Worked Design Checks
Complete strength checks of a bracket, a shaft, a column and a bolted joint, from loads to safety factor.
The method every check follows
A strength check is the same six steps whether the part is a bracket or a bridge: 1. find the loads (and the worst combination); 2. find the internal forces and moments; 3. compute the stresses at the critical point with the section properties; 4. choose the failure criterion that matches the failure mode (yield, buckling, fatigue, bolt proof strength); 5. compute the safety factor n = strength ÷ stress; 6. check deflection and the second failure modes you have not yet examined. The four checks below use the formulas of the other pages of this guide: Stress, Strain and Safety Factors, Beam Bending and Deflection, Columns and Buckling, Torsion, Shear and Combined Loads and Bolted and Welded Joints. Steel is taken as E = 210 000 MPa.
Check 1: a cantilever bracket arm
- Given. A flat steel arm (S235, Sy = 235 MPa), 10 mm wide and 40 mm deep, sticks out 300 mm from a wall and carries 1 200 N at its tip, with the 40 mm side vertical.
- Moment. The worst section is at the wall: M = P L = 1 200 × 300 = 360 000 N·mm.
- Section. S = b h² ÷ 6 = 10 × 40² ÷ 6 = 2 667 mm³ and I = b h³ ÷ 12 = 53 333 mm⁴.
- Stress and safety factor. σ = M ÷ S = 360 000 ÷ 2 667 = 135 MPa, so n = 235 ÷ 135 = 1.74 against yield.
- Deflection. δ = P L³ ÷ (3 E I) = 1 200 × 300³ ÷ (3 × 210 000 × 53 333) = 0.96 mm, about L ÷ 311, inside a limit of L ÷ 180 = 1.67 mm.
- Lessons. Turning the arm flat (40 mm side horizontal) would reduce S to 667 mm³ and quadruple the stress; and if the load is applied thousands of times, the weld or bolts at the wall must be checked for fatigue (Fatigue and Stress Concentration).
Check 2: a shaft under bending and torque
- Given. A 30 mm shaft of AISI 1045 (Sy = 530 MPa) carries M = 120 N·m and T = 90 N·m (see Shafts, Keys and Couplings).
- Stresses. σ = 32 M ÷ (π d³) = 45.3 MPa and τ = 16 T ÷ (π d³) = 17.0 MPa.
- von Mises. σ′ = √(σ² + 3 τ²) = √(45.3² + 3 × 17.0²) = 54.0 MPa, so the static safety factor is n = 530 ÷ 54.0 = 9.8.
- But a rotating shaft is a fatigue problem. The same shaft with a shoulder fillet, machined finish and 99 percent reliability has a fatigue safety factor of only about 2.2 (the full calculation is in Fatigue and Stress Concentration). Static strength alone would have hidden that: always run the fatigue check on rotating parts.
Check 3: a tubular column
- Given. A pinned-pinned steel tube 48.3 × 3.2 mm (S235) is 1 500 mm long and carries 8 kN of compression.
- Section. Inside diameter 41.9 mm; A = 453 mm²; I = 115 857 mm⁴; radius of gyration r = √(I ÷ A) = 16.0 mm.
- Slenderness. L ÷ r = 1 500 ÷ 16.0 = 94. The limit between short and long columns for this steel is π√(2E ÷ Sy) = 133, so this is a long column and Euler’s formula applies.
- Buckling load. Pcr = π² E I ÷ L² = π² × 210 000 × 115 857 ÷ 1 500² = 106.7 kN, a safety factor of 106722.7 ÷ 8 000 = 13.3 against buckling.
- Crushing check. σ = P ÷ A = 17.6 MPa, only 8 percent of the yield strength: buckling, not crushing, governs. With a fixed end the effective length falls to about 0.7 L and Pcr roughly doubles; with a free end (a flagpole) it is 2 L and Pcr falls to a quarter.
Check 4: a preloaded bolted joint
- Given. An M10 class 8.8 bolt (tensile-stress area 58 mm², proof strength 600 MPa) is tightened to a preload Fi = 26.1 kN and then carries an external tension P = 10 kN. The joint stiffness ratio is C = kb ÷ (kb + km) = 0.25 (the bolt is soft compared with the clamped plates).
- Bolt load. Fb = Fi + C P = 26.1 + 0.25 × 10 = 28.6 kN: the bolt feels only a quarter of the external load.
- Proof check. σ = Fb ÷ At = 28 600 ÷ 58 = 493 MPa, so n = 600 ÷ 493 = 1.22 on proof strength.
- Clamping. The clamp force left is Fm = Fi − (1 − C) P = 18.6 kN, still compressing the plates. The joint would separate at P = Fi ÷ (1 − C) = 34.8 kN, a factor of 3.48 above the working load.
- Remember. These numbers are the same as in Bolted and Welded Joints. A bolt in dynamic tension also needs the fatigue check, where the preload and the small stiffness ratio keep the alternating stress low.
Summary and typical safety factors
| Check | Governing mode | Result | Typical required n |
|---|---|---|---|
| Bracket arm | Bending yield; tip deflection | n = 1.74; δ = 0.96 mm | 1.5 to 2 for static loads; deflection limit set by function |
| Shaft | Fatigue (not static yield) | static n = 9.8; fatigue n ≈ 2.2 | 1.5 to 2.5 on fatigue |
| Column | Euler buckling | n = 13.3 | 2 to 3 or more; a higher value for slender or uncertain columns |
| Bolted joint | Proof load; separation | n = 1.22; separation 3.48 | 1.2 to 1.5 on proof when preload is controlled; 2 or more on separation |
A safety factor is a judgment about how well the loads, the material and the analysis are known, not a property of the part. Use the low end for well-known loads, tested materials and controlled production, and the high end where a failure endangers people, loads are uncertain or inspection is difficult. Codes (building, pressure vessel, lifting) fix the factors for their field: follow them.
Related
- Structures and Strength of Materials: the overview of this guide.
- Beam Bending and Deflection, Section Properties and Structural Profiles and Columns and Buckling: the formulas used here.
- Material Properties: yield and ultimate strengths for the material you choose.
- Bolt Torque and Thread Engagement tool: the tightening torque for the preload in Check 4.
FAQ
Which safety factor should I use?
Use the one your code or standard requires. Where none applies, start from 1.5 to 2 for well-understood static loads and raise it for uncertain loads, brittle materials, hazards to people and parts that cannot be inspected.
Why check deflection as well as stress?
A part can be strong enough and still too flexible: a machine bracket that moves a millimeter ruins the alignment. Stiffness limits often govern, especially for long, light members.
When do I need finite element analysis?
When the geometry or the loading is too complex for the formulas: irregular shapes, contact, local stress concentrations. Hand calculations are still the way to check that the FEA result is believable.
References
- Shigley, J. E. and Mischke, C. R., Mechanical Engineering Design, McGraw-Hill.
- Beer, F. P., Johnston, E. R., Mechanics of Materials, McGraw-Hill; Gere, J. M., Mechanics of Materials.
- Young, W. C. and Budynas, R. G., Roark’s Formulas for Stress and Strain, McGraw-Hill.
Last reviewed: October 2026. These are worked illustrations, not a design for any particular application: check the loads, material and code for your own part.