Rc04 bending2

Strength of Rectangular Section in Bending

44

� Location of Reinforcement

� Behavior of Beam under Load

� Beam Design Requirements

� Working Stress Design (WSD)

� Practical Design of RC Beam

Reinforced Concrete DesignReinforced Concrete Design

Asst.Prof.Dr.Mongkol JIRAVACHARADET

S U R A N A R E E

UNIVERSITY OF TECHNOLOGY

INSTITUTE OF ENGINEERING

SCHOOL OF CIVIL ENGINEERING

Location of ReinforcementLocation of Reinforcement

•• Simply supported beamSimply supported beam

Concrete cracks due to tension, and as a result, reinforcement iConcrete cracks due to tension, and as a result, reinforcement is requireds required

where flexure, axial loads, or shrinkage effects cause tensile swhere flexure, axial loads, or shrinkage effects cause tensile stresses.tresses.

tensile stresses and cracks are

developed along bottom of the beam

BMD

longitudinal reinforcement is placed

closed to the bottom side of the beam

Positive

Moment

Location of ReinforcementLocation of Reinforcement

• Cantilever beam

- Top bars

- Ties and anchorage

to support

•• Continuous beamContinuous beam

Location of ReinforcementLocation of Reinforcement

Location of ReinforcementLocation of Reinforcement

• Continuous beam with 2 spans

Figure B-13 : Reinforcement Arrangement for Suspended Beams

Figure B-14 : Reinforcement Arrangement for

Suspended Cantilever Beams

Behavior of Beam under Load

Working Stress Condition

sε

cf f ′<cε

T = As fs

C

w

L

cε 2.0r cf f f ′< =

cf f ′<Elastic Bending (Plain Concrete)

cε

s yε ε<T = As fs

C

Brittle failure mode

s yf f<

s yε ε≥T = As fs

C

Ductile failure mode

s yf f=

Crushing

εcu= 0.003

εc < 0.003

Beam Design Requirements

1) Minimum Depth (for deflection control)

oneway

slabL/24 L/28 L/10L/20

BEAM L/18.5 L/21 L/8L/16

2) Temperature Steel (for slab)

As

t

bSR24: As = 0.0025 bt

SD30: As = 0.0020 bt

SD40: As = 0.0018 bt

fy > 4,000 ksc: As = 0.0018� 4,000 bt

fy

3) Minimum Steel (for beam)

AsAs min = 14 / fy

To ensure that steel not fail before first crack

5) Bar Spacing

> 4/3 max. aggregate size

4) Concrete Covering

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stirrup

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Durability and Fire protection

WSD of Beam for Moment

Assumptions:

1) Section remains plane

2) Stress proportioned to Strain

3) Concrete not take tension

4) No concrete-steel slip

Modular ratio (n):

62.04 10 134

15,100

s

c c c

En

E f f

×= = ≈

′ ′

Effective Depth (d) : Distance from compression face to centroid of steel

d

d

compression face

b

kd

cε

sε s s

s s s

T A f

f E ε

=

=

N.A.

c c cf E ε=C

jd

Cracked transformed section

strain condition force equilibrium

Equilibrium ΣFx= 0 :

Compression = Tension

1

2c s sf b kd A f=

Compression in concrete:1

2cC f b kd=

Tension in steel:s sT A f=

s s

s s s

T A f

f E ε

=

=

N.A.

c c cf E ε=C

jd

kd

Reinforcement ratio: /sA bdρ =

2c

s

f

f k

ρ= 1

Strain compatibility:

1

/

/ 1

1

c

s

c c

s s

c

s

kd k

d kd k

f E k

f E k

f kn

f k

εε= =

− −

=−

=−

2

d

kd

cε

sε

Analysis: know ρ find k 1 2 ( )22k n n nρ ρ ρ= + −

Design: know fc , fs find k 21

1

c

sc s

c

n fk

fn f f

n f

= =+ +

Example 3.1: = 150 ksc , fs = 1,500 ksccf ′

13410.94 10 (nearest integer)

150

0.375(150) 56 ksc

10.2515

1,5001

9(56)

c

n

f

k

= = ⇒

= =

= =+

Allowable Stresses

20.33 60 kg/cmc cf f ′= ≤

Plain concrete:

SR24: fs = 0.5(2,400) = 1,200 ksc

SD30: fs = 0.5(3,000) = 1,500 ksc

SD40, SD50: fs = 1,700 ksc

Steel:

20.375 65 kg/cmc cf f ′= ≤

Reinforced concrete:

Resisting Moment

M

T = As fs

jd

kd/3

1

2cC f k b d=

Moment arm distance : j d

3

kdjd d= −

13

kj = −

Steel:s sM T jd A f jd= × =

Concrete: 2 21

2cM C jd f k j b d R b d= × = =

1

2cR f k j=

Design Step: known M, fc, fs, n

1) Compute parameters

1

1 s c

kf n f

=+

1 / 3j k= −1

2cR f k j=

45

50

55

60

65

fc

(kg/cm2)

R (kg/cm2)

fs=1,200(kg/cm2)

fs=1,500(kg/cm2)

fs=1,700(kg/cm2)

