Seun Sambath-ACI 318-05 Mathcad in Concrete Structures (2010)

CONSTRUCTION MANAGEMENT
Concept-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 1
lbf
in
2

ksi = kilopound per square inch 1ksi 1
kip
in
2

psf = pound per square foot 1psf 1
lbf
ft
2

1psi 6.895 kPa 1kPa 0.145 psi 1Pa 1
N
m
2

1ksi 6.895 MPa 1MPa 0.145 ksi 1kPa 1
kN
m
2

1MPa 1
N
mm
2

1psf 0.048
kN
m
2
 1
kN
m
2
20.885 psf
1psf 4.882
kgf
m
2
 1
kgf
m
2
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 1.915
kN
m
2
 200
kgf
m
2
40.963 psf
100psf 4.788
kN
m
2
 4.80
kN
m
2
100.25 psf
Page 2

Density
1pcf 1
lbf
ft
3

125pcf 19.636
kN
m
3
 145pcf 22.778
kN
m
3

Moments
1ft kip 1.356 kN m
User setting
Riels 1
USD 4165Riels
124USD 516460 Riels
200000Riels 48.019 USD
CostSteel 665
USD
1tonnef

670kgf CostSteel 445.55 USD
Page 3

2. Simple Calculation
1 3
2
3







3
23.472
a 2 b 4 c 1
Δ b
2
4 a c
x1
b Δ
2 a
0.225
x2
b Δ
2 a
2.225
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
Concrete ultimate strain
εu 0.003
Cubic and cylinder compression strength
fcube
f'c
0.85
=
f'c 20MPa fcube
f'c
0.85
23.529 MPa
f'c
0.85
29.412 MPa
f'c
0.85
41.176 MPa
Modulus of rupture (tensile strength)
fr 7.5 f'c= (in psi)
Metric coefficient 7.5psi
f'c
psi
 7.5MPa
psi
MPa

f'c
MPa
MPa
psi
= C MPa
f'c
MPa
=
C 7.5
psi
MPa
MPa
psi
 0.623
fr 0.623 f'c= (in MPa)
Page 5

Modulus of elasticity
Ec 33 wc
1.5
 f'c= (in psi)
wc is a unit weight of concrete (in pcf)
Metric coefficient 33psi
kN
m
3
pcf








1.5

MPa
psi
 44.011 MPa
Ec 44 wc
1.5
 f'c= (in MPa)
wc in kN/m3
Example 3.1
Concrete compression strength f'c 25MPa
Unit weight of concrete wc 24
kN
m
3
 145pcf 22.778
kN
m
3

Modulus of rupture
fr 7.5psi
f'c
psi
 3.114 MPa
fr 0.623MPa
f'c
MPa
 3.115 MPa
Ec 33psi
wc
pcf






1.5

f'c
psi
 2.587 10
4
 MPa
Ec 44MPa
wc
kN
m
3










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 10
5
 MPa
Es 2 10
5
MPa
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


































 D
no
8
in 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
A
π 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


































cm
2

Weight of steel reinforcements
W A 7850
kgf
m
3
 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 (cm
2
) 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
A
π D
2

4
 W A 7850
kgf
m
3

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 (cm
2
) 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 0.005if
0.70 εt 0.002if
1.7 200 εt
3
otherwise
= 0.7
0.9 0.7
0.005 0.002
εt 0.002 
0.7
200
3
εt 0.002 
1.7 200 εt
3
=
For other members
ϕ 0.9 εt 0.005if
0.65 εt 0.002if
1.45 250 εt
3
otherwise
=
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
tmin
Perimeter
180
111.111 mm
t
1
30
1
50






La 133.333 80( ) mm
t 110mm
Section of beam B1
L 6m
h
1
10
1
15






L 600 400( ) mm h 500mm
b 0.3 0.6( ) h 150 300( ) mm b 250mm
For girders h
1
8
1
10






L=
For two-way slab beams h
1
10
1
15






L=
For floor beams h
1
15
1
20






L=
Section of beam B2
L 4m
h
1
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 22
kN
m
3
1.1
kN
m
2

RC slab Slab t 25
kN
m
3
2.75
kN
m
2

Ceiling Ceiling 0.40
kN
m
2

M & E Mechanical 0.20
kN
m
2

Partition Partition 1.00
kN
m
2

Dead load DL Cover Slab Ceiling Mechanical Partition 5.45
kN
m
2

Live load for Lab LL 60psf 2.873
kN
m
2

Factored load wu 1.2 DL 1.6 LL 11.137
kN
m
2

C. Loads of Wall
Void 30mm 30 mm 190 mm 4
wbrick.hollow 90mm 90 mm 190 mm Void( ) 20
kN
m
3
1.744 kgf
wbrick.solid 45mm 90 mm 190 mm 20
kN
m
3
1.569 kgf
ρbrick.hollow
wbrick.hollow
90mm 90 mm 190 mm
11.111
kN
m
3

Brickhollow.10 120mm Void
55
1m
2






20
kN
m
3
1.648
kN
m
2

110
1m
2






20
kN
m
3
2.895
kN
m
2

Page 13

D. Loads on Beam B1
Self weight SW 25cm 50cm 110mm( ) 25
kN
m
3
2.438
kN
m

Wall wwall Brickhollow.10 3.5m 50cm( ) 4.943
kN
m

Slab α
4m
2
6m
0.333 k 1 2 α
2
 α
3
 0.815
wD.slab DL
4m
2
 2 k 17.763
kN
m

wL.slab LL
4m
2
 2 k 9.363
kN
m

Dead load wD SW wwall wD.slab 25.143
kN
m

Live load wL wL.slab 9.363
kN
m

Factored load wu 1.2 wD 1.6 wL 45.153
kN
m

E. Loads on Beam B2
Self weight SW 20cm 30cm 110mm( ) 25
kN
m
3
0.95
kN
m

Wall wwall Brickhollow.10 3.5m 30cm( ) 5.272
kN
m

Slab α
4m
2
4m
0.5 k 1 2 α
2
 α
3
 0.625
wD.slab DL
4m
2
 2 k 13.625
kN
m

wL.slab LL
4m
2
 2 k 7.182
kN
m

Dead load wD SW wwall wD.slab 19.847
kN
m

Live load wL wL.slab 7.182
kN
m

kN
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( ) 25
kN
m
3
L 14.625 kN
PB2 20cm 30cm 110mm( ) 25
kN
m
3
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 kN
Pu
PD PL
1.299
Control
PD PL
n B L
11.622
kN
m
2
 (Ref. 10
kN
m
2
18
kN
m
2
 )
PL
PD PL
24.72 % (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)
L
6m
2
3 m
tmin
L
28
107.143 mm
t
1
25
1
35






L 120 85.714( ) mm
t 120mm
Page 16

Section of floor beam B1
L 8m
h
1
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
h
1
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 22
kN
m
3
1.1
kN
m
2

Slab t 25
kN
m
3
3
kN
m
2

Ceiling 0.40
kN
m
2

Mechanical 0.20
kN
m
2

Partition 1.00
kN
m
2

DL Cover Slab Ceiling Mechanical Partition 5.7
kN
m
2

LL 60psf 2.873
kN
m
2

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
55
1m
2






20
kN
m
3
1.648
kN
m
2

SW 25cm 50cm 120mm( ) 25
kN
m
3
2.375
kN
m

wwall Brickhollow.10 3.5m 50cm( ) 4.943
kN
m

wD.slab DL 3 m 17.1
kN
m

wL.slab LL 3 m 8.618
kN
m

wD SW wwall wD.slab 24.418
kN
m

wL wL.slab 8.618
kN
m

wu 1.2 wD 1.6 wL 43.091
kN
m

D. Loads on Girder B2
SW 30cm 60cm 120mm( ) 25
kN
m
3
3.6
kN
m

wwall Brickhollow.10 3.5m 60cm( ) 4.778
kN
m

PB1 25cm 50cm 120mm( ) 25
kN
m
3
8m 4m
2
 14.25 kN
Pwall Brickhollow.10 3.5m 50cm( )
8m 4m
2
 29.657 kN
PD.slab DL 3 m
8m 4m
2
 102.6 kN
PL.slab LL 3 m
8m 4m
2
 51.711 kN
Page 18

Factored loads
wD SW wwall 8.378
kN
m

wL 0
wu 1.2 wD 1.6 wL 10.054
kN
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 G
3.5m
12
291.667 mm
H
3.8m
24
158.333 mm G 2 H 60.833 cm
Slope angle α atan
H
G






28.496 deg
Loads on Waist Slab
Thickness of waist slab t 120mm
Step cover Cover 50mm H G( ) 22
kN
m
3
1m
G 1 m
 1.697
kN
m
2

Page 20

SW of step Step
G H
2
24
kN
m
3
1m
G 1 m
 1.9
kN
m
2

SW of waist slab Slab t 25
kN
m
3
1m
2
1m
2
cos α( )
 3.414
kN
m
2

Renderring Renderring 0.40
kN
m
2
1m
2
1m
2
cos α( )
 0.455
kN
m
2

Handrail Handrail 0.50
kN
m
2

Total dead load DL Cover Step Slab Renderring Handrail
DL 7.966
kN
m
2

Live load for public staircase LL 100psf 4.788
kN
m
2

Factored load wu 1.2 DL 1.6 LL 17.22
kN
m
2

Loads on Landing Slab
Cover 50mm 22
kN
m
3
1.1
kN
m
2

Slab 150mm 25
kN
m
3
3.75
kN
m
2

Renderring 0.40
kN
m
2

Handrail 0.50
kN
m
2

DL Cover Slab Renderring Handrail 5.75
kN
m
2

LL 100psf 4.788
kN
m
2

wu 1.2 DL 1.6 LL 14.561
kN
m
2

Page 21

8. Loads on Roof
Slope angle α atan
3m
10m
2














30.964 deg
Srokalinh tile Tile 30mm 20
kN
m
3
0.6
kN
m
2

Purlins w20x20x1.0 20mm 20 mm 18mm 18 mm( ) 7850
kgf
m
3
0.597
kgf
1m

Purlin w20x20x1.0
1m
1m 100 mm cos α( )
 0.068
kN
m
2

Rafters w40x80x1.6 40mm 80 mm 36.8mm 76.8 mm( ) 7850
kgf
m
3

w40x80x1.6 2.934
kgf
1m

Rafter w40x80x1.6
1m
750mm1 m cos α( )
 0.045
kN
m
2

Page 23

Roof beam Beam 20cm 30 cm 25
kN
m
3
1m
1m
10m
4

 0.6
kN
m
2

Roof column Column 20cm 20 cm
3m
2
 25
kN
m
3
1
10m
4
4 m





 0.15
kN
m
2

Total dead load wD Tile Purlin Rafter Beam Column 1.463
kN
m
2

Live load wL 1.00
kN
m
2

kN
m
2

Page 24

9. ASCE Wind Loads
Basic wind speed V 120
km
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 0.613
N
m
2
V
m
s








2
 0.681
kN
m
2

Velocity pressure coefficients
zg 274m α 9.5 (For exposure C)
Kz z( ) 2.01
max z 4.6m( )
zg






2
α
 Kz 10m( ) 1.001
Velocity wind pressure
qz z( ) qs Kz z( ) Kzt I Kd qz 10m( ) 0.667
kN
m
2

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
λ
L
B
1.111
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 a
i
1
i
k
H
k

H
i
2

a
n 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
Pwindwardi
ai
ai 1
zpz z Cp.windward 



d Bwindward
Pleewardi
pz h Cp.leeward  a
i 1
a
i
  Bleeward
Psidei
pz h Cp.side  a
i 1
a
i
  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
b
i
1
i
k
H
k


Prectanglei
pz b
i
Cp.windward  a
i 1
a
i
  Bwindward
Ptrapeziumi
pz a
i
Cp.windward  pz a
i 1
Cp.windward 
2
a
i 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 c
a
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 α
c
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
γ
α
β1
=
0.72
0.85
0.847
0.68
0.80
0.85
0.64
0.75
0.853
0.60
0.70
0.857
0.56
0.65
0.862
Page 28

Conclusion: γ 0.85=
β1 0.85 f'c 4000psiif
0.65 f'c 8000psiif
0.85 0.05
f'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 d
a
2






= T d
a
2






=
Mn 0.85 f'c a b d
a
2






= (2.1)
Mn As fs d
a
2






= (2.2)
Conditions of strain compatibility
εs
εu
d c
c
=
εs εu
d c
c
= or εt εu
dt c
c
= (3.1)
c d
εu
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 β1
f'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 β1
f'c
fy

εu
εu εy
= 0.85 β1
f'c
fy

600MPa
600MPa fy
=
εu 0.003 Es 2 10
5
MPa εu Es 600 MPa
Maximum steel ratio
ACI 318-99 ρmax 0.75 ρb=
ACI 318-02 and later ρmax 0.85 β1
f'c
fy

εu
εu εt
= with εt 0.004
For fy 390MPa εs
fy
Es
0.002
For εt 0.004
ρmax
ρb
εu εy
εu 0.004
=
5
7
= 0.714=
For εt 0.005
ρmax
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
a
As fy
0.85 f'c b
= c
a
β1
=
Mn As fy d
a
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 5
π 16mm( )
2

4
 10.053 cm
2

d h 30mm 6mm 16mm
40mm
2






 278 mm
dt h 30mm 6mm
16mm
2






 306 mm
Materials f'c 25MPa fy 390MPa
Solution
Checking for steel ratios
β1 0.65 max 0.85 0.05
f'c 27.6MPa
6.9MPa







min 0.85






0.85
εu 0.003
ρmax 0.85 β1
f'c
fy

εu
εu 0.004
 0.02
ρmin max
0.249MPa
f'c
MPa

fy
1.379MPa
fy











0.00354
ρ
As
b d
0.018
Steel_Reinforcement "is Enough" ρ ρminif
"is not Enough" otherwise

Steel_Reinforcement "is Enough"
As min ρ ρmax  b d 10.053 cm
2

Calculation of flexural strength
a
As fy
0.85 f'c b
92.252 mm c
a
β1
108.532 mm
Mn As fy d
a
2






 90.911 kN m
Page 32

εt εu
dt c
c
 0.00546
ϕ 0.65 max
1.45 250 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
w
a
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'c
1=
Flexural resistance factor
R
Mn
b d
2

=
As fy d
a
2







b d
2

=
As
b d
fy
d
a
2

d
= ρ fy 1
1
2
w





=
R ρ fy 1
ρ fy
1.7 f'c







= 0.85 f'c w 1
1
2
w





=
Quadratic equation relative w
R
0.85 f'c
w 1
1
2
w





=
w
2
2 w 2
R
0.85 f'c
 0=
w1 1 1 2
R
0.85 f'c
 1= w2 1 1 2
R
0.85 f'c
 1=
w 1 1 2
R
0.85 f'c
=
ρ 0.85
f'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
R
Mn
b d
2

=
ρ 0.85
f'c
fy
 1 1 2
R
0.85 f'c







=
(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
a
As fy
0.85 f'c b
= (As is a provided steel area)
Mn As fy d
a
2






