381 

© 2024 The Author(s). Published by College of Education for Pure Science (Ibn Al-Haitham), 

University of Baghdad. This is an open-access article distributed under the terms of the Creative Commons 

Attribution 4.0 International License 

 

 

Ibn Al-Haitham Journal for Pure and Applied Sciences 

Journal homepage: jih.uobaghdad.edu.iq 

PISSN: 1609-4042, EISSN: 2521-3407 

 IHJPAS. 2024, 37(4) 

 

 Analysis of Loaded Beam, Cantilever, and Elongated Vertical Column via 

Integral Rohit Transform  

Rohit Gupta1,* , Shivam Sharma2  and Anikait Gupta3  

1Applied Sciences, Faculty of Physics, Yogananda College of Engineering and Technology (YCET), Jammu, India. 

2, 3Civil Engineering, Yogananda College of Engineering and Technology (YCET), Jammu, India. 

*Corresponding Author. 

 

  Received: 9 February 2024                     Accepted: 21 April 2024                Published: 20 October 2024  

 doi.org/10.30526/37.4.3928 

 
Abstract  

Structural analysis is a branch of solid mechanics which utilizes straight forward models for solids. 

The main objective of Structural analysis is to find out the effect of loads on the physical structures 

and their components. A beam is a structure with a constant cross-section and is described by its 

significant length in comparison to its thickness and width. A cantilever, on the other hand, is a 

slender beam with a uniform cross-sectional shape that is fixed horizontally at one end and 

subjected to a load at the other end. Columns, which serve as vertical compression members in 

building frames, are susceptible to buckling and failure when subjected to relatively small axial 

loads. The analysis of loaded beams, cantilevers, and elongated vertical columns is typically 

carried out using the principles of calculus. However, this paper introduces the integral Rohit 

transform for the analysis of loaded beam supported at ends, cantilever, and elongated columns 

with low buckling axial loads. It is found that the depression grows as the cantilever and beam 

lengths that are loaded in the middle and supported at both ends rise. An attempt has been made to 

analyze the elongated column with low axial buckling loads and derive the Euler's formula for 

buckling load. The obtained solutions are graphically represented, and the results demonstrate 

accuracy, capability and effectiveness of the integral Rohit transform technique when compared 

to existing methods in the literature. The Rohit transform involves simple formulation and less 

computational work compared to other methods available in the literature. 

Keywords: Loaded Beam, Cantilever, Elongated Columns, Integral Rohit Transform 

 

 

 

 

https://creativecommons.org/licenses/by/4.0/
https://creativecommons.org/licenses/by/4.0/
https://orcid.org/0000-0002-9744-5131
mailto:guptarohit565@gmail.com
https://orcid.org/0009-0004-6302-6096
mailto:srsharma.sharma001@gmail.com
https://orcid.org/0000-0003-2250-8588
mailto:aniketgupta321@gmail.com


IHJPAS. 2024, 37(4 ) 

382 
 

1. Introduction 

A beam is a structural element with a uniform cross-section. It is described by its significant length 

compared to its width and thickness. In such structures, the shearing stress across any cross-section 

is considered to be negligibly small [1]. A cantilever, on the other hand, refers to a slender and 

uniform beam that is horizontally fixed at one end and subjected to loading at the opposite end. 

Beams are commonly employed in the construction of bridges or for the purpose of supporting 

heavy loads, often found in the structure of multistoried buildings [2]. In engineering design, the 

elastic behavior of materials assumes a critical role in various applications such as the construction 

of buildings, bridges, automobiles, and rope-ways. The property of elasticity in beam materials 

leads to the generation of a restoring couple when subjected to deforming forces, which acts in 

equilibrium and is equal in magnitude but opposite in direction to the bending couple. This 

restoring moment is known as the bending moment [3]. Building frames employ columns as one 

of their vertical compression components, and they are susceptible to buckling and failing at mild 

axial stresses. These buckles when the axial load reaches a threshold value known as the critical 

buckling load because they are significantly longer than their lateral dimensions [4]. One of the 

failures of a structure supporting a load is buckling. Because they are thin, columns buckle when 

the axial load reaches a threshold amount called the critical buckling load. They also deflect 

laterally when compressed. It has been shown that low buckling axial loads cause the columns to 

fail. According to Euler's Theory of Columns, a column behaves to resist buckling. Buckling is 

influenced by the end condition of the columns and flexural rigidity. Commonly, standard methods 

like the calculus technique are used to analyze loaded beams supported at their ends, cantilevers, 

and elongated columns with low buckling axial loads [5–8]. Moreover, Euler's Theory of Columns 

is used to find out the buckling load of the column. This study presents the analysis of the loaded 

beam supported at its ends, the cantilever, and the elongated columns with modest buckling axial 

loads via the integral Rohit transform. This integral transform [9] has been proposed by the author 

Rohit Gupta in the year 2020. It has been applied to solve initial value problems in science and 

engineering [10-12]. In contrast to the calculus method, the suggested method presents an alternate 

approach for the analysis of loaded beams supported at their ends, cantilevers, and elongated 

columns with low buckling axial loads.  

