flat slab
DESCRIPTION
this is the spreed sheet prepared in excell in checking shear action in flat slab desingTRANSCRIPT
SLAB DESIGN WORKSHEETTwo-Way Post-Tensioned Flat Plate (slab without beams)Project:Designer:Date:
All units are in Newtons and meters unless otherwise indicated
The following are the assumptions of this worksheet:1. All spans (long and short direction) are uniform length.2. All columns have uniform dimensions (long and short direction).
5. Slab is of uniform thickness.6. Direct design method is applicable.
Fixed inputs:
3. There may or may not be any overhang (Dlong and/or Dshort may or may not be zero)
4. There may or may not be any wall load (Wwall may or may not be zero).
Span in long direction, Llong =
Llong
CornerSlab
EdgeSlab 1
L sho
rtL s
hort
Dlong
Dsh
ort
Shor
tD
irecti
on
A
1
2
3
Floor to floor height, H =
426.409.3.2.1
426.409.3.2.3
Superimposed dead load:
426.409.2.1
Design inputs:Slab thickness, t =
204-1
Clear span in long direction Lnlong = Llong-clong =
Overhang in long direction, Dlong =
Span in short direction, Lshort =
Clear span in short direction, Lnshort = Lshort-cshort =
Overhang in short direction, Dshort =
Column width in long direction, clong =
Column width in short direction, cshort =
Capacity reduction factor for flexure fb =
Capacity reduction factor for shear fv =
Live load, wLL =
Floor finish, waterproofing, etc., wDL1 =
Other dead loads, wDL2 =
Exterior wall load, Wwall (N/m) =
Load Factor for Dead Load, LFDL =
Load Factor for Live Load, LFLL =
Unit weight of concrete gconc =
Concrete strength, f'c =
Steel yield strength, fy =
m= fy/0.85*f'c =
Page 1 of 6
All units are in Newtons and meters unless otherwise indicated
1. All spans (long and short direction) are uniform length.2. All columns have uniform dimensions (long and short direction).
7.200
long and/or Dshort may or may not be zero)
wall may or may not be zero).
Llong
EdgeSlab 2
InteriorSlab
LongDirection
cshort
clong
B C
6.750
0.000
7.200
6.750
0.000
0.450
0.4504.500
0.900
0.850
4,800
3,120
1,200
7,800
1.40
1.70
0.180
23,600
12,860,000
309,000,000
28.268
nlong = Llong-clong =
nshort = Lshort-cshort =
SLAB DESIGN WORKSHEETTwo-Way Post-Tensioned Flat Plate (slab without beams)Project:Designer:Date:
409.6.3.2 Minimum slab thickness:
Table 409-3 Minimum slab thicknessYield Strength Exterior Panels Interior Panels280,000,000 0.205 0.188309,000,000 0.209 0.191415,000,000 0.225 0.205
N.A. 0.000 0.000520,000,000 0.241 0.218
Uniform load on slab:Dead Load:
Live Load =
Direct design method criteria:413.7.1.1 1. There are a minimum of three continous spans in each direction.413.7.1.2 2. Panels are rectangular, with ratio of longer to shorter span center-to-center support within a
panel not greater than 2.413.7.1.3 3. Successive span lengths center-to-center supports in each direction do not differ by more than
one-third the longer span.413.7.1.4 4. Offsets of columns are a minimum of 10% of the span in direction of offset from either axis
between center lines of successive columns.413.7.1.5 5. All loads shall be due to gravity loads only and distributed over the entire panel.413.7.1.5 6. Live load shall not exceed two times the dead load.413.7.1.6 7. The relative stiffness of beams in two perpendicular directions
shall not be less than 0.2 nor greater than 5.0.
Minimum thickness, exterior panel, tminext =
Minimum thickness, interior panel, tminint =
Minimum thickness, tmin = smaller of tminext or tminint =
Capacity/Demand Ratio for Slab Thickness = t/tmin =
Slab weight, wDLslab = t*gconc =
Floor finish, waterproofing, etc., wDL1 =
Other dead loads, wDL2 =
wDL = wDLslab+wDLothers =
Total load, wtotal = LFLL*wLL+LFDL*wDL =
(af1*L22)/af2*L1
2)
Page 2 of 6
0.209
0.191
0.191
0.94
4,248
3,120
1,200
8,5684,800
20,155
1. There are a minimum of three continous spans in each direction.2. Panels are rectangular, with ratio of longer to shorter span center-to-center support within a
