STRUTS AND COLUMNS

Introduction

Collapse of long structural members under compressive axial loads is a

common occurrence. Through an efficient design of strut or column

members in a structure, the collapse must be prevented; otherwise, in

addition to damage to structure, there will be loss of human lives.

Euler has developed a buckling theory for long column members; however,

for short columns, the theory gives absurd results. However, he provided

the basic formula for buckling of columns. Rankine modified the Euler’s

buckling load formula by combining the crushing load for a short column

and the buckling load (provided by Euler) for a long column. In Rankine’s

formula, there are two experimentally determined constants. This formula is

applicable for long, as well as medium-long columns.

Many a times, load is eccentrically applied on a column and buckling takes

place at a smaller load than the load given by Euler or Rankine. For

eccentrically loaded columns, expressions for maximum stress have been

derived in the form of a secant formula. Many researchers have modified the

secant formula for eccentrically loaded columns.

Higher-order differential equations have been used for determining the

buckling loads of columns with different end conditions. Moreover, energy

approach is briefly discussed to determine the buckling loads of columns,

taking various shape functions for the deflection curve.

Euler’s Theory of Buckling

Euler has developed a theory of buckling for columns or struts, but

considered the column/strut already buckled under a particular load while

developing the theory of buckling. He has made the following assumptions:

1.material of the column is homogeneous and isotropic,

2.compressive load acting on the column is fully axial,

3.column has failed only due to buckling,

4.weight of the column is negligible,

5.column is initially straight and buckles suddenly at a particular load,

6.fixed ends are rigid and

7.hinged ends are frictionless.

Example 9.1

Consider a column AB with one end fixed and the other

end free, buckled under the load P as shown in the figure.

Length of the column is L. Deflection at free end B is a as

shown, and the buckling load is P. Consider a section of

the column at a distance x from end A.

Column fixed at one end and free at the other end

Bending moment at section XX = P(a − y), or

Solution of this differential equation is y = A sin mx + B cos mx + a where and A and B are constants.

y

x

EIPa

EI

Py

dx

yd

yaPdx

ydEI

2

2

2

2

EIP

m

End conditions:

y = 0, x = 0, dy/dx = 0 at x = 0, at end A.

y = a, x = L, at end B

so, 0 = A sin 0 + B cos 0 + a

0 = 0 + B + a

or constant, B = −a

dy/dx = A m cos mx − B m sin mx

0 at x = 0

0 = A m × cos 0 − B m × sin 0

0 = A m

m # 0, because , then the load P would become zero,

if m # 0, then constant A = 0.

y

x

y = A sin mx + B cos mx + a

EIPm /

Finally, y = −a cos mx + a

At end B, x = L, y = a

a = −a cos mL + a or,

a cos mL = 0

a # 0, because if a = 0, then no buckling takes place.

so,

mL=p/2 (minimum significant value)

Euler’s buckling load, Pe = p2EI/4L2.

y

x

Example 9.2

Let us take another case of a column hinged at one end

and fixed at the other end as shown in the figure. The

column is under buckled condition. P is the buckling

load, H is the horizontal reaction at B. Let’s consider a

section at a distance x from the fixed end A as shown.

Column with one end fixed, other end hinged

Bending moment at section XX

Solution of this differential equation is

H

End conditions x = 0, y = 0

Constant, B = −HL/P Then,

So, Am = H/P

Constant, A = H/mP

Putting the values of A and B in the equation

H

At the end B, x = L, y = 0,

so

or tan mL = mL

H

Exercise 9.1

A column of length L and flexural rigidity EI is hinged at both

the ends. Derive the expression of Euler’s buckling load.

[Hint: M = −Py, y = A sin mx + B cos mx,

End conditions x = 0, y = 0, x = L, y = 0]

Exercise 9.2

Derive the expressions for Euler’s buckling load of a column fixed at both the

ends. Length of column L, buckling load P . Fixing moments at fixed ends are

MA and MB, respectively.

M at section shown = –Py + MA

y = A sin mx+ B cos mx + MA/ P

End conditions x = 0, y = 0, dy/dx = 0

x = L, y = 0, dy/dx = 0

and (d) a column with one end fixed and

Equivalent Length

Buckling load of a column is given by Euler’s formula and it depends on the end conditions of a column. However, to express the buckling load formula in a simplified manner, equivalent length of column is determined that is equivalent to the length of column with both ends hinged. This length can be obtained by completing the bending curve of the column with different end conditions similar to the bending curve of a column with both the ends hinged as shown in Figure 1.

