Steel and composite clamping members share load according to stiffness, seating and permitted movement. Model their contact state with the plate response to establish which path carries the required force.
Calculation basis. Project-specific values and gates in the illustrative screens below are assumptions; cited catalogue values retain their source test conditions.
Identify the topology first. Members compressed in parallel between common plates can share force. A tie rod and the stack it clamps are parts of a different joint relationship; they should not be treated as two parallel areas merely because both appear in the same section.
A parallel-path illustration
Consider fictional linear compression springs between rigid plates: steel stiffness kₛ = 100 kN/mm and composite stiffness kc = 20 kN/mm. With both paths in contact and total force 12 kN, common plate movement is δ = 12/(100 + 20) = 0.100 mm. The steel takes 10 kN and the composite 2 kN.
Now give only the composite path an initial 0.100 mm gap. After it contacts, forces are Fₛ = 100δ and Fc = 20(δ − 0.100), in kN. Solving Fₛ + Fc = 12 gives δ = 0.116667 mm, Fₛ = 11.6667 kN and Fc = 0.3333 kN. The computed movement exceeds the gap, so the assumed contact state is consistent.
These are illustrative spring values, not steel or transformerwood material data. The example shows why equal area or an unchanged material modulus cannot establish load sharing when contact geometry changes.
Map the actual assembly
- Member compliance
- Identify axial, bending and local compression contributions using the relevant directional material properties.
- Contact engagement
- Record initial gaps, seating sequence and surfaces that can separate or slide.
- Plate and clamp movement
- Check whether the supposedly common displacement is altered by bending or unequal supports.
- Temperature and time
- Evaluate the material state that changes stiffness, free dimensions and relaxation during the load history.
For nonlinear or time-dependent members, one constant spring value may describe only a limited state. A composite contact can densify or relax while a steel path remains comparatively stable. That can redistribute force even if the total assembly load is maintained.
Thermal movement can change contact before it changes the nominal stiffness. Use a defined reference state and directional free movement. Retain the named grade, conditioning and test direction from the densified-wood data sheet when supplying material inputs to the joint model. [1]
Verify the distribution using suitable force or deformation observations at representative paths. A total load cell cannot establish each member’s share. Instrumentation should account for its own effect on contact and stiffness.
When one path appears underloaded, check seating and plate movement before increasing its cross-section. The correct intervention may be a contact-geometry change, while added area leaves the gap mechanism intact.
Derive the seating tolerance from the required load share
Keep the named wood grade and test-method boundary separate from the assumed spring stiffnesses [2]. Actual joint load sharing depends on compliance and contact [3]. Continue the 100 and 20 kN/mm parallel springs under 12 kN, and assume the composite must carry at least 1.0 kN.
With a composite gap g in mm after contact, Fc = 20(12−100g)/120 kN. Requiring Fc ≥ 1 gives g ≤ 0.060 mm. The earlier 0.100 mm gap leaves only 0.333 kN and fails that assumed participation requirement, despite both members being present.
Control actual seating, plate bending and preload before changing composite area. Replace the spring assumptions with conditioned material and assembled-path evidence.
References
- Röchling Industrial, Lignostone L I/2-E3-HQ (P2R), Technical Data Sheet, version 2.0, released 20 September 2023.
- IEC, IEC 61061-2:1992, Specification for non-impregnated, densified laminated wood for electrical purposes — Part 2: Methods of test.
- R. T. Barrett, Fastener Design Manual, NASA Reference Publication 1228, 1990.