CON4332 REINFORCED CONCRETE DESIGN
│CHAPTER 2│
Design Formulae for Bending
Learning Objectives
 Appreciate the stress-strain properties of concrete and steel for
R.C. design
 Appreciate the derivation of the design formulae for bending
 Apply the formulae to determine the steel required for bending
CONTENTS
2.1
2.1.1
2.1.2
2.1.3
2.2
2.2.1
2.2.2
2.2.3
2.2.4
2.2.5
2.2.6
2.2.7
2.2.8
2.2.9
2.3
2.3.1
1.3.2
2.4
2.4.1
2.4.2
Material Stress-strain Relations
Concrete
Reinforcement
Example – Design Ultimate Capacity for Axial Compression
Design Formulae for Bending
Limit to Neutral Axis
Examples – Effective Depth
Simplified Stress Block
Design Formulae for Singly Reinforced Section
Limits of the Lever Arm
The Balanced Section
Examples – Singly Reinforced Section
Design Formulae for Doubly Reinforced Section
Examples – Doubly Reinforced Section
Flanged Section
Effective Flange Width
Examples – Flanged Section
Limits to Bar Spacing and Steel Ratio
Bar Spacing
Maximum and Minimum Percentage of Steel
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2.1
Material Stress-strain Relations
The stress-strain curve of a material describes the deformation of the
material in response to the load acted upon it. In generalized terms,
deformation is presented in terms of change in length per unit length, i.e.
strain, while load is in terms of force per unit area, i.e. stress.
Strain, Ɛ =
Stress, f =
Deformation
Length
Force
Area
(Dimensionless)
(in N/mm2 or MPa)
The stress-strain curves of concrete and steel provides the fundamental
knowledge to understand the behaviour of the R.C. composite under loads
and for deriving the necessary formulae for R. C. design.
2.1.1
Concrete
Concrete is comparatively very weak in tension. Its tensile strength is
about 1/10 of the compressive strength. It is usually ignored in the design.
Therefore, the stress-strain curve of concrete is usually referring to concrete
under compression. Typical stress-strain curves of concrete are shown in
Figure 2.1 below.
The shape of stress-strain curves varies with the strength of the concrete.
The elastic modulus, i.e. the slope of the initial part of the curves, is higher
for higher strength concrete. In addition, the higher is the strength of the
concrete, the more sudden the failure of the concrete, i.e. more brittle.
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CON4332 REINFORCED CONCRETE DESIGN
Stress
(MPa)
120
80
40
0
0.001
0.002
0.003
0.004
0.005
Strain
Figure 2.1 – Typical Stress-strain Curves of Concrete
In order to facilitate the derivation of design formulae, an idealized
stress-strain curve in parabolic-rectangular shape is adopted in the design
code as shown in Figure 2.2 below.
Figure 2.2 – Idealized Stress-strain Curve of Concrete for Design
(Figure 3.8 of HKCP-2013)
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Take note of the following points:
(a)
The design ultimate strength of concrete is:
0.67 fcu / m = 0.67 fcu / 1.5  0.45 fcu
where
 0.67 is to account for the differences between the testing
condition and the actual effect on concrete in the structure.
 1.5 is the partial factor of safety for the strength of concrete
under bending or axial load.
(b)
The ultimate compressive strain of concrete is:1
Ɛcu = 0.0035
The concrete crushes when it deforms to this value and the failure is
brittle and sudden. This value defines the ultimate limit state, ULS,
of R. C structure.
2.1.2
Reinforcement
Steel is much stronger and more ductile than concrete as illustrated in the
typical stress-strain curves of steel in Figure 2.3.
The initial part of the curve is linear and the slope, i.e. the elastic module, is
constant disregard of the strength. The following value of elastic modulus
is adopted in R.C. design.
Es = 200 kN/m2
1
or 200 000 N/m2
This value is for concrete not higher than Grade C60. Concrete becomes more brittle when its strength is higher,
and therefore the value of Ɛcu is lower.
Details refer to the design code.
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Stress
(MPa)
800
700
0.2% Proof
Stress
600
500
400
Yield Stress
300
200
A typical stress-strain
curve of concrete
100
0
0.002
0.04
0.08
0.12
0.16
0.20
Strain
Figure 2.3 – Typical Stress-strain Curves of Steel
Most of the grades of steel exhibit a definite yield point at which strain
increases suddenly without increase in stress. The stress at this point, i.e.
yield stress, fy, is adopted for design. For steel without yield, 0.2% proof
stress is adopted. Beyond this point, the strength of steel continues to
increase but with substantial increase in strain.
Steel deforms substantially before rupture, and the ultimate strength
increases by more than 8% above the yield, and the ultimate strain is more
than 0.05. This strength hardening and ductility properties render steel a
good structural material.
The tensile and compressive strength properties of steel are the same.
The design code provides an idealized stress-strain curve for design as
given in Figure 2.4.
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Ɛy
Figure 2.4 – Idealized Stress-strain Curve
of Reinforcement for Design
(Figure 3.9 of HKCP-2013)
Take note of the following points:
(a)
The design yield strength is:
fy / m = fy / 1.15 = 0.87 fy
(b)
Within the elastic range, i.e. before yield:
Stress = Es Ɛs
= 200 000 Ɛs
(c)
The yield strain, i.e. beyond which the stress of steel is 0.87fy, is:
Ɛy = 0.87fy / 200 000
For grade 500 steel,
Ɛy = 0.87 x 500 / 200 000
= 0.002175
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2.1.3
Example – Design Ultimate Capacity for Axial Compression
The design ultimate capacity of a concrete section subject to axial
compression is given by Eqn 6.5 of HKCP-20132:
Nuz = Compression Resistance of Concrete
+ Compression Resistance of Steel
= 0.45 fcu Anc + 0.87 fy Asc
where
Anc = Net cross-sectional area of concrete
Asc = Area of steel in compression
Question
Determine the design ultimate capacity for axial compression of the following concrete
section:
Concrete :
Dimensions :
Rebars :
C40
400mm x 400mm
4T25 vertical bars fully restrained by links
Solution
Asc =
=
Anc =
=
Nuz =
2
4 x 491
1 964 mm2
400 x 400 – 1 964
158 036 mm2
0.45 fcu Anc + 0.87 fy Asc
=
(0.45 x 40 x 158036 + 0.87 x 500 x 1964) x 10-3
=
2845 + 854
=
3 699 kN
The application of this equation is subject to the following conditions: (i) the column is subject to axial load only,
without eccentricity and moment, (ii) the rebars restrained from buckling, and (iii) the column is not slender.
Design of column will be discussed in another chapter.
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2.2
Design Formulae for Bending
When a beam is under downward bending as shown in Figure 2.5 below, the
upper part of the beam is in compression and the lower part is in tension. If
the plane section remains plane after deformation as shown in figure (a), the
strain distribution will be linear as shown in figure (b), with zero strain at the
neutral axis and increasing linearly outward towards the top and bottom
fibres of the section.3
a
d
b
c
Elevation of a Beam under Load
Ɛcc
a
d
x
M
M
Neutral
Axis
d
Rebar
c
b
Ɛst
(a) Deformation of a-b-c-d
(b) Strain
(c) Stress (Elastic)
(d) Stress (Plastic)
Figure 2.5 – Distribution of Stresses and Strains
across a Beam Section
When the load is small and the material is still linear elastic, the stress will be
in linear proportion to the strain. The distribution of the compressive stress
above the neutral axis is then in triangular shape as shown in figure (c).
The concrete below the neutral axis is assumed unable to take up any
tensile stress, and rebars are provided to take up the tension.
3
The "plane section remains plane" assumption is usually valid in beam design, except under some
circumstances, for examples, deep beam with span-to-depth ratio is smaller than 4, at section under very high
shear force, etc. It is out of the scope of this chapter.
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If the load is further increased until the section become plastic, the
compressive stress block of concrete will become parabolic in shape and the
tension steel become yielded as shown in figure (d).
2.2.1
Limits to the Neutral Axis
Considering the compatibility of strain in figure (b) above, and assuming
there no slip at the interface of concrete and steel bar, the relationship of the
maximum strain of concrete compression, Ɛcc, and the strain of steel in
tension, Ɛst, is given by:
Ɛst =
Ɛcc (d – x)
x
where
d = The effective depth of the section. It is the depth
measured from the top of the section (for sagging
moment) to the centroid of the tension reinforcement.
x = The depth of the neutral axis, above which (for sagging
moment) the section is in compression while below
which the section cracks under tension.
In order to ensure ductility, it is desirable to have the tension reinforcement
yielded before the concrete crushes. That is
Ɛst ≥ 0.002175
Therefore,
when
0.0035 (d – x)
x
Re-arranging, it becomes
Ɛcc = 0.0035
≥ 0.002175
x ≤ 0.617 d
In other words, the section should be designed such that depth of neutral
axis should not exceed the limit to ensure ductility.
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CON4332 REINFORCED CONCRETE DESIGN
HKCP-2013 limits the depth of neutral axis to:4
x ≤ 0.5 d
[2.1]
If moment redistribution is more than 10%, i.e. βb < 0.9, the depth of neutral
axis is limited to:
x≤
(βb - 0.4) d
[2.1a]
For example, if βb = 0.8, the limiting neutral axis is 0.4d.
2.2.2
Examples – Effective Depth
Question A
Determine the effective depth of the following section:
Overall beam depth, h =
Concrete cover =
Size of link:
Bottom bars:
500 mm
40 mm
d
h
10
3T32 in one layer
3T32
Solution
Effective Depth, d =
=
500 – 40 – 10 – 32/2
434 mm
d
Bar size / 2
Link size
Cover
Question B
Determine the effective depth of the following section:
Overall beam depth, h =
Concrete cover =
Size of link:
Bottom bars:
650 mm
45 mm
12
h
d
2T32
2T40 + 3T32 in two layers
2T40 + T32
4
This limit is for concrete not higher than Grade C45.
For higher grade of concrete the limit is more stringent.
Details refer to the design code.
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CON4332 REINFORCED CONCRETE DESIGN
Solution
The clear spacing between two layers of bars should not be less than (vide Cl.8.2 of
HKCP-2013):
(a)
maximum bar size
(b)
aggregate size + 5 mm
(c)
20 mm
In this case, the maximum bar size controls, i.e. 40mm.
Distance to the bottom T40 =
650 – 45 – 12 – 20 = 573 mm
Distance to the bottom T32 =
650 – 45 – 12 – 16 = 577 mm
Distance to the 2nd layer T32 =
Effective Depth, d =
=
650 – 45 – 12 – 40 – 40 - 16 = 497 mm
1257 x 2 x 573 + 804 x 577 + 804 x 2 x 497
1257 x 2 + 804 x 3
549 mm
Alternatively, the effective depth is simply taken to the "center", instead of the centroid, of
the two layers of rebars as follows.
Effective depth, d =
=
650 – 45 – 12 – 40 – 20
533 mm (the deviation is about 2.9% only)
Unless rigorous checking is required, this method is in general acceptable for manual
calculation in design office.
In fact, during the initial design stage, the amount steel
required is unknown. Assumption has to be made on the bar size, based on which to
estimate the effective depth for calculating the steel required and then the number and size
of bars. Once the bar size is known, the initial assumption on effective depth has to
verified. If the initial assumption is on conservative side and does not deviate too much
from actual value, the result will then be treated as acceptable and the calculation would not
be re-done.
?
Q.1 – Q.4
2.2.3
Simplified Stress Block
After the steel has yielded, the beam continues to deform until the top
concrete crushes at the ultimate strain, Ɛcu, and the distribution of
compressive stress in the compression zone, i.e. above the neutral axis, will
then be in the shape of rectangular-parabolic as shown in (b) of Figure 2.6
below. In order to make it more manageable in deriving the design formula
for bending, a simplified rectangular stress block as shown figure (c) of
Figure 2.6 is adopted (Figure 6.1 of HKCP-2013).
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Ɛcu
0.45fcu
0.45fcu
s/2
s
x
Neutral
Axis
d
z
Ɛst
Section
(a)
Strain at Ultimate
Limit State
(b)
Parabolic Stress
Block
(c)
Simplified Rectangular
Stress Block
Figure 2.6 – Stress and Strain Distribution at Ultimate Limit State
In the simplified stress block, a uniform compressive stress of
0.67 fcu / m = 0.45fcu is adopted over a depth of:5
s = 0.9 x
and, the lever arm, z, between the centroid of the compression force in the
concrete and the tension force of rebars is:
z = d – s/2
Rearranging,
2.2.4
s = 2(d – z)
Design Formulae for Singly-Reinforced Section
The objective of the design formulae is to determine the steel area, As, with
the following information given:
The design ultimate moment: M
5
It is for concrete not higher than Grade C45. The value of "s" is smaller for higher grade of concrete.
Details
refer to the design code.
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CON4332 REINFORCED CONCRETE DESIGN
Grade of concrete : fcu
Grade of steel : fy
Breadth of section : b
Effective depth : d
0.45fcu
s = 0.9 x
b
x
d
Neutral
Axis
s/2
Fcc
z
M
Fst
As
Simplified Rectangular
Stress Block
Section
Figure 2.7 – Simplified Stress Block for R.C. Design
Compression in concrete
Fcc = 0.45 fcu (b s)
= 0.9 fcu b (d - z)
Tension in the rebar
Fst = 0.87 fy As
Take moment about the rebar, and by equilibrium of moment:
M = Fcc z
= 0.9 fcu b (d - z) z
Divide both sides by (bd2fcu), and
let
K = M/(bd2fcu)
[2.2]
the equation becomes:
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CON4332 REINFORCED CONCRETE DESIGN
(z/d)2 – (z/d) + K/0.9 = 0
Solve for z/d, the level arm factor:
z/d = [0.5 + (0.25 – K/0.9)0.5]
z = [0.5 + (0.25 – K/0.9)0.5] d
and then, the level arm:
[2.3]
This is Eqn 6.10 of HKCP-2013.
Take moment about the centroid of the compression force, and by
equilibrium of moment:
M = Fst z
= 0.87 fy As z
Rearranging,
As = M / (0.87 fy z)
[2.4]
This is Eqn 6.12 of HKCP-2013.
2.2.5
Limits of the Lever Arm, z
The limit to the depth of neutral axis, i.e. x ≤ 0.5 d, imposes a lower limit to
the lever arm6:
z ≥ d – 0.9(0.5d) / 2
z ≥ 0.775 d
[2.5]
If moment redistribution is more than 10%, i.e. βb < 0.9
z ≥ (1.18 – 0.45βb) d
6
This limit is for concrete not higher than Grade 45.
[2.5a]
This limit is more stringent for higher grade of concrete.
Refer to the design code for details.
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In addition, the design code also provides an upper limit to the lever arm (Cl.
6.1.2.4 (c) of HKCP2013):
z ≤ 0.95 d
When putting z = 0.95d into the equation of lever arm [2.3], we can find that
the corresponding value of K is 0.0428. In other words,
If K ≤ 0.0428
2.2.6
z = 0.95 d
[2.6]
The Balanced Section, K'
If the amount of reinforcement is provided such that the depth of the neutral
axis is just at the limit of 0.5d, the section will then fail by crushing of the
concrete immediately after the steel has yielded. This beam section is
called a balanced section. The corresponding level arm is 0.775d.
Putting this value into the equation of equilibrium of moment about the
tension force, the moment of resistance of the balanced section is:
Mbal = Fcc z
= 0.9 fcu b (d – 0.775d) 0.775d
Therefore,
and,
Mbal = 0.157 fcu bd2
K' = Mbal / (bd2 fcu) = 0.157
If the design moment of a section is larger than Mbal, it will fail by crushing of
concrete before yielding of steel no matter how much tension steel is
provided, that is undesirable; unless compression steel is provided as
discussed in 2.2.8 below.
In other words, if a section is over-reinforced, the neutral axis will exceed the
upper limit leading to failure without ductility, though the moment capacity is
increased, as illustrated in following figure.
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Moment
CON4332 REINFORCED CONCRETE DESIGN
As > As,bal (w/o compression steel)
Mbal
As  As,bal
As < As,bal
0
Deflection
HKCP-2013 specifies the K-value for the balanced section as:7
K' = 0.156
(for βb ≥ 0.9)
[2.7]
If moment redistribution is more than 10%, i.e. βb < 0.9, the value of K' is
reduced to:
K' = 0.402(βb – 0.4) - 0.18(βb – 0.4)2
[2.7a]
For example, if βb = 0.8, K' = 0.132.
2.2.7
Examples – Singly Reinforced Section
In summary, the procedures to determine the area of tension steel are:
7
1. Calculate the K value
K = M / (bd2 fcu)
2. Check balanced section
K < K'
3. Calculate the lever arm
z = [0.5 + (0.25 – K/0.9)0.5] d
It is for concrete not higher than Grade C45.
The value of K' is smaller for higher grade of concrete.
Details
refer to the design code.