=
c
a
β1
= εt εu
dt c
c
= ϕ ϕ εt =
FS
Mu
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 18mm
40mm
2






 424 mm
Page 34

dt h 30mm 8mm
18mm
2






 453 mm
Solution
Steel ratios
β1 0.65 max 0.85 0.05
f'c 27.6MPa
6.9MPa







min 0.85






0.85
εu 0.003
ρmax 0.85 β1
f'c
fy

εu
εu 0.004
 0.02
ρmin max
0.249MPa
f'c
MPa

fy
1.379MPa
fy











0.00354
Assume ϕ 0.9
Mn
Mu
ϕ
170 kN m
Steel area
R
Mn
b d
2

4.728 MPa
ρ 0.85
f'c
fy
 1 1 2
R
0.85 f'c







 0.014
ρmin ρ ρmax 1
As ρ b d 11.783 cm
2

As 6
π 16mm( )
2

4
 12.064 cm
2

Checking for flexural strength
a
As fy
0.85 f'c b
110.702 mm c
a
β1
130.238 mm
Mn As fy d
a
2






 173.444 kN m
Page 35

εt εu
dt c
c
 0.00743
ϕ 0.65 max
1.45 250 εt
3






min 0.9






0.9
FS
Mu
ϕ Mn
0.98
The_beam "is safe" FS 1if
"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 β1
f'c
fy

εu
εu εt
= R ρ fy 1
ρ fy
1.7 f'c







=
ϕ ϕ εt = Mn
Mu
ϕ
=
bd
2
Mn
R
=
Option 1: b
Mn
R
d
2
=
Option 2: d
Mn
R
b
=
Option 3: k
b
d
= d
3
Mn
R
k
= b k d=
Step 2. Calculation of steel area
R
Mn
b d
2

=
Page 36

ρ 0.85
f'c
fy
 1 1 2
R
0.85 f'c







=
As max ρ ρmin  b d=
Step 3. Checking for flexural strength
a
As fy
0.85 f'c b
= c
a
β1
=
Mn As fy d
a
2






=
εt εu
dt c
c
= ϕ ϕ εt =
FS
Mu
ϕ Mn
=
Example 10.3
Solution
Steel ratios
β1 0.65 max 0.85 0.05
f'c 27.6MPa
6.9MPa







min 0.85






0.85
εu 0.003
ρmax 0.85 β1
f'c
fy

εu
εu 0.004
 0.02
ρmin max
0.249MPa
f'c
MPa

fy
1.379MPa
fy











0.00354
Assume εt 0.007
ρ 0.85 β1
f'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 max
1.45 250 εt
3






min 0.9






0.9 Mn
Mu
ϕ
777.778 kN m
Concrete dimension
k
b
d
= k
400
600
 Cover 30mm 10mm 25mm
40mm
2

Cover 85 mm
d
3
Mn
R
k
627.231 mm b k d 418.154 mm
h Round d Cover 50mm( ) 700 mm b Round b 50mm( ) 400 mm
d h Cover 615 mm
b
h






400
700






mm
Steel area
R
Mn
b d
2

5.141 MPa
ρ 0.85
f'c
fy
 1 1 2
R
0.85 f'c







 0.015
As max ρ ρmin  b d 37.741 cm
2

As 8
π 25mm( )
2

4
 39.27 cm
2
 dt h 30mm 10mm
25mm
2







dt 647.5 mm
Checking for flexural strength
a
As fy
0.85 f'c b
180.18 mm c
a
β1
211.976 mm
Mn As fy d
a
2






 803.914 kN m
εt εu
dt c
c
 0.00616
ϕ 0.65 max
1.45 250 εt
3






min 0.9






0.9
FS
Mu
ϕ Mn
96.749 %
Page 38

11. Design of Doubly Reinforced Beams
(with tensile reinforcements only)
(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 d
a
2






= 0.85 f'c a b d
a
2






=
Mn2 T Cs  d
a
2






= As fy A's f's  d
a
2






=
Page 39

εs
εu
d c
c
= (3.1)
εs εu
d c
c
= or εt εu
dt c
c
=
c d
εu
εu εs
= or c dt
εu
εu εt
=
ε's
εu
c d'
c
= (3.2)
ε's εu
c 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 β1
f'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 β1
f'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'
A's
b d
= ρt.max ρmax ρ'=
ρ ρt.max= As ρ b d=
Step 2. Determination of compression parameters
2.1. Assume
f's fy=
2.2. Calculate
a
As fy A's f's
0.85 f'c b
= c
a
β1
=
ε's εu
c 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=
a
As 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 εu
c d'
c
=
As fy 0.85 f'c a b A's Es εu
β1 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 a
2
 b A's f1 As fy  a A's f1 β1 d' 0=
0.85 f'c a
2
 b A's f1 As fy  a A's f1 β1 d'
0.85 f'c d
2
 b
0=
a
d






2 ρ' f1 ρ fy
0.85 f'c
a
d

ρ' f1 β1
0.85 f'c
d'
d
 0=
w
2
2 p w q 0= p
1
2
ρ' f1 ρ fy
0.85 f'c
= q
ρ' f1 β1
0.85 f'c
d'
d
=
w1 p p
2
q 0= w2 p p
2
q 0=
a d p p
2
q = c
a
β1
=
ε's εu
c 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  d
a
2






=
εt εu
dt c
c
= ϕ ϕ εt =
ϕMn ϕ Mn1 Mn2 =
Page 42

Example 11.1
π 20mm( )
2

4
 25.133 cm
2

d h 30mm 10mm 16mm 20mm
40mm
2







d 454 mm
dt h 30mm 10mm 16mm
20mm
2






 484 mm
A's 4
π 20mm( )
2

4
 12.566 cm
2

d' 30mm 10mm
20mm
2
 50 mm
Solution
Steel ratios
β1 0.65 max 0.85 0.05
f'c 27.6MPa
6.9MPa







min 0.85






0.85
εu 0.003
ρmax 0.85 β1
f'c
fy

εu
εu 0.005
 0.0174
ρmin max
0.249MPa
f'c
MPa

fy
1.379MPa
fy











0.00354
Checking for singly reinforced beam
ρ
As
b d
0.0185
The_beam "is singly reinforced" ρ ρmaxif
"is doubly reinforced" otherwise

The_beam "is doubly reinforced"
ρ'
A's
b d
9.226 10
3
 ρt.max ρmax ρ' 0.027
Page 43

Concrete "is enough" ρ ρt.maxif
"is not enough" otherwise

Concrete "is enough"
ρ min ρ ρt.max  0.0185
As ρ b d 25.133 cm
2

Determination of compression parameters
Es 2 10
5
MPa f1 Es εu 600 MPa
εy
fy
Es
1.95 10
3

ρcy 0.85 β1
f'c
fy

εu
εu εy

d'
d
 ρ' 0.024
Compression_steel "will yield" ρ ρcyif
"will not yield" otherwise

Compression_steel "will not yield"
a
As fy A's fy
0.85 f'c b
Compression_steel "will yield"=if
p
1
2
ρ' f1 ρ fy
0.85 f'c

q
ρ' f1 β1
0.85 f'c
d'
d

d p p
2
q 
otherwise

a 90.825 mm c
a
β1
106.853 mm
f's fy Compression_steel "will yield"=if
ε's εu
c 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  d
a
2






 236.576 kN m
εt εu
dt c
c
 0.011
ϕ 0.65 max
1.45 250 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 β1
f'c
fy

