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**Chapter 2 Beams (Part 3)**

** (b) Doubly reinforced section**

If depth and width are restricted.

If this beam has to support a BM more than Mr of the balanced section.

Ther are 2 options.

• Provide an over-reinforced section, or

• Provide a doubly Reinforcement-section

Over re-inforced section has many disadvantage like brittle failure so always a doubly reinforced

section is a better option.

Properties

• Steel is provided on both side of NA

• Permissible stress for compression steel = 1.5 mC

where C' = stress in concrete around compression steel.

• Equivalent area of steel in terms of concrete for compression steel = 1.5 mAsc For tension steel

equivalent area = mAst.

(i) Actual Depth of Neutral Axis, (Xa)

Equating moment of area on both sides of NA

( ) sc ( a c ) st ( a )

2a

1.5m –1 A X d mA d X

2

BX + - = -

Here, Xa = Actual depth of Neutral axis

(ii) Critical Depth of Neutral Axis, (Xc)

d .

t mc

X mc

c +

=

(iii) Moment of Resistance (Mr)

Mr = C1 (LA1) + C2 LA2

Mr = BXa 2

Ca (1.5m –1)

3

d Xa + ÷ø

ö

çè

æ - Asc . C' (d–dc)

C' can be calcualted using similar triangles

For a balanced section

Xa = Xc

Ca = cbc s

Mr = Q.B.d2 + (1.5 m – 1) Asc. C' (d – dc )

Mr = Mr1 + Mr2

Where, Mr1 = QBd2 = moment of resistance of singly reinforced balanced section

The above doubly reinforced Beam can be assumed to be made-up of two Beams A and B, as shown in

the figure.

Moment of resistance from tension side

For balanced section

ta = st

Design Steps

1. Given values : B, D.

2. If load is given BM =

8

wl2

design moment = BM

3. Find moment of resistance of the balanced section (singly reinforced)

Mr(bal) = Q.B.d2 =Mrl

4. If BM > Mr (balance) then only, a dubly reinforced section is required.

5. Area of steel for singly reinforced balanced section from Beam-A

Mr1 = C1 × LA1 = T1 × LA1

6. Area of steel for remaining BM (Part - II)

Mr2 = (BM – Mr1)

Mr2 = (1.5 m –1) Asc C' (d – dc)

= A (d d

7. From eq. (ii)

or equating moment of area on both sides of N.A. in Beams–B.

(1.5 m – 1)Asc (Xa – dc) = mAst2 (d – Xa)

Limitation in Working Stress Method of Design

1. The shrinkage and the creep of concrete have a considerable effect of stresses due to service loads and

these can not be estimated easily (Shrinkage usually means the drying shrinkage, i.e. decrease in all the 3

dimension due to drying).

Shrinkage strain of concrete, e 0.03% cs »

Elastic strain under service loads = 0.035% which is of the same order as that shrinkage. So shrinkage has

considerable effect. Creep occurs (in case of steel) only when stress > 0.5 fy creep is a time dependent

strain under sustained stress. Creep strain has a considerable effect when it is comparable to eci and it is

so. We define creep coefficient, cc = 1.6

e

e

ci

cc » (approx).

eci

Its value varies from 1.1 to 2.2 depending upon the age of loading, the greater value corresponding to

lesser age of loading.

2. Owing to the presence of cracks, the elastic theory for the analysis of stresses due to shear can not be

extended to R.C. beams and the design for shear must be based on the behaviour at ultimate state. The

actual margin of safety is not equal to the 'factory of sately' because stress-strain relationship is not linear

upto collapse. So' limit state method is a bettter method of design.

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