 

2. Rohit Transform and Its Properties 

The integral Rohit transform, also written as integral RT, [9] is defined for a function of 

exponential order by the integral Equations as 

R{h(t)} =  𝑞3 ∫ 𝑒−𝑞𝑡∞

0
ℎ(𝑡)𝑑𝑡, 𝑡 ≥ 0 , 𝑞1 ≤ 𝑞 ≤  𝑞2. 

The variable q is used to factor the variable t in the argument of the function h. 

The Rohit transforms of unidentified functions [10] are given by  

 𝑅 {𝑡𝑛} =
𝑛!

𝑞𝑛−2 

 𝑅 {𝑠𝑖𝑛𝑏𝑡} =
𝑏 𝑞3

𝑞2+𝑏2 

 𝑅 {𝑐𝑜𝑠𝑏𝑡} =
 𝑞4

𝑞2+𝑏2 

 𝑅 {𝑒𝑏𝑡} =
 𝑞3

𝑞−𝑏
 

 



IHJPAS. 2024, 37(4 ) 

383 
 

The Rohit transforms (RT) of some derivatives are [11] given by  

𝑅 {𝑔′(𝑡)} = 𝑞𝐺(𝑞) −  𝑞3𝑔(0), 

𝑅{𝑔′′(𝑡)} = 𝑞2𝐺(𝑞) −  𝑞4𝑔(0) −  𝑞3𝑔′(0), 

𝑅{𝑔′′′(𝑡)} = 𝑞3𝐺(𝑞) − 𝑞5𝑔(0) −  𝑞4𝑔′(0) −  𝑞3𝑔′′(0). 

In general, 𝑅{𝑔n(𝑡)} = 𝑞n𝑅{𝑔(𝑡)} − ∑ 𝑞𝑛−𝑘+3𝑛
𝑘=1 𝑔k−1(0). 

A unit step function is written as 𝑈(𝑡 − 𝑎) = 0 𝑓𝑜𝑟 𝑡 < 𝑎 𝑎𝑛𝑑 1 𝑓𝑜𝑟 𝑡 ≥ a. 

The Rohit transform of a unit step function is given by 

R{𝑈(𝑡 − 𝑎)} =  𝑞3 ∫ 𝑒−𝑞𝑡
∞

0

𝑈(𝑡 − 𝑎)𝑑𝑡, 

R{𝑈(𝑡 − 𝑎)} =  𝑞3 ∫ 𝑒−𝑞𝑡
∞

𝑎

𝑑𝑡, 

R{𝑈(𝑡 − 𝑎)} =  𝑞2𝑒−𝑞𝑎 . 

 

Shifting property of Rohit transform 

Let  R{g(t)} = 𝐺(𝑞), 𝑡ℎ𝑒𝑛 𝑅[𝑔(𝑡 − 𝑎)𝑈(𝑡 − 𝑎)] = 𝑒−𝑞𝑎𝐺(𝑞). 

Proof: The Rohit transform of [𝑔(𝑡 − 𝑎)𝑈(𝑡 − 𝑎)] is given by 

𝑅[𝑔(𝑡 − 𝑎)𝑈(𝑡 − 𝑎)] =  𝑞3 ∫ 𝑒−𝑞𝑡
∞

0

𝑔(𝑡 − 𝑎)𝑈(𝑡 − 𝑎)𝑑𝑡, 

𝑅[𝑔(𝑡 − 𝑎)𝑈(𝑡 − 𝑎]) =  𝑞3 ∫ 𝑒−𝑞𝑡
∞

𝑎

𝑔(𝑡 − 𝑎)𝑑𝑡, 

𝑅[𝑔(𝑡 − 𝑎)𝑈(𝑡 − 𝑎)] =  𝑞3 ∫ 𝑒−𝑞(𝑣+𝑎)
∞

0

𝑔(𝑣)𝑑𝑣,     𝑤ℎ𝑒𝑟𝑒 𝑣 = 𝑡 − 𝑎, 

𝑅[𝑔(𝑡 − 𝑎)𝑈(𝑡 − 𝑎)] = 𝑒−𝑞(𝑎) 𝑞3 ∫ 𝑒−𝑞(𝑣)
∞

0

𝑔(𝑣)𝑑𝑣, 

𝑅[𝑔(𝑡 − 𝑎)𝑈(𝑡 − 𝑎)] = 𝑒−𝑞(𝑎) 𝑞3 ∫ 𝑒−𝑞(𝑡)
∞

0

𝑔(𝑡)𝑑𝑡, 

𝑅[𝑔(𝑡 − 𝑎)𝑈(𝑡 − 𝑎)] = 𝑒−𝑞(𝑎)𝐺(𝑞). 

 

3. Algorithm for Proposed Method 

The algorithm for the proposed method is as follows: 

Firstly, brief information of the integral Rohit transform and its attributes is provided.  

Secondly, the analysis of loaded beam supported at ends, cantilever, and elongated columns with 

low buckling axial loads is done via the integral Rohit transform.   

Thirdly, the obtained solutions are graphically represented, and the results obtained are compared 

to existing methods in the literature. 

Finally, the conclusions of the study are presented. 

 

 



IHJPAS. 2024, 37(4 ) 

384 
 

4.  Material and Method 

In this section, the analysis of loaded beam supported at its ends, cantilevers, and elongated 

columns with low buckling axial loads is carried out via the integral Rohit transform. 

4.1 Analysis of Loaded Beam Supported at Its Ends 

In this study, consider a beam supported on the two knife edges A and B and loaded in the middle 

with a load W vertically downwards. Let L be the length of the beam between the points A (at x = 

0) and B (at x = L). The bending moment [1], [12] at the section X is given by the differential 

Equation: 

�̈�(x) +  
W

2YI
𝑥 = 0                                                                                                                           (1) 

where ‘I’ is the geometrical moment of inertia and Y is the Young’s modulus. Here y is the 

depression of the beam at the section X at the distance x from the end A.  

On taking the Rohit transform of Equation (1), we have 

𝑅{�̈�(x)} +  
W

2YI
𝑅{𝑥} = 0, 

 𝑞3 ∫ 𝑒−𝑞𝑥
∞

0

�̈�(x) 𝑑𝑥 +
W

2YI
𝑞 = 0, 

 𝑞3 [∫ 𝑒−𝑞𝑥
𝐿

0

�̈�(x) 𝑑𝑥 + ∫ 𝑒−𝑞𝑥
∞

𝐿

�̈�(x) 𝑑𝑥] +
W

2YI
𝑞 = 0. 

As 0 < x < L, therefore, ∫ 𝑒−𝑞𝑥∞

𝐿
�̈�(x) 𝑑𝑥 = 0. 

Thus,  

 𝑞3 ∫ 𝑒−𝑞𝑥
𝐿

0

�̈�(x) 𝑑𝑥 +
W

2YI
𝑞 = 0, 

 𝑞3 [𝑒−𝑞𝐿𝑦′(𝐿) − 𝑦′(0) + 𝑞 ∫ 𝑒−𝑞𝑥
𝐿

0

y′(𝑥)𝑑𝑥] +
W

2YI
𝑞 = 0, 

 𝑞3 [𝑒−𝑞𝐿𝑦′(𝐿) − 𝑦′(0) + 𝑞 𝑒−𝑞𝐿𝑦(𝐿) − 𝑞 𝑦(0) + 𝑞2 ∫ 𝑒−𝑞𝑥
𝐿

0

y(x) 𝑑𝑥] +
W

2YI
𝑞 = 0, 

 𝑞3𝑒−𝑞𝐿𝑦′(𝐿) −  𝑞3𝑦′(0) +  𝑞4 𝑒−𝑞𝐿𝑦(𝐿) −  𝑞4 𝑦(0) + 𝑞2𝑅{𝑦(𝑥)} +
W

2YI
𝑞 = 0                             (2) 

 Applying initial conditions: 𝑦 (0) = y(L) = 0, ẏ(0) = 𝐶 𝑎𝑛𝑑  ẏ(L) = 𝐷. Equation (2) becomes 

 𝑞3𝑒−𝑞𝐿𝐷 −  𝑞3𝐶 + 𝑞2𝑅{𝑦(𝑥) +
W

2YI
𝑞 = 0, 

𝑅{𝑦(𝑥)} = 𝑞𝐶 − 𝑞𝑒−𝑞𝐿𝐷 +
W

2YI

1

𝑞
 . 