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CON4332 REINFORCED CONCRETE DESIGN
4. Check
z / d ≤ 0.95
5. Calculate the steel area
As = M / (0.87 fy z)
Question A
Determine the rebars for the following beam section:
Design ultimate moment, M =
350 kN-m
Breadth, b =
350 mm
Effective depth, d =
480 mm
Concrete, fcu =
35 MPa
Steel, fy =
βb =
1.0
500 MPa
Solution
K=
βb = 1.0
M / (bd2 fcu)
=
350 x 106 / (350 x 4802 x 35)
=
0.124
<
0.156
Lever arm, z =
[0.5 + (0.25 – K/0.9)0.5] d
=
[0.5 + (0.25 – 0.124/0.9)0.5] x 480
=
0.835 x 480
=
401 mm
Tension steel req'd, As =
(Singly reinforced)
M / (0.87 fy z)
=
350 x 106 / (0.87 x 500 x 401)
=
2 006 mm2
(Provide 2T32 + 1T25)
As,pro =
=
2 x 804 + 491
2099 mm2
Question B
For the section in Example A, find the moment of resistance of the balanced section and the
corresponding amount of steel.
Solution
Mbal =
K' fcu bd2
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=
0.156 x 35 x 350 x 4802 x 10-6
=
440 kN-m
z=
0.775 d
=
0.775 x 480
=
372 mm
As =
M / (0.87 fy z)
=
440 x 106 / (0.87 x 500 x 372)
=
2719 mm2
Question C
Determine the ultimate moment of resistance of the following beam section:
Breadth, b
= 350 mm
Effective depth, d
= 480 mm
Area of steel provided, As
Concrete, fcu
Steel, fy
= 2412 mm2 (i.e. 3T32)
= 35 MPa
= 500 MPa
Solution
(The design formulae in the design
0.45fcu
the determination of steel area from
x
the design moment. This example,
however, requires you to work back
N.A.
section from the steel area provided.)
For equilibrium of compression and tension forces:
0.45 fcu (b s) =
s=
Simplified Rectangular
Stress Block
Fst
0.87 fy As
(assuming steel has yielded)
0.87 fy As / (0.45 fcu b)
=
190.3 mm
s / 0.9
190.3 / 0.9
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M
Fst
0.87 x 500 x 2412 / (0.45 x 35 x 350)
=
Fcc
z
=
Depth of neutral axis, x =
s/2
d
the ultimate moment capacity of the
Fcc =
s = 0.9 x
code are given in the form to facilitate
18
CON4332 REINFORCED CONCRETE DESIGN
=
211.5 mm
<
0.5 x 480 = 240 mm
Lever arm, z =
(steel has yielded as assumed)
d – s/2
=
480 – 190.3 / 2
=
384.9 mm
Take moment about the centroid of compression zone:
Moment of resistance, M =
Fst z
=
0.87 fy As z
=
0.87 x 500 x 2412 x 384.9 x 10-6
=
404 kN-m
?
Q.5 – Q.8
2.2.8
Design Formulae for Doubly Reinforced Section
When the design moment becomes so large that
K > K',
compression reinforcement is required to provide additional compressive
resistance in the compression zone of the section. If the area of
compression steel, A's, is located at d' from the top of the section as shown
in Figure 2.8, the formulae for determining As and A's are derived as follows.
The neutral axis cannot be further lowered.
to retain ductility. Therefore,
It remains at the limiting depth
z = (1.18 – 0.45βb) d
Fcc z = K' fcu bd2
For βb < 0.9,
z = 0.775 d
Fcc z = 0.156 fcu bd2
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CON4332 REINFORCED CONCRETE DESIGN
Ɛcu
0.45fcu
Ɛsc
d'
s
x
M
z = d – s/2
d
Fcc
Neutral
Axis
As
d – d'
As '
Fsc
Fst
Ɛst
Section
(a)
Strain at Ultimate Limit
State
(b)
Simplified Rectangular
Stress Block
Figure 2.8 – Stress and Strain Distribution
for Doubly Reinforced Section
Compression in the compression reinforcement:
Fsc = fsc A's
where
fsc = stress in the compression steel
Take moment about the tension steel, and by equilibrium of moment:
M = Fcc z + Fsc (d - d')
= 0.156 fcu bd2 + fsc A's (d - d')
Rearranging, it becomes
A's =
M - 0.156 fcu bd2
fsc (d - d')
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CON4332 REINFORCED CONCRETE DESIGN
or
A's =
(K – K') fcu bd2
[2.8a]
fsc (d - d')
By equilibrium of forces
Fst = Fcc + Fsc
0.87 fy As = 0.156 fcu bd2 / z + fsc As'
Rearranging, it becomes
As =
or
As =
0.156 fcu bd2
0.87 fy z
K' fcu bd2
0.87 fy z
+
As'
+
As'
fsc
0.87 fy
fsc
0.87 fy
[2.9a]
The value of fsc can be determined from the strain distribution in Figure 2.8(a)
above, that is
Ɛsc / (x - d') = Ɛcu / x
Rearranging,
d' / x = 1 - Ɛsc / Ɛcu
If Ɛsc > 0.002175 at Ɛcu = 0.0035, the compression steel has yielded at the
ultimate limit state, i.e. fsc = 0.87 fy :
In other words, if
d' / x < 1 – 0.002175 / 0.0035 = 0.38
A's=
and,
As =
(K – K') fcu bd2
[2.8]
0.87 fy (d - d')
K' fcu bd2
0.87 fy z
[2.10]
+
A's
[2.9]
These are Eqns 6.14 and 6.15 in HKCP-2013.
