

In [1]:

    
#Pkg.add("SymPy") # Run this line the first time to install the SymPy package
using SymPy

Ex 1

$$U_S(x) = \int \frac{M(x)}{2EI} dx$$



In [2]:

    
@syms x, q, R1, R2, l, i, E
M1(x) = R1*x-(q*x^(2))/(2)
M2(x) = R1*x+R2*(x-l/(2))-(q*x^(2))/(2)









    Out[2]:





M2 (generic function with 1 method)

Using Castigliano's second theorem, that states $$\frac{\partial U}{\partial Q_i}=q_i$$

Where $Q_i$ is genneralized forces and $q_i$ is genneralized displacements. Since there is no deflection in the two supports then the differentiated with repect to the reaction force set to zero



In [3]:

    
U = integrate(M1(x)^2/(2*E*i),x,0,l/2) + integrate(M2(x)^2/(2*E*i),x,l/2,l)









    Out[3]:





$$\frac{R_{1}^{2} l^{3}}{48 e i} - \frac{R_{1} l^{4} q}{128 e i} + \frac{R_{2}^{2} l^{3}}{16 e i} + \frac{l^{5} q^{2}}{40 e i} - \frac{15 l^{4}}{128 e i} \left(R_{1} q + R_{2} q\right) + \frac{7 l^{3}}{96 e i} \left(2 R_{1}^{2} + 4 R_{1} R_{2} + 2 R_{2}^{2} + R_{2} l q\right) - \frac{3 l^{2}}{16 e i} \left(R_{1} R_{2} l + R_{2}^{2} l\right)$$



In [4]:

    
eq1 = diff(U,R1)









    Out[4]:





$$\frac{R_{1} l^{3}}{24 e i} - \frac{3 R_{2} l^{3}}{16 e i} - \frac{l^{4} q}{8 e i} + \frac{7 l^{3}}{96 e i} \left(4 R_{1} + 4 R_{2}\right)$$



In [5]:

    
eq2= diff(U,R2)









    Out[5]:





$$\frac{R_{2} l^{3}}{8 e i} - \frac{15 l^{4} q}{128 e i} + \frac{7 l^{3}}{96 e i} \left(4 R_{1} + 4 R_{2} + l q\right) - \frac{3 l^{2}}{16 e i} \left(R_{1} l + 2 R_{2} l\right)$$



In [6]:

    
solve([eq1,eq2], [R1, R2])









    Out[6]:





\begin{equation*}\begin{cases}R_{1} & \text{=>} &\frac{11 l}{56} q\\R_{2} & \text{=>} &\frac{4 l}{7} q\\\end{cases}\end{equation*}

Super, return the same as maple

ex2

Since we would like to find the displacement in point A, we call it P1 and then later apply the relation between P and P1



In [7]:

    
@syms x, q, P, P1, l, i, E
M1(x) = ((l-x)*2)*P
M2(x) = ((l-x)*2)*P-P1*((1/2)*l-x)

U = integrate(M2(x)^2/(2*E*i),x,0,l/2) + integrate(M1(x)^2/(2*E*i),x,l/2,l)









    Out[7]:





$$\frac{P^{2} l^{3}}{12 e i} + \frac{l^{3}}{48 e i} \left(4 P^{2} - 4 P P_{1} + P_{1}^{2}\right) - \frac{0.25 l^{2}}{e i} \left(2.0 P^{2} l - 1.5 P P_{1} l + 0.25 P_{1}^{2} l\right) + \frac{0.5 l}{e i} \left(2.0 P^{2} l^{2} - 1.0 P P_{1} l^{2} + 0.125 P_{1}^{2} l^{2}\right)$$

differentiating with respect to P1



In [8]:

    
eq1 = diff(U, P1)









    Out[8]:





$$\frac{l^{3}}{48 e i} \left(- 4 P + 2 P_{1}\right) - \frac{0.25 l^{2}}{e i} \left(- 1.5 P l + 0.5 P_{1} l\right) + \frac{0.5 l}{e i} \left(- 1.0 P l^{2} + 0.25 P_{1} l^{2}\right)$$

Remember that P1 is equal P and substitue in the equation. So the displacement in point A isequal to



In [9]:

    
eq1(P1 => P) # Substitute P1 with P









    Out[9]:





$$- \frac{0.166666666666667 P l^{3}}{e i}$$

ex 3



In [10]:

    
M1(x) = x*P
M2(x) = x*P-(x-l/(2))*3/(2)*P
U = integrate(M1(x)^2/(2*E*i),x,0,l/2) + integrate(M2(x)^2/(2*E*i), x, l/2, l*3/2)
eq1 = diff(U, P)









