16.21 Techniques of structural analysis and design

Home assignment 4

Question 1: Note that we require the complete elasticity solution: \(TraditionalForm\`\(u\_\(\(i\)\(,\)\)\) ϵ\_ij, \ σ\_ij\)

Show[Import["block.eps",  "EPS"]];

a) plane stress condition:

(Subscript[σ, 33]) ≠  0, 
No distortions:
\(TraditionalForm\`ϵ\_23, \ ϵ\_13, \ ϵ\_12\)
\(TraditionalForm\`σ\_23, \ σ\_13, \ σ\_12\ are\ all\ zero\)

(Subscript[ϵ, 23]) = (Subscript[ϵ, 13]) = (Subscript[ϵ, 12]) = 
            0;
(Subscript[σ, 23]) = (Subscript[σ, 13]) = (Subscript[σ, 12]) = 0;

From equilibrium:
(∂_((Subscript[x, 1])))(Subscript[σ, 11]) = 0 => (Subscript[σ, 11]) =  C, b.c.: (Subscript[σ, 11])(((-w)/2), (Subscript[x, 2]), (Subscript[x, 3])) = 0  => C = 0
⟹

(Subscript[σ, 11]) = 0;

Similarly (from the plane stress condition)

(Subscript[σ, 22]) = 0;

(∂_((Subscript[x, 3])))(Subscript[σ, 33]) = 0 

Constitutive:
\(TraditionalForm\`ϵ\_33 = \ \(1\/E\)[σ\_33 - ν\ \((0 + 0)\)]\)

Then:
\(TraditionalForm\`∂\_\(x\_3\)\((E\ ϵ\_33)\)\  = \ \(0\ ⟹\ ϵ\_33 = \ \(∂\_\(x\_3\)u\_3 = c\)\)\)

(Subscript[u, 3]) = c (Subscript[x, 3]) + d;

The bondary conditions are:

(Subscript[u, 3])[(L/2)]  =  c (L/2) + d  =  0
(Subscript[u, 3])[((-L)/2)]  =  -c (L/2) + 
            d  =  δ

rules  =   Solve[{c (L/2) + d  == 
                0,  -c (L/2) + d  == δ}, {c, d}] // 
          Flatten;
sol  =  (Subscript[u, 3]) /.  rules;   
(Subscript[u, 3])  =  sol;

(Subscript[ϵ, 33]) =  (∂_((Subscript[x, 3])))(Subscript[u, 3]);
(Subscript[σ, 33]) =  EE (Subscript[ϵ, 33]);

    
(Subscript[ϵ, 11]) =  (Subscript[ϵ, 22]) =  -ν (Subscript[ϵ, 33]);

\(TraditionalForm\`ϵ\_11 = \ \(∂\_\(x\_1\)u\_1\  = \ \(-ν\)\ ϵ\_33\)\)
(Subscript[u, 1]) =  (Subscript[ϵ, 11]) (Subscript[x, 1]) +  d
The boundary condition is:
(Subscript[u, 1])  =  0 at (Subscript[x, 1]) =  0 => d = 0

(Subscript[u, 1]) =  (Subscript[ϵ, 11]) (Subscript[x, 1]);

\(TraditionalForm\`ϵ\_22 = \ \(∂\_\(x\_2\)u\_2\  = \ \(-ν\)\ ϵ\_22\)\)
(Subscript[u, 2]) =  (Subscript[ϵ, 22]) (Subscript[x, 2]) +  d
The boundary condition is:
(Subscript[u, 2])  =  0 at (Subscript[x, 2]) =  0 => d = 0

(Subscript[u, 2]) =  (Subscript[ϵ, 22]) (Subscript[x, 2]);

{(Subscript[u, 1]), (Subscript[u, 2]), (Subscript[u, 3])} // MatrixForm
(({{(Subscript[ϵ, 11]), (Subscript[ϵ, 12]), (Subscript[ϵ, 12])}, {(Subscript[ϵ, 12]), (Subscript[ϵ, 22]), (Subscript[ϵ, 23])}, {(Subscript[ϵ, 13]), (Subscript[ϵ, 23]), (Subscript[ϵ, 33])}}))//MatrixForm
(({{(Subscript[σ, 11]), (Subscript[σ, 12]), (Subscript[σ, 12])}, {(Subscript[σ, 12]), (Subscript[σ, 22]), (Subscript[σ, 23])}, {(Subscript[σ, 13]), (Subscript[σ, 23]), (Subscript[σ, 33])}}))//MatrixForm

b) Plane strain conditions:
The only things that change are:

(Subscript[σ, 22])≠0, (Subscript[ϵ, 22]) =  0

Use this in the constitutive equation in compliance form for (Subscript[ϵ, 22]):

