Concrete Design 1

1. Slab design

Building Location

Chatswood, NSW

Exposure Classification: A2, AS3600…. Table 4.3

Use f ‘c = 32 MPa, Ec = 30100 Mpa AS3600 ... Table 3.1.2

Slab Reinforcement N12, fsy =500 MPa

Slab thickness D s

AS3600 standard requirements

1. Fire Resistance Periods (FRP) for InsulationTime Required Cover 60 min 10mm90 min 100mm120 min 120mm

2. FRPs for Structural Adequacy for Slab Axis distance (ax - mm) Time L y/Lx <= 1.5 1.5<= L y/Lx<=260 min 10mm 15mm90 min 15mm 20mm120 min 20mm 25mm

3. Concrete Cover requirementsExposure Classification is A2, f ‘c = 32MPaConcrete Cover = 25 mm, AS3600….. Table 4.10.3.2

Serviceability Design

Initial assumption for Slab thickness (Dsl): 200mm

AS3600 Deemed to comply span-to-depth ratio for reinforced slabs

Lef

d≤ K3 K4[ ( ∆

Lef

)×1000 × Ec

Fd . ef]

13

Coefficients

Lef = 7500mm,

∆Lef

= 1250 AS3600 ... Table 2.3.2, total deflection – all members,

Fd . ef=( 1+ K cs ) G+(ψs+K cs ψ L) Q

Kcs = 2.0… No compressive reinforcement,

ψs = 0.7, ψL = 0.4, AS1170.0…Table 4.1

Slab Loads G,Q

G = [(0.2*25) + 0.5 +1.0 + 0.5) = 7.0KPa

Q = 4.0KPa

Fd.ef = [(1.0 + 2.0)*7] + [(0.7 + (2.0*0.4))*4] = 27 KPa,

K3 = 1.0, AS3600…. 9.3.4.2 (a),

K4 obtained from table 9.3.4.2 (according to slab’s supports condition)

Lef

d=7500

d≤ K3 K 4∗[ ( ∆

Lef)∗1000∗Ec

Fd . ef]

13

≤ K 4∗1.0∗[ 0.004∗1000∗3010027 ]

13 ≤ 16. 459 K4

K4 16.459K4 d (mm)2.95 48.55 154.52.10 35.564 2112.5 41 1823.38 55.6 135

Use d= 211mm,

Dsl = 211 + ½ (12) + 25 = 248.5mm,

Use Slab thickness (Dsl) = 250mm.

Strength Design

Design Bending Moment, AS3600….. 6.10.3.2

M x¿=βx Fd Lx

2

M y¿=β y Fd Lx

2

Capacity Reduction (Φ)

ΦMu = M*, Φ = 0.8, AS3600….. Table 2.2.2

Design Reinforcement, AS3600…….9.1.1(b)

Pmin=0.19∗( D

d )2

∗f ct .f'

f sy

Slab A (Two adjacent edges discontinuous)

Design Load Fd for 1m wide

Fd = 1.2G *1 + 1.5Q * 1

= (1.2 * (0.25*25 + 2) *1) + (1.5 * 4 *1) = 15.9 KN/m

Lx = 7.5m, Ly = 7.5m

AS3600..Table 6.10.3.2(A) for Ly/Lx = 1,

βx = 0.035, βy = 0.035

dx = 250 – ( ½ *12) – 25 = 219mm,

dy = 219 – db = 219 – 12 = 207mm.

The flexural lever arm jd ~ 0.95d

X direction

(+ve mid-span) M*x =0.035*15.9*(7.5)2 = + 31.30 KN.m,

(-ve continuous edge) M*x = 1.33*27.2 = - 41.63 KN.m,

(-ve discontinuous edge) M*x = 0.5*27.2 = - 15.65 KN.m,

Y direction

(+ve mid-span) M*y = 0.035*14.4*(7.5)2 = + 31.30 KN.m,

(-ve continuos edge) M*y = 1.33*27.2 = - 41.63 KN.m,

(-ve discontinuos edge) M*y = 0.5*27.2 = - 15.65 KN.m,

Details Discontinuous Edge Mid-Span Continuous Edge

Short Span "X" direction  

M*x (KN.m) -15.65 31.30 -41.63

M* / Φ (KN.m) -19.56 39.13 -52.04d (mm) 219 219 219

jdx (mm) 208 208 208Ast (mm2)/ 1m wide =

M*ux/fsy* jd

188 376 500

P Actual 0.00086 0.00172 0.00228

Pmin 0.00168 0.00168 0.00168

# N12 (113mm2)Spacing (mm)

Long Span "Y" Direction 

M*y (KN.m) -15.65 31.30 -41.63

M* / Φ (KN.m) -19.56 39.13 -52.04d 207 207 207

jdy (mm) 197 197 197Ast (mm2) 199 398 529

P Actual 0.00096 0.00192 0.00256Pmin 0.00188 0.00188 0.00188

# N12 (113mm2)Spacing (mm)

Slab B, C and F (One Long edge discontinuous)

Design Load Fd for 1m wide

Fd = 1.2G *1 + 1.5Q * 1

= (1.2 * (0.25*25 + 2) *1) + (1.5 * 4 *1) = 15.9 KN/m

Lx = 7.5m, Ly = 15m

AS3600..Table 6.10.3.2(A) for Ly/Lx = 2, βx = 0.066, βy = 0.028

Details Discontinuous Edge Mid-Span Continuous Edge

Short Span "X" direction  

M*x (KN.m) 0.00 59.03 -78.51

M* / Φ (KN.m) 0.00 73.79 -98.14d (mm) 0.00 219 219

jdx (mm) 0.00 208 208

Ast (mm2) 0.00 709 943

P Actual 0.00 0.00324 0.00431

Pmin 0.00 0.00168 0.00168

# N12 (113mm2) 0.00Spacing (mm) 0.00

Long Span "Y" Direction 

M*y (KN.m) -12.52 25.04 -33.31

M* / Φ (KN.m) -15.65 31.3 -41.63d 207 207 207

jdy (mm) 197 197 197

Ast (mm2) 159 318 423

P Actual 0.00077 0.00154 0.00205

Pmin 0.00188 0.00188 0.00188

# N12 (113mm2)Spacing (mm)

Slab D and E (Four edges Continuous)

Design Load Fd for 1m wide

Fd = 1.2G *1 + 1.5Q * 1

= (1.2 * (0.25*25 + 2) *1) + (1.5 * 4 *1) = 15.9 KN/m

Lx = 7.5m, Ly = 15m

AS3600..Table 6.10.3.2(A) for Ly/Lx = 2, βx = 0.048, βy = 0.024

Details Discontinuous Edge Mid-Span Continuous Edge

Short Span "X" direction  

M*x (KN.m) 0.00 42.93 -57.10

M* / Φ (KN.m) 0.00 53.66 -71.37d (mm) 0.00 219 219

jdx (mm) 0.00 208 208

Ast (mm2 0.00 516 686

P Actual 0.00 0.00236 0.00313

Pmin 0.00 0.00168 0.00168

# N12 (113mm2) 0.00Spacing (mm) 0.00

Long Span "Y" Direction 

M*y (KN.m) 0.00 21.47 -28.55

M* / Φ (KN.m) 0.00 26.83 -35.69d 0.00 207 207

jdy (mm) 0.00 197 197

Ast (mm2) 0.00 273 363

P Actual 0.00 0.00132 0.00175

Pmin 0.00 0.00188 0.00188

# N12 (113mm2) 0.00Spacing (mm) 0.00

Slab G (Four edges Continuous)

Design Load Fd for 1m wide

Fd = 1.2G *1 + 1.5Q * 1

= (1.2 * (0.25*25 + 2) *1) + (1.5 * 4 *1) = 15.9 KN/m

Lx = 13.5m, Ly = 15m

AS3600..Table 6.10.3.2(A) for Ly/Lx = 1.11,

βx = 0.028, βy = 0.024

Details Discontinuous Edge Mid-Span Continuous Edge

Short Span "X" direction  

M*x (KN.m) 0.00 81.14 -107.91

M* / Φ (KN.m) 0.00 101.42 -134.89d (mm) 0.00 219 219

jdx (mm) 0.00 208 208

Ast (mm2) 0.00 975 1297

P Actual 0.00 0.00445 0.00592

Pmin 0.00 0.00168 0.00168

# N12 (113mm2) 0.00Spacing (mm) 0.00

Long Span "Y" Direction 

M*y (KN.m) 0.00 69.55 -92.50

M* / Φ (KN.m) 0.00 86.93 -115.62d 0.00 207 207

jdy (mm) 0.00 197 197

Ast (mm2) 0.00 884 1176

P Actual 0.00 0.00427 0.00568

Pmin 0.00 0.00188 0.00188

# N12 (113mm2) 0.00Spacing (mm) 0.00

Beam Design

1. Edge Beam: Spandrel Beam, Dt = 1000mm, w = 450mm,

f’c = 32 MPa fct.f = 0.6√32 = 3.39 Mpa

bef=bw+0.1 a …………. a=0.7 L ,bw=450 mm.

