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CON4332 REINFORCED CONCRETE DESIGN

Chapter 2 1

HD in Civil Engineering (Aug 2014)

│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 Material Stress-strain Relations 2.1.1 Concrete

2.1.2 Reinforcement

2.1.3 Example – Design Ultimate Capacity for Axial Compression

2.2 Design Formulae for Bending 2.2.1 Limit to Neutral Axis

2.2.2 Examples – Effective Depth

2.2.3 Simplified Stress Block

2.2.4 Design Formulae for Singly Reinforced Section

2.2.5 Limits of the Lever Arm

2.2.6 The Balanced Section

2.2.7 Examples – Singly Reinforced Section

2.2.8 Design Formulae for Doubly Reinforced Section

2.2.9 Examples – Doubly Reinforced Section

2.3 Flanged Section 2.3.1 Effective Flange Width

1.3.2 Examples – Flanged Section

2.4 Limits to Bar Spacing and Steel Ratio 2.4.1 Bar Spacing

2.4.2 Maximum and Minimum Percentage of Steel


Chapter 2 2


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 = Deformation

(Dimensionless) Length

Stress, f = Force

(in N/mm2 or MPa) Area

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.


Chapter 2 3


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)

Strain

Stress (MPa)

0.001 0.002 0.003 0.004 0.005

40

80

120

0


Chapter 2 4


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 or 200 000 N/m2

1 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.


Chapter 2 5


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.

0.04 0.08 0.12 0.16 0.20

100

200

300

400

500

600

700

800

Stress (MPa)

Strain 0

0.002

0.2% Proof

Stress

A typical stress-strain

curve of concrete

Yield Stress


Chapter 2 6


Figure 2.4 – Idealized Stress-strain Curve

of Reinforcement for Design


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

Ɛy


Chapter 2 7


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 : C40

Dimensions : 400mm x 400mm

Rebars : 4T25 vertical bars fully restrained by links

Solution Asc = 4 x 491

= 1 964 mm2

Anc = 400 x 400 – 1 964

= 158 036 mm2

Nuz = 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

2 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.


Chapter 2 8


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

b

d

c

M M

a

b

d

c

Elevation of a Beam under Load

Rebar

d

x

Ɛcc

Ɛst

(a) Deformation of a-b-c-d (b) Strain (c) Stress (Elastic) (d) Stress (Plastic)

Neutral Axis

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.


Chapter 2 9


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 when Ɛcc = 0.0035

Therefore, 0.0035 (d – x) ≥ 0.002175

x

Re-arranging, it becomes 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.


Chapter 2 10


3T32

h d

2T40 + T32

h d

2T32

d

Bar size / 2 Link size Cover

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 = 500 mm

Concrete cover = 40 mm

Size of link: 10

Bottom bars: 3T32 in one layer

Solution Effective Depth, d = 500 – 40 – 10 – 32/2

= 434 mm

Question B Determine the effective depth of the following section:

Overall beam depth, h = 650 mm

Concrete cover = 45 mm

Size of link: 12

Bottom bars: 2T40 + 3T32 in two layers

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.


Chapter 2 11


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 = 650 – 45 – 12 – 40 – 40 - 16 = 497 mm

Effective Depth, d =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.

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).

?Q.1 – Q.4


Chapter 2 12


Ɛcu

Ɛst

(a) Strain at Ultimate

Limit State

(b) Parabolic Stress

Block

Section

0.45fcu

d

x

0.45fcu

s s/2

z

(c) Simplified Rectangular

Stress Block

Neutral Axis

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, s = 2(d – z)

2.2.4 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.


Chapter 2 13


Grade of concrete : fcu

Grade of steel : fy

Breadth of section : b

Effective depth : d

Section

d

x

0.45fcu

s =

0.9

x

s/2

z

Simplified Rectangular Stress Block

Neutral Axis

As

Fcc

Fst

M

b

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:


Chapter 2 14


(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]

and then, the level arm: z = [0.5 + (0.25 – K/0.9)0.5] d [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 [2.5a]

6 This limit is for concrete not higher than Grade 45. This limit is more stringent for higher grade of concrete.

Refer to the design code for details.