3. Successive span lengths center-to-center supports in each direction do not differ by more than
4. Offsets of columns are a minimum of 10% of the span in direction of offset from either axis
5. All loads shall be due to gravity loads only and distributed over the entire panel.
7. The relative stiffness of beams in two perpendicular directions
SLAB DESIGN WORKSHEETTwo-Way Post-Tensioned Flat Plate (slab without beams)Project:Designer:Date:
All units are in Newtons and meters unless otherwise indicated
Moment analysis:Slab strips in long direction:
Slab strips in short direction:
Column strip
Half middle strip
Column strip
3
2
1
Hal
f mid
dle
strip
Colu
mn
strip
Colu
mn
strip
3
2
1
A B
A B
Half middle strip
Half middle strip
Shor
tD
irecti
on
Hal
f mid
dle
strip
Hal
f mid
dle
strip
Colu
mn
strip
Colu
mn
strip
3
Shor
tD
irecti
on
Hal
f mid
dle
strip
Page 1 of 6
All units are in Newtons and meters unless otherwise indicated
Column strip
Half middle strip
Column strip
C
C
0.50*Lshort+0.5*cshort+Dshort
Half middle strip
Half middle strip
L sho
rt
LongDirection
Hal
f mid
dle
strip
Dsh
ort
0.5*
L sho
rt0.5*cshort
Hal
f mid
dle
strip
SLAB DESIGN WORKSHEETTwo-Way Post-Tensioned Flat Plate (slab without beams)Project:Designer:Date:
413.7.2.2 Total factored static moment:Slab strips in long direction:
Slab strips in short direction:
Wall load:
413.7.3.3 Calculation of longitudinal moments
Frame 1
439,072
-M at exterior support 114,159+M at exterior span 0.52*Mo 228,318-M at first interior support 0.70*Mo 307,351-M at typical interior support 0.65*Mo 285,397+M at typical interior support 0.35*Mo 153,675
413.7.4.2 Percentage of exterior negative moment going to column strip:
Frames 1 and 2:
0.500 1.000 1.000
0 100.00% 100.00%0.09 99.07% 99.07% 99.07%
2.5 75.00% 75.00% 75.00%
Column strip %, exterior negative moment
Frames 1 & 2 1.000 0.000 0.000Frames A & B 1.000 0.000 0.000
Torsional constant C:In long direction:
In short direction:
Slab strip 1: Mo1 = wtotal*(0.5*Lshort+0.5*cshort+Dshort)*Lnlong2/8 =
Slab strip 2: Mo2 = wtotal*Lshort*Lnlong2/8 =
Slab strip A: MoA = wtotal*(0.5*Llong+0.5*clong*Dlong)*Lnshort2/8 =
Slab strip B: MoB = wtotal*Llong*Lnshort2/8 =
In long direction: Mow1 = LFDL*Wwall*Lnlong2/8 =
In short direction: MowA = LFDL*Wwall*Lnshort2/8 =
Mo1
Mo
0.26*Mo
L2/L1
bt =
bt >
L2/L1 a1 a1*(L2/L1)
Clong = (1-0.63*t/cshort)*(t3*cshort/3) =
Cshort = (1-0.63*t/clong)*(t3*clong/3) =
cshort t
cclong t
413.7.4.1 Percentage of interior negative moments going to column strip =
413.7.4.4 Percentage of positive moment going to column strip =
Moments in column strip and middle strip slabs:
Totalmoment
Frame 1 Exterior span -Mext 114,159+M 228,318
-Mint 307,351Interior span -M 285,397
+M 153,675Frame 2 Exterior span -Mext 214,887
+M 429,774-Mint 578,542
Interior span -M 537,218+M 289,271
Frame A Exterior span -Mext 114,159+M 228,318
-Mint 307,351Interior span -M 285,397
+M 153,675Frame B Exterior span -Mext 214,887
+M 429,774-Mint 578,542
Interior span -M 537,218+M 289,271
439,072.37
826,489.17
439,072.37
826,489.17
62,192.81
62,192.81
Frame 2 Frame A Frame B Wall 1 Wall A
826,489 439,072 826,489 62,193 62,193
214,887 114,159 214,887 16,170 16,170429,774 228,318 429,774 32,340 32,340578,542 307,351 578,542 43,535 43,535537,218 285,397 537,218 40,425 40,425289,271 153,675 289,271 21,767 21,767
Frames A and B:
1.000 1.000 2.000
0 100.00% 100.00%0.09 99.07% 99.07% 99.07%
2.5 75.00% 75.00% 75.00%
Column strip %, exterior negative moment
C %0.00065 0.00350 0.094 99.07%0.00065 0.00350 0.094 99.07%
0.000654
0.000654
short+Dshort)*Lnlong2/8 =
*Dlong)*Lnshort2/8 =
Mo2 MoA MoB MoW1 MoWA
L2/L1
bt =
bt >
Is bt
= (1-0.63*t/cshort)*(t3*cshort/3) =
= (1-0.63*t/clong)*(t3*clong/3) =
Percentage of interior negative moments going to column strip = 75.00%
60.00%
% momnt Moment in Moment Momentto column column strip in column in middlestrip slab slab wall strip slab strip slab99.07% 113,091 16,170 129,262 1,06760.00% 136,991 32,340 169,331 91,32775.00% 230,513 43,535 274,048 76,83875.00% 214,048 40,425 254,473 71,34960.00% 92,205 21,767 113,973 61,47099.07% 212,878 212,878 2,00960.00% 257,865 257,865 171,91075.00% 433,907 433,907 144,63675.00% 402,913 402,913 134,30460.00% 173,563 173,563 115,70899.07% 113,091 16,170 129,262 1,06760.00% 136,991 32,340 169,331 91,32775.00% 230,513 43,535 274,048 76,83875.00% 214,048 40,425 254,473 71,34960.00% 92,205 21,767 113,973 61,47099.07% 212,878 212,878 2,00960.00% 257,865 257,865 171,91075.00% 433,907 433,907 144,63675.00% 402,913 402,913 134,30460.00% 173,563 173,563 115,708
SLAB DESIGN WORKSHEETTwo-Way Post-Tensioned Flat Plate (slab without beams)Project:Designer:Date:
Shear analysis:Edge column B1:
411.13.2.1
a1 = 1.06*Llong =a2 = Dshort+0.5*cshort+0.44*Lshort =a3 = clong+t =a4 = Dshort+cshort+0.5*t =a5 = (Aac*0.5*a4+Abd*0.5*a4)/(Acd+Aac+Abd) =
Aac = Abd = a4*t =Acd = a3*t =
