Example Calc s

EXAMPLE CALCULATION Mechanical Anchors (ACI 318 App. D/ICC-ES AC193) Example calculation for a single Strong-Bolt™ anchor using USD: Determine if a single ¹⁄₂" diameter Strong-Bolt ™ torque-controlled expansion anchor with a minimum 5" embedment (h ( h ef = 4¹⁄₂ inches) installed 4" from the edge of a 12" deep spandrel beam is adequate for a service tension load of 1,000 lb. for wind and a reversible service shear load of 350 lb. for wind. The anchor will be in the tension zone, away from other anchors in ƒ'c = 3,000 psi normal-weight concrete.

1000 lb. 350 lb. 4¹⁄₂"

4"

Reference the appropriate tables in this catalog for Strong-Bolt anchor performance values as determined from testing in accordance with ACI 355.2 and ICC-ES AC193.

CALCULATIONS CALCULATION S AND DISCUSSION

REFERENCE

Note: Calculations are performed in accordance with ACI 318-05. 1. Determine the factored tension and shear design loads:

ACI 318, 9.2.1

N ua = 1.6W 1.6 W = = 1.6 x 1,000 = 1,600 lb.

Note: Rebar not shown for clarity.

CALCULATIONS CALCULATIO NS AND DISCUSSION

4. Concrete breakout capacity under tension loading:

D.5.2

φN cb ≥ N ua N cb =

V ua = 1.6W 1.6 W = = 1.6 x 350 = 560 lb.

REFERENCE

Eq. (D-1)

ANc ψ ψ ψ N ANco ed,N c,N cp,N b

Eq. (D-4);

where: 2. Design considerations:

D.4.1.2

Eq. (D-7)

substituting:

This is a combined tension and shear interaction problem where values for both φN n and φV n need to be determined. φN n is the lesser of the design tension strength controlled by: steel ( φN sa ), concrete breakout (φN cb ), or pull-out ( φnN pn ). φV n is the lesser of the design shear strength controlled by: steel ( φV sa ), concrete breakout (φV cb ), or pryout ( φV cp ).

3. Steel capacity under tension Loading:

N b = k c √ƒ ' 'c h ef 1.5

φN cb = φ ANc ψ ed,N ψ c,N ψ cp,N k c √ƒ ' 'c h ef 1.5 ANco

where:

D.5.1

k c = k cr = 17 (Anchor is installed in a tension zone, therefore, cracking is assumed at service loads

This catalog

ψ cp,N = 1.0

D.5.2.7 Eq. (D-11)

φN sa ≥ N ua

Eq. (D-1)

ψ = 0.7 + 0.3 c a,min when c ed,N a,min < 1.5 h ef 1.5h 1.5 h ef

N sa = 13,500 lb.

This catalog

by observation, c a,min = 4 < 1.5h 1.5 h ef

φ = 0.75

This catalog

ψ ed,N = 0.7 + 0.3

n = n = 1 (single anchor)

(4) 1.5(4.5)

= 0.88

Calculating for φN sa :

ψ c,N = 1.0 assuming cracking

φN sa = 0.75 x 1 x 13,500 = 10,125 lb. > 1,600 lb. – OK

at service loads (ft > fr)

D.5.2.6

φ = 0.65 for Condition B

This catalog

∴ ¹⁄₂ in. diameter anchor is adequate

(no supplementary reinforcement provided) ANco = 9h 9h ef 2 = 9(4.5) 2 = 182.25 in. 2

Eq. (D-6)

ANc = (ca 1 + 1.5h 1.5 h ef )(2 x 1.5h 1.5h ef ) = (4 + 1.5(4.5))(2 x 1.5(4.5)) = 145.13 in. 2

Fig. RD.5.2.1(a)

ANc 145.13 = = 0.8 ANco 182.25 Calculating for φN cb : Would you like help with these c alculations? Visit www.simpsonanchors.com to download the Simpson Strong-Tie ® Anchor Designer ™ software.

24

φN cb = 0.65 x 0.8 x 1.0 x 0.88 x 1.0 x 17 x √3,000 x (4.5) 1.5 = 4,067 lb. > 1,600 lb. – OK

Continued on next page.