Pe = Euler’s buckling load = p2EImin/Le2

Figure 1 (a) A column with both ends hinged, Le = L; (b) a column with both ends fixed, Le = L/2; (c) a

column with one end fixed and the other end hinged,

the other end free, Le = 2L.

Example 9.3 An allowable axial load for a 3-m-long column with hinged ends is 30 kN. Another column of the same material, same cross-section and same length, but with one end fixed and the other end hinged, suffers buckling; what is the buckling load for the column?

Euler’s buckling load,

where Le = L, both ends hinged.

other column with one end fixed and the other end hinged

Limitations of Euler’s Theory of Buckling

Euler’s theory of buckling is applicable for long columns. If there is a limit on the

slenderness ratio of the column and it is less than this limiting value, then Euler’s

formula gives the value of buckling load that is greater than the crushing load,

which is not possible.

Euler’s buckling load,

where,

E = Young’s modulus

Imin = Ak2min, where A is the area of cross-section

Le = equivalent length, kmin = the minimum radius of gyration

so

Pe /A < σc (the crushing stress of a short column)

Le/Kmin = the slenderness ratio of a column

For mild steel

E = 208 GPa = 208,000 N/mm2

σc = crushing strength

= 320 N/mm2

For mild steel column, the slenderness ratio should be greater than 80, so

that Euler’s formula can be used to predict the buckling load of a column.

Example 9.4

For what length of a mild steel bar of 60 mm in diameter used as a strut, the

Euler’s theory is applicable, if the ultimate compressive strength is 0.33 kN/mm2

and E = 210 kN/mm2, when one end of the strut is hinged and the other end is

fixed?

End conditions: one end hinged and the other end fixed

Equivalent length,

kmin = minimum radius of gyration

Length of the column,

Length should be greater than 1.68 m.

4

16

4

64

2

2

42

dk

d

d

d

A

Ik

p

p

Higher-Order Differential Equation

For any end conditions, the buckling load of a column/strut can be determined by

a higher order differential equation as follows:

Solution of this differential equation is

y = C1sin kx + C2cos kx + C3x + C4 Constants C1, C2, C3 and C4 are determined using the end conditions.

Example 9.5 Let us consider a column with both the ends fixed as shown in the figure.

Column AB of length L is fixed at ends A and B. At the

ends, both the slope and deflection are zero, that is,

the boundary conditions of the column are:

At end A, x = 0, y = 0, dy/dx = 0

At end B, x = L, y = 0, dy/dx = 0

Differentiating y = C1sin kx + C2cos kx + C3x + C4

with respect to x, we get

dy/dx = C1k cos kx − C2k sin kx + C3

Taking x = 0, y = 0, x = L, y = 0, we get,

0 = C1 sin 0 + C2cos 0 + C3 × 0 + C4

0 = 0 + C2 + 0 + C4

0 = C1sin kL + C2cos kL + C2L + C4

Taking dy/dx = 0 at x = 0, and x = L we get,

0 = C1k cos 0 − C2k sin 0 + C3

0 = C1k − 0 + C3

0 = C1k cos kL − C2 ksin kL + C3

Using these four equations, the following matrix is constructed:

The solution of this matrix yields:

Let us take 1 − cos kL = 0,

cos kL = 0

kL = 0, 2p, 4p - - -

So, kL = 2p (minimum significant value)

or,

Squaring both sides,

Buckling load,

Moreover, sin kL = sin 2p = 0 (also), which will also give P = 4p2EI/L2.

p2LEI

P

22 4pLEI

P

2

24

L

EIP

p

Example 9.6

Let us take another case of a column fixed at one end

and hinged at the other end as shown in the figure.

End A is fixed, end B is hinged.

Boundary conditions

at x = 0, end A, y = 0, dy/dx = 0

at x = L, end B, y = 0, d2y/dx2 = 0

(moment at hinged end is zero).

Equation of deflection,

y = C1sin kx + C2cos kx + C3x + C4

At x = 0, y = 0

0 = C1 × sin 0 + C2cos 0 + C3 × 0 + C4

= 0 + C2 + 0 + C4

At x = 0, dy/dx = 0

0 = C1kcos 0 − C2ksin 0 + C3

0 = C1 k + C3

At x = L, y = 0

0 = C1sin kL + C2cos kL + C3L + C4

At x = L, d2y/dx2 = 0, that is, the bending moment is

zero at the hinged end

0 = −C2k2sin kL − C2k

2cos kL

Using these set of equations, the following matrix is constructed:

or 0 = kL coskL − sin kL

tan kL = kL

tan 0 = θ

Buckling load,

Rankine Gordon Formula

Euler has developed a theory for the buckling load of long columns, but short

columns get crushed under the compressive load and crushing stress (σc) is

much greater than the stress (σe) given by Euler’s buckling load. Some

columns are of medium length and cannot be classified as short columns.

