1. What are the assumptions made in limit state of collapse (flexure)?

Ans:

Design of limit state of collapse is based on the following assumptions.
1. Plane sections normal to the axis remains plane after bending.
2. The maximum strain in concrete at the outermost compression fiber taken as 0.0035 in bending
3. The relationship between the compressive stress distribution in concrete and the strain in concrete maybe assumed to rectangle, trapezoidal or parabolic or any other shape which results in prediction of strength in substantial agreement with the results of test.
4. The tensile strength is ignored.
5. Partial safety factor equal to 1.15 shall be applied for design purpose.
6. The maximum strain in the tension reinforcement in the section at failure shall not be less than;

f_y/1.15Es + 0.002
Where, fy = characteristic strength of steel
     Es = Modulus of elasticity of steel.

2. When is beam said to be under reinforced section? Give the formula for calculating moment of resistance.

Ans:

When Xu < Xu,max or pt < pt lim the section is said to be under reinforced.
Moment of resistance (Mu) is calculated by
Mu = Tu×z
Where, z = (d - 0.42 Xu)
Mu = 0.87 fy Ast (d – 0.42 Xu)

3.What is the limiting value of Mu for Fe 500 steel?

Ans:

Limiting value of moment of resistance is calculated by different formulas for different steels.
Mu,lim = Ru,max .bd² (N-mm)
For Fe 500 it is given by
Mu,lim = 0.133 fck bd²

4.What are the steps followed for analysis of the beam when you are given with fck, fy, b, d, Ast?

Ans:

Mu and allowable load on given span.
Steps:
A.Calculate Xu from Xu= 0.87.fy.Ast/(0.36 fck b)
B.Determine depth of balanced neutral axis Xu,max = ku,max.d
C.Compare Xu with Xu,max
    If Xu < Xu,max the section is under reinforced
    If Xu = Xu,max the section is balanced
    If Xu > Xu,max the section is over reinforced.
D.Calculate Mu, Mu = Tu.z = 0.87 fy.Ast (d – 0.42Xu)
E.Obtain Md in terms of wd and span L and equate it with Mu to get unknown Wd and hence working load.

5.Which type of section is not permitted in designing of beam as per IS code?

Ans:

Providing over reinforced section of the beam is not allowable as per the IS code and hence in such cases Xu = Xu,max.

6.How will you find out the required depth?
    1. If you are given with the value of b.
    2. If you are given with the value of ratio b/d.

Ans:

When b is given, d = √Md/(Ru,max × b)
When ratio b/d is given, d = ³√Md/(Ru,max × b/d)

7.What is doubly reinforced beam?

Ans:

The RCC beam in which the reinforcements are provided on both tension and compression side is known as doubly reinforced beam.

8.What are the conditions for provision of doubly reinforced beam?

Ans:

Conditions for provision can be stated as:

   1. When the applied moment exceeds the moment resisting capacity of a singly reinforced beam
   2. When the dimensions b and d of the section are restricted due to architectural, structural and/or constructional purposes.
   3. When the sections are subjected to reversal of bending moment. For example piles underground water tanks, winds braces, etc.
   4. In continuous T beam where the portion of beam over middle support has to be designed as doubly reinforced.
   5. When the beams are subjected to eccentric loading, shocks or impact loads.

9.In case of doubly reinforced beam the depth of neutral axis is greater than the maximum depth of neutral axis then the section is of which type. Give the equation for calculating moment of resistance.

Ans:

If Xu > Xu,max, then section is over reinforced.
Mu = q.fck.bd²+(fsc - fcc)Asc(d - d’)