖𝑝𝑠 1000

Design Strength = 𝜙𝑁𝑠𝑎 ≥ 𝑁𝑢𝑎,𝑖 𝜙 = 0.75 for ductile steel element

…Table 17.3.1.1 …17.3.3

𝜙𝑁𝑠𝑎 = 0.75 × 8.236 = 6.18 𝑘𝑖𝑝𝑠 Tension on most stressed anchors = 2 𝑘𝑖𝑝𝑠 2

Strength utilization = 6.18 = 0.32

4.2. Concrete breakout strength of anchor in tension (17.4.2) For concrete breakout in tension if 𝑠 < 3ℎ𝑒𝑓 , then we consider all the anchors as a group. For our case; 𝑠max = 8𝑖𝑛 < 3 × 12 = 36 𝑖𝑛 , so the anchors act as a group. ...17.2.1.1 Edge distance to the anchors are 4 𝑖𝑛, 4 𝑖𝑛 and 28 𝑖𝑛 from three sides. 1.5 ℎ𝑒𝑓 = 1.5 × 12 = 18 𝑖𝑛 Since only 2 edges are less than 1.5 ℎ𝑒𝑓 , full ℎ𝑒𝑓 will be taken

...17.4.2.3

For a group of anchors 𝐴

𝑁𝑐𝑏𝑔 = 𝐴 𝑁𝑐 𝜓𝑒𝑐,𝑁 𝜓𝑒𝑑,𝑁 𝜓𝑐,𝑁 𝜓𝑐𝑝,𝑁 𝑁𝑏

…Eq. 17.4.2.1b

2 𝐴𝑁𝑐𝑜 = 9ℎ𝑒𝑓 = 9 × 122 = 1296 𝑖𝑛2

…Eq. 17.4.2.1c

𝑁𝑐𝑜

a) Calculate 𝐴𝑁𝑐𝑜

Fig. R17.4.2.1 3

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b) Calculate 𝐴𝑁𝑐

𝐴𝑁𝑐 = (1.5ℎ𝑒𝑓 + 𝑠1 + 𝑐𝑎1 ) × (1.5ℎ𝑒𝑓 + 𝑠2 + 𝑠2 + 𝑐𝑎2 ) = (1.5 × 12 + 8 + 4) × (1.5 × 12 + 6 + 6 + 4) = 1020 𝑖𝑛2

c) 𝜓𝑒𝑐,𝑁 : modification factor for anchor groups loaded eccentrically in tension. …17.4.2.4 1 𝜓𝑒𝑐,𝑁 = ≤1 2𝑒 ′ 1+ 𝑛 3ℎ𝑒𝑓 Since, for our case all the anchors are in tension, the eccentricity will remain same which can also be calculated as below: 𝑁 𝑁 𝑁 5𝑁 𝑁 𝑁 (12𝑢 − 6𝑢 + 8𝑢 − 24𝑢 + 6𝑢 − 4𝑢 ) 𝑒𝑥 = × 4 = −1 𝑁𝑢 𝑁𝑢 𝑁𝑢 𝑁𝑢 5𝑁𝑢 𝑁𝑢 12 + 8 + 6 + 6 + 24 + 4

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For eccentric loading in both axis 𝜓𝑒𝑐,𝑁 = 𝜓𝑒𝑐,𝑁𝑥 × 𝜓𝑒𝑐,𝑁𝑦 𝜓𝑒𝑐,𝑁𝑥 = 𝜓𝑒𝑐,𝑁𝑦 =

1 = 0.947 2×1 1 + 3 × 12

𝜓𝑒𝑐,𝑁 = 0.947 × 0.947 = 0.897 d) 𝜓𝑒𝑑,𝑁 : modification factor for edge effects for single anchor or anchor groups loaded in tension. …17.4.2.5 𝑐𝑎,𝑚𝑖𝑛 = min(4, 4, 20) = 4𝑖𝑛 Since 𝑐𝑎,𝑚𝑖𝑛 = 4 < 1.5 × 12 = 18 𝑐𝑎,𝑚𝑖𝑛 𝜓𝑒𝑑,𝑁 = 0.7 + 0.3 ( ) = 0.77 1.5ℎ𝑒𝑓 e) 𝜓𝑐,𝑁 : modification factor for cracking Where analysis indicates cracking at service loads, 𝜓𝑐,𝑁 = 1

…17.4.2.6

f) 𝜓𝑐𝑝,𝑁 : modification factor for post-installed anchors designed for uncracked concrete without supplementary reinforcement to control splitting. …17.4.2.7 𝜓𝑐𝑝,𝑁 = 1 g) 𝑁𝑏 : basic concrete breakout strength.

