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Daniel
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PROJECT :
PAGE :
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T. LiCLIENT :
DESIGN BY :
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JOB NO. : DATE :
REVIEW BY :
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Grade Beam Design Based on ACI 318-99
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INPUT DATA & DESIGN SUMMARY
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CONCRETE STRENGTH
fc' =3ksi
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REBAR YIELD STRESS
fy =60ksi
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SQUIRE PAD SIZE
B =8ft
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T =24in
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GRADE BEAM SIZE
W =60in
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D =24in
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COLUMN DISTANCE
L =22ft
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GRADE BEAM EXTENSION
Le =5ft
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AXIAL LOADS
PD,1 =25kipsPD,2 =25kips
(Dead Load)
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PL,1 =15kipsPL,2 =15kips(Live Load)
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PE,1 =-30kipsPE,2 =30kips
(Seismic Load)
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SEISMIC SHEAR LOADS
VE,1 =50kipsVE,2 =30kips
(Seismic Load)
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SEISMIC MOMENTS
ME,1 =50ft-kipsME,2 =50ft-kips
(Seismic Load)
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ALLOW SOIL PRESSURE
Qa =2.5ksf
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PAD REINFORCING
10#8
@ 9 o.c., Each Way, Bottom.
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GRADE BEAM LONGITUDINAL REINFORCING
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TOP7#8
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( d =19.88in )
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( 1 Layer)
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BOTTOM7#8
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( d =19.88in )
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( 1 Layer)
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GRADE BEAM HOOPS (ACI 21.3.3)
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LOCATIONAT ENDAT SPLICE
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LENGTH48in55in
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( 2h )
MAX{0.075fyabgdb/[(fc')0.5(c+Ktr)/db], 12}
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BAR4Legs #54Legs #5
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(Legs to alternate long bars supported, ACI 7.10.5.3)
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SPACING@ 4 in o.c.@ 4 in o.c.
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MIN(d/4, 8db, 24dt, 12)
MIN(d/4, 4)
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THE GRADE BEAM DESIGN IS ADEQUATE.
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ANALYSIS
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CHECK OVERTURNING AT CENTER BOTTOM OF PAD 2
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MO =
ME,1 + ME,2 + (VE,1+VE,2)(D+T) - PE,1L
=1080.0ft-kips
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MR =
(PD,1 + gconc B2T) L + 0.5gconc(L + 2Le) L D W =
2028.4>MO / 0.9 =1200ft-kips
[Satisfactory]
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CHECK SOIL BEARING CAPACITY
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2.46ksf<4/3 Qa
[Satisfactory]
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whereW0 =2.2
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(CBC 1633A.2.12)
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gconc =0.15 kcf
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gsoil =0.11 kcf
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CHECK PAD FLEXURAL REINFORCING
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0.0011<
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rprovd =0.0041
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whered =20.00in
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Qu,max = 1.5 Qmax =
3.69ksf
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Mu = 0.125 (B-W)2 B Qu,max =
181ft-kips
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rmax =0.0160
(ACI 10.2.7.3 & 10.3.3)
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rmin =0.0018
(ACI 7.12.2.1)
[Satisfactory]
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(cont'd)
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CHECK PAD ONE WAY SHEAR CAPACITY
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Vu<fVn
[Satisfactory]
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where
Vu = 0.5 (B - W) B Qu,max - d B Qu,max =
0.0kips
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fVn = f 2 d B (fc')0.5 =
178.8kips
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f = 0.85
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CHECK GB SECTION REQUIREMENTS (ACI 21.3.1)
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Pu = VE,1 - VE,2 =
20kips<0.1Agfc' =432kips
[Satisfactory]
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Ln=L - B =14.00ft>4 d =6.63ft
[Satisfactory]
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W / D =2.50>0.3
[Satisfactory]
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W =60in>10in
[Satisfactory]
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<B+1.5D = 132in
[Satisfactory]
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CHECK GB FLEXURAL REQUIREMENTS
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(ACI 21.3.2.1)
rtop =0.005>
rmin=MIN[3(fc')0.5/fy, 200/fy ]=
0.003
[Satisfactory]
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<rmax =0.025
[Satisfactory]
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rbot =0.005>rmin =0.003
[Satisfactory]
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<rmax =0.025
[Satisfactory]
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(ACI 21.3.2.2)
Mn,top>
(1/2)Mn,bot
[Satisfactory]
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where
Mn,bot = rbot bd2fy (1 - 0.588rbot fy/fc') =
520ft-kips>Mu,bot / f
[Satisfactory]