Mathcad in Concrete Structures (ACI 318-05) 6th Edition (CM)

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  1.1 m

1mi  200m  1.124  mi D  6in  15.24  cm

Size of standard concrete cylinder

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

1psi  1 

= pound per square inch

lbf 2

in ksi

1ksi  1 

= kilopound per square inch

kip 2

in psf

1psf  1 

= pound per square foot

lbf ft

1psi  6.895  kPa

1kPa  0.145  psi

2

1Pa  1 

N 2

m 1ksi  6.895  MPa

1MPa  0.145  ksi

1kPa  1 

kN 2

m 1MPa  1 

N 2

mm 1psf  0.048 

kN

1

2

m 1psf  4.882 

kN 2

 20.885 psf

m

kgf

1

2

m

kgf 2

 0.205  psf

m

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

200

2

m 100psf  4.788 

kgf 2

 40.963 psf

m

kN 2

m

4.80

kN 2

m

Page 2

 100.25 psf

Density 1pcf  1 

lbf ft

3

125pcf  19.636

kN 3

145pcf  22.778

m

3

m

Moments 1ft kip  1.356  kN m

User setting Riels  1 USD  4165Riels 124USD  516460 Riels 200000Riels  48.019 USD

CostSteel  665

kN

USD 1tonnef

670kgf  CostSteel  445.55 USD

Page 3

2. Simple Calculation

3

 2 1  3    23.472 3  

a  2

b  4

c  1

2

Δ  b  4  a c

x 1  x 2 

b 

Δ

 0.225

Δ

 2.225

2 a b  2 a

Page 4

3. Materials 1. Concrete Standard concrete cylinder

D  6in  15.24  cm

D  150mm

H  12in  30.48  cm

H  300mm

Concrete compression strength f'c  3000psi  20.7 MPa

f'c  20MPa  2900.8 psi

f'c  4000psi  27.6 MPa

f'c  25MPa  3625.9 psi

f'c  5000psi  34.5 MPa

f'c  35MPa  5076.3 psi

Concrete ultimate strain ε u  0.003 Cubic and cylinder compression strength f'c fcube = 0.85 f'c  20MPa

f'c fcube   23.529 MPa 0.85

f'c  25MPa

f'c fcube   29.412 MPa 0.85

f'c  35MPa

f'c fcube   41.176 MPa 0.85

Modulus of rupture (tensile strength) fr = 7.5 f'c

(in psi)

Metric coefficient

7.5psi

f'c psi

C  7.5 fr = 0.623  f'c

= 7.5MPa 

psi MPa



(in MPa) Page 5

MPa psi

psi MPa



 0.623

f'c MPa



MPa psi

= C MPa 

f'c MPa

Modulus of elasticity

Ec = 33 wc wc

1.5

 f'c

(in psi)

is a unit weight of concrete (in pcf)

 kN  3 m 33psi   pcf

Metric coefficient

Ec = 44 wc wc

1.5

 f'c

   

1.5



MPa psi

 44.011 MPa

(in MPa)

in kN/m3

Example 3.1 Concrete compression strength

f'c  25MPa

Unit weight of concrete

wc  24

kN 3

m Modulus of rupture f'c fr  7.5psi  3.114  MPa psi f'c fr  0.623MPa   3.115  MPa MPa Modulus of elasticity

 wc  Ec  33psi    pcf 

1.5

 wc  Ec  44MPa    kN   m3   



f'c

4

psi

1.5



 2.587  10  MPa

f'c MPa

4

 2.587  10  MPa

Page 6

145pcf  22.778

kN 3

m

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

5

Es  29000000psi  1.999  10  MPa 5

Es  2  10 MPa Steel yield strength

εy =

fy Es

US Steel Reinforcements

Bar No. (#)

 3  4 5 6   7 8 no    9  10   11     12   13   14   

Bar diameter

D 

Page 7

no 8

in

 9.5   12.7   15.9   19     22.2   25.4  D    mm  28.6   31.8   34.9     38.1   41.3   44.4   

Steel area

 0.71   1.27   1.98   2.85     3.88   5.07  2 A    cm  6.41   7.92   9.58     11.4   13.38   15.52   

2

A 

π D 4

Weight of steel reinforcements

W  A 7850

kgf 3

m

 0.559   0.994   1.554   2.237     3.045   3.978  kgf W    5.034  m  6.215   7.52     8.95   10.503   12.182   

ORIGIN  1 n  1  9

n Area  A n

Page 8

2

Bar #

Diameter (mm)

1

3

9.5

0.71

1.43

2.14

2.85

3.56

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

Area of cross section (cm ) for the number of bars is equal to 2 3 4 5 6 7 8 4.28

4.99

5.70

9

Weight (kgf/m)

6.41

0.559

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

Metric Steel Reinforcements T

D  ( 6 8 10 12 14 16 18 20 22 25 28 32 36 40 ) mm 2

A 

π D

W  A 7850

4

kgf 3

m n Area  A n

n  1  9

2

Diameter (mm)

1

6

0.28

0.57

0.85

1.13

1.41

1.70

1.98

8

0.50

1.01

1.51

2.01

2.51

3.02

10

0.79

1.57

2.36

3.14

3.93

4.71

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 40

Area of cross section (cm ) for the number of bars is equal to 2 3 4 5 6 7 8

9

Weight (kgf/m)

2.26

2.54

0.222

3.52

4.02

4.52

0.395

5.50

6.28

7.07

0.617

10.18 20.36 30.54 40.72 50.89 61.07 71.25 81.43 91.61 12.57 25.13 37.70 50.27 62.83 75.40 87.96 100.53 113.10

Page 9

7.990 9.865

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 ( ε t  0.005)

ϕ = 0.9

Compression-controlled ( ε t  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 if ε t  0.005

0.7 

0.70 if ε t  0.002 1.7  200  ε t 3

0.7  otherwise

For other members ϕ=

0.9 if ε t  0.005 0.65 if ε t  0.002 1.45  250  ε t 3

otherwise

Page 11

0.9  0.7 0.005  0.002 200





 ε t  0.002



 ε t  0.002 = 3



1.7  200  ε t 3

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  t 

Perimeter 180

 111.111  mm

 1 1   L  ( 133.333 80 )  mm   a  30 50 

t  110mm

Section of beam B1 L  6m h 

 1 1   L  ( 600 400 )  mm    10 15 

h  500mm

b  ( 0.3 0.6 )  h  ( 150 300 )  mm

b  250mm

For girders

h=

 1 1  L    8 10 

For two-way slab beams

h=

 1 1  L    10 15 

For floor beams

h=

 1 1  L    15 20 

Section of beam B2 L  4m h 

 1 1   L  ( 400 266.667 )  mm    10 15 

b  ( 0.3 0.6 )  h  ( 90 180 )  mm Page 12

h  300mm b  200mm

B. Loads on Slab Floor cover

kN

Cover  50mm 22

 1.1

3

m Slab  t 25

RC slab

kN

kN 2

m

Ceiling  0.40

Ceiling

2

m

 2.75

3

kN

m kN 2

m

Mechanical  0.20

M&E

kN 2

m Partition  1.00

Partition

kN 2

m Dead load

DL  Cover  Slab  Ceiling  Mechanical  Partition  5.45

kN 2

m Live load for Lab

LL  60psf  2.873 

kN 2

m Factored load

kN wu  1.2 DL  1.6 LL  11.137 2 m

C. Loads of Wall Void  30mm 30mm 190 mm 4 wbrick.hollow  ( 90mm 90mm 190 mm  Void)  20

kN 3

 1.744  kgf

m wbrick.solid  45mm 90mm 190 mm 20

kN 3

 1.569  kgf

m

wbrick.hollow kN ρbrick.hollow   11.111 3 90mm 90mm 190 mm m Brickhollow.10   120mm  Void

55  kN kN  20  1.648    2 3 2 1m  m m  110  kN kN Brickhollow.20   220mm  Void  20  2.895    2 3 2 1m  m m 

Page 13

D. Loads on Beam B1 Self weight

SW  25cm ( 50cm  110mm)  25

kN 3

 2.438 

m Wall

kN m

kN wwall  Brickhollow.10  ( 3.5m  50cm)  4.943  m 4m

Slab

α 

2

6m

 0.333

2

3

k  1  2  α  α  0.815

4m kN wD.slab  DL  2  k  17.763 2 m 4m kN wL.slab  LL  2  k  9.363  2 m Dead load

kN wD  SW  wwall  wD.slab  25.143 m

Live load

kN wL  wL.slab  9.363  m

Factored load

kN wu  1.2 wD  1.6 wL  45.153 m

E. Loads on Beam B2 Self weight

SW  20cm ( 30cm  110mm)  25

kN 3

 0.95

m Wall

kN m

kN wwall  Brickhollow.10  ( 3.5m  30cm)  5.272  m 4m

Slab

α 

2

4m

 0.5

2

4m kN wD.slab  DL  2  k  13.625 2 m 4m kN wL.slab  LL  2  k  7.182  2 m Dead load

kN wD  SW  wwall  wD.slab  19.847 m

Live load

kN wL  wL.slab  7.182  m

Factored load

kN wu  1.2 wD  1.6 wL  35.308 m

Page 14

3

k  1  2  α  α  0.625

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 PB1  25cm ( 50cm  110mm)  25

Beam loads

kN 3

 L  14.625 kN

m PB2  20cm ( 30cm  110mm)  25

kN 3

 B  3.8 kN

m

Pwall.1  Brickhollow.10  ( 3.5m  50cm)  L  29.657 kN

Wall loads

Pwall.2  Brickhollow.10  ( 3.5m  30cm)  B  21.089 kN SW = ( 5%  7%)  PD

SW of column

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

Pu  1.2 PD  1.6 PL  2535.958 kN

PD  PL

 1.299

Control PD  PL n B L PL PD  PL

 11.622

kN

(Ref. 10

kN 2

2

m

m

 18

kN m

(Ref. 15%  35% )

 24.72  %

Page 15

2

)

06 . Loads on Structures (Case of One-Way Slabs)

A. Preliminary Design Thickness of one-way slab (both ends continue) L 

6m 2

tmin  t 

 3m L

28

 107.143  mm

 1 1   L  ( 120 85.714 )  mm    25 35 

t  120mm

Page 16

Section of floor beam B1 L  8m h 

 1 1   L  ( 533.333 400 )  mm    15 20 

b  ( 0.3 0.6 )  h  ( 150 300 )  mm

h  500mm b  250mm

Section of girder B2 L  6m

 1 1   L  ( 750 600 )  mm    8 10 

h  600mm

b  ( 0.3 0.6 )  h  ( 180 360 )  mm

b  300mm

h 

B. Loads on Slab kN


 1.1

3


kN

 3

3

m

Ceiling  0.40

kN 2

m

kN 2

m kN 2

m


kN 2


kN 2

m


kN 2

m LL  60psf  2.873 

kN 2

m

kN m

wu  1.2 DL  1.6 LL  11.437 1m

Page 17

C. Loads on Beam B1 Void  30mm 30mm 190 mm 4 Brickhollow.10   120mm  Void

 

SW  25cm ( 50cm  120mm)  25

  20 kN  1.648  kN 2 3 2 1m  m m 55

kN 3

 2.375 

kN

m

m

kN wwall  Brickhollow.10  ( 3.5m  50cm)  4.943  m kN wD.slab  DL 3 m  17.1 m kN wL.slab  LL 3 m  8.618  m kN wD  SW  wwall  wD.slab  24.418 m kN wL  wL.slab  8.618  m kN wu  1.2 wD  1.6 wL  43.091 m

D. Loads on Girder B2 SW  30cm ( 60cm  120mm)  25

kN 3

 3.6

m

kN m

kN wwall  Brickhollow.10  ( 3.5m  60cm)  4.778  m PB1  25cm ( 50cm  120mm)  25

kN 8m  4m   14.25  kN 3 2 m

Pwall  Brickhollow.10  ( 3.5m  50cm) 

PD.slab  DL 3 m PL.slab  LL 3 m

8m  4m 2

8m  4m 2

8m  4m

 102.6  kN  51.711 kN

Page 18

2

 29.657 kN

Factored loads kN wD  SW  wwall  8.378  m wL  0 kN wu  1.2 wD  1.6 wL  10.054 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

Slope angle

G 

3.5m

H 

3.8m

12

24

α  atan

 291.667  mm  158.333  mm

G  2  H  60.833 cm

H

  28.496 deg  G

Loads on Waist Slab Thickness of waist slab

t  120mm

Step cover

Cover  50mm ( H  G)  22

kN

1m

3 G 1 m

m

Page 20



 1.697 

kN 2

m

Step 

SW of step

G H 2

Slab  t 25

SW of waist slab

kN

 24

3 G 1 m

kN 2

m

2

kN

1m

3



2

kN 2

m Handrail  0.50

 3.414 

1m  cos( α)

Renderring  0.40

Handrail

 1.9

m

m Renderring

1m



kN 2

m 2

1m



2

1m  cos( α)

 0.455 

kN 2

m

kN 2

m

DL  Cover  Step  Slab  Renderring  Handrail

Total dead load

DL  7.966 

kN 2

m

LL  100psf  4.788 

Live load for public staircase

kN 2

m

kN wu  1.2 DL  1.6 LL  17.22  2 m

Factored load

Loads on Landing Slab kN


3

 1.1

m

kN

Slab  150mm 25

3

kN 2

m  3.75

m

kN 2

m

kN

Renderring  0.40

2


kN 2

m

DL  Cover  Slab  Renderring  Handrail  5.75

kN 2

m LL  100psf  4.788 

kN 2

m

kN wu  1.2 DL  1.6 LL  14.561 2 m

Page 21

8. Loads on Roof

Slope angle

α  atan

  30.964 deg    2 

Srokalinh tile

Tile  30mm 20

3m

  10m 

kN 3

 0.6

m Purlins

kN 2

m

w20x20x1.0  ( 20mm 20mm  18mm 18mm)  7850

kgf 3

 0.597 

m 1m kN Purlin  w20x20x1.0   0.068  2 1m 100 mm cos( α) m Rafters

w40x80x1.6  ( 40mm 80mm  36.8mm 76.8mm)  7850

kgf 3

m kgf w40x80x1.6  2.934  1m 1m kN Rafter  w40x80x1.6  0.045 2 750mm 1m cos ( α) m

Page 23

kgf 1m

Roof beam

kN

Beam  20cm 30cm 25

3

m Roof column

Total dead load

Live load

Column  20cm 20cm

3m 2

1m



1m  25

10m

kN 2

m

4

kN 3

m



1

 10m  4m    4 

 0.15

kN 2

m

kN wD  Tile  Purlin  Rafter  Beam  Column  1.463  2 m wL  1.00

kN 2

m Factored load

 0.6

kN wu  1.2 wD  1.6 wL  3.356  2 m

Page 24

9. ASCE Wind Loads V  120

Basic wind speed

km

V  33.333

hr

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

q s  0.613

N 2

m

 

 

V m s

m

V  74.565mph 

s

2

  0.681 kN  2 m  

Velocity pressure coefficients zg  274m

α  9.5

(For exposure C) 2

Kz( z)  2.01 

max( z4.6m) 



zg

α

Kz( 10m)  1.001

 

Velocity wind pressure q z( z)  q s  Kz( z)  Kzt I Kd

q z( 10m)  0.667

kN 2

m Design wind pressure





p z zCp  q z( 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

 0   0.5      0.5    1     Cp.leeward  linterp 2   0.3  λ  0.478  4   0.2         40   0.2   Cp.side  0.7