. C N I Y N A P M O C E I T G N O R T S N O S P M I S 9 0 0 2 © 9 0 0 2 S A S C

EXAMPLE CALCULATION Mechanical Anchors (ACI 318 App. D/ICC-ES AC193) Continued from previous page. CALCULATIONS AND DISCUSSION

REFERENCE

5. Pullout capacity:

D.5.3

CALCULATIONS AND DISCUSSION

REFERENCE

9. Concrete pryout strength:

D.6.3

Pullout capacity, N pn,cr , is established by reference tests in cracked concrete by the reliability test of ACI 355.2. Data from the anchor prequaliﬁcation testing must be used. Reference Strong-Bolt ™ anchor "characteristic tension design values" table for the 5 percent fractile value, N pn,cr.

φnV cp ≥ V ua

Eq. (D-2)

V cp = k cp N cb

Eq. (D-29)

φN pn ≥ N ua

n = 1

Eq. (D-1)

N pn,cr = 2,995 x

3,000 ( 2,500 )

0.5

= 3,281 lb.

This catalog

k cp = 2.0 and φ = 0.70 4,067 = 12,514 lb. k cp N cb = 2.0 x 0.65

D.4.1.2

φnV cp = 0.70 x 1 x 12,514 = 8,760 lb. > 560 lb.

This catalog

φ = 0.65 φN pn = 0.65 x 3,281 = 2,133 lb. > 1,600 lb.

–

OK

6. Check all failure modes under tension loading:

where:

Summary: Steel capacity = 10,125 lb. Concrete breakout capacity = 4,067 lb. Pullout capacity = 2,133 lb. ← Controls ∴ φN n = 2,133 lb. as pullout capacity controls

This catalog D.6.3.1 –

OK

10. Check all failure modes under shear Loading:

D.4.1.2

Summary: Steel capacity

= 4,264 lb.

D.6.1

φV sa ≥ V ua

Concrete breakout capacity = 2,301 lb. ← Controls

Eq. (D-2)

Pryout capacity

V sa = 6,560 lb.

This catalog

∴ φV n = 2,301 lb. as concrete breakout capacity controls

7. Steel capacity under shear loading:

φ = 0.65

This catalog

= 8,760 lb.

Calculating for φV sa :

φV sa = 0.65 x 6,560 = 4,264 lb. > 560 l b.

–

11. Check interaction of tension and shear forces:

OK

∴ ¹⁄₂ in. diameter anchor is adequate 8. Concrete breakout capacity under shear loading:

φV cb ≥ V ua V cb =

AVc ψ ψ ed,V c,V V b AVco

D.6.2

If 0.2 φV n ≥ V ua, then the full tension design strength is permitted.

Eq. (D-2)

By observation, this is not the case.

Eq. (D-21)

If 0.2 φN n ≥ N ua , then the full shear design strength is permitted


(d o e )

D.7.1

D.7.2


where: V b = 7

D.7

0.2

√d o √ƒ 'c c a11.5

Eq. (D-24)

substituting: φV cb = φ AVc ψ ed,V ψ c,V 7 e AVco d o

( )

Therefore: N ua V + ua ≤ 1.2 φN n φV n

0.2

√d o √ƒ 'c c a11.5

Eq. (D-31)

1,600 560 + = 0.75 + 0.24 = 0.99 < 1.2 2,133 2,301

where:

–

OK

φ = 0.70 for Condition B (no supplementary reinforcement provided)

D4.4(c)(i)

AVco = 4.5ca 12 = 4.5(4) 2 ∴ AVco = 72 in. 2

Eq. (D-23)

AVc = 2(1.5c a 1)(1.5ca 1) = 2(1.5(4))(1.5(4)) ∴ AVc = 72 in. 2 72 AVc = =1 AVco 72

Fig. RD.6.2.1(a)

ψ ed,V = 1.0 since c a 2 > 1.5c a 1

Eq. (D-27)

12. Summary A single ¹⁄₂" diameter Strong-Bolt anchor at a 5" embedment depth is adequate to resist the applied service tension and shear loads of 1,000 lb. and 350 lb., respectively. ™

D.6.2.1

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

D.6.2.7

d o = 0.5 in. e = 8d 0 = 8 (0.5) = 4"

D.6.2.2

ca 1 = 4 in.

φV cb = 0.70 x 1 x 1.0 x 1.0 x 7 x

( 0.54 )

0.2

x √0.5

x √3,000 x (4) 1.5 = 2,301 lb. > 560 lb.

