Compute the shear center of channel, Z, T, angle and I sections in real time. Visualise the shear flow distribution q = VQ/I and understand intuitively how torsion arises.
Section Presets
Dimensions
Flange width b (mm)
mm
Web height h (mm)
mm
Flange thickness tf (mm)
mm
Web thickness tw (mm)
mm
Live Values
--
e: SC offset (mm)
--
I_z (mm⁴ ×10⁴)
--
Load eccentricity d (mm)
--
Torque T=V·d (N·m)
--
Twist angle φ (schematic)
--
q_max (N/mm @ V=1kN)
Apply the Load — No Twist When It Passes Through the Shear Center
Load position d (distance from the shear center)0 mm
Shear flow q(s) distribution (V = 1 kN)
Shear Flow q Distribution
Theory & Key Formulas
Second moment of area:
$$I_z \approx \frac{t_w h^3}{12} + 2 b t_f \left(\frac{h}{2}\right)^2$$
Shear-center offset (C-section: distance from the web centerline; 0 for Z and I; for T and L the distance between the centroid and the intersection of the leg centerlines):
$$e = \frac{b^2 h^2 t_f}{4 I_z}$$
Shear flow:
$$q(s) = \frac{V \cdot Q(s)}{I_z}$$
Torque of the applied load (eccentricity d from the shear center):
$$T = V \cdot d \qquad (d = 0 \Rightarrow \text{no twist})$$
What Is the Shear Center Calculator?
🙋
What is the shear center? It is described as "the point where no twist occurs", but how is it different from the centroid?
🎓
Roughly speaking, it is the line of action a sideways force must follow so that the section "deforms straight without rotating". Try it with the C-section (channel) in this tool: increase "Flange width b" at the top. Can you see the shear center (red SC) moving further and further outside the web? The centroid (green G) moves the other way, towards the flanges, so the gap between the two keeps widening. At the default b=60 the SC is 23.7 mm outside the web centerline; at b=100 it is 43.1 mm. That is why a force applied at the centroid produces an unintended twist.
🙋
Really? Then what do the "shear flow" arrows drawn on the figure represent?
🎓
Those arrows show the magnitude and direction of the "flow of shear force" running through the walls of the section, the $q$ in the formula $q = VQ/I$. Change the web thickness $t_w$ and you should see the density of the flow in the web change. In practice, the position of the shear center is found by integrating this flow distribution.
🙋
I see! But how is this used in real design? When does a gap between the centroid and the shear center cause trouble?
🎓
A very important point. For example, when a load is hung from a crane girder (a channel section) in a factory, getting the load path wrong makes the girder "wring" unexpectedly, causing deformation and vibration. Select the "Angle (L)" section in this tool and move the parameters. Can you see that the shear center (SC) sits at the corner point where the centerlines of the two legs meet, while the centroid (G) lies inside, away from the corner? When applying forces to such members you must keep this point in mind.
Frequently Asked Questions
The shear center is the point at which a shear force produces no torsion. For an asymmetric section the resultant of the shear flow does not pass through the centroid, so the shear center is offset from it. The offset is especially large for C and L sections, and to avoid torsion the load must act through the shear center.
Aligning the support points or load application points of beams and columns with the shear center suppresses torsional deformation. When a channel is used as a beam, for example, apply the load at the distance e from the web centerline, or restrain the twist with bracing or stiffeners.
It gives an intuitive picture of the distribution and direction of shear stress within the section. The more asymmetrically the shear flow runs, the further the shear center lies from the centroid and the more readily torsion occurs. The visualisation shows which flange or web carries concentrated stress, which helps when deciding where to reinforce.
For an ordinary symmetric I-section the shear center coincides with the centroid, but for an asymmetric I-section whose top and bottom flanges differ in width or thickness it does not. The I-section in this tool covers only the symmetric case with identical flanges and shows e=0. For an asymmetric I-section the shear center divides the depth in proportion to the second moments of area of the two flanges, I₁ and I₂ (e₁ = h·I₂/(I₁+I₂) from the top flange centerline), so compute it separately.
Real-World Applications
Building and civil structures: Light-gauge Z and C sections (lipped channels) are widely used as roof and wall purlins and girts. When wind or snow applies transverse loads to these members, unless they are supported and connected so that the load passes through the shear center, torsional vibration causes local buckling and noise.
Aircraft and spacecraft structures: Thin-walled open sections made of skin and stringers carry large shear forces such as wing lift. Knowing the shear center accurately and designing the load path accordingly is crucial to prevent unintended wing twist (such as aileron reversal) and ensure flight stability.
Cranes and conveyors: Crane girders in factories and warehouses are typically thin-walled I or box sections. If the travelling trolley (the load-carrying device) applies the load away from the shear center, the girder wrings, impeding smooth travel and shortening fatigue life.
Automotive bodies: Even for closed-section members such as the side sills and pillars of a monocoque body, the shear-flow concept matters when considering the transmission path of crash loads. CAE analyses evaluate the relationship between the shear center of these members and the actual load input points to optimise the torsional stiffness of the body.
Common Misconceptions and Points to Note
First, grasp that "the shear center is independent of material properties and Young's modulus". Notice that although you change thickness and width in this tool, there is no material input. The shear center is a purely "geometric property" determined by the shape and dimensions of the section alone, so steel and aluminium with the same section shape have the same shear center. Conversely, "confusing the centroid with the shear center" is the biggest pitfall. Play with the L-section in the tool: the centroid lies inside, away from the corner, while the shear center always stays at the corner point where the centerlines of the two legs meet (for L60×60×6 the two are 20.1 mm apart). Apply the load at the centroid and twisting appears at once. In practice, supports and load points must be designed to coincide with this corner point.
One more thing: pay attention to the realism of the parameter settings. For instance, try setting the flange thickness \(t_f\) extremely thin compared with the web thickness \(t_w\) (say 1/10 or less). Does the shear flow distribution look odd? That is because the thin-wall assumption, "the thickness is small compared with the other dimensions", breaks down and the accuracy of the formulas themselves drops. In real design it is common practice to keep the thickness ratio within a certain range (roughly 1/2 to 2).
Choose a section preset (C, Z, T, L or I). Typical dimensions are filled in when you select one
Enter the flange width b (20–150 mm), web height h (40–300 mm) and the flange and web thicknesses tf, tw (2–20 mm) with the sliders or numeric fields. Values update as you type
The shear force is fixed at V = 1 kN; the shear flow distribution q = VQ/I (N/mm) and q_max are displayed. Moving the "Load position d" slider shifts the load line and changes the torque T = V·d; at d = 0 ("Through shear center") the twist disappears
The red SC marker is the shear center and the green G marker the centroid. For the C-section e is the distance from the web centerline to the shear center, for T and L the distance from the centroid to the shear center, and for Z and I it is 0
Worked Example
Example: for a channel with b=80 mm, h=100 mm, tf=6 mm, tw=5 mm and V=1 kN, Iz=2.82×10^6 mm^4, e=34.1 mm, Qmax=30,250 mm^3 and qmax=10.7 N/mm.
Notes
Check case: for L60×60×6 the shear center is at the intersection of the leg centerlines SC=(3.0, 57.0) mm, the centroid is G=(17.2, 42.8) mm and the offset is 20.1 mm. For the T-section the SC is shown near the intersection of the web and flange centerlines. The shear force is fixed at V = 1 kN and the twist angle φ is a schematic display without material or length. The L-section is asymmetric with inclined principal axes, so q = VQ/I is an approximation about the horizontal axis.