…17.4.2.2

𝑁𝑏 = 𝑘𝑐 𝜆𝑎 √𝑓𝑐′ ℎ1.5 𝑒𝑓

…Eq. 17.4.2.2a

𝑘𝑐 = 24; for cast in anchor Since, ℎ𝑒𝑓 = 12 𝑖𝑛 and when 11 𝑖𝑛 ≤ ℎ𝑒𝑓 ≤ 25 𝑖𝑛 5

Then, 𝑁𝑏 =

3 16𝜆𝑎 √𝑓𝑐′ ℎ𝑒𝑓

𝜆𝑎 = 1, (modification factor for light weight concrete) 5 √4000 𝑁𝑏 = 16 × 1 × × 123 = 63.65 1000

…Eq. 17.4.2.2b …17.2.6

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Nominal concrete group breakout strength in tension, 𝑁𝑐𝑏𝑔 =

1020 × 0.897 × 0.77 × 1 × 1 × 63.65 = 34.6 𝑘𝑖𝑝𝑠 1296

Design strength for concrete breakout in tension 𝜙𝑁𝑐𝑏𝑔 ≥ 𝑁𝑢𝑎,𝑔 𝜙 = 0.7 for condition B where supplementary reinforcement is not present.

...Table 17.3.1.1 …17.3.3c(i)

The anchor design tensile strength for resisting earthquake forces shall be determined by 0.75𝜙𝑁𝑐𝑏𝑔

…17.2.3.4.4b

0.75 × 0.7 × 34.6 = 18.165 8

Strength utilization = 18.165 = 0.44

4.3. Pullout strength if cast-in, post installed expansion and undercut anchors in tension (17.4.3) Nominal pullout strength of single cast-in, post installed expansion and post installed undercut anchor in tension, 𝑁𝑝𝑛 , shall not exceed. 𝑁𝑝𝑛 = 𝜓𝑐,𝑃 𝑁𝑝 a) 𝜓𝑐,𝑝 : modification factor for cracking Where there is cracking under service load. 𝜓𝑐,𝑝 = 1

…17.4.3.6

b) 𝑁𝑃 : basic pullout strength in tension of a single headed stud or headed bolt 𝑁𝑝 = 8𝐴𝑏𝑟𝑔 𝑓𝑐′

…Eq. 17.4.3.4

= 8 × 0.291 × 4 = 9.3120 𝑘𝑖𝑝𝑠 So, 𝑁𝑝𝑛 = 1 × 9.3120 = 9.3120 𝑘𝑖𝑝𝑠 Design strength = 0.75𝜙𝑁𝑝𝑛 for a single anchor, or the most highly stressed individual anchor in a group of anchors. …17.2.3.4.4c 𝜙 = 0.7 , as Condition B applies for pullout strength

…17.3.3c (ii) 6

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= 0.75 × 0.7 × 9.312 = 4.89 Tension in most highly stressed anchor as calculated before = 2 𝑘𝑖𝑝𝑠 2

Strength utilization = 4.89 = 0.41

4.4. Concrete side-face blowout strength of a headed anchor in tension. (17.4.3) Since, for our case ℎ𝑒𝑓 = 12 𝑖𝑛 > 2.5 × 𝑐𝑎1 = 2.5 × 4 = 10 So, our case falls under deep embedment close to an edge, and side-face blowout should be checked. …17.4.4.1 For a single headed anchor 𝑁𝑠𝑏 = 160𝑐𝑎1 √𝐴𝑏𝑟𝑔 𝜆𝑎 √𝑓𝑐′ = 160 × 4 × √0.29 × 1 × √4000 = 21.8 𝑘𝑖𝑝𝑠 This will be same along both axis as 𝑐𝑎1 = 4 𝑖𝑛, along both the axis. For multiple headed anchors with deep embedment close to an edge and with anchor spacing 8𝑖𝑛 (𝑚𝑎𝑥 𝑠𝑝𝑎𝑐𝑖𝑛𝑔) < 6𝑐𝑎1 = 24 𝑖𝑛. …17.4.4.2 𝑁𝑠𝑏𝑔 = (1 +

