\section*{Problem 1 - 051211}

A steel angle (equal-leg L-angle) has the cross-section shown in the figure. Here, \(b=50 \mathrm{~mm}, d=5 \mathrm{~mm}, r=2 r_{1}=7 \mathrm{~mm}\).

a) What is its mass for a length of \(5 \mathrm{~m}\)? (Density of steel \(\rho=7.85 \frac{\mathrm{g}}{\mathrm{cm}^{3}}\))?

b) What is the percentage error if, for the sake of simplification, the rounding is neglected and it is assumed that the cross-sectional area consists of two rectangular areas?

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1800.jpg?height=569&width=596&top_left_y=732&top_left_x=730)

c) What is the maximum percentage error that occurs in the approximate calculation described in b) if \(b=50 \mathrm{~mm}\) and \(r=7 \mathrm{~mm}\) are constant and \(d\) ranges from 5 \(\mathrm{mm}\) to \(9 \mathrm{~mm}\)?
}

a) For the mass of the body, one first needs the volume and for that, the area of the cross-section (since the length \(5 \mathrm{~m}\) is known). The cross-sectional area is composed of the areas \(A_{1}\) to \(A_{7}\), as shown in the diagram, where \(A_{1}=A_{2}, A_{3}=A_{4}\), and \(A_{5}=A_{6}\). The partial areas are as follows:

\[
\begin{array}{r}
A_{1}=d \cdot\left(b-r-r_{1}-d\right)=d \cdot\left(b-3 r_{1}-d\right)=b d-3 r_{1} d-d^{2} \\
A_{2}=\left(d-r_{1}\right) \cdot r_{1}=r_{1} d-r_{1}^{2} \\
A_{3}=\frac{1}{4} \cdot \pi \cdot r_{1}^{2}
\end{array}
\]

\[
A_{7}=(d+r)^{2}-\frac{1}{4} \cdot \pi \cdot r^{2}=d^{2}+4 r_{1}^{2}+4 r_{1} d-\pi r_{1}^{2}
\]

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1800.jpg?height=557&width=602&top_left_y=2006&top_left_x=727)

\[
A=2 \cdot\left(b d-3 r_{1} d-d^{2}\right)+2 \cdot\left(r_{1} d-r_{1}^{2}\right)+2 \cdot\left(\frac{1}{4} \cdot \pi \cdot r_{1}^{2}\right)+d^{2}+4 r_{1}^{2}+4 r_{1} d-\pi r_{1}^{2}
\]

\[
\begin{gathered}
=2 b d-d^{2}+2 r_{1}^{2}-\frac{1}{2} \cdot \pi \cdot r_{1}^{2}=480,258 \mathrm{~mm}^{2}=4,80258 \mathrm{~cm}^{2} \\
\Rightarrow m=\rho \cdot V=\rho \cdot A \cdot l=7,85 \frac{\mathrm{g}}{\mathrm{cm}^{3}} \cdot 4,80258 \mathrm{~cm}^{2} \cdot 500 \mathrm{~cm}=18850 \mathrm{~g}=18,85 \mathrm{~kg}
\end{gathered}
\]

b) The percentage error is obtained by dividing the simplified value by the exact value, multiplying by \(100 \%\), and then finding the difference from \(100 \%\).

Since the length and density are the same and would cancel out, it suffices to work with the areas:

The simplified area is \(A^{*}=b d+(b-d) d=2 b d-d^{2}\). From this, the error is:

\[
\Delta=\left(1-\frac{2 b d-d^{2}}{2 b d-d^{2}+2 r_{1}^{2}-\frac{1}{2} \cdot \pi \cdot r_{1}^{2}}\right) \cdot 100 \%=\left(\frac{2 r_{1}^{2}-\frac{1}{2} \cdot \pi \cdot r_{1}^{2}}{2 b d-d^{2}+2 r_{1}^{2}-\frac{1}{2} \cdot \pi \cdot r_{1}^{2}}\right) \cdot 100 \%=1,09 \%
\]

c) In the formula found in b), the numerator remains independent of \(d\). Therefore, only the denominator needs to be examined. Since \(b\) is much larger than \(d\), the term \(2 b d-d^{2}\) increases with increasing \(d\). Thus, the denominator becomes larger, and with a constant numerator, the fraction becomes smaller.

The percentage error will also become smaller. For values of \(d\) in the range of 5 to \(9 \mathrm{~mm}\), the percentage error is largest at \(d=5 \mathrm{~mm}\) with a value of \(1,09 \%\).