Two - Way Slab Design - 2

Prepared by : Arnel H. Sinconigue, C.ENSCP 2010 STRUCTURAL ENGINEERING DESIGN Date : Oct. 13, 2011Specification and CONSTRUCTION SUPERVISION PRC # : 116853Section and TIN # : 406-365-953-000

Equations No. Sheet Content : Two-Way Slab Design 2S-3 Issued On: 3/24/2011Date : Nov. 30, 2011 Issued At: Quezon City

Project Title : Three Storey Resedential BuildingLocation : Brgy. Banale, Pagadian City, Zamboanga Del Sur

Client : Engr. Arnel H. SinconiegueAddress : Dagat-Dagatan, Caloocan City, Metro Manila

Design Refference : NSCP Volume 1, Fourth Edition 2010

: Fundamentals of Reinforced Concrete (Using USD Method, NSCP 2001) by Besavilla

A. Design Criteria

Materials :

Compressive strenght for slab,fc' = 28.00 MpaCompressive strenght for beam, fc' = 28.00 MpaYeild strenght, fy = 413.00 Mpa

24870.00 Mpa

24870.00 Mpa

24.00

410.3.7.3 0.85409.4.2.1 Capacity Reduction factor, Ф = 0.90

Dead Load :

1. Floor Finish

Ceramic Tile = 0.80 KpaWater Proofing = 0.10 Kpa

2. Ceilling

Suspended Steel Channel = 0.20 KpaElectrical Wirings = 0.30 Kpa

3.ConcreteHollow Blocks

CHB 4" = 2.10 KpaCHB 6" = 2.73 Kpa

4. Partition Load

Parttion = 1.00 Kpa

Live Load :

Residential = 2.00 Kpa

B. Design Data

Long Span, L = 3.00 mShort Span, S = 3.00 mBeam width, b = 250.00 mmBeam Deep, d = 400.00 mm

C. Design of Thickness

3.00

SIDE A = DISCONTINOUS EDGESIDE C = DISCONTINOUS EDGE

3.00

: Design of Reinforce Concrete, 2ND Edition by Jack C. McCormac

Modulus of Elasticity for beam, Ebb =

Modulus of Elasticity forslab, Ebs =

Unit Weight of Concrete, Ȣc = KN/m3

Compressive block deep, β1 =

A

C

D

B

250 x 400

250

x 40

0

250

x 40

0

250 X 400

C-1. Type of Slab

when S/L = 1.00 Two-Way Slab

C-2. Calculation for slab thickness

Assumed Thickness for Design Purposes, h = 100.00 mm

Beam A

Dimension :

h = 100.00 mma = 300.00 mmb = 250.00 mmd = 400.00 mme = 1625.00 mm

Centroid :

30000.00

100000.00

130000.00

By Varignon's Theorem

y = 165.38 mm

Calculation of the Moment of Inertia

1877564102.56

e*h^3/12 135416666.67

Beam B

Check if αm is greater than 2

A1 = mm2

A2 = mm2

AT = mm2

AT(y) = A1(h/2) + A2(d/2)

Ib = ((a*h^3/12 + A1*(y-h/2)^2)) + ((b*d^3/12 + A2*(d/2 -y)^2))

Ib = mm4

Is = mm4

d

h

a

a b

y

x

A1

A2

h

e

A3

A1 A2h

a

a b a

d

h

e

Dimension :


Centroid :

30000.00

30000.00

100000.00

160000.00


y = 143.75 mm

Calculation of Moment of Inertia

2227083333.33

250000000.00

Beam C

Dimension :


Centroid :

30000.00

100000.00

130000.00

By Varignon's Therem

y = 165.38 mm


1877564102.56

135416666.67

A1 = mm2

A2 = mm2

A3= mm2

AT = mm2

AT(y)= A1(h/2) + A2(h/2) + A3(d/2)

Ib = ((a*h3/12) + A1(y-h/2)2) + ((a*h3/12) + A2(y-h/2)2) + ((b*d3/12) + A3(d/2 - y)2)

Ib = mm4

Is = e*h^3/12 mm4

A1 = mm2

A2 = mm2

AT = mm2

AT(y) = A1(h/2) + A2(d/2)

Ib = ((a*h^3/12) + A1*(y-h/2)^2)) + ((a*h^3/12) + A2*(d/2 - y)^2))

Ib = mm4

Is = e*h^3/12 mm4

d

h

a

a b

y

x

A1

A2

h

e

Beam D

Dimension :


Centroid :

30000.00

30000.00

100000.00

160000.00


y = 143.75 mm


2227083333.33

250000000.00

409.1 Calculation of ratio of flexural stiffness of beam section to flexural stiffness of a widthof a slab bounded laterally by center line of adjacent panel (if any) on each side of beam.