Chapter 2 15


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 z = 0.95 d [2.6]

2.2.6 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, Mbal = 0.157 fcu bd2

and, 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.


Chapter 2 16


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:

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

7 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.

Deflection

Mom

ent

0

Mbal

As < As,bal

As A

s,bal

As > A

s,bal (w/o compression steel)


Chapter 2 17


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 βb = 1.0

Breadth, b = 350 mm

Effective depth, d = 480 mm

Concrete, fcu = 35 MPa

Steel, fy = 500 MPa

Solution K = M / (bd2 fcu)

= 350 x 106 / (350 x 4802 x 35)

= 0.124

βb = 1.0 < 0.156 (Singly reinforced)

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 = 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


Chapter 2 18


Simplified Rectangular Stress Block

d

x

0.45fcu

s =

0.9

x

s/2

z

N.A.

Fcc

Fst

M

= 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 = 2412 mm2 (i.e. 3T32)


Steel, fy = 500 MPa

Solution (The design formulae in the design

code are given in the form to facilitate

the determination of steel area from

the design moment. This example,

however, requires you to work back

the ultimate moment capacity of the

section from the steel area provided.)

For equilibrium of compression and tension forces:

Fcc = Fst

0.45 fcu (b s) = 0.87 fy As (assuming steel has yielded)

s = 0.87 fy As / (0.45 fcu b)

= 0.87 x 500 x 2412 / (0.45 x 35 x 350)

= 190.3 mm

Depth of neutral axis, x = s / 0.9

= 190.3 / 0.9


Chapter 2 19


= 211.5 mm

< 0.5 x 480 = 240 mm (steel has yielded as assumed)

Lever arm, z = 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

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. It remains at the limiting depth

to retain ductility. Therefore,

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

?Q.5 – Q.8


Chapter 2 20


Section

Ɛcu

Ɛst

(a) Strain at Ultimate Limit

State

d

x

0.45fcu

s

z =

d –

s/2

(b) Simplified Rectangular

Stress Block

Neutral Axis

d'

Ɛsc

d –

d'

As

As'

Fst

Fcc

Fsc

M

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')


Chapter 2 21


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 =0.156 fcu bd2

+ As' fsc

0.87 fy z 0.87 fy

or As =K' fcu bd2

+ As' fsc

[2.9a]0.87 fy z 0.87 fy

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 [2.10]

A's=(K – K') fcu bd2

[2.8]0.87 fy (d - d')

and, As =K' fcu bd2

+ A's [2.9]0.87 fy z

These are Eqns 6.14 and 6.15 in HKCP-2013.

On the other hand,


Chapter 2 22


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 to ensure rebar has yielded

5. Calculate compression steel A's=(K – K') fcu bd2

0.87 fy (d - d')

6. Calculate tension steel As =K' fcu bd2

+ A's 0.87 fy z

Question A Determine the steel required for the following beam section:


Breadth, b = 350 mm

Effective depth of tension steel, d = 480 mm

Effective depth to comp'n steel, d' = 70 mm


Steel, fy = 500 MPa


Chapter 2 23


Solution K = M / (bd2 fcu)

= 500 x 106 / (350 x 4802 x 35)

= 0.177

βb =1.0 > 0.156 (Compression steel is required)

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

= 0.29 < 0.38 (fsc = 0.87fy)

Compression steel req'd, A's =(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 = 2 x 201

= 402 mm2

Tension steel req'd, As =K' fcu bd2

+ As' 0.87 fy z

=0.156 x 35 x 350 x 4802

+ 332 0.87 x 500 x 372

= 2721 + 332

= 3053 mm2

(Provide 4 T32 Bottom Bars)

As, pro = 4 x 804

= 3216 mm2


Chapter 2 24


d

x

Neutral Axis

d' Ɛ

sc

Ɛcu

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

Solution For doubly reinforced section, x = 0.5 d

= 0.5 x 350

= 175 mm

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

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

?Q.9 – Q.13


Chapter 2 25


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)


Chapter 2 26


beff

bw

T – Section

hd

bw

beff

L – Section

hf

Compression

Tension

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.