Taking inverse Rohit transform, we have  

 y (x)  = 𝐶𝑥 − 𝐷(𝑥 − 𝐿) 𝑈(𝑡 − 𝐿) −  
W

2YI

𝑥3

3!
.                                                                                  (3) 

Now, for x <  L, 𝑈(𝑥 − 𝐿) = 0. 

Thus, y (x)  = 𝐶𝑥 −
W

2YI

𝑥3

3!
                                                                                                             (4) 

At x =
L

2
 and ẏ (

L

2
) = 0. Therefore, using Equation (4) and solving for C, we get 

𝐶 =
3W𝐿2

32YI
                                                                                                                                        (5) 

 



IHJPAS. 2024, 37(4 ) 

385 
 

Using Equation (5) in Equation (4), we have 

y (x) = (
3W𝐿2

32YI
𝑥 −

W

2YI

𝑥3

3!
), where 0 < x < L                                                                                    (6) 

Taking, for example,  
W

YI
= 32 𝑎𝑛𝑑 𝐿 = 10, the graph of y(x) is shown in the Figure 1. 

 

Figure 1. Numerical solution of Equation (1). 

 

At the middle, the total depression is given by 

y (L/2)  =  
1

3!
 [

3W𝐿2

8YI
𝐿/2 −  

W

2YI
(𝐿/2)3], 

⇒ y (L/2)  =  
W𝐿3

48YI
                                                                                                                       (7) 

For a beam of circular cross-section [4], [13], we have 

I = 𝜋𝑟4/4 

Hence, from Equation (7), we have 

y(L/2) =  
W𝐿3

12Y𝜋𝑟4
 

4.2 Analysis of Cantilever Beam 

In this study, consider a horizontal beam AB of length L attached at end A (at x = 0) and loaded 

with a load W vertically downward from the free end B (at x = L). The bending moment [3], [4] 

at point X is obtained from the differential Equation: 

�̈�(x)  +  
W

YI
(L −  x) = 0                                                                                                                 (8) 

where I is the geometrical moment of inertia and Y is the Young’s modulus. Here, y is the 

depression of the beam at the section X at the distance x from fixed end A. 

On taking the Rohit transform [10], [11] of Equation (8), we get 

 𝑞3 ∫ 𝑒−𝑞𝑥
∞

0

�̈�(x) 𝑑𝑥 +
W

YI
(𝐿𝑞2 − 𝑞) = 0, 

 𝑞3 [∫ 𝑒−𝑞𝑥
𝐿

0

�̈�(x) 𝑑𝑥 + ∫ 𝑒−𝑞𝑥
∞

𝐿

�̈�(x) 𝑑𝑥] +
W

YI
(𝐿𝑞2 − 𝑞) = 0. 

As 0 < x < L, therefore, ∫ 𝑒−𝑞𝑥∞

𝐿
�̈�(x) 𝑑𝑥 = 0. 

Thus,  

 𝑞3 ∫ 𝑒−𝑞𝑥
𝐿

0

�̈�(x) 𝑑𝑥 +
W

YI
(𝐿𝑞2 − 𝑞) = 0, 

 𝑞3 [𝑒−𝑞𝐿𝑦′(𝐿) − 𝑦′(0) + 𝑞 ∫ 𝑒−𝑞𝑥
𝐿

0

y′(𝑥)𝑑𝑥] +
W

YI
(𝐿𝑞2 − 𝑞) = 0, 



IHJPAS. 2024, 37(4 ) 

386 
 

 𝑞3 [𝑒−𝑞𝐿𝑦′(𝐿) − 𝑦′(0) + 𝑞 𝑒−𝑞𝐿𝑦(𝐿) − 𝑞 𝑦(0) + 𝑞2 ∫ 𝑒−𝑞𝑥
𝐿

0

y(x) 𝑑𝑥] +
W

YI
(𝐿𝑞2 − 𝑞) = 0, 

 𝑞3𝑒−𝑞𝐿𝑦′(𝐿) −  𝑞3𝑦′(0) +  𝑞4 𝑒−𝑞𝐿𝑦(𝐿) −  𝑞4 𝑦(0) + 𝑞2𝑅{𝑦(𝑥)} +
W

YI
(𝐿𝑞2 − 𝑞) = 0          (9) 

 Applying initial conditions: 𝑦 (0) = 0, ẏ(0) = 0, 𝑦′(𝐿) = 𝐷 𝑎𝑛𝑑 𝑦(𝐿) = 𝐶. Equation (9) 

becomes,  

 𝑞3𝑒−𝑞𝐿𝐷 +  𝑞4 𝑒−𝑞𝐿𝐶 + 𝑞2𝑅{𝑦(𝑥)} +
W

YI
(𝐿𝑞2 − 𝑞) = 0, 

𝑅{𝑦(𝑥)} = −𝑞𝑒−𝑞𝐿𝐷 −  𝑞2 𝑒−𝑞𝐿𝐶 −
W

YI
(𝐿 −

1

𝑞
)                                                                        (10) 

Taking inverse Rohit transform, we have 

y(x) = −D (𝑥 − 𝐿)𝑈(𝑥 − 𝐿) − C𝑈(𝑥 − 𝐿) + 
W

YI
 (𝐿𝑥2 −

𝑥3

3!
)                                                    

Now, for x <  L, 𝑈(𝑥 − 𝐿) = 0. 

Therefore,  

y(x) =
W

YI
(𝐿𝑥2 −

𝑥3

3!
)                                                                                                                   (11) 

Taking, for example,  
W

YI
= 32 𝑎𝑛𝑑 𝐿 = 10, the graph of y(x) is shown in the Figure 2. 

 

Figure 2. Numerical solution of Equation (8). 

 

4.3 Analysis of Elongated Vertical Column  

In this study, consider an elongated vertical column AB (A is at top and B is at bottom) of 

length 'L' and of uniform cross-section. Let "y" be the lateral deflection of the column section at 

height "x". We now consider three different cases: 

Case-I: When both ends A and B of the column are pinned or hinged 

In this case, the bending moment [1], [13] at the section is given by  

�̈�(𝑥) + 𝑘2𝑦(𝑥) = 0,                                                                                                                     (12) 

 where k =  √
𝑃

𝑌I
 .  

Taking Rohit transform of Equation (12), we get 

 𝑞3 ∫ 𝑒−𝑞𝑥
∞

0

�̈�(x) 𝑑𝑥 + 𝑘2𝑅{𝑦(𝑥)} =  0, 

 𝑞3 [∫ 𝑒−𝑞𝑥𝐿

0
�̈�(x) 𝑑𝑥 + ∫ 𝑒−𝑞𝑥∞

𝐿
�̈�(x) 𝑑𝑥] + 𝑘2𝑅{𝑦(𝑥)} = 0                                                     (13) 

As 0 < x < L, therefore, ∫ 𝑒−𝑞𝑥∞

𝐿
�̈�(x) 𝑑𝑥 = 0. 



IHJPAS. 2024, 37(4 ) 

387 
 

Thus,  

 𝑞3 ∫ 𝑒−𝑞𝑥
𝐿

0

�̈�(x) 𝑑𝑥 + 𝑘2𝑅{𝑦(𝑥)} = 0, 

 𝑞3 [𝑒−𝑞𝐿𝑦′(𝐿) − 𝑦′(0) + 𝑞 ∫ 𝑒−𝑞𝑥
𝐿

0

y′(𝑥)𝑑𝑥] 𝑘2𝑅{𝑦(𝑥) = 0, 

 𝑞3 [𝑒−𝑞𝐿𝑦′(𝐿) − 𝑦′(0) + 𝑞 𝑒−𝑞𝐿𝑦(𝐿) − 𝑞 𝑦(0) + 𝑞2 ∫ 𝑒−𝑞𝑥
𝐿

0

y(x) 𝑑𝑥] + 𝑘2𝑅{𝑦(𝑥) = 0, 

 𝑞3𝑒−𝑞𝐿𝑦′(𝐿) −  𝑞3𝑦′(0) +  𝑞4 𝑒−𝑞𝐿𝑦(𝐿) −  𝑞4 𝑦(0) + 𝑞2𝑅{𝑦(𝑥)} + 𝑘2𝑅{𝑦(𝑥)} = 0      (14) 

Applying initial conditions: 𝑦(0) = 0, y(L) = 0, 𝑦′(0) = 𝐴, 𝑎𝑛𝑑 𝑦′(𝐿) = 𝐵. Equation (14) becomes 

 𝑞3𝑒−𝑞𝐿𝐵 −  𝑞3𝐴 + 𝑞2𝑅{𝑦(𝑥)} + 𝑘2𝑅{𝑦(𝑥)} = 0, 

𝑅{𝑦(𝑥)}(𝑞2 + 𝑘2) = − 𝑞3𝑒−𝑞𝐿𝐵 +  𝑞3𝐴, 

{𝑦(𝑥)} =
− 𝑞3𝑒−𝑞𝐿𝐵

𝑞2+𝑘2 +
 𝑞3𝐴

𝑞2+𝑘2                                                                                                         (15)  R 

Taking inverse Rohit transform of Equation (15), we get 

𝑦(x)  =  −
𝐵

𝑘
 sin k (x − L)U(x − L) +  

𝐴

𝑘
 sin (k x).      

