Voided Biaxial Slab - Structural Design Softwareengineering-international.com/VoidedBiaxialSlabs.xls · 2017-09-28Voided Biaxial Slab - Structural Design Software

PROJECT : PAGE :

CLIENT : DESIGN BY : JOB NO. : DATE : REVIEW BY :

Voided Two-Way Slab Design Based on ACI 318-14

DESIGN CRITERIA1. The voided sphere or ellipse bubbles within slab can reduce concrete weight, so both seismic mass (ASCE 7 12.7.2) and gravity

loads reduced. And the long-term deflection (3 DL + LL) limits may not govern the two-way slab design (ACI 318 8.3).2. The entire slab bottom formwork can be flat, without girder, beam, drop panel or cap, but the punching area (ACI 318 8.4 & 8.5),

or lateral frame diaphragm area (ACI 318 12.5.3), may need to be solid as normal concrete shear transfer.3. The section forces of voided slab can be determined by a two-way finite element method or by ACI 318 8, but PT slab

can only be designed by one way method because the secondary moment of PT slab is one way concept. Also, the voidedtwo-way slab is better for depressed floor,or irregular opening, than PT slab.

4. The bottom two direction rebar can bedistributed as a regular solid two-way slab,without Waffle slab or hollow core plank limits.

INPUT DATA & DESIGN SUMMARYCONCRETE STRENGTH

5 ksi, (34 MPa)REBAR YIELD STRESS

60 ksi, (414 MPa)

TOTAL SLAB THICKNESSt = 40 in, (1016 mm)

TOP & BOTTOM SOLID THICKNESS

5 in, (127 mm)VOIDED BUBBLE HORIZONTAL DIAMETER

D = 48 in, (1219 mm)

COLUMN SPACING EACH WAYL = 69 ft, (21.03 m)B = 69 ft, (21.03 m)

COLUMN SIZE (SHORT EDGE)50 in, (1270 mm)

SUPERIMPOSED DEAD LOAD, ASD

20 psf, (1.0 kPa)LIVE LOAD

LL = 60 psf, (2.9 kPa)

TOP BARS AT COLUMNS EACH WAY41 # 10 @ 5 in. o.c. (127 mm o.c.)

x 23.0 ft. long, with 1 in. cover(7.0 m) (25 mm)

(All top bars to column strip suggested, if column strip & middle strip used.)

BOTTOM LAYER BOTTOM BARS# 9 @ 12 in. o.c. (305 mm o.c.) THE DESIGN IS ADEQUATE.

BOTTOM LAYER TOP BARS# 9 @ 12 in. o.c. (305 mm o.c.)

with 1 in. (25 mm), bottom concrete cover(75% total bottom bars to middle strip & 25% to column strip suggested, if column strip & middle strip used.)

ANALYSISDETERMINE SECTION PROPERTY & DEAD LOAD

5 in > 0.75 + 2.54 + 1 = 4.29 in, top solid min thk> 0.75 + 2.26 + 1 = 4.01 in, bot solid min thk

(inside cover) (2 rebar thick) (top & bot cover) [Satisfactory]

D = 48 in > 30 in, height of voided sphere or ellipse bubble [Satisfactory]

150 pcf, (ACI 318-14 19.2.2.1) V = 36191

Wt = 323 psf, self weight reduced 35% DL = 343 psf

269333 205716

4287 ksi, (ACI 318-14 19.2.2.1) 3274 ksi, for Finite Element Method

0.63 t = 25.2 in, for Slab only (ACI 318-14 8.3.2, 24.2 & 6)

fc' =

fy =

tsolid =

c =

DLsup =

tsolid =

wc = in3, volume of a voided sphere or ellipse bubble

DLsup + Wt =

Isoild = in4 Ig = in4

Ec = wc1.5 33 f'c0.5 = ( Ig / Isoild ) Ec =

te = ( Ie / Ig )1/3 t = (0.25 Ig / Ig )1/3 t = (0.25)1/3 t =

(cont'd)DETERMINE SECTION FORCE AND SLAB DEFLECTION USING FINITE ELEMENT METHOD

JointNumber in kips Bending

1 0 603.61 Section ft-k/ft2 1.05 1 - 2 463.53 1.69 2 - 3 -21.24 1.05 1 - 6 463.55 0 603.61 6 - 11 -21.26 1.05 3 - 8 -9.97 1.50 8 - 13 -118.18 1.91 11 - 12 -9.99 1.50 12 - 13 -118.1

10 1.0511 1.6912 1.9113 2.1414 1.91 DETERMINE FACTORED LOAD (ACI 318-14 5.3)

