Courses

# Beams (Part - 4) Civil Engineering (CE) Notes | EduRev

## Civil Engineering (CE) : Beams (Part - 4) Civil Engineering (CE) Notes | EduRev

The document Beams (Part - 4) Civil Engineering (CE) Notes | EduRev is a part of Civil Engineering (CE) category.
All you need of Civil Engineering (CE) at this link: Civil Engineering (CE)

Chapter 2 (Part 4) Beams

(II) Limit State Method
According to this method a structure is so designed that the probability of reaching the limit state is
acceptably low during the design life of the structure.
Limit state is the state in which the structure becomes unfit for use.
(a) Limit States
It can be classified into two categories
1. Limit state of serviceability
2. Limit state of collapse.
1. Limit Stateof Serviceability
These relate to the satisfactory performance of the structure under service loads.
These includes the limit states of deflection, cracking vibration etc.
2. Limit State of Collapse
These provide adequate margin of safety for normal overloads. There include limit states of overturning
sliding, failure of one or more critical sections, failure due to buckling etc.
We consider all the relevant limit states during design.
3. Difference b/w Serviceability & Collapse
If a structure raches to the limit state of serviceability. it recovers when the loads are removed but at the
limit state of collapse, the structure cannot recovery.
Flexure
Flexural members are those members subjected to bending.
In working stress method, the Stress-strain relation of the concrete and steel are assumed to the lines within the
range of permissible stresses.
In limit state method, the design stress strain curves concrete and steel are assumed as follows:

• Design value of strength
For concrete

Here, mc g Partial factor of safety for concrete
= 1.5
fd = desgin value of strength
For steel

Apart from other common assumption in flexutre we take here an important assumption i.e. On the
tension side, tension is borne entirely by steel. Though in real, this is not true as there is a certain area below
Neutral axis which carries tensile stresses (of small magnitude) and remains uncracked.Force carried is small
and its resultant is also small. So it is ignored and it is the justification of the assumption.
(b) Assumption in LSM
1. Plane section before bending remain plane after bending.
2. Maximum strain in concrete at outermost compression fibre = 0.0035
3. Tensile strength of concrete is ignored
4. Stress in reinforcement are derived from stress-strain curve. For design purpose partial factor of
safety = 1.15 shall be applied
Permissible stress in steel = 0.87 fy
5. Maximum stress is tension reinforcement at failure shall not be less than

A. Analysis
(i) Analysis of stress block parameter

The above stress diagram can be divided into two portions rectangular portion ABED and parabolic
portion BCE.
Compressive force C1 = Width of cross-section x Area of rectangular portion

Now depth of CG of rectangular portion (y1)

Now Compressive force C2 = Width of cross-section × Area of parabolic portion

= 0.1714 fck BXu
Now, depth of CG of parabolic portion (y2)

Now, Total compressive force
C = C1 +C2 = (0.1928 + 0.1714) fck BXu
C = 0.36fck BXu
Now depth of CG of stress diagram from top

= 0.42 Xu
(ii) Limiting depth of N.A. (Xu(lim))
The limiting values of depth of neutral axis Xu(lim) for different grades of steel can be obtained from the
strain diagram.

From similar triangles

The value of Xu(lim) for three grades of steel are given in table.
fy (N/mm2) Xulimit/d
250 0.53
415 0.48
500 0.46

(iii) Actual depth of N.A.
The actual depth of neutral axis can be obtained by considering the equilibrium of the normal forces, that
is
Resultant force of compression = Average stress × area = 0.36 fck BXu
Resultant force of tension = 0.87fy Ast
Force of compression should be equal to force of tension
0.36 fck BXu = 0.87 fy Ast

(iv) Ultimate moment of resistance (Mu(lim))
Since the maximum depth of neutral axis is limited, the maximum value of the moment of resistance is also
limited.
Mu(lim) with respect of concrete
= 0.36 fck BXu(lim) × lever arm
= 0.36 fck BXu(lim) × lever arm
Lever arm = d – 0.42 Xu(lim)
Mu(lim) = 0.36 fck BXu(lim) × (d–0.42 Xu(lim))
Mu(lim) with respect to steel
= 0.87 fy Ast × Lever arm
= 0.87 fy Ast × (d – 0.42 Xu(lim))
For a rectangular beam section, the limiting value of Mu(lim) depends on the type of concrete mix and the
grade of steel. The values of limiting moment of resistance Mu(lim) with respect to concrete for different
grades of concrete and steel are given in table.

In general moment of resistance of balance section = QBd2
where Q = moment of resistance coefficient
= 0.148 fck for Fe 250
= 0.138 fck for Fe 415
= 0.133 fck for Fe 500

Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;