HAC_Technical-Guide
382 383
Cast-In Anchor Channel Product Guide, Edition 1 • 02/2019
1. Anchor
Channel Systems
2. HAC
Portfolio
3. HAC
Applications
4. Design
Introduction
5. Base material 6. Loading
7. Anchor Channel
Design Code
8. Reinforcing
Bar Anchorage
9. Special Anchor
Channel Design
10. Design
Software
11. Best
Practices
12. Instructions
for Use
13. Field Fixes
14. Design
Example
Code Discussion Calculations
Step 1: Load Combination and Bracket dimension
ASCE 7-10
2.3.2
There load combination that is evaluated:
1.2DL + WL/0.6
Slab depth: 8”
Pocket height: 2”
Front edge distance : 5”
Side edge distance : 6”
T-bolt spacing: 6”
Bracket tolerance: 1.5” in/out
Worst position: Bracket out
Wind Load:
WL(ASD)=40psfx10.5’x5’=2100lbs
WL(Strength level)=2100lbs/0.6=3500lbs
DL(ASD)=14psfx10.5’x5’=735lbs
DL(Strength level)=1.2x735lbs=882lbs
Slab depth deducting pocket: 8”-2”=6”
Available tolerance along the channel length: (13.8”-(2x0.984”)-6”)/2=2.916”
Bracket dimension in y-direction: 8.75”
Bracket dimension in x-direction: 1.5”
Figure 14.1.5— Design example – Profis anchor channel - 3D front View Figure 14.1.6— Design example – Profis anchor channel - 3D Back View
Step 2: Determination of T-bolt Tension and shear forces
Determine the tension loads on the T-bolts.
Assume the fixture is rigid. PROFIS Anchor
Channel assumes Ec = 30,000 Mpa
(2) HBC-C 8.8F M16x60 are used.
A
T-bolts
= (2)(tensile stress area of T-bolt) = (2)(157 mm
2
)[1in
2
/ (25.4 mm)
2
] = 0.487in
2
E
T-bolts
= (modulus of elasticity for T-bolt) = 29,000,000 lb/in
2
E
concrete
= (modulus of elasticity for concrete) = 30,000 MPa = 4,351,200 lb/in
2
Discussion Calculations
The application is statically indeterminate. Use
compatibility equations and statics equations to
solve for the location of the neutral axis (x). Once
x is known, the tension force on each T-bolt can
be calculated. Using the compatibility equations,
define the tension force acting on both T-bolts
(FT-bolts) in terms of the concrete compressive
stress under the fixture (σc).
εs
εc
x
5”-x
γ
γ
( )( )
( )( )
T bolts
T bolts T bolts
FL
PL
AE A E
-
--
®
tan( )
(5 )
(5 )
.
.
.
. . .(5 )
: ""
( ).( ).( ).(5 )
(
cs
cs
T bolts
s
T bolts T bolts
c
c
c
c T bolts
c T bolts T bolts
T bolts T bolts c
T bolts
c
is so small that
xx
xx
FL
EA
L
E
F
Ex E A x
Compatility equations assume L unity
AE x
F
E
ee
gg
ee
e
s
e
s
s
-
--
-
--
--
-
==
-
=
-
=
=
=
-
=
-
=
).( )x
Using statics equations, define the concrete
stress under the fixture (σc) in terms of the
lcompressive ocation of the neutral axis (x).
Solve for x via trial and error. Once x is known,
the values for σc and FT-bolts can be calculated.
Once FT-bolts is known, the tension force on
each T-bolt can be determined. The applied shear
force is distributed equally on each T-bolt.
0 M =
å
22
2
2
2
(0.487 ).(29000000 / ).( ).(5 )
(4351200 / ).( )
(3.246 ).( ).(5 )
0
(1).( ).( ).( )
882 0
2
(3.246 ).( ).(5 ) (1).(
882
c
T bolts
c
T bolts
c
T bolts
cc
in lb in x
F
lb in x
in x
F
x
Sum the tension forces and moments
F
xb
F lb
in x
x
s
s
s
ss
-
-
-
-
=
-
=
+ =
+ - =
-
+ -
å
2
2
2
2
).( ).( )
0
2
(1).( ).(10) (3.246 ).(5 )
. . 882
2
(5).( ) 3.246 16.23
. 882
882
1
(5).( ) 3.246 16.23
(3.246 ).( ).(5 )
882
(3.246).
(
cc
c
c
c
T bolts
T bolts
xb
x in x
x
xx
x
x
Eqn
xx
in x
F
x
x
F
ss
s
s
s
-
-
=
éù
éù
-
éù
- =
êú
êú
êú
ëû
ëû
ëû
éù
+ -
=
êú
ëû
éù
= ®®®
êú
+ -
ëû
-
=
=
[ ]
[ ]
2
2
2
.(5 )
5).( ) 3.246 16.23
14314.86 2862.972
2
(5).( ) 3.246 16.23
(882). 2 (3500). 1.5 ( ). 5 0
33
14314.86 2862.972
(882). 2 (3500). 1.5
3 (5).( )
T bolts
T bolts
x
xx
x
x
F Eqn
xx
xx
F
xx
x
-
-
éù
-
êú
+ -
ëû
éù
-
= ®®®
êú
+ -
ëû
éù éù
++ --=
êú êú
ëû ëû
-
éù
++ -
êú
+
ëû
.5 0 3
3.246 16.23 3
x
Eqn
x
éù
éù
- = ®®®
êú
êú
-
ëû
ëû
After trial and error x=1.779993152175” the solution to Eqn3 leads to 0.0000000221,
which is approximately equal to zero.
Substituting the value for x into equation 2 will lead to FT-bolts= 1700lbs.
Tension force on each t-bolt = 1700lbs/2=850lbs.
Shear Force per each t-bolt = Vy=3500lbs/2=1750lbs per t-bolt
Figure 14.1.7 — Design example – Section View - Force Resultant Figure 14.1.8 — Design example – Plan View