Two - Way Slab Design - 2

Prepared by : Date : PRC # : TIN # : 2S-3 Issued On: Issued At:

STRUCTURAL ENGINEERING DESIGN and CONSTRUCTION SUPERVISION

NSCP 2010 Specification Section and Equations No.

Sheet Content : Two-Way Slab Design Date : Nov. 30, 2011

Arnel H. Sinconigue, C.E Oct. 13, 2011 116853 406-365-953-000 3/24/2011 Quezon City

Project Title : Three Storey Resedential Building Location : Brgy. Banale, Pagadian City, Zamboanga Del Sur Client : Address :

Engr. Arnel H. Sinconiegue Dagat-Dagatan, Caloocan City, Metro Manila

Design Refference : NSCP Volume 1, Fourth Edition 2010 : Design of Reinforce Concrete, 2ND Edition by Jack C. McCormac : Fundamentals of Reinforced Concrete (Using USD Method, NSCP 2001) by Besavilla A. Design Criteria Materials : 28.00 28.00 413.00 24870.00

Modulus of Elasticity forslab, Ebs =

24870.00 Mpa 3 24.00 KN/m

Unit Weight of Concrete, Ȣc = Compressive block deep, β1 = Capacity Reduction factor, Ф =

Mpa Mpa Mpa Mpa

0.85 0.90

Dead Load : 1. Floor Finish Ceramic Tile = Water Proofing =

0.80 Kpa 0.10 Kpa

2. Ceilling Suspended Steel Channel = Electrical Wirings =

0.20 Kpa 0.30 Kpa

3.ConcreteHollow Blocks CHB 4" = CHB 6" =

2.10 Kpa 2.73 Kpa

4. Partition Load Parttion =

1.00 Kpa

Live Load : Residential =

2.00 Kpa

B. Design Data Long Span, L = Short Span, S = Beam width, b = Beam Deep, d =

3.00 3.00 250.00 400.00

m m mm mm

C. Design of Thickness

C

B

250 x 400

250 X 400 A

250 x 400

410.3.7.3 409.4.2.1

Compressive strenght for slab,fc' = Compressive strenght for beam, fc' = Yeild strenght, fy = Modulus of Elasticity for beam, Ebb =

3.00

D

250 x 400 3.00

SIDE A = DISCONTINOUS EDGE SIDE C = DISCONTINOUS EDGE

C-1.

Type of Slab when S/L =

C-2.

1.00

Two-Way Slab

Calculation for slab thickness Assumed Thickness for Design Purposes, h =

100.00 mm

Check if αm is greater than 2 Beam A

y x h

A1

d a A2

a

b

h

e

Dimension : h= a= b= d= e=

100.00 300.00 250.00 400.00 1625.00

mm mm mm mm mm

Centroid :

A2 =

2 30000.00 mm 2 mm 100000.00

AT =

2 130000.00 mm

A1 =

By Varignon's Theorem AT(y) = A1(h/2) + A2(d/2) y=

165.38 mm

Calculation of the Moment of Inertia Ib = ((a*h^3/12 + A1*(y-h/2)^2)) + ((b*d^3/12 + A2*(d/2 -y)^2)) 4 1877564102.56 mm

Ib = Is =

4 135416666.67 mm

e*h^3/12

Beam B

h

A1

A2

d

a A3

a

b

a

h e

Dimension :

h= a= b= d= e=

100.00 300.00 250.00 400.00 3000.00

mm mm mm mm mm

Centroid : 2 30000.00 mm 2 30000.00 mm

A1 = A2 = A3=

2 100000.00 mm

AT =

2 160000.00 mm

By Varignon's Theorem AT(y)= A1(h/2) + A2(h/2) + A3(d/2) y=

143.75 mm

Calculation of Moment of Inertia 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) 4 2227083333.33 mm

Ib =

4 250000000.00 mm

Is = e*h^3/12 Beam C

y x h

A1

d a A2 a

b

h

e Dimension : h= a= b= d= e=

100.00 300.00 250.00 400.00 1625.00

mm mm mm mm mm

Centroid :

