Static Analysis of 3 bars under thermal and mechanical loads, connected to a rigid beam (Method 2)

Aims and expectations at this course

Our expectation after you study:

1- Modifying an existing model to create a new model.

2- Modeling wire type deformable and rigid parts separately

3- Using MPC constraint to connect two points together.

As the Figure shows, a rigid beam, AB, hanged via 3 vertical bars with the same lengths and cross sectional areas (0.1 in2). The middle bar is made from steel while the others are from copper. The temperature of bars increases 10 oF and a load of Q=4000 lb is applied on middle of the rigid beam. We are aiming to calculate the stresses in all bars, firstly and then, compare them with the results given in reference [1]. The material properties have been presented in Table 1.

Figure 1: A rigid mass hanged from 3 vertical bars


Table 1: Bars material properties


Creating the model:

Open the model you created in Example 5, firstly. Rename the model to hanfing slab-MPC. Save the model, then, with name of three-bar-method2. Now we need to edit existing sketch, hence, right click on Section Sketch, as Figure 2 shows, in model tree and select the item Edit. You will enter Sketch module by doing this.


Figure 2: The steps for going to sketch module with aim of modification


Click on icon (Delete) and choose the horizontal line which was created to model the rigid body. This leads to eliminating the line and only deformable (vertical) lines remain. Now, exit Sketch module by pressing Done in prompt area. Select OK when facing the message shown in Figure 3 appears.


Figure 3: Warning that tells you it is needed to regenerate the sketch


As the message guided you, select the items below, from the menu bar:

Feature > Regenerate

This causes that the viewport updates the geometry by deleting the horizontal line.

In order to create the rigid part, separately, click on icon (Create Part) and configure the window opened like illustrated in Figure 4. After all, click on Continue button.


Figure 4: Create Part window and its configurations to create a discrete rigid part


Note selecting Discrete rigid item as the Type of the part. Unlike prior Example, by doing this, Abaqus considers the part rigid, since beginning. Utilizing Create Lines tool, as shown in Figure 5, draw a horizontal line with length of 20 inch, divided in half. Then, put dimensions and fixing constraints on the part to define it completely.


Figure 5: 2 segments of lines related to the horizontal rigid beam


After that, exit Sketch module. The Reference Point related to the rigid part must be defined in either this module or the Property one. Like you did in previous Example, before creating Reference Point, you need to define Datum, firstly. Therefore, from the menu bar choose the tool of creating Datum as following:

Tools > Datum

Like what seen in Figure 6, select the items that are pointed out.


Figure 6: The items selected for defining a Datum offset from a point


Now, pick the middle point of line, specified in Figure 7.


Figure 7: The point from which the Datum offsets


In prompt area, insert coordinate of (0,-1,0) and press Enter key to create a Datum Point with distance of 1 in from the middle of rigid part and in negative direction of Y. Next, through the steps below, make a Reference Point by selecting the Datum Point you have created.

Tools > Reference Point

After that, next to the Reference Point an expression of RP emerges. Unlike prior Example, in Interaction module, we were able to define infinite quantity of Reference Points, but, here in Part module or the Property, we can only assign one Reference Point to each part.

Editing Sections assigned to the parts:

Enter Property module and, as shown in Figure 8, select Part-1 containing 3 vertical bars.


Figure 8: Selecting Part-1 in Property module


Click on icon of Section Assignment Manager, displayed in Figure 9.


Figure 9: The tool used for modifying Sections


Click on the last Section, displayed in Figure 10 which was assigned to the rigid part. Then, press Delete and Yes buttons to eliminate it from the model.


Figure 10: Deleting the Section of rigid beam defined in previous Example


The job in this step is done so that it is not needed to assign any Section to Discrete Rigid Part. In the next step we insert the rigid part into assembly module.

Inserting parts in Assembly module:

Enter Assembly module. At this moment, a message like, what shown in Figure 11, appears in Message Area, informing you that some meshes are deleted in some regions of an instance with name of Part-1-1 (The first instance of Part-1). This is for reason of deleting those two lines in previous steps.


Figure 11: The message warning about some meshes elimination


Because we inserted Part-1 prior to this and we deleted the horizontal rigid part above, only 3 vertical bars can be seen in viewport. Consequently, we need only to insert the rigid part to Assembly module. So, as Figure 12 shows do the inserting of the rigid part. Take notice that we selected the Independent option again.


Figure 12: Inserting the part named rigid into Assembly module


If we had created both parts exactly in the right places as the bottom endpoints of each vertical bar laid over middle point and right-hand and left-hand endpoints of the rigid beam, it would have been difficult to select the linked points correctly. Fortunately, we did not concern about the position of parts as shown in Figure 13. Leave the parts in the way they are without any translation for the moment.


Figure 13: The model and its parts in Assembly module


As you see, there remained Reference Point of RP-1 and its relevant Datum. From the path displayed in Figure 14, find and delete them. Select Yes when facing coming messages.


Figure 14: Deleting the Datum and Reference Point that we do not need them anymore


Deleting Rigid body constraint applied in previous Example and connecting related points of parts:

Enter Interaction module and click on Constraint Manager, which its icons is displayed in Figure 15.


