16.21 Techniques of structural analysis and design

Home assignment 7
Solution Set

Question 1: Problem 5.6 from textbook

Show[{Import["reddy-5.6-1.jpg", "JPG"], 
        Import["reddy-5.6-2.jpg", "JPG"]}]

Question 2: Problem 7.22-7.29 from textbook

The unknown displacement field for this problem is the deflection of the beam's neutral axis \(TraditionalForm\`\((w\_0)\) . \ 
          We\ approximate\ this\ field\ using\ a\ Ritz\ approximation\ of\ the\ \(type : 
w\_0\ ~\ c\_i  ϕ\_i\)\)

The essential boundary conditions at the clamped end (x=0) of this cantilever beam require:
(Subscript[ϕ, i])(0)=0, (Subscript[ϕ, i])'(0) = 0
A good candidate family of functions (Subscript[ϕ, i])would be: (Subscript[ϕ, i]) =  (((x/L))^(i + 1)),  i = 1, n. Note (x^1) does not satisfy  the zero-slope boundary condition so it's been eliminated.

Clear["Global`*"]; 
    Off[General::spell, General::spell1];

1) approximation functions:

Ck[n_] :=  Table[ck[i], {i, n}]
ϕk[n_]  :=  
      Table[((((x/L)))^(i + 1)), {i, n}]
w[n_]  :=  Ck[n] . ϕk[n]
w[2]

2) approximate potential Π(Ck) for this problem

Π[
        n_]  := (EI/2) (∫_0^L) ((((∂_(x, x))w[
                          n]))^2) \[DifferentialD]x  -  (∫_0^L) (Subscript[q, 0]) w[n] \[DifferentialD]x // Expand

Π[2]

3) Minimize the potential to find the value of the coefficients that best approximate this problem

gradΠ[n_]  :=  
      Table[D[Π[n], ck[i]], {i, n}]

gradΠ[2]

solution[n_]  :=  
      Solve[gradΠ[n]  ==  0 ,  Ck[n]] // 
        Flatten

solution[5]

wapprox[n_]  :=  w[n]  /.  solution[n]

wapprox[2]

Plot the solution

values  =  {EI  ->  70 (10^9) ((0.05 (0.3^3))/12), 
          L ->  3, (Subscript[q, 0]) ->  1000 };
Plot[Evaluate[wapprox[4] /.  values], {x, 0, L /.  values},  
        PlotRange -> All, 
        PlotStyle ->  {Thickness[0.01]}];

Question 3: Problem 7.24-7.31 from textbook

The unknown displacement field for this problem is the deflection of the beam's neutral axis \(TraditionalForm\`\((w\_0)\) . \ 
          We\ approximate\ this\ field\ using\ a\ Ritz\ approximation\ of\ the\ \(type : 
w\_0\ ~\ c\_i  ϕ\_i\)\)The essential boundary conditions at the clamped end require:
(Subscript[ϕ, i])(0)=0, (Subscript[ϕ, i])'(0) = 0
at the pinned end:
(Subscript[ϕ, i])(L)  =  0
A good candidate family of functions (Subscript[ϕ, i])would be: (Subscript[ϕ, i]) =  (((x/L))^(i + 1)) (1 - 
              (x/L)),  i = 1, n. Note (x^1) does not satisfy  the zero-slope boundary condition so it's been eliminated

Clear["Global`*"]; 
    Off[General::spell, General::spell1];

Beam clamped on one end and simply supported on the other end with concentrated load at x=L/2

1) approximation functions:

Ck[n_] :=  Table[ck[i], {i, n}]
ϕk[n_]  :=  
      Table[((((x/L)))^(i + 1)) (1 - (x/L)), {i, 
          n}]
w[n_]  :=  Ck[n] . ϕk[n]
w[2]

L = 1;
Plot[Evaluate[ϕk[4], {x, 0, L}]];
L =. ;

2) approximate potential Π(Ck) for this problem

Δ[
        n_]  :=  (w[n]  /.  
          x ->  (L/2))
Π[
        n_]  := (EI/2) (∫_0^L) ((((∂_(x, x))w[
                          n]))^2) \[DifferentialD]x  - (∫_0^L) (Subscript[q, 0]) 
              w[n] \[DifferentialD]x -  (Subscript[F, 0]) Δ[n] // 
        Expand

Π[2]

3) Minimize the potential to find the value of the coefficients that best approximate this problem

gradΠ[n_]  :=  
      Table[D[Π[n], ck[i]], {i, n}]

gradΠ[2]

solution[n_]  :=  
      Solve[gradΠ[n]  ==  0 ,  Ck[n]] // 
        Flatten

solution[4]

Note (Subscript[q, 0])only appears in ck[i] with i<=2

wapprox[n_]  :=  w[n]  /.  solution[n]

wapprox[2]

Plot the solution

values  =  {EI ->  70 (10^9) ((0.05 (0.3^3))/12), 
          L ->  3, (Subscript[F, 0]) ->  10000 , (Subscript[q, 0]) -> 0};
Plot[Evaluate[wapprox[4] /.  values], {x, 0, L /.  values},  
        PlotRange -> All, 
        PlotStyle ->  {Thickness[0.01]}];

Question 4: Problem 7.25-7.32 from textbook

The unknown displacement field for this problem is the deflection of the beam's neutral axis \(TraditionalForm\`\((w\_0)\) . \ 
          We\ approximate\ this\ field\ using\ a\ Ritz\ approximation\ of\ the\ \(type : 
w\_0\ ~\ c\_i  ϕ\_i\)\)
The essential boundary conditions at both ends require:
(Subscript[ϕ, i])(0)=0, 
(Subscript[ϕ, i])(L)  =  0
A good candidate family of functions (Subscript[ϕ, i])would be: (Subscript[ϕ, i]) =  (((x/L))^i) (1 - (x/L)),  
      i = 1, n.

