1. Anchor
Channel Systems
Code
2. HAC
Portfolio
3. HAC
Applications
Discussion
4. Design
Introduction
5. Base material
6. Loading
8. Reinforcing
Bar Anchorage
9. Special Anchor
Channel Design
Calculations Discussion
There load combination that is evaluated: Wind Load: 1.2DL + WL/0.6 WL(ASD)=40psfx10.5’x5’=2100lbs Slab depth: 8” WL(Strength level)=2100lbs/0.6=3500lbs 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).
Pocket height: 2” DL(ASD)=14psfx10.5’x5’=735lbs Front edge distance : 5” DL(Strength level)=1.2x735lbs=882lbs Side edge distance : 6” Slab depth deducting pocket: 8”-2”=6” T-bolt spacing: 6” Available tolerance along the channel length: (13.8”-(2x0.984”)-6”)/2=2.916” Bracket tolerance: 1.5” in/out Bracket dimension in y-direction: 8.75” Worst position: Bracket out Bracket dimension in x-direction: 1.5”
Step 1: Load Combination and Bracket dimension
ASCE 7-10
2.3.2
7. Anchor Channel
Design Code
x
ε s
γ
g is so small that tan( g ) =
e c
5”-x
e c
x
=
e s
(5 - x )
e s
=
(5 - x )
F T - bolts . L
e s =
E T - bolts . A T - bolts
e c =
s c
E c . x
E c
F T - bolts
E T - bolts . A T - bolts .(5 - x )
Compatility equations : assume L = " unity "
F T - bolts =
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 A T-bolts = ( 2)(tensile stress area of T-bolt) = (2)(157 mm 2 )[1in 2 / (25.4 mm) 2 ] = 0.487in 2
(2) HBC-C 8.8F M16x60 are used. E concrete = ( modulus of elasticity for concrete) = 30,000 MPa = 4,351,200 lb/in 2
E T-bolts = (modulus of elasticity for T-bolt) = 29,000,000 lb/in 2
F T - bolts =
13. Field Fixes
14. Design
Example
(0.487 in 2 ).(29000000 lb / in 2 ).( s c ).(5 - x )
(4351200 lb / in 2 ).( x )
(3.246 in 2 ).( s c ).(5 - x )
x
Sum the tension forces and moments
F T - bolts =
å F + = 0
(1).( s c ).( x ).( b )
= 0
2
2
(3.246 in ).( s c ).(5 - x )
(1).( s c ).( x ).( b )
= 0
+ 882 -
x
2
é (3.246 in 2 ).(5 - x ) ù
é é (1).( x ).(10) ù ù
- s c . ê
s c . ê ê
ú = 882
ú
ú
2
x
û û
ë ë
ë
û
é (5).( x 2 ) + 3.246 x - 16.23 ù
ú = 882
x
ë
û
s c . ê
é
ù
882 x
ú ®®® Eqn 1
2
+
-
(5).(
x
)
3.246
x
16.23
ë
û
s c = ê
( A T - bolts ).( E T - bolts ).( s c ).(5 - x )
( E c ).( 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.
(3.246 in 2 ).( s c ).(5 - x )
x
é
ù
882 x
(3.246). ê
ú .(5 - x )
2
+
-
(
5).(
x
)
3.246
x
16.23
ë
û
=
x
é 14314.86 - 2862.972 x ù
= ê
ú ®®® Eqn 2
2
ë (5).( x ) + 3.246 x - 16.23 û
F T - bolts =
s c . L
=
12. Instructions
for Use
F T - bolts + 882 lb -
γ
ε c
11. Best
Practices
Calculations
( F T - bolts )( L )
PL
®
AE
( A T - bolts )( E T - bolts )
x
10. Design
Software
F T - bolts
F T - bolts
å M = 0
x ù
x ù
é
é
(882). ê 2 + ú + (3500). [ 1.5 ] - ( F T - bolts ). ê 5 - ú = 0
3 û
ë
ë 3 û
é 14314.86 - 2862.972 x ù é
x ù
x ù
é
(882). ê 2 + ú + (3500). [ 1.5 ] - ê
ú . ê 5 - ú = 0 ®®® Eqn 3
2
3 û
ë
ë (5).( x ) + 3.246 x - 16.23 û ë 3 û
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
382
Figure 14.1.8 — Design example – Plan View
Cast-In Anchor Channel Product Guide, Edition 1 • 02/2019
383