a6 = 0.5*a4-a5 =
Shear capacity f*Vc:bo = 2*a4+a3 =Vc1 = (1/6)*[1+2/(clong/cshort)]*sqrt(f'c/1,000,000)*bo*t*1,000,000 =Vc2 = (1/12)*(as*t/b0+2)*sqrt(f'c/1,000,000)*bo*t*1,000,000 =
For edge column, as = 30Vc3 = (1/3)*sqrt(f'c/1,000,000)*bo*t*1,000,000 =Vc = smallest of Vc1, Vc2 or Vc3 =fv*Vc =fv*vc = fv*Vc/(bo*t) =
C
0.56*Llong
a1 = 1.06*Llong
Z
W
W
a3 = clong+t
Centroid ofshear perimeter
a
c
Shor
t dire
ction
B
1
Moment about axis Z-Z to be transferred to the column:413.7.3.6
413.6.3.2
413.6.3.3
Capacity/Demand Ratio for flexure, moment about axis Z-Z:
413.6.3.2
Slab adequate in bending if C/D > 1.0Shear stress due to moment about axis Z-Z transferred by shear:
Moment about axis W-W to be transferred to the column (from column line 1 with LL on span A-B only):
413.6.3.2
413.6.3.3
Capacity/Demand Ratio for flexure, moment about axis W-W:
Direct shear due to slab and wall loads, Vudirect:Vuslab = wtotal*[(a1*a2-a3*a4] =Vuwall = LFDL*Wwall*(a1-a3) =Vudirect = Vuslab+Vuwall =Vudirect/(fv*Vc) =
Direct shear stress due to slab and wall loads, vudirect:vudirecta = vudirectb = vudirectc = vudirectd = Vudirect/(bo*t) =
The gravity load moment to be transferred between slab and edge column shall be 0.3*Mo.MZZ = 0.30*MoB =The fraction of the unbalanced moment given by gf*Mu shall be considered to be transferred by flexuregfZZ1 = 1/[1+(2/3)*sqrt(a4/a3)]For edge columns with unbalanced moments about an axis parallel to the edge, gf = 1.0 provided thatVu at an edge support does not exceed 0.75*fv*Vc
gfZZ = gfaZZ1 if Vudirect/(fv*Vc)>0.75; gfZZ = 100% if Vudirect/(fv*Vc)<0.75
Moment be transferred by flexure MZZb = gfZZ*Mzz =Effective width for flexure = beff = clong+3*t =Moment capacity Mcap = beff*MCSshort =C/D Ratio in bending = Mcap/MZZb =
Moment to be transferred by shear MZZv = (1-gfZZ)*Mzz =vuZZc = vuZZd = MZZv*a5/JZZ =vuZZa = vuzzb = -Mzzv*(a4-a5)/JZZ =
JZZ = Jac+Jbd+Jcd = Jac = Jbd = Ixac+Iyac =
Ixac = a4*t3/12 =Iyac = a4
3*t/12+(a4*t)*a62 =
Jcd = (a3*t)*a52 =
Unbalanced moment from slab, MWWslab:Negative moment at first interior support M1 = 0.70*Mo1 =Negative moment at typical interior support M2 = 0.65*M01 =
Negative moment at typical interior support, dead load only, M3 = (LFDL*wDL/wtotal)*M2 =MWWslab = M1-M3 =
Unbalanced moment from wall, MWWwall:Negative moment at first interior support M1 = 0.70*Mow1 =Negative moment at typical interior support M2 = 0.65*Mow1 =MWWwall = M1-M2 =
MWW = MWWslab+MWWwall =The fraction of the unbalanced moment given by gf*Mu shall be considered to be transferred by flexuregfaWW = 1/[1+(2/3)*sqrt(a3/a4)]For edge columns with unbalanced moments about an axis transverse to the edge, increase gf to as much as1.25 times the value but not more than 1.0 provided that Vu at the support does not exceed 0.40*fv*Vc.Percentage of moment transferred by flexure, gfWW = gfaWW if Vudirect/Vc>0.40; gf = 1.25*gfaWW if Vudirect/Vc<0.40
Moment be transferred by flexure MWWb = gfWW*MWW =
Slab adequate in bending if C/D > 1.0Shear stress due to moment about axis W-W transferred by shear:
Shear stresses:
Critical shear stress:
Capacity/Demand Ratio for shear:
Slab adequate in shear if C/D ratio > 1.0
Effective width for flexure = beff = shorter of cshort+3*t or cshort+1.5*t+Dshort
cshort+3*t =cshort+1.5*t+Dshort =
Moment capacity Mcap = beff*MCSlong =C/D Ratio = Mcap/MWWb =
Moment to be transferred by shear MWWv = (1-gfWW)*MWW =vuWWa = vuWWc = MWWv*0.5*a3/JWW =vuWWb = vuWWd = -vuWWa =
Jww = Jcd+Jac+Jbd =Jcd = Ixcd+Iycd =
Ixcd = a3*t3/12 =Iycd = a3
3*t/12 =Jac = Jbd = a4*t*(0.5*a3)2 =
vua = vudirecta+vuZZa+vuWWa =vub = vudirectb+vuZZv+vuWWb =vuc = vudirectc+vuZZc+vuWWc =vud = vudirectd+vuZZd+vuWWd =
vu = largest of absolute values of vua, vub, vuc or vud =
C/D Ratio = fv*vc/vu =
Page 6 of 6
7.6323.3930.6300.5400.1710.0970.1130.099
1.71551,898.36474,438.94
367,932.24367,932.24312,742.40
1,016,057.19
/1,000,000)*bo*t*1,000,000 =/1,000,000)*bo*t*1,000,000 =
0.50*Llong
a1 = 1.06*Llong
Dsh
ort
0.5*cshort
0.44
*Lsh
ort
a 2 =
0.4
4*L s
hort+0
.5*c
shor
t+D
shor
t
Z
W
W
a3 = clong+t
a 4 =
0.5
*t+c
shor
t+D
shor
t
0.5*
a 4
a 5a 6
b
d
Long direction
B
515,069.6876,461.84
591,531.521.89
1,921,804.82Moment about axis Z-Z to be transferred to the column:
247,946.75
61.83%
61.83%Capacity/Demand Ratio for flexure, moment about axis Z-Z:
153,317.270.99
0.00Not adequate
Shear stress due to moment about axis Z-Z transferred by shear:94,629.48
1,541,246.89-3,339,368.25
0.010470.003590.000260.003320.00330
Moment about axis W-W to be transferred to the column (from column line 1 with LL on span A-B only):
307,351285,397169,852137,499
43,53540,425
3,110140,609
58.14%
58.14%Capacity/Demand Ratio for flexure, moment about axis W-W:
81,745.29
Direct shear stress due to slab and wall loads, vudirect:
The gravity load moment to be transferred between slab and edge column shall be 0.3*Mo.
