5

5.4 BEAM in PLANE WITH BEAMS NO.13

Copy the example file B4_X.DXF to Z88X.DXF.

B4_X.DXF ---> Z88X.DXF input file for CAD converter Z88X

CAD:

Import Z88X.DXF into your CAD program and look at it. Usually you would have designed this example in a CAD system (makes not much sense because this example is extremely simple) and then exported as Z88X.DXF.

Z88: (in reduced form, more detailed instructions cf. examples 5.1, 5.2 and 5.3)

Z88X, conversion, "from Z88X.DXF to Z88I*.TXT"

Z88P, looking at structure, structure file Z88I1.TXT

Z88F calculates deflections

Z88D calculates stresses

Z88E calculates nodal forces

Z88P, plot FE structure, now also deflected (FUX, FUY, FUZ per 10.)

This example deals with a beam, fixed on both sides, and loaded with 1,648 N in the middle in downward direction. This mechanical problem is covered in every mechanical and civil engineering handbook. Geometry: Length 1,000 mm, cross-cut 50 x 10 mm. Thus: A = 500 mm**2, Izz = 4,167 mm**4, ezz = 5 mm.

The deflection curve has inflection points, we therefore take 4 beams No.13. Nodes 1 and 5 will be fixed and node 3 is loaded.

You would calculate analytically:

f in the middle: F*L**3/(192*E*I) = 10 mm

f in the inflection points: fw = f/2 = 5 mm

The bending moments on the left, middle, on the right: F*L/8 = 206,000 Nmm

The slope angle in the inflection points: Phi = atan (3*f/L) = 0.029991 rad

When interpreting the results of Z88O2.TXT (deflections) and Z88O4.TXT (nodal forces and moments) refer to the sign definition of chapter 3.13. Especially Z88O4.TXT, node 3: The force F(2) = force in Y direction is the sum of the forces of elements 2 and 3, due to extrinsic force. The force F(3) = bending moment is not a summary of elements 2 and 3, because it is an intrinsic moment, not an extrinsic load ! Also the signs of the load F(3) at node 1 and F(3) at node 5 are correct, refer to chapter 4.13. Keep in mind that the classical mechanical science sometimes uses different conventions.

5.4.1 Input:

This example shows that a FEA basically needs nodes in all locations where you want to get results. As the beam is fixed left and right, the maximum of displacements appears in the middle for x = L/2, but the bending curve features two inflection points for x = L/4 and x = 3L/4. To calculate results for this locations, the structure must be subdivided with nodes in x = 0, x = L/4, x = L/2 and x = 3L/4.

Only the file input is shown here because CAD use is not worth here.

Z88I1.TXT so becomes:

2 5 4 15 1 0 1 0 (2-D,5 nodes,4 ele, 5 DOF,1 mat info, KFLAG 0, IBFLAG , IPFLAG 01)

1 3 0 0 (1.node, 3 DOF, X and Y coordinate)

2 3 250 0

3 3 500 0

4 3 750 0

5 3 1000 0

1 13 (1. element, type beam in plane No.13)

1 2 (coincidence for 1. element)

2 13

2 3

3 13

3 4

4 13

4 5

1 4 206000 0.3 1 500 0 0 4167 5 0 0 (mat info for ele 1 to 4, Young's, Poisson's, INTORD (any), QPARA = area, Ixx=0, exx=0, Izz, ezz, It=0, Wt=0)

The node 1 is fixed in all degrees of freedom at the boundary conditions. It is important to fix especially the DOF 1 = displacement in X direction so that the structure cannot move. Node 5 is fixed in DOF 2 = displacement in Y direction and DOF 3 = rotation around Z axis. You could also fix DOF 1 for node 5, if you wish. But in reality one of the bearings or supports will allow for thermal expansion. This was taken into account in Z88I2.TXT.

Here Z88I2.TXT:

6 (6 Boundary conditions)

1 1 2 0 (Node 1, DOF 1 gets a displacment of 0 = DOF 1 fixed)

1 2 2 0 (Node 1, DOF 2 fixed)

1 3 2 0 (Node 1, DOF 3 fixed (restraining moment)

3 2 1 -1648 (Node 3, DOF 2 gets load of -1,648 N)

5 2 2 0

5 3 2 0

The parameter file for the stress processor Z88I3.TXT can have any content (cf. sections 3.5 and 4.13), because Gauss points, radial and tangential stresses as well as calculation of the von Mises stresses has no significance for Beams No.13.

5.4.2 Results

The Cholesky solver Z88F provides the following output files:

Z88O0.TXT stores the processed structure data. For documentation purposes.

Z88O1.TXT stores the processed boundary conditions: For documentation purposes.

Z88O2.TXT, the displacements, the main task and solution of the FEA problem.

The stress processor Z88D internally uses the calculated displacements from Z88F and stores Z88O3.TXT, the calculated stresses. The results in Z88O3.TXT do not depend on the header parameters in Z88I3.TXT for Beams No.13.

The nodal force processor Z88E internally uses the calculated deflections of Z88F and stores Z88O4.TXT, the computed nodal forces.

The following picture of the plot program shows the deflected structure for FUX, FUY and FUZ = 10 each (magnifications of the deflections).

Attention to the results of the nodal force calculation: Node 3: The force F(2) = force in Y direction is the sum of the forces of elements 2 and 3, due to extrinsic force. The force F(3) = bending moment is not a summary of elements 2 and 3, because it is an intrinsic moment, not an extrinsic load ! Also the signs of the load F(3) at node 1 and F(3) at node 5 are correct, refer to chapter 4.13. Keep in mind that the classical mechanical science sometimes uses different conventions.

Additional remark: Such simple examples are well suitable to become aware of the sign omen definitions. Experiment with this example and calculate other bend cases from good handbooks. Frameworks with Beams No.2 are calculated accordingly. However, a real spatial structure then must be available: At least one Z coordinate must not equal 0.

View of undeflected and deflected structure

Take into account: The plot program Z88P connects the nodes with straight lines, although the deflection curve represents a cubic parable in the case of a Beam No.13 or No.2. This means: Z88P shows the deformations correctly for the node, but straight lines are between the nodes. Therefore, no deflection curve is shown. If you want to plot a real nice deflection curve with Z88P, then use basically more nodes, e.g. 15 to 20 nodes for this example (the cubic bending curve is then featured by a couple of straight lines).