Equivalent Stiffness (Springs in Series)
Calculate the equivalent stiffness of two springs in series, 1 ÷ k_eq = 1/k₁ + 1/k₂, from the individual stiffnesses k₁ and k₂. The result, in the same unit (N/m), is always smaller than the smallest stiffness — springs in series are more flexible, since each deforms under the same force and the displacements add. It is the fundamental calculation to reduce suspension systems, isolators and structures with elastic elements in sequence to a single-degree-of-freedom model, the basis for finding the natural frequency. Enter the two stiffnesses.
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Rigidez equivalente — molas em série
Quando duas molas (ou elementos elásticos) são ligadas em série, uma após a outra, a mesma força atravessa as duas, mas cada uma se deforma, e os deslocamentos se somam. O resultado é um conjunto mais flexível que qualquer das molas isoladas, com rigidez equivalente dada por 1 ÷ k_eq = 1/k₁ + 1/k₂ (a mesma forma da associação de resistores em paralelo, ou de capacitores em série — é o padrão das grandezas que se somam pelo inverso). Para duas molas iguais de 100 N/m, a equivalente é 50 N/m — metade. A regra geral: a rigidez equivalente em série é sempre menor que a menor das rigidezes. Esse cálculo é o primeiro passo para analisar qualquer sistema vibratório real, que quase nunca tem uma única mola: pense numa suspensão automotiva (mola + pneu + bucha em série), num isolador empilhado, ou na flexibilidade combinada de um eixo e seus mancais. Reduzindo todas as molas a uma rigidez equivalente única, o sistema complexo vira um modelo simples de massa-mola de um grau de liberdade, do qual se calcula a frequência natural f_n = (1/2π)√(k_eq/m) — a informação mais importante para prever ressonâncias. Informe as duas rigidezes.
Related Tools
Equivalent Stiffness (Springs in Parallel)
Calculate the equivalent stiffness of two springs in parallel, k_eq = k₁ + k₂, by adding the individual stiffnesses. The result, in the same unit (N/m), is always larger than the largest stiffness — springs in parallel are stiffer, since they share the load under the same displacement and the forces add. This is the case of mounts, isolators and supports placed side by side carrying the same component. Reducing spring assemblies to an equivalent stiffness is the first step to compute a vibrating system's natural frequency. Enter the two stiffnesses.
Natural Frequency from Static Deflection
Calculate a system's natural frequency from its static deflection, f_n = (1 ÷ 2π)·√(g ÷ δ), where δ is the static deflection caused by self-weight and g the gravitational acceleration (9.81 m/s²). The result, in Hz, is a practical and elegant way to estimate the natural frequency without separately knowing mass and stiffness — you just measure how much the system sags under its own weight. Larger deflections (more flexible systems) give lower natural frequencies, desirable in vibration isolators. It is widely used in spring and mount design. Enter the static deflection (in metres).
Bolt Load under External Load
Calculate the total tensile force in the bolt when an external load is applied to the joint, F_b = F_i + C·P, from the preload F_i (N), the joint stiffness constant C and the external tensile load P (N). This is one of the most important — and most surprising to the uninitiated — relations of bolted joints: when you apply an external load P trying to 'separate' the parts, the bolt tension does NOT rise from F_i to F_i + P (as intuition suggests), but only to F_i + C·P, where C is typically 0.2-0.4. That is, the bolt only 'feels' a FRACTION of the external load! The reason: most of the external load (1−C)·P merely RELIEVES the compression between the parts (which were compressed by the preload), rather than stretching the bolt more. This is the genius of the preloaded joint — it 'hides' the external load from the bolt. So a well-tightened joint, under a CYCLIC external load (causing fatigue), exposes the bolt to a very small stress variation (proportional to C·ΔP, not ΔP), making it extremely fatigue-resistant. This formula holds while the joint does NOT separate (P below the separation load); above that, the bolt carries the whole load. Enter the preload, stiffness constant and external load.
Bolt Stiffness
Calculate a bolt's stiffness (spring constant), k_b = (A_t·E) ÷ L, from the tensile area A_t (mm²), the material elastic modulus E (MPa) and the grip length L (mm, the effective length under tension between head and nut). When tensioned by the preload, the bolt behaves as a very stiff SPRING: it stretches an amount proportional to the force (Hooke's law), and its stiffness is force per unit elongation. This stiffness is one of two essential ingredients of bolted-joint analysis — the other is the stiffness of the clamped PARTS (members). The ratio between these two stiffnesses (the joint stiffness constant C) determines how an external load splits between the bolt and the parts. Typically the parts (massive, with large effective compression area) are MUCH stiffer than the bolt (thin and long), which is the DESIRED situation: stiff parts absorb most of the external load, protecting the bolt from stress variation and fatigue. Long, thin bolts have low stiffness (good for sharing load), while short, thick bolts are stiff. Knowing k_b is the starting point of fatigue and joint-separation analysis. Enter the tensile area, elastic modulus and grip length.
Joint Stiffness Constant
Calculate a bolted joint's stiffness constant, C = k_b ÷ (k_b + k_m), from the bolt stiffness k_b (N/mm) and the members' (clamped parts) stiffness k_m (N/mm). The constant C (also called bolt load fraction) is the heart of bolted-joint analysis: it tells what FRACTION of an external tensile load is carried by the BOLT, the rest (1−C) being carried by decompression of the MEMBERS. The value of C reveals the elegant, protective behavior of a preloaded joint: since the (massive) members are usually much stiffer than the (thin) bolt, k_m >> k_b, so C is SMALL (typically 0.2-0.4). This means that when an external load P is applied, only a small portion C·P adds to the bolt tension — most of the load (1−C)·P is absorbed by RELIEF of the compression between the parts. That is why the bolt stress variation is small (good fatigue resistance) and why preload is so beneficial. The smaller C (stiff parts, flexible bolt), the better the bolt protection. The constant C appears in all subsequent formulas: bolt load, residual member force, separation load and fatigue safety factor. Enter the bolt stiffness and the members' stiffness.
Amplification Factor (Q)
Calculate the resonance amplification factor (quality factor Q), Q = 1 ÷ (2·ζ), from the damping ratio ζ. The dimensionless result shows how many times the vibration amplitude at resonance exceeds the equivalent static deflection: lightly damped systems (small ζ) have high Q and sharp, dangerous resonance peaks; well-damped systems have low Q and smooth response. It is central to designing structures, machines and instruments to avoid destructive amplification and to characterizing the selectivity of filters and resonators. Enter the damping ratio.
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