Reinforced Concrete Design

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

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

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

• Continuous beam
Location of Reinforcement

Location 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
w
Working Stress Condition
L
Elastic Bending (Plain Concrete)
ε s
ε f < f = 2.0 f ′
c
εc
εc
f < f ′ c
r c
f < f ′ c
C
T = A s f s

Brittle failure mode
Crushing
Ductile failure mode
εs < εy
εs ≥εy
ε cu = 0.003
fs < fy
ε c < 0.003
fs = fy
C
T = A s f s
C
T = A s f s

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

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

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):
6
Es
2.04× 10 134
n = = ≈
E 15,100 f ′ f ′
c c c

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

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

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

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

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

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

Design Parameter k and j
f c
(kg/cm 2 )
45
50
55
60
65
n
12
12
11
11
10
f s =1,200
(kg/cm 2 )
k j
0.310
0.333
0.335
0.355
0.351
0.897
0.889
0.888
0.882
0.883
f s =1,500
(kg/cm 2 )
k j
0.265
0.286
0.287
0.306
0.302
0.912
0.905
0.904
0.898
0.899
f s =1,700
(kg/cm 2 )
k j
0.241
0.261
0.262
0.280
0.277
0.920
0.913
0.913
0.907
0.908
1) For greater f s , 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 bd 2
Such that resisting moment of concrete M c = R b d 2 ≥ Required M
Usually b ≈ d / 2 : b = 10 cm, 20 cm, 30 cm, 40 cm, . . .
3) Determine steel area
From
d = 20 cm, 30 cm, 40 cm, 50 cm, . . .
M = A f jd → A =
s s s
4) Select steel bars and Detailing
M
f j d
s

ก.1 ก,
. 2
Bar Dia.
RB6
RB9
DB10
DB12
DB16
DB20
DB25
Number of Bars
1 2 3 4 5 6
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

ก.3 ก
ACI
Member
One-way slab
Beam
L = span length
Simple
supported
L/20
L/16
One-end
continuous
L/24
L/18.5
Both-ends
continuous Cantilever
L/28
L/21
L/10
For steel with f y not equal 4,000 kg/cm 2 multiply with 0.4 + f y /7,000
L/8

Example 3.2: Working Stress Design of Beam
5.0 m
w = 4 t/m
Concrete: f c = 65 kg/cm 2
Steel: f s = 1,700 kg/cm 2
From table: n = 10, R = 8.161 kg/cm 2
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:
M c = R b d 2 = 8.161 (30) (45) 2
= 495,781 kg-cm
TRY section 40 x 80 cm d = 75 cm
Steel area:
A
s
=
= 4.96 t-m < 12.5 t-m NG
M c = R b d 2 = 8.161 (40) (75) 2
M
f jd
s
= 1,836,225 kg-cm
= 18.36 t-m > 12.5 t-m OK
=
1,
700
12.
5
×
× 10
0.
908
5
× 75
Select steel bar 4DB20 (A s = 12.57 cm 2 )
=
10.
8
cm
2

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

Revised Design due to Self Weight
From selected section 40 x 80 cm
Beam weight w bm = 0.4 × 0.8 × 2.4(t/m 3 ) = 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
30 cm
Column width 30 cm
4.7 m clear span
5.0 m span
30 cm
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.
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 ± = 8.62 cm 2 (2DB25)
V = 5.75 t (RB9@0.20 St.)
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 /10 5 = 8.02 t-m
Vc = 0.29(173) 1/2 (30)(55)/10 3
= 6.29 t
As ± = 6.39×10 5 /(1,500×0.899×55)
= 8.62 cm 2

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

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

Analysis of RC Beam
Given: Section A s , b, d Materials f c , f s
Find: M allow = Moment capacity of section
STEP 1 : Locate Neutral Axis (kd)
2 ( ρn
) n
k = 2ρn
+ − ρ
j = 1 − k
A s
/
3
where ρ =
bd
= Reinforcem ent ratio
Es
n =
E
6
2.04× 10 134
= ≈
15,100 f ′ f ′
c c c

STEP 2 : Resisting Moment
Concrete:
1
f
2
Mc = c
Steel: A f j d
Ms s s =
If M c > M s , Over reinforcement M allow = M s
If M c < M s , Under reinforcement M allow = M c
k
Under reinforcement is preferable because steel is weaker
than concrete. The RC beam would fail in ductile mode.
j
b
d
2

