UNIVERSIDAD AUTÓNOMA DE YUCATÁN

FACULTAD DE INGENIERÍA

MASTER IN STRUCTURAL ENGINEERING

ADVANCED STRUCTURAL ANALYSIS

UNIDAD 2. FRAMES

ADA 3 - ANALYZE FRAMES WITH THE STIFFNESS METHOD

STUDENT: I.C. LOURDES ELENA PECH CRUZ

PROFESOR: DR. LUIS ENRIQUE FERNÁNDEZ BAQUEIRO

AUGUST - DECEMBER 2023

 ADA 3 - PROCESS
UNIT 2. FRAMES
ADVANCED STRUCTURAL ANALYSIS
Learning outcome

Analyze frames with the Stiffness Method.

Activity

Individually, determine the internal forces and reactions of the following frame structure using the
Stiffness Method. If you want, you can neglect the axial deformation of the elements.

Determine the stiffness matrix of each element.

• Assemble the stiffness matrix of the structure.

• Determine the external force vector.
• Determine the fixed end force vector.
• Determine the displacements.
• Determine the internal forces at the ends of the elements.
• Draw the free-body diagram.
• Verify that the structure is in equilibrium with the reactions.
• Draw the internal forces diagrams.

The frame is made of reinforced concrete and has a modulus of elasticity 𝐸 = 1.7 × 105 𝑘𝑔/𝑐𝑚2.

The columns have a square cross section of 40 × 40 𝑐𝑚.

The beams have a rectangular cross section of 40 × 80 𝑐𝑚.

There is a vertical displacement beneath column 1 equal to 2 𝑐𝑚.

 Fig. 1. Geometry and loads definition.

Compare your results with those obtained from the commercial program. Show the main images of the
procedure to input data in the commercial program and images of the results: deformed shape, internal
forces diagrams and reactions.

Write your conclusions comparing the results and presenting your main learnings.

Indicate the references you used to do your homework.

 ADA 3

1. Select the units system of units and determine the parameters with that system of units.
The frame will be analyzed considering the units in tons and meters.

Material properties
𝐸 = 1.7 × 105 𝑘𝑔/𝑐𝑚2 = 1.7 × 106 𝑡𝑜𝑛/𝑚2

Geometric properties
• Column: 𝐴 = 0.16 𝑚2 and : 𝐼 = 2.133 × 10−3 𝑚4
• Beam: 𝐴 = 0.32 𝑚2 and : 𝐼 = 0.0170 𝑚4

2. Identify nodes and elements. Define the local axis.

Fig. 2. Numbering of nodes and elements. Local axes x'.

3. Identify known and unknown degrees of freedom.

There are 6 unknown degrees of freedom:

(𝑢1 = 𝜑1 = 𝑢2 = 𝑣2 = 𝜑2 = 0, 𝑣1 = 0.02 𝑚)

There are 18 unknown degrees of freedom:

( 𝑢3, 𝑣3, 𝜑3, 𝑢4, 𝑣4, 𝜑4, 𝑢5, 𝑣5, 𝜑5, 𝑢6, 𝑣6, 𝜑6, 𝑢7, 𝑣7, 𝜑7, 𝑢8, 𝑣8, 𝜑8).

4. Identify the degrees of freedom of the structure organized into unknown and known:

𝑢3, 𝑣3, 𝜑3, 𝑢4, 𝑣4, 𝜑4, 𝑢5, 𝑣5, 𝜑5, 𝑢6, 𝑣6, 𝜑6, 𝑢7, 𝑣7, 𝜑7, 𝑢8, 𝑣8, 𝜑8
𝑢1, 𝑣1, 𝜑1, 𝑢2, 𝑣2, 𝜑2

5. Determine the stiffness matrix of each element.

6. Determine the stiffness matrix of the structure.
7. Determine the submatrices of the stiffness matrix of the structure: [𝐾𝑢𝑢], [𝐾𝑢𝑢∗], [𝐾𝑢∗𝑢] y [𝐾𝑢∗𝑢∗].
 ̅̅̅̅
8. Determine the fixed ends forces for each element {𝐹}
Elements 1, 2, 3, 4, 5, 8 and 9 don’t have loads.
Element 6. Nodes 3-4. Point load: 𝑃 = 6 𝑡𝑜𝑛 at 3 meters.

Element 7. Nodes 4-5. Uniform load: 𝑤 = 2.5 𝑡𝑜𝑛/𝑚.