13.87

8.91

13.87

8.91

409.1

11.39 Greater than 2, use Eqn.(409-13)

409.6.3.3 Since the average value is greater than 2, the thickness shall not be less than

Eqn. (409-13) 66.92 Non-Compliant

ln = L-b 2750.00

1.00

therefore used thickness.h = 90.00 mm

A1 = mm2

A2 = mm2

A3 = mm2

AT = mm2

AT(y) = A1(h/2) + A2(h/2) + A3(d/2)

Ib = ((a*h3/12) + A1(y-h/2)2) + ((a*h3/12) + A2(y-h/2)2) + ((b*d3/12) + A3(d/2 - y)2)

Ib = mm4

Is = e*h^3/12 mm4

αA = Eb*Ib/Es*Is

αB = Eb*Ib/Es*Is

αC = Eb*Ib/Es*Is

αD = Eb*Ib/Es*Is

Average value for α for all beams on edges of a panel.

αm = (αA + αB + αC + αD)/4

hmin = ((ln(0.8+fy/1400))/(36 + 9β)

β = L/s

A3

A1 A2h

a

a b a

d

h

e

D. Design of Reinforcement by (COEFFICIENT METHOD) Remarks

D-1.Calculation of LoadingsConsideirng 1m strip, b = 1.00 m

Dead Loads :

Partition = 1.00 KpaFloor Finish = 0.90 KpaCeiling = 0.50 KpaC.H.B wall = 0.00 KpaSelfweigth = 2.16 Kpa

DL = 4.56 KpaLive Load :LL = 2.00 Kpa

426.409.2.1 6.38 Kpa426.409.2.1 LLU = 1.7 LL 3.40 Kpa

Ultimate Load (considering 1m strip)

9.78 KN/m

D-2. Type of Slab

m = Ls/Lb 1.00 Two-Way Slab

D-3. Case Number : Case 4

Coefficient for Negative Moment in Slab

0.05000 Short Direction

0.05000 Long Direction

Coefficients for Dead-Load Pos. Moments in Slab



Coefficient for Live-Load Pos. Moment in Slab



D - 4. Negative Moment at Continous Edges

Ms = Ca. neg*Wu*Ls^2 4.40 KN.m

Mb = Cb. Neg*Wu*Lb^2 4.40 KN.m

D - 5. Positive Moment along Short Direction

Ms. DL = Ca.DL* DLu * Ls^2 1.55 KN.m

Ms. LL = Ca.LL * LLu * Ls^2 0.98 KN.m

2.53 KN.m

D - 6. Positive Moment along Long Direction

Mb. DL = Cb. DL*Dlu*Lb^2 1.55 KN.m

Mb. LL = Cb. LL*LL.u*Lb^2 0.98 KN.m

2.53 KN.m

D - 7. Design of reinf. Spacing along short direction (mid-span)

where : Mu = 4.40 KN.mcc = 20.00 mm

12.00 mmds = h - cc - ø/2 64.00 mmb = 1000.00 mm

0.0723 = 0By Quadratic Equation :

DLU = 1.4DL

WU = DLU + LLU

Ca neg. =

Cb neg. =

Ca. DL =

Cb. DL =

Ca. LL =

Cb.LL =

MTs =

MTb =

Mu = Ф*fc'*b*ds^2*ω*(1-0.59ω)

