cdm.ing.unimore.it

Hi, this is a discussion board. Once logged in, you may post here questions about the obscure lesson points that need clarification. The teacher will treat those point in dedicated addenda.

Q&A of a student with respect to the beam theory, if it may be useful to someone else.

Q: How does the user decide the orientation of vector v, which is used to define the local axis system?

A: It depends on the specific kind of profile you want to represent. Consider the following example. We have an elliptic, thin walled cross section, and, in the cross sectional view, the first local axis 1 is aligned with the minor ellipse semi-axis.

Now, in a global xyz reference system, the profile axis is aligned with the global x axis. If we set v=(0,0,-1), the profile cross sections are oriented such that their minor semi-axis is parallel to the global z, whereas their major semi-axis is aligned with the global y. If we set, instead, v=(0,1,0) the cross sections are globally rotated in space by 90° around the profile axis with respect to the previous configuration.

Q: What does the force F and I represent , why are they called auxiliary/fictitious ?

A: Since the Castigliano second theorem allows the deflection evaluation only in correspondence of an external force, an external force has to be added at points whose deflection is under scrutiny, even if the original load case does not consider such a load.

Obviously, the magnitude of such a spuriously added force has to be imposed zero.

Those forces, which were added just to perform the deflection calculations, are hence named “auxiliary”, and are “fictitious”, or “spurious” in nature.

Questions of a student with respect to the ladder frame chassis maxima worksheet, and answers, if they may be useful to someone else.

What does the term compliance mean technically?

In general, it's the inverse of the stiffness. It is usually defined as the magnitude ratio between a deflection and the applied external action that induces it.

Why do you specifically decide to choose W_a as parametric , Can you keep W_d as parametric ?

It was an arbitrary choice. A few worksheet versions ago ThetaD was the internal action component kept parametric. Such a choice might only affect the complexity of the algebraic expressions, think e.g. how would they appear if the parametric internal action component were the vertical shear force at the midspan of the modeled profile portion.

In line %i12 (linsolve command) , why don’t you make the equation equal to zero ? Like command %i13,%i14

The linsolve command treats its first argument as an equation; in the absence of an explicit “=” sign, the supplied expression is assumed to be equated to zero. You may think that a “=0” suffix is appended in the absence of an “=” sign.

In Line %i24 , can you tell me what is the significance of chi ?

Those (chi1, chi2, chi12) terms are corrective factors for the shear stress/strain contribution to the internal energy lineic density. They account for the deviation between the pointwise shear stress value and its [shear force]/[cross. sec. area] average. In the last two live streaming lessons we evaluated those factors for the “almost closed” OTW profile analyzed in torsion.

Why haven't we defined the “F” (Force applied ) in our code? ( I think it's because , we are dealing with ratios)

You're right. Such a term is simplified away in the “ts” definition at %i21.

Result Analysis in which we make alpha components zero:
In this , I'm assuming that alpha_axial is not taken into consideration as N=0, correct?

Yes. The “N” internal action component is symmetric with respect to the plane the structure lies on, and it is zero for purely skew-symmetrical load cases.

I actually forgot the concept of how shear stresses can cause bending , can you give me an intuitive example please?

Shear stresses and the first variation of the bending moment along the beam axis are both caused by the presence of a nonzero shear force. On the pointwise scale, the equilibrium equation for the infinitesimal cube also relates the first variation in z of the sigma_z stresses with the first variation in y (or, x) of the tau_yz (or, tau_zx) stresses. The (arm-weighted) accumulation of the sigma_z stresses along the cross section generates the bending moment, and the accumulation of the tau_yz (tau_xz) stresses generates the Q_y (Q_x) shear forces. Nonuniform bending is observable in straight beam in (and only in) the presence of nonzero shear force.

How are the bending and shear stiffness connected to each other and why can't they act alone ? (My intuition tries to tell me that , if there is no shear stiffness ,it becomes path of least resistance and thus the whole force acts through this weak link not providing any torsional rigidity, something like a paper)

The same vertical shift “2f” of the two profile ends may be obtained both through an S shaped purely flexural deformation, e.g. with a transverse deflection v(z)=-(f*z*(z^2-3*a^2))/(2*a^3), z in [-a,+a] (try to plot it), or through a constant slope, card deck like shift of the cross sections, which is due to pure shear. The least resistance path between those two options is actually followed.

Can you give me an intuitive explanation why the results of line i%48 and i%52 are like that? (For i%48 , I'm thinking that, because the shear stiffness is soo much , it resists the bending of the structure and thus the overall increase)

sorry but I have so many slightly different versions of the worksheet that I lost which passages are actually numbered %i48 and %i52 in yours. In general, if one of the sources of compliance – namely, the torsional compliance, the bending compliance, the shear compliance – is amplified in magnitude, or inhibited, the overall compliance follows such a variation according to a scaling factor which is in the between 0 and 1; such a factor may be discussed based on the proposed “spring in series / in parallel” sketch, and on the relative weigth of the single sources of compliance along the various elastic support paths.