16.21 Techniques of structural analysis and design

Home Assignment 1

Question 1:
a) The style in this notebook
b)

m = (({{a, b, c}, {d, e, f}, {g, h, i}}));
v =(({{j}, {k}, {l}}));
w  =  
        m . v;
MatrixForm[w]

m = (({{1, -2, 3}, {-4, 5, -6}, {7, -8, 9}}));
v = (({{10}, {11}, {12}}));
w  =  
        m . v;
MatrixForm[w]

matA  =  Table[i + j, {i, 5}, {j, 5}];
MatrixForm[matA]

v  =  
        Table[i ( i + 1),  {i, 5}];
MatrixForm[v]

w  =  matA . v;
MatrixForm[w]

c)Differential equation:

lhs[w_] := EI D[w, {x, 4}]  +  q0 Sin[π (x/L)]

i)(Subscript[w, 1])(x) = A+B x + C x^2 + D x^3

(Subscript[w, 1]) = 
        A  +  B x  +  C (x^2) +  D (x^3);
lhs[(Subscript[w, 1])]

This is different from "0"  for all  0<x<L regardless of the value of D, therefore it is not the solution

ii) (Subscript[w, 2])(x) = A Sin[π (x/L)]

(Subscript[w, 2])  = A Sin[π (x/L)];
lhs[(Subscript[w, 2])]

Note that this equation is satisfied for all 0<x<L if  "A" is chosen such that:

constantA  =  Solve[lhs[(Subscript[w, 2])]  ==  0,  A]

get rid of extra layer of braces that Solve adds with Flatten and use "rules" to assign constant value to variable "AA"

w2  =  (Subscript[w, 2])/.  Flatten[
          constantA]

Boundary conditions: this possible solution satisfies the following boundary conditions:

w2 /.  x ->  0
w2  /.  x ->  L
D[w2, x]  /.  x ->  0
D[w2, x]  /.  x ->  L
D[w2, {x, 2}]  /.  x ->  0
D[w2, {x, 2}]  /.  x ->  L

This is, the beam is pinned at both ends

iii) (Subscript[w, 3])(x) = A Cos[π (x/L)] + B Sin[π (x/L)]

(Subscript[w, 3]) = A Cos[π (x/L)] + B Sin[π (x/L)];
LHS  = 
      lhs[(Subscript[w, 3])]

The warning message can be eliminated with the following command:

Off[General::spell, General::spell1];

Note that this equation is satisfied for all 0<x<L if A=0 and B is chosen as in example ii):

Question 2

Potential energy

Π[
        w_] :=  (∫_0^L) (EI/2) ((((∂_(x, x))w))^2) \[DifferentialD]x  +  (∫_0^L) q0 w \[DifferentialD]x // 
        Simplify

Case 1)

w1  =  
      a  q0 ((L^4)/EI) (x/L) (1 - 
            (x/L)); Π1  =  Π[
          w1]  /.  {EI ->  1,  q0 ->  1,  
          L ->  1}
sol1  =  
      Solve[(∂_a) Π1 == 0, a] // 
        Flatten
a1  =  a  /.  sol1
Plot[Π1, {a, -2 a1, 3  a1}]

w3  =  (( q0 (L^4))/( EI)) (a (x/L) (1 -  (x/L)) + b  (x/L) (((1 - (x/L)))^2) +  
              c (x/L) (((1 - (x/L)))^3));
Π3  =  Π[w3]  /.  {EI ->  1,  
          q0 ->  1,  L ->  1}

a =. ; b =. ; c =. ;
sol3  =  
      Solve[{(∂_a) Π3 == 
              0,  (∂_b) Π3 == 
              0,  (∂_c) Π3 == 0}, {a, b, c}] // 
        Flatten
Π3v  =  Π3  /. 
        sol3[[{2, 3}]]
a3  =  a  /.  sol3[[1]]
Plot[{Π1, Π3v}, {a, -1 a3, 3  a3}, 
        PlotStyle -> {RGBColor[1, 0, 0], 
            RGBColor[0, 1, 0]}];

Comments: clearly, the second solution with three terms gives a lower energy. This can be explained by the fact that the solution with fewer terms, i.e. with fewer deformation modes and degrees of freedom gives a stiffer response.

