Theory of structures

THEORY

OF

STRUCTURES

Theory of Structures

Revision of De Saint Venant's Problem 14-09-20

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

Def: 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 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 solve 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 stress tensor, once we have the stress fields we can easily compute the strain field (ε_j, ε_t).

Now, we want to 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 last equation of 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 summarised in a single differential equation

From boundary conditions on Γ we obtain: equilibrium on Γ

We can obtain Σ_zx, Σ_yp 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 it to compute CONSTANT C

cos: 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 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 due to the fact that boundary condition is not affected.

We can add to y_G a constant that is that 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 / J

J_w = 4/3 ∫ b³ ds

From the stress distribution we observed that the maximum shear is in correspondence of the boundary for m maximum and m = m₁ ± b₂.

When 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

For KINEMATICS an important parameter is β ( M_t / 4GJ )

L 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 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 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 modelled 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 tangent vector is orthogonal to 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 to the certain point.

compatibility equation.

Start from a point and do the whole loop of the parameter, and arrive to the certain point.

OSS:

By comparing two found expression of

we can easily compute

Torsional inertia for a closed section

from the basis of geometry

Summary: