KIT - Klausuren - TM2

TM 2 Sammlung

TUHH-TM2

Totale Differenzierbarkeit

Mittelwertsatz

Mathe I

euklidischer Algorithmus

Brückenschaltung

Lineare Algebra 

QR-Zerlegung

Cholesky-Zerlegung

Stützmpment

Simplex Algorithmus

Optimierung in mehreren Variablen + Envelop Theorem 

Exterma Untersuchung

Basiswechselmatrix

Differenzenverstärker

übertragungsfunktion

Darstellungsmatrizen in neuer Basis darstellen

Basis wechsel, Transformationsmatrizen

Monotieverhalten

Cauchy-Produktformel

physik

elastischer stoss

lösbarkeit einer Matrix

kinematik

impuls

kugelkoordinaten une zylinderkoordinaten 

Matrix

Integralgleichung

kettenregel

Lösbarkeit einer Matrix

Glechgewicht

Schiefe Ebene

Cauchy Verdichtungssatz 

Schaltvorgänge

Singulärwertzerlegung

Spann

Lineares Gleichungssystem

spannung

Basis bestimmen

Fourier-Reihen

Unterschied In/Homogen

Gruppen

lineare abbildung 

quadratische ergänzung

basis 

Bild & Urbild

Zweigstromanalyse

Transistor als Thermometer

Investment

Teilfolgen

Epsilon Delta Kriterium

prinzip der virtuellen verrückung 

Linear Kombination

Cholesky Zerlegung

Polynomfunktionen beschreiben

Teilchenphysik und Festkörperphysik

test

Regel von de L'Hopital

KRAFT

Grenzwert Funktion 

Folgenkriterium

ϵ-δ Kriterium 

Beweis Stetigkeit 

Basiswechselsatz

Vorderrad fahradgabel lagerkräfte berechnen 

Spannungsregelung

Test

Induktion 

Jordan-Normalform

Inverse einer Matrix

Gleichmäßige Stetigkeit

addition

Stationäre Stoffübertragung

komplexe Ungleichungen 

Rekursive Folgen

Exponentialreihe

restglied

Newton Verfahren

Randwertaufgabe

Numerische Integration: Trapez- und Simpsonregel

dualräume

Dualraum

Komplexe Nullstellen

Chemie

Folgedefinition des Funktionslimes

Grenzwerte und Stetigkeit 

Komplexe Zahlen

Direkte Summe 

Exponentielle Verzinsung

Tilgungsrechnung

Abschreibungsrechnung

Zins- und Barwertrechnung

Grenzkosten

Grenzerlös, Grenzgewinn und Grenzproduktivität

Ökonomische Anwendung

\[ F_{A}=F_{A 0}+X_{1} F_{A 1}+X_{2} F_{A 2}=\frac{2}{5} q_{0} a \]
\[ F_{B}=F_{B 0}+X_{1} F_{B 1}+X_{2} F_{B 2}=\frac{11}{10} q_{0} a \]
\[F_{C}=F_{C 0}+X_{1} F_{C 1}+X_{2} F_{C 2}=\frac{11}{10} q_{0} a \]
\[ F_{D}=F_{D 0}+X_{1} F_{D 1}+X_{2} F_{D 2}=\frac{2}{5} q_{0} a \]

Der Balken ist zweifach unbestimmt gelagert. Eine Möglichkeit die Lagerkräfte zu ermitteln ist das Auslösen von zwei Bindungen. Hier ein Vorschlag zur Berechnung der Lagerreaktionen durch das einführen von Gelenken an den mittleren beiden Loslagern. Ein alternativer Lösungsweg wäre z.B. die Verwendung von Biegelinientafeln und die Superposition von den Verschiebungen

"0"-System (Statisch bestimmtes System)

"1"-System(Statisch bestimmtes System)

"2" -System (Statisch bestimmtes System)

Die Verschiebungen in den Systemen können mit dem Prinzip der virtuellen Arbeit bestimmt werden.
Verschiebungen $ \delta_{01}, \delta_{02}: $
\[ \quad \mathrm{EI} \delta_{\mathrm{i} 0}=\int \mathrm{M}_{\mathrm{i}} \mathrm{M}_{0} \mathrm{~d} \mathrm{x} \]
Verschiebungen $ \delta_{11}, \delta_{12}, \delta_{21}, \delta_{22} $ :
\[ \quad \mathrm{EI} \delta_{\mathrm{ik}}=\int \mathrm{M}_{\mathrm{i}} \mathrm{M}_{\mathrm{k}} \mathrm{dx} \]

