Many conventional structural shapes are very vulnerable to torsional loading. In doubly-symmetric shapes, like W-shapes, we try to keep imposed loads acting through the centroid. Shear loads that act eccentrically produce torsion in the member.
The shear center is the point on a cross section to which any shear load may be applied without causing torsion in the section. In doubly-symmetric sections, the shear center and centroid are the same.
Finding the shear center by hand can be tricky business, but a few tips can help along the way. Any section with all linear flanges converging on one point will have the shear center in that point. Clever selection of coordinate systems and careful consideration of which point to take moments about can also help.
When dealing with more complex shapes, it becomes necessary to resort to computer tools, or make simplifying assumptions, like the thin-sections assumption demonstrated in Example 3.
Worksheet and Solutions Download
In the videos below I introduce the concept of a shear center, then work through three example problems. Grab the worksheet packet ahead of time, and fill it in as you watch the videos, or just give them a try on your own, then check yourself against the answer key below.
Printable PDF Shear Center Worksheet Packet Download
Printable PDF Shear Center Answer Key Download
Video Option: Explanation Video & Three Worked Examples
If you prefer video to article format, check out the playlist below, in which I introduce the concept of the shear center, then solve the three increasingly difficult problems available in the printable handouts above.
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Background Mechanics
The shear center is all about equilibrium, informed by symmetry and our knowledge that the shear stress at any free boundary must be zero.
Continuity tells us that at any corner of a cross-section, the magnitude of the shear stress on one side of the joint must be equal to the magnitude of the shear stress on the other side of the joint.
Using these pieces of information, finding the shear center becomes a statics problem of equilibrium, informed by the knowledge that the resultant shear force of each section is the integral of the shear stress over the area, and summing moments about an aptly-chosen point.
The resultant shear force over any area is:
And tau is known from introductory Mechanics of Materials courses to be:
Where:
- V is the applied shear
- Q is the “first moment of area” (Area of the portion of the section beyond the cut in question times the distance from the centroid of the overall shape to the centroid of same area)
- I is the moment of inertia of the overall section
- t is the thickness of the flange at the location in question
Resultant forces are computed for each portion of the section, and moments summed about some convenient point to solve for the eccentricity at which the applied shear force must be imposed to create moment equilibrium, thus avoiding twist on the beam.
Taking Advantage of Symmetry
Any section that is at least doubly-symmetric will have a shear center located at the intersection of the axes of symmetry, which is also where the centroid of such shapes can be found.
Shapes that feature only one axis of symmetry still have half as much work as asymmetric shapes, as the shear center will necessarily be found on that axis of symmetry. This leaves only one coordinate to solve for.
Choosing an Easy Moment Summation Point
The resultant force from each rectangular flange making up a built-up shape acts along the axis of that shape. Choosing as a moment summation point the intersection of as many flanges as possible minimizes the number of resultant shear forces that must be computed.
For example, in the second worked problem (a C-shape with asymmetric flanges) finding the x-coordinate of the shear center is made easy by selecting as the moment summation point the intersection of the web and lower flange.
When Hand Solutions are Difficult
In a section composed of rectangular areas, the shear stress distribution in elements oriented normal to the external applied shear force varies linearly, while the stress in parallel elements varies quadratically.
Integrating that shear force over the perpendicular elements is easily done, and the quadratic makes it only slightly more difficult. The real difficulty in integration arises when the parallel elements have perpendicular elements attached outboard of the centroid.
The quadratic stress distribution mixes in with truly ugly expressions for the first moment of area Q, caused by trying to work a weighted average with variables to find the centroid-to-centroid distance needed in Q. While typically solvable, the process becomes so long and arduous as to lend itself to human error in calculation.
Thin Sections Approximation
When I was introduced to the shear center, thin sections were the only sort we investigated, and it left me very confused as to the underlying principles.
Much of the mathematical complexity of computing shear centers can be eliminated by use of the “thin sections approximation”, wherein it is assumed that the thickness of the member is negligible compared with the other dimensions of the section.
Just as in Euler-Bernoulli Beam Theory’s small angle assumption, the thin sections assumption lets us do away with higher-order terms, simplifying computation tremendously. If thickness t is small, then the thickness squared or cubed must be negligible. Higher-order terms crop up in both the overall section moment of inertia, I, and first moment of area, Q. Neglecting these terms results in some error (small for thin sections) but greatly simplifies the math.
Most appropriate for computation of shear centers in things like light-gauge steel studs, even here an appreciable amount of error is produced. The results wouldn’t be something I would rely on for design, but they can help a student to build intuition as to the probable location of the shear center in unfamiliar shapes.
See Example 3 in the PDF worksheet download above, or the Example 3 video of the Shear Center playlist embedded above for a full demonstration of the application of this simplifying assumption.
Computer Tools
For actual design based on the location of the shear center, such as figuring torsional loads on C-shapes with the load applied to their flanges, I would strongly recommend the use of a section analysis tool. My office uses ShapeBuilder from IES, though the well-known structural analysis tool company RISA has also begun marketing their own RISASection tool.
Both of those section analysis tools take AutoCAD .dxf files as inputs and numerically solve for various section properties, many more than the built-in AutoCAD utility offers.
If trying to solve the more complex shapes by hand, try using WolframAlpha to do the heavy lifting for you, or at least to check your work. I lean on Wolfram for checking any long derivation I carry out, and it proved exceptionally useful a few months back as I was deriving expressions for tension-only panels under lateral loading.
Conclusion
Shear centers can be tricky things to compute by hand, but working at least a few can give students the intuition to understand when a computer tool is kicking out a bad result.
If you haven’t already, download the Printable PDF Shear Center Worksheet Packet and give it a go on your own, then check your work against the Printable PDF Shear Center Answer Key.
If you have any questions, topics you’d like to see covered, or other feedback, please comment below or email me at eric@structed.org.
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