ø =

4.40*1000*1000 = 0.90*28*1000*74^2*ω(1-0.59ω)ω2 - 1.69ω +

a = 1.00b = -1.69c = 0.0723

0.0439

0.00298 Non-Compliant

0.00339 Use for design

0.02901

0.02176 Compliant

0.00339

216.95

Calculation for reinf. Spacing

Use steel dia. D = 12.00 mm

521.00 mm

Spacing limits for Slab Reinforcement

s > h 90.00 mm 407.7.5 s < 3*h 270.00 mm407.7.5 s < 450 450.00 mm

Design Spacing, S = 270.00 mm

D - 8. Design of reinf. Spacing along short direction (continous-edges)

where :Mu = 2.53 KN.mcc = 20.00 mm

12.00 mmds = h - cc - ø/2 64.00 mmb = 1000.00 mm


1.00a = -1.69b = 0.0416c =

0.0250


0.00339 Use for Design

0.02901

0.02176 Compliant

0.00339

ω = (- b ± SQRT(b2 - 4ac))/2a

Calculation of actual steel ratio, ρact

ρact = ω*fc'/fy

Calculation of min. steel ratio, ρmin

ρmin = 1.4/fy

Calculation of max. steel ratio, ρmax

ρb = 0.85*β1*fc'*600/fy(fy+600)

ρmax = 0.75*ρb

Therefore used, ρact =

Calculation of steel area, As

As = ρact*b*d mm2

S = 1000*π*D2/(4*As)

Mu = Ф*fc'*b*ds^2*ω*(1-0.59ω)

ø =

2.53*1000*1000 = 0.90*28*1000*74^2*ω(1-0.59ω)ω2 - 1.69ω +

ω = (- b ± SQRT(b2 - 4ac))/2a


ρact = ω*fc'/fy


ρmin = 1.4/fy


ρb = 0.85*β1*fc'*600/fy(fy+600)

ρmax = 0.75*ρb


216.95



521.00 mm


s > h 90.00 mm 407.7.5 s < 3*h 270.00 mm407.7.5 s < 450 450.00 mm


D - 9. Design of reinf. Spacing along long direction (mid-span)


12.00 mmds = h - cc - Ø - Ø/2 52.00 mmb = 1000.00 mm


a = 1.00b = -1.69c = 0.10951

0.06750

0.00458 Compliant

0.00339 Use actual steel ratio for Design

0.02901

0.02176 Compliant

0.00458

237.96



475.00 mm


s > h 90.00 mm 407.7.5 s < 3*h 270.00 mm407.7.5 s < 450 450.00 mm


D - 10. Design of reinf. Spacing along long direction (continous-edges)


12.00 mmds = h - cc - Ø/2 64.00 mmb = 1000.00 mm


As = ρact*b*d mm2

S = 1000*π*D2/(4*As)

Mu = Ф*fc'*b*ds^2*ω*(1-0.59ω)

ø =

4.40*1000*1000 = 0.90*28*1000*74^2*ω(1-0.59ω)ω2 - 1.69ω +

ω = (- b ± SQRT(b2 - 4ac))/2a


ρact = ω*fc'/fy


ρmin = 1.4/fy


ρb = 0.85*β1*fc'*600/fy(fy+600)

ρmax = 0.75*ρb



As = ρact*b*d mm2

S = 1000*π*D2/(4*As)

Mu = Ф*fc'*b*ds^2*ω*(1-0.59ω)

ø =

2.53*1000*1000 = 0.9*28*1000*74^2*ω*(1-0.59ω)


a = 1.00b = -1.69c = 0.04155

0.02496


0.00339 Use for Design

0.02901

0.02176 Compliant

0.00339

216.95



521.00 mm


s > h 90.00 mm 407.7.5 s < 3*h 270.00 mm407.7.5 s < 450 450.00 mm


E . Design Summary

For short direction the reinforcement spacing : Ø12mm spaced @ 270.00 mm

For long direction the reinforcement spacing : Ø12mm spaced @ 270.00 mm

ω2 - 1.69ω +

ω = (- b ± SQRT(b2 - 4ac))/2a


ρact = ω*fc'/fy


ρmin = 1.4/fy


ρb = 0.85*β1*fc'*600/fy(fy+600)

ρmax = 0.75*ρb



As = ρact*b*d mm2

S = 1000*π*D2/(4*As)

Two - Way Slab Design - 2

Documents

actual steel

steel ratio

long direction

short direction

dl cb

steel dia

steel area

slab ca