UNIVERSIDAD AUTÓNOMA DE YUCATÁN

FACULTAD DE INGENIERÍA

MASTER IN STRUCTURAL ENGINEERING

ADVANCED STRUCTURAL ANALYSIS

UNIDAD 3.

ADA 6H - ANALYZE GRIDS WITH THE STIFFNESS METHOD

STUDENT: I.C. LOURDES ELENA PECH CRUZ

PROFESOR: DR. LUIS ENRIQUE FERNÁNDEZ BAQUEIRO

AUGUST - DECEMBER 2023

 ADA 6 - PROCESS
UNIT 2. FRAMES
ADVANCED STRUCTURAL ANALYSIS
Learning outcome

Analyze grids with the Stiffness Method.

Activity

Individually, determine the internal forces and reactions of the following frame structure using the
Stiffness Method.

• Determine the stiffness matrix of each element.

• Assemble the stiffness matrix of the structure.
• Determine the displacements.
• Determine the internal forces at the ends of the elements
• Draw the free-body diagram in global axes.
• Draw the free-body diagram in local axes.
• Draw the internal forces diagrams.

The grid is made of reinforced concrete and has a modulus of elasticity 𝐸 = 1.8 × 105 𝑘𝑔/𝑐𝑚2.

The three elements have a rectangular cross section of 60 cm x 110 cm and the same length (4 m).
Do not consider the shear deformation.

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.
 ADA 6H

1. Select the units system and determine the parameters.

The grid will be analyzed considering the units in tons and meters.

Geometric and material properties

• Cross section: 𝑏 = 0.6 𝑚, ℎ = 1.1 𝑚
• Area: 𝐴 = 𝑏ℎ = 0.66 𝑚2
𝑏ℎ3
• Inertia: 𝐼 = = 0.067 𝑚4
12
16 𝑏 𝑏𝑎4
• Polar moment of inertia: 𝐽 = 𝑎𝑏 3 [ 3 − 3.36 𝑎 (1 − 12𝑎4 ] = 0.052 𝑚4

• Modulus of elasticity: 𝐸 = 1.8 × 105 𝑘𝑔/𝑐𝑚2 = 1.8 × 106 𝑡𝑜𝑛/𝑚2

𝐸
• Shear modulus: 𝐺 = 2(1+𝑣) = 750,000 𝑡𝑜𝑛/𝑚2

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

Fig. 3. Numbering of nodes and elements. Local axes x'.

3. Identify known and unknown degrees of freedom.

There are 6 known degrees of freedom:

𝜑𝑥3 = 𝜑𝑦3 = 𝑢𝑧3 = 𝜑𝑥4 = 𝜑𝑦4 = 𝑢𝑧4 = 0

There are 6 unknown degrees of freedom:

𝜑𝑥1 , 𝜑𝑦1 , 𝑢𝑧1 , 𝜑𝑥2 , 𝜑𝑦2 , 𝑢𝑧2
4. Identify the degrees of freedom of the structure organized into unknown and known:
𝜑𝑥1 , 𝜑𝑦1 , 𝑢𝑧1 , 𝜑𝑥2 , 𝜑𝑦2 , 𝑢𝑧2 𝜑𝑥3 , 𝜑𝑦3 , 𝑢𝑧3 , 𝜑𝑥4 , 𝜑𝑦4 , 𝑢𝑧4

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. In this case, there are no
prescribed displacements {𝑢∗ } = 0. Therefore, it is only necessary to determine: [𝐾𝑢𝑢], [𝐾𝑢∗𝑢].

8. Determine the fixed ends forces of each element.

9. Determine the fixed ends forces vector of the structure (know a unknow degrees of freedom).

10. Determine the external forces vector {𝐹𝐸𝑥𝑡}. In this case there are not external forces.}

11. Determine the unknown displacements.

{u} = [𝐾𝑢𝑢 ]−1 [{ 𝐹 𝐸𝑥𝑡 2 } − [𝐾𝑢𝑢∗ ]{𝑢∗ }]
{ 𝐹 𝐸𝑥𝑡 2 } = { 𝐹 𝐸𝑥𝑡 } − { 𝐹 𝑒𝑚𝑝 𝑑 }

In this case, there are no prescribed displacements.

{u} = [𝐾𝑢𝑢 ]−1 [{ 𝐹 𝐸𝑥𝑡 2 }

Finally, determine the unknown displacements: {u}

The known displacements

12. Determine the displacement vector considering all degrees of freedom.

13. Determine the reactions: 𝑀𝑥3 , 𝑀𝑦3 , 𝐹𝑧3 , 𝑀𝑥4 , 𝑀𝑦4 , 𝐹𝑧4
{R} = [𝐾𝑢∗𝑢 ]{u} + [𝐾𝑢∗𝑢∗ ]{u∗ }

14. Verify the equilibrium of the structure:

15. Determine the internal forces at the ends of the elements:
{𝐹 𝐼𝑛𝑡 } = [K]{u} + 𝐹 𝑒𝑚𝑝
16. Draw the free body diagram in global axes.

