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2.0 Design of Slab
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document.xls One Way Slab
Design of One Way SlabReference: NSCP 2001 Volume 1
Design of Reinforced Concrete by J.C. McCormac
Dimension: Reinforcement: Loads:
cc = 20 mm = 12 mm = 2.400 kPa
h = 100 mm = 12 mm = 2.000 kPa
b = 1000 mm Material Strength: = 1.900 kPa
d = 74 mm fc' = 21 Mpa = 9.390 kPaLoads: fy = 276 Mpa
= 2.6709 kN-m = 0.9
= 3.0048 kN-m
Check if One Way Slab s / l < 0.50s = 1.600 ml = 4.400 ms
= 0.364 one way slab!l
when fy = 415 when fy = 415s / 20 simply supported multiply by:s / 24 one end continuous
0.40 +fy
= 1.07s / 28 two end continuous 415s / 10 cantilever
=s
= 57.14 ok!28
> 0.65 NSCP Section 410.3.7.3
= 0.85 - 0.05( fc' - 30 ) = 0.85
7
= 0.914use:
= 0.85 > 0.65 ok!
NSCP Section 410.4.3 NSCP Section 410.6.1
==
fc'> =
1.4
= 600 4 fy fy
fy 600 + fy = 0.0042 < = 0.0051
= 0.0377 use:= 0.0282 = 0.0051
Main Reinforcements Top Bars Bot Bars units Spcg Limitations:m = 0.847 fc' / fy 0.0644 0.0644 - a. Main Reinf.
0.5419 0.6097 Mpa NSCP Section 407.7.5
0.0020 0.0022 - s < 3h = 300
0.0051 0.0051 - s < 450
375.36 375.36 b.Temp Reinf.
113.10 113.10 NSCP Section 407.13
301.30 301.30 pcs s < 5h = 500300.00 300.00 pcs s < 450
Temperature / Shrinkage ReinforcementsNSCP Section 407.13
= 0.002bh
= 200
= 565.49 mmProvide = 450 mm
Summary:Top Bars 12 @ 300 mmBot Bars 12 @ 300 mmTemp Bars 12 @ 450 mm
main bar q WSLAB
temp bar q WDL
WLL
WU
Mutop fM
Mubot
Solve for hmin
hmin
hmin
Solve fo b1 :a.) if fc' > 30 Mpa ; b.) if fc' < 30 Mpa ;
b1b1
b1
b1
Solve for rmax : Solve for rmin :
rmax 0.75 rbalrmin rmin
rbal
0.85 b1 fc'
rmin rmin
rbal
rmax rmin
R = Mu / fbd2
rreqd = m - m2 - 2Rm/fy
rsupplied
Asreqd = rbd mm2
As main bar q mm2
spcgreqd
Provide (spcg)
Astemp
Astemp mm2
spcgreqd
document.xls Two Way Slab
Analysis of Two-Way Slab with BeamsReference: NSCP 2001 Volume 1
Design of Reinforced Concrete by J.C. McCormac
NSCP Section 409.6.3.3
= > 125 mm
= > 90 mm
where:
=a = ratio of flexural stiffness of beam section to flexural
stiffness of a width of slab bounded laterally by centerlines of adjacent panels on each side of the beam
a =
= length of clear span in long direction of two-wayconstruction, measured face to face of supports
b = ratio of clear span in long to short direction of two-way slabs
B. Total Factored Static Moment for a Span NSCP Section 413.7.2.2
=8
where:
= length of span in direction that moments are being determined, measured c to c of supports
=of supports
= length of span in direction that moments are being determined, measured face to face of supports
C. Negative & Positive Factored Moments NSCP Section 413.7.3
Negative Factored Moments = 0.65Positive Factored Moments = 0.35
0.75 0.70 0.70 0.70 0.65
0.63 0.57 0.52 0.50 0.35
0.00 0.16 0.26 0.30 0.65
D. Factored Moments in Column Strips NSCP Section 413.7.4
0.50 1.00 2.00
75 75 75
90 75 45
A. Computation of hmin
1. For am greater than 0.20 but not greater than 2.0
hmin
ln ( 0.80 + fy / 1500 )
36 + 5b ( am - 0.20 )
2. For am greater than 2.0
hminln ( 0.80 + fy / 1500 )
36 + 9b
am ave. of a
ECBIB
ECSIS
ln
Mowu l2 ln2
l1
l2 length of span transverse to l1, measured c to c
ln
1. Interior Spans
2. Exterior Spans
Exterior Edge Unrestrained
Slabs With Beams
Between All Supports
Slab Without Beams Between Interior Supports Exterior Edge
Fully RestrainedWithout Edge
BeamsWith Edge
Beams
Interior Negative Factored Moment
Positive Factored Moment
Exterior Negative Factored Moment
1. Interior Negative Factored Moments (%)
l2 / l1
a1 l1 / l2 = 0
a1 l1 / l2 > 1
document.xls Two Way Slab
0.50 1.00 2.00
100 100 100
75 75 75
100 100 100
90 75 45
0.50 1.00 2.00
60 60 60
90 75 45
Note: NSCP Section 413.7.5
if: > 1.0 85 % - beam15 % - slab
< 1.0 interpolate from 0% to 85%where:
=
= ratio of torsional dtiffness of edge beam section toflexural stiffness of a width od slab equal to span lengthof beam, c to c of supports
=
C = cross-sectional constant to define torsional properties
C = S 1 - 0.63xy 3
E. Slab Reinforcements NSCP Section 413.4
Spacing of reinforcement at critical sections shallnot exceed two times the slab thickness
F. Column Strip and Middle Strip
`
Middle Strip
2. Exterior Negative Factored Moments (%)
l2 / l1
a1 l1 / l2 = 0bt = 0
bt > 2.5
a1 l1 / l2 > 1bt = 1
bt > 2.5
3. Positive Factored Moments (%)
l2 / l1
a1 l1 / l2 = 0
a1 l1 / l2 > 1
a1 l2 / l1
a1 l2 / l1
a1 a in direction of l1
bt y2
x2
bt
ECBCy1
2ECSIS y1 - x2
x3y x1
l2
Column Strip
Column Strip
l1
0.25 l1 or 0.25 l2, which is smaller
document.xls
Design of Two-Way Slab with BeamsReference: NSCP 2001 Volume 1
Design of Reinforced Concrete by J.C. McCormacDesign of Concrete Structures by Arthur H. Nilson
= 21 Mpa
= 21 Mpafy = 276 Mpa
= 0.90 4.00 m
= 4.80 kPa b x h
= 2.50 kPa 250 x 400
= 10.92 kPa A
= 4.25 kPa= 15.17 kPa 5.00 m
= 12 mm 300 x 500 C D 300 x 500h = 125 mm
= 93 mm B
= 0.779250 x 400
= 3.00 kPa5.00 m
5.00 m 4.00 m 3.00 m
Side A
= ==
= 21675.58 MPa = 21675.58 MPa
= = = 1.82
= 1.33E+09 = 7.32E+08Side B
= ==
= 21675.58 MPa = 21675.58 MPa
= = = 1.64
= 1.33E+09 = 8.14E+08Side C
= ==
= 21675.58 MPa = 21675.58 MPa
= = = 4.27
= 3.13E+09 = 7.32E+08Side D
= ==
= 21675.58 MPa = 21675.58 MPa
= = = 5.49
= 3.13E+09 = 5.70E+08
=4
= 3.30
B. Check Minimum Thickness = 4.750 m b = 1.284
= > 125 mm
= 125.00 mm
fc'CS
fc'CB
fM
WDL
WLL
WuDL
WuLL
Wu
qbar
deff
la
lb
WSLAB
A. Solve for am
ECB 4730 fc'CB ECS 4730 fc'CSaA
ECBICB
ECB ECS ECSICS
ICB bh3 / 12 ICS bh3 / 12 aA
ICB mm4 ICS mm4
ECB 4730 fc'CB ECS 4730 fc'CSaB
ECBICB
ECB ECS ECSICS
ICB bh3 / 12 ICS bh3 / 12 aB
ICB mm4 ICS mm4
ECB 4730 fc'CB ECS 4730 fc'CSaC
ECBICB
ECB ECS ECSICS
ICB bh3 / 12 ICS bh3 / 12 aC
ICB mm4 ICS mm4
ECB 4730 fc'CB ECS 4730 fc'CSaD
ECBICB
ECB ECS ECSICS
ICB bh3 / 12 ICS bh3 / 12 aD
ICB mm4 ICS mm4
amaA + aB + aC + aD
am
ln
1. For am greater than 0.20 but not greater than 2.0
hmin
ln ( 0.80 + fy / 1500 )
36 + 5b ( am - 0.20 )
hmin
document.xls
= > 90 mm
= 100.00 mm
Therefore use: = 100.00 mm ok!