Floor heights

 3.5m     3.5m   3.5m  H   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

i  1  n



a  i

H H  k

k1

a

n 1



i

2

H

Bwindward  6m

Bleeward  6m

Bside  4m a

 i 1 Pwindward   p z z Cp.windward dz Bwindward i a





i







Pleeward  p z h Cp.leeward  a  a  Bleeward i 1 i i







Pside  p z h Cp.side  a  a  Bside i 1 i i

Page 26

 24.5   22.75     19.25   15.75  reverse ( a)   m  12.25   8.75   5.25     1.75 

 5.70   11.13  10.71 reverse  augment Pwindward Pleeward Pside    10.21   9.61  8.81  8.10 

3.43 3.35 



6.86 6.71  6.86 6.71 

6.86 6.71   kN



6.86 6.71  6.86 6.71 



6.86 6.71 

Alternative ways i  1  n i

b  i



H

k

k 1







Prectangle  p z b Cp.windward  a  a  Bwindward i i 1 i i

Ptrapezium  i







p z a Cp.windward  p z a Cp.windward i i 1 2

 5.70   11.13  10.71 reverse  augment Pwindward Prectangle Ptrapezium    10.21   9.61  8.81  8.10 

Page 27

5.75 11.13 10.71 10.22 9.62 8.83 8.08

  a  a B  i1 i windward

  11.12  10.70  10.20  kN  9.59  8.78   8.20  5.70

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 β c =

Location

a 2

C = α f'c c b = γ f'c a b

Resultant

a = 2  β c = β1  c

Thus,

γ = α

c a

=

where

β1 = 2  β

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

0.72

β1

0.85

 0.847

0.68 0.80

 0.85

0.64 0.75

 0.853

Page 28

0.60 0.70

 0.857

0.56 0.65

 0.862

Conclusion:

γ = 0.85 β1 =

0.85 if f'c  4000psi

4000psi  27.6 MPa

0.65 if f'c  8000psi

8000psi  55.2 MPa

0.85  0.05

f'c  4000psi 1000psi

otherwise

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 M n = C  d 

  = T  d  2    a  M n = 0.85 f'c a b   d   2  a M n = As fs  d   2  a

a



2 (2.1) (2.2)

Conditions of strain compatibility εs dc = c εu εs = εu c = d

dc c εu

εu  εs

dt  c

or

εt = εu

or

εu c = d t εu  εt

Unknowns = 3

a As fs

Equations = 2

X = 0

M = 0

Additional condition

fs = fy

(From economic criteria)

Page 29

(3.1)

c (3.2)

C. Steel Ratios

ρ=

As b d

=

As fy b  d  fy

ρ = 0.85 β1 

f'c



=

0.85 f'c a b b  d  fy

εu

fy ε u  ε s

f'c c f'c c d t = 0.85 β1   = 0.85 β1    fy d fy d t d

= 0.85 β1 

f'c

εu





dt

fy ε u  ε t d

Balanced steel ratio fc = f'c

fs = fy

εs = εy =

fy Es

f'c εu f'c 600MPa ρb = 0.85 β1   = 0.85 β1   fy ε u  ε y fy 600MPa  fy 5

ε u  0.003

Es  2  10 MPa

ε u  Es  600  MPa

Maximum steel ratio ACI 318-99

ρmax = 0.75 ρb

ACI 318-02 and later

f'c εu ρmax = 0.85 β1   fy ε u  ε t

For fy  390MPa For ε t  0.004 For ε t  0.005

ε s 

fy Es

ρmax ρb ρmax ρb

=

=

Minimum steel ratio

ρmin =

3  f'c



200

fy fy 0.249  f'c 1.379 ρmin =  fy fy

(in psi)

(in MPa)

Page 30

with ε t  0.004

 0.002 εu  εy ε u  0.004 εu  εy ε u  0.005

=

=

5 7 5 8

= 0.714

= 0.625

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

Step 2. Calculation of flexural strength a=

As fy 0.85 f'c b

c=

a β1

a M n = As fy   d   2  εt = εu

dt  c c

 

ϕ = ϕ εt

The design flexural strength is ϕ M n

Example 10.1

Page 31

As = ρ b  d

Concrete dimension

b  200mm

Steel reinforcements

As  5 

h  350mm

π ( 16mm)

2

4

 10.053 cm

2

d  h   30mm  6mm  16mm 

40mm 

  278  mm 2   16mm  d t  h   30mm  6mm    306  mm 2   f'c  25MPa

Materials

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  

ε u  0.003 f'c εu ρmax  0.85 β1    0.02 fy ε u  0.004 f'    0.249MPa c  MPa 1.379MPa   ρmin  max     0.00354 fy fy   ρ 

As b d

 0.018

Steel_Reinforcement 

"is Enough" if ρ  ρmin "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 M n  As fy   d    90.911 kN m 2  Page 32

a β1

 108.532  mm

dt  c

ε t  ε u 

c

 0.00546

 1.45  250  ε t    min 0.9  0.9 3   

ϕ  0.65 max 

The design flexural strength is ϕ M n  81.82  kN m

E. Determination of Steel Area Given:

M u 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

=

a As fy   d   2 

2

b d



R = ρ fy   1 



=

2

As b d

a

d  fy 

2

d

1 = ρ fy   1   w 2  

 1    = 0.85 f'c w  1   w 1.7 f'c 2    ρ fy

Quadratic equation relative w R 0.85 f'c

= w  1 



w1 = 1 

ρ = 0.85

1  2

fy

 w



0.85 f'c

1  2

f'c

2

R

2

w  2 w  2

w=1

1

=0

R 0.85 f'c

1

w2 = 1 

R 0.85 f'c

 w = 0.85

f'c fy



1 



1  2

  0.85 f'c 

Page 33

R

1  2

R 0.85 f'c

1

Step 1. Assume ϕ = 0.9 Mu

Mn =

ϕ

Step 2. Calculation of steel area Mn

R=

b d

2

ρ = 0.85

f'c fy



1 

1  2



  0.85 f'c  R

ρ  ρmax

: the beam is doubly reinforced (concrete is not enough)

ρ  ρmax






As = max ρ ρmin  b  d

(this is a required steel area)

Step 3. Checking for flexural strength a=

As fy

( As is a provided steel area)

0.85 f'c b

a M n = As fy   d   2  c=

a

εt = εu

β1

FS =

Mu ϕ M n

dt  c

 

ϕ = ϕ εt

c

(usage percentage)

FS  1

: the beam is safe

FS  1

: the beam is not safe

Example 10.2 Required strength

M u  153kN m

Concrete section

b  200mm

h  500mm

d  h   30mm  8mm  18mm 



Page 34

40mm  2

  424  mm 

d t  h   30mm  8mm 



f'c  25MPa

Materials

18mm 

  453  mm 

2

fy  390MPa

Solution Steel ratios

 

β1  0.65 max  0.85  0.05

f'c  27.6MPa 

  min 0.85  0.85  

6.9MPa

ε u  0.003 f'c εu ρmax  0.85 β1    0.02 fy ε u  0.004 f'    0.249MPa c  MPa 1.379MPa   ρmin  max     0.00354 fy fy   ϕ  0.9

Assume M n 

Mu ϕ

 170  kN m

Steel area R 

Mn b d

2

ρ  0.85

 4.728  MPa f'c fy



1 

1  2



  0.014  0.85 f'c  R

ρmin  ρ  ρmax  1 As  ρ b  d  11.783 cm As  6 

π ( 16mm) 4

2

2

 12.064 cm

2

Checking for flexural strength a 

As fy 0.85 f'c b

 110.702  mm

c 

a M n  As fy   d    173.444  kN m 2 

Page 35

a β1

 130.238  mm

dt  c

ε t  ε u 

c

 0.00743

 1.45  250  ε t    min 0.9  0.9 3   

ϕ  0.65 max  Mu

FS 

ϕ M n

The_beam 

 0.98 "is safe" if FS  1

The_beam  "is safe"

"is not safe" otherwise

F. Determination of Concrete Dimension and Steel Area Given:

M u f'c fy

Find:

b  d As

Step 1. Determination of concrete dimension Assume ε t  0.004

(Usually ε t  0.005) ρ fy   R = ρ fy   1   1.7 f'c   Mu Mn = ϕ

f'c εu ρ = 0.85 β1   fy ε u  ε t

 

ϕ = ϕ εt 2

bd =

Mn R Mn

Option 1:

R

b=

d

2 Mn

Option 2:

R

d=

b 3

Option 3:

k=

b

d=

d

Step 2. Calculation of steel area

R=

Mn b d

2 Page 36

Mn R

k

b = k d

ρ = 0.85

f'c fy



1 

1  2





  0.85 f'c  R



As = max ρ ρmin  b  d

Step 3. Checking for flexural strength

a=

As fy

c=

0.85 f'c b

a β1

a M n = As fy   d   2  εt = εu FS =

dt  c

 

ϕ = ϕ εt

c

Mu ϕ M n


M u  700kN m

Materials

f'c  25MPa

fy  390MPa


 

β1  0.65 max  0.85  0.05

f'c  27.6MPa  6.9MPa

  min 0.85  0.85  

ε u  0.003 f'c εu ρmax  0.85 β1    0.02 fy ε u  0.004 f'    0.249MPa c  MPa 1.379MPa   ρmin  max     0.00354 fy fy   Assume

ε t  0.007 f'c εu ρ  0.85 β1    0.014 fy ε u  ε t

Page 37

ρ fy   R  ρ fy   1    4.728  MPa 1.7 f'c  

 1.45  250  ε t    min 0.9  0.9 3   

ϕ  0.65 max 

M n 

Mu ϕ

 777.778  kN m

Concrete dimension k=

b

k 

d 3

400

Cover  30mm  10mm  25mm 

600

40mm 2

Cover  85 mm Mn R

d 

 627.231  mm

k

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    400   mm      h   700 

Steel area Mn

R 

b d

 5.141  MPa

2

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 As  8 

π ( 25mm) 4

2

 39.27  cm

2

2

d t  h   30mm  10mm 



d t  647.5  mm Checking for flexural strength a 

As fy

 180.18 mm

0.85 f'c b

c 

a M n  As fy   d    803.914  kN m 2  ε t  ε u 

dt  c c

 0.00616

 1.45  250  ε t    min 0.9  0.9 3   

ϕ  0.65 max  FS 

Mu ϕ M n

 96.749 %

Page 38

a β1

 211.976  mm

25mm  2

 

11. Design of Doubly Reinforced Beams ρ  ρmax

: the beam is singly reinforced (with tensile reinforcements only)

ρ  ρmax

: the beam is doubly reinforced (with tensile and compression reinforcements)

A. Strength Analysis

Equilibrium in forces

X = 0 T = C  Cs

(1)

T = As fs = As fy C = 0.85 f'c a b Cs = A's f's


M = 0 M n = M n1  M n2

(2)

M n1 = T ( d  d') = A's f's ( d  d') M n2 = C  d 

a

a   = 0.85 f'c a b  d   2 2   a M n2 =  T  Cs   d   =  As fy  A's f's   d  2  

Page 39

a



2

Conditions of strain compatibility εs εu

=

dc

(3.1)

c dc

εs = εu c = d ε's εu

=

εt = εu

or

c εu

c εu

c = d t εu  εt

or

εu  εs

dt  c

c  d'

(3.2)

c ε's = ε u  c = d'

c  d' c εu

ε u  ε's

B. Steel Ratios Compression steel ratio A's ρ' = 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

f'c c f's = 0.85 β1    ρ' fy d fy

Maximum tensile steel ratio f's ρt.max = ρmax  ρ' fy ρ  ρt.max

: concrete is enough

ρ  ρt.max

: concrete is not enough

Minimum tensile steel ratio f'c c f's f'c εu f's d' ρ = 0.85 β1    ρ' = 0.85 β1     ρ' fy d fy fy ε u  ε's d fy fy f's = fy ε's = ε y = Es

Page 40

f'c εu d' ρcy = 0.85 β1     ρ' fy ε u  ε y d 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 d t d' As A's f'c fy

Find:

ϕMn

Step 1. Checking for singly reinforced beam ρ=

ρ' =

As b d ρ  ρmax


ρ  ρmax

: the beam is doubly reinforced

A's

ρt.max = ρmax  ρ'

b d

ρ  ρt.max


ρ  ρt.max

: concrete is not enough ρ = ρt.max

As = ρ b  d

Step 2. Determination of compression parameters 2.1. Assume f's = fy 2.2. Calculate As fy  A's f's a= 0.85 f'c b ε's = ε u  If

c  d' c

f's.revised  f's

c=

a β1

f's.revised = Es ε's  fy 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 ε u  As fy = 0.85 f'c a b  A's Es ε u 

β1  c  β1  d' β1  c

c  d' c

= 0.85 f'c a b  A's f1 

a  β1  d' a

where f1 = Es ε u = 600MPa 2





2





0.85 f'c a  b  A's f1  As fy  a  A's f1  β1  d' = 0 0.85 f'c a  b  A's f1  As fy  a  A's f1  β1  d'

=0

2

0.85 f'c d  b 2  a   ρ' f1  ρ fy  a  ρ' f1 β1  d' = 0   d 0.85 f'c 0.85 f'c d d 2

w  2 p w  q = 0 2

w1 = p 

p q0



a = d  p  ε's = ε u 

p=

2

p q



w2 = p  c=

c  d'

1 ρ' f1  ρ fy  2 0.85 f'c

a β1

f's = Es ε's  fy

c

Step 3. Calculation of flexural strength M n1 = A's f's ( d  d') a M n2 = As fy  A's f's   d   2  dt  c εt = εu ϕ = ϕ εt c





 



2

p q0



ϕMn = ϕ M n1  M n2

Page 42

q=

ρ' f1  β1 d'  0.85 f'c d

Example 11.1 Concrete dimension

b  300mm


As  8 

h  550mm

π ( 20mm)

2

 25.133 cm

4

2

d  h   30mm  10mm  16mm  20mm 



40mm  2

 

d  454  mm d t  h   30mm  10mm  16mm 



A's  4 

π ( 20mm)

2

4

 12.566 cm

d'  30mm  10mm 

20mm

f'c  25MPa

Materials

2

2

 50 mm

fy  390MPa


 

β1  0.65 max  0.85  0.05

f'c  27.6MPa  6.9MPa

  min 0.85  0.85  

ε u  0.003 f'c εu ρmax  0.85 β1    0.0174 fy ε u  0.005 f'    0.249MPa c  MPa 1.379MPa   ρmin  max     0.00354 fy fy   Checking for singly reinforced beam ρ 

As b d

 0.0185

The_beam 

"is singly reinforced" if ρ  ρmax "is doubly reinforced" otherwise

The_beam  "is doubly reinforced" ρ' 

A's b d

3

 9.226  10

ρt.max  ρmax  ρ'  0.027

Page 43

20mm  2

  484  mm 

Concrete 

"is enough" if ρ  ρt.max "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 5

Es  2  10 MPa ε y 

fy Es

f1  Es ε u  600  MPa

 1.95  10

3

f'c εu d' ρcy  0.85 β1     ρ'  0.024 fy ε u  ε y d Compression_steel 

"will yield" if ρ  ρcy "will not yield" otherwise

Compression_steel  "will not yield" a 

As fy  A's fy

if Compression_steel = "will yield"

0.85 f'c b otherwise p

1 ρ' f1  ρ fy  2 0.85 f'c

q

ρ' f1  β1 d'  0.85 f'c d



d  p 

2

p q

a  90.825 mm f's 

 c 

a β1

 106.853  mm

fy if Compression_steel = "will yield" otherwise ε's  ε u 



c  d' c

min Es ε's fy



f's  319.24 MPa Flexural strength Page 44

M n1  A's f's ( d  d')  162.072  kN m a M n2  As fy  A's f's   d    236.576  kN m 2  dt  c ε t  ε u   0.011 c





