集中荷重をかけた支承付き片持ち梁

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途中Qの符号を間違えて未知数が消えてしまい、悩んでいた

$ M=\min\{f(x),(P-V)(l-x)\}

$ f=ax+b

$ f(0)=\frac12(P-2V)l=f(\frac12l)-\frac12Vl<0

$ f(\frac12l)=\frac12(P-V)l>0

$ \begin{pmatrix}0&1\\\frac12l&1\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix}=\frac12l\begin{pmatrix}P-2V\\P-V\end{pmatrix}

$ \iff \begin{pmatrix}a\\b\end{pmatrix}=\frac1{-\frac12l}\begin{pmatrix}1&-1\\-\frac12l&0\end{pmatrix}\cdot\frac12l\begin{pmatrix}P-2V\\P-V\end{pmatrix}

$ = \begin{pmatrix}-1&1\\\frac12l&0\end{pmatrix}\begin{pmatrix}P-2V\\P-V\end{pmatrix}

$ = \begin{pmatrix}V\\f(0)\end{pmatrix}

$ \therefore f=Vx+\frac12(P-2V)l

$ = \min\{Vx+\frac12(P-2V)l,(P-V)(l-x)\}

$ = \min\{Vx+\frac12(P-2V)l,Vx-Px+(P-V)l\}

$ =Vx+\min\{\frac12(P-2V)l,-Px+(P-V)l\}

$ =Vx+\min\{\frac12Pl-Vl,-Px+Pl-Vl\}

$ =Vx+\frac12Pl-Vl+\min\{0,-Px+\frac12Pl\}

$ =Vx+\frac12(P-2V)l+P(\frac12l-x)\llbracket x\ge\frac12l\rrbracket

check

$ v=\frac16\frac{V}{EI}x^3+\frac14(\frac{P}{EI}-2\frac{V}{EI})lx^2-\frac16\frac{P}{EI}(\frac12l-x)^3\llbracket x\ge\frac12l\rrbracket+Cx

2023-04-17 21:09:15 左端はたわみ角が0になることを忘れてた

$ v'(0)=0+C=0\therefore C=0

$ v(l)=\frac16\frac{V}{EI}l^3+\frac14(\frac{P}{EI}-2\frac{V}{EI})l^3+\frac16\cdot\frac18\frac{P}{EI}l^3=0

$ \implies 8V+12(P-2V)+P=0

$ \iff (8-24)V+13P=0

$ \iff -16V+13P=0

$ \iff V=\frac{13}{16}P

check:

$ M=\frac{13}{16}Px-\frac12\frac58Pl+P(\frac12l-x)\llbracket x\ge\frac12l\rrbracket

$ =\frac{13}{16}P(x-\frac{5}{13}l)+P(\frac12l-x)\llbracket x\ge\frac12l\rrbracket

$ v=\frac16\frac{13}{16}\frac{P}{EI}x^3-\frac14\frac58\frac{P}{EI}lx^2-\frac16\frac{P}{EI}(\frac12l-x)^3\llbracket x\ge\frac12l\rrbracket

$ v'=\frac{13}{32}\frac{P}{EI}x^2-\frac{10}{32}\frac{P}{EI}lx+\frac{16}{32}\frac{P}{EI}(\frac12l-x)^2\llbracket x\ge\frac12l\rrbracket

$ = \frac{13}{32}\frac{P}{EI}(x-\frac{10}{13}l)x+\frac{16}{32}\frac{P}{EI}(\frac12l-x)^2\llbracket x\ge\frac12l\rrbracket

$ v'(\frac{5}{13}l)=-\frac{25}{32\cdot13}\frac{P}{EI}l^2

$ v'(l)=\frac{3}{32}\frac{P}{EI}l^2+\frac{4}{32}\frac{P}{EI}l^2=\frac{7}{32}\frac{P}{EI}l^2

$ v'=0\impliedby (13x-10l)x+16(\frac12l-x)^2=0

$ \iff(13x-10l)x+4(l-2x)^2=0

$ \iff 13x^2-10lx+16x^2-16lx+4l^2=0

$ \iff 27x^2-26lx+4l^2=0

↑計算間違えている

正しくは

$ v=-\frac{11}{96}\frac{P}{EI}x^3+\frac{3}{32}\frac{P}{EI}lx^2+\frac16\frac{P}{EI}(x-\frac12l)^3\llbracket x\ge\frac12l\rrbracket

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