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CONSTRUCTION MANAGEMENTConcept-Develop-Execute-Finish
__________________________________________________________________________________
Seun Sambath, PhD
Concrete Structures(ACI 318-05)
in Mathcad
Sixth Edition
Phnom Penh 2010___________________________________________________________________________________
Address: #M41, St.308, Sangkat Tonle Basac, Khan Chamkarmon,Phnom Penh, Cambodia.
Tel: 012 659 848.
Table of Contents
1. Unit conversion ....................................................................................................... 1
2. Simple calculation ................................................................................................... 4
3. Materials ................................................................................................................ 5
4. Safety provision ....................................................................................................... 10
5. Loads on structures (case of two-way slab) ................................................................ 12
6. Loads on structures (case of one-way slab) ................................................................ 16
7. Loads on staircase .................................................................................................. 20
8. Loads on tile roof ..................................................................................................... 23
9. ASCE wind loads .................................................................................................... 25
10. Design of singly reinforced beams ............................................................................ 28
11. Design of doubly reinforced beams ......................................................................... 39
12. Design of T beams ................................................................................................. 52
13. Shear design ......................................................................................................... 60
14. Column design ...................................................................................................... 68
15. Design of footings .................................................................................................. 99
16. Design of pile caps ............................................................................................... 106
17. Slab design .......................................................................................................... 115
18. Design of staircase ............................................................................................... 142
19. Deflections ........................................................................................................... 148
20. Development lengths ............................................................................................. 158
Reference .................................................................................................................. 162
1. Unit Conversion
Length
in = inch
ft = foot
yd = yard
mi = mile
1in 25.4 mm 1cm 0.394 in
1ft 0.305 m 1m 3.281 ft 1ft 12 in
1yd 0.914 m 1m 1.094 yd 1yd 3 ft
1mi 1.609 km 1km 0.621 mi 1mi 1760 yd
1mi 5280 ft
1m 100mm 1.1m 1mi 200m 1.124 mi
Size of standard concrete cylinder D 6in 15.24 cm
H 12in 30.48 cm
Force
lbf = pound force
kip = kilopound force
1lbf 4.448 N 1lbf 0.454 kgf 1kgf 9.807 N
1N 0.225 lbf 1kgf 2.205 lbf 1N 0.102 kgf
1kip 4.448 kN 1kip 0.454 tonnef 1tonf 0.907 tonnef
1kN 0.225 kip 1tonnef 2.205 kip 1tonnef 1.102 tonf
1tonnef 1000 kgf 1tonf 2000 lbf
Page 1
Stress
psi = pound per square inch 1psi 1lbf
in2
ksi = kilopound per square inch 1ksi 1kip
in2
psf = pound per square foot 1psf 1lbf
ft2
1psi 6.895 kPa 1kPa 0.145 psi 1Pa 1N
m2
1ksi 6.895 MPa 1MPa 0.145 ksi 1kPa 1kN
m2
1MPa 1N
mm2
1psf 0.048kN
m2
1kN
m2
20.885 psf
1psf 4.882kgf
m2
1kgf
m2
0.205 psf
Concrete compression strength
3000psi 20.7 MPa 20MPa 2900.8 psi
4000psi 27.6 MPa 25MPa 3625.9 psi
5000psi 34.5 MPa 35MPa 5076.3 psi
8000psi 55.2 MPa 55MPa 7977.1 psi
Steel yield strength
60ksi 413.7 MPa 390MPa 56.6 ksi
75ksi 517.1 MPa 490MPa 71.1 ksi
Live loads
40psf 1.915kN
m2
200kgf
m2
40.963 psf
100psf 4.788kN
m2
4.80kN
m2
100.25 psf
Page 2
Density
1pcf 1lbf
ft3
125pcf 19.636kN
m3
145pcf 22.778kN
m3
Moments
1ft kip 1.356 kN m
User setting
Riels 1
USD 4165Riels
124USD 516460 Riels
200000Riels 48.019 USD
CostSteel 665USD
1tonnef
670kgf CostSteel 445.55 USD
Page 3
2. Simple Calculation
1 32
3
3
23.472
a 2 b 4 c 1
Δ b2
4 a c
x1b Δ
2 a0.225
x2b Δ
2 a2.225
Page 4
3. Materials
1. Concrete
Standard concrete cylinder D 6in 15.24 cm D 150mm
H 12in 30.48 cm H 300mm
Concrete compression strength
f'c 3000psi 20.7 MPa f'c 20MPa 2900.8 psi
f'c 4000psi 27.6 MPa f'c 25MPa 3625.9 psi
f'c 5000psi 34.5 MPa f'c 35MPa 5076.3 psi
Concrete ultimate strain
εu 0.003
Cubic and cylinder compression strength
fcube
f'c
0.85=
f'c 20MPa fcube
f'c
0.8523.529 MPa
f'c 25MPa fcube
f'c
0.8529.412 MPa
f'c 35MPa fcube
f'c
0.8541.176 MPa
Modulus of rupture (tensile strength)
fr 7.5 f'c= (in psi)
Metric coefficient 7.5psif'c
psi 7.5MPa
psi
MPa
f'c
MPa
MPa
psi= C MPa
f'c
MPa=
C 7.5psi
MPa
MPa
psi 0.623
fr 0.623 f'c= (in MPa)
Page 5
Modulus of elasticity
Ec 33 wc1.5
f'c= (in psi)
wc is a unit weight of concrete (in pcf)
Metric coefficient 33psi
kN
m3
pcf
1.5
MPa
psi 44.011 MPa
Ec 44 wc1.5
f'c= (in MPa)
wc in kN/m3
Example 3.1
Concrete compression strength f'c 25MPa
Unit weight of concrete wc 24kN
m3
145pcf 22.778kN
m3
Modulus of rupture
fr 7.5psif'c
psi 3.114 MPa
fr 0.623MPaf'c
MPa 3.115 MPa
Modulus of elasticity
Ec 33psiwc
pcf
1.5
f'c
psi 2.587 10
4 MPa
Ec 44MPawc
kN
m3
1.5
f'c
MPa 2.587 10
4 MPa
Page 6
2. Steel Reinforcements
Steel yield strength of deformed bar (DB) fy 390MPa 56.565 ksi
Steel yield strength of round bar (RB) fy 235MPa 34.084 ksi
Modulus of elasticity Es 29000000psi 1.999 105
MPa
Es 2 105MPa
Steel yield strength εy
fy
Es=
US Steel Reinforcements
Bar No. (#) Bar diameter
no
3
4
5
6
7
8
9
10
11
12
13
14
Dno
8in D
9.5
12.7
15.9
19
22.2
25.4
28.6
31.8
34.9
38.1
41.3
44.4
mm
Page 7
Steel area
Aπ D
2
4 A
0.71
1.27
1.98
2.85
3.88
5.07
6.41
7.92
9.58
11.4
13.38
15.52
cm2
Weight of steel reinforcements
W A 7850kgf
m3
W
0.559
0.994
1.554
2.237
3.045
3.978
5.034
6.215
7.52
8.95
10.503
12.182
kgf
m
ORIGIN 1
n 1 9 Area n A n
Page 8
1 2 3 4 5 6 7 8 9
3 9.5 0.71 1.43 2.14 2.85 3.56 4.28 4.99 5.70 6.41 0.559
4 12.7 1.27 2.53 3.80 5.07 6.33 7.60 8.87 10.13 11.40 0.994
5 15.9 1.98 3.96 5.94 7.92 9.90 11.88 13.86 15.83 17.81 1.554
6 19.1 2.85 5.70 8.55 11.40 14.25 17.10 19.95 22.80 25.65 2.237
7 22.2 3.88 7.76 11.64 15.52 19.40 23.28 27.16 31.04 34.92 3.045
8 25.4 5.07 10.13 15.20 20.27 25.34 30.40 35.47 40.54 45.60 3.978
9 28.6 6.41 12.83 19.24 25.65 32.07 38.48 44.89 51.30 57.72 5.034
10 31.8 7.92 15.83 23.75 31.67 39.59 47.50 55.42 63.34 71.26 6.215
11 34.9 9.58 19.16 28.74 38.32 47.90 57.48 67.06 76.64 86.22 7.520
12 38.1 11.40 22.80 34.20 45.60 57.00 68.41 79.81 91.21 102.61 8.950
13 41.3 13.38 26.76 40.14 53.52 66.90 80.28 93.66 107.04 120.42 10.503
14 44.5 15.52 31.04 46.55 62.07 77.59 93.11 108.63 124.14 139.66 12.182
Bar #Diameter
(mm)Area of cross section (cm2) for the number of bars is equal to Weight
(kgf/m)
Metric Steel Reinforcements
D 6 8 10 12 14 16 18 20 22 25 28 32 36 40( )T mm
Aπ D
2
4 W A 7850
kgf
m3
n 1 9 Area n A n
1 2 3 4 5 6 7 8 9
6 0.28 0.57 0.85 1.13 1.41 1.70 1.98 2.26 2.54 0.222
8 0.50 1.01 1.51 2.01 2.51 3.02 3.52 4.02 4.52 0.395
10 0.79 1.57 2.36 3.14 3.93 4.71 5.50 6.28 7.07 0.617
12 1.13 2.26 3.39 4.52 5.65 6.79 7.92 9.05 10.18 0.888
14 1.54 3.08 4.62 6.16 7.70 9.24 10.78 12.32 13.85 1.208
16 2.01 4.02 6.03 8.04 10.05 12.06 14.07 16.08 18.10 1.578
18 2.54 5.09 7.63 10.18 12.72 15.27 17.81 20.36 22.90 1.998
20 3.14 6.28 9.42 12.57 15.71 18.85 21.99 25.13 28.27 2.466
22 3.80 7.60 11.40 15.21 19.01 22.81 26.61 30.41 34.21 2.984
25 4.91 9.82 14.73 19.63 24.54 29.45 34.36 39.27 44.18 3.853
28 6.16 12.32 18.47 24.63 30.79 36.95 43.10 49.26 55.42 4.834
32 8.04 16.08 24.13 32.17 40.21 48.25 56.30 64.34 72.38 6.313
36 10.18 20.36 30.54 40.72 50.89 61.07 71.25 81.43 91.61 7.990
40 12.57 25.13 37.70 50.27 62.83 75.40 87.96 100.53 113.10 9.865
Diameter(mm)
Area of cross section (cm2) for the number of bars is equal to Weight(kgf/m)
Page 9
4. Safety Provision
Required_Strength Design_Strength
U ϕSn
where
U = required strength (factored loads)
ϕSn = design strength
Sn = nominal strength
ϕ = strength reduction factor
a. Load Combinations
Basic combination U 1.2 D 1.6 L=
Roof combination U 1.2 D 1.6 L 1.0 Lr=
Wind combination U 1.2 D 1.6 W 1.0 L 0.5 Lr=
where
D = dead load
L = live load
Lr = roof live load
W = wind load
Page 10
b. Strength Reduction Factor
Strength Condition Strength reduction factor ϕ
Tension-controlled members ( εt 0.005 ) ϕ 0.9=
Compression-controlled ( εt 0.002 )
Spirally reinforced ϕ 0.70=
Other ϕ 0.65=
Shear and torsion ϕ 0.75=
where
εt = net tensile strain
For spirally reinforced members
ϕ 0.9 εt 0.005if
0.70 εt 0.002if
1.7 200 εt
3otherwise
= 0.70.9 0.7
0.005 0.002εt 0.002
0.7200
3εt 0.002
1.7 200 εt
3=
For other members
ϕ 0.9 εt 0.005if
0.65 εt 0.002if
1.45 250 εt
3otherwise
=
Page 11
5. Loads on Structures (Case of Two-Way Slabs)
Slab dimension
Short side La 4m
Long side Lb 6m
A. Preliminary Design
Thickness of two-way slab
Perimeter La Lb 2
tminPerimeter
180111.111 mm
t1
30
1
50
La 133.333 80( ) mm
t 110mm
Section of beam B1
L 6m
h1
10
1
15
L 600 400( ) mm h 500mm
b 0.3 0.6( ) h 150 300( ) mm b 250mm
For girders h1
8
1
10
L=
For two-way slab beams h1
10
1
15
L=
For floor beams h1
15
1
20
L=
Section of beam B2
L 4m
h1
10
1
15
L 400 266.667( ) mm h 300mm
b 0.3 0.6( ) h 90 180( ) mm b 200mm
Page 12
B. Loads on Slab
Floor cover Cover 50mm 22kN
m3
1.1kN
m2
RC slab Slab t 25kN
m3
2.75kN
m2
Ceiling Ceiling 0.40kN
m2
M & E Mechanical 0.20kN
m2
Partition Partition 1.00kN
m2
Dead load DL Cover Slab Ceiling Mechanical Partition 5.45kN
m2
Live load for Lab LL 60psf 2.873kN
m2
Factored load wu 1.2 DL 1.6 LL 11.137kN
m2
C. Loads of Wall
Void 30mm 30 mm 190 mm 4
wbrick.hollow 90mm 90 mm 190 mm Void( ) 20kN
m3
1.744 kgf
wbrick.solid 45mm 90 mm 190 mm 20kN
m3
1.569 kgf
ρbrick.hollow
wbrick.hollow
90mm 90 mm 190 mm11.111
kN
m3
Brickhollow.10 120mm Void55
1m2
20kN
m3
1.648kN
m2
Brickhollow.20 220mm Void110
1m2
20kN
m3
2.895kN
m2
Page 13
D. Loads on Beam B1
Self weight SW 25cm 50cm 110mm( ) 25kN
m3
2.438kN
m
Wall wwall Brickhollow.10 3.5m 50cm( ) 4.943kN
m
Slab α
4m
2
6m0.333 k 1 2 α
2 α
3 0.815
wD.slab DL4m
2 2 k 17.763
kN
m
wL.slab LL4m
2 2 k 9.363
kN
m
Dead load wD SW wwall wD.slab 25.143kN
m
Live load wL wL.slab 9.363kN
m
Factored load wu 1.2 wD 1.6 wL 45.153kN
m
E. Loads on Beam B2
Self weight SW 20cm 30cm 110mm( ) 25kN
m3
0.95kN
m
Wall wwall Brickhollow.10 3.5m 30cm( ) 5.272kN
m
Slab α
4m
2
4m0.5 k 1 2 α
2 α
3 0.625
wD.slab DL4m
2 2 k 13.625
kN
m
wL.slab LL4m
2 2 k 7.182
kN
m
Dead load wD SW wwall wD.slab 19.847kN
m
Live load wL wL.slab 7.182kN
m
Factored load wu 1.2 wD 1.6 wL 35.308kN
m
Page 14
F. Loads on Column
Tributary area B 4m L 6m
Slab loads PD.slab DL B L 130.8 kN
PL.slab LL B L 68.948 kN
Beam loads PB1 25cm 50cm 110mm( ) 25kN
m3
L 14.625 kN
PB2 20cm 30cm 110mm( ) 25kN
m3
B 3.8 kN
Wall loads Pwall.1 Brickhollow.10 3.5m 50cm( ) L 29.657 kN
Pwall.2 Brickhollow.10 3.5m 30cm( ) B 21.089 kN
SW of column SW 5% 7%( ) PD=
Total loads for number of floors n 7
PD PD.slab PB1 PB2 Pwall.1 Pwall.2 1.05 n 1469.787 kN
PL PL.slab n 482.633 kN
Pu 1.2 PD 1.6 PL 2535.958 kNPu
PD PL1.299
Control
PD PL
n B L11.622
kN
m2
(Ref. 10kN
m2
18kN
m2
)
PL
PD PL24.72 % (Ref. 15% 35% )
Page 15
06 . Loads on Structures (Case of One-Way Slabs)