 1.45  250  ε t    min 0.9  0.9 3   

ϕ  0.65 max 





ϕMn  ϕ M n1  M n2  358.783  kN m

D. Design of Doubly Reinforced Beam Given:

M u b  d d t d' f'c fs

Find:

As A's

Step 1. Assume ε t f'c εu ρ = 0.85 β1   fy ε u  ε t

 

ϕ = ϕ εt

Step 2. Cheching for singly reinforced beam As = ρ b  d a=

As fy 0.85 f'c b

a M n = As fy   d   2  M u  ϕMn


M u  ϕMn

: the beam is doubly reinforced

Step 3. Case of doubly reinforced beam M n2 = M n M n1 =

Mu ϕ

 M n2

Page 45

c=

a

f's = Es ε u 

β1

A's = As =

c  d' c

M n1

 fy ρ' =

f's ( d  d') 0.85 f'c a b  A's f's

ρ=

fy

A's

ρt.max = ρmax  ρ'

b d As b d

ρ  ρt.max


ρ  ρt.max



M u  1350kN m

Concrete dimension

b  400mm

h  800mm

d  h   30mm  12mm  25mm  25mm 



40mm  2

 

d  688  mm d t  h   30mm  12mm  25mm 



d'  30mm  12mm  f'c  25MPa

Materials

25mm 2

fy  390MPa

Steel ratios

 

f'c  27.6MPa  6.9MPa

  min 0.85  0.85  

ε u  0.003 f'c εu ρmax  0.85 β1    0.0174 fy ε u  0.005 f'    0.249MPa c  MPa 1.379MPa   ρmin  max     0.00354 fy fy   Assume

ε t  0.0092 f'c εu ρ  0.85 β1    0.011 fy ε u  ε t

Page 46

2

 54.5 mm

Solution

β1  0.65 max  0.85  0.05

25mm 

  720.5  mm 

 1.45  250  ε t    min 0.9  0.9 3   

ϕ  0.65 max 

Checking for singly reinforced beam As  ρ b  d  31.342 cm a 

As fy 0.85 f'c b

2

 143.803  mm

c 

a M n2  As fy   d    753.074  kN m 2  The_beam 

a β1

 169.18 mm

"is singly reinforced" if M u  ϕ M n2 "is doubly reinforced" otherwise

The_beam  "is doubly reinforced"

Case of doubly reinforced beam M n1 

Mu ϕ

 M n2  746.926  kN m

f's  min Es ε u 

c  d'



A's 

M n1

f's ( d  d') A's ρ'   0.011 b d As 

c

fy  390  MPa

 30.232 cm

2

ρt.max  ρmax  ρ'  0.028

0.85 f'c a b  A's f's

Concrete 



fy

 61.574 cm

2

ρ 

"is enough" if ρ  ρt.max

As

 0.022

b d


"is not enough" otherwise A's  30.232 cm

Compression steel

As  61.574 cm

Tensile steel

400mm  ( 12mm  30mm)  2  28mm 5 4

Page 47

2

2

 44 mm

5

π ( 28mm)

10

2

 30.788 cm

4 π ( 28mm) 4

2

2

 61.575 cm

2

E. Determination of Tensile Steel Area Given:

M u b  d d t 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

M n1 = A's f's ( d  d') M n2 = M n  M n1 R=

M n2 b d

2

ρ = 0.85

f'c fy



1 



1  2

  0.85 f'c  R

As = ρ b  d a=

As fy

c=

0.85 f'c b

f's.revised = Es ε u  εt = εu

c  d' c

dt  c

a β1

 fy

 

ϕ = ϕ εt

c f's.revised  f's

: Goto 1.2

M u  ϕ M n

: Goto 1.2

Step 2. Calculation of tensile steel area As = ρ' =

0.85 f'c a b  A's f's A's b d

ρ=

fy

As b d

ρt.max = ρmax  ρ' ρ  ρt.max


ρ  ρt.max


Page 48


M u  1350kN m

Concrete dimension

b  400mm

h  800mm

d  h   30mm  12mm  25mm  28mm 



40mm  2

 

d  685  mm d t  h   30mm  12mm  25mm 



d'  30mm  12mm 

Compression reinforcements

A's  5 

π ( 28mm)

2

 56 mm

2

4

f'c  25MPa

Materials

28mm

 30.788 cm

2

fy  390MPa


 

β1  0.65 max  0.85  0.05

f'c  27.6MPa  6.9MPa

  min 0.85  0.85  

ε u  0.003 f'c εu ρmax  0.85 β1    0.0174 fy ε u  0.005 f'    0.249MPa c  MPa 1.379MPa   ρmin  max     0.00354 fy fy   Compression parameters

Page 49

28mm  2

  719  mm 

Compression ( ε ) 

f's  fy ϕ  0.9 for i  0  99 Mn 

Mu ϕ

M n1  A's f's ( d  d') M n2  M n  M n1 M n2

R

b d

2

ρ  0.85

f'c fy



1 



1  2

R 0.85 f'c

  

As  ρ b  d a c

As fy 0.85 f'c b a β1

f's.revised  min Es ε u 



c  d' c

fy



  f's   fy    ϕ   i Z  a    d    f's.revised   fy    εt  εu

dt  c c

 1.45  250  εt    min 0.9 3     f's.revised  f's  ( break) if   ε   M u  ϕ M n  f's  

ϕ  0.65 max 

f's  f's.revised

 T

reverse Z

Z  Compression( 0.000001)

Page 50

a  Z

 d  142.792  mm

c 

0 2

a β1

 167.99 mm

f's  Z  fy  390  MPa 0 0 Tensile steel area ρ'  As 

A's b d

 0.011

0.85 f'c a b  A's f's

Concrete 

fy

ρt.max  ρmax  ρ'  0.029  61.909 cm

2

ρ 

"is enough" if ρ  ρt.max

As b d

 0.023


"is not enough" otherwise Tensile steel

As  61.909 cm

Page 51

2

10

π ( 28mm) 4

2

 61.575 cm

2

12. Design of T Beams 12.1. Effective Flange Width

For symmetrical T beam: b

L

b  b w  16h f

4

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 or

where

a  hf

M u  ϕMnf

or

hf   M nf = 0.85 f'c h f  b   d   2  

Page 52

M u  ϕMnf

X = 0

Equilibrium in forces T = C1  C2

T = As fs = As fy





C1 = 0.85 f'c h f  b  b w = Asf  fy C2 = 0.85 f'c a b w = T  C1 = As fy  Asf  fy

M = 0

Equilibrium in moments M n = M n1  M n2

hf  hf    M n1 = C1   d   = Asf  fy  d   2  2    a a M n2 = C2   d   = 0.85 f'c a b w  d   2 2   a a M n2 = T  C1   d   = As fy  Asf  fy   d   2 2  









Condition of strain compatibility εs εu

=

dc

εt

or

εs

c εs = εu c = d

dc

=

dt  c

εt = εu

c εu

c dt  c c εu

c = d t εu  εt

εu  εs

12.3. Steel Ratios As As fy 0.85 f'c a b w  Asf  fy ρw = = = b w d b w d  fy b w d  fy f'c c ρw = 0.85 β1    ρf fy d

where

Page 53

Asf ρf = b w d

Maximum steel ratio ρw.max = ρmax  ρf ρw  ρw.max


ρw  ρw.max


12.4. Determination of Moment Capacity Given:

b w  d b d t h f As f'c fy

Find:

ϕMn

Step 1. Checking for rectangular beam a=

As fy 0.85 f'c b a  hf a  hf

: the beam is rectangular : the beam is tee

Step 2. Case of T beam Asf =





0.85 f'c h f  b  b w fy

hf   M n1 = Asf  fy   d   2   a=

As fy  Asf  fy

c=

0.85 f'c b w

a β1

a M n2 = As fy  Asf  fy   d   2 



εt = εu



dt  c

 

ϕ = ϕ εt

c





ϕMn = ϕ M n1  M n2

Page 54

Example 12.1

b  28in  711.2  mm

Concrete dimension

h f  6in  152.4  mm b w  10in  254  mm h  30in  762  mm

d  26in  660.4  mm

d t  27.5in  698.5  mm π  As  6 


10

in

2

8  4

f'c  3000psi  20.684 MPa

Materials

fy  60ksi  413.685  MPa


 

β1  0.65 max  0.85  0.05

f'c  4000psi 

  min 0.85  0.85 1000psi  

ε u  0.003 f'c εu ρmax  0.85 β1    0.014 fy ε u  0.005 f'    3psi c  psi 200psi   ρmin  max   0.00333  fy fy   

Checking for rectangular beam

Page 55

2

As  7.363  in

a 

As fy

 157.162  mm

0.85 f'c b

The_beam 

"is rectangular" if a  h f

The_beam  "is T"

"is T" otherwise Case of T beam Asf 





0.85 f'c h f  b  b w fy

 29.613 cm

2

Asf ρf   0.018 b w d As ρw   0.028 b w d Concrete 

ρw.max  ρmax  ρf  0.031

"is enough" if ρw  ρw.max


"is not enough" otherwise





2

As  min ρw ρw.max  b w d

As  7.363  in

hf   M n1  Asf  fy   d    715.669  kN m 2   a 

As fy  Asf  fy 0.85 f'c b w

 165.734  mm

a M n2  As fy  Asf  fy   d    427.446  kN m 2 



ε t  ε u 



dt  c c

 0.008

  1.45  250  ε t   min 0.9  0.9 3   

ϕ  0.65 max 





ϕMn  ϕ M n1  M n2  1028.803 kN m

Page 56

c 

a β1

 194.981  mm

12.5. Determination of Steel Area Given:

M u b w  d d t b h f f'c fy

Find:

As

Step 1. Checking for rectangular beam hf   M nf = 0.85 f'c h f  b   d   2   ϕ = 0.9 M u  ϕMn

: the beam is rectangular

M u  ϕMn

: the beam is tee

Step 2. Case of T beam





0.85 f'c h f  b  b w

Asf =

Asf ρf = b w d

fy

hf   M n1 = Asf  fy   d   2   M n2 =

Mu ϕ

 M n1

M n2

R=

b w d ρ = 0.85

2

f'c fy



1 



1  2

  0.85 f'c  R

As2 = ρ b w d a=

As2  fy 0.85 f'c b w

As =

0.85 f'c a b w  Asf  fy

As ρw = b w d

fy

ρw.max = ρmax  ρf ρw  ρw.max


ρw  ρw.max


Page 57

Example 12.2

h f  3in  76.2 mm

Concrete dimension

L  24ft  7.315 m

s  47in  1.194  m

b w  11in  279.4  mm

d  20in  508  mm

Required strength

M u  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

b w  16 h f s

b  1193.8 mm



Steel ratios

 

β1  0.65 max  0.85  0.05

f'c  4000psi 

  min 0.85  0.85 1000psi  

ε u  0.003 f'c εu ρmax  0.85 β1    0.014 fy ε u  0.005 f'    3psi c  psi 200psi   ρmin  max   0.00333  fy fy    Checking for rectangular beam ϕ  0.9 hf   M nf  0.85 f'c h f  b   d    751.538  kN m 2   The_beam 

"is rectangular" if M u  ϕ M nf "is tee" otherwise

Case of T beam

Page 58

The_beam  "is tee"





0.85 f'c h f  b  b w

Asf 

 29.613 cm

fy

Asf ρf   0.021 b w d

2

hf   M n1  Asf  fy   d    575.646  kN m 2   M n2 

Mu ϕ

M n2

R 

b w d ρ  0.85

2

f'c fy

 M n1  227.801  kN m  3.159  MPa



1 



1  2

R 0.85 f'c

As2  ρ b w d  12.042 cm a 

As2  fy 0.85 f'c b w

As 

  8.484  10 3  

2

 101.408  mm

0.85 f'c a b w  Asf  fy fy

c 

 41.655 cm

2

a β1

 119.304  mm

As ρw   0.029 b w d

ρw.max  ρmax  ρf  0.034 Concrete 

"is enough" if ρw  ρw.max


"is not enough" otherwise As.min  ρmin b w d





As  max As As.min  41.655 cm

2

Page 59

6

π ( 32mm) 4

2

 48.255 cm

2

13. Shear Design Safety provision Vu  ϕVn where

Vu

= required shear strength

Vn

= nominal shear strength

ϕ = 0.75 ϕVn

is a strength reduction factor for shear

= 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 b w d

(in psi)

Vc = 0.166  f'c b w d

(in MPa)

Steel shear strength Av  fy  d Vs = s where = area of stirrup

Av fy

= yield strength of stirrup

s

= spacing of stirrup

No required stirrups

Vu  ϕVc 2

ϕVc

: no stirrup is required

2  Vu  ϕVc

Vu  ϕVc

: stirrup is minimum : stirrup is required

Minimum stirrups b w s b w s Av.min = 0.75 f'c  50 fy fy b w s b w s Av.min = 0.062  f'c  0.345  fy fy

Page 61

(in psi)

(in MPa)

Maximum spacing of stirrup smax = smax =

Case

Av  fy 0.75 f'c b w Av  fy



0.062  f'c b w

Av  fy

(in psi)

50 b w



Av  fy

(in MPa)

0.345  b w

Vs  2  Vc d smax =  24in = 600mm 2

Case

2  Vc  Vs  4  Vc d smax =  12in = 300mm 4

Case

Vs  4  Vc Concrete is not enough

Example 13.1

Materials

f'c  25MPa

fy  390MPa

Page 62

LL  6.00

Live load for garage

kN 2

m Loads on slab kN

Hardener  8mm 24

3

 0.192 

m Slab  200mm 25

kN 3

2

m

 5

m Mechanical  0.30

kN

kN 2

m

kN 2

m

DL  Hardener  Slab  Mechanical  5.492 

kN 2

m LL  6 

kN 2

m Loads on beam

wbeam  30cm ( 60cm  200mm)  25

kN 3

 3

m

kN m

kN

wD.slab  DL 3.5m  19.222 m kN wL.slab  LL 3.5m  21 m

kN wD  wbeam  wD.slab  22.222 m kN wL  wL.slab  21 m kN wu  1.2 wD  1.6 wL  60.266 m Shear L  8m wu  L V0   241.066  kN 2 V( x )  V0  wu  x Concrete shear strength b w  300mm

d  600mm   40mm  10mm 



f'c Vc  0.166MPa   b  d  134.46 kN MPa w ϕ  0.75

Page 63

20mm  2

  540  mm 

Location of no stirrup zone V0  wu  x =

ϕVc

V0 

x 

2

ϕ Vc

wu

2

 3.163 m

Minimum stirrup Av  2 

π ( 10mm)

2

4

 1.571  cm Av  fy

2

fy  390MPa Av  fy

   591.894  mm 0.345MPa  b w  f'c  b  0.062MPa  MPa w   

smax  min







smax  Floor smax 50mm  550  mm Av  fy  d

Vs.min 

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 

Vu Vs   Vc  170.89 kN ϕ Concrete 

"is enough" if Vs  4  Vc


"is not enough" otherwise s 

Av  fy  d Vs

 193.581  mm

smax.1  smax  550  mm smax.2 

d min 600mm if Vs  2  Vc 2 

smax.2  270  mm

d min 300mm otherwise 4 

 





s  Floor min s smax.1 smax.2 50mm

Page 64

s  150  mm

Example 13.2 Design of shear in support and midspan zones.

