Theories of columns

THEORIES OF COLUMNS

BY JISHNU V

ENGINEER BHEL

A member of a structure which carries an axial compressive load is called a strut

A vertical strut which is liable for failure due to buckling or bending is called a column

Failure of a STRUT occurs due to any one of the following stresses set up in the column

Direct stress- Due to compressive stress (for short columns)

Buckling stress or crippling stress- Due to lateral bending of columns

Combined direct compressive stress and buckling stresses

7.1 COLUMNS & STRUTS

Column is initially straight and the load is applied axially

Cross section of the column is uniform through out its length

The column material is isotropic and homogenous

Length of the column is very large compared to its lateral dimensions

Direct stress (compressive stress) is very small compared to buckling stress

The column will fail by buckling alone Self weight of the column is negligible

7.2 ASSUMPTIONS MADE IN EULER’S COLUMN THEORY

7.3 EULER’S CRITICAL LOAD FOR A COLUMN WHEN BOTH THE ENDS ARE HINGED


M=EI d²y/dx² EI d²y/dx²= -Pyd²y/dx² + (P/EI).y=0

Solution of the differential equation, d²y/dx² + m²y=0 is y= Acosmx + Bsinmx

Hence solution of DE d²y/dx² + (P/EI).y=0 is y=Acos√(P/EI)x + Bsin√(P/EI)x


Boundary conditions are (i) When x=0, y=0 (ii) When x=L, y=0 Using BC (i), A=0 Hence y= Bsin √(P/EI)x Using BC (ii), B sin√(P/EI)L=0√(P/EI)L=nΠ, where n=1,2,3...... Simplest case is when n=1, √(P/EI)L=Π P=Π²EI/L²

7.3EULER’S CRITICAL LOAD FOR A COLUMN WHEN BOTH THE ENDS ARE HINGED


y= Bsin √(P/EI)x; This means y is a sin function of x. Hence, we can plot various column failure modes as follows:

7.3EULER’S CRITICAL LOAD FOR A COLUMN WHEN BOTH THE ENDS ARE HINGED

Consider a column AB of length L and uniform cross sectional area, fixed at A and free at B.

The free end will sway sideways when the load is applied and the curvature will be similar to that of upper half of the column subjected to compressive load whose both ends are hinged.

Let P be the critical load, ie crippling load at which the column starts buckling

7.4 EULER’S CRITICAL LOAD FOR A COLUMN WHEN ONE END IS FIXED & OTHER FREE


Let y be the deflection at a section which is x distance away from A.

Moment due to crippling load at the section, M= P(a-y)

M=EI d²y/dx² EI d²y/dx²= -P(y-a)d²y/dx² + (P/EI).y - Pa=0 (1)

Let λ = (P/EI).y – Pa; d λ/dx= (P/EI).dy/dx


d² λ/dx²= (P/EI).d²y/dx²d²y/dx²= (EI/P). d² λ/dx²

Eqn (1) becomes (EI/P). d² λ/dx² + λ=0 d² λ/dx² + λ.(P/EI)=0

λ= Acos√(P/EI)x + Bsin√(P/EI)x (P/EI).y – Pa= Acos√(P/EI)x + Bsin√(P/EI)x y=(EI/P){ Acos√(P/EI)x + Bsin√(P/EI)x+Pa}


Boundary conditions:I) At x=0, y=0II) At x=L, dy/dx=0 Using condition (i), y=0=A + Pa; A=-Pa y= (EI/P){-Pacos√(P/EI)x + Bsin√(P/EI)x+Pa} Using condition (ii), dy/dx=0= B√(P/EI).cos

√(P/EI).L √(P/EI).L=(2n-1) Π/2, n=1,2,3.... Simplest case is when n=1; √(P/EI).L= Π/2 P=Π²EI/4L²


Consider a column AB of length L and uniform cross sectional area, fixed at both A and B. Let P be the critical load, ie crippling load at which the column starts buckling

7.4 EULER’S CRITICAL LOAD FOR A COLUMN WHEN BOTH THE ENDS ARE FIXED

Let y be the deflection at a section which is x distance away from B.

Moment due to crippling load at the section, M= Mo-Py

M=EI d²y/dx²


EI d²y/dx²= Mo-Pyd²y/dx² + (P/EI).y – Mo/EI=0 (1)

Let λ = (P/EI).y – Mo/EI; d λ/dx= (P/EI).dy/dxd² λ/dx²= (P/EI).d²y/dx²d²y/dx²= (EI/P). d² λ/dx²Eqn (1) becomes (EI/P). d² λ/dx² + λ=0


d² λ/dx² + λ.(P/EI)=0 λ= Acos√(P/EI)x + Bsin√(P/EI)x(P/EI).y – Mo= Acos√(P/EI)x + Bsin√(P/EI)x y=(EI/P){ Acos√(P/EI)x + Bsin√(P/EI)x+Mo} Boundary conditions:a) At x=0, y=0b) At x=L, y=0c) At x=0, dy/dx=0 Using condition (a), y=0=A + Mo; A=-Mo


y= (EI/P){-Mocos√(P/EI)x + Bsin√(P/EI)x+Mo}

Using condition (c),B=0 y= (EI/P){-Mocos√(P/EI)x + Mo) Using condition (b), -Mocos√(P/EI)L + Mo=0cos√(P/EI)L =1√(P/EI)L=2nΠ, n=1,2,3,..... Simplest case is when n=1, √(P/EI)L=2Π P= 4Π²EI/L²


Consider a column AB of length L and uniform cross sectional area, fixed at end A and hinged at end B

Let P be the critical load, ie crippling load at which the column starts buckling

At the fixed end of the column, there will be a fixed end moment=Mo

This Mo will tend to make the slop of deflection at fixed end zero.