εu
εu εt
=
ϕ ϕ εt =
Step 2. Cheching for singly reinforced beam
As ρ b d=
a
As fy
0.85 f'c b
=
Mn As fy d
a
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
ϕ
Mn2=
Page 45

c
a
β1
= f's Es εu
c d'
c
 fy=
A's
Mn1
f's d d'( )
= ρ'
A's
b d
As
0.85 f'c a b A's f's
fy
= ρ
As
b d
=
Example 11.2
d h 30mm 12mm 25mm 25mm
40mm
2







d 688 mm
25mm
2






 720.5 mm
d' 30mm 12mm
25mm
2
 54.5 mm
Solution
Steel ratios
β1 0.65 max 0.85 0.05
f'c 27.6MPa
6.9MPa







min 0.85






0.85
εu 0.003
ρmax 0.85 β1
f'c
fy

εu
εu 0.005
 0.0174
ρmin max
0.249MPa
f'c
MPa

fy
1.379MPa
fy











0.00354
Assume εt 0.0092
ρ 0.85 β1
f'c
fy

εu
εu εt
 0.011
Page 46

ϕ 0.65 max
1.45 250 εt
3






min 0.9






0.9
Checking for singly reinforced beam
As ρ b d 31.342 cm
2

a
As fy
0.85 f'c b
143.803 mm c
a
β1
169.18 mm
Mn2 As fy d
a
2






 753.074 kN m
The_beam "is singly reinforced" Mu ϕ Mn2if
"is doubly reinforced" otherwise

The_beam "is doubly reinforced"
Case of doubly reinforced beam
Mn1
Mu
ϕ
Mn2 746.926 kN m
f's min Es εu
c d'
c
 fy





390 MPa
A's
Mn1
f's d d'( )
30.232 cm
2

ρ'
A's
b d
0.011 ρt.max ρmax ρ' 0.028
As
0.85 f'c a b A's f's
fy
61.574 cm
2
 ρ
As
b d
0.022
 Concrete "is enough"
Compression steel A's 30.232 cm
2
 5
π 28mm( )
2

4
 30.788 cm
2

Tensile steel As 61.574 cm
2
 10
π 28mm( )
2

4
 61.575 cm
2

400mm 12mm 30mm( ) 2 28mm 5
4
44 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=
R
Mn2
b d
2

=
ρ 0.85
f'c
fy
 1 1 2
R
0.85 f'c







=
As ρ b d=
a
As fy
0.85 f'c b
= c
a
β1
=
f's.revised Es εu
c 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
Page 48

Example 11.3
d h 30mm 12mm 25mm 28mm
40mm
2







d 685 mm
28mm
2






 719 mm
d' 30mm 12mm
28mm
2
 56 mm
Compression reinforcements A's 5
π 28mm( )
2

4
 30.788 cm
2

Solution
Steel ratios
β1 0.65 max 0.85 0.05
f'c 27.6MPa
6.9MPa







min 0.85






0.85
εu 0.003
ρmax 0.85 β1
f'c
fy

εu
εu 0.005
 0.0174
ρmin max
0.249MPa
f'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
R
Mn2
b d
2


ρ 0.85
f'c
fy
 1 1 2
R
0.85 f'c








As ρ b d
a
As fy
0.85 f'c b

c
a
β1

f's.revised min Es εu
c d'
c
 fy






Z
i 
f's
fy
ϕ
a
d
f's.revised
fy























εt εu
dt c
c

ϕ 0.65 max
1.45 250 εt
3






min 0.9







break( )
f's.revised f's
f's
ε






Mu ϕ Mn if
f's f's.revised
i 0 99for
reverse Z
T 

Z Compression 0.000001( )
Page 50

a Z
0 2
d 142.792 mm c
a
β1
167.99 mm
f's Z
0 0
fy 390 MPa
Tensile steel area
ρ'
A's
b d
0.011 ρt.max ρmax ρ' 0.029
As
0.85 f'c a b A's f's
fy
61.909 cm
2
 ρ
As
b d
0.023
Tensile steel As 61.909 cm
2
 10
π 28mm( )
2

4
 61.575 cm
2

Page 51

12. Design of T Beams
12.1. Effective Flange Width
For symmetrical T beam:
b
L
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 d
hf
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 d
hf
2







= Asf fy d
hf
2







=
Mn2 C2 d
a
2






= 0.85 f'c a bw d
a
2






=
Mn2 T C1  d
a
2






= As fy Asf fy  d
a
2






=
Condition of strain compatibility
εs
εu
d c
c
= or
εt
εs
dt c
c
=
εs εu
d c
c
= εt εu
dt c
c
=
c d
εu
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 β1
f'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
a
As 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 d
hf
2







=
a
As fy Asf fy
0.85 f'c bw
= c
a
β1
=
Mn2 As fy Asf fy  d
a
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
π
10
8
in





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 0.05
f'c 4000psi
1000psi







min 0.85






0.85
εu 0.003
ρmax 0.85 β1
f'c
fy

εu
εu 0.005
 0.014
ρmin max
3psi
f'c
psi

fy
200psi
fy











0.00333
Checking for rectangular beam
Page 55

a
As fy
0.85 f'c b
157.162 mm
The_beam "is rectangular" a hfif
"is T" otherwise
 The_beam "is T"
Case of T beam
Asf
0.85 f'c hf b bw 
fy
29.613 cm
2

ρf
Asf
bw d
0.018 ρw.max ρmax ρf 0.031
ρw
As
bw d
0.028
Concrete "is enough" ρw ρw.maxif
As min ρw ρw.max  bw d As 7.363 in
2

Mn1 Asf fy d
hf
2







 715.669 kN m
a
As fy Asf fy
0.85 f'c bw
165.734 mm c
a
β1
194.981 mm
Mn2 As fy Asf fy  d
a
2






 427.446 kN m
εt εu
dt c
c
 0.008
ϕ 0.65 max
1.45 250 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 d
hf
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 d
hf
2







=
Mn2
Mu
ϕ
Mn1=
R
Mn2
bw d
2

=
ρ 0.85
f'c
fy
 1 1 2
R
0.85 f'c







=
As2 ρ bw d=
a
As2 fy
0.85 f'c bw
=
As
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 min
L
4
bw 16 hf s





 b 1193.8 mm
Steel ratios
β1 0.65 max 0.85 0.05
f'c 4000psi
1000psi







min 0.85






0.85
εu 0.003
ρmax 0.85 β1
f'c
fy

εu
εu 0.005
 0.014
ρmin max
3psi
f'c
psi

fy
200psi
fy











0.00333
Checking for rectangular beam
ϕ 0.9
Mnf 0.85 f'c hf b d
hf
2







 751.538 kN m
The_beam "is rectangular" Mu ϕ Mnfif
"is tee" otherwise
 The_beam "is tee"
Case of T beam
Page 58

Asf
0.85 f'c hf b bw 
fy
29.613 cm
2
 ρf
Asf
bw d
0.021
Mn1 Asf fy d
hf
2







 575.646 kN m
Mn2
Mu
ϕ
Mn1 227.801 kN m
R
Mn2
bw d
2

3.159 MPa
ρ 0.85
f'c
fy
 1 1 2
R
0.85 f'c







 8.484 10
3

As2 ρ bw d 12.042 cm
2

a
As2 fy
0.85 f'c bw
101.408 mm c
a
β1
119.304 mm
As
fy
41.655 cm
2
 ρw
As
bw d
0.029
ρw.max ρmax ρf 0.034
Concrete "is enough" ρw ρw.maxif
As.min ρmin bw d
As max As As.min  41.655 cm
2
 6
π 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
2
Vu ϕVc : stirrup is minimum
Vu ϕVc : stirrup is required
Minimum stirrups
Av.min 0.75 f'c
bw s
fy
 50
bw s
fy
= (in psi)
Av.min 0.062 f'c
bw 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
smax
d
2
24in= 600mm=
Case 2 Vc Vs 4 Vc
smax
d
4
12in= 300mm=
Case Vs 4 Vc
Concrete is not enough
Example 13.1
Page 62

