Mexico DF Ejemplos Anclajes

1-1

Example 1 - Single Headed Anchor in Tension Away from Edges

Check the capacity of a single anchor, 1 in. diameter, F1554 Grade 36 headed bolt with heavy-hex head installed in the top of a foundation without edge effects to resist a factored load of 17,000 lb tension (determined from ACI 318 9.2.1). The foundation is located in an area of high seismic risk. Assume normal weight concrete and that a crack forms in the plane of the anchor. Note: Foundation reinforcement not shown for clarity

Calculations and Discussion

Code Reference

1. The factored design load is: Nua = 17,000 lb

2. General requirement for anchor strength

φ Nn ≥ Nua where φ Nn is the lowest design strength from all appropriate failure modes in tension as determined from consideration of φ Nsa , φ Npa , φ Nsb , and φ Ncb. In regions of moderate or high seismic risk, the design strength of anchors is limited to 0.75 φ Nn . 0.75 φ Nn ≥ Nua

3. Check strength of steel anchor and requirements for ductility

Since the anchor is located in a region of moderate or high seismic risk, the anchor strength shall be governed by tensile or shear strength of a ductile steel element, ACI 318, D.3.3.4 unless the other parts of the attachment are designed to yield (D.3.3.5). ASTM F1554 Grade 36 bolt material meets the requirements of the ductile steel element definition in D.1 [tensile elongation of at least 14 percent and reduction of area of at least 30 percent]. F1554 material has 23 % elongation in 2 in. of length with 40 % reduction in area (See Table A). The basic design strength for the steel anchor in regions of moderate or high seismic risk is: 0.75 φ Nsa ≥ Nua and φ Nsa < capacity of concrete breakout, pullout, side-face blowout, or splitting failure in tension to satisfy ductility requirements, where:

D.4.1.2

Eq. (D-1)

D.4.1.2

D.3.3.3

D.3.3.4

17,000 lb

hef = 8 in.

8 in.

psi 6,000='cf

BORRADOR

1-2


Code Reference

φ = 0.75 and the steel bolt strength is:

D.4.4(a)

Nsa = n Ase (futa)

where the minimum value for futa = 58,000 psi is given for F1554, Grade 36 material. See Table A of this design guide. Note, futa shall not be taken greater than the smaller of 1.9 fya and 125,000 psi (See D.5.1.2).

n = 1 (single anchor) ANSI / ASME B1.1 defines Ase:

2

9743.04 ⎜⎜⎝

⎛⎟⎟⎠

⎞−=

tose n

dA π

substituting 1 in. for do and 8 (number of threads) for nt, Ase = 0.606 in.2 Nsa = (1) (0.606 in.2) (58,000 psi) = 35,148 lb φ Nsa = (0.75) 35,148 lb = 26,361 lb Reduction for regions of moderate or high seismic risk 0.75 (φ Nsa) = 0.75 (26,361 lb) = 19,771 lb 19,771 lbs > 17,000 lbs (requirement of 0.75φ Nsa > Nua is met)

4. Using the embedment depth, (hef = 8 in.), check the various capacities of the concrete

The basic capacity for nominal concrete breakout strength in tension for a single anchor in regions of moderate or high seismic risk is: 0.75 φ Ncb ≥ Nua and φ Ncb > φ Nsa (to ensure that the steel failure mode governs) where:

φ = 0.70 (for concrete breakout), φ = 0.75 (for ductile steel in tension) Determine the concrete breakout strength assuming condition B (no supplementary reinforcement has been provided)

bNcpNcNedNco

Nccb N

AA

N ,,, ψψψ=

Eq. (D-3)

RD.5.1.2

D.3.3.3

D.4.4

Eq. (D-4)

BORRADOR

1-3


Code Reference

where:

the Nco

Nc

AA

= 1.0 for single anchor remote from edges (a complete breakout prism

is developed and no reduction is required) ψed,N = 1.0 since anchor is remote from edges, (ca,min ≥ 1.5 hef) ψc,N = 1.0 since for this example it has been assumed cracking occurs in the concrete at service load levels ψcp,N = 1.0 (for cast-in anchors) The basic concrete breakout strength of a single anchor in tension is:

5.1'efccb hfkN =

where kc = 24 for cast-in anchors For this example hef = 8 in. which is less than 11 in. so equation D-8,

( 3/5'16 efcb hfN = ), does not apply

Substituting:

5.1)8(000,624=bN = 42,065 lb and φ Ncb = 0.7(42,065 lb) = 29,445 lb Reduction for regions of moderate or high seismic risk 0.75 φ Ncb = 0.75 (29,445 lb) = 22,084 lb 22,084 lb ≥ Nua and 22,084 lb > 0.75φ Nsa (or 19,771 lb) therefore the anchor strength remains governed by ductile steel failure

5. Check the anchor bolt pullout strength

To prevent a pullout failure mode, the requirement for pullout strength checks the bearing stress under bolt head. This failure mode is initiated by crushing of the concrete: 0.75 φ Npn ≥ Nua and φ Npn > φ Nsa (to ensure ductile steel failure mode governs) where: φ = 0.70

RD.5.2.1

D.5.2.5

D.5.2.6

D.5.2.7

(D-7)

D.3.3.3

D.4.4(c)(ii)

BORRADOR

1-4


Code Reference

pPcpn NN ,ψ=

where:

cbrgp fAN ′=8 ψc,P = 1.0 considering concrete cracking at service loads For 1 in. diameter heavy hex bolt head the bearing area of the head, Abrg = 1.501 in.2 (See Table B1 of this design guide) Substituting:

lbpsiinN p 048,72)000,6().501.1(8 2 == and φ Npn = 0.7 (72,048 lb) = 50,433 lb Reduction for regions of moderate or high seismic risk 0.75 φ Npn = 0.75 (50,433 lb) = 37,825 lb 37,825 lb ≥ Nua and 37,825 lb > 0.75φ Nsa (or 19,771 lb) therefore the anchor strength remains governed by ductile steel failure

6. Evaluate concrete side-face blowout failure mode

Since this anchor is remote from a free edge of concrete (ca1 ≥ 0.4 hef), concrete side-face blowout failure mode is not applicable.

7. Evaluate splitting failure: This type of failure can occur in thin slabs where the anchors have torque applied. The minimum edge distance for cast-in anchors with torque applied is 6 (do) where do is the bolt diameter. This failure mode is not applicable since the anchor is remote from an edge.

8. Summary:

Steel 0.75φ Nsa 19,771 lb (Controls) Concrete Breakout 0.75 φ Ncb 22,084 lb Concrete Pullout 0.75 φ Npn 37,825 lb

Side-Face Blowout NA The lowest design strength for considering all failure modes 19,771 lb for steel failure (controlled by the seismic reduction design capacity equation 0.75φ Nsa ); therefore the ASTM F1554 Grade 36, 1-in. diameter anchor bolt with a heavy-hex head and an 8-in. embedment is adequate to resist the 17,000 lb tension factored load in a moderate or high seismic region.

Eq. (D-14)

Eq. (D-15)

D.5.3.6

D.3.3.3

D.5.4

D.8.2

BORRADOR

1-5


Code Reference

BORRADOR

2-1

Example 2 – Single Hooked Anchor in Tension Away from Edges

Check the capacity of a single anchor, 1 in. diameter, F1554 Grade 36 standard hooked bolt (L-bolt) installed in the top of a foundation without edge effects to resist a factored load of 17,000 lb tension (determined from ACI 318 9.2.1). The foundation is located in an area of high seismic risk. Assume normal weight concrete and that a crack forms in the plane of the anchor. Note: Foundation reinforcement not shown for clarity


Code Reference

1. The factored design load is:

Nua = 17,000 lb

2. General requirement for anchor strength:

φ Nn ≥ Nua

where φ Nn is the lowest design strength from all appropriate failure modes in tension as determined from consideration of φ Nsa , φ Npa , φ Nsb , and φ Ncb. In regions of moderate or high seismic risk, the design strength of anchors is limited to 0.75 φ Nn.

3. Check strength of steel anchor and requirements for ductility

Since the anchor is located in a region of moderate or high seismic risk, the anchor strength shall be governed by tensile or shear strength of a ductile steel element, ACI 318 D.3.3.4. ASTM F1554 Grade 36 material meets the requirements of the ductile steel element definition in D.1 [tensile elongation of at least 14 percent and reduction of area of at least 30 percent]. F1554 material has 23 % elongation in 2 in. of length with 40% reduction in area (See Table A). The basic design strength for the steel anchor in regions of moderate or high seismic risk is: 0.75 φ Nsa ≥ Nua and φ Nsa < the capacity of concrete breakout, pullout, side-face blowout, or splitting failure in tension to satisfy ductility requirements, where: φ = 0.75 and the steel bolt strength is:

D.4.1.2 Eq. (D-1)

D.4.1.2

D.3.3.3

D.3.3.4

D.4.4(a)

hef = 8 in.

17,000 lb

eh = 3 in. 8 in.

psi 6,000='cf

BORRADOR

2-2


Code Reference

Nsa = n Ase (futa)

Eq. (D-3)

where the minimum value for futa = 58,000 psi is given for F1554 Grade 36 material. See Table A of this design guide. Note, futa shall not be taken greater than the smaller of 1.9 fya and 125,000 psi (See D.5.1.2). n = 1 (single anchor) ANSI / ASME B1.1 defines Ase:

29743.0

4 ⎜⎜⎝

⎛⎟⎟⎠

⎞−=

tose n

dA π.

Substituting 1 in. for do and 8 (number of threads) for nt Ase = 0.606 in.2 Nsa = (1) (0.606 in.2) (58,000 psi) = 35,148 lb φ Nsa = (0.75)35,148 lb = 26,361 lb Reduction for regions of moderate or high seismic risk 0.75 (φ Nsa ) = 0.75 (26,361 lb) = 19,771 lb 19,771 lbs > 17,000 lbs (requirement of 0.75φ Nsa ≥ Nua is met)

4. Using the embedment depth (hef = 8 in.) check the various capacities of the concrete

The basic capacity for nominal concrete breakout strength in tension for a single anchor in regions of moderate or high seismic risk is: 0.75 φ Ncb ≥ Nua and φ Ncb > φ Nsa (to ensure that the steel failure mode governs) where: φ = 0.70 (for concrete breakout), φ = 0.75 (for ductile steel in tension) Determine the concrete breakout strength assuming condition B (no supplementary reinforcement has been provided)

bNcpNcNedNco

Nccb N

AA

N ,,, ψψψ=

where:

the Nco

Nc

AA

= 1.0 for single anchor remote from edges (a complete breakout prism is

developed and no reduction is required)

RD.5.1.2

D.3.3.3

D.4.4

Eq. (D-4)

RD.5.2.1

BORRADOR

2-3


Code Reference

ψed,N = 1.0 since anchor is remote from edges, (ca,min ≥ 1.5 hef) ψc,N = 1.0 since for this example it has been assumed cracking occurs in the concrete at service load levels ψcp,N = 1.0 (for cast-in anchors) The basic concrete breakout strength of a single anchor in tension is:

5.1'efccb hfkN =

where kc = 24 for cast-in anchors For this example hef = 8 in. which is less than 11

in. so equation D-8, ( 3/5'16 efcb hfN = ), does not apply Substituting:

5.1)8(000,624=bN = 42,065 lb and φ Ncb = 0.7 (42,065 lb)= 29,445 lb Reduction for regions of moderate or high seismic risk 0.75 (φ Ncb) = 0.75 (29,445 lb) = 22,084 lb 22,084 lb ≥ Nua and 22,084 lb > 0.75φ Nsa (or 19,771 lb) therefore the anchor strength at this point remains governed by ductile steel failure.

5. Check the anchor bolt pullout strength To prevent a pullout failure mode, the requirement for pullout strength of the hooked bolt checks the bearing stress inside the hook. This failure mode is initiated by crushing of the concrete: 0.75 φ Npn ≥ Nua and φ Npn > φ Nsa (to insure ductile steel failure mode governs) where: φ = 0.70

pPcpn NN ,ψ= where:

ohcp defN '9.0=

D.5.2.5

D.5.2.6

D.5.2.7

Eq. (D-7)

D.3.3.3

D.5.3

D.4.4(c)(ii)

Eq. (D-14)

Eq. (D-16)

BORRADOR

2-4


Code Reference

ψc,P = 1.0 considering concrete cracking at service loads.

he = 3 in., od = 1 in.

and the hook length falls within the range 3 4 .5≤ ≤o h od e d Substituting:

( )( )( ) lbpsiN p 200,16"1"360009.0 == and φ Npn = 0.7 (16,200 lb) = 11,340 lb Reduction for regions of moderate or high seismic risk 0.75 (11,340 lb) = 8,505 lb 0.75 φ Npn < Nua and φ Npn < φ Nsa strength requirements are not met. Anchor pullout strength is less than the factored load and less than the steel capacity due to pullout capacity of the hook. Extending the hook will increase the capacity but the hook length is limited to 4.5 times anchor diameter or 4 ½ in. (D.5.3.5) Try extending hook to the maximum allowed amount. Substituting 4.5 in. for he , recalculate pN and φ Npn :

( )( )( ).9 6000 psi 4.5" 1" 24,300 lbpN = =

and φ Npn = 0.7 (24,300 lb) = 17,010 lb Reduction for regions of moderate or high seismic risk 0.75 (φ Npn) = 0.75 (17,010 lb) = 12,757 lb Anchor pullout strength with seismic reduction is less than factored load, 17,000 lb, and less than the steel capacity, 19,771 lb.

6. Evaluate concrete side-face blowout failure mode

Since this anchor is located far from a free edge of concrete (ca1 ≥ 0.4 hef), concrete side-face blowout failure mode is not applicable, however, Section D.5.4 for side-face blowout applies only to headed anchors.

7. Evaluate splitting failure: This type of failure can occur in thin slabs where the anchors have torque applied. The minimum edge distance for cast-in anchors with torque applied is 6 (do) where do is the bolt diameter. This failure mode is not applicable since the anchor is remote from an edge.

D.5.3.6

D.5.3.5

D.3.3.3

D.3.3.3

D.5.4

D.8.2

BORRADOR

2-5


Code Reference

8. Summary:

Steel 0.75φ Nsa 19,771 lb

Concrete Breakout 0.75 φ Ncb 22,084 lb Concrete Pullout 0.75 φ Npn 12,757 lb Controls

Side-Face Blowout NA The ASTM F1554 Grade 36, 1 in. diameter hooked bolt with an 8 in. embedment using the maximum allowed effective hook length of 4 ½ in. with the 0.75 seismic reduction is inadequate to resist the 17,000 lb tension factored load and cannot be used in a moderate to high seismic region since the ductile steel failure mode does not govern the capacity. In a non-seismic or low seismic region the hooked bolt with 4 ½ in. hook is adequate to resist the factor load of 17,000 lbs. with the pullout strength (17,010 lb) being the governing mode of failure.

BORRADOR

3-1

Example 3 – Single Post-Installed Anchor in Tension Away from Edges

Determine the minimum diameter post-installed torque-controlled expansion anchor for installation in the bottom of an 8-in. slab with a concrete compressive strength of cf ′ = 4,000 psi to support a 3,000 lb service dead load. The anchor will be in the tension zone (cracking at service load level is assumed), away from edges and other anchors in normal weight concrete. See Table C of this document for sample anchor installation and performance data. The data in Table C is not from any specific anchor and should not be used for design in accordance with ACI 318-05, Appendix D. However it is similar to what would be expected from testing and an evaluation report prepared by an independent testing and evaluation agency for the manufacturer in accordance with ACI 355.2-04. Example 3 Calculations and Discussion

Code Reference

1. Determine factored design load: Nua = 1.4 D Nua = 1.4 (3,000) = 4,200 lb

9.2.1

2. Determine fastener material:

Select a qualified post-installed torque-controlled expansion anchor with a bolt conforming to the requirements of ASTM F 1554 Grade 55. Design information resulting from anchor prequalification testing according to ACI 355.2 for the selected torque-controlled expansion anchor is given in Table C of this document. For ASTM F 1554 Grade 55: futa = 75,000 psi fya = 55,000 psi Elongation at 2 in. = 21 % min. Reduction of area = 30 % min.

ACI 318 Appendix D requires 14 % min. elongation and 30 % min. reduction of area to qualify as a ductile steel anchor.