Stirrups in Support Zone Vu  V0  wu 

Required shear strength

400mm 2

 229.012  kN

Concrete shear strength f'c Vc  0.166MPa   b  d  134.46 kN MPa w ϕ  0.75 Stirrup 

"is minimum" if Vu  ϕ Vc

Stirrup  "is required"

"is required" otherwise Required steel shear strength Vu Vs   Vc  170.89 kN ϕ Concrete 



"is not enough" otherwise Spacing of stirrup Av  2  s 

π ( 10mm)

Av  fy  d Vs

4

2

 1.571  cm

2

fy  390MPa

 193.581  mm



   591.894  mm 0.345MPa  b w  f'c  b  0.062MPa  MPa w  

smax.1  min

Av  fy

Page 65



Av  fy

d min 600mm if Vs  2  Vc 2 

smax.2 

smax.2  270  mm

d min 300mm otherwise 4 

 





s  Floor min s smax.1 smax.2 50mm  150  mm

Stirrups in Midspan Zone L Vu  V0  wu   120.533  kN 4

Required shear strength Stirrup 

"is minimum" if Vu  ϕ Vc

Stirrup  "is required"

"is required" otherwise Required steel shear strength Vu Vs   Vc  26.25  kN ϕ Concrete 



"is not enough" otherwise Spacing of stirrup Av  2  s 

π ( 10mm) 4

Av  fy  d Vs

2

 1.571  cm

2

fy  390MPa

 1260.208 mm



Av  fy

Av  fy

   591.894  mm 0.345MPa  b w  f'c  b  0.062MPa  MPa w  

smax.1  min

smax.2 



d min 600mm if Vs  2  Vc 2  d min 300mm otherwise 4 

 





s  Floor min s smax.1 smax.2 50mm  250  mm

Page 66

smax.2  270  mm

Page 67

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 M u 2.2. Long (slender) columns (with buckling) Pu M u  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  

where

with

ϕ = 0.70

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 Ast ρg = = 1%  8% Ag

Page 69

Determination of Concrete Section Pu 0.80 ϕ

Ag = 0.85 f'c 1  ρg  fy  ρg





Determination of Steel Area Pu Ast =

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 2

m f'c  25MPa

Materials

fy  390MPa

Solution Loads on slab kN


3

m

kN

Slab  120mm 25

3

m Ceiling  0.40

kN 2

m


kN 2


kN 2

m


kN 2

m LL  3 

kN 2

m

Page 70

Reduction of live load 2

Tributary area

AT  B L  24 m

For interior column

KLL  4

Influence area

AI  KLL AT  96 m

Live load reduction factor

αLL  0.25 

2

4.572

 0.717

AI 2

m

kN LL0  LL αLL  2.15 2 m

Reduced live load

Loads of wall Void  30mm 30mm 190 mm 4 Brickhollow.10   120mm  Void

  20 kN  1.648  kN 2 3 2 1m  m m

Brickhollow.20   220mm  Void

  20 kN  2.895  kN 2 3 2 1m  m m

   

55

110

Loads on column PD.slab  DL B L  136.8  kN PL.slab  LL B L  72 kN PB1  25cm ( 50cm  120mm)  25

kN 3

 L  14.25  kN

m PB2  20cm ( 35cm  120mm)  25

kN 3

 B  4.6 kN

m

Pwall.1  Brickhollow.10  ( 3.5m  50cm)  L  29.657 kN Pwall.2  Brickhollow.10  ( 3.5m  35cm)  B  20.76  kN n  6

Number of floors





PD  PD.slab  PB1  PB2  Pwall.1  Pwall.2  1.05 n  1298.219 kN PL  PL.slab n  432  kN PD  PL B  L n

 12.015

kN 2

m

SW = ( 5%  7%)  PD PL PD  PL Page 71

 24.968 %

Pu  1.2 PD  1.6 PL  2249.063 kN

Determination of column section ρg  0.03

Assume

k=

b

k 

h

300 500

ϕ  0.65 Pu 0.80 ϕ

Ag  0.85 f'c 1  ρg  fy  ρg



h 

Ag k

Ag  1338.529 cm



 472.322  mm

2

b  k  h  283.393  mm

h  Ceil( h 50mm)  500  mm

b  Ceil( b 50mm)  300  mm

 b    300   mm      h   500 

Ag  b  h  1500 cm

2

Determination of steel area Pu Ast 

0.80 ϕ

 0.85 f'c Ag

Ast  30.851 cm

0.85 f'c  fy 6

π ( 20mm) 4

2

 6

Stirrups Main bars

D  20mm

Stirrup dia.

Dv  10mm

Spacing of tie

s  min 16 D 48 Dv b  300  mm





Page 72

π ( 16mm) 4

2

2

 30.913 cm

2

2. Short Columns

Safety provision Pu  ϕPn M u  ϕMn Equilibrium in forces

X = 0

Pn = C  Cs  T Pn = 0.85 f'c a b  A's f's  As fs Equilibrium in moments

M = 0

a h h h M n = Pn  e = C     Cs   d'  T  d   2 2 2 2   a h h h M n = Pn  e = 0.85 f'c a b      A's f's   d'  As fs  d   2 2 2 2   Conditions of strain compatibility εs εu

=

dc c

εs = εu

dc c

fs = Es ε s = Es ε u  ε's εu

=

c  d' c

ε's = ε u 

dc c

c  d' c

f's = Es ε's = Es ε u  Page 73

c  d' c

Unknowns = 5 :

a As A's fs f's

Equations = 4 :

X = 0

M = 0

Case of symmetrical columns:

As = A's

Case of unsymmetrical columns:

fs = fy

2 conditions of strain compatibility

A. Interaction Diagram for Column Strength Interaction diagram is a graph of parametric function, where Abscissa :

M n ( a)

Ordinate:

Pn ( a)

B. Determination of Steel Area Given:

M u Pu b  h f'c fy

Find:

As = A's

Answer:

As = AsN( a) = AsM ( a) Pu  0.85 f'c a b ϕ AsN( a) = f's  fs Mu AsM( a) =

h a  0.85 f'c a b     ϕ 2 2 h  d'  fs  d   2 2  

f's 

f's( a) = Es ε u  fs( a) = Es ε u 

h

c  d' c dc c

 fy  fy

Page 74

Example 14.2 Construction of interaction diagram for column strength. b  500mm

Concrete dimension

As  5 


h  200mm

π ( 16mm)

2

4

 10.053 cm

A's  As  10.053 cm

2

d'  30mm  6mm 

16mm 2

2

 44 mm

d  h  d'  156  mm f'c  25MPa

Materials

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 



c( a) 

6.9MPa

  min 0.85  0.85  

a β1 5

Es  2  10 MPa fs( a)  min Es ε u 

ε u  0.003

d t  d

d  c( a)

fy c( a)   c ( a )  d' f's( a)  min Es ε u  fy c ( a )   ϕ( a) 

εt  εu

d t  c( a) c( a)

 1.45  250  ε t    min 0.90 3   

ϕ  0.65 max 

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  



a  0 

h 100

h

2

a

h     A's f's( a)    d'  As fs( a)   d  2 2  



 h

Interaction diagram for column strength 1500

1250

1000 ϕPn( a)

750

kN 500

250

0

0

20

40

60

ϕMn( a) kN m

Example 14.3 Determination of steel area. Required strength

Pu  1152.27kN M u  42.64kN m

Concrete dimension

b  500mm

Materials

f'c  25MPa

h  200mm

fy  390MPa Concrete cover to main bars

cc  30mm  6mm 

Page 76

16mm 2

h 



2 

Solution Location of steel re-bars d'  cc  44 mm d  h  cc  156  mm Case of axially loaded column Ag  b  h ϕ  0.65 Pu  0.85 f'c Ag ϕ  0.65 0.80 ϕ 2 Ast   2.465  cm 0.85 f'c  fy Case of eccentric column



β1  0.65 max  0.85  0.05

f'c  27.6MPa 



6.9MPa

  min 0.85  0.85  

a

c( a) 

β1 5

Es  2  10 MPa

ε u  0.003

fs( a)  min Es ε u 

d  c( a)

d t  d

fy

  c ( a )  d' f's( a)  min Es ε u  fy c( a)   ϕ( a) 

εt  εu

c( a)

d t  c( a) c( a)

 1.45  250  ε t    min 0.90 3   

ϕ  0.65 max 

Graphical solution Pu AsN( a) 

ϕ( a)

 0.85 f'c a b

f's( a)  fs( a) Mu

AsM( a) 

a1  134.2mm a  a1 a1 

a h  0.85 f'c a b     ϕ( a) 2 2 h h f's( a)    d'  fs( a)   d   2 2  

a2  134.25mm

a2  a1 50

 a2

Page 77

8.735 10

4

8.73 10

4

8.725 10

4

A sN( a) A sM( a)

8.72 10

4

4

8.715 10 0.13418

0.1342

0.13422

0.13424

0.13426

a

a  134.23mm AsN( a)  8.722  cm

2

AsM( a)  8.725  cm As 

2

AsN( a)  AsM( a) 2

 8.724  cm

2

5

Page 78

π ( 16mm) 4

2

 10.053 cm

2

Analytical solution ORIGIN  1 Asteel( No) 

k1 for a  cc cc 

h No

 h

f  f's( a)  fs( a) ( continue ) if f = 0 Pu AsN 

ϕ( a)

 0.85 f'c a b f

( continue ) if AsN  0 h h fd  f's( a)    d'  fs( a)   d   2 2   ( continue ) if fd = 0 Mu

AsM 

a h  0.85 f'c a b     ϕ( a) 2 2 fd

( continue ) if AsM  0 a  h   AsN  Ag  k Z  AsM  Ag   A A sM  sN  Ag

          

kk1



T

csort Z 4



Z  Asteel( 5000) a  Z

rows( Z)  2046

 h  134.24 mm

1 1

AsN  Z  Ag  8.719  cm 1 2

As 

AsN  AsM 2

2

 8.723  cm

AsM  Z  Ag  8.728  cm 1 3 2

Page 79

2

C. Case of Distributed Reinforcements

Pn

b

e

a

d1 dn

0.85 f c

h Tn

T1 C s,1 u s,n

c dn

rUb 3>1> ssrcakp©it EdlmanEdkBRgayeRcInCYr CMuvijmuxkat;ebtug X = 0

Equilibrium in forces n

Pn = C 



n

 As if s i

T = 0.85 f'c a b  i

i 1


i 1

M = 0

a h M n = Pn  e = C     2 2

n



i 1 n

a h M n = 0.85 f'c a b      2 2

T   d  h   i i  2   



i 1

A  f   d   s i s i  i  

Page 80

h 



2 

Conditions of strain compatibility ε

s i

εu

d c i

=

c d c ε

= εu s i

i

c d c

f

= Es ε = Es ε u  s i s i

i

c

Example 14.4 Checking for column strength. Pu  13994.6kN

Required strength

M u  57.53kN m f'c  35MPa

Materials

fy  390MPa

Solution Determination of Concrete Section Case of axially loaded column ϕ  0.65 ρg  0.04

Assume

Pu 0.80 ϕ

Ag   6094.36  cm 0.85 f'c 1  ρg  fy  ρg





Aspect ratio of column section λ = h 

Ag λ

b h

 780.664  mm

2

λ  1 b  λ h  780.664  mm

h  Ceil( h 50mm)

b  Ceil( b 50mm)

 b    800   mm      h   800 

Ag  b  h  6400 cm

Page 81

2

Steel area Pu Ast 

0.80 ϕ

 0.85 f'c Ag

Ast  218.534  cm

0.85 f'c  fy

( 4  7 4) 

π ( 25mm)

2

 ( 4  5 4) 

4

π ( 20mm) 4

800mm  50mm 2

Spacing

2

 232.478  cm

2

2

 87.5 mm

8

Interaction Diagram for Column Strength Distribution of reinforcements

 25  25  25   25 Bars   25  25   25  25  25

25 25 25 25 25 25 25 25 



20 20 20 20 20 20 20 25  20 0

0

0

0

0 20 25 

20 0

0

0

0

0 20 25

20 0

0

0

0

0 20

20 0

0

0

0

0 20

20 0

0

0

0

0 20

20 20 20 20 20 20 20 25 25 25 25 25 25 25

n  cols( Bars)  9

Number of reinforcement rows Steel area As0 

 π Bars

2

4

i  1  n Ast 

  25  mm 25   25  25  25 

As  i

i

 As0

 As

Ast  232.478  cm

2

Location of reinforcement rows Cover  30mm  10mm  40 mm

Concrete cover Bars d  Cover  1

1 n

2

i  2  n

h  d 2  52.5 mm d  d i

T

i 1

ΔS 

1

n1

 86.875 mm

 ΔS

reverse ( d )  ( 747.5 660.63 573.75 486.88 400 313.13 226.25 139.38 52.5 )  mm Case of axially loaded column Page 82





ϕPn.max  0.80 ϕ 0.85 f'c Ag  Ast  fy  Ast   ϕPn.max  14255.808  kN Case of eccentric column

 

β1  0.65 max  0.85  0.05 c( a) 

f'c  27.6MPa 

  min 0.85  0.796  

6.9MPa

a β1 d  c( a)

fs( i a) 

εs  εu

 

i

c( a)



sign ε s  min Es ε s fy



d t  max( d )  747.5  mm ϕ( a) 

εt  εu

d t  c( a) c( a)

 1.45  250  ε t    min 0.9 3   

ϕ  0.65 max 

n       ϕPn ( a)  min ϕ( a)  0.85 f'c a b   Asi fs( i a)  ϕPn.max      i 1    





ϕMn ( a)  ϕ( a)  0.85 f'c a b  

 

a  0 

h 100

h

2

 h

Page 83



a



2

n





i 1

A  f ( i a)   d   si s  i  

h 

 

2 

10000 ϕPn( a) kN Pu kN 5000

0

0

1000

2000 ϕMn( a) kN m



3000 Mu

kN m

Page 84

D. Design of Circular Columns

Symbols ns

= number of re-bars

Dc Ds

= column diameter = diameter of re-bar circle

Location of steel re-bar

 

d = rc  rs cos α i s i α

s i

=

2 π ns

Dc rc = 2

 ( i  1)

Page 85

Ds rs = 2

Depth of compression concrete  rc  a  α = acos   rc  Area and centroid of compression concrete





1 1 2 Asector =  Radius  Arch =  rc  rc 2  α = rc  α 2 2 x1 =

2 3

 rc

sin( α) α 1

1 2 Atriangle =  Base  Height =  2  rc sin( α)  rc cos( α) = rc  sin( α)  cos( α) 2 2 x2 =

2

 r  cos( α) 3 c 2

Ac = Asegment = Asector  Atringle = rc  ( α  sin( α)  cos( α) )

xc =

xc =

Asector x 1  Atrinagle x 2 Ac 2  rc 3



sin( α)

r  3 c

=

sin( α)  sin( α)  cos( α)

2

α  sin( α)  cos( α)

3

α  sin( α)  cos( α)

X = 0

Equilibrium in forces ns

Pn = C 

2



ns

 As if s i

T = 0.85 f'c Ac  i

i 1


i 1

M = 0 ns

M n = Pn  e = C x c 



i 1

Dc    Ti  di   2    ns

M n = Pn  e = 0.85 f'c Ac x c 



i 1

As i f s i  d i  rc

Conditions of strain compatibility ε

s i

εu

d c =

i

c d c f

= Es ε = Es ε  s i s i u

i

with c

Page 86

f

s i

 fy

Example 14.5 Pu  3437.31kN

Required strength

M u  42.53kN m f'c  20MPa

Materials

fy  390MPa

Solution Determination of concrete dimension ϕ  0.70 ρg  0.02

Assume

Pu 0.85 ϕ

Ag   2361.812 cm 0.85 f'c 1  ρg  fy  ρg



 Ag

Dc  Ceil

 

