Torsion & twist calculator

Only for the circle and circular tube are the two equal, because these sections do not warp under torsion. For all other sections (rectangle, open and closed profiles) the I_t that governs the twist is smaller than I_p. Using I_p for a rectangle or hollow section significantly overestimates the stiffness - the calculator always uses the correct I_t.

They apply to thin-walled closed hollow sections. The first Bredt formula gives the shear stress that is constant across the wall, τ = T/(2·A_m·t), with A_m the area enclosed by the wall centreline. The second Bredt formula gives the twist via I_t = 4·A_m²/(∫ds/t), for constant wall thickness I_t = 4·A_m²·t/U. Note that A_m is the centreline area, not the outer contour.

You do not - you enter the raw dimensions. For the closed section choose the input via outside dimensions, enter outside width b, outside height h and wall thickness t, and the calculator derives the wall centreline itself: A_m = (b − t)·(h − t) and U = 2·((b − t) + (h − t)). That is the standard route because A_m and U describe the same centreline and, entered as two free numbers, can be detuned against each other unnoticed. Using the outer contour flatters the result: for an 80 × 40 × 4 rectangular tube it gives A_m = 3200 instead of 2736 mm², the shear stress comes out 15 percent too low and the twist 22 percent too small. Direct entry of A_m and U remains for sections that are not rectangular tubes; there the calculator at least checks whether the perimeter you entered can enclose the area you entered.

Yes, open thin-walled sections are a cross-section type of their own: I, U, L, T, cruciform, slit tube or flat bar. What you enter is a named profile shape with a fixed number of parts - flat bar one part, L, T and cruciform two, U, I and IPB three - and no longer a freely extendable list of loose rectangles: the summation formula holds for simply bounded, that is one-piece, cross-sections (Hütte, 33rd ed., p. E 81). The difference from a closed section remains dramatic all the same. In a closed section the shear flow runs as a loop around the cavity and produces a large internal moment (Bredt); an open section cannot form that loop, every rectangle has to close the shear flow within itself. For a slit tube with centreline diameter d_m and wall thickness s the closed tube carries (3/2)·(d_m/s) times the torque at the same stress and is (3/4)·(d_m/s)² times stiffer - at d_m/s = 20 that is 30 times stronger and 300 times stiffer. A slit one millimetre wide is enough, because it interrupts the loop completely.

It appears in I_t = (η/3)·Σ h_i·t_i³ and is the allowance for the rigid shear connection between the parts - not, tempting as that reading is, an allowance for the extra material in the fillets at the junctions. The summation formula treats the individual rectangles as independent and assumes that no shear stresses are transferred between them (Wriggers/Nackenhorst/Beuermann/Spiess/Löhnert, Technische Mechanik kompakt, 2nd ed. 2006, p. 235). A rolled section violates exactly that: its parts are rigidly connected and can no longer warp freely, and the restrained warping increases the torsion constant (ibid., table 13.3). The numbers are test values for rolled sections (Hirschfeld, Baustatik, 5th ed., p. 127, eq. (22), from tests by Föppl; likewise Dubbel and Hütte): 0.99 for L, 1.12 for U and T, 1.17 for cruciform, 1.29 for IPB/HEB and 1.31 for I. You do not pick the factor separately but through the profile shape - it follows from it. For the flat bar η = 1.00 applies, and there that is not merely conservative but correct: a single rectangle has no neighbouring part that could restrain its warping. For welded or folded sections, pick the shape of the same family; the calculator does not accept a free sum of several flat bars, because the formula presupposes a one-piece cross-section. Bear in mind that η comes from tests on ROLLED sections: for a welded or sharply folded inner corner without a fillet it is on the unsafe side - the rolled-series factor is too favourable there.

Not where the table formula puts it. W_t = I_t/t_max gives the nominal value in the middle of the long side of the thickest rectangle; the true maximum sits at the re-entrant corner, that is where web and flange meet. The flow analogy shows it clearly (Dubbel, fig. 36 c, on an angle section; Läpple, fig. 11.7), and Hütte on p. E 91 explicitly names the re-entrant corners of a cross-section contour. It is quantified there on p. E 83 as τ_max = τ·[c + √(1+c²)] with c = t/(4·r), where r is the fillet radius: at t/r = 1 the factor is 1.28, at r = t/2 already 1.62, at r = t/4 then 2.41, and for a sharp corner without any fillet it grows without bound. For a rolled IPE with t_web = 5.6 mm and r = 12 mm it stays at 1.12. The calculator does not apply this factor - it reports the nominal stress. For rolled sections with a generous radius the increase is small; for welded or sharply folded inner corners it belongs in the calculation, either via this formula or via a notch-effect check.