ACI 355.2-04, Table 4.2

D.1

3,000 lb

8-in. hef

BORRADOR

Example 3 (cont’d) Calculations and Discussion

Code Reference

3-2

3. Steel strength requirement under tension loading:

φNn ≥ Nua = 4,200 lb

where φNn is the lowest of φNsa, φNpn , φNsb, and φNcb

Nsa = nAse futa

futa ≤ the smaller of 1.9fya and 125,000 psi.

φ nAse futa ≥ Nua = 4,200 lb For ductile steel, φ = 0.75 n = 1 (single anchor) Substituting and solving for Ase:

uta

se fA

φ200,4

≥

000757502004

,.,Ase ×

≥

Ase ≥ 0.0747 in.2 (minimum effective cross-sectional area of anchor) ∴ Bolt diameter ≥ 3/8-in. (Ase = 0.0775 in.2)

D.5.1

D.4.1.1 and

D.4.1.2

Eq. (D-3)

D.5.1.2

D.4.4

4. Concrete breakout strength requirement for tension loading: φNcb ≥ Nua = 4,200 lb

bNcpNcNedNco

Nccb N

AAN ,,, ψψψ=

where: 5.1'

efccb hfkN =

substituting:

lbhfkAA

efccNcpNcNedNco

Nc 200,45.1,,, =′⋅ψψψφ

where:

1=Nco

Nc

AA

for a single anchor

kc = 17 for a post-installed anchor where cracking is expected ψed,N = 1 for no edge effects ψcp,N = 1 assuming cracking at service loads for a category 1 anchor, φ = 0.65 (condition B, no supplemental reinforcement) for a category 2 anchor, φ = 0.55 (condition B, no supplemental reinforcement)

D.5.2

Eq. (D-1)

Eq. (D-4)

Eq. (D-7)

D.5.2.5D.5.2.6

D.4.4(c)iiD.4.4(c)ii

BORRADOR


Code Reference

3-3

[Note that an anchor category is established independently for each anchor diam-eter from the reliability tests of ACI 355.2. Smaller diameter anchors are, in general, more sensitive to the conditions established for reliability tests, including reduced torque, variations in hole diameter and cracks in the concrete.] Assuming a category 1 anchor, solve for the required embedment, hef, corresponding to the desired concrete breakout strength:

32

000,41711165.0200,4

⎥⎦

⎤⎢⎣

⎡

×××××≥efh = 3.3 in.

If we instead assume a category 2 anchor, the required embedment becomes: 3

2

000,41711155.0200,4

⎥⎦

⎤⎢⎣

⎡

×××××≥efh = 3.7 in.

for a category 2 anchor for breakout strength. Substituting the data from anchor categories and the embedment depths from Table C into the combined expression:

φNcb = 5.1',,, efccNcpncNed

Nco

Nc hfkAA

ψψψφ yields the following table;

Acceptable anchor diameters and embedment depths for concrete breakout strength in tension (for this example) are: Diameter (in.) Suitable embedments (in.) Category Minimum hef (in.)

3/8 4.5 2 3.7 ½ 5.5 2 3.7

5/8 4.5, 6.5 1 3.3 3/4 3.5, 5,8 1 3.3

5. Pullout strength: For static loading, pullout strength is established by reference tests in cracks and by the crack movement reliability test of ACI 355.2. For post-installed anchors, data from the anchor prequalification testing must be used. φNpn ≥ Nua = 4,200 lb

ppcpn NN ,ψ= where: ψc.p = 1 φ = 0.65 for a Category 1 and Condition B φ = 0.55 for a Category 2 and Condition B From the evaluation report, the following 5 percent fractile capacities were established for the embedments that satisfy concrete breakout strength using the above equation:

D.5.3

Eq. (D-1)Eq. (D-14)

D.5.3.6

D.4.4(c)iiD.4.4(c)ii

BORRADOR


Code Reference

3-4

diameter (in.) embedment (in.) anchor pullout capacity (lb) 3/8 4.5 φNpn = 0.55 × 5,583 = 3,070 < 4,200 lb – N.G. 1/2 5.5 φNpn = 0.55 × 7,544 = 4,149 < 4,200 lb – N.G. 5/8 4.5 φNpn = 0.65 × 8,211 = 5,337 > 4,200 lb – OK 5/8 6.5 φNpn = 0.65 × 14,254 = 9,265 > 4,200 lb – OK 3/4 3.5 φNpn = 0.65 × 5,632 = 3,661 < 4,200 lb – N.G. 3/4 5 φNpn = 0.65 × 9,617 = 6,251 > 4,200 lb – OK 3/4 8 φNpn = 0.65 × 19,463 = 12,651 > 4,200 lb – OK Try a 5/8 in. diameter anchor with hef = 4.5 in. from the selected anchor system.

6. Check all failure modes: For a post-installed torque-controlled expansion anchor with a 5/8 in. diameter and embedment depth of 4.5 in.: a) For steel strength: φNsa = φ nAse futa ≥ Nua = 4,200 lb 0.75 × 1 × 0.2260 × 75,000 = 12,712 lb > 4,200 lb – OK b) For concrete breakout strength:

φNcb = 5.1',, efccNcpNed

Nco

Nc hfkAA ψψφ ≥ Nua = 4,200 lb

φNcb = 0.65 × 1 × 1 × 1 × 17 × 51540004 .., × = 6,671 lb > 4,200 lb – OK

c) For pullout strength: φNpn ≥ Nua = 4,200 lb (value obtained from Table C, manufacturer’s evaluation report) 0.65 × 8,211 = 5,337 lb > 4,200 lb – OK

D.5.1

D.5.2

D.5.3

7. Check for minimum concrete thickness:

In order to prevent splitting, the thickness of the concrete in which the anchor is embedded must be at least 1.5 hef. 1.5 × 4.5 = 6-3/4 in. < 8 in. – OK

D.8.5

8. Summary:

Strength mode Strength (lb) Steel strength φNsa 12,712 Concrete breakout strength φNcb 6,671 Concrete pullout strength φNpn 5,337

Therefore pullout strength controls.

BORRADOR


Code Reference

3-5

9. Final Recommendation: Use a post-installed torque-controlled expansion anchor with a 5/8-in. diameter and an embedment depth of 4.5-in. meeting the requirements of ASTM F 1554, Grade 55.

BORRADOR

4-1

Example 4 - Group of Headed Studs in Tension Near an Edge

Design a group of four welded, AWS D1.1 Type B welded headed studs spaced 6-in. on center each way and concentrically loaded with a 10,000 lb service dead load. The anchor group is to be installed in the bottom of an 8-in. thick normal weight concrete slab made with the centerline of the connection 6-in. from a free edge of the slab. Note: Reinf. not shown for clarity


Code Reference

1. Determine factored design load Nua = 1.4 (10,000) = 14,000 lb

2. Determine anchor diameter

Using AWS D1.1 Type B welded, headed studs. Assume tension steel failure controls. The basic requirement for the anchor steel is: φ Nsa ≥ Nua where: φ = 0.75

Note: Per the Ductile Steel Element definition in D.1, AWS D1.1 Type B studs qualify as a ductile steel element (20% minimum elongation in 2-in. which is greater than the 14% required and a minimum reduction in area of 50% that is greater than the 30% required. See Table A for this information.

Nsa = n Ase futa For design purposes, Eq. (D-1) with Eq. (D-3) may be rearranged as:

uta

uase nf

NA

φ≥

9.2

D.5.1

Eq. (D-1)

D.4.1.2

D.4.4(a)i

Eq. (D-3)

psi 4,000='cf

6 in.

hef

6 in.

6 in.

½ in. plate

8 in.

10,000 lb

BORRADOR

4-2

Example 4 (cont’d) Calculations and Discussion Code

Reference where: Nua = 14,000 lb φ = 0.75 n = 4 futa = 65,000 psi

Note: Per Section D5.1.2, futa shall not be taken greater than 1.9 fya or 125,000 psi. For AWS D1.1 headed studs, 1.9 fya = 1.9 (51,000 psi) = 96,900 psi, therefore use the specified minimum futa of 65,000 psi.

Substituting:

( )( ) 072.0000,65475.0

000,14=≥seA in2

Per Table B1, ½-in. diameter welded, headed studs will satisfy this requirement (Ase = 0.196 in.2). Note: Per AWS D1.1 Table 7.1, Type B welded studs applies to studs in ½- in., 5/8-in., ¾-in., 7/8-in., and 1 in. diameters. Although individual manufacturers may list smaller diameters they are not explicitly covered by AWS D1.1 The total design steel strength of four ½-in. headed studs:

( )( )( ) 220,38196.0000,65475.0 ==saNφ lb.

3. Determine the required embedment length (hef) based on concrete breakout in tension:

Two different equations are given for calculating concrete breakout strength; for single anchors Eq. (D-4) applies, and for anchor groups Eq. (D-5) applies. An “anchor group” is defined as:

“a number of anchors of approximately equal effective embedment depth with each anchor spaced at less than three times its embedment depth from one or more adjacent anchors.”

Since the spacing between anchors is 6-in., the anchors must be treated as a group if the embedment depth exceeds 2-in.. Although the embedment depth is unknown, at this point it will be assumed that the provisions for an anchor group will apply.

The basic requirement for embedment of a group of anchors is: φ Ncbg ≥ Nua where:

D.5.1.2

D.5.2

D.1

Eq. (D-1)

D.4.1.2

BORRADOR

4-3


Code Reference

φ = 0.70, for anchors governed by concrete breakout and Condition B applies since no supplementary reinforcement has been provided (e.g., hairpin type reinforcement that ties the failure prism into the structural member).

, , , ,Nc

cbg ec N ed N c N cp N bNco

AN NA

ψ ψ ψ ψ=

Since this connection is likely to be affected by both group effects and edge effects, the embedment length hef is difficult to solve for directly. In this case, an embedment length must be assumed at the outset and then be proven to satisfy the requirement of Eq. (D-1). Note: Welded studs are generally available in fixed lengths. Available lengths may be determined from manufacturers' catalogs. For this example, one manufacturers’ product line indicates a length of 4-in. (after welding) for a standard ½- in. headed stud. The effective embedment depth, hef = 4-in. + 0.5-in. = 4.5-in. Note: The effective embedment length, hef , for the welded stud anchor is the shank length of the stud (4-in.) plus the thickness of the embedded plate (0.5-in.) as shown in the following figure.

Evaluate the terms in Eq. (D-5) with hef = 4.5 in. Determine ANc and ANco for the anchor group: ANc is the projected area of the failure surface as approximated by a rectangle with edges bounded by 1.5hef (1.5 x 4.5 = 6.75 in. in this case) and free edges of the concrete from the centerlines of the anchors. ANc shall not be taken greater than n ANco where n = number of anchors in a group.

D.4.4(c)ii RD.4.4

Eq. (D-5)

D.5.2.1

6.75"

6"

6.75"

6.75"6 "3 "

A Nc

hef = 4.5 in.

0.5 in.

4 in.

BORRADOR

4-4


Code Reference

ANc = (3 + 6 + 6.75) (6.75 + 6 + 6.75) = 307 in.2 ANco = 9 hef

2 = 9 (4.5)2 = 182 in.2 (projected area of failure for one anchor without edge effects) Check: ANc ≤ nANco 307 in.2 < 4(182 in.2) = 728 in.2 - OK Determine ψec,N: ψec,N = 1.0 (no eccentricity in the connection) Determine ψed,N since ca1 = ca,min < 1.5hef (note: ca1 = 3 in.)

,min, 0.7 0.3

1.5a

ed Nef

ch

ψ = +

,3.00.7 0.3 0.83

1.5(4.5)ed Nψ = + =

Determine ψc,N: ψc,N = 1.0 for locations where concrete cracking is likely to occur (i.e., the bottom

of the slab) and cracking is controlled by flexural or confining reinforcement.

Determine ψcp,N: Note, this factor only applies to post-installed anchors. Therefore: ψcp,N = 1.0 for cast-in-place anchors. Determine Nb:

5.1'efccb hfkN =

where kc = 24 for cast-in anchors

( ) 490,145.4000,424 5.15.1' === efccb hfkN lb Substituting into Eq. (D-5):

20,287(14,490)(1.0)(0.83)(1.0)182307

=⎥⎦⎤

⎢⎣⎡=cbgN lb

Check if: φ Ncbg ≥ Nua

Eq. (D-6)

D.5.2.1

D.5.2.4

D.5.2.5

Eq. (D-11)

D.5.2.6

D.5.2.7

D.5.2.2

Eq. (D-7)

Eq. (D-1)

BORRADOR

4-5


Code Reference

Substituting: φ Ncbg =(0.70) (20,287) = 14,201 lb 14,201 lb > 14,000 lb – OK Specify a 4-in. length for the welded, headed studs with the ½- in. thick base plate.

4. Determine if welded stud head size is adequate for pullout

φ Npn ≥ Nua where: φ = 0.70 Note: Condition B applies in all cases when pullout strength governs.

,pn c P pN Nψ= where:

'8 cbrgp fAN = ψc,P = 1.0 for locations where concrete cracking is likely to occur (i.e., the bottom

of the slab) and cracking is controlled by flexural or confining reinforcement.

For design purposes Eq. (D-1) with Eq. (D-14) and Eq. (D-15) may be rearranged as:

cPc

uabrg f

NA

'8,φψ≥

For the group of four studs the individual factored tension load Nua on each stud is:

500,34000,14

==uaN lb

Substituting:

0.156(4,000)(8)(1.0)0.70

3,500)( =≥requiredbrgA in2

D.5.3

Eq. (D-1)

D.4.1.2

D.4.4(c)ii

Eq. (D-14)

Eq. (D-15)

D.5.3.6

BORRADOR

4-6


Code Reference

The bearing area of welded, headed studs should be determined from manufacturers' catalogs. Figure 7.1 from AWS D1.1 lists the diameter of the head for a ½-in. diameter stud is 1 in. AWS states that alternate head configurations may be used with proof of full-strength development of design. Procedures in ACI 318 Appendix D can be used to confirm alternate head configurations.

( ) ( )( ) 589.05.00.14

22)( =−=

πprovidedbrgA in2 > 0.156 in2 – OK

The total design pullout strength of four ½-in. headed studs:

( )( )( )( ) 774,52)000,4(589.00.1847.0 ==pnnNφ lb

5. Evaluate side-face blowout

Side-face blowout needs to be considered when the edge distance from the center-line of an anchor to the nearest free edge is less than 0.4 hef. For this example: 0.4 hef = 0.4 (4.5) = 1.8 in. < 3 in. actual edge distance ∴The side-face blowout failure mode is not applicable.

6. Required edge distances, spacing, and thickness to preclude splitting failure

Since a welded, headed anchor is not torqued the minimum cover requirements of ACI 318, Section 7.7 apply. Per Section 7.7, the minimum clear cover for a ½-in. bar not exposed to earth or weather is ¾-in. which is less than the 2 ¾-in. cover provided - OK

7. Summary of connection strength:

Steel saNφ 38,220 lb.

Concrete breakout

cbgNφ 14,201 lb. ← Controls

Pullout pnnNφ 52,774 lb. Side-face blowout

sbNφ Not Applicable

Use four ½-in. diameter welded studs meeting AWS D1.1 Type B with an effective embedment of 4 ½- in.

D.5.4

D.8

7.7.1(c)

BORRADOR

51

Example 5 Single Headed Bolt in Shear Near an Edge

Determine the reversible service wind load shear capacity for a single ½in. diameter castin hex headed boltmeeting ASTM F1554 Grade 36. The headed bolt is installed in a normal weight continuous concretefoundation with a 7 in. embedment and a 2¾ in. edge distance. No supplemental reinforcing is present.

Note: This is the minimum anchorage requirement at the foundation required by IBC 2003 Section 2308.6for conventional lightframe wood construction. The 2¾in edge distance represents a typical connection atthe base of framed walls using wood 2 6 sill members.

Note: Foundation reinforcement not shown for clarity

Calculations and DiscussionCode

Reference

1. This problem provides the anchor diameter, embedment depth, and anchormaterial type. The designer is required to compute the maximum reversibleservice shear load due to wind. In this case, it is best to first determine thecontrolling shear design strength, Vn, based on the smaller of the steel strengthand/or concrete strength. Step 6 of this example provides the conversion of thecontrolling factored shear load, Vua , to a service load due to wind using factoredloads of Section 9.2.

