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For me, it's a bit hard to imagine where shear center can be, and what purpose it serves. The definition I found online is very vague:

Shear center is defined as the point on the beam section where load is applied and no twisting is produced.

Given the general profile of a section, is there a proper mathematical derivation/formula that shows how the shear center is defined/derived?

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  • $\begingroup$ Explanation here. It's a bit complicated and I don't have the time to post a proper answer now, but this should get you started. $\endgroup$
    – grfrazee
    Dec 30, 2015 at 13:59
  • $\begingroup$ I also found this explanation helpful $\endgroup$
    – CableStay
    Dec 30, 2015 at 21:11
  • $\begingroup$ It's all about the integrals :-) $\endgroup$ Dec 30, 2015 at 21:19

2 Answers 2

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SHEAR CENTER

Why do we care?

Because we need to know if the beam is subjected to torsion in addition to flexure.

What is it?

The point in the cross section (or outside the cross section) where we can apply load and produce beam bending without twisting. Loads applied anywhere other than the shear center will produce both bending (moment) and twisting (torsion).

Generally when we think about a beam in bending, we only think about, well...the bending. But depending on how that load is applied to the cross section (i.e. is the load through the shear center), there may also be twisting.

I-Beam Deflections

Here are shear center locations for some typical structural sections (from AISC Design Guide 9).

Structural Section Shear Centers

Note that for doubly symmetric sections, the shear center is coincident with the centroid. In other words, the shear center is at the intersection of the axes of symmetry. In singly symmetric sections, the shear center lies somewhere along the axis of symmetry. Lucky for us, in typical civil engineering applications we very often deal with symmetric sections. (Aero folks get saddled with the harder math.)

Intuitive Approach to Estimating Shear Center

For starters, you can probably look at the two I-beam load cases above and intuitively sense that applying a load offset from the centerline will produce beam twisting.

A more quantifiable procedure is to visualize the shear flows and estimate the point about which the net moment produced by the shear flow will be zero.

Shear Flows

For the Mathematically Inclined

Well, I could go bonanzas with the LaTex (and I may do so a bit later), but since my lunch break is coming to an end, I'll refer you to this explanation.

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    $\begingroup$ The link is broken $\endgroup$
    – Graviton
    Jan 13, 2018 at 13:12
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I hope answering this after all this time is not a huge faux pass, but this question comes up on search engines when looking for shear center calculations.

First, some preliminaries.

  • This applies to simply connected cross sections, like channels, I-shapes, L-shapes, etc., but not things like pipes or HSS boxes. -The cross section is defined by the $y$ and $z$ coordinates, with the $x$ coordinate being the axis of the element.

With this out of the way, the following picture illustrates the set-up for this. Cross section centered on centroid

In the following I refer to the cross sectional domain as $\Omega$ and its boundary as $\partial\Omega$. The first step is to solve for the auxiliary warping function $\omega_{0}$, by solving

$$ \begin{aligned} div(grad(\omega_{0}))&=0& \qquad &\textrm{in } \Omega \\ grad(\omega_{0})\cdot n&=\frac{1}{2} \frac{d}{ds}(y^2+z^2)& \qquad &\textrm{on } \partial\Omega \end{aligned} $$

This is a Neumann problem, so it's only solvable if $$ \oint_{\partial\Omega} \frac{d}{ds}(y^2+z^2)\,ds=0, $$ luckily this is automatic in our case, as this expression reduces to the expression inside the derivative evaluated at the starting point and endpoint of the path. However, since the path is a closed loop this is automatically 0. All Neumann problems are solved uniquely up to a constant, so fix any single point to an arbitrary value (usually a corner to 0) and you are golden.

Now that you have $\omega_{0}$ the coordinates of the shear center are given by $$ y_s=-\frac{I_zI_{y\omega}-I_{yz}I_{z\omega}}{I_zI_y-I^2_{yz}} \qquad \qquad z_s=\frac{I_yI_{z\omega}-I_{yz}I_{z\omega}}{I_zI_y-I^2_{yz}} $$ where $I_y$, $I_z$ and $I_{yz}$ are the second moments of area (usually called moments of inertia), given by $$ I_y=\int_{\Omega}z^2\,dydz \quad I_z=\int_{\Omega}y^2\,dydz \quad I_{yz}=\int_{\Omega}yz\,dydz, $$ and $$ I_{y\omega}=\int_{\Omega}z\omega_{0}\,dydz \quad I_{z\omega}=\int_{\Omega}y\omega_{0}\,dydz $$

I derived this from an old book I had, but you can find an in-depth explanation and derivation in this thesis, the shear center expressions are eqs (2.98) and (2.99) but the whole chapter 2 is worth a read.

As an illustration, some examples. The following pictures show the values of $\omega_0$ (in the colorbar, units are not important) and the $y-z$ dimensions (in mm) for different simply connected shapes.

Centroid and shear center for I beam Here's the solution for an I-beam. As expected, the centroid (circle) and shear center (cross) are coincident.

Centroid and shear center for an L-shape Next we have an uneven L-shape cross section. The centroid is somewhere outside the cross-section, while the shear center, as you'll usually find in the literature, is at the intersection of both legs.

Centroid and shear center for a channel Finally we have a channel (or C shape or U shape). As expected, the centroid is somewhere to the right of the web (between the flanges) while the shear center is to the left of the web (away from the flanges).

As a bonus of doing all this, $\omega_{0}$ lets you get the torsional properties of the cross section for almost no extra work. First you construct the warping function from the auxiliary warping function by doing $$ \omega=\omega_0-z_sy+y_sz, $$ then the warping constant $C_{\omega}$ is given by $$ C_{\omega}=\int_{\Omega}\omega^2\,dydz $$

And the torsional constant of the cross section, $J$ is given by $$ J=I_s-W_0 \\ I_s=\int_{\Omega}(y-y_s)^2+(z-z_s)^2\, dydz \qquad W_0=\int_{\Omega}\left(\frac{\partial\omega}{\partial y}\right)^2+\left(\frac{\partial\omega}{\partial z}\right)^2 dydz $$

PS: Any tips on how to better format images are welcome.

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