–

OK

25

EXAMPLE CALCULATION Adhesive Anchors (ICC-ES AC308) Example calculation for a single SET-XP™ epoxy adhesive anchor using USD: Determine if a single ¹⁄₂" diameter ASTM A193 Grade B7 anchor rod in SET-XP ™ epoxy adhesive anchor with a minimum 4 ¹⁄₂" embedment (h ef = 4¹⁄₂") installed 1³⁄₄" from the edge of a 12" deep spandrel beam is adequate for a service tension load of 560 lb. for wind and a reversible service shear load of 425 lb. for wind. The anchor will be in uncracked dry concrete, away from other anchors in ƒ'c = 3,000 psi normal-weight concrete. The anchor will be subjected to a maximum short-term temperature of 110˚F and a maximum long-term temperature of 75 ˚F. Continuous inspection will be provided

560 lb. 425 lb. 4¹⁄₂"

1³⁄₄"


Reference the appropriate tables in this catalog f or SET-XP epoxy adhesive anchor performance values as determined from testing in accordance with ICC-ES AC308. CALCULATIONS AND DISCUSSION

REFERENCE

Note: Calculations are performed in accordance with ICC-ES AC308 and ACI 318-05. 1. Determine the factored tension and shear design loads:


REFERENCE


D.5.2

φN cb ≥ N ua ACI 318, 9.2.1 N cb =

N ua = 1.6W = 1.6 x 560 = 900 lb. V ua = 1.6W = 1.6 x 425 = 680 lb.

Eq. (D-1)

ANc ψ ψ ψ N ANco ed,N c,N cp,N b

where: N b = k c √ƒ 'c h ef 1.5

2. Design considerations:

D.4.1.2

This is a combined tension and shear interaction problem where values for both φN n and φV n need to be determined. φN n is the lesser of the design tension strength controlled by: steel ( φN sa ), concrete breakout (φN cb ), or adhesive ( φN a ). φV n is the lesser of the design shear strength controlled by: steel ( φV sa ), concrete breakout (φV cb ), or pryout ( φV cp ).

3. Steel capacity under tension loading:

Eq. (D-4);

Eq. (D-7)

substituting:

φN cb = φ ANc ψ ed,N ψ c,N ψ cp,N k c √ƒ 'c h ef 1.5 ANco

where: k c = k uncr = 24

This catalog

c a,min = 1.75" = c min – OK

D.5.1

c ac = 3h ef = 3(4.5) = 13.5" ψ cp,N = c a,min = 0.13 ≥ 1.5h ef = 0.5 c ac c ac

φN sa ≥ N ua

Eq. (D-1)

ψ cp,N = 0.5

N sa = 17,750 lb.

This catalog

ψ

φ = 0.75

This catalog

c a,min

when c a,min < 1.5 h ef ed,N = 0.7 + 0.3 1.5h ef

n = 1 (single anchor)

by observation, c a,min < 1.5h ef

Calculating for φN sa :

ψ ed,N = 0.7 + 0.3 1.75 = 0.78

φN sa = 0.75 x 1 x 17,750 = 13,313 lb. > 900 lb. – OK

This catalog Eq. (D-13)

Eq. (D-11)

1.5(4.5)

ψ c,N = 1.0 since k c = k uncr = 24

D.5.2.6


This catalog

(no supplementary reinforcement provided)

Would you like help with these calculations? Visit www.simpsonanchors.com to download the Simpson Strong-Tie ® Anchor Designer ™ software.

ANco = 9h ef 2 = 9(4.5) 2 = 182.25 in. 2

Eq. (D-6)

ANc = (ca 1 + 1.5h ef )(2 x 1.5h ef ) = (1.75 + 1.5(4.5))(2 x 1.5(4.5)) = 114.75 in. 2 ANc 114.75 = = 0.63 ANco 182.25

Fig. RD.5.2.1(a)

Calculating for φN cb :

φN cb = 0.65 x 0.63 x 1.0 x 0.78 x 0.5 x 24 x √3,000 x (4.5) 1.5 = 2,004 lb. > 900 lb. – OK

26



EXAMPLE CALCULATION Adhesive Anchors (ICC-ES AC308) Continued from previous page. CALCULATIONS AND DISCUSSION

REFERENCE


REFERENCE

5. Adhesive anchor capacity under tension loading:

AC308 Section 3.3

8. Concrete breakout capacity under shear loading:

D.6.2

φN a ≥ N ua ANa ψed,Na ψ p,Na N ao N a = ANao

Eq. (D-1)

N ao = τk,uncr πdh ef = 2,422π(0.5)(4.5) = 17,120 lb.

