# Rám¶

Vypracoval: Filip Horák, 2017/2018

U uzavřeného prutu dle obrázku vypočtěte posunutí bodu B.

## Řešení¶

Ze začátku je nutné naimportovat potřebné knihovny.

In [1]:

    import sympy as sp
    import numpy as np
    from IPython.core.display import Image
    sp.init_printing()

Vyvoláme obrázek úlohy.

In [2]:

    Image(filename='06-o-ram-1.png', width=250)

Out[2]:

Využitím symetrie provedeme částečné uvolnění a sestavíme deformační podmínky: $$\begin{align} u_{Bx} & = 0 \\ \varphi_B & = 0 \end{align}$$

In [3]:

    Image(filename='06-o-ram-2.png', width=200)

Out[3]:

Zavedeme potřebné proměnné jako symboly.

In [4]:

    F,a,R=sp.symbols('F a R')
    fi1,x2,x3=sp.symbols('varphi_1 x2 x3')
    FBx,FBz,MB=sp.symbols('F_Bx F_Bz M_B')
    E,J=sp.symbols('E J')
    wBz=sp.symbols('w_Bz')

Vyjádříme momentové vnitřní účinky jednotlivých úseků.

In [5]:

    Moy1=-MB+FBz*R*sp.sin(fi1)-FBx*R*(1-sp.cos(fi1))
    Moy2=-MB+FBz*R-FBx*(R+x2)
    Moy3=-MB+FBz*(R-x3)-FBx*(R+a)
    Moy1,Moy2,Moy3

Out[5]:
$$\left ( - F_{Bx} R \left(- \cos{\left (\varphi_{1} \right )} + 1\right) + F_{Bz} R \sin{\left (\varphi_{1} \right )} - M_{B}, \quad - F_{Bx} \left(R + x_{2}\right) + F_{Bz} R - M_{B}, \quad - F_{Bx} \left(R + a\right) + F_{Bz} \left(R - x_{3}\right) - M_{B}\right )$$

Derivujeme jednotlivé úseky podle síly $F_{Bx}$ a momentu $M_B$.

In [6]:

    dMoy1dM=Moy1.diff(MB)
    dMoy1dFBx=Moy1.diff(FBx)
    dMoy2dM=Moy2.diff(MB)
    dMoy2dFBx=Moy2.diff(FBx)
    dMoy3dM=Moy3.diff(MB)
    dMoy3dFBx=Moy3.diff(FBx)
    dMoy1dM,dMoy2dM,dMoy3dM,dMoy1dFBx,dMoy2dFBx,dMoy3dFBx

Out[6]:
$$\left ( -1, \quad -1, \quad -1, \quad - R \left(- \cos{\left (\varphi_{1} \right )} + 1\right), \quad - R - x_{2}, \quad - R - a\right )$$

Sestavíme rovnice dle deformačních podmínek a vyjádříme sílu $F_{Bz}$.

In [7]:

    eqn1=sp.Integral(Moy1*dMoy1dFBx,[fi1,0,sp.pi/2])+ \
         sp.Integral(Moy2*dMoy2dFBx,[x2,0,a])+ \
         sp.Integral(Moy3*dMoy3dFBx,[x3,0,R])
    eqn2=sp.Integral(Moy1*dMoy1dM,[fi1,0,sp.pi/2])+ \
         sp.Integral(Moy2*dMoy2dM,[x2,0,a])+ \
         sp.Integral(Moy3*dMoy3dM,[x3,0,R])
    eqn3=FBz-F/2
    eqn1,eqn2,eqn3

Out[7]:
$$\left ( \int_{0}^{R} \left(- R - a\right) \left(- F_{Bx} \left(R + a\right) + F_{Bz} \left(R - x_{3}\right) - M_{B}\right)\, dx_{3} + \int_{0}^{a} \left(- R - x_{2}\right) \left(- F_{Bx} \left(R + x_{2}\right) + F_{Bz} R - M_{B}\right)\, dx_{2} + \int_{0}^{\frac{\pi}{2}} \left(- R \left(- \cos{\left (\varphi_{1} \right )} + 1\right) \left(- F_{Bx} R \left(- \cos{\left (\varphi_{1} \right )} + 1\right) + F_{Bz} R \sin{\left (\varphi_{1} \right )} - M_{B}\right)\right)\, d\varphi_{1}, \quad \int_{0}^{R} \left(F_{Bx} \left(R + a\right) - F_{Bz} \left(R - x_{3}\right) + M_{B}\right)\, dx_{3} + \int_{0}^{a} \left(F_{Bx} \left(R + x_{2}\right) - F_{Bz} R + M_{B}\right)\, dx_{2} + \int_{0}^{\frac{\pi}{2}} \left(F_{Bx} R \left(- \cos{\left (\varphi_{1} \right )} + 1\right) - F_{Bz} R \sin{\left (\varphi_{1} \right )} + M_{B}\right)\, d\varphi_{1}, \quad - \frac{F}{2} + F_{Bz}\right )$$

Spočteme soustavu rovnic, čímž získáme $F_{Bx}$, $F_{Bz}$ a $M_{B}$.

In [8]:

    eqn1_,eqn2_=eqn1.doit(),eqn2.doit()
    eqn1_=sp.expand(eqn1_)
    eqn2_=sp.expand(eqn2_)
    sol=sp.linsolve([eqn1_,eqn2_,eqn3],[MB,FBx,FBz])
    MB_solved,FBx_solved,FBz_solved=next(iter(sol))
    MB_solved,FBx_solved,FBz_solved

Out[8]:
$$\left ( \frac{F R \left(3 \pi R^{3} + 6 R^{2} a^{2} + 36 R^{2} a - 36 R^{2} + 12 \pi R^{2} + 10 R a^{3} - 6 \pi R a^{2} + 54 R a^{2} + 2 a^{4} + 8 a^{3}\right)}{2 \left(6 \pi R^{3} + 6 \pi R^{2} a + 48 R^{2} a - 24 R^{2} + 3 \pi^{2} R^{2} + 8 R a^{3} + 24 R a^{2} + 12 \pi R a^{2} + 2 a^{4} + 4 \pi a^{3}\right)}, \quad - \frac{3 F R \left(R a^{2} - \pi R a + 2 R a - 4 R + \pi R - \pi a^{2} + 2 a^{2}\right)}{6 \pi R^{3} + 6 \pi R^{2} a + 48 R^{2} a - 24 R^{2} + 3 \pi^{2} R^{2} + 8 R a^{3} + 24 R a^{2} + 12 \pi R a^{2} + 2 a^{4} + 4 \pi a^{3}}, \quad \frac{F}{2}\right )$$

Zavedeme číselné hodnoty veličin. $$\begin{align} a & = 3000\, \mathrm{mm} \\ R & = 1500\, \mathrm{mm} \\ D & = 50\, \mathrm{mm} \\ F & = 1000\, \mathrm{N} \\ E & = 2.1\cdot 10^5\, \mathrm{MPa} \\ J & = \frac{\pi d^4}{64} \approx 306796\, \mathrm{mm^4}\\ \end{align}$$

In [9]:

    a_,R_=3000.,1500.
    F_=1000.
    D_=50.
    E_=2.1e5
    J_=sp.pi*D_**4/64.

Čísla dosadíme do rovnic $F_{Bx}$, $F_{Bz}$ a $M_{B}$.

In [10]:

    FBx_=FBx_solved.subs({F:F_,a:a_,R:R_})
    FBz_=FBz_solved.subs({F:F_,a:a_,R:R_})
    MB_=MB_solved.subs({F:F_,a:a_,R:R_})
    sp.N(FBx_,6),sp.N(FBz_,6),sp.N(MB_,6)

Out[10]:
$$\left ( -124.425, \quad 500.0, \quad 1.06031 \cdot 10^{6}\right )$$

Jednitlivé úseky zderivujeme podle $F_{Bz}$.

In [11]:

    dMoy1dFBz=Moy1.diff(FBz)
    dMoy2dFBz=Moy2.diff(FBz)
    dMoy3dFBz=Moy3.diff(FBz)
    dMoy1dFBz,dMoy2dFBz,dMoy3dFBz

Out[11]:
$$\left ( R \sin{\left (\varphi_{1} \right )}, \quad R, \quad R - x_{3}\right )$$

Sestrojíme rovnici pro posuv ve směru $y$ a vyjádříme z rovnice $w_{Bz}$.

In [12]:

    eqn=1/(E*J)*(sp.Integral(Moy1*dMoy1dFBz,[fi1,0,sp.pi/2])+sp.Integral(Moy2*dMoy2dFBz,[x2,0,a])+sp.Integral(Moy3*dMoy3dFBz,[x3,0,R]))-wBz
    eqn_=eqn.doit()
    sol_wBz=sp.solve(eqn_,wBz)
    sol_wBz[0]

Out[12]:
$$\frac{R}{12 E J} \left(- 6 F_{Bx} R^{2} - 18 F_{Bx} R a - 6 F_{Bx} R - 6 F_{Bx} a^{2} + 4 F_{Bz} R^{2} + 12 F_{Bz} R a + 3 \pi F_{Bz} R - 6 M_{B} R - 12 M_{B} a - 12 M_{B}\right)$$

Dosadíme číselné hodnoty do rovnice a získáme $w_{Bz}$ v [mm].

In [13]:

    wBz_=sol_wBz[0].subs({F:F_,a:a_,R:R_,FBx:FBx_,FBz:FBz_,MB:MB_,E:E_,J:J_})
    sp.N(wBz_,4)

Out[13]:
$$4.382$$