A2 =

2 30000.00 mm 2 100000.00 mm

AT =

2 130000.00 mm

A1 =

By Varignon's Therem AT(y) = A1(h/2) + A2(d/2) y=

165.38 mm

Calculation of Moment of Inertia Ib = ((a*h^3/12) + A1*(y-h/2)^2)) + ((a*h^3/12) + A2*(d/2 - y)^2)) Ib = Is = e*h^3/12

4 1877564102.56 mm 4 135416666.67 mm

Beam D

h

A1

A2

d a A3

a

b

a

h e Dimension : h= a= b= d= e=

100.00 300.00 250.00 400.00 3000.00

mm mm mm mm mm

Centroid :

A2 =

2 30000.00 mm 2 30000.00 mm

A3 =

2 100000.00 mm

AT =

2 160000.00 mm

A1 =

By Varignon's Theorem AT(y) = A1(h/2) + A2(h/2) + A3(d/2) y=

143.75 mm

Calculation of Moment of Inertia 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 = Is = e*h^3/12

409.1

409.1

Eqn. (409-13)

4 250000000.00 mm

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

αA

= Eb*Ib/Es*Is

13.87

αB

= Eb*Ib/Es*Is

8.91

αC

= Eb*Ib/Es*Is

13.87

αD

= Eb*Ib/Es*Is

8.91

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

αm = (αA + αB + αC + αD)/4 409.6.3.3

4 2227083333.33 mm

11.39

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

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

hmin = ((ln(0.8+fy/1400))/(36 + 9β) ln = L-b β = L/s

therefore used thickness.h =

66.92 2750.00 1.00

90.00 mm

Non-Compliant

D. Design of Reinforcement by (COEFFICIENT METHOD)

Remarks

D-1.Calculation of Loadings Consideirng 1m strip, b =

1.00 m

Dead Loads :

426.409.2.1 426.409.2.1

Partition = Floor Finish = Ceiling = C.H.B wall = Selfweigth =

1.00 0.90 0.50 0.00 2.16

Kpa Kpa Kpa Kpa Kpa

DL = Live Load : LL =

4.56 Kpa

DLU = 1.4DL LLU = 1.7 LL

6.38 Kpa 3.40 Kpa

2.00 Kpa

Ultimate Load (considering 1m strip) W U = DLU + LLU

9.78 KN/m

D-2. Type of Slab m = Ls/Lb

1.00

Two-Way Slab

Coefficient for Negative Moment in Slab Ca neg. =

0.05000

Short Direction

Cb neg. =

0.05000

Long Direction

0.02700 0.02700

Short Direction Long Direction

Ca. LL =

0.03200

Short Direction

Cb.LL =

0.03200

Long Direction

D-3. Case Number : Case 4

Coefficients for Dead-Load Pos. Moments in Slab Ca. DL = Cb. DL = 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

MTs =

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

MTb =

2.53 KN.m

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

Mu = Ф*fc'*b*ds^2*ω*(1-0.59ω) where : Mu = cc = ø= ds = h - cc - ø/2 b= 4.40*1000*1000 = 0.90*28*1000*74^2*ω(1-0.59ω) ω2 - 1.69ω + 0.0723 = 0 By Quadratic Equation :

4.40 20.00 12.00 64.00 1000.00

KN.m mm mm mm mm

a= b= c=

1.00 -1.69 0.0723

ω = (- b ± SQRT(b2 - 4ac))/2a

0.0439

Calculation of actual steel ratio, ρact ρact = ω*fc'/fy

0.00298

Non-Compliant

0.00339

Use for design

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)

0.02901

ρmax = 0.75*ρb

0.02176

Therefore used, ρact =

0.00339

Compliant

Calculation of steel area, As 2 216.95 mm

As = ρact*b*d Calculation for reinf. Spacing Use steel dia. D = 2

S = 1000*π*D /(4*As)