𝑠 )𝑁 6𝑐𝑎1 𝑠𝑏

where ‘𝑠’ is the distance between the outer anchors along the edge, and 𝑁𝑠𝑏 is obtained from before without modification for a perpendicular edge distance. 𝑠 = 8 𝑖𝑛, outer edge distance along x-axis and 12 𝑖𝑛, along y-axis 𝑁𝑠𝑏𝑔,𝑦 = (1 +

8 ) 21.8 = 29.07 𝑘𝑖𝑝𝑠 6×4

𝑁𝑠𝑏𝑔,𝑥 = (1 +

12 ) 21.8 = 32.7 𝑘𝑖𝑝𝑠 6×4

Along x-axis, the right column i.e. anchors 2, 4, and 6 are resisting side-face blowout. Total 𝑁 𝑁 5𝑁 15𝑁 tension shared by this column = 6𝑢 + 4𝑢 + 24𝑢 = 24𝑢 24

So, the nominal side-face blowout strength along x-axis = 32.7 × 15 = 52.3 𝑘𝑖𝑝𝑠

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Along y-axis, the bottom most row, i.e. anchors 1 and 2, resist side-face blowout. Total 𝑁 𝑁 5𝑁 tension shared by this row = 6𝑢 + 4𝑢 = 12𝑢 So, the nominal side-face blowout strength along y-axis = 29.07 ×

12 5

= 69.77 𝑘𝑖𝑝𝑠

Governing nominal side-face blowout strength for the group 𝑁𝑐𝑏𝑔 = 52.3 𝑘𝑖𝑝𝑠 Design strength = 0.75𝜙𝑁𝑠𝑏𝑔 for a group of anchors

…17.2.3.4.4d

𝜙 = 0.7 , as Condition B applies where supplementary reinforcement is not present …17.3.3c (ii) = 0.75 × 0.7 × 52.3 = 27.46 𝑘𝑖𝑝𝑠 Tension applied on the group =8 𝑘𝑖𝑝𝑠 8

Strength utilization = 27.46 = 0.29

4.5. Concrete breakout strength of anchor in shear

…17.5.2

For anchors to be considered in a group𝑠 ≤ 3𝑐𝑎1 Along direction ‘X’ there is no shear and along ‘Y’ 𝑐𝑎1 = 4 𝑖𝑛; 3𝑐𝑎1 = 12 𝑖𝑛. Since 6 𝑖𝑛 ≤ 12 𝑖𝑛 , so the anchors shall be considered as a group.

…17.2.1.1

For a group 𝐴

𝑉𝑐𝑏𝑔 = 𝐴 𝑣𝑐 𝜓𝑒𝑐,𝑉 𝜓𝑒𝑑,𝑉 𝜓𝑐,𝑉 𝜓ℎ,𝑉 𝑉𝑏 𝑣𝑐𝑜

…17.5.2.1(b)

We have to check failure modes governing from different edges.

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4.5.1 Failure from first row of anchors nearest to the edge

Since 𝑠𝑦 = 6 𝑖𝑛 ≥ 𝑐𝑎1,1 = 4 𝑖𝑛; so we can distribute the shear force on each of the anchors. Forces acting at the centroid of the plate will be the shear acting in negative ‘Y’ direction and its corresponding torsion due to the eccentricity in positive ‘X’ direction. Due to Shear acting at the centroid, the calculations are straightforward