15 1.69 1.2 DL + 1.6 LL = 0.507 ksf16 1.0517 1.50 DETERMINE FLEXURE CAPACITY (ACI 318-14 7.6.1.1, 8.6.1.1, & 22)18 1.9119 1.50 Top Bar Bot. Layer Bot. Bot. Layer Top 20 1.05 41 # 10 @ 5" o.c. 9 @ 12" o.c. 9 @ 12" o.c.21 0 603.61 37.10 38.44 37.31

22 1.05 3.05 1.00 1.00

23 1.69 1.57 0.86 0.8624 1.05 4.59 1.52 1.52

25 0 603.61 483.2 170.2 165.1

CHECK FLEXURE CAPACITY

463.5 ft-k/ft < 483.2 ft-k/ft [Satisfactory]



CHECK LIVE LOAD DEFLECTION (ACI 318-14 Table 24.2.2)

0.25 in < L / 360 = 2.30 in[Satisfactory]

CHECK LONG-TERM DEFLECTION (ACI 318-14 24.2.4)


CHECK COLUMN PUNCHING CAPACITY (ACI 318-14 8.4 & 8.5)

2414.5 kips (See Punching.xls Software for More Information.)

2741.42 kips > [Satisfactory]

where 0.75 (ACI 318-14 21.2)

1.00d = 37.1 in

4 c + 4 d = 348.4 in

12923.2

y = 2.0

Du Ru

Mu

wu =

d (in)

As (in2/ft)

As, min (in2/ft)c (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 =

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

'

2'

'

2 0.85, 29000

0.85 2 , 0

0.85 ,

,,

C

CC

C

S

fksiEso Ec

c c forf c of oo

forf c oforEss s t

f forf s ty

PROJECT : PAGE :


Voided Section Design Based on ACI 318-08


5 ksiREBAR YIELD STRESS

60 ksi

TOTAL SLAB THICKNESSt = 40 in


5 inVOIDED HORIZONTAL DIAMETER

D = 48 in, (input 30 for Sphere/Circle) 483.2 ft-kips / ft

VERTICAL WEB THICKNESS 17.9 kips / ftw = 2.5 in

THE DESIGN IS INADEQUATE, SEE ANALYSIS BELOWSECTION BARS # 10 @ 5 o.c.

with 2.27 in. bottom concrete cover

ANALYSISCHECK SECTION LIMITATIONS

5 in < 0.75 + 2.54 + 2.27 = 5.56 in, solid min thk(inside cover) (2 rebar thick) (top & bot cover) [Unsatisfactory]

D = 48 in > 30 in, voided height [Satisfactory]

w = 2.5 in > 2.5 in [Satisfactory]

DETERMINE FLEXURE CAPACITY (ACI 318-08 7.12.2.1, 10.2, 10.5.1)

150 pcf, (ACI 318-08 8.5.1) 4287 ksi, (ACI 318-08 8.5.1)

d = 37.10 in c = 4.59 in

3.05 > 0.86 [Satisfactory]

483.2 ft-kips / ft, (by pure math method) 0.9 ,(ACI 318-08 Fig R9.3.2)

DETERMINE ONE WAY SHEAR CAPACITY (ACI 318-08 11.1.3.1, & 11.2)

17.9 kips / ft 0.75 ,(ACI 318-08 9.3.2.3)

169

fc' =

fy =

tsolid =

fMn =

fVn =

tsolid =

wc = Ec = wc1.5 33 f'c0.5 =

As = in2 / ft Amin = in2 / ft

fMn = f =

fVn = f 2Ac (fc')0.5 = f =

Ac = in2 / ft

'

2'

'

2 0.85, 29000

0.85 2 , 0

0.85 ,

,,

C

CC

C

S

fksiEso Ec

c c forf c of oo

forf c oforEss s t

f forf s ty

PROJECT : PAGE :





60 ksi






THE DESIGN IS ADEQUATE.SECTION BARS # 9 @ 12 o.c.

with 1 in. bottom concrete cover


5 in > 0.75 + 2.26 + 1 = 4.01 in, solid min thk(inside cover) (2 rebar thick) (top & bot cover) [Satisfactory]




150 pcf, (ACI 318-08 8.5.1) 4287 ksi, (ACI 318-08 8.5.1)

d = 38.44 in c = 1.52 in




19.6 kips / ft 0.75 ,(ACI 318-08 9.3.2.3)

185

fc' =

fy =

tsolid =

fMn =

fVn =

tsolid =

wc = Ec = wc1.5 33 f'c0.5 =


fMn = f =

fVn = f 2Ac (fc')0.5 = f =

Ac = in2 / ft

'