On the other hand,
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CON4332 REINFORCED CONCRETE DESIGN
if
d' / x > 0.38
the compression bars are so close to the neutral axis that they has not
yielded and the stress in the compression bars has to be calculated by:
fsc = Es Ɛsc
2.2.9
Examples – Doubly Reinforced Section
In summary, the procedures to determine the steel areas for doubly
reinforced section are (for βb > 0.9):
1. Provide comp'n steel, if
K > 0.156
2. Calculate the lever arm
z = 0.775d
3. Calculate the neutral axis
x = 0.5d
4. Check
d' / x ≤ 0.38
5. Calculate compression steel
6. Calculate tension steel
A's=
As =
to ensure rebar has yielded
(K – K') fcu bd2
0.87 fy (d - d')
K' fcu bd2
0.87 fy z
+
A's
Question A
Determine the steel required for the following beam section:
Design ultimate moment, M =
500 kN-m
Breadth, b =
350 mm
Effective depth of tension steel, d =
480 mm
Effective depth to comp'n steel, d' =
70 mm
Concrete, fcu =
Steel, fy =
βb =
1.0
35 MPa
500 MPa
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Solution
K=
βb =1.0
M / (bd2 fcu)
=
500 x 106 / (350 x 4802 x 35)
=
0.177
>
0.156
Lever arm, z =
0.775 d
=
0.775 x 480
=
372 mm
Depth to neutral axis, x =
0.5 d
=
0.5 x 480
=
240 mm
Check d' / x =
70 / 240
=
Compression steel req'd, A's =
=
=
(Compression steel is required)
0.29 < 0.38
(fsc = 0.87fy)
(K – K') fcu bd2
0.87 fy (d - d')
(0.177 – 0.156) x 35 x 350 x 4802
0.87 x 500 x (480 – 70)
332 mm2
(Provide 2T16 Top Bars)
A's,pro =
=
Tension steel req'd, As =
=
2 x 201
402 mm2
K' fcu bd2
0.87 fy z
+
As'
0.156 x 35 x 350 x 4802
0.87 x 500 x 372
=
2721 + 332
=
3053 mm2
+
332
(Provide 4 T32 Bottom Bars)
As, pro =
=
4 x 804
3216 mm2
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CON4332 REINFORCED CONCRETE DESIGN
Question B
Determine the strain and stress of the compression reinforcement for a doubly reinforced
concrete section with the following information:
d=
350 mm
d' =
70 mmm
Ɛcu
Ɛsc
d'
x
Solution
For doubly reinforced section,
x=
0.5 d
=
0.5 x 350
=
175 mm
d
Neutral
Axis
d' / x =
=
70 / 175
0.40 > 0.38
(Compression steel has not yielded)
Refer to the strain diagram,
the relationship of the strain of concrete and the strain of steel:
Ɛsc / (x - d') =
Ɛcu / x
Ɛsc =
0.0035 x (1 – d'/x)
=
0.0035 x (1- 0.40)
=
0.0021
Stress of comp'n steel, fsc =
Es Ɛs
=
200 000 x 0.0021
=
420 N/mm2
?
Q.9 – Q.13
2.3
Flanged Section
Reinforced concrete beams are usually constructed monolithically with the
floor slab, and therefore they will act in integral to resist sagging moment as
shown in the following figure. The slab acts as the top flange of the beam
to share the flexural compressive stress. As the slab is much wider than
the breadth of the beam, the compressive zone can be achieved by a much
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CON4332 REINFORCED CONCRETE DESIGN
shallower neutral axis, which, in most circumstances, falls within the flange
without trespassing into the web of the beam. On the other hand, the
flange does not assist in resisting the hogging moment at the supports,
where the compression zone is at the bottom of the section.
Figure 2.9 – Slab Acting as the Flange of a Beam (Isometric View)
(For clarity of illustration, the main beams supporting beam A-B are not shown)
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CON4332 REINFORCED CONCRETE DESIGN
beff
beff
Compression
hf
d
Tension
bw
bw
T – Section
L – Section
h
Figure 2.10 – Flanged Sections
2.3.1
Effective Flange Width
The flexural compressive stress in the flange is assumed to be uniformly
distributed over an effective width, beff, which depends on the dimensions of
the beam & slabs and the length of the sagging moment, lpi, which may be
obtained from the following figures.
Notes:
(a)
The length of the cantilever, l3, should be less than half the adjacent span.
(b)
The ratio of the adjacent spans should lie between 2/3 and 1.5.
Figure 2.11 – Definition of lpi
for Calculation of Effective Flange Width
(Figure 5.1 of HKCP-2013)
For simply-supported beam, the whole span is under sagging moment, and
therefore, lpi = effective span of the simply-supported beam (L).
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CON4332 REINFORCED CONCRETE DESIGN
c/c distance of adjacent slabs
Figure 2.12 – Effective Flange Width
(Figure 5.2 of HKCP-2013)
The design of flanged section for bending can be simply treated as the
design of rectangular section by putting:
b = beff
provided that the neutral axis is within the flange, i.e.
x ≤ hf
or
(d – z) / 0.45≤ hf
[2.11]
where hf is the thickness of the flange, i.e. the slab.
2.3.2
Examples – Flanged Section
Question A
Determine the effective flange width for an interior span of a continuous beam with
approximately equal spans with the following information:
Breadth, bw =
350 mm
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CON4332 REINFORCED CONCRETE DESIGN
Effective span, L =
Clear spacing btw adjacent beams =
6 700 mm
2 500mm (same on both sides)
Solution
Internal span of ctu beam
lpi =
0.7 x 6700
=
4 690 mm
b 1 = b2 =
2 500 / 2
=
1250 mm
beff,1 = beff,2 =
Min (0.2x1250+0.1x4690 or 0.2x4690 or 1250)
=
Min (719 or 938 or 1250)
=
719 mm
beff =
=
2 x 719 + 350
1788 mm
Note: As 719 > 0.1 x 4690, the shear stress between the web and flange has to be checked,
i.e. Note 1 of Figure 2.12, which is outside the scope of this chapter and is ignored for the
purpose of this course.