\(TraditionalForm\`ϵ\_\(\(22\)\(\ \)\) = \ \(0\  = \ \(1\/EE\) \((σ\_22 - \ ν\ \((σ\_11 + \ σ\_33)\))\)\)\) => \(TraditionalForm\`σ\_22 = \ ν\ \((\ σ\_11 + \ σ\_33)\)\)

(Subscript[σ, 11]) =  0, as before

replace (Subscript[σ, 22])in constitutive equation for (Subscript[ϵ, 33]):

\(TraditionalForm\`ϵ\_33 = \ \(\(1\/EE\) \((σ\_33 - \ ν\ σ\_22)\)\  = \ \(\(1\/EE\) \((σ\_33 - \ \(ν\^2\) σ\_33)\)\  = \ \(\((1 - ν\^2)\)\/EE\) σ\_33\)\)\)

We note that this gives a stiffer behavior, since the factor in parenthesis is smaller than "1". 
Note that from here on the solution is the same as in a) but with
(Subscript[ϵ, 33])= (((1 - (ν^2)))/EE) (Subscript[σ, 33]), instead of (Subscript[ϵ, 33])= (1/EE) (Subscript[σ, 33])
So we can infer the solution by replacing E with (E/(1 - (⋁^2)))

(Subscript[σ, 33  ]) = (EE/(1 - (ν^2))) (Subscript[ϵ, 33]);
(Subscript[σ, 22]) =  ν (Subscript[σ, 33]);

(Subscript[ϵ, 11]) =  
      Simplify[((-ν)/EE) ((Subscript[σ, 22]) + (Subscript[σ, 33]))];  
    (Subscript[u, 1]) =  (Subscript[ϵ, 11]) (Subscript[x, 1]); (Subscript[σ, 11]) =  0;

    (Subscript[u, 2]) =  0; (Subscript[ϵ, 22]) =  
      0 ;

{(Subscript[u, 1]), (Subscript[u, 2]), (Subscript[u, 3])} // MatrixForm
(({{(Subscript[ϵ, 11]), (Subscript[ϵ, 12]), (Subscript[ϵ, 12])}, {(Subscript[ϵ, 12]), (Subscript[ϵ, 22]), (Subscript[ϵ, 23])}, {(Subscript[ϵ, 13]), (Subscript[ϵ, 23]), (Subscript[ϵ, 33])}}))//MatrixForm
(({{(Subscript[σ, 11]), (Subscript[σ, 12]), (Subscript[σ, 12])}, {(Subscript[σ, 12]), (Subscript[σ, 22]), (Subscript[σ, 23])}, {(Subscript[σ, 13]), (Subscript[σ, 23]), (Subscript[σ, 33])}}))//MatrixForm

Question 2: Problem 4.4 from textbook

For bar:
(Subscript[U, 0]) =  (1/2) σϵ = (1/2) E (ϵ^2)= (1/2) E ((((du/dx)))^2)
(Subscript[U, bar]) = (∫_V) (Subscript[U, 0]) dV= ((∫_0)^L) EA (1/2) ((((du/dx)))^2)dx

For spring:
(Subscript[U, spring]) =  (1/2) k ((u(L))^2)

U = (Subscript[U, bar]) +  (Subscript[U, spring])= ((∫_0)^L) EA (1/2) ((((du/dx)))^2)dx + (1/2) k ((u(L))^2)

External work:

(Subscript[W, ext]) =  
        P u(L)  +  (∫_V)(( f(x))/A) u(x) dV= P u(L) + (A/A) ((∫_0)^L) f(x) u(
              x) dx

(Subscript[W, ext])= P u(L) + (A/A) ((∫_0)^L) f(x) u(
              x) dx

Question 3: Problem 4.6 from textbook

U = \(TraditionalForm\`∫\_V\( 
              1\/2\) \(σ\_ij\) \(ϵ\_ij\) dV\)

We consider the contributions from bending :
\(TraditionalForm\`σ\_11 = \ \(-\(M(x\_1)\)\)\/I\ x\_3, \ ϵ\_11 = \ σ\_11\/E\)= ((-M((Subscript[x, 1])))/EI) (Subscript[x, 3])
(U^b) =  (1/2)\(TraditionalForm\`∫\_V\( 
                1\/2\) \(σ\_11\) \(ϵ\_11\) dV\  = \ \(\(1\/2\) \(∫\_V\ \(x\_3\^2\) \(\(M(x\_1)\)\^2\/EI\^2\) 
                dV\)\  = \ \(1\/2\) \(∫\_0\%L\(∫\_A\( x\_3\^2\) dA\ \(\(M(x\_1)\)\^2\/EI\^2\) dx\_1\)\)\)\)

