Calculation Criteria


Material: Steel S 235 (former St 37)
circular cross section, radius r = 2 mm
usual steel in machine industrie and steel construction with moderate load
Elastic modulus E E = 210000 N/mm2
Area moment IF, circular cross section, d = 4mm IF = p · r4 / 4 = 12,566 mm4
Tensile strength Rm Rm = 340 N/mm2
Yield point Rp 0,2
after total discharge remains a plastic
strech of 0,2%
Rp 0,2 = 235 N/mm2
Equivalent factors for steel k1 = 0,5,   k2 = 1,4     for bending
k1 = 0,3,   k2 = 0,58   for torsion
Bending moment Mb for one-sided bearing Mb = F · l
Section modulus W
for circular cross section, r = 2 mm
W = p · r3 / 4 = 6,283 mm3
Bending stress sb sb = Mb / W
Bending moment Mb for double sided movable bearing Mb = F · l / 4
Torsional moment MtMt = F · r
Polar section modulus Wp
for circular cross section, r = 2 mm
Wp = p · r3 / 2 = 12,566 mm3
Torsional stress tb tt = Mt / Wp
Equivalent stress sv sv = Ö[sb2 + 3(a0 · tt) 2]   mit a0 » 0,7
Assumed limits for bending stress sb zul
according to the fatigue strength diagram
sb zul = k2 · Rp 0,2 = 330 N/mm2 for static bending
sb zul =      ssch     = 255 N/mm2 for swelling bending
sb zul = k1 · Rm     = 170 N/mm2 for alternating bending
Assumed limits for torsional stress tt zul
according to the fatigue strength diagram
tt zul = k2 · Rp 0,2 = 135 N/mm2 for static torsion
tt zul =      tsch     = 135 N/mm2 for swelling torsion
tt zul = k1 · Rm     = 100 N/mm2 for alternating torsion
Required torque M to accelerate a vehicle
with the mass m on a flat bottom from zero speed
to a velocity v within the time t by means of wheels
with radius R
M = F · R = m · a · R = (m · v · R) / t
7kB (37kB)
Fatigue strength diagram for bending
6kB (32kB)
Fatigue strength diagram for torsion