Flexibility describes the elastic property of a restraint and can be linear or rotational. PASS/START-PROF calculates expansion joint flexibility values using catalog data, industry standards, or structural mechanics methods for complex configurations.
Calculate expansion joint flexibility using Start-Elements: expansion joint flexibility.
Linear Flexibility (C) is the displacement (Δ) caused by a unit force (P) of 1 kg: C = Δ/P. Units: mm/kg.
Rotational Flexibility (C) is the rotation angle (φ in radians) caused by a unit moment (M) of 1 kg·m: C = φ/M. Units: rad/(kg·m).
Stiffness (K) is the inverse of flexibility: K = 1/C.
Linear Stiffness (K) is the force (P) required to produce a unit displacement (Δ) of 1 mm: K = P/Δ. Units: kg/mm.
Rotational Stiffness (K) is the moment (M) required to produce a unit rotation (φ) of 1 radian: K = M/φ. Units: kg·m/rad.
To model infinite flexibility (zero stiffness), input:
Rotational flexibility = 1 rad/(kg·m)
Linear flexibility = 1000 mm/kg
Examples:
A rotational bellows allows 3° (0.05236 rad) rotation under 120 kg·m moment. Rotational flexibility C = φ/M = 0.05236/120 = 0.000436 rad/(kg·m).
An axial expansion joint allows 25 mm displacement under 1500 kg force. Linear flexibility C = Δ/P = 25/1500 = 0.0167 mm/kg.
Expansion joint catalogs typically specify stiffness values. To convert stiffness to flexibility: C = 1/K. For K = 50 kg/mm, axial flexibility C = 1/50 = 0.02 mm/kg.
When the number of waves is unknown, model the expansion joint as perfectly flexible using linear flexibility = 1000 mm/kg and rotational flexibility = 1 rad/(kg·m). After preliminary analysis, select the appropriate expansion joint from catalog data and perform final verification analysis.
A spiral coil spring per OST 108.764.01-80 (Rev. 09) allows 70 mm deflection under 3225 kg load. Spring flexibility C = Δ/P = 70/3225 = 0.0217 mm/kg.