Behavior of Beam under Load. ▫ Beam Design Requirements. ▫ Working Stress
Design (WSD). ▫ Practical Design of RC Beam. Reinforced Concrete Design.
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 UNIVERSITY OF TECHNOLOGY
INSTITUTE OF ENGINEERING SCHOOL OF CIVIL ENGINEERING
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
Location of Reinforcement • Cantilever beam - Top bars - Ties and anchorage to support
Location of Reinforcement • Continuous beam
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
L εc
Elastic Bending (Plain Concrete)
εc
εc
Working Stress Condition
f < f c′
f < f r = 2.0 f c′
f < f c′
C
T = As fs εs
Brittle failure mode
εcu= 0.003 C
Crushing
T = As fs
εs 4,000 ksc:
As = 0.0018 4,000 bt fy
b t As
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
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′
Effective Depth (d) : Distance from compression face to centroid of steel d
Cracked transformed section strain condition
εc
compression face kd
force equilibrium f c = Ecε c C N.A.
d
εs b
T = As f s f s = Es ε s
jd
Compression in concrete:
1 C = f c b kd 2
kd
N.A.
T = As f s
Tension in steel:
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: fc 2 ρ = fs k
f c = Ecε c C
ρ = As / bd 1
jd
Strain compatibility: εc
εc kd k = = ε s d − kd 1 − k
kd
f c / Ec k = f s / Es 1 − k
d
fc k n = fs 1 − k
εs Analysis: know ρ find k
Design: know fc , fs find k
1
2
2
2
k = 2n ρ + ( n ρ ) − n ρ 2
k=
n fc 1 = n fc + fs 1 + fs n fc
Allowable Stresses Steel:
Plain concrete:
f c = 0.33 f c′ ≤ 60 kg/cm 2 Reinforced concrete:
f c = 0.375 f c′ ≤ 65 kg/cm 2
SR24: fs = 0.5(2,400)
= 1,200 ksc
SD30: fs = 0.5(3,000)
= 1,500 ksc
SD40, SD50: fs
= 1,700 ksc
Example 3.1: f c′ = 150 ksc , fs = 1,500 ksc n=
134 = 10.94 ⇒ 10 (nearest integer) 150
f c = 0.375(150) = 56 ksc k=
1 = 0.2515 1,500 1+ 9(56)
Resisting Moment Moment arm distance : j d
kd/3
C=
M
1 fc k b d 2
jd T = As fs
kd jd = d − 3 k j = 1− 3
Steel:
M = T × jd = As f s jd
Concrete:
1 M = C × jd = f c k j b d 2 = R b d 2 2 1 R = fc k j 2
Design Step: known M, fc, fs, n 1) Compute parameters
k=
1 1 + fs n fc
j = 1− k / 3
R=
1 fc k j 2
R (kg/cm2)
fc (kg/cm2)
n
45
fs=1,200 (kg/cm2)
fs=1,500 (kg/cm2)
fs=1,700 (kg/cm2)
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
Design Parameter k and j fc (kg/cm2)
n
fs=1,200 (kg/cm2)
fs=1,500 (kg/cm2)
fs=1,700 (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.
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 From
M = As f s jd
→
M As = fs j d
4) Select steel bars and Detailing
.1 , .2 Number of Bars Bar Dia.
1
2
3
RB6
0.283
0.565
0.848
RB9
0.636
1.27
DB10
0.785
DB12
4
5
6
1.13
1.41
1.70
1.91
2.54
3.18
3.82
1.57
2.36
3.14
3.93
4.71
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
.3 ! " #$%! & ACI Member
Simple One-end Both-ends supported continuous continuous
Cantilever
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
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
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 Steel area:
OK
M 12 . 5 × 10 5 As = = = 10 . 8 cm 2 f s jd 1,700 × 0 . 908 × 75
Select steel bar 4DB20 (As = 12.57 cm2)
Alternative Solution: From Mc = R b d2 = required moment M
bd
2
=
M R
⇒
d =
M Rb
For example M = 12.5 t-m, R = 8.161 ksc, b = 40 cm d =
12 . 5 × 10 5 = 61 . 88 cm 8 . 161 × 40
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 30 cm
Column width 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
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 wall slab w
0.43 0.63 1.24 2.30
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
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) V = 5.75 t (
[email protected] St.)
