ICC ESR-1545 for HSL-3 Heavy Duty Expansion Anchors

ESR-1545
|
Most Widely Accepted and Trusted Page 12 of 14
Given:
Two HSL-3 M10 anchors under static tension load
as shown.
h
ef
= 2.76 in.
Normal weight concrete with f '
c
= 3,000 psi.
No supplementary reinforcement.
Condition B (ACI 318-14 17.3.3(c) or ACI 318-11
D.4.3(c))
Assume uncracked normal-weight concrete.
Needed:
Using Allowable Stress Design (ASD) Calculate
the allowable tension load for this configuration.
Calculation per ACI 318-14 Chapter 17, ACI 318-11 Appendix D and this report.
ACI 318-14
Ref.
ACI
318-11
Ref.
Report
Ref.
Step 1. Calculate steel strength of anchor in tension lbxx 880,20000,116090.02 
utase,Nsa
n fAN
17.4.1.2 D.5.1.2 Table 3
Step 2. Calculate steel capacity lb600,15=880,20x75.0=
sa
ΦN
17.3.3(a) D.4.3(a) Table 3
Step 3. Calculate concrete breakout strength of anchor in tension
bNcp,Nc,Ned,Nec,
Nco
Nc
ψψψψ=
cbg
N
A
A
N
17.4.2.1 D.5.2.1 §4.1.3
Step 4. Verify minimum spacing and edge distance:
Table 4 Case A: h
min
= 5-1/2 in. < 6 in. okay
9.5 - 2.75
slope = = -3.0
2.75 - 5
⇒
.in4=For
min
c
okay in. 6<in. 5.75=)]2.75)(-3.0-[(4-9.5=
∴
min
s
17.7 D.8
Table 3
Table 4
Step 5. Calculate A
Nco
and A
Nc
for the anchorage:
222
efNco
in6.68=)76.2(9=9= hA
okayin
∴
Nco
2
efef
sc AhhA
Nc
22.116]6)76.2(3][4)76.2(5.1[)3)(5.1( 
17.4.2.1 D.5.2.1 Table 3
Step 6. Calculate
lb
a
027,6)76.2(000,3)0.1(24  
1.51.5
ef
'
cuncrb
hfkN 
17.4.2.2 D.5.2.2 Table 3
Step 7. Modification factor for eccentricity → no eccentricity
1 . 0 = 0 =
∴
Nec,
'
N
Ψe
17.4.2.4 D.5.2.4 -
Step 8. Modification factor for edge
Ned,ef
Ψc ∴>.in13.4=)76.2(5.1=5.1 h must be calculated:
99.0
)76.2(5.1
4
3.07.0 
Ned,
Ψ
17.4.2.5 D.5.2.5 Table 3
Step 9. Modification factor for cracked concrete, k = 24 used in D.5.2.2 → 1.0Ψ
Nc,
=
(see Step10)
17.4.2.6 D.5.2.6 Table 3
Step 10. Splitting Modification factor
94.0=
375.4
)76.2(5.1
=
|5.1|max
=
ac
ef
Ncp,
c
Ψ
hc
⇔
17.4.2.7 D.5.2.7 Table 4
Step 11. Calculate
lb500,9=027,6x94.0x0.1x99.0x0.1x
6.68
2.116
=
cbg
N
17.4.2.1 D.5.2.1 -
Step 12. Check pullout strength in Table 3 →
uncrp,
N does not govern
17.4.3.2 D.5.3.2
§ 4.1.4
Table 3
Step 13. controls<lb175,6=500,9x65.0=
cbgscbg
ΦΦ NN ∴NΦ
17.3.3(c) D.4.3(c) Table 3
Step 14. To convert to ASD, assume U = 1.2D + 1.6L: T
allow
=
,
.
= 4,172 lb.
- - § 4.2
s
min
(5, 2.75)
c
min
(2.75, 9.5)
5.75
4
2.76ʺ
FIGURE 6—EXAMPLE CALCULATION