Bending in Beam 3

� Strength Design Method (SDM)

� Beam behavior under increasing load

� Nominal Moment Strength (Mn)

� Design Procedure

Reinforced Concrete DesignReinforced Concrete Design

Mongkol JIRAVACHARADET

S U R A N A R E E INSTITUTE OF ENGINEERING

UNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING

Ultimate Stress Design (USD)

Advantage of SDM over WSD:

1) Consider mode of failure

2) Nonlinear behavior of concrete

3) More realistic F.S.

4) Ultimate load prediction ≅ 5%

5) Saving (lower F.S.)

Strength Design Method (SDM)Strength Design Method (SDM)

Working Stress Design (WSD): early 1900s → early 1960s

Strength Design Method (SDM) is more realistic for safety and reliability at the strength limit state.

Strength Design Method (SDM)Strength Design Method (SDM)

Design Strength ≥ Required Strength (U)

Design Strength = Strength Reduction Factor (φ) × Nominal Strength (N)

Strength Reduction Factor ( φφφφ) accounts for

(1) Variations in material strengths and dimensions

(2) Inaccuracies in the design equations

(3) Ductility & reliability of the members

(4) Importance of the member in the structure

Nominal Strength = Strength of a member calculated by using SDM

Required Strength = Load factors × Service Load

REQUIRED STRENGTH (U)REQUIRED STRENGTH (U)

Load Combinations:

U = 1.4(D + F)

U = 1.2(D + F + T) + 1.6(L + H) + 0.5(Lr or S or R)

U = 1.2D + 1.6(Lr or S or R) + (1.0L or 0.8W)

U = 1.2D + 1.6W + 1.0L + 0.5(Lr or S or R)

U = 1.2D + 1.0E + 1.0L + 0.2S

U = 0.9D + 1.6W + 1.6H

U = 0.9D + 1.0E + 1.6H

where

D = dead load, E = earthquake, F = fluid pressure, H = lateral soil pressure,

L = live load, Lr = roof live load, R = rain load, S = snow load, T = temp. load,

U = required strength to resist factored load, and W = wind load

Required Strength for Simplified Load CombinationsRequired Strength for Simplified Load Combinations

Loads

Dead (D) and Live (L)

Required Strength

1.4D

1.2D + 1.6L + 0.5Lr

Dead, Live and Wind (W) 1.2D + 1.6Lr + 1.0L

1.2D + 1.6Lr + 0.8W

1.2D + 1.6W + 1.0L + 0.5Lr

0.9D + 1.6W

Dead, Live and Earthquake (E) 1.2D + 1.0L + 1.0E

0.9D + 1.0E

In Thailand, we still use: U = 1.4D + 1.7L

Strength Reduction Factors Strength Reduction Factors φφφφφφφφ in the Strength Design Methodin the Strength Design Method

0.90Tension-controlled sections

0.700.65

Compression-controlled sectionsMembers with spiral reinforcementOther reinforced members

0.75Shear and torsion

0.65Bearing on concrete

0.85Post-tensioned anchorage zones

0.75Struts, ties, nodal zones and shearing areas

Behavior of Concrete Beam under increasing load

As

h

b

As

d

cε

sε

ctεfs

fc

fct

Before crack

As

After crack

cε

sε fs

fc

Working Stress State

cuε

sε fs

fc

Strength Limit State

Overload

Ultimate strain

Nominal Moment Strength ( Mn )

b

d

As

εcu = crushing strain

εs

N.A.

fc’

x

T = As fy ( for εs > εy )

C = k fc’ bx

T = As fy

C = 0.85 fc’ a b

0.85 fc’

a = β1xa/2

d – a/2

Equivalent Stress Distribution(Whitney stress block)

[ΣFx = 0] C = T

0.85 fc’ a b = As fy

ρ= =

′ ′0.85 0.85s y y

c c

A f f da

f b f

Equivalent Stress Distribution(Whitney stress block)

