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PROJECT : PAGE :
CLIENT : DESIGN BY : JOB NO. : DATE : REVIEW BY :
Two-Way Slab Design Based on ACI 318-14 using Finite Element Method
INPUT DATA & DESIGN SUMMARYCONCRETE STRENGTH
3 ksi, (21 MPa)REBAR YIELD STRESS
60 ksi, (414 MPa)COLUMN SPACING EACH WAY
L = 24 ft, (7.32 m)B = 24 ft, (7.32 m)
SLAB THICKNESSt = 9.5 in, (241 mm)
BENDING DROP PANEL THK. (12.0 ft x 12.0 ft)
2.5 in, (64 mm)PUNCHING CAP THICKNESS
0 in, (0 mm)COLUMN SIZE (SHORT EDGE)
24 in, (610 mm)DEAD LOAD & SELF WT
DL = 150 psf, (7.2 kPa)LIVE LOAD
LL = 70 psf, (3.4 kPa)
TOP BARS AT COLUMNS EACH WAY6 # 6 @ 12 in, (305 mm), o.c.x 8.0 ft. long, with 0.75 in, (19 mm), cover
(All top bars to column strip suggested, if column strip & middle strip used.)
BOTTOM LAYER BOTTOM BARS# 5 @ 18 in, (457 mm), o.c. THE DESIGN IS ADEQUATE.
BOTTOM LAYER TOP BARS# 5 @ 18 in, (457 mm), o.c.
with 0.75 in, (19 mm), bottom concrete cover(75% total bottom bars to middle strip & 25% to column strip suggested, if column strip & middle strip used.)
ANALYSIS150 pcf, (ACI 318-14 19.2.2.1) 3321 ksi, (ACI 318-14 19.2.2.1)
0.63 t = 6.0 in, for Slab only (ACI 318-14 8.3.2, 24.2 & 6)8.5 in, for Slab & Drop Panel
JointNumber in kips Bending
1 0 42.05 Section ft-k/ft2 0.29 1 - 2 12.93 0.53 2 - 3 -0.24 0.29 1 - 6 12.95 0 42.05 6 - 11 -0.26 0.29 3 - 8 -0.57 0.46 8 - 13 -6.08 0.64 11 - 12 -0.59 0.46 12 - 13 -6.0
10 0.2911 0.5312 0.6413 0.7514 0.64 DETERMINE FACTORED LOAD (ACI 318-14 5.3)
15 0.53 1.2 DL + 1.6 LL = 0.292 ksf16 0.2917 0.46 DETERMINE FLEXURE CAPACITY (ACI 318-14 7.6.1.1, 8.6.1.1, & 22)18 0.6419 0.46 Top Bar Bot. Layer Bot. Bot. Layer Top 20 0.29 6 # 6 @ 12" o.c. 5 @ 18" o.c. 5 @ 18" o.c.21 0 42.05 10.13 8.44 7.81
22 0.29 0.44 0.21 0.21
23 0.53 0.41 0.21 0.2124 0.29 0.86 0.41 0.41
25 0 42.05 19.2 7.7 7.1
CHECK FLEXURE CAPACITY
12.9 ft-k/ft < 19.2 ft-k/ft [Satisfactory]
6.0 ft-k/ft < 7.7 ft-k/ft [Satisfactory]
6.0 ft-k/ft < 7.1 ft-k/ft [Satisfactory]
fc' =
fy =
tdrop =
tcap =
c =
wc = Ec = wc1.5 33 f'c0.5 =
te = ( Ie / Ig )1/3 t = (0.25 Ig / Ig )1/3 t = (0.25)1/3 t =
Du Ru
Mu
wu =
d (in)
As (in2/ft)
As, min (in2/ft)a (in)
fMn (ft-k/ft)
Mu,Top = Max( Mu,1-2 , Mu,1-6 ) = fMn =
Mu,Bot,Bot = - Min( Mu,8-13 , Mu,12-13 ) = fMn =
Mu,Bot,Top = - Max( Mu,8-13 , Mu,12-13 ) = fMn =
(cont'd)CHECK LIVE LOAD DEFLECTION (ACI 318-14 Table 24.2.2)
0.18 in < L / 360 = 0.80 in[Satisfactory]
CHECK LONG-TERM DEFLECTION (ACI 318-14 24.2.4)
1.33 in < L / 180 = 1.60 in[Satisfactory]
CHECK COLUMN PUNCHING CAPACITY (ACI 318-14 8.4 & 8.5)
168.2 kips (See Punching.xls Software for More Information.)