Ag 

π





50mm  550  mm

4

π Dc

2

 

2

4

 2375.829 cm

2

Determination of steel area Pu Ast 

0.85 ϕ

 0.85 f'c Ag

 46.597 cm

0.85 f'c  fy

Ds  Dc   30mm  10mm 



 π Ds  n s  ceil   15  100mm  Ast  n s As0  47.124 cm

20mm 

  2  450  mm 

2

As0  2

2

s 

π ( 20mm)

π Ds ns

4

2

 3.142  cm

 94.248 mm

Interaction diagram for column strength





ϕPn.max  0.85 ϕ 0.85 f'c Ag  Ast  fy  Ast  3448.996 kN  

Page 87

2

 

β1  0.65 max  0.85  0.05 5

Es  2  10 MPa c( a) 

f'c  27.6MPa  6.9MPa

  min 0.85  0.85  

ε u  0.003

a β1

2 π αs   (i  1) i ns

i  1  n s

d 

Dc

i

2



Ds 2

 cos αs 



i

d t  max( d )  495.083  mm ϕ( a) 

εt  εu

d t  c( a) c( a)

 1.7  200  εt    min 0.9 3   

ϕ  0.70 max 

d  c( a) fs( i a) 

εs  εu

 

i

c( a)



sign ε s  min Es ε s fy Dc rc  2



 rc  a    rc 

α( a)  acos x c( a) 

2  rc 3



sin( α( a) )

3

α( a)  sin( α( a) )  cos( α( a) )

2

Ac( a)  rc  ( α( a)  sin( α( a) )  cos( α( a) ) ) ns        ϕPn ( a)  min ϕ( a)  0.85 f'c Ac( a)  As0 fs(i a)  ϕPn.max   i 1    



ns     ϕMn ( a)  ϕ( a)  0.85 f'c Ac( a)  x c( a)  As0 fs( i a)   di  rc  i 1  



a  0 

Dc 100

 Dc

Page 88

Interaction diagram for column strength

3000 ϕPn( a) kN Pu

2000

kN

1000

0

0

100

200 ϕMn( a) kN  m



300 Mu kN  m

3. Long (Slender) Columns Stability index Q=

ΣPu  Δ0 Vu  Lc

where ΣPu Vu Δ0

= total vertical force and story shear

Lc

= center-to-center length of column

Q  0.05

: Frame is nonsway (braced)

Q  0.05

: Frame is sway (unbraced)

= relative deflection between column ends

Page 89

Braced Frame

Shear Wall

Unbraced Frame

Braced Frame Brick Wall

Ties

Slenderness of column The column is short, if k  Lu

In nonsway frame:

r k  Lu

In sway frame:

r



M1



M2

 min 34  12  22

where



M 1 = min M A M B





M 2 = max M A M B



= 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



40



 ψ=



 EIc     Lc   EIb  L   b

ψ=0

: column is fixed

ψ=∞

: column is pinned

Moments of inertia

Ig

For column

Ic = 0.70Ig

For beam

Ib = 0.35Ig

= 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

2

  k

ψA  ψB 2

π   k 1  π  tan   k Page 91

 2 tan π      2 k  = 1  π  k 

For unbraced frames: ψA ψB 

π

2

  36 k =



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 M c = M 2  δns  M 2.min δns where M 2.min = Pu  ( 15mm  0.03h ) Moment magnification factor

Page 92

Cm

δns = 1

1

Pu 0.75 Pc

Euler's critical load 2

Pc =

EI =

π  EI

 k  Lu 

2

0.4 Ec Ig 1  βd

1.2 PD βd = 1.2 PD  1.6 PL Coefficient Cm = 0.6  0.4

M1 M2

 0.4


Pu  6402.35kN

 PD    PL 

M A  77.75kN m

M B  122.68kN m

Length of column

Lc  7.8m

Upper and lower columns

 ba     h a   L   a

Upper and lower beams

Materials

 60cm   60cm     3.6m 

 bb     h b   L   b

 4273.41kN     796.25kN 

 65cm   65cm     1.5m 

 b a1     h a1   L   a1 

 30cm   50cm     6m 

 b a2     h a2   L   a2 

 30cm   50cm     6m 

 b b1     h b1   L   b1 

 30cm   50cm     6m 

 b b2     h b2   L   b2 

 30cm   50cm     6m 

f'c  30MPa fy  390MPa Page 93

Solution Determination of concrete dimension ϕ  0.65 ρg  0.03

Assume

Pu 0.80 ϕ

Ag   3379.226 cm 0.85 f'c 1  ρg  fy  ρg





Proportion of column section h 

Ag

k=

 581.311  mm

k

b

2

k 

h

60 60

b  k  h  581.311  mm

h  Ceil( h 50mm)  600  mm

 b    600   mm      h   600 

b  Ceil( b 50mm)  600  mm

Determination of steel area 3

2

Ag  b  h  3.6  10  cm Pu  0.85 f'c Ag 0.80 ϕ 2 Ast   85.932 cm 0.85 f'c  fy

1 1  1 Bars   1 1  1

20

π ( 25mm)

2

 98.175 cm

4

2

1 1 1 1 1

0 0 0 0 1





0 0 0 0 1 0 0 0 0 1

 25mm

As0 

π Bars

 0 0 0 0 1  1 1 1 1 1

 

i  1  cols As0

As1  i Ast 

 

n s  rows As

i

 As0



As

ns  6

Cover  40mm  10mm 

25mm 2

 62.5 mm

Page 94

2

4

As  As1 Ast  98.175 cm

2

Ast Ag

 0.027

h  Cover 2

d1  Cover

Δs 

i  2  n s

d1  d1

1

ns  1 i 1

i

 95 mm

 62.5   157.5    252.5   d  mm  347.5   442.5     537.5 

 Δs

d  d1





ϕPn.max  0.80 ϕ 0.85 f'c Ag  Ast  fy  Ast  6634.405 kN  

Slenderness of column Stability index

Q  0

Radius of gyration

r 

h

 0.173 m

12 Modulus of elasticity wc  24

kN 3

m

 wc  Ec  44MPa    kN   m3   

1.5



f'c MPa

4

 2.834  10  MPa

Degree of end restraint 3

Ia1  0.35

b a1 h a1

Ica  0.70

Ia2  0.35

12 b b1 h b1

Ib1  0.35

3

12 b a h a

3

Ib2  0.35

3

Icb  0.70

12

b a2 h a2 12

b b2 h b2

3

12 bb hb

3

12

3

b h Ic  0.70 12 Σi ca 

Σi ba  ψA 

Ec Ica La Ec Ia1 La1

Σi ca Σi ba





Ec Ic Lc Ec Ia2 La2

 8.418

Σi cb 

Σi bb  ψB 

Page 95

Ec Icb Lb Ec Ib1 Lb1

Σi cb Σi bb





Ec Ic Lc Ec Ib2 Lb2

 21.699

Effective length factor k  0.6 Given ψA ψB 4

 

π

ψA  ψB

2

  k

k  0.5

2 k  1.0

k  Find( k )

π   k 1  π  tan   k

 2 tan π   2 k     =1  π  k 

k  0.969

Checking for long column M 1 

MA  MB

M A if

M 2 

M B otherwise

M A otherwise

M 1  77.75  kN m Lu  Lc  k  Lu r

M 2  122.68 kN m







max h a1 h a2  max h b1 h b2

k  Lu

"is short" if

r

The_column  "is long"

Case of long column 1.2 PD βd   0.801 1.2 PD  1.6 PL 3

Ig  12

EI 

0.4 Ec Ig 1  βd

2

π  EI

 k  Lu 

2



M1



M2

 13409.955  kN



M1



M2

Cm  max 0.6  0.4



M1



M2

 min 34  12

"is long" otherwise

Pc 

 7.3 m min 34  12

 40.834

b h



2

The_column 

MA  MB

M B if



0.4  0.4



Page 96



40





40  40







Cm

δns  max

Pu   1  0.75 P c 

1  1.101

  

M 2.min  Pu  ( 15mm  0.03 h )  211.278  kN m M c 



δns  max M 2 M 2.min



max M 2 M 2.min





if The_column = "is long"

otherwise

Interaction diagram for column strength c( a) 

a β1

d t  max( d )  537.5  mm ϕ( a) 

εt  εu

d t  c( a) c( a)

 1.45  250  ε t    min 0.90 3   

ϕ  0.65 max 

d  c( a) fs( i a) 

εs  εu

 

i

c( a)



sign ε s  min Es ε s fy



ns         ϕPn ( a)  min ϕ( a)  0.85 f'c a b   Asi fs( i a)  ϕPn.max     i 1    



ns   h a h      ϕMn ( a)  ϕ( a)  0.85 f'c a b      Asi fs( i a)   d i    2  2 2   i 1  



a  0 

h 100

 h

Page 97

Interaction diagram for column strength 7000

6000

5000

ϕPn( a)

4000

kN Pu kN

3000

2000

1000

0

0

200

400

600 ϕMn( a) kN  m



Page 98

800 Mc kN  m

1000

1200

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

q e = q a  20

kN 3

H

m

= allowable bearing capacity of soil with FS = 2.5  3

qa 20

kN

= average density of soil and concrete

3

m

= depth of foundation

H

Checking for maximum stress of soil under footing q max  q u P

q max =

B L

  1 



6 e 

L

 if e  L  6

4P 3  B  ( L  2  e)

if e 

L 6

where qu

= design bearing capacity of soil 1.2PD  1.6 PL

q u = q a P

PD  PL

= axial load on footing





P = 1.2 PD  P0  1.6 PL P0 = 20

kN 3

 H B L

m e

= eccentricity of load

Page 99

e= L B

M P

= long and width of footing

B. Determination of Depth of Footing Checking for Punching Vu  ϕVc where Vu

= punching shear

Vc

= punching shear strength is a strength reduction factor for shear

ϕ = 0.75

Punching shear



Vu = q u  A  A0



A = B L





A0 = b c  d  h c  d



Punching shear strength Vc = 4  f'c b 0  d

(in psi)

Page 100

Vc = 0.332  f'c b 0  d b0 =

(in MPa)

 b c  d   h c  d  2

Checking for Beam Shear Vu1  ϕVc1 Vu2  ϕVc2 where Vu1 Vu2 Vc1 Vc2

= beam shears = beam shear strength

Beam shears

 L hc  Vu1 = q u  B    d 2 2   B bc  Vu2 = q u  L    d 2 2  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 M u1 =

L1 =

q 1  L1

L 2



hc 2

2

2

Design section: rectangular singly reinforced beam of B  d

Steel re-bars in short direction Required strength q2 = qu L M u2 =

L2 =

q 2  L2

B 2



bc 2

2

2

Design section: rectangular singly reinforced beam of L  d


PD  484.71kN PL  228.56kN

PL PD  PL

 0.32

M u  5.03kN m Dimension of column stub

b c  350mm

Depth of foundation

H  2.0m q a  178.33

Allowable bearing capacity of soil Materials

h c  350mm

f'c  25MPa fy  390MPa Page 102

kN 3 kN   213.996  2 2.5 2 m m

Solution

Determination of Dimension of Footing Effective bearing capacity of soil q e  q a  20

kN 3

 H  173.996 

m

kN 2

m

Required area of footing Areq 

PD  PL qe

Footing proportion k = Areq

L 

k

2

 4.099 m

B

2

k 

L

 2.075  m

2.1

B  k  L  1.976 m

L  Ceil( L 50mm)  2.1 m

B  Ceil( B 50mm)  2 m

B   2  m      L   2.1  Design bearing capacity of soil q u  q a

1.2 PD  1.6 PL PD  PL

 284.224 

kN 2

m

Checking for maximum stress of soil Pu  1.2  PD  B L H 20 



e 

Mu Pu

q max 

3

m



 1.6 PL  1148.948 kN

 4.378  mm Pu B L

  1 



6 e 

L

 if e  L  6

4Pu 3  B ( L  2  e) Soil 

kN 

otherwise

q max  276.981  q max

 0.975

Soil  "is safe"

"is not safe" otherwise

Page 103

2

m qu

"is safe" if q max  q u

kN

Determination of depth of footing Punching shear





A0 ( d )  b c  d  h c  d



Vu ( d )  q u  A  A0 ( d )





A  B L Vu ( 320mm)  1066.154 kN

Punching shear strength b 0 ( d ) 

 b c  d   h c  d  2

ϕ  0.75 f'c ϕVc( d )  ϕ 0.332 MPa   b ( d)  d MPa 0

ϕVc( 320mm)  1067.712 kN

Beam shears

 L hc  Vu1( d )  q u  B    d 2 2 

Vu1( 300mm)  326.858  kN

 B bc  Vu2( d )  q u  L    d 2 2 

Vu2( 300mm)  313.357  kN

Beam shear strength f'c ϕVc1( d )  ϕ 0.166 MPa   B d MPa

ϕVc1( 300mm)  373.5  kN

f'c ϕVc2( d )  ϕ 0.166 MPa   L d MPa

ϕVc2( 300mm)  392.175  kN

c  50mm  20mm 

20mm 2

 80 mm

d min  150mm  c  70 mm d 

d  d min



 

 

while Vu ( d )  ϕVc( d )  Vu1( d )  ϕVc1( d )  Vu2( d )  ϕVc2( d ) d  d  50mm d d  320  mm

h  d  c  400  mm

Page 104



Steel reinforcements ρshrinkage 

( return 0.0020) if fy  50ksi ( return 0.0018) if fy  60ksi

 return max 0.0018 60ksi 0.0014  otherwise    fy    ρshrinkage  0.0018

Re-bars in long direction b  B

Ln  wu  Ln

M u 

b d

hc

 0.875 m

2

wu  q u  b

 217.609  kN m

M n 

f'c fy



1 



1  2



s  150mm

b

kN m 1m

 20.012 cm

2

n  floor

D  14mm

2

4

0.9

 108.805 

  0.00312  0.85 f'c 



n

Mu

R

As  max ρ b  d ρshrinkage  b  h  19.944 cm

π D

Mu

 1.181  MPa

2

ρ  0.85

2



2

2

Mn

R 

L



b  75mm 2  D  s

  1  13 

2

Re-bars in short direction b  L

Ln  wu  Ln

M u 

b d

bc

 0.825 m

2

wu  q u  b

 203.123  kN m

M n 

f'c fy



1 



1  2

s  160mm



b

 20.012 cm

2

n  floor

D  14mm

2

4

0.9

 96.725

  0.00276  0.85 f'c 



n

Mu

R

As  max ρ b  d ρshrinkage  b  h  18.554 cm

π D

Mu

 1.05 MPa

2

ρ  0.85

2



2

2

Mn

R 

B



2

Page 105

b  75mm 2  D  s

  1  13 

kN m 1m

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 kN 2 Qe = Qa  20  ( 3 D)  H 3 m

