Download - Bending in Beam 3
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)