2026-09-08

VE0062 | Gravity Loading

Description

A rod with a square cross-section is fixed on the top end according to the following figure. The rod is loaded by its self-weight. For comparison, the example is also modeled with the concentrated force load of which the value is equal to the gravity. The aim of this verification example is to show the difference between these types of loading, although the total loading force is equal. The problem is described by the following set of parameters.

Material	Steel	Modulus of Elasticity	E	210000.000	MPa
		Poisson's Ratio	ν	0.296	-
		Density	ρ	7850.000	kg/m³
Geometry	Rod	Length	L	10.000	m
Geometry	Rod	Cross-Section Width	w	20.000	mm
Load		Gravitational Acceleration	g	10.000	m/s²

Gravity Loading

The self-weight is neglected in the first case. Determine the maximum deformation of the rod u_z,max.

Analytical Solution

The equation of equilibrium in one-dimensional case can be written as:

\frac{d σ_{z}}{d z} + f_{z} = 0

σ_z	Axial stress
f_z	Volume force

Equation of Equilibrium for One-Dimensional Case

The axial stress σ_z is defined by Hooke's law.

σ_{z} = E ε_{z}

Hooke's Law

Concentrated Force

At first, the loading by the concentrated force is considered. The concentrated force is equal to the gravity.

G = g ρ A L = 314.000 N

Gravitational Force

The volume force f_z is equal to zero in this case. Then the strain ε_z remains constant

\frac{d σ_{z}}{d z} = \frac{d (E ε_{z})}{d z} = \frac{E d ε_{z}}{d z} = 0

Equation of Equilibrium for Concentrated Force

where the strain ε_z is defined as:

ε_{z} = \frac{d u_{z}}{d z}

Axial strain

Then the final differential equation is:

\frac{d^{2} u_{z}}{d z^{2}} = 0

Final Differential Equation for Concentrated Force

The solution of this differential equation has the form of the following linear function.

u_{z} (z) = C_{1} z + C_{2}, C_{1} \in ℝ, C_{2} \in ℝ

General Solution of the Differential Equation

The integration constants C₁, C₂ can be obtained from the boundary conditions.

\frac{d u_{z}}{d z} = ε_{z} = \frac{G}{E A},

u_{z} (L) = 0 .

Boundary Conditions

Hence,

C_{1} = \frac{G}{E A},

C_{2} = - \frac{G L}{E A} .

Integration Constants

Finally, the formula for the displacement u_z can be completed and the maximum displacement u_z,max calculated.

u_{z} (z) = \frac{G}{E A} (L - z),

u_{z, \max} = \frac{G L}{E A} = - 0.037 mm .

Maximum Displacement

Self-weight

In case of self-weight, the volume force is defined as:

f_{z} = ρ g

Volume Force

Using the equation of equilibrium mentioned above again, the differential equation of equilibrium can be rewritten as follows:

\frac{d^{2} u_{z}}{d z^{2}} = - \frac{ρ g}{E}

Equation of Equilibrium for Self-Weight

Please note that the strain ε_z is not constant. The deformation u_z is then the following:

u_{z} (z) = - \frac{ρ g}{2 E} z^{2} + C_{1} x + C_{2}, C_{1} \in ℝ, C_{2} \in ℝ

General Solution of the Differential Equation

The boundary conditions in this case are the following:

\frac{d u_{z} (0)}{d z} = 0,

u_{z} (L) = 0 .

Boundary Conditions

Hence,

C_{1} = 0,

C_{2} = - \frac{ρ g}{2 E} L^{2} .

Integration Constants

Finally, the formula for the displacement u_z can be completed and the maximum displacement u_z,max calculated.

u_{z} (z) = \frac{ρ g}{2 E} (L^{2} - z^{2}),

u_{z, \max} = \frac{ρ g L^{2}}{2 E} = - 0.019 mm .

Maximum Displacement

RFEM Settings

Modeled in RFEM 6.14 and RFEM 5.06
The element size is l_FE = 0.100 m
Isotropic linear elastic material model is used

Results

Type of Loading	Analytical Solution [mm]	RFEM 6 [mm]	Ratio [-]	RFEM 5 [mm]	Ratio [-]
Concentrated force	−0.037	−0.037	1.000	−0.037	1.000
Self-weight	−0.019	−0.019	1.000	−0.019	1.000

Verification Example 000062 | 1

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