D. Check Slab on Shear ( Coefficient Method )
Va = by linear interpolation Vu Mu(-)
Vb = 0.750 0.7600 1.00 0.500 0.045 0.018 0.027
= 0.7311 kN 0.779 x 0.7311 0.95 0.550 0.050 0.020 0.030
= 0.2689 kN 0.800 0.7100 0.90 0.600 0.055 0.022 0.034Va = 194.91 kN 0.85 0.660 0.060 0.024 0.037Vb = 71.70 kN 0.750 0.2400 0.80 0.710 0.065 0.026 0.041
= 0.779 x 0.2689 0.75 0.760 0.069 0.028 0.045= 362.25 kN ok! 0.800 0.2900 0.70 0.810 0.074 0.030 0.049
0.65 0.850 0.077 0.032 0.053D. Determine Design Moments ( Coefficient Method ) 0.60 0.890 0.081 0.034 0.058
Continuous Edge 0.55 0.920 0.084 0.035 0.062Negative Moments: by linear interpolation 0.50 0.940 0.086 0.037 0.066
Ma = 0.750 0.0690
Mb = 0.779 x 0.0667 1.00 0.500 0.045 0.018 0.027
= 0.0667 kN 0.800 0.0650 0.95 0.450 0.041 0.016 0.025
= 0.0249 kN 0.90 0.400 0.037 0.014 0.022
Ma = 13.85 kN-m 0.750 0.0220 0.85 0.340 0.031 0.012 0.019Mb = 8.52 kN-m 0.779 x 0.0249 0.80 0.290 0.027 0.011 0.017
0.800 0.0270 0.75 0.240 0.022 0.009 0.0140.70 0.190 0.017 0.007 0.012
Positive Moments: by linear interpolation 0.65 0.150 0.014 0.006 0.010
= 0.750 0.0280 0.60 0.110 0.010 0.004 0.007
= 0.779 x 0.0268 0.55 0.080 0.007 0.003 0.006
= 0.0268 kN 0.800 0.0260 0.50 0.060 0.006 0.002 0.004
= 0.0427 kN
= 4.01 kN-m 0.750 0.0450
= 2.48 kN-m 0.779 x 0.0427
= 6.50 kN-m 0.800 0.0410
by linear interpolation
= 0.750 0.0090
= 0.779 x 0.0102
= 0.0102 kN 0.800 0.0110
= 0.0157 kN
= 2.50 kN-m 0.750 0.0140
= 1.51 kN-m 0.779 x 0.0157
= 4.01 kN-m 0.800 0.0170
E. Design Reinforcements:
> 0.65 NSCP Section 410.3.7.3
= 0.85 - 0.05( fc' - 30 ) = 0.85
7
= 0.914use:
= 0.85 > 0.65 ok!
NSCP Section 410.4.3 NSCP Section 410.6.1
==
fc'> =
1.4
= 600 4 fy fy
fy 600 + fy = 0.0042 < = 0.0051
= 0.0377 use:
2. For am greater than 2.0
hminln ( 0.80 + fy / 1500 )
36 + 9b
hmin
hmin
la/lb
Ca
CaWulalb Mu(+)DL Mu(+)LL
CbWulalb Ca
Ca
Cb
Cb
fVc 0.85 fc'CS(1000)(deff)fVc
CaWula2 Ca la/lb Cb
CbWulb2
Ca
Cb
Cb
MaDL CaDLWuDLla2 CaDL
MaLL CaLLWuLLla2
CaDL
CaLL
MaDL CaLL
MaLL
MaTOT
MbDL CbDLWuDLlb2 CbDL
MbLL CbLLWuLLlb2
CbDL
CbLL
MbDL CbLL
MbLL
MbTOT
Solve fo b1 :a.) if fc' > 30 Mpa ; b.) if fc' < 30 Mpa ;
b1b1
b1
b1
Solve for rmax : Solve for rmin :
rmax 0.75 rbalrmin rmin
rbal
0.85 b1 fc'