20

kN

= average density of soil and concrete

3

m 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

Checking for pile reaction

R = i

P n

M x 

M y

y i

n

 xk 

 2

k1

x i

n

 yk 

 Qu 2

k1

where P

= load on pile cap

M xM y

= moments on pile cap

Qu

= design bearing capacity of pile

Qu = Qa

1.2 PD  1.6 PL PD  PL

Page 106

 200mm

2. Depth of Pile Cap Case of punching Vu  ϕ Vc where Vu

= punching shear Vu =

Vc

 Routside = Qu noutside

= punching shear strength Vc = 0.332  f'c b 0  d b0 =

 b c  d   h c  d  2

Case of beam shear Vu1  ϕVc1 Vu2  ϕVc2 where Vu1 Vu2

= beam shears

Vc1 Vc2

= beam shear strength Rleft  Rright    

Vu1 = max

Rbottom  Rtop    

Vu2 = max

Vc1 = 0.166  f'c B d Vc2 = 0.166  f'c L d

Page 107

3. Determination of Steel Reinforcements In long direction



Design section:



 Rleft xleft  

M u1 = max

Required moment:

hc 



 Rright xright 

  2 

h c  2

 

Rectangular singly reinforced of B  d

In short direction





M u2 = max

Required moment: Design section:



 

Rbottom  x bottom 

bc 



  2 

 

Rtop   x top 

Rectangular singly reinforced of L  d

Example 16.1 Pile size

D  300mm

Allowable bearing capacity of pile

Qa  351.5kN

Loads on pile cap

PD  1769.88kN

PL  417.11kN

M y  33.92kN m

M x  56.82kN m

Depth of foundation

H  1.5m

Column stub

b c  350mm

Materials

f'c  25MPa fy  390MPa

Diameters of main bar

 D1    D2 

Concrete cover

c  75mm

Depth of concrete crack

h shrinkage  200mm

Diameter of shrinkage rebar

Dshrinkage  12mm

Solution Design of pile Required strength of pile concrete Page 108

 16mm     16mm 

Lp  9m

h c  500mm

b c  2

 

Ag  D D Qa f'c.pile   15.622 MPa 1 A 4 g

f'c.pile  20MPa

Use

Steel re-bars Ast  0.005  Ag  4.5 cm

2

4

π ( 16mm) 4

2

 8.042  cm

2

Dimension of pile cap Effective bearing capacity of pile Qe  Qa  20

kN 3

2

 ( 3  D)  H  327.2  kN

m Number of piles n 

PD  PL Qe

 6.684

ceil( n )  7

Required number of piles

Location of pile

 1m   0     1m   0.5m  X     0.5m   1m   0     1m 

Number of piles

 0.8m   0.8m     0.8m   0  Y     0   0.8m   0.8m     0.8m 

n  rows( X)

n8

Dimension of pile cap B  max( Y)  min( Y)   min

D

200mm 

D

 2  2  2 D D L  max( X)  min( X)   min 200mm    2  2  2 Checking for pile reactions

Page 109

B  2.2 m L  2.6 m

Qu  Qa P0  20

1.2 PD  1.6 PL PD  PL

kN 3

 448.616  kN

 H B L  171.6  kN

m





Pu  1.2 PD  P0  1.6 PL  2997.152 kN i  1  n Ru  i

ORIGIN  1

Pu n



My X

i

n

  k X

k1

 1    1  L Xcap   1    1  2    1 

Mx Y

i



n

 Yk

2

2

k 1

 1     1  B Ycap   1   1  2    1 

i  1  n

 1  1   D i Xpile  X   1   i  1  2    1 

 1   1    D i Ypile  Y   1   i 1 2    1 

2

1 Ycap Ypile Y

2

1

0

1

2

1

2 Xcap Xpile X

Page 110

 0.845   0.861     0.878  Ru  0.827    Qu  0.844   0.792   0.809     0.826 

Determination of Depth of Pile Cap Punching shear   hc bc  d  d  Outside ( d )   X    Y    2  2  2 2  Vu ( d )  Ru  Outside ( d )

Vu ( 700mm)  2247.864 kN

Punching shear strength ϕ  0.75

 h c  d   b c  d  2

b 0 ( d ) 

f'c ϕVc( d )  ϕ 0.332 MPa   b ( d)  d MPa 0

ϕVc( 700mm)  3921.75  kN

Beam shears

    hc  Left( d )  X    d  2      bc  Bottom( d )  Y    d  2 

   hc  Right( d )  X    d  2     bc  Top( d )  Y    d  2 



Vu1( d )  max Ru  Left( d ) Ru  Right( d )





Vu2( d )  max Ru  Bottom( d ) Ru  Top( d )

Vu1( 700mm)  764.364  kN



Vu2( 700mm)  0 N

Beam shear strength f'c ϕVc1( d )  ϕ 0.166 MPa   B d MPa

ϕVc1( 700mm)  958.65 kN

f'c ϕVc2( d )  ϕ 0.166 MPa   L d MPa

ϕVc2( 700mm)  1132.95  kN

Depth of pile cap D2 Cover  c  D1   99 mm 2 d 

d  300mm  Cover



 

 

while Vu ( d )  ϕVc( d )  Vu1( d )  ϕVc1( d )  Vu2( d )  ϕVc2( d ) d  d  50mm d d  651  mm

h  d  Cover  750  mm Page 111



Steel Reinforcements ρshrinkage 


 return max 0.0018 60ksi 0.0014  otherwise    fy    In long direction b  B M u1 

M u2 

    hc  Left( 0)  Ru    X  643.378  kN m 2







  h c    Right( 0 )  Ru  X    667.876  kN m 2











M u  max M u1 M u2  667.876  kN m R 

Mn b d

f'c fy



1 



1  2

0.9

  0.00208  0.85 f'c  R



As  max ρ b  d ρshrinkage  b  h As1 

Mu

 0.796  MPa

2

ρ  0.85

M n 

π D1

2



As  29.797 cm

 As    15  As1  D1     b  c   2 2     s1  Floor 5mm  145  mm   n1  1   n 1  ceil

4

In short direction b  L M u1 

M u2 

   b c  Bottom( 0 )  Ru   Y  680.262  kN m 2







 b c    Top( 0 )  Ru  Y    724.653  kN m 2

 

2







M u  max M u1 M u2  724.653  kN m

Page 112

M n 

Mu 0.9

R 

Mn b d

 0.731  MPa

2

ρ  0.85

f'c fy



1 



1  2

  0.00191  0.85 f'c  R



As  max ρ b  d ρshrinkage  b  h As2 

π D2

2



As  35.1 cm

2

 As    18  As2  D2     b  c   2 2     s2  Floor 5mm  140  mm   n2  1   n 2  ceil

4

Shrinkage reinforcement b  1m

h shrinkage 

h if h shrinkage = 0 h shrinkage otherwise

As  ρshrinkage  b  h shrinkage  3.6 cm As0 

π Dshrinkage 4

2

n 

2

As As0

b sshrinkage  Floor 5mm  310  mm n  D1      2  L  c   s D 1 1 2     n1 mm mm m     D  2 Table   B   c   2  D2 s2 2     n2 m mm mm    Dshrinkage sshrinkage    "N/A" "N/A" mm mm  

Page 113

Dimension of pile cap

Depth of pile cap

B=

2.20

m

L=

2.60

m

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 La Lb

 0.5

: the slab in one-way

 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  1.8 m

Live load

LL  12

kN 2

m

f'c  20MPa

Materials

fy  390MPa

Solution Thickness of one-way slab tmin 

Ln 28

 64.286 mm t  100mm

Use Loads on slab

kN


 1.1

3


kN

 2.5

3

m

kN 2

m kN 2

m kN

Ceiling  0.40

2

m


kN 2

m kN

Partition  1.00

2

m


kN 2

m kN wu  1.2 DL  1.6 LL  25.44  2 m Bending moments M support  M midspan 

1 11

2

 wu  Ln  7.493 

1 16

2

kN m

 wu  Ln  5.152 

1m kN m 1m


Page 116

 

β1  0.65 max  0.85  0.05

f'c  27.6MPa 

  min 0.85  0.85  

6.9MPa

ε u  0.003 f'c εu ρmax  0.85 β1    0.014 fy ε u  0.005 f'    0.249MPa c  MPa 1.379MPa   ρmin  max     0.00354 fy fy   ρshrinkage 


 return max 0.0018 60ksi 0.0014  otherwise    fy    ρshrinkage  0.0018 Top rebars d  t   20mm 

b  1m



10mm 

  75 mm 

2

M u  M support b  7.493  kN m M n 

Mu 0.9

Mn

R 

b d

 1.48 MPa

2

ρ  0.85

 8.326  kN m

f'c fy



1 



1  2



  0.004  0.85 f'c  R



As  max ρ b  d ρshrinkage  b  t  2.982  cm As0 

π ( 10mm)

2

n 

4

ρ  ρmax  1 2

As As0

b s  min Floor 10mm smax  260  mm  n   Bottom rebars M u  M midspan b  5.152  kN m M n 

Mu 0.9

 5.724  kN m

Page 117

smax  min( 3  t 450mm)

R 

Mn b d

 1.018  MPa

2

ρ  0.85

f'c fy



1 



1  2

R 0.85 f'c



  0.003  




π ( 10mm)

2

n 

4

ρ  ρmax  1 2

As As0

smax  min( 3  t 450mm)

b s  min Floor 10mm smax  300  mm  n   Link rebars As  ρshrinkage  b  t  1.8 cm As0 

π ( 10mm)

2

2

4

n 

As As0

b s  min Floor 10mm smax  430  mm  n  

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

smax  min( 5  t 450mm)

(1) Load Distribution Method Principle: Equality of deflection in short and long directions fa = fb αa

wa La

4

= αb 

EI

wb  Lb

4

EI

Case αa = αb wa wb

=

Lb La

4

1

=

4

λ=

4

λ

La Lb

wa  wb = wu From which,

wa = wu 

wb = wu 

For

λ  1

1 1λ λ

4

1λ

λ  0.8

λ  0.6

λ  0.5

λ  0.4

4

1 1λ

For

4

1 1λ

For

4

1 1λ

For

4

1 1λ

For

4

4

1 1λ

4

 0.5

λ

4

1λ  0.709

λ

λ

λ

λ

 0.291

4

 0.115

4

1λ  0.975

4

4

1λ  0.941

 0.5

4

1λ  0.885

4

4

 0.059

4

1λ

4

Page 119

 0.025

Example 17.2 La  4.3m

Slab dimension

Lb  5.5m kN

LL  2.00

Live load

2

m f'c  20MPa

Materials

fy  390MPa

Solution Thickness of two-way slab





Perimeter  La  Lb  2 tmin  t 

Perimeter

 108.889  mm

180

 1 1   L  ( 143.333 86 )  mm   a  30 50  t  120mm

Use Loads on slab

SDL  50mm 22

kN 3

kN

 0.40

2

m DL  SDL  t 25

kN 3

m

kN

2

m

kN 2

m

kN wu  1.2 DL  1.6 LL  9.8 2 m Load distribution λ 

La Lb

wa 

wb 

 0.782 1

1λ λ

4

 wu  7.134 

4

 wu  2.666 

1λ

kN 2

m

4

2

m

kN

 5.5

m LL  2 

 1.00

kN 2

m

Page 120

 2.5

kN 2

m

Bending moments M a.neg  M a.pos  M b.neg  M b.pos 

1 11 1 16 1 11 1 16

2

 wa La  11.992 2

 wa La  8.245  2

 wb  Lb  7.33 2

 wb  Lb  5.04

kN m 1m

kN m 1m

kN m 1m kN m 1m


 

β1  0.65 max  0.85  0.05

f'c  27.6MPa  6.9MPa

  min 0.85  0.85  

ε u  0.003 f'c εu ρmax  0.85 β1    0.014 fy ε u  0.005 f'    0.249MPa c  MPa 1.379MPa   ρmin  max     0.00354 fy fy   ρshrinkage 



Top rebar in short direction d  t   20mm  10mm 

b  1m



M u  M a.neg b  11.992 kN m

M n  R 

Mu 0.9

Mn b d

2

 13.325 kN m  1.844  MPa

Page 121

10mm  2

  85 mm 

ρ  0.85

f'c fy



1 

1  2



  0.005  0.85 f'c  R





As  max ρ b  d ρshrinkage  b  t  4.265  cm π ( 10mm)

As0 

2

n 

4

ρ  ρmax  1 2

As

smax  min( 2  t 450mm)

As0

b s  min Floor 10mm smax  180  mm  n   Bottom rebar in short direction M u  M a.pos b  8.245  kN m M n  R 

Mu 0.9

Mn b d

 1.268  MPa

2

ρ  0.85

 9.161  kN m

f'c fy



1 



1  2

  0.003  0.85 f'c  R






π ( 10mm)

2

n 

4

ρ  ρmax  1 2

As

smax  min( 2  t 450mm)

As0

b s  min Floor 10mm smax  240  mm  n   Top rebar in long direction M u  M b.neg  b  7.33 kN m

M n  R 

Mu 0.9

Mn b d

 1.127  MPa

2

ρ  0.85

 8.145  kN m

f'c fy





1 



1  2

  0.003  0.85 f'c  R



As  max ρ b  d ρshrinkage  b  t  2.544  cm

Page 122

ρ  ρmax  1 2

π ( 10mm)

As0 

2

n 

4

As

smax  min( 2  t 450mm)

As0

b s  min Floor 10mm smax  240  mm  n   Bottom rebar in long direction M u  M b.pos b  5.04 kN m M n 

Mu 0.9

Mn

R 

b d

 0.775  MPa

2

ρ  0.85

 5.599  kN m

f'c fy



1 



1  2

  0.002  0.85 f'c  R





As  max ρ b  d ρshrinkage  b  t  2.16 cm As0 

π ( 10mm)

2

n 

4

ρ  ρmax  1

2

As As0

smax  min( 2  t 450mm)


Shrinkage rebars b  1m As  ρshrinkage  b  t  2.16 cm As0 

π ( 10mm) 4

2

2

n 

As As0


Page 123

smax  min( 5  t 450mm)

(2) Moment Coefficient Method Negative moments M a.neg = Ca.neg wu  La

2

M b.neg = Cb.neg  wu  Lb

2

Positive moments 2

M a.pos = Ca.pos.DL wD La  Ca.pos.LL wL La 2

2

M b.pos = Cb.pos.DL  wD Lb  Cb.pos.LL wL Lb

2

Ca.neg Cb.neg Ca.pos.DL Ca.pos.LL Cb.pos.DL Cb.pos.LL

where

are tabulated moment coefficients wD = 1.2 DL

wL = 1.6 LL

wu = 1.2  1.6 LL

Example 17.3 La  5.0m  25cm  4.75 m

Slab dimension

Lb  5.5m  20cm  5.3 m LL  2.40

Live load for office

kN 2

m f'c  20MPa

Materials

fy  390MPa

Boundary conditions in short and long directions

 Simple      Continuous 

0   1

Short 

 Continuous     Continuous 

Long 

 Continuous     Continuous 

Solution Thickness of two-way slab





Perimeter  La  Lb  2

Page 124

tmin  t 

Perimeter 180

 111.667  mm

 1 1   L  ( 158.333 95 )  mm    30 50  a

Use

t  120mm

Loads on slab kN


 1.1

3

m kN

Slab  t 25

 3

3

m

Ceiling  0.40

kN 2

m

kN 2

m kN 2


kN 2

m

SDL  Cover  Ceiling  Partition  2.5

kN 2

m DL  SDL  Slab  5.5

kN

wD  1.2 DL

2

m LL  2.4

kN

wL  1.6 LL

2

m

kN wu  1.2 DL  1.6LL  10.44  2 m Moment coefficients

Page 125

Table 12.3a Coefficients for negative moments in short direction of slab m Case 1 Case 2 Case 3 Case 4 Case 5 1.00 0.000 0.045 0.000 0.050 0.075 0.95 0.000 0.050 0.000 0.055 0.079 0.90 0.000 0.055 0.000 0.060 0.080 0.85 0.000 0.060 0.000 0.066 0.082 0.80 0.000 0.065 0.000 0.071 0.083 0.75 0.000 0.069 0.000 0.076 0.085 0.70 0.000 0.074 0.000 0.081 0.086 0.65 0.000 0.077 0.000 0.085 0.087 0.60 0.000 0.081 0.000 0.089 0.088 0.55 0.000 0.084 0.000 0.092 0.089 0.50 0.000 0.086 0.000 0.094 0.090