𝑉𝑢 6 = = 1 𝑘𝑖𝑝𝑠 6 6 Due to Torsion of 𝑉𝑢𝑦 × 𝑒𝑥 = 6 × 1 = 6 𝑘𝑖𝑝𝑠 − 𝑖𝑛 9

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θ

𝑠

Where, 𝑐𝑜𝑠𝜃 = 2𝑑𝑥 and from static equilibrium we know 2 × 𝑅 × 2 × 𝑑 + 𝑅𝑦 × 𝑠𝑥 = 𝑉𝑢 …Eq-1 𝑅 × 𝑐𝑜𝑠𝜃 = 𝑅𝑦 …Eq-2 From all the above equations it can be derived that 𝑠𝑥 𝑉𝑢 𝑅𝑦 = 𝑉𝑢 × 2 = = 0.1 𝑘𝑖𝑝𝑠 2 8𝑑 + 𝑠𝑥 60 Horizontal component of R will generate forces in horizontal direction over the anchors 1, 2, 5 and 6. 𝑠𝑦 𝑅𝑥 = 𝑅𝑦 × 𝑡𝑎𝑛𝜃 = 𝑅𝑦 × 2 × = 0.15 𝑘𝑖𝑝𝑠 𝑠𝑥

So, neglecting ‘X’ direction forces, and adding up forces in ‘Y’ direction for both the effects, 10

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The revised eccentricity for the redistributed forces will be: 𝑠𝑥 𝑒𝑟𝑒𝑣,𝑥 = 0.1 × 3 × = 0.4 𝑖𝑛 𝑉𝑢 2 a) 𝐴𝑣𝑐0 = 4.5𝑐𝑎1,1 = 4.5 × 42 = 72 𝑖𝑛2

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b) 𝐴𝑣𝑐 𝐴𝑣𝑐

ℎ𝑎 = 18 𝑖𝑛 > 1.5 𝑐𝑎1,1 = 6 𝑖𝑛 …OK = 1.5𝑐𝑎1,1 × (1.5𝑐𝑎1,1 + 𝑠𝑥 + 𝑐𝑎1,2 ) = 6 × (6 + 8 + 4) = 6 × 18 = 108 𝑖𝑛2

c) Calculate 𝑉𝑏 , basic concrete breakout strength in shear.

…17.5.2.2

𝑙𝑒 0.2 𝑉𝑏 = min {7 ( ) √𝑑𝑎 𝜆𝑎 √𝑓𝑐′ (𝑐𝑎1 )1.5 , 9𝜆𝑎 √𝑓𝑐′ (𝑐𝑎1 )1.5 } 𝑑𝑎 …Eq.17.5.2.2.a & b where, 𝑙𝑒 = min(ℎ𝑒𝑓 , 8𝑑𝑎 ) = 4 𝑖𝑛 and 𝜆𝑎 = 1 𝑉𝑏 = 3.8 𝑘𝑖𝑝𝑠 d) 𝜓𝑒𝑐,𝑉 : modification factor for eccentric loading.

…17.5.2.5

1 )≤1 2𝑒𝑉′ 1 + 3𝑐 𝑎1 Taking the revised eccentricity based on the distribution of shear and torsion in ‘Y’ direction. 𝑒𝑟𝑒𝑣,𝑥 = 0.4 𝑖𝑛 𝜓𝑒𝑐,𝑉 = 0.9375 𝜓𝑒𝑐,𝑉 = (

e) 𝜓𝑒𝑑,𝑉 : modification factor for edge effect for a single anchor or a group of anchors loaded in shear. …17.5.2.6 𝑐𝑎2 𝜓𝑒𝑑,𝑉 = 0.7 + 0.3 ( ) = 0.9 1.5𝑐𝑎1

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f) 𝜓𝑐,𝑉 = 1 for cracked concrete without supplementary reinforcement or with edge reinforcement smaller than No.4 bar. …17.5.2.7 g) 𝜓ℎ,𝑉 = 1 as ℎ𝑎 > 1.5𝑐𝑎1 108 𝑉𝑐𝑏𝑔 = × 0.9375 × 0.9 × 1 × 1 × 3.8 = 4.81 𝑘𝑖𝑝𝑠 72 Concrete breakout strength in shear , 𝜙𝑉𝑐𝑏𝑔 𝜙 = 0.7 for condition B for shear 𝜙𝑉𝑐𝑏𝑔 = 0.7 × 4.81 = 3.37 𝑘𝑖𝑝𝑠

…17.5.2.8

…17.3.3(c)(i)

