1. Determine the reinforcement ratio ρ. 2. Calculate the modular ratio n based on concrete and steel properties. 3. Use an iterative process to locate the neutral axis depth kd by solving for the parameter k. 4. With k determined, calculate the moment arm j. 5. Compute the moment capacity as Mallow = R * b * d^2, where R is the resisting stress block parameter dependent on k and j.

1. 4 Reinforced Concrete Design
Strength of Rectangular Section in Bending
Location of Reinforcement
Behavior of Beam under Load
Beam Design Requirements
Working Stress Design (WSD)
Practical Design of RC Beam
Asst.Prof.Dr.Mongkol JIRAVACHARADET
SURANAREE INSTITUTE OF ENGINEERING
UNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING

2. Location of Reinforcement
Concrete cracks due to tension, and as a result, reinforcement is required
where flexure, axial loads, or shrinkage effects cause tensile stresses.
• Simply supported beam
tensile stresses and cracks are
developed along bottom of the beam
Positive
Moment
BMD
longitudinal reinforcement is placed
closed to the bottom side of the beam

3. Location of Reinforcement
• Cantilever beam
- Top bars
- Ties and anchorage
to support

4. Location of Reinforcement
• Continuous beam

5. Location of Reinforcement
• Continuous beam with 2 spans

6. Figure B-13 : Reinforcement Arrangement for Suspended Beams

7. Figure B-14 : Reinforcement Arrangement for
Suspended Cantilever Beams

8. Behavior of Beam under Load
w
L
εc f < f c′
Elastic Bending (Plain Concrete)
εc f < f r = 2.0 f c′
εc f < f c′
Working Stress Condition
C
T = As fs
εs

9. Brittle failure mode
εcu= 0.003
Crushing C
T = As fs
εs <εy fs < fy
Ductile failure mode
εc < 0.003
C
T = As fs
εs ≥εy fs = fy

10. Beam Design Requirements
1) Minimum Depth (for deflection control)
oneway
L/20 L/24 L/28 L/10
slab
BEAM L/16 L/18.5 L/21 L/8
2) Temperature Steel (for slab)
SR24: As = 0.0025 bt b
SD30: As = 0.0020 bt t
SD40: As = 0.0018 bt As
fy > 4,000 ksc: As = 0.0018 4,000 bt
fy

11. 3) Minimum Steel (for beam)
As min = 14 / fy As
To ensure that steel not fail before first crack
4) Concrete Covering
stirrup
Durability and Fire protection
> 4/3 max. aggregate size
5) Bar Spacing

12. 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):
Es 2.04 × 106 134
n= = ≈
Ec 15,100 f c′ f c′

13. Effective Depth (d) : Distance from compression face to centroid of steel
d
Cracked transformed section
strain condition force equilibrium
compression face εc f c = Ecε c
C
kd
d N.A.
jd
εs T = As f s
b f s = Es ε s

14. f c = Ecε c
1
Compression in concrete: C = f c b kd kd
C
2
N.A.
jd
Tension in steel: T = As f s
T = As f s
f s = Es ε s
Equilibrium ΣFx= 0 :
Compression = Tension
1
f c b kd = As f s
2
Reinforcement ratio: ρ = As / bd
fc 2 ρ
= 1
fs k

15. Strain compatibility:
εc εc kd k
= =
ε s d − kd 1 − k
kd
d f c / Ec k
=
f s / Es 1 − k
fc k
εs n = 2
fs 1 − k
k = 2n ρ + ( n ρ ) − n ρ
2
Analysis: know ρ find k 1 2
n fc 1
Design: know fc , fs find k 2 k= =
n fc + fs 1 + fs
n fc

16. Allowable Stresses
Plain concrete: Steel:
f c = 0.33 f c′ ≤ 60 kg/cm 2 SR24: fs = 0.5(2,400) = 1,200 ksc
Reinforced concrete: SD30: fs = 0.5(3,000) = 1,500 ksc
f c = 0.375 f c′ ≤ 65 kg/cm 2 SD40, SD50: fs = 1,700 ksc
Example 3.1: f c′ = 150 ksc , fs = 1,500 ksc
134
n= = 10.94 ⇒ 10 (nearest integer)
150
f c = 0.375(150) = 56 ksc
1
k= = 0.2515
1,500
1+
9(56)

17. Resisting Moment
kd/3 Moment arm distance : j d
1 kd
C= fc k b d jd = d −
M 2 3
jd
k
j = 1−
T = As fs 3
Steel: M = T × jd = As f s jd
1
Concrete: M = C × jd = f c k j b d 2 = R b d 2
2
1
R = fc k j
2

18. Design Step: known M, fc, fs, n
1) Compute parameters
1 1
k= j = 1− k / 3 R= fc k j
1 + fs n fc 2
R (kg/cm2)
fc
n
(kg/cm2) fs=1,200 fs=1,500 fs=1,700
(kg/cm2) (kg/cm2) (kg/cm2)
45 12 6.260 5.430 4.988
50 12 7.407 6.463 5.955
55 11 8.188 7.147 6.587
60 11 9.386 8.233 7.608
65 10 10.082 8.835 8.161

19. Design Parameter k and j
fs=1,200 fs=1,500 fs=1,700
fc (kg/cm2) (kg/cm2) (kg/cm2)
n
(kg/cm2)
k j k j k j
45 12 0.310 0.897 0.265 0.912 0.241 0.920
50 12 0.333 0.889 0.286 0.905 0.261 0.913
55 11 0.335 0.888 0.287 0.904 0.262 0.913
60 11 0.355 0.882 0.306 0.898 0.280 0.907
65 10 0.351 0.883 0.302 0.899 0.277 0.908
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.

20. 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
M
From M = As f s jd → As =
fs j d
4) Select steel bars and Detailing

21. .1 F F , .2
Number of Bars
Bar Dia.
1 2 3 4 5 6
RB6 0.283 0.565 0.848 1.13 1.41 1.70
RB9 0.636 1.27 1.91 2.54 3.18 3.82
DB10 0.785 1.57 2.36 3.14 3.93 4.71
DB12 1.13 2.26 3.53 4.52 5.65 6.79
DB16 2.01 4.02 6.03 8.04 10.05 12.06
DB20 3.14 6.28 9.42 12.57 15.71 18.85
DB25 4.91 9.82 14.73 19.63 24.54 29.45

22. .3 F ACI
Simple One-end Both-ends
Member Cantilever
supported continuous continuous
One-way slab L/20 L/24 L/28 L/10
Beam L/16 L/18.5 L/21 L/8
L = span length
For steel with fy not equal 4,000 kg/cm2 multiply with 0.4 + fy/7,000

23. Example 3.2: Working Stress Design of Beam
w = 4 t/m Concrete: fc = 65 kg/cm2
Steel: fs = 1,700 kg/cm2
5.0 m 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

24. 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
M 12 . 5 × 10 5
Steel area: As = = = 10 . 8 cm 2
f s jd 1,700 × 0 . 908 × 75
Select steel bar 4DB20 (As = 12.57 cm2)

25. Alternative Solution:
From Mc = R b d2 = required moment M
M M
bd 2
= ⇒ d =
R Rb
For example M = 12.5 t-m, R = 8.161 ksc, b = 40 cm
12 . 5 × 10 5
d = = 61 . 88 cm
8 . 161 × 40
USE section 40 x 80 cm d = 75 cm

26. 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
Column width 30 cm
30 cm 30 cm Required moment:
M = (4.768) (4.7)2 / 8
= 13.17 t-m
4.7 m clear span
5.0 m span

27. 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
Load
w = 2.30 t/m
dl 0.43
k = 0.302, j = 0.899, R = 8.835 ksc
wall 0.63
slab 1.24
b = 30 cm, d = 60 - 5 = 55 cm
5.00 w 2.30
Mc = 8.835(30)(55)2/105 = 8.02 t-m
M± = (1/9)(2.3)(5.0)2 = 6.39 t-m
Vc = 0.29(173)1/2(30)(55)/103
As± = 8.62 cm2 (2DB25)
= 6.29 t
V = 5.75 t (RB9@0.20 St.)
As± = 6.39×105/(1,500×0.899×55)
= 8.62 cm2

28. B2 40x80 Mc = 19.88 t-m, Vc = 11.44 t.
w = 2.64 t/m w = 2.64 t/m
8.00 5.00
8.54 9.83
SFD 12.58 3.37
+13.81 +2.15
BMD -16.17
13.65 15.99 2.13
As 3DB25 4DB25 2DB25

29. GRASP Version 1.02
B11-B12
Membe
Mz.i [T-m] Mz.pos [T-m] Mz.j [T-m] Fy.i [Ton] Fy.j [Ton]
r
1 0 39.03 -53.42 33.04 -50.84
2 -53.42 17.36 -37.97 44.52 -39.36
3 -37.97 20.75 -46.35 40.54 -43.34
4 -46.35 25.88 -28.26 44.96 -38.92
5 -28.26 6.59 -92.25 31.27 -52.61
6 -92.25 81.47 0.00 69.70 -47.73

