Stress Calculation
Tension · Compression · Shear · Bending · Torsion · Equivalent Stress
Stress Calculator
Zugkraft: σ = +F/A (positiv)
Druckkraft: σ = −F/A (negativ)
Positive = tension, negative = compression
IPE 270: A = 45.9 cm² | HEB 200: A = 78.1 cm² | Rectangle b×h
Shaft / TubeFormulas & Stress Types
Normal Stress (Tension / Compression):
σ
σ = F / A [N/mm²]
F [N], A [mm²] | Tension: σ > 0, Compression: σ < 0
Check: |σ| ≤ f_y / γ_M
σ = F / A [N/mm²]
F [N], A [mm²] | Tension: σ > 0, Compression: σ < 0
Check: |σ| ≤ f_y / γ_M
Shear Stress (Direct Shear):
τ
τ = F / (n · A_s) [N/mm²]
n = number of shear planes | A_s = shear area
Bolt ∅d: A_s = π·d²/4 | Check: τ ≤ f_y / (√3 · γ_M)
τ = F / (n · A_s) [N/mm²]
n = number of shear planes | A_s = shear area
Bolt ∅d: A_s = π·d²/4 | Check: τ ≤ f_y / (√3 · γ_M)
Bending Stress (Navier):
σ_b
σ_b = M / W_y = M · e / I_y [N/mm²]
M [N·mm] | W_y [mm³] = I_y / e
Check: σ_b ≤ f_y / γ_M
σ_b = M / W_y = M · e / I_y [N/mm²]
M [N·mm] | W_y [mm³] = I_y / e
Check: σ_b ≤ f_y / γ_M
Torsional Stress (Circular Section):
τ_t
τ_t = M_t / W_p [N/mm²]
W_p (solid) = π · d³ / 16 | W_p (tube) = π · (D⁴−d⁴) / (16·D)
τ_t = M_t / W_p [N/mm²]
W_p (solid) = π · d³ / 16 | W_p (tube) = π · (D⁴−d⁴) / (16·D)
Equivalent Stress – von Mises Criterion:
σ_v
σ_v = √(σ² + 3·τ²) [N/mm²]
Combined normal and shear stress
Check: σ_v ≤ f_y / γ_M
σ_v = √(σ² + 3·τ²) [N/mm²]
Combined normal and shear stress
Check: σ_v ≤ f_y / γ_M
Allowable Stress (Code Check):
σ_allow = f_y / γ_M [N/mm²]
τ_allow = f_y / (√3 · γ_M) [N/mm²]
Utilisation factor η = σ_act / σ_allow ≤ 1.0
σ_allow = f_y / γ_M [N/mm²]
τ_allow = f_y / (√3 · γ_M) [N/mm²]
Utilisation factor η = σ_act / σ_allow ≤ 1.0
Material Strength Values
| Material | f_y [N/mm²] | σ_allow (γ=1) | τ_allow (γ=1) |
|---|---|---|---|
| Steel S235 | 235 | 235 | 136 |
| Steel S355 | 355 | 355 | 205 |
| Steel S420 | 420 | 420 | 242 |
| Timber GL28h | 28 | 22.4 (γ=1.25) | – |
Stress Analysis – Fundamentals of Structural Engineering
What is mechanical stress?
Mechanical stress is an internal force per unit area that arises when a structural member is subjected to external loads. If the stress exceeds the material's limit (yield strength, ultimate tensile strength), the member fails by yielding, cracking or fracture. Stress calculation is the cornerstone of structural analysis and design.
Stress Types at a Glance
| Type | Symbol | Cause |
|---|---|---|
| Normal stress | σ | Tension / Compression |
| Shear stress | τ | Shear force / direct shear |
| Bending stress | σ_b | Bending moment |
| Torsional stress | τ_t | Torque |
| Equivalent stress | σ_v | Combined loading |
Unit Conversions
| Quantity | Conversion |
|---|---|
| σ = F/A | 1 kN / cm² = 10 N/mm² |
| F in kN, A in cm² | σ[N/mm²] = F[kN]×100/A[cm²] |
| M in kN·m, W in cm³ | σ[N/mm²] = M×10⁶/W×10³ |
| M_t in N·m, W_p in cm³ | τ[N/mm²] = M_t×10³/W_p×10³ |
Detailed Derivation of Stress Formulas
1. Normal Stress σ = F / A
Uniformly distributed tensile force F over cross-section A:
σ = F / A [N/mm²]
Valid for centrically loaded cross-sections (no bending moment).
Tension: σ > 0 (fibres elongate) | Compression: σ < 0 (fibres shorten)
Example: Tension rod ∅30 mm, F = 50 kN → A = π×15² = 706.9 mm² → σ = 50000/706.9 = 70.7 N/mm²
σ = F / A [N/mm²]
Valid for centrically loaded cross-sections (no bending moment).
Tension: σ > 0 (fibres elongate) | Compression: σ < 0 (fibres shorten)
Example: Tension rod ∅30 mm, F = 50 kN → A = π×15² = 706.9 mm² → σ = 50000/706.9 = 70.7 N/mm²
2. Shear Stress τ = F / A_s
Shear force F acts parallel to shear area A_s:
τ = F / (n · A_s) [N/mm²]
Bolted connection ∅20 mm, F = 80 kN, single shear (n=1):
A_s = π×10² = 314.2 mm² → τ = 80000/314.2 = 254.6 N/mm²
Check S355: τ_allow = 355/√3 = 205 N/mm² → ✗ Increase bolt diameter!