2. Determine Vua as controlled by the steel strength of the anchor in shear

Vsa Vua

where:= 0.65

Per the Ductile Steel Element definition in D.1, ASTM F1554 Grade 36 steelqualifies as a ductile steel element(See Table A).

Vsa = n 0.6 Ase futa

To determine Vua for the steel strength, Eq. (D2) can be combined withEq. (D20) to give:

0.6ua sa se utaV V n A f

where:= 0.65

n = 1Ase = 0.142 in.2 for a 1/2 in. threaded bolt (See Table B1)futa = 58,000 psi

D.6.1

Eq. (D2)D.4.1.1

D.4.4

Eq. (D20)

psi4,000'cf

2.75in

7in

Vservice

18 in

BORRADOR

52

Example 5 (cont’d) Calculations and DiscussionCode

Reference

Per ASTM F1554, Grade 36 has a specified minimum yield strength (fya)of 36ksi and a specified tensile strength (futa)of 58 ksi (see Table A). For designpurposes, the minimum tensile strength of 58 ksi should be used.

Note: Per Section D.6.1.2, futa shall not be taken greater than 1.9 fya or125,000 psi. For ASTM F1554 Grade 36, 1.9 fya = 1.9 (36,000) = 68,400 psi,therefore use the specified minimum futa of 58,000 psi.

Substituting, Vua as controlled by steel strength is:

212,3000,58142.06.0165.0saua VV lb

3. Determine Vua governed by concrete breakout strength with shear directed towarda free edge.

Vcb Vua.

where:= 0.70 Condition B no supplementary reinforcement has been provided.

, ,Vc

cb ed V c V bVco

AV VA

where:

23(2.75)(1.5)(2.75) 1.0

4.5(2.75)Vc

Vco

AA

ed,V = 1.0

In this instance, the member thickness is greater than 1.5 ca1, and the distance toan orthogonal edge, ca2, is greater than 1.5 ca1.

c,V = 1.0 for locations where concrete cracking is likely to occur (i.e., the edgeof the foundation is susceptible to cracks) and no supplementalreinforcement is provided or edge reinforcement is smaller than a No. 4bar.

5.11

2.0

0

'7 acoe

b cfdd

V

where:e = load bearing length of the anchor for shear, not to exceed 8d0

e = 8 do = 8 (0.5) = 4.0 in. < 7.0 in

For this problem 8do will control since the embedment depth hef is 7 in.

To determine Vua for the given embedment depth governed by concrete breakoutstrength, Eq. (D2) can be combined with Eq. (D21) and Eq. (D23) to give:

D.6.1.2

D.6.2

Eq. (D2)D.4.1.2

D.4.4(c)i

Eq. (D21)

D.6.2.1

D.6.2.6

D.6.2.7

Eq. (D24)

D.6.2.2

BORRADOR

53


Reference

5.11

2.0

0,, '7 aco

eVcVed

Vco

Vccbua cfd

dAAVV

Substituting, Vua as controlled by concrete breakout strength is:

514,175.2000,45.05.0

470.10.10.170.0 5.12.0

cbua VV lb

4. Determine Vua governed by concrete pryout strength

Note: The pryout failure mode typically controls only for stiff anchors installed atshallow embedments where the concrete breakout occurs behind the anchor in adirection opposite the shear force (See Figure RD.4.1(ii)). For this example,where the shear may be directed either toward the free edge or away from the freeedge, the small edge distance may be the controlling value for pryout strength.

Vcp Vua

where:= 0.70 Condition B applies in all cases when pryout strength governs

cbcpcp NkV

where:kcp = 2.0 for hef 2.5 in.

bNcpNcNedNco

Nccb N

AAN ,,,

Evaluate the terms of Eq. (D4) for this problem:

ANc is the projected concrete failure area on the surface as approximated by arectangle with edges bounded by 1.5hef (1.5(7)= 10.5 in. in this case) in adirection perpendicular to the shear force and the free edge of the concrete fromthe centerline of the anchor.

ANc = (2.75 + 10.5)(10.5 + 10.5) = 278 in.2

ANco = 9 hef2 = 9 (7.0)2 = 441 in.2

D.6.3

Eq. (D2)D.4.1.2

D.4.4(c)i

Eq. (D29)

Eq. (D4)

Eq. (D6)

BORRADOR

54


Reference

Determine ed,N:

,min, 0.7 0.3

1.5a

ed Nef

ch

78.000.75.1

75.23.07.0,Ned

Determine c,N:

c,N = 1.0 for locations where concrete cracking is likely to occur (i.e., the edge ofthe foundation is susceptible to cracks) with cracking controlled by flexuralreinforcement or confining reinforcement.

Determine cp,N:

cp,N = 1.0 for castinplace anchors.

Determine Nb for the anchor:

5.1'efccb hfkN

where, kc = 24 for castin anchors

28,1127.0400024 1.5bN lb

Substituting into Eq. (D4):

13,822(28,112))0.1((1.0)(0.78)441278

cbN lb

To determine Vua for the given embedment depth governed by pryout strength,

D.5.2.5

Eq. (D11)

D.5.2.6

D.5.2.7

D.5.2.2

Eq. (D7)

10.5in

10.5in

10.5in2.75in

ANc

BORRADOR

55


Reference

Eq. (D2) can be combined with Eq. (D29) to give:

ua cp cp cbV V k N

Substituting, Vua for the embedment length governed by pryout strength is:

350,19)822,13)(0.2(7.0cpua VV lb

5. Required edge distances, spacings and thickness to preclude splitting failure:Since a headed bolt used to attach wood frame construction is not likely to betorqued significantly, the minimum cover requirements of Section 7.7 apply.

Per Section 7.7 the minimum clear cover for a 1/2 in. bar is 1 1/2 in. whenexposed to earth or weather. The clear cover provided for the bolt is 2 1/2 in. (23/4 in. to bolt centerline less onehalf bolt diameter). Note that the bolt head willhave slightly less cover (2 3/16 in. for a hex head) – OK

6. Summary:

The factored shear load (Vua = Vn ) based on the governing strength (steel,concrete breakout, and concrete pryout) can be summarized as:

Steel strength Vsa 3,212 lbConcrete Breakout Vcb 1,514 lb Controls

Concrete pryout Vcp 19,350 lb

In accordance with Section 9.2 the load factor for wind load is 1.6:

9466.1

514,16.1cb

serviceVV lb

The reversible service load shear strength from wind load of the IBC 2003Section 2308.6 minimum foundation connection for conventional woodframeconstruction (1/2 in. diameter bolt embedded 7 in.) is 946 lb per bolt. Thestrength of the attached member (i.e., the 2x6 sill plate) also needs to beevaluated.

Note that this embedment strength is only related to the anchor being installed inconcrete with a specified compressive strength of 4,000 psi. In many cases, concrete used in foundations such as this is specified at 2,500 psi, the minimumstrength permitted by the code. Since the concrete breakout strength controlledthe strength of the connection, a revised strength based on using 2,500 psi concrete rather than the 4,000 psi concrete used in the example can be determined asfollows:

747000,4500,2

946500,2', psifservice cV lb

D.8

7.7

9.2

BORRADOR

61

Example 6 Single Headed Bolt in Tension and Shear Near an Edge

Determine if a single ½in. diameter castin hexheaded bolt installed with a 7 in. embedment depth and a 2¾in. edge distance in a normal weight, continuous concrete foundation is adequate for a service tension loadfrom wind of 1,000 lb and reversible service shear load from wind of 600 lb. No supplemental reinforcing ispresent.

Note: This is an extension of Example 5 that includes a tension load on the anchor as well as a shear load.

f c = 4,000 psi

ASTM F1554 Grade 36 hex head anchor bolt material

Note: Foundation reinforcement not shown for clarity


Reference

1. Determine the factored design loads

Nua = 1.6 (1,000) = 1,600 lbVua = 1.6 (600) = 960 lb

2. This is a tension/shear interaction problem where values for both the designtensile strength ( Nn) and design shear strength ( Vn) will need to be determined.

Nn is the smallest of the design tensile strengths as controlled by steel ( Nsa),concrete breakout ( Ncb), pullout ( Npn), and sideface blowout ( Nsb). Vn isthe smallest of the design shear strengths as controlled by steel ( Vsa), concretebreakout ( Vcb), and concrete pryout ( Vcp).

3. Determine the design tensile strength ( Nn)

a. Steel strength, ( Nsa):

Nsa = n Ase futa

where:

= 0.75

n = 1 (for one anchor)

Per the Ductile Steel Element definition in D.1, ASTM F1554 Grade 36 steelqualifies as a ductile steel element.

9.2

D.4.1.2

D.5

D.5.1

Eq. (D3)

D.4.4(a)i

7in.

600 lb

18in.

1,000 lb

2.75in.

BORRADOR

62


Reference

Ase = 0.142 in.2 (for ½” anchor bolt, see Table B1)

futa = 58,000 psi (see Table A)

Note: Per Section D.6.1.2, futa shall not be taken greater than 1.9 fya or125,000 psi. For ASTM F1554 Grade 36, 1.9 fya = 1.9 (36,000) = 68,400 psi,therefore use the specified minimum futa of 58,000 psi.

Substituting:

Nsa = 0.75 (1) (0.142) (58,000) = 6,177 lb

b. Concrete breakout strength ( Ncb):

Since no supplementary reinforcement has been provided, = 0.70

In the process of calculating the pryout strength for this fastener in ExampleNo. 5, Step 4, Ncb for this anchor was found to be 13,822 lb. Substituting:

Ncb = 0.70 (13,822) = 9,675 lb

c. Pullout strength ( Npn)

Npn = c,P Np

where:

= 0.70 – Condition B applies in all cases when pullout strength governs

c,P = 1.0, cracking may occur at the edges of the foundation

cbrgP fAN '8

Abrg = 0.291 in.2, for a ½in. hexhead bolt (see Table B1)

Pullout strength ( Npn):

0.7(1.0)(8)(0.291)(4,000) 6,518pnN lb

d. Concrete sideface blowout strength ( Nsb)

The sideface blowout failure mode must be investigated when the edge distance ca1 is less than 0.4 hef

0.4 hef = 0.4 (7) = 2.80 in. > 2.75 in., therefore, the sideface blowoutstrength must be determined

'a1 brg c160 c A fsbN

D.5.2

D.4.4(c)ii

D.5.3

Eq. (D14)

D.4.4(c)ii

D.5.3.6

Eq. (D15)

D.5.4

D.5.4.1

Eq. (D17)

BORRADOR

63


Reference

where: = 0.70, no supplementary reinforcement has been provided

ca1 = 2.75 in.

Abrg = 0.291 in.2, for a ½in. hexhead bolt (see Table B1)

Substituting:

508,1040000.291(2.75)1600.70sbN lb

e. Summary:

Steel strength Nsa 6,177 lb ControlsConcrete breakout strength Ncb 9,675 lbConcrete pullout strength Npn 6,518 lb

Sideface blowout strength Nsb 10,508

Check if Nn Nua:

6,177 lb > 1,600 lb OK

Therefore:

Nn = 6,177 lb

4. Determine the design shear strength ( Vn)

Summary of steel strength, concrete breakout strength, and pryout strength foranchor in shear from Example Problem No. 5:

Steel strength Vsa 3,212 lbConcrete Breakout Vcb 1,514 lb Controls

Concrete pryout Vcp 19,350 lb

Check Vn Vua: 1,514 lb > 960 lb OK

Therefore:

Vn = 1,514 lb

5. Check tension and shear interaction

If Vua 0.2 Vn then the full tension design strength is permitted

Vua = 960 lb

0.2 Vn = 0.2 (1,514) = 302 lb < 960 lb

D.4.4(c)ii

D.5.1D.5.2D.5.3D.5.4

Eq. (D1)

D.6

D.6.1D.6.2D.6.3

Eq. (D2)

D.7

D.7.1

BORRADOR

64


Reference

Vua exceeds 0.2 Vn, so the full tension design strength is not permitted

If Nua 0.2 Nn then the full shear design strength is permitted

Nua = 1,600 lb0.2 Nn = 0.2 (6,177) = 1,235 lb < 1,600 lb

Nua exceeds 0.2 Nn , so the full shear design strength is not permitted

The interaction equation must be used

1.2ua ua

n n

N VN V

1,600 960 0.26 0.63 0.89 1.26,177 1,514

OK

6. Required edge distances, spacings, and thickness to preclude splitting failure:

Since a headed bolt used to attach wood frame construction is not likely to betorqued significantly, the minimum cover requirements of Section 7.7 apply.

Per Section 7.7 the minimum clear cover for a 1/2 in. bar is 1 ½in. when exposedto earth or weather. The clear cover provided for the bolt is 2 ½in. (2 ¾in. tobolt centerline less one half bolt diameter). Note that the bolt head will haveslightly less cover (2 3/16in. for a hex head) OK

7. Summary

A single ½in. diameter castin hexheaded bolt installed with a 7 in. embedmentdepth and a 2 ¾in. edge distance in a concrete foundation is adequate for aservice tension load from wind of 1,000 lb and reversible service shear load fromwind of 600 lb.

D.7.2

D.7.3

Eq. (D31)

D.8

7.7

BORRADOR

71

Example 7 Single PostInstalled Anchor in Tension and Shear Near an Edge

Determine if a single 1/2 inch diameter postinstalled torquecontrolled expansion anchor with a minimum5 1/2 inch effective embedment installed 3 inches from the edge of a continuous normalweight concretefooting (cast against the earth) is adequate for a service tension load of 1,000 lb for wind and a reversibleservice shear load of 350 lb for wind. The anchor will be installed in the tension zone and the concrete isassumed to be cracked.

1,000 lb

350 lb

Note: Reinf. not shown for clarity.

See Table C for a sample table of postinstalled anchor data from manufacturer (fictitious for examplepurposes only) as determined from testing in accordance with ACI 355.204.


Reference

1. Determine the factored tension and shear design loads

Nua = 1.6W = 1.6 1,000 = 1,600 lb.Vua = 1.6W = 1.6 350 = 560 lb.

2. Design considerations

This is a tension/shear interaction problem where values for both Nn and Vn

need to be determined. Nn is the lesser of the design tension strength controlledby: Steel ( Nsa), Concrete Breakout ( Ncb), Concrete SideFace Blowout ( Nsb), orConcrete PullOut ( Npn ). Vn is the lesser of the design shear strength controlledby: Steel ( Vsa ), Concrete Breakout ( Vcb), or Pryout ( Vcp ). Concrete SideFaceBlowout requirements usually apply to castin anchors (see RD.5.4) and are notconsidered here.

3. Evaluate anchor material:

For this example consider that the postinstalled torquecontrolled expansionanchor stud is manufactured from carbon steel material conforming to the materialrequirements of ASTM F 1554 grade 55 which is a headed bolt ASTMspecification. The data from anchor prequalification testing according toACI 355.2 is shown in Table C.

9.2

D.4.1.2

51/2 in.3 in.

f’c =3,000 psi

BORRADOR

72


Reference

For ASTM F1554 Grade 55 material (See Table A):

futa = 75,000 psi

fya = 55,000 psi

Elongation at 2 in. = 21 % min. with a reduction of area = 30 % min.

Appendix D requires 14 % min. elongation and 30 % min. reduction of area for ananchor to be considered as a ductile steel element.

Anchor steel is ductile.

4. Steel strength under tension loading

Nsa Nua

Nsa = nAsefuta

nAsefuta Nua = 1,600 lb

For ductile steel as controlling failure mode, = 0.75

n = 1 (single anchor)

Calculating for Nsa:

000,75142.0175.0saN = 7,988 lb > 1,600 lb – OK

1/2 in. diameter anchor steel strength is adequate under tension loading.

D.1

D.5.1

D.4.1.1

Eq. (D3)

D.4.4(a)i

5. Minimum Edge Distance Requirements

The minimum edge distance for postinstalled anchors shall be based on thegreater of the minimum cover requirements in 7.7, minimum edge distancerequirements for the products as determined by tests in accordance withACI 355.2 (See Table C), and shall not be less than 2 times the maximumaggregate size.

ca,min = (3 in. – maximum cover requirements for concrete cast permanentlyagainst the earth;2.5 in. – Product requirement; or 2(0.75) = 1.5 in.) –Assuming ¾in. maximum aggregate size)

ca,min = 3 in.