Figure 15: Constraint Manager icon to eliminate Rigid body constraint


In opened window, using Delete button displayed in Figure 16, delete the Constraint-1 which was for defining Rigid body. Answer the warning by Yes.


Figure 16: Constraint Manager window


After that, in order to connect the points of vertical bars and rigid part, while the Constraint Manager window is open, click on Create button and configure the items shown in Figure 17, then, Continue.


Figure 17: The MPC Constraint for bonding the points of separated parts


In prompt area, you are asked to choose the MPC Control Point. This point specifies the leader side of connection while the other side must move and lay on this one. In other words, the first point is considered as the master and some of its behaviors will be followed by the other one like the slave. Select the point number 1, labeled in Figure 18, as the control point.


Figure 18: Labels of the points for using MPC Constraint


Thereafter, select the second point, as the Slave node of this connection and press Done in prompt area. In the coming window (Edit Constraint), as shown in Figure 19, select the item Pin from the list of MPC Type, then, hit OK. A green expression (MPC Pin) and a yellow line connecting the points selected as the sides of the connection appears in viewport which symbolize existence of the MPC Constraint with type of Pin between these points.


Figure 19: The item Pin selected to constraint and connect two points


Take notice that this constraint only relates the displacements of the points and cannot consider rotations.


Note 1

In the case we are solving, we selected Pin Type constraint for two reasons:

1- In each constraints both points lay on each other (Main reason for not utilizing the Link Type constraint).

2- The Truss element includes only displacement type degrees of freedom (Main reason for not using the Beam Type constraint).

Note that we could have used either Beam or Link MPC Constraints. But, the Link Type are usually used when connecting points with different positions. Moreover, Beam MPC Constraint bonds rotational degrees of freedom of both nodes as well as the translational ones. So, using this type seems rational when we have rotational degrees of freedom. Otherwise, as Figure A shows, we will see a warning, in solving step, that tells us there activate extra degrees of freedom in the model.


Figure A: The warning appears when using Beam Constraint instead of the Pin Type


Define two other connections linking middle and the right-hand bars with name of mid bar and right bar. Consider points of 3 and 5 as the control points of these constraints and the points 4 and 6 as the slave ones, respectively. If you pick wrong points, an error will comes up when submitting the relevant job. After all, the Constraint Manager must be like what seen in Figure 20.


Figure 20: Constraint Manager window and the MPC Constraints


Translating the Rigid part in the right place considering vertical bars:

Enter Assembly module and click on icon (Translate Instance). Select horizontal (Rigid) bar as the instance which is going to translate, then, Done. Next, pick the points 1 and 2, shown in Figure 21, in sequence, to place Rigid beam on its right position. Press OK in prompt area to finish the action.


Figure 21: The points defined for start (1) and end (2) of translation vector


Modifying loads and boundary conditions:

Enter Load module. As vertical bars have not edited, their boundary conditions and thermal conditions are kept with no change. Although, we can ensure whether this is true or not by clicking on the icons (Manager) and Edit buttons, related to these sorts of loads and boundary conditions. On the other hand, we eliminated the point on which the load is applied, so, it needs to be redefined. Hence, click on icon (Load Manager) and selecting the load item in Step-1, press Edit button. There will appear a message, shown in Figure 22, warning you that the position of applying load has been missed. Actually, this is what we are willing to modify.


Figure 22: The massage making you aware of missing force position


Answer the warning by Yes. This brings up the Edit Load window. Click on the arrow icon pointed out as pointed out in Figure 23.


Figure 23: Edit Region icon in Edit Load window


After this, pick the Reference Point you have just created and is known with name of RP in viewport, then, press Done in prompt area. Finally, click on OK button to save the changes and close Edit Load window.

Meshing the parts:

At this stage, enter Mesh module. No changes happened in the vertical bars, so, their meshes still exist. The only action we need to take, here, is meshing the rigid beam. Using the icon (Seed Edges) divide the both parts of rigid beam to 1 division. This assigns 2 elements to the beam. Click on icon (Mesh Part Instance) and mesh the rigid beam. To specify the elements type, click on icon (Assign Element Type) and dragging the mouse choose both horizontal elements, then, press Done button. As seen in Figure 24 select the items Discrete Rigid Element and Rigid link. This sets the element type on R2D2, a 2D rigid link having two nodes. At the end, press OK button.


Figure 24: Element Type window and choosing type of rigid element for rigid beam


Analyzing the problem:

Enter Job module and click on icon (Create Job). Create a new job with name of bar3-slab-MPC, then, submit it in order to analyze the model. Note that the previous job with name of bar3-slab must be deleted because its related model (hanging slab) does not exist anymore. Pressing the Results button, when the job has completed, go to Visualization module. Click on icon (Plot Contours on Deformed Shape) to observe deformed shape of the system. Using one of the methods explained in previous Examples, extract the stress magnitudes in steel and copper bars. Table 2 compares the results obtained from prior and current Examples (Method 1 and Method 2), however, no difference exists.


Table 2: Stresses in vertical bars obtained using methods 1 and 2



[1] Timoshenko, S. P., “Strength of Materials, Part l, Elementary Theory and Problems,” 3rd Ed., D. Van Nostrand Co., Inc., l956, p. 30.

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