Eq. (D-16f)

where:

Eq. (D-16d)

V b = 7

s cr,Na = 20d

√

τk,uncr ≤

1,450

s cr,Na = (20)(0.5)

√

Eq. (D-16a)

3h ef

AVc ψ ψ V AVco ed,V c,V b

V cb =

(d o e )

Eq. (D-21)

0.2

√d o √ƒ 'c c a11.5

Eq. (D-24)

( )

Eq. (D-16e)

0.2

√d o √ƒ 'c c a11.5

where:


D4.4(c)(i)

ANa = (c a1 + c cr,Na )(s cr,Na ) = (1.75 + 6.46)(12.92) = 106.07" 2 ψed,Na = (0.7 + 0.3 c a,min ) ≤ 1.0 Since c a,min < c cr,Na Eq. (D-16m) c cr,Na

AVco = 4.5ca 12 = 4.5(1.75) 2 ∴ AVco = 13.78 in. 2

Eq. (D-23)

ψed,Na = (0.7 + 0.3 c a,min ) = (0.7 + 0.3 1.75 ) = 0.78

AVc = 2(1.5c a 1)(1.5ca 1) = 2(1.5(1.75))(1.5(1.75)) ∴ AVc = 13.78 in. 2 AVc 13.78 = =1 AVco 13.78

Fig. RD.6.2.1(a)

ψ ed,V = 1.0 since ca 2 > 1.5ca 1

Eq. (D-27)

ψ c,V = 1.4 for uncracked concrete

D.6.2.7

2

2

ANao = (s cr,Na ) = (12.92) = 166.93"

2

c cr,Na

Eq. (D-16c)

6.46

ψp,Na = max [c a,min ; c cr,Na ] when c a,min < c ac c ac

Eq. (D-16p)

ψp,Na = max [1.75; 6.46] = 6.46 = 0.48 13.5 13.5 φ = 0.65 for dry concrete

This catalog

Calculating for φN a : φN a = 0.65x 106.07 x0.78x0.48x17,120 = 2,647 lb. > 900 lb. – OK 166.93 6. Check all failure modes under tension loading:


Eq. (D-2)


2,422 = 12.92" ≤ 3h = 13.5" ef 1,450

s cr,Na = 12.92" s 12.92 c cr,Na = cr,Na = = 6.46" 2 2

φV cb ≥ V ua

D.4.1.2

Summary: Steel capacity = 13,313 lb. Concrete breakout capacity = 2,004 lb. ← Controls Adhesive capacity = 2,647 lb. ∴ φN n = 2,004 lb. as concrete breakout capacity controls

D.6.2.1

d o = 0.5 in. e = 8d 0 = 8 (0.5) = 4"

D.6.2.2

ca 1 = 1.75 in.

φV cb = 0.70 x 1 x 1.0 x 1.4 x 7 x

( 0.54 )

0.2

x √0.5

x √3,000 x (1.75) 1.5 = 932 lb. > 680 lb. – OK 9. Concrete pryout capacity per AC308 V cp = min[k cp N a ; k cp N cb ]

Eq. (D-30a)

D.6.1

φV sa ≥ V ua

k cp = 2.0 for h ef ≥ 2.5"

Eq. (D-2)

N a = 4,072 lb. from adhesive-capacity calculation without φ factor

V sa = 10,650 lb.

This catalog

7. Steel capacity under shear loading:

φ = 0.65


φV sa = 0.65 x 10,650 = 6,923 lb. > 680 lb. – OK

This catalog

N cb = 3,083 lb. from concrete-breakout calculation without φ factor V cp = (2.0)(3,083) = 6,166 lb. controls

φ = 0.7

This catalog

φV cp = (0.7)(6,166) = 4,316 lb. > 680 lb. – OK


27

EXAMPLE CALCULATION Adhesive Anchors (ICC-ES AC308) Continued from previous page. CALCULATIONS AND DISCUSSION

REFERENCE

10. Check all failure modes under shear loading:

D.4.1.2

Summary: Steel capacity

= 6,923 lb.

Concrete breakout capacity = 932 lb. ← Controls Pryout capacity

= 4,316 lb.

∴ φV n = 932 lb. as concrete breakout capacity controls

11. Check interaction of tension and shear forces: If 0.2 φV n ≥ V ua, then the full tension design strength is permitted.

