Semi-Monocoque Theory



In [1]:

    
from pint import UnitRegistry
import sympy
import networkx as nx
import numpy as np
import matplotlib.pyplot as plt
import sys
%matplotlib inline
from IPython.display import display

Import Section class, which contains all calculations



In [2]:

    
from Section import Section

Initialization of sympy symbolic tool and pint for dimension analysis (not really implemented rn as not directly compatible with sympy)



In [3]:

    
ureg = UnitRegistry()
sympy.init_printing()

Define sympy parameters used for geometric description of sections



In [4]:

    
A, A0, t, t0, a, b, h, L = sympy.symbols('A A_0 t t_0 a b h L', positive=True)

We also define numerical values for each symbol in order to plot scaled section and perform calculations



In [5]:

    
values = [(A, 450 * ureg.millimeter**2),(A0, 250  * ureg.millimeter**2),(a, 130 * ureg.millimeter), \
          (b, 300 * ureg.millimeter),(h, 150 * ureg.millimeter),(L, 650 * ureg.millimeter)]
datav = [(v[0],v[1].magnitude) for v in values]

Multiconnected Section

Define graph describing the section:

1) stringers are nodes with parameters:

x coordinate
y coordinate
Area

2) panels are oriented edges with parameters:

thickness
lenght which is automatically calculated



In [6]:

    
stringers = {1:[(sympy.Integer(0),h),A],
             2:[(sympy.Integer(0),sympy.Integer(0)),A],
             3:[(a,sympy.Integer(0)),A],
             4:[(a+b,sympy.Integer(0)),A],
             5:[(a+b,h),A],
             6:[(a,h),A]}

panels = {(1,2):t,
          (2,3):t,
          (3,4):t,
          (4,5):t,
          (5,6):t,
          (6,1):t,
          (3,6):t}

Define section and perform first calculations



In [7]:

    
S3 = Section(stringers, panels)

As we need to compute $x_{sc}$, we have to perform

$$A \cdot q_{ext} = T$$

where:

A is a matrix with number of nodes + number of loops rows and number of edges +1 columns (it is square)
q is a column vector of unknowns: #edges fluxes and shear center coordinate
T is the vector of known terms: $-\frac{T_y}{J_x} \cdot S_{x_i}$ or $-\frac{T_x}{J_y} \cdot S_{y_i}$ for n-1 nodes and the rest are 0

Expression of A



In [8]:

    
sympy.simplify(S3.A)









    Out[8]:





$$\left[\begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & -1 & 0\\-1 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\0 & -1 & 1 & 1 & 0 & 0 & 0 & 0\\0 & 0 & -1 & 0 & 1 & 0 & 0 & 0\\0 & 0 & 0 & 0 & -1 & 1 & 0 & 0\\\frac{h}{3} \left(2 a + b\right) & \frac{a h}{2} & \frac{b h}{2} & \frac{h}{3} \left(a - b\right) & \frac{h}{3} \left(a + 2 b\right) & \frac{b h}{2} & \frac{a h}{2} & -1\\\frac{h}{t} & \frac{a}{t} & 0 & \frac{h}{t} & 0 & 0 & \frac{a}{t} & 0\\0 & 0 & \frac{b}{t} & - \frac{h}{t} & \frac{h}{t} & \frac{b}{t} & 0 & 0\end{matrix}\right]$$

Expression of T



In [9]:

    
sympy.simplify(S3.T)









    Out[9]:





$$\left[\begin{matrix}- \frac{1}{3 h}\\\frac{1}{3 h}\\\frac{1}{3 h}\\\frac{1}{3 h}\\- \frac{1}{3 h}\\0\\0\\0\end{matrix}\right]$$

Resulting fluxes and coordinate



In [10]:

    
sympy.simplify(S3.tempq)









    Out[10]:





$$\left[\begin{matrix}- \frac{1}{3 h}\\0\\0\\\frac{1}{3 h}\\\frac{1}{3 h}\\0\\0\\0\end{matrix}\right]$$

Plot of S3 section in original reference frame



In [11]:

    
start_pos={ii: [float(S3.g.node[ii]['ip'][i].subs(datav)) for i in range(2)] for ii in S3.g.nodes() }



In [12]:

    
plt.figure(figsize=(12,8),dpi=300)
nx.draw(S3.g,with_labels=True, arrows= True, pos=start_pos)
plt.arrow(0,0,20,0)
plt.arrow(0,0,0,20)
#plt.text(0,0, 'CG', fontsize=24)
plt.axis('equal')
plt.title("Section in starting reference Frame",fontsize=16);

Expression of Inertial properties wrt Center of Gravity in with original rotation



In [13]:

    
S3.Ixx0, S3.Iyy0, S3.Ixy0, S3.α0









    Out[13]:





$$\left ( \frac{3 A}{2} h^{2}, \quad \frac{4 A}{3} \left(a^{2} + a b + b^{2}\right), \quad 0, \quad 0\right )$$

Plot of S3 section in inertial reference Frame

Section is plotted wrt center of gravity and rotated (if necessary) so that x and y are principal axes. Center of Gravity and Shear Center are drawn



In [14]:

    
positions={ii: [float(S3.g.node[ii]['pos'][i].subs(datav)) for i in range(2)] for ii in S3.g.nodes() }



In [15]:

    
x_ct, y_ct = S3.ct.subs(datav)

plt.figure(figsize=(12,8),dpi=300)
nx.draw(S3.g,with_labels=True, pos=positions)
plt.plot([0],[0],'o',ms=12,label='CG')
plt.plot([x_ct],[y_ct],'^',ms=12, label='SC')
#plt.text(0,0, 'CG', fontsize=24)
#plt.text(x_ct,y_ct, 'SC', fontsize=24)
plt.legend(loc='lower right', shadow=True)
plt.axis('equal')
plt.title("Section in pricipal reference Frame",fontsize=16);

Expression of inertial properties in principal reference frame



In [16]:

    
S3.Ixx, S3.Iyy, S3.Ixy, S3.θ









    Out[16]:





$$\left ( \frac{3 A}{2} h^{2}, \quad \frac{4 A}{3} \left(a^{2} + a b + b^{2}\right), \quad 0, \quad 0\right )$$

Shear center expression



In [17]:

    
S3.ct









    Out[17]:





$$\left[\begin{matrix}0\\0\end{matrix}\right]$$

Loops detection



In [18]:

    
S3.cycles









    Out[18]:





$$\left [ \left [ 3, \quad 4, \quad 5, \quad 6, \quad 3\right ], \quad \left [ 2, \quad 3, \quad 6, \quad 1, \quad 2\right ]\right ]$$

Compute axial loads in Stringers in S3

Set loads on the section:

Example 1: shear in y direction and bending moment in x direction



In [19]:

    
Tx, Ty, Nz, Mx, My, Mz, F, ry, ry, mz = sympy.symbols('T_x T_y N_z M_x M_y M_z F r_y r_x m_z')
S3.set_loads(_Tx=0, _Ty=Ty, _Nz=0, _Mx=Mx, _My=0, _Mz=0)

Compute axial loads in stringers and shear flows in panels



In [20]:

    
S3.compute_stringer_actions()
S3.compute_panel_fluxes();

Expression of matrix A:

second to last row: Mz equilibrium
last row: equivalence of rotations in the form: $\dot{\theta_1} - \dot{\theta_2} = 0$



In [21]:

    
sympy.simplify(S3.A)









    Out[21]:





$$\left[\begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & -1\\-1 & 1 & 0 & 0 & 0 & 0 & 0\\0 & -1 & 1 & 1 & 0 & 0 & 0\\0 & 0 & -1 & 0 & 1 & 0 & 0\\0 & 0 & 0 & 0 & -1 & 1 & 0\\\frac{h}{3} \left(2 a + b\right) & \frac{a h}{2} & \frac{b h}{2} & \frac{h}{3} \left(a - b\right) & \frac{h}{3} \left(a + 2 b\right) & \frac{b h}{2} & \frac{a h}{2}\\- \frac{1}{2 a t} & - \frac{1}{2 h t} & \frac{1}{2 h t} & - \frac{a + b}{2 a b t} & \frac{1}{2 b t} & \frac{1}{2 h t} & - \frac{1}{2 h t}\end{matrix}\right]$$

Expression of T



In [22]:

    
sympy.simplify(S3.T)









    Out[22]:





$$\left[\begin{matrix}- \frac{T_{y}}{3 h}\\\frac{T_{y}}{3 h}\\\frac{T_{y}}{3 h}\\\frac{T_{y}}{3 h}\\- \frac{T_{y}}{3 h}\\0\\0\end{matrix}\right]$$

Axial loads



In [23]:

    
S3.N









    Out[23]:





$$\left \{ 1 : \frac{M_{x}}{3 h}, \quad 2 : - \frac{M_{x}}{3 h}, \quad 3 : - \frac{M_{x}}{3 h}, \quad 4 : - \frac{M_{x}}{3 h}, \quad 5 : \frac{M_{x}}{3 h}, \quad 6 : \frac{M_{x}}{3 h}\right \}$$

Shear flows



In [24]:

    
sympy.simplify(S3.q)









    Out[24]:





$$\left \{ \left ( 1, \quad 2\right ) : - \frac{T_{y}}{3 h}, \quad \left ( 2, \quad 3\right ) : 0, \quad \left ( 3, \quad 4\right ) : 0, \quad \left ( 3, \quad 6\right ) : \frac{T_{y}}{3 h}, \quad \left ( 4, \quad 5\right ) : \frac{T_{y}}{3 h}, \quad \left ( 5, \quad 6\right ) : 0, \quad \left ( 6, \quad 1\right ) : 0\right \}$$

Example 2: twisting moment in z direction



In [25]:

    
S3.set_loads(_Tx=0, _Ty=0, _Nz=0, _Mx=0, _My=0, _Mz=Mz)
S3.compute_stringer_actions()
S3.compute_panel_fluxes();