rmin rmin
rbal
document.xls
= 0.0282 = 0.0051
Middle StripShort Span Long Span
Top Bars Bot Bars Top Bars Bot BarsDesign Moments 13.85 6.50 8.52 4.01
Bar Diameter 12 12 12 12m = 0.847 fc' / fy 0.0644 0.0644 0.0644 0.0644
1.7791 0.8346 1.0946 0.5154
0.0068 0.0031 0.0041 0.0019
0.0068 0.0051 0.0051 0.0051
632.90 471.74 471.74 471.74113.10 113.10 113.10 113.10
178.70 239.75 239.75 239.75175.00 225.00 225.00 225.00
Column Strip Top Bars Bot Bars Top Bars Bot Bars
262.50 337.50 337.50 337.50250.00 250.00 250.00 250.00
E. Temperature / Shrinkage Reinforcements:As = 0.002bh b.Temp Reinf.
As = 250.00 NSCP Section 407.13
= 12.00 mm s < 5h = 625
= 113.10 s < 450spcg = 450.00 mm
rmax rmin
R = Mu / fbdeff2
rreqd = m - m2 - 2Rm/fy
rsupplied
Asreqd = rbdAs main bar q
spcgreqd
Provide (spcg)
spcgreqd = 3/2(spcgMID-STRIP)Provide (spcg)
mm2
qbar
Asbar mm2
document.xls
Design of Two-Way Slab with BeamsReference: NSCP 2001 Volume 1
Design of Reinforced Concrete by J.C. McCormacDesign of Concrete Structures by Arthur H. Nilson
= 21 Mpa b x h
= 21 Mpa 250 x 400fy = 276 Mpa A
= 0.90
= 4.35 kPa
= 1.92 kPa 4.00 m 200 x 400 C D 200 x 400
= 9.45 kPa
= 3.26 kPa B= 12.71 kPa 200 x 400
= 12 mmh = 100 mm
= 68 mm 4.00 m
= 0.477
= 2.40 kPa
2.00 m 2.00 m
Side A
= ==
= 21675.58 MPa = 21675.58 MPa
= = = 8.00
= 1.33E+09 = 1.67E+08Side B
= ==
= 21675.58 MPa = 21675.58 MPa
= = = 3.20
= 1.07E+09 = 3.33E+08Side C
= ==
= 21675.58 MPa = 21675.58 MPa
= = = 6.40
= 1.07E+09 = 1.67E+08Side D
= ==
= 21675.58 MPa = 21675.58 MPa
= = = ###
= 1.07E+09 = 8.33E+07
=4
= 7.60
B. Check Minimum Thickness = 3.775 m b = 2.097
= > 125 mm
= 125.00 mm
= > 90 mm
= 90.00 mm
fc'CS
fc'CB
fM
WDL
WLL
WuDL
WuLL
Wu
qbar
deff
la
lb
WSLAB
A. Solve for am
ECB 4730 fc'CB ECS 4730 fc'CSaA
ECBICB
ECB ECS ECSICS
ICB bh3 / 12 ICS bh3 / 12 aA
ICB mm4 ICS mm4
ECB 4730 fc'CB ECS 4730 fc'CSaB
ECBICB
ECB ECS ECSICS
ICB bh3 / 12 ICS bh3 / 12 aB
ICB mm4 ICS mm4
ECB 4730 fc'CB ECS 4730 fc'CSaC
ECBICB
ECB ECS ECSICS
ICB bh3 / 12 ICS bh3 / 12 aC
ICB mm4 ICS mm4
ECB 4730 fc'CB ECS 4730 fc'CSaD
ECBICB
ECB ECS ECSICS
ICB bh3 / 12 ICS bh3 / 12 aD
ICB mm4 ICS mm4
amaA + aB + aC + aD
am
ln
1. For am greater than 0.20 but not greater than 2.0
hmin
ln ( 0.80 + fy / 1500 )
36 + 5b ( am - 0.20 )
hmin
2. For am greater than 2.0
hminln ( 0.80 + fy / 1500 )
36 + 9b
hmin
document.xls
Therefore use: = 90.00 mm ok!hmin
document.xls
D. Check Slab on Shear ( Coefficient Method )
Va = by linear interpolation Vu Mu(-)
Vb = 0.750 0.7600 1.00 0.500 0.050 0.027 0.032
= 1.0332 kN 0.477 x 1.0332 0.95 0.550 0.055 0.030 0.035
= -0.0332 kN 0.800 0.7100 0.90 0.600 0.060 0.033 0.039Va = 89.26 kN 0.85 0.660 0.066 0.036 0.043Vb = -2.87 kN 0.750 0.2400 0.80 0.710 0.071 0.039 0.048
= 0.477 x -0.0332 0.75 0.760 0.076 0.043 0.052= 264.87 kN ok! 0.800 0.2900 0.70 0.810 0.081 0.046 0.057
0.65 0.850 0.085 0.050 0.062D. Determine Design Moments ( Coefficient Method ) 0.60 0.890 0.089 0.053 0.067
Continuous Edge 0.55 0.920 0.092 0.056 0.072Negative Moments: by linear interpolation 0.50 0.940 0.094 0.059 0.077
Ma = 0.750 0.0760
Mb = 0.477 x 0.1033 1.00 0.500 0.050 0.027 0.032
= 0.1033 kN 0.800 0.0710 0.95 0.450 0.045 0.024 0.029
= -0.0033 kN 0.90 0.400 0.040 0.022 0.026
Ma = 4.26 kN-m 0.750 0.0240 0.85 0.340 0.034 0.019 0.023Mb = -0.60 kN-m 0.477 x -0.0033 0.80 0.290 0.029 0.016 0.020
0.800 0.0290 0.75 0.240 0.024 0.013 0.0160.70 0.190 0.019 0.011 0.014
Positive Moments: by linear interpolation 0.65 0.150 0.015 0.009 0.011
= 0.750 0.0430 0.60 0.110 0.011 0.007 0.009
= 0.477 x 0.0649 0.55 0.080 0.008 0.005 0.007
= 0.0649 kN 0.800 0.0390 0.50 0.060 0.006 0.004 0.005
= 0.0739 kN
= 1.99 kN-m 0.750 0.0520
= 0.78 kN-m 0.477 x 0.0739
= 2.77 kN-m 0.800 0.0480
by linear interpolation
= 0.750 0.0130
= 0.477 x -0.0034
= -0.0034 kN 0.800 0.0160
= -0.0059 kN
= -0.46 kN-m 0.750 0.0160
= -0.27 kN-m 0.477 x -0.0059
= -0.73 kN-m 0.800 0.0200
E. Design Reinforcements:
> 0.65 NSCP Section 410.3.7.3
= 0.85 - 0.05( fc' - 30 ) = 0.85
7
= 0.914use:
= 0.85 > 0.65 ok!
NSCP Section 410.4.3 NSCP Section 410.6.1
==
fc'> =
1.4
= 600 4 fy fy
fy 600 + fy = 0.0042 < = 0.0051
= 0.0377 use:= 0.0282 = 0.0051
la/lb
Ca
CaWulalb Mu(+)DL Mu(+)LL
CbWulalb Ca
Ca
Cb
Cb
fVc 0.85 fc'CS(1000)(deff)fVc
CaWula2 Ca la/lb Cb
CbWulb2
Ca
Cb
Cb
MaDL CaDLWuDLla2 CaDL
MaLL CaLLWuLLla2
CaDL
CaLL
MaDL CaLL
MaLL
MaTOT
MbDL CbDLWuDLlb2 CbDL
MbLL CbLLWuLLlb2
CbDL
CbLL
MbDL CbLL
MbLL
MbTOT
Solve fo b1 :a.) if fc' > 30 Mpa ; b.) if fc' < 30 Mpa ;
b1b1
b1
b1
Solve for rmax : Solve for rmin :
rmax 0.75 rbalrmin rmin
rbal
0.85 b1 fc'
rmin rmin
rbal
rmax rmin
document.xls
REINFORCEMENTSShort Span Long Span
Cont. Discont. Cont. Discont.
