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2025-08-22

VE0021 | Plastic Bending with Zero Tensile Strength

Description

A cantilever is fully fixed on the left end (x = 0) and subjected to a transverse force F and an axial force F_a on the right end. The tensile strength is zero and the behavior in the compression remains elastic. The problem is described by the following set of parameters. Small deformations are considered and the self-weight is neglected in this example. Determine the maximum deflection u_z,max.

Material	Elastic-Plastic	Modulus of Elasticity	E	210000.000	MPa
		Poisson's Ratio	ν	0.000	-
		Shear Modulus	G	105000.000	MPa
		Tensile Plastic Strength	f_t	0.000	MPa
Geometry	Cantilever	Length	L	2.000	m
		Width	w	0.005	m
		Thickness	t	0.005	m
Load		Transverse Force	F	4.000	N
Load		Axial Force	F_a	5000.000	N

Plastic Bending with Zero Tensile Strength

Analytical Solution

The transverse force F together with the axial force F_a causes the elastic-plastic state of the cantilever according to the following sketch. The elastic-plastic zone length is described by the parameter x_p. The stress σ_x is composed of the bending stress σ_b and the compressive stress σ_c, and it is defined according to the following formula:

Plastic Bending with Zero Tensile Strength - Stress Distribution

σ_{x} (x, z) = - κ (x) E (z - z_{0} (x))

Normal Stress

where κ(x) is the curvature and the parameter z₀(x) is defined so that σ_x(x,z₀)=0. The first yield occurs when the bending stress on the top surface at the fixed end reaches the value of the compressive stress.

\frac{M (0)}{I_{y}} \frac{t}{2} = \frac{F_{a}}{A}

Equality of Bending Stress and Compressive Stress

The transverse force results F=2.083 N. Thus, the the cantilever under transverse force F=4.000 N is in an elastic-plastic state. In the elastic-plastic zone (x < x_p) the equilibrium between bending moments and axial forces has to be satisfied.

M_{ep} = \int_{z_{0}}^{t / 2} - κ_{p} E (z - z_{0}) z w d z = M

Equilibrium of Internal and External Bending Moments

N_{ep} = \int_{z_{0}}^{t / 2} - κ_{p} E (z - z_{0}) w d z = - F_{a}

Equilibrium of Internal and External Normal Force

Solving these equations, the curvature in the elastic-plastic zone κ_p and the parameter z₀ results as follows.

κ_{p} (x) = \frac{8 F_{a}^{3}}{9 E w (F_{a} t + 2 M)^{2}}

Result Curvature

z_{0} (x) = \frac{t}{2} - \frac{3}{2} \frac{F_{a} t + 2 M}{F_{a}}

Result Parameter z₀

The elastic-plastic zone length x_p can be obtained from the last equation under the condition z₀(x_p)=-t/2.

x_{p} = L - \frac{t F_{a}}{6 F} \approx 958.333 m m

Elastic-Plastic Zone Length

The curvature κ_e in the elastic zone (x > x_p) is described by the Bernoulli-Euler formula:

κ_{e} = - \frac{M}{E I_{y}}

Curvature in Elastic Zone

The maximum deflection u_z,max can finally be calculated according to the following formula

u_{z, \max} = \int_{0}^{x_{p}} κ_{p} (L - x) d x + \int_{x_{p}}^{L} κ_{e} (L - x) d x \approx 1.232 m

Result Maximum Deflection

RFEM Settings

Modeled in RFEM 5.26 and RFEM 6.11
The element size is l_FE= 0.020 m
Geometrically linear analysis is considered
The number of increments is 10
Shear stiffness of the members is neglected

Results

Material Model	Analytical Solution	RFEM 6		RFEM 5
Material Model	u_z,max [m]	u_z,max [m]	Ratio [-]	u_z,max [m]	Ratio [-]
Isotropic Nonlinear Elastic 1D	1.232	1.231	0.999	1.231	0.999
Isotropic Masonry 2D		1.237	1.004	1.237	1.004
Nonlinear Elastic 2D/3D, Mohr - Coulomb		1.237	1.004	1.237	1.004
Nonlinear Elastic 2D/3D, Drucker - Prager		1.237	1.004	1.237	1.004
Isotropic Plastic 2D/3D, Mohr - Coulomb		1.237	1.004	1.237	1.004
Isotropic Plastic 2D/3D, Drucker - Prager		1.237	1.004	1.236	1.003

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Verification Example 0021 | 1

References

Lubliner, J. (1990). Plasticity Theory. New York: Macmillan.

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