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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 :
h = 100.00 mma = 300.00 mmb = 250.00 mmd = 400.00 mme = 3000.00 mm
Centroid :
30000.00
30000.00
100000.00
160000.00
By Varignon's Theorem
y = 143.75 mm
Calculation of Moment of Inertia
2227083333.33
250000000.00
Beam C
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 Therem
y = 165.38 mm
Calculation of Moment of Inertia
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 :
h = 100.00 mma = 300.00 mmb = 250.00 mmd = 400.00 mme = 3000.00 mm
Centroid :
30000.00
30000.00
100000.00
160000.00
By Varignon's Theorem
y = 143.75 mm
Calculation of Moment of Inertia
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
0.02700 Short Direction
0.02700 Long Direction
Coefficient for Live-Load Pos. Moment in Slab
0.03200 Short Direction
0.03200 Long Direction
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
0.04155 = 0By Quadratic Equation :
1.00a = -1.69b = 0.0416c =
0.0250
0.00169 Non-Compliant
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
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 =
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 - 9. Design of reinf. Spacing along long direction (mid-span)
where :Mu = 4.40 KN.mcc = 20.00 mm
12.00 mmds = h - cc - Ø - Ø/2 52.00 mmb = 1000.00 mm
0.10951 = 0By Quadratic Equation :
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
Calculation for reinf. Spacing
Use steel dia. D = 12.00 mm
475.00 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 - 10. Design of reinf. Spacing along long 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
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 =
4.40*1000*1000 = 0.90*28*1000*74^2*ω(1-0.59ω)ω2 - 1.69ω +
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 = 0.9*28*1000*74^2*ω*(1-0.59ω)
0.04155 = 0By Quadratic Equation :
a = 1.00b = -1.69c = 0.04155
0.02496
0.00169 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
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
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)