Closed: $q = \dfrac{T}{2A_{enc}}$
$\tau = \dfrac{T}{2A_{enc}\,t}$
Open: $\tau_{max}= \dfrac{T\,t}{\frac{1}{3}\sum b_i t_i^3}$
Apply Bredt-Batho theory to closed (box, circle) and open (C-channel, I-beam) sections. Animated shear flow arrows, max shear stress, twist angle per unit length, and closed-vs-open stiffness comparison — all in real time.
Closed: $q = \dfrac{T}{2A_{enc}}$
$\tau = \dfrac{T}{2A_{enc}\,t}$
Open: $\tau_{max}= \dfrac{T\,t}{\frac{1}{3}\sum b_i t_i^3}$
The governing theory for closed, thin-walled sections under pure torsion is Bredt-Batho theory. It states that the shear flow (shear force per unit length) is constant around the entire closed perimeter. This leads to a simple formula relating applied torque to shear flow.
$$q = \frac{T}{2 A_{enc}}$$$q$ is the constant shear flow (Force/Length, e.g., N/mm).
$T$ is the applied torque (Force·Length, e.g., N·m).
$A_{enc}$ is the area enclosed by the median line of the wall section.
From the constant shear flow, we can find the shear stress in each wall segment. The stress depends on the local wall thickness. For open sections (like a C-channel), a different formula applies, where stress is proportional to thickness and torque, and inversely proportional to the torsional constant.
$$\tau_{closed}= \frac{q}{t}= \frac{T}{2 A_{enc} t}\quad \quad \tau_{open,max}= \frac{T \cdot t}{J}$$$\tau$ is the shear stress (Force/Area, e.g., MPa).
$t$ is the local wall thickness.
$J$ is the torsional constant. For an open section, $J \approx \frac{1}{3} \sum b_i t_i^3$, where $b_i$ and $t_i$ are the length and thickness of each wall segment.
Aerospace Fuselages & Wings: Aircraft fuselages are classic thin-walled closed sections (monocoque construction). Engineers use Bredt-Batho theory to ensure the skin and frames can handle torsional loads from maneuvers or uneven lift, optimizing material use for minimum weight.
Automotive Chassis & Roll Cages: The cabin of a car or the tubular frame of a race car roll cage forms a closed box section. Analyzing shear flow helps engineers place reinforcements and understand how torque from the wheels is transmitted through the structure, impacting handling and safety.
Bridge Box Girders: Many modern bridges use large, thin-walled steel or concrete box girders. These closed sections provide excellent torsional stiffness to resist wind loads and eccentric traffic loads, preventing unwanted twisting of the deck.
Industrial Conveyor & Machinery Frames: Long, thin-walled beams that support rotating machinery are subject to torsion. Choosing between an open I-beam (poor in torsion) or a closed tubular section (excellent in torsion) is a direct application of this theory to prevent excessive twist and vibration.
When you start using this tool, there are a few points that beginners in CAE often stumble over. First is the definition of "thin-walled". As the tool's name implies, the Bredt-Batho theory assumes "thin-walled" sections. A good rule of thumb is a plate thickness less than 1/10th of the section's representative dimension (e.g., the side length of a box section). For example, using a 15mm plate thickness for a 100mm wide box section pushes it into the "moderately thick" or thicker category, where the theory's accuracy decreases. In practice, this is an area requiring detailed verification with FEA.
Next is input errors for material constants. The shear modulus G varies significantly by material—around 80 GPa for steel and 27 GPa for aluminum. Getting this wrong will yield a calculated twist angle far from reality. For instance, inputting steel's G value for an aluminum member would lead you to mistakenly judge it as three times stiffer (with a smaller twist angle) than it actually is.
Finally, the pitfall of "closed sections". While selecting a box section in the tool might seem strong, in real structures, welded seams or bolt holes can become weak points. Even if theoretically closed, if the seam cannot fully transmit shear forces, it effectively becomes an "open section," potentially leading to failure much earlier than the calculated value. The mark of a professional is not blindly trusting simulation results but constantly being mindful of the actual load path through real joints.
Rectangular steel box section: width 100 mm, height 80 mm, wall thickness 3 mm, shear force V = 15 kN. Enclosed area A_enc = 100 × 80 = 8000 mm². Second moment I ≈ 5.1 × 10⁶ mm⁴. Shear flow q = (15000 × 8000)/(5.1 × 10⁶ × 3) ≈ 7.8 N/mm. Maximum shear stress τ = 7.8/3 ≈ 2.6 MPa (well below 250 MPa yield). For applied torque 50 kN·mm with G = 80 GPa and J ≈ 1.15 × 10⁶ mm⁴, twist angle θ/L = 50000/(80000 × 1.15 × 10⁶) ≈ 0.00054 rad/m.