Diese Verschiebungen werden mit Hilfe der Koppeltafel ermittelt:

\[ \mathrm{EI} \delta_{01}=\int \mathrm{M}_{0} \mathrm{M}_{1} \mathrm{~d} \mathrm{x}=\frac{1}{3} \cdot \frac{\mathrm{q}_{0} \mathrm{a}^{2}}{8} \cdot 1 \cdot \mathrm{a}+\frac{1}{3} \cdot \frac{\mathrm{q}_{0} \mathrm{a}^{2}}{8} \cdot 1 \cdot \mathrm{a}=\frac{\mathrm{q}_{0} \mathrm{a}^{3}}{12} \]
\[ \mathrm{EI} \delta_{02}=\int \mathrm{M}_{0} \mathrm{M}_{2} \  \mathrm{dx}=\frac{1}{3} \cdot \frac{\mathrm{q}_{0} \mathrm{a}^{2}}{8} \cdot 1 \cdot \mathrm{a}+\frac{1}{3} \cdot \frac{\mathrm{q}_{0} \mathrm{a}^{2}}{8} \cdot 1 \cdot \mathrm{a}=\frac{\mathrm{q}_{0} \mathrm{a}^{3}}{12} \]

\[ \mathrm{EI} \delta_{11}=\int \mathrm{M}_{1} \mathrm{M}_{1} \mathrm{~d} \mathrm{x}=\frac{1}{3} \cdot 1.1 \cdot \mathrm{a}+\frac{1}{3} \cdot 1.1 \cdot \mathrm{a}=\frac{2}{3} \mathrm{a} \]
\[ \mathrm{EI} \delta_{12}=\int \mathrm{M}_{1} \mathrm{M}_{2} \mathrm{~d} \mathrm{x}=\frac{1}{6} \cdot 1.1 \cdot \mathrm{a}=\frac{1}{6} \mathrm{a} \]

\[ \mathrm{EI} \delta_{22}=\int \mathrm{M}_{2} \mathrm{M}_{2} \mathrm{dx}=\frac{1}{3} \cdot 1.1 . \mathrm{a}+\frac{1}{3} \cdot 1.1 . \mathrm{a}=\frac{2}{3} \mathrm{a} \]

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\mathrm{EI} \delta_{22}=\int \mathrm{M}_{2} \mathrm{M}_{2} \mathrm{dx}=\frac{1}{3} \cdot 1.1 . \mathrm{a}+\frac{1}{3} \cdot 1.1 . \mathrm{a}=\frac{2}{3} \mathrm{a}
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Aufgrund der Symmetrie gilt zudem:
\[ \quad \delta_{12}=\delta_{21}=\frac{2}{3} \mathrm{a} \]

Aufgrund der Symmetrie gilt zudem:
<mjx-container class="MathJax" jax="SVG" display="true" justify="left"><svg alt="
\quad \delta_{12}=\delta_{21}=\frac{2}{3} \mathrm{a}
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Nutze den Zusammenhang\[ \ {\left[\begin{array}{c} \text { Gesuchte Grösse im } \\ \text { endgültigen Zustand } \end{array}\right]=\left[\begin{array}{l} \text { Grösse im } \\ \text { Nullzustand } \end{array}\right]+\mathrm{X}_{1}\left[\begin{array}{c} \text { Grösse im } \\ \text { Einheitszustand } \\ \mathrm{X}_{1}=1 \end{array}\right]+\mathrm{X}_{2}\left[\begin{array}{c} \text { Grösse im } \\ \text { Einheitszustand } \\ \mathrm{X}_{2}=1 \end{array}\right]} \]

\[ \Rightarrow\left[\begin{array}{l} 0 \\ 0 \end{array}\right] =\left[\begin{array}{l} \delta_{01} \\ \delta_{02} \end{array}\right]+\mathrm{X}_{1} \left[\begin{array}{l} \delta_{11} \\ \delta_{21} \end{array}\right] +\mathrm{X}_{2} \left[\begin{array}{l} \delta_{12} \\ \delta_{22} \end{array}\right]\]

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\Rightarrow\left[\begin{array}{l}
0 \\
0
\end{array}\right] =\left[\begin{array}{l}
\delta_{01} \\
\delta_{02}
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\delta_{11} \\
\delta_{21}
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\delta_{12} \\
\delta_{22}
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\[ \Rightarrow\left[\begin{array}{l} \mathrm{X}_{1} \\ \mathrm{X}_{2} \end{array}\right]=-\left[\begin{array}{ll} \delta_{11} &amp; \delta_{12} \\ \delta_{21} &amp; \delta_{22} \end{array}\right]^{-1}\left[\begin{array}{l} \delta_{01} \\ \delta_{02} \end{array}\right] \]