τ = F / (n · A_s) [N/mm²]
Bolted connection ∅20 mm, F = 80 kN, single shear (n=1):
A_s = π×10² = 314.2 mm² → τ = 80000/314.2 = 254.6 N/mm²
Check S355: τ_allow = 355/√3 = 205 N/mm² → ✗ Increase bolt diameter!
3. Bending Stress σ_b = M / W_y
σ_b = M / W_y = M · e / I_y
Peak tensile stress at the outermost tension fibre, equal and opposite compression at the top.
Example: M = 50 kN·m, IPE 270: W_y = 429 cm³
σ_b = 50×10⁶ / (429×10³) = 116.5 N/mm² < 235 N/mm² ✓
Peak tensile stress at the outermost tension fibre, equal and opposite compression at the top.
Example: M = 50 kN·m, IPE 270: W_y = 429 cm³
σ_b = 50×10⁶ / (429×10³) = 116.5 N/mm² < 235 N/mm² ✓
4. Torsional Stress τ_t = M_t / W_p
τ_t = M_t / W_p
W_p (solid) = π·d³/16 | W_p (tube) = π·(D⁴−d⁴)/(16·D)
Shaft ∅50 mm, M_t = 500 N·m:
W_p = π×50³/16 = 24 544 mm³
τ_t = 500×10³ / 24 544 = 20.4 N/mm²
W_p (solid) = π·d³/16 | W_p (tube) = π·(D⁴−d⁴)/(16·D)
Shaft ∅50 mm, M_t = 500 N·m:
W_p = π×50³/16 = 24 544 mm³
τ_t = 500×10³ / 24 544 = 20.4 N/mm²
5. Equivalent Stress – von Mises Criterion
σ_v = √(σ² + 3·τ²)
Applicable for plane stress state (σ_x, τ_xy). For pure shear (σ = 0):
σ_v = √3·τ → yield condition τ_y = f_y/√3 ≈ 0.577·f_y
Example: σ = 120 N/mm², τ = 60 N/mm²:
σ_v = √(120² + 3×60²) = √(14400 + 10800) = √25200 = 158.7 N/mm²
Applicable for plane stress state (σ_x, τ_xy). For pure shear (σ = 0):
σ_v = √3·τ → yield condition τ_y = f_y/√3 ≈ 0.577·f_y
Example: σ = 120 N/mm², τ = 60 N/mm²:
σ_v = √(120² + 3×60²) = √(14400 + 10800) = √25200 = 158.7 N/mm²
Practical Example: Weld Throat Check
Task:
Fillet weld (a = 5 mm, weld length l = 200 mm) carries simultaneously:
– Normal force F = 40 kN (tensile stress component)
– Shear force V = 30 kN (shear stress component)
Check for steel S235, γ_M = 1.25 (welds, EC3).
Solution:
- Weld throat area A_w = a × l = 5 × 200 = 1 000 mm²
- σ_⊥ = F / A_w = 40 000 / 1 000 = 40 N/mm²
- τ_∥ = V / A_w = 30 000 / 1 000 = 30 N/mm²
- Equivalent stress: σ_v = √(40² + 3×30²) = √(1600+2700) = √4300 = 65.6 N/mm²
- σ_allow = f_u / (β_w × γ_M2) ≈ 360 / (0.8 × 1.25) = 360 N/mm² → ✓ Check satisfied
Frequently Asked Questions
σ (sigma) is normal stress – it acts perpendicular to the cut plane
(tension = positive, compression = negative).
τ (tau) is shear stress – it acts parallel to the cut plane
(direct shear, torsion, transverse forces).
A bar under pure axial load F has only σ = F/A.
A bolt under transverse force has τ = F/A.
In practice both usually occur simultaneously and are combined using the von Mises criterion.
The von Mises yield criterion σ_v = f_y reduces for pure shear
(σ = 0, only τ) to:
σ_v = √(0 + 3·τ²) = √3·τ = f_y → τ_yield = f_y / √3 ≈ 0.577·f_y.
For steel S235: τ_allow = 235/√3 ≈ 135.7 N/mm².
The alternative Tresca criterion gives τ = f_y / 2 = 0.5·f_y (more conservative).
EC3 adopts the von Mises (distortion energy) criterion.
When tension members contain holes (bolts, pins) the load-bearing section is reduced:
A_net = A_gross − n × d_0 × t
(n = number of holes, d_0 = hole diameter, t = plate thickness).
According to EC3 the net cross-section uses γ_M2 = 1.25 (rather than γ_M0 = 1.0)
because tensile fracture at the net section is brittle-like.
Whenever σ and τ act simultaneously and the two stress states cannot be
checked independently. Typical cases:
weld seams (normal + shear), shafts (bending + torsion),
pressure vessels, connection plates under combined loading.
For pure axial force or pure bending the von Mises check is not required –
use σ ≤ f_y directly.
According to Eurocode EC3 (steel):
- γ_M0 = 1.0 – cross-section resistance (plastic)
- γ_M1 = 1.0 – member stability (buckling, lateral torsional buckling)
- γ_M2 = 1.25 – net section fracture, bolts and welds
Summary
Normal / Shear Stress
σ = F/A [N/mm²]
τ = F/(n·A_s) [N/mm²]
Bending / Torsional Stress
σ_b = M/W_y
τ_t = M_t/W_p
Equivalent Stress
σ_v = √(σ²+3τ²)
σ_v ≤ f_y / γ_M
Typical Applications
- Tension bars and hangers: σ = F/A, net section for holed members
- Bolts and pins: shear stress τ = F/A_s (single or double shear)
- Beams: bending stress σ_b = M/W_y at extreme fibres
- Drive shafts: torsional stress τ_t = M_t/W_p (combined with bending)
- Weld seams: von Mises check from normal and shear stress
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