IntroductionIn the first task of this CAE, a symmetric model of a tank and a connected pipe is analyses. The second part of the report deals with four simple geometries. The main theme of these analyses is applying the correct load and finding and evaluating the reaction force.Part one

A symmetric model was analysed. Four different approaches were used to model the loading and boundary conditions. Physical/mechanical constants and applied pressure are common between all four models and are as follows.Young modulus of elasticity (E) : 200 GPaPoisson ratio ():0.3Pressure:1.6 MPaTank nozzle load (1/2 full load):402.1 kNPipe nozzle load (1/4 full load):12.444 kNElement type:SHELL181Tanks Plate thickness:30 mmPipes plate thickness:63.5 mmRemote loadAs the first approach, nozzle load (in the cut section of the tank and the pipe) was applied as a remote force on the geometry (Edges on section A and section B).

Figure 1 Nozzle loads were applied as remote forces

Figure 2 Symmetry was applied on edges

Figure 3 Tangential displacement of edges in section A and B were constrained (y component in respective cylindrical coordinate systems). All rotation were also constrained

Figure 4 DeformationRigidDeformable

Figure 5 Reaction force in the section A RigidDeformable

Figure 6 Reaction force in the section BAs expected the value of reaction forces under rigid remote force is higher than deformable. That is because rigid force does not update itself with the deformed shape.

Edge loadIn this part, nozzle load was applied as an edge load. Everything else is almost similar to the previous approach. Results are presented below

It could be seen that the results are close to the first section (Remote force) provided that the deformable option is chosen.Using nodal loadsUnfortunately results for this section are not reasonable and do not conform with previous results and engineering sense but they are nevertheless presented for the sake of troubleshooting.Named selection were exclusively used in this section to apply constraints and symmetry boundary conditions. As shown below for the symmetry on x, those nodes which are located on pipe opening and tank opening were excluded (since these nodes should be oriented in the local cylindrical coordinate systems)

Figure 7 named selection for symmetry in x direction

Figure 8 named selection for Symmetry in z directionNodal orientation of all nodes on the sections A and B were converted to the local cylindrical coordinates. The nodal triads are shown for nodes on the section A below.

Figure 9 Nodal triads for the section A (tank opening)Nodal rotation and nodal displacement constraints were later applied on these oriented nodes.For the application of nozzle loads, the extreme nodes (on the symmetry plane) and interior nodes were dealt differently. Whereas a full load was applied on the interior nodes, just a half-load was applied on the extreme nodes (due to the fact that these extreme nodes are common between two symmetric halves and therefore just half the load should be applied on each symmetric half). To do this at first a full load was applied on all section nodes, then a half load in opposite direction was applied on the extreme nodes. Figure 10 Nozzle load on the pipe opening applied as the superposition of the shown loadsThe total deformation is shown below. It is quite clear that the model is not correct. While he tangential displacement in the section A (y direction in the local coordinate) is set to zero seemingly the model ended up with a y constrained in the global coordinate. Moreover all rotation in the section A is set to zero but from deformation it is obvious that the section A is rotating!

Part 2A pressure of 10 Pa was applied on different surfaces and the reaction force was retrieved from Ansys. These reaction forces then were compared to the analytical calculations. SHELL181 was used as the element type for all these surfaces. The plate thickness for all geometries were considered to be 1 mm.Plane shell

Figure 11 Total deformation for the plane shellThe reaction forces are shown below:

Figure 12 Reaction force for the plane shellIt is clear that the applied force on the shell equals pressure by projected area which lead to 10 N.Half cylinder

Figure 13 Reaction force for the half cylinderAgain as expected the reaction force equals projected are multiplied by pressure:

cylinder

Figure 14 Reaction force for the 1/4 cylinderIt is quite clear that load has two equal element in x and y direction. The projected area in both x and y direction equals

Half sphere

Figure 15 Reaction force for the half sphereThe projected are equals. Therefor the reaction force would be 31.4 N which fits with FE results.