<mjx-container class="MathJax" jax="SVG" display="true" justify="left"><svg alt=" \Rightarrow\left[\begin{array}{l}
\mathrm{X}_{1} \\
\mathrm{X}_{2}
\end{array}\right]=-\left[\begin{array}{ll}
\delta_{11} & \delta_{12} \\
\delta_{21} & \delta_{22}
\end{array}\right]^{-1}\left[\begin{array}{l}
\delta_{01} \\
\delta_{02}
\end{array}\right]
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Das Einsetzen der Werte führt zu:

\[ {\left[\begin{array}{l} \mathrm{X}_{1} \\ \mathrm{X}_{2} \end{array}\right]=-\left[\begin{array}{ll} 2 &amp; \frac{1}{2} \\ \frac{1}{2} &amp; 2 \end{array}\right]^{-1}\left[\begin{array}{l} 1 \\ 1 \end{array}\right] \frac{\mathrm{q}_{0} \mathrm{a}^{2}}{4}} \]

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{\left[\begin{array}{l}
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\[ \Rightarrow \mathrm{X}_{1}=\mathrm{X}_{2}=-\frac{\mathrm{q}_{0} \mathrm{a}^{2}}{10} \]

<mjx-container class="MathJax" jax="SVG" display="true" justify="left"><svg alt="
\Rightarrow \mathrm{X}_{1}=\mathrm{X}_{2}=-\frac{\mathrm{q}_{0} \mathrm{a}^{2}}{10}
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Mit den bekannten Reaktionskräften aus den statisch bestimmten Systemen "0”, "1", "2" ergeben sich die gesuchten Lagerreaktionen zu

\[ F_{A}=F_{A 0}+X_{1} F_{A 1}+X_{2} F_{A 2}=\frac{2}{5} q_{0} a \]
\[ F_{B}=F_{B 0}+X_{1} F_{B 1}+X_{2} F_{B 2}=\frac{11}{10} q_{0} a \]
\[F_{C}=F_{C 0}+X_{1} F_{C 1}+X_{2} F_{C 2}=\frac{11}{10} q_{0} a \]
\[ F_{D}=F_{D 0}+X_{1} F_{D 1}+X_{2} F_{D 2}=\frac{2}{5} q_{0} a \]

Da keine Belastung in Horizontaler Richtung auftritt, ist die horizontale Lagerkraft im Festlager A gleich null:
\[ \quad \mathrm{F}_{\mathrm{AH}}=0 \]

Da keine Belastung in Horizontaler Richtung auftritt, ist die horizontale Lagerkraft im Festlager A gleich null:
<mjx-container class="MathJax" jax="SVG" display="true" justify="left"><svg alt="
\quad \mathrm{F}_{\mathrm{AH}}=0
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Ein Balken ($\mathrm{EI}$, Länge $3a$) ist mit einem Festlager und drei Loslagern gelagert und mit einer Streckenlast $ q_{0} $ belastet (siehe Skizze) Bestimmen Sie für alle Lager $ (\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}) $ die Lagerreaktionen.
<img src="https://storage.googleapis.com/max-academy-image-bucket/607d0d88b29296cf2249c4db.png" alt="Abbildung" />
Gegeben: $ a,\ \mathrm{E I},\ \mathrm{q_{0}} $