6. Concrete breakout strength under tension loading

Ncb Nua

D.8.3

D.5.2

D.4.1.1

BORRADOR

73


Reference

bNcpNcNedNco

Nccb N

AAN ,,,

where:

5.1'efccb hfkN

Substituting:

5.1,,, ' efccNcpNcNed

Nco

Nccb hfk

AAN Nua =1,600 lb

where:

kc = kcr (See Table C) = 17

ed,N =ef

a

hc

5.13.07.0 min, when ca,min < 1.5hef

by observation, ca,min < 1.5hef

ed,N = 81.05.55.1

33.07.0

c,N= 1.0 assuming cracking at service loads (ft > fr)

cp,N= 1.0 assuming cracking at service loads (ft > fr)

for a category 2 anchor, = 0.55 (No supplemental reinforcement provided)

ANco = 9 hef2

= 9 (5.5)2

ANco = 272.25 in.2

ANc = (ca1 + 1.5hef ) (2 x 1.5hef ) = (3.0 + 1.5(5.5)) (2 x (1.5)(5.5))

ANc = 185.63 in.2

25.27263.185

Nco

Nc

AA = 0.68

Calculating Ncb:

Ncb = 5.15.5000,3170.10.181.068.0 = 6,615 lb

Ncb = 0.55(6,615 lb) = 3,638 lb > Nua =1,600 lb – OK

Eq. (D4)

Eq. (D7)

Eq. (D11)

D.5.2.6

D.5.2.7

D.4.4(c)ii

Eq. (D6)

D.5.2.1Fig. RD.5.2.1(b)

BORRADOR

74


Reference

7. Pullout Strength

Pullout strength, Np, for postinstalled anchors is established by reference tests incracked and uncracked concrete in accordance with ACI 355.2. Data from theanchor prequalification testing must be used. Np = Np,cr = 7,544 lb (See

Table C ).

Npn Nua

Npn = pPc N,

for a category 2 anchor, = 0.55 (Condition B applies in all cases when pulloutstrength governs)

Pc, = 1.0 assuming cracking at service loads (ft > fr)

Npn= 0.55 7,544 = 4,149 > 1,600 lb – OK

8. Check All Failure Modes under Tension Loading

Summary:

Steel saN 7,988 lb.Concretebreakout cbN 3,638 lb. Controls

Pullout pnN 4,149 lb.

Nn = 3,638 lb as concrete breakout strength controls.

9. Steel strength under shear loading

Vsa Vua

Vsa = n0.6Asefuta (Anchor does not have sleeve to extend through shear plane)

n0.6Asefuta Vua = 560 lb

For ductile steel as controlling failure mode, = 0.65

n = 1 (single anchor)

Calculating for Vsa:

000,75142.06.065.0saV = 4,153 lb > 560 lb – OK

1/2 in. diameter anchor steel strength is adequate under shear loading.

D.5.3.2

D.4.1.1

Eq. (D14)

D.4.4(c)ii

D.5.3.6

D.4.1.2

D.6.1

D.4.1.1

Eq. (D20)

D.4.4(a)ii

BORRADOR

75


Reference

10. Concrete breakout strength under shear loading

Vcb Vua

bVcVedVco

Vccb V

AAV ,,

where:

5.11

'2.0

7 acoo

eb cfd

dV

Substituting:

uaacoo

eVcVed

Vco

Vccb Vcfd

dAA

V 5.11

'2.0

,, 7 = 560 lb

where:

= 0.7 for anchors governed by concrete breakout due to shear (Condition B nosupplemental reinforcement is provided)

Vco

Vc

AA = 1.0, and ca1 = 3 in. (edge distance)

ca2 is the distance from the center of the anchor to the edge of concrete in thedirection orthogonal to ca1 (not specified in this example but consider this distancegreater than 1.5 hef)

ed,V = 1.0 since ca2 1.5ca1

c,V = 1.0 assuming cracking at service loads (ft > fr)

do = 0.5 in.

e = hef = 5.5 in., but oe d8 ; 8do = 8(0.5) = 4 in.

e = 4 in.

5.12.0

3000,35.05.0

470.10.10.17.0cbV

= 1,495 lb > 560 lb – OK

11. Concrete Pryout Strength:

Vcp VuaVcp = kcpNcb

D.6.2

D.4.1.1

Eq. (D21)

Eq. (D24)

D.4.4(c)ii

Eq. (D27)

D.6.2.7

D.6.2.2

D.6.3

D.4.1.1

BORRADOR

76


Reference

where:

kcp = 2.0, since hef = 5.5 in. > 2.5 in.Ncb = 6,615 lb (See section 6 of this example)

= 0.7 (Condition B applies)

615,60.270.0cpV = 9,261 lb > 560 lb – OK

12. Check All Failure Modes under Shear Loading:

Summary:

Steel saV 4,153 lb.Concretebreakout

cbV 1,495 lb. Controls

Pryout cpV 9,261 lb.

Vn = 1,495 lb from concrete breakout strength controls

13. Check Interaction of Tension and Shear Forces

If Vua < 0.2 Vn , then the full strength in tension is permitted: Nn Nua

0.2 Vn = 0.2(1,495 lb) = 299 lb < 560 lb – Requirement not met

If Nua < 0.2 Nn , then the full strength in shear is permitted: Vn Vua

0.2 Nn = 0.2(3,638 lb) = 728 lb < 1,600 lb – Requirement not met

Since Vua > 0.2 Vn and Nua > 0.2 Nn, then:

2.1n

ua

n

ua

VV

NN

495,1560

638,3600,1

= 0.44 + 0.37 = 0.81 < 1.2 – OK

14. Summary

The PostInstalled Torque Controlled Expansion Anchor, 1/2in. diameter at a5 1/2 in. effective embedment depth is adequate to resist the applied servicetension and shear loads of 1,000 lb and 350 lb, respectively.

Eq. (D29)

D.6.3.1

D.4.4(c)i

D.4.1.2

D.7

D.7.1

D.7.2

D.7.3

Eq. (D31)BORRADOR

8-

1

Example 8 – Group of Anchors in Tension and Shear with Two Free Edges, and Supplemental Reinforcement

Check the capacity of a fastener group with four ¾” diameter ASTM F1554 Grade 55 cast-in anchor rods embedded 12” using hex nuts into the 2’-0” thick, fc’= 3000 thickened slab made of normal-weight concrete to support a factored shear of 4 kips shear and simultaneous uplift of 12 kips. The plate is symmetrically placed at the corner. Seismic forces are not a consideration. 60 ksi reinforcement.

Calculation and Discussion Code

Reference Because both shear and tension are to be considered the capacity of this detail will be based on the interaction equation given in D.7.

2”

8”

2”

4,000 lb

¾”x 1’-0”x1’-0” Base Plate with oversized holes

1 ½ " non-shrink, min 3000psi grout

Hairpins each way- see example text

#4 @12 slab edge reinforcement.

12,000 lb

#5 continuous perimeter bar

#5 continuous perimeter bar

4” 8”

2’-0

” 12”

BORRADOR

8-

2

Example 8 (cont’d) Calculation and Discussion Code

Reference

1. Factored design load

Nua = 12,000 lb Vua = 4,000 lb

2. Design considerations

The differences between Condition A and Condition B, and the associated requirements for supplemental reinforcement are the subject of different interpretations. The disagreement centers around whether Condition A supplemental reinforcement must be proportioned, detailed, and developed to carry the full design load, thereby essentially eliminating the breakout failure mode, or if Condition A is an intermediate condition between Condition B and a condition in which the supplementary reinforcement is proportioned, detailed, and developed to completely restrain the breakout prism and carry the full load.

In the first interpretation, Condition A supplemental reinforcement must be proportioned, detailed, and developed to carry the full design load. In the second interpretation, Condition A supplemental reinforcement is not required to be proportioned, detailed, and developed to carry the full load. This second interpretation is consistent with three possible conditions: Condition A – Supplemental reinforcement is present and is oriented and detailed to help restrain the breakout prism, but is not specifically designed to completely restrain the breakout prism. Because this reinforcement increases the ductility of the system a small increase in the phi factor is provided by the code. Condition B – Either reinforcement is present near the anchor but is not oriented to aid in restraining the breakout prism or no reinforcement is present. Fully Developed Condition – Reinforcement is proportioned, detailed, and developed to hold the breakout prism in place and to support a design load greater than the breakout strength. The authors of this problem consider the second interpretation correct and have considered Condition B for tension and Condition A for shear. It is hoped that this issue will be clarified in ACI318-08.

In the following calculations, the supplementary reinforcement (#4s, #5s, hairpin) is not considered to directly restrain the shear breakout prism. It is considered to enhance the ductility of the system in shear as noted in the explanation of the second interpretation above for Condition A. Condition B of the second interpretation is assumed for all tension capacity calculations. The anchor system must be designed as follows: φ Nn ≥ Nua where φ Nn is the lowest design strength in tension from all appropriate failure

Eq. (D-1)

BORRADOR

8-

3


Reference

modes. φ Vn ≥ Vua Where φ Vn is the lowest design strength in shear from all appropriate failure modes.

3. Anchor material

ASTM F1554 Grade 55 rod material meets the requirements of the ductile steel element definition in D.1 (i.e. elongation of at least 14% and reduction in area of at least 30%. F1554 material has 21 % elongation in 2" of length with 30 % reduction in area (see Table A of this design guide).

Tensile Capacity Calculations

1. Determine the steel strength of anchors in tension, φ Nsa .

φ Nsa = φ n Ase futa φ = 0.75 Ase = 0.334 See Table B1 of this design guide.

0.75(4) (0.334) (75,000) 75,150 lbse utanA fφ = =

2. Determine the breakout capacity of the anchor group in tension, φNcbg .

ec,N ed,N c,N cp,NNc

cbg bNco

AN NA

φ φ= ψ ψ ψ ψ

(ѱcp,N = 1.0 for cast-in anchors) φ = 0.70 We select Condition B, no supplementary reinforcement is provided to resist tension breakout. There is no reinforcement provided which is oriented to restrain the tension breakout prism. The reinforcement present is oriented in such a way that it will aid in enhancing the ductility of the system in shear – more on that in the shear section (See Section 7 in shear calculations) – and not the tension breakout prism. 1.5hef=18”

Check the basic concrete breakout strength.

AN c = (4 + 8 + 18)(4 + 8 + 18) = 900 in.2

Eq. (D-2)

D.1

D.5.1.2

Eq. (D-3)

D.4.4

D.5.2

Eq. (D-5)

D.5.2.7

D.4.4

RD.5.2.1

BORRADOR

8-

4


Reference

ANco = 9 2efh = 9(12)2 = 1,296 in.2

ψec,N = 1.0 as no eccentricity is exists

Check to see if the edge distance modifier, ψed,N , is other than 1.0

ca,min = 4″

ca,min < 1.5 hef = 18”, therefore ψed,N will be less than one.

ed,N4.00.7 0.3 0.7 0.3 0.767

1.5 1.5(12)a,min

ef

ch

ψ = + = + =

Check if the cracked concrete, c,Nψ , modifier is other than one. Unless an analysis is done to show no cracking at service loads cracking must be assumed, therefore c,N 1=ψ .0.

hef = 12" > 11", therefore Eq. (D-8) shall be applied

5/ 316 16 3,000 (12) 55,120 lb' 5/3b c efN f h= = =

9000.7 (1.0)(0.767)(1.0)(55,120) 20,551 lb

1,296cbgNφ = =

3. Determine the pullout strength in tension, φNpn .

φ Npn = φ ψc,P Np

Eq. (D-6)

D.5.2.4

D.5.2.5

Eq. (D-11)

D.5.2.6

D.5.2.2

Eq. (D-8)

D.5.3

Eq. (D-14)

D4.4

4”

8”

18”

4” 8” 18”

Hatched Area Is ANC

BORRADOR

8-

5


Reference

φ = 0.70 Unless an analysis is done to show no cracking is present at service loads cracking must be assumed, therefore ψc,P = 1.0.

8 (8)0.654 (3,000) 15,696 lb'p brg cN A f= = =

See Table B1 for bearing area of hex nut. There are four rods, thus the total pullout capacity is (4)15,696 62,784lb.=

4. Determine concrete side-face blowout capacity, φ Nsbg .

Evaluate if side-face blowout is a consideration. The smallest edge distance, ca1, must be less than 0.4 hef = 0.4(12) = 4.8″ and the spacing of the anchors must be less than 6ca1 = 6(4) = 24″. Both requirements are met, so side face blowout must be considered.

1

16sbg sb

a

sN Nc

φ φ⎛ ⎞

= +⎜ ⎟⎝ ⎠

φ = 0.70 s = 8”

160 160(4) 0.654 3,000 28,348 lb'sb a1 brg cN c A f= = =

80.70 1 28,348 26,458 lb

6(4)⎛ ⎞

= + =⎜ ⎟⎝ ⎠

φ sbgN

5. Summary

φ Nsa = 75,150 lb (steel strength) φ Ncbg = 20,551 lb (concrete breakout) φ Npn = 62,784 lb (concrete pullout) φ Nsbg = 26,458 lb (side face blowout) Concrete breakout controls the tension capacity at a capacity of 20.55 kips.

D.5.3.6

Eq. (D-15)

D.5.4

D.5.4.2

Eq. (D-18)

D.4.4

Eq. (D-17)

BORRADOR

8-

6


Reference

Shear Capacity Calculation 6. Determine the steel strength of anchor in shear, saVφ .

sa se utaV n0.6A fφ = φ

65.0=φ

seA 0.334= See Table B1 of this design guide The steel strength of each anchor is:

0.6 0.65 (0.6)(0.334)(75,000) 9.77kipsse utan A fφ = = Where anchors are used with built-up grout pads, the design strength must be multiplied by 0.8. φ Vsa = 0.8 (9.77) = 7.82 kips per anchor

7. Determine the breakout strength in shear, cbgVφ .

ec,V ed,V c,V= ψ ψ ψφ φ Vccbg b

Vco

AV VA

φ = 0.75 Condition A. The combination of the continuous #5 perimeter bars and #4 bent slab bars qualify as supplementary reinforcement for resisting a shear breakout, as do the hairpins. The hairpins could potentially be designed to directly restrain the concrete breakout thereby eliminating breakout as a failure mode. However, at this time the combination of the continuous #5 perimeter bars and #4 bent slab bars and hairpins are considered as supplementary reinforcement in terms of Condition A (in other words, the hairpin is not considered to directly restrain the breakout prism and does not eliminate breakout as a failure mode. We are acknowledging the fact that there is reinforcement present that will help retrain the breakout prism and improve ductility allowing an increase in the capacity by changing the phi factor to the Condition A value).

0.2' 1.5e

b o c a1o

V 7 d f (c )d

⎛ ⎞= ⎜ ⎟

⎝ ⎠

e can not exceed "6)4/3(88 ==od

ca1 can be either from the first line of fasteners or the second. The base plate will be fabricated with oversized holes. It is therefore possible the two fasteners closest to the edge will engage before those at the line farthest from the edge. Un-less steps are taken to ensure that the fasteners farthest from the edge will be

D.6.1.2

Eq. (D-20)

D.4.4

D.6.1.3

D.6.2

Eq. (D-22)

D.4.4

Eq. (D-24)

D.6.2.2

RD6.2.1 Fig RD.6.2.1(b)

BORRADOR

8-

7


Reference

engaged (such as welding plate washers with standard holes to the base plate in the field – which will be considered later in this example), ca1 should conservatively be taken as the distance from the free edge to the first line of fasteners. ca1 = 4″

AVc = (ca2 +s+1.5 ca1)(1.5 ca1 )=(4 + 8 + 6)(6) = 108 in.2

AVco = 4.5 (ca1 )2 = 4.5(4)2 = 72 in.2

0.2

1.567 0.75 3,000 (4) 4,026 lb0.75bV ⎛ ⎞= =⎜ ⎟

⎝ ⎠

ψec,V = 1.0 no eccentric shear is applied.

ca2 = 4″ therefore ca2 < 1.5ca1 and ψ ed.V must be calculated

ed,V4.00.7 0.3 0.7 0.3 0.9

1.5 1.5(4.0)a2

a1

cc

ψ = + = + =

A continuous #5 bar is present around the edge of the slab. Therefore ψ c,V = 1.2

1080.75 (1.0)(0.9)(1.2)(4,026) 4.89 kips72cbgVφ = =

The breakout capacity of the detail is limited. The designer may design the hairpins in the slab to directly restrain concrete breakout per D.4.2.1. and thereby eliminate breakout as a failure mode. Side Note: Consider the effect on the capacity if plate washers with standard holes were welded to the base plate at each anchor. In such a case the breakout prism could be assumed to form from the back anchors per RD.6.2.2 ca1 = 12″ 1.5ca1 = 18″

AVc = (4 + 8 + 18)(18) = 540 in.2

AVco = 4.5 (ca1 )2 = 4.5(12)2 = 648 in.2

( )0.2

1.5b

6V 7 0.75 3,000 12 20,921 lb0.75

⎛ ⎞= =⎜ ⎟⎝ ⎠

ca2 = 4″ therefore ca2 < 1.5ca1 and ψ ed.V must be calculated ψed,V = 0.767

5400.75 (1)(0.767)(1.2)(20,921) 12.0kips648cbgVφ = =

Thus, the addition of plate washers with standard holes welded to the base plate would allow the designer to assume that the breakout would occur from the back row of fasteners and to realize a substantial increase in breakout capacity. If this assumption is made all of the load is carried by the rear fasteners and the steel strength of the rear fasteners alone must be sufficient to support the full design load. Additionally, the bearing capacity of the thin edge of the steel washer against the bolt should be checked.

