Stability Takehome

Stability of Structures

Problem S1

For the beam depicted below, the student must evaluate the critical load and the associated critical mode shape by adopting two cubic finite elements; discuss the results depending upon the parameter β.

Problem S2

Evaluate the critical load of the internally continuous frame below by means of:

• (a) Newmark's formula;
• (b) One finite element on the right column;
• (c) Two finite elements on the right column.

Problem S3

A column with section in the figure has flexural-torsional clamped conditions at each edge. Evaluate its critical load as a function of the length.

Data (N, mm): h = 150, b = 150, t_f = t_w = 10, E = 2.0E+5, G = 8.0E+4, σ_n = 240

PROBLEM S4

The following finite elements and degrees of freedom were proposed for the solution of the problem:

The degrees of freedom of each element are represented by the vector:

q_e = {q_e1 q_e2 q_e3 q_e4}^T

For the problem, the incidence vectors that connect the global and local degrees of freedom will be:

N₁ = [0 0 1 2]

N₂ = [1 2 0 0]

We recall the shape functions:

N₁ = 1 - 3ξ² + 2ξ³

N₂ = ξ - 2ξ² + ξ³

N₃ = 3ξ² - 2ξ³      where   ξ = x/L

N₄ = -ξ² + ξ³

The local displacement model can be written as:

Y_e(x) = N₁(x)q_e1 + N₂(x)q_e2 + N₃(x)q_e3 + N₄(x)q_e4 = N(ξ)^Tq_e

The total potential energy is described by the following expression:

V_e,p(q_e) = 1/2 ∫₀^L E I v'² dx - ∫₀^L p_e v dx

= 1/2 q^T _e (K_e = N_e K_ea) q_e

Considering the previous shape functions we have:

_{KE = EI/L3} symmetr_ic

• 12      6L_e -12      6L_e
• 4L_e2 -6L_e      2L_e2
•     12 -6L_{e symme}
•     4L_e2

Kea= ⅓/30L_e

• 36     3L_e -36     3L_e
• 4L_e2 -3L_e L_e2
• 36 -3L_e
• 4L_e2

PROBLEM S2

Evaluate the critical load of the internally continuous frame below by means of:

1. Newmark's formula
2. One finite element on the right column
3. Two finite elements on the right column

a) The solution of a simply supported beam with elastical constants (springs) at the opposite extremities is generally given by the Newmark's formulation:

P_E = cπ²EI/L

where c =

In this problem we have: k_c = EI/l.

While k₂ is computed by the displacement approach on the frame. So we distinguish two situations:

1. u₂ = 1 and u₂ = 0

   K₁₁ = ^4(2EI)/_l + ^8EI/_l = ^12EI/_l

   K₂₁ = ^2EI/_l

2. u₂ = 0 and u₂ = 1

   K₂₂ = ^4EI/_e

   K₁₂ = ^2EI/_e

Taking into account that is present a moment in node c due to rotational spring, we are able to build the system:

[

[K₁₁ K₁₂] [u₁] [0]

[K₂₁ K₂₂] [u₂] = [M_C]

] + [

[^12EI/_e ^2EI/_e] [u₁] = [0]

[^2EI/_e ^4EI/_e] [u₂] = [M_C]

]

PROBLEM S3

h=150 mmb=150 mmtf, tw=10 mmE=2.10⁵ MPaG=8.10⁴ MPaQ₀=240 N/mm

We compute some geometrical parameters in order to determine the centroid. Taking into account the symmetry of the section, the following initial reference system was proposed:

We compute the area:A=10(150+150)=3000 mm²

The moment of inertia:S_xG=0 - due to symmetryS_yG=150.10-(¹⁵⁰/₂)=-112500 mm³

The center of gravity:x_G=^S_yG/_A=-37,5 mmy_G=0

The moment of inertia:I_x=^hb3/₁₂=10(¹⁵⁰/₃)² 2181.10⁶ mm⁴I_y=^t_fb3/₁₂+b(37,5)²+t_h(37,5-^b/₂)²=7,03.10⁶ mm⁴ with x_CFT=-37,5 mm

The center of shear is computed by using the symmetry so:x_C=-37,5 mmy_C=0

while the torsional inertia:J=^t3/₃(b_h)=400000=1.10⁵ mm⁴

so:I_G=I_x+I_y=1045.10⁶ mm⁴I_C=I_G+A.d_CG2²=141,77.10⁶ mm⁴

and the interaction factor is given by:k=^I_C/_IG=1,43