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l º P

AB, BC and CD are rigid bars The system has two degrees

connected by rotational springs of freedom: q₁, q₂

of stiffness K₁, K₂

Assuming that q₁ and q₂ are small (sin q₁ » q₁, sin q₂ » q₂, cos q₁ » 1, cosq₂» 1) the equilibrium equations

lead to the following system of equations, written in matrix form

The corresponding iteration scheme is

where the R.H.S. represents the internal forces (bending moments in B and C). After each iteration, by equating {q}_i+1 to {q}_i it is possible to obtain a lower bound, an upper bound and an estimate of P_cr. We consider now two cases:

(i) K₁ = K₂ = K

\ 0.923 £ P_cr £ 1.071 = 0.992

{q}_I »

Exact results: P_cr = 1 {q}_I =

(ii) K₁ = 2K₂ = K

\ 0.6 £ P_cr £ 0.75 = 0.667 {q}_I =

Exact results: P_cr = 0.634 {q}_I =

NOTE: The estimates of the critical load were obtained by averaging the values of {q}_i+1/{1}_i.

The critical buckling load of an axially loaded column with arbitrary boundary conditions is the lowest eigenvalue of the following eigenvalue-eigenfunction problem:

(1)

where the axial force N is assumed to depend on a single load parameter l . Here we will be concerned only with problems in which N is piecewise constant, which means that equation (1) becomes:

(2)

One observes that equation (2) is equivalent to:

(3)

where the bending moment M (x, l ) has the form

M (x, l ) = N (l ) y (x) + A (l ) . x + B (l ) (4)

and the values of A (l ) and B (l ) may be determined from the boundary conditions. It should be noticed that the application of Vianello's method requires the calculation of M (x,l ) in each iteration (step (ii)) which, particularly in the case of statically indeterminate columns, may involve considerable effort. Here, only the very simple case of a simple supported column under uniform compression will be considered in order to illustrate the application of the method to a continuous system.

Since, in this case, A (l ) = B (l ) = 0, the differential equation of equilibrium is given by

and leads to the iteration scheme

Assume the initial estimate of the deflected configuration of the column as y₁ (x)= (L - x), noticing that it must satisfy the kinematic boundary conditions y₁ (0) = y₁ = 0. Then, multiplying by P to obtain the corresponding bending moment, integrating twice and imposing the boundary conditions one is led to:

Now define

and obtain an upper bound and a lower bound of P_cr by equating [Q₁(x)]_max and [Q₁(x)]_min to the unity:

To calculate estimates of the critical load it is necessary to define criteria of similarity between y₁(x) and y₂(x). Two possible criteria are:

(a) To equate the average values

(y₁)_av =

(y₂)_av =

(b) To equate y₁ (L/2) and y₂ (L/2)

Repeating the procredure with the initial estimate

y₂(x) = x

(same shape as the result of interation 1)

one obtains

y₃(x) =

Q₂ (x) =

=

=

(a) (y₂)_av = (y₃)_av Þ = 9.882

(b) y₂ (L/2) = y₃ (L/2) Þ = 9.836

Estimate of buckling mode: y₁(x) »

Exact results: P_cr = p ² y_I(x) = sen

3.3 Illustrative Example

Consider the beam represented in Figure 2, which is divided into 6 equal segments of length 1.

FIGURE 2

i) Distributed loads (q)

0 -1 -2 0 0 0 0

ii) Equivalent concentrated loads (q_disc)

Linear disc. -1 -6 -5/0 0 0 0 0

(-5)

iii) Concentrated Loads (Q)

0 0 0 0 -6 0 -3

iv) Total concentrated loads (F_disc)

-1 -6 -5 0 -6 0 -3

v) Uncorrected average shears (V¢ )

The calculation starts from the right, so that the average shears are known on the basis of only one arbitrary value, which is R_E = 0

20 14 9 9 3 3

vi) Uncorrected bending moments (M¢ )

Same direction as in V¢ . Influence of R_E missing. Includes a jump in pointC due to the concentrated moment.

-64 -44 -30 -21/-15 -6 -3 0

vii) Bending moment correction (M_corr)

The correction is due to R_E and is based on the fact that M_B = 0. It is linear between E and A.

50 40 30 20 10 0 0

viii) Average shear correction (V_corr)

It is the value of R_E, i.e. the slope of M_corr

-10 -10 -10 -10 -10 0

ix) Average shears (V) v) + viii)

10 4 -1 -1 -7 3

x) Bending moments (M) vi) + viii)

-14 -4 0 -1/5 4 -3 0

xi) Curvatures (c )

-7 -2 0 -0.5/5 4 -3 0

xii) Equivalent concentrated curvatures (c _disc)

In point C, due to the jump in curvature the formula for end stations must be used twice.