Now, for x <  L, 𝑈(𝑥 − 𝐿) = 0. 

Thus, 

 𝑦(𝑥) =
𝐴

𝑘
𝑠𝑖𝑛(𝑘𝑥)                                                                                                               (16) 

As 𝑦(L) = 0, therefore, Equation (16) gives  

sin (k L) =  0,   

where n is an integer greater than equal to zero. 

𝑘𝐿 = 𝑛 𝜋, 

k = 
n𝜋

𝐿
                                                                                                                                  (17)          

The least practical value of n is 1, therefore, considering n = 1, we have  

𝑘 =
𝜋

𝐿
 , 

√
P

YI
=  

π

L
 , 

 P = 
π2YI

L2                                                                                                                               (18) 

The Euler's formula for the critical buckling load of the elongated column with pins at both ends 

is found in Equation (18). 

Case-II: When the bottom end B of the column is fixed and the upper end A is hinged 

In this case, the bending moment [4] at the section is given by          

�̈�(𝑥) + 𝑘2𝑦(𝑥)  =  𝐻(𝐿 − 𝑥)                                                                                                      (19)  

where H =
H0

EI
. Here, H0 is horizontal force at the fixed end 𝐵. 

Taking Rohit transform of Equation (19), we get 

 𝑞3 ∫ 𝑒−𝑞𝑥
∞

0

�̈�(x) 𝑑𝑥 + 𝑘2𝑅{𝑦(𝑥)} =  𝐻(𝐿 𝑞2 − 𝑞), 

 𝑞3 [∫ 𝑒−𝑞𝑥
𝐿

0

�̈�(x) 𝑑𝑥 + ∫ 𝑒−𝑞𝑥
∞

𝐿

�̈�(x) 𝑑𝑥] + 𝑘2𝑅{𝑦(𝑥)} = 𝐻(𝐿 𝑞2 − 𝑞). 

As 0 < x < L, therefore, ∫ 𝑒−𝑞𝑥∞

𝐿
�̈�(x) 𝑑𝑥 = 0. 

 



IHJPAS. 2024, 37(4 ) 

388 
 

Thus,  

 𝑞3 ∫ 𝑒−𝑞𝑥
𝐿

0

�̈�(x) 𝑑𝑥 + 𝑘2𝑅{𝑦(𝑥)} = 𝐻(𝐿 𝑞2 − 𝑞), 

 𝑞3 [𝑒−𝑞𝐿𝑦′(𝐿) − 𝑦′(0) + 𝑞 ∫ 𝑒−𝑞𝑥
𝐿

0

y′(𝑥)𝑑𝑥] + 𝑘2𝑅{𝑦(𝑥) = 𝐻(𝐿 𝑞2 − 𝑞), 

 𝑞3 [𝑒−𝑞𝐿𝑦′(𝐿) − 𝑦′(0) + 𝑞 𝑒−𝑞𝐿𝑦(𝐿) − 𝑞 𝑦(0) + 𝑞2 ∫ 𝑒−𝑞𝑥
𝐿

0

y(x) 𝑑𝑥] + 𝑘2𝑅{𝑦(𝑥) = 𝐻(𝐿 𝑞2 − 𝑞), 

𝑞3𝑒−𝑞𝐿𝑦′(𝐿) − 𝑞3𝑦′(0) +  𝑞4𝑒−𝑞𝐿𝑦(𝐿) − 𝑞4 𝑦(0) + 𝑞2𝑅{𝑦(𝑥)} + 𝑘2𝑅{𝑦(𝑥)} = 𝐻(𝐿 𝑞2 − 𝑞)                                 

                                                                                                                                                     (20) 

Applying initial conditions: 𝑦(0) = 0, y(L) = 0, 𝑦′(0) = 0, 𝑎𝑛𝑑 𝑦′(𝐿) = 𝐵. Equation (20) becomes 