6.260

7.407

8.188

9.386

10.082

n

12

12

11

11

10

5.430

6.463

7.147

8.233

8.835

4.988

5.955

6.587

7.608

8.161

Design Parameter k and j

45

50

55

60

65

fc

(kg/cm2)

fs=1,200(kg/cm2)

fs=1,500(kg/cm2)

fs=1,700(kg/cm2)

0.310

0.333

0.335

0.355

0.351

n

12

12

11

11

10

k j

0.897

0.889

0.888

0.882

0.883

0.241

0.261

0.262

0.280

0.277

k j

0.920

0.913

0.913

0.907

0.908

k j

0.265

0.286

0.287

0.306

0.302

0.912

0.905

0.904

0.898

0.899

1) For greater fs , k becomes smaller → smaller compression area

2) j ≈ 0.9 → moment arm j d ≈ 0.9d can be used in approximation

design.

2) Determine size of section bd2

Such that resisting moment of concrete Mc = R b d 2 ≥ Required M

Usually b ≈ d / 2 : b = 10 cm, 20 cm, 30 cm, 40 cm, . . .

d = 20 cm, 30 cm, 40 cm, 50 cm, . . .

3) Determine steel area

s s s

s

MM A f jd A

f j d= → =From

4) Select steel bars and Detailing

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Bar Dia.Number of Bars

1 2 3 4 5 6

RB6

RB9

DB10

DB12

DB16

DB20

DB25

0.283

0.636

0.785

1.13

2.01

3.14

4.91

0.565

1.27

1.57

2.26

4.02

6.28

9.82

0.848

1.91

2.36

3.53

6.03

9.42

14.73

1.13

2.54

3.14

4.52

8.04

12.57

19.63

1.41

3.18

3.93

5.65

10.05

15.71

24.54

1.70

3.82

4.71

6.79

12.06

18.85

29.45

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Member

One-way slab

Beam

Simplesupported

One-endcontinuous

Both-endscontinuous

Cantilever

L/20

L/16

L/24

L/18.5

L/28

L/21

L/10

L/8

L = span length

For steel with fy not equal 4,000 kg/cm2 multiply with 0.4 + fy/7,000

Example 3.2: Working Stress Design of Beam

w = 4 t/m

5.0 m

Concrete: fc = 65 kg/cm2

Steel: fs = 1,700 kg/cm2

From table: n = 10, R = 8.161 kg/cm2

Required moment strength M = (4) (5)2 / 8 = 12.5 t-m

Recommended depth for simple supported beam:

d = L/16 = 500/16 = 31.25 cm

USE section 30 x 50 cm with steel bar DB20

d = 50 - 4(covering) - 2.0/2(bar) = 45 cm

Moment strength of concrete:

Mc = R b d2 = 8.161 (30) (45)2

= 495,781 kg-cm

= 4.96 t-m < 12.5 t-m NG

TRY section 40 x 80 cm d = 75 cm

Mc = R b d2 = 8.161 (40) (75)2

= 1,836,225 kg-cm

= 18.36 t-m > 12.5 t-m OK

Steel area: 25

cm 8.1075908.0700,1

105.12=

×××

==jdf

MA

s

s

Select steel bar 4DB20 (As = 12.57 cm2)

Alternative Solution:

From Mc = R b d2 = required moment M

bR

Md

R

Mdb =⇒=2

For example M = 12.5 t-m, R = 8.161 ksc, b = 40 cm

cm 88.6140161.8

105.12 5

=××

=d

USE section 40 x 80 cm d = 75 cm

Revised Design due to Self Weight

From selected section 40 x 80 cm

Beam weight wbm = 0.4 × 0.8 × 2.4(t/m3) = 0.768 t/m

Required moment M = (4 + 0.768) (5)2 / 8 = 14.90 < 18.36 t-m OK

Revised Design due to Support width

5.0 m span

30 cm30 cmColumn width 30 cm

4.7 m clear span

Required moment:

M = (4.768) (4.7)2 / 8

= 13.17 t-m

Practical Design of RC Beam

B1 30x60 Mc = 8.02 t-m, Vc = 6.29 t.fc = 65 ksc, fs = 1,500 ksc, n = 10

k = 0.302, j = 0.899, R = 8.835 ksc

b = 30 cm, d = 60 - 5 = 55 cm

Mc = 8.835(30)(55)2/105 = 8.02 t-m

Vc = 0.29(173)1/2(30)(55)/103

= 6.29 t

w = 2.30 t/m

5.00

Load

dl 0.43

wall 0.63

slab 1.24

w 2.30

M± = (1/9)(2.3)(5.0)2 = 6.39 t-m

As± = 6.39×105/(1,500×0.899×55)

= 8.62 cm2

As± = 8.62 cm2 (2DB25)

V = 5.75 t ([email protected] St.)