Case 6 0.071 0.075 0.079 0.083 0.086 0.088 0.091 0.093 0.095 0.096 0.097

Case 7 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000

Case 8 0.033 0.038 0.043 0.049 0.055 0.061 0.068 0.074 0.080 0.085 0.089

Case 9 0.061 0.065 0.068 0.072 0.075 0.078 0.081 0.083 0.085 0.086 0.088

Table 12.3b Coefficients for negative moments in long direction of slab m Case 1 Case 2 Case 3 Case 4 Case 5 1.00 0.000 0.045 0.076 0.050 0.000 0.95 0.000 0.041 0.072 0.045 0.000 0.90 0.000 0.037 0.070 0.040 0.000 0.85 0.000 0.031 0.065 0.034 0.000 0.80 0.000 0.027 0.061 0.029 0.000 0.75 0.000 0.022 0.056 0.024 0.000 0.70 0.000 0.017 0.050 0.019 0.000 0.65 0.000 0.014 0.043 0.015 0.000 0.60 0.000 0.010 0.035 0.011 0.000 0.55 0.000 0.007 0.028 0.008 0.000 0.50 0.000 0.006 0.022 0.006 0.000

Case 6 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000

Case 7 0.071 0.067 0.062 0.057 0.051 0.044 0.038 0.031 0.024 0.019 0.014

Case 8 0.061 0.056 0.052 0.046 0.041 0.036 0.029 0.024 0.018 0.014 0.010

Case 9 0.033 0.029 0.025 0.021 0.017 0.014 0.011 0.008 0.006 0.005 0.003

ORIGIN  1 Index 

1 2   2 3

I  Index

 Short11 Short21  3

J  Index

1 7 3 Table   6 4 8    5 9 2

 Long11 Long21  3

Case  Table

I J

2

Vλ  reverse ( Vλ)



Case Vaneg  reverse Taneg



  Case  VaposLL  reverse  TaposLL Case VaposDL  reverse TaposDL

λ 

La Lb



Case Vbneg  reverse Tbneg



  Case  VbposLL  reverse  TbposLL Case VbposDL  reverse TbposDL

 0.896

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 kN m 2 M a.neg  Ca.neg wu  La  13.043 1m kN m 2 M b.neg  Cb.neg  wu  Lb  10.729 1m kN m 2 2 M a.pos  Ca.pos.DL  wD La  Ca.pos.LL wL La  6.266  1m kN m 2 2 M b.pos  Cb.pos.DL  wD Lb  Cb.pos.LL wL Lb  4.911  1m


 

β1  0.65 max  0.85  0.05

f'c  27.6MPa  6.9MPa

  min 0.85  0.85  

ε u  0.003 f'c εu ρmax  0.85 β1    0.014 fy ε u  0.005 f'    0.249MPa c  MPa 1.379MPa   ρmin  max     0.00354 fy fy  

ρshrinkage 



Top rebars in short direction b  1m

d  t   20mm  10mm 



Page 127

10mm  2

  85 mm 

M u  M a.neg b  13.043 kN m

M n 

Mu

Mn

R 

 14.492 kN m

0.9

b d

 2.006  MPa

2

ρ  0.85

f'c fy



1 

1  2



  0.005  0.85 f'c 



R




π ( 10mm)

As0 

2

n 

4

ρ  ρmax  1 2

As

smax  min( 2  t 450mm)

As0


Bottom rebars in short direction M u  M a.pos b  6.266  kN m M n  R 

Mu 0.9

Mn b d

 0.964  MPa

2

ρ  0.85

 6.963  kN m

f'c fy



1 



1  2



  0.003  0.85 f'c  R




π ( 10mm)

2

n 

4

ρ  ρmax  1 2

As As0


Top rebars in long direction M u  M b.neg  b  10.729 kN m

M n 

Mu 0.9

 11.921 kN m

Page 128

smax  min( 2  t 450mm)

Mn

R 

b d

 1.65 MPa

2

ρ  0.85

f'c fy



1 

1  2



R 0.85 f'c



  0.004  



As  max ρ b  d ρshrinkage  b  t  3.79 cm

π ( 10mm)

As0 

2

n 

4

ρ  ρmax  1

2

As

smax  min( 2  t 450mm)

As0


Bottom rebars in long direction M u  M b.pos b  4.911  kN m M n  R 

Mu 0.9

Mn b d

 0.755  MPa

2

ρ  0.85

 5.457  kN m

f'c fy



1 



1  2

R 0.85 f'c



  0.002  




π ( 10mm)

2

n 

4

ρ  ρmax  1

2

As As0

smax  min( 2  t 450mm)


Shrinkage rebars b  1m As  ρshrinkage  b  t  2.16 cm As0 

π ( 10mm) 4

2

2

n 

As As0


Page 129

smax  min( 5  t 450mm)

(3) Direct Design Method (DDM) Total static moment

M0 =

wu  L2  Ln

2

8

Longitudinal distribution of moments M neg = Cneg M 0 M pos = Cpos M 0

Lateral distribution of moments M neg.col = Cneg.col M neg M neg.mid = Cneg.mid M neg M pos.col = Cpos.col M pos M pos.mid = Cpos.mid  M pos

Page 130

Example 17.4 Slab dimension

La  4m

Live load for hospital

LL  3.00

Lb  6m kN 2

m

Materials

f'c  25MPa

fy  390MPa

Page 131

Solution Section of beam in long direction L  Lb  6 m 1 1  h     L  ( 600 400 )  mm  10 15 

h  500mm

b  ( 0.3 0.6 )  h  ( 150 300 )  mm

 b b    h b 

b  250mm

b   h

Section of beam in short direction L  La  4 m 1  1 h     L  ( 400 266.667 )  mm  10 15 

h  300mm

b  ( 0.3 0.6 )  h  ( 90 180 )  mm

 b a    h a 

b  200mm

b   h

Determination of slab thickness





Perimeter  La  Lb  2 t min 

Perimeter

 111.111  mm

180

t  120mm

Assume In long direction

b w  b b

h  h b

h f  t

h w  h  h f





b  min b w  2  h w b w  8  h f  1.01 m h A1  b w h x 1  2





A2  b  b w  h f x c 

I1  I2 

x 2 

x 1  A1  x 2  A2 A1  A2 b w h 12

hf 2

 169.852  mm

3



 A1  x 1  x c

b  bw hf 12



2

3



 A2  x 2  x c Page 132



2

5

Ib  I1  I2  4.617  10  cm Is 

La  h f

3

12 3

 wc  Ec  44MPa    kN   m3   

 8.016

αb  α

kN

wc  24

m

α 

Ec Ib Ec Is

4

1.5



In short direction b w  b a

h  h a

h f  t

h w  h  h f





b  min b w  2  h w b w  8  h f  0.56 m h A1  b w h x 1  2





A2  b  b w  h f

x 2 

x 1  A1  x 2  A2

x c 

A1  A2 b w h

I1 

 112.326  mm



 A1  x 1  x c

b  bw hf 3 12

2



 A2  x 2  x c 4

Ib  I1  I2  7.053  10  cm

Is 

α 

Lb  h f 12 Ec Ib Ec Is

4

3 4

 8.64  10  cm  0.816

4

αa  α

Required thickness of slab αm  β 

αa 2  αb  2

Lb La

2

3

12

I2 

hf

4

 4.416

 1.5

Ln  Lb  20cm  5.8 m

Page 133

2 Iba  Ib

f'c MPa

4

 2.587  10  MPa

fy      Ln   0.8  200ksi     h f  max 5in if 0.2  αm  2.0  36  5  β  αm  0.2    f    y    Ln  0.8   200ksi    3.5in if 2.0  αm  5.0 max 36  9  β   "DDM is not applied" otherwise h f  126.876  mm

Loads on slab DL  50mm 22

kN 3

 t  25

m LL  3 

kN 3

 0.40

m

kN

 1.00

2

m

kN 2

 5.5

m

kN 2

m

kN 2

m

kN wu  1.2 DL  1.6 LL  11.4 2 m

In long direction L1  Lb  6 m

Ln  L1  b a  5.8 m

L2  La  4 m

α1  αb  8.016

Total static moment M 0 

wu  L2  Ln 8

2

 191.748  kN m

Longitudinal distribution of moments M neg  0.65 M 0  124.636  kN m M pos  0.35 M 0  67.112 kN m

Lateral distribution of moments L2 k 1   0.667 L1

L2 k 2  α1   5.344 L1

linterp2 ( VX VY M x y ) 

for j  1  rows( VY)



j V  linterp VX M x j

linterp( VY V y )

Page 134



 0   0.5   0.75 0.75 0.75          Cneg.col  linterp2 1  1.0  0.90 0.75 0.45 k 2 k 1  0.85         10   2.0   0.90 0.75 0.45   Cneg.mid  1  Cneg.col  0.15

 0   0.5   0.60 0.60 0.60          Cpos.col  linterp2 1  1.0  0.90 0.75 0.45 k 2 k 1  0.85         10   2.0   0.90 0.75 0.45   Cpos.mid  1  Cpos.col  0.15 M neg.col  Cneg.col M neg  105.941  kN m M neg.mid  Cneg.mid M neg  18.695 kN m M pos.col  Cpos.col M pos  57.045 kN m M pos.mid  Cpos.mid  M pos  10.067 kN m

 0   0   Ccol.beam  linterp 1   0.85  k 2  0.85       10   0.85   Ccol.slab  1  Ccol.beam  0.15 M neg.col.beam  Ccol.beam M neg.col  90.05  kN m M neg.col.slab  Ccol.slab M neg.col  15.891 kN m M pos.col.beam  Ccol.beam M pos.col  48.488 kN m M pos.col.slab  Ccol.slab M pos.col  8.557  kN m

b col 



min L1 L2 4



2  2 m

b mid  L2  b col  2 m

Top rebars in column strip b  b col

d  t   20mm  10mm 



M u  M neg.col.slab  15.891 kN m

M n 

Mu 0.9

 17.657 kN m

Page 135

10mm  2

  85 mm 

Mn

R 

b d

 1.222  MPa

2

ρ  0.85

f'c fy



1 

1  2



R 0.85 f'c



  0.003  




π ( 10mm)

As0 

2

n 

4

ρ  ρmax  1 2

As

smax  min( 2  t 450mm)

As0


Bottom rebars in column strip M u  M pos.col.slab  8.557  kN m M n  R 

Mu 0.9

Mn b d

 0.658  MPa

2

ρ  0.85

 9.508  kN m

f'c fy



1 



1  2

R 0.85 f'c



  0.002  




π ( 10mm)

2

n 

4

ρ  ρmax  1

2

As As0

smax  min( 2  t 450mm)


Top rebars in middle strip b  b mid M u  M neg.mid  18.695 kN m M n  R 

Mu 0.9

Mn b d

2

ρ  0.85

 20.773 kN m  1.438  MPa

f'c fy



1 



1  2

  0.004  0.85 f'c  R

Page 136

ρ  ρmax  1






π ( 10mm)

As0 

2

n 

4

2

As

smax  min( 2  t 450mm)

As0


Bottom rebars in middle strip M u  M pos.mid  10.067 kN m M n  R 

Mu

 11.185 kN m

0.9

Mn b d

 0.774  MPa

2

ρ  0.85

f'c fy



1 



1  2

  0.002  0.85 f'c  R






π ( 10mm)

2

n 

4

2

As As0


In short direction L1  La  4 m

Ln  L1  b b  3.75 m

L2  Lb  6 m

α1  αa  0.816

Total static moment M 0 

wu  L2  Ln 8

2

 120.234  kN m

Longitudinal distribution of moments M neg  0.65 M 0  78.152 kN m M pos  0.35 M 0  42.082 kN m

Lateral distribution of moments

Page 137

ρ  ρmax  1

smax  min( 2  t 450mm)

k 1 

L2 L1

L2 k 2  α1   1.224 L1

 1.5

 0   0.5   0.75 0.75 0.75          Cneg.col  linterp2 1  1.0  0.90 0.75 0.45 k 2 k 1  0.6         10   2.0   0.90 0.75 0.45   Cneg.mid  1  Cneg.col  0.4

 0   0.5   0.60 0.60 0.60          Cpos.col  linterp2 1  1.0  0.90 0.75 0.45 k 2 k 1  0.6         10   2.0   0.90 0.75 0.45   Cpos.mid  1  Cpos.col  0.4 M neg.col  Cneg.col M neg  46.891 kN m M neg.mid  Cneg.mid M neg  31.261 kN m M pos.col  Cpos.col M pos  25.249 kN m M pos.mid  Cpos.mid  M pos  16.833 kN m

 0   0   Ccol.beam  linterp 1   0.85  k 2  0.85       10   0.85   Ccol.slab  1  Ccol.beam  0.15 M neg.col.beam  Ccol.beam M neg.col  39.858 kN m M neg.col.slab  Ccol.slab M neg.col  7.034  kN m M pos.col.beam  Ccol.beam M pos.col  21.462 kN m M pos.col.slab  Ccol.slab M pos.col  3.787  kN m b col 



min L1 L2 4



2  2 m

b mid  L2  b col  4 m

Top rebars in column strip b  b col M u  M neg.col.slab  7.034  kN m M n 

Mu 0.9

 7.815  kN m

Page 138

Mn

R 

b d

 0.541  MPa

2

ρ  0.85

f'c fy



1 

1  2



R 0.85 f'c



  0.001  




π ( 10mm)

As0 

2

n 

4

ρ  ρmax  1

2

As

smax  min( 2  t 450mm)

As0


Bottom rebars in column strip M u  M pos.col.slab  3.787  kN m M n  R 

Mu 0.9

Mn b d

 0.291  MPa

2

ρ  0.85

 4.208  kN m

f'c fy



1 



1  2

R 0.85 f'c



  0.001  




π ( 10mm)

2

n 

4

ρ  ρmax  1

2

As As0

smax  min( 2  t 450mm)


Top rebars in middle strip b  b mid M u  M neg.mid  31.261 kN m M n  R 

Mu 0.9

Mn b d

2

ρ  0.85

 34.734 kN m  1.202  MPa

f'c fy



1 



1  2

  0.003  0.85 f'c 

Page 139

R

ρ  ρmax  1






π ( 10mm)

As0 

2

n 

4

2

As

smax  min( 2  t 450mm)

As0


Bottom rebars in middle strip M u  M pos.mid  16.833 kN m M n  R 

Mu 0.9

Mn b d

 0.647  MPa

2

ρ  0.85

 18.703 kN m

f'c fy



1 



1  2

  0.002  0.85 f'c 



R




π ( 10mm)

2

4

n 

2

As As0


Page 140

ρ  ρmax  1

smax  min( 2  t 450mm)

Page 141

18. Design of Staircase

Step dimension

G 

3.1m

H 

1.8m

11 11

 281.818  mm  163.636  mm

G  2  H  60.909 cm

Number of steps

Reference: G  2  H = 60cm  64cm H = 150mm  190mm

n  11

Loads on waist slab Slope angle

α  atan

H

Thickness of waist slab

twaist  120mm

Step cover

Cover  50mm ( H  G)  22

  30.141 deg  G

kN



1m

3 1m G

 1.739 

m Concrete step

RC slab

Step 

G H 2

 24

kN

3 1m G

 1.964 

m

Slab  twaist 25

kN 3

m

Page 142

1m



kN 2

2



2

1m  cos( α)

