Question

To generate the $j^{\text{th}}$ column of the flexibility matrix

• A. A unit force is applied at coordinate $j$ and the displacements are calculated at all coordinates
• B. A unit displacement is applied at co-ordinate $j$ and the forces are calculated at all coordinates
• C. A unit force is applied at coordinate $j$ and the forces are calculated at all coordinates
• D. A unit displacement is applied at co-ordinate $j$ and the displacements are calculated at all co-ordinates

Hint:

Flexibility matrix → apply unit {force}, find {displacements}. Stiffness matrix → apply unit {displacement}, find {forces}.

Answer 1

The Correct Option is A

Step 1: Recall the definition of flexibility coefficients.
Flexibility coefficients represent the displacement at one coordinate due to a unit force applied at another coordinate.
Step 2: Meaning of the $j^{\text{th}}$ column.
The $j^{\text{th}}$ column of the flexibility matrix consists of displacements at all coordinates caused by a unit force applied at coordinate $j$.
Step 3: Analyze the options.
(A) Correct, this exactly matches the definition of flexibility coefficients.
(B) Incorrect, this describes stiffness matrix generation.
(C) Incorrect, flexibility matrix relates forces to displacements, not forces to forces.
(D) Incorrect, again related to stiffness, not flexibility.
Step 4: Conclusion.
Hence, option (A) is correct.

Answer 2

We can also arrive at this by looking directly at the matrix equation that defines the flexibility matrix, rather than only restating the coefficient definition in words.

The flexibility matrix relates the vector of coordinate displacements \(\{\delta\}\) to the vector of coordinate forces \(\{F\}\) as \(\{\delta\} = [f]\{F\}\). If we choose the force vector to be a unit vector with a 1 in position j and 0 everywhere else, i.e. apply a unit force only at coordinate j and nothing elsewhere, then multiplying out \([f]\{F\}\) simply picks out the \(j^{\text{th}}\) column of \([f]\), and the resulting displacement vector \(\{\delta\}\) gives exactly that column's entries.

1. A unit force is applied at coordinate j and the displacements are calculated at all coordinates: This is exactly the operation of substituting a unit vector at position j into \(\{\delta\} = [f]\{F\}\), which by matrix multiplication produces the \(j^{\text{th}}\) column of \([f]\) as the resulting displacement vector.
2. A unit displacement is applied at co-ordinate j and the forces are calculated at all coordinates: This describes instead substituting a unit displacement vector into the inverse relationship \(\{F\} = [K]\{\delta\}\), which extracts a column of the stiffness matrix \([K]\), not the flexibility matrix.
3. A unit force is applied at coordinate j and the forces are calculated at all coordinates: Forces cannot be "calculated" as an output when a force is already the specified input; the output of the flexibility relation for a force input is always a displacement vector, not another force vector.
4. A unit displacement is applied at co-ordinate j and the displacements are calculated at all co-ordinates: Applying a displacement and reading off displacements does not correspond to either of the standard matrix relations \([f]\) or \([K]\); one of the two quantities in any such relation must be the applied unit input and the other the computed output of a different physical type.

Substituting a unit force vector directly into the defining matrix equation confirms which operation generates a column of the flexibility matrix.

Therefore, the correct answer is a unit force is applied at coordinate j and the displacements are calculated at all coordinates.