The document discusses column buckling and failure modes. It defines a strut and column, and describes failure due to direct stress, buckling stress, or a combination. It provides differential equations to model column buckling based on the column's end conditions, including fixed-fixed, fixed-free, and fixed-hinged. Solutions provide the critical buckling load. The document also discusses Rankine's formula which relates crushing load and buckling load for a column's total capacity.

1. BY
JISHNU V
ENGINEER
BHEL

2. 
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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

3. 
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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

6. M=EI d²y/dx²
 EI d²y/dx²= -Py
d²y/dx² + (P/EI).y=0

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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

7. 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


9. 
y= Bsin √(P/EI)x; This means y is a sin
function of x. Hence, we can plot various
column failure modes as follows:

10. 
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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

11. 
Let y be the deflection at a section which is x
distance away from A.

12. 
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


13. d² λ/dx²= (P/EI).d²y/dx²
d²y/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}


14. Boundary conditions:
I) At x=0, y=0
II) 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


15. 
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

16. 
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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

17. EI d²y/dx²= Mo-Py
d²y/dx² + (P/EI).y – Mo/EI=0 (1)

Let λ = (P/EI).y – Mo/EI;
 d λ/dx= (P/EI).dy/dx
d² λ/dx²= (P/EI).d²y/dx²
d²y/dx²= (EI/P). d² λ/dx²
Eqn (1) becomes (EI/P). d² λ/dx² + λ=0


18. 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=0
b) At x=L, y=0
c) At x=0, dy/dx=0
 Using condition (a), y=0=A + Mo; A=-Mo


19. 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=0
cos√(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


20. 
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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

21. 
Let y be the deflection at a section which is x
distance away from A.

22. 
Hence the moment at section at a distance x
from fixed end, Moment M=-Py + H(L-x)

23. 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 + H
d ²λ/dx²=(P/EI)d²y/dx²
 d²y/dx²= (EI/P) d ²λ/dx²


24. 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=0
b) At x=L, y=0
c) At x=0, dy/dx=0


25. 
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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

26. 
Tan √(P/EI)L=√(P/EI)L
Solution of tanθ= θ is θ= 4.5 radian
√(P/EI)L=4.5
P/EI=20.25/L²; 2π²=20.25
 P=2π²EI/L²


27. 
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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

29. 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)/Le²
P= π²E (A)/(Le/k)²
P/A= π²E /(Le/k)²
 Le/k = Equivalent length/least radius of
gyration=Slenderness ratio


30. 
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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

31. 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


32. 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

33. 
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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.

34. 
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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

35. 
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= 0
λ= 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}

36. 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


37. 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}
