2026-08-11

VE0096 | Euler Buckling

Description

A strut with circular cross-section is supported according to four basic cases of Euler buckling and it is subjected to pressure force P according to the following figure. Determine the critical load P_cr. The problem is described by the following parameters.

Material	Steel	Modulus of Elasticity	E	210000.000	MPa
Material	Steel	Poisson's Ratio	ν	0.300	-
Geometry	Strut	Length	L	2.000	m
Geometry	Strut	Diameter	d	30.000	mm
Load	Force	Force	P	10.000	kN

Euler Buckling

Analytical Solution

The governing differential equation for all buckling cases is following:

\frac{d^{2}}{d x^{2}} y = - \frac{M}{E I},

y	Coordinate in buckling direction
M	Bending moment related to the cross-section
I	Moment of inertia

Governing Differential Equation for Buckling

Buckling Case 1

The strut is fixed on one end and free on the other one. The governing equation can then be rewritten as:

\frac{d^{2}}{d x^{2}} y = - \frac{P}{E I} (δ - y)

δ	Maximum lateral deflection of the tip of the strut

Governing equation for Case 1 (Fixed-Free)

Using the substitution

α = \sqrt{\frac{P}{E I}}

Substitution

the solution of the equation above reads as

y = C_{1} \cos (α x) + C_{2} \sin (α x) + δ

General solution y for Case 1

Integration constants C₁ and C₂ can be obtained from the following boundary conditions

y (0) = 0

\frac{d}{d x} y (0) = 0

Boundary Conditions for Case 1

The resulting deflection then reads

y = δ (1 - \cos (α x)))

Resulting Deflection y for Case 1

For nonzero δ and for the tip of the strut, the following condition has to hold

\cos (α L) = 0 \Rightarrow α L = k \frac{π}{2}, k = 1, 3, 5, . . .

Condition for Case 1

The minimum value corresponds to k = 1 and the critical force P_cr is then, by substituting into the α relation,

P_{c r} = \frac{π^{2} E I}{4 L^{2}} \approx 5.151 kN

Critical Force for Case 1

Buckling Case 2

The strut is supported by means of pinned joints, where one is, furthermore, sliding. The governing equation reads as

\frac{d^{2}}{d x^{2}} y = - \frac{P}{E I} y

Governing Equation for Case 2 (Pinned-Pinned Sliding)

the solution of this differential equation is of the form

y = C_{1} \cos (α x) + C_{2} \sin (α x)

General Solution for Case 2

Integration constants C₁ and C₂ can be obtained from the following boundary conditions

y (0) = 0,

y (L) = 0

Boundary Conditions for Case 2

From the boundary conditions follows

C_{2} \sin (α L) = 0

Condition for Case 2

For nonzero constant C₂ it can be written

\sin (α L) = 0 \Rightarrow α L = k π, k = 1, 2, 3, . . .

Condition for Case 2

The minimum value corresponds to k = 1 and the critical force P_cr is then

P_{c r} = \frac{π^{2} E I}{L^{2}} \approx 20.602 k N

Critical Force for Case 2

Buckling Case 3

The strut is fixed on one end and supported by means of a sliding pinned joint on the other one. The governing equation can then be rewritten as follows

\frac{d^{2}}{d x^{2}} y = - \frac{P}{E I} (y - \frac{H}{P} (L - x))

Governing Equation for Case 3 (Fixed-Sliding Pinned)

where H is the horizontal reaction force in the pinned joint. The solution of this differential equation reads as

y = C_{1} \cos (α x) + C_{2} \sin (α x) + \frac{H}{P} (L - x)

General Solution for Case 3

Integration constants C₁ and C₂ can be obtained from the same boundary conditions as in Case 1. The resulting deflection is then

y = \frac{H}{P} (\frac{1}{α} \sin (α x) - L \cos (α x) + (L - x))

Resulting Deflection for Case 3

This equation must also satisfy the boundary condition for the second end of the strut, after substitution

0 = \frac{H}{P} (\frac{1}{α} \sin (α L) - L \cos (α L))

Boundary Condition Substitution for Case 3

For nonzero horizontal reaction force H, the following equation has to hold

\tan (α L) = α L

Transcendental Condition

This transcendental equation can be solved numerically or graphically, with approximate solution

P_{c r} \approx 42.147 kN

Critical Force for Case 3

Buckling Case 4

The strut is fixed on both ends and one end is sliding. The governing equation takes the form

\frac{d^{2}}{d x^{2}} y = - \frac{P}{E I} (y - \frac{M_{1}}{P})

M₁	Reaction moment in the fixed support

Governing Equation for Case 4 (Fixed-Fixed Sliding)

where M₁ is the reaction moment in the fixed support. The solution of this differential equation reads as

y = C_{1} \cos (α x) + C_{2} \sin (α x) + \frac{M_{1}}{P}

General Solution for Case 4

Integration constants C₁ and C₂ can be obtained from the boundary conditions of Case 1. The general solution then reads

y = \frac{M_{1}}{P} (1 - \cos (α x))

General Solution (Simplified) for Case 4

This equation must also satisfy the boundary condition for the second end of the strut, after substitution

0 = \frac{M_{1}}{P} (1 - \cos (α L))

Boundary Condition Substitution for Case 4

For nonzero reaction moment M₁, the following equation

\cos (α L) = 1 \Rightarrow α L = 2 k π, k = 1, 2, 3, . . .

Condition for Case 4

yields a minimum value for k = 1 and the critical force P_cr is

P_{c r} = \frac{4 π^{2} E I}{L^{2}} \approx 82.409 kN

Critical Force for Case 4

RFEM and RSTAB Settings

Modeled in RFEM 6.13, RSTAB 9.12 and RFEM 5.16 and RSTAB 8.16
Element size l_FE = 0.1 m
Isotropic linear elastic material is used
Lanczos method is used for eigenvalue analysis

Results

Buckling Case	Analytical Solution P_cr [kN]	RFEM 6 P_cr [kN]	Ratio [-]	RSTAB 9 P_cr [kN]	Ratio [-]	RFEM 5 – RF-STABILITY P_cr [kN]	Ratio [-]	RSTAB 8 – RSBUCK P_cr [kN]	Ratio [-]
Case 1	5.151	5.150	1.000	5.150	1.000	5.150	1.000	5.153	1.000
Case 2	20.602	20.594	1.000	20.594	1.000	20.596	1.000	20.749	1.007
Case 3	42.147	42.109	0.999	42.109	0.999	42.107	0.999	43.190	1.025
Case 4	82.409	82.289	0.999	82.285	0.998	82.270	0.998	83.324	1.011

Results in RFEM / RSTAB - First Eigenvector

Verification Example 0096 | 1

;