 3.469 

2

m

m

1m

kN

kN 2

m

Ceiling  0.40

Ceiling

kN 2

2

1m



2

1m  cos( α)


Handrail

 0.463 

kN 2

m

kN 2

m

DL  Cover  Step  Slab  Ceiling  Handrail

Dead load

DL  8.134 

kN 2

m LL  4.80

Live load

kN 2

m

kN wwaist  1.2 DL  1.6 LL  17.441 2 m

Factored load

Loads on landing slab Thickness of landing slab

tlanding  150mm

Slab cover


kN 3

 1.1

m Slab  tlanding 25

RC slab

kN 3

kN

 3.75

m Ceiling  0.40

Ceiling

2

m

kN 2

m

kN 2


Handrail

kN 2

m

DL  Cover  Slab  Ceiling  Handrail  5.75

Dead load

kN 2

m LL  4.80

Live load

kN 2

m

kN wlanding  1.2 DL  1.6 LL  14.58  2 m

Factored load

Analysis of Staircase Concrete modulus of elasticity f'c  25MPa wc  24

kN 3

m

 wc  Ec  44MPa    kN   m3    Page 143

1.5



f'c MPa

4

 2.587  10  MPa

Geometry of staicase L0  3.1m

L2  1.9m

2

L1 

h  1.8m

2

L0  h  3.585 m

t1  twaist  120  mm

t2  tlanding  150  mm

Flexural stiffness b  1m b  t1

3

b  t2

EI1  Ec 12

3

EI2  Ec 12

Loads on staircase kN w1  wwaist b  17.441 m kN w2  wlanding b  14.58  m Coefficients EI1

EI2

r11  4   3 L1 L2

R1p 

w1  L0 12

Angular rotation Z1 

R1p

4

 4.723  10

r11

φB  Z1

Bending moments

M A  2 

EI1 L1

M BA  4 

M BC  3 

 Z1 

EI1 L1

EI2 L2

w1  L0

 14.949 kN m

12

 Z1 

 Z1 

2

w1  L0

2

 12.004 kN m

12

w2  L2

2

8

 12.004 kN m

M C  0 Shears kN 2 w0  w1  cos( α)  13.043 m VAB 

M BA  M A L1



w0  L1 2

 24.199 kN

Page 144

2



w2  L2 8

2

VBA  VAB  w0  L1  22.557 kN VBC 

M C  M BC L2



w2  L2 2

 20.169 kN

VCB  VBC  w2  L2  7.533  kN Positive moments x 1 

VAB w0

 1.855 m

M max.AB  M A  VAB x 1 

x 2 

VBC w2

w0  x 1

2

 7.5 kN m

2

 1.383 m

M max.BC  M BC  VBC x 2 

w2  x 2

2

2

 1.946  kN m

Design of Staircase Materials f'c  25MPa

fy  390MPa

ε u  0.003

 

β1  0.65 max  0.85  0.05

f'c  27.6MPa 

  min 0.85  0.85  

6.9MPa

f'c εu ρmax  0.85 β1    0.017 fy ε u  0.005 f'    0.249MPa c  MPa 1.379MPa   ρmin  max    fy fy   ρshrinkage 

return 0.0020 if fy  50ksi

ρshrinkage  0.0018

return 0.0018 if fy  60ksi return max 0.0018



60ksi fy

0.0014 otherwise

Top rebars in waist slab

Page 145



d  twaist   30mm 

b  1m







M u  max M A  M BA R 

Mn b d

  82 mm 

2

 14.949 kN m

M n 

f'c fy



1 



1  2

R 0.85 f'c



200mm



0.9

  0.00675  



As  max ρ b  d ρshrinkage  b  twaist  5.537  cm b

Mu

 2.47 MPa

2

ρ  0.85

16mm 

π ( 14mm)

2

 7.697  cm

4

2

2

Top rebars in landing slab d  tlanding   30mm 

b  1m





M u  max M BC  M C R 

Mn b d

  112  mm 

2

 12.004 kN m

M n 

f'c fy



1 



1  2

  0.0028  0.85 f'c 



200mm



0.9

R



As  max ρ b  d ρshrinkage  b  twaist  3.134  cm b

Mu

 1.063  MPa

2

ρ  0.85



16mm 

π ( 10mm) 4

2

 3.927  cm

2

2

Bottom rebars in waist slab d  twaist   30mm 

b  1m



16mm  2

  82 mm 

M u  M max.AB  7.5 kN m R 

Mn b d

 1.239  MPa

2

ρ  0.85

M n 

f'c fy





1 



1  2

  0.00328  0.85 f'c  R



As  max ρ b  d ρshrinkage  b  twaist  2.687  cm

Page 146

2

Mu 0.9

b 200mm



π ( 10mm)

2

 3.927  cm

4

2

Bottom rebars in landing slab d  tlanding   30mm 

b  1m



16mm 

  112  mm 

2

M u  M max.BC  1.946  kN m R 

Mn b d

 0.172  MPa

2

ρ  0.85

M n 

f'c fy



1 



1  2

  0.00044  0.85 f'c  R





As  max ρ b  d ρshrinkage  b  twaist  2.16 cm b 200mm



π ( 10mm)

2

 3.927  cm

4

2

2

Link rebars



b  1m

As  ρshrinkage  b  t  2.7 cm b 250mm



t  max twaist tlanding  150  mm



π ( 10mm) 4

2

2

 3.142  cm

2

Page 147

Mu 0.9

19. Deflection A. Immediate (Initial) Deflection Initial or short-term deflection in midspan of continuous beam 5  M a  Ln

2

Δi = K 48 Ec Ie where

Ma

= support moment for cantilever or midspan moment for simple or cotinuous beam

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  0.2

4. Fixed-hinged beam Midspan deflection

K = 0.8

Max. deflection using max. moment

K = 0.74

5. Fixed-fixed beam

where

M0 =

w Ln

K = 0.60

2

8

Effective moment of inertia of cracked section 3

 Mcr  Ie = Icr      Ig  Icr  Ig  Ma  where Icr

= moment of inertia of cracked transformed section

Ig

= moment of inertia of gross section

M cr

= cracking moment Page 148

M0 Ma

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

fs = εs = Es Ec Es fs =  f = n  fc Ec c

From which

where

n=

Es

is a modulus ratio

Ec

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 M cr =

Iut fr

(exact expression)

yt Page 149

M cr =

Ig  fr

(simplied expression)

yt

where Iut

= moment of inertia of uncracked transformed section At

Ig

= moment of inertia of gross section Ag

fr

= modulus of rupture f'c f'c fr = 7.5psi = 0.623MPa  psi MPa

yt

= distance from neutral axis to the tension face

Moment of inertia of cracked section

Condition of strain compatibity εs εu

=

dx x

εs = εu

dx x

Equilibrium in forces C=T fc x 2

 b = As fs

Ec  ε u  b  x 2

= As Es ε s = As Es ε u 

dx x

2

b  x = 2  n  As ( d  x )

Page 150

b x

2

b d

2

2  n  As ( d  x )

=

b d

2

2

 x  = 2 n  ρ  1  x      d d  2

 x   2 n  ρ x  2 n  ρ = 0   d d x d

= n  ρ 

2

( n  ρ)  2  n  ρ

Moment of inertia of cracked section Icr =

b x

3

3

 n  As ( d  x )

2

B. Long-Term Deflection Long-term deflection due to combined effect of creep and shrinkage Δt = λ Δi where

ρ' =

ξ

λ=

ξ 1  50 ρ'

A's b d = 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  9.2 m

Beam section

b  300mm

Bending moments

M D.neg  419.34kN m

M D.pos  319.33kN m

M L.neg  223.09kN m

M L.pos  176.58kN m

Steel re-bars

As.sup  6 

h  750mm

π ( 25mm)

π ( 25mm) π ( 25mm)

2

2

 14.726 cm

4

A's.mid  3  Materials

 29.452 cm

4

A's.sup  3  As.mid  5 

2

2

 24.544 cm

4 π ( 25mm)

2

2

2

4

 14.726 cm

2

f'c  25MPa fy  390MPa

Solution Modulus of rupture f'c fr  0.623MPa   3.115  MPa MPa Modulus ratio 5

Es  2  10 MPa kN

wc  24

3

m n 

Es Ec

 7.732

 wc  Ec  44MPa    kN   m3   

1.5



f'c MPa

4

 2.587  10  MPa

Support section Centroid of uncracked section h

A1  b  h

y 1 

As  As.sup

d  h   30mm  10mm  20mm  25mm 

2



Page 153

40mm  2

  645  mm 

A2  n  As y c 

y 2  d

A1  y 1  A2  y 2

 399.815  mm

A1  A2

Moment of inertia of gross section 3

b h I1  12



Ig  I1  A1  y 1  y c



2



 A2  y 2  y c



2

6

 1.205  10  cm

4

Cracking moment y t  h  y c  350.185  mm Ig M cr  fr  107.228  kN m yt Location of neutral axis of cracked section C=T fc x 2

 b = As fs

Ec ε u  x  b = 2  As Es ε s ε u  x  b = 2  As

Es Eu

 εu

dx x

2

b  x = 2  As n  ( d  x ) 2

 x  = 2 ρ n  1  x      d d 

As

ρ 

b d

 0.015

2

 x   2 ρ n x  2 ρ n = 0   d d 

x  d  ρ n 



2

( ρ n )  2  ρ n  246.092  mm

Moment of inertia of cracked section Icr 

b x 3

3

2

5

 A2  ( d  x )  5.114  10  cm

4

Effective moment of inertia of cracked section M neg  M D.neg  M L.neg  642.43 kN m

  Ie1  min Icr   

M a  M neg

3   Mcr   5 4     Ig  Icr Ig  5.146  10  cm  Ma  

Page 154

Ie2  Ie1

Midspan section Centroid of uncracked section h

A1  b  h

y 1 

As  As.mid

d  h   30mm  10mm  25mm 

A2  n  As

y 2  d

y c 

2



A1  y 1  A2  y 2

40mm  2

  665  mm 

 397.557  mm

A1  A2

Moment of inertia of gross section 3

b h I1  12



Ig  I1  A1  y 1  y c

2  A2 y2  yc2  1.202  106 cm4

Cracking moment y t  h  y c  352.443  mm Ig M cr  fr  106.225  kN m yt Location of neutral axis of cracked section C=T fc x 2

 b = As fs

Ec ε u  x  b = 2  As Es ε s ε u  x  b = 2  As

Es Eu

 εu

dx x

2

b  x = 2  As n  ( d  x ) 2

 x  = 2 ρ n  1     d 

x

 d

ρ 

As b d

 0.012

2

 x   2 ρ n x  2 ρ n = 0   d d 

x  d  ρ n 

2



( ρ n )  2  ρ n  233.616  mm

Page 155

Moment of inertia of cracked section b x

Icr 

3

3

2

5

 A2  ( d  x )  4.806  10  cm

4

Effective moment of inertia of cracked section M pos  M D.pos  M L.pos  495.91 kN m

  Im  min Icr   

M a  M pos

3   Mcr   5 4  I  I  I    g cr g  4.877  10  cm Ma   

Calculation of deflection Effective moment of inertia





5

Ie  0.70 Im  0.15 Ie1  Ie2  4.958  10  cm Initial deflection due to dead and live loads M a  495.91 kN m M 0  M neg  M pos  1138.34  kN m K  1.2  0.2

M0 Ma

 0.741

5  M a  Ln

2

ΔD+L  K  25.259 mm 48 Ec Ie Long-term deflection due to dead load ξ  2 A's  A's.mid λ 

ξ 1  50 ρ'

ρ' 

A's b d

 1.461

M D.pos ΔD  λ ΔD+L  23.761 mm M pos Long-term deflection due to sustained live load Δ0.20L  ΔD+L

0.20 M L.pos M pos

 1.799  mm

Short-term deflection due to live load Page 156

4

Δ0.80L  ΔD+L

0.80 M L.pos M pos

 7.195  mm

Total deflection Δ  ΔD  Δ0.20L  Δ0.80L  32.755 mm Permisible deflection Ln 480

Ln

 19.167 mm

360

Page 157

 25.556 mm

20. Development Lengths A. Development length of deformed bar in tension Diameter of deformed bar

d b  20mm

Steel yield strength

fy  390MPa

Concrete compression strength

f'c  25MPa

Depth of concrete below development length

H  350mm

Reinforcement coating

Type of concrete

Concrete cover

c  1  d b

Clear spacing of re-bars

s  2  d b

ψt 

ψt  1.3

1.3 if H  300mm 1.0 otherwise

ψe 

ψe  1

1.0 if Coating = "Uncoated" otherwise 1.5 if c  3d b  s  6d b 1.2 otherwise

ψs 

0.8 if d b  20mm

ψs  0.8

1.0 otherwise λ 

λ1

1.3 if Concrete = "Lightweight" 1.0 otherwise

Ktr  0

(for a design simplification)

db s  db     cb  1.5d b max min c    min 2.5db   30 mm 2 2    







Development of tension bar in tension Page 158



cb  Ktr db

 1.5



Ld  max

fy



ψt ψe ψs λ

f'c  cb  Ktr     1.107MPa MPa  d b   

Ld db



 d b 300mm  977.055  mm

  



 48.853



Ceil Ld 10mm  980  mm

B. Splice length in tension Splice class

Lst 

1.0 Ld if Class = "Class A"

Lst  1270.172 mm

1.3 Ld otherwise



Lst db

 63.509



Ceil Lst 10mm  1280 mm

C. Development length of deformed bar in compression Diameter of development bar

d b  32mm


fy  390MPa


f'c  25MPa

Development length in compression



Ldc  max

fy

f'c   4.152MPa  MPa 

Ldc db

 d b 



fy 22.983MPa

 d b 200mm  601.156  mm

  



 18.786



Ceil Ldc 10mm  610  mm

Page 159

D. Development length of standard hook in tension

Diameter of development bar

d b  10mm


fy  390MPa


f'c  25MPa

Reinforcement coating

Type of concrete

Side cover

cside  65mm

Cover beyond hook

cbeyond  50mm

ψe 

ψe  1

1.0 if Coating = "Uncoated" otherwise 1.5 if c  3d b  s  6d b 1.2 otherwise

λ 

λ1

1.3 if Concrete = "Lightweight" 1.0 otherwise Page 160

α 

Modified factor



 



0.8 if cside  65mm  cbeyond  50mm 0.7 otherwise

α  0.7

Development length of standard hook in tension



Ldh  α max

ψe λ fy



f'c   4.152MPa MPa 

Ldh db

 d b 8d b 150mm  131.503  mm

  



 13.15



Ceil Ldh 10mm  140  mm

E. Lap Splice Length in Compression Diameter of splice bar

d b  25mm


fy  390MPa

Lap splice length in compression

Lsc 

fy    d b 300mm if fy  60ksi  13.79MPa  f    y  24  d b 300mm otherwise max  7.661MPa  

max

Lsc  707.034  mm Lsc db

 28.281

Page 161





Ceil Lsc 10mm  710  mm

Reference 1.

Reinforced Concrete / A Fundamental Approch. 5th Edition. Edward G. Nawy. - Pearson Prentice 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. - 3rd Edotion. 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

Mathcad in Concrete Structures (ACI 318-05) 6th Edition (CM)

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