Fig. RD.6.2.1(a)

D.6.2.1

Eq. (D-23)

Eq. (D-24)

D.6.2.5

D.6.2.6

Eq. (D-28)

D.6.2.7

Eq. (D-22)

BORRADOR

8-

8


Reference

8. Determine the concrete pryout strength, φ Vcp .

cpg cp cbgV k Nφ φ=

0.7φ =

2.0cpk = when "5.2≥efh

ed,N c,N900 (0.767)(1)(55,120) 29.3 kips

1,296Nc

cbg bNco

AN NA

ψ ψ= = =

0.7(2)(29.3) 41 kipscpVφ = =

9. Summary

7.82 kipssaVφ = per anchor

4.89 kipscbgVφ = assuming no welded plate washers are provided.

41 kipscpVφ = Side Note: In the case where plate washers were welded to the farthest line of fasteners a substantial increase in concrete breakout capacity is realized. Then the steel strength of these two anchors must be compared to the concrete breakout capacity engaged by these anchors. The concrete breakout capacity of 12 kips with two anchors effective is less than the steel capacity 2(7.82) = 15.6 kips. Furthermore, the pryout capacity of the two rear fasteners alone must be checked:

cpg cp cbgV k Nφ φ=

0.7φ =

2.0cpk = when "5.2≥efh

2660)84)(1884( inANc =+++= for the rear two anchors.

kipsNcbg 5.21)120,55)(1)(767.0(296,1

660==

kipsVcp 30)5.21)(2(7.0 =−φ

Concrete Breakout controls.

Shear breakout severely limits the capacity of this connection.

D.6.3

Eq. (D-30)

D.4.4

D.6.3.1

Eq. (D-4)

BORRADOR

8-

9


Reference

Interaction of Tensile and Shear Forces 10. First the provisions of D.7.1 and D.7.2 are checked.

4 kips 0.2 0.2(4.89) 0.98 kipsua nV Vφ= > = = (assumes no plate washers welded to base plate and no hairpins provided)

12 kips 0.2 0.2(20.6) 4.12 kipsua nN Nφ= > = =

Therefore the interaction of the two must be checked

12 4 1.4 1.220.6 4.89

ua ua

n n

N VN Vφ φ

+ = + = > No Good

Add field welded plate washers and thus increase φ Vn to 12.0 kips (see above, concrete breakout failure of the two rear anchors governs). Vua = 4 kips > 0.2 φ Vn = 0.2 (12.0) = 2.4 kips Therefore the interaction must still be checked,

12 4 0.92 1.220.6 12.0

u u

na na

N VN Vφ φ

+ = + = ≤ OK

The four anchor rod group is adequate to resist the applied loading assuming plate washers are welded to the base plate at all anchor rod locations.

D.7

D.7.1

D.7.2

Eq. (D-31)

BORRADOR

Example 10 - Multiple Anchor Connection Subjected to Moment and Shear

Design an embed plate beam to support the end reaction and end bending moment of a steel beam using a group of eight welded headed studs spaced as shown in Figure 1. The factored design bending moment (Mua) is 30,000 ft-lb and the factored shear load coming from the beam end reaction (Vua) is 20,000 lb factored shear load (ACI 318-05 Section 9.2 load combinations used). The connection is located on the side face of a reinforced concrete spandrel girder (see Figure 1).

Figure 1 – Beam Support Consisting of Embed Plate with Welded Studs


Code Reference

1. It is often difficult to simultaneously determine the required anchor diameter and embedment length simultaneously due to the complexity of the load distribution to each anchor and the interaction of tensile and shear forces. Thus, the general approach from a practical design perspective is to assume some design values in advance and verify their acceptability in the design calculation. In this example, the connection detail shown in Figure 1 as based on preliminary design is to be checked for conformance with code requirements.

2. Design basis:

A. basic assumptions

- cracked concrete

- elastic design

B. materials

10-1/2" 10-1/2" 6" 1-1/2" typ.

10-1

/2"

6"

1 2 3

6 7 8

4 5 y

x z

1-1/2"

1'-6"

hef = 9" 7/8" dia. welded headed studs

W14x53

Mua Vua

#5 stirrups @ 4" o.c.

Normal wt. concrete f'c = 4,500 psi

4 - #10

Bottom soffit of spandrel girder

#6 typ.

1-1/2" x 2'-0" sq. embed plate

10-1

/2"

Girder beyond

Vua

BORRADOR

10-2

Example 10 (cont’d)

Calculations and Discussion Code

Reference

- embed plate: ASTM A 36 1-1/2" x 24" x 24" (tp = 1-1/2 in)

- anchors:

7/8" x 8-3/16" AWS D1.1 Type B mild steel welded headed stud (standard

length shear stud)

specified yield strength of anchor steel fya = 51 ksi

specified tensile strength of anchor steel futa = 65 ksi

stud diameter d = 0.875 in.

stud head diameter dh = 1.375 in.

stud head thickness th = 3/8 in.

reduction in stud length due to welding ~ 3/16 in.

effective anchor embedment:

hef = 8-3/16" – 3/8" – 3/16" + 1-1/2" = 9.125" ≈ 9 in.

Figure 2 – Welded Headed Stud

3. Determine anchor reactions: D.3.1

A. Tension: The determination of the anchor reactions is not trivial. While the use of finite element modeling may provide the closest approximation, it is possible to use conservative assumptions and statics to derive an acceptable set of anchor forces.

Simplified statically determinant analysis: Assume that the compression reaction is located directly beneath the toe of the W14 beam (conservative) and that the embed plate exhibits rigid-body rotation (see Figure 3).

dh

t h

stud

leng

th

h ef

d

BORRADOR

10-3



Reference

Figure 3 – Tension Load Distribution

TT = tension reaction of top anchors (6, 7 and 8)

TM = tension reaction of middle anchors (4 and 5)

( ) ( ). T MM 0 17 5 T 7 T 360 0= ⇒ + − =∑ (1)

Assume rigid-body rotation of embed plate and determine ratio of anchor reactions

(K = single anchor elastic stiffness, Δ = anchor elastic displacement):

. .Δ= ⇒ Δ = Δ

ΔT

T MM

17 5 2 57

(2)

Tension reaction of back 3 anchors: T TT 3 K= Δ (3)

Substituting Eq. (2) into Eq. (3): T MT 7.5 K= Δ (4)

Tension reaction of middle 2 anchors: M MT 2 K= Δ (5)

Substituting Eq. (5) into Eq. (4): T MT 3.75T= (6)

Substituting into Eq. (1) above:

. . / .. . / .

M 4 5

T 6 7 8

T 4 95k T T 4 95 2 2 48 kT 18 6k T T T 18 6 3 6 20 k

= ∴ = = == ∴ = = = =

Total tension reaction: ( )( ) ( )( )uaN = 3 6.20 + 2 2.48 = 23.6 k

Check assumption of plate rigidity using results from analysis above:

The assumption that the compression reaction is located at the compression flange of the steel beam as shown in Figure 3 is rational based on observations of tested assemblies. While the load distribution to the tension-loaded anchors will also be affected by their proximity to the connected wide flange shape, such effects are likely to occur at larger rotations. Per D.3.1, the analysis may be based on elastic

10.5

" 7"

TT

TM

C

360 in - k

a

BORRADOR

10-4



Reference

anchor response. On the tension side, checking the moment at the beam face and compare it to the moment in the plate:

( )22 24 1.5bhM = S f = × f = × 36 ksi = 324 in - ky,pl x y,pl y,pl6 6

( ) ( )TM = T × a = 3 6.20 3.5 = 65.1in - k 324 in - kface ∑ ≤ ∴ ok

B. Shear: The anchors are welded to the attachment. Per RD.6.2.1 for anchors welded to a common plate the shear is carried by the back anchor row.

Figure 4 – Assumption of Critical Anchor Row for Shear Loading

RD.6.2.1

4. Determine design tensile strength, φ Nn , as follows:

Nominal steel strength in tension, Nsa : D.5.1.1

check: uta yaf = 65 ksi < 1.9 f = (1.9)(51)= 96.9 k < 125 ksi∴ o.k. D.5.1.2

Effective cross-sectional area of anchor:

2 22

seπ× d π×0.875A = = = 0.601 in

4 4

Tension, sa se utaN = A × f = (0.601)(65)= 39.07 k for a single stud Eq.(D-3)

Concrete breakout strength in tension, Ncbg: D.5.2

Nccbg ec ,N ed ,N c,N cp ,N b

Nco

AN NA

ψ ψ ψ ψ= Eq.(D-5)

Determine ANc for the group of tensioned anchors, T4, T5, T6, T7 and T8 and ANco for single anchor:

1 2 3

6 7 8

4 5

y

x z

27"

10.5

" 6"

20 k c a1

6" ca2

critical anchor row

BORRADOR

10-5



Reference

The projected concrete failure area, ANc, is shown shaded in Figure 5 and is calculated as follows:

hef = 9 in.

N c ef a2 x ef y

2

A = (1.5×h +c + 2× s ) (2×1.5×h + s )

= (1.5×9 + 6 + 2×10.5) (2×1.5×9 + 10.5)

= 1,519 in.

D.5.2.1

Figure 5 – Determination of Projected Failure Area and e'N for Tension-Loaded Anchors

Determine ANco for the single anchor:

( )2 2 2Nco efA 9 h 9 9 729 in.= × = = Eq.(D-6)

Determine the eccentricity modification factor ψ ec ,N :

ec ,N

N

ef

1 1.02 e13 h

ψ = ≤⎛ ⎞′⋅+⎜ ⎟⎜ ⎟⋅⎝ ⎠

Eq.(D-9)

Determine the eccentricity, ′Ne , of the resultant tension load with respect to the centroid of the tension-loaded anchors.

Fig. RD.5.2.4(b)

The location of the geometric centroid of tension-loaded anchors (T4 - T8) as measured from axis a-a (see Figure 5) is given by y(2 5 )s . Summing moments about a-a, the location of the resultant of the tension-loaded anchors is given

1 2 3

8

4 5

y

x z

37.5

"

13.5

" 13

.5"

10.5

"

40.5"

NcA

a a

s y

e'N

centroid of tension-loaded anchors (T4 - T8)

resultant of tension-loaded anchors (T4 - T8)

y2 s5

ca2

7 6

sx

BORRADOR

10-6



Reference

by 4 5 y

ua

(T +T )sN

. The eccentricity ′Ne between the centroid of the tension-loaded

anchors and the tension resultant is thus given as:

y 4 5 yN

ua

2×s (T +T )s (2)(10.5) (2.48+ 2.48)(10.5)e - = - = 1.99 in.5 N 5 23.6

′ =

ec,N1ψ = = 0.87

2(1.99)1+3(9)

⎛ ⎞⎜ ⎟⎝ ⎠

a,min efc = 6 in.< 1.5h = 13.5in.

Determine the near-edge modification factor ψ ed ,N :

a,mined,N

ef

c 6ψ = 0.7 +0.3 = 0.7 +0.3 = 0.831.5 h 1.5(9)

Eq.(D-11)

Determine the modification factor for cracked/uncracked concreteψ c ,N :

Per D.5.2.6, unless an analysis is performed to show no cracking at service loads, the concrete is assumed to be cracked for the purposes of the determining the anchor design strength: c,Nψ = 1.0 for cracked concrete.

D.5.2.6

Determine the splitting modification factorψ cp ,N :

=ψ cp ,N 1.0 for cast-in anchors. D.5.2.7

Determine the basic concrete breakout tensile strength for a single anchor, Nb : D.5.2.2

' 1.5b c c efN = k f h Note: = ⇒efh 9 in.< 11 in. Use of Eq. (D-7) required. Eq.(D-7)

ck = 24

1.5bN = (24) 4,500 (9) = 43,469 lb = 43.5 k

Substituting into Eq.(D-5): cbg1,519N = (0.87)(0.83)(1.0)(1.0)(43.5)= 65.5 k729

Determine the pullout strength of anchors in tension, Npn : D.5.3

pn c ,P pN Nψ= Eq.(D-14)

where: ψc,P = 1.0 cracking assumed D.5.3.6

BORRADOR

10-7



Reference

Pullout strength of single anchor in tension, Np D.5.3.4

'p brg cN = 8 A f Eq.(D-15)

Determine bearing area of the head of a single stud:

2 2 2 22h

brgπ(d - d ) π(1.375 -0.875 )A = = = 0.884 in.

4 4

pN = (8)(0.884)(4,500)= 31,824lb = 31.8 k∴

Substituting into Eq. (D-14):

pnN = (1.0)(31.8)= 31.8 k for a single anchor

Determine the side-face blowout tensile strength, Nsb : D.5.4

a,min efc = 6 in. 0.4h = 3.6 in.≥ ∴side-face blowout failure mode not applicable D.5.4.1

The design tensile strengths are calculated with the calculated nominal strengths and strength reduction factors of D.4.4.

Steel strength, = 0.75φ D.4.4

Per AWS D1.1 Type B studs qualify as a ductile steel element (20% minimum elongation in 2", see Table A.)

( )saN = (0.75) 39.07 = 29.3 kφ for a single anchor. D.4.1.2

Strength reduction factor for concrete breakout, Condition B: = 0.70φ D.4.4

Note: Although the cantilever beam contains significant amounts of beam reinforcing located in the anchor vicinity, this reinforcing is not configured to provide the necessary restraint to the tension-induced concrete breakout surface to warrant the use of the strength reduction factor associated with Condition A. Therefore, Condition B is assumed (supplementary reinforcement not provided). D.4.4

cbgN = (0.7)(65.5)= 45.9 kφ for the anchor group comprised of anchors T4 - T8. D.4.1.2

Strength reduction factor for pullout = 0.70φ D.4.4

pn pnN = N = (0.70)(31.8)= 22.3kφ φ for a single anchor. D.5.3

Note: Since this is an elastic analysis, the check for the controlling strength is dependent on the nature of the load distribution. In this case the anchor loads are not uniform, so the steel and pullout checks must be performed on the most

BORRADOR

10-8



Reference

critically-loaded anchor. Since the CCD method predicts the concrete breakout resistance of the group, the concrete breakout check is always performed for the group of tension-loaded anchors.

Summary of connection strength – tension: D.4.1.2

Failure mode

Critical anchor(s)

Associated strength, nNφ

Factored Load, uaN Ratio ua

n

NNφ

Steel 6, 7 or 8 saNφ 29.3 k 6.4 k 0.22

Concrete breakout group 4 - 8 cbgNφ 45.9 k 23.6 k 0.51←controls

Concrete pullout 6, 7 or 8 pnNφ 22.3 k 6.4 k 0.29

Side-face blowout 8 φ sbN not applicable

5. Determine design shear strength, φ Vn, as follows:

Nominal steel strength in shear for cast-in headed studs, Vsa : D.6.1.2

sa se utaV = nA f = (1)(0.601)(65)= 39.1 k Eq.(D-19)

For a single anchor: saV = (1)(0.601)(65)= 39.1 k

Concrete breakout strength in shear, Vcbg : D.6.2.1

Vccbg ec ,V ed ,V c ,V b

Vco

AV VA

ψ ψ ψ= Eq.(D-22)

Note: Per RD.6.2.1 for anchors welded to a common plate the shear is carried by the back anchor row (anchors 6, 7 and 8, see Figure 4) The shear load per anchor is therefore uaV 20k / 3 6.7k= = . Note also that the contribution to the shear capacity of the fastening by the concrete located in front of the embedded plate will be neglected.

