Leaf Spring

Calculate bending stress and free-end deflection for rectangular, trapezoidal, and triangular leaf springs.

Inputs

Spring Profile

W

Applied Load

l

Free Length

b

Width (fixed end)

t

Thickness

E

Elastic Modulus

GPa

Steel ≈ 210 GPa; Stainless steel ≈ 193 GPa; Phosphor bronze ≈ 110 GPa

Formula Interpretation

① Bending Stress

\sigma = \dfrac{6Wl}{bt^2} \quad \text{(MPa)}

Bending stress $\sigma$ is computed using the cantilever beam model. $W$ is the applied load, $l$ is the free length, $b$ is the width at the fixed end, and $t$ is the thickness. This formula applies to both rectangular and trapezoidal leaf springs.

② Rectangular Deflection

\delta = \dfrac{4Wl^3}{bt^3 E} \quad \text{(mm)}

For a rectangular leaf spring (constant width), $k$ = 1. Deflection scales cubically with length $l$ and inversely with modulus $E$ and the cube of thickness $t$ .

③ Trapezoidal Deflection

\delta = k\,\dfrac{4Wl^3}{bt^3 E} \quad \text{(mm)}

For a trapezoidal leaf spring, deflection is multiplied by the correction factor $k$ , which depends on the width ratio b₁/b. When b₁/b = 1 (rectangular), $k$ = 1; when b₁/b = 0 (triangular), $k$ = 1.5.

④ Triangular Deflection

\delta = \dfrac{6Wl^3}{bt^3 E} \quad \text{(mm)}

A triangular leaf spring is a special case of the trapezoidal spring where the free-end width b₁ = 0, giving $k$ = 1.5. The deflection formula simplifies to $\delta = \dfrac{6Wl^3}{bt^3 E}$ .

Knowledge Points

The bending stress of a leaf spring can be determined using the cantilever beam model with a rectangular cross-section. For a rectangular leaf spring, the bending stress varies along the length — it is highest at the fixed end. For a trapezoidal leaf spring, the bending stress is uniform along the entire length, making it a more efficient use of material.

Worked Example

A rectangular leaf spring has width $b = 12\,\text{mm}$ , thickness $t = 2\,\text{mm}$ , free length $l = 100\,\text{mm}$ . A load of $W = 50\,\text{N}$ is applied at the free end. The elastic modulus is $E = 210\,\text{GPa}$ . Find the bending stress and deflection.

Step 1 — Bending stress σ (Formula ①)

\sigma = \dfrac{6 \times 50 \times 100}{12 \times 2^2} = \dfrac{30\,000}{48} = 625 \text{ MPa}

Step 2 — Free-end deflection δ (Formula ②)

\delta = \dfrac{4 \times 50 \times 100^3}{12 \times 2^3 \times 210 \times 10^3} = \dfrac{2 \times 10^8}{2.016 \times 10^7} \approx 9.92 \text{ mm}

Therefore: bending stress $\sigma = 625\,\text{MPa}$ , free-end deflection $\delta \approx 9.92\,\text{mm}$ .

Extended Knowledge

•A rectangular leaf spring is a special case of a trapezoidal spring where the free-end width equals the fixed-end width (b₁/b = 1, k = 1).
•A triangular leaf spring is a special case where the free-end width is zero (b₁/b = 0, k = 1.5). Its deflection formula simplifies to δ = 6Wl³/(bt³E).
•Trapezoidal leaf springs have uniform bending stress along their length, resulting in more efficient material utilization than rectangular springs.
•Leaf springs are widely used in automotive suspensions, railroad bogies, and heavy machinery where compact, high-load spring elements are required.
•Multi-leaf springs stack several leaves of decreasing length. The effective stiffness is approximately n times that of a single leaf, where n is the number of leaves.