The fraction of the unbalanced moment given by gf*Mu shall be considered to be transferred by flexure
For edge columns with unbalanced moments about an axis parallel to the edge, gf = 1.0 provided that
= 100% if Vudirect/(fv*Vc)<0.75
= (1-gfZZ)*Mzz =
Negative moment at first interior support M1 = 0.70*Mo1 =Negative moment at typical interior support M2 = 0.65*M01 =
Negative moment at typical interior support, dead load only, M3 = (LFDL*wDL/wtotal)*M2 =
Negative moment at first interior support M1 = 0.70*Mow1 =Negative moment at typical interior support M2 = 0.65*Mow1 =
The fraction of the unbalanced moment given by gf*Mu shall be considered to be transferred by flexure
For edge columns with unbalanced moments about an axis transverse to the edge, increase gf to as much as1.25 times the value but not more than 1.0 provided that Vu at the support does not exceed 0.40*fv*Vc.Percentage of moment transferred by flexure, gfWW = gfaWW if Vudirect/Vc>0.40; gf = 1.25*gfaWW if Vudirect/Vc<0.40
fWW*MWW =
0.7200.9900.720
0.00Not adequate
Shear stress due to moment about axis W-W transferred by shear:58,863.33
794,216.20-794,216.20
0.023350.004060.000310.003750.00964
-623,347.23-2,211,779.644,257,267.912,668,835.50
4,257,267.91
0.24Not adequate
= shorter of cshort+3*t or cshort+1.5*t+Dshort
= (1-gfWW)*MWW =
uc or vud =
474,438.94
-22117804257267.91
Criti
cal s
ectio
n
0.50
*Lsh
ort
Criti
cal s
ectio
n
0.56*Llong-0.5*clong-dave0.56
*Lsh
ort
0.50
*Lsh
ort
clong+dave0.50*Lshort
Llong
0.50*Llong0.56*Lshort
clong+dave
cshort+daveCriticalSection
0.50*Lshort
0.56*Llong
SLAB DESIGN WORKSHEETTwo-Way Post-Tensioned Flat Plate (slab without beams)Project:Designer:Date:
Edge column A2:
411.13.2.1
a1 = 1.06*Lshort =a2 = Dlong+0.5*clong+0.44*Llong =a3 = cshort+t =a4 = Dlong+clong+0.5*t =a5 = (Aac*0.5*a4+Abd*0.5*a4)/(Acd+Aac+Abd) =
Aac = Abd = a4*t =Acd = a3*t =
a6 = 0.5*a4-a5 =
Shear capacity f*Vc:bo = 2*a4+a3 =Vc1 = (1/6)*[1+2/(clong/cshort)]*sqrt(f'c/1,000,000)*bo*t*1,000,000 =Vc2 = (1/12)*(as*t/b0+2)*sqrt(f'c/1,000,000)*bo*t*1,000,000 =
Dlong 0.44*Llong
a2 = Dlong+0.5*clong+0.44*Llong
ZZ
W W
a4 = Dlong+clong+0.5*t
a 3 =
csh
ort+
t
Cent
roid
of
shea
r per
imet
er
Shor
t di
recti
onA
2
0.5*clong
0.5*a4
a6 a5
ab
cd
Moment about axis Z-Z to be transferred to the column:413.7.3.6
413.6.3.2
413.6.3.3
Capacity/Demand Ratio for flexure, moment about axis Z-Z:
413.6.3.2
Slab adequate in bending if C/D > 1.0Shear stress due to moment about axis Z-Z transferred by shear:
Moment about axis W-W to be transferred to the column (from column line A with LL on span 1-2 only):
413.6.3.2
For edge column, as = 30Vc3 = (1/3)*sqrt(f'c/1,000,000)*bo*t*1,000,000 =Vc = smallest of Vc1, Vc2 or Vc3 =fv*Vc =fv*vc = fv*Vc/(bo*t) =
Direct shear due to slab and wall loads, Vudirect:Vuslab = wtotal*[(a1*a2-a3*a4] =Vuwall = LFDL*Wwall*(a1-a3) =Vudirect = Vuslab+Vuwall =Vudirect/(fv*Vc) =
Direct shear stress due to slab and wall loads, vudirect:vudirecta = vudirectb = vudirectc = vudirectd = Vudirect/(bo*t) =
The gravity load moment to be transferred between slab and edge column shall be 0.3*Mo.MZZ = 0.30*Mo2 =The fraction of the unbalanced moment given by gf*Mu shall be considered to be transferred by flexuregfZZ1 = 1/[1+(2/3)*sqrt(a4/a3)]For edge columns with unbalanced moments about an axis parallel to the edge, gf = 1.0 provided thatVu at an edge support does not exceed 0.75*fv*Vc
gfZZ = gfaZZ1 if Vudirect/(fv*Vc)>0.75; gfZZ = 100% if Vudirect/(fv*Vc)<0.75