9. Determine the fixed ends forces vector of the structure.

10. Determine the external forces vector {𝐹𝐸𝑥𝑡}.

11. Determine the unknown displacements.

{u} = [𝐾𝑢𝑢 ]−1 [{ 𝐹 𝐸𝑥𝑡 } − [𝐾𝑢𝑢∗ ]{𝑢∗ }]

Determine the prescribed displacements vector: {𝑢∗ }

Determine the forces vector: 𝐹 𝐸𝑥𝑡 − 𝐹𝑒𝑚𝑝 − [𝐾𝑢𝑢∗ ]{𝑢∗ } = 𝐹𝑒𝑥𝑡2

Finally, determine the unknown displacements: {u}
12. Determine the reactions: 𝑅𝑥1 , 𝑅𝑦1 , 𝑀𝑧1 , 𝑅𝑥2 , 𝑅𝑦2 , 𝑀𝑧2 ,

{R} = [𝐾𝑢∗𝑢 ]{u} + [𝐾𝑢∗𝑢∗ ]{u∗ }

13. Verify the equilibrium of the structure:

14. Determine the internal forces at the ends of the elements:

{𝐹 𝐼𝑛𝑡 } = [K]{u} + {F̅}
15. Draw the free body diagram.

Fig. 3. Free body diagram of internal forces in elements.

 Fig. 4. Free body diagram of reactions and internal forces at nodes.

The following table shows that equilibrium is verify in each node of the frame according to the free body
diagram (Fig.4). This is because when adding the internal forces in each nodenode, the result is zero.

Table 1. Verification the equilibrium

Internal forces in echa node

Node Fx Fy Mz
1 1.1705-1.1705=0 12.037-12.037=0. 3.52-3.52=0
2 1.1705-1.1705=0 15.4628-15.4628=0 0.3863-0.3863=0

3 2.6372+1.1705-3.808=0 12.037-5.6273-6.4099=0 5.7091+1.15-6.8676=0

4 2.9394-0.3032-2.6372=0 1.13+0.4099-1.5407=0 0.513+12.259-12.772=0

5 4.11-1.1705-2.9394=0 15.4628-4.504-10.9593=0 11.2875-5.0685-6.219=0
6 3.8078-3.8078=0 5.6273-5.6273=0 5.7141-5.71=0
7 3.8078+0.3032-4.11=0 5.6273+4.5035-1.13-9=0 0.3936+16.40-16.8052=0
8 4.11-4.11=0 4.504-4.504=0 6.11-6.11=0
16. Obtain the internal forces diagrams.

Fig. 5. Internal forces diagram (axial).

Fig. 6. Internal forces diagram (shear).

Fig. 7. Internal forces diagram (bending moment).
 STRUCTURAL ANALYSIS - COMMERCIAL PROGRAM

1. Create a new file in the STAAD Pro program and configure it with metric units.
2. Enter the coordinates of the nodes and join them to form the elements of the frame.

3. Create and assign the cross-sections of the frame elements.

4. Assign the type of material: concrete with modulus of elasticity of 𝐸 = 1.7×105 <𝑘𝑔/𝑐𝑚2.
5. Create the fixed supports.

6. Create a new state of load.

7. Create the loads ítems:
Point load: −6 𝑡𝑜𝑛 = −58.8399 𝑘𝑁

Point load: -9 𝑡𝑜𝑛 = −88.2599 𝑘𝑁

Uniform load: −2.5 𝑡𝑜𝑚 = −24.5166 𝑘𝑁

8. Create the displacement of 0.02 m to the support of node 1..

9. Assign the loads and the displacement to the corresponding nodes!

10. Perform the configuration to run the analysis.

11. Run the program to analyze the framework and verify that there are no errors.

12. After running the program, perform the configuration to observe the values of the results obtained.
The results obtained from the STADD Pro program are presented below.

• Displacements

Fig. 8. Displacements results in each node of the frame.

• Reactions

Fig. 9. Reactions results.

• Internal forces

Fig. 10. Internal forces results.

Fig. 11. Internal forces diagram (axial).

Fig. 12. Internal forces diagram (shear)

Fig. 13. Internal forces diagram (bending moment)

.
Fig. 14. Deformed shape.
Comparing the results between Stiffness Method and STAAD Pro program:

Table 2. Comparison of reactions results.

Reactions (ton)
Reaction Stiffness STAAD Pro
Method
𝑅𝑥1 1.1705 1.167
𝑅𝑦1 12.0372 12.037
𝑀𝑧1 -3.5288 -3.54
𝑅𝑥2 -1.1705 -1.167
𝑅𝑦2 15.4628 15.463
𝑀𝑧2 -0.3863 -0.407

Table 3. Comparison of displacements results in each node of the truss.