2'

'

2 0.85, 29000

0.85 2 , 0

0.85 ,

,,

C

CC

C

S

fksiEso Ec

c c forf c of oo

forf c oforEss s t

f forf s ty

PROJECT : PAGE :





60 ksi






THE DESIGN IS INADEQUATE, SEE ANALYSIS BELOWSECTION BARS # 9 @ 12 o.c.

with 2.128 in. bottom concrete cover


5 in < 0.75 + 2.26 + 2.128 = 5.13 in, solid min thk(inside cover) (2 rebar thick) (top & bot cover) [Unsatisfactory]




150 pcf, (ACI 318-08 8.5.1) 4287 ksi, (ACI 318-08 8.5.1)

d = 37.31 in c = 1.52 in




18.2 kips / ft 0.75 ,(ACI 318-08 9.3.2.3)

172

fc' =

fy =

tsolid =

fMn =

fVn =

tsolid =

wc = Ec = wc1.5 33 f'c0.5 =


fMn = f =

fVn = f 2Ac (fc')0.5 = f =

Ac = in2 / ft

'

2'

'

2 0.85, 29000

0.85 2 , 0

0.85 ,

,,

C

CC

C

S

fksiEso Ec

c c forf c of oo

forf c oforEss s t

f forf s ty

PROJECT : PAGE :


Two-Way Slab Design Based on ACI 318-08 using Finite Element Method



60 ksiCOLUMN SPACING EACH WAY

L = 69 ftB = 69 ft

SLAB THICKNESSt = 40 in

BENDING DROP PANEL THK. (0 for no drap panel)

0 inPUNCHING CAP THICKNESS

0 inDEAD LOAD & SELF WT

DL = 342.60981 psf, (1.0 kPa)LIVE LOAD

LL = 60 psf

TOP BARS AT COLUMNS EACH WAY17 # 6 @ 12 o.c.

x 23.0 ft. long, with 2 in. cover(All top bars to column strip suggested, if column strip & middle strip used.)

BOTTOM LAYER BOTTOM BARS# 5 @ 18 o.c. THE DESIGN IS INADEQUATE, SEE ANALYSIS BELOW

BOTTOM LAYER TOP BARS# 5 @ 18 o.c.

with 0.75 in. bottom concrete cover(75% total bottom bars to middle strip & 25% to column strip suggested, if column strip & middle strip used.)

ANALYSIS150 pcf, (ACI 318-08 8.5.1) 4287 ksi, (ACI 318-08 8.5.1)

0.63 t = 25.2 in, for Slab only (ACI 318-08 9.5.3.4 & 10.11.1)25.2 in, for Slab & Drop Panel

JointNumber in Bending

1 0 Section ft-k/ft2 0.90 1 - 2 414.73 1.31 2 - 3 -39.14 0.66 3 - 4 -21.15 0 4 - 5 225.76 0.98 1 - 6 430.97 1.30 6 - 11 -22.18 1.41 12 - 13 -118.49 0.68 13 - 14 -112.410 0 7 - 12 -131.511 1.58 8 - 13 -107.412 1.66 9 - 14 -46.413 1.5114 0.70 DETERMINE FACTORED LOAD (ACI 318-08 9.2.1)

15 0 1.2 DL + 1.6 LL = 0.507 ksf16 0.9817 1.30 DETERMINE FLEXURE CAPACITY (ACI 318-08 7.12.2.1, 10.2, 10.5.1)18 1.4119 0.68 Top Bar Bot. Layer Bot. Bot. Layer Top 20 0 17 # 6 @ 12" o.c. 5 @ 18" o.c. 5 @ 18" o.c.21 0 36.88 38.94 38.31

22 0.90 0.44 0.21 0.21

23 1.31 [Unsatisfactory] ==> 1.56 0.86 0.8624 0.66 0.52 0.24 0.24

25 0 72.5 36.1 35.5

CHECK FLEXURE CAPACITY

430.9 ft-k/ft > 72.5 ft-k/ft [Unsatisfactory]

131.5 ft-k/ft > 36.1 ft-k/ft [Unsatisfactory]

46.4 ft-k/ft

> 35.5 ft-k/ft [Unsatisfactory]

CHECK LIVE LOAD DEFLECTION (ACI 318-08 Table 9.5b)


CHECK LONG-TERM DEFLECTION (ACI 318-08 9.5.2.5)


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) =

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Voided Biaxial Slab - Structural Design Softwareengineering-international.com/VoidedBiaxialSlabs.xls · 2017-09-28Voided Biaxial Slab - Structural Design Software