ρρ

= − ′

2 11.7

yn y

c

fM f bd

f

T = As fy

C = 0.85 fc’ a b

0.85 fc’

a = β1xa/2

d – a/2

ρ = − = − ′

( / 2)2(0.85)

yn s y

c

f dM T d a A f d

f

Flexural resistance factor Rn : = 2n nM R bd

ρρ

= − ′

11.7

yn y

c

fR f

f

=′0.85

y

c

fm

fModular ratio: ρ ρ

= −

11

2n yR f m

for fc’ ≤ 280 ksc, β =1 0.85

for fc’ > 280 ksc, β′ −

= − ≥

1

2800.85 0.05 0.65

70cf

Reinforcement Ratio ρρρρ :

ρ ρ = −

11

2n yR f mFrom ρ ρ− + =2 2 2 0y y nm f f R

ρ

= − −

211 1 n

y

mRm f

1β

cf ′280

0.85

0

0.65

560

210 0.85

cf ′ 1β

240 0.85

280 0.85

320 0.82

350 0.80

Balance Steel Ratio ( ρρρρb )

��������ก�� ��������������ก�������ก���ก����� ��� !����ก��

T = As fy

d

baf85.0C c′=

εcu = 0.003

εs = εy = fy / Es

cd

Concrete crushing

Steel yielding

From strain condition,

From force equilibrium, [ΣFx = 0]

cu

cu y

cd

ε=

ε + εcu

cu y

c d ε

= ε + ε

C = T

c s y0.85 f ab A f′ = c 1 y0.85 f cb f bd′ β = ρ

����"� εcu = 0.003

c cub 1

y cu y

0.85 ff

′ ερ = β ε + ε

Balance Steel Ratio:

εy = fy / 2.04×106��

cb 1

y y

0.85 f 6,120f 6,120 f

′ρ = β +

��������ก�� ��#���$%��&��������ก����� ρρρρ = As/bd

ρρρρ = ρρρρb

���������balance condition

εcu

εy

������� ก��กOver RC

ρρρρ > ρρρρbεcu

εs < εy

������� ก����Under RC

εcu

ρρρρ < ρρρρb

εs > εy

��$�'"�� ก����$%��&��������ก����� (Under RC)

30 cm

50 c

m

2DB25

cf 240 ksc′ =

β1 = 0.85

fy = 4,000 ksc

b0.85 240 6,120

0.854,000 6,120 4,000

× ρ = × × +

= 0.0262

As = 2×4.91 = 9.82 cm2

sA 9.820.0073

bd 30 45ρ = = =

×ρ < ρb Under RC

Steel Yield : fs = fy

������� ก����Under RC

εcu

ρρρρ < ρρρρb

εεεεs > εεεεy

C = T

0.85×240×0.85×c×30 = 9.82×4,000

c = 7.55 cm

c 1 s y0.85 f cb A f′ β =

Strain Condition:

d

c

εcu = 0.003

εεεεs = ?

s

cu

d cc

ε −=

ε

εs = (45 – 7.55) / 7.55 × 0.003 = 0.0149

εy = fy / Es = 4,000 / 2.04e6 = 0.00196

εεεεs > εεεεy Steel Yield : fs = fy Under RC

ACI 318-08 Section 10.5 : Minimum reinforcement of flexural members

10.5.1 – At every section of a flexural member where tensile reinforcement is required, As provided shall not be less than that given by

cs, min

y

0.8 fA b d

f

′=

and not less than 14 b d / f y To prevent concrete first crack

��$�'"�� ก����$%��&��������ก����� (Over RC)

cf 240 ksc′ =

β1 = 0.85

fy = 4,000 ksc

b0.85 240 6,120

0.854,000 6,120 4,000

× ρ = × × +

= 0.0262

As = 8×4.91 = 39.28 cm2

ρ > ρb Over RC

Steel NOT Yield : fs < fy

C = T

0.85×240×0.85×c×30 = 39.28 fs

2 unknowns: c and fs ?