A. Preliminary Design
Thickness of one-way slab (both ends continue)
L6m
23 m
tminL
28107.143 mm
t1
25
1
35
L 120 85.714( ) mm
t 120mm
Page 16
Section of floor beam B1
L 8m
h1
15
1
20
L 533.333 400( ) mm h 500mm
b 0.3 0.6( ) h 150 300( ) mm b 250mm
Section of girder B2
L 6m
h1
8
1
10
L 750 600( ) mm h 600mm
b 0.3 0.6( ) h 180 360( ) mm b 300mm
B. Loads on Slab
Cover 50mm 22kN
m3
1.1kN
m2
Slab t 25kN
m3
3kN
m2
Ceiling 0.40kN
m2
Mechanical 0.20kN
m2
Partition 1.00kN
m2
DL Cover Slab Ceiling Mechanical Partition 5.7kN
m2
LL 60psf 2.873kN
m2
wu 1.2 DL 1.6 LL 11.437
kN
m
1m
Page 17
C. Loads on Beam B1
Void 30mm 30 mm 190 mm 4
Brickhollow.10 120mm Void55
1m2
20kN
m3
1.648kN
m2
SW 25cm 50cm 120mm( ) 25kN
m3
2.375kN
m
wwall Brickhollow.10 3.5m 50cm( ) 4.943kN
m
wD.slab DL 3 m 17.1kN
m
wL.slab LL 3 m 8.618kN
m
wD SW wwall wD.slab 24.418kN
m
wL wL.slab 8.618kN
m
wu 1.2 wD 1.6 wL 43.091kN
m
D. Loads on Girder B2
SW 30cm 60cm 120mm( ) 25kN
m3
3.6kN
m
wwall Brickhollow.10 3.5m 60cm( ) 4.778kN
m
PB1 25cm 50cm 120mm( ) 25kN
m3
8m 4m
2 14.25 kN
Pwall Brickhollow.10 3.5m 50cm( )8m 4m
2 29.657 kN
PD.slab DL 3 m8m 4m
2 102.6 kN
PL.slab LL 3 m8m 4m
2 51.711 kN
Page 18
Factored loads
wD SW wwall 8.378kN
m
wL 0
wu 1.2 wD 1.6 wL 10.054kN
m
PD PB1 Pwall PD.slab 146.507 kN
PL PL.slab 51.711 kN
Pu 1.2 PD 1.6 PL 258.545 kN
Page 19
7. Loads on Staircase
Run and rise of step G3.5m
12291.667 mm
H3.8m
24158.333 mm G 2 H 60.833 cm
Slope angle α atanH
G
28.496 deg
Loads on Waist Slab
Thickness of waist slab t 120mm
Step cover Cover 50mm H G( ) 22kN
m3
1m
G 1 m 1.697
kN
m2
Page 20
SW of step StepG H
224
kN
m3
1m
G 1 m 1.9
kN
m2
SW of waist slab Slab t 25kN
m3
1m2
1m2
cos α( ) 3.414
kN
m2
Renderring Renderring 0.40kN
m2
1m2
1m2
cos α( ) 0.455
kN
m2
Handrail Handrail 0.50kN
m2
Total dead load DL Cover Step Slab Renderring Handrail
DL 7.966kN
m2
Live load for public staircase LL 100psf 4.788kN
m2
Factored load wu 1.2 DL 1.6 LL 17.22kN
m2
Loads on Landing Slab
Cover 50mm 22kN
m3
1.1kN
m2
Slab 150mm 25kN
m3
3.75kN
m2
Renderring 0.40kN
m2
Handrail 0.50kN
m2
DL Cover Slab Renderring Handrail 5.75kN
m2
LL 100psf 4.788kN
m2
wu 1.2 DL 1.6 LL 14.561kN
m2
Page 21
8. Loads on Roof
Slope angle α atan3m
10m
2
30.964 deg
Srokalinh tile Tile 30mm 20kN
m3
0.6kN
m2
Purlins w20x20x1.0 20mm 20 mm 18mm 18 mm( ) 7850kgf
m3
0.597kgf
1m
Purlin w20x20x1.01m
1m 100 mm cos α( ) 0.068
kN
m2
Rafters w40x80x1.6 40mm 80 mm 36.8mm 76.8 mm( ) 7850kgf
m3
w40x80x1.6 2.934kgf
1m
Rafter w40x80x1.61m
750mm1 m cos α( ) 0.045
kN
m2
Page 23
Roof beam Beam 20cm 30 cm 25kN
m3
1m
1m10m
4
0.6kN
m2
Roof column Column 20cm 20 cm3m
2 25
kN
m3
1
10m
44 m
0.15kN
m2
Total dead load wD Tile Purlin Rafter Beam Column 1.463kN
m2
Live load wL 1.00kN
m2
Factored load wu 1.2 wD 1.6 wL 3.356kN
m2
Page 24
9. ASCE Wind Loads
Basic wind speed V 120km
hr V 33.333
m
s V 74.565mph
Exposure category Expoure C=
Importance factor I 1.15
Topograpic factor Kzt 1.0
Gust factor G 0.85
Wind directionality factor Kd 0.85
Static wind pressure qs 0.613N
m2
V
m
s
2
0.681kN
m2
Velocity pressure coefficients
zg 274m α 9.5 (For exposure C)
Kz z( ) 2.01max z 4.6m( )
zg
2
α
Kz 10m( ) 1.001
Velocity wind pressure
qz z( ) qs Kz z( ) Kzt I Kd qz 10m( ) 0.667kN
m2
Design wind pressure
pz z Cp qz z( ) G Cp
Dimension of building in plan
B 6m 3 18 m
L 4m 5 20 m
λL
B1.111
External pressure coefficients
Cp.windward 0.8
Page 25
Cp.leeward linterp
0
1
2
4
40
0.5
0.5
0.3
0.2
0.2
λ
0.478
Cp.side 0.7
Floor heights
H
3.5m
3.5m
3.5m
3.5m
3.5m
3.5m
3.5m
H reverse H( ) h H 24.5 m
ORIGIN 1
n rows H( ) 7
Wind forces
i 1 n ai
1
i
k
Hk
Hi
2
an 1 H reverse a( )
24.5
22.75
19.25
15.75
12.25
8.75
5.25
1.75
m
Bwindward 6m Bleeward 6m
Bside 4m
Pwindwardiai
ai 1
zpz z Cp.windward
d Bwindward
Pleewardipz h Cp.leeward a
i 1 ai
Bleeward
Psideipz h Cp.side a
i 1 ai
Bside
Page 26
reverse augment Pwindward Pleeward Pside
5.70
11.13
10.71
10.21
9.61
8.81
8.10
3.43
6.86
6.86
6.86
6.86
6.86
6.86
3.35
6.71
6.71
6.71
6.71
6.71
6.71
kN
Alternative ways
i 1 n
bi
1
i
k
Hk
Prectangleipz b
iCp.windward a
i 1 ai
Bwindward
Ptrapeziumi
pz ai
Cp.windward pz ai 1 Cp.windward
2ai 1 a
i Bwindward
reverse augment Pwindward Prectangle Ptrapezium
5.70
11.13
10.71
10.21
9.61
8.81
8.10
5.75
11.13
10.71
10.22
9.62
8.83
8.08
5.70
11.12
10.70
10.20
9.59
8.78
8.20
kN
Page 27
10. Design of Singly Reinforced Beams
A. Concrete Stress Distribution
In actual distribution
Resultant C α f'c c b=
Location β c
In equivalent distribution
Location β ca
2=
Resultant C α f'c c b= γ f'c a b=
Thus, a 2 β c= β1 c= where β1 2 β=
γ αc
a=
α
β1=
f'c 4000psi 5000psi 6000psi 7000psi 8000psi
α 0.72 0.68 0.64 0.60 0.56
β 0.425 0.400 0.375 0.350 0.325
β1 2 β= 0.85 0.80 0.75 0.70 0.65
γα
β1=
0.72
0.850.847
0.68
0.800.85
0.64
0.750.853
0.60
0.700.857
0.56
0.650.862
Page 28
Conclusion: γ 0.85=
β1 0.85 f'c 4000psiif
0.65 f'c 8000psiif
0.85 0.05f'c 4000psi
1000psi otherwise
= 4000psi 27.6 MPa
8000psi 55.2 MPa
1000psi 6.9 MPa
B. Strength Analysis
Equilibrium in forces
X 0=
C T=
0.85 f'c a b As fs= (1)
Equilibrium in moments
M 0=
Mn C da
2
= T da
2
=
Mn 0.85 f'c a b da
2
= (2.1)
Mn As fs da
2
= (2.2)
Conditions of strain compatibility
εs
εu
d c
c=
εs εud c
c= or εt εu
dt c
c= (3.1)
c dεu
εu εs= or c dt
εu
εu εt= (3.2)
Unknowns = 3 a As fs
Equations = 2 X 0= M 0=
Additional condition fs fy= (From economic criteria)
Page 29
C. Steel Ratios
ρ
As
b d=
As fy
b d fy=
0.85 f'c a b
b d fy= 0.85 β1
f'c
fy
c
d= 0.85 β1
f'c
fy
c
dt
dt
d=
ρ 0.85 β1f'c
fy
εu
εu εs= 0.85 β1
f'c
fy
εu
εu εt
dt
d=
Balanced steel ratio
fc f'c= fs fy= εs εy=fy
Es=
ρb 0.85 β1f'c
fy
εu
εu εy= 0.85 β1
f'c
fy
600MPa
600MPa fy=
εu 0.003 Es 2 105MPa εu Es 600 MPa
Maximum steel ratio
ACI 318-99 ρmax 0.75 ρb=
ACI 318-02 and later ρmax 0.85 β1f'c
fy
εu
εu εt= with εt 0.004
For fy 390MPa εs
fy
Es0.002
For εt 0.004ρmax
ρb
εu εy
εu 0.004=
5
7= 0.714=
For εt 0.005ρmax
ρb
εu εy
εu 0.005=
5
8= 0.625=
Minimum steel ratio
ρmin
3 f'c
fy
200
fy= (in psi)
ρmin
0.249 f'c
fy
1.379
fy= (in MPa)
Page 30
D. Determination of Flexural Strength
Given: b d As f'c fy
Find: ϕMn
Step 1. Checking for steel ratio
ρ
As
b d=
ρ ρmin : Steel reinforcement is not enough
ρmin ρ ρmax : the beam is singly reinforced
ρ ρmax : the beam is doubly reinforced
ρ ρmax= As ρ b d=
Step 2. Calculation of flexural strength
aAs fy
0.85 f'c b= c
a
β1=
Mn As fy da
2
=
εt εu
dt c
c= ϕ ϕ εt =
The design flexural strength is ϕ Mn
Example 10.1
Page 31
Concrete dimension b 200mm h 350mm
Steel reinforcements As 5π 16mm( )
2
4 10.053 cm
2
d h 30mm 6mm 16mm40mm
2
278 mm
dt h 30mm 6mm16mm
2
306 mm
Materials f'c 25MPa fy 390MPa
Solution
Checking for steel ratios
β1 0.65 max 0.85 0.05f'c 27.6MPa
6.9MPa
min 0.85
0.85
εu 0.003
ρmax 0.85 β1f'c
fy
εu
εu 0.004 0.02
ρmin max
0.249MPaf'c
MPa
fy
1.379MPa
fy
0.00354
ρ
As
b d0.018
Steel_Reinforcement "is Enough" ρ ρminif
"is not Enough" otherwise
Steel_Reinforcement "is Enough"
As min ρ ρmax b d 10.053 cm2
Calculation of flexural strength
aAs fy
0.85 f'c b92.252 mm c
a
β1108.532 mm
Mn As fy da
2
90.911 kN m
Page 32
εt εu
dt c
c 0.00546
ϕ 0.65 max1.45 250 εt
3
min 0.9
0.9
The design flexural strength is ϕ Mn 81.82 kN m
E. Determination of Steel Area
Given: Mu b d f'c fy
Find: As
Relative depth of compression concrete
wa
d=
0.85 f'c a b
0.85 f'c b d=
As fy
0.85 f'c b d=
ρ fy
0.85 f'c1=
Flexural resistance factor
RMn
b d2
=
As fy da
2
b d2
=
As
b dfy
da
2
d= ρ fy 1
1
2w
=
R ρ fy 1ρ fy
1.7 f'c
= 0.85 f'c w 11
2w
=
Quadratic equation relative w
R
0.85 f'cw 1
1
2w
=
w2
2 w 2R
0.85 f'c 0=
w1 1 1 2R
0.85 f'c 1= w2 1 1 2
R
0.85 f'c 1=
w 1 1 2R
0.85 f'c=
ρ 0.85f'c
fy w= 0.85
f'c
fy 1 1 2
R
0.85 f'c
=
Page 33
Step 1. Assume ϕ 0.9=
Mn
Mu
ϕ=
Step 2. Calculation of steel area
RMn
b d2
=
ρ 0.85f'c
fy 1 1 2
R
0.85 f'c
=
ρ ρmax : the beam is doubly reinforced(concrete is not enough)
ρ ρmax : the beam is singly reinforced
As max ρ ρmin b d= (this is a required steel area)
Step 3. Checking for flexural strength
aAs fy
0.85 f'c b= (As is a provided steel area)
Mn As fy da
2
=
ca
β1= εt εu
dt c
c= ϕ ϕ εt =
FSMu
ϕ Mn= (usage percentage)
FS 1 : the beam is safe
FS 1 : the beam is not safe
Example 10.2
Required strength Mu 153kN m
Concrete section b 200mm h 500mm
d h 30mm 8mm 18mm40mm
2
424 mm
Page 34
dt h 30mm 8mm18mm
2
453 mm
Materials f'c 25MPa fy 390MPa
Solution
Steel ratios
β1 0.65 max 0.85 0.05f'c 27.6MPa
6.9MPa
min 0.85
0.85
εu 0.003
ρmax 0.85 β1f'c
fy
εu
εu 0.004 0.02
ρmin max
0.249MPaf'c
MPa
fy
1.379MPa
fy
0.00354
Assume ϕ 0.9
Mn
Mu
ϕ170 kN m
Steel area
RMn
b d2
4.728 MPa
ρ 0.85f'c
fy 1 1 2
R
0.85 f'c
0.014
ρmin ρ ρmax 1
As ρ b d 11.783 cm2
As 6π 16mm( )
2
4 12.064 cm
2
Checking for flexural strength
aAs fy
0.85 f'c b110.702 mm c
a
β1130.238 mm
Mn As fy da
2
173.444 kN m
Page 35
εt εu
dt c
c 0.00743
ϕ 0.65 max1.45 250 εt
3
min 0.9
0.9
FSMu
ϕ Mn0.98
The_beam "is safe" FS 1if
"is not safe" otherwise
The_beam "is safe"
F. Determination of Concrete Dimension and Steel Area
Given: Mu f'c fy
Find: b d As
Step 1. Determination of concrete dimension
Assume εt 0.004 (Usually εt 0.005 )
ρ 0.85 β1f'c
fy
εu
εu εt= R ρ fy 1
ρ fy
1.7 f'c
=
ϕ ϕ εt = Mn
Mu
ϕ=
bd2 Mn
R=
Option 1: b
Mn
R
d2
=
Option 2: d
Mn
R
b=
Option 3: kb
d= d
3Mn
R
k= b k d=
Step 2. Calculation of steel area
RMn
b d2
=
Page 36
ρ 0.85f'c
fy 1 1 2
R
0.85 f'c
=
As max ρ ρmin b d=
Step 3. Checking for flexural strength
aAs fy
0.85 f'c b= c
a
β1=
Mn As fy da
2
=
εt εu
dt c
c= ϕ ϕ εt =
FSMu
ϕ Mn=
Example 10.3
Required strength Mu 700kN m
Materials f'c 25MPa fy 390MPa
Solution
Steel ratios
β1 0.65 max 0.85 0.05f'c 27.6MPa
6.9MPa
min 0.85
0.85
εu 0.003
ρmax 0.85 β1f'c
fy
εu
εu 0.004 0.02
ρmin max
0.249MPaf'c
MPa
fy
1.379MPa
fy
0.00354
Assume εt 0.007
ρ 0.85 β1f'c
fy
εu
εu εt 0.014 R ρ fy 1
ρ fy
1.7 f'c
4.728 MPa
Page 37
ϕ 0.65 max1.45 250 εt
3
min 0.9
0.9 Mn
Mu
ϕ777.778 kN m
Concrete dimension
kb
d= k
400
600 Cover 30mm 10mm 25mm
40mm
2
Cover 85 mm
d
3Mn
R
k627.231 mm b k d 418.154 mm
h Round d Cover 50mm( ) 700 mm b Round b 50mm( ) 400 mm
d h Cover 615 mmb
h
400
700
mm
Steel area
RMn
b d2
5.141 MPa
ρ 0.85f'c
fy 1 1 2
R
0.85 f'c
0.015
As max ρ ρmin b d 37.741 cm2
As 8π 25mm( )
2
4 39.27 cm
2 dt h 30mm 10mm
25mm
2
dt 647.5 mm
Checking for flexural strength
aAs fy
0.85 f'c b180.18 mm c
a
β1211.976 mm
Mn As fy da
2
803.914 kN m
εt εu
dt c
c 0.00616
ϕ 0.65 max1.45 250 εt
3
min 0.9
0.9
FSMu
ϕ Mn96.749 %
Page 38
11. Design of Doubly Reinforced Beams
ρ ρmax : the beam is singly reinforced(with tensile reinforcements only)
ρ ρmax : the beam is doubly reinforced(with tensile and compression reinforcements)
A. Strength Analysis
Equilibrium in forces
X 0=
T C Cs= (1)
T As fs= As fy=
C 0.85 f'c a b=
Cs A's f's=
Equilibrium in moments
M 0=
Mn Mn1 Mn2= (2)
Mn1 T d d'( )= A's f's d d'( )=
Mn2 C da
2
= 0.85 f'c a b da
2
=
Mn2 T Cs da
2
= As fy A's f's da
2
=
Page 39
Conditions of strain compatibility
εs
εu
d c
c= (3.1)
εs εud c
c= or εt εu
dt c
c=
c dεu
εu εs= or c dt
εu
εu εt=
ε's
εu
c d'
c= (3.2)
ε's εuc d'
c=
c d'εu
εu ε's=
B. Steel Ratios
Compression steel ratio
ρ'A's
b d=
Tensile steel ratio
ρ
As
b d=
As fy
b d fy=
0.85 f'c a b A's f's
b d fy= 0.85 β1
f'c
fy
c
d ρ'
f's
fy=
Maximum tensile steel ratio
ρt.max ρmax ρ'f's
fy=
ρ ρt.max : concrete is enough
ρ ρt.max : concrete is not enough
Minimum tensile steel ratio
ρ 0.85 β1f'c
fy
c
d ρ'
f's
fy= 0.85 β1
f'c
fy
εu
εu ε's
d'
d ρ'
f's
fy=
f's fy= ε's εy=fy
Es=
Page 40
ρcy 0.85 β1f'c
fy
εu
εu εy
d'
d ρ'=
ρ ρcy : compression steel will yield f's fy=
ρ ρcy : compression steel will not yield f's fy
C. Determination of Flexural Strength
Given: b d dt d' As A's f'c fy
Find: ϕMn
Step 1. Checking for singly reinforced beam
ρ
As
b d=
ρ ρmax : the beam is singly reinforced
ρ ρmax : the beam is doubly reinforced
ρ'A's
b d= ρt.max ρmax ρ'=
ρ ρt.max : concrete is enough
ρ ρt.max : concrete is not enough