Rankine has combined the two loads, that is, the crushing load for short

columns and the Euler’s buckling load for long columns to determine the

buckling load for any long or short column.

Rankine’s load, PR is

Rankine’s load,

where σc = crushing stress,

Rankine has taken σc/p2E a constant that is determined experimentally.

Rankine’s load,

both σc and a are determined experimentally.

Le = equivalent length of column

kmin = minimum radius of gyration of the section of the column

Le/Kmin is known as slenderness ratio of the column, an important

parameter, and the buckling load depends on the slenderness ratio;

the greater the slenderness ratio, the less is the buckling load. The

table below gives the values of σc and a for various materials.

Rankine’s constants for different materials:

Example 9.7

A cast iron column of a hollow circular section with an external diameter of

250 mm and a wall thickness of 45 mm is subjected to an axial compressive

load. The column is 7 m long with both ends hinged. Taking factor of safety

(FOS) as 8, determine safe value of P.

Rankine’s constants are, σc = 560 N/mm2, a = 1/1,600

External diameter, D = 250mm

Wall thickness, t = 45mm

Internal diameter, d = 250 − 2 × 45 = 160 mm

Area of cross-section,

Length, L = 7 m = 7,000 mm

End condition: both the ends are hinged

Rankine’s load,

Safe load, P′R = 286.6/8 = 35.83 kN

A

Ik 2

Johnson’s Parabolic Formula

We have learnt so far that buckling loads depend upon the slenderness ratio

(Le /kmin) of the column. As the slenderness ratio increases, the buckling

required for the column decreases. Based on this principle, Johnson has

suggested a formula:

Working stress, σw = σc′[1 − φ(Le/k)] as function of Le/k, slenderness ratio

Taking φ(Le/k) = b(Le2/k2), where b is a constant

σw = σc′[1 − b(L2e/k

2)], an equation of a parabola.

σ’c = allowable stress in compression taking into account the FOS (factor of

safety)

= 110 N/mm2 for mild steel.

Constant, b = 0.00003 for hinged ends

= 0.00002 for fixed ends.

Straight line formula

σw′ = σc′[1 − C(Le/k)], applicable for columns for which the slenderness

ratio is greater than 90.

For structural steel,

σc′ = 140 N/mm2 in compression

C = 0.0054 for hinged ends

= 0.0038 for riveted ends.

Example 9.8

A strut is built-up of two 100 mm × 45 mm channels placed back-to-back at a

distance of 100 mm apart and riveted to two flange plates each 200 mm × 10

mm symmetrically, properties of one channel are:

Area, A = 7.41 cm2, Ixx = 123.8 cm4, Iyy = 14.9 cm4, and = 14 mm =1.4 cm

(distance of CG from the outer edge of the web). If the effective length is 5

m, calculate the working load for the strut using Johnson’s parabolic formula:

where σw = σc′[1 − b(L2e/k

2)], and constant b = 0.00003 for hinged ends and

σc

′ = 110 N/mm2.

Area of cross-section of built-up section, A

= 2 × 200 ×10 + 2 × 741

= 4,000 + 1,482 = 5,482 mm2

Ixx of built-up section = 2 ×123.8 ×104 + 2 × 200 ×10 × (55)2

= 247.6 ×104 +1210 ×104

= 1457.6 ×104 mm4

Iyy of built-up section (figure) Iyy = 2 ×14.9 ×104 + 2 × 741× (50 +14)2 + 2 ×10 ×

(2002/12)

= 29.6 ×104 + 607 ×104 +1,333.33×104

= 1,970 ×104 mm4 Now,

Imin = Ak2min

Eccentric Loading of Columns

A column AB of length L loaded eccentrically e (eccentricity

with respect to axis of column) has buckled under load P

as shown in Figure 2. Say, deflection at end B is a, when

the column has buckled. Consider a section at a distance x

from end A, where the deflection is y.

Figure 2 Eccentric Loaded Column

Bending moment at the section,

M = P(a + e − y)

or, EI(d2y/dx2) = P(a + e − y)

or,

y = A sin mx + Bcos mx + (a + e)

where m= A and B are constants

At end A, x = 0, y = 0 dy/dx = 0

Solution of this differential equation is,

At x = 0, y = 0

0 = Asin 0 + Bcos 0 + (a + e)

0 = 0 + B + (a + e)

or constant, B = −(a + e)

Moreover, dy/dx = Am cos mx − Bm sin mx

dy/dx = 0, at x = 0

0 = Am cos 0 − Bm sin 0

0 = Am

m ≠ 0, because m =

therefore, constant A = 0.