\(TraditionalForm\`U\^b = \(1\/2\) \(∫\_0\%L\ \(\(M(x\_1)\)\^2\/EI\) \(\(dx\_1\)\(\ \)\)\)\)

and the contribution from shear:

from equilibrium:
(Subscript[σ, 11, 1]) +  (Subscript[σ, 31, 3]) =  
        0
\(TraditionalForm\`σ\_\(31, 3\) = \(M' \((x\_1)\)\)\/I\ x\_3\), also from equilibrium M' ((Subscript[x, 1]))= V((Subscript[x, 1)]), the shear force.
(Subscript[σ, 31])= ((V((Subscript[x, 1])))/(2  I)) (((Subscript[x, 3])^2) -  
            (h^(2 ))), where we have also applied the boundary condition (Subscript[σ, 31])= 0 
at (Subscript[x, 3]) =  h (assume h is half the height of the rectangular beam of dimensions b,2h)

(U^s)= \(TraditionalForm\`\(1\/2\) \(∫\_V\((\(σ\_13\) σ\_13\/\(2  G\) + \ \(σ\_31\) σ\_31\/\(2  G\))\)\ dV\)\  = \(\(1\/2\) \(∫\_V\( σ\_13\) σ\_13\/G\ dV\)\  = \ \(1\/2\) \(∫\_V\( 
                    1\/G\) \((\(V(x\_1)\)\/\(2  I\)\ \((x\_3\^2 - \ h\^\(\(2\)\(\ \)\))\))\)\^2\ dV\)\)\)
= \(TraditionalForm\`\(1\/2\) \(∫\_0\%L\(\( V(x\_1)\)\^2\/\(4  
                  G\)\) \(1\/I\^2\) \(∫\_\(-h\)\%h 
                b\ \(\((x\_3\^2 - \ 
                        h\^\(\(2\)\(\ \)\))\)\^2\) dx\_\(\(3\)\(\ \)\)\ dx\_1\)\)\)
In this expression:

Ii  =  (∫_(-h)^h) 
          b (x^2) \[DifferentialD]x
geomFactor  =  (∫_(-h)^h) 
            b (((((Subscript[x, 3])^2) -  
                    (h^(2 ))))^2) \[DifferentialD](Subscript[x, 3 ])
(geomFactor/(Ii^2))

that is:
(U^s)= (1/2) (∫_0^L)((( V((Subscript[x, 1])))^2)/(4  
                  G)) (12/(5 A)) d(Subscript[x, 1])
where A = bh is the area of the cross section.
Therefore:
\(TraditionalForm\`U\^s = \ \(1\/2\) \(∫\_0\%L\( 12 \( V(x\_1)\)\^2\)\/\(20\ GA\)\ dx\_1\)\)

For this specific problem:
M((Subscript[x, 1])) = (2/3)P (Subscript[x, 1])    if 0  < (Subscript[x, 1]) <  0.5 L
	=(1/3) P (1.5  L  - 
            (Subscript[x, 1]))   if 0.5 L  < (Subscript[x, 1]) <  1.5 L
	This can be conveniently expressed in Mathematica as:

(Subscript[F, 0]) =  12000;
L  =  1;
Eb  =  205 (10^9);
νb =  0.3;
Gb  =  (Eb/(2 (1 + νb)))
Ib  =  0.2604 (10^(-6));
Ab  =  (3/50) (10^4) Ib;    
M =  
      Which[0  <  x  <  0.5 L, (2/3) (Subscript[F, 0]) x, 
        0.5 L  <  x  <  1.5 L,  (1/3) (Subscript[F, 0]) (1.5 L  -  x), 
        True, 0];
(∂_x) M

V  =  
        Which[0 < x < 0.5, ((2 (Subscript[F, 0]))/3), 
          0.5 L < x < 1.5 L, -((Subscript[F, 0])/3), True, 
          0];
Plot[{M, V}, {x, 0, 1.5  L}];

Ub  =  (1/(2 Eb Ib)) ((∫_0^(0.5 L)) ((((2/3) (Subscript[F, 0]) 
                        x))^2) \[DifferentialD]x  + (∫_(0.5 L)^(1.5 L)) ((((1/3) (Subscript[F, 0]) (1.5 L - 
                            x)))^2) \[DifferentialD]x)

This value matches the solution given in the textbook.