30. Analysis of RC Beam
Given: Section As , b, d Materials fc , fs
Find: Mallow = Moment capacity of section
STEP 1 : Locate Neutral Axis (kd)
k = 2 ρn + (ρn ) − ρn
2
j =1−k / 3
As
where ρ = = Reinforcem ent ratio
bd
Es 2.04 ×106 134
n= = ≈
Ec 15,100 f c′ f c′

31. STEP 2 : Resisting Moment
1
Concrete: Mc = f c k j b d 2
2
Steel: M s = As f s j d
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.

32. Example 3.3 Determine the moment strength of beam
40 cm fc = 65 ksc, fs = 1,700 ksc,
n = 10, d = 75 cm
As 12 . 57
80 cm ρ= = = 0 . 00419 , ρ n = 0 . 0419
bd 40 × 75
k = 2 × 0 . 0419 + ( 0 . 0419 ) 2 − 0 . 0419
4 DB 20
= 0 . 251 → j = 1 − 0 . 251 / 3 = 0 . 916
As = 12.57 cm2
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)

33. Double Reinforcement
- Increase steel area
- Enlarge section
When Mreq’d > Mallow
- Double RC
only when no choice
A’s εc T’ = A’s f’s
d’ ε’s 1
M C = 2 fc k b d
As As1 fs
T = As fs
εs As2 fs

34. F F F
T’ = A’s f’s
1 T’ = A’s f’s
C=
1
f kbd C = 2 fckbd
2 c
jd d-d’
T = As fs T1 = As1 fs T2 = As2 fs
1 M2 = M − Mc
Moment strength M 1 = M c = f c kjbd 2
2
M = M1 + M2 = As 2 f s (d − d ′)
= As1 f s jd
= As′ f s′(d − d ′)
Mc M − Mc
Steel area As = As1 = + As 2 =
f s jd f s (d − d ′)

35. Compatibility Condition
d’ εc
εs d − kd
=
kd ε’s ε s′ kd − d ′
d From Hook’s law: εs = Es fs, ε’s = Es f’s
Es f s fs d − kd
= =
Es f s′ f s′ kd − d ′
εs
k − d′ d
f s′ = f s
1− k
. . . F F k − d′ d
f s′ = 2 f s
1− k

36. ( A’s )
T’ = A’s f’s Force equilibrium [ ΣFx=0 ]
T’ = T2
d-d’
A’s f’s = As2 fs
T2 = As2 fs k − d′ d
Substitute f s′ = 2 f s
1− k
1 1− k
As′ = As 2
2 k − d′ d

37. F (k ) Compression = Tension
d’ εc Cc + Cs′ = T
1
f c b kd + As′ f s′ = As f s
kd ε’ s 2
d k − d′ d As′
Substitute f s′ = 2 f s , ρ′ =
1− k bd
1− k As
εs f s = n fc , ρ=
k bd
 d′  2
′  + n ( ρ + 2 ρ ′) − n ( ρ + 2 ρ ′)
2
k = 2n  ρ + 2 ρ
 d

38. Example 3.4 Design 40x80 cm beam using double RC
w = 6 t/m fc = 65 ksc, fs = 1,700 ksc,
n = 10, d = 75 cm
5.0 m k = 0.277, j = 0.908, R = 8.161 ksc
Beam weight wbm = 0.4 × 0.8 × 2.4(t/m3) = 0.768 t/m
Required M = (6.768) (5)2 / 8 = 21.15 t-m
Mc = Rbd2 = 8.161(40)(75)2/105 = 18.36 t-m < req’d M Double RC
Mc 18.36 × 105
As1 = = = 15.86 cm 2
f s jd 1, 700 × 0.908 × 75
M − Mc (21.15 − 18.36) ×105
As 2 = = = 2.34 cm 2
f s (d − d ′) 1, 700 × (75 − 5)

39. Tension steel As = As1 + As2 = 15.86 + 2.34 = 18.20 cm2
USE 6DB20 (As = 18.85 cm2)
Compression steel
1 1− k 1 1 − 0.277
As′ = As 2 = × 2.34 × = 4.02 cm 2
2 k − d′ d 2 0.277 − 5 / 75
USE 2DB20 (As = 6.28 cm2)
2DB20
0.80 m
6DB20
0.40 m

40. F 175
3 F 3 F