Shear force on the first row of anchors = 1.1 + 0.9 = 2 𝑘𝑖𝑝𝑠 𝑉𝑢𝑎𝑦

Strength utilization = 𝜙𝑉

𝑐𝑏𝑔

2

= 3.37 = 0.59

Shear parallel to the edge a) 𝐴𝑣𝑐0 = 72 𝑖𝑛2 b) 𝐴𝑣𝑐 = (4 + 6 + 6 + 6) × 6 = 132 𝑖𝑛2 c) 𝜓𝑒𝑐,𝑉 , is not calculated for parallel to an edge case. d) 𝜓𝑒𝑑,𝑉 = 1, for shear parallel to an edge. e) 𝜓𝑐,𝑉 = 1 f) 𝜓ℎ,𝑉 = 1 g)

𝑉𝑏 = 𝑙

min {7 ( 𝑒 )

0.2

𝑑𝑎

√𝑑𝑎 𝜆𝑎 √𝑓𝑐′ (𝑐𝑎1 )1.5 , 9𝜆𝑎 √𝑓𝑐′ (𝑐𝑎1 )1.5 } =

3.8 𝑘𝑖𝑝𝑠

𝑉𝑐𝑏𝑔 =

132 × 72

1 × 1 × 1 × 3.8 = 6.96 𝑘𝑖𝑝𝑠

Concrete breakout strength for shear parallel to an edge is 2 × 𝜙𝑉𝑐𝑏𝑔 = 2 × 0.75 × 6.96 = 10.44 𝑘𝑖𝑝𝑠 Total shear in ‘Y’ for concerned edge 1.1 × 3 = 3.3 Strength utilization =

𝑉𝑢𝑎𝑦 𝜙𝑉𝑐𝑏𝑔

=

3.3 10.44

= 0.32 13

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4.5.2 Failure from second row of anchors

Since, 𝑠𝑦 < 𝑐𝑎2,1 (6 < 10) ; So it would not be conservative to consider breakout against the load distributed to all the active anchors. So, we shall check for concrete breakout by applying the entire load on the second row of anchors (CASE 3) …Fig. R17.5.2.1b

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Distribution of forces due to shear acting along ‘Y’ direction and torsion

The eccentricity will remain the same for this case.

a) 𝐴𝑣𝑐0 = 4.5 × 102 = 450 𝑖𝑛2 b) 𝐴𝑣𝑐 = (15 + 8 + 4) × 15 = 405 𝑖𝑛2 c) 𝜓𝑒𝑐,𝑉 = (

1+

1 2𝑒′𝑉 3𝑐𝑎1

) ≤ 1 = 0.94

𝑐

𝑎2 d) 𝜓𝑒𝑑,𝑉 = 0.7 + 0.3 (1.5𝑐 ) = 0.78 𝑎1

e) 𝜓𝑐,𝑉 = 1, for cracked concrete without supplementary reinforcement f) 𝜓ℎ,𝑉 = 1 0.2

𝑙

g) 𝑉𝑏 = min {7 (𝑑𝑒 ) 𝑎

√𝑑𝑎 𝜆𝑎 √𝑓𝑐′ (𝑐𝑎1 )1.5 , 9𝜆𝑎 √𝑓𝑐′ (𝑐𝑎1 )1.5 } = 15.05 𝑘𝑖𝑝𝑠

𝑉𝑐𝑏𝑔 =

405 × 0.94 × 0.78 × 1 × 1 × 15.05 = 9.93 𝑘𝑖𝑝𝑠 450

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ANCHOR DESIGN | ACI 318-14

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Concrete breakout strength in shear, 𝜙𝑉𝑐𝑏𝑔 = 0.7 × 9.93 = 6.95 𝑘𝑖𝑝𝑠 Shear force on the first row of anchors = 6 𝑘𝑖𝑝𝑠 Strength utilization 𝑉𝑢𝑎𝑦 6 = = 0.86 𝜙𝑉𝑐𝑏𝑔 6.95 Shear parallel to the edge a) 𝐴𝑣𝑐0 = 72 𝑖𝑛2 b) 𝐴𝑣𝑐 = (3 × 4) × 6 = 72 𝑖𝑛2 c) 𝜓𝑒𝑐,𝑉 , is not calculated for parallel to an edge case. d) 𝜓𝑒𝑑,𝑉 = 1, for shear parallel to an edge. e) 𝜓𝑐,𝑉 = 1 f) 𝜓ℎ,𝑉 = 1 g) 𝑉𝑏 = 𝑙

