Sa1 09-column-struts-euler-theory

BY:BY:BHAVIN THAKRANIBHAVIN THAKRANINITESH VARSANINITESH VARSANI

STRUCTURAL ANALYSIS-1

Columns and StrutsColumns and Struts

STRUT:STRUT:

A structural member subjected to axial compressive force is called strut.

Strut may be vertical, horizontal or inclined.

The cross – sectional dimensions of strut are small.

Normally, struts carry smaller compressive loads.

Struts are used in roof truss and bridge trusses.

COLUMN:COLUMN:When strut is vertical it is known

as column.The cross – sectional dimensions

of column are large.Normally, columns carry heavy

compressive loads.Columns are used in concrete

and steel buildings.

Q. Compare Column and StrutsA.

Radius of Gyration(k) , k = √(I/A) or I = AK²

K=radius of gyrationI = Moment of Inertia (mm4)A = Area of Section (mm2)

Slenderness ratio, Slenderness Ratio = effective length of

column/Minimum radius of gyration λ = le/kmin

LONG COLUMN :LONG COLUMN :When length of column is more as

compared to its c/s dimension, it is called long column.

Long Column Le/kmin > 50

Or,Le/d > 15 for Long

Where,Le = effective length of columnd = least lateral dimension of

column.Kmin = Minimum radius of gyration

SHORT COLUMN :SHORT COLUMN :When length of column is less as

compared to its c/s dimension, it is called Short column.

Short Column Le/kmin <50

Or, Le/d < 15

Crushing Load : The load at which, short column fails by crushing is called crushing load.

CRIPPLING LOAD OR CRIPPLING LOAD OR BUCKLINGBUCKLINGLOADLOADThe load at which, long column

starts buckling(bending) is called buckling load or crippling load.

Buckling of column depends upon the following factors.

1. Amount of load. 2. Length of column 3. End condition of column 4. C/s dimensions of column 5. Material of column.

COLUMN END CONDITION COLUMN END CONDITION AND EFFECTIVE LENTH :AND EFFECTIVE LENTH : 1.Both end hinged. 2.Both end fixed. 3.One end fixed and other

hinged. 4.One end fixed and other free.

Effective length (le)

Where l is actual length

COLUMNS HAVING VARIOUS TYPES OF SUPPORTSCOLUMNS HAVING VARIOUS TYPES OF SUPPORTS

Effective length

12

Euler’s Formula

Euler’s Crippling Load, PE = ∏²EI /le²

Where, E is Modulus of Elasticity (Mpa) I is MOI or 2nd Moment of area

(mm4) Le is Effective length (mm)

Also known as Critical Buckling Load

Eccentric LoadingEccentric Loading

Short Columnσmax = P/A + P.e/Z = P/A (1 + eyc/k2)

Z = Ak2/ yc

Long Column◦ Rankine’s Formula σc= P/A (1 + eyc/k2) (1 + αle/k)

◦ Euler’s Formula σmax = P/A + Pe v /Z

σmin = P/A – Pe v /Z

v = sec {(le/2) /√[P/(EI)]}

Euler’s Formula for both end of the columns are hingedEuler’s Formula for both end of the columns are hinged

The Bending Moment at the Section is given by

General solution is

Since x = 0 at y = 0, then C2 = 0.Since x = l at y = 0, then

02

2

yEI

P

dx

yd

x

EI

PCx

EI

PCy sincos 21

15

0sin2

L

EI

PC

IDEAL COLUMN WITH HINGED SUPPORTSIDEAL COLUMN WITH HINGED SUPPORTS

for C1 = 0, we get

Which is satisfied if

or2

2

L

EIP

LEI

P

16

0sin

LEIP

Rankine’s Formula1/P = 1/PC + 1/PE

Where, P is Rankine’s crippling Load

PC is Crushing Load

PE is Euler’s crippling Load

If A is the Cross section area of columnPC = fC . A

PE = ∏²EI /le²

I = Ak2

Where Rankine’s Constant, α = fc/(∏²E)Thus, P = PR = (fC . A) / (1 + α λ)

If a dice is thrown what is the probability of an even number or a “3”

Since three is not an even number (!) P(Even or “3”) = P(Even) + P(3)

= 3/6 + 1/6 = 4/6