ACI-349 Approval for HSL-3 Heavy-Duty Expansion Anchor

Report WC 11-02 Hilti HSL-3 Compliance September 30, 2011
13
FIGURE 5—EXAMPLE CALCULATION
Given:
(2) HSL-3 M10 anchors under static tension load
as shown.
h
ef
= 2.76 in.
Slab on grade with f '
c
= 3,000 psi.
No supplementary reinforcing.
Assume cracked normal-weight concrete.
Calculate the design strength in tension for this
configuration.
Calculation per ACI 349-01 Appendix B and this document. ACI 349-01 Guide Ref.
Step 1. Calculate steel strength of anchor in tension
lb,,x.x 8802000011609002 ===
utNse,s
n fAN
B.5.1.2 Table 1
Step 2. Calculate steel capacity lb,,x. 704168802080 ==
s
ΦN
B.4.4 a Table 1
Step 3. Calculate concrete breakout strength of anchor in tension
b321
No
N
ψψψ=
cbg
N
A
A
N
B.5.2.1
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
B.8 Table 1
Table 2
Step 5. Calculate A
No
and A
N
for the anchorage:
222
ef
No
in..)in.( 66876299 === hA
okayin[.]).(][).(.[))(.( ] ∴
∴∴
∴
No
2
efef
sc AhhA
N
2211667623476251351 <=++=++=
B.5.2.1 Table 1
Step 6. Calculate ]lb[,).(, 2745762000321 ===
1.51.5
ef
'
ccrb
hfkN
B.5.2.2 Table 1
Step 7. Modification factor for eccentricity → no eccentricity 1.0
1.01.0
1.0=
==
=0
00
0=
==
= ∴
∴∴
∴
1
'
N
Ψ e
B.5.2.4 -
Step 8. Modification factor for edge
2ef
Ψc ∴ 1347625151 >== .in..)in.(.. h
must be calculated
990
76251
4
3070 .
).(.
.. =+=
2
Ψ
B.5.2.5 Table 1
Step 9. Modification factor for cracked concrete
1.0Ψ
3
=
==
=
B.5.2.2
B.5.2.8
Table 1
Step 10. Calculate
lb,lb,x.x.x.x
.
.
844827450199001
668
2116
==
cbg
N
B.5.2.1 -
Step 11. φ
φφ
φ N
cbg
= 0.75x8,844 lb = 6,633 lb controls
B.4.4.c
Table 1
Step 12. Calculate pullout strength: φ
φφ
φ N
p,cr
φ
φφ
φ N
p,cr
= 0.75 x 4,496 lb = 3,372 lb/anchor x2 anchors = 6,744 lb
B.5.3.2 Table 1
Step 13. Ductility check according to B.3.6.1
For tension φ
φφ
φ min [ N
cbg
; N
p,cr
] ≥ A
se
f
uta
where φ = 0.85
0.85 (8,844 lb) < 20,880 lb => ductility not met
B.3.6.3 requires an additional reduction factor of 0.6 for non-ductile anchors
φ
φφ
φ N
cbg
= 0.6 x 6,633 = 3,980 lb
Note: According to B.3.6.2, alternatively the attachment / base plate can be designed to yield at a load
level of 75% of the anchor design strength. In this case, the 0.6 factor would not have to be applied.
B.3.6.1
B.3.6.3
B.4.1
Table 1
s
min
(5, 2.75)
c
m
(2.75, 9.5)
5.7
4