Fig. 4. Free body diagram of internal forces in elements (global axes).

 Fig. 5. Free body diagram of reactions and internal forces at nodes (global axes).

The following table shows that equilibrium is verify in each node of the frame according to the free body
diagram (Fig.5). This is because when adding the internal forces in each node, the result is zero.

Table 1. Verification the equilibrium

Internal forces in each node

Node Mx My Fz
1 0.2-0.2=0 1.322-1.322=0 5.526-5.526=0
2 0.2-0.2=0 0.577-0.577=0 6.475-6.475=0
3 25.698-25.698=0 0.577-0.577=0 6.475-6.475=0
4 28.002-28.002=0 14.73-14.73=0 15.526-15.526=0
17. Draw the free body diagram in local axes.

Fig. 5. Free body diagram of reactions and internal forces in elements (local axes).

18. Obtain the internal forces diagrams.

Fig. 6. Internal forces diagram (torsion).

 Fig. 7. Internal forces diagram (shear).

Fig. 8. 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 grid.

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

4. Assign the type of material: concrete.
6. Create the fixed supports.

7. Create a new state of load.

8. Create the loads items:
Point load: −10 𝑡𝑜𝑛 = −99.6402 𝑘𝑁

Uniform load: -3 𝑡𝑜𝑛/𝑚 = −29.41 𝑘𝑁/𝑚

9. Assign the loads to the corresponding nodes and elements.

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. 9. Displacements results in each node of the frame.

• Reactions

Fig. 10. Reactions results.

• Internal forces

Fig. 11. Internal forces results.

Fig. 12. Internal forces diagram (torsion).

Fig. 13. Internal forces diagram (shear)

Fig. 14. Internal forces diagram (bending moment)

.
Fig. 15. Deformed shape.
Comparing the results between Stiffness Method and STAAD Pro program:

Table 2. Comparison of reactions results.

Reactions (ton)
Reaction Stiffness Method STAAD Pro
𝑀𝑥3 25.698 25.706
𝑀𝑦3 0.577 0.582
𝑅𝑧3 6.475 6.4759
𝑀𝑥4 28.002 28.119
𝑀𝑦4 -14.731 -14.790
𝑅𝑧4 15.526 15.6805

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

Displacements
Stiffness Method STAAD Pro
Node Φx Φy Horizontal z Φx Φy Horizontal z
(rad) (rad) (m) (rad) (rad) (m)
1 -0.00041 0.00009 -0.0011 -0.00041 0.00009 -0.00117
2 -0.00043 -0.00006 -0.00114 -0.00043 -0.00006 -0.0012
3 0 0 0 0 0 0
4 0 0 0 0 0 0

Table 4. Comparison of internal forces results.

Internal forces
Stiffness Method STAAD Pro
Element Node
Local axes Local axes
Mx My Fz Mx My Fz
4 1.245 -31.616 15.526 1.245 -31.616 15.6805
1
1 -1.245 -0.488 -5.526 -1.245 -0.488 -5.5201
2 0.577 25.698 6.475 0.582 25.709 6.4759
2
4 -0.577 0.2 -6.475 -0.582 0.198 -6.4759
3 0.2 1.321 5.525 0.198 1.3330 5.5201
3
5 -0.2 0.577 6.475 -0.198 0.582 6.4759
 CONCLUSIONS

In this ADA 6, a grid 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 present differences that are
minimal and, therefore, could be neglected and it’s verified that the structural analysis is correct.

As part of my learning, I concluded that both methods are equally efficient, however, using a computer
program to analyze a grid can be faster and equally accurate process than applying the Stiffness
Method.

In this unit we learned about the grids, which is like the topic of frames, with the difference that in the
grids, the loads are applied in a normal direction to the plane and mechanical elements of shear force,
bending moment and torsional moment are produced.

In previous activities, when there was an inclined support, certain considerations were taken; however,
in this activity, by having an inclined and fixed support, its three degrees of freedom are known, which
are equal to zero. And therefore, having it at 90° or at any angle, the effects and results will be the
same. However, if the support were inclined but not fixed (for example pinned support), you would have
to be very careful and take the necessary considerations regarding the angle.

One of the biggest challenges in this activity was that one of the elements of the grid had an angle of
inclination and fixed ends forces. This is because the fixed ends forces were calculated in local axes
and had to be transformed to global axes (Femp), and this was done by multiplying it by a new
transformation matrix (Temp), and these Femp in global axes were used to determine the internal forces
of the elements in global axes.

And the other challenge came when we had to draw the diagrams of mechanical elements, this
because the internal forces had to be calculated in local axes and this was done by multiplying Fint
vector to the transformation matrix [T].