ANALYSIS OF THE

SECTIONS

Introduction

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A satisfactory and economic of a concrete structure rarely depends on a complex theoretical analysis.

Wherever possible the analysis should be kept simple, yet it should be based on the observed and tested behavior of reinforced concrete members.

Assumption

3

The flexural strength at the ultimate limit state is

determined assuming the following conditions as

given in clause 3.4.4.1 BS8110 Part 1:

Plane sections remain plane.

The compressive stresses in the concrete may be

derived from the stress-strain curve in Figure 2.1

The stresses in the reinforcement are derived

from the stress-strain curve in Figure 2.2.

The tensile strength of the concrete is ignored.

Sizing a concrete beam (BS 8110)

Table 1 : Span /effective depth ratio for initial design

Support Condition Span/Effective Depth

Cantilever 6

Simply supported 12

Continuous 15

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Example 1

A simply supported beam has an effective span

8 m and supported characteristic dead (gk) and

live (qk) loads of 15 kN/m and 10 kN/m

respectively. Determine suitable dimensions for

the effective depth and width of the beam.

6

Solution

Ratio= Span/ Depth

= 12

d = 8000/12 = 666.67 mm

Take d = 670 mm

Total ultimate load = (1.4 gk + 1.6 qk) 8

= 296 kN

Design shear force (V) = 296/2 = 148 kN

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Solution

Design shear stress v = V/bd

= 148 x 103/(670 x b)

Assuming v is equal to 1.2 N/mm2 ,

gives width of beam,b,of

Check shear stress in beam

b =V/dv = 148 x 103/(670 x 1.2)

≈ 185 mm

RECTANGULAR SECTIONS Singly Reinforced Sections

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b = breadth

h = overall depth

x = depth of neutral axis

As = area of tension reinforcement

d = effective depth

εcu = max concrete strain

εs = max steel strain

fcu = characteristic strength of concrete

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(0.67/m)fcu εcu=0.0035 b

As

T

C

PN

εs

Fc

Fs

0.45x

Z=d-0.45x

0.9x x

h

Fig 2.1:Strain Block Fig 2.2:Stress Block

d

• Fc = total compressive force in

concrete

• Fs = total tensile force in steel

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1. Consider the moment of the compressive force about the line of action of Ft:

Ultimate concrete moment, M = FcZ

= [0.45fcub(0.9x)]Z

= 0.405fcubxZ

from the stress block: Z = d-0.45x

then,

M = 0.405fcubx(d-0.45x)

M = {0.405[x/d]}{1-0.45[x/d]}fcubd2 -------------(1)

M = kfcubd2 @ k = M/fcubd2

2. To enable balance failure to occur (other failures occur without warning), x is limited to ≤ 0.5d, then equation 1 becomes:

Mu = 0.156fcubd2 ---------------- (2)

= k’fcubd2 where k’=0.156

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3. Equation 2 represents the moment capacity of the section based on the

concrete area in compression.

4. The value of 0.156 is the limiting value of k and is given the symbol k’.

5. To obtain the required area of reinforcement, As :

Consider the moment of the tensile force about the line of action of Fc :

M = FsZ

M = 0.95fyAsZ

As = M/0.95fyZ

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6. From (1);

M = {0.405[x/d]}{1-0.45[x/d]}fcubd2

M/fcubd2 = {0.405[x/d]}{1-0.45[x/d]}

From the stress block;

lever arm,

Z = d-0.45x---- x = (d-z)/0.45

(Z/d)2 – (Z/d) + 1.111k = 0

Z = d{0.5 + √[0.25 – (k/0.9)]} ≤ 0.95d

where k = M/fcubd2

M = max design moment

NB: the value of Z is limited to ≤ 0.95d

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Example 2

Design moment 185kNm. Determine As. Given fy = 460N/mm2 and fcu = 30N/mm2.

d = 440

b = 260

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Solution

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k = M/bd2fcu = 185x106/(260x4402x30) = 0.122

z = d{0.5 + √[0.25 – (k/0.9)]}

= 440{0.5+√[0.25-(0.122/0.9)]} = 369mm

As = M/0.95fyz = 185x106/(0.95x460x369) = 1147.3mm2

NB: As is the area of reinforcement required to resist the excess design

moment.

