Column and Strut: Common Difference of Strut & Column

By BYJU'S Exam Prep

Updated on: September 25th, 2023

Columns and Struts are any structural members experiencing axial compression due to axial or longitudinal loads.  A column is a vertical member which is used in structures for supporting the horizontal load over it. A strut can be a horizontal/ vertical/ inclined member that is subjected to axial compression load. It is comparatively a shorter column.

In this article, we will try to learn about columns and struts in detail, the basic differences between them, and the classification of columns. We will also see how columns are further divided in terms of their slenderness ratio.

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Difference Between Columns and Struts

The basic difference between columns and struts is that a column is a vertical member designed to carry axial or longitudinal loads. In contrast, a strut is a horizontal/inclined/vertical member designed to take axial or longitudinal loads. Columns are supported by fixed supports at both ends, while struts are supported by hinged or pinned joints at both ends, as struts are generally used in a truss.

In columns, failure is generally due to buckling, while in struts, failure is generally due to crushing. Columns have a more slenderness ratio, while struts have a less slenderness ratio. Columns can carry only both compressive and tensile loads.

Effective Length of Columns and Struts

The load-carrying capacity of columns and struts, i.e., compression members, depends on the length of the member, the cross-sectional area of the member, and the type of support. The effective length of a column is the distance between two points on a column where the bending moment is zero. It is written as the product of effective length factor and unsupported length.

Effective Length = KL


  • K – Effective length factor
  • L – Unsupported Length

Both Ends Fixed


Both ends are restrained against rotation and translation/lateral movement or held in position and direction.

K = 0.5

Leff = L/2

IS Code recommendation , K = 0.65

One End Fixed One End Pinned


Held in position at both ends but held in direction at one end only or Restricted against translation at both ends, rotation is allowed in one end only.

K= 12=0.707

Leff= L2

IS Code recommendation , K = 0.8

Both Ends Pinned


Held in position at both ends or restrained against lateral movement at both ends.

K = 1

Leff = L

IS Code recommendation , K = 1

One End Fixed and Other End Free


Held in position and direction at one end only or Restricted against rotation and translation at one end only.

K = 2

Leff = 2L

Slenderness Ratio

The slenderness ratio is the effective length ratio to the least lateral dimension.

λ = Leff/b


  • λ = Slenderness ratio
  • Leff = Effective Length
  • B = Least Lateral Dimension


  • Least lateral dimension in the case of figure 1 is ‘d’.
  • Least lateral dimension in the case of figure 2 is a min of (b,d).

Limits of Slenderness Ratio

Both ends restrained

L/b <60= L <60b

One end restrained

L < 100b2/d

Classification of Columns based on Slenderness Ratio

Short Column

It is that column in which the ratio of effective length to least lateral dimension ratio is less than 12.

3 < λ ≤12

Shorts columns always fail because of direct compression.

Long / Slender Columns

It is that column in which the ratio of the effective length to least lateral dimension is more than 12.

λ >12

Elastic Instability

Columns might fail by crushing or buckling.


Where Pcr – Buckling Load / Crippling

Columns and Struts which fail by buckling are analyzed by Euler’s theory.

Assumptions for Euler’s Equation for Elastic Instability

Material is homogenous, isotropic, and elastic. The cross-section area of the column remains uniform from top to bottom. The line of thrust coincides exactly with the axis of the column. The column is initially straight, loading must be axial and self-weight is negligible.

Euler’s Theory of Buckling

Moment at x-x axis

M = Py

As per the double integration method,


d2y/dx2+ (P/EI)y=0

For second-order differential equation solutions are:

y= c1 cos√(P/EI) x+ c2 sin√(P/EI) x

Boundary conditions at x=0, y=0. Therefore, c1 = 0

At x=L, y=0

√P/EI L=nπ

P= π2EI/L2 (Both sides pinned joint)

Generalized Euler’s Formula

Pcr= π2EI/KL2


KL = Effective Length

K = Effective Length Factor

L = Actual Length

Rankine’s Theory For Columns

Euler’s formula doesn’t provide accurate results for columns and struts of medium length. Rankine’s theory can be used for both long and short columns. Rankine’s theory assumes combined modes of failure due to buckling and crushing. Rankine gave an empirical formula :

1/P= 1/Pc+ 1/Pe


P = Combined Load

Pc = Crushing Load

Pe = Euler’s Buckling Load

P= fc A/(1+ αλ2)

Where α= fc2E Rankine Constant

λ = Slenderness Ratio

fc = Crushing Strength

For Long Column,

The value of λ >> λc

Pe will be small.

1Pe> 1Pc

Therefore, P = Pe

Long Columns fail under buckling.

For Short Column,

The value of λ << λc

Pe will be small.

1Pe< 1Pc

Therefore, P = Pc

Short Columns fail under crushing.

Columns and Struts Solved Problems

Problem 1: A rectangular column of dimension 40mm x 60mm and length of 3m has the end conditions as given:

“Restrained against rotation and translation at both ends”. Find the buckling load if E = 2 x 105 N/mm2.

Solution- Restrained against rotation and translation at both ends refers to the support system where both ends are fixed.

The effective length factor K = 0.5.

Leff = 0.5 x L

= 0.5 x 3000 = 1500mm

Pcr= π2EImin/leff2

= 280.75kN

Problem 2: There are two columns, A and B, and both the columns have the same E, L, and I. The effective length factor is 2 and 0.707 for columns A and B respectively. Determine the ratio of the buckling load of column B to column A.

Solution- P= π2EI/KL2

PA= π2EI/4L2

PB= π2EI/0.5L2


Problem 3: Given the Euler’s Buckling load for a column as 100kN and crushing load as 120kN. Find the Rankine Load (in kN).

Solution – Given, Pcr = 1000kN, Pc = 1200kN

1/P= 1/Pc+ 1/Pcr

P= PcPcr/(Pc+ Pcr)= 1200 ×1000/(1200+1000) = 545kN

Problem 4: A column of 1.5m long (diameter 50mm) is free at one end and fixed at another end. Crushing stress = 600 MPa. Find safe failure load if the factor of safety = 3. α = 1/1600.

Solution – K = 2

Leff = KL = 2 x 1.5 = 3000mm

fc = 600MPa, α = 1/1600

Safe Failure Load= Rankine Load (P)/ FOS

Rankine’s Load=P= fc A/(1+ αλ2)

λ= leff/ Rmin

Rmin= √Imin/A


λ= 3000/12.5=240


Safe Load= 31.8/3=10.61kN

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