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009954
2024-01-31

VE0054 | Influence of Normal Force on Torsion

Description

Member with the given boundary conditions is loaded with the moment M and the axial force Fx. Neglecting it's self-weight, determine beam's maximum torsional deformation φx,max as well as its inner torsional moment MT defined as a sum of a primary torsional moment MTpri and torsional moment caused by the normal force MTN. Provide a comparison of those values while assuming or neglecting the influence of the normal force. The verification example is based on the example introduced by Gensichen and Lumpe (see the reference).

Material Steel Modulus of Elasticity E 210000.000 MPa
Shear Modulus G 81000.000 MPa
Geometry Beam Length L 3.000 m
Height h 0.400 m
Width b 0.180 m
Web Thickness s 0.010 m
Flange Thickness t 0.014 m
Load Torsional Moment M 1.200 kNm
Axial Force Fx 500.000 kN

Analytical Solution

Assuming that the relative torsion φ' is constant and no secondary torsional moment acts on the structure, beam's torsional moment MT can be obtained as a sum of a primary torsional moment MTpri and torsional moment caused by the normal force MTN.

where

Torsional moment equals to the acting moment (MT=M) and the normal force has the opposite value of the acting force (N=-Fx), beam's relative torsion φ' can be expressed as follows:

Maximum torsional deformation at the end of the beam φx,max can be calculated as follows:

RFEM Settings

  • Modeled in version RFEM 5.03 and RFEM 6.01
  • The element size is lFE= 0.300 m
  • The number of increments is 1
  • The element type is member
  • Isotropic linear elastic material model is used
  • Shear stiffness of members is activated

Results

φx,max [rad] Analytical Solution RFEM 6 Ratio RFEM 5 - RF-FE-LTB Ratio
N = 0 kN 0.101 0.101 1.000 0.101 1.000
N = -500 kN 0.166 0.165 0.994 0.165 0.994
MTpri [kNm] Analytical Solution RFEM 6 Ratio RFEM 5 - RF-FE-LTB Ratio
N = 0 kN 1.200 1.200 1.000 1.200 1.000
N = -500 kN 1.972 1.966 0.997 1.966 0.997
MTN [kNm] Analytical Solution RFEM 6 Ratio RFEM 5 - RF-FE-LTB Ratio
N = 0 kN 0.000 0.000 - 0.000 -
N = -500 kN -0.772 -0.766 0.992 -0.766 0.992
MT [kNm] Analytical Solution RFEM 6 Ratio RFEM 5 - RF-FE-LTB Ratio
N = 0 kN 1.200 1.200 1.000 1.200 1.000
N = -500 kN 1.200 1.200 1.000 1.200 1.000

References
  1. LUMPE, G. and GENSITEN, V. Evaluation of Linear and Nonlinear Member Analysis in Theory and Software: Test examples, causes of failure, detailed theory. Ernest.