Displacements
Stiffness Method STAAD Pro
Node Horizontal x Vertical y Rotacional rZ Horizontal x Vertical y Rotacional rZ
(m) (m) (rad) (m) (m) (rad)
1 0 -0.02 0 0 -0.02 0
2 0 0 0 0 0 0
3 -0.0043 -0.0202 0.0013 -0.004 -0.020 0.001
4 -0.0043 -0.0145 0.002 -0.004 -0.015 0.002
5 -0.0043 -0.0002 0.003 -0.004 -0.000 0.003
6 -0.0106 -0.0202 0.0013 -0.011 -0.020 0.001
7 -0.0107 -0.0145 0.0021 -0.01 -0.015 0.002
8 -0.107 -0.0003 0.003 -0.11 0.000 0.003
 Table 4. Comparison of internal forces results.

Internal forces
Stiffness Method STAAD Pro
Element Node
Global coordinates Local coordinates
Fx Fy Mz Fx Fy Mz

1 1 1.1705 12.0372 -3.5288 12.037 1. 167 -3.508

(Column) 3 -1.1705 -12.0372 -1.1534 -12.037 -1.167 -1.16

2 2 -1.1705 15.4628 -0.3863 15.463 -1.167 -0.408

(Column) 5 1.1705 -15.4628 5.0685 -15.463 1.167 5.075

3 3 3.8078 5.6273 -5.7142 5.626 3.741 -5.617

(Column) 6 -3.8078 -5.6273 -5.7091 -5.626 -3.741 -5.609

4 4 0.3022 -1.1308 -0.5131 -1.129 0.298 -0.508

(Column) 7 -0.3022 1.1308 -0.3936 1.129 -0.298 -0.387

5 5 -4.11 4.5035 6.219 4.503 -4.039 6.112

(Column) 8 4.11 -4.5035 6.1109 -4.503 4.039 6.006

6 3 -2.6372 6.4099 6.8676 -2.575 6.411 6.774

(Beam) 4 2.6372 -6.4099 12.772 2.575 -6.411 12.870

7 4 -2.9394 1.5407 -12.9394 -2.873 1.54 -12.362

(Beam) 5 2.9394 10.9593 -11.2875 2.873 10.960 -11.187

8 6 3.8078 5.6273 5.7091 3.741 5.626 5.6809

(Beam) 7 -3.8078 -5.6273 -5.6273 -3.741 -5.626 16.896

9 7 4.11 -4.5035 -16.4065 4.039 -4.503 -16.509

(Beam) 8 -4.11 4.5035 -6.1109 -4.039 4.503 -6.006
 CONCLUSIONS

In this ADA 3, a frame was analyzed using the Stiffness method, the values of the displacements at
the nodes, the reactions and the internal forces of the elements were obtained. Subsequently, the same
structure was analyzed in a computer program called STAAD Pro.

As could be seen in the tables previously presented, the results of the displacements, reactions and
internal forces obtained with the Stiffness Method and the STAAD Pro program present differences
that are minimal and, therefore, could be neglected and it’s verified that the structural analysis is
correct.

The values of internal forces of the elements, in Mathcad are obtained in the direction of global
coordinates and, on the contrary, in STADD Pro they are presented in global coordinates. For this
reason, the forces result Fx and Fy of the columns are presented the opposite of those obtained in
Mathcad. However, with the correct interpretation of the results with both methods, it is observed that
they are the same results.
The same way, for this activity all degrees of freedom were considered for each node; however, if the
frame analysis had been intended to be simplified, the unknow degrees of freedom would have been
reduced, neglecting the axial deformations of the elements. So that:

𝑢1 = 𝑢4 = 𝑢5

𝑢6 = 𝑢7 = 𝑢8

𝑣1 = 𝑣3 = 𝑣6

𝑣2 = 𝑣5 = 𝑣8 = −0.02

𝑣4 = 𝑣7

According to the results obtained, it is verified that the displacements in x direction (u) and in the y
direction (v) are very similar for the nodes in which a similar displacement would have been assumed.
Therefore, it is correct to be able to neglect axial deformations.

As part of my learning, I concluded that both methods are equally efficient, however, using a computer
program to analyze a frame can be a faster and equally accurate process than applying the Stiffness
Method.
Únicamente hay que tener cuidado al momento de introducir la información del marco como lo son:
los nodos, los elementos, las propiedades de los materiales, fuerzas externas, tipos de apoyos,
asentamientos y cualquier otro tipo de consideración, pues el equivocarse podría arrojar resultados
erróneos. Y finalmente, los conceptos, ejemplos y todo lo aprendido durante las sesiones me sirvieron
de base para poder aplicar mis conocimientos y poder realizar esta actividad de aprendizaje.
You just must be careful when introducing the frame information such as: nodes, elements, material
properties, external forces, supports, settlements and any other type of consideration, because making
a mistake could give wrong results. And finally, the concepts, examples and everything learned during
the sessions served as a basis to apply my knowledge and carry out this learning activity.