Theory of structures

Theory of structures

Revision of De Saint Venant's Problem

Consider an elastic node with any possible shape section that must be constant over the length. DSV considers a reference system centered in the geometric centroid of the section (⟶ we need a geometric analysis to find it).

Properties of centre of gravity:

S_x = ∫ y dA = 0

S_y = ∫ x dA = 0

First order moments

In order to have a principal reference system we need that I_xy = ∫ xy dA = 0 (Mixed moment of inertia).

In addition, we define also:

I_x = ∫ y² dA > 0

I_y = ∫ x² dA > 0 (Moments of inertia).

Properties of DSV's Problem

1. Linear-elastic homogeneous isotropic material
2. Small strain & displacement
3. No thermal strain (⟶ no inelastic effects)
4. No body forces (e.g., gravity) ⟶ body forces are distributed over the volume
5. No forces on the lateral surface ⟶ we can only have forces on initial and end sections
6. No extreme constraints ⟶ body is completely free ⟶ global equilibrium must be neglected

⟶ We know the resultant actions on the extreme sections.

Definition: Cauchy's Stress Components

G_xx, G_xy, G_xz ⟶ direct components

Normal stress = δ_xx

Shear stress = τ_xy, τ_xz - refer to 1:2 x-axis

⟶ Triangular components (equilibrium of a small volume element) refers to the 1:2 y-axis and 1:2 z-axis.

Balance Equations:

∫ G_xa dA = N

∫ x G_ya dA = M_x

∫ y G_xa dA = M_y

N.B. For an open section: we do not consider shear center.

Consider x_c - y_c dA = M + T_x - T_y

Note: External constraint is a combination of displacements, something reduction in increments. DSV solves the problem in the absence of any external constraint on Γ boundary conditions.

Initial Hypothesis (G_x, G_y, Z_xy = 0) are ok because all equations are satisfied. Knowing the stress tensor, once we have the stress fields we can easily compute the strain field (ε_j, ε_t).

Focus on the problem of Pure Torsion

DSV hypotheses (G_x, G_y, Z_xy) are zero; then M = only torque → Σ_i, Σ_ip = 0.

To have an analytical solution, we have to consider what happens to Σ_zx, Σ_yp. We take into account the last equation of the independent equation system m, which is → equilibrium equation in A, and we manipulate the two remaining compatibility equations m which E + E_y - σ_y = 0. They can be summarized in a single differential equation.

From boundary conditions on Γ we obtain: equilibrium on Γ. We can obtain Σ_zx, Σ_yp by solving the linear differential problem of 2 equations and 2 unknowns.

Equilibrium in A

Compatibility in A + boundary condition in strong form on Γ + boundary condition in weak form.

Next, compute CONSTANT Ccos: it is a well-posed problem that admits a unique solution. Applying the constitutive law:

χ_xx=⟨χ_xx⟩_C=Gβ∂⟨y_G⟩_C/∂x∂y

χ_yy=⟨χ_yy⟩_C=Gβ∂⟨y_G⟩_C/∂y²

Let's see now if they are in agreement with governing equations:

Compatibility Equation:

∂²χ_xx/∂x = ∂²y_G/∂y = C⇒ GP ( ∂²φ/∂x∂y + ... + 1) = C

Equilibrium Equation:

∂²χ_xx/∂x + ∂²y_G/∂y = 0⇒ GP ( ∂²χ_xx/∂x² + ∂²χ_yy/∂y² ) = 0

We obtained the well-known LAPLACE OPERATION: ∇²y_G = 0. If I take also the harmonic problem, I can compute the warping function.

Boundary Condition:

m STRONG FORM ⟨χ_xx⟩_m + ⟨χ_yy⟩_m M_y = 0 ⇒ (∂y_G/∂x) M_x + (∂y_G/∂y) M_y = 0 y_G = y_mx - x_my ∇²y_G = 0 ∂y_G/∂m = y_mx - x_my

Neumann-Dirichlet Problem

HR is a boundary condition on a normal derivative. It admits a solution, but the solution is not unique. This is due to the fact that the boundary condition is expressed in terms of the derivative of something (y_G). So, we can always add a constant to y_G without altering the governing equation. This is because the boundary condition is not affected. We can add to y_G a constant that represents the rigid body rotation. The solution is not unique because of these reasons, BUT the dimensions of φ are unique.

Continue with thin-walled open sections

We have seen STATICS.C_2B = 2 M_t, m / JJ_w = 4/3 ∫ b³ ds. From the stress distribution, we observed that the maximum shear is in correspondence with the boundary for m maximum and m = m₁ ± b₂. When the section is not uniform along the mean line, the maximum tangential stress is where the thickness b is the greatest. τ_c^max = 2 M_t b(s) / J.

Kinematics

An important parameter is β ( M_t / 4GJ )L, which describes the way in which torsional rotation increases along the axis, but it is not sufficient to describe the whole kinematic field. Remember that we defined (if the centroid is the reference point):

S_X = βxy

S_Y = βxz

Once we have computed β, we can obtain S_X, S_Y. To have an approximate solution, we assume γ₀ constant across the thickness of the profile, because the thickness is very small. We focus our attention on the warping function on the mean line only γ₆(s).

For computing the warping function on the mean line, we can consider the following procedure: Try to compute the moment of the warping function along the mean line: dγ₆ / ds. From differential geometry, we can compute it as:

dγ₆ = ∂γ∂x . ∂x∂s + ∂γ∂y . ∂y∂s

We are assuming that the mean line is a PARAMETRICAL curve - the points are modeled through the curve as a function of parameters, and then we have x(s), y(s). If we know the shape of the mean line, we can compute the two terms ∂x/∂s, ∂y/∂s that are functions of s. Remember that the tangent vector to the mean line is given by t_s = d²d¹ (from differential geometry). Remember also that the tangent vector is orthogonal to the normal vector t_n . m.

Angle warping function for a closed profile is given by: Consider an axial path. b(s) ds. From a point and do the whole loop of the parameter, and arrive at a certain point. Compatibility equation. Start from a point and do the whole loop of the parameter, and arrive at a certain point.

OSS: By comparing two found expressions, we can easily compute torsional inertia for a closed section from the basis of geometry.

Summary: