commit ccec96d0dbfbf53391958b0cadc18d893d31920b
Author: Simon Schurti <simon.schurti@stud.fhgr.ch>
Date:   Mon Nov 25 14:39:41 2024 +0100

    Upload files to "TM1"

diff --git a/TM1/trussSolver2d.py b/TM1/trussSolver2d.py
new file mode 100644
index 0000000..56ec9b4
--- /dev/null
+++ b/TM1/trussSolver2d.py
@@ -0,0 +1,125 @@
+import numpy as np
+
+
+class truss:
+    n: int
+    s: int
+    nodes: np.ndarray
+    externalForces: list[list]
+    beams: np.ndarray
+    beamForces: np.ndarray
+
+    # n: number nodes
+    # s: number beams
+    # externalForces: list of externalForces each force is list with [Fx, Fy, NodeIndex] so a single force in y direction on node 3 would look like: [[0,F,3]] (zero indexed!!)
+    # nodeCoordinates: noarray of nodes x and y coordinates: [[x1,y1],[x2,y2]]
+    # beams: ndarray of beams every beam contains node indices it connects. e.g. beam0 connnects nodes 0 and 1 -> beams= [[0,1]]
+    def __init__(
+        self,
+        s: int,
+        n: int,
+        externalForces: list,
+        nodeCoordinates: np.ndarray,
+        beams: np.ndarray,
+    ):
+        self.n = n
+        self.s = s
+        self.externalForces = externalForces
+        self.nodes = nodeCoordinates
+        self.beams = beams
+
+    def solve(self):
+        # coeficient matrices A in x and y dir. with  #beams rows and #nodes cols
+        # Ax,Ay is padded because numpy requires square coefficient matrix
+        # Ax and Ay are concatenated efter lineaer system is constructed
+        Ax = np.zeros((self.n, 2 * self.n))  # lhsx
+        Ay = np.zeros((self.n, 2 * self.n))  # lhsy
+
+        # rhs vectors all zero or external force
+        bx = np.zeros(self.n)
+        by = np.zeros((self.n))
+        for ef in self.externalForces:
+            bx[ef[2]] = ef[0]
+            by[ef[2]] = ef[1]
+
+        # construct lgs
+        for ni in range(len(self.nodes)):  # for every node index ni
+            for b in np.where(self.beams == ni)[
+                0
+            ]:  # for every beam connected to node[ni]
+                (x1, y1) = self.nodes[self.beams[b][0]]
+                (x2, y2) = self.nodes[self.beams[b][1]]
+
+                # find deltaX and deltaY of beam to calculate force components
+                vy, vx = 0, 0
+                if (
+                    b < (2 * ni) and ni > 0
+                ):  # find force direction for beams on left side of node
+                    vy = y1 - y2
+                    vx = x1 - x2
+                else:  # different force direction for nodes on right side of node
+                    vy = y2 - y1
+                    vx = x2 - x1
+
+                lv = np.sqrt(vx**2 + vy**2)  # |v|
+
+                ux = ((1 / lv) if lv != 0 else 0) * vx  # (v^)x unit vector X
+                uy = ((1 / lv) if lv != 0 else 0) * vy  # (v^)y unit vector Y
+
+                Ax[ni, b] = ux  # koeffizient stabkraft F[b] an node[ni] in x richtung
+                Ay[ni, b] = uy  # koeffizient stabkraft F[b] an node[ni] in y richtung
+
+        A = np.concatenate((Ax, Ay), axis=0)
+        b = np.concatenate((bx, by), axis=0)
+
+        self.beamForces = np.linalg.lstsq(A, b)[0]
+
+    def print(self):
+        [
+            print(f"F{i+1}: {n:#.3g} kN")
+            for i, n in enumerate(self.beamForces)
+            if n != 0
+        ]  # ignore padding collumns
+
+
+### sample useage
+## TM1 Aufgabe 164
+if __name__ == "__main__":
+    n = 15  # nr. nodes
+    s = 2 * n - 3  # 27 beams(0-26) + F[27]+ Fa[28] + Fb[29]
+
+    F = 28
+    Fa = (((n - 3) / 2) * F) / 2
+    Fb = Fa
+
+    ## Node x and y coordinates
+    nodes = np.array([[i, 0 if i % 2 == 0 else 2] for i in range(n)])
+
+    # ExternalForces (x,y,nodeIndex)
+    externalForces = [
+        [0, Fa, 0],
+        [0, -F, 2],
+        [0, -F, 4],
+        [0, -F, 6],
+        [0, -F, 8],
+        [0, -F, 10],
+        [0, -F, 12],
+        [0, Fb, 14],
+    ]
+
+    ## beams each row represents a beam and contains the node indices it connects to
+    beams = np.array(
+        [
+            [
+                x / 2 if x % 2 == 0 else (x - 1) / 2,
+                x / 2 + 1 if x % 2 == 0 else (x - 1) / 2 + 2,
+            ]
+            for x in range(s)
+        ],
+        dtype=int,
+    )
+
+    t = truss(s, n, externalForces, nodes, beams)
+
+    t.solve()
+    t.print()