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

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



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


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


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


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



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


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


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



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


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


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= 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



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


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


As


fy= ρ

As

b d=



Example 11.2




2

d 688 mm


2

720.5 mm

d' 30mm 12mm25mm

2 54.5 mm


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


fy61.574 cm

2 ρ

As

b d0.022




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


fy= ρ

As

b d=

ρ'A's




Page 48

Example 11.3




2

d 685 mm


2

719 mm

d' 30mm 12mm28mm

2 56 mm

Compression reinforcements A's 5π 28mm( )

2

4 30.788 cm

2


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


fy61.909 cm

2 ρ

As

b d0.023




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



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


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


fy41.655 cm

2 ρw

As

bw d0.029

ρw.max ρmax ρf 0.034

Concrete "is enough" ρw ρw.maxif



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


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


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



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


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




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




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

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


Solution

Loads on slab

Cover 50mm 22kN

m3

Slab 120mm 25kN

m3

Ceiling 0.40kN

m2

Mechanical 0.20kN

m2

Partition 1.00kN

m2


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



1m2

20kN

m3

1.648kN

m2


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


Pn C Cs T=

Pn 0.85 f'c a b A's f's As fs=


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

=


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.



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


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


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


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

rUb 3>1> ssrcakp©it EdlmanEdkBRgayeRcInCYrCMuvijmuxkat;ebtug

b

dn

d1

Pn e

Tn T1

C

h

a

cf 85.0

c

u

s,1

s,n dn


Pn C

1

n

i

Ti

= 0.85 f'c a b

1

n

i

As i f

s i

=


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


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.


Mu 57.53kN m

Materials f'c 35MPa

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 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


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 0.05f'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( )=


Pn C

1

ns

i

Ti

= 0.85 f'c Ac

1

ns

i

As i f

s i

=


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

=


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


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


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



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( )


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


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


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


Slenderness of column

Stability index Q 0

Radius of gyration rh

120.173 m


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


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( )


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


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


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


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


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


MPa B d ϕVc1 700mm( ) 958.65 kN


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



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


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


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



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


As0π 10mm( )

2

4 n

As


s min Floorb

n10mm

smax

300 mm

Link rebars

As ρshrinkage b t 1.8 cm2

As0π 10mm( )

2

4 n

As


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


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


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



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


As0π 10mm( )

2

4 n

As


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


As0π 10mm( )

2

4 n

As


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


Page 122

As0π 10mm( )

2

4 n

As


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


As0π 10mm( )

2

4 n

As


s min Floorb

n10mm

smax

240 mm

Shrinkage rebars

b 1m


As0π 10mm( )

2

4 n

As


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


Boundary conditions in short and long directions

Simple

Continuous

0

1

ShortContinuous

Continuous

LongContinuous

Continuous

Solution


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


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



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


As0π 10mm( )

2

4 n

As


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


As0π 10mm( )

2

4 n

As


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


As0π 10mm( )

2

4 n

As


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


As0π 10mm( )

2

4 n

As


s min Floorb

n10mm

smax

240 mm

Shrinkage rebars

b 1m


As0π 10mm( )

2

4 n

As


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


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


Mneg 0.65 M0 124.636 kN m

Mpos 0.35 M0 67.112 kN m


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


As0π 10mm( )

2

4 n

As


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


As0π 10mm( )

2

4 n

As


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


As0π 10mm( )

2

4 n

As


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


As0π 10mm( )

2

4 n

As


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


Mneg 0.65 M0 78.152 kN m

Mpos 0.35 M0 42.082 kN m


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


As0π 10mm( )

2

4 n

As


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


As0π 10mm( )

2

4 n

As


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


As0π 10mm( )

2

4 n

As


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


As0π 10mm( )

2

4 n

As


s min Floorb

n10mm

smax

240 mm

Page 140

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


m2

1m2

1m2

cos α( ) 0.463

kN

m2


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


m2


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


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


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


b

200mm

π 10mm( )2

4 3.927 cm

2

Link rebars

b 1m t max twaist tlanding 150 mm


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=


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=


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


Icrb x

3

3A2 d x( )

2 5.114 10

5 cm

4


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


Icrb x

3

3A2 d x( )

2 4.806 10

5 cm

4


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


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



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



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


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

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

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