Middle Strip Top Bars Bot Bars Top Bars Top Bars Bot Bars Top BarsDesign Moments 4.26 2.77 0.92 -0.60 -0.73 -0.24
Bar Diameter 12 12 12 12 12 12m = 0.847 fc' / fy 0.0644 0.0644 0.0644 0.0644 0.0644 0.0644
1.0227 0.6648 0.2216 -0.1444 -0.1752 -0.0584
0.0038 0.0025 0.0008 -0.0005 -0.0006 -0.0002
0.0051 0.0051 0.0051 0.0051 0.0051 0.0051
344.93 344.93 344.93 344.93 344.93 344.93113.10 113.10 113.10 113.10 113.10 113.10
327.89 327.89 327.89 327.89 327.89 327.89Provide (spcg) 200.00 200.00 200.00 200.00 200.00 200.00
Column Strip Top Bars Bot Bars Top Bars Top Bars Bot Bars Top Bars
300.00 300.00 300.00 300.00 300.00 300.00Provide (spcg) 200.00 200.00 200.00 200.00 200.00 200.00
E. Temperature / Shrinkage Reinforcements:As = 0.002bh b.Temp Reinf.
As = 200.00 NSCP Section 407.13
= 12.00 mm s < 5h = 500
= 113.10 s < 450spcg = 450.00 mm
R = Mu / fbdeff2
rreqd = m - m2 - 2Rm/fy
rsupplied
Asreqd = rbdAs main bar q
spcgreqd
spcgreqd = 3/2(spcgMID-STRIP)
mm2
qbar
Asbar mm2
document.xls
Design of Two-Way Slab with BeamsReference: NSCP 2001 Volume 1
Design of Reinforced Concrete by J.C. McCormacDesign of Concrete Structures by Arthur H. Nilson
= 21 Mpa b x h
= 21 Mpa 250 x 400fy = 276 Mpa A
= 0.90 long span
= 4.80 kPa
= 2.50 kPa 3.70 m 300 x 500 C D 300 x 500
= 10.92 kPa
= 4.25 kPa B= 15.17 kPa 250 x 400
= 12 mmh = 125 mm 4.00 m 4.00 m
= 93 mm
= 0.932
= 3.00 kPa
Side A
= ==
= 21675.58 MPa = 21675.58 MPa
= = = 4.43
= 1.33E+09 = 3.01E+08Side B
= ==
= 21675.58 MPa = 21675.58 MPa
= = = 4.43
= 1.33E+09 = 3.01E+08Side C
= ==
= 21675.58 MPa = 21675.58 MPa
= = = 4.80
= 3.13E+09 = 6.51E+08Side D
= ==
= 21675.58 MPa = 21675.58 MPa
= = = 9.60
= 3.13E+09 = 3.26E+08
=4
= 5.81
B. Check Minimum Thickness = 3.700 m b = 1.072
= > 125 mm
= 125.00 mm
= > 90 mm
= 90.00 mm
Therefore use: = 90.00 mm ok!
fc'CS
fc'CB
fM
WDL
WLL
WuDL
WuLL
Wu
qbar
deff
la
lb
WSLAB
A. Solve for am
ECB 4730 fc'CB ECS 4730 fc'CSaA
ECBICB
ECB ECS ECSICS
ICB bh3 / 12 ICS bh3 / 12 aA
ICB mm4 ICS mm4
ECB 4730 fc'CB ECS 4730 fc'CSaB
ECBICB
ECB ECS ECSICS
ICB bh3 / 12 ICS bh3 / 12 aB
ICB mm4 ICS mm4
ECB 4730 fc'CB ECS 4730 fc'CSaC
ECBICB
ECB ECS ECSICS
ICB bh3 / 12 ICS bh3 / 12 aC
ICB mm4 ICS mm4
ECB 4730 fc'CB ECS 4730 fc'CSaD
ECBICB
ECB ECS ECSICS
ICB bh3 / 12 ICS bh3 / 12 aD
ICB mm4 ICS mm4
amaA + aB + aC + aD
am
ln
1. For am greater than 0.20 but not greater than 2.0
hmin
ln ( 0.80 + fy / 1500 )
36 + 5b ( am - 0.20 )
hmin
2. For am greater than 2.0
hminln ( 0.80 + fy / 1500 )
36 + 9b
hmin
hmin
document.xls
D. Check Slab on Shear ( Coefficient Method )
Va = by linear interpolation Vu Mu(-)
Vb = 0.750 0.8800 1.00 0.710 0.071 0.033 0.035
= 0.8070 kN 0.932 x 0.8070 0.95 0.750 0.075 0.036 0.038
= 0.1930 kN 0.800 0.8600 0.90 0.790 0.079 0.039 0.042Va = 156.28 kN 0.85 0.830 0.083 0.042 0.046Vb = 37.37 kN 0.750 0.1200 0.80 0.860 0.086 0.045 0.051
= 0.932 x 0.1930 0.75 0.880 0.088 0.048 0.055= 362.25 kN ok! 0.800 0.1400 0.70 0.910 0.091 0.051 0.060
0.65 0.930 0.093 0.054 0.064D. Determine Design Moments ( Coefficient Method ) 0.60 0.950 0.095 0.056 0.068
Continuous Edge 0.55 0.960 0.096 0.058 0.073Negative Moments: by linear interpolation 0.50 0.970 0.097 0.061 0.078
Ma = 0.750 0.0880
= 0.0807 kN 0.932 x 0.0807 1.00 0.290 - 0.027 0.032Ma = 14.57 kN-m 0.800 0.0860 0.95 0.250 - 0.024 0.029
0.90 0.210 - 0.021 0.025Positive Moments: by linear interpolation 0.85 0.170 - 0.017 0.022
= 0.750 0.0480 0.80 0.140 - 0.015 0.019
= 0.932 x 0.0371 0.75 0.120 - 0.012 0.016
= 0.0371 kN 0.800 0.0450 0.70 0.090 - 0.009 0.013
= 0.0404 kN 0.65 0.070 - 0.007 0.010
= 4.82 kN-m 0.750 0.0550 0.60 0.050 - 0.006 0.008
= 2.04 kN-m 0.932 x 0.0404 0.55 0.040 - 0.004 0.006
= 6.86 kN-m 0.800 0.0510 0.50 0.030 - 0.003 0.005
by linear interpolation
= 0.750 0.0120
= 0.932 x 0.0229
= 0.0229 kN 0.800 0.0150
= 0.0269 kN
= 3.43 kN-m 0.750 0.0160
= 1.57 kN-m 0.932 x 0.0269
= 5.00 kN-m 0.800 0.0190
E. Design Reinforcements:
> 0.65 NSCP Section 410.3.7.3
= 0.85 - 0.05( fc' - 30 ) = 0.85
7
= 0.914use:
= 0.85 > 0.65 ok!