𝑞3𝑒−𝑞𝐿𝐵 + 𝑞2𝑅{𝑦(𝑥)} + 𝑘2𝑅{𝑦(𝑥)} = 𝐻(𝐿 𝑞2 − 𝑞), 

𝑅{𝑦(𝑥)}(𝑞2 + 𝑘2) = − 𝑞3𝑒−𝑞𝐿𝐵 + 𝑘2𝑑 𝑞2, 

𝑅{𝑦(𝑥)} = −
 𝑞3𝑒−𝑞𝐿𝐵

𝑞2 + 𝑘2
+

𝐻𝐿 𝑞2

𝑞2 + 𝑘2
−

𝐻𝑞

𝑞2 + 𝑘2
 , 

𝑅{𝑦(𝑥)} = −
 𝑞3𝑒−𝑞𝐿𝐵

𝑞2 + 𝑘2
+

𝐻𝐿

𝑘2
[𝑞2 −

𝑞4

𝑞2 + 𝑘2
] −

𝐻

𝑘2
[𝑞 −

𝑞3

𝑞2 + 𝑘2
]                                             (21) 

Taking inverse Rohit transform of Equation (21), we get 

𝑦(x) = −
𝐵

𝑘
 sin k (x − L)U(x − L) +  H [

𝐿

𝑘2 −
𝐿

𝑘2 cos (k x)] − H [
𝑥

𝑘2 −
sin 𝑘𝑥

𝑘3  ]                           (22)  

Now for x <  L, 𝑈(𝑥 − 𝐿) = 0. 

Thus, from Equation (22), we have 

𝑦(x) = H [
𝐿

𝑘2 −
𝐿

𝑘2 cos (k x) −
𝑥

𝑘2 +
sin 𝑘𝑥

𝑘3  ]                                                                                         (23)  

Applying the condition: 𝑦(L) = 0. Equation (23) gives 

H  [
𝐿

𝑘2  −
𝐿

𝑘2 cos (k L) − 
𝐿

𝑘2  + 
sin 𝑘𝐿

𝑘3  ] = 0, 

[−
𝐿

𝑘2  cos (k L)  +  
sin 𝑘𝐿

𝑘3  ]  = 0,  

𝐿

𝑘2
cos (k L) =

sin 𝑘𝐿

𝑘3
 

tan (k L) =  kL                                                                                                                                           (24) 

and solving, we get 𝑘𝐿 power of thupto 5 tan 𝑘𝐿 On expanding 

 𝑘𝐿 = 4.5 radians, 

 √
P

YI
𝐿 = 4.5 radians 

 P = 
20.25YI

𝐿2 , 

 P = 
2π2YI

𝐿2                                                                                                                                       (25) 

The Equation (25) is Euler's formula for the critical buckling load of the elongated column with a 

fixed lower end and a fixed upper end. 

Case-III: When both the ends A and B of the column are fixed 

In this case, the bending moment at the section is given by 

 

  �̈�(𝑥) + 𝑘2𝑦(𝑥) = M                                                                                                                  (26) 

where M = 
M0

EI
. Here, M0is the restraint moment at each end. 

 



IHJPAS. 2024, 37(4 ) 

389 
 

Taking Rohit transform of Equation (26), we get 

 𝑞3 ∫ 𝑒−𝑞𝑥
∞

0

�̈�(x) 𝑑𝑥 + 𝑘2𝑅{𝑦(𝑥)} =  𝑀 𝑞2, 

 𝑞3 [∫ 𝑒−𝑞𝑥
𝐿

0

�̈�(x) 𝑑𝑥 + ∫ 𝑒−𝑞𝑥
∞

𝐿

�̈�(x) 𝑑𝑥] + 𝑘2𝑅{𝑦(𝑥)} = 𝑀 𝑞2. 

As 0 < x < L, therefore, ∫ 𝑒−𝑞𝑥∞

𝐿
�̈�(x) 𝑑𝑥 = 0. 