= 6.29 t As± = 6.39×105/(1,500×0.899×55) = 8.62 cm2
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 +13.81
3.37 +2.15
-16.17 13.65
15.99
2.13
3DB25
4DB25
2DB25
GRASP Version 1.02 B11-B12
Membe r
Mz.i [T-m]
Mz.pos [T-m]
Mz.j [T-m]
Fy.i [Ton]
Fy.j [Ton]
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
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 = Reinforcem ent ratio bd Es 2.04 ×106 134 n= = ≈ Ec 15,100 f c′ f c′
where ρ =
STEP 2 : Resisting Moment Concrete:
1 Mc = f c k j b d 2
Steel:
M s = As f s j d
2
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 40 cm
fc = 65 ksc, fs = 1,700 ksc, n = 10, d = 75 cm
As 12 . 57 ρ= = = 0 . 00419 , ρ n = 0 . 0419 bd 40 × 75
80 cm
k = 2 × 0 . 0419 + ( 0 . 0419 ) 2 − 0 . 0419 4 DB 20 As = 12.57 cm2
= 0 . 251 → j = 1 − 0 . 251 / 3 = 0 . 916
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 - Increase steel area - Enlarge section
When Mreq’d > Mallow
- Double RC only when no choice εc
A’s d’ M
As
εs
ε’s
T’ = A’s f’s 1
C = 2 fc k b d
T = As fs
As1 fs As2 fs
T’ = A’s f’s C=
1
C = 2 fckbd
1 f kbd 2 c
d-d’
jd T = As fs
T1 = As1 fs
Moment strength M = M1 + M2
Steel area
As =
T’ = A’s f’s
T2 = As2 fs
1 M 1 = M c = f c kjbd 2 2 = As1 f s jd Mc As1 = f s jd
+
M2 = M − Mc = As 2 f s (d − d ′) = As′ f s′(d − d ′) M − Mc As 2 = f s (d − d ′)
Compatibility Condition d’ kd
εc
εs d − kd = ε s′ kd − d ′
ε’s
From Hook’s law: εs = Es fs, ε’s = Es f’s
d
εs
. . .
Es f s fs d − kd = = Es f s′ f s′ kd − d ′ k − d′ d f s′ = f s 1− k k − d′ d f s′ = 2 f s 1− k
( A’s ) T’ = A’s f’s
Force equilibrium [ ΣFx=0 ] T’ = T2
d-d’
A’s f’s = As2 fs T2 = As2 fs
Substitute
k − d′ d f s′ = 2 f s 1− k
1 1− k As′ = As 2 2 k − d′ d
( k ) d’ kd d
εc
Compression = Tension
Cc + Cs′ = T
ε’ s Substitute
1 f c b kd + As′ f s′ = As f s 2 k − d′ d As′ f s′ = 2 f s , ρ′ = 1− k bd
εs
As 1− k f s = n fc , ρ= k bd
k =
d′ 2 2 ′ ′ 2n ρ + 2 ρ + n ( ρ + 2 ρ ) − n ( ρ + 2 ρ ′) d
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)
Tension steel As = As1 + As2 = 15.86 + 2.34 = 18.20 cm2 USE 6DB20 (As = 18.85 cm2) Compression steel
As′ =
1 1− k 1 1 − 0.277 = × 2.34 × = 4.02 cm 2 As 2 2 k − d′ d 2 0.277 − 5 / 75
USE 2DB20 (As = 6.28 cm2)
0.80 m
2DB20
6DB20 0.40 m
175 3 3