NSCP Section 410.4.3 NSCP Section 410.6.1
==
fc'> =
1.4
= 600 4 fy fy
fy 600 + fy = 0.0042 < = 0.0051
= 0.0377 use:= 0.0282 = 0.0051
la/lb
Ca
CaWulalb Mu(+)DL Mu(+)LL
CbWulalb Ca
Ca
Cb
Cb
fVc 0.85 fc'CS(1000)(deff)fVc
CaWula2 Ca la/lb Cb
Ca
MaDL CaDLWuDLla2 CaDL
MaLL CaLLWuLLla2
CaDL
CaLL
MaDL CaLL
MaLL
MaTOT
MbDL CbDLWuDLlb2 CbDL
MbLL CbLLWuLLlb2
CbDL
CbLL
MbDL CbLL
MbLL
MbTOT
Solve fo b1 :a.) if fc' > 30 Mpa ; b.) if fc' < 30 Mpa ;
b1b1
b1
b1
Solve for rmax : Solve for rmin :
rmax 0.75 rbalrmin rmin
rbal
0.85 b1 fc'
rmin rmin
rbal
rmax rmin
document.xls
REINFORCEMENTSShort Span Long Span
Cont. Discont. Discont.
Middle Strip Top Bars Bot Bars Top Bars Bot Bars Top BarsDesign Moments 14.57 6.86 2.29 5.00 1.67
Bar Diameter 12 12 12 12 12m = 0.847 fc' / fy 0.0644 0.0644 0.0644 0.0644 0.0644
1.8720 0.8813 0.2938 0.6421 0.2140
0.0072 0.0033 0.0011 0.0024 0.0008
0.0072 0.0051 0.0051 0.0051 0.0051
668.01 471.74 471.74 471.74 471.74113.10 113.10 113.10 113.10 113.10
169.31 239.75 239.75 239.75 239.75Provide (spcg) 150.00 225.00 225.00 225.00 225.00
Column Strip Top Bars Bot Bars Top Bars Bot Bars Top Bars
225.00 337.50 337.50 337.50 337.50Provide (spcg) 225.00 250.00 250.00 250.00 250.00
E. Temperature / Shrinkage Reinforcements:As = 0.002bh b.Temp Reinf.
As = 250.00 NSCP Section 407.13
= 12.00 mm s < 5h = 625
= 113.10 s < 450spcg = 450.00 mm
R = Mu / fbdeff2
rreqd = m - m2 - 2Rm/fy
rsupplied
Asreqd = rbdAs main bar q
spcgreqd
spcgreqd = 3/2(spcgMID-STRIP)
mm2
qbar
Asbar mm2
document.xls
Design of Two-Way Slab with BeamsReference: NSCP 2001 Volume 1
Design of Reinforced Concrete by J.C. McCormacDesign of Concrete Structures by Arthur H. Nilson
= 21 Mpa b x h
= 21 Mpa 250 x 400fy = 276 Mpa A
= 0.90
= 4.80 kPa 5.00 m
= 2.50 kPa 300 x 500 C D 300 x 500
= 10.92 kPa
= 4.25 kPa B= 15.17 kPa 250 x 400
= 12 mmh = 125 mm
= 93 mm 6.00 m short span
= 0.779
= 3.00 kPa
4.00 m
Side A
= ==
= 21675.58 MPa = 21675.58 MPa
= = = 3.28
= 1.33E+09 = 4.07E+08Side B
= ==
= 21675.58 MPa = 21675.58 MPa
= = = 1.49
= 1.33E+09 = 8.95E+08Side C
= ==
= 21675.58 MPa = 21675.58 MPa
= = = 9.60
= 3.13E+09 = 3.26E+08Side D
= ==
= 21675.58 MPa = 21675.58 MPa
= = = 9.60
= 3.13E+09 = 3.26E+08
=4
= 5.99
B. Check Minimum Thickness = 4.750 m b = 1.284
= > 125 mm
= 125.00 mm
= > 90 mm
= 100.00 mm
fc'CS
fc'CB
fM
WDL
WLL
WuDL
WuLL
Wu
qbar
deff
la
lb
WSLAB
A. Solve for am
ECB 4730 fc'CB ECS 4730 fc'CSaA
ECBICB
ECB ECS ECSICS
ICB bh3 / 12 ICS bh3 / 12 aA
ICB mm4 ICS mm4
ECB 4730 fc'CB ECS 4730 fc'CSaB
ECBICB
ECB ECS ECSICS
ICB bh3 / 12 ICS bh3 / 12 aB
ICB mm4 ICS mm4
ECB 4730 fc'CB ECS 4730 fc'CSaC
ECBICB
ECB ECS ECSICS
ICB bh3 / 12 ICS bh3 / 12 aC
ICB mm4 ICS mm4
ECB 4730 fc'CB ECS 4730 fc'CSaD
ECBICB
ECB ECS ECSICS
ICB bh3 / 12 ICS bh3 / 12 aD
ICB mm4 ICS mm4
amaA + aB + aC + aD
am
ln
1. For am greater than 0.20 but not greater than 2.0
hmin
ln ( 0.80 + fy / 1500 )
36 + 5b ( am - 0.20 )
hmin
2. For am greater than 2.0
hminln ( 0.80 + fy / 1500 )
36 + 9b
hmin
document.xls
Therefore use: = 100.00 mm ok!hmin
document.xls
D. Check Slab on Shear ( Coefficient Method )
Va = by linear interpolation Vu Mu(-)
Vb = 0.750 0.5600 1.00 0.290 - 0.027 0.032
= 0.5195 kN 0.779 x 0.5195 0.95 0.330 - 0.031 0.036
= 0.4805 kN 0.800 0.4900 0.90 0.380 - 0.035 0.040Va = 138.50 kN 0.85 0.430 - 0.040 0.045Vb = 128.11 kN 0.750 0.4400 0.80 0.490 - 0.045 0.051
= 0.779 x 0.4805 0.75 0.560 - 0.051 0.056= 362.25 kN ok! 0.800 0.5100 0.70 0.620 - 0.058 0.063
0.65 0.690 - 0.065 0.070D. Determine Design Moments ( Coefficient Method ) 0.60 0.760 - 0.073 0.077
Continuous Edge 0.55 0.810 - 0.081 0.085Negative Moments: by linear interpolation 0.50 0.860 - 0.089 0.092
Mb = 0.750 0.0440
= 0.0481 kN 0.779 x 0.0481 1.00 0.710 0.071 0.033 0.035Mb = 16.45 kN-m 0.800 0.0510 0.95 0.670 0.067 0.031 0.032
0.90 0.620 0.062 0.028 0.029Positive Moments: by linear interpolation 0.85 0.570 0.057 0.025 0.026
= 0.750 0.0510 0.80 0.510 0.051 0.022 0.023
= 0.779 x 0.0475 0.75 0.440 0.044 0.020 0.020
= 0.0475 kN 0.800 0.0450 0.70 0.380 0.038 0.017 0.017
= 0.0531 kN 0.65 0.310 0.031 0.014 0.014
= 7.10 kN-m 0.750 0.0560 0.60 0.240 0.024 0.012 0.011
= 3.09 kN-m 0.779 x 0.0531 0.55 0.190 0.019 0.009 0.009
= 10.19 kN-m 0.800 0.0510 0.50 0.140 0.014 0.007 0.007
by linear interpolation
= 0.750 0.0200
= 0.779 x 0.0212
= 0.0212 kN 0.800 0.0220
= 0.0217 kN
= 5.21 kN-m 0.750 0.0200
= 2.08 kN-m 0.779 x 0.0217
= 7.30 kN-m 0.800 0.0230
E. Design Reinforcements:
> 0.65 NSCP Section 410.3.7.3
= 0.85 - 0.05( fc' - 30 ) = 0.85
7
= 0.914use:
= 0.85 > 0.65 ok!