Figure 2.11 – Definition of lpi

for Calculation of Effective Flange Width


For simply-supported beam, the whole span is under sagging moment, and

therefore, lpi = effective span of the simply-supported beam (L).

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.


Chapter 2 27


Figure 2.12 – Effective Flange Width


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

c/c distance of adjacent slabs


Chapter 2 28


Effective span, L = 6 700 mm

Clear spacing btw adjacent beams = 2 500mm (same on both sides)

Solution Internal span of ctu beam lpi = 0.7 x 6700

= 4 690 mm

b1 = 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:



Clear spacing btw adjacent beams = 2 000mm (same on both sides)

Solution Simply supported beam lpi = 9000

b1 = b2 = 2000 / 2

= 1000 mm

beff,1 = beff,2 = 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


Chapter 2 29


Question C Determine the steel required for the following beam section:

Design ultimate moment, M = 500 kN-m (sagging)


Slab thickness, hf = 150 mm

Effective flange width, beff = 1780 mm

Effective depth of tension steel, d = 480 mm


Steel, fy = 500 MPa

Solution b = beff = 1780mm

K = M / (bd2 fcu)

= 500 x 106 / (1780 x 4802 x 35)

= 0.0348

< 0.156 (No compression steel required)

K < 0.0428 Lever arm, z = 0.95d

= 0.95 x 480

= 456 mm

Check x = (480 – 456) / 0.45

= 53 < 150 mm (N.A. is within the flange)

Tension steel req'd, As = M / (0.87 fy z)

= 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


Chapter 2 30


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. maximum bar size

ii. aggregate size + 5 mm

iii. 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.


Chapter 2 31


Example

For a beam of 400 mm wide, provide 3 to 4 bars at the outer layer depending on the size

of the bars.

For slab, under most circumstances, limit the spacing of main bars to not

more than 2h or 250 mm whichever is lesser.

2.4.2 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

4

Flexural tension steel

Rectangular section 0.13

Flanged section (bw/b < 0.4) 0.18

Flexural compression steel

Rectangular section 0.20

Column 0.80 6

Wall

4 Vertical bars 0.40

Horizontal bars 0.25

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.


Chapter 2 32


Overall depth, h,

instead of effective

depth, d, is used to

check steel ratio.

For rectangular section, Ac = bh

For flanged section, Ac = bwh

Example

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


Chapter 2 33


│Key Concepts/Terms│

Design Ultimate Strength of Concrete 0.45 fcu

Ultimate Strain of Concrete Ɛcu = 0.0035

Design Yield Strength of Steel 087fy

Effective Depth d

Depth of Neutral Axis and Its Limit 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


Chapter 2 34


│Self-Assessment Questions│

Q.1 Find the effective depth, d, of the following beam section.

Overall beam depth, h = 600 mm Referred size of link: 10

Concrete cover = 40 mm Preferred size of main bars: 40

A. 510 mm

B. 530 mm

C. 265 mm

D. 550 mm

Q.2 Find the effective depth, d, of the following beam section.


Concrete cover = 40 mm Main bars: 2 layers of 32

A. 557 mm

B. 541 mm

C. 525 mm

D. 312.5 mm

Q.3 Find the effective depth, d, of the following slab section.

Overall slab thickness, h = 175 mm


A. 159 mm

B. 138 mm

C. 144 mm

D. 150 mm


Chapter 2 35


Q.4 Determine the allowable depth of the neutral axis, x, of the following rectangular beam

section.



βb = 1.0

A. 322 mm

B. 300 mm

C. 539 mm

D. 269.5 mm

Q.5 Determine the K value for the following rectangular beam section.


