Springs, fasteners, clutches and component functions

A spring law needs a direction and working range

For an ideal linear spring, the force magnitude is k times extension from its specified reference state. Its restoring direction opposes the displacement. The stiffness k is in N/m or N/mm, which differ by a factor of 1000. Stored elastic energy is kx²/2 for a spring initially unstressed at x=0.

Real springs can have preload, nonlinear response, solid-height or travel limits, and fatigue or relaxation. A supplied linear relation should therefore carry its range and reference. A component’s stiffness is not its maximum load. An intentionally compliant element can accommodate motion, but its displacement still must remain compatible with the rest of the assembly.

Connection changes which quantity is shared

For ideal massless springs in series under a static axial load, the same force passes through both and their extensions add. With stiffnesses k 1 and k 2, the equivalent compliance is 1/keq=1/k 1+1/k 2. This assumes no other load or support at their intermediate connection.

For parallel springs attached between the same two rigid endpoints, the extension is common and the forces add, so keq=k 1+k 2. These are connection rules rather than a statement that one arrangement is universally better. Endpoint geometry, preload and unequal alignment can change how real components share load.

Clamping and coupling introduce other interactions

A preloaded threaded joint stretches the fastener while compressing the clamped members. Subsequent external load changes both according to the joint’s stiffness and geometry. Tightening torque is not a direct measurement of clamp force: friction, process variation and relaxation matter. No torque setting or preload target follows from this lesson.

A clutch establishes or interrupts a torque-transmission path. Some types permit relative slip while transmitting torque; others rely on positive engagement under appropriate conditions. In a slipping friction clutch, the power difference associated with equal opposing torque magnitude τ and angular-speed difference Δω is τΔω. This transfer can become heat and affect the operating limits, so coupling behaviour must be part of the duty model.

Follow a separating load through a preloaded joint

Imagine one straight bolt clamping two members. Before any external separating load is applied, bolt tension and member compression have the same magnitude P0. The bolt is stretched; the members are shortened relative to their unloaded states. This stored elastic deformation is preload. The forces act through different parts of the joint rather than being two additional external loads on the whole assembly.

Consider a deliberately simple axial model: load is centred on the bolt axis and applied at the two outer ends of the clamped stack, the bolt and members stay linearly elastic, and the contact remains closed. Ignore bending, prying and additional load paths. Let bolt axial stiffness be kb and effective member compression stiffness be kc. If the distance between the bolt head and nut increases by a small amount δ, bolt tension increases by kbδ while member compression decreases by kcδ as the members recover some of their initial shortening.

A free-body balance at the loaded end gives bolt tension minus remaining member compression equal to the external separating load P. Starting from the balanced preload, P=(kb+kc)δ. Hence the bolt receives the fraction Cj=kb/(kb+kc) of the added load: Fb=P0+CjP, while remaining compression is Fc=P0−(1−Cj)P. Cj is a dimensionless load-sharing fraction for this particular idealisation, not a universal joint constant. Changing where the load enters, the member shape or the stiffness can change the relationship.

For an original toy case, supply P0=100 N, kb=1000 N/mm and kc=4000 N/mm. Then Cj=0.20. Adding P=50 N gives δ=50/5000=0.010 mm, bolt tension Fb=110 N and remaining compression Fc=60 N. Their difference is 50 N, as equilibrium requires. Adding the whole 50 N directly to bolt tension would give 150 N and miss the members’ recovery in this maintained-contact model.

As P rises in this model, compression reaches zero at P=100/0.80=125 N, with δ=0.025 mm and bolt tension 125 N. This is the first-separation boundary of the model, not an allowable working load. Continuing the same formula to P=150 N would produce Fc=−20 N. The nonadhesive contact cannot supply that tensile clamping force, so the maintained-contact model must stop and a changed contact/load-path model is needed. Stiffness and preload alone do not establish bolt strength, fatigue life or a permissible load.

A real joint may have eccentric load entry, local flange bending, several fasteners, nonlinear contact or changing preload. The simple ledger teaches why bolt tension can increase while clamp force decreases; it does not determine a tightening torque or installation method. The earlier torque-preload warning still applies, and any claim about a particular joint needs its actual geometry, material and loading evidence.

Compare the same two springs in two arrangements

Take k 1=200 N/m and k 2=300 N/m. In series, keq=(200×300)/(200+300)=120 N/m. A 6 N load extends the springs 6/200=0.030 m and 6/300=0.020 m, a total 50 mm. The equivalent energy is½×120×0.050²=0.15 J.

In parallel, the combined stiffness is 500 N/m and a total 6 N produces 6/500=0.012 m, or 12 mm. The forces are 2.4 N and 3.6 N, and total energy½×500×0.012²=0.036 J. Equal total force does not mean equal displacement or stored energy across different arrangements. These are ideal supplied-range cases, not spring ratings.

Try the reasoning

1. Which quantity is shared by the two series springs?

   Show the answer and reasoning

   Force is shared in this ideal static, massless intermediate-connection model; extensions add.

2. Does a larger k automatically mean a larger permissible load?

   Show the answer and reasoning

   No. Stiffness and component capacity require different information.

3. Why is applied tightening torque insufficient as proof of clamp force?

   Show the answer and reasoning

   Friction, process variation, joint response and relaxation affect the actual preload.

4. In the toy maintained-contact joint, the external separating load becomes 75 N. Find bolt tension, remaining compression and their difference.

   Show the answer and reasoning

   Cj=0.20 remains the supplied model value. Bolt tension is 100+0.20×75=115 N, and member compression is 100−0.80×75=40 N. The difference 115−40=75 N checks the end-force balance. The corresponding added gap is 75/5000=0.015 mm.

5. Someone continues Fc=100−0.80P to P=150 N and interprets −20 N as a useful extra clamp. What is wrong?

   Show the answer and reasoning

   A negative compression result signals loss of the assumed closed nonadhesive contact. The model reached zero compression at 125 N and cannot continue unchanged. It does not predict a negative useful clamp or establish that the actual bolt/assembly can carry either load.

What to carry forward

Describe the intended component function and the relationship its connection preserves.