NSCP Section 410.4.3 NSCP Section 410.6.1
==
fc'> =
1.4
= 600 4 fy fy
fy 600 + fy = 0.0042 < = 0.0051
= 0.0377 use:= 0.0282 = 0.0051
la/lb
Ca
CaWulalb Mu(+)DL Mu(+)LL
CbWulalb Ca
Ca
Cb
Cb
fVc 0.85 fc'CS(1000)(deff)fVc
CbWulb2 Cb la/lb Cb
Cb
MaDL CaDLWuDLla2 CaDL
MaLL CaLLWuLLla2
CaDL
CaLL
MaDL CaLL
MaLL
MaTOT
MbDL CbDLWuDLlb2 CbDL
MbLL CbLLWuLLlb2
CbDL
CbLL
MbDL CbLL
MbLL
MbTOT
Solve fo b1 :a.) if fc' > 30 Mpa ; b.) if fc' < 30 Mpa ;
b1b1
b1
b1
Solve for rmax : Solve for rmin :
rmax 0.75 rbalrmin rmin
rbal
0.85 b1 fc'
rmin rmin
rbal
rmax rmin
document.xls
REINFORCEMENTSShort Span Long Span
Discont. Cont. Discont.
Middle Strip Bot Bars Top Bars Top Bars Bot Bars Top BarsDesign Moments 10.19 3.40 16.45 7.30 2.43
Bar Diameter 12 12 12 12 12m = 0.847 fc' / fy 0.0644 0.0644 0.0644 0.0644 0.0644
1.3097 0.4366 2.1129 0.9375 0.3125
0.0049 0.0016 0.0082 0.0035 0.0011
0.0051 0.0051 0.0082 0.0051 0.0051
471.74 471.74 760.17 471.74 471.74113.10 113.10 113.10 113.10 113.10
239.75 239.75 148.78 239.75 239.75Provide (spcg) 225.00 225.00 125.00 225.00 225.00
Column Strip Bot Bars Top Bars Top Bars Bot Bars Top Bars
337.50 337.50 187.50 337.50 337.50Provide (spcg) 250.00 250.00 175.00 250.00 250.00
E. Temperature / Shrinkage Reinforcements:As = 0.002bh b.Temp Reinf.
As = 250.00 NSCP Section 407.13
= 12.00 mm s < 5h = 625
= 113.10 s < 450spcg = 450.00 mm
R = Mu / fbdeff2
rreqd = m - m2 - 2Rm/fy
rsupplied
Asreqd = rbdAs main bar q
spcgreqd
spcgreqd = 3/2(spcgMID-STRIP)
mm2
qbar
Asbar mm2
document.xls
Design of Two-Way Slab with BeamsReference: NSCP 2001 Volume 1
Design of Reinforced Concrete by J.C. McCormacDesign of Concrete Structures by Arthur H. Nilson
= 21 Mpa
= 21 Mpafy = 276 Mpa
= 0.90 6.00 m
= 4.80 kPa b x h
= 2.50 kPa 300 x 500
= 10.92 kPa A
= 4.25 kPa long span= 15.17 kPa 6.00 m
= 12 mm 300 x 500 C D 300 x 500h = 125 mm
= 93 mm B
= 1.000300 x 500
= 3.00 kPa6.00 m
6.00 m 5.70 m
Side A
= ==
= 21675.58 MPa = 21675.58 MPa
= = = 3.20
= 3.13E+09 = 9.77E+08Side B
= ==
= 21675.58 MPa = 21675.58 MPa
= = = 3.20
= 3.13E+09 = 9.77E+08Side C
= ==
= 21675.58 MPa = 21675.58 MPa
= = = 6.40
= 3.13E+09 = 4.88E+08Side D
= ==
= 21675.58 MPa = 21675.58 MPa
= = = 3.28
= 3.13E+09 = 9.52E+08
=4
= 4.02
B. Check Minimum Thickness = 5.700 m b = 1.000
= > 125 mm
= 101.79 mm
fc'CS
fc'CB
fM
WDL
WLL
WuDL
WuLL
Wu
qbar
deff
la
lb
WSLAB
A. Solve for am
ECB 4730 fc'CB ECS 4730 fc'CSaA
ECBICB
ECB ECS ECSICS
ICB bh3 / 12 ICS bh3 / 12 aA
ICB mm4 ICS mm4
ECB 4730 fc'CB ECS 4730 fc'CSaB
ECBICB
ECB ECS ECSICS
ICB bh3 / 12 ICS bh3 / 12 aB
ICB mm4 ICS mm4
ECB 4730 fc'CB ECS 4730 fc'CSaC
ECBICB
ECB ECS ECSICS
ICB bh3 / 12 ICS bh3 / 12 aC
ICB mm4 ICS mm4
ECB 4730 fc'CB ECS 4730 fc'CSaD
ECBICB
ECB ECS ECSICS
ICB bh3 / 12 ICS bh3 / 12 aD
ICB mm4 ICS mm4
amaA + aB + aC + aD
am
ln
1. For am greater than 0.20 but not greater than 2.0
hmin
ln ( 0.80 + fy / 1500 )
36 + 5b ( am - 0.20 )
hmin
document.xls
= > 90 mm
= 124.64 mm
Therefore use: = 124.64 mm ok!
