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2026-03-19

VE0087 | Curved Beam with Distributed Loading

Description

A curved beam consists of two perpendicular beams of length L and rectangular cross-section w × h. It is loaded by a distributed loading p. While neglecting self-weight, the goal is to determine the maximum stress σ_x,max on the top surface of the horizontal beam.

Material	Isotropic Linear Elastic	Modulus of Elasticity	E	210000.000	MPa
Material	Isotropic Linear Elastic	Poisson's Ratio	ν	0.296	-
Geometry		Length	L	1.000	m
		Cross-section Width	w	25.000	mm
		Cross-section Height	h	50.000	mm
Load		Distributed Loading	p	10.000	N/mm

Curved Beam with Distributed Loading

Analytical Solution

The equations of equilibrium yields that the given structure is statically indeterminate. To complete the set of equations, further constraint has to be found.

x : A_{x} - B_{x} = 0,

z : p L - A_{z} - B_{z} = 0,

M_{A} : \frac{p L^{2}}{2} - B_{z} L - B_{x} L = 0

Equations of Equilibrium

where A_x, A_z, B_x, B_z are the corresponding reaction forces. The missing equation is defined by means of the condition of zero deflection at point B in z-direction:

u_{z B} = 0

Condition of Zero Deflection

Curved Beam with Distributed Loading - Free Body Diagram

The general deflection v of beams and curved beams can conveniently be determined by Maxwell-Mohr integral:

v = \frac{1}{E I_{y}} \int_{L} M (x) m (x) d x

Maxwell-Mohr Integral

where I_y is the second moment of area, M(x) is the bending moment caused by the outer forces and m(x) is the bending moment caused by the unitary force. The following formulas define these bending moments in two regions with coordinate x₁:

x_{1} \in [0, L] :

M (x_{1}) = \frac{p x_{1}^{2}}{2} - B_{z} x_{1},

m (x_{1}) = x_{1}

Bending Moment in Horizontal Beam

and coordinate x₂:

x_{2} \in [0, L] :

M (x_{2}) = \frac{p L^{2}}{2} - B_{x} x_{2} - B_{z} L,

m (x_{2}) = L - x_{2}

Bending Moment in Vertical Beam

The deflection of the point B is then equal to:

u_{z B} = \frac{1}{E I_{y}} (\int_{0}^{L} M (x_{1}) m (x_{1}) d x_{1} + \int_{0}^{L} M (x_{2}) m (x_{2}) d x_{2}) = 0

Deflection of the Point B Calculation

Considering the equations of equilibrium and the deflection condition, the reaction forces are equal to:

A_{x} = \frac{1}{16} p L,

A_{z} = \frac{9}{16} p L,

B_{x} = \frac{1}{16} p L,

B_{z} = \frac{7}{16} p L

Result Reaction Forces

The maximum stress occurs at the point with maximum bending moment M_max. This point is on the horizontal beam at distance:

x_{1} = \frac{7}{16} L

Position of the Maximum Bending Moment

The horizontal beam is also loaded by the axial reaction force B_x. The maximum stress σ_x,max on the top surface is composed of the maximum bending stress and the pressure stress caused by the axial reaction force B_x, hence:

σ_{x, \max} = σ_{b, \max} + σ_{a} = \frac{6 M_{\max}}{w h^{2}} + \frac{- B_{x}}{w h} = - 92.375 MPa

Maximum Stress

RFEM Settings

Modeled in RFEM 5.39 and RFEM 6.13

Element size l_FE = 0.050 m

Number of increments: 10

Isotropic linear elastic material is used

Shear stiffness of the members is deactivated

Kirchhoff bending theory for plates is used

Results

Entity	Theory σ_x,max [MPa]	RFEM 6 σ_x,max [MPa]	Ratio [-]	RFEM 5 σ_x,max [MPa]	Ratio [-]
Member	- 92.375	- 91.774	0.993	- 91.774	0.993
Plate, horizontal		- 92.422	1.001	- 92.422	1.001
Plate, vertical		- 91.619	0.992	-	-
Solid		- 91.176	0.987	-	-

Remark: The shear effect are neglected in case of beam and horizontal plate. In case of vetical plate and solid these effects are included.

RFEM results – Stress distribution along the curved beam

RFEM Results – Stress Distribution Along Curved Beam

RFEM results – bending moment behavior along the curved beam

RFEM Results – Bending Moment Behavior Along Curved Beam

Model 000188 | Verification Example 000087 | 1

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Verification Example 000087 | 1

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