Columns lecture#3

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

LECTURE # 3

1. TRIAL SIZE OF COLUMN:

Approximate formula;

i. For tied columns,

ii. For spiral columns,

2. CONCENTRICALLY LOADED SHORT COLUMN:

Column is said to be a short column if it satisfies the following condition;

…………………………………………...

(1)

For rectangular section, r = 0.3 hmin

where, hmin is the least column dimension.

And for concentrically loaded column, k = 0.75 and

Putting these values in equation (1),

or ……………. (1)

Now, for circular section, r = 0.25 D

where, D is the column diameter.

Putting these values in equation (1),

or …………………… (2)

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If above conditions in equation (1) and (2) are satisfied then the column is said to be a short column.3. DESIGN PROBLEM:

An axially loaded short column has a length of 4m and carries a factored load (Pu)

of 2000 kN. Other data is as follows;

fc’ = 28 MPa

fy = 420 MPa

Maximum aggregate size = 20 mm

Using US Customary Bars design the column as;

i. Square tied column.

ii. Circular spiral column.

Solution:

i. Design of Square Tied Column:

STEP 1: Area Calculations:

Trial size of column = mm

Say, trial size of column is,

mm

STEP 2: Check for the type of column:

Therefore, it is a short column.

STEP 3: Reinforcement calculations:

For a tied column,

Pu = Pn = ( 0.65 x 0.8 ) [ 0.85 fc’Ag + ( fy – 0.85 fc’ )Ast ]

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( 2000 x 1000 ) = (0.65 x 0.8)

Ast = 1260.2 mm2

STEP 4: Reinforcement Ratio check:

= = 0.009 < 0.01 ( Not Satisfied )

Therefore,

= min = 1% = 0.01

So,

Ast = 0.01 x (375)2 ≈ 1407 mm2

Selecting,

8 # 16 Bars

STEP 5: Design of ties:

As maximum diameter of the longitudinal bars is < 32 mm, so,

Diameter of the ties = 10 mm

Spacing of the ties = Minimum of;

a. 16 x 16 = 352 mm.

b. 48 x 10 = 480 mm.

c. Least column dimension = 375 mm.

d. 300 mm. ( Governs )

Therefore,

Spacing of the ties, S = 300 mm.

Note: In this step shape of the ties is not given. In step 6, shape of the ties is decided

during detailing.

STEP 6: Detailing:

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ii. Design of Circular Spiral Column:

STEP 1: Area Calculations:

Trial diameter of column, D = 374 mm

Say, trial diameter of column is,

D = 375 mm

STEP 2: Check for the type of column:

Therefore, it is a short column.

STEP 3: Reinforcement calculations:

For a spiral column,

Pu = Pn = ( 0.7 x 0.85 ) [ 0.85 fc’Ag + ( fy – 0.85 fc’ )Ast ]

( 2000 x 1000 ) = (0.7 x 0.85)

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Ast = 1849.4 mm2

STEP 4: Reinforcement Ratio check:

= = 0.0167 > 0.01 ( Satisfied )

Therefore,

Ast ≈ 1850 mm2

Selecting,

4 # 16 , 4 # 19 Bars

STEP 5: Design of Spiral:

Diameter of the spiral = dsp = 10 mm

Minimum pitch of the spiral = Larger of;

a. 1.5 x 20 = 30 mm. ( Governs )

b. 25 mm.

Therefore,

Smin = 30 mm.

Smax = .

Now, Dc = 375 – ( 40 x 2 ) = 295 mm

Ac = = 68349.2 mm2

Ag = = 110446.6 mm2

Therefore,

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Smax = = 57.63 mm < 75 mm ( O.K. )

Selecting pitch between Smin and Smax ,

S = 45 mm

STEP 6: Detailing:

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4. INTERACTION CURVE:

It is a graph between load and moment. In order to draw an interaction curve we

require the coordinates of different points.

Fig. Interaction Curve

5. PLASTIC CENTROID:

It is a point through which the resultant of all the internal forces should pass, in

pure axially loaded column case, when no moment is acting on the column at ultimate

stage of failure.

6. SYMMETRICAL SECTIONS:

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Those sections in which plastic centroid coincide with the geometrical centroid

are called symmetrical sections.

7. UNSYMMETRICAL SECTIONS:

Those sections in which plastic centroid do not coincide with the geometrical

centroid are called unsymmetrical sections. Plastic centroid of unsymmetrical sections

can be found out in the following manner;

Fig. Section unsymmetrical about x-axis

R = Cc + F1 + F2 + F3

Lets suppose plastic centroid is at a distance ‘ ’ from the bottom face. Taking moment

about bottom face,

R = + F1 L1+ F2 L2+ F3 L3

= ………………………….. (1)

Equation (1) is used to find out location of plastic centroid.

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