ρ ρt.max= As ρ b d=
Step 2. Determination of compression parameters
2.1. Assume
f's fy=
2.2. Calculate
aAs fy A's f's
0.85 f'c b= c
a
β1=
ε's εuc d'
c= f's.revised Es ε's fy=
If f's.revised f's then f's f's.revised=
Goto 2.2
Page 41
Direct calculation
Case f's fy=
aAs fy A's fy
0.85 f'c b=
Case f's fy
As fy 0.85 f'c a b A's f's= 0.85 f'c a b A's Es εuc d'
c=
As fy 0.85 f'c a b A's Es εuβ1 c β1 d'
β1 c= 0.85 f'c a b A's f1
a β1 d'
a=
where f1 Es εu= 600MPa=
0.85 f'c a2
b A's f1 As fy a A's f1 β1 d' 0=
0.85 f'c a2
b A's f1 As fy a A's f1 β1 d'
0.85 f'c d2
b0=
a
d
2 ρ' f1 ρ fy
0.85 f'c
a
d
ρ' f1 β1
0.85 f'c
d'
d 0=
w2
2 p w q 0= p1
2
ρ' f1 ρ fy
0.85 f'c= q
ρ' f1 β1
0.85 f'c
d'
d=
w1 p p2
q 0= w2 p p2
q 0=
a d p p2
q = ca
β1=
ε's εuc d'
c= f's Es ε's fy=
Step 3. Calculation of flexural strength
Mn1 A's f's d d'( )=
Mn2 As fy A's f's da
2
=
εt εu
dt c
c= ϕ ϕ εt =
ϕMn ϕ Mn1 Mn2 =
Page 42
Example 11.1
Concrete dimension b 300mm h 550mm
Steel reinforcements As 8π 20mm( )
2
4 25.133 cm
2
d h 30mm 10mm 16mm 20mm40mm
2
d 454 mm
dt h 30mm 10mm 16mm20mm
2
484 mm
A's 4π 20mm( )
2
4 12.566 cm
2
d' 30mm 10mm20mm
2 50 mm
Materials f'c 25MPa fy 390MPa
Solution
Steel ratios
β1 0.65 max 0.85 0.05f'c 27.6MPa
6.9MPa
min 0.85
0.85
εu 0.003
ρmax 0.85 β1f'c
fy
εu
εu 0.005 0.0174
ρmin max
0.249MPaf'c
MPa
fy
1.379MPa
fy
0.00354
Checking for singly reinforced beam
ρ
As
b d0.0185
The_beam "is singly reinforced" ρ ρmaxif
"is doubly reinforced" otherwise
The_beam "is doubly reinforced"
ρ'A's
b d9.226 10
3 ρt.max ρmax ρ' 0.027
Page 43
Concrete "is enough" ρ ρt.maxif
"is not enough" otherwise
Concrete "is enough"
ρ min ρ ρt.max 0.0185
As ρ b d 25.133 cm2
Determination of compression parameters
Es 2 105MPa f1 Es εu 600 MPa
εy
fy
Es1.95 10
3
ρcy 0.85 β1f'c
fy
εu
εu εy
d'
d ρ' 0.024
Compression_steel "will yield" ρ ρcyif
"will not yield" otherwise
Compression_steel "will not yield"
aAs fy A's fy
0.85 f'c bCompression_steel "will yield"=if
p1
2
ρ' f1 ρ fy
0.85 f'c
qρ' f1 β1
0.85 f'c
d'
d
d p p2
q
otherwise
a 90.825 mm ca
β1106.853 mm
f's fy Compression_steel "will yield"=if
ε's εuc d'
c
min Es ε's fy
otherwise
f's 319.24 MPa
Flexural strength
Page 44
Mn1 A's f's d d'( ) 162.072 kN m
Mn2 As fy A's f's da
2
236.576 kN m
εt εu
dt c
c 0.011
ϕ 0.65 max1.45 250 εt
3
min 0.9
0.9
ϕMn ϕ Mn1 Mn2 358.783 kN m
D. Design of Doubly Reinforced Beam
Given: Mu b d dt d' f'c fs
Find: As A's
Step 1. Assume εt
ρ 0.85 β1f'c
fy
εu
εu εt=
ϕ ϕ εt =
Step 2. Cheching for singly reinforced beam
As ρ b d=
aAs fy
0.85 f'c b=
Mn As fy da
2
=
Mu ϕMn : the beam is singly reinforced
Mu ϕMn : the beam is doubly reinforced
Step 3. Case of doubly reinforced beam
Mn2 Mn=
Mn1
Mu
ϕMn2=
Page 45
ca
β1= f's Es εu
c d'
c fy=
A's
Mn1
f's d d'( )= ρ'
A's
b d= ρt.max ρmax ρ'=
As
0.85 f'c a b A's f's
fy= ρ
As
b d=
ρ ρt.max : concrete is enough
ρ ρt.max : concrete is not enough
Example 11.2
Required strength Mu 1350kN m
Concrete dimension b 400mm h 800mm
d h 30mm 12mm 25mm 25mm40mm
2
d 688 mm
dt h 30mm 12mm 25mm25mm
2
720.5 mm
d' 30mm 12mm25mm
2 54.5 mm
Materials f'c 25MPa fy 390MPa
Solution
Steel ratios
β1 0.65 max 0.85 0.05f'c 27.6MPa
6.9MPa
min 0.85
0.85
εu 0.003
ρmax 0.85 β1f'c
fy
εu
εu 0.005 0.0174
ρmin max
0.249MPaf'c
MPa
fy
1.379MPa
fy
0.00354
Assume εt 0.0092
ρ 0.85 β1f'c
fy
εu
εu εt 0.011
Page 46
ϕ 0.65 max1.45 250 εt
3
min 0.9
0.9
Checking for singly reinforced beam
As ρ b d 31.342 cm2
aAs fy
0.85 f'c b143.803 mm c
a
β1169.18 mm
Mn2 As fy da
2
753.074 kN m
The_beam "is singly reinforced" Mu ϕ Mn2if
"is doubly reinforced" otherwise
The_beam "is doubly reinforced"
Case of doubly reinforced beam
Mn1
Mu
ϕMn2 746.926 kN m
f's min Es εuc d'
c fy
390 MPa
A's
Mn1
f's d d'( )30.232 cm
2
ρ'A's
b d0.011 ρt.max ρmax ρ' 0.028
As
0.85 f'c a b A's f's
fy61.574 cm
2 ρ
As
b d0.022
Concrete "is enough" ρ ρt.maxif
"is not enough" otherwise
Concrete "is enough"
Compression steel A's 30.232 cm2
5π 28mm( )
2
4 30.788 cm
2
Tensile steel As 61.574 cm2
10π 28mm( )
2
4 61.575 cm
2
400mm 12mm 30mm( ) 2 28mm 5
444 mm
Page 47
E. Determination of Tensile Steel Area
Given: Mu b d dt d' A's f'c fy
Find: As
Step 1. Calculation of compression parameters
1.1. Assume f's fy= ϕ 0.9=
1.2. Calculate
Mn
Mu
ϕ=
Mn1 A's f's d d'( )=
Mn2 Mn Mn1=
RMn2
b d2
=
ρ 0.85f'c
fy 1 1 2
R
0.85 f'c
=
As ρ b d=
aAs fy
0.85 f'c b= c
a
β1=
f's.revised Es εuc d'
c fy=
εt εu
dt c
c= ϕ ϕ εt =
f's.revised f's : Goto 1.2
Mu ϕ Mn : Goto 1.2
Step 2. Calculation of tensile steel area
As
0.85 f'c a b A's f's
fy= ρ
As
b d=
ρ'A's
b d= ρt.max ρmax ρ'=
ρ ρt.max : concrete is enough
ρ ρt.max : concrete is not enough
Page 48
Example 11.3
Required strength Mu 1350kN m
Concrete dimension b 400mm h 800mm
d h 30mm 12mm 25mm 28mm40mm
2
d 685 mm
dt h 30mm 12mm 25mm28mm
2
719 mm
d' 30mm 12mm28mm
2 56 mm
Compression reinforcements A's 5π 28mm( )
2
4 30.788 cm
2
Materials f'c 25MPa fy 390MPa
Solution
Steel ratios
β1 0.65 max 0.85 0.05f'c 27.6MPa
6.9MPa
min 0.85
0.85
εu 0.003
ρmax 0.85 β1f'c
fy
εu
εu 0.005 0.0174
ρmin max
0.249MPaf'c
MPa
fy
1.379MPa
fy
0.00354
Compression parameters
Page 49
Compression ε( ) f's fy
ϕ 0.9
Mn
Mu
ϕ
Mn1 A's f's d d'( )
Mn2 Mn Mn1
RMn2
b d2
ρ 0.85f'c
fy 1 1 2
R
0.85 f'c
As ρ b d
aAs fy
0.85 f'c b
ca
β1
f's.revised min Es εuc d'
c fy
Z i
f's
fy
ϕ
a
d
f's.revised
fy
εt εu
dt c
c
ϕ 0.65 max1.45 250 εt
3
min 0.9
break( )f's.revised f's
f'sε
Mu ϕ Mn if
f's f's.revised
i 0 99for
reverse ZT
Z Compression 0.000001( )
Page 50
a Z0 2 d 142.792 mm c
a
β1167.99 mm
f's Z0 0 fy 390 MPa
Tensile steel area
ρ'A's
b d0.011 ρt.max ρmax ρ' 0.029
As
0.85 f'c a b A's f's
fy61.909 cm
2 ρ
As
b d0.023
Concrete "is enough" ρ ρt.maxif
"is not enough" otherwise
Concrete "is enough"
Tensile steel As 61.909 cm2
10π 28mm( )
2
4 61.575 cm
2
Page 51
12. Design of T Beams
12.1. Effective Flange Width
For symmetrical T beam:
bL
4 b bw 16hf b s
where
L = span length of beam
s = spacing of beam
12.2. Strength Analysis
Design as rectangular section Design as T section
a hf a hf
or Mu ϕMnf or Mu ϕMnf
where Mnf 0.85 f'c hf b dhf
2
=
Page 52
Equilibrium in forces X 0=
T C1 C2=
T As fs= As fy=
C1 0.85 f'c hf b bw = Asf fy=
C2 0.85 f'c a bw= T C1= As fy Asf fy=
Equilibrium in moments M 0=
Mn Mn1 Mn2=
Mn1 C1 dhf
2
= Asf fy dhf
2
=
Mn2 C2 da
2
= 0.85 f'c a bw da
2
=
Mn2 T C1 da
2
= As fy Asf fy da
2
=
Condition of strain compatibility
εs
εu
d c
c= or
εt
εs
dt c
c=
εs εud c
c= εt εu
dt c
c=
c dεu
εu εs= c dt
εu
εu εt=
12.3. Steel Ratios
ρw
As
bw d=
As fy
bw d fy=
0.85 f'c a bw Asf fy
bw d fy=
ρw 0.85 β1f'c
fy
c
d ρf= where ρf
Asf
bw d=
Page 53
Maximum steel ratio
ρw.max ρmax ρf=
ρw ρw.max : concrete is enough
ρw ρw.max : concrete is not enough
12.4. Determination of Moment Capacity
Given: bw d b dt hf As f'c fy
Find: ϕMn
Step 1. Checking for rectangular beam
aAs fy
0.85 f'c b=
a hf : the beam is rectangular
a hf : the beam is tee
Step 2. Case of T beam
Asf
0.85 f'c hf b bw
fy=
Mn1 Asf fy dhf
2
=
aAs fy Asf fy
0.85 f'c bw= c
a
β1=
Mn2 As fy Asf fy da
2
=
εt εu
dt c
c= ϕ ϕ εt =
ϕMn ϕ Mn1 Mn2 =
Page 54
Example 12.1
Concrete dimension b 28in 711.2 mm
hf 6in 152.4 mm
bw 10in 254 mm
h 30in 762 mm d 26in 660.4 mm
dt 27.5in 698.5 mm
Steel reinforcements As 6
π10
8in
2
4 As 7.363 in
2
Materials f'c 3000psi 20.684 MPa
fy 60ksi 413.685 MPa
Solution
Steel ratios
β1 0.65 max 0.85 0.05f'c 4000psi
1000psi
min 0.85
0.85
εu 0.003
ρmax 0.85 β1f'c
fy
εu
εu 0.005 0.014
ρmin max
3psif'c
psi
fy
200psi
fy
0.00333
Checking for rectangular beam
Page 55
aAs fy
0.85 f'c b157.162 mm
The_beam "is rectangular" a hfif
"is T" otherwise
The_beam "is T"
Case of T beam
Asf
0.85 f'c hf b bw
fy29.613 cm
2
ρf
Asf
bw d0.018 ρw.max ρmax ρf 0.031
ρw
As
bw d0.028
Concrete "is enough" ρw ρw.maxif
"is not enough" otherwise
Concrete "is enough"
As min ρw ρw.max bw d As 7.363 in2
Mn1 Asf fy dhf
2
715.669 kN m
aAs fy Asf fy
0.85 f'c bw165.734 mm c
a
β1194.981 mm
Mn2 As fy Asf fy da
2
427.446 kN m
εt εu
dt c
c 0.008
ϕ 0.65 max1.45 250 εt
3
min 0.9
0.9
ϕMn ϕ Mn1 Mn2 1028.803 kN m
Page 56
12.5. Determination of Steel Area
Given: Mu bw d dt b hf f'c fy
Find: As
Step 1. Checking for rectangular beam
Mnf 0.85 f'c hf b dhf
2
=
ϕ 0.9=
Mu ϕMn : the beam is rectangular
Mu ϕMn : the beam is tee
Step 2. Case of T beam
Asf
0.85 f'c hf b bw
fy= ρf
Asf
bw d=
Mn1 Asf fy dhf
2
=
Mn2
Mu
ϕMn1=
RMn2
bw d2
=
ρ 0.85f'c
fy 1 1 2
R
0.85 f'c
=
As2 ρ bw d=
aAs2 fy
0.85 f'c bw=
As
0.85 f'c a bw Asf fy
fy= ρw
As
bw d=
ρw.max ρmax ρf=
ρw ρw.max : concrete is enough
ρw ρw.max : concrete is not enough
Page 57
Example 12.2
Concrete dimension hf 3in 76.2 mm
L 24ft 7.315 m s 47in 1.194 m
bw 11in 279.4 mm d 20in 508 mm
Required strength Mu 6400in kip 723.103 kN m
Materials f'c 3000psi 20.684 MPa fy 60ksi 413.685 MPa
Solution
Effective flange width
b minL
4bw 16 hf s
b 1193.8 mm
Steel ratios
β1 0.65 max 0.85 0.05f'c 4000psi
1000psi
min 0.85
0.85
εu 0.003
ρmax 0.85 β1f'c
fy
εu
εu 0.005 0.014
ρmin max
3psif'c
psi
fy
200psi
fy
0.00333
Checking for rectangular beam
ϕ 0.9
Mnf 0.85 f'c hf b dhf
2
751.538 kN m
The_beam "is rectangular" Mu ϕ Mnfif
"is tee" otherwise
The_beam "is tee"
Case of T beam
Page 58
Asf
0.85 f'c hf b bw
fy29.613 cm
2 ρf
Asf
bw d0.021
Mn1 Asf fy dhf
2
575.646 kN m
Mn2
Mu
ϕMn1 227.801 kN m
RMn2
bw d2
3.159 MPa
ρ 0.85f'c
fy 1 1 2
R
0.85 f'c
8.484 103
As2 ρ bw d 12.042 cm2
aAs2 fy
0.85 f'c bw101.408 mm c
a
β1119.304 mm
As
0.85 f'c a bw Asf fy
fy41.655 cm
2 ρw
As
bw d0.029
ρw.max ρmax ρf 0.034
Concrete "is enough" ρw ρw.maxif
"is not enough" otherwise
Concrete "is enough"
As.min ρmin bw d
As max As As.min 41.655 cm2
6π 32mm( )
2
4 48.255 cm
2
Page 59
13. Shear Design
Safety provision
Vu ϕVn
where Vu = required shear strength
Vn = nominal shear strength
ϕ 0.75= is a strength reduction factor for shear
ϕVn = design shear strength
Required shear strength
Page 60
Nominal shear strength
Vn Vc Vs=
where
Vc = concrete shear strength
Vs = steel shear strength
Concrete shear strength
Vc 2 f'c bw d= (in psi)
Vc 0.166 f'c bw d= (in MPa)
Steel shear strength
Vs
Av fy d
s=
where
Av = area of stirrup
fy = yield strength of stirrup
s = spacing of stirrup
No required stirrups
Vu
ϕVc
2 : no stirrup is required
ϕVc
2Vu ϕVc : stirrup is minimum
Vu ϕVc : stirrup is required
Minimum stirrups
Av.min 0.75 f'cbw s
fy 50
bw s
fy= (in psi)
Av.min 0.062 f'cbw s
fy 0.345
bw s
fy= (in MPa)
Page 61
Maximum spacing of stirrup
smax
Av fy
0.75 f'c bw
Av fy
50 bw= (in psi)
smax
Av fy
0.062 f'c bw
Av fy
0.345 bw= (in MPa)
Case Vs 2 Vc
smaxd
224in= 600mm=
Case 2 Vc Vs 4 Vc
smaxd
412in= 300mm=
Case Vs 4 Vc
Concrete is not enough
Example 13.1
Materials f'c 25MPa fy 390MPa
Page 62
Live load for garage LL 6.00kN
m2
Loads on slab
Hardener 8mm 24kN
m3
0.192kN
m2
Slab 200mm 25kN
m3
5kN
m2
Mechanical 0.30kN
m2
DL Hardener Slab Mechanical 5.492kN
m2
LL 6kN
m2
Loads on beam
wbeam 30cm 60cm 200mm( ) 25kN
m3
3kN
m
wD.slab DL 3.5 m 19.222kN
m
wL.slab LL 3.5 m 21kN
m
wD wbeam wD.slab 22.222kN
m
wL wL.slab 21kN
m
wu 1.2 wD 1.6 wL 60.266kN
m
Shear
L 8m
V0
wu L
2241.066 kN
V x( ) V0 wu x
Concrete shear strength
bw 300mm d 600mm 40mm 10mm20mm
2
540 mm
Vc 0.166MPaf'c
MPa bw d 134.46 kN
ϕ 0.75
Page 63
Location of no stirrup zone
V0 wu xϕVc
2= x
V0
ϕ Vc
2
wu3.163 m
Minimum stirrup
Av 2π 10mm( )
2
4 1.571 cm
2 fy 390MPa
smax minAv fy
0.062MPaf'c
MPa bw
Av fy
0.345MPa bw
591.894 mm
smax Floor smax 50mm 550 mm
Vs.min
Av fy d
smax60.147 kN
Location of minimum stirrup zone
V0 wu x ϕ Vc Vs.min = xV0 ϕ Vc Vs.min
wu1.578 m
Required spacing of stirrup
Vu V0 wu400mm
2
229.012 kN
Vs
Vu
ϕVc 170.89 kN
Concrete "is enough" Vs 4 Vcif
"is not enough" otherwise
Concrete "is enough"
sAv fy d
Vs193.581 mm
smax.1 smax 550 mm
smax.2 mind
2600mm
Vs 2 Vcif
mind
4300mm
otherwise
smax.2 270 mm
s Floor min s smax.1 smax.2 50mm s 150 mm
Page 64
Example 13.2
Design of shear in support and midspan zones.
