Prolegomena

We need

a short list of sympy functions,
the ability to pretty-print sympy's objects,
some additional display functions from IPython.
some names for some symbols to be used in symbolic expressions.



In [1]:

    
from sympy import symbols, Rational as rat, Eq, latex
%load_ext sympyprinting
from IPython.core.display import Image, display, Math, Latex
m, k, p, L, u, a, delta = symbols('m k p_0 L u_G a_G delta')
d = delta*u

Rigid Bodies



In [2]:

    
Image(filename='figures/rigid.png')









    Out[2]:

Centres of Instantaneous Rotation

The CIR of left body, $\Omega_1$, is at the intersection of an horizontal line in A and a vertical line in C, hence the coordinates of $\Omega_1$ are $(L,L)$ if we put the origin in B.

The CIR of the right body, $\Omega_2$, is at the intersection of an horizontal line in G and a $135^o$ line going from D, internal hinge, to $\Omega_1$, hence the coordinates of $\Omega_2$ are $(0,2L)$.

The rotation $\theta_2$ is given in terms of $u_G$ by $\theta_2\\,3L=u_G$, the rotation $\theta_1$ is $\theta_1=2\theta_2$.

Objects

In our problem, we have two type of objects, 1) the rotation of a rigid body characterized by the coordinates of the centre of rotation and the angle of rotation and 2) the significant points on a rigid body, characterized by the rotation, the attached mass, the attached spring etc.

We define two simple classes that model the behaviour of our objects.



In [3]:

    
class Centre:
    def __init__(self, x, y, rot):
        self.x = x ; self.y = y ; self.r = rot
        
class Point:
    def __init__(self, x, y, centre, mass=0, kx=0, ky=0, px=0, py=0, length=0):
        if kx*ky: raise NameError('Only one stiffness can be different from zero.')
        j_r = mass*length**2/12
        ux = -(y-centre.y)*centre.r
        uy = +(x-centre.x)*centre.r
        self.mvw = -a*d*(mass*(ux*ux+uy*uy)+j_r*centre.r**2)
        self.kvw = -u*d*(kx*ux**2+ky*uy**2)
        self.pvw = +d*(px*ux+py*uy)

Specifying the centres of rotation

The centres of rotation and the rotations, for the left and the right rigid body, are defined as follows (the origin of coordinates is in B, note that the rotations are divided by the unit displacement):



In [4]:

    
w1 = Centre( L, L,   rat(2,3)/L)
w2 = Centre( 0, 2*L, rat(1,3)/L)

Specifying the relevant points of the rigid bodies

The significant points are A (external force), B (point mass), D (spring), F (both spring and center of mass of the vertical segment of DEFG bar) and GDE (centre of mass of the horizontal segment).

A Point object is initialized for each significant point, with the relevant quantities explicited.



In [5]:

    
A = Point( 0, L, w1, py=-p)
B = Point( 0, 0, w1, mass=3*m)
D = Point( 2*L, 0, w1, ky=k)
F = Point( 3*L, L, w2, kx=k, mass=2*m, length=2*L)
GDE = Point( rat(5,2)*L, 0, w2, mass=m, length=L)

Computing the work of different forces for a virtual displacement

We need the sum of virtual works for the different types of forces in the equation of the virtual displacements, here it is an handy function to do it...



In [6]:

    
def sum_vw(which):
    return sum([getattr(point,which) for point in (A,B,D,F,GDE)])

We use our handy function to compute the individual contributions to the virtual work.



In [7]:

    
vw_inertial_f = sum_vw('mvw')
display(Math(r'\delta W_I = '+latex(vw_inertial_f/d)+'\\,'+latex(d)+','))
vw_elastic_f  = sum_vw('kvw')
display(Math(r'\delta W_S = '+latex(vw_elastic_f/d)+'\\,'+latex(d)+','))
vw_external_f = sum_vw('pvw')
display(Math(r'\delta W_E = '+latex(vw_external_f/d)+'\\,'+latex(d)+'.'))









    





$$\delta W_I = - \frac{55}{9} a_{G} m\,\delta u_{G},$$






    





$$\delta W_S = - \frac{5}{9} k u_{G}\,\delta u_{G},$$






    





$$\delta W_E = \frac{2}{3} p_{0}\,\delta u_{G}.$$

The equation of dynamic equilibrium

The sum of the individual contribution to the virtual work, i.e., the total virtual work, is zero for a rigid body in equilibrium or, equivalently,

$-\delta W_I -\delta W_S = \delta W_E$

Simplyfing $\delta u_g$ we have



In [8]:

    
display(Eq( -vw_inertial_f/d - vw_elastic_f/d, vw_external_f/d))









    





$$\frac{55}{9} a_{G} m + \frac{5}{9} k u_{G} = \frac{2}{3} p_{0}$$

The generalized parameters



In [9]:

    
display(Latex(r'The generalized mass is $m^* = %s$, '%latex(-vw_inertial_f/d/a) +
              r'the generalized stiffness is $k^* = %s$, '%latex(-vw_elastic_f/d/u) +
              r'the generalized load is $p^* = %s\,\mathrm{f}(t)$.'%latex(vw_external_f/d)))









    





The generalized mass is $m^* = \frac{55}{9} m$, the generalized stiffness is $k^* = \frac{5}{9} k$, the generalized load is $p^* = \frac{2}{3} p_{0}\,\mathrm{f}(t)$.