DESIGN & ANALYSIS OF DOUBLY REINFORCED BEAM | IS 456 | Limit State Design | Mumbai University

C = Cc + Cs

Cc = 0.36 fck b xu – fcc Asc

= 0.36 fck b xu

.

Cs = fsc Asc and T = 0.87 fy Ast

.

C = T gives

0.36 fck b xu + fsc Asc = 087 fy Ast

.

From it xu may be found.

* The strain at the level of compression steel

esc = 0.0035

The corresponding stress in compression steel may be obtained from stress-strain curve. Then

fsc = esc Es = 0.0045 × 2 × 10

5

= 700 , subject to maximum of 0.87 fy

.

Mu = Cc

(d – 0.42 xu

) + Cs

(d – d¢)

= 0.36 fck b xu

(d – 0.42 xu

) + fck Asc

(d – d¢).

Strength of R.C. Section in Shear, Torsion and Bond

* Design procedure is given by IS: 456–2000 based on average shear stress across the section.

Average shear stress Tv =

where Vu = factored (design) shear force

b = bredth

d = effective depth

In case of varying depth,

tv =

where Mu = factored bending moment

b = angle between the top and bottom edges.

* Table 19 in IS 456–2000 gives design shear strength (tc

) for various mixes for different percentage

of reinforcements.

* Shear strength Vu can be enhanced by providing shear reinforcement in any one of the following:

1. Vertical stirrups

2. Bent up bars

3. Inclined stirrups.

If a is the angle between the inclined stirrup or bent up bar and the axis of the member, then total

vertical shear force resisted is given by

1. Vertical stirrups: Vus = Asvd

2. Bent up bars: Vus =

3. Inclined stirrups: Vus =

However, code specifies wherever bent up bars are provided, their contribution towards shear

resistance shall not be more than half that of the total shear reinforcement. Code also specifies that

to avoid compression failure of the section in shear, under no circumstances, even with shear

reinforcement nominal shear stress tv exceeds tc max

, given in Table 7.3.

Torsional Shear Strength: IS code recommends the effect of torsional moment Tu

, may be split into

a bending moment and a shear force as given below:

Mu = Tu

Vu = 1.6

where b = bredth of beam or breadth of web in case of flanged sections

D = overall depth of beam

\ If a beam is subjected to Mus

, Tu and Vu

Mequivalent = Mu + Tu

Vequivalent = Vu + 1.6 .

* Bond strength:

Tbf =

where dj

is lever arm.

\ Bond length Ld

is given by

Ld = where f = Diameter of bar.

The values for different grade of concrete are as given in Table 7.4.

Table 7.4 Design bond stress in plain bars in tension

Equivalent Development Lengths: For standard hooks and bends the equivalent development bends

may be taken as given below:

1. U-type bends: 16 × diameter of bar

2. For standard bends, it shall be 4 times the diameter of the bar for each 45° bend subject to a

maximum of 16 times the diameter of the bar.