D. Check Slab on Shear ( Coefficient Method )
Va = by linear interpolation Vu Mu(-)
Vb = 1.000 0.3300 1.00 0.330 0.033 0.020 0.028
= 0.3300 kN 1.000 x 0.3300 0.95 0.380 0.038 0.022 0.031
= 0.6700 kN 0.950 0.3800 0.90 0.430 0.043 0.025 0.035Va = 162.65 kN 0.85 0.490 0.049 0.029 0.040Vb = 330.23 kN 1.000 0.6700 0.80 0.550 0.055 0.032 0.044
= 1.000 x 0.6700 0.75 0.610 0.061 0.036 0.049= 362.25 kN ok! 0.950 0.6200 0.70 0.680 0.068 0.040 0.054
0.65 0.740 0.074 0.044 0.059D. Determine Design Moments ( Coefficient Method ) 0.60 0.800 0.080 0.048 0.065
Continuous Edge 0.55 0.850 0.085 0.052 0.070Negative Moments: by linear interpolation 0.50 0.890 0.089 0.056 0.076
Ma = 1.000 0.0330
Mb = 1.000 x 0.0330 1.00 0.670 0.061 0.023 0.030
= 0.0330 kN 0.950 0.0380 0.95 0.620 0.056 0.021 0.027
= 0.0610 kN 0.90 0.570 0.052 0.019 0.024
Ma = 16.26 kN-m 1.000 0.0610 0.85 0.510 0.046 0.017 0.022Mb = 30.07 kN-m 1.000 x 0.0610 0.80 0.450 0.041 0.015 0.019
0.950 0.0560 0.75 0.390 0.036 0.013 0.0160.70 0.320 0.029 0.011 0.014
Positive Moments: by linear interpolation 0.65 0.260 0.024 0.009 0.011
= 1.000 0.0360 0.60 0.200 0.018 0.007 0.009
= 1.000 x 0.0360 0.55 0.150 0.014 0.005 0.007
= 0.0360 kN 0.950 0.0320 0.50 0.110 0.010 0.004 0.005
= 0.0490 kN
= 12.77 kN-m 1.000 0.0490
= 6.77 kN-m 1.000 x 0.0490
= 19.54 kN-m 0.950 0.0440
by linear interpolation
= 1.000 0.0130
= 1.000 x 0.0130
= 0.0130 kN 0.950 0.0150
= 0.0160 kN
= 4.61 kN-m 1.000 0.0160
= 2.21 kN-m 1.000 x 0.0160
= 6.82 kN-m 0.950 0.0190
E. Design Reinforcements:
> 0.65 NSCP Section 410.3.7.3
= 0.85 - 0.05( fc' - 30 ) = 0.85
7
= 0.914use:
= 0.85 > 0.65 ok!
NSCP Section 410.4.3 NSCP Section 410.6.1
==
fc'> =
1.4
= 600 4 fy fy
fy 600 + fy = 0.0042 < = 0.0051
= 0.0377 use:
2. For am greater than 2.0
hminln ( 0.80 + fy / 1500 )
36 + 9b
hmin
hmin
la/lb
Ca
CaWulalb Mu(+)DL Mu(+)LL
CbWulalb Ca
Ca
Cb
Cb
fVc 0.85 fc'CS(1000)(deff)fVc
CaWula2 Ca la/lb Cb
CbWulb2
Ca
Cb
Cb
MaDL CaDLWuDLla2 CaDL
MaLL CaLLWuLLla2
CaDL
CaLL
MaDL CaLL
MaLL
MaTOT
MbDL CbDLWuDLlb2 CbDL
MbLL CbLLWuLLlb2
CbDL
CbLL
MbDL CbLL
MbLL
MbTOT
Solve fo b1 :a.) if fc' > 30 Mpa ; b.) if fc' < 30 Mpa ;
b1b1
b1
b1
Solve for rmax : Solve for rmin :
rmax 0.75 rbalrmin rmin
rbal
0.85 b1 fc'
rmin rmin
rbal
document.xls
= 0.0282 = 0.0051rmax rmin
document.xls
REINFORCEMENTSShort Span Long Span
Cont. Discont. Cont.
Middle Strip Top Bars Bot Bars Top Bars Top Bars Bot BarsDesign Moments 16.26 19.54 6.51 30.07 6.82
Bar Diameter 12 12 12 12 12m = 0.847 fc' / fy 0.0644 0.0644 0.0644 0.0644 0.0644
2.0895 2.5101 0.8367 3.8624 0.8764
0.0081 0.0098 0.0031 0.0160 0.0033
0.0081 0.0098 0.0051 0.0160 0.0051
751.14 915.74 471.74 1485.57 471.74113.10 113.10 113.10 113.10 113.10
150.57 123.50 239.75 76.13 239.75Provide (spcg) 150.00 100.00 225.00 75.00 225.00
Column Strip Top Bars Bot Bars Top Bars Top Bars Bot Bars
225.00 150.00 337.50 112.50 337.50Provide (spcg) 225.00 150.00 250.00 100.00 250.00
