(a) An aluminium tensile specimen has a diameter of 30·50 mm and a gauge length 275 mm. If the force of 17·50 × 10⁴ N elongates…

(a) Aluminium specimen: E, ν and change in diameter

Given: d = 30.50 mm, L = 275 mm, P = 17.50 × 10⁴ N, ΔL = 1.28 mm, G = 22 GPa, σ_y = 435 MPa.

A = π(30.50)²/4 = 730.6 mm². Stress σ = 175 000/730.6 