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Example 3

A simply supported rectangular beam of 7 m carries characteristics dead (including self weight of beam), gk and imposed, qk, loads of 12 kN/m and 8 kN/m respectively. The beam dimensions are breadth, b, 275 mm and effective depth, d, 450 mm. Assuming the following material strengths, calculate the area of reinforcement required. Given fy = 460N/mm2 and fcu = 30N/mm2.

d = 450

b = 275

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7 m

qk = 8 kN/m

gk = 12 kN/m

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Solution

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Ultimate load (w) = 1.4 gk + 1.6 qk

= 1.4 x 12 + 1.6 x 8 =29.6 kN/m

Design moment (M) = wl2/8 = (29.6 x 72)/8 = 181.3 kNm

Ultimate moment of resistance (Mu) = 0.156fcubd2

= 0.156 x 30 x 275 x 4502 x 10-6 = 260.6 kNm

Since Mu > M design as a singly reinforced beam

k = M/fcubd2 = 181.3 x106/(30x275x4502) = 0.1085

z = d{0.5 + √[0.25 – (k/0.9)]}

= 450{0.5+√[0.25-(0.1085/0.9)]}

= 386.8 mm 0.95 d (= 427.5) OK

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Solution

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As = M/0.95fyz = 181.3 x 106/(0.95 x 460 x 386.8)

= 1073 mm2

NB: As is the area of reinforcement required to resist the excess design

moment.

Hence provide 4T20 (T refers to high yield steel bars, fy =460 N/mm2)

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Doubly Reinforced Sections

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• εs’ = strain in the compression reinforcement

• Fs’ = force in the compression reinforcement

• d’ = depth from the compression face of the concrete to the centroid

of the compression reinforcement.

•Happens when the applied design bending moments exceeds the concrete

capacity.

d’

(0.67/m)fcu εcu=0.0035 b

As

T

C

PN

εs

Fc

Fs

0.45x

Z=d-0.45x

0.9x x

h

Stress Block Strain Block

d

As’ εs’

Fs’

d-d’

20

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Consider the moment of the compressive forces about the line of action of Ft:

M = FcZ + Fs’(d-d’)

= 0.45fcub(0.9x)(d-0.45x) + 0.95fyAs‘(d-d’)

consider balance failure, x/d=0.5

M = 0.156fcubd2 + 0.95fyAs’(d-d’)

the equation can be rewritten as:

As’ = (M-0.156fcubd2)/0.95fy(d-d’)

The required area of tension reinforcement can be determined by equating the compressive and the tensile forces acting on the cross section:

Fs = Fc + Fs’

0.95fyAs = 0.20fcubd + 0.95fyAs’

As = (0.20fcubd/0.95fy) + As’

As = (0.156fcubd2/0.95fyZ) + As’

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Example 4

The reinforced concrete beam shown in Figure above has an effective span of 9 m carries uniformly distributed dead (including self weight of beam), gk and imposed, qk, loads of 4 kN/m and 5 kN/m respectively. Design the bending reinforcement assuming the following: fy = 460N/mm2 and fcu = 30N/mm2. Cover to main steel = 40 mm

h = 370

b = 230

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9 m

qk = 5 kN/m

gk = 4 kN/m

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Solution:

l) Design Moment, M

Ultimate design load (W) = (1.4 gk + 1.6 qk) span

= (1.4 x 4 + 1.6 x 5) 9

= 122.4 kN

Max design moment = Wl/8 = (122.4 x 9)/8

= 137.7 kNm

II) Ultimate moment of resistance, Mu

Effective depth, d

Assume diameter of tension bars() = 25 mm

d = h- /2 – cover

= 370 – 25/2 – 40

= 317 mm

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Solution:

Ultimate moment resistance Mu = 0.156fcubd2

Mu = 0.156 x 30x 230x 3172)

= 108.2 x 106 Nmm = 108.2 kNm

Since M > Mu compression reinforcement is required

III) Compression Reinforcement

Assume diameter of compression bars () = 16 mm

d’ = cover + /2 = 40 + 16/2 = 48 mm

As’ = M - Mu/ 0.95fy(d-d’)

= (137.7 – 108.2)(106) /0.95(460)(317 - 48)

= 251 mm2

Provide 2T 16 ( As’ = 402 mm2)

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IV) Tension Reinforcement

z = d{0.5 + √(0.25-k’/0.9)}

= d{0.5 + √(0.25-0.156/0.9)}

= 246 mm

As = [Mu / 0.95fyz] + As’

= 108.2 x 106)/0.9 5(460)(246) + 251

= 1258 mm2

Provide 3T 25 ( As = 1470 mm2 )

2 T 16

3 T 25

d = 48 mm

d = 317 mm