Example: Line Segment 2-D

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Example: Line Segment 2-D

$\def \half {\frac{1}{2}} \def \kwart {\frac{1}{4}} \def \hieruit {\quad \Longrightarrow \quad} \def \EN {\quad \mbox{and} \quad}$ The equation of a line segment between

$(x_1,y_1)$ and

$(x_2,y_2)$ is:

$\left\{ \begin{array}{c} x = x_1 + \xi (x_2 - x_1) \\ y = y_1 + \xi (y_2 - y_1) \end{array} \right. \qquad \mbox{where:} \qquad 0 \le \xi \le 1$ It is assumed that weights are uniformly distributed across this line segment:

$w(\xi) = 1$ . The midpoint of the line segment is:

$\overline{x} = \int_0^1 x \, dx = x_1 + (x_2-x_1) \int_0^1 \xi \, d\xi = x_1 + (x_2-x_1) \half = \half (x_1 + x_2)$ Quite analogously for

$\overline{y}$ . Therefore:

$\overline{x} = \half (x_1 + x_2) \EN \overline{y} = \half (y_1 + y_2)$ The second moments of inertia are:

$\overline{x^2} = \int_0^1 x^2 \, dx = x_1^2 + 2 x_1 (x_2-x_1) \int_0^1 \xi \, d\xi + (x_2-x_1)^2 \int_0^1 \xi^2 \, d\xi =$

$x_1^2 + 2 x_1 (x_2-x_1) \half + (x_2-x_1)^2 \frac{1}{3} = x_1^2 + x_1 x_2 - x_1^2 + x_2^2/3 - 2 x_1 x_2 / 3 + x1^2/3 =$

$\frac{1}{3} \left( x_1^2 + x_2^2 + x_1 x_2 \right) \hieruit$

$\overline{x^2} - \overline{x}^2 = \frac{1}{3} \left( x_1^2 + x_2^2 + x_1 x_2 \right) - \kwart \left( x_1^2 + x_2^2 + 2 x_1 x_2 \right) =$

$\frac{1}{12} \left( x_1^2 + x_2^2 - 2 x_1 x_2 \right) = \frac{1}{12} \left( x_1 - x_2 \right)^2$ Quite analogously for

$\overline{y^2} - \overline{y}^2$ . Therefore:

$\sigma_{xx} = \frac{1}{12} \left( x_1 - x_2 \right)^2 \EN \sigma_{yy} = \frac{1}{12} \left( y_1 - y_2 \right)^2$ Cross moments of intertia are a somewhat different piece of cake. MAPLE has been employed here for the purpose of being error proof:

$\overline{x y} - \overline{x} \, \overline{y} =$

  simplify(x_1*y_1 + x_1*(y_2-y_1)/2 + y_1*(x_2-x_1)/2
   + (x_2-x_1)*(y_2-y_1)/3 - (x_2+x_1)*(y_2+y_1)/4);

Giving:

    1/12 x_1 y_1 - 1/12 x_1 y_2 - 1/12 y_1 x_2 + 1/12 x_2 y_2

$\hieruit \sigma_{xy} = \overline{x y} - \overline{x} \, \overline{y} = \frac{1}{12} (x_2 - x_1) (y_2 - y_1)$