worked-example

name	worked-example
description	Generate a fully worked hand-calculation example for any mechanical engineering problem. Given/Find/Solution format, all steps shown, unit checks, verification, design interpretation.
metadata	{"priority":7,"promptSignals":{"phrases":["worked example","solve this","step by step","hand calculation","walk me through","show the work"],"minScore":4}}

Worked Example Generator — Complete Skill

Format (Always Follow This)

Problem Statement

Restate the problem clearly with all given information and what to find.

Given Information

• List every given value with units
• State material properties (E, Sy, Sut, ν, etc.) — look up if not given
• State geometry (dimensions, cross-section type)
• State loading (forces, moments, pressures, temperatures)

Assumptions

List ALL assumptions made:

• Loading is static (or fatigue, specify R)
• Material is homogeneous, isotropic, linear elastic
• Small deformation theory applies
• Which failure criteria to use (von Mises, Tresca, Goodman)
• Stress concentrations included/excluded and why
• Plane stress or plane strain (if applicable)

Free Body Diagram

Describe (or draw ASCII) the FBD:

• Support reactions
• Applied loads
• Internal forces at critical section

Solution Steps (Numbered)

Step 1: Find reactions (if statically indeterminate, state approach) Show equilibrium equations explicitly: ΣFx = 0: ... ΣFy = 0: ... ΣM_A = 0: ...

Step 2: Identify critical section Where is M, T, or P maximum? Calculate or identify: M_max = ... at location x = ... V_max = ... at location x = ...

Step 3: Section properties A = ... mm² or in² I = ... mm⁴ or in⁴ c = ... mm (to extreme fiber) S = I/c = ... mm³ J = ... mm⁴ (if torsion)

Step 4: Calculate stresses σ = M×c/I = ()()/() = ___ MPa [show substitution fully] τ = T×c/J = ()()/() = ___ MPa [show substitution fully]

Step 5: Combined stress σ_von = √(σ² + 3τ²) = √(()² + 3()²) = ___ MPa

Step 6: Factor of safety n = Sy/σ_von = / = ___ Is n ≥ n_design? → YES/NO

Step 7: Deflection (if required) δ = FL³/(3EI) = ()(³)/(()()) = ___ mm

Unit Check

Always verify units at each step: [N][m]/[m⁴] = [N/m²] = [Pa] ✓ [N·m][m]/[m⁴] = [N/m²] = [Pa] ✓

Verification

Check result makes physical sense:

• Is stress less than Sy? (elastic assumption valid)
• Does deflection seem reasonable for geometry?
• Does factor of safety match expected value for this application?
• Sanity check: if P doubles, does σ double? (linearity check)

Design Interpretation

• Is the design adequate? (n ≥ n_design = ___)
• What is the governing failure mode?
• What changes would most effectively increase safety factor?
  • Double section modulus: σ halved
  • Change material: Sy increased
  • Reduce load: σ proportionally reduced
• Is there a simpler/lighter design that achieves the same safety factor?

Example: Simply Supported Beam Under UDL

Given: W12×26 beam, L=4.57m (15ft), w=14.6kN/m (1000 lb/ft), Fy=248MPa (36ksi) Find: max stress, deflection, safety factor

Solution: Step 1 — Reactions (by symmetry): R_A = R_B = wL/2 = 14.6×4.57/2 = 33.4 kN

Step 2 — Max moment (at midspan): M_max = wL²/8 = 14.6×4.57²/8 = 38.1 kN·m = 38.1×10⁶ N·mm

Step 3 — Section properties (W12×26): Ix = 8.49×10⁶ mm⁴, c = 155mm [d/2], S_x = 54,810 mm³

Step 4 — Bending stress: σ = M/S = 38.1×10⁶/54,810 = 695 N/mm² ... [Wait — units wrong. W12×26 Sx = 33.4 in³ = 547,500 mm³] σ = 38.1×10⁶ N·mm / 547,500 mm³ = 69.6 MPa ✓

Step 5 — Safety factor: n = 248/69.6 = 3.56 ✓ (adequate for Fy=36ksi)

Step 6 — Deflection: δ = 5wL⁴/(384EI) = 5×14.6×4570⁴/(384×200,000×84.9×10⁶) = 6.2mm = L/738 < L/360 (limit) ✓

Conclusion: Design adequate. Bending governs. Deflection well within limit.