Moment be transferred by flexure MZZb = gfZZ*Mzz =Effective width for flexure = beff = clong+3*t =Moment capacity Mcap = beff*MCSshort =C/D Ratio in bending = Mcap/MZZb =
Moment to be transferred by shear MZZv = (1-gfZZ)*Mzz =vuZZc = VuZZd = MZZv*a5/JZZ =vuZZa = VuZZb = -Mzzv*(a4-a5)/JZZ =
JZZ = Jac+Jbd+Jcd = Jac = Jbd = Ixac+Iyac =
Ixac = a4*t3/12 =Iyac = a4
3*t/12+(a4*t)*a62 =
Jcd = (a3*t)*a52 =
Unbalanced moment from slab, MWWslab:Negative moment at first interior support M1 = 0.70*MoA =Negative moment at typical interior support M2 = 0.65*M0A =
Negative moment at typical interior support, dead load only, M3 = (LFDL*wDL/wtotal)*M2 =MWWslab = M1-M3 =
Unbalanced moment from wall, MWWwall:Negative moment at first interior support M1 = 0.70*MowA =Negative moment at typical interior support M2 = 0.65*MowA =MWWwall = M1-M2 =
MWW = MWWslab+MWWwall =The fraction of the unbalanced moment given by gf*Mu shall be considered to be transferred by flexuregfaWW = 1/[1+(2/3)*sqrt(a3/a4)]
413.6.3.3
Capacity/Demand Ratio for flexure, moment about axis W-W:
Slab adequate in bending if C/D > 1.0Shear stress due to moment about axis W-W transferred by shear:
Shear stresses:
Critical shear stress:
Capacity/Demand Ratio for shear:
Slab adequate in shear if C/D ratio > 1.0
For edge columns with unbalanced moments about an axis transverse to the edge, increase gf to as much as1.25 times the value but not more than 1.0 provided that Vu at the support does not exceed 0.40*fv*Vc.Percentage of moment transferred by flexure, gfWW = gfaWW if Vudirect/Vc>0.40; gf = 1.25*gfaWW if Vudirect/Vc<0.40
Moment be transferred by flexure MWWb = gfWW*MWW =Effective width for flexure = beff = shorter of cshort+3*t or cshort+1.5*t+Dshort
cshort+3*t =cshort+1.5*t+Dshort =
Moment capacity Mcap = beff*MCSlong =C/D Ratio = Mcap/MWWb =
Moment to be transferred by shear MWWv = (1-gfWW)*MWW =vuWWb = vuWWd = MWWv*0.5*a3/JWW =vuWWa = vuWWc = -MWWv*0.5*a3/JWW =
Jww = Jcd+Jac+Jbd =Jcd = Ixcd+Iycd =
Ixcd = a3*t3/12 =Iycd = a3
3*t/12 =Jac = Jbd = a4*t*(0.5*a3)2 =
vua = vudirecta+vuZZa+vuWWa =vub = vudirectb+vuZZb+vuWWb =vuc = vudirectc+vuZZc+vuWWc =vud = vudirectd+vuZZd+vuWWd =
vu = largest of vua, vub, vuc or vud =
C/D Ratio = fv*vc/vu =
/1,000,000)*bo*t*1,000,000 =*t*1,000,000 =
0.56
*Lsh
ort
0.50
*Lsh
ort
a 1 =
1.0
6*L s
hort
Long direction
Capacity/Demand Ratio for flexure, moment about axis Z-Z:
Shear stress due to moment about axis Z-Z transferred by shear:
Moment about axis W-W to be transferred to the column (from column line A with LL on span 1-2 only):
The gravity load moment to be transferred between slab and edge column shall be 0.3*Mo.
The fraction of the unbalanced moment given by gf*Mu shall be considered to be transferred by flexure
For edge columns with unbalanced moments about an axis parallel to the edge, gf = 1.0 provided that
udirect/(fv*Vc)<0.75
Negative moment at first interior support M1 = 0.70*MoA =Negative moment at typical interior support M2 = 0.65*M0A =
Negative moment at typical interior support, dead load only, M3 = (LFDL*wDL/wtotal)*M2 =
Negative moment at first interior support M1 = 0.70*MowA =Negative moment at typical interior support M2 = 0.65*MowA =
The fraction of the unbalanced moment given by gf*Mu shall be considered to be transferred by flexure
Capacity/Demand Ratio for flexure, moment about axis W-W:
Shear stress due to moment about axis W-W transferred by shear:
For edge columns with unbalanced moments about an axis transverse to the edge, increase gf to as much as1.25 times the value but not more than 1.0 provided that Vu at the support does not exceed 0.40*fv*Vc.
