Mathematical Background

(a) Symmetry and positive definiteness

$K_{12} = K_{21} = -2$, $K_{23} = K_{32} = -1$, $K_{13} = K_{31} = 0$, so $[K]$ is symmetric.

For positive definiteness, all leading principal minors must be positive:

• $\Delta_1 = 5 > 0$
• $\Delta_2 = 5(4) - (-2)(-2) = 16 > 0$
• $\Delta_3 = \det[K] = 5(12-1) + 2(-6-0) = 55 - 12 = 43 > 0$

All are positive, so $[K]$ is positive definite and a unique solution exists.

(b) Gauss elimination

Augmented matrix:

$$\left[\begin{array}{ccc|c} 5 & -2 & 0 & 10 \\ -2 & 4 & -1 & 0 \\ 0 & -1 & 3 & 6 \end{array}\right]$$

Step 1: $R_2 \leftarrow R_2 + 0.4R_1$ gives $[0,\ 3.2,\ -1\,|\,4]$.

Step 2: $R_3 \leftarrow R_3 + \frac{1}{3.2}R_2 = R_3 + 0.3125R_2$ gives $[0,\ 0,\ 2.6875\,|\,7.25]$.

Back substitution:

$$\begin{aligned} u_3 &= \frac{7.25}{2.6875} = 2.698\ \text{mm} \\ u_2 &= \frac{4 + 2.698}{3.2} = 2.093\ \text{mm} \\ u_1 &= \frac{10 + 2(2.093)}{5} = 2.837\ \text{mm} \end{aligned}$$

Check: row 3 gives $-2.093 + 3(2.698) = 6.00$ kN. Also the product of pivots $5 \times 3.2 \times 2.6875 = 43 = \det[K]$.

Why positive definite: strain energy $U = \tfrac{1}{2}\{u\}^T[K]\{u\}$ must be positive for every non-zero displacement. If a rigid-body motion remains (supports missing), $[K]$ is singular and the equations cannot be solved. Positive pivots also guarantee that elimination runs without row interchange.

Answer: $u_1 = 2.837$ mm, $u_2 = 2.093$ mm, $u_3 = 2.698$ mm.

Differences

Classification

$EI\,w'''' = q$ is a linear, fourth-order, ordinary differential equation (one independent variable $x$). For a fourth-order equation $m = 2$, so $w$ and $w'$ are primary variables (essential), while $EIw''$ (moment) and $EIw'''$ (shear) are secondary variables (natural).

Cantilever beam (fixed at $x=0$, free at $x=L$)

• Essential: $w(0) = 0$, $\dfrac{dw}{dx}(0) = 0$
• Natural: $EI\dfrac{d^2w}{dx^2}(L) = 0$ (zero moment), $EI\dfrac{d^3w}{dx^3}(L) = 0$ (zero shear)