30 cm

50 c

m

8DB25

39.280.0312

30 42ρ = =

×

������� ก��กOver RC

ρρρρ > ρρρρbεcu

εεεεs < εεεεy

c 1 s s0.85 f cb A f′ β =

1

Strain Condition:

dc

εcu = 0.003

εεεεs = fs/Es

s

cu

d cc

ε −=

ε

ss cu

s

f d cE c

− ε = = ε

s42 c

f 6,120c− =

1

5,202 c2 + 240,393.6 c – 10,096,531.2 = 0

>> roots([5202 240393.6 -10096531.2]) ans = -72.8530 26.6412

MATLAB:

c = 26.6 cm

fs = 3,543 ksc

425,202 39.28 6,120

− = ×

cc

c

fs < fy

Steel NOT Yield Over RC

ρρρρ = 0.5 ρρρρmax = 0.375 ρρρρb

ACI 318-08: Section 10.3 – General principles and re quirements

10.3.5 – For flexural members, a net tensile strain εεεεt in extreme tension steel shall not be less than 0.004.

For conservative design, we may use

ACI Code before 2002, ρρρρmax = 0.75 ρρρρb

From

2n nM R bd=

yn y

c

fR f 1

1.7 f

ρ = ρ − ′

If we use ρρρρmax Rn,max Mn,max

where Mn,max is the maximum moment

capacity of the section

�������� ก.5 ��������ก��������"����������*�+������� �

f’c(ksc)

ρmin ρb ρmax m Rn,max(ksc)

180

210

240

280

320

350

0.0035

0.0035

0.0035

0.0035

0.0035

0.0035

0.0197

0.0229

0.0262

0.0306

0.0338

0.0360

0.0147

0.0172

0.0197

0.0229

0.0253

0.0270

26.1

22.4

19.6

16.8

14.7

13.4

47.62

55.55

63.49

74.07

82.46

88.36

������� fy = 4,000 ก.ก./��.2

0 0.005 0.01 0.015 0.02 0.0250

10

20

30

40

50

60

70

80

Strength Curve (Rn vs. ρρρρ) for SD40 Reinforcement

Coe

ffici

ent o

f res

ista

nce

Rn

(kg/

cm2 )

Reinforcement ratio ρ = As /bd

f’c = 180 ksc

f’c = 210 ksc

f’c = 240 ksc

f’c = 280 ksc

Upper limit at 0.75ρb

��������ก������������ก��

�ก���ก��ก�������

��������ก ρρρρb �� !�����ก"!� 1/m ����(�$��ก�$%�) �����'�

Required moment from load = Mu

Design Moment Strength = MnuM

=φ

2nR bd= u

n 2

MR

bd=

φ

From n y1

R f 1 m2

= ρ − ρ

where y

c

fm

0.85 f=

′

�()�� 2y y nmf 2f 2R 0ρ − ρ + =

2y y n y

y

2f 4 f 8mR f

2mf

± −ρ = n

y

2mR11 1

m f

= ± −

n

y

2mR11 1

m f

ρ = − −

STEP 1 ���ก�������� ก�����*������(��+��,�-!"� min maxρ ρ ρ≤ ≤

min max

1

1

14 / 0.75

0.85 6,120

6,120

0.85 ; 280 ksc

2800.85 0.05 ;280 560 ksc

70

0.65 ; 560 ksc

y b

cb

y y

c

cc

c

f

f

f f

f

ff

f

ρ ρ ρ

ρ β

β

= =

′= +

′ ≤ ′− ′= − < ≤

′ >

Conservative design select ρρρρ = 0.5= 0.5= 0.5= 0.5ρρρρmax = 0.375 ρρρρb

,�-����ก����ก�&&������ ������ก����������ก

STEP 2

onewayslab L/24 L/28 L/10L/20

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

���ก.��� ��*��%���ก�� b d 2

��ก 2n nM R bd= 2 n

n

Mbd

R= u

n

MR

=φ

n y1

R f 1 m2

= ρ − ρ

����� y

c

fm

0.85 f=

′

STEP 3 ���ก d ��ก "��/ก h *����(����0�����ก������12��ก����!�%�"

���ก b ≈≈≈≈ d/2

ก�����ก.�������%�� ���(,-��.�% ������. �-!� 30x50 ��.

STEP 4 ������� !� ρρρρ %��.�������%��*�����ก (b, d)

��ก 2n nM R bd= n

n 2

MR

bd= u

2

M

bd=

φ

n

y

2mR11 1

m f

ρ = − −

STEP 5 %�"������������ ก"!���$!,�-!"� ρρρρmin < ρρρρ < ρρρρmax ����)�! ?