fWW = gfaWW if Vudirect/Vc>0.40; gf = 1.25*gfaWW if Vudirect/Vc<0.40
+3*t or cshort+1.5*t+Dshort
fWW)*MWW =
Page 6 of 6
7.6323.3930.6300.5400.1710.0970.1130.099
1.71551,898.36474,438.94 474,438.94
30367,932.24367,932.24312,742.40
1,016,057.19
515,069.6876,461.84
591,531.521.89
1,921,804.82
247,946.75
61.83%
61.83%
153,317.270.99
0.00Not adequate
9,038,176.19147,206,361.56
-318,947,116.710.010470.003590.000260.003320.00330
307,351285,397169,852137,499
43,53540,425
3,110140,609
58.14%
58.14%
81,745.290.7200.9900.720
0.00Not adequate
58,863.33794,216.20
-794,216.200.023350.004060.000310.003750.00964
-317,819,528.09-316,231,095.69 -317819528148,333,950.18 -317819528149,922,382.58
-317,819,528.09
0.00Not adequate
0.56
*Lsh
ort
0.50
*Lsh
ort
Criti
cal s
ectio
n
0.56*Llong-0.5*clong-dave0.56
*Lsh
ort
0.50
*Lsh
ort
Llong
0.50*Llong0.56*Lshort
clong+dave
cshort+daveCriticalSection
0.50*Lshort
0.56*Llong
Llong
SLAB DESIGN WORKSHEETTwo-Way Post-Tensioned Flat Plate (slab without beams)Project:Designer:Date:
Interior column B2:
411.13.2.1
a1 = 1.06*Llong =a2 = 1.06*Lshort =a3 = clong+t =a4 = cshort+t =
Shear capacity f*Vc:bo = 2*(a3+a4) =Vc1 = (1/6)*[1+2/(clong/cshort)]*sqrt(f'c/1,000,000)*bo*t*1,000,000 =Vc2 = (1/12)*(as*t/b0+2)*sqrt(f'c/1,000,000)*bo*t*1,000,000 =
For interior column as = 40Vc3 = (1/3)*sqrt(f'c/1,000,000)*bo*t*1,000,000 =Vc = smallest of Vc1, Vc2 or Vc3 =fv*Vc =fv*vc = fv*Vc/(bo*t) =
Direct shear due to slab loads, Vudirect:Vudirect = wtotal*[(a1*a2-a3*a4] =Vudirect/(fv*Vc) =
Direct shear stress due to slab and wall loads, vudirect:
0.56*Llong
a1 = 1.06*Llong
a3 = clong+t
W
W
Z Z
a b
c d
Shor
t dire
ction
B
2
Moment about axis Z-Z to be transferred to the column (from column line B with LL on span 1-2 only):
413.6.3.2
Capacity/Demand Ratio for flexure, moment about axis Z-Z:
413.6.3.2
Slab adequate in bending if C/D > 1.0Shear stress due to moment about axis Z-Z transferred by shear:
Moment about axis W-W to be transferred to the column (from column line 2 with LL on span A-B only):
413.6.3.2Capacity/Demand Ratio for flexure, moment about axis W-W:
Slab adequate in bending if C/D > 1.0Shear stress due to moment about axis W-W transferred by shear:
Shear stresses:
vudirecta = vudirectb = vudirectc = vudirectd = Vudirect/(bo*t) =
Negative moment at first interior support, M1 = 0.70*MoB =Negative moment at typical interior support, M2 = 0.65*MoB =
Negative moment at typical interior support, dead load only, M3 = (LFDL*wDL/wtotal)*M2 =MZZ = M1-M3 =The fraction of the unbalanced moment given by gf*Mu shall be considered to be transferred by flexuregfZZ = 1/[1+(2/3)*sqrt(a4/a3)]
Moment be transferred by flexure MZZf = gfZZ*Mzz =Effective width for flexure = beff = clong+3*t =Moment capacity Mcap = beff*MCSshort =Capacity/Demand Ratio = Mcap/MZZb =
Moment to be transferred by shear MZZv = (1-gfZZ)*Mzz =vuZZa = vuZZb = MZZv*0.5*a4/JZZ =vuZZc = vuZZd = -vuZZa =
JZZ = Jac+Jbd+Jab+Jcd = Jac = Jbd = Ixac+Iyac =
Ixac = a4*t3/12 =Iyac = a4
3*t/12 =Jab = Jcd = (a3*t)*(0.5*a4)2 =
Negative slab moment at first interior support M1 = 0.70*Mo2 =Negative slab moment at typical interior support M2 = 0.65*Mo2 =
Negative slab moment at typical interior support, dead load only M3 = (LFDL*wDL/wtotal)*M2 =MWW = M1-M3 =gfWW = 1/[1+(2/3)*sqrt(a3/a4)]
Moment be transferred by flexure MWWb = gfWW*MWW =Effective width for flexure = beff = shorter of cshort+3*t or cshort+1.5*t+Dshort
cshort+3*t =cshort+1.5*t+Dshort =
Moment capacity Mcap = beff*MCSlong =Capacity/Demand Ratio = Mcap/MWWb =
Moment to be transferred by shear MWWv = (1-gfWW)*MWW =vuWWa = vuWWc = MWWv*0.5*a3/JWW =vuWWb = vuWWd = -vuWWa =
JWW = Jac+Jbd+Jab+Jcd =Jac = Jbd = (a4*t)*(0.5*a3)2 =Jab = Jcd = Ixab+Iyab =
Ixab = a3*t3/12 =Iyab = a3
3*t/12 =
vua = vudirecta+vuZZa+vuWWa =
Critical shear stress:
Capacity/Demand Ratio for shear:
Slab adequate in shear if C/D ratio > 1.0