Stirrups in Support Zone
Required shear strength Vu V0 wu400mm
2 229.012 kN
Concrete shear strength
Vc 0.166MPaf'c
MPa bw d 134.46 kN
ϕ 0.75
Stirrup "is minimum" Vu ϕ Vcif
"is required" otherwise
Stirrup "is required"
Required steel shear strength
Vs
Vu
ϕVc 170.89 kN
Concrete "is enough" Vs 4 Vcif
"is not enough" otherwise
Concrete "is enough"
Spacing of stirrup
Av 2π 10mm( )
2
4 1.571 cm
2 fy 390MPa
sAv fy d
Vs193.581 mm
smax.1 minAv fy
0.062MPaf'c
MPa bw
Av fy
0.345MPa bw
591.894 mm
Page 65
smax.2 mind
2600mm
Vs 2 Vcif
mind
4300mm
otherwise
smax.2 270 mm
s Floor min s smax.1 smax.2 50mm 150 mm
Stirrups in Midspan Zone
Required shear strength Vu V0 wuL
4 120.533 kN
Stirrup "is minimum" Vu ϕ Vcif
"is required" otherwise
Stirrup "is required"
Required steel shear strength
Vs
Vu
ϕVc 26.25 kN
Concrete "is enough" Vs 4 Vcif
"is not enough" otherwise
Concrete "is enough"
Spacing of stirrup
Av 2π 10mm( )
2
4 1.571 cm
2 fy 390MPa
sAv fy d
Vs1260.208 mm
smax.1 minAv fy
0.062MPaf'c
MPa bw
Av fy
0.345MPa bw
591.894 mm
smax.2 mind
2600mm
Vs 2 Vcif
mind
4300mm
otherwise
smax.2 270 mm
s Floor min s smax.1 smax.2 50mm 250 mm
Page 66
Page 67
14. Column Design
Type of columns (by design method)
1. Axially loaded columns
eM
P= 0=
2. Eccentric columns
eM
P= 0
2.1. Short columns (without buckling)
Pu Mu
2.2. Long (slender) columns (with buckling)
Pu Mu δns
1. Axially Loaded Columns
Safety provision
Pu ϕPn.max
where Pu = axial load on column
ϕPn.max = design axial strength
For tied columns
ϕPn.max 0.80 ϕ 0.85 f'c Ag Ast fy Ast =
with ϕ 0.65=
For spirally reinforced columns
ϕPn.max 0.85 ϕ 0.85 f'c Ag Ast fy Ast =
with ϕ 0.70=
where Ag = area of gross section
Ast = area of steel reinforcements
Ag Ast Ac= is an area of concrete section
Page 68
For tied columns
Diameter of tie
Dv 10mm= for D 32mm
Dv 12mm= for D 32mm
Spacing of tie
s 48Dv s 16D s b
For spirally reinforced columns
Diameter of spiral Dv 10mm
Clear spacing 25mm s 75mm
Column steel ratio
ρg
Ast
Ag= 1%= 8%
Page 69
Determination of Concrete Section
Ag
Pu
0.80 ϕ
0.85 f'c 1 ρg fy ρg=
Determination of Steel Area
Ast
Pu
0.80 ϕ0.85 f'c Ag
0.85 f'c fy=
Example 14.1
Tributary area B 4m L 6m
Thickness of slab t 120mm
Section of beam B1 b 250mm h 500mm
Section of beam B2 b 200mm h 350mm
Live load for lab LL 3.00kN
m2
Materials f'c 25MPa fy 390MPa
Solution
Loads on slab
Cover 50mm 22kN
m3
Slab 120mm 25kN
m3
Ceiling 0.40kN
m2
Mechanical 0.20kN
m2
Partition 1.00kN
m2
DL Cover Slab Ceiling Mechanical Partition 5.7kN
m2
LL 3kN
m2
Page 70
Reduction of live load
Tributary area AT B L 24 m2
For interior column KLL 4
Influence area AI KLL AT 96 m2
Live load reduction factor αLL 0.254.572
AI
m2
0.717
Reduced live load LL0 LL αLL 2.15kN
m2
Loads of wall
Void 30mm 30 mm 190 mm 4
Brickhollow.10 120mm Void55
1m2
20kN
m3
1.648kN
m2
Brickhollow.20 220mm Void110
1m2
20kN
m3
2.895kN
m2
Loads on column
PD.slab DL B L 136.8 kN
PL.slab LL B L 72 kN
PB1 25cm 50cm 120mm( ) 25kN
m3
L 14.25 kN
PB2 20cm 35cm 120mm( ) 25kN
m3
B 4.6 kN
Pwall.1 Brickhollow.10 3.5m 50cm( ) L 29.657 kN
Pwall.2 Brickhollow.10 3.5m 35cm( ) B 20.76 kN
Number of floors n 6
PD PD.slab PB1 PB2 Pwall.1 Pwall.2 1.05 n 1298.219 kN
PL PL.slab n 432 kN SW 5% 7%( ) PD=
PD PL
B L n12.015
kN
m2
PL
PD PL24.968 %
Page 71
Pu 1.2 PD 1.6 PL 2249.063 kN
Determination of column section
Assume ρg 0.03 kb
h= k
300
500
ϕ 0.65
Ag
Pu
0.80 ϕ
0.85 f'c 1 ρg fy ρg Ag 1338.529 cm
2
hAg
k472.322 mm b k h 283.393 mm
h Ceil h 50mm( ) 500 mm b Ceil b 50mm( ) 300 mm
b
h
300
500
mm Ag b h 1500 cm2
Determination of steel area
Ast
Pu
0.80 ϕ0.85 f'c Ag
0.85 f'c fy Ast 30.851 cm
2
6π 20mm( )
2
4 6
π 16mm( )2
4 30.913 cm
2
Stirrups
Main bars D 20mm
Stirrup dia. Dv 10mm
Spacing of tie s min 16 D 48 Dv b 300 mm
Page 72
2. Short Columns
Safety provision
Pu ϕPn
Mu ϕMn
Equilibrium in forces X 0=
Pn C Cs T=
Pn 0.85 f'c a b A's f's As fs=
Equilibrium in moments M 0=
Mn Pn e= Ch
2
a
2
Csh
2d'
T dh
2
=
Mn Pn e= 0.85 f'c a bh
2
a
2
A's f'sh
2d'
As fs dh
2
=
Conditions of strain compatibility
εs
εu
d c
c= εs εu
d c
c=
fs Es εs= Es εud c
c=
ε's
εu
c d'
c= ε's εu
c d'
c=
f's Es ε's= Es εuc d'
c=
Page 73
Unknowns = 5 : a As A's fs f's
Equations = 4 : X 0= M 0= 2 conditions of strain compatibility
Case of symmetrical columns: As A's=
Case of unsymmetrical columns: fs fy=
A. Interaction Diagram for Column Strength
Interaction diagram is a graph of parametric function, where
Abscissa : Mn a( )
Ordinate: Pn a( )
B. Determination of Steel Area
Given: Mu Pu b h f'c fy
Find: As A's=
Answer: As AsN a( )= AsM a( )=
AsN a( )
Pu
ϕ0.85 f'c a b
f's fs=
AsM a( )
Mu
ϕ0.85 f'c a b
h
2
a
2
f'sh
2d'
fs dh
2
=
f's a( ) Es εuc d'
c fy=
fs a( ) Es εud c
c fy=
Page 74
Example 14.2
Construction of interaction diagram for column strength.
Concrete dimension b 500mm h 200mm
Steel reinforcements As 5π 16mm( )
2
4 10.053 cm
2
A's As 10.053 cm2
d' 30mm 6mm16mm
2 44 mm
d h d' 156 mm
Materials f'c 25MPa
fy 390MPa
Solution
Case of axially loaded column
Ag b h
Ast As A's
ϕ 0.65
ϕPn.max 0.80 ϕ 0.85 f'c Ag Ast fy Ast 1490.536 kN
Case of eccentric column
β1 0.65 max 0.85 0.05f'c 27.6MPa
6.9MPa
min 0.85
0.85
c a( )a
β1
Es 2 105MPa εu 0.003 dt d
fs a( ) min Es εud c a( )
c a( ) fy
f's a( ) min Es εuc a( ) d'
c a( ) fy
ϕ a( ) εt εu
dt c a( )
c a( )
ϕ 0.65 max1.45 250 εt
3
min 0.90
Page 75
ϕPn a( ) min ϕ a( ) 0.85 f'c a b A's f's a( ) As fs a( ) ϕPn.max
ϕMn a( ) ϕ a( ) 0.85 f'c a bh
2
a
2
A's f's a( )h
2d'
As fs a( ) dh
2
a 0h
100 h
0 20 40 600
250
500
750
1000
1250
1500
Interaction diagram for column strength
ϕPn a( )
kN
ϕMn a( )
kN m
Example 14.3
Determination of steel area.
Required strength Pu 1152.27kN
Mu 42.64kN m
Concrete dimension b 500mm h 200mm
Materials f'c 25MPa
fy 390MPa
Concrete cover to main bars cc 30mm 6mm16mm
2
Page 76
Solution
Location of steel re-bars
d' cc 44 mm
d h cc 156 mm
Case of axially loaded column
Ag b h ϕ 0.65
ϕ 0.65Ast
Pu
0.80 ϕ0.85 f'c Ag
0.85 f'c fy2.465 cm
2
Case of eccentric column
β1 0.65 max 0.85 0.05f'c 27.6MPa
6.9MPa
min 0.85
0.85
c a( )a
β1
Es 2 105MPa εu 0.003 dt d
fs a( ) min Es εud c a( )
c a( ) fy
f's a( ) min Es εuc a( ) d'
c a( ) fy
ϕ a( ) εt εu
dt c a( )
c a( )
ϕ 0.65 max1.45 250 εt
3
min 0.90
Graphical solution
AsN a( )
Pu
ϕ a( )0.85 f'c a b
f's a( ) fs a( )
AsM a( )
Mu
ϕ a( )0.85 f'c a b
h
2
a
2
f's a( )h
2d'
fs a( ) dh
2
a1 134.2mm a2 134.25mm
a a1 a1a2 a1
50 a2
Page 77
0.13418 0.1342 0.13422 0.13424 0.134268.715 10
4
8.72 104
8.725 104
8.73 104
8.735 104
AsN a( )
AsM a( )
a
a 134.23mm
AsN a( ) 8.722 cm2
AsM a( ) 8.725 cm2
As
AsN a( ) AsM a( )
28.724 cm
2 5
π 16mm( )2
4 10.053 cm
2
Page 78
Analytical solution
ORIGIN 1
Asteel No( ) k 1
f f's a( ) fs a( )
continue( ) f 0=if
AsN
Pu
ϕ a( )0.85 f'c a b
f
continue( ) AsN 0if
fd f's a( )h
2d'
fs a( ) dh
2
continue( ) fd 0=if
AsM
Mu
ϕ a( )0.85 f'c a b
h
2
a
2
fd
continue( ) AsM 0if
Z k
a
h
AsN
Ag
AsM
Ag
AsN AsM
Ag
k k 1
a cc cch
No hfor
csort ZT 4
Z Asteel 5000( ) rows Z( ) 2046
a Z1 1 h 134.24 mm
AsN Z1 2 Ag 8.719 cm
2 AsM Z
1 3 Ag 8.728 cm2
As
AsN AsM
28.723 cm
2
Page 79
C. Case of Distributed Reinforcements
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b
dn
d1
Pn e
Tn T1
C
h
a
cf 85.0
c
u
s,1
s,n dn
Equilibrium in forces X 0=
Pn C
1
n
i
Ti
= 0.85 f'c a b
1
n
i
As i f
s i
=
Equilibrium in moments M 0=
Mn Pn e= Ch
2
a
2
1
n
i
Ti
di
h
2
=
Mn 0.85 f'c a bh
2
a
2
1
n
i
As i f
s i di
h
2
=
Page 80
Conditions of strain compatibility
εs i
εu
di
c
c=
εs i εu
di
c
c=
fs i Es ε
s i= Es εud
ic
c=
Example 14.4
Checking for column strength.