Question 3

\(\(\(Show[Import["springs.eps", "EPS"]];\)\(
\)
    \)\)

(a) Vector approach: the equations for the displacements (Subscript[u, 1]), (Subscript[u, 2]), (Subscript[u, 3])are obtained by applying equilibrium of the nodes as follows:

(Subscript[loadbalance, 1])  =  
      (Subscript[P, 1])  -  ((Subscript[k, d])  +  (Subscript[k, e])) ((Subscript[u, 1]) - (Subscript[u, 2]))  -  
        (Subscript[k, b]) ((Subscript[u, 1]) - (Subscript[u, 3]))
(Subscript[loadbalance, 2]) =  
      (Subscript[P, 2])  -  (Subscript[k, c]) ((Subscript[u, 2]) - (Subscript[u, 3]))  +  ((Subscript[k, d]) + 
              (Subscript[k, e])) ((Subscript[u, 1]) - (Subscript[u, 2]))
(Subscript[loadbalance, 3])  =  
      (Subscript[P, 3]) -  (Subscript[k, a]) (Subscript[u, 3])  +  (Subscript[k, b]) ((Subscript[u, 1]) - (Subscript[u, 3]))  +  
        (Subscript[k, c]) ((Subscript[u, 2]) - (Subscript[u, 3]))

(b) Energy approach

U  =  (1/2) ((Subscript[k, a]) ((Subscript[u, 3])^2) +  (Subscript[k, b]) ((((Subscript[u, 1]) - (Subscript[u, 3])))^2) + (Subscript[k, c]) ((((Subscript[u, 2]) - (Subscript[u, 3])))^2) +  ((Subscript[k, d]) + 
                  (Subscript[k, e])) ((((Subscript[u, 1]) - (Subscript[u, 2])))^2))

W  =  (Subscript[P, 1]) (Subscript[u, 1]) +  (Subscript[P, 2]) (Subscript[u, 2]) +  (Subscript[P, 3]) (Subscript[u, 3])

Π  =  U  -  W

gradΠ  =  Table[D[Π, (Subscript[u, i])], {i, 3}] // 
          Expand // Simplify

Let's check that the two sets of equations are the same:

Table[gradΠ[[i]] + (Subscript[loadbalance, i]), {i, 3}] // 
      Simplify

Let's find the solution

solution  =  Solve[
            gradΠ == 0, {(Subscript[u, 1]), (Subscript[u, 2]), (Subscript[u, 3])}] // Simplify // 
        Flatten

(c) Special solutions
	i.)

solution  /.  {(Subscript[k, a]) ->  k,  (Subscript[k, b]) ->  k,  
        (Subscript[k, c]) ->  k,  (Subscript[k, d]) ->  k,  (Subscript[k, e]) ->  k, 
        (Subscript[P, 1]) ->  0,  (Subscript[P, 2]) ->  0 , (Subscript[P, 3]) -> P}

This corresponds to the case of all the springs being equal and the only load being applied at point 3. The displacement of the spring (Subscript[k, a]) is only affected by this load and the result is (Subscript[u, 3 ]) =  (P/k)the rest of the springs don't deform, they displace rigidly.
	ii.)

solution  /.  {(Subscript[k, a]) ->  k,  (Subscript[k, b]) ->  k,  
        (Subscript[k, c]) ->  k,  (Subscript[k, d]) ->  k,  (Subscript[k, e]) ->  k, 
        (Subscript[P, 1]) ->  P,  (Subscript[P, 2]) ->  0 , (Subscript[P, 3]) -> 0}

This cooresponds to the case of all the springs being equal and the only load being applied at point 1. The displacements are the result of the relative stiffnesses provided by the equivalent springs attached to them.

Question 4

(Subscript[σ, n])  =  (t^((n))) .  
            n  =  (Subscript[n, i]) (Subscript[σ, ij]) (Subscript[n, j])
(Subscript[σ, s])  =  (t^((n))) .  
            s  =  (Subscript[n, i]) (Subscript[σ, ij]) (Subscript[s, j])

n  =  {Cos[θ],  
          Sin[θ]};
s  =  {-Sin[θ],  Cos[θ]};

sigma  =  Table[(Subscript[σ, i, j]), {i, 2}, {j, 2}]

(Subscript[σ, n]) = n . sigma . 
              n  /.  {(Subscript[σ, 1, 2]) ->  σ12,  (Subscript[σ, 2, 1]) -> σ12} // Expand // TrigReduce

(Subscript[σ, s])  =  n . sigma . 
              s /.  {(Subscript[σ, 1, 2]) ->  σ12,  (Subscript[σ, 2, 1]) -> σ12} // Expand // TrigReduce

From this equation we can obtain the principal directions, i.e., those directions for which the shear stresses are zero:
((sin(2  θ))/cos)[2  θ) =  
        2 σ(12/((Subscript[σ, 11]) - (Subscript[σ, 22])))

Principal stresses

Solve[
      Det[sigma  -  λ IdentityMatrix[2]] == 
        0,  λ]