Clear["Global`*"]; 
    Off[General::spell, General::spell1];

1) approximation functions:

Ck[n_] :=  Table[ck[i], {i, n}]
ϕk[n_]  :=  
      Table[((((x/L)))^i) (1 - (x/L)), {i, 
          n}]
w[n_]  :=  Ck[n] . ϕk[n]
w[2]

L = 1;
Plot[Evaluate[ϕk[3], {x, 0, L}]];
L =. ;

2) approximate potential Π(Ck) for this problem

Δ[
        n_]  :=  (w[n]  /.  
          x ->  (L/2))
Π[
        n_]  := (EI/2) (∫_0^L) ((((∂_(x, x))w[
                          n]))^2) \[DifferentialD]x  + 
          k Δ[n]^2 - (∫_0^L) (Subscript[q, 0]) 
              w[n] \[DifferentialD]x -  (Subscript[F, 0]) Δ[n] // 
        Expand

Π[2]

3) Minimize the potential to find the value of the coefficients that best approximate this problem

gradΠ[n_]  :=  
      Table[D[Π[n], ck[i]], {i, n}]

gradΠ[2]

solution[n_]  :=  
      Solve[gradΠ[n]  ==  0 ,  Ck[n]] // 
        Flatten

solution[3]

wapprox[n_]  :=  w[n]  /.  solution[n]

wapprox[3] // Simplify

Plot the solution

values  =  {EI ->  70 (10^9) ((0.05 (0.3^3))/12), 
          L ->  3, (Subscript[F, 0]) ->  0 , (Subscript[q, 0]) -> 10, 
          k ->  (10^4)};
Plot[Evaluate[wapprox[4] /.  values], {x, 0, L /.  values},  
        PlotRange -> All, 
        PlotStyle ->  {Thickness[0.01]}];

Question 5 a: Problem 7.36 a from textbook

Clear["Global`*"]; Off[General::spell, General::spell1];

Simply supported beam on an elastic foundation under uniform distributed load.

1) Seek approximate solutions of the form:

Ck[n_] :=  Table[ck[i], {i, n}]
ϕk[n_]  :=  
      Table[(x^i) (L - x), {i, n}]
w[n_]  :=  Ck[n] . ϕk[n]

w[2]

2) Define the approximate potential Π(Ck) for this problem

Π[n_] :=((ym im)/2) (∫_0^L) ((((∂_(x, x))w[n]))^2) \[DifferentialD]x  +  (k/2) (∫_0^L) w[
                        n]^2 \[DifferentialD]x  -  
            q0 (∫_0^L) w[n] \[DifferentialD]x // 
          Expand

Π[2]

3) Minimize the potential to find the value of the coefficients that best approximate this problem

gradΠ[n_]  :=  
      Table[D[Π[n], ck[i]], {i, n}]

gradΠ[2]

solution[n_]  :=  
      Solve[gradΠ[n]  ==  0 ,  Ck[n]] // 
        Flatten

solution[2]

wapprox[n_]  :=  w[n]  /.  solution[n]

wapprox[2]

Plot the solution

values  =  {ym  ->  70 (10^9), L ->  3, 
          im ->  ((0.05 (0.3^3))/12), q0 ->  -(10^5),  
          k  ->   (10^6)};
Plot[Evaluate[wapprox[2] /.  values], {x, 0, L /.  values},  
        PlotRange -> All, 
        PlotStyle ->  {Thickness[0.01]}];

Question 5 b: Problem 7.36 b from textbook

Clear["Global`*"]; Off[General::spell, General::spell1];

Simply supported beam on an elastic foundation under uniform distributed load.

1) Seek approximate solutions of the form:

Ck[n_] :=  Table[ck[i], {i, n}]
ϕk[n_]  :=  
      Table[Sin[(2 i - 1) π (x/L)], {i, 
          n}]
w[n_]  :=  Ck[n] . ϕk[n]

w[2]

2) Define the approximate potential Π(Ck) for this problem

Π[n_] :=((ym im)/2) (∫_0^L) ((((∂_(x, x))w[n]))^2) \[DifferentialD]x  +  (k/2) (∫_0^L) w[
                        n]^2 \[DifferentialD]x  -  
            q0 (∫_0^L) w[n] \[DifferentialD]x // 
          Expand

Π[2]

3) Minimize the potential to find the value of the coefficients that best approximate this problem

gradΠ[n_]  :=  
      Table[D[Π[n], ck[i]], {i, n}]

gradΠ[2]

solution[n_]  :=  
      Solve[gradΠ[n]  ==  0 ,  Ck[n]] // 
        Flatten

solution[2]

wapprox[n_]  :=  w[n]  /.  solution[n]

wapprox[2]

Plot the solution

values  =  {ym  ->  70 (10^9), L ->  3, 
          im ->  ((0.05 (0.3^3))/12), q0 ->  -(10^5),  
          k  ->   (10^6)};
Plot[Evaluate[wapprox[2] /.  values], {x, 0, L /.  values},  
        PlotRange -> All, 
        PlotStyle ->  {Thickness[0.01]}];