min {7 ( 𝑒 ) 𝑑𝑎

0.2

√𝑑𝑎 𝜆𝑎 √𝑓𝑐′ (𝑐𝑎1 )1.5 , 9𝜆𝑎 √𝑓𝑐′ (𝑐𝑎1 )1.5 } =

3.8 𝑘𝑖𝑝𝑠

𝑐𝑎1 = 4 𝑖𝑛 𝑉𝑐𝑏𝑔 =

72 × 1 × 1 × 1 × 3.8 = 3.8 𝑘𝑖𝑝𝑠 72

Concrete breakout strength for shear parallel to an edge is 2 × 𝜙𝑉𝑐𝑏𝑔 = 2 × 0.7 × 3.8 = 5.32 𝑘𝑖𝑝𝑠 Total shear in ‘Y’ for concerned edge = 3.75 𝑘𝑖𝑝𝑠 Strength utilization 𝑉𝑢𝑎𝑦 3.75 = = 0.70 𝜙𝑉𝑐𝑏𝑔 5.32

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ANCHOR DESIGN | ACI 318-14

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4.5.3 Failure from third row of anchors This case is not considered as already Case 3 has been considered in failure from second row anchors.

4.6. Steel strength of anchor in shear

…17.5.1

For cast in headed bolt anchors 𝑉𝑠𝑎 = 0.6 𝐴𝑠𝑐,𝑉 𝑓𝑢𝑡𝑎 …17.5.1.2b 𝐴𝑠𝑒,𝑉 = 0.142 𝑖𝑛2 𝑓𝑢𝑡𝑎 ≤ min(1.9𝑓𝑦𝑎 , 125000) 𝑓𝑢𝑡𝑎 = 58000 𝑉𝑠𝑎 = 0.6 × 0.142 ×

58000 = 4.94 𝑘𝑖𝑝𝑠 1000

Number of active anchors for concrete failure in shear = 4 anchors To calculate maximum shear in 4 anchors, we have distribute the shear force and torsion due to eccentricity. Due to Torsion

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2𝑅 × 2𝑑 = 6 ; 𝑅 =

6 4𝑑

;𝑅 =

6 4×√32 +42

𝑅𝑦 = 𝑅 × 𝑐𝑜𝑠𝜃 = 𝑅 ×

4 = 0.24 𝑑

𝑅𝑥 = 𝑅 × 𝑠𝑖𝑛𝜃 = 𝑅 ×

3 = 0.18 𝑑

=0.3

6

Due to shear all the anchors will carry a force of 4 = 1.5 𝑘𝑖𝑝𝑠 The total force on all the active anchors will be

2×0.24×8

Revised eccentricity = = 0.64 6 Design strength of the most stressed anchor = 𝜙𝑉𝑠𝑎 = 0.65 × 4.94 = 3.21 𝑘𝑖𝑝𝑠 …17.3.3(a) Strength utilization for most stressed anchor 𝑉𝑢𝑎𝑦 1.74 = = 0.54 𝜙𝑉𝑠𝑎 3.21

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ANCHOR DESIGN | ACI 318-14

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4.7 Concrete pryout strength of anchor in shear …17.5.3 For a group of anchors 𝑉𝑐𝑝𝑔 = 𝑘𝑐𝑝 × 𝑁𝑐𝑝𝑔 For cast-in anchors 𝑁𝑐𝑝𝑔 = 𝑁𝑐𝑏𝑔 , concrete breakout strength in tension.

…Eq. 17.5.3.1b …Eq. 17.4.2.1b

In the direction along ‘Y’ 𝐴

𝑁𝑐𝑏𝑔 = 𝐴 𝑁𝑐 𝜓𝑒𝑐𝑦,𝑁 𝜓𝑒𝑑,𝑁 𝜓𝑐,𝑁 𝜓𝑐𝑝,𝑁 𝑁𝑏 𝑁𝑐𝑜

…Eq. 17.4.2.1b

From 6.1 we know that the revised eccentricity for shear along ‘Y’ due to combined action of shear and torsion is = 0.4 𝑖𝑛. So, 𝜓𝑒𝑐𝑦,𝑁 =