RD.6.2.1

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10-9



Reference

Figure 6 – Determination of Projected Failure Area for Shear Loading

Figure 7 – Determination of Projected Failure Surface for Shear Loading – View of Girder Soffit

The edge distance from the edge to the farthest anchor row = 27 in. (see Figure 6).

Determine the projected concrete failure areas, AVco and AVc:

Note that anchors are influenced by three edges (including the beam width), and are located less than 1.5 hef, therefore the value of ca1 used in Eq. D-21 through D-28 will be reduced therefore check:

x,maxa2 aa1

sc h 6 18 21c = max ; ; = max ; ; = 12 in.1.5 1.5 3 1.5 1.5 3

D.6.2.4

AVc

45"

18"

#5 stirrups @4" o.c.

40.5" 10.5" 10.5" 6"

27"

1 2 3

6 7 8

4 5

c a1 =

12"

BORRADOR

10-10



Reference

2 2 2Vco a1A = 4.5× c = 4.5(12) = 648 in. Eq.(D-23)

2Vc a1 x a2A = (1.5× c + 2× s + c )(h)= (45)(18)= 810 in for anchors 6, 7 and 8 D.6.2.1

Determine the eccentricity modification factor ψ ec,V :

ψec,V = 1.0 for no eccentricity in the connection D.6.2.5

Determine the near-edge modification factor ,ψ ed V :

a2ed,V

a1

c 6ψ = 0.7 +0.3 = 0.7 +0.3 = 0.801.5c 1.5 12

⎛ ⎞⎜ ⎟×⎝ ⎠

Eq.(D-28)

Determine the modification factor for cracked/uncracked concrete ,ψ c V :

Per D.6.2.7: ψc,V = 1.4 for cracked concrete with supplementary reinforcement of a No. 4 bar or greater between the anchor and the edge, and with the supplementary reinforcement enclosed within stirrups spaced at not more than 4 inches

The #10 girder flexural reinforcement substantially exceeds the requirement of a No. 4 bar, and the associated stirrups for this example will be considered effective in meeting the requirements of D.6.2.7. One issue arises from the fact that the “edge” reinforcing, in this case a #10 bar, does not extend beyond the failure surface on the right-hand side. While the language of D.6.2.7 does not specifically require this, it is a reasonable expectation that the edge reinforcing is intended to act as a tension tie across the face of the failure prism. The assumption is made here that the beam reinforcing is adequate to justify the increase provided by D.6.2.7. A check for development of the #5 stirrups in the assumed breakout surface as shown below should be made.

D.6.2.7

Refer to Figure 8. Per Section 12.2.2, for #6 and smaller straight deformed bars, the development length is given by:

⎛ ⎞⋅ ⋅ ⋅ ⎛ ⎞= ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

y t ed b

c

f 60,000×1×1×1d = 0.625=22.3in.25 f ' 25 4,500ψ ψ λ

( )≈d ,provided 2×10.5 +6 - 2 - 2 = 23in.

12.2.2

The #5 stirrups in this case are also closed loop stirrups anchored by their interaction with the #10 longitudinal steel, so say ok.

For this example the #5 stirrups will be considered as supplemental reinforcement in meeting the requirements of tying the shear failure prism to the structure.

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10-11



Reference

Figure 8 – Elevation of Girder Reinforcement at End

Determine basic concrete breakout shear strength, Vb , for cast-in anchors continuously welded to a steel attachment having a minimum thickness equal to the greater of 3/8 in. or half of the anchor diameter:

D.6.2.3

0.2' 1.5e

b o c a1o

V 8 d f cd

⎛ ⎞= ⎜ ⎟⎝ ⎠

Eq.(D-25)

where:

e = 8 do = (8)(0.875) = 7 in. (8 do controls) D.6.2.2

Substituting into Eq.(D-25)

0.21.5

b7V = (8) 0.875 4,500 (12) = 31,629lb = 31.6 k

0.875⎛ ⎞⎜ ⎟⎝ ⎠

Substituting into Eq.(D-22)

⎛ ⎞⎜ ⎟⎝ ⎠

cbg810V = (1.0)(0.80)(1.4)(31.6)= 44.2 k648

Determine the concrete pryout strength cpgV D.6.3

deve

lopm

ent l

engt

h ≈

23 in

. de

velo

pmen

t len

gth

#5 stirrups #10

assumed failure surface

BORRADOR

10-12



Reference

cpg cp cbgV = k N Eq.(D-30)

Check the pryout capacity associated with the back three anchors. This is conservative.

Determine the concrete breakout strength of rear three anchors (6, 7 and 8):

Nccbg ec,N ed,N c,N b

Nco

AN = ψ ψ ψ NA

Eq.(D-5)

Figure 9 – Determination of Projected Failure Area for Pryout

Nc ef ef x a2A = (3h )(1.5h + 2× s +c )= (27)(13.5+ 21+6)= 1,094

cpk = 2.0 for efh 2.5 in.≥

cbg1,094N = (1.0)(0.83)(1.0)(43.5)= 54.1 k729

⎛ ⎞⎜ ⎟⎝ ⎠

Note: For the calculation of the pryout capacity, the tension resistance of the anchor group is calculated assuming uniform load distribution to the anchors, therefore ec ,N 1.0ψ = .

cpgV = (2.0)(54.1)= 108.3 k

The design shear strengths are calculated with the calculated nominal strengths and strength reduction factors of D.4.4.

10.5" 10.5" 6" 13.5"

13.5

" 13

.5"

ANc

1 2 3

6 7 8

4 5

BORRADOR

10-13



Reference

Per AWS D1.1 Type B studs qualify as a ductile steel element (20% minimum elongation in 2”) D.1

Determination of the number of anchors contributing to the steel shear capacity is dependent on the assumption for concrete edge failure. The contributions of anchors 1, 2, 3, 4 and 5 are neglected for the steel shear calculation, since these anchors are contained within the assumed breakout surface (see Figure 10). This is reflected in the calculation of Vua = 6.7k.

Figure 10 – Determination of No. of Anchors Contributing to Vsa

Strength reduction factor for steel strength, = 0.65φ D.4.4

saV = (0.65)(39.1)= 25.4 kφ D.6.1

Strength reduction factor for concrete breakout, Condition A: = 0.75φ D.4.4

Note: For the concrete breakout strength of the anchors in shear, the # 5 stirrups in this example will be considered as supplemental reinforcement (Condition A), since the girder reinforcement will substantially engage the breakout surface. D.4.4

cbgV = (0.75)(44.2)= 33.2 kφ D.6.2

Strength reduction factor for pryout, Condition B: = 0.70φ D.4.4

cpgV = (0.70)(108.3)=75.8 kφ D.6.3

Note: As with the tension case, it is important to ensure that the most critically-loaded anchor is identified for the steel check. Since the CCD method predicts the concrete breakout resistance of the group, the concrete breakout and pryout checks are performed for the anchor group only.

1 2 3

6 7 8

4 5

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10-14


Summary of connection strength – shear: D.4.1.2

Failure mode

Critical anchor(s)

Associated strength, nVφ

Factored Load, uaV

Ratio ua

n

VVφ

Steel 6, 7 or 8 saVφ 25.4 k 6.7 k 0.26

Concrete breakout group 4 - 8 cbgVφ 33.2 k 20 k 0.60 ←controls

Concrete pryout group 6 - 8 cpgVφ 75.8 k 20 k 0.26

6. Check tension and shear interaction (concrete breakout): D.7

Nua = 23.6 k > 0.2 φ Nn (= 9.18 k) D.7.2

Vua = 20 k > 0.2 φ Vn (= 6.64 k) D.7.1

ua ua

n n

N V 23.6 20 + = 1.12 1.2N V 45.9 33.2φ φ

∴ + = < ∴⋅ ⋅

ok Eq.(D-31)

Interaction for steel failure ok by inspection.

BORRADOR

Example 12 - Multiple Anchor Connection Subjected to Seismic Moment and Shear

Check the design of an embedded plate with a group of eight welded headed studs spaced as shown to support a factored reversible bending moment, Mua of 30,000 lb-ft (360 kip-in ) and a factored reversible end reaction, Vua of 60,000 lbs (60 kips) (based on elastic analysis, see note below ). The loading results from an earthquake in a region of moderate or high seismic risk. The connection is located far away from any edges of the concrete member; assume it to be in a shear wall (see Figure 1).

Figure 1 – Beam Support Consisting of Embed Plate with Welded Studs


Code Reference

1. Note : As discussed in the ACI 318-05 Commentary R21.2.1, the design forces may be less than those corresponding to linear response at the anticipated earthquake intensity. The design loads for this connection should be established consistent with the requirements of a code such as ASCE 7. Any reduction factor, Rw, should consider the type of construction and the behavior required for the steel beam to concrete connection.

2. The connections shown in Figure 1 is to be checked for the given loading and ductility requirements of Appendix D for resisting seismic loads in a region of moderate or high seismic risk.

3. Design basis:

(1) basic assumptions

- cracked concrete

- plastic design

2" x 1'-0" x 1'-8" embed plate

4" 4" 2" typ.

8"

8"

1 2

5 6

3 4

y

x z

2"

2'-0"

hef = 14"

welded headed studs W14x53

Mua Vua

normal wt. concrete f'c = 6,000 psi #6 @12"

o.c., typ.

BORRADOR

12-2



Reference

(2) materials

- embed plate: ASTM A 36, 2” x 12” x 20” (tp = 2 in.)

- anchors:

3/4" diameter x 13-3/16" AWS D1.1 Type B mild steel welded headed stud

specified yield strength of anchor steel fya = 51 ksi

specified tensile strength of anchor steel futa = 65 ksi

stud head diameter dh = 1.25 in.

stud head thickness th = 3/8 in.

reduction in stud length due to welding ~ 3/16 in.

effective anchor embedment:

hef = 13-3/16 – 3/8 – 3/16 + 2 = 14.63 Assume hef = 14 in.

4. Requirements for design:

Connection must be proportioned to satisfy D.3.3, i.e., either the anchor design is governed by steel strength or the attachment will undergo ductile yielding before the anchors reach design strength.

Checking the plastic capacity of the attached beam (assume Fy = 50 ksi):

Z = 87.1 in3, Mpl = 87.1 x 50 = 4,355 kip-in.

D.3.3.5

Clearly, it is not practical to develop the plastic strength of the beam. In this case, the anchor design must be controlled by steel strength of the anchors. D.3.3.4

5. Determine anchor forces: D.3.1

A. Tension: Anchor forces are determined on the basis of plastic response.

Assume a simplified elastic statically determinate analysis: Assume that the compression reaction is located directly beneath the toe of the W14 beam (conservative) and that the embedded plate is stiff enough to exhibit rigid-body rotation (see Figure 2).

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12-3



Reference

Figure 2 – Tension Load Distribution

TT = tension force on top anchors (5 and 6) TM = tension force on middle anchors (3 and 4)

( ) ( ) 360272150 =−+⇒=∑ MT TTM 30 TT + 14 TM = 360 Because plastic behavior is assumed, all tension anchors have equal force:

MT TT = Substituting into the above equation:

( ) kN

kinkipTT

ua

MT

72.3218.84

18.844

.360

==∴

=

−

==

Check assumption of plate rigidity:

The 2-in. embed plate has a nominal yield moment capacity of:

( ) .288366

0.2126

22

inkipksifbhfSM yyxy −=×=×==

However, the plastic moment capacity of the plate is given by:

.432.2885.15.1 inkipinkipMMfM yyplpl −=−×===

Since the yield capacity of the embed plate at the toe of the connected shape is close to the applied moment, the stiffness of the plate should minimize the prying forces in this connection. The assumption of the compression reaction at the toe is reasonable and conservative. While the load distribution to the tension-loaded anchors will also be affected by their proximity to the connected wide flange shape, such effects are likely to occur at larger rotations. Per D.3.1, plastic analysis is permitted. On the tension side, checking the moment in the plate at the top face of the beam flange:

( )( )∑ ∴−≤−==×= okinkipinkipinkaTM Tface .288.36.16.118.82

6. Determine design tensile strength of the anchors, φ Nn :

Calculate the nominal steel strength in tension, Nsa : D.5.1.2

check: OKksiksifksif yuta 1259.96519.19.165 <=×=≤= D.5.1.2

Effective cross-sectional area of anchor:

8"

7"TT

TM

C

360 in - k

a

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12-4



Reference

2 22

seπ× d π×0.75A = = = 0.442 in

4 4

Nominal Steel Tensile Strength

( )( ) kipsfAN utasesa 73.2865442.0 ==×= for a single anchor

Eq.(D-3)

Calculate the nominal concrete breakout strength in tension, Ncbg: D.5.2

)( ,,,,e bNcpNcNedNcNco

Nccbg N

AAN ψψψψ= Eq.(D-5)

Determine the ratio of ANc for the tension anchor group, T3, T4, T5 and T6, and ANco for a single anchor.

The projected concrete failure area, ANc , shown as a shaded area on the figure below is calculated as follows (see Figure 3):

hef = 14 in.

D.5.2.1

ANc = (1.5hef + sx + 1.5hef)(1.5hef + sy + 1.5hef)

= (21 + 8 + 21)(21 + 8 + 21)

= 2,500 in2

BORRADOR

12-5



Reference

Figure 3 – Determination of Projected Failure Area and e'N for Tension-Loaded

Anchors

Determine ANco for single anchor

ANco = 9(hef)2 = 9(14)2 = 1,764 in2 for a single anchor.

Eq.(D-6)

Determine the modification factors ψ ec ,N , ed ,Nψ , c ,Nψ and cp ,Nψ :

Determine the eccentricity modification factor, ψ ec ,N

0.1

321

1', ≤

+

=

ef

N

Nec

he

ψ Eq.(D-9)

Because plastic analysis is being used, all tension anchors have equal force, and the resultant of the four tension anchors is concentric with respect to the center of gravity of the anchors group, and the eccentricity '

Ne is zero.

0.1, =Necψ Fig. RD.5.2.4(b)

Determine the edge modification factor, ed ,Nψ

Connection is remote from edges: 0.1, =Nedψ

Determine the modification factor for cracked concrete, c ,Nψ

Per D.5.2.6, unless an analysis is performed to show no cracking at service loads, the concrete is assumed to be cracked for the purposes of the determining

D.5.2.6

1 2

3

6

4

5

50"

21"

21"

8"

25"

a a

s y

resultant of tension-loaded anchors (T3, T4, T5 and T6)

ys2

25"

ANc

sx

BORRADOR

12-6



Reference

the anchor design strength (this is a prudent assumption for most seismic design cases): 0.1, =Ncψ

Determine the modification factor for post-installed anchor in uncracked concrete, cp ,Nψ

Since it is not a post-installed anchor, 1.0cp ,Nψ =

Determine the concrete breakout strength, Nb, for a single anchor in tension :

D.5.2.2

5.1'efccb hfkN = Eq.(D-7)

hef = 14 in.

16=ck for cast-in anchors, .25.11 inhin ef ≤≤

( ) kipsNb 8.10014000,616 3/5 ==

Substituting into Eq.(D-5) for the group capacity:

( ) ( ) ( ) ( ) kipsNcbg 9.1428.100)0.1(0.10.10.1764,1500,2

==

Determine the pullout strength of anchors in tension, Npn : D.5.3

pPcpn NN ,ψ= Eq.(D-14)

where: Pc,ψ = 1.0 cracked concrete assumed D.5.3.6

Determine pullout capacity of single anchor in tension, Np D.5.3.4

'8 cbrgp fAN =

Eq.(D-15)

Determine bearing area of the head of a single anchor :

2 2 2 22h

brgπ(D - d ) π(1.25 - 0.75 )A = = = 0.785 in.

