Columns lecture#4

Engr. Ayaz Waseem ( Lecturer/Lab Engr., CED) Columns

LECTURE # 4

1. COLUMNS UNDER ECCENTRIC LOADING:

Let’s suppose a load acts on a column at a distance ‘e’ from plastic centroid as

shown in figure 3. In the figure,

As′ = Area of compressive steel.

As = Area of tensile steel.

fs = Stress in tensile steel.

fs′ = Stress in compressive steel.

T = Tensile force in steel.

Cc = Compressive force in concrete.

Cs = Compressive force in steel.

Cc = 0.85 fc′ b a

Cs = As′ fs′

T = As fs

Now,

Pn = Cc + Cs − T

Pn ………….. (1)

Equation (1) is called ‘Load Equation’.

To find out moment carrying capacity of the

column we take moment about plastic centroid.

x e = Mn = Cc. + Cs. + T.

Let, d′′ =

and e′ = d′′ + e

Considering moment about center line of tension steel,

x e′ = Mn = Cc. + Cs.

Now, considering moment about plastic centroid in terms of d′′,

1


Pn x e = Mn = Cc. + Cs.[ d − d′′ − d′ ]+ T.

Mn = 0.85 fc’ b a. + As’ fs’.[ d − d′′ − d′ ]+ As fs. ……………….. (2)

Equation (2) is called ‘Moment Equation’.

Case 1: PURE AXIAL LOAD/CRUSHING FAILURE:

We know that,

Pn = Cc + Cs − T

But in this case there is pure compression,

so, no tension steel is present

Ast = As + As′ and Cs = C1 + C2

Pn

Pn

Pn = 0.85 fc′ Ag + ( fy – 0.85 fc′ ) Ast

Mn = 0

Case 2: BALANCE FAILURE:

For balanced condition,

a = ab = 1 cb ……………………..…………. (1)

Fitst we need to find cb, in figure comparing c f g and

a b c we get,

If we want to express cb in terms of fy then,

2

LOAD EQUATION

MOMENT EQUATION


Using Es = 200, 000 MPa and rearranging,

……………………………..... (3)

Using equation (2) in equation (1) we get,

1

Now, in figure comparing c f g and c d e, we get,

If εs′ < εy , then fs′ = Es. εs′

If εs′ ≥ εy , then fs′ = fy

and for balanced failure we already know that fs = fy

So,

Pn

If As = As′ and fs′ = fy , then

Pn = 0.85 fc′ b

and

3

LOAD EQUATIONS

MOMENT

EQUATION


Case 3: PURE FLEXURAL FAILURE:

In this case no axial load is acting is

acting on the column and column just

behaves like a beam. Therefore,

Pn = 0

In figure comparing a b c and c f g, we

get,

………………………. (1)

Note: In equation (1) value of ‘c’ can not be found out because even though we know

that s >> y but we don’t know the exact value of s . Whereas in case of balance failure

‘c’ could be found out as the equation involved y rather than s .

Now, in figure comparing a b c and c d e, we get,

……………… (2)

Now,

For pure flexural failure we know that, fs = fy and Pn = 0, so

…………. (3)

4

LOAD EQUATION


Assuming, fs′ = fy and As′ = As , we get

This means that concrete is not taking any load which is not possible. So, our assumption

is wrong. Therefore, compression steel can not yield at pure flexural failure point.

Now, as εs′ < εy , so

fs′ = Es. εs′

Using equation (2) in above equation, we get,

………………….. (4)

Now, using equation (4) in (3), we get,

………………….. (5)

Equation (5) results in a quadratic equation. We solve above equation for the value of ‘a’

and then using equation (4) we find fs′.

Now, using the values of a and fs′ in moment equation we can easily find value of Mn ,

whereas, value of Pn is already known i.e., Pn = 0.

5

Es = 200,000 MPa

MOMENT

EQUATION

Columns lecture#4

Design

Transcript of Columns lecture#4