Inorder to balance the moment a reaction force H will be generated at B

7.5 EULER’S CRITICAL LOAD FOR A COLUMN WITH ONE END FIXED AND OTHER END HINGED


Let y be the deflection at a section which is x distance away from A.

Hence the moment at section at a distance x from fixed end, Moment M=-Py + H(L-x)


M=EI d²y/dx² EI d²y/dx²= -Py + H (L-x)d²y/dx² + (P/EI).y –H (L-x)=0 (1)

Let λ = (P/EI).y –H (L-x);d λ/dx=(P/EI)dy/dx + Hd ²λ/dx²=(P/EI)d²y/dx² d²y/dx²= (EI/P) d ²λ/dx²


Hence eqn (1) becomes (EI/P) d ²λ/dx² + λ=0

λ= A cos √(P/EI) x + B sin√(P/EI) x (P/EI).y –H (L-x)= A cos √(P/EI) x + B

sin√(P/EI) x

Boundary conditions:a) At x=0, y=0b) At x=L, y=0c) At x=0, dy/dx=0


Using condition (a) A = -HL

Using condition (b), 0= -HL cos {√(P/EI) L} + B sin{√(P/EI) L} (2)

Using condition (c), -H= √(P/EI)B (3)

Using condition eqns (2) and (3),- √(P/EI)Lcos{√(P/EI)L} + sin√(P/EI)L=0


Tan √(P/EI)L=√(P/EI)L

Solution of tanθ= θ is θ= 4.5 radian√(P/EI)L=4.5P/EI=20.25/L²; 2π²=20.25 P=2π²EI/L²


The effective length (Le) of a given column with given end conditions is the length of an equivalent column of same material and cross section hinged at its either ends and having the value of Euler’s crippling load equal to that of the column.

Crippling load for any type of end conditions is given by, P=π²EI/Le²

The moment of inertia in the equation in Euler’s equation is the least among Ixx and Iyy

7.6 EQUIVALENT LENGTH OF A COLUMN

7.6 EQUIVALENT LENGTH OF A COLUMN

Moment of inertia I =Ak², where A is the cross sectional area and k is the least radius of gyration.

Crippling load P=π²EI/Le²; Substituting I=Ak², P= π²E (Ak²)/Le²P= π²E (A)/(Le/k)²P/A= π²E /(Le/k)² Le/k = Equivalent length/least radius of

gyration=Slenderness ratio

7.7 CRIPPLING STRESS IN TERMS OF SLENDERNESS RATIO

It is an empirical formula which is applicable to all columns whether short or long

1/P=1/Pc + 1/PE- P= Crippling load- Pc= Crushing load= σc x A- PE= Euler’s crippling load= π²EI/ Le²- σc= Ultimate crushing stress (=330

MPa for Mild steel)- A= Cross sectional area

7.8 RANKINE’S FORMULA

P=Pc.PE/(Pc + PE) OR P=Pc/{(Pc/PE) + 1} P= σc.A/( σc.A.Le²/Π²EI + 1)

P= σc.A/( σc /Π²E x (Le/k)² + 1) σc /Π²E=α=Rankine’s constant Hence P= σc.A/( α x (Le/k)² + 1) Case-1:- If a column is short Pc>>Pe; Hence P≈Pc Case-2:- If a column is long

Pe>>Pc; Hence P≈PE

7.8 RANKINE’S FORMULA

SL No Material Crushing stress in MPa

α

1 Wrought iron 250 1/9000

2 Cast iron 550 1/1600

3 Mild steel 330 1/7500

4 Timber 50 1/750

7.8 RANKINE FORMULA

Consider a column AB of length L and uniform cross section fixed at A and free at B subjected to a compressive load P at an eccentricity of amount e

Let a be the lateral deflection at free end

Consider a section at a distance x from A where the lateral displacement is y.

7.9 COLUMNS WITH ECCENTRIC LOADING

Bending moment at the section, M= P(a+e-y) EI d²y/dx²= P(a+e-y) d²y/dx² + (P/EI) x y – P (a + e)/EI=0


Let λ=(P/EI) x y – P (a + e)/EI

Then d² λ/dx²=(P/EI) x d² y/dx²(EI/P) d² λ/dx² + λ= 0λ= A cos {√(P/EI)x} + B sin {√(P/EI)x}

(P/EI).y – P(a+e)/EI= A cos {√(P/EI)x} + B sin {√(P/EI)x}


Boundary conditions are A) x=0, y=0 B) x=0,y’=0 C) x=L, y=a Condition (A) A= -P(a+e)/EI Condition (B) B=0


Condition (c)e= (a + e) x cos{√(P/EI)L} (a+e)= e x sec {√(P/EI)L} Maximum stress induced at a section Maximum stress at any section= crushing stress

+ bending stress Crushing stress= P/A Bending stress= M/Z; where Z is the section

modulus M= P (a+e)= P x e x sec {√(P/EI)L} Hence σmax= P/A + P x e x sec {√(P/EI)L}


Theories of columns

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