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E.10 Simple Constraints

E.10.1 Simple Constraint 1

An analysis was performed on the structure shown in Figure E.10.1 with applied constraints. For all load cases, the constraint \(\small UX4 = UX3\) was imposed. Case 1 examines the constraint equation applied to a degree of freedom where a load is applied. Case 2 considers the case where a constraint is applied to a degree of freedom with an imposed displacement. Case 3 involves applying a load to a slave DOF connected to a degree of freedom with an imposed displacement. Essentially, all load cases represent the same problem and should yield identical displacement responses for the degrees of freedom, although the reaction forces at the nodes may be displayed differently.

Figure E.10.1 Analysis Model

Table E.10.1 Analysis Results (Case 1)

(a) Node Results

Node Displacement External Load Internal Load Remark
1 0.0000 -3.0 -3.0
2 0.0015 1.0 1.0
3 0.0025 1.0 2.0
4 0.0025 1.0 0.0

(b) Element Results

Element Axial Force Remark
1 3.0
2 2.0
3 0.0

Table E.10.2 Analysis Results (Case 2)

(a) Node Results

Node Displacement External Load Internal Load Remark
1 0.0000 -3.0 -3.0
2 0.0015 1.0 1.0
3 0.0025 2.0 (반력) 2.0
4 0.0025 0.0 0.0

(b) Element Results

Element Axial Force Remark
1 3.0
2 2.0
3 0.0

Table E.10.3 Analysis Results (Case 3)

(a) Node Results

Node Displacement External Load Internal Load Remark
1 0.0000 -3.0 -3.0
2 0.0015 1.0 1.0
3 0.0025 1.0 (반력) 2.0
4 0.0025 1.0 0.0

(b) Element Results

Element Axial Force Remark
1 3.0
2 2.0
3 0.0

E.10.2 Simple Constraint 2

Figure E.10.2 Analysis Model

Table E.10.4 Analysis Results (Case 1)

(a) Node Results

Node Displacement External Load Internal Load Remark
1 0.0000 -3.0 -3.0
2 0.0015 1.0 1.0
3 0.0025 1.0 2.0
4 0.0025 1.0 0.0

(b) Element Results

Element Axial Force Remark
1 3.0
2 2.0
3 -

Table E.5.5 Analysis Results (Case 2)

(a) Node Results

Node Displacement External Load Internal Load Remark
1 0.0000 -3.0 -3.0
2 0.0015 1.0 1.0
3 0.0025 2.0 (Reaction) 2.0
4 0.0025 0.0 0.0

(b) Element Results

Element Axial Force Remark
1 3.0
2 2.0
3 -

Table E.10.6 Analysis Results (Case 3)

(a) Node Results

Node Displacement External Load Internal Load Remark
1 0.0000 -3.0 -3.0
2 0.0015 1.0 1.0
3 0.0025 1.0 (Reaction) 2.0
4 0.0025 1.0 0.0

(b) Element Results

Element Axial Force Remark
1 3.0
2 2.0
3 -

Input File

  • sc1.inp

  • sc2.inp

  • sc3.inp

  • sc4.inp

E.10.3 Unrealistic Constraint

An analysis was performed for the case where each node of the upper and lower truss members is constrained in parallel along the X direction, as shown in the figure. The solution can be derived analytically, and the results are presented in the table.

Figure E.10.3 Analysis Model

Table E.10.7 Analysis Results (Case 1)

(a) Node Results

Node Displacement External Load Internal Load Remark
1 0.0000 -2.0 -0.4
2 0.0020 0.0 0.4
11 0.0000 0.0 -0.8
12 0.0040 1.0 0.8

(b) Element Results

Element Axial Force Remark
1 0.4
2 0.8

Table E.10.8 Analysis Results (Case 2)

(a) Node Results

Node Displacement External Load Internal Load Remark
1 0.0000 -2.0 -0.4
2 0.0020 0.0 0.0
3 0.0040 0.0 0.0
4 0.0060 0.0 0.0
5 0.0080 0.0 0.4
11 0.0000 0.0 -0.8
12 0.0040 0.0 0.0
13 0.0080 0.0 0.0
14 0.0012 0.0 0.0
15 0.0016 1.0 0.8

(b) Element Results

Element Axial Force Remark
1 0.4
2 0.4
3 0.4
4 0.4
5 0.8
6 0.8
7 0.8
8 0.8

□ Remark

A hand calculation of the solution is presented. The equation of motion is formulated as follows.