In [26]:

    
S3.N









    Out[26]:





$$\left \{ 1 : 0, \quad 2 : 0, \quad 3 : 0, \quad 4 : 0, \quad 5 : 0, \quad 6 : 0\right \}$$



In [27]:

    
S3.q









    Out[27]:





$$\left \{ \left ( 1, \quad 2\right ) : \frac{M_{z} \left(2 a b + 2 a h + b h\right)}{4 h \left(a^{2} b + a^{2} h + a b^{2} + a b h + b^{2} h\right)}, \quad \left ( 2, \quad 3\right ) : \frac{M_{z} \left(2 a b + 2 a h + b h\right)}{4 h \left(a^{2} b + a^{2} h + a b^{2} + a b h + b^{2} h\right)}, \quad \left ( 3, \quad 4\right ) : \frac{M_{z} \left(2 a b + a h + 2 b h\right)}{4 h \left(a^{2} b + a^{2} h + a b^{2} + a b h + b^{2} h\right)}, \quad \left ( 3, \quad 6\right ) : \frac{M_{z} \left(a - b\right)}{4 a^{2} b + 4 a^{2} h + 4 a b^{2} + 4 a b h + 4 b^{2} h}, \quad \left ( 4, \quad 5\right ) : \frac{M_{z} \left(2 a b + a h + 2 b h\right)}{4 h \left(a^{2} b + a^{2} h + a b^{2} + a b h + b^{2} h\right)}, \quad \left ( 5, \quad 6\right ) : \frac{M_{z} \left(2 a b + a h + 2 b h\right)}{4 h \left(a^{2} b + a^{2} h + a b^{2} + a b h + b^{2} h\right)}, \quad \left ( 6, \quad 1\right ) : \frac{M_{z} \left(2 a b + 2 a h + b h\right)}{4 h \left(a^{2} b + a^{2} h + a b^{2} + a b h + b^{2} h\right)}\right \}$$

Set loads on the section:

Example 3: shear in x direction and bending moment in y direction



In [28]:

    
S3.set_loads(_Tx=Tx, _Ty=0, _Nz=0, _Mx=0, _My=My, _Mz=0)
S3.compute_stringer_actions()
S3.compute_panel_fluxes();



In [29]:

    
S3.N









    Out[29]:





$$\left \{ 1 : - \frac{3 M_{y} \left(- \frac{2 a}{3} - \frac{b}{3}\right)}{4 a^{2} + 4 a b + 4 b^{2}}, \quad 2 : - \frac{3 M_{y} \left(- \frac{2 a}{3} - \frac{b}{3}\right)}{4 a^{2} + 4 a b + 4 b^{2}}, \quad 3 : - \frac{3 M_{y} \left(\frac{a}{3} - \frac{b}{3}\right)}{4 a^{2} + 4 a b + 4 b^{2}}, \quad 4 : - \frac{3 M_{y} \left(\frac{a}{3} + \frac{2 b}{3}\right)}{4 a^{2} + 4 a b + 4 b^{2}}, \quad 5 : - \frac{3 M_{y} \left(\frac{a}{3} + \frac{2 b}{3}\right)}{4 a^{2} + 4 a b + 4 b^{2}}, \quad 6 : - \frac{3 M_{y} \left(\frac{a}{3} - \frac{b}{3}\right)}{4 a^{2} + 4 a b + 4 b^{2}}\right \}$$



In [30]:

    
S3.q









    Out[30]:





$$\left \{ \left ( 1, \quad 2\right ) : 0, \quad \left ( 2, \quad 3\right ) : \frac{T_{x} \left(2 a + b\right)}{4 a^{2} + 4 a b + 4 b^{2}}, \quad \left ( 3, \quad 4\right ) : \frac{T_{x} \left(a + 2 b\right)}{4 a^{2} + 4 a b + 4 b^{2}}, \quad \left ( 3, \quad 6\right ) : 0, \quad \left ( 4, \quad 5\right ) : 0, \quad \left ( 5, \quad 6\right ) : - \frac{T_{x} \left(a + 2 b\right)}{4 a^{2} + 4 a b + 4 b^{2}}, \quad \left ( 6, \quad 1\right ) : - \frac{T_{x} \left(2 a + b\right)}{4 a^{2} + 4 a b + 4 b^{2}}\right \}$$

Verify that $$q_i \cdot l_i = T_x$$



In [31]:

    
sympy.simplify(S3.q[(2,3)]*a+S3.q[(3,4)]*b-S3.q[(5,6)]*b-S3.q[(6,1)]*a)









    Out[31]:





$$T_{x}$$

Torsional moment of inertia



In [32]:

    
S3.compute_Jt()
S3.Jt









    Out[32]:





$$\left[\begin{matrix}\frac{8 h^{2} t \left(a^{2} b + a^{2} h + a b^{2} + a b h + b^{2} h\right)}{4 a b + 4 a h + 4 b h + 3 h^{2}}\end{matrix}\right]$$



In [ ]: