## 片持ち等分布†

### 境界値問題†

$$q=EIv''''_{(z)}$$ $$EIv''''_{(z)}=q$$ $$EIv'''_{(z)}=qz+A$$ $$EIv''_{(z)}=\frac{qz^2}{2}+Az+B$$ $$EIv'_{(z)}=\frac{qz^3}{6}+\frac{Az^2}{2}+Bz+C$$ $$EIv_{(z)}=\frac{qz^4}{24}+\frac{Az^3}{6}+\frac{Bz^2}{2}+Cz+D$$ $$M_{(z)}=-EI(v''_{(z)}+\frac{q}{kGA})$$ $$S_{(z)}=-EIv'''_{(z)}$$ $$\theta_{(z)}=\frac{M'_{(z)}}{kGA}-v'_{(z)}$$ $$v_{(0)}=0$$ $$D=0$$ $$S_{(\ell)}=0=-EIv'''_{(\ell)}$$ $$EIv'''_{(z)}=q\ell+A$$ $$-q\ell-A=0$$ $$A=-q\ell$$ $$M_{(\ell)}=0=-EI(v''_{(\ell)}-\frac{q}{kGA})$$ $$EIv''_{(\ell)}=\frac{q\ell^2}{2}-q\ell^2+B$$ $$0=-\frac{q\ell^2}{2}+q\ell^2-B+\frac{qEI}{kGA}$$ $$B=\frac{q\ell^2}{2}+\frac{qEI}{kGA}$$ $$\theta_{(0)}=\frac{M'_{(0)}}{kGA}-v'_{(0)}$$ $$M'_{(0)}=q\ell$$ $$-v'_{(0)}=-\frac{C}{EI}$$ $$\theta_{(0)}=0=\frac{q\ell}{kGA}-\frac{C}{EI}$$ $$C=\frac{q\ell EI}{kGA}$$ $$EIv_{(z)}=\frac{qz^4}{24}-\frac{q\ell z^3}{6}+\frac{q\ell^2 z^2}{4}-\frac{qEIz^2}{2kGA}+\frac{q\ell EIz}{kGA}$$ $$v_{(z)}=\frac{q}{24EI}(z^4-4\ell z^3+6\ell^2 z^2)+\frac{q}{2kGA}(2\ell z-z^2)$$ $$v_{(\ell)}=\frac{q}{24EI}(\ell^4-4\ell^4+6\ell^4)+\frac{q\ell^2}{2kGA}$$ $$v_{(\ell)}=\frac{q\ell^4}{8EI}+\frac{q\ell^2}{2kGA}$$

## 単純梁等分布†

### 境界値問題†

$$q=EIv''''_{(z)}$$ $$EIv''''_{(z)}=q$$ $$EIv'''_{(z)}=qz+A$$ $$EIv''_{(z)}=\frac{qz^2}{2}+Az+B$$ $$EIv'_{(z)}=\frac{qz^3}{6}+\frac{Az^2}{2}+Bz+C$$ $$EIv_{(z)}=\frac{qz^4}{24}+\frac{Az^3}{6}+\frac{Bz^2}{2}+Cz+D$$ $$M_{(z)}=-EI(v''_{(z)}+\frac{q}{kGA})$$ $$S_{(z)}=-EIv'''_{(z)}$$ $$\theta_{(z)}=\frac{M'_{(z)}}{kGA}-v'_{(z)}$$ $$v_{(0)}=0$$ $$D=0$$ $$M_{(0)}=0$$ $$EIv''_{(0)}=B$$ $$-EIv''_{(0)}=-B$$ $$M_{(0)}=-B-\frac{qEI}{kGA}$$ $$B=-\frac{qEI}{kGA}$$ $$S_{(\frac{\ell}{2})}=0$$ $$EIv'''_{(\frac{\ell}{2})}=\frac{q\ell}{2}+A$$ $$0=-\frac{q\ell}{2}-A$$ $$A=-\frac{q\ell}{2}$$ $$v_{(\ell)}=0$$ $$EIv_{(z)}=\frac{qz^4}{24}-\frac{q\ell z^3}{12}-\frac{qEIz^2}{2kGA}+\frac{q\ell^3 z}{24}+\frac{qEI\ell z}{2kGA}$$ $$v_{(z)}=\frac{q}{24EI}(z^4-2\ell z^3+q\ell^3 z)+\frac{q\ell^2}{2kGA}(-z^2+\ell z)$$ $$v_{(\frac{\ell}{2})}=\frac{q\ell^4}{24EI}(\frac{1}{16}-\frac{1}{4}+\frac{1}{2})+\frac{q\ell^2}{2kGA}(-\frac{1}{4}+\frac{1}{2})$$ $$v_{(\frac{\ell}{2})}=\frac{5q\ell^4}{384EI}+\frac{q\ell^2}{8kGA}$$

Last-modified: 2020-01-20 (月) 10:38:39