D.7 D.7.1

By observation, this is not the case. If 0.2 φN n ≥ N ua , then the full shear design strength is permitted

D.7.2

By observation, this is not the case. Therefore: N ua V + ua ≤ 1.2 φN n φV n

Eq. (D-31)

900 680 + = 0.45 + 0.73 = 1.18 < 1.2 – OK 2,004 932

12. Summary

A single ¹⁄₂" diameter ASTM A193 Grade B7 anchor rod in SET-XP epoxy adhesive at a 4 ¹⁄₂" embedment depth is adequate to resist the applied service tension and shear loads of 560 lb. and 425 lb., respectively. ™


28

EXAMPLE CALCULATION Torque Controlled Adhesive Anchors (ICC-ES AC308) Example calculation for a single IXP™ anchor with SET-XP™ epoxy using USD: Determine if a single ¹⁄₂" diameter IXP™ torque-controlled adhesive anchor with SET-XP™ epoxy with a minimum 4 ⁷⁄₈" embedment (h ef = 4⁷⁄₈") installed 4⁷⁄₈" from the edge of a 12" deep spandrel beam is adequate for a service tension load of 2,500 lb. for wind and a reversible service shear load of 450 lb. for wind. The anchor will be in the tension zone, away from other anchors in ƒ'c = 3,000 psi normal-weight concrete. The anchor will be subjected to a maximum short-term temperature of 180 ˚F and a maximum long-term temperature of 75 ˚F. Continuous inspection will be provided.

2,500 lb. 450 lb. 4⁷⁄₈"

4⁷⁄₈"

Reference the appropriate tables in this catalog for IXP anchor with SET-XP epoxy performance values as determined from test ing in accordance with ICC-ES AC308.


REFERENCE

Note: Calculations are performed in accordance with ICC-ES AC308 and ACI 318-05. 1. Determine the factored tension and shear design loads:



REFERENCE


D.5.2

φN cb ≥ N ua ACI 318, 9.2.1 N cb =

N ua = 1.6W = 1.6 x 2,500 = 4,000 lb. V ua = 1.6W = 1.6 x 450 = 720 lb.

Eq. (D-1)

ANc ψ ψ ψ ed,N c,N cp,N N b ANco

where: N b = k c √ƒ 'c h ef 1.5

2. Design considerations:

D.4.1.2

This is a combined tension and shear interaction problem where values for both φN n and φV n need to be determined. φN n is the lesser of the design tension strength controlled by: steel ( φN sa ), concrete breakout (φN cb ), or pull-out ( φnN pn ). φV n is the lesser of the design shear strength controlled by: steel ( φV sa ), concrete breakout (φV cb ), or pryout ( φV cp ).


3. Steel capacity under tension loading:

Eq. (D-4);

Eq. (D-7)

substituting:

φN cb = φ ANc ψ ed,N ψ c,N ψ cp,N k c √ƒ 'c h ef 1.5 ANco

where:

D.5.1

φN sa ≥ N ua

Eq. (D-1) .

N sa = 11,075 lb.

This catalog

φ = 0.75

This catalog

n = 1 (single anchor) Calculating for φN sa :

φN sa = 0.75 x 1 x 11,075 = 8,306 lb. > 4,000 lb. – OK ∴ ¹⁄₂ in. diameter anchor is adequate

k c = k cr = 17 (Anchor is installed in a tension zone, therefore, cracking is assumed at service loads)

This catalog

ψ cp,N = 1.0

D.5.2.7

ψ = 0.7 + 0.3 c a,min when c ed,N a,min < 1.5 h ef 1.5h ef

Eq. (D-11)

by observation, c a,min = 4⁷⁄₈ < 1.5h ef

ψ ed,N = 0.7 + 0.3

(4⁷⁄₈) 1.5(4⁷⁄₈)

= 0.90

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

D.5.2.6


This catalog

(no supplementary reinforcement provided) ANco = 9h ef 2 = 9(4 ⁷⁄₈)2 = 213.89 in. 2

Eq. (D-6)

ANc = (ca 1 + 1.5h ef )(2 x 1.5h ef ) = (4⁷⁄₈ + 1.5(4 ⁷⁄₈))(2 x 1.5(4 ⁷⁄₈)) = 178.24 in. 2

Fig. RD.5.2.1(a)

ANc 178.24 = = 0.83 ANco 213.89 Calculating for φN cb : Would you like help with these c alculations? Visit www.simpsonanchors.com to download the Simpson Strong-Tie ® Anchor Designer ™ software.