4�� ρρρρ < ρρρρmin ,��,-� ρρρρ = ρρρρmin

4�� ρρρρ > ρρρρmax ,���0���.�������%�� ��" ���"�,��!

STEP 6 ���"�0�'�*���� ก As = ρρρρbd ��"���ก.����(����"��� ก�����

STEP 7 %�"����ก��������%�� y2n y

c

fM f bd 1

1.7 f

ρ = ρ − ′

uM≥

φ

����bf85.0

fAa

c

ys

′=

−=

2a

dfAM ysnuM

≥φ

c

y

f85.0

fm

′=�����

�������� ก.4 � ��ก ����������������(��.)

����ก����� �������ก����������� ����� ����

���������2 3 4 5 6 7 8

DB12

DB16

DB20

DB25

DB28

16.9

17.3

17.7

18.2

18.8

A B C D

20.6

21.4

22.2

23.2

24.4

24.3

25.5

26.7

28.2

30.0

28.0

29.6

31.2

33.2

35.6

31.7

33.7

35.7

38.2

41.2

35.4

37.8

40.2

43.2

46.8

39.1

41.9

44.7

48.2

52.4

3.7

4.1

4.5

5.0

5.6

A = 4 ��. �(�����ก9�" ��ก��%4/��� ก��ก

B = 9 ��. �� ก��กC = 1.9 ��.D = -!��"!���(�"!���� ก = db ���� 2.5 ��.

Example 2.5 Design B1 in the floor plan shown below.

Slab thickness = 12 cm

LL = 300 kg/m2

= 280 kg/cm2

Steel: SD40

cf ′B1

B2

8.00

4.00

2.00

5.00 3.00

Reaction at B2’s ends = wL/2 = (2,331+504)(4)/2 = 5,670 kg

Slab DL = 0.12(2,400) = 288 kg/m2

Ultimate load = 1.4(288) + 1.7(300) = 913.2 kg/m2

2913.2(4) 913.2(3) 3 0.752,331 kg/m

3 3 2

−+ =

Load on B2 =

B2 weight (assume section 30× 50 cm) = 1.4(0.3)(0.5)(2,400) = 504 kg/m

Load on B1:

B1

B2:5,670 kg

2,350 kg/m 1,826 kg/m

5,670 kg

5.00 m 3.00 m

5,670 kg

913.2 kg/m913.21,437 kg/m

B1 weight: simply support min. depth = 800/16 = 50 cm

Try section 30 × 60 cm, wu = 1.4(0.3)(0.6)(2400) = 605 kg/m

Mmax = 2,431(8.0)2/8

= 19,448 kg-m

1

max

524(5)(5 / 2)819 kg

8819(3) 2,456 kg-m

R

M

= =

= =

1,826 + 605 = 2,431 kg/m

8.00

5.00

2,350-1,826=524 kg/m

3.00R1

5.00

5,670 kg

3.00

max

5,670(5.0)(3.0)

810,631 kg-m

M =

=

Mu = 19,448 + 2,456 + 10,631 = 32,535 kg-m

Max. moment on B1:

USE DB20: d = 60 - 4 - 2.0/2 - 0.9 = 54 cm

0.85(280) 6120(0.85) 0.0306

4,000 6120 4000bρ = = +

ρmax= 0.75ρb = 0.75(0.0306) = 0.0230

2 2

4,00016.81

0.85 0.85(280)

32,535(100)41.32

0.9(30)(54)

y

c

un

fm

f

MR

bdφ

= = =′

= = =

ρmin = 14/fy = 14/4,000 = 0.0035

21Required 1 1

1 2(16.81)(41.32)1 1

16.81 (4,000)

0.0114

n

y

m R

m fρ

= − −

= − −

=

ρρρρmin = 0.0035 < ρρρρ = 0.0114 < ρρρρmax = 0.0230 OK

0.60

0.30

6DB20BUT 6DB20 need bmin = 35.7 cm NG

Home work: redesign section

As = ρbd = 0.0114(30)(54) = 18.51 cm2

USE 6DB20 (As = 18.85 cm2)