𝑁𝑐𝑏𝑔 =

1 = 0.98 2 × 0.4 1 + 3 × 12

1020 × 0.98 × 0.77 × 1 × 1 × 63.65 = 37.8 𝑘𝑖𝑝𝑠 1296

Since, ℎ𝑒𝑓 = 12 𝑖𝑛 ≥ 2.5 𝑖𝑛 ; 𝑘𝑐𝑝 = 2 𝑉𝑐𝑝𝑔 = 2 × 37.8 = 75.6 𝑘𝑖𝑝𝑠 𝜙, for pryout strength: Condition B applies, so 𝜙 = 0.70 …17.3.3(c)(i) Design strength of the most stressed anchor = 𝜙𝑉𝑐𝑝𝑔 = 0.7 × 75.6 = 52.92 𝑘𝑖𝑝𝑠 …17.3.3(a) Strength utilization for most stressed anchor 𝑉𝑢𝑎𝑦 6 = = 0.11 𝜙𝑉𝑐𝑝𝑔 52.92

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ANCHOR DESIGN | ACI 318-14

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5 Check for ductility Anchors and their attachments shall satisfy one of options (a) through (d)

…17.2.3.4.3

a) For a group of anchors; the ratio of tensile load on the most highly stressed anchor to the steel strength of that anchor shall be equal to or greater than the ratio of tensile load on tension loaded anchors to the concrete governed strength of those anchors. Steel Strength = 1.2𝑁𝑠𝑎 = 1.2 × 8.236 = 9.88 𝑘𝑖𝑝𝑠 Tensile load on most highly stressed anchor (Anchor No. 2) =

…17.2.3.4.3a (i) 𝑁𝑢 4

= 2 𝑘𝑖𝑝𝑠

2

Utilization ratio = 9.88 = 0.2024 Number of tension loaded anchors, which is all in our case; so Tension load = 8 𝑘𝑖𝑝𝑠 Concrete governed strength will be the nominal strength considering pullout, sideface blowout and concrete breakout. …17.2.3.4.3a (ii) (i) Pullout: For consideration of pullout in groups, the ratio shall be calculated for the most highly stressed anchor. 𝑁𝑝𝑛 = 9.312 Tensile load on most highly stressed anchor (Anchor No. 2) = 2 𝑘𝑖𝑝𝑠 2

Utilization ratio for pullout = 9.312 = 0.215

(ii) Side-face blowout 𝑁𝑠𝑏𝑔 = 52.3 𝑘𝑖𝑝𝑠 𝑁𝑢 = 8 𝑘𝑖𝑝𝑠 8

Utilization ratio for side-face blowout = 52.3 = 0.1530 (iii) Concrete Breakout 𝑁𝑐𝑏𝑔 = 34.6 𝑘𝑖𝑝𝑠 𝑁𝑢 = 8 𝑘𝑖𝑝𝑠 8

Utilization ratio for concrete breakout = 34.6 = 0.2312 Governing concrete utilization ratio is for concrete breakout= 0.2312. This is greater than the utilization ratio for steel failure= 0.2024. Hence, our mode of failure is brittle. So we need to check other options b,c, and d

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ANCHOR DESIGN | ACI 318-14

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6 Summary S.No.

Loading Type

Failure Mode

Utilization Ratio

1

Steel strength of anchor in tension

0.32

2

Concrete breakout strength of anchor in tension

0.44

3

Pullout strength

0.41

4

Concrete side-face blowout

0.29

5

Concrete breakout strength of anchor in shear

0.86

Steel strength of anchor in shear

0.54

Concrete pryout strength of anchor in shear

0.11

Tension

6

Shear

7

Governing Tension Load 0.44 Governing Shear Load 0.86

7 Interaction of tensile and shear forces Since,

𝑉𝑢𝑎 𝜙𝑉𝑛

> 0.2 for the governing strength in shear, and

𝑁𝑢𝑎 𝜙𝑁𝑛

…17.6 > 0.2 for the governing strength in

tension, then

…17.6.3 𝑁𝑢𝑎 𝑉𝑢𝑎 + ≤ 1.2 𝜙𝑁𝑛 𝜙𝑉𝑛 0.44 + 0.86 = 1.3 > 1.2, … 𝑁𝑂𝑇 𝑂𝐾

All the results have been verified using Quick Anchor, software by S K Ghosh Associates, which provides detailed calculations at each and every step. The numbers got are within the same range as that from the software.

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