vub = vudirectb+vuZZb+vuWWb =vuc = vudirectc+vuZZc+vuWWc =vud = vudirectd+vuZZd+vuWWd =
vu = largest of absolute values of vua, vub, vuc or vud =
C/D Ratio for shear = fv*vc/vu =
Page 6 of 6
7.6327.6320.6300.630
2.52813,323.90658,405.06
40542,215.93542,215.93460,883.54
1,016,057.19
1,165,988.882.53
/1,000,000)*bo*t*1,000,000 =*t*1,000,000 =
0.56
*Lsh
ort
0.50
*Lsh
ort
a 2 =
1.0
6*L s
hort
0.50*Llong
a 4 =
csh
ort+
t
Z
Long Direction
2,570,522.23Moment about axis Z-Z to be transferred to the column (from column line B with LL on span 1-2 only):
578,542.42537,217.96319,720.81338,751.81
60.00%Capacity/Demand Ratio for flexure, moment about axis Z-Z:
203,251.090.990
0.00Not adequate
Shear stress due to moment about axis Z-Z transferred by shear:135,500.72
1,394,040.37-1,394,040.37
0.030618000.004060.000310.003750.01125
Moment about axis W-W to be transferred to the column (from column line 2 with LL on span A-B only):578,542537,218319,721258,82260.00%
Capacity/Demand Ratio for flexure, moment about axis W-W:155,292.97
0.7200.9900.720
0.00Not adequate
Shear stress due to moment about axis W-W transferred by shear:103,528.64
1,065,109.50-1,065,109.50
0.030620.011250.004060.000310.00375
5,029,672.10
= 0.70*MoB = = 0.65*MoB =
Negative moment at typical interior support, dead load only, M3 = (LFDL*wDL/wtotal)*M2 =
The fraction of the unbalanced moment given by gf*Mu shall be considered to be transferred by flexure
= 0.70*Mo2 =Negative slab moment at typical interior support M2 = 0.65*Mo2 =
Negative slab moment at typical interior support, dead load only M3 = (LFDL*wDL/wtotal)*M2 =
+3*t or cshort+1.5*t+Dshort
fWW)*MWW =
2,899,453.102,241,591.35
111,372.35
5,029,672.10
0.20Not adequate
658,405.06
233805.4
5029672.15029672.1
Criti
cal s
ectio
n
0.56*Llong-0.5*clong-dave0.56
*Lsh
ort
0.50
*Lsh
ort
Llong
0.50*Llong0.56*Lshort
clong+dave
cshort+daveCriticalSection
0.50*Lshort
0.56*Llong
SLAB DESIGN WORKSHEETTwo-Way Post-Tensioned Flat Plate (slab without beams)Project:Designer:Date:
Corner column A1:
Two-way shear:
411.13.2.1
a1 = Dlong+0.5*clong+0.44*Llong =a2 = 0.44*Lshort+0.5*cshort+Dshort =a3 = Dlong+clong+0.5*t =a4 = 0.5*t+cshort+Dshort =a5 = a3*t*0.5*a3/(a3*t+a4*t) =a6 = 0.5*a3-a5 =a7 = a4*t*0.5*a4/(a3*t+a4*t) =a8 = 0.5*a4-a7 =a9 = Dlong+clong+1.414*t =a10 = Dshort+cshort+1.414*t =a11 = Dlong+clong+a10 =a12 = Dshort+cshort+a9 =
Shear capacity fv*Vc and maximum shear stress fv*vc:bo = a3+a4 =Vc1 = (1/6)*[1+2/(clong/cshort)]*sqrt(f'c/1,000,000)*bo*t*1,000,000 =
0.44*Llong
a1 = Dlong+0.5*clong+0.44*Llong
a3 = Dlong+clong+0.5*t
a4 = 0.5*t+cshort
+Dshort
W
W
Z
a b
c d
Shor
t dire
ction
A
1
Dlong 0.5*clong
a 7a 8
0.5*
a 4
Centroid ofshear perimeter
0.5*a3
a5a6
Z
Two-way shear
Moment about axis Z-Z to be transferred to the column (from column line A):413.7.3.6
413.6.3.2
Capacity/Demand Ratio for flexure, moment about axis Z-Z:
413.6.3.2
Slab adequate in bending if C/D > 1.0Shear stress due to moment about axis Z-Z transferred by shear:
Moment about axis W-W to be transferred to the column (from column line 1):413.7.3.6
413.6.3.2
Capacity/Demand Ratio for flexure, moment about axis W-W:
Slab adequate in bending if C/D > 1.0
Vc2 = (1/12)*(as*t/b0+2)*sqrt(f'c/1,000,000)*bo*t*1,000,000 =For corner column as = 20:
Vc3 = (1/3)*sqrt(f'c/1,000,000)*bo*t*1,000,000 =Vc = smallest of Vc1, Vc2 or Vc3 =Shear capacity, fv*Vc =Maximum shear stress, fv*vc = fv*Vc/(bo*t) =
Direct shear due to slab and wall loads, Vudirect:Vuslab = wtotal*[(a1*a2-a3*a4] =Vuwall = LFDL*Wwall*(a1+a2-a3-a4) =Vudirect = Vuslab+Vuwall =Vudirect/(fv*Vc) =
Direct shear stress due to slab and wall loads, vudirect:vudirectb = vudirectc = vudirectd = Vudirect/(bo*t) =
The gravity load moment to be transferred between slab and edge column shall be 0.3*Mo.MZZ = 0.30*MoA =The fraction of the unbalanced moment given by gf*Mu shall be considered to be transferred by flexuregfZZ = 1/[1+(2/3)*sqrt(a4/a3)]