Thus,  

 𝑞3 ∫ 𝑒−𝑞𝑥
𝐿

0

�̈�(x) 𝑑𝑥 + 𝑘2𝑅{𝑦(𝑥)} = 𝑀 𝑞2, 

 𝑞3 [𝑒−𝑞𝐿𝑦′(𝐿) − 𝑦′(0) + 𝑞 ∫ 𝑒−𝑞𝑥
𝐿

0

y′(𝑥)𝑑𝑥] + 𝑘2𝑅{𝑦(𝑥) = 𝑀 𝑞2, 

 𝑞3 [𝑒−𝑞𝐿𝑦′(𝐿) − 𝑦′(0) + 𝑞 𝑒−𝑞𝐿𝑦(𝐿) − 𝑞 𝑦(0) + 𝑞2 ∫ 𝑒−𝑞𝑥
𝐿

0

y(x) 𝑑𝑥] + 𝑘2𝑅{𝑦(𝑥) = 𝑀 𝑞2, 

 𝑞3𝑒−𝑞𝐿𝑦′(𝐿) −  𝑞3𝑦′(0) +  𝑞4 𝑒−𝑞𝐿𝑦(𝐿) −  𝑞4 𝑦(0) + 𝑞2𝑅{𝑦(𝑥)} + 𝑘2𝑅{𝑦(𝑥)} = 𝑀 𝑞2    (27) 

Applying initial conditions: 𝑦(0) = 0, y(L) = 0, 𝑦′(0) = 0, 𝑎𝑛𝑑 𝑦′(𝐿) = 𝐵. Equation (27) becomes 

 𝑞3𝑒−𝑞𝐿𝐵 + 𝑞2𝑅{𝑦(𝑥)} + 𝑘2𝑅{𝑦(𝑥)} = 𝑀 𝑞2, 

𝑅{𝑦(𝑥)}(𝑞2 + 𝑘2) = − 𝑞3𝑒−𝑞𝐿𝐵 + 𝑘2𝑑 𝑞2, 

𝑅{𝑦(𝑥)} = −
 𝑞3𝑒−𝑞𝐿𝐵

𝑞2 + 𝑘2
+

𝑀 𝑞2

𝑞2 + 𝑘2
 , 

𝑅{𝑦(𝑥)} = −
 𝑞3𝑒−𝑞𝐿𝐵

𝑞2 + 𝑘2
+

𝑀

𝑘2
[𝑞2 −

𝑞4

𝑞2 + 𝑘2
]                                                                                     (28) 

Taking inverse Rohit transform of Equation (28), we get 

y(x) = −
𝐵

𝑘
 sin k (x − L)U(x − L) +

𝑀

𝑘2 [1- cos (k x)]                                                                 (29) 

Now, for x <  L, 𝑈(𝑥 − 𝐿) = 0. 

Thus, from Equation (29), we have 

y(x) = 
𝑀

𝑘2 [1- cos (k x)]                                                                                                                 (30) 

Applying the condition: 𝑦(L) = 0. Equation (30) gives 

   
𝑀

𝑘2 −
𝑀

𝑘2 cos (k L) = 0, 

 cos (k L)  = 1, 

  𝑘 =
2n𝜋

𝐿
                                                                                                                                      (31) 

The least practical value of n is 1, therefore, considering n = 1, we have  

𝑘 =
2𝜋

𝐿
, 

√
P

YI
=  

2π

L
 , 

 P = 
4π2YI

L2                                                                                                                                     (32) 

The Equation (32) is the Euler’s formula for critical buckling load for the elongated column 

whose both ends are fixed. 

 

5. Discussion 

The integral Rohit transform has effectively handled the analysis of a beam supported at both ends 



IHJPAS. 2024, 37(4 ) 

390 
 

and loaded in the middle, as well as cantilevers and elongated columns with modest buckling axial 

stress. Figures 1 and 2 make it abundantly evident that the depression grows as the cantilever and 

beam lengths that are loaded in the middle and supported at both ends rise. An attempt has been 

made to provide an example of the Rohit transform in order to analyze the elongated column with 

low axial buckling loads and derive the buckling load Euler's formula. It is found that the critical 

buckling load for elongated columns subjected to axial loads is inversely related to the square of 

length of the column in all of the cases that were studied. 

 

6. Conclusion 

According to the calculus approach described in the literature [14-21], the results obtained by the 

integral Rohit transform are accurate. This demonstrates the efficacy and ability of the method to 

analyze beams supported at both ends and loaded in the middle, as well as cantilevers and 

elongated columns with minimal axial buckling loads. In contrast to the calculus method, the 

suggested method presents an alternate approach for the analysis of loaded beams supported at 

their ends, cantilevers, and elongated columns with low buckling axial loads. The dominance of 

integral Rohit transform over other methods available in the literature is in terms of simplicity, 

speed, and accuracy. It involves simple formulation and less computational work compared to 

other methods available in the literature. 

 

Acknowledgment 

The authors would like to thank Prof. Dinesh Verma for his guidance. 

 

Conflict of Interest  

 The authors declare that they have no conflicts of interest. 

Funding 

There is no financial support in preparation for the publication. 

 

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