Displacement Calculations – Various Load Cases (3)

Welcome to the third part of the article on displacement calculations in beams. In this section, I will cover everything you need to know about calculating displacements, deflections, and angles of deflection (rotation), as well as twist angles.

The article will be somewhat theoretical—but I guarantee that it will clarify certain issues before you tackle the tasks. In this part of the article,

I will cover the following topics:

• Displacements in tensile bars
• Twist angle in freely twisting bars
• Przemieszczenia przy zginaniu prostym

Displacements in tensile bars – these forces are kind of “abnormal”…

Actually, if you’ve read the previous part of the article, there’s not much more I can add here. The equations for calculating displacements (extension/contraction) in bars have already been derived in detail. I’ll just briefly review them to refresh your memory.

We usually calculate displacements based on two approaches:

• using differential equations derived from fundamental physical or mechanical relationships:

u'(x) = ε_x → u' (x)={N(x)\over E(x)⋅A(x)}→u(x)= ∫{N(x)\over E(x)⋅A(x)} dx

• Using equations describing the elastic energy of the system:

u=∫_L{N⋅N_F\over{EA}}dx

where: N – normal forces, A – cross-sectional area,  E – Young’s modulus of the material,  N_F– normal forces caused by a unit force applied at the point where the displacement is calculated.

In the first case, the result is a displacement function that allows you to calculate displacements at any point along the bar. However, when using the formula, be mindful of whether the quantities in the equation are constant (which simplifies the work) or functions that must actually be integrated.

If you use general formulas, remember to include the integration constants and calculate them from the boundary conditions (this is how EquiBeam calculates in the general method).

If the system’s parameters are constant, you can directly calculate the total extension using the formula:

ΔL={NL\over EA}

If you are analyzing a system where the quantities change, you can divide it into smaller segments. Then the total extension is calculated as the sum of the extensions from each segment.

In the second case, the calculation results in the displacement only at the point where we introduce a fictitious (virtual) load. Additionally, using energy methods requires calculating the normal forces—for both the actual load and the fictitious load (unless you use Castigliano’s method, which handles this in one step).

As in the first method, if the system consists of multiple segments where the internal force functions change, the integral is broken down into a sum of integrals over all segments. These integrals can be solved either analytically (don’t worry—most cases involve simple polynomial integrals) or using graphical methods, which are easier for some and mind-bending for others (I’ll probably write an article about that one day ????).

As you can see, calculating displacements for axial tension/compression is not too complicated—solving a few problems in the EquiBeam application will likely clear up any doubts.

Let’s move on to the second strength case: free torsion.

Twist angle in freely twisting bars – tricky shafts…

Free torsion refers to a type of twisting where cross-sectional warping can occur.

This means that points on the cross-section can move in the direction parallel to the bar’s axis during twisting. This behavior is natural for twisted bars whose cross-sections are not “blocked,” and the theory describing this phenomenon was formulated by Saint-Venant.

As a curiosity, warping does not occur in circular cross-sections (or round tubes), which is why torsion calculations for these cross-sections are quite “pleasant.”

If someone wants to analyze constrained torsion in bars, they will have to deal with bimoments, bending-torsional moments, and strange additional stresses in the bar (I wouldn’t recommend it in general…).

Although torsion is a completely different case from tension, all the formulas for calculating the twist angle are exactly analogous—we again have two basic methods (I won’t describe them in detail), leading to two types of formulas:

• the differential equation:

θ' (x)={M_s(x)\over G(x)⋅J(x)}→θ(x)= ∫{M_s(x)\over G(x)⋅J(x)} dx

• the elastic energy-based equation:

θ=∫_L{M_s⋅M_{sK}\over{GJ}}dx

where:  M_s– twisting moments, J – torsional stiffness of the cross-section,  G – shear modulus (Kirchhoff’s modulus) of the material, M_sK – twisting moments caused by a unit moment applied at the point where the twist angle is calculated.

As you can see, the formulas for torsion analysis are essentially the same. What changes?

Normal forces N are replaced by twisting moments M_s​, Young’s modulus E is replaced by Kirchhoff’s modulus G, and the cross-sectional area A is replaced by the torsional stiffness J—and this is where most of the complications arise…

While we calculate the cross-sectional area in elementary school, the torsional stiffness coefficient can only be easily calculated for circular cross-sections (and round tubes), as it equals their polar moments of inertia:

• for a circular bar:

J=I_O={πD^4\over32}

• for a round tube:

J=I_O={π(D^4-d^4)\over32}

For all other cross-sections, the calculations are either “more” complicated or even impossible to perform analytically.

Typical cross-sections where such calculations are performed include rectangular cross-sections (there are tables for this), thin-walled open sections (such as I-beams), or closed sections (such as square tubes). For each of these cross-sections, the torsional stiffness is calculated differently, and the stresses are calculated differently as well (I’ll write an article on this someday ????).