Live load for garage LL 6.00
kN
m
2

Loads on slab
Hardener 8mm 24
kN
m
3
0.192
kN
m
2

Slab 200mm 25
kN
m
3
5
kN
m
2

Mechanical 0.30
kN
m
2

DL Hardener Slab Mechanical 5.492
kN
m
2

LL 6
kN
m
2

Loads on beam
wbeam 30cm 60cm 200mm( ) 25
kN
m
3
3
kN
m

wD.slab DL 3.5 m 19.222
kN
m

wL.slab LL 3.5 m 21
kN
m

wD wbeam wD.slab 22.222
kN
m

wL wL.slab 21
kN
m

wu 1.2 wD 1.6 wL 60.266
kN
m

Shear
L 8m
V0
wu L
2
241.066 kN
V x( ) V0 wu x
bw 300mm d 600mm 40mm 10mm
20mm
2






 540 mm
Vc 0.166MPa
f'c
MPa
 bw d 134.46 kN
ϕ 0.75
Page 63

Location of no stirrup zone
V0 wu x
ϕVc
2
= x
V0
ϕ Vc
2

wu
3.163 m
Minimum stirrup
Av 2
π 10mm( )
2

4
 1.571 cm
2
 fy 390MPa
smax min
Av fy
0.062MPa
f'c
MPa
 bw
Av fy
0.345MPa bw











591.894 mm
smax Floor smax 50mm  550 mm
Vs.min
Av fy d
smax
60.147 kN
Location of minimum stirrup zone
V0 wu x ϕ Vc Vs.min = x
V0 ϕ Vc Vs.min 
wu
1.578 m
Required spacing of stirrup
Vu V0 wu
400mm
2






 229.012 kN
Vs
Vu
ϕ
Vc 170.89 kN
Concrete "is enough" Vs 4 Vcif
s
Av fy d
Vs
193.581 mm
smax.1 smax 550 mm
smax.2 min
d
2
600mm





Vs 2 Vcif
min
d
4
300mm





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 wu
400mm
2
 229.012 kN
Vc 0.166MPa
f'c
MPa
 bw d 134.46 kN
ϕ 0.75
Stirrup "is minimum" Vu ϕ Vcif
"is required" otherwise
 Stirrup "is required"
Required steel shear strength
Vs
Vu
ϕ
Vc 170.89 kN
Spacing of stirrup
Av 2
π 10mm( )
2

4
 1.571 cm
2
 fy 390MPa
s
Av fy d
Vs
193.581 mm
smax.1 min
Av fy
0.062MPa
f'c
MPa
 bw
Av fy
0.345MPa bw











591.894 mm
Page 65

smax.2 min
d
2
600mm





Vs 2 Vcif
min
d
4
300mm





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 wu
L
4
 120.533 kN
Stirrup "is minimum" Vu ϕ Vcif
"is required" otherwise
 Stirrup "is required"
Required steel shear strength
Vs
Vu
ϕ
Vc 26.25 kN
Spacing of stirrup
Av 2
π 10mm( )
2

4
 1.571 cm
2
 fy 390MPa
s
Av fy d
Vs
1260.208 mm
smax.1 min
Av fy
0.062MPa
f'c
MPa
 bw
Av fy
0.345MPa bw











591.894 mm
smax.2 min
d
2
600mm





Vs 2 Vcif
min
d
4
300mm





otherwise
 smax.2 270 mm
s Floor min s smax.1 smax.2  50mm  250 mm
Page 66

14. Column Design
Type of columns (by design method)
1. Axially loaded columns
e
M
P
= 0=
2. Eccentric columns
e
M
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 ϕ
0.85 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 3.00
kN
m
2

Solution
Loads on slab
Cover 50mm 22
kN
m
3

Slab 120mm 25
kN
m
3

Ceiling 0.40
kN
m
2

Mechanical 0.20
kN
m
2

Partition 1.00
kN
m
2

kN
m
2

LL 3
kN
m
2

Page 70

Reduction of live load
Tributary area AT B L 24 m
2

For interior column KLL 4
Influence area AI KLL AT 96 m
2

Live load reduction factor αLL 0.25
4.572
AI
m
2
 0.717
Reduced live load LL0 LL αLL 2.15
kN
m
2

Loads of wall
Void 30mm 30 mm 190 mm 4
55
1m
2






20
kN
m
3
1.648
kN
m
2

110
1m
2






20
kN
m
3
2.895
kN
m
2

Loads on column
PD.slab DL B L 136.8 kN
PL.slab LL B L 72 kN
PB1 25cm 50cm 120mm( ) 25
kN
m
3
L 14.25 kN
PB2 20cm 35cm 120mm( ) 25
kN
m
3
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 n
12.015
kN
m
2

PL
PD PL
24.968 %
Page 71

Pu 1.2 PD 1.6 PL 2249.063 kN
Determination of column section
Assume ρg 0.03 k
b
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

h
Ag
k
472.322 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 cm
2

Determination of steel area
Ast
Pu
0.80 ϕ
0.85 f'c Ag
0.85 f'c fy
 Ast 30.851 cm
2

6
π 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
 0=
Pn C Cs T=
Pn 0.85 f'c a b A's f's As fs=
 0=
Mn Pn e= C
h
2
a
2






 Cs
h
2
d'





 T d
h
2






=
Mn Pn e= 0.85 f'c a b
h
2
a
2






 A's f's
h
2
d'





 As fs d
h
2






=
εs
εu
d c
c
= εs εu
d c
c
=
fs Es εs= Es εu
d c
c
=
ε's
εu
c d'
c
= ε's εu
c d'
c
=
f's Es ε's= Es εu
c 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
ϕ
0.85 f'c a b
f's fs
=
AsM a( )
Mu
ϕ
0.85 f'c a b
h
2
a
2







f's
h
2
d'





 fs d
h
2







=
f's a( ) Es εu
c d'
c
 fy=
fs a( ) Es εu
d c
c
 fy=
Page 74

Example 14.2
Construction of interaction diagram for column strength.
π 16mm( )
2

4
 10.053 cm
2

A's As 10.053 cm
2

d' 30mm 6mm
16mm
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 0.05
f'c 27.6MPa
6.9MPa







min 0.85






0.85
c a( )
a
β1

Es 2 10
5
MPa εu 0.003 dt d
fs a( ) min Es εu
d c a( )
c a( )
 fy






f's a( ) min Es εu
c a( ) d'
c a( )
 fy






ϕ a( ) εt εu
dt c a( )
c a( )

ϕ 0.65 max
1.45 250 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 b
h
2
a
2






 A's f's a( )
h
2
d'





 As fs a( ) d
h
2













a 0
h
100
 h
0 20 40 60
0
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
fy 390MPa
Concrete cover to main bars cc 30mm 6mm
16mm
2

Page 76

Solution
Location of steel re-bars
d' cc 44 mm
d h cc 156 mm
Ag b h ϕ 0.65
ϕ 0.65
Ast
Pu
0.80 ϕ
0.85 f'c Ag
0.85 f'c fy
2.465 cm
2

β1 0.65 max 0.85 0.05
f'c 27.6MPa
6.9MPa







min 0.85






0.85
c a( )
a
β1

Es 2 10
5
MPa εu 0.003 dt d
fs a( ) min Es εu
d c a( )
c a( )
 fy






f's a( ) min Es εu
c a( ) d'
c a( )
 fy






ϕ a( ) εt εu
dt c a( )
c a( )