Us =  (3/(10  
              Gb Ab)) ((∫_0^(0.5 L)) ((((2/3) 
                        (Subscript[F, 0])))^2) \[DifferentialD]x  + (∫_(0.5 L)^(1.5 L)) ((((1/3) (Subscript[F, 0]) (-x)))^2) \[DifferentialD]x)

The result in the textbook is given in term of a shape factor (see page 92-94)
(Subscript[f, s]) =  (A/((I^2) 
                (b^2))) (∫_A)(( Q(
                    (Subscript[x, 3])))^2) d(Subscript[x, 3]),
where
\(TraditionalForm\`Q(
          z)\  = \ ∫\_\(\_z\)\%h 
            b\ \(x\_3\) dx\_3\)
which gives the same expression for the shear energy

Q  = (∫_((Subscript[x, 3]))^h) 
          b z \[DifferentialD]z
(Subscript[f, s]) =  ((b h)/(((((2/3) b (h^3)))^(2 )) 
              (b^2))) 
        b (∫_(-h)^h) 
                (Q^2) \[DifferentialD](Subscript[x, 3])

Finally, let's note that the complementary strain energy has the same value as the strain energy, as this is a linear elastic material

Question 4: Problem 4.7  from textbook

For linear elastic beam:

\(TraditionalForm\`U\_0 = \ \(∫\_0\^ϵ\_ij\( σ\_ij\) dϵ\_ij = \ \(\(1\/2\) \(σ\_11\) ϵ\_11 = \ \(1\/2\) E\ ϵ\_11\^2\)\), \ ϵ\_11 = \ \(-x\_3\)\ du\_3\/dx\_1\)
(Subscript[U, 0]) =  (1/(2 )) 
          E ((((Subscript[x, 3])^2)(((Subscript[du, 3])/(Subscript[dx, 1]))))^2)
(Subscript[U, beam]) = (∫_V) (Subscript[U, 0]) dV= ((∫_0)^L) (1/2) E (((((Subscript[du, 3])/(Subscript[dx, 1]))))^2)(∫_A) ((Subscript[x, 3])^2) dA (Subscript[dx, 1])

(Subscript[U, beam]) = (1/2) ((∫_0)^L) 
              E I (((((Subscript[du, 3])/(Subscript[dx, 1]))))^2) (Subscript[dx, 1])
(Subscript[U, spring]) =  (Subscript[((∫_0)^w), 0]) (Subscript[F, s]) 
                      dw = \(∫\_0\^\(\(u\_3\)(L)\)k\ w\_0\^n\ dw\_0\)=k\(\(\(w\_0\^\(n + 1\)\/\(n + 
                            1\)\)\(\( | \_0\)\^\(\(u\_3\) 
                            L\)\)\)\(=\)\(\ \)\(k\)\(\ \)\)((((Subscript[u, 3])(L))^(n + 1))/(n + 
                      1)) 

(Subscript[U, beam]) +  (Subscript[U, spring])= (1/2) ((∫_0)^L) 
            E I (((((Subscript[du, 3])/(Subscript[dx, 1]))))^2) (Subscript[dx, 1]) + k((((Subscript[u, 3])(L))^(n + 1))/(n + 1))

External work:

(Subscript[W, ext])= (∫_S) (Subscript[t, i]) (Subscript[u, i]) dS
The only external force in this case is (Subscript[q, 0])which is a force per unit length distributed uniformly over the top surface of the beam. Then:
(Subscript[W, ext]) =  b ((∫_0)^L)((q(x))/b)(Subscript[u, 3])((Subscript[x, 1]))(Subscript[dx, 1])
(Subscript[W, ext]) =  ((∫_0)^L)q(x)(Subscript[u, 3])((Subscript[x, 1]))(Subscript[dx, 1])

Question 5:

Clear["Global`*"]
values  =  {L  ->  0.1,  w ->  0.1,  
          d ->  0.001,  
          EE ->  70 (10^9),  ν ->  0.3,  δ ->  
            0.002};

a) Plane stress

ϵ=(({{((δ ν)/L), 0, 0}, {0, ((δ ν)/L), 0}, {0, 0, -(δ/L)}}));
σ=(({{0, 0, 0}, {0, 0, 0}, {0, 0, -((EE δ)/(L ))}}));
(Subscript[U, 0]) =  (1/2) (∑_(i = 1)^3)(∑_(j = 1)^3) σ[[i, j]] ϵ[[i, 
                  j]]
V  =  
        d w L;
U =  (Subscript[U, 0]) V    
U /. values

In the case of plane stress, the strain energy is 14000 J, about the amount of energy consumed by a 60 W light bulb in:

(140/60) // N

in seconds.

b) Plane strain

ϵ=(({{((δ ν)/(L - L ν)), 0, 0}, {0, 0, 0}, {0, 0, -(δ/L)}}));
σ=(({{0, 0, 0}, {0, -((EE δ ν)/(L (1 - (ν^2)))), 0}, {0, 0, -((EE δ)/(L (1 - (ν^2))))}}));
(Subscript[U, 0]) =  (1/2) (∑_(i = 1)^3)(∑_(j = 1)^3) σ[[i, j]] ϵ[[i, 
                  j]]
V  =  
        d w L;
U =  (Subscript[U, 0]) V    
U  /.  
      values

In the case of plane stress the energy is higher, as the plate is stiffer
Note that in both cases the only component of stress contributing to the strain energy is (Subscript[σ, 33]), as either the 11 and 22 components of stress or the strain vanish in both cases.