Bracket Material Choice When Deflection Governs
A cantilever support bracket is a solid rectangular bar of width b = 40 mm and depth h = 80 mm, with length 1.2 m, loaded by a 2.0 kN point load at the free end. Two materials are stocked: hot-rolled steel with E = 200 GPa, ρ = 7850 kg/m³ and σ_allow = 160 MPa; and 6061-T6 aluminium with E = 69 GPa, ρ = 2700 kg/m³ and σ_allow = 110 MPa. The bracket must satisfy a tip deflection limit of L/250 and a stress limit with a factor of safety of 1.5 on the stated allowable. A designer proposes aluminium to save weight, arguing that the bending stress is only about 56 MPa and therefore safe. The mounting is assumed rigid and the load is static. The team wants the lightest feasible prismatic bar of the same 2:1 depth-to-width ratio, using only the stocked materials.
1. Check the proposed 40 mm × 80 mm section in both materials against the stress and deflection requirements. Is the designer's stress-only argument sufficient?
Read the worked answer
The second moment is I = b h³/12 = 0.040 × 0.080³/12 = 1.707e-6 m⁴. The maximum bending moment is M = PL = 2000 × 1.2 = 2400 N·m. The maximum stress is σ = M c/I = 2400 × 0.040/1.707e-6 = 56.3 MPa. With a factor of safety of 1.5, the design allowables are 160/1.5 = 106.7 MPa for steel and 110/1.5 = 73.3 MPa for aluminium. The stress is acceptable in both materials. The deflection limit is L/250 = 1.2/250 = 4.8 mm. For steel, δ = PL³/(3EI) = 2000 × 1.2³/(3 × 200e9 × 1.707e-6) = 3.37 mm, which passes. For aluminium, δ = 2000 × 1.2³/(3 × 69e9 × 1.707e-6) = 9.78 mm, which fails. The stress-only argument is therefore insufficient: the aluminium bar is strong enough but too flexible, and deflection governs the design.
2. Find the minimum depth h (keeping h/b = 2) for aluminium to satisfy the deflection limit, and compare its mass with the steel bar.
Read the worked answer
For a 2:1 section, b = h/2 and I = b h³/12 = h⁴/24. The required I for aluminium is I_req = PL³/(3Eδ_allow) = 2000 × 1.728/(3 × 69e9 × 0.0048) = 3.48e-6 m⁴. Then h⁴ = 24 × 3.48e-6 = 8.35e-5 m⁴, so h = (8.35e-5)^(1/4) = 0.0956 m = 95.6 mm and b = 47.8 mm. The aluminium volume is 0.0478 × 0.0956 × 1.2 = 5.49e-3 m³, giving mass 5.49e-3 × 2700 = 14.8 kg. The original steel bar has volume 0.040 × 0.080 × 1.2 = 3.84e-3 m³ and mass 3.84e-3 × 7850 = 30.1 kg. The resized aluminium bar is therefore feasible and about half the mass of the steel bar. It also has lower stress because the section is larger. The correct comparison is not the original aluminium bar versus steel, but the resized aluminium bar versus steel.
3. If only the existing 40 mm × 80 mm section can be used, which material meets both limits, and what is the mass penalty relative to the aluminium proposal?
Read the worked answer
With the fixed 40 × 80 section, steel passes both checks: stress 56.3 MPa < 106.7 MPa and deflection 3.37 mm < 4.8 mm. Aluminium fails the deflection check at 9.78 mm > 4.8 mm, even though its stress is acceptable. The steel bar has mass 30.1 kg. The original aluminium proposal has volume 3.84e-3 m³ and mass 3.84e-3 × 2700 = 10.4 kg. The mass penalty for choosing the feasible steel option is therefore 30.1 - 10.4 = 19.7 kg, roughly a factor of 2.9. This penalty is real only if the section cannot be changed. If the section can be resized, the aluminium bar at 95.6 mm × 47.8 mm is feasible at 14.8 kg and is the lighter option. The judgement is to identify which constraint governs before selecting the material.