Required strength Pu 13994.6kN
Mu 57.53kN m
Materials f'c 35MPa
fy 390MPa
Solution
Determination of Concrete Section
Case of axially loaded column
ϕ 0.65
Assume ρg 0.04
Ag
Pu
0.80 ϕ
0.85 f'c 1 ρg fy ρg6094.36 cm
2
Aspect ratio of column section λb
h= λ 1
hAg
λ780.664 mm b λ h 780.664 mm
h Ceil h 50mm( ) b Ceil b 50mm( )
b
h
800
800
mm Ag b h 6400 cm2
Page 81
Steel area
Ast
Pu
0.80 ϕ0.85 f'c Ag
0.85 f'c fy Ast 218.534 cm
2
4 7 4( )π 25mm( )
2
4 4 5 4( )
π 20mm( )2
4 232.478 cm
2
Spacing 800mm 50mm 2
887.5 mm
Interaction Diagram for Column Strength
Distribution of reinforcements
Bars
25
25
25
25
25
25
25
25
25
25
20
20
20
20
20
20
20
25
25
20
0
0
0
0
0
20
25
25
20
0
0
0
0
0
20
25
25
20
0
0
0
0
0
20
25
25
20
0
0
0
0
0
20
25
25
20
0
0
0
0
0
20
25
25
20
20
20
20
20
20
20
25
25
25
25
25
25
25
25
25
25
mm
Number of reinforcement rows n cols Bars( ) 9
Steel area
As0π Bars
2
4
i 1 n AsiAs0
i
Ast As Ast 232.478 cm2
Location of reinforcement rows
Concrete cover Cover 30mm 10mm 40 mm
d1
CoverBars
1 n
2 52.5 mm ΔS
h d1
2
n 186.875 mm
i 2 n di
di 1 ΔS
reverse d( )T 747.5 660.63 573.75 486.88 400 313.13 226.25 139.38 52.5( ) mm
Case of axially loaded column
Page 82
ϕPn.max 0.80 ϕ 0.85 f'c Ag Ast fy Ast
ϕPn.max 14255.808 kN
Case of eccentric column
β1 0.65 max 0.85 0.05f'c 27.6MPa
6.9MPa
min 0.85
0.796
c a( )a
β1
fs i a( ) εs εu
di
c a( )
c a( )
sign εs min Es εs fy
dt max d( ) 747.5 mm
ϕ a( ) εt εu
dt c a( )
c a( )
ϕ 0.65 max1.45 250 εt
3
min 0.9
ϕPn a( ) min ϕ a( ) 0.85 f'c a b
1
n
i
Asifs i a( )
ϕPn.max
ϕMn a( ) ϕ a( ) 0.85 f'c a bh
2
a
2
1
n
i
Asifs i a( ) d
ih
2
a 0h
100 h
Page 83
0 1000 2000 30000
5000
10000ϕPn a( )
kN
Pu
kN
ϕMn a( )
kN m
Mu
kN m
Page 84
D. Design of Circular Columns
Symbols
ns = number of re-bars
Dc = column diameter
Ds = diameter of re-bar circle
Location of steel re-bar
di
rc rs cos αs i = rc
Dc
2= rs
Ds
2=
αs i
2 π
nsi 1( )=
Page 85
Depth of compression concrete
α acosrc a
rc
=
Area and centroid of compression concrete
Asector1
2Radius Arch=
1
2rc rc 2 α = rc
2α=
x12
3rc
sin α( )
α=
Atriangle1
2Base Height=
1
22 rc sin α( ) rc cos α( )= rc
2sin α( ) cos α( )=
x22
3rc cos α( )=
Ac Asegment= Asector Atringle= rc2
α sin α( ) cos α( )( )=
xc
Asector x1 Atrinagle x2
Ac=
2
3rc
sin α( ) sin α( ) cos α( )2
α sin α( ) cos α( )=
xc
2 rc
3
sin α( )3
α sin α( ) cos α( )=
Equilibrium in forces X 0=
Pn C
1
ns
i
Ti
= 0.85 f'c Ac
1
ns
i
As i f
s i
=
Equilibrium in moments M 0=
Mn Pn e= C xc
1
ns
i
Ti
di
Dc
2
=
Mn Pn e= 0.85 f'c Ac xc
1
ns
i
As i f
s i di
rc
=
Conditions of strain compatibility
εs i
εu
di
c
c=
fs i Es ε
s i= Es εu
d
ic
c= with f
s i fy
Page 86
Example 14.5
Required strength Pu 3437.31kN
Mu 42.53kN m
Materials f'c 20MPa
fy 390MPa
Solution
Determination of concrete dimension
ϕ 0.70
Assume ρg 0.02
Ag
Pu
0.85 ϕ
0.85 f'c 1 ρg fy ρg2361.812 cm
2
Dc CeilAg
π
4
50mm
550 mm
Ag
π Dc2
42375.829 cm
2
Determination of steel area
Ast
Pu
0.85 ϕ0.85 f'c Ag
0.85 f'c fy46.597 cm
2
Ds Dc 30mm 10mm20mm
2
2 450 mm
ns ceilπ Ds
100mm
15 As0π 20mm( )
2
43.142 cm
2
Ast ns As0 47.124 cm2
sπ Ds
ns94.248 mm
Interaction diagram for column strength
ϕPn.max 0.85 ϕ 0.85 f'c Ag Ast fy Ast 3448.996 kN
Page 87
β1 0.65 max 0.85 0.05f'c 27.6MPa
6.9MPa
min 0.85
0.85
Es 2 105MPa εu 0.003
c a( )a
β1
i 1 ns αsi
2 π
nsi 1( ) d
i
Dc
2
Ds
2cos αsi
dt max d( ) 495.083 mm
ϕ a( ) εt εu
dt c a( )
c a( )
ϕ 0.70 max1.7 200 εt
3
min 0.9
fs i a( ) εs εu
di
c a( )
c a( )
sign εs min Es εs fy
rc
Dc
2
α a( ) acosrc a
rc
xc a( )2 rc
3
sin α a( )( )3
α a( ) sin α a( )( ) cos α a( )( )
Ac a( ) rc2
α a( ) sin α a( )( ) cos α a( )( )( )
ϕPn a( ) min ϕ a( ) 0.85 f'c Ac a( )
1
ns
i
As0 fs i a( )
ϕPn.max
ϕMn a( ) ϕ a( ) 0.85 f'c Ac a( ) xc a( )
1
ns
i
As0 fs i a( ) di
rc
a 0Dc
100 Dc
Page 88
0 100 200 3000
1000
2000
3000
Interaction diagram for column strength
ϕPn a( )
kN
Pu
kN
ϕMn a( )
kN m
Mu
kN m
3. Long (Slender) Columns
Stability index
QΣPu Δ0
Vu Lc=
where
ΣPu Vu = total vertical force and story shear
Δ0 = relative deflection between column ends
Lc = center-to-center length of column
Q 0.05 : Frame is nonsway (braced)
Q 0.05 : Frame is sway (unbraced)
Page 89
Unbraced Frame Braced Frame
She
ar W
all
Braced Frame
Brick Wall
Ties
Slenderness of column
The column is short, if
In nonsway frame:k Lu
rmin 34 12
M1
M2 40
In sway frame:k Lu
r22
where
M1 min MA MB = M2 max MA MB =
= minimum and maximum moments at the ends of column
Lu = unsuppported length of column
r = radius of gyration
rI
A=
I A = moment of inertia and area of column section
k = effective length factor
k k ψA ψB =
ψA ψB = degree of end restraint (release)
Page 90
ψ
EIc
Lc
EIb
Lb
=
ψ 0= : column is fixed
ψ ∞= : column is pinned
Moments of inertia
For column Ic 0.70Ig=
For beam Ib 0.35Ig=
Ig = moment of inertia of gross section
Determination of effective length factor
Way 1. Using graph
Way 2. Using equations
For braced frames:
ψA ψB
4
π
k
2
ψA ψB
21
π
k
tanπ
k
2 tanπ
2 k
π
k
1=
Page 91
For unbraced frames:
ψA ψBπ
k
2
36
6 ψA ψB
π
k
tanπ
k
=
Way 3. Using approximate relations
In nonsway frames:
k 0.7 0.05 ψA ψB 1.0=
k 0.85 0.05 ψmin 1.0=
ψmin min ψA ψB =
In sway frames:
Case ψm 2
k20 ψm
201 ψm=
Case ψm 2
k 0.9 1 ψm=
ψm
ψA ψB
2=
Case of column is hinged at one end
k 2.0 0.3 ψ=
ψ is the value in the restrained end.
Moment on column
Mc M2 δns M2.min δns=
where
M2.min Pu 15mm 0.03h( )=
Moment magnification factor
Page 92
δns
Cm
1Pu
0.75 Pc
1=
Euler's critical load
Pcπ
2EI
k Lu 2=
EI0.4 Ec Ig
1 βd=
βd
1.2 PD
1.2 PD 1.6 PL=
Coefficient
Cm 0.6 0.4M1
M2 0.4=
Example 14.6
Required strength Pu 6402.35kNPD
PL
4273.41kN
796.25kN
MA 77.75kN m MB 122.68 kN m
Length of column Lc 7.8m
Upper and lower columns
ba
ha
La
60cm
60cm
3.6m
bb
hb
Lb
65cm
65cm
1.5m
Upper and lower beams ba1
ha1
La1
30cm
50cm
6m
ba2
ha2
La2
30cm
50cm
6m
bb1
hb1
Lb1
30cm
50cm
6m
bb2
hb2
Lb2
30cm
50cm
6m
Materials f'c 30MPa
fy 390MPa
Page 93
Solution
Determination of concrete dimension
ϕ 0.65
Assume ρg 0.03
Ag
Pu
0.80 ϕ
0.85 f'c 1 ρg fy ρg3379.226 cm
2
Proportion of column section kb
h= k
60
60
hAg
k581.311 mm b k h 581.311 mm
h Ceil h 50mm( ) 600 mm
b Ceil b 50mm( ) 600 mmb
h
600
600
mm
Determination of steel area
Ag b h 3.6 103
cm2
Ast
Pu
0.80 ϕ0.85 f'c Ag
0.85 f'c fy85.932 cm
2 20
π 25mm( )2
4 98.175 cm
2
Bars
1
1
1
1
1
1
1
0
0
0
0
1
1
0
0
0
0
1
1
0
0
0
0
1
1
0
0
0
0
1
1
1
1
1
1
1
25 mm As0π Bars
2
4
i 1 cols As0 As1iAs0
i As As1
Ast As Ast 98.175 cm2
Ast
Ag0.027
ns rows As ns 6
Cover 40mm 10mm25mm
2 62.5 mm
Page 94
d11
Cover Δsh Cover 2
ns 195 mm
i 2 ns d1i
d1i 1 Δs
d d1 d
62.5
157.5
252.5
347.5
442.5
537.5
mm
ϕPn.max 0.80 ϕ 0.85 f'c Ag Ast fy Ast 6634.405 kN
Slenderness of column
Stability index Q 0
Radius of gyration rh
120.173 m
Modulus of elasticity
wc 24kN
m3
Ec 44MPawc
kN
m3
1.5
f'c
MPa 2.834 10
4 MPa
Degree of end restraint
Ia1 0.35ba1 ha1
3
12 Ia2 0.35
ba2 ha23
12
Ib1 0.35bb1 hb1
3
12 Ib2 0.35
bb2 hb23
12
Ica 0.70ba ha
3
12 Icb 0.70
bb hb3
12
Ic 0.70b h
3
12
Σica
Ec Ica
La
Ec Ic
Lc Σicb
Ec Icb
Lb
Ec Ic
Lc
Σiba
Ec Ia1
La1
Ec Ia2
La2 Σibb
Ec Ib1
Lb1
Ec Ib2
Lb2
ψA
Σica
Σiba8.418 ψB
Σicb
Σibb21.699
Page 95
Effective length factor
k 0.6
Given
ψA ψB
4
π
k
2
ψA ψB
21
π
k
tanπ
k
2 tanπ
2 k
π
k
1=
k 0.5 k 1.0
k Find k( ) k 0.969
Checking for long column
M1 MA MA MBif
MB otherwise
M2 MB MA MBif
MA otherwise
M1 77.75 kN m M2 122.68 kN m
Lu Lc
max ha1 ha2 max hb1 hb2
2 7.3 m
k Lu
r40.834 min 34 12
M1
M2 40
40
The_column "is short"k Lu
rmin 34 12
M1
M2 40
if
"is long" otherwise
The_column "is long"
Case of long column
βd
1.2 PD
1.2 PD 1.6 PL0.801
Igb h
3
12 EI
0.4 Ec Ig
1 βd
Pcπ
2EI
k Lu 213409.955 kN
Cm max 0.6 0.4M1
M2 0.4
0.4
Page 96
δns maxCm
1Pu
0.75 Pc
1
1.101
M2.min Pu 15mm 0.03 h( ) 211.278 kN m
Mc δns max M2 M2.min The_column "is long"=if
max M2 M2.min otherwise
Interaction diagram for column strength
c a( )a
β1
dt max d( ) 537.5 mm
ϕ a( ) εt εu
dt c a( )
c a( )
ϕ 0.65 max1.45 250 εt
3
min 0.90
fs i a( ) εs εu
di
c a( )
c a( )
sign εs min Es εs fy
ϕPn a( ) min ϕ a( ) 0.85 f'c a b
1
ns
i
Asifs i a( )
ϕPn.max
ϕMn a( ) ϕ a( ) 0.85 f'c a bh
2
a
2
1
ns
i
Asifs i a( ) d
ih
2
a 0h
100 h
Page 97
0 200 400 600 800 1000 12000
1000
2000
3000
4000
5000
6000
7000
Interaction diagram for column strength
ϕPn a( )
kN
Pu
kN
ϕMn a( )
kN m
Mc
kN m
Page 98
15. Footing Design
A. Determination of Footing Dimension
Required area of footing
Areq
PD PL
qe=
where
PD PL = dead and live loads on footing
qe = effective bearing capacity of soil
qe qa 20kN
m3
H=
qa = allowable bearing capacity of soil with FS 2.5= 3
20kN
m3
= average density of soil and concrete
H = depth of foundation
Checking for maximum stress of soil under footing
qmax qu
qmaxP
B L1
6 e
L
eL
6if
4P
3 B L 2 e( )e
L
6if
=
where
qu = design bearing capacity of soil
qu qa
1.2PD 1.6 PL
PD PL=
P = axial load on footing
P 1.2 PD P0 1.6 PL=
P0 20kN
m3
H B L=
e = eccentricity of load
Page 99
eM
P=
L B = long and width of footing
B. Determination of Depth of Footing
Checking for Punching
Vu ϕVc
where
Vu = punching shear
Vc = punching shear strength
ϕ 0.75= is a strength reduction factor for shear
Punching shear
Vu qu A A0 =
A B L=
A0 bc d hc d =
Punching shear strength
Vc 4 f'c b0 d= (in psi)
Page 100
Vc 0.332 f'c b0 d= (in MPa)
b0 bc d hc d 2=
Checking for Beam Shear
Vu1 ϕVc1
Vu2 ϕVc2
where
Vu1 Vu2 = beam shears
Vc1 Vc2 = beam shear strength
Beam shears
Vu1 qu BL
2
hc
2 d
=
Vu2 qu LB
2
bc
2 d
=
Beam shear strength
Vc1 0.166 f'c B d=
Vc2 0.166 f'c L d=
C. Determination of Steel Area
Page 101
Steel re-bars in long direction
Required strength
q1 qu B= L1L
2
hc
2=
Mu1
q1 L12
2=
Design section: rectangular singly reinforced beam of B d
Steel re-bars in short direction
Required strength
q2 qu L= L2B
2
bc
2=
Mu2
q2 L22
2=
Design section: rectangular singly reinforced beam of L d
Example 15.1
Required strength PD 484.71kN
PL 228.56kNPL
PD PL0.32
Mu 5.03kN m
Dimension of column stub bc 350mm hc 350mm
Depth of foundation H 2.0m
Allowable bearing capacity of soil qa 178.33kN
m2
3
2.5 213.996
kN
m2
Materials f'c 25MPa
fy 390MPa
Page 102
Solution
Determination of Dimension of Footing
Effective bearing capacity of soil
qe qa 20kN
m3
H 173.996kN
m2
Required area of footing
Areq
PD PL
qe4.099 m
2
Footing proportion kB
L= k
2
2.1
LAreq
k2.075 m B k L 1.976 m
L Ceil L 50mm( ) 2.1 m B Ceil B 50mm( ) 2 m
B
L
2
2.1
m
Design bearing capacity of soil
qu qa
1.2 PD 1.6 PL
PD PL 284.224
kN
m2
Checking for maximum stress of soil
Pu 1.2 PD B L H 20kN
m3
1.6 PL 1148.948 kN
eMu
Pu4.378 mm
qmax
Pu
B L1
6 e
L
eL
6if
4Pu
3 B L 2 e( )otherwise
qmax 276.981kN
m2
qmax
qu0.975
Soil "is safe" qmax quif
"is not safe" otherwise
Soil "is safe"
Page 103
Determination of depth of footing
Punching shear