227.10 kips > [Satisfactory]
where 0.75 (ACI 318-14 21.2)
1.00d = 10.1 in
4 c + 4 d = 136.5 in
1382.1
y = 2.0
DLL = Du,Max LL / (1.2 DL + 1.6 LL) =
D3DL + LL = Du,Max (3DL + LL) / (1.2 DL + 1.6 LL) =
Pu = 4 Ru,max =
Pu
f =
bc =
b0 =
Ap = b0 d = in2
MIN(2 , 4 / bc , 40 d / b0) =
'2 y fV Apn cf f
PROJECT : PAGE :
CLIENT : DESIGN BY : JOB NO. : DATE : REVIEW BY :
Two-Way Slab Design Based on ACI 318-14 using Finite Element Method
INPUT DATA & DESIGN SUMMARYCONCRETE STRENGTH
3 ksi, (21 MPa)REBAR YIELD STRESS
60 ksi, (414 MPa)COLUMN SPACING EACH WAY
L = 24 ft, (7.32 m)B = 24 ft, (7.32 m)
SLAB THICKNESSt = 9.5 in, (241 mm)
BENDING DROP PANEL THK. (12.0 ft x 12.0 ft)
2.5 in, (64 mm)PUNCHING CAP THICKNESS
0 in, (0 mm)DEAD LOAD & SELF WT
DL = 150 psf, (7.2 kPa)LIVE LOAD
LL = 70 psf, (3.4 kPa)
TOP BARS AT COLUMNS EACH WAY6 # 6 @ 12 in, (305 mm), o.c.x 8.0 ft. long, with 2 in, (51 mm), cover
(All top bars to column strip suggested, if column strip & middle strip used.)
BOTTOM LAYER BOTTOM BARS# 5 @ 18 in, (457 mm), o.c. THE DESIGN IS ADEQUATE.
BOTTOM LAYER TOP BARS# 5 @ 18 in, (457 mm), o.c.
with 0.75 in, (19 mm), bottom concrete cover(75% total bottom bars to middle strip & 25% to column strip suggested, if column strip & middle strip used.)
ANALYSIS150 pcf, (ACI 318-14 19.2.2.1) 3321 ksi, (ACI 318-14 19.2.2.1)
0.63 t = 6.0 in, for Slab only (ACI 318-14 8.3.2, 24.2 & 6)8.5 in, for Slab & Drop Panel
JointNumber in Bending
1 0 Section ft-k/ft2 0.30 1 - 2 13.33 0.56 2 - 3 -0.64 0.32 3 - 4 -2.35 0 4 - 5 12.96 0.30 1 - 6 13.27 0.47 6 - 11 -0.38 0.62 12 - 13 -6.99 0.33 13 - 14 -6.810 0 7 - 12 -7.311 0.54 8 - 13 -5.812 0.64 9 - 14 -2.813 0.6814 0.34 DETERMINE FACTORED LOAD (ACI 318-14 5.3)
15 0 1.2 DL + 1.6 LL = 0.292 ksf16 0.3017 0.47 DETERMINE FLEXURE CAPACITY (ACI 318-14 7.6.1.1, 8.6.1.1, & 22)18 0.6219 0.33 Top Bar Bot. Layer Bot. Bot. Layer Top 20 0 6 # 6 @ 12" o.c. 5 @ 18" o.c. 5 @ 18" o.c.21 0 8.88 8.44 7.81
22 0.30 0.44 0.21 0.21
23 0.56 0.36 0.21 0.2124 0.32 0.86 0.41 0.41
25 0 16.7 7.7 7.1
CHECK FLEXURE CAPACITY
13.3 ft-k/ft < 16.7 ft-k/ft [Satisfactory]
7.3 ft-k/ft < 7.7 ft-k/ft [Satisfactory]
2.8 ft-k/ft
< 7.1 ft-k/ft [Satisfactory]
CHECK LIVE LOAD DEFLECTION (ACI 318-14 Table 24.2.2)
0.16 in < L / 360 = 0.80 in[Satisfactory]
CHECK LONG-TERM DEFLECTION (ACI 318-14 24.2.4)
1.20 in < L / 180 = 1.60 in[Satisfactory]
fc' =
fy =
tdrop =
tcap =
wc = Ec = wc1.5 33 f'c0.5 =
te = ( Ie / Ig )1/3 t = (0.25 Ig / Ig )1/3 t = (0.25)1/3 t =
Du
Mu
wu =
d (in)
As (in2/ft)
As, min (in2/ft)a (in)
fMn (ft-k/ft)
Mu,Top = Max( Mu,1-5 , Mu,1-11 ) = fMn =
Mu,Bot,Bot = - Min( Mu ) = fMn =
Mu,Bot,Top = - Max( Mu,12-14 , Mu,7-12 , Mu,8-13 , Mu,9-14 ) =
fMn =
DLL = Du,Max LL / (1.2 DL + 1.6 LL) =
D3DL + LL = Du,Max (3DL + LL) / (1.2 DL + 1.6 LL) =
