Laminated Leaf Spring

Calculate bending stress, deflection, and required number of leaves for laminated (multi-leaf) springs.

Inputs

W

Central Load

l

Span

b

Leaf Width

t

Leaf Thickness

E

Elastic Modulus

GPa

Steel ≈ 210 GPa; Stainless steel ≈ 193 GPa

n

Number of Leaves

Optional Corrections

e

Clamp Width

Width of the central clamp/band; effective span becomes l' = l − 0.6e

\mu

Friction Coefficient

Inter-leaf friction coefficient, typically 0.1 – 0.3

Formula Interpretation

① Bending Stress

\sigma = \dfrac{3Wl}{2nbt^2} \quad \text{(MPa)}

Bending stress $\sigma$ for equal-thickness laminated leaves. $W$ is the central load, $l$ is the span, $n$ is the number of leaves, $b$ is the leaf width, and $t$ is the thickness.

② Deflection

\delta = \dfrac{3Wl^3}{8nbt^3 E} \quad \text{(mm)}

Free-end deflection $\delta$ is proportional to the cube of the span $l$ and inversely proportional to $n$ , thickness cubed, and elastic modulus $E$ .

③ Clamp-corrected Span

l' = l - 0.6\,e \quad \text{(mm)}

When a clamp (band) of width $e$ is used, the effective span is reduced. Use $l' = l - 0.6e$ in place of $l$ for stress and deflection calculations.

④ Friction-corrected Deflection

\delta_1 = \dfrac{5(1 \pm \mu)}{5 + \mu}\,\delta \quad \text{(mm)}

Inter-leaf friction reduces deflection during loading and increases it during unloading. $\mu$ is the friction coefficient; use '−' for loading and '+' for unloading.

Knowledge Points

A laminated leaf spring can be thought of as a spring steel plate cut into multiple long, narrow triangular strips stacked together. After stacking, the assembly is secured by a center bolt or a clamp band. Laminated leaf springs are used in automobiles and trains as cushioning devices. For rectangular-section leaves of equal thickness, the bending stress varies along the length — it is greatest at the center where the load is applied.

Worked Example

A laminated leaf spring has span $l = 600\,\text{mm}$ , width $b = 45\,\text{mm}$ , thickness $t = 6\,\text{mm}$ , and allowable stress $\sigma_{\text{allow}} = 400\,\text{MPa}$ . When a load of $W = 2000\,\text{N}$ is applied at the center, find the required number of leaves and deflection (ignoring friction). Also find the deflection considering inter-leaf friction with $\mu = 0.15$ . Given: $E = 210\,\text{GPa}$ .

Step 1 — Number of leaves n (Formula ①)

n = \dfrac{3Wl}{2bt^2\sigma} = \dfrac{3 \times 2000 \times 600}{2 \times 45 \times 6^2 \times 400} = \dfrac{3\,600\,000}{1\,296\,000} = 2.78

Step 2 — Deflection δ (Formula ②)

\delta = \dfrac{3 \times 2000 \times 600^3}{8 \times 3 \times 45 \times 6^3 \times 210 \times 10^3} = \dfrac{1.296 \times 10^{12}}{4.898 \times 10^{10}} \approx 26.5 \text{ mm}

Step 3 — Friction-corrected deflection δ₁ (Formula ④)

\delta_1 = \dfrac{5(1 - 0.15)}{5 + 0.15} \times 27 = \dfrac{4.25}{5.15} \times 27 \approx 22.3 \text{ mm}

Therefore: required $n = 3$ leaves, deflection $\delta \approx 27\,\text{mm}$ ; with friction correction the loading deflection is approximately $\delta_1 \approx 23\,\text{mm}$ .

Extended Knowledge

•A laminated leaf spring is essentially multiple triangular cantilever springs stacked. The stacking multiplies the load capacity while sharing the stress among all leaves.
•Center-bolt fixation is simpler but produces a stress concentration at the bolt hole. Clamp-band fixation distributes the clamping force more evenly.
•The inter-leaf friction effect creates hysteresis: deflection under load is less than calculated, while deflection on unloading is greater. This hysteresis provides additional damping.
•Laminated leaf springs are widely used in heavy truck rear suspensions, railway bogies, and agricultural trailers where high load capacity and robust construction are required.
•Modern parabolic leaf springs use leaves with variable thickness to achieve uniform stress, reducing weight by up to 50% compared to constant-thickness multi-leaf designs.