12.00 mm 521.00 mm

Spacing limits for Slab Reinforcement

407.7.5 407.7.5

s>h s < 3*h s < 450

90.00 mm 270.00 mm 450.00 mm

Design Spacing, S =

270.00 mm

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

Mu = Ф*fc'*b*ds^2*ω*(1-0.59ω) where : Mu = cc = ø= ds = h - cc - ø/2 b=

2.53 20.00 12.00 64.00 1000.00

KN.m mm mm mm mm

2.53*1000*1000 = 0.90*28*1000*74^2*ω(1-0.59ω) ω2 - 1.69ω + 0.04155 = 0 By Quadratic Equation : a= b= c=

ω = (- b ± SQRT(b2 - 4ac))/2a

1.00 -1.69 0.0416

0.0250

Calculation of actual steel ratio, ρact ρact = ω*fc'/fy

0.00169

Non-Compliant

0.00339

Use for Design

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)

0.02901

ρmax = 0.75*ρb

0.02176

Therefore used, ρact =

0.00339

Compliant

Calculation of steel area, As 2 216.95 mm

As = ρact*b*d Calculation for reinf. Spacing Use steel dia. D = 2

S = 1000*π*D /(4*As)

12.00 mm 521.00 mm

Spacing limits for Slab Reinforcement

407.7.5 407.7.5

s>h s < 3*h s < 450

90.00 mm 270.00 mm 450.00 mm

Design Spacing, S =

270.00 mm

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

Mu = Ф*fc'*b*ds^2*ω*(1-0.59ω) where : Mu = cc = ø= ds = h - cc - Ø - Ø/2 b=

4.40 20.00 12.00 52.00 1000.00

KN.m mm mm mm mm

4.40*1000*1000 = 0.90*28*1000*74^2*ω(1-0.59ω) ω2 - 1.69ω + 0.10951 = 0 By Quadratic Equation : a= b= c= ω = (- b ± SQRT(b2 - 4ac))/2a

1.00 -1.69 0.10951 0.06750

Calculation of actual steel ratio, ρact ρact = ω*fc'/fy

0.00458

Compliant

0.00339

Use actual steel ratio for Design

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)

0.02901

ρmax = 0.75*ρb

0.02176

Therefore used, ρact =

0.00458

Compliant

Calculation of steel area, As As = ρact*b*d

2 237.96 mm

Calculation for reinf. Spacing Use steel dia. D = S = 1000*π*D2/(4*As)

12.00 mm 475.00 mm

Spacing limits for Slab Reinforcement

407.7.5 407.7.5

s>h s < 3*h s < 450

90.00 mm 270.00 mm 450.00 mm

Design Spacing, S =

270.00 mm

D - 10. Design of reinf. Spacing along long direction (continous-edges) Mu = Ф*fc'*b*ds^2*ω*(1-0.59ω) where : Mu = cc = ø= ds = h - cc - Ø/2 b=

2.53 20.00 12.00 64.00 1000.00

KN.m mm mm mm mm

2.53*1000*1000 = 0.9*28*1000*74^2*ω*(1-0.59ω) ω2 - 1.69ω + 0.04155 = 0 By Quadratic Equation : a= b= c=

1.00 -1.69 0.04155

ω = (- b ± SQRT(b2 - 4ac))/2a

0.02496

Calculation of actual steel ratio, ρact ρact = ω*fc'/fy

0.00169

Non-Compliant

0.00339

Use for Design

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)

0.02901

ρmax = 0.75*ρb

0.02176

Therefore used, ρact =

0.00339

Compliant

Calculation of steel area, As 2 216.95 mm

As = ρact*b*d Calculation for reinf. Spacing Use steel dia. D = 2

S = 1000*π*D /(4*As)

12.00 mm 521.00 mm

Spacing limits for Slab Reinforcement

407.7.5 407.7.5

s>h s < 3*h s < 450

90.00 mm 270.00 mm 450.00 mm

Design Spacing, S =

270.00 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