Finally, y = −(a + e) cos mx + (a + e)

At the end B, x = L, y = a, putting this boundary condition

a = −(a + e)cos mL + (a + e)

or, −e = −(a + e) cos mL

(a + e) = e sec mL

Maximum bending moment occurs at fixed end A.

Mmax = P(a + e) = Pe sec mL

σb = maximum stress due to bending,

= P e sec mL/ , secant formula

= section modulus = I /yc,

I = moment of inertia

yc = distance of extreme layer in compression from neutral layer

σc = direct compressive stress

where

Resultant stress at fixed end, = P/A

I

Myb

In this case, we have considered that one end of the column is fixed and the other

end is free. In this case, Le = 2L, that is, equivalent length, Le = 2L. Secant

formula can be generalized for any column with any type of end conditions as

follows:

σr = resultant maximum stress

where Le is the equivalent length.

Example 9.9

A steel tube of 80 mm outer diameter, 50 mm inner diameter and 3 m long is

used as a strut with both ends hinged. The load is parallel to the axis of the strut

but is eccentric. Find the maximum value of eccentricity so that the crippling load

on strut is equal to 50 per cent of the Euler’s crippling load.

Yield strength = 320 N/mm2, E = 210 kN/mm2.

End conditions: both ends hinged

Length, Le = L = 3 m

E = 210 kN/mm2

Steel tube

Euler’s buckling load,

Crippling load due to eccentricity, P = Pe /2 = 1,96,184.7 N

Eccentricity, e = (320 - 64.05)/10.37 = 24.68 mm

Long Columns with Eccentricity in Geometry

A column AB of length L with both the ends hinged has

an initial eccentricity e′ in the centre. Assuming deflection

curve to be sinusoidal (Figure 3),

eccentricity at any section,

Figure 3

e′ = maximum initial eccentricity at centre.

Taking a section at a distance of x from end A, let’s

consider the column buckled under load P.

Final deflection = y

Change in deflection = y-y′ (as shown)

BM at the section at a distance x from A as shown

′

′

as given

Putting this value in:

Let us assume the solution as

Putting this value in,

Putting the assumed value of y

However, p2EI/L2 = Pe, Euler’s buckling load

Finally, the deflection curve equation will be,

Maximum deflection occurs at the centre, where

Maximum bending moment,

σmax = maximum compressive stress at central section of column

where

k = radius of gyration

yc = distance of extreme layer

However, Pe = σeA, Euler’s load.

P = σeA, putting these values; load applied.

Example 9.10

A 5-m-long hollow circular steel strut having an outside diameter of 120 mm and

an inside diameter of 80 mm with both the ends hinged is initially bent. Assume

that the centre line of the strut as sinusoidal with maximum deviation of 6 mm.

Determine the maximum stress developed due to an axial load of 100 kN.

E = 210 kN/mm2

Length, L = 5 m

Eccentricity, e′ = 6 mm

Area of cross-section, A = (p /4)(1202 - 802) = 6283.2mm2

Moment of inertia, I = (p /64)(1204 — 804)

= 816.816 × 104 mm4

Distance, yc = 120/2 = 60 mm

e′ = 6 mm

Lateral Loading of Strut with Point Load

A strut or column may buckle under the combined action of axial thrust P and

transverse load (at centre) as shown in Figure 4. Lateral loading produces

deflection in the strut and axial thrust produces a bending moment on account of

deflection. The figure shows a column AB of length L, hinged at both the ends and

carrying a central transverse load W as shown. Due to W at C, there are

horizontal support reactions, W / 2 each.

Consider a section at a distance x from end A

Bending moment at the section, M = −Py − (Wx/2) or

Figure 4 Lateral point load on a strut

or

Solution of this differential equation is

where

Boundary conditions are x = 0, y = 0,

0 = A cos 0 + B sin 0 − 0

0 = A + 0 − 0

or constant, A = 0

Moreover,

Equation of deflection curve becomes

Maximum deflection occurs at centre at x = L/2

Maximum bending moment occurs at the centre, x = L / 2

putting the value of m

yc = distance of extreme layer in compression from neutral layer

or,

Example 9.11

A 4-m-long horizontal pin-ended strut is formed from a standard T-section of 150

mm × 100 mm × 12.5 mm. The axial compressive load is 60 kN. A lateral

concentrated load of 6 kN acts at the centre of the strut. Find the maximum stress

developed if the xx axis is horizontal and the table of the T-section forms the

compressive face. The centre of gravity is 24 mm away from the edge of the

table.