Lc.c = 7500mm Le = 7500 – 450 = 7050mm,

bef=450+(0.1∗0.7∗7050 )=944 mm

y ' (From tension extreme fiber to N.A)

y '=((250∗494∗125 )+(1000∗450∗500 ))

( (250∗494 )+(1000∗450))=419 mm

I g=[( 494∗2503

12 )+(494∗250∗(419−125 )2)]+[( 450∗10003

12 )+(450∗1000∗(500−419 )2)] I g=5.177∗1010 mm4

M uo ,min± ve =1.2∗Z∗f ct . f

' …………… AS 3600 …8.1 .4 .1

Hogging Moment (-ve moment)

M uo ,min−ve =1.2∗5.177∗1010

419∗3.39=502.6 KN . m

** Assume using N28, fy = 500 MPa,

d = 1000 – 25 – 12 – 28/2 =949 mm,

A st .min=502.6∗106

0.95∗949∗500=1115 mm2

a=A st∗f y

α2∗f ' c∗bef

= 1232∗5000.85∗32∗944

=24 mm,

M u=A st∗f y∗(d−a2 )=1232∗500∗(949−24

2 )=577.2 KN .m,

φM u=0.8∗577.2=461.6 KN .m ,

M uo ,max−ve =1700 KN .m ,¿moment envelope

A st ,max=( 1700461.6 )∗1232=4537 mm2

# Use 2N28 = 2 * 616 = 1232 mm2

d = 1000 – 25 – 12 -16 = 947 mm,

a=A st∗f y

α2∗f ' c∗bef

= 4824∗5000.85∗32∗944

=94 mm,

M u=A st∗f y∗(d−a2 )=4824∗500∗(947−94

2 )=2170.8 KN . m,

φM u=0.8∗2170.8=1736.6 KN .m>1700 KN . m ………OK

2. Mid Beam – L Beam

bef=bw+0.2 a …………. a=0.7 L ,bw=450 mm.

Lc.c = 7500mm Le = 7500 – 450 = 7050 mm,

bef=450+(0. 1∗0.7∗7050 )=944 mm

y ' (From compression extreme fiber to N.A)

y '=419 mm

I g=5.177∗1010 mm4

Sagging Moment (+ve moment)

M uo ,min+ve =1.2∗Z∗f ct . f

'

M uo ,min+ve =1.2∗5.177∗1010

(419)∗3.39=502.6 KN . m

** Assume using N28, fy = 500 MPa,

d = 1000 – 25 – 12 – 28/2 =949 mm,

A sc .min=5 02 .6∗106

0.95∗949∗500=1115 mm2

# Use 6N32 = 6 * 804 = 4824 mm2

# Use 2N28 = 2 * 616 = 1232 mm2

a=A st∗f y

α2∗f ' c∗bef

= 1232∗5000.85∗32∗944

=24 mm,

M u=A st∗f y∗(d−a2 )=1232∗500∗(949−24

2 )=5 77.2 KN .m,

φM u=0.8∗577.2=461.6 KN .m ,

M uo ,max+ve =720 KN . m ,¿ moment envelope

A sc , max=( 720461 .6 )∗1232=19 22 mm2

d = 1000 – 25 – 12 – 28/2 =949 mm,

a=A st∗f y

α2∗f ' c∗bef

= 2 464∗5000.85∗32∗944

=48 mm,

M u=A st∗f y∗(d−a2 )=2464∗500∗(94 9−48

2 )=1139 KN . m,

φM u=0.8∗1139=911 KN . m>720 KN . m……OK

Hogging Moment (-ve moment)

M uo ,min−ve =1.2∗Z∗f ct . f

'

M uo ,min−ve =1.2∗5.177∗1010

581∗3.39=362.5 KN . m≪ Muo , min

−ve at edgebeam=502. 6 KN .m

a=A st∗f y

α2∗f ' c∗bef

= 1232∗5000.85∗32∗944

=24 mm,

M u=A st∗f y∗(d−a2 )=1232∗500∗(949−24

2 )=577.2 KN .m,

# Use 4 N28 = (4 * 616) = 2464 mm2

# Use 2N28 for the whole section

φM u=0.8∗577.2=461.6 KN .m ,

M uo ,max−ve =1500 KN .m ,¿moment envelope

A st ,max=( 1500461.6 )∗1232=400 4 mm2≪ A st , max for the dge beam=4537 mm2

d = 1000 – 25 – 12 -16 = 947 mm,

a=A st∗f y

α2∗f ' c∗bef

= 4824∗5000.85∗32∗944

=94 mm,

M u=A st∗f y∗(d−a2 )=4824∗500∗(947−94

2 )=2170.8 KN . m,

φM u=0.8∗2170.8=1736.6 KN .m>1500 KN . m ………OK

Stress Development

Lsy .tb=0.5 K1 K3 f sy db

K2√ f ' c

≥ 29 K1 db ………13.1.2 .2

#N28

K1=1.3 , K2=1.04 , K3=1.0 ,

Lsy .tb=0.5∗1.3∗1.0∗500∗28

1.0 4∗√32=1550 mm ≥29∗1.3∗28=1056 mm

#N32

K1=1.3 , K2=1.04 , K3=1.0 ,

Lsy .tb=0.5∗1.3∗1.0∗500∗32

1.0∗√32=1840 mm≥ 29∗1.3∗32=1206.5 mm

Beam Shear Design

# Use 6N32 for the whole section