Post-Installed Reinforcing Bar Guide
2.0 HOW ARE THEY DESIGNED ?
2.6.3 DESIGN EXAMPLE — DEVELOPMENT LENGTH IN A SPECIAL MOMENT FRAME
Requirement : Provide post-installed reinforcement for an addition to a special moment frame in a structure assigned to Seismic Design Category D ( high seismic ). in Figure 11 .
For 4 - # 9 top bars , probable moment strength :
4 ( 1.0 ) ρ - = = 0.0162
14 x 17.6
K pr
-
= 1.25 ρ - f y 1 - 0.735 ρ - f y f l c
60000 = 1.25 x 0.0162 x 60000 1-0.735 x 0.0162 = 998 psi
4000
( E ) special moment frame
( N ) beam
-
K pr b · d 2 998 x 14 x ( 17.6 ) 2
-
M pr
= = = 361 ft-k 12000 12000
20 in .
For 2 - # 9 top bars :
( N ) drilled-in dowels roughened surface , 1 / 4 in . amplitude
2 ( 1.0 ) ρ + = = 0.0081
14 x 17.6
24 in .
60000
+
K pr
= 1.25 x 0.0081 x 60000 1-0.735 x 0.0081 4000
= 553 psi
Figure 15 — Addition to a special moment frame .
-
K pr b · d 2 553 x 14 x ( 17.6 ) 2
+
M pr
= = = 200 ft-k 12000 12000
Step 1 : Establish requirements for new bars . Existing construction : Special moment frame with 18 x 24- inch columns , 14 x 20-inch beams , 20-ft . bays , 4 ksi normal weight concrete , A706 Gr . 60 bar , 4 - # 9 top bars , 2 - # 9 bottom bars , # 3 hoops and cross ties @ 3-1 / 2 in . on center typical at joint . Factored w u
= D + L = 3.6 k / ft
New construction : ( N ) 14 x 20-inch beam as shown , 4 ksi normal weight concrete , ASTM A706 Gr . 60 reinforcement , 4 - # 9 top bars , 2 - # 9 bottom bars , # 3 hoops and cross ties at 3-1 / 2 in . on center
Determine shear force associated with plastic hinges at beam ends per ACI 318-19 Section 18.6.5.1 and 18.8.2.1
Effective depth of bars in beams : d = 20 - 1.5 - 0.375 - 1.128 / 2 = 17.6 in .
Shear associated with formation of plastic hinges :
M pr1
+ M pr2
W n l n
361 + 200 3.6 ( 20 )
V c
= ± = ± = 64 k l n
2 20 2
Step 2 : Check ability of bottom bars to carry shear at joint face :
V
A vf
= u
V
= u
64000 = = 1.42 in 2 2- # 9 = 2.0 .. · ok ϕf y μ ϕf y
1.0λ 0.75 x 60000 x 1.0
Step 3 : Calculate the required embedment for the new top and bottom bars using ACI 318-19 Eq . ( 18.8.5.1 ) and 18.8.5.3 ( a ):
f y d b
60000 x 1.128 l d
= = 2.5x = 41 in . 65 65√4000
Per 18.8.5.4 , portion of development length not within confined core = 41-24 = 17in .
l d
= 24 + 1.6 x 17 = 51 in .
10 November 2022