Ein Balken (<mjx-container class="MathJax" jax="SVG"><svg alt="\mathrm{EI}" style="vertical-align: 0" xmlns="http://www.w3.org/2000/svg" width="2.357ex" height="1.545ex" role="img" focusable="false" viewBox="0 -683 1042 683" xmlns:xlink="http://www.w3.org/1999/xlink"><defs><path id="MJX-1-TEX-N-45" d="M128 619Q121 626 117 628T101 631T58 634H25V680H597V676Q599 670 611 560T625 444V440H585V444Q584 447 582 465Q578 500 570 526T553 571T528 601T498 619T457 629T411 633T353 634Q266 634 251 633T233 622Q233 622 233 621Q232 619 232 497V376H286Q359 378 377 385Q413 401 416 469Q416 471 416 473V493H456V213H416V233Q415 268 408 288T383 317T349 328T297 330Q290 330 286 330H232V196V114Q232 57 237 52Q243 47 289 47H340H391Q428 47 452 50T505 62T552 92T584 146Q594 172 599 200T607 247T612 270V273H652V270Q651 267 632 137T610 3V0H25V46H58Q100 47 109 49T128 61V619Z"></path><path id="MJX-1-TEX-N-49" d="M328 0Q307 3 180 3T32 0H21V46H43Q92 46 106 49T126 60Q128 63 128 342Q128 620 126 623Q122 628 118 630T96 635T43 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-22Q158 -22 100 21T42 130Q42 158 60 175T105 193Q133 193 151 175T169 130Q169 119 166 110T159 94T148 82T136 74T126 70T118 67L114 66Q165 21 238 21Q293 21 321 74Q338 107 338 175V195Q338 290 274 322Q259 328 213 329L171 330L168 332Q166 335 166 348Q166 366 174 366Q202 366 232 371Q266 376 294 413T322 525V533Q322 590 287 612Q265 626 240 626Q208 626 181 615T143 592T132 580H135Q138 579 143 578T153 573T165 566T175 555T183 540T186 520Q186 498 172 481T127 463Z"></path><path id="MJX-2-TEX-I-1D44E" d="M33 157Q33 258 109 349T280 441Q331 441 370 392Q386 422 416 422Q429 422 439 414T449 394Q449 381 412 234T374 68Q374 43 381 35T402 26Q411 27 422 35Q443 55 463 131Q469 151 473 152Q475 153 483 153H487Q506 153 506 144Q506 138 501 117T481 63T449 13Q436 0 417 -8Q409 -10 393 -10Q359 -10 336 5T306 36L300 51Q299 52 296 50Q294 48 292 46Q233 -10 172 -10Q117 -10 75 30T33 157ZM351 328Q351 334 346 350T323 385T277 405Q242 405 210 374T160 293Q131 214 119 129Q119 126 119 118T118 106Q118 61 136 44T179 26Q217 26 254 59T298 110Q300 114 325 217T351 328Z"></path></defs><g stroke="currentColor" fill="currentColor" stroke-width="0" transform="scale(1,-1)"><g data-mml-node="math"><g data-mml-node="mn"><use data-c="33" xlink:href="#MJX-2-TEX-N-33"></use></g><g data-mml-node="mi" transform="translate(500,0)"><use data-c="1D44E" xlink:href="#MJX-2-TEX-I-1D44E"></use></g></g></g></svg></mjx-container>) ist mit einem Festlager und drei Loslagern gelagert und mit einer Streckenlast <mjx-container class="MathJax" jax="SVG"><svg alt=" q_{0} " style="vertical-align: -0.439ex" xmlns="http://www.w3.org/2000/svg" width="1.997ex" height="1.439ex" role="img" focusable="false" viewBox="0 -442 882.6 636" xmlns:xlink="http://www.w3.org/1999/xlink"><defs><path id="MJX-3-TEX-I-1D45E" d="M33 157Q33 258 109 349T280 441Q340 441 372 389Q373 390 377 395T388 406T404 418Q438 442 450 442Q454 442 457 439T460 434Q460 425 391 149Q320 -135 320 -139Q320 -147 365 -148H390Q396 -156 396 -157T393 -175Q389 -188 383 -194H370Q339 -192 262 -192Q234 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data-mjx-texclass="ORD"><g data-mml-node="mn"><use data-c="30" xlink:href="#MJX-3-TEX-N-30"></use></g></g></g></g></g></svg></mjx-container> belastet (siehe Skizze) Bestimmen Sie für alle Lager <mjx-container class="MathJax" jax="SVG"><svg alt=" (\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}) " style="vertical-align: -0.566ex" xmlns="http://www.w3.org/2000/svg" width="11.439ex" height="2.262ex" role="img" focusable="false" viewBox="0 -750 5056 1000" xmlns:xlink="http://www.w3.org/1999/xlink"><defs><path id="MJX-4-TEX-N-28" d="M94 250Q94 319 104 381T127 488T164 576T202 643T244 695T277 729T302 750H315H319Q333 750 333 741Q333 738 316 720T275 667T226 581T184 443T167 250T184 58T225 -81T274 -167T316 -220T333 -241Q333 -250 318 -250H315H302L274 -226Q180 -141 137 -14T94 250Z"></path><path id="MJX-4-TEX-N-41" d="M255 0Q240 3 140 3Q48 3 39 0H32V46H47Q119 49 139 88Q140 91 192 245T295 553T348 708Q351 716 366 716H376Q396 715 400 709Q402 707 508 390L617 67Q624 54 636 51T687 46H717V0H708Q699 3 581 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Dies ist eine interaktive Aufgabe zu: TUHH-TM2 mit praktischen Tipps zum Lösen und einer Zusammenfassung der nötigen Theorie

Aufgabenstellung:

Lösungsweg:

Lösung: