Question 1.4: (a) Derive the stiffness matrix for the chain of springs sho......

(a) The (3 × 3) stiffness matrix can be found by setting each of the coordinates x_{1}, x_{2} and x_{3} to 1 in turn, with the others at zero, and writing down the forces required at each node to maintain equilibrium:

when : \begin{aligned}&\begin{array}{lll} x_{1}=1 &&& x_{2} = 0 &&& x_{3} = 0\end{array}\end{aligned}

then :

when : \begin{aligned}&\begin{array}{lll} x_{1} = 0 &&& x_{2} = 1 &&& x_{3} = 0\end{array}\end{aligned}

then :

when : \begin{aligned}&\begin{array}{lll} x_{1} = 0 &&& x_{2} = 0 &&& x_{3} = 1 \end{array}\end{aligned}

then :

Equations (A_{1}) now give the first column of the stiffness matrix, Eqs (A_{2}) the second column, and so on, as follows:

\begin{Bmatrix} f_{1} \\ f_{2} \\ f_{3} \end{Bmatrix} = \begin{bmatrix} (k_{1}+k_{2}) & -k_{2} & 0 \\ -k_{2} & (k_{2}+k_{3}) & -k_{3} \\ 0 & -k_{3} & k_{3} \end{bmatrix} \begin{Bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{Bmatrix}                        (B)

The required stiffness matrix is given by Eq. (B).

(b) The flexibility matrix can be found by setting f_{1}, f_{2} and f_{3} to 1, in turn, with the others zero, and writing down the displacements x_{1}, x_{2} and x_{3}:

Equations (C_{1}) give the first column of the flexibility matrix, Eqs (C_{2}) the second column, and so on:

(D)

Equation (D) gives the required flexibility matrix.

(c) To prove that the flexibility matrix is the inverse of the stiffness matrix, and vice versa, we multiply them together, which should produce a unit matrix, as is, indeed, the case.

(E)