Parab.disc. -61 -54 -5 -1.5/62 84 -52 -22

(60.5)

xiii) Uncorrected average slopes (q ¢ )

The calculation starts from the left, so that the average slopes are known on the basis of only one arbitrary value, which is the discontinuity in slope in point B.

-61 -115 -120 -59.5 24.5 -27.5

xiv) Uncorrected deflections (y¢ )

Same directions as in q ¢ . Influence of _q _B missing.

0 -61 -176 -296 -355.5 -331 -358.5

xv) Deflection correction (y_corr)

The correction is due to _q _B and is based on the fact that y_E = 0. It is linear between B and F.

0 0 0 110.33 220.67 331 441.33

xvi) Average slope correction (q _corr)

It is the value of _q _B, i.e. the slop of y_corr

0 0 110.33 110.33 110.33 110.33

xvii) Average slopes (q ) xiv) + xv)

-61 -115 -9.67 50.83 134.83 82.83

xviii) Deflections (y) xiv) + xv)

0 -61 -176 -185.67 -134.83 0 82.83

1. 0 1 2 1 0 y₀ initial guess

i = 1

2. 0 1 2 1 0 y₀
3. 1 1 1 1 P N

-1 -1 1 1 __y

-1 -1 1 1 P _M

1. 0 -1 -2 -1 0 P M¢ º M
2. 0 -1 -2 -1 0 c

-4 -24 -44 -24 -4 c _disc

-4 -28 -72 -96 q ¢

1. 0 -4 -32 -104 -200 y¢
2. 0 50 100 150 200 y_corr
3. 0 46 68 46 0 yº y₁

Now it is necessary to equate y₁ to y₀:

- 46 34 46 0 y₁/y₀

Considering the smallest and largest values of y₁/y₀ one obtains an upper bound and a lower bound of P_cr:

By considering the average of the values of y₁/y₀ one obtains and estimate of P_cr:

1. 0 1 1.478 1 0 y₁

-1 -0.478 0.478 1 _y

-1 -0.478 0.478 1 P _M

2. 0 -1 -1.478 -1 0 P M¢ º M
3. 0 -1 -1.478 -1 0 c

-4.522 -22.956 -33.56 -22.956 -4.522 c _disc

-4.522 -27.478 -61.038 -83.994 -4.522 q ¢

1. 0 -4.522 -32 -93.038 -177.032 y¢
2. 0 44.258 88.516 132.744 117.032 y_corr
3. 0 39.736 56.516 39.736 0 y º y₂

- 39.736 38.238 39.736 - y₂/y₁

9.664 £ p_cr £ 10.042 = 9.787

Exact result: P_cr = p ² = 9.870

1. 0 0.5 1 2 4 y₀ initial guess

i = 1

2. 0 0.5 1 2 4 y₀
3. 3 3 1 1 P N

-0.5 -0.5 -1 -2 _y

-1.5 -1.5 -1 -2 P _M

1. 6 4.5 3 2 0 P M¢ º M

-3 2.25 1.5/3 2 0 c

2. 33 54 112.5/33 46 9 c _disc

(145.5)

3. 33 87 232.5 278.5 q ¢ º q
4. 0 33 120 352.5 631 y¢ º yº y₁

- 66 120 176.25 157.75 y₁/y₀

2.179 £ P_cr £ 5.818

1. 0 0.28 1 2.94 5.26 y₁

-0.28 -0.72 -1.94 -2.32 _y

-0.84 -2.16 -1.94 -2.32 P _M

2. 6.26 5.42 4.26 2.32 0 P M
3. 3.13 2.71 2.13/4.262 2.32 0 c
4. 36.04 64.72 28.04/43.74 54.92 9.66 c _disc

(71.71)

1. 36.04 100.76 172.54 227.46 q
2. 0 36.04 136.8 309.34 536.8 y º y₂

- 128.71 136.8 105.22 102.05 y₂/y₁

2.807 £ P_cr £ 3.763 = 3.249

Exact result: P_cr = 3.217

Since the column is statically indeterminate it is necessary to use the force method. The basis system is defined by suppressing the support at C.