Example 3.3 Determine the moment strength of beam
80 cm
40 cm f c = 65 ksc, f s = 1,700 ksc,
4 DB 20
A s = 12.57 cm 2
n = 10, d = 75 cm
12.
57
ρ = = = 0.
00419 , n
bd 40 × 75
k
=
=
A s ρ
2
×
0.
251
0.
0419
→
j
=
+
1
( 0.
0419
−
0.
251
)
/
2
3
−
=
0.
0419
=
0.
916
M c = 0.5(65)(0.251)(0.916)(40)(75) 2 /10 5 = 16.81 t-m
0.
0419
M s = (12.57)(1,700)(0.916)(75)/10 5 = 14.68 t-m (control)

Double Reinforcement
M
When M req’d > M allow
A’ s
A s
d’
ε s
ε c
- Increase steel area
- Enlarge section
- Double RC
only when no choice
ε’ s
T’ = A’ s f’ s
1
C = fc k b d
2
T = A s f s
A s1 f s
A s2 f s

กก
T’ = A’ s f’ s
1
C = fckbd 2
T = A s f s
Moment strength
M = M 1 + M 2
jd
Steel area As = s1
1
C = f
2 ckbd
T 1 = A s1 f s
1
M1 = M c = fckjbd 2
= A f jd
A
s1 s
M
c = + As
2
fs jd
2
2
d-d’
M = M − M
T’ = A’ s f’ s
T 2 = A s2 f s
= As 2 fs ( d − d′
)
= A′ f ′ ( d − d′
)
=
s s
c
M − M c
f ( d − d′
)
s

Compatibility Condition
d
d’
ε c
kd ε’ s
ε s
ε s d − kd
=
ε′
kd − d′
s
From Hook’s law: ε s = E s f s, ε’ s = E s f’ s
Es fs fs d − kd
= =
E f ′ f ′ kd − d′
s s s
f ′ = f
s s
... ก ′ = 2
f f
s s
k − d′ d
1−
k
k − d′ d
1−
k

ก ( A’ s )
d-d’
T’ = A’ s f’ s
T 2 = A s2 f s
′ =
Force equilibrium [ ΣF x=0 ]
T’ = T 2
A’ s f’ s = A s2 f s
f ′ = f
Substitute 2
1 1−
k
As As 2
2 k −
d′ d
s s
k − d′ d
1−
k

ก ( k )
d
d’
ε c
kd ε’ s
ε s
1
2
Compression = Tension
C + C′ = T
c s
f bkd + A′ f ′ = A f
c s s s s
k − d′ d A′
s
Substitute f ′ s = 2 fs
, ρ′
=
1−
k b d
1−
k As
fs = n fc
, ρ =
k b d
⎛ d′
⎞ 2
k = 2n⎜ ρ + 2ρ′ ⎟ + n ρ + 2ρ′ − n ρ + 2ρ′
⎝ d ⎠
2
( ) ( )

Example 3.4 Design 40x80 cm beam using double RC
5.0 m
w = 6 t/m
f c = 65 ksc, f s = 1,700 ksc,
n = 10, d = 75 cm
Required M = (6.768) (5) 2 / 8 = 21.15 t-m
k = 0.277, j = 0.908, R = 8.161 ksc
Beam weight w bm = 0.4 × 0.8 × 2.4(t/m 3 ) = 0.768 t/m
M c = Rbd 2 = 8.161(40)(75) 2 /10 5 = 18.36 t-m < req’d M Double RC
A
A
s1
s2
5
M c 18.36× 10
= = = 15.86 cm
f jd 1,700 × 0.908× 75
s
5
M − M c (21.15 − 18.36) × 10
= = = 2.34 cm
f ( d − d′
) 1,700 × (75 −
5)
s
2
2

Tension steel A s = A s1 + A s2 = 15.86 + 2.34 = 18.20 cm 2
USE 6DB20 (A s = 18.85 cm 2 )
Compression steel
1 1− k 1 1− 0.277
′ = = × 2.34× = 4.02 cm
2 − 2 0.277 − 5/ 75
As As 2
k d′ d
USE 2DB20 (A s = 6.28 cm 2 )
0.80 m
0.40 m
2DB20
6DB20
2

175
3 3