# Columns and Struts | Difference Between Columns and Struts

By Aina Parasher|Updated : May 26th, 2022

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.

## 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

Where, 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, 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 defined as the ratio of effective length to the least lateral dimension.

λ = Leff/b

Where, 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 the top to the 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,

EId2y/dx2 =-Py

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

Where

KL = Effective Length

K = Effective Length Factor

L = Actual Length

## Rankine’s Theory For Columns

For columns and struts of medium length, Euler’s formula doesn’t provide accurate results. 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

Where

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 & Struts 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

PB/PA=8

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= Pc Pcr/(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 other end. Crushing stress = 600 MPa. Find safe failure load if factor of safety = 3. α = 1/1600.

Solution – K = 2

Leff = KL = 2 x 1.5 = 3000mm

fc = 600MPa, α = 1/1600

λ= leffRmin

Rmin= √Imin/A

D/4=0.5mm

λ= 3000/12.5=240

P=31.8kN

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## FAQs on Columns and Struts

• “Effectively held in position at both ends” – The translation is restrained at both ends.

“Restrained against rotation at one end” – Rotation is restricted at one end and rotation is allowed at another end.

The above conditions are true for columns whose one end is fixed and another end is hinged.

k= 1/2=0.707

• Column is a vertical member that is designed to carry longitudinal loads whose both ends are provided with fixed supports. It is designed to carry different types of loads including gravity loads. The failure of the column occurs by buckling and columns are analyzed using Euler’s theory.

Strut is a horizontal/vertical/inclined member which is usually used in the structure to bring stability to the structure. Failure of the strut is due to crushing.

• A strut is a short column that can be a horizontal/vertical/inclined member that is designed to carry axial compression as well as tensile load. Struts are used to bring stability to the structure.

• It is a type of failure of a column when axial/longitudinal load is applied to a column. The column undergoes a lateral displacement under which the column might fail. This type of failure is called buckling in a column and the load responsible for buckling is called buckling or crippling load.

• A column is a vertical compression member which is provided for supporting beams, ceilings, and the horizontal load coming over it. GradeStack Learning Pvt. Ltd.Windsor IT Park, Tower - A, 2nd Floor, Sector 125, Noida, Uttar Pradesh 201303 help@byjusexamprep.com