0 0.5 1 0.8 0 y₀ initial guess

i) Calculation of the flexibility

0 0 0 0 1 F_disc

-1 -1 -1 -1 V

4 3 2 1 0 M

4 3 2 1 0 c

Lin.disc. 11 18 12 6 1 c _disc

11 29 41 47 q

0 11 40 81 128 yº

ii) Iteration procedure

i = 1

Basic system:

0 0.5 1 0.8 0 y₀

-0.5 -0.5 0.2 0.8 _y

-1 -1 -1/0 0 0 p P

lin.disc. -3 -6 -3/0 0 0 p_disc

(-3)

18 0 0 0 -6 P

15 -6 -3 0 -6 F_disc

15 9 6 6 N

-7.5 -4.5 1.2 4.8 _M

6 -1.5 -6 -4.8 0 M

6 -1.5 -6 -4.8 0 c

39 -30 -132.6 -108 -22.8 c _disc

39 9 -123.6 -231.6 q

0 39 48 -75.6 -307.2 y

Calculation of R_c:

-307.2 + 128 R_c = 0 Þ R_c = 01. p

Calculation of y₁:

y₁ = y + 0.1 p

0 65.4 144 118.8 0 y₁

- 130.8 144 148.5 - y₁/y₀

62.061 £ (pl)_cr £ 70.459

i = 2

Basic system:

0 0.454 1 0.825 0 y₁

-0.454 -0.546 0.175 0.825 _y

15 9 9 9 N

-6.81 -4.914 1.05 4.95 _M

5.724 -1.086 -6 -4.95 0 M

5.724 -1.086 -6 -4.95 0 c

39.55 -22.27 -132.07 -111 -23.7 c _disc

39.55 17.28 -114.74 -225.74 q

0 39.55 56.83 -57.96 -283.75 y

R_c = 0.09237 p

0 63.93 145.50 121.60 0 y₂

- 140.81 145.50 147.39 - y₂/y₁

62.528 £ (pl)_cr £ 65.81 = 63.744

Exact result: (pl)_cr = 59.81

One observes that the method converges to a value which is larger than the exact result. This is mainly due to the error introduced by the discretisation of p and c . In fact, solving this problem with 8 segments and an initial guess of

1. 0 0.217 0.435 0.718 1 0.92 0.84 0.42 0 leads,

after 3 iterations, to 59.80 £ (pl)_cr £ 60.28

P = ⁵ = 0.507P_cr

q = =

i) Problem I

-1 -1 -1 -1 -1 q

-1 -2 -2 -2 -1 q_{disc º} F_disc

-1 -3 -5 -7 V¢

0 -1 -4 -9 -16 M¢

0 4 8 12 16 M_corr

0 3 4 3 0 M

0 3 4 3 0 c

7 34 46 34 7 c _disc

7 41 87 121 q ¢

0 7 48 135 256 y¢

0 -64 -128 -192 -256 y_corr

0 -57 -80 -57 0 y º y_I

ii) Iteration procedure

i = 1

Using the first iteration in example 4.1, which corresponds to the solution of Problem II, one obtains

0 46 68 46 0 y_II

Calculation of _

_ (0 + 1 + 2 + 1 + 0) = (0 + 46 + 68 + 46 + 0) + (0 - 57 - 80 - 57 -0)

\ _ = -0.00824 L

Calculation of y₁ (x):

y₁ (x) = y_I (x) + y_II (x) [_ = -0.008324 L]

0 -0.00957 -0.01381 -0.00957 0 L y₁

i = 2

Problem II

0 -1 -1.443 -1 0 _ y₁

0 1 1.443 1 0 M

0 1 1.443 1 0 c

2.279 11.443 16.43 11.443 2.279 c _disc

2.279 13.722 30.152 41.595 q ¢

0 2.279 16.001 46.153 87.748 _ y¢

0 -21.937 -43.874 -65.811 -87.748 _ y_corr

0 -19.658 -27.874 -19.658 0 _ y

Calculation of _:

_ (-3.443) = (-67.14) + Þ _ = 0.00932 L

Calculation of y₂ (x):

0 -0.00941 -0.01328 -0.00941 0 L y₂

Exact result: y (L/2) = -0.01326 L

If the initial imperfection consists of an eccentricity e₀ of all the applied loads, then y_I (x) is the solution of (N piecewise constant):

= - N e₀ (x) + Ax + B (9)

Finally, it should be mentioned that the method will diverge if the axial loading parameter l is larger than the corresponding critical value l _cr.

Assuming that q₁ and q₂ are small (cos q₁ = , cos q₂ = and cose (q₁-1₂) = ) the potential energy of the system is given by

V=U+W = K₁(2q₁-q₂)² + K₂ (2q₁ - q₁)²-

= (2 K₁+-P1)+-(2 K₁+2 K₂-P1) q₁-q₂

the conditions for a stationary potential energy = 0 and = 0 lead to the following system of equations, written in matrix form

and to the corresponding iteration scheme is

K₂

One should notice that this procedure always leads to a symmetric matrix, which was not that case in the previous solution. Obviously, both lead to the same results.

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