4 4

( )( ) kipsN p 6.37000,6785.08 ==

Substituting into Eq. (D-14)

BORRADOR

12-7



Reference

Npn = (1.0)(37.6) = 37.6 kips for a single anchor

Determine side-face blowout tensile capacity, Nsb (not applicable since there are no near edges) D.5.4

Determine strength reduction factors applicable for the conditions

The design tensile strengths are determined with the calculated nominal strengths and strength reduction factors of D.4.4. Determine the design steel tensile strength:

Strength reduction factor for ductile steel in tension, = 0.75φ

D.4.4.a.i

Per AWS D1.1 Type B studs qualify as a ductile steel element (20% minimum elongation in 2”)

( ) ( ) kipsNnN satsa 2.8673.28475.0 === φφ D.4.1.1

Determine design concrete breakout strength in tension:

Strength reduction factor for concrete breakout, Condition B: = 0.70φ

D.4.4.c.ii

Note: The wall element in which the connection is embedded contains normal orthogonal reinforcement, which will not significantly interact with the tension failure surface of the tension-loaded anchors. Therefore, Condition B is assumed (supplementary reinforcement not provided). D.4.4

( ) kipsNcbg 03.1009.1427.0 ==φ D.4.1.1

Determine design concrete pullout strength in tension :

Strength reduction factor for pullout = 0.70φ

D.4.4.c.ii

( ) ( )pn t pnN = n N = 0.70 4 37.6 = 105.3kφ φ ips D.4.1.1

Summary of design anchor strength – tension: D.4.1.2

Steel saNφ 86.2 kips ←controls OK D.3.3.4

Concrete breakout

cbgNφ 100.03 kips

Concrete pullout

pnNφ 105.3 kips

BORRADOR

12-8



Reference

Note: The design strength of steel is exceeded by the design strength of concrete breakout and concrete pullout. For plastic design, it may be prudent to require that the nominal strength of steel be exceeded by the design strength of concrete breakout and concrete pullout. This would require an increase in embedment from 14 in. to about 17 in.

7. Design Strength for Tension in a region of moderate or high seismic risk :

( ) kipsNn 65.642.8675.075.0 ==φ D.3.3.3

8. Determine design shear strength, φ Vn,

The contribution to the shear capacity of the anchorage by the concrete located in front of the embedded plate is neglected.

Nominal steel strength in shear, Vsa : for six anchors D.6.1.2

( ) ( ) kipsfAnV utasesa 4.17265442.06 === Eq.(D-19)

Nominal Concrete breakout strength in shear, Vcbg :

Not applicable, since there are no proximate edges. D.6.2.1

Determine the anchor group nominal pryout strength , Vcpg

Concrete pryout strength Vcpg D.6.3

cpg cp cbgV = k N Eq.(D-30)

Concrete breakout strength of four anchors (3,4,5 and 6):

Note : As all six anchors are in shear, technically Ncbg of all six anchors need to be determined. However, conservatively, Ncbg for the anchors in tension are taken for the following calculation.

Ncbg = 142.9 kips Eq.(D-5)

cpk = 2.0 for hef ≥ 2.5 in

Vcpg = 2.0 x 142.9 = 285.8 kips

Determine design shear strength of the anchors

The design shear strength is determined with the calculated nominal strengths and strength reduction factors of D.4.4.

Per AWS D1.1 Type B studs qualify as a ductile steel element (20% minimum elongation in 2”) D.1

BORRADOR

12-9



Reference

Strength reduction factor for ductile steel in shear,

= 0.65φ

D.4.4.a.i

( ) kipsVsa 0.1123.17265.0 ==φ Eq.(D-19)

Strength reduction factor for concrete pryout

= 0.70φ

D.4.4.c.i

cpgVφ = 0.70(285.8) = 200.0 kips D.4.1.1

Summary of anchor strength – shear: D.4.1.2

Steel saVφ 112.0 kips ←controls OK D.3.3.4

Concrete pryout

cpgVφ 200.0 kips

9. Design strength for Shear in a region of moderate or high seismic risk :

( )φ n0.75 V = 0.75 112.0 = 84.0 k ips

D.3.3.3

10. Check tension and shear interaction: D.7

Nua = 32.7 kips > 0.2 φ Nn = 0.2 (64.65) = 12.93 kips D.7.2

Vua = 60 kips > 0.2 φ Vn = 0.2(84) = 16.8 kips D.7.1

2.122.10.84

6065.647.32

75.075.0>=+=

⋅+

⋅∴

n

ua

n

ua

VV

NN

φφ

Interaction is not met and design strength must be increased. It can be achieved by using larger diameter anchors.

Eq.(D-31)

BORRADOR

16-1

Example 16 - Group of Tension Anchors on a Pier with Shear Lug

Design the anchors and shear lug for the steel column located atop the concrete pedestal shown in Figure A.1. Tension is resisted by the anchors, and all the shear is resisted by the shear lug. This is a common design situation encountered in industrial facilities. The pedestal, in this example, could just as easily have been a wall or pilaster. The challenge for this application is to design all of the elements to work properly together while making certain that the design is constructible. This example goes beyond the provisions of Appendix D and uses other provisions such as ACI 349 and AISC Column Base Plate Design Guide. These other provisions are used to evaluate the shear lug. The reader is cautioned to review these other documents to fully understand their limitations.

Given:

• Pedestal geometry shown in Figure A.1 • Pedestal vertical reinforcement 16 - #8 bars (designed per ACI 318 for

tension plus bending at the top of footing) • Pedestal transverse tie sets of #4 bars arranged as shown • Concrete strength f’c = 4000 psi, reinforcement strength fy = 60,000 psi • W12x58 column with 16” square base plate 1-½” thick • All plate steel is ASTM A36 • Use 4 anchors, as shown in Figure A.1, are ASTM F 1554 Grade 55

material • Use anchors with double heavy hex nut at the embedded end • Anchors are to be a ductile design

• Service Dead Load (DL) = 200 kips • Service Wind Load (WL) = 230 kips uplift and 40 kips shear (acting as

shown in Figure A.1)

BORRADOR


Code Reference

16-2

Figure A.1

Design Ultimate Loads

Load combinations are from Section 9.2 of ACI 318. Combination (9-6) controls for tension and shear on the anchors.

Tension

Tu = 0.9D + 1.6W + 1.6H = 0.9 (200 kip) + 1.6 (-230 kip) = -188 kips (upward)

Shear

Vu = 0.9D + 1.6W + 1.6H = 1.6 (40 kip) = 64 kips

Anchor Design

1. Select the anchor diameter to resist the factored load using Section D.5.1 of ACI 318. Try 1-¼ in. diameter anchor.

ASTM F 1554 Grade 55 is ductile steel. Check the assumed anchor diameter by checking the tensile capacity.

Therefore, φ equals 0.75 for tension D.4.4

utasesa fAnN φφ = Eq. (D-3)

2969.0 inAse = From Table B1 of this design guide

ksifuta 75= From Table A of this design guide

kipksiinNsa 218)75)(969.0)(4(75.0 2 ==φ > 188 kips (looks ok)

2. Make Initial Estimate of Embedment Depth:

Estimate the required embedded length hef at 12 diameters. Given the 7 in. anchor spacing and 30 in. pier, 12 diameters equals hef of ~15 in., which will include the entire area of the pier in the projected failure area ANc. Note from Section RD.5.2 that the failure surface projects outward 1.5 units per every 1 unit of depth (~35 degrees to the concrete surface).

Since the tension force in this example places the entire concrete pier section in tension it is doubtful that the concrete breakout capacity will be sufficient to resist the applied load (this will be verified by calculation in Section 3). In this case, the load path must be from the anchors to the vertical pier reinforcement. The transfer of force from anchors to reinforcement requires development of the reinforcing bars into the concrete breakout cone. The #8 vertical bars have a

BORRADOR


Code Reference

16-3

tension development length of approximately 48 in. For now, provide hef such that a #6 bar with hook could be developed into the failure cone as shown in Figure A.2within a distance of Ld.

Figure A.2

Allowing 10 in. to develop a #6 hooked bar (including the length reduction factor of 0.7 from ACI 318, 12.5.3(a)) and 2 in. for cover at the top of pier:

inhef 67.1967.7210 =++=

Try 20 in. as the embedment depth for the anchors

3. Determine Concrete Breakout Capacity of Anchor Group:

Use Section D.5.2 of ACI 318 to calculate the concrete breakout capacity. Note that this check is included in this example for illustrative purposes only, since it is apparent that the concrete breakout load will have to be resisted by supplemental reinforcement. According to Section D.4.4.a.i and D.4.4.c.i of ACI 318, the strength reduction factor φ will be 0.75 for shear and tension since Condition A applies.

The group tension capacity is:

bNcpNcNedNecNco

Nccbg N

AAN ,,,, ψψψψφφ = Eq. (D-5)

Determine the maximum embedment depth allowed:

This determination is required because if the actual embedment depth is used the calculation is overly conservative.

BORRADOR


Code Reference

16-4

ca,max = (30 - 7) / 2 = 11.5 in. ; 1.5hef = 1.5(20) = 30 in.

inhinc efa 305.15.11max, =<= ∴ Per ACI 318, D.5.2.3, limit the embedment depth used in EQ. D-4 through D-11.

hef > smax/3 = 7/3 =2.33 in

inch aef 67.75.1/5.115.1/max, === For use in D-4 through D-11

hef maximum of ca,max/1.5 or smax/3

hef = 7.67 in

Determine the concrete breakout areas to account for spacing

- Determine concrete breakout area for anchor group

22 900)30( ininANc == Since 1.5hef = 30 in, by inspection the projected area includes the entire pier area.

- Determine the concrete breakout area for a single anchor

29 efNco hA = Eq. (D-6)

22 529)67.7(9 ininANco ==

Determine the single anchor tension breakout capacity

5.1efccb hfkN ′= Note: Although the anchors have an embedment depth of about 20 in.,

D5.2.3 reduces the embedment depth because of the multiple side effects which requires the use of Eq. D-7 instead of D-8. Eq. ( D-7)

24=ck For cast-in anchors

lbNb 32243)67.7(400024 5.1 ==

Determine the eccentricity modification factor, Ψec,N

0.1, =Necψ Eq. (D-9)

Factor equals 1, since there is no eccentric load.

Determine the edge distance modification factor, Ψed,N

ef

aNed h

c5.1

3.07.0 min,, +=ψ Eq. (D-11)

BORRADOR


Code Reference

16-5

Factor applies due to close proximity of edges.

0.1)67.7(5.1

5.113.07.0, =+=Nedψ

Determine the cracking modification factor, Ψc,N

25.1, =Ncψ D.5.2.6

By inspection of the magnitude and direction of the applied factored loads (axial tension and shear only) it is concluded that at service loads there will be no cracking along the axis of the anchors. Since there is no cracking that would affect the concrete breakout capacity, The cracking modification factor equals 1.25.

Calculate the anchor group concrete tension breakout capacity

kipslbNcbg 4.5151427)32243)(25.1)(0.1)(0.1(52990075.0 ===φ

Clearly, the concrete tension breakout capacity is not adequate to resist the factored tension load. Supplemental reinforcement is required.

4. Determine the anchor pullout capacity. D.5.3

pPcpn NnnN ,ψφφ = Eq. (D-14)

cbrgp fAN ′= 8 Eq. (D-15)

2237.2 inAbrg = See Table B1 of this design guide

Determine the cracking modification factor, Ψc,N 4.1, =Pcψ D.5.3.6

Factor equals 1.4, since there is no cracking that would affect bearing at the embedded head of the anchor. Calculate the single anchor pullout capacity, Np

kipslbN p 5.7171584)4000()237.2(8 ===

Calculate the anchor group pullout capacity, Npg

kipslbNn pn 7.300300652)71584)(4.1)(4(75.0 ===φ

5. Check side-face blowout. D.5.4

ca1 = 11.5 in. Edge distance

BORRADOR


Code Reference

16-6

0.4hef = 0.4(20) = 8 in. < ca1 = 11.5 in. ∴ Tension capacity is not affected by side-face blowout.

Summary of anchor capacity for tension load:

Anchor steel capacity = 218 kips Concrete breakout capacity = 51.4 kips Pullout capacity = 300.7 kips Side-face blowout capacity = not applicable

Since the concrete breakout capacity controls and is less than the (RD.4.2.1) factored tension of 188 kips, reinforcement is required to be anchored on both sides of the breakout planes. In addition, a ductile design is required as part of the problem statement. A ductile design requires that overall capacity be limited by the anchor’s steel capacity, which is 218 kips.

Reinforcement Design

1. There are many possible options for designing the reinforcement. The challenge is to make certain that the design is constructible. Often, the design will have to be sketched to scale in order to check for congestion and interferences. It is also advisable to coordinate with a member of the construction team.

This example requires vertical reinforcement to constrain the concrete failure prism. The reinforcement must be developed on both sides of the failure plane. As previously discussed, the #8 vertical bars have an approximate tension development length of 48 in. (ACI 318, Section 12.2). Since the effective embedment hef is equal to 20 in., a straight #8 bar cannot be developed. One option, which will not be examined in this example, would be to make the anchors long enough to develop the #8 bars; if there were an excess area of #8 bars, then the development length could be reduced in accordance with ACI 318, Section 12.2.5. Several other options will be evaluated.

The following is the area of reinforcement required to develop the anchor capacity of 218 kips:

204.4)60(9.0

218 inksi

kipsAs == 9.3.2.1

#8 bars equivalent to As = 4.04/0.79 = 5.11 => 6 bars

#6 bars equivalent to As = 4.04/0.44 = 9.18 => 10 bars

BORRADOR


Code Reference

16-7

Figure A.3

Figure A.3 shows three reinforcement options. One option not considered in this example is the use of headed reinforcing bars for the #6 and #8 bar options. A general discussion of the three options considered follows:

A. Option (A) requires a 180 degree hook at the top of each #8 bar. The development length for a standard hook on a #8 bar is approximately 14 in. (including the length reduction factor of 0.7 from ACI 318, Section 12.5.3(a)). Since there is an excess of #8 bars for developing the anchors (6 bars required, 16 bars in the pier), the development length may be reduced according to ACI 318, Section 12.5.3 (d), as follows:

ininprovidedArequiredAinl

s

sdh 47.4

)79.0(1604.4)14()14( === 12.5.3 (d)

However, ldh shall not be less than 8db or 6 in.

ininldh 8)0.1(8 == Since 10 in. was previously allotted for bar development, the #8 bars with hooks will work.

The “J” dimension, or width of the hook, is 8 in., which puts the hooks in close proximity to the anchors. Also, the hooks would have to be rotated in order to avoid interfering with one another at the corners of the pier. As shown in Figure A.1, the #8 bars are continuous into the foundation below the pier; therefore, once the #8 bars are cast into the foundation, it will be difficult to adjust the hooks at the top in order to accommodate the anchors and the shear lug blockout. For this example, Option (A) is ruled out.

BORRADOR


Code Reference

16-8

B. Option (B) requires #6 U-bars (or hairpins) to be installed adjacent to the anchors. Since 10 - #6 bars are required to develop the anchors, 5 U-bars would have to be placed.

This option allows the #6 U-bars bars to be placed at the same time as the anchors and shear lug blockout. Unfortunately, the U-bars will have to compete for space with the anchors, shear key blockout, and lateral tie sets. In addition, construction will have to take additional measures to secure the U-bars during the casting of concrete. For this example, Option (B) is ruled out.

C. Option (C) is the solution chosen for this example. This option makes use of hooked #6 bars that are lap spliced with the #8 pier reinforcement. The required number of #6 bars is 10, but 12 bars will be provided for convenience because there are three #8 bars between corners in the pedestal (3 on each side of the pier). The hooked bars may be placed at the same time as the anchors and shear key blockout, which will permit adjustment of the hooks to avoid congestion and interferences. Also, the hooked bars will not interfere with the interior legs of the lateral tie sets.

The lap splice for the #6 bars must be Class B, and the required lap length is approximately 37 in., according to ACI 318, Section 12.15. Therefore, the overall length of the hooked bars is 10 in. (the development length) plus 37 in. (the lap splice length) plus 7.67 in due to slope of shear cone (See Figure A.3.c), which equals 54 in. The 4 ft-6 in. (54 in.) pier height is sufficient to accommodate the hooked bars.

2. ACI 318, Section 7.10.5.6, requires lateral reinforcement to encompass the anchors in the top of a pier/column. As shown in Figure A.1, 2 sets of #4 ties will be located within the top 5 in. of the pier.

Shear Lug Design

A single cantilever type shear lug is proposed to transfer the entire shear load to the top of the concrete pier. For convenience, the lug will use the same 1-½”-thick plate used for the column base plate. The methodology used in this example is based on ACI 349-06, Appendix D, and Column Base Plates: Steel Design Guide Series 1, American Institute of Steel Construction, Inc., Chicago, Illinois, 1990.