Question B
Determine the effective flange width for the following simply-supported beam:
Breadth, bw =
Effective span, L =
Clear spacing btw adjacent beams =
300 mm
9 000 mm
2 000mm (same on both sides)
Solution
Simply supported beam
lpi =
b 1 = b2 =
=
beff,1 = beff,2 =
9000
2000 / 2
1000 mm
Min (0.2x1000+0.1x9000 or 0.2x9000 or 1000)
=
Min (1100 or 1800 or 1000)
=
1000 mm
beff =
=
2 x 1000 + 300
2300 mm
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CON4332 REINFORCED CONCRETE DESIGN
Question C
Determine the steel required for the following beam section:
Design ultimate moment, M =
500 kN-m (sagging)
Breadth, bw =
350 mm
Slab thickness, hf =
150 mm
Effective flange width, beff =
1780 mm
Effective depth of tension steel, d =
480 mm
Concrete, fcu =
35 MPa
Steel, fy =
500 MPa
b = beff =
1780mm
Solution
K=
K < 0.0428
M / (bd2 fcu)
=
500 x 106 / (1780 x 4802 x 35)
=
0.0348
<
0.156
Lever arm, z =
0.95d
Check
=
0.95 x 480
=
456 mm
x=
(No compression steel required)
(480 – 456) / 0.45
=
53 < 150 mm
Tension steel req'd, As =
M / (0.87 fy z)
(N.A. is within the flange)
=
500 x 106 / (0.87 x 500 x 456)
=
2 521 mm2
(Provide 2T40 + 1T20)
As,pro =
=
2 x 1257 + 314
2828 mm2
This example is similar to Question A in 2.2.9 except that it is a flanged section. As the
value of b in this case is 5 times that of the rectangular section in the previous example, the
value of K is therefore reduced also by almost 5 times and becomes much smaller than K'
and, as a result, the upper bound value of z is adopted. When the total steel area required
(2521 mm2) is compared with that required for rectangular section (322 + 3043 = 3365 mm2),
?
there is a saving of 25%.
Q.14 – Q.17
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CON4332 REINFORCED CONCRETE DESIGN
2.4
Limits to Bar Spacing and Steel Ratio
There are lower and upper limits to the amount of steel and the spacing
between bars in reinforced concrete. The lower bound is to prevent
unsightly cracking due to shrinkage, temperature effect, restrained action
and brittle failure. On the other hand, the upper bound is to prevent
congestion of reinforcement bars that would affect the proper compaction of
concrete.
2.4.1
Bar Spacing
Adequate clear spacing should be provided
between bars such that concrete can be placed
and compacted satisfactorily around the bars.
The clear distance (horizontal and vertical)
between individual or horizontal layers of parallel
bars should not be less than (Cl.8.2 of
HKCP-2013):
i.
ii.
iii.
maximum bar size
aggregate size + 5 mm
20 mm
On the other hands, reinforcement bars cannot be placed too far apart; they
have to be placed close enough to distribute the cracks on the surface of the
concrete element. The maximum spacing of the bars is determined by the
service stress in the rebars, their distance from the concrete surface and the
thickness of the concrete element. Detailed requirements for beams and
slabs can be found in Cl.9.2.1.4 and Cl.9.3.1.1 of HKCP-2013. They will not
be covered in this chapter. For simplicity, the following rules of thumb can
be adopted for preliminary design:

For beam, the maximum bar spacing requirement can in general be
complied with by providing one bar for every 100 to 150 mm width of the
beam.
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CON4332 REINFORCED CONCRETE DESIGN
Example
For a beam of 400 mm wide, provide 3 to 4 bars at the outer layer depending on the size
of the bars.

2.4.2
For slab, under most circumstances, limit the spacing of main bars to not
more than 2h or 250 mm whichever is lesser.
Maximum and Minimum Percentage of Steel
The maximum and minimum limits for Grade 500 steel commonly used in
R.C. design are summarized in the table below:8
Elements
Minimum (%)
Maximum (%)
Beam
Flexural tension steel
Rectangular section
0.13
Flanged section (bw/b < 0.4)
0.18
4
Flexural compression steel
Rectangular section
Column
0.20
0.80
6
Vertical bars
0.40
4
Horizontal bars
0.25
Wall
Table 2.1 – Minimum and Maximum Percentage
of Reinforcement (Grade 500)
In calculating the steel ratio for the above table, the gross area of the
concrete, Ac, is adopted.
8
There are more specific requirements on the steel ratios, like compression steel in the flange of flanged beam,
steel area at the lapping of rebars, cantilever slab, ductility requirements for members resisting lateral load, etc.
Refer to the design code for details.
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CON4332 REINFORCED CONCRETE DESIGN
For rectangular section,
For flanged section,
Ac = bh
Ac = bwh
Overall depth, h,
instead of effective
depth, d, is used to
Example
check steel ratio.
For a beam of 600 (h) x 300 (b), the minimum flexural tension
steel is 0.13 x 600 x 300 / 100 = 234 mm2.
?
Q.18 – Q.19
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CON4332 REINFORCED CONCRETE DESIGN
│Key Concepts/Terms│
Design Ultimate Strength of Concrete
Ultimate Strain of Concrete
Design Yield Strength of Steel
Effective Depth
Depth of Neutral Axis and Its Limit
0.45 fcu
Ɛcu = 0.0035
087fy
d
x < 0.5d
K and Balanced Section
K < 0.156
Lever Arm and Its Limits
0.775d < z < 0.95d
Effective Flange Width
beff
Maximum and Minimum Steel Ratios
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CON4332 REINFORCED CONCRETE DESIGN
│Self-Assessment Questions│
Q.1
Find the effective depth, d, of the following beam section.
Overall beam depth, h =
Concrete cover =
Q.2
A.
510 mm
B.
530 mm
C.
265 mm
D.
550 mm
40 mm
Referred size of link:
10
Preferred size of main bars:
40
Find the effective depth, d, of the following beam section.
Overall beam depth, h =
Concrete cover =
Q.3
600 mm
A.
557 mm
B.
541 mm
C.
525 mm
D.
312.5 mm
625 mm
Referred size of link:
40 mm
Main bars:
12
2 layers of 32
Find the effective depth, d, of the following slab section.
Overall slab thickness, h =
Concrete cover =
A.
159 mm
B.
138 mm
C.
144 mm
D.
150 mm
175 mm
25 mm
Preferred size of main bars:
12
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CON4332 REINFORCED CONCRETE DESIGN
Q.4
Determine the allowable depth of the neutral axis, x, of the following rectangular beam
section.