B2 40x80 Mc = 19.88 t-m, Vc = 11.44 t.

w = 2.64 t/m

8.00 5.00

w = 2.64 t/m

SFD8.54 9.83

12.58 3.37

BMD+13.81

-16.17

+2.15

As13.65

4DB25

2.13

3DB25

15.99

2DB25

GRASP Version 1.02

B11-B12

-47.7369.700.0081.47-92.256

-52.6131.27-92.256.59-28.265

-38.9244.96-28.2625.88-46.354

-43.3440.54-46.3520.75-37.973

-39.3644.52-37.9717.36-53.422

-50.8433.04-53.4239.0301

Fy.j [Ton]Fy.i [Ton]Mz.j [T-m]Mz.pos [T-m]Mz.i [T-m]Membe

r

Analysis of RC Beam

Given: Section As , b, d Materials fc , fs

Find: Mallow = Moment capacity of section

STEP 1 : Locate Neutral Axis (kd)

( ) nnnk ρρρ −+= 22

3/1 kj −=

where ratioent Reinforcem==bd

Asρ

62.04 10 134

15,100

s

c c c

En

E f f

×= = ≈

′ ′

STEP 2 : Resisting Moment

Concrete: 2

2

1dbjkfM cc =

Steel: djfAM sss =

If Mc > Ms , Over reinforcement Mallow = Ms

If Mc < Ms , Under reinforcement Mallow = Mc

Under reinforcement is preferable because steel is weaker

than concrete. The RC beam would fail in ductile mode.

Example 3.3 Determine the moment strength of beam

4 DB 20

As = 12.57 cm2

80 cm

40 cm fc = 65 ksc, fs = 1,700 ksc,

n = 10, d = 75 cm

0419.0,00419.07540

57.12==

×== n

bd

As ρρ

916.03/251.01251.0

0419.0)0419.0(0419.02 2

=−=→=

−+×=

j

k

Mc = 0.5(65)(0.251)(0.916)(40)(75)2/105 = 16.81 t-m

Ms = (12.57)(1,700)(0.916)(75)/105 = 14.68 t-m (control)

Double Reinforcement

When Mreq’d > Mallow

- Increase steel area- Enlarge section

- Double RConly when no choice

εs

εc

M

T = As fs

C = fc k b d1

2

As

A’s

ε’sd’T’ = A’s f’s

As1 fs

As2 fs

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T = As fs

C = fckbd1

2

T’ = A’s f’s

T1 = As1 fs

C = fckbd1

2

T2 = As2 fs

T’ = A’s f’s

jd d-d’

2

1

1

1

2c c

s s

M M f kjbd

A f jd

= =

=

2

2 ( )

( )

c

s s

s s

M M M

A f d d

A f d d

= −

′= −

′ ′ ′= −

Steel area As =1

cs

s

MA

f jd= +

2( )

cs

s

M MA

f d d

−=

′−

M = M1 + M2

Moment strength

Compatibility Condition

εc

d

d’

kd ε’s

εs

s

s

d kd

kd d

εε

−=

′ ′−

From Hook’s law: εs = Es fs, ε’s = Es f’s

s s s

s s s

E f f d kd

E f f kd d

−= =

′ ′ ′−

1s s

k d df f

k

′−′ =

−

.�.�. ������������ 21

s s

k d df f

k

′−′ =

−

������� ����������������� ( A’s )

T2 = As2 fs

T’ = A’s f’s

d-d’

Force equilibrium [ ΣFx=0 ]

T’ = T2

A’s f’s = As2 fs

Substitute 21

s s

k d df f

k

′−′ =

−

2

1 1

2s s

kA A

k d d

−′ =

′−

���� � ��������� ( k )

εc

d

d’

kd ε’s

εs

Compression = Tension

c sC C T′+ =

1

2c s s s sf bkd A f A f′ ′+ =

Substitute 2 ,1

ss s

Ak d df f

k b dρ

′′−′ ′= =

−

1, s

s c

Akf n f

k b dρ

−= =

( ) ( )222 2 2 2d

k n n nd

ρ ρ ρ ρ ρ ρ′ ′ ′ ′= + + + − +

Example 3.4 Design 40x80 cm beam using double RC

fc = 65 ksc, fs = 1,700 ksc,

n = 10, d = 75 cm

k = 0.277, j = 0.908, R = 8.161 ksc

w = 6 t/m

5.0 m

Required M = (6.768) (5)2 / 8 = 21.15 t-m

Beam weight wbm = 0.4 × 0.8 × 2.4(t/m3) = 0.768 t/m

Mc = Rbd2 = 8.161(40)(75)2/105 = 18.36 t-m < req’d M Double RC

52

1

18.36 1015.86 cm

1,700 0.908 75

cs

s

MA

f jd

×= = =

× ×

52

2

(21.15 18.36) 102.34 cm

( ) 1,700 (75 5)

cs

s

M MA

f d d

− − ×= = =

′− × −

Tension steel As = As1 + As2 = 15.86 + 2.34 = 18.20 cm2

USE 6DB20 (As = 18.85 cm2)

Compression steel

2

2

1 1 1 1 0.2772.34 4.02 cm

2 2 0.277 5 / 75s s

kA A

k d d

− −′ = = × × =

′− −

USE 2DB20 (As = 6.28 cm2)

6DB20

2DB200

.80

m

0.40 m

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