E. Temperature / Shrinkage Reinforcements:As = 0.002bh b.Temp Reinf.
As = 250.00 NSCP Section 407.13
= 12.00 mm s < 5h = 625
= 113.10 s < 450spcg = 450.00 mm
R = Mu / fbdeff2
rreqd = m - m2 - 2Rm/fy
rsupplied
Asreqd = rbdAs main bar q
spcgreqd
spcgreqd = 3/2(spcgMID-STRIP)
mm2
qbar
Asbar mm2
document.xls
Design of Two-Way Slab with BeamsReference: NSCP 2001 Volume 1
Design of Reinforced Concrete by J.C. McCormacDesign of Concrete Structures by Arthur H. Nilson
= 21 Mpa b x h
= 21 Mpa 250 x 400 short spanfy = 276 Mpa A
= 0.90
= 4.80 kPa
= 2.50 kPa 300 x 500 C D 300 x 500 5.00 m
= 10.92 kPa
= 4.25 kPa= 15.17 kPa
= 12 mm Bh = 125 mm 250 x 400
= 93 mm 5.00 m
= 0.779
= 3.00 kPa
5.00 m 4.00 m 3.00 m
Side A
= ==
= 21675.58 MPa = 21675.58 MPa
= = = 3.28
= 1.33E+09 = 4.07E+08Side B
= ==
= 21675.58 MPa = 21675.58 MPa
= = = 1.64
= 1.33E+09 = 8.14E+08Side C
= ==
= 21675.58 MPa = 21675.58 MPa
= = = 4.27
= 3.13E+09 = 7.32E+08Side D
= ==
= 21675.58 MPa = 21675.58 MPa
= = = 5.49
= 3.13E+09 = 5.70E+08
=4
= 3.67
B. Check Minimum Thickness = 4.750 m b = 1.284
= > 125 mm
= 125.00 mm
= > 90 mm
= 100.00 mm
fc'CS
fc'CB
fM
WDL
WLL
WuDL
WuLL
Wu
qbar
deff
la
lb
WSLAB
A. Solve for am
ECB 4730 fc'CB ECS 4730 fc'CSaA
ECBICB
ECB ECS ECSICS
ICB bh3 / 12 ICS bh3 / 12 aA
ICB mm4 ICS mm4
ECB 4730 fc'CB ECS 4730 fc'CSaB
ECBICB
ECB ECS ECSICS
ICB bh3 / 12 ICS bh3 / 12 aB
ICB mm4 ICS mm4
ECB 4730 fc'CB ECS 4730 fc'CSaC
ECBICB
ECB ECS ECSICS
ICB bh3 / 12 ICS bh3 / 12 aC
ICB mm4 ICS mm4
ECB 4730 fc'CB ECS 4730 fc'CSaD
ECBICB
ECB ECS ECSICS
ICB bh3 / 12 ICS bh3 / 12 aD
ICB mm4 ICS mm4
amaA + aB + aC + aD
am
ln
1. For am greater than 0.20 but not greater than 2.0
hmin
ln ( 0.80 + fy / 1500 )
36 + 5b ( am - 0.20 )
hmin
2. For am greater than 2.0
hminln ( 0.80 + fy / 1500 )
36 + 9b
hmin
document.xls
Therefore use: = 100.00 mm ok!hmin
document.xls
D. Check Slab on Shear ( Coefficient Method )
Va = by linear interpolation Vu Mu(-)
Vb = 0.750 0.8600 1.00 0.670 0.061 0.023 0.030
= 0.8426 kN 0.779 x 0.8426 0.95 0.710 0.065 0.024 0.032
= 0.1574 kN 0.800 0.8300 0.90 0.750 0.068 0.026 0.036Va = 224.66 kN 0.85 0.790 0.072 0.028 0.039Vb = 41.96 kN 0.750 0.1400 0.80 0.830 0.075 0.029 0.042
= 0.779 x 0.1574 0.75 0.860 0.078 0.031 0.046= 362.25 kN ok! 0.800 0.1700 0.70 0.890 0.081 0.033 0.050
0.65 0.920 0.083 0.034 0.054D. Determine Design Moments ( Coefficient Method ) 0.60 0.940 0.085 0.036 0.059
Continuous Edge 0.55 0.950 0.086 0.037 0.063Negative Moments: by linear interpolation 0.50 0.970 0.088 0.038 0.067
Ma = 0.750 0.0780
Mb = 0.779 x 0.0763 1.00 0.330 0.033 0.020 0.028
= 0.0763 kN 0.800 0.0750 0.95 0.290 0.029 0.017 0.025
= 0.0157 kN 0.90 0.250 0.025 0.015 0.022
Ma = 15.84 kN-m 0.750 0.0140 0.85 0.210 0.021 0.013 0.020Mb = 5.39 kN-m 0.779 x 0.0157 0.80 0.170 0.017 0.010 0.017
0.800 0.0170 0.75 0.140 0.014 0.007 0.0130.70 0.110 0.011 0.006 0.011
Positive Moments: by linear interpolation 0.65 0.080 0.008 0.005 0.009
= 0.750 0.0310 0.60 0.060 0.006 0.004 0.007
= 0.779 x 0.0298 0.55 0.050 0.005 0.003 0.006
= 0.0298 kN 0.800 0.0290 0.50 0.030 0.003 0.002 0.004
= 0.0437 kN
= 4.46 kN-m 0.750 0.0460
= 2.54 kN-m 0.779 x 0.0437
= 7.00 kN-m 0.800 0.0420
by linear interpolation
= 0.750 0.0070
= 0.779 x 0.0087
= 0.0087 kN 0.800 0.0100
= 0.0153 kN
= 2.15 kN-m 0.750 0.0130
= 1.47 kN-m 0.779 x 0.0153
= 3.62 kN-m 0.800 0.0170
E. Design Reinforcements:
> 0.65 NSCP Section 410.3.7.3
= 0.85 - 0.05( fc' - 30 ) = 0.85
7
= 0.914use:
= 0.85 > 0.65 ok!
NSCP Section 410.4.3 NSCP Section 410.6.1
==
fc'> =
1.4
= 600 4 fy fy
fy 600 + fy = 0.0042 < = 0.0051
= 0.0377 use:
= 0.0282 = 0.0051
la/lb
Ca
CaWulalb Mu(+)DL Mu(+)LL
CbWulalb Ca
Ca
Cb
Cb
fVc 0.85 fc'CS(1000)(deff)fVc
CaWula2 Ca la/lb Cb
CbWulb2
Ca
Cb
Cb
MaDL CaDLWuDLla2 CaDL
MaLL CaLLWuLLla2
CaDL
CaLL
MaDL CaLL
MaLL
MaTOT
MbDL CbDLWuDLlb2 CbDL
MbLL CbLLWuLLlb2
CbDL
CbLL
MbDL CbLL
MbLL
MbTOT
Solve fo b1 :a.) if fc' > 30 Mpa ; b.) if fc' < 30 Mpa ;
b1b1
b1
b1
Solve for rmax : Solve for rmin :
rmax 0.75 rbalrmin rmin
rbal
0.85 b1 fc'
rmin rmin
rbal
rmax rmin
document.xls
REINFORCEMENTSShort Span Long Span
Cont. Cont. Discont.
Middle Strip Top Bars Bot Bars Top Bars Bot Bars Top BarsDesign Moments 15.84 7.00 5.39 3.62 1.21
Bar Diameter 12 12 12 12 12m = 0.847 fc' / fy 0.0644 0.0644 0.0644 0.0644 0.0644
2.0347 0.8996 0.6920 0.4652 0.1551
0.0079 0.0033 0.0026 0.0017 0.0006
0.0079 0.0051 0.0051 0.0051 0.0051
730.06 471.74 471.74 471.74 471.74113.10 113.10 113.10 113.10 113.10
154.91 239.75 239.75 239.75 239.75150.00 225.00 225.00 225.00 225.00
Column Strip Top Bars Bot Bars Top Bars Bot Bars Top Bars
225.00 337.50 337.50 337.50 337.50225.00 250.00 250.00 250.00 250.00
E. Temperature / Shrinkage Reinforcements:As = 0.002bh b.Temp Reinf.
As = 250.00 NSCP Section 407.13
= 12.00 mm s < 5h = 625
= 113.10 s < 450spcg = 450.00 mm
R = Mu / fbdeff2
rreqd = m - m2 - 2Rm/fy
rsupplied
Asreqd = rbdAs main bar q
spcgreqd
Provide (spcg)
spcgreqd = 3/2(spcgMID-STRIP)Provide (spcg)
mm2
qbar
Asbar mm2