φN cb = 0.65 x 0.83 x 1.0 x 0.90 x 1.0 x 17 x √3,000 x (4 ⁷⁄₈)1.5 = 4,866 lb. > 4,000 lb. – OK


29

EXAMPLE CALCULATION Torque Controlled Adhesive Anchors (ICC-ES AC308) Continued from previous page. CALCULATIONS AND DISCUSSION

REFERENCE

5. Pullout capacity:

D.5.3

Pullout capacity, N pn,cr , is established by reference tests in cracked concrete by the reliability test of AC 308. Pullout capacity does not need to be calculated for IXP with SET-XP epoxy since steel failure controls over pullout capacity. ™

™

6. Check all failure modes under tension loading: Summary: Steel capacity

D.4.1.2

= 8,306 lb.

Concrete breakout capacity = 4,866 lb. ← Controls Pullout capacity = N/A ∴ φN n = 4,866 lb. as concrete breakout capacity controls 7. Steel capacity under shear loading:

D.6.1

REFERENCE

9. Concrete pryout strength:

D.6.3

φnV cp ≥ V ua

Eq. (D-2)

V cp = k cp N cb

Eq. (D-29)

where: n = 1 k cp = 2.0 and φ = 0.70 4,866 = 14,972 lb. k cp N cb = 2.0 x 0.65

This catalog D.6.3.1

φnV cp = 0.70 x 1 x 14,972 = 10,480 lb. > 720 lb.

–

OK

10. Check all failure modes under shear loading:

D.4.1.2

φV sa ≥ V ua

Eq. (D-2)

Summary:

V sa = 10,450 lb.

This catalog

Steel capacity

This catalog

Concrete breakout capacity = 3,096 lb. ← Controls

φ = 0.65


φV sa = 0.65 x 10,450 = 6,793 lb. > 720 lb.

–

Pryout Capacity

OK

8. Concrete breakout capacity under shear loading:

Eq. (D-2)

AVc ψ ψ ed,V c,V V b AVco

Eq. (D-21)

(d o e )

0.2

√d o √ƒ 'c c a11.5

Eq. (D-24)


If 0.2 φV n ≥ V ua, then the full tension design strength is permitted.

D.7 D.7.1

If 0.2 φN n ≥ N ua , then the full shear design strength is permitted

D.7.2


( )

0.2

√d o √ƒ 'c c a11.5

Therefore:

where:

N ua V + ua ≤ 1.2 φN n φV n


D4.4(c)(i)

2

AVco = 4.5ca 1

= 4.5(4 ⁷⁄₈)

11. Check interaction of tension and shear forces:


where: V b = 7

= 10,480 lb.

D.6.2

φV cb ≥ V ua V cb =

= 6,793 lb.

∴ φV n = 3,096 lb. as concrete breakout capacity controls

∴ ¹⁄₂ in. diameter anchor is adequate

Eq. (D-23)

4,000 720 + = 0.82 + 0.23 = 1.05 < 1.2 4,866 3,096

Eq. (D-31)

–

OK

2

∴ AVco = 106.95 in. 2

12. Summary

AVc = 2(1.5c a 1)(1.5ca 1)

Fig. RD.6.2.1(a)

= 2(1.5(4 ⁷⁄₈))(1.5(4⁷⁄₈))

∴ AVc = 106.95 in. 2 AVc = 106.95 = 1 AVco 106.95

D.6.2.1

ψ ed,V = 1.0 since c a 2 > 1.5ca 1

Eq. (D-27)

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

D.6.2.7

d o = 0.5 in. e = 8d 0 = 8 (0.5) = 4"

D.6.2.2

ca 1 = 4⁷⁄₈ in.

φV cb = 0.70 x 1 x 1.0 x 1.0 x 7 x 30


( 0.54 )

0.2

x √0.5

x √3,000 x (4 ⁷⁄₈)1.5 = 3,096 lb. > 720 lb.

–

OK

A single ¹⁄₂" diameter IXP anchor with SET-XP epoxy at a 4 ⁷⁄₈" embedment depth is adequate to resist the applied service tension and shear loads of 2,500 lb. and 450 lb., respectively.


Example Calc s

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