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 Dg t = 15394 mm2
E = 210000 N/mm2
Ib = 4.622*108 mm4
W = Ib/0.5Dg = 1.8865*106 mm3

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:
ΔXB = 0 mm
ΔZC = 1006.6 mm
Theoretical result:
ΔZC = 5/384 *q l4/EIb = 1006.1 mm

Geometrically non-linear results of Ple4Win:
ΔXB = -50 mm
ΔZC = 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 l2 = 3.75*108 Nmm at point C.
The maximum axial and Von Mises stress σ = M / W = 198.7 N/mm2 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 = σF=0 + F200/A = 198.7 + 13 = 211.7 N/mm2 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 ΔZB = 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 = ΔZ0 / (1 + F/FE)

where:
ΔZ    =  the resulting vertical displacement at the free end point
ΔZ0 =  the vertical displacement at the free end point when pull force F is zero
FE     =  the critical buckling load = π2 EIb / 4 l2

In our case:
ΔZ0   =  ΔZB    =  1004.6 mm
FE     =  π2 EIb / 4 l2 =  383 kN
F       =   200 kN
ΔZ    =  1004.6 / (1 + 200/383) = 1006.4 / 1.522 = 660 mm (= ΔZC)

The vertical support force in end point B SB = 30 kN and equal to the total deadweight DCB of beam section C-B. The resulting bending moment in C:

MC = SB * 25000 – DCB * 12500 – F * 660 = 7.5*108 – 3.75 * 108 –1.32 * 108 = 2.43 * 108.
The maximum axial and Von Mises stress σ = M / W + F/A = 128.8 + 13 = 141.8  N/mm2 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.