Overall beam depth, h =
Concrete cover =
βb =
Q.5
A.
322 mm
B.
300 mm
C.
539 mm
D.
269.5 mm
600 mm
35 mm
Referred size of link:
10
Preferred size of main bars:
32
1.0
Determine the K value for the following rectangular beam section.
Design ultimate moment, M =
A.
0.142
B.
0.108
C.
1.415
D.
0.124
350 kN-m
Breadth, b =
300 mm
Effective depth, d =
454 mm
Concrete, fcu =
40 MPa
βb =
Overall depth, h =
1.0
520mm
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CON4332 REINFORCED CONCRETE DESIGN
Q.6
Determine the amount of tension steel, As, required for the following rectangular beam
section.
Design ultimate moment, M =
Q.7
A.
3212 mm2
B.
2320 mm2
C.
1856 mm2
D.
2134 mm2
422 kN-m
βb =
Breadth, b =
325 mm
Effective depth, d =
Concrete, fcu =
40 MPa
Steel, fy =
1.0
534 mm
500 MPa
Determine the amount of tension steel, As, required for the following rectangular beam
section.
Design ultimate moment, M =
Q.8
A.
701 mm2
B.
2320 mm2
C.
1856 mm2
D.
693 mm2
153 kN-m
βb =
Breadth, b =
350 mm
Effective depth, d =
Concrete, fcu =
30 MPa
Steel, fy =
1.0
534 mm
500 MPa
Determine the amount of tension steel, As, required for the following rectangular beam
section.
Design ultimate moment, M =
Breadth, b =
Concrete, fcu =
A.
372 mm2
B.
366 mm2
C.
327 mm2
D.
701 mm2
26 kN-m
1000 mm
30 MPa
βb =
Effective depth, d =
Steel, fy =
0.8
169 mm
500 MPa
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CON4332 REINFORCED CONCRETE DESIGN
Q.9
Determine the lever arm, z, for the following rectangular beam section.
K=
Q.10
A.
462 mm
B.
426 mm
C.
317 mm
D.
491 mm
0.178
βb =
Breadth, b =
350 mm
Effective depth, d =
Concrete, fcu =
30 MPa
Steel, fy =
1.0
634 mm
500 MPa
Determine the compression steel, A's, required for the following rectangular beam section.
K=
A.
3617 mm2
B.
642 mm2
C.
4776 mm2
D.
491 mm2
0.188
βb =
Breadth, b =
350 mm
Overall depth, h =
Effective depth, d =
634 mm
d' =
Concrete, fcu =
35 MPa
Steel, fy =
1.0
700 mm
70 mm
500 MPa
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CON4332 REINFORCED CONCRETE DESIGN
Q.11
Determine the steel required for the following rectangular beam section.
Design ultimate moment, M =
Q.12
βb =
Breadth, b =
350 mm
Overall depth, h =
Effective depth, d =
505 mm
d' =
Concrete, fcu =
40 MPa
Steel, fy =
A.
A's = 563 mm2 and As = 3525 mm2
B.
A's = 563 mm2 and As = 4483 mm2
C.
A's = 666 mm2 and As = 3272 mm2
D.
A's = 666 mm2 and As = 3938 mm2
1.0
575 mm
70 mm
500 MPa
Determine the steel required for the following rectangular beam section.
Design ultimate moment, M =
Q.13
683 kN-m
766 kN-m
βb =
Breadth, b =
400 mm
Overall depth, h =
Effective depth, d =
475 mm
d' =
Concrete, fcu =
40 MPa
Steel, fy =
A.
A's = 1151 mm2 and As = 4668 mm2
B.
A's = 1151 mm2 and As = 3516 mm2
C.
A's = 666 mm2 and As = 3272 mm2
D.
A's = 666 mm2 and As = 3938 mm2
1.0
545 mm
70 mm
500 MPa
Determine the lever arm, z, for the following rectangular beam section.
K=
A.
462 mm
B.
520 mm
C.
317 mm
D.
491 mm
0.150
βb =
Breadth, b =
350 mm
Effective depth, d =
Concrete, fcu =
30 MPa
Steel, fy =
0.8
634 mm
500 MPa
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CON4332 REINFORCED CONCRETE DESIGN
Q.14
Determine the effective flange width, beff, for the end span of a continuous beam with
approximately equal spans with the following information:
Breadth, bw =
Effective span, L =
Clear distance between adjacent beams =
Q.15
A.
2918 mm
B.
2600 mm
C.
2129 mm
D.
1829 mm
300 mm
7 700 mm
2 600mm (same on both sides)
Determine the effective flange width, beff, for the following simply-supported beam:
Breadth, bw =
Effective span, L =
c/c distance between adjacent beams =
A.
1395 mm
B.
2290 mm
C.
2540 mm
D.
2750 mm
250 mm (same for adjacent beams)
8 700 mm
3000mm (same on both sides)
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CON4332 REINFORCED CONCRETE DESIGN
Q.16
Determine the lever arm, z, for the following flanged beam section.
Design ultimate moment, M =
Q.17
A.
457 mm
B.
355 mm
C.
376 mm
D.
461 mm
βb =
Breadth, bw =
400 mm
Effective flange width, beff =
Effective depth, d =
485 mm
hf =
Concrete, fcu =
40 MPa
Steel, fy =
1.0
2150 mm
200 mm
500 MPa
Determine the steel area required for the following flanged beam section.
Design ultimate moment, M =
Q.18
666 kN-m (sagging)
668 kN-m (sagging)
βb =
Breadth, bw =
350 mm
Effective flange width, beff =
Effective depth, d =
486 mm
hf =
Concrete, fcu =
40 MPa
Steel, fy =
A.
A's = 0 mm2 and As = 3326 mm2
B.
A's = 0 mm2 and As = 4077 mm2
C.
A's = 823 mm2 and As = 4900 mm2
D.