As a matter of good practice, the shear lug should be embedded a minimum of 2 in. into the concrete pier. The 1-½ in. thick grout pad between the base plate and top of pier is considered to be ineffective for transfer of shear. The shear lug must therefore be a minimum of 3

BORRADOR


Code Reference

16-9

½ in. deep. Try 2 in. of concrete embedment a lug length of 12 in. (see sketch in Figure A.4)

Figure A.4

1. Check the lug for concrete bearing. (Note that φ factors From ACI 349 are different than those used in ACI 318 and that ACI 349 φ factors will be used for this design)

lugcn AfP ′= 3.1φφ Bearing capacity according to ACI 349, Section D.4.6.2

kipslbPn 4.8787360)122)(4000)(3.1(7.0 ==×=φ > Vu = 64 kips ∴ O.K.

2. Check for shear acting towards the edge of the pier. The methodology of ACI 349, Section D.11.2, is applied. The effective stress area is found by projecting 45-degree planes from the bearing edges of the shear lug as shown in Figure A.4.

25.463)212()25.1630( inAeff =×−×= Note, the area of the lug must be deducted.

effcn AfV ′= 4φφ

8.0=φ

kipslbVn 8.9393803)5.436(4000)4(8.0 ===φ > Vu = 64 kips ∴ O.K.

BORRADOR


Code Reference

16-10

3. Check the shear lug for bending and shear stresses. A uniform bearing pressure is assumed over the 2 in. of embedded depth into the top of the pedestal. The maximum moment will occur where the lug attaches to the baseplate. The plate capacity is based on the Load and Resistance Factor Design (LRFD) capacity for a plate in bending and shear.

Determine the bending capacity of shear lug

inkipM u ⋅=+= 160)225.1)(64( Applied factored moment at the weld of

the lug to the baseplate, based on the shear load being applied at mid-depth of lug.

yn ZFM φφ = Plastic moment capacity of the shear lug plate AISCLRFD F1.1

inkipM n ⋅== 7.218)36(4

)5.1)(12()9.0(2

φ > Mu = 160 kip-in ∴ O.K.

Determine the shear capacity of the shear lug

ygn FAV 6.0φφ = Shear capacity of the shear lug plate AISC LRFD J5.3

kipsVn 350)36)(5.112)(6.0(9.0 =×=φ > Vu = 64 kips ∴ O.K.

Figure A.5

4. Design the shear lug-to-baseplate weld using E70XX electrodes. The resultant forces on the welds are as shown in Figure A.5. The shear force is shared equally between the two fillets. The moment is resisted by a force

BORRADOR


Code Reference

16-11

couple at the two fillets. The vertical force couple with distance “s” is taken between the centroids of the two fillet welds. For 1-½ in. plate, the minimum allowable fillet size is 5/16 in.

inkipfv /67.2)212/()64( =×= Shear force per in. of weld

inkipft /8.7)]16

53

12(5.1)[12(160

=××+

= Tension force per in. of weld

inkipffR tv /24.822 =+= Total force per in. of weld

)sin50.00.1(6.0 5.1 θφφ += EXXw FF Weld capacity AISC LRFD

Appendix J2.4

θ = 90 degrees since the force is perpendicular to the axis of the weld.

φ = 0.75

Solve for the required weld leg size, a :

ina 25.0)707.0))(90(sin5.01)(70)(6.0(75.0

24.85.1 =

+= < 5/16 in.

∴ Use the minimum allowable size of 5/16 in.

Design Summary

Anchors: 1-¼ in. diameter ASTM 1554 Grade 55 w/ heavy hex nuts

20 in. embedment depth (top of concrete to top of embedded head/nut)

Supplementary reinforcement: 12 - #6 bars with standard 180 hooks at the top, lap splice with #8 pier reinforcement

Shear lugs: 12 in. x 3-½ in. x 1-½ in. thick plate ASTM A36

5/16 in. fillet weld E70XX on each of shear lug side to baseplate

BORRADOR

Table A

MATERIALS FOR HEADED ANCHORS AND THREADED RODS

Yield Strength, Min. Elongation, Min. Material

Grade or

Type

Diameter (in.)

Tensile Strength,

Min. (ksi)

(ksi) Method (%) Length

Reduction of Area,

Min. (%)

Comments

Welded Studs AWS D1.1:2004 ASTM A29-05 /

A 108-03

B 1010 1020

½ to 1 65 51 0.2% 20 2 in. 50 Structural Welding Code – Steel, Section 7, covers welded headed or welded bent studs. AWS D1.1 requires studs to be made from cold drawn bar stock conforming to the requirements of ASTM A108.

36 ¼ to 4 58 36 0.2% 23 2 in. 40

55 ≤2* 75 55 0.2% 21 2 in. 30 ASTM F1554-

04 (H,HD, T)1

105 ¼ to 3 125 105 0.2% 15 2 in. 45

ASTM F1554, Standard Specification for Anchor Bolts, Steel, 36, 55, and 105-ksi Yield Strength is the preferred material specification for anchors. *Diameters larger than 2 in. (up to 4 in) are available, but the reduction of area will vary for Grade 55.

≤2-½ 125 105 0.2% 16 4D 50

2-½ to 4 115 95 0.2% 16 4D 50 ASTM A193-05 (H,T) B7

Over 4 to 7 100 75 0.2% 18 4D 50

ASTM A193, Standard Specification for Alloy-Steel and Stainless Steel Bolting Materials for High-Temperature Service: Grade B7 is an alloy steel for use in high-temperature service.

A ¼ to 4 60 … … 18 2 in. … ASTM A307-04 (Gr. A: HD) (Gr. C: H, T) C ¼ to 4 58 36 … 23 2 in. …

ASTM A307, Standard Specification for Carbon Steel Bolts and Studs, 60000 PSI Tensile Strength: ACI 349 specifies that elements meeting ASTM A307 shall be considered ductile. Note that Grade C conforms to tensile properties for ASTM A36.

ASTM A36-05 (H, T) - To 8 58 36 … 23 2 in. …

ASTM A36, Standard Specification for Carbon Structural Steel: Since ACI 318 considers ASTM A307 to be ductile, A36 will also qualify since it is the basis for ASTM A307 Grade C.

¼ to 1 120 92 0.2% 14 4D 35

Over 1 to 1-½ 105 81 0.2% 14 4D 35 ASTM A449-

04b (H, HD, T)

1

Over 1-½ to 3 90 58 0.2% 14 4D 35

ASTM A449, Standard Specification for Quenched and Tempered Steel Bolts and Studs: This specification is for general high strength applications.

Notes:

1 Anchor type availabilities are denoted as follows: H = hooked bolt, HD = headed bolt, and T = threaded rod. BORRADOR

Table B1Gross Area (AD), Effective Area (Ase), and Bearing Area (Abrg) for cast-in-place anchors, threaded rod with nuts, and threaded

rods with nuts and washers

do (in.) AD (in.2) Ase (in.2)

0.250 0.049 0.032 0.092 0.117 - 0.142 0.201 0.117 0.167

0.375 0.110 0.078 0.206 0.164 - 0.280 0.362 0.164 0.299

0.500 0.196 0.142 0.366 0.291 0.467 0.464 0.569 0.291 0.467

0.625 0.307 0.226 0.572 0.454 0.671 0.693 0.822 0.454 0.671

0.750 0.442 0.334 0.824 0.654 0.911 0.824 1.121 0.654 0.911

0.875 0.601 0.462 1.121 0.891 1.188 1.121 1.465 0.891 1.188

1.000 0.785 0.606 1.465 1.163 1.501 1.465 1.855 1.163 1.501

1.125 0.994 0.763 1.854 1.472 1.851 1.854 2.291 1.472 1.851

1.250 1.227 0.969 2.288 1.817 2.237 2.288 2.773 1.817 2.237

1.375 1.485 1.16 2.769 2.199 2.659 2.769 3.300 2.199 2.659

1.500 1.767 1.41 3.295 2.617 3.118 3.295 3.873 2.617 3.118

1.750 2.405 1.90 - 3.562 4.144 - - - 4.1442.000 3.142 2.50 - 4.653 5.316 - - - 5.316

Notes:1 Dimensions taken from AISC Steel Design Manual2 Ase = 0.7854(d0-(0.9743/n))2

3 AH = F2 (Square Head Bolt/Nut) or AH = 1.5F2tan30 (Hex/Heavy Hex Head Bolt/Nut)

4 Abrg = AH-AD

Gross Cross-Sectional Area

Nominal Diameter

Table B1 - Dimensional Characteristics of Cast-In-Place Bolts & Threaded Rods with Nuts

Abrg (in.2)

Square Head Bolt

Hex Head Bolt

Heavy Hex Head Bolt

Threaded Rod w/

Square Nut

Threaded Rod w/ Heavy

Square Nut

Threaded Rod w/ Hex

Nut

Threaded Rod w/

Heavy Hex Nut

Effective Cross-Sectional Area

Note: All washers need to meet the minimum thickness requirements in Table B2, or a reduced Abrg. needs to be calculated according to D.5.2.8.

BORRADOR

Stud Gross AreaDiameter (do) of stud (AD) Width (F) Gross Area (AH) Abrg Width (F) Gross Area (AH) Abrg Width (F) Gross Area (AH) Abrg

(in.) (in.2) (in.) (in.) (in.)

0.250 0.049 0.375 0.141 0.092 0.438 0.166 0.117 N.A. N.A. N.A.

0.375 0.110 0.563 0.316 0.206 0.563 0.274 0.164 N.A. N.A. N.A.

0.500 0.196 0.750 0.563 0.366 0.750 0.487 0.291 0.875 0.663 0.467

0.625 0.307 0.938 0.879 0.572 0.938 0.761 0.454 1.063 0.978 0.671

0.750 0.442 1.125 1.266 0.824 1.125 1.096 0.654 1.250 1.353 0.911

0.875 0.601 1.313 1.723 1.121 1.313 1.492 0.891 1.438 1.790 1.188

1.000 0.785 1.500 2.250 1.465 1.500 1.949 1.163 1.625 2.287 1.501

1.125 0.994 1.688 2.848 1.854 1.688 2.466 1.472 1.813 2.845 1.851

1.250 1.227 1.875 3.516 2.288 1.875 3.045 1.817 2.000 3.464 2.237

1.375 1.485 2.063 4.254 2.769 2.063 3.684 2.199 2.188 4.144 2.659

1.500 1.767 2.250 5.063 3.295 2.250 4.384 2.617 2.375 4.885 3.118

1.750 2.405 N.A. N.A. N.A. 2.625 5.967 3.562 2.750 6.549 4.144

2.000 3.142 N.A. N.A. N.A. 3.000 7.794 4.653 3.125 8.457 5.316



4 Abrg = AH-AD

5 Ref. Figures of Table B1.

(in.2) (in.2)

TABLE B2 - DIMENSIONAL PROPERTIES OF BOLTS FOR DETERMINING BEARING AREA, Abrg

Square Head Bolt Hex Head Bolt Heavy Hex Head Bolt

(in.2)

BORRADOR

Stud Gross AreaDiameter (do) of stud (AD) Width (F) Gross Area (AH) Abrg Width (F) Gross Area (AH) Abrg Width (F) Gross Area (AH) Abrg Width (F) Gross Area (AH) Abrg

(in.) (in.2) (in.) (in.) (in.) (in.)

0.250 0.049 0.438 0.191 0.142 0.500 0.250 0.201 0.438 0.166 0.117 0.500 0.217 0.167

0.375 0.110 0.625 0.391 0.280 0.688 0.473 0.362 0.563 0.274 0.164 0.688 0.409 0.299

0.500 0.196 0.813 0.660 0.464 0.875 0.766 0.569 0.750 0.487 0.291 0.875 0.663 0.467

0.625 0.307 1.000 1.000 0.693 1.063 1.129 0.822 0.938 0.761 0.454 1.063 0.978 0.671

0.750 0.442 1.125 1.266 0.824 1.250 1.563 1.121 1.125 1.096 0.654 1.250 1.353 0.911

0.875 0.601 1.313 1.723 1.121 1.438 2.066 1.465 1.313 1.492 0.891 1.438 1.790 1.188

1.000 0.785 1.500 2.250 1.465 1.625 2.641 1.855 1.500 1.949 1.163 1.625 2.287 1.501

1.125 0.994 1.688 2.848 1.854 1.813 3.285 2.291 1.688 2.466 1.472 1.813 2.845 1.851

1.250 1.227 1.875 3.516 2.288 2.000 4.000 2.773 1.875 3.045 1.817 2.000 3.464 2.237

1.375 1.485 2.063 4.254 2.769 2.188 4.785 3.300 2.063 3.684 2.199 2.188 4.144 2.659

1.500 1.767 2.250 5.063 3.295 2.375 5.641 3.873 2.250 4.384 2.617 2.375 4.885 3.118

1.750 2.405 - - - - - - - - - 2.750 6.549 4.144

2.000 3.142 - - - - - - - - - 3.125 8.457 5.316



4 Abrg = AH-AD

5 Ref. Figures of Table B1.

(in.2) (in.2) (in.2) (in.2)

TABLE B3 - DIMENSIONAL PROPERTIES OF NUTS FOR DETERMINING BEARING AREA, Abrg

Heavy Hex NutSquare Nut Heavy Square Nut Hex Nut

BORRADOR

Table C—Sample data for a post-installed torque-controlled mechanical expansion anchor

[Note: This table was created for illustrating the use of test data as would be developed from a test program according to ACI 355.2-04 for use with the design procedures of ACI 318-05 Appendix D. This data is purely fictional and does not represent any specific anchor system. This data should not be used for actual design purposes. For actual anchor design data, obtain data that has been tested, developed, and certified to be in accordance with ACI 355.2-04.] Anchor system is qualified for use in both cracked and uncracked concrete in accordance with test program of Table 4.2 of ACI 355.2-04. The material, ASTM F1554 grade 55, meets the ductile steel element requirements of ACI 318-05 Appendix D (tensile test elongation of at least 14 percent and reduction in area of at least 30 percent).

Characteristic Symbol Units Nominal anchor diameter Installation information

Outside diameter do in. 3/8 ½ 5/8 ¾ 1.75 2.5 3 3.5 2.75 3.5 4.5 5 Effective embedment depth hef in. 4.5 5.5 6.5 8

Installation torque Tinst ft-lb 30 65 100 175 Minimum edge distance cmin in. 1.75 2.5 3 3.5

Minimum spacing smin in. 1.75 2.5 3 3.5 Minimum concrete thickness hmin in. 1.5hef 1.5hef 1.5hef 1.5hef Critical edge distance @ hmin cac in. 2.1 3.0 3.6 4.0

Anchor data Anchor material ASTM F 1554 Grade 55 (meets ductile steel element requirements) Category number 1, 2, or 3 — 2 2 1 1

Yield strength of anchor steel fya psi 55,000 55,000 55,000 55,000 Ultimate strength of anchor steel futa psi 75,000 75,000 75,000 75,000

Effective tensile stress area Ase in.2 0.0775 0.142 0.226 0.334 Effective shear stress area Ase in.2 0.0775 0.142 0.226 0.334

Effectiveness factor for uncracked concrete

kuncr — 24 24 24 24

Effectiveness factor for cracked concrete used for ACI 318 design kcr* — 17 17 17 17

ψc,N for ACI 318 design in cracked concrete ψc,N* — 1.0 1.0 1.0 1,0

ψc,N = kuncr/kcr for ACI 318 design in uncracked concrete ψc,N* — 1.4 1.4 1.4 1.4

hef Np hef Np hef Np hef Np 1.75 1,354 2.5 2,312 3 4,469 3.5 5,632 2.75 2,667 3.5 3,830 4.5 8,211 5 9,617

Pullout or pull-through resistance from tests Np lb

4.5 5,583 5.5 7,544 6.5 14,254 8 19,463 1.75 903 2.5 1,541 3 2,979 3.5 3,755 Tension resistance of single anchor for

seismic loads Neq lb 4.5 3,722 5.5 5,029 6.5 9,503 8 12,975 Shear resistance of single anchor for seismic

loads Veq lb 2,906 5,321 8,475 12,543

Axial stiffness in service load range β lb/in. 55,000 57,600 59,200 62,000 Coefficient of variation for axial stiffness in service load range. ν % 12 11 10 9

*These are values used for kc and ψc,N in ACI 318 for anchors qualified for use only in both cracked and uncracked concrete.

BORRADOR

Mexico DF Ejemplos Anclajes

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