152x

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 F_x. Neglecting it's self-weight, determine beam's maximum torsional deformation φ_x,max as well as its inner torsional moment M_T defined as a sum of a primary torsional moment M_Tpri and torsional moment caused by the normal force M_TN. 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
Material	Steel	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
Load		Axial Force	F_x	500.000	kN

Influence of Normal Force on Torsion

Analytical Solution

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

M_{T} = M_{Tpri} + M_{TN}

Torsional moment

where

M_{Tpri} = GJφ'

J	Torsional moment of inertia

Primary Torsional Moment

M_{TN} = N {i_{p}}^{2} φ'

i_p	Polar radius of inertia

Torsional Moment Caused by Normal Force

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

φ' = \frac{M_{T}}{GJ + N (i_{p})^{2}}

Relative Torsion

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

φ_{x, \max} = φ (x = L) = φ' L

Maximum Torsional Deformation

RFEM Settings

Modeled in version RFEM 5.03 and RFEM 6.01
The element size is l_FE= 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

φ_x,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

M_Tpri [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

M_TN [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

M_T [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

417x

12x

Verification Example 000054 | 1

References

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

Gensichen

;