ϕ 0.65 max
1.45 250 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
2
d'





 fs a( ) d
h
2








a1 134.2mm a2 134.25mm
a a1 a1
a2 a1
50
 a2
Page 77

0.13418 0.1342 0.13422 0.13424 0.13426
8.715 10
4

8.72 10
4

8.725 10
4

8.73 10
4

8.735 10
4

AsN a( )
AsM a( )
a
a 134.23mm
AsN a( ) 8.722 cm
2

AsM a( ) 8.725 cm
2

As
AsN a( ) AsM a( )
2
8.724 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 0if
fd f's a( )
h
2
d'





 fs a( ) d
h
2







continue( ) fd 0=if
AsM
Mu
ϕ a( )
0.85 f'c a b
h
2
a
2







fd

continue( ) AsM 0if
Z
k 
a
h
AsN
Ag
AsM
Ag
AsN AsM
Ag

























k k 1
a cc cc
h
No
 hfor
csort Z
T
4 

Z Asteel 5000( ) rows Z( ) 2046
a Z
1 1
h 134.24 mm
AsN Z
1 2
Ag 8.719 cm
2
 AsM Z
1 3
Ag 8.728 cm
2

As
AsN AsM
2
8.723 cm
2

Page 79

C. Case of Distributed Reinforcements
rUb 3>1> ssrcakp©it EdlmanEdkBRgayeRcInCYr
CMuvijmuxkat;ebtug
b
dn
d1
Pne
Tn T1
C
h
a
cf 85.0
c
u
s,1
s,n
dn
 0=
Pn C
1
n
i
T
i

= 0.85 f'c a b
1
n
i
A
s i
f
s i
 

=
 0=
Mn Pn e= C
h
2
a
2







1
n
i
T
i
d
i
h
2













=
Mn 0.85 f'c a b
h
2
a
2







1
n
i
A
s i
f
s i
 d
i
h
2













=
Page 80

ε
s i
εu
d
i
c
c
=
ε
s i
εu
d
i
c
c
=
f
s i
Es ε
s i
= Es εu
d
i
c
c
=
Example 14.4
Checking for column strength.
Mu 57.53kN m
fy 390MPa
Solution
Determination of Concrete Section
ϕ 0.65
Assume ρg 0.04
Ag
Pu
0.80 ϕ
0.85 f'c 1 ρg  fy ρg
6094.36 cm
2

Aspect ratio of column section λ
b
h
= λ 1
h
Ag
λ
780.664 mm b λ h 780.664 mm
h Ceil h 50mm( ) b Ceil b 50mm( )
b
h






800
800






mm Ag b h 6400 cm
2

Page 81

Steel area
Ast
Pu
0.80 ϕ
0.85 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
8
87.5 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
As0
π Bars
2

4


i 1 n Asi
As0
i 

Ast As Ast 232.478 cm
2

Location of reinforcement rows
Concrete cover Cover 30mm 10mm 40 mm
d
1
Cover
Bars
1 n
2
 52.5 mm ΔS
h d
1
2
n 1
86.875 mm
i 2 n d
i
d
i 1
ΔS
reverse d( )
T
747.5 660.63 573.75 486.88 400 313.13 226.25 139.38 52.5( ) mm
Page 82

ϕPn.max 0.80 ϕ 0.85 f'c Ag Ast  fy Ast 
ϕPn.max 14255.808 kN
β1 0.65 max 0.85 0.05
f'c 27.6MPa
6.9MPa







min 0.85






0.796
c a( )
a
β1

fs i a( ) εs εu
d
i
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 max
1.45 250 εt
3






min 0.9








ϕPn a( ) min ϕ a( ) 0.85 f'c a b
1
n
i
Asi
fs i a( )













 ϕPn.max









ϕMn a( ) ϕ a( ) 0.85 f'c a b
h
2
a
2







1
n
i
Asi
fs i a( ) d
i
h
2























a 0
h
100
 h
Page 83

0 1000 2000 3000
0
5000
10000
ϕPn 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
d
i
rc rs cos α
s i = rc
Dc
2
= rs
Ds
2
=
α
s i
2 π
ns
i 1( )=
Page 85

Depth of compression concrete
α acos
rc a
rc






=
Area and centroid of compression concrete
Asector
1
2
Radius Arch=
1
2
rc rc 2 α = rc
2
α=
x1
2
3
rc
sin α( )
α
=
Atriangle
1
2
Base Height=
1
2
2 rc sin α( ) rc cos α( )= rc
2
sin α( ) cos α( )=
x2
2
3
rc cos α( )=
Ac Asegment= Asector Atringle= rc
2
α sin α( ) cos α( )( )=
xc
Asector x1 Atrinagle x2
Ac
=
2
3
rc
sin α( ) sin α( ) cos α( )
2

α sin α( ) cos α( )
=
xc
2 rc
3
sin α( )
3
α sin α( ) cos α( )
=
 0=
Pn C
1
ns
i
T
i

= 0.85 f'c Ac
1
ns
i
A
s i
f
s i
 

=
 0=
Mn Pn e= C xc
1
ns
i
T
i
d
i
Dc
2















=
Mn Pn e= 0.85 f'c Ac xc
1
ns
i
A
s i
f
s i
 d
i
rc  

=
ε
s i
εu
d
i
c
c
=
f
s i
Es ε
s i
= Es ε
u

d
i
c
c
= with f
s i
fy
Page 86

Example 14.5
Mu 42.53kN m
fy 390MPa
Solution
Determination of concrete dimension
ϕ 0.70
Ag
Pu
0.85 ϕ
0.85 f'c 1 ρg  fy ρg
2361.812 cm
2

Dc Ceil
Ag
π
4
50mm








550 mm
Ag
π Dc
2

4
2375.829 cm
2

Ast
Pu
0.85 ϕ
0.85 f'c Ag
0.85 f'c fy
46.597 cm
2

Ds Dc 30mm 10mm
20mm
2






2 450 mm
ns ceil
π Ds
100mm






15 As0
π 20mm( )
2

4
3.142 cm
2

Ast ns As0 47.124 cm
2
 s
π Ds
ns
94.248 mm
Page 87

β1 0.65 max 0.85 0.05
f'c 27.6MPa
6.9MPa







min 0.85






0.85
Es 2 10
5
MPa εu 0.003
c a( )
a
β1

i 1 ns αsi
2 π
ns
i 1( ) d
i
Dc
2
Ds
2
cos αsi





dt max d( ) 495.083 mm
ϕ a( ) εt εu
dt c a( )
c a( )

ϕ 0.70 max
1.7 200 εt
3






min 0.9








d
i
c a( )
c a( )


rc
Dc
2

α a( ) acos
rc a
rc







xc a( )
2 rc
3
sin α a( )( )
3
α a( ) sin α a( )( ) cos α a( )( )

Ac a( ) rc
2
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( ) d
i
rc  











a 0
Dc
100
 Dc
Page 88

0 100 200 300
0
1000
2000
3000
ϕPn a( )
kN
Pu
kN
ϕMn a( )
kN m
Mu
kN m

3. Long (Slender) Columns
Stability index
Q
ΣPu 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
ShearWall
Braced Frame
Brick Wall
Ties
Slenderness of column
The column is short, if
In nonsway frame:
k Lu
r
min 34 12
M1
M2
 40







In sway frame:
k Lu
r
22
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
r
I
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
2
1
π
k
tan
π
k


















2 tan
π
2 k







π
k
 1=
Page 91

For unbraced frames:
ψA ψB
π
k






2
 36
6 ψA ψB 
π
k
tan
π
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
k
20 ψm
20
1 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
1
Pu
0.75 Pc

