## Geometrically non-linear effects (string theory)

Consider a horizontal beam supported at one end (A) by a hinged support and at the other end (B) by a roller support. The beam is subdivided in 100 elements with length 500 mm and loaded by its deadweight q = 1.2 N/mm. The diameter is 500 mm and the wallthickness 10 mm.

Further derived quantities:

Cross-sectional area_{ }^{ }

Young’s modulus^{ }

Moment of inertia_{ }^{ }

Moment of resistance_{ }^{ }

A = π D_{g} t = 15394 mm^{2}

E = 210000 N/mm^{2}

I_{b} = 4.622*10^{8} mm^{4}

W = I_{b}/0.5D_{g} = 1.8865*10^{6} mm^{3}

The beam calculations have been performed both with the geometrically linear model option and the geometrically non-linear model option. The difference between both options is that in the linear option equilibrium is obtained for the undeformed beam and in case of the non-linear option for the deformed beam. In the latter option the so-called ‘stress-stiffening’ method – an iterative method – is applied in Ple4Win resulting in stiffer elements.

Now we look at the horizontal displacement at end B and the vertical displacement at midpoint C.

Geometrically **linear** results of Ple4Win:

ΔX_{B} = 0 mm

ΔZ_{C} = 1006.6 mm

Theoretical result:

ΔZ_{C }= 5/384 *q l^{4}/EI_{b} = 1006.1 mm

Geometrically **non-linear** results of Ple4Win:

ΔX_{B} = -50 mm

ΔZ_{C} = 1004.6 mm

Despite the negative X-displacement at end B the vertical displacement in point C is (somewhat) less compared to the linear calculation. This is due to the stress-stiffening.

The resulting bending moment M = 1/8 q l^{2} = 3.75*10^{8} Nmm at point C.

The maximum axial and Von Mises stress σ = M / W = 198.7 N/mm^{2} at point C.

The following step is to apply an additional horizontal pull force at end B with magnitude varying from 50 to 300 kN with a force interval of 50 kN. Next we consider the case of 200 kN.

In case of a **linear **calculation the vertical displacement at point C does not change because of the equilibrium on the undeformed beam. At the roller end B a small positive horizontal displacement results caused by the axial tensile force: ΔX = F l / EA. For F = 200 kN ΔX = 3.09 mm.

Then the maximum axial and Von Mises stress σ = σ_{F=0} + F_{200}/A = 198.7 + 13 = 211.7 N/mm^{2} at point C.

Now the **non-linear** calculation:

We use the symmetry of the vertical displacements with respect to midpoint C and consider the right half of the beam as fixed at point C. Then the end B has a vertical displacement ΔZ_{B} = 1004.6 mm upward if the horizontal pull force F = 0 (see calculation above).

From literature the following equation is known for an one-end fixed beam loaded by a vertical load and a horizontal pull force F at the other free end:

ΔZ = ΔZ_{0} / (1 + F/F_{E})

where:

ΔZ = the resulting vertical displacement at the free end point

ΔZ_{0} = the vertical displacement at the free end point when pull force F is zero

F_{E} = the critical buckling load = π^{2} EI_{b} / 4 l^{2}

In our case:

ΔZ_{0 } = ΔZ_{B } = 1004.6 mm

F_{E} = π^{2} EI_{b} / 4 l^{2} = 383 kN

F = 200 kN

ΔZ = 1004.6 / (1 + 200/383) = 1006.4 / 1.522 = 660 mm (= ΔZ_{C})

The vertical support force in end point B S_{B} = 30 kN and equal to the total deadweight D_{CB} of beam section C-B. The resulting bending moment in C:

M_{C} = S_{B} * 25000 – D_{CB} * 12500 – F * 660 = 7.5*10^{8} – 3.75 * 10^{8 }–1.32 * 10^{8} = 2.43 * 10^{8}.

The maximum axial and Von Mises stress σ = M / W + F/A = 128.8 + 13 = 141.8 N/mm^{2} at point C.

Ple4Win obtains the same results by means of the stress-stiffening approach.

For all calculated pull forces in end point B the results are shown in the picture below.