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2023-08-31

VE0048 | Uniaxial Bending with Pressure

Description

A structure made of I-profile is fully fixed on the left end and embedded into the sliding support on the right end. The structure consists of two segments according to the following sketch. The self-weight is neglected in this example. Determine the maximum deflection of the structure u_z,max, the bending moment M_y on the fixed end, the rotation φ_2,y of the segment 2 and the reaction force R_Bz by means of the geometrically linear analysis and the second-order analysis. 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	Poisson's Ratio	ν	0.300	-
Geometry	Structure	Segment Length 1	L₁	6.000	m
	Structure	Segment Length 2	L₂	1.200	m
	Cross-section	Height	h	400.000	mm
		Width	b	180.000	mm
		Web Thickness	s	10.000	mm
		Flange Thickness	t	14.000	mm
Load		Axial Force	F_x	100.000	kN
Load		Transverse Force	F_z	0.500	kN

01 Uniaxial Bending with Pressure

Analytical Solution

Geometrically Linear Analysis

Geometrically linear analysis is carried out at first. In this case, the axial force F_x is not taken into account. The problem can be then solved as well as a cantilever of the length L₁ loaded only by the transverse force F_z. The maximum deflection u_z,max can be calculated using Mohr's integral and results into expression:

u_{z, \max} = \frac{F_{z} L_{1}^{3}}{3 {EI}_{y}} = 0.743 mm

F<sub>y</sub>

Quadratic moment of the cross-section with respect to the y-axis

Maximum Deflection – Geometrically Linear Analysis

The bending moment on the fixed end can be calculated according to the following formula:

M_{y} (0) = F_{z} L_{1} = 3.000 kNm

Maximum Bending Moment – Geometrically Linear Analysis

The rotation of the segment 2 φ_2,y is calculated from the geometric condition as follows:

φ_{2, y} = \arctan (\frac{u_{z, \max}}{L_{2}}) = 0.619 mrad

Rotation of Segment 2 – Geometrically Linear Analysis

The reaction force in the sliding joint R_Bz considering the zero effect of the axial force F_x can be obtained from the free body diagram shown in the following sketch.

R_{B z} = - \frac{F_{x} u_{z, \max}}{L_{2}} = 0.000

Reaction Force in Sliding Joint – Geometrically Linear Analysis

02 Free Structure Design

Second-order Analysis

Because of the nonnegligible effect of the axial force F_x the second-order analysis should be considered. Thus the axial force F_x is taken into account and produces another contribution to the bending moment. The problem can be described by the free body diagram of the segments according to the sketch. The unknown reaction forces can be obtained from the equilibrium equations and then the bending moment formula can be written.

M_{y} = - R_{A z} (L_{1} - x) - R_{A x} (u_{z, \max} - u_{z} (x))

Bending Moment in Segment 1 – Second-Order Analysis

The solution can be found by the Euler-Bernoulli differential equation.

\frac{d^{2} u_{z}}{{dx}^{2}} = - \frac{M_{y}}{{EI}_{y}}

Euler-Bernoulli Differential Equation

Considering boundary conditions the solution on the differential equation can be found and maximum deflection of the structure calculated.

u_{z} (x) = \frac{F_{z} L_{2} [- \cos ({αL}_{1}) αx + \cos ({αL}_{1}) \sin (αx) - \sin ({αL}_{1}) \cos (αx) + \sin ({αL}_{1})]}{F_{x} [{αL}_{1} \cos ({αL}_{1}) + {αL}_{2} \cos ({αL}_{1}) - \sin ({αL}_{1})]}

Deflection Function for Segment 1 – Second-Order Analysis

u_{z, \max} = \frac{F_{z} L_{2} [{αL}_{1} \cos ({αL}_{1}) - \sin ({αL}_{1})]}{F_{x} [αcos ({αL}_{1}) (L_{1} + L_{2}) - \sin ({αL}_{1})]} = 0.878 mm

Maximum Deflection of Structure – Second-Order Analysis

The bending moment on the fixed end can be calculated according to the following formula:

M_{y} (0) = R_{A z} L_{1} + R_{A x} u_{z, \max} = 3.527 kNm

Maximum Bending Moment – Second-Order Analysis

The rotation of the segment 2 φ_2,y is calculated from the geometric condition as follows:

φ_{2, y} = \arctan (\frac{u_{z, \max}}{L_{2}}) = 0.732 mrad

Rotation of Segment 2 – Second-Order Analysis

The reaction force in the sliding joint R_Bz results:

R_{B z} = - \frac{F_{x} u_{z, \max}}{L_{2}} = - 0.073 kN

Reaction Force in Sliding Joint – Second-Order Analysis

It is obvious that the influence of the axial force F_x is considerable. The total deflection of the structure under the prescribed loading in case of the second-order analysis is approximately 18 % greater than in case of geometrically linear analysis. The comparison of the geometrically linear analysis and the second-order analysis is shown in the graph, considering the ratio of the loading forces F_z = F_x/200. It is obvious that the difference between these analysis is more considerable when the loading is grater. The second-order analysis solution is approaching the horizontal asymptote. The numerical solution gives the value of the horizontal asymptote F_x,cr = 650.873 kN.

03 Comparison of Geometrically Linear Analysis (Dashed Line) and Second-Order Analysis (Solid Line)

RFEM and RSTAB Settings

Modeled in RFEM 5.05 and RSTAB 8.05 and RFEM 6.01, RSTAB 9.01
The number of elements is 2 (one element per member)
The number of increments is 5
Isotropic linear elastic material model is used
The structure is modeled using members
Shear stiffness of the members is neglected

Results

Geometrically Linear Analysis	Analytical Solution	RFEM 6	Ratio	RSTAB 9	Ratio
u_z,max [mm]	0.743	0.743	1.000	0.743	1.000
M_y(0) [kNm]	3.000	3.000	1.000	3.000	1.000
φ_2,y [mrad]	0.619	0.619	1.000	0.619	1.000
R_Bz [kN]	0.000	0.000	-	0.000	-

Geometrically Linear Analysis	Analytical Solution	RFEM 5	Ratio	RSTAB 8	Ratio
u_z,max [mm]	0.743	0.743	1.000	0.743	1.000
M_y(0) [kNm]	3.000	3.000	1.000	3.000	1.000
φ_2,y [mrad]	0.619	0.619	1.000	0.619	1.000
R_Bz [kN]	0.000	0.000	-	0.000	-

Second-order Analysis	Analytical Solution	RFEM 6	Ratio	RSTAB 9	Ratio
u_z,max [mm]	0.878	0.878	1.000	0.878	1.000
M_y(0) [kNm]	3.527	3.527	1.000	3.527	1.000
φ_2,y [mrad]	0.732	0.732	1.000	0.732	1.000
R_Bz [kN]	-0.073	-0.073	1.000	-0.073	1.000

Second-order Analysis	Analytical Solution	RFEM 5	Ratio	RSTAB 8	Ratio
u_z,max [mm]	0.878	0.878	1.000	0.878	1.000
M_y(0) [kNm]	3.527	3.527	1.000	3.527	1.000
φ_2,y [mrad]	0.732	0.732	1.000	0.732	1.000
R_Bz [kN]	-0.073	-0.073	1.000	-0.073	1.000

499x

14x

Verification Example 000048 | 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.

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