1=
Euler's critical load
Pc
π
2
EI
k Lu 
2
=
EI
0.4 Ec Ig
1 βd
=
βd
1.2 PD
1.2 PD 1.6 PL
=
Coefficient
Cm 0.6 0.4
M1
M2
 0.4=
Example 14.6
PD
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









fy 390MPa
Page 93

Solution
Determination of concrete dimension
ϕ 0.65
Ag
Pu
0.80 ϕ
0.85 f'c 1 ρg  fy ρg
3379.226 cm
2

Proportion of column section k
b
h
= k
60
60

h
Ag
k
581.311 mm b k h 581.311 mm
h Ceil h 50mm( ) 600 mm
b Ceil b 50mm( ) 600 mm
b
h






600
600






mm
Ag b h 3.6 10
3
 cm
2

Ast
Pu
0.80 ϕ
0.85 f'c Ag
0.85 f'c fy
85.932 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 As0
π Bars
2

4


i 1 cols As0  As1i
As0
i 
 As As1
Ast As Ast 98.175 cm
2

Ast
Ag
0.027
ns rows As  ns 6
Cover 40mm 10mm
25mm
2
 62.5 mm
Page 94

d1
1
Cover Δs
h Cover 2
ns 1
95 mm
i 2 ns d1
i
d1
i 1
Δs
d d1 d
62.5
157.5
252.5
347.5
442.5
537.5
















mm
Slenderness of column
Stability index Q 0
Radius of gyration r
h
12
0.173 m
wc 24
kN
m
3

Ec 44MPa
wc
kN
m
3










1.5

f'c
MPa
 2.834 10
4
 MPa
Degree of end restraint
Ia1 0.35
ba1 ha1
3

12
 Ia2 0.35
ba2 ha2
3

12

Ib1 0.35
bb1 hb1
3

12
 Ib2 0.35
bb2 hb2
3

12

Ica 0.70
ba ha
3

12
 Icb 0.70
bb hb
3

12

Ic 0.70
b 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
Σiba
8.418 ψB
Σicb
Σibb
21.699
Page 95

Effective length factor
k 0.6
Given
ψA ψB
4
π
k






2

ψA ψB
2
1
π
k
tan
π
k


















2 tan
π
2 k







π
k
 1=
k 0.5 k 1.0
k Find k( ) k 0.969
Checking for long column
M1 MA MA MBif
MB otherwise
 M2 MB MA MBif
MA otherwise

M1 77.75 kN m M2 122.68 kN m
Lu Lc
max ha1 ha2  max hb1 hb2 
2
 7.3m
k Lu
r
40.834 min 34 12
M1
M2
 40






40
The_column "is short"
k Lu
r
min 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 PL
0.801
Ig
b h
3

12
 EI
0.4 Ec Ig
1 βd

Pc
π
2
EI
k Lu 
2
13409.955 kN
Cm max 0.6 0.4
M1
M2
 0.4






0.4
Page 96

δns max
Cm
1
Pu
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

c a( )
a
β1

dt max d( ) 537.5 mm
ϕ a( ) εt εu
dt c a( )
c a( )

ϕ 0.65 max
1.45 250 εt
3






min 0.90








d
i
c a( )
c a( )


ϕPn a( ) min ϕ a( ) 0.85 f'c a b
1
ns
i
Asi
fs i a( )













 ϕPn.max









ϕMn a( ) ϕ a( ) 0.85 f'c a b
h
2
a
2







1
ns
i
Asi
fs i a( ) d
i
h
2























a 0
h
100
 h
Page 97

0 200 400 600 800 1000 1200
0
1000
2000
3000
4000
5000
6000
7000
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 20
kN
m
3
H=
qa = allowable bearing capacity of soil with FS 2.5= 3
20
kN
m
3
= average density of soil and concrete
H = depth of foundation
Checking for maximum stress of soil under footing
qmax qu
qmax
P
B L
1
6 e
L






 e
L
6
if
4P
3 B L 2 e( )
e
L
6
if
=
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 20
kN
m
3
H B L=
e = eccentricity of load
Page 99

e
M
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 B
L
2
hc
2
 d






=
Vu2 qu L
B
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= L1
L
2
hc
2
=
Mu1
q1 L1
2

2
=
Design section: rectangular singly reinforced beam of B d
Steel re-bars in short direction
Required strength
q2 qu L= L2
B
2
bc
2
=
Mu2
q2 L2
2

2
=
Design section: rectangular singly reinforced beam of L d
Example 15.1
Required strength PD 484.71kN
PL 228.56kN
PL
PD PL
0.32
Mu 5.03kN m
Dimension of column stub bc 350mm hc 350mm
Depth of foundation H 2.0m
Allowable bearing capacity of soil qa 178.33
kN
m
2
3
2.5
 213.996
kN
m
2

fy 390MPa
Page 102

Solution
Determination of Dimension of Footing
Effective bearing capacity of soil
qe qa 20
kN
m
3
H 173.996
kN
m
2

Required area of footing
Areq
PD PL
qe
4.099 m
2

Footing proportion k
B
L
= k
2
2.1

L
Areq
k
2.075 m B k L 1.976 m
L Ceil L 50mm( ) 2.1m 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
m
2

Checking for maximum stress of soil
Pu 1.2 PD B L H 20
kN
m
3






 1.6 PL 1148.948 kN
e
Mu
Pu
4.378 mm
qmax
Pu
B L
1
6 e
L






 e
L
6
if
4Pu
3 B L 2 e( )
otherwise
 qmax 276.981
kN
m
2

qmax
qu
0.975
Soil "is safe" qmax quif
"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
b0 d( ) bc d  hc d   2
ϕ 0.75
ϕVc d( ) ϕ 0.332 MPa
f'c
MPa
 b0 d( ) d ϕVc 320mm( ) 1067.712 kN
Beam shears
Vu1 d( ) qu B
L
2
hc
2
 d






 Vu1 300mm( ) 326.858 kN
Vu2 d( ) qu L
B
2
bc
2
 d






 Vu2 300mm( ) 313.357 kN
Beam shear strength
ϕVc1 d( ) ϕ 0.166 MPa
f'c
MPa
 B d ϕVc1 300mm( ) 373.5 kN
ϕVc2 d( ) ϕ 0.166 MPa
f'c
MPa
 L d ϕVc2 300mm( ) 392.175 kN
c 50mm 20mm
20mm
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 50ksiif
0.0018return( ) fy 60ksiif
max 0.0018
60ksi
fy
 0.0014





return





otherwise

ρshrinkage 0.0018
Re-bars in long direction
b B Ln
L
2
hc
2
 0.875 m wu qu b
Mu
wu Ln
2

2
217.609 kN m Mn
Mu
0.9

Mu
b
108.805
kN m
1m

R
Mn
b d
2

1.181 MPa
ρ 0.85
f'c
fy
 1 1 2
R
0.85 f'c







 0.00312
As max ρ b d ρshrinkage b h  19.944 cm
2

s 150mm D 14mm n floor
b 75mm 2 D
s






1 13
n
π D
2

4
 20.012 cm
2

Re-bars in short direction
b L Ln
B
2
bc
2
 0.825 m wu qu b
Mu
wu Ln
2

2
203.123 kN m Mn
Mu
0.9

Mu
b
96.725
kN m
1m

R
Mn
b d
2

1.05 MPa
ρ 0.85
f'c
fy
 1 1 2
R
0.85 f'c







 0.00276
As max ρ b d ρshrinkage b h  18.554 cm
2

s 160mm D 14mm n floor
b 75mm 2 D
s






1 13
n
π D
2

4
 20.012 cm
2

Page 105

16. Design of Pile Caps
1. Determination of Pile Cap
Number of required piles
n
PD PL
Qe
=
where
PD PL = dead and live loads on pile cap
Qe = effective bearing capacity of pile
Qe Qa 20
kN
m
3
3 D( )
2
 H=
20
kN
m
3
= 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
2
200mm
Checking for pile reaction
R
i
P
n
M
y
x
i