Breadth, b = 300 mm

Effective depth, d = 454 mm Overall depth, h = 520mm


A. 0.142

B. 0.108

C. 1.415

D. 0.124


Chapter 2 36


Q.6 Determine the amount of tension steel, As, required for the following rectangular beam

section.


Breadth, b = 325 mm Effective depth, d = 534 mm

Concrete, fcu = 40 MPa Steel, fy = 500 MPa

A. 3212 mm2

B. 2320 mm2

C. 1856 mm2

D. 2134 mm2


section.




A. 701 mm2

B. 2320 mm2

C. 1856 mm2

D. 693 mm2


section.




A. 372 mm2

B. 366 mm2

C. 327 mm2

D. 701 mm2


Chapter 2 37


Q.9 Determine the lever arm, z, for the following rectangular beam section.

K = 0.178 βb = 1.0



A. 462 mm

B. 426 mm

C. 317 mm

D. 491 mm

Q.10 Determine the compression steel, A's, required for the following rectangular beam section.

K = 0.188 βb = 1.0

Breadth, b = 350 mm Overall depth, h = 700 mm

Effective depth, d = 634 mm d' = 70 mm


A. 3617 mm2

B. 642 mm2

C. 4776 mm2

D. 491 mm2


Chapter 2 38


Q.11 Determine the steel required for the following rectangular beam section.





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

Q.12 Determine the steel required for the following rectangular beam section.





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

Q.13 Determine the lever arm, z, for the following rectangular beam section.

K = 0.150 βb = 0.8



A. 462 mm

B. 520 mm

C. 317 mm

D. 491 mm


Chapter 2 39


Q.14 Determine the effective flange width, beff, for the end span of a continuous beam with

approximately equal spans with the following information:



Clear distance between adjacent beams = 2 600mm (same on both sides)

A. 2918 mm

B. 2600 mm

C. 2129 mm

D. 1829 mm

Q.15 Determine the effective flange width, beff, for the following simply-supported beam:

Breadth, bw = 250 mm (same for adjacent beams)


c/c distance between adjacent beams = 3000mm (same on both sides)

A. 1395 mm

B. 2290 mm

C. 2540 mm

D. 2750 mm


Chapter 2 40


Q.16 Determine the lever arm, z, for the following flanged beam section.

Design ultimate moment, M = 666 kN-m (sagging) βb = 1.0

Breadth, bw = 400 mm Effective flange width, beff = 2150 mm

Effective depth, d = 485 mm hf = 200 mm


A. 457 mm

B. 355 mm

C. 376 mm

D. 461 mm

Q.17 Determine the steel area required for the following flanged beam section.

Design ultimate moment, M = 668 kN-m (sagging) βb = 1.0




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

Q.18 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


Chapter 2 41


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?

A. 4T20

B. 3T25

C. 4T25

D. 12T32

Q.20 Which of the following is the most appropriate rebars for the flanged beam under sagging

moment as described below?



Overall depth, h = 400 mm

The amount of bottom steel required to resist the sagging moment = 380 mm2

A. 2T20

B. 2T16

C. 4T12

D. 5T12


Chapter 2 42


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.


Chapter 2 43


│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:


Breadth, b = 600 mm Concrete cover = 35mm

Overall depth, h = 400 mm Preferred Link size: 10

Concrete, fcu = 40 MPa Preferred Bar size: 32

Steel, fy = 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:


Breadth, bw = 600 mm Concrete cover = 35mm

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 = 500 MPa

c/c distance btw adjacent beams = 2 500mm Concrete, fcu = 40 MPa

Concrete cover = 35mm

Determine the following:

(a) The effective depth

(b) The effective flange width

(c) The K value

(d) The lever arm, z


Chapter 2 44


(e) The steel area required

(f) Number and size of reinforcement bars, and steel area provided

(g) Check if the steel provided comply with the max. & min. limits

AQ3 Identify the assumptions that have been made in deriving the formulae for

bending.

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