\section*{Task 2 - 011232}

From aluminum sheet with a thickness of \(2 \mathrm{~mm}\), 10000 workpieces are to be punched according to the attached drawing 1. (All internal angles are the same, \(a=34 \mathrm{~mm}, b=8 \mathrm{~mm}\).)
![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1736.jpg?height=268&width=922&top_left_y=437&top_left_x=570)

a) How long and how wide must the strip of sheet metal be from which the pieces are punched? It should be noted that the web width (distance between the parts or from the edge) \(s_{1}=2 \mathrm{~mm}\) must be maintained.

How many square meters of sheet metal will be consumed? How many square meters of waste will there be?

b) A suggestion is made to punch according to drawing 2 to save material. How long and how wide must the strip of sheet metal now be? How many square meters of sheet metal will be consumed? How many square meters of waste will there be?

What percentage of material savings is achieved compared to the method specified in a)? (Web width here \(s_{2}=3 \mathrm{~mm}\).)
}

Since the workpiece is a hexagon with equal interior angles, each interior angle is \(\frac{1}{6}(6-2) \cdot 180^{\circ} = 120^{\circ}\).

We will now calculate the length (in the longitudinal direction of the sheet metal strip) and width (in the transverse direction) of a workpiece, and from this, the required dimensions. a) (Figure a) The length is

\[
b \cdot \cos \left(30^{\circ}\right) + a \cdot \cos \left(30^{\circ}\right) = \frac{\sqrt{3}}{2}(a + b) = 36.373 \text{ mm}
\]

for 10000 workpieces, this is \(363.730 \text{ m}\) plus \(20.002 \text{ m}\) for 10001 gaps \(s_{1}\). Thus, the sheet metal strip must be \(383.732 \text{ m}\) long. The width of the workpiece is

\[
a + 2 b \sin \left(30^{\circ}\right) = a + b = 42 \text{ mm}
\]

together with two gaps \(s_{1}\), the sheet metal strip must be \(46 \text{ mm}\) wide. Its area is \(383.732 \text{ m} \cdot 0.046 \text{ m} = 17.652 \text{ m}^2\).

The area of a workpiece is equal to the area of an equilateral triangle with side length \(a + 2b\) minus the area of three equilateral triangles with side length \(b\), thus

\[
\frac{\sqrt{3}}{4}(a + 2b)^2 - 3 \frac{\sqrt{3}}{4} b^2 = 999.4 \text{ mm}^2
\]

The 10000 workpieces thus require an area of \(9.994 \text{ m}^2\), from which the waste is calculated to be \(7.658 \text{ m}^2\).

b) (Figure b) Since the workpieces are now rotated by \(90^{\circ}\), the width is equal to the length from a), together with the double flange width \(s_{2}\), the width of the sheet metal strip is \(42.4 \text{ mm}\).

To calculate the length, \(s_{2} = 0\) is initially assumed. The centers of two workpieces are then

\[
\frac{b}{2} + \frac{b}{2} + \frac{a}{2} + \frac{b}{2} = \frac{1}{2}(a + 3b) = 29.0 \text{ mm}
\]

apart. The total length of the sheet metal strip is here

\[
9999 \cdot \left(29.0 \text{ mm} + s_{2} \cos \left(30^{\circ}\right)\right) + 2 \cdot 21.0 \text{ mm} + 2 s_{2} = 315.997 \text{ m}
\]

the consumption is \(13.398 \text{ m}^2\) and the waste is \(3.404 \text{ m}^2\). The material savings compared to a) is \(24.1 \%\). Problem solved by Eckard Specht