    Out[10]:





$$\frac{P l^{3}}{8 e i}$$

Displacement in point P is found

ex 4



In [11]:

    
M1(x) = x*P+(q*x^(2))/(2)









    Out[11]:





M1 (generic function with 1 method)



In [12]:

    
U = integrate(M1(x)^2/(2*E*i),x,0,l)
eq1 = diff(U, P)









    Out[12]:





$$\frac{P l^{3}}{3 e i} + \frac{l^{4} q}{8 e i}$$

Rember that P is equal 0, the displacement in x=0 is



In [13]:

    
eq1(P=>0)









    Out[13]:





$$\frac{l^{4} q}{8 e i}$$

ex 5



In [14]:

    
@syms M
M1(x) = M+(q*x^(2))/(2)
M2(x) = M+(q*l^(2))/(2)
U = integrate(M1(x)^2/(2*E*i),x,0,l) + integrate(M2(x)^2/(2*E*i),x,l,2l)
eq1 = diff(U, M)
# M is 0
eq1(M=>0)









    Out[14]:





$$\frac{2 l^{3} q}{3 e i}$$

ex 6



In [15]:

    
@syms q1,q2, A, Q1, E_mod
θ = 30*((1/180)*pi)
ɛ1(x) = q1/l
ɛ2(x) = 2*q2/l
ɛ3(x) = (cos((1/4)*pi)*q1+cos((1/4)*pi)*q2)/l
U = integrate((E_mod*A*ɛ1(x)^2)/2,x,0,l) + integrate((E_mod*A*ɛ2(x)^2)/2,x,0,l/2) + integrate((E_mod*A*ɛ3(x)^2)/2,x,0,l) - Q1*q1*cos(θ)-Q1*q2*sin(θ)
eq2 = diff(U,q2)
eq1 = diff(U,q1)
solve([eq1, eq2], [q1, q2])









    Out[15]:





\begin{equation*}\begin{cases}q_{1} & \text{=>} &\frac{0.547161002703171 Q_{1} l}{A E_{mod}}\\q_{2} & \text{=>} &\frac{0.0905677994593659 Q_{1} l}{A E_{mod}}\\\end{cases}\end{equation*}

ex 7



In [16]:

    
@syms R1
M1(x) = P1*x
M2(x) = P1*l+R1*x
M3(x) = P1*l+R1*x-2*P*(x-l)
M4(x) = P1*(l-x)+R1*l*(3)/(2)-2*P*(l)/(2)
U = integrate(M1(x)^2/(2*E*i),x,0,l) + integrate(M2(x)^2/(2*E*i),x,0,l) + 
    integrate(M3(x)^2/(2*E*i),x,l,3*l/2) + integrate(M4(x)^2/(2*E*i),x,0,l/2)
eq1 = diff(U,P1)(P1 => P)
eq2 = diff(U,R1)(P1 => P)
r1 = solve(eq2, R1)[1]









    Out[16]:





$$- \frac{29 P}{108}$$



In [17]:

    
δ_p = eq1(R1=>r1)









    Out[17]:





$$\frac{67 P l^{3}}{64 e i}$$

ex 8



In [18]:

    
@syms Mb, R1,R2
M1(x) = P*x
M2(x) = P*x-R1*(x-l/(2))+Mb
M3(x) = P*x-R1*(x-l/(2))-R2*(x-(3* l)/(2))+Mb
U = integrate(M1(x)^2/(2*E*i),x,0,l/2) + integrate(M2(x)^2/(2*E*i),x,l/2,l*3/2) + 
    integrate(M3(x)^2/(2*E*i),x,3/2 * l,5*l/2)
#Reaction forces is determined from eq1 and eq2
eq1 = subs(diff(U,R1),Mb, 0)
eq2 = subs(diff(U,R2),Mb, 0)
res = solve([eq1, eq2], [R1, R2])









    Out[18]:





\begin{equation*}\begin{cases}R_{1} & \text{=>} &1.64285714285714 P\\R_{2} & \text{=>} &- 0.857142857142857 P\\\end{cases}\end{equation*}



In [19]:

    
eq3 = diff(U,P)(Mb => 0 ,res...)









    Out[19]:





$$\frac{0.113095238095238 P l^{3}}{e i}$$



In [20]:

    
eq4 = diff(U,Mb)(Mb => 0 ,res...)









    Out[20]:





$$\frac{0.142857142857143 P l^{2}}{e i}$$



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