A0 d( ) bc d hc d A B L
Vu d( ) qu A A0 d( ) Vu 320mm( ) 1066.154 kN
Punching shear strength
b0 d( ) bc d hc d 2
ϕ 0.75
ϕVc d( ) ϕ 0.332 MPaf'c
MPa b0 d( ) d ϕVc 320mm( ) 1067.712 kN
Beam shears
Vu1 d( ) qu BL
2
hc
2 d
Vu1 300mm( ) 326.858 kN
Vu2 d( ) qu LB
2
bc
2 d
Vu2 300mm( ) 313.357 kN
Beam shear strength
ϕVc1 d( ) ϕ 0.166 MPaf'c
MPa B d ϕVc1 300mm( ) 373.5 kN
ϕVc2 d( ) ϕ 0.166 MPaf'c
MPa L d ϕVc2 300mm( ) 392.175 kN
c 50mm 20mm20mm
2 80 mm
dmin 150mm c 70 mm
d d dmin
d d 50mm
Vu d( ) ϕVc d( ) Vu1 d( ) ϕVc1 d( ) Vu2 d( ) ϕVc2 d( ) while
d
d 320 mm h d c 400 mm
Page 104
Steel reinforcements
ρshrinkage 0.0020return( ) fy 50ksiif
0.0018return( ) fy 60ksiif
max 0.001860ksi
fy 0.0014
return
otherwise
ρshrinkage 0.0018
Re-bars in long direction
b B LnL
2
hc
2 0.875 m wu qu b
Mu
wu Ln2
2217.609 kN m Mn
Mu
0.9
Mu
b108.805
kN m
1m
RMn
b d2
1.181 MPa
ρ 0.85f'c
fy 1 1 2
R
0.85 f'c
0.00312
As max ρ b d ρshrinkage b h 19.944 cm2
s 150mm D 14mm n floorb 75mm 2 D
s
1 13
nπ D
2
4 20.012 cm
2
Re-bars in short direction
b L LnB
2
bc
2 0.825 m wu qu b
Mu
wu Ln2
2203.123 kN m Mn
Mu
0.9
Mu
b96.725
kN m
1m
RMn
b d2
1.05 MPa
ρ 0.85f'c
fy 1 1 2
R
0.85 f'c
0.00276
As max ρ b d ρshrinkage b h 18.554 cm2
s 160mm D 14mm n floorb 75mm 2 D
s
1 13
nπ D
2
4 20.012 cm
2
Page 105
16. Design of Pile Caps
1. Determination of Pile Cap
Number of required piles
nPD PL
Qe=
where
PD PL = dead and live loads on pile cap
Qe = effective bearing capacity of pile
Qe Qa 20kN
m3
3 D( )2
H=
20kN
m3
= average density of soil and concrete
D = pile size
H = depth of foundation
Distance between piles = 2 D 4 D
Distance from face of pile to face of pile cap = D
2200mm
Checking for pile reaction
Ri
P
n
My
xi
1
n
k
xk 2
M
xy
i
1
n
k
yk 2
Qu=
where
P = load on pile cap
Mx My = moments on pile cap
Qu = design bearing capacity of pile
Qu Qa
1.2 PD 1.6 PL
PD PL=
Page 106
2. Depth of Pile Cap
Case of punching
Vu ϕ Vc
where
Vu = punching shear
Vu Routside= Qu noutside=
Vc = punching shear strength
Vc 0.332 f'c b0 d=
b0 bc d hc d 2=
Case of beam shear
Vu1 ϕVc1
Vu2 ϕVc2
where
Vu1 Vu2 = beam shears
Vc1 Vc2 = beam shear strength
Vu1 max Rleft Rright
=
Vu2 max Rbottom Rtop
=
Vc1 0.166 f'c B d=
Vc2 0.166 f'c L d=
Page 107
3. Determination of Steel Reinforcements
In long direction
Required moment: Mu1 max Rleft xleft
hc
2
Rright xright
hc
2
=
Design section: Rectangular singly reinforced of B d
In short direction
Required moment: Mu2 max Rbottom xbottom
bc
2
Rtop xtop
bc
2
=
Design section: Rectangular singly reinforced of L d
Example 16.1
Pile size D 300mm Lp 9m
Allowable bearing capacity of pile Qa 351.5kN
Loads on pile cap PD 1769.88kN PL 417.11kN
My 33.92kN m Mx 56.82kN m
Depth of foundation H 1.5m
Column stub bc 350mm hc 500mm
Materials f'c 25MPa
fy 390MPa
Diameters of main barD1
D2
16mm
16mm
Concrete cover c 75mm
Depth of concrete crack hshrinkage 200mm
Diameter of shrinkage rebar Dshrinkage 12mm
Solution
Design of pile
Required strength of pile concrete
Page 108
Ag D D
f'c.pile
Qa
1
4Ag
15.622 MPa Use f'c.pile 20MPa
Steel re-bars
Ast 0.005 Ag 4.5 cm2
4π 16mm( )
2
4 8.042 cm
2
Dimension of pile cap
Effective bearing capacity of pile
Qe Qa 20kN
m3
3 D( )2
H 327.2 kN
Number of piles
nPD PL
Qe6.684
Required number of piles ceil n( ) 7
Location of pile
X
1 m
0
1m
0.5 m
0.5m
1 m
0
1m
Y
0.8m
0.8m
0.8m
0
0
0.8 m
0.8 m
0.8 m
Number of piles n rows X( ) n 8
Dimension of pile cap
B max Y( ) min Y( ) minD
2200mm
D
2
2 B 2.2m
L max X( ) min X( ) minD
2200mm
D
2
2 L 2.6m
Checking for pile reactions
Page 109
Qu Qa
1.2 PD 1.6 PL
PD PL 448.616 kN
P0 20kN
m3
H B L 171.6 kN
Pu 1.2 PD P0 1.6 PL 2997.152 kN
i 1 n ORIGIN 1
Rui
Pu
n
My Xi
1
n
k
Xk 2
Mx Y
i
1
n
k
Yk 2
Ru
Qu
0.845
0.861
0.878
0.827
0.844
0.792
0.809
0.826
Xcap
1
1
1
1
1
L
2 Ycap
1
1
1
1
1
B
2
i 1 n
Xpile i Xi
1
1
1
1
1
D
2 Ypile i Y
i
1
1
1
1
1
D
2
2 1 0 1 2
2
1
1
2
Ycap
Ypile
Y
Xcap Xpile X
Page 110
Determination of Depth of Pile Cap
Punching shear
Outside d( ) Xhc
2
d
2
Ybc
2
d
2
Vu d( ) Ru Outside d( ) Vu 700mm( ) 2247.864 kN
Punching shear strength
ϕ 0.75
b0 d( ) hc d bc d 2
ϕVc d( ) ϕ 0.332 MPaf'c
MPa b0 d( ) d ϕVc 700mm( ) 3921.75 kN
Beam shears
Left d( ) Xhc
2d
Right d( ) Xhc
2d
Bottom d( ) Ybc
2d
Top d( ) Ybc
2d
Vu1 d( ) max Ru Left d( ) Ru Right d( ) Vu1 700mm( ) 764.364 kN
Vu2 d( ) max Ru Bottom d( ) Ru Top d( ) Vu2 700mm( ) 0 N
Beam shear strength
ϕVc1 d( ) ϕ 0.166 MPaf'c
MPa B d ϕVc1 700mm( ) 958.65 kN
ϕVc2 d( ) ϕ 0.166 MPaf'c
MPa L d ϕVc2 700mm( ) 1132.95 kN
Depth of pile cap
Cover c D1D2
2 99 mm
d d 300mm Cover
d d 50mm
Vu d( ) ϕVc d( ) Vu1 d( ) ϕVc1 d( ) Vu2 d( ) ϕVc2 d( ) while
d
d 651 mm h d Cover 750 mm
Page 111
Steel Reinforcements
ρshrinkage 0.0020return( ) fy 50ksiif
0.0018return( ) fy 60ksiif
max 0.001860ksi
fy 0.0014
return
otherwise
In long direction
b B
Mu1 Left 0( ) Ruhc
2X
643.378 kN m
Mu2 Right 0( ) Ru Xhc
2
667.876 kN m
Mu max Mu1 Mu2 667.876 kN m Mn
Mu
0.9
RMn
b d2
0.796 MPa
ρ 0.85f'c
fy 1 1 2
R
0.85 f'c
0.00208
As max ρ b d ρshrinkage b h As 29.797 cm2
As1
π D12
4 n1 ceil
As
As1
15
s1 Floor
b cD1
2
2
n1 15mm
145 mm
In short direction
b L
Mu1 Bottom 0( ) Rubc
2Y
680.262 kN m
Mu2 Top 0( ) Ru Ybc
2
724.653 kN m
Mu max Mu1 Mu2 724.653 kN m Mn
Mu
0.9
Page 112
RMn
b d2
0.731 MPa
ρ 0.85f'c
fy 1 1 2
R
0.85 f'c
0.00191
As max ρ b d ρshrinkage b h As 35.1 cm2
As2
π D22
4 n2 ceil
As
As2
18
s2 Floor
b cD2
2
2
n2 15mm
140 mm
Shrinkage reinforcement
b 1m hshrinkage h hshrinkage 0=if
hshrinkage otherwise
As ρshrinkage b hshrinkage 3.6 cm2
As0
π Dshrinkage2
4 n
As
As0
sshrinkage Floorb
n5mm
310 mm
Table
L cD1
2
2
m
B cD2
2
2
m
"N/A"
D1
mm
D2
mm
Dshrinkage
mm
n1
n2
"N/A"
s1
mm
s2
mm
sshrinkage
mm
Page 113
Dimension of pile cap B= 2.20 m
L= 2.60 m
Depth of pile cap h= 750 mm
Direction Length (mm) Dia. (mm) NOS Spacing (mm)
Long 2.43 16 15 145
Short 2.03 16 18 140
Top N/A 12 N/A 310
Page 114
17. Slab Design
A. Design of One-Way Slabs
La = length of short side
Lb = length of long side
La
Lb0.5 : the slab in one-way
La
Lb0.5 : the slab is two-way
Thickness of one-way slab
Simply supportedLn
20
One end continuousLn
24
Both ends continuousLn
28
CantileverLn
10
Analysis of one-way slab
Design scheme: continuous beam
Determination of bending moments: using ACI moment coefficients
Design of one-way slab
Design section: rectangular section of 1m x h
Type section: singly reinforced beam
Page 115
Example 17.1
Span of slab Ln 2m 20cm 1.8 m
Live load LL 12kN
m2
Materials f'c 20MPa
fy 390MPa
Solution
Thickness of one-way slab
tmin
Ln
2864.286 mm
Use t 100mm
Loads on slab
Cover 50mm 22kN
m3
1.1kN
m2
Slab t 25kN
m3
2.5kN
m2
Ceiling 0.40kN
m2
Mechanical 0.20kN
m2
Partition 1.00kN
m2
DL Cover Slab Ceiling Mechanical Partition 5.2kN
m2
wu 1.2 DL 1.6 LL 25.44kN
m2
Bending moments
Msupport1
11wu Ln
2 7.493
kN m
1m
Mmidspan1
16wu Ln
2 5.152
kN m
1m
Steel reinforcements
Page 116
β1 0.65 max 0.85 0.05f'c 27.6MPa
6.9MPa
min 0.85
0.85
εu 0.003
ρmax 0.85 β1f'c
fy
εu
εu 0.005 0.014
ρmin max
0.249MPaf'c
MPa
fy
1.379MPa
fy
0.00354
ρshrinkage 0.0020return( ) fy 50ksiif
0.0018return( ) fy 60ksiif
max 0.001860ksi
fy 0.0014
return
otherwise
ρshrinkage 0.0018
Top rebars
b 1m d t 20mm10mm
2
75 mm
Mu Msupport b 7.493 kN m
Mn
Mu
0.98.326 kN m
RMn
b d2
1.48 MPa
ρ 0.85f'c
fy 1 1 2
R
0.85 f'c
0.004 ρ ρmax 1
As max ρ b d ρshrinkage b t 2.982 cm2
As0π 10mm( )
2
4 n
As
As0 smax min 3 t 450mm( )
s min Floorb
n10mm
smax
260 mm
Bottom rebars
Mu Mmidspan b 5.152 kN m
Mn
Mu
0.95.724 kN m
Page 117
RMn
b d2
1.018 MPa
ρ 0.85f'c
fy 1 1 2
R
0.85 f'c
0.003 ρ ρmax 1
As max ρ b d ρshrinkage b t 2.019 cm2
As0π 10mm( )
2
4 n
As
As0 smax min 3 t 450mm( )
s min Floorb
n10mm
smax
300 mm
Link rebars
As ρshrinkage b t 1.8 cm2
As0π 10mm( )
2
4 n
As
As0 smax min 5 t 450mm( )
s min Floorb
n10mm
smax
430 mm
B. Design of Two-Way Slabs
Design methods:- Load distribution method- Moment coefficient method- Direct design method (DDM)- Equivalent frame method- Strip method- Yield line method
Page 118
(1) Load Distribution Method
Principle: Equality of deflection in short and long directions
fa fb=
αa
wa La4
EI αb
wb Lb4
EI=
Case αa αb=
wa
wb
Lb4
La4
=1
λ4
= λ
La
Lb=
wa wb wu=
From which, wa wu1
1 λ4
=
wb wuλ
4
1 λ4
=
For λ 11
1 λ4
0.5
λ4
1 λ4
0.5
For λ 0.81
1 λ4
0.709
λ4
1 λ4
0.291
For λ 0.61
1 λ4
0.885
λ4
1 λ4
0.115
For λ 0.51
1 λ4
0.941
λ4
1 λ4
0.059
For λ 0.41
1 λ4
0.975
λ4
1 λ4
0.025
Page 119
Example 17.2
Slab dimension La 4.3m
Lb 5.5m
Live load LL 2.00kN
m2
Materials f'c 20MPa
fy 390MPa
Solution
Thickness of two-way slab
Perimeter La Lb 2
tminPerimeter
180108.889 mm
t1
30
1
50
La 143.333 86( ) mm
Use t 120mm
Loads on slab
SDL 50mm 22kN
m3
0.40kN
m2
1.00kN
m2
2.5kN
m2
DL SDL t 25kN
m3
5.5kN
m2
LL 2kN
m2
wu 1.2 DL 1.6 LL 9.8kN
m2
Load distribution
λ
La
Lb0.782
wa1
1 λ4
wu 7.134
kN
m2
wbλ
4
1 λ4
wu 2.666
kN
m2
Page 120
Bending moments
Ma.neg1
11wa La
2 11.992
kN m
1m
Ma.pos1
16wa La
2 8.245
kN m
1m
Mb.neg1
11wb Lb
2 7.33
kN m
1m
Mb.pos1
16wb Lb
2 5.04
kN m
1m
Steel reinforcements
β1 0.65 max 0.85 0.05f'c 27.6MPa
6.9MPa
min 0.85
0.85
εu 0.003
ρmax 0.85 β1f'c
fy
εu
εu 0.005 0.014
ρmin max
0.249MPaf'c
MPa
fy
1.379MPa
fy
0.00354
ρshrinkage 0.0020return( ) fy 50ksiif
0.0018return( ) fy 60ksiif
max 0.001860ksi
fy 0.0014
return
otherwise
ρshrinkage 0.0018
Top rebar in short direction
b 1m d t 20mm 10mm10mm
2
85 mm
Mu Ma.neg b 11.992 kN m
Mn
Mu
0.913.325 kN m
RMn
b d2
1.844 MPa
Page 121
ρ 0.85f'c
fy 1 1 2
R
0.85 f'c
0.005 ρ ρmax 1
As max ρ b d ρshrinkage b t 4.265 cm2
As0π 10mm( )
2
4 n
As
As0 smax min 2 t 450mm( )
s min Floorb
n10mm
smax
180 mm
Bottom rebar in short direction
Mu Ma.pos b 8.245 kN m
Mn
Mu
0.99.161 kN m
RMn
b d2
1.268 MPa
ρ 0.85f'c
fy 1 1 2
R
0.85 f'c
0.003 ρ ρmax 1
As max ρ b d ρshrinkage b t 2.875 cm2
As0π 10mm( )
2
4 n
As
As0 smax min 2 t 450mm( )
s min Floorb
n10mm
smax
240 mm
Top rebar in long direction
Mu Mb.neg b 7.33 kN m
Mn
Mu
0.98.145 kN m
RMn
b d2
1.127 MPa
ρ 0.85f'c
fy 1 1 2
R
0.85 f'c
0.003 ρ ρmax 1
As max ρ b d ρshrinkage b t 2.544 cm2
Page 122
As0π 10mm( )
2
4 n
As
As0 smax min 2 t 450mm( )
s min Floorb
n10mm
smax
240 mm
Bottom rebar in long direction
Mu Mb.pos b 5.04 kN m
Mn
Mu
0.95.599 kN m
RMn
b d2
0.775 MPa
ρ 0.85f'c
fy 1 1 2
R
0.85 f'c
0.002 ρ ρmax 1
As max ρ b d ρshrinkage b t 2.16 cm2
As0π 10mm( )
2
4 n
As
As0 smax min 2 t 450mm( )
s min Floorb
n10mm
smax
240 mm
Shrinkage rebars
b 1m
As ρshrinkage b t 2.16 cm2
As0π 10mm( )
2
4 n
As
As0 smax min 5 t 450mm( )
s min Floorb
n10mm
smax
360 mm
Page 123
(2) Moment Coefficient Method
Negative moments
Ma.neg Ca.neg wu La2
=
Mb.neg Cb.neg wu Lb2
=
Positive moments
Ma.pos Ca.pos.DL wD La2
Ca.pos.LL wL La2
=
Mb.pos Cb.pos.DL wD Lb2
Cb.pos.LL wL Lb2
=
where Ca.neg Cb.neg Ca.pos.DL Ca.pos.LL Cb.pos.DL Cb.pos.LL
are tabulated moment coefficients
wD 1.2 DL= wL 1.6 LL=
wu 1.2 1.6 LL=
Example 17.3
Slab dimension La 5.0m 25cm 4.75 m
Lb 5.5m 20cm 5.3 m
Live load for office LL 2.40kN
m2
Materials f'c 20MPa fy 390MPa
Boundary conditions in short and long directions
Simple
Continuous
0
1
ShortContinuous
Continuous
LongContinuous
Continuous
Solution
Thickness of two-way slab
Perimeter La Lb 2
Page 124
tminPerimeter
180111.667 mm
t1
30
1
50
La 158.333 95( ) mm
Use t 120mm
Loads on slab
Cover 50mm 22kN
m3
1.1kN
m2
Slab t 25kN
m3
3kN
m2
Ceiling 0.40kN
m2
Partition 1.00kN
m2
SDL Cover Ceiling Partition 2.5kN
m2
DL SDL Slab 5.5kN
m2
wD 1.2 DL
LL 2.4kN
m2
wL 1.6 LL
wu 1.2 DL 1.6LL 10.44kN
m2
Moment coefficients
Page 125
Table 12.3aCoefficients for negative moments in short direction of slab
m Case 1 Case 2 Case 3 Case 4 Case 5 Case 6 Case 7 Case 8 Case 91.00 0.000 0.045 0.000 0.050 0.075 0.071 0.000 0.033 0.0610.95 0.000 0.050 0.000 0.055 0.079 0.075 0.000 0.038 0.0650.90 0.000 0.055 0.000 0.060 0.080 0.079 0.000 0.043 0.0680.85 0.000 0.060 0.000 0.066 0.082 0.083 0.000 0.049 0.0720.80 0.000 0.065 0.000 0.071 0.083 0.086 0.000 0.055 0.0750.75 0.000 0.069 0.000 0.076 0.085 0.088 0.000 0.061 0.0780.70 0.000 0.074 0.000 0.081 0.086 0.091 0.000 0.068 0.0810.65 0.000 0.077 0.000 0.085 0.087 0.093 0.000 0.074 0.0830.60 0.000 0.081 0.000 0.089 0.088 0.095 0.000 0.080 0.0850.55 0.000 0.084 0.000 0.092 0.089 0.096 0.000 0.085 0.0860.50 0.000 0.086 0.000 0.094 0.090 0.097 0.000 0.089 0.088
Table 12.3bCoefficients for negative moments in long direction of slab
m Case 1 Case 2 Case 3 Case 4 Case 5 Case 6 Case 7 Case 8 Case 91.00 0.000 0.045 0.076 0.050 0.000 0.000 0.071 0.061 0.0330.95 0.000 0.041 0.072 0.045 0.000 0.000 0.067 0.056 0.0290.90 0.000 0.037 0.070 0.040 0.000 0.000 0.062 0.052 0.0250.85 0.000 0.031 0.065 0.034 0.000 0.000 0.057 0.046 0.0210.80 0.000 0.027 0.061 0.029 0.000 0.000 0.051 0.041 0.0170.75 0.000 0.022 0.056 0.024 0.000 0.000 0.044 0.036 0.0140.70 0.000 0.017 0.050 0.019 0.000 0.000 0.038 0.029 0.0110.65 0.000 0.014 0.043 0.015 0.000 0.000 0.031 0.024 0.0080.60 0.000 0.010 0.035 0.011 0.000 0.000 0.024 0.018 0.0060.55 0.000 0.007 0.028 0.008 0.000 0.000 0.019 0.014 0.0050.50 0.000 0.006 0.022 0.006 0.000 0.000 0.014 0.010 0.003
ORIGIN 1
Index1
2
2
3
I IndexShort1 1 Short2 1 3
J IndexLong1 1 Long2 1 3
Table
1
6
5
7
4
9
3
8
2
Case TableI J 2
Vλ reverse Vλ( )
Vaneg reverse Taneg Case Vbneg reverse Tbneg Case
VaposDL reverse TaposDL Case VbposDL reverse TbposDL Case
VaposLL reverse TaposLL Case VbposLL reverse TbposLL Case
λ
La
Lb0.896
vs1 pspline Vλ Vaneg( ) Ca.neg interp vs1 Vλ Vaneg λ( ) 0.055
Page 126
vs2 pspline Vλ Vbneg( ) Cb.neg interp vs2 Vλ Vbneg λ( ) 0.037
vs3 pspline Vλ VaposDL( ) Ca.pos.DL interp vs3 Vλ VaposDL λ( ) 0.022
vs4 pspline Vλ VbposDL( ) Cb.pos.DL interp vs4 Vλ VbposDL λ( ) 0.014
vs5 pspline Vλ VaposLL( ) Ca.pos.LL interp vs5 Vλ VaposLL λ( ) 0.034
vs6 pspline Vλ VbposLL( ) Cb.pos.LL interp vs6 Vλ VbposLL λ( ) 0.022
Bending moments
Ma.neg Ca.neg wu La2
13.043kN m
1m
Mb.neg Cb.neg wu Lb2
10.729kN m
1m
Ma.pos Ca.pos.DL wD La2
Ca.pos.LL wL La2
6.266kN m
1m
Mb.pos Cb.pos.DL wD Lb2
Cb.pos.LL wL Lb2
4.911kN m
1m
Steel reinforcements
β1 0.65 max 0.85 0.05f'c 27.6MPa
6.9MPa
min 0.85
0.85
εu 0.003
ρmax 0.85 β1f'c
fy
εu
εu 0.005 0.014
ρmin max
0.249MPaf'c
MPa
fy
1.379MPa
fy
0.00354
ρshrinkage 0.0020return( ) fy 50ksiif
0.0018return( ) fy 60ksiif
max 0.001860ksi
fy 0.0014
return
otherwise
ρshrinkage 0.0018
Top rebars in short direction
b 1m d t 20mm 10mm10mm
2
85 mm
Page 127
Mu Ma.neg b 13.043 kN m
Mn
Mu
0.914.492 kN m
RMn
b d2
2.006 MPa
ρ 0.85f'c
fy 1 1 2
R
0.85 f'c
0.005 ρ ρmax 1
As max ρ b d ρshrinkage b t 4.666 cm2
As0π 10mm( )
2
4 n
As
As0 smax min 2 t 450mm( )
s min Floorb
n10mm
smax
160 mm
Bottom rebars in short direction
Mu Ma.pos b 6.266 kN m
Mn
Mu
0.96.963 kN m
RMn
b d2
0.964 MPa
ρ 0.85f'c
fy 1 1 2
R
0.85 f'c
0.003 ρ ρmax 1
As max ρ b d ρshrinkage b t 2.163 cm2
As0π 10mm( )
2
4 n
As
As0 smax min 2 t 450mm( )
s min Floorb
n10mm
smax
240 mm
Top rebars in long direction
Mu Mb.neg b 10.729 kN m
Mn
Mu
0.911.921 kN m
Page 128
RMn
b d2
1.65 MPa
ρ 0.85f'c
fy 1 1 2
R
0.85 f'c
0.004 ρ ρmax 1
As max ρ b d ρshrinkage b t 3.79 cm2
As0π 10mm( )
2
4 n
As
As0 smax min 2 t 450mm( )
s min Floorb
n10mm
smax
200 mm
Bottom rebars in long direction
Mu Mb.pos b 4.911 kN m
Mn
Mu
0.95.457 kN m
RMn
b d2
0.755 MPa
ρ 0.85f'c
fy 1 1 2
R
0.85 f'c
0.002 ρ ρmax 1
As max ρ b d ρshrinkage b t 2.16 cm2
As0π 10mm( )
2
4 n
As
As0 smax min 2 t 450mm( )
s min Floorb
n10mm
smax
240 mm
Shrinkage rebars
b 1m
As ρshrinkage b t 2.16 cm2
As0π 10mm( )
2
4 n
As
As0 smax min 5 t 450mm( )
s min Floorb
n10mm
smax
360 mm
Page 129
(3) Direct Design Method (DDM)
Total static moment
M0
wu L2 Ln2
8=
Longitudinal distribution of moments
Mneg Cneg M0=
Mpos Cpos M0=
Lateral distribution of moments
Mneg.col Cneg.col Mneg=
Mneg.mid Cneg.mid Mneg=
Mpos.col Cpos.col Mpos=
Mpos.mid Cpos.mid Mpos=
Page 130
Example 17.4
Slab dimension La 4m Lb 6m
Live load for hospital LL 3.00kN
m2
Materials f'c 25MPa fy 390MPa
Page 131
Solution
Section of beam in long direction
L Lb 6 m
h1
10
1
15
L 600 400( ) mm h 500mm
b 0.3 0.6( ) h 150 300( ) mm b 250mm
bb
hb
b
h
Section of beam in short direction
L La 4 m
h1
10
1
15
L 400 266.667( ) mm h 300mm
b 0.3 0.6( ) h 90 180( ) mm b 200mm
ba
ha
b
h
Determination of slab thickness
Perimeter La Lb 2
tminPerimeter
180111.111 mm
Assume t 120mm
In long direction
bw bb h hb hf t
hw h hf
b min bw 2 hw bw 8 hf 1.01 m
A1 bw h x1h
2
A2 b bw hf x2
hf
2
xc
x1 A1 x2 A2
A1 A2169.852 mm
I1
bw h3
12A1 x1 xc 2
I2
b bw hf3
12A2 x2 xc 2
Page 132
Ib I1 I2 4.617 105
cm4
Is
La hf3
12
wc 24kN
m3
Ec 44MPawc
kN
m3
1.5
f'c
MPa 2.587 10
4 MPa
α
Ec Ib
Ec Is8.016 αb α
In short direction
bw ba h ha hf t
hw h hf
b min bw 2 hw bw 8 hf 0.56 m
A1 bw h x1h
2
A2 b bw hf x2
hf
2
xc
x1 A1 x2 A2
A1 A2112.326 mm
I1
bw h3
12A1 x1 xc 2
I2
b bw hf3
12A2 x2 xc 2
Ib I1 I2 7.053 104
cm4
Iba Ib
Is
Lb hf3
128.64 10
4 cm
4
α
Ec Ib
Ec Is0.816 αa α
Required thickness of slab
αm
αa 2 αb 2
44.416
β
Lb
La1.5
Ln Lb 20cm 5.8 m
Page 133
hf max
Ln 0.8fy
200ksi
36 5 β αm 0.2 5in
0.2 αm 2.0if
max
Ln 0.8fy
200ksi
36 9 β3.5in
2.0 αm 5.0if
"DDM is not applied" otherwise
hf 126.876 mm
Loads on slab
DL 50mm 22kN
m3
t 25kN
m3
0.40kN
m2
1.00kN
m2
5.5kN
m2
LL 3kN
m2
wu 1.2 DL 1.6 LL 11.4kN
m2
In long direction
L1 Lb 6 m Ln L1 ba 5.8 m
L2 La 4 m α1 αb 8.016
Total static moment
M0
wu L2 Ln2
8191.748 kN m
Longitudinal distribution of moments
Mneg 0.65 M0 124.636 kN m
Mpos 0.35 M0 67.112 kN m
Lateral distribution of moments
k1
L2
L10.667 k2 α1
L2
L1 5.344
linterp2 VX VY M x y( )
Vj
linterp VX M j x
j 1 rows VY( )for
linterp VY V y( )
Page 134
Cneg.col linterp2
0
1
10
0.5
1.0
2.0
0.75
0.90
0.90
0.75
0.75
0.75
0.75
0.45
0.45
k2 k1
0.85
Cneg.mid 1 Cneg.col 0.15
Cpos.col linterp2
0
1
10
0.5
1.0
2.0
0.60
0.90
0.90
0.60
0.75
0.75
0.60
0.45
0.45
k2 k1
0.85
Cpos.mid 1 Cpos.col 0.15
Mneg.col Cneg.col Mneg 105.941 kN m
Mneg.mid Cneg.mid Mneg 18.695 kN m
Mpos.col Cpos.col Mpos 57.045 kN m
Mpos.mid Cpos.mid Mpos 10.067 kN m
Ccol.beam linterp
0
1
10
0
0.85
0.85
k2
0.85
Ccol.slab 1 Ccol.beam 0.15
Mneg.col.beam Ccol.beam Mneg.col 90.05 kN m
Mneg.col.slab Ccol.slab Mneg.col 15.891 kN m
Mpos.col.beam Ccol.beam Mpos.col 48.488 kN m
Mpos.col.slab Ccol.slab Mpos.col 8.557 kN m
bcol
min L1 L2 4
2 2 m
bmid L2 bcol 2 m
Top rebars in column strip
b bcol d t 20mm 10mm10mm
2
85 mm
Mu Mneg.col.slab 15.891 kN m
Mn
Mu
0.917.657 kN m
Page 135
RMn
b d2
1.222 MPa
ρ 0.85f'c
fy 1 1 2
R
0.85 f'c
0.003 ρ ρmax 1
As max ρ b d ρshrinkage b t 5.489 cm2
As0π 10mm( )
2
4 n
As
As0 smax min 2 t 450mm( )
s min Floorb
n10mm
smax
240 mm
Bottom rebars in column strip
Mu Mpos.col.slab 8.557 kN m
Mn
Mu
0.99.508 kN m
RMn
b d2
0.658 MPa
ρ 0.85f'c
fy 1 1 2
R
0.85 f'c
0.002 ρ ρmax 1
As max ρ b d ρshrinkage b t 4.32 cm2
As0π 10mm( )
2
4 n
As
As0 smax min 2 t 450mm( )
s min Floorb
n10mm
smax
240 mm
Top rebars in middle strip
b bmid
Mu Mneg.mid 18.695 kN m
Mn
Mu
0.920.773 kN m
RMn
b d2
1.438 MPa
ρ 0.85f'c
fy 1 1 2
R
0.85 f'c
0.004 ρ ρmax 1
Page 136
As max ρ b d ρshrinkage b t 6.494 cm2
As0π 10mm( )
2
4 n
As
As0 smax min 2 t 450mm( )
s min Floorb
n10mm
smax
240 mm
Bottom rebars in middle strip
Mu Mpos.mid 10.067 kN m
Mn
Mu
0.911.185 kN m
RMn
b d2
0.774 MPa
ρ 0.85f'c
fy 1 1 2
R
0.85 f'c
0.002 ρ ρmax 1
As max ρ b d ρshrinkage b t 4.32 cm2
As0π 10mm( )
2
4 n
As
As0 smax min 2 t 450mm( )
s min Floorb
n10mm
smax
240 mm
In short direction
L1 La 4 m Ln L1 bb 3.75 m
L2 Lb 6 m α1 αa 0.816
Total static moment
M0
wu L2 Ln2
8120.234 kN m
Longitudinal distribution of moments
Mneg 0.65 M0 78.152 kN m
Mpos 0.35 M0 42.082 kN m
Lateral distribution of moments
Page 137
k1
L2
L11.5 k2 α1
L2
L1 1.224
Cneg.col linterp2
0
1
10
0.5
1.0
2.0
0.75
0.90
0.90
0.75
0.75
0.75
0.75
0.45
0.45
k2 k1
0.6
Cneg.mid 1 Cneg.col 0.4
Cpos.col linterp2
0
1
10
0.5
1.0
2.0
0.60
0.90
0.90
0.60
0.75
0.75
0.60
0.45
0.45
k2 k1
0.6
Cpos.mid 1 Cpos.col 0.4
Mneg.col Cneg.col Mneg 46.891 kN m
Mneg.mid Cneg.mid Mneg 31.261 kN m
Mpos.col Cpos.col Mpos 25.249 kN m
Mpos.mid Cpos.mid Mpos 16.833 kN m
Ccol.beam linterp
0
1
10
0
0.85
0.85
k2
0.85
Ccol.slab 1 Ccol.beam 0.15
Mneg.col.beam Ccol.beam Mneg.col 39.858 kN m
Mneg.col.slab Ccol.slab Mneg.col 7.034 kN m
Mpos.col.beam Ccol.beam Mpos.col 21.462 kN m
Mpos.col.slab Ccol.slab Mpos.col 3.787 kN m
bcol
min L1 L2 4
2 2 m
bmid L2 bcol 4 m
Top rebars in column strip
b bcol
Mu Mneg.col.slab 7.034 kN m
Mn
Mu
0.97.815 kN m
Page 138
RMn
b d2
0.541 MPa
ρ 0.85f'c
fy 1 1 2
R
0.85 f'c
0.001 ρ ρmax 1
As max ρ b d ρshrinkage b t 4.32 cm2
As0π 10mm( )
2
4 n
As
As0 smax min 2 t 450mm( )
s min Floorb
n10mm
smax
240 mm
Bottom rebars in column strip
Mu Mpos.col.slab 3.787 kN m
Mn
Mu
0.94.208 kN m
RMn
b d2
0.291 MPa
ρ 0.85f'c
fy 1 1 2
R
0.85 f'c
0.001 ρ ρmax 1
As max ρ b d ρshrinkage b t 4.32 cm2
As0π 10mm( )
2
4 n
As
As0 smax min 2 t 450mm( )
s min Floorb
n10mm
smax
240 mm
Top rebars in middle strip
b bmid
Mu Mneg.mid 31.261 kN m
Mn
Mu
0.934.734 kN m
RMn
b d2
1.202 MPa
ρ 0.85f'c
fy 1 1 2
R
0.85 f'c
0.003 ρ ρmax 1
Page 139
As max ρ b d ρshrinkage b t 10.792 cm2
As0π 10mm( )
2
4 n
As
As0 smax min 2 t 450mm( )
s min Floorb
n10mm
smax
240 mm
Bottom rebars in middle strip
Mu Mpos.mid 16.833 kN m
Mn
Mu
0.918.703 kN m
RMn
b d2
0.647 MPa
ρ 0.85f'c
fy 1 1 2
R
0.85 f'c
0.002 ρ ρmax 1
As max ρ b d ρshrinkage b t 8.64 cm2
As0π 10mm( )
2
4 n
As
As0 smax min 2 t 450mm( )
s min Floorb
n10mm
smax
240 mm
Page 140
Page 141
18. Design of Staircase
Step dimension G3.1m
11281.818 mm
H1.8m
11163.636 mm
G 2 H 60.909 cm Reference: G 2 H 60cm= 64cmH 150mm= 190mm
Number of steps n 11
Loads on waist slab
Slope angle α atanH
G
30.141 deg
Thickness of waist slab twaist 120mm
Step cover Cover 50mm H G( ) 22kN
m3
1m
1m G 1.739
kN
m2
Concrete step StepG H
224
kN
m3
1m
1m G 1.964
kN
m2
RC slab Slab twaist 25kN
m3
1m2
1m2
cos α( ) 3.469
kN
m2
Page 142
Ceiling Ceiling 0.40kN
m2
1m2
1m2
cos α( ) 0.463
kN
m2
Handrail Handrail 0.50kN
m2
Dead load DL Cover Step Slab Ceiling Handrail
DL 8.134kN
m2
Live load LL 4.80kN
m2
Factored load wwaist 1.2 DL 1.6 LL 17.441kN
m2
Loads on landing slab
Thickness of landing slab tlanding 150mm
Slab cover Cover 50mm 22kN
m3
1.1kN
m2
RC slab Slab tlanding 25kN
m3
3.75kN
m2
Ceiling Ceiling 0.40kN
m2
Handrail Handrail 0.50kN
m2
Dead load DL Cover Slab Ceiling Handrail 5.75kN
m2
Live load LL 4.80kN
m2
Factored load wlanding 1.2 DL 1.6 LL 14.58kN
m2
Analysis of Staircase
Concrete modulus of elasticity
f'c 25MPa
wc 24kN
m3
Ec 44MPawc
kN
m3
1.5
f'c
MPa 2.587 10
4 MPa
Page 143
Geometry of staicase
L0 3.1m L2 1.9m h 1.8m
L1 L02
h2
3.585 m
t1 twaist 120 mm t2 tlanding 150 mm
Flexural stiffness
b 1m
EI1 Ec
b t13
12 EI2 Ec
b t23
12
Loads on staircase
w1 wwaist b 17.441kN
m
w2 wlanding b 14.58kN
m
Coefficients
r11 4EI1
L1 3
EI2
L2 R1p
w1 L02
12
w2 L22
8
Angular rotation
Z1
R1p
r114.723 10
4 φB Z1
Bending moments
MA 2EI1
L1 Z1
w1 L02
12 14.949 kN m
MBA 4EI1
L1 Z1
w1 L02
12 12.004 kN m
MBC 3EI2
L2 Z1
w2 L22
8 12.004 kN m
MC 0
Shears
w0 w1 cos α( )2
13.043kN
m
VAB
MBA MA
L1
w0 L1
2 24.199 kN
Page 144
VBA VAB w0 L1 22.557 kN
VBC
MC MBC
L2
w2 L2
2 20.169 kN
VCB VBC w2 L2 7.533 kN
Positive moments
x1
VAB
w01.855 m
Mmax.AB MA VAB x1w0 x1
2
2 7.5 kN m
x2
VBC
w21.383 m
Mmax.BC MBC VBC x2w2 x2
2
2 1.946 kN m
Design of Staircase
Materials
f'c 25MPa fy 390MPa
εu 0.003
β1 0.65 max 0.85 0.05f'c 27.6MPa
6.9MPa
min 0.85
0.85
ρmax 0.85 β1f'c
fy
εu
εu 0.005 0.017
ρmin max
0.249MPaf'c
MPa
fy
1.379MPa
fy
ρshrinkage 0.0020return fy 50ksiif
0.0018return fy 60ksiif
max 0.001860ksi
fy 0.0014
return otherwise
ρshrinkage 0.0018
Top rebars in waist slab
Page 145
b 1 m d twaist 30mm16mm
2
82 mm
Mu max MA MBA 14.949 kN m Mn
Mu
0.9
RMn
b d2
2.47 MPa
ρ 0.85f'c
fy 1 1 2
R
0.85 f'c
0.00675
As max ρ b d ρshrinkage b twaist 5.537 cm2
b
200mm
π 14mm( )2
4 7.697 cm
2
Top rebars in landing slab
b 1 m d tlanding 30mm16mm
2
112 mm
Mu max MBC MC 12.004 kN m Mn
Mu
0.9
RMn
b d2
1.063 MPa
ρ 0.85f'c
fy 1 1 2
R
0.85 f'c
0.0028
As max ρ b d ρshrinkage b twaist 3.134 cm2
b
200mm
π 10mm( )2
4 3.927 cm
2
Bottom rebars in waist slab
b 1 m d twaist 30mm16mm
2
82 mm
Mu Mmax.AB 7.5 kN m Mn
Mu
0.9
RMn
b d2
1.239 MPa
ρ 0.85f'c
fy 1 1 2
R
0.85 f'c
0.00328
As max ρ b d ρshrinkage b twaist 2.687 cm2
Page 146
b
200mm
π 10mm( )2
4 3.927 cm
2
Bottom rebars in landing slab
b 1 m d tlanding 30mm16mm
2
112 mm
Mu Mmax.BC 1.946 kN m Mn
Mu
0.9
RMn
b d2
0.172 MPa
ρ 0.85f'c
fy 1 1 2
R
0.85 f'c
0.00044
As max ρ b d ρshrinkage b twaist 2.16 cm2
b
200mm
π 10mm( )2
4 3.927 cm
2
Link rebars
b 1m t max twaist tlanding 150 mm
As ρshrinkage b t 2.7 cm2
b
250mm
π 10mm( )2
4 3.142 cm
2
Page 147
19. Deflection
A. Immediate (Initial) Deflection
Initial or short-term deflection in midspan of continuous beam
Δi K5 Ma Ln
2
48 Ec Ie=
where Ma = support moment for cantilever or midspan moment for simple or cotinuousbeam
Ln = span length of beam
Ec = concrete modulus of elasticity
Ie = effective moment of inertia of cracked section
K = deflection coefficient for uniform distributed load w
1. Cantilever K 2.40=
2. Simple beam K 1.0=
3. Continuous beam K 1.2 0.2M0
Ma=
4. Fixed-hinged beam
Midspan deflection K 0.8=
Max. deflection usingmax. moment
K 0.74=
5. Fixed-fixed beam K 0.60=
where M0
w Ln2
8=
Effective moment of inertia of cracked section
Ie Icr
Mcr
Ma
3
Ig Icr Ig=
where
Icr = moment of inertia of cracked transformed section
Ig = moment of inertia of gross section
Mcr = cracking moment
Page 148
Case of continuous beams
According ACI Code 9.5.2 for continuous span
Ie 0.50 Im 0.25 Ie1 Ie2 =
Beams with both ends continuous
Ie 0.70 Im 0.15 Ie1 Ie2 =
Beams with one end continuous
Ie 0.85 Im 0.15 Icont.end=
where
Im = midspan section Ie
Ie1 Ie2 = Ie for the respective beam ends
Icont.end = Ie of continuous end
Transformed Section
εc
fc
Ec= εs=
fs
Es=
From which fs
Es
Ecfc= n fc=
where nEs
Ec= is a modulus ratio
Axial force
P Ac fc As fs= Ac n As fc= At fc=
where At Ac n As= Ag n 1( ) As=
is a transformed section
Ag = area of gross section
Cracking Moment
Mcr
Iut fr
yt= (exact expression)
Page 149
Mcr
Ig fr
yt= (simplied expression)
where
Iut = moment of inertia of uncracked transformed section At
Ig = moment of inertia of gross section Ag
fr = modulus of rupture
fr 7.5psif'c
psi= 0.623MPa
f'c
MPa=
yt = distance from neutral axis to the tension face
Moment of inertia of cracked section
Condition of strain compatibity
εs
εu
d x
x= εs εu
d x
x=
Equilibrium in forces
C T=
fc x
2b As fs=
Ec εu b x
2As Es εs= As Es εu
d x
x=
b x2
2 n As d x( )=
Page 150
b x2
b d2
2 n As d x( )
b d2
=
x
d
2
2 n ρ 1x
d
=
x
d
2
2 n ρx
d 2 n ρ 0=
x
dn ρ n ρ( )
22 n ρ=
Moment of inertia of cracked section
Icrb x
3
3n As d x( )
2=
B. Long-Term Deflection
Long-term deflection due to combined effect of creep and shrinkage
Δt λ Δi= λξ
1 50 ρ'=
where ρ'A's
b d=
ξ = time-dependent coefficient
Sustained load duration Value ξ________________________________________________
5 year and more 2.0
12 months 1.4
6 months 1.2
3 months 1.0
Page 151
C. Minimum Depth-Span Ratio
D. Permisible Deflection
Page 152
Example
Span length Ln 9.8m 60cm 9.2m
Beam section b 300mm h 750mm
Bending moments MD.neg 419.34kN m MD.pos 319.33kN m
ML.neg 223.09kN m ML.pos 176.58kN m
Steel re-bars As.sup 6π 25mm( )
2
4 29.452 cm
2
A's.sup 3π 25mm( )
2
4 14.726 cm
2
As.mid 5π 25mm( )
2
4 24.544 cm
2
A's.mid 3π 25mm( )
2
4 14.726 cm
2
Materials f'c 25MPa
fy 390MPa
Solution
Modulus of rupture
fr 0.623MPaf'c
MPa 3.115 MPa
Modulus ratio
Es 2 105MPa
wc 24kN
m3
Ec 44MPawc
kN
m3
1.5
f'c
MPa 2.587 10
4 MPa
nEs
Ec7.732
Support section
Centroid of uncracked section
A1 b h y1h
2
As As.sup d h 30mm 10mm 20mm 25mm40mm
2
645 mm
Page 153
A2 n As y2 d
yc
A1 y1 A2 y2
A1 A2399.815 mm
Moment of inertia of gross section
I1b h
3
12
Ig I1 A1 y1 yc 2 A2 y2 yc 2 1.205 106
cm4
Cracking moment
yt h yc 350.185 mm
Mcr fr
Ig
yt 107.228 kN m
Location of neutral axis of cracked section
C T=
fc x
2b As fs=
Ec εu x b 2 As Es εs=
εu x b 2 AsEs
Eu εu
d x
x=
b x2
2 As n d x( )=
x
d
2
2 ρ n 1x
d
= ρ
As
b d0.015
x
d
2
2 ρ nx
d 2 ρ n 0=
x d ρ n ρ n( )2
2 ρ n 246.092 mm
Moment of inertia of cracked section
Icrb x
3
3A2 d x( )
2 5.114 10
5 cm
4
Effective moment of inertia of cracked section
Mneg MD.neg ML.neg 642.43 kN m Ma Mneg
Ie1 min Icr
Mcr
Ma
3
Ig Icr Ig
5.146 105
cm4
Page 154
Ie2 Ie1
Midspan section
Centroid of uncracked section
A1 b h y1h
2
As As.mid d h 30mm 10mm 25mm40mm
2
665 mm
A2 n As y2 d
yc
A1 y1 A2 y2
A1 A2397.557 mm
Moment of inertia of gross section
I1b h
3
12
Ig I1 A1 y1 yc 2 A2 y2 yc 2 1.202 106
cm4
Cracking moment
yt h yc 352.443 mm
Mcr fr
Ig
yt 106.225 kN m
Location of neutral axis of cracked section
C T=
fc x
2b As fs=
Ec εu x b 2 As Es εs=
εu x b 2 AsEs
Eu εu
d x
x=
b x2
2 As n d x( )=
x
d
2
2 ρ n 1x
d
= ρ
As
b d0.012
x
d
2
2 ρ nx
d 2 ρ n 0=
x d ρ n ρ n( )2
2 ρ n 233.616 mm
Page 155
Moment of inertia of cracked section
Icrb x
3
3A2 d x( )
2 4.806 10
5 cm
4
Effective moment of inertia of cracked section
Mpos MD.pos ML.pos 495.91 kN m Ma Mpos
Im min Icr
Mcr
Ma
3
Ig Icr Ig
4.877 105
cm4
Calculation of deflection
Effective moment of inertia
Ie 0.70 Im 0.15 Ie1 Ie2 4.958 105
cm4
Initial deflection due to dead and live loads
Ma 495.91 kN m
M0 Mneg Mpos 1138.34 kN m
K 1.2 0.2M0
Ma 0.741
ΔD+L K5 Ma Ln
2
48 Ec Ie 25.259 mm
Long-term deflection due to dead load
ξ 2
A's A's.mid ρ'A's
b d
λξ
1 50 ρ'1.461
ΔD λ ΔD+LMD.pos
Mpos 23.761 mm
Long-term deflection due to sustained live load
Δ0.20L ΔD+L
0.20 ML.pos
Mpos 1.799 mm
Short-term deflection due to live load
Page 156
Δ0.80L ΔD+L
0.80 ML.pos
Mpos 7.195 mm
Total deflection
Δ ΔD Δ0.20L Δ0.80L 32.755 mm
Permisible deflection
Ln
48019.167 mm
Ln
36025.556 mm
Page 157
20. Development Lengths
A. Development length of deformed bar in tension
Diameter of deformed bar db 20mm
Steel yield strength fy 390MPa
Concrete compression strength f'c 25MPa
Depth of concrete below development length H 350mm
Reinforcement coating
Type of concrete
Concrete cover c 1 db
Clear spacing of re-bars s 2 db
ψt 1.3 H 300mmif
1.0 otherwise
ψt 1.3
ψe 1.0 Coating "Uncoated"=if
1.5 c 3db s 6dbif
1.2 otherwise
otherwise
ψe 1
ψs 0.8 db 20mmif
1.0 otherwise
ψs 0.8
λ 1.3 Concrete "Lightweight"=if
1.0 otherwise
λ 1
Ktr 0 (for a design simplification)
cb 1.5db max min cdb
2
s db
2
min 2.5db
30 mmcb Ktr
db1.5
Development of tension bar in tension
Page 158
Ld maxfy
1.107MPaf'c
MPa
ψt ψe ψs λ
cb Ktr
db
db 300mm
977.055 mm
Ld
db48.853 Ceil Ld 10mm 980 mm
B. Splice length in tension
Splice class
Lst 1.0 Ld Class "Class A"=if
1.3 Ld otherwise
Lst 1270.172 mmLst
db63.509
Ceil Lst 10mm 1280 mm
C. Development length of deformed bar in compression
Diameter of development bar db 32mm
Steel yield strength fy 390MPa
Concrete compression strength f'c 25MPa
Development length in compression
Ldc maxfy
4.152MPaf'c
MPa
dbfy
22.983MPadb 200mm
601.156 mm
Ldc
db18.786 Ceil Ldc 10mm 610 mm
Page 159
D. Development length of standard hook in tension
Diameter of development bar db 10mm
Steel yield strength fy 390MPa
Concrete compression strength f'c 25MPa
Reinforcement coating
Type of concrete
Side cover cside 65mm
Cover beyond hook cbeyond 50mm
ψe 1.0 Coating "Uncoated"=if
1.5 c 3db s 6dbif
1.2 otherwise
otherwise
ψe 1
λ 1.3 Concrete "Lightweight"=if
1.0 otherwise
λ 1
Page 160
Modified factor α 0.8 cside 65mm cbeyond 50mm if
0.7 otherwise
α 0.7
Development length of standard hook in tension
Ldh α maxψe λ fy
4.152MPaf'c
MPa
db 8db 150mm
131.503 mm
Ldh
db13.15 Ceil Ldh 10mm 140 mm
E. Lap Splice Length in Compression
Diameter of splice bar db 25mm
Steel yield strength fy 390MPa
Lap splice length in compression Lsc maxfy
13.79MPadb 300mm
fy 60ksiif
maxfy
7.661MPa24
db 300mm
otherwise
Lsc 707.034 mm Ceil Lsc 10mm 710 mm
Lsc
db28.281
Page 161
Reference
1. Reinforced Concrete / A Fundamental Approch. 5th Edition. Edward G. Nawy. - PearsonPrentice Hall, 2005.
2. Design of Concrete Structures / Arthur Nilson, David Darwin, Charles W/ Dolan - 13 ed.McGraw-Hill, 2003.
3. Structural Concrete: Theory and Design / M. Nadim Hassoun, Akthem Al-Manaseer. - 3rdEdotion. John Wiley and Sons, 2005.
4. Reinforced Concrete: Mechanics and Design / James McGregor, James K. Wight. - 4th Edition.Pearson Education, 2005.
5. ACI Code 318-05
Page 162