Given Ixx = 250 × 104 mm4, A = 3,100 mm2, E = 200 GPa

Axial load, P = 60 kN

Lateral load, W = 6 kN

Ixx = 250 × 104 mm2

A = 3,100 mm2

Length, L = 4 m

T-section, yc = 24 mm

Strut with a Uniformly Distributed Lateral Load

A strut subjected to axial thrust P and lateral load w per unit length is shown in

Figure 5. The length of the strut is L and it is hinged at both the ends A and B.

Consider a section at a distance x from end A. The bending moment at the

section,

Figure 5

Solution of this differential equation, y = complementary function + particular

integral.

Complementary function = Acos mx + Bsin mx, where

Particular integral,

or constant, A = w EI / P2

Then,

dy/dx = 0 at x = L/2, at the centre of the beam due to symmetrical loading about

centre C,

However, A = w EI / P2

So constant,

Finally, the equation of y

At the centre, deflection is maximum, therefore, taking

Bending moment at the centre,

Maximum stress,

Example 9.12

A circular rod of diameter 50 mm is supported horizontally through pin

joints at its ends and carries a uniformly distributed load of 1 kN/m running

throughout its length and an axial thrust of 25 kN. If its length is 2.4 m,

estimate the maximum stress induced in the rod, E = 200 GPa.

Axial thrust, P = 25,000 N

Diameter, d = 50 mm

Area, A = (p/4)d2 = 1963.5 mm2

w = 1 kN/m = 1 N/mm

E = 200,000 N/mm2

yc = 25 mm

Length, L = 2.4 m = 2400 mm

Energy Approach

To determine the critical load at which a column ceases to be in a stable

equilibrium, energy criterion can be used. Moreover, in a situation where the exact

solution of a differential equation is not possible or difficult to obtain, energy

approach can be used to get a good approximate solution. However, this approach

converges to the exact solution if a large number of terms are taken to represent

the deflection curve.

Energy stored in a system = work done by external loads

δU + δVe = 0

δVe = external work done due to virtual displacement.

δ(U+Ve) = 0

δ(Kp) = 0

U + Ve = Kp = a constant referred to as the total potential of the system.

Energy stored,

External work done,

For small values of P, Kp is positive for any non-trivial admissible function y(x).

The critical load can be obtained by assuming suitable admissible function.

Critical condition is reached when constant Kp = 0

Example 9.13

Let us consider a case of column that is fixed at one end

and free at the other end, with boundary conditions.

Deflection, y = 0 and x = 0

slope, dy/dx = 0, x = 0

Moment at end B is zero, that is,

Let us take a shape function,

Putting these values in the eq.

Pcr = 3EI / L2, but the exact value given by Euler’s theory is

p2 EI / 4L2 = 2.467(EI /L2)

Let us take more terms in the

deflection shape function

Putting these values in the equation of total potential Kp,

Example 9.14

A long strut AB of length L is of uniform section throughout. A thrust P is applied

at the ends eccentrically on the same side of the centre line with eccentricity at

the end B twice than that at end A. Show that the maximum bending moment

occurs at a distance x from end A, where

The strut is buckled under the thrust load P.

P at two ends produce a couple P(2e − e) (cw)

that is to be balanced by anticlockwise couple

of horizontal reactions F each at A and B, as

shown F × L (ccw).

Force F is unknown. so 2Pe − Pe − FL = 0 or,

Consider a section at a distance x from end A,

Bending moment at the section (considering top part),

M = –P(2e + y) + F(L – x) or

Solution of this differential equation is,

where A and B are constants, and

End conditions

At end A, x = 0, y = 0, putting this in,

0 = A cos 0 + B cos 0 – e – 0

Constant, A = e

Moreover, at end B, k = L, y = 0

0 = Acos kL+ Bsin kL – 2e

However, A = e

so, 0 = e cos kL + Bsin kL – 2e

Constant,

Finally, the equation of deflection becomes,

Bending moment at the section,

For maximum bending moment, dMx /dx = 0

Example 9.15

A strut of length L is fixed at its lower end, its upper end is eccentrically

supported against a lateral deflection (through a spring), so that the resisting

force is k times the end deflection, k is the spring constant. Show that the

crippling load P is given by 1 – (P/kL) = tan mL/mL.

where, .

The figure shows a strut AB of length L, fixed at

end A and free at end B. At the end B, the

horizontal reaction is H = ka, given by a spring

and P is the crippling load. Consider a section, at

a distance x from end A.

Bending moment at the section (top part),

M = P(a − y) − ka(L − x)

Solution of the differential equation

At end A, fixed end, x = 0, y = 0

Then,

or constant, B = –(ka/mP)

Finally, the equation of deflection curve is

At the end B,

or