1
n
k
x
k 
2



M
x
y
i

1
n
k
y
k 
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
fy 390MPa
Diameters of main bar
D1
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
4
Ag
15.622 MPa Use f'c.pile 20MPa
Steel re-bars
Ast 0.005 Ag 4.5 cm
2
 4
π 16mm( )
2

4
 8.042 cm
2

Dimension of pile cap
Effective bearing capacity of pile
Qe Qa 20
kN
m
3
3 D( )
2
 H 327.2 kN
Number of piles
n
PD PL
Qe
6.684
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( ) min
D
2
200mm





D
2






2 B 2.2m
L max X( ) min X( ) min
D
2
200mm





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 20
kN
m
3
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 X
i

1
n
k
X
k 
2



Mx Y
i

1
n
k
Y
k 
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 
X
i
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( ) X
hc
2
d
2







Y
bc
2
d
2
















Vu d( ) Ru Outside d( ) Vu 700mm( ) 2247.864 kN
ϕ 0.75
b0 d( ) hc d  bc d   2
ϕVc d( ) ϕ 0.332 MPa
f'c
MPa
 b0 d( ) d ϕVc 700mm( ) 3921.75 kN
Beam shears
Left d( ) X
hc
2
d














 Right d( ) X
hc
2
d















Bottom d( ) Y
bc
2
d














 Top d( ) Y
bc
2
d















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 MPa
f'c
MPa
 B d ϕVc1 700mm( ) 958.65 kN
ϕVc2 d( ) ϕ 0.166 MPa
f'c
MPa
 L d ϕVc2 700mm( ) 1132.95 kN
Depth of pile cap
Cover c D1
D2
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
max 0.0018
60ksi
fy
 0.0014





return





otherwise

In long direction
b B
Mu1 Left 0( ) Ru
hc
2
X














 643.378 kN m
Mu2 Right 0( ) Ru X
hc
2















 667.876 kN m
Mu max Mu1 Mu2  667.876 kN m Mn
Mu
0.9

R
Mn
b d
2

0.796 MPa
ρ 0.85
f'c
fy
 1 1 2
R
0.85 f'c







 0.00208
As max ρ b d ρshrinkage b h  As 29.797 cm
2

As1
π D1
2

4
 n1 ceil
As
As1






15
s1 Floor
b c
D1
2







2
n1 1
5mm










145 mm
In short direction
b L
Mu1 Bottom 0( ) Ru
bc
2
Y














 680.262 kN m
Mu2 Top 0( ) Ru Y
bc
2















 724.653 kN m
Mu max Mu1 Mu2  724.653 kN m Mn
Mu
0.9

Page 112

R
Mn
b d
2

0.731 MPa
ρ 0.85
f'c
fy
 1 1 2
R
0.85 f'c







 0.00191
As max ρ b d ρshrinkage b h  As 35.1 cm
2

As2
π D2
2

4
 n2 ceil
As
As2






18
s2 Floor
b c
D2
2







2
n2 1
5mm










140 mm
Shrinkage reinforcement
b 1m hshrinkage h hshrinkage 0=if
hshrinkage otherwise

As ρshrinkage b hshrinkage 3.6 cm
2

As0
π Dshrinkage
2

4
 n
As
As0

sshrinkage Floor
b
n
5mm





310 mm
Table
L c
D1
2







2
m
B c
D2
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
Lb
0.5 : the slab in one-way
La
Lb
0.5 : the slab is two-way
Thickness of one-way slab
Simply supported
Ln
20
One end continuous
Ln
24
Both ends continuous
Ln
28
Cantilever
Ln
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 1.8m
Live load LL 12
kN
m
2

fy 390MPa
Solution
Thickness of one-way slab
tmin
Ln
28
64.286 mm
Use t 100mm
Loads on slab
Cover 50mm 22
kN
m
3
1.1
kN
m
2

Slab t 25
kN
m
3
2.5
kN
m
2

Ceiling 0.40
kN
m
2

Mechanical 0.20
kN
m
2

Partition 1.00
kN
m
2

kN
m
2

wu 1.2 DL 1.6 LL 25.44
kN
m
2

Bending moments
Msupport
1
11
wu Ln
2
 7.493
kN m
1m

Mmidspan
1
16
wu Ln
2
 5.152
kN m
1m

Steel reinforcements
Page 116

β1 0.65 max 0.85 0.05
f'c 27.6MPa
6.9MPa







min 0.85






0.85
εu 0.003
ρmax 0.85 β1
f'c
fy

εu
εu 0.005
 0.014
ρmin max
0.249MPa
f'c
MPa

fy
1.379MPa
fy











0.00354
max 0.0018
60ksi
fy
 0.0014





return





otherwise

ρshrinkage 0.0018
Top rebars
b 1m d t 20mm
10mm
2






 75 mm
Mu Msupport b 7.493 kN m
Mn
Mu
0.9
8.326 kN m
R
Mn
b d
2

1.48 MPa
ρ 0.85
f'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 cm
2

As0
π 10mm( )
2

4
 n
As
As0
 smax min 3 t 450mm( )
s min Floor
b
n
10mm





smax





260 mm
Bottom rebars
Mu Mmidspan b 5.152 kN m
Mn
Mu
0.9
5.724 kN m
Page 117

R
Mn
b d
2

1.018 MPa
ρ 0.85
f'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 cm
2

As0
π 10mm( )
2

4
 n
As
As0
 smax min 3 t 450mm( )
s min Floor
b
n
10mm





smax





300 mm
Link rebars
As ρshrinkage b t 1.8 cm
2

As0
π 10mm( )
2

4
 n
As
As0
 smax min 5 t 450mm( )
s min Floor
b
n
10mm





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 La
4

EI
 αb
wb Lb
4

EI
=
Case αa αb=
wa
wb
Lb
4
La
4
=
1
λ
4
= λ
La
Lb
=
wa wb wu=
From which, wa wu
1
1 λ
4

=
wb wu
λ
4
1 λ
4

=
For λ 1
1
1 λ
4

0.5
λ
4
1 λ
4

0.5
For λ 0.8
1
1 λ
4

0.709
λ
4
1 λ
4

0.291
For λ 0.6
1
1 λ
4

0.885
λ
4
1 λ
4

0.115
For λ 0.5
1
1 λ
4

0.941
λ
4
1 λ
4

0.059
For λ 0.4
1
1 λ
4

0.975
λ
4
1 λ
4

0.025
Page 119

Example 17.2
Slab dimension La 4.3m
Lb 5.5m
Live load LL 2.00
kN
m
2

fy 390MPa
Solution
Thickness of two-way slab
Perimeter La Lb  2
tmin
Perimeter
180
108.889 mm
t
1
30
1
50






La 143.333 86( ) mm
Use t 120mm
Loads on slab
SDL 50mm 22
kN
m
3
0.40
kN
m
2
 1.00
kN
m
2
 2.5
kN
m
2

DL SDL t 25
kN
m
3
 5.5
kN
m
2

LL 2
kN
m
2

wu 1.2 DL 1.6 LL 9.8
kN
m
2

Load distribution
λ
La
Lb
0.782
wa
1
1 λ
4

wu 7.134
kN
m
2

wb
λ
4
1 λ
4

wu 2.666
kN
m
2

Page 120

Seun Sambath-ACI 318-05 Mathcad in Concrete Structures (2010)

Seun Sambath-ACI 318-05 Mathcad in Concrete Structures (2010)

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