Stress Calculations
Stress Formulas & Solved Examples
Explore categorized stress formulas in solid mechanics with solved examples for better understanding.
1. Normal Stress
σ = F / A
Definition: Axial stress perpendicular to a surface.
σ: Normal stress [MPa] — F: Force [N] — A: Area [mm²]
Example:
F = 1500 N, A = 300 mm² ⇒ σ = 1500 / 300 = 5 MPa
2. Tensile & Compressive Stress
σt/c = Ft/c / A
Ft: Tensile force [N] — Fc: Compressive force [N] — A: Area [mm²]
Example:
Ft = 1200 N, A = 200 mm² ⇒ σt = 6 MPa
3. Shear Stress
τ = F / A
Definition: Stress parallel to the surface due to shear force.
τ: Shear stress [MPa] — F: Shear force [N] — A: Shear area [mm²]
Example:
F = 500 N, A = 250 mm² ⇒ τ = 2 MPa
4. Double Shear Stress
τ = F / 2A
Definition: Shear force divided over two planes.
F: Force [N] — A: Area of one shear plane [mm²]
Example:
F = 800 N, A = 200 mm² ⇒ τ = 800 / (2 × 200) = 2 MPa
5. Bearing (Contact) Stress
σ = F / (d × l)
Definition: Stress between two contacting surfaces.
F: Load [N] — d: Diameter [mm] — l: Length of contact [mm]
Example:
F = 1000 N, d = 10 mm, l = 20 mm ⇒ σ = 1000 / (10×20) = 5 MPa
6. Bending Stress
σ = (M × y) / I
M: Bending moment [N·mm] — y: Distance from neutral axis [mm] — I: Moment of inertia [mm⁴]
Example:
M = 10000 N·mm, y = 10 mm, I = 5000 mm⁴ ⇒ σ = (10000×10)/5000 = 20 MPa
7. Torsional Shear Stress
τ = (T × r) / J
T: Torque [N·mm] — r: Radius [mm] — J: Polar moment of inertia [mm⁴]
Example:
T = 3000 N·mm, r = 5 mm, J = 200 mm⁴ ⇒ τ = 75 MPa
8. Thermal Stress
σ = E × α × ΔT
E: Young's modulus [MPa] — α: Thermal expansion coefficient — ΔT: Temperature change [°C]
Example:
E = 200000 MPa, α = 12e-6, ΔT = 50°C ⇒ σ = 200000×12e-6×50 = 120 MPa
9. Hoop Stress
σh = (p × r) / t
p: Pressure [MPa] — r: Inner radius [mm] — t: Thickness [mm]
Example:
p = 2 MPa, r = 100 mm, t = 10 mm ⇒ σh = (2×100)/10 = 20 MPa
10. Radial Stress
σr = [...] / (ro² − ri²)
Note: Use full equation for thick-walled pressure vessels.
Example with assumed values requires longer derivation – shown in extended section.
11. Normal Stress on Inclined Plane
σθ = (P / A₀) × cos²θ
Example:
P = 1000 N, A₀ = 200 mm², θ = 45° ⇒ σθ = (1000 / 200) × cos²45° ≈ 2.5 MPa
12. Shear Stress on Inclined Plane
τθ = (P / A₀) × sinθ × cosθ
Example:
P = 1000 N, A₀ = 200 mm², θ = 45° ⇒ τθ ≈ 2.5 MPa
13. Principal Stresses (2D)
σ1,2 = (σx + σy) / 2 ± √[((σx − σy) / 2)² + τxy²]
Example:
σx = 10, σy = 4, τxy = 3 ⇒ σ1,2 ≈ 7 ± 3.6 → σ₁ = 10.6, σ₂ = 3.4 MPa
14. Von Mises Stress (2D)
σvm = √[σx² − σxσy + σy² + 3τxy²]
Example:
σx = 10, σy = 4, τxy = 3 ⇒ σvm ≈ 9.49 MPa
15. Von Mises Stress (3D)
σvm = √[½((σx − σy)² + (σy − σz)² + (σz − σx)² + 6(τxy² + τyz² + τzx²))]
Example:
σx = 100, σy = 50, σz = 0, τxy = 20, τyz = 0, τzx = 0 ⇒ σvm ≈ 79.06 MPa
16. Maximum Shear Stress
τmax = (σ1 − σ3) / 2
Example:
σ1 = 100, σ3 = 20 ⇒ τmax = (100−20)/2 = 40 MPa
17. Mean (Hydrostatic) Stress
σm = (σ1 + σ2 + σ3) / 3
Example:
σ1 = 100, σ2 = 60, σ3 = 40 ⇒ σm = 66.67 MPa