In two places, and the calculator points out both. First, the summation formula assumes slender rectangles. Dankert/Dankert, Technische Mechanik, 4th ed. 2006, p. 360, puts the deviation at a thickness ratio t/h = 0.1, that is exactly at h/t = 10, at about 6 percent for I_t as well as for W_t; the approximation overestimates the stiffness by 6 percent and underestimates the stress by 10 percent, so for the stress it is on the unsafe side. A flange 100 mm wide and 10 mm thick is thus exactly the borderline case, and anything stockier is below it - a rolled flange like that of an IPE 200 at 100 × 8.5 mm sits just above at h/t = 11.8, while the legs of channel and angle sections regularly fall below. Second, the calculator covers only Saint-Venant torsion with free warping. Open sections warp strongly, and as soon as that warping is restrained - a fixed end, a welded end plate, load introduction through a gusset - warping torsion contributes as well. For short open members held at their ends it can take the larger share, and this calculator cannot estimate it; that needs a check per warping torsion theory with the warping constant C_M. The warping note also appears for the rectangle, because it follows from the section not being circular, not from the section being open.

You choose it. The default is the distortion energy hypothesis (von Mises) with τ_F = R_e/√3 ≈ 0.577·R_e; the shear stress hypothesis (Tresca) with τ_F = R_e/2 can be selected instead, which is 15.5 percent lower and therefore on the safe side. Läpple sizes torsion with R_e/2 throughout. Which hypothesis was used is stated in the input list of the report. From it follow the allowable shear stress τ_allow = τ_F/S_req and the available safety factor S = τ_F/τ, which is compared with the required safety factor S_req.

On offer are 1.2 / 1.35 / 1.5 / 1.7 / 2.0 and a custom value; the default is 1.5, the single best-documented value (Roloff/Matek, Böge, Tabellenbuch Metall). The grading follows the consequences of failure and how well the load is known: 1.2 for a well-known load with minor consequences, 1.35 for a well-known load with severe consequences, 1.7 for preliminary sizing with the section still open, 2.0 where people are at risk or the load assumption is uncertain. There is no grading by static, pulsating or alternating load - the load case sits in the material property, not in the safety factor. Cyclic loading needs a fatigue check (DIN 743 shaft calculator), not a larger factor. The series applies to ductile materials against the yield strength; brittle materials are checked against tensile strength with 4.0 to 9.0 (Läpple, table 2.2).

They are published guide values from the product standards: EN 10025-2 for the structural steels S235 and S355 as well as E295, EN 10083-2/-3 for C45 and 42CrMo4, EN 10088-2 for the stainless X5CrNi18-10 and EN 755-2 for the aluminium alloys. The yield strengths apply to product thicknesses up to 16 mm; thicker material is lower, for S355 per EN 10025-2 for example 345 N/mm² up to 40 mm and 335 N/mm² up to 63 mm. The shear modulus is taken as 81000 N/mm² for steel, 77000 N/mm² for the austenitic steel and 26000 N/mm² for aluminium. If you have an inspection certificate, a data sheet or a different material, pick the custom entry in the list and enter G and R_e yourself.

It is the angle of twist per metre of bar length, φ' = T/(G·I_t), reported in degrees per metre. Roloff/Matek (18th edition, p. 358) gives 0.25 to 0.5 degrees per metre for gearbox and machine shafts and treats the stiffness check explicitly as the second step after the stress check. Three levels are on offer - 0.25 for gearbox and machine-tool shafts, 0.35 for general mechanical engineering, 0.5 for secondary shafts - plus any value of your own. It is a serviceability guide value and not a code check; what matters is what the function tolerates. Where the twist itself is disturbing - indexing of a rotary table, angular position of a measuring system, torsional vibration in a drive train - your own, usually stricter value applies. Because φ' falls with the fourth power of the diameter while the stress falls only with the third, stiffness almost always governs on long shafts, not strength. For an open section the traffic light is therefore red with the starting values, and that is not a mistake: an IPE 200 twists by 5.2 degrees per metre under 500 Nm, a U 200 by 3.1 - the shaft guide value of 0.25 degrees per metre is the wrong yardstick for such a section. If you load an open section in torsion, check what the function tolerates and set the guide value accordingly; where torsional stiffness matters, a closed section belongs there anyway.

Via T = 9550·P/n with P in kW, n in revolutions per minute and T in Nm. The number is 30000/π = 9549.3 and follows from P = T·ω with ω = 2·π·n/60. Example: 7.5 kW at 1450 rpm gives 49.4 Nm. Downstream of a gearbox use the output speed, otherwise the motor speed; the torque, speed & power calculator does the conversion including units for you. Mind the difference between rated and peak torque: what counts for a check against the yield strength is the largest torque that occurs - starting, braking, stalling - not the continuous torque in service.

The calculator rates two things separately. For the shear check the utilisation counts, that is the required safety factor divided by the available one. Red means: the check is not satisfied - the torsional shear stress exceeds τ_allow = τ_F/S_req. Amber means: satisfied, but with less than 10 percent reserve. An available safety factor below the required one is therefore red and not amber, because a check that is not satisfied is not borderline, it has failed. The specific twist follows the same direction, only against a guide value instead of a yield limit: red means the guide value you set is exceeded, amber means it is used up by more than 90 percent. If you want more reserve, raise S_req or tighten the guide value instead of interpreting the light.

Straight, prismatic bars under pure torsion (Saint-Venant) with free warping, linear-elastic. Not included are restrained warping (e.g. at fixed ends of open sections), superimposed bending, notch effects at shoulders and keyways, and local buckling of thin-walled sections. For rotating, fatigue-loaded shafts an additional fatigue check per DIN 743 is required.