For engineering purposes, these calculations are usually done numerically (e.g., using FEA)—and note that EquiBeam can calculate these values both analytically and numerically for predefined cross-sections!

Returning to the calculation of the twist angle—the methodology is the same as for tensile systems, so I won’t elaborate…

An important point to emphasize is that (again…) if the parameters are constant, you can quickly calculate the twist angle over a given section of the bar using the formula:

Δθ={M_sL \over GJ}

That’s about it for twisted bars: check out EquiBeam for examples. Now, let’s move on to a slightly different strength case—bending.

Displacements in simple bending – it gets worse with curves…

To start, we should say that, as in the previous strength cases, the calculation methods for bending will be divided into the same two basic groups.

In the energy method group, nothing really changes. The forms of the equations are almost the same, except that the set of internal forces, cross-sectional parameters, and material properties change. In general, they will look like this:

y=∫_L{M_g ⋅M_{gF} \over{EI}} dx  + ∫_L{T ⋅T_F \over{kGA}} dx,

φ=∫_L{M_g ⋅M_{gK} \over{EI}} dx  + ∫_L{T ⋅T_K \over{kGA}} dx

gdzie: M_g – bending moments, T – shear forces, I – central moment of inertia of the cross-section, k – shear stiffness modifier for the cross-section. The other data has already been explained earlier.

As you can see—if we want to calculate the deflection yyy, the formulas analyze the influence of a concentrated (unit) force applied at the point of displacement analysis; if we want to calculate the angle of deflection, we analyze the influence of a concentrated (unit) moment applied at that point.

Additionally—often in analysis (especially for long and slender beams), the integrals related to shear forces are not considered—they have little effect on deflection and stiffness. However, don’t be fooled—shear can have a significant impact on results for short beams—reactions, internal forces, etc. For now, I won’t go into the details of bending analysis using energy methods—I’ll cover those in the next part of the article, where I’ll explain their modifications in each method.

I think it’ll be more interesting to now present the derivation of the differential equation for the deflected axis—especially since scientists really did an impressive job here

The beginning of this derivation can be based on assumptions derived (again) from constitutive equations, geometric conditions, and equilibrium equations (I’ll skip this part—it has too many derivatives and integrals…).

After performing these operations, we arrive at a very pleasant equation:

{M_g \over EJ}={1 \over ρ}

where  ρ is the radius of curvature of the deflected beam at a given point

And now the fun begins, because mathematicians created a great formula for the radius of curvature containing quite a few nice derivatives, which looks something like this:

ρ= {\big(1+y^{'2} (x)\big)^{3/2} \over |y^{''} (x)|}

If we substitute this monster into the deflection equation, we get the relationship:

{M_g \over EJ}={|y^{''} (x)| \over \big(1+y^{'2} (x)\big)^{3/2}}

Now comes the climax—you can’t solve the above equation easily, so I suspect that the physicists had something like this conversation:

Thanks to this conversation, they proposed the differential equation for the deflected axis in the form:

y'' (x)= \stackrel{+}{-}{M_g(x)\over EJ}→y (x) =  \stackrel{+}{-}∫\bigg(∫{M_g(x)\over EJ} dx\bigg)dx

The plus and minus signs essentially depend on the assumed direction of the vertical axis.

Now, if we know the bending moment functions, we can try to integrate this equation twice, yielding the deflection function with two constants of integration (which can be calculated from the beam’s support conditions—i.e., boundary conditions).

A final curiosity concerns the determination of the angle of deflection. We can infer its value from the above illustration. It shows that the tangent of the deflection angle equals the derivative of the deflection function.

But these angles are small, aren’t they? So, let’s do a little trick:

\tgφ(x) = y'(x) → \tg⁡φ(x) ≅ φ(x) → φ(x) = y'(x)

It turns out that once we have the deflection, all we need is to differentiate it to get the deflection angles…

Hmm… Didn’t we just assume earlier, to derive the differential equation for the deflected axis, that they must be equal to zero for everything to check out? Never mind… Who cares, since it matches the experiment quite well ????.

So—important info—the above equations describe systems characterized by small displacements. They are also simplified by excluding displacements caused by shear forces—these are the so-called Euler-Bernoulli beams. The methods for calculating displacements using these formulas (e.g., the general method—differential equation of the deflected axis, or Clebsch’s method) will be presented in future articles ????.

OK, to summarize the second part of the article—for different strength cases, we use different material properties, cross-sectional parameters, or internal forces, which affect how we perform displacement analyses. The calculations themselves are actually done in a “fairly” similar way (especially in energy methods), and the main difference—or problem—is appropriately formulating the equations or drawing the graphs of the relevant internal forces.

In future articles, I will focus on methods for solving displacements in bending beams, which usually have the most complex procedure (due to the analysis of the bending moment), although some of them can also be used for tensile or torsional systems.