By applying the constraint

a condensed system can be constructed as follows:

Applying the boundary conditions yields \(\small u_{2} = 0.0002\) and \(\small P_{1} = -2\):

The internal forces in each element are 0.4 for element 1 and 0.8 for element 2 (tension).
When the element internal forces are assembled into nodal internal forces, the result is:

The external force P and the internal force F satisfy equilibrium with respect to the condensed system. That is, the condensed force for each is \(\small \begin{bmatrix} -2 & +2 \end{bmatrix}^{T}\). While the internal force F sums to zero over all components, the external force P does not. This is because an unrealistic constraint condition was imposed.

Input File

  • urc1.inp : Input file for Case 1

  • urc4.inp : Input file for Case 2

E.10.4 Constraint of Space Frame

An analysis was performed for the space frame shown in the figure, with arbitrary constraints applied, and the results were compared with those from SAP2000. The constraints defined by the *Constraint command are handled using the transformation method. In the transformation method, there is one slave DOF corresponding to each constraint equation, and the following limitations are generally known:

  • A slave DOF that has been used once may be used as a master DOF in another constraint equation, but it cannot be used again as a slave DOF.
  • A slave DOF cannot have displacement boundary conditions (zero DOF or prescribed DOF). However, it can be subjected to external loads (nodal loads).

In hfAnalyzer, these conditions are managed through a special handling method, so the user does not need to be concerned about them. The following is a test example.

  • Case 1: UX1 - UX4 = 0, UX2 - UX4 = 0, UX3 - UX4=0

  • Case 2: UX1 - UX4 = 0, UX2 - UX1 = 0, UX3 - UX1=0

  • Case 3: UX1 - UX4 = 0, UX2 - UX1 = 0, UX3 - UX2=0

  • Case 4: UX1 - UX2 = 0, UX2 - UX3 = 0, UX3 - UX4=0, UX4-UX1=0(over)

Figure E.10.4 Analysis Model

Table E.10.9 Analysis Results (Case 1)

Classification Node Degree of Freedoms Remark
X Y Z RX RY RZ
Displacement 11134.7-2481.6-126.47151.6583.2560.01039
21134.7-1759.8-379.39151.6583.2560.01039
31134.7-1820.9126.46181.3783.2560.017191
41134.7-3015379.4181.3883.2560.017191
Reaction 5-2.55.99722.0172-34.384-14.91412.555
6-2.53.48556.0513-21.825-14.91412.555
7-2.53.1809-2.0171-21.164-14.91420.773
8-2.57.3364-6.0514-41.942-14.91420.773

Input File

  • fc.inp

E.10.5 NodeToSurface Constraint

A NodeToSurface constraint refers to a constraint that connects a slave node to a master surface. When specified in the input as slave surface – master surface, the first-listed slave surface is converted into slave nodes corresponding to the nodes contained in that surface.

Figure E.10.5 Verification Model (Cantilever with 20\*4\*1 m)

Figure E.10.6 Verifcation Models

Figure E.10.7 Verifcation Models – Deformed Shape

Table E.10.10 Analysis Results

oneC oneF twoF2C twoC2F

leftToRight

(fineToCoarse)

rightToLeft

(coarseToFine)

leftToRight

(coarseToFine)

rightToLeft

(fineToCoarse)

Static analysis

(end load)

 19599 19460 19503 19475 19556 19551

Frequency analysis

(f1, Hz)

 0.158249 0.15777 0.157817 0.157717 0.158231 0.158191

In general, in a slave–master relationship, the master should be defined on the stiffer or coarser side (ABAQUS recommendation). However, in this example, since a sufficiently fine mesh is used, it is found that selecting the mesh on the left—whose behavior has a greater influence on the overall response—results in a solution closer to the exact one.

References

  1. R. D. Cook, D. S. Malkus, M. E. Plesha, and R. J. Witt (2002) “Concepts and Applications of Finite Element Analysis", Forth Edition pp. 491