A's = 823 mm2 and As = 3326 mm2
1.0
2050 mm
190 mm
500 MPa
In the design of a 600mm (h) x 400mm (b) reinforced concrete beam, if the amount of steel
required for resisting the design moment by the rectangular section is found to be 210 mm2,
which of the following reinforcement is most appropriate?
A.
2T12
B.
2T16
C.
4T12
D.
1T20
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CON4332 REINFORCED CONCRETE DESIGN
Q.19
In the design of a 400mm x 400mm reinforced concrete column, if the amount of steel
required for resisting the design axial force is found to be 1250 mm2, which of the following
reinforcement is most appropriate?
Q.20
A.
4T20
B.
3T25
C.
4T25
D.
12T32
Which of the following is the most appropriate rebars for the flanged beam under sagging
moment as described below?
Breadth, bw =
600 mm
Effective flange width, beff =
Effective depth, d =
350 mm
hf =
Overall depth, h =
400 mm
The amount of bottom steel required to resist the sagging moment =
A.
2T20
B.
2T16
C.
4T12
D.
5T12
2050 mm
150 mm
380 mm2
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CON4332 REINFORCED CONCRETE DESIGN
Answers:
Q1
B
d = 600 – 40 – 10 – 40/2 = 530 mm
Q2
C
d = 625 – 40 – 12 – 32 – 32/2 = 525 mm
Q3
C
d = 175 – 25 – 12/2 = 144 mm
Q4
D
max x = 0.5d = 0.5 x (600 – 35 -10 – 32/2) = 269.5 mm
Q5
A
K = 350 x 106 / (300 x 4542 x 40) = 0.142
Q6
D
K = 0.114; z = 0.851 x 534 = 454.7mm; As = 422 x 106 / (0.87 x 500 x 454.7) = 2134 mm2
Q7
A
K = 0.0511; z = 0.940 x 534 = 501.7mm; As = 153 x 106 / (0.87 x 500 x 501.7) = 701 mm2
Q8
A
K = 0.0303; z = 0.95 x 169 = 160.6mm, As = 26 x 106 / (0.87 x 500 x 160.6) = 372 mm2
Q9
D
K = 0.178 > 0.156, therefore, z = 0.775 x 634 = 492 mm
Q10
B
As' = (0.188-0.156)x35x350x6342 / (0.87x500x(634-70)) = 643 mm2
Q11
D
K = 0.1913; As' = (0.1913-0.156)x40x350x5052 / (0.87x500x(505-70)) = 666 mm2; As =
0.156x40x350x5052 / (0.87x500x0.775x505) + 666 = 3938 mm2
Q12
A
K = 0.2122; As' = (0.2122-0.156)x40x400x4752 / (0.87x500x(475-70)) = 1151 mm2; As =
0.156x40x400x4752 / (0.87x500x0.775x475) + 1151 = 4668 mm2
Q13
B
βb = 0.8 < 0.9, max x = (βb – 0.4)d = 0.4 x 634 = 253.6; z = 634 – 0.45 x 253.6 = 520mm
Q14
C
bi = 2600/2 = 1300mm; lpi = 0.85x7700 = 6545mm; beff = 2 x min(0.2x1300+0.1x6545, 0.2x6545,
1150) + 300 = 2129 mm
Q15
C
bi = (3000-250)/2 = 1375mm; lpi = 8700mm; beff = 2 x min(0.2x1375+0.1x8700, 0.2x8700, 1150) +
250 = 2540 mm
Q16
D
K = 0.0329 < 0.0428; z = 0.95 x 485 = 461 mm
Q17
A
K = 0.0345 < 0.0428; z = 0.95 x 486 = 461.7 mm; As = 668 x 106 / (0.87 x 500 x 461.7) = 3326 mm2
Q18
C
Min As = 0.13x600x400/100 = 312 > 210 mm2; preferably provide 4 bars over 400mm; therefore
use 4T12, As = 452 mm2
Q19
C
(A) 100As/bh = 0.785 < 0.8 unacceptable; (B) At least 4 bars; (D) 100As/bh = 6.03 > 3 unacceptable
Q20
D
Min As = 0.18x600x400/100 = 432 > 380; although the steel area 4T12 is ok, it is preferable to
provide at least 5 – 6 bars over the width of 600mm.
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CON4332 REINFORCED CONCRETE DESIGN
│Tutorial Questions│
[Present your answers with detailed working steps in a neat, tidy and logical manner.]
AQ1
Given the following information of a rectangular beam section:
Design ultimate moment, M =
425 kN-m
βb =
1.0
Breadth, b =
600 mm
Concrete cover =
Overall depth, h =
400 mm
Preferred Link size:
10
Concrete, fcu =
40 MPa
Preferred Bar size:
32
Steel, fy =
35mm
500 MPa
Determine the following:
(a)
The effective depth
(b)
The K value
(c)
The lever arm, z
(d)
The steel area required
(e)
Number and size of reinforcement bars, and steel area provided
(f)
Check if the steel provided comply with the max. & min. limits
AQ2
Given the following information of a flanged beam section of a simply
supported beam:
Design ultimate moment, M =
425 kN-m
βb =
1.0
Breadth, bw =
600 mm
Concrete cover =
Overall depth, h =
400 mm
Preferred Link size:
10
Slab thickness, hf =
150 mm
Preferred Bar size:
32
Effective span, L =
7 700 mm
Steel, fy =
c/c distance btw adjacent beams =
2 500mm
Concrete, fcu =
Concrete cover =
35mm
500 MPa
40 MPa
35mm
Determine the following:
(a)
The effective depth
(b)
The effective flange width
(c)
The K value
(d)
The lever arm, z
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CON4332 REINFORCED CONCRETE DESIGN
(e)
(f)
(g)
AQ3
The steel area required
Number and size of reinforcement bars, and steel area provided
Check if the steel provided comply with the max. & min. limits
Identify the assumptions that have been made in deriving the formulae for
bending.
Chapter 2
HD in Civil Engineering (Aug 2014)
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