Moment be transferred by flexure MZZf = gfZZ*Mzz =Effective width for flexure = beff = shorter of cshort+3*t or cshort+1.5*t+Dshort
clong+3*t =clong+1.5*t+Dlong =
Moment capacity Mcap = beff*MCSshort =Capacity/Demand Ratio = Mcap/MZZb =
Moment to be transferred by shear MZZv = (1-gfZZ)*Mzz =vuZZc = vuZZd = MZZv*a7/JZZ =vuZZb = -MZZv*(a4-a7)/JZZ =JZZ = Jbd+Jcd =
Jbd = Ixbd+Iybd =Ixbd = a4*t3/12 =Iybd = a4
3*t/12+a4*t*a82 =
Jcd = (a3*t)*a72 =
The gravity load moment to be transferred between slab and edge column shall be 0.3*Mo.MWW = 0.30*Mo1 =The fraction of the unbalanced moment given by gf*Mu shall be considered to be transferred by flexuregfWW = 1/[1+(2/3)*sqrt(a3/a4)]
Moment be transferred by flexure MWWb = gfWW*MWW =Effective width for flexure = beff = shorter of cshort+3*t or cshort+1.5*t+Dshort
cshort+3*t =cshort+1.5*t+Dshort =
Moment capacity Mcap = beff*MCSlong =Capacity/Demand Ratio = Mcap/MWWb =
Shear stress due to moment about axis W-W transferred by shear:
Shear stresses:
Design shear stress:
Capacity/Demand Ratio for two-way shear:
Slab adequate in shear if C/D ratio > 1.0One-way shear:
Shear capacity:
411.4.1.1Direct shear due to slab and wall load:
Capacity/Demand Ratio for one-way shear:
Slab adequate in shear if C/D ratio > 1.0
Moment to be transferred by shear MWWv = (1-gfWW)*MWW =vuWWb = vuWWd = MWWv*a5/JWW =vuWWc = -MWWv*(a3-a5)/JWW =
JWW = Jbd+Jcd =Jbd = (a4*t)*a5
2 =Jcd = Ixcd+Iycd =
Ixcd = a3*t3/12 =Iycd = a3
3*t/12+(a3*t)*a62 =
vub = vudirectb+vuZZb+vuWWb =vuc = vudirectc+vuZZc+vuWWc =vud = vudirectd+vuZZd+vuWWd =
vu = largest of absolute values of vub, vuc or vud =
C/D Ratio for two-way shear = fv*vc/vu =
bo = sqrt(a112+a12
2) =fv*Vc = fv*0.17*sqrt(f'c)*bo*t =
Vuslab = wtotal*(a1*a2-0.5*a11*a12) =Vuwall = LFDL*Wwall*(a1+a2-a11-a12) =Vu = Vuslab+Vuwall =
C/D Ratio for one-way shear = fv*Vc/Vu =
and maximum shear stress fv*vc:
/1,000,000)*bo*t*1,000,000 =
0.5*cshort
t
a 10 =
Dsh
ort+
c sho
rt+1
.414
*t
a9 = Dlong+clong+1.414*t
a11 = Dlong+clong+a10
a 12 =
Dsh
ort+
c sho
rt+a
9
One-way shear
45o
Moment about axis Z-Z to be transferred to the column (from column line A):
Capacity/Demand Ratio for flexure, moment about axis Z-Z:
Shear stress due to moment about axis Z-Z transferred by shear:
Moment about axis W-W to be transferred to the column (from column line 1):
Capacity/Demand Ratio for flexure, moment about axis W-W:
/1,000,000)*bo*t*1,000,000 =
Direct shear stress due to slab and wall loads, vudirect:
The gravity load moment to be transferred between slab and edge column shall be 0.3*Mo.
The fraction of the unbalanced moment given by gf*Mu shall be considered to be transferred by flexure
= shorter of cshort+3*t or cshort+1.5*t+Dshort
= (1-gfZZ)*Mzz =
The gravity load moment to be transferred between slab and edge column shall be 0.3*Mo.
The fraction of the unbalanced moment given by gf*Mu shall be considered to be transferred by flexure
fWW*MWW = = shorter of cshort+3*t or cshort+1.5*t+Dshort
Shear stress due to moment about axis W-W transferred by shear: = (1-gfWW)*MWW =
Page 6 of 6
3.3933.3930.5400.5400.1350.1350.1350.1350.7050.7051.1551.155
1.08348,567.39
a 2 =
0.4
4*L s
hort+0
.5*c
s hor
t+D
shor
t
Long DirectionDshort
0.5*cshort
0.44
*Lsh
ort
309,837.68 309,837.68
232,378.26232,378.26197,521.52
1,016,057.19
226,158.4662,309.52
288,467.981.46
1,483,888.76
131,721.71
60.00%
79,033.030.7200.9900.720
0.00Not adequate
52,688.681,153,329.06
-3,459,987.170.006170.004400.000260.004130.00177
131,722
60.00%
79,033.030.7200.9900.720
0.00Not adequate
52,688.681,153,329.06
-3,459,987.170.006170.001770.004400.000260.00413
-822,769.35-822,769.35 -822769.35
3,790,546.88
3,790,546.88
0.27Not adequate
1.633152,292.07
218,603.1134,920.29
253,523.40
0.601Not adequate
0.56*Llong-0.5*clong-dave0.56
*Lsh
ort
0.50
*Lsh
ort
Criti
cal s
ectio
n
0.56*Llong-0.5*clong-dave
Llong
0.50*Llong0.56*Lshort
clong+dave
cshort+daveCriticalSection
0.50*Lshort
0.56*Llong
Llong