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diff --git a/tutorials/module_1/array.md b/tutorials/module_1/array.md index 449135a..3c0290e 100644 --- a/tutorials/module_1/array.md +++ b/tutorials/module_1/array.md @@ -118,27 +118,27 @@ Due to the nature of zero-based indexing. If we want to call the value `4.054973 ## Exercise Let's solve a statics problem given the following problem -A simply supported bridge of length L=20L = 20L=20 m is subjected to three point loads: +A simply supported bridge of length L = 20 m is subjected to three point loads: -- $P1=10P_1 = 10P1=10 kN$ at $x=5x = 5x=5 m$ -- $P2=15P_2 = 15P2=15 kN$ at $x=10x = 10x=10 m$ -- $P3=20P_3 = 20P3=20 kN$ at $x=15x = 15x=15 m$ +- $P_1 = 10 kN$ at $x = 5 m$ +- $P_2 = 15 kN$ at $x = 10 m$ +- $P_3 = 20 kN$ at $x = 15 m$ The bridge is supported by two reaction forces at points AAA (left support) and BBB (right support). We assume the bridge is in static equilibrium, meaning the sum of forces and sum of moments about any point must be zero. ##### Equilibrium Equations: 1. **Sum of Forces in the Vertical Direction**: - $RA+RB−P1−P2−P3=0R_A + R_B - P_1 - P_2 - P_3 = 0RA+RB−P1−P2−P3=0$ + $R_A + R_B - P_1 - P_2 - P_3 = 0$ 2. **Sum of Moments About Point A**: - $5P1+10P2+15P3−20RB=05 P_1 + 10 P_2 + 15 P_3 - 20 R_B = 05P1+10P2+15P3−20RB=0$ + $5 P_1 + 10 P_2 + 15 P_3 - 20 R_B = 0$ 3. **Sum of Moments About Point B**: - $20RA−15P3−10P2−5P1=020 R_A - 15 P_3 - 10 P_2 - 5 P_1 = 020RA−15P3−10P2−5P1=0$ + $20 R_A - 15 P_3 - 10 P_2 - 5 P_1 = 0$ ##### System of Equations: - -{RA+RB−10−15−20=05(10)+10(15)+15(20)−20RB=020RA−5(10)−10(15)−15(20)=0\begin{cases} R_A + R_B - 10 - 15 - 20 = 0 \\ 5(10) + 10(15) + 15(20) - 20 R_B = 0 \\ 20 R_A - 5(10) - 10(15) - 15(20) = 0 \end{cases}⎩ - +$$ +\begin{cases} R_A + R_B - 10 - 15 - 20 = 0 \\ 5(10) + 10(15) + 15(20) - 20 R_B = 0 \\ 20 R_A - 5(10) - 10(15) - 15(20) = 0 \end{cases} +$$ ### Solution ```python |