# Giorgio Serino » 8.Matrix Methods of Structural Analysis... Basics

### TECNICA DELLE COSTRUZIONI II

Matrix Methods of Structural Analysis… Basics

Dr. (Eng) Chandrasekaran

### Structures are REAL, PHYSICAL entities

Concrete, Steel, Cables, Nuts, Bolts…it seems a mess!!!

How do you handle this mathematically???

### We require an “IDEALIZATION”

Means that we assume that they are made up of some simple ‘elements’ whose behavior can be predicted using principles of mechanics (of course under the assumption that the structure is indeed made up of ‘elements’).

Accuracy of our results depend on the accuracy of our idealization.

### Example 1 – SI

SI is often associated with removal of ‘redundant’ or ‘extra’ supports, which will make the structure statically determinate at the same time maintaining its stability.

We have 3 equilibrium equations for this beam (vertical & horizontal force equilibrium and moment equilibrium). So, the structure will be determinate if we have only 3 unknown actions. But, we have 4 reaction forces here.

Hence, in order to make the structure determinate, we must remove ONE action. This can be done by removing either of the roller supports.

It can be easily seen that removing a reaction from the hinge support will make the structure unstable.

### Example 1 – KI

It follows from the definition that KI is the number of independent displacements (rotations or translations) which can the structural system can have at its joints.

Since all the supports are hinge or roller type, therefore the only possible independent displacements could be the ‘rotational’ displacements as shown in pink arrows.

It should be noted that despite the presence of roller supports, the system cannot have a horizontal translation because of the presence of one hinge supports which denies any such displacement.

Same goes with the vertical displacement.

### Example 2 – SI

We have 3 equilibrium equations for this frame (vertical & horizontal force equilibrium and moment equilibrium). So, the structure will be determinate if we have only 3 unknown actions. But, we have 6 reaction forces here.

Hence, in order to make the structure determinate, we must remove THREE action. This can be done in two ways:

1. By making one of the fixed supports a hinged support and other, a roller support.
2. By keeping one support as fixed and making the other free.

### Example 2 – KI

It can be easily seen from the figure that ‘two’ rotations are possible at the joints and ‘one’ translation (or sway) is possible.

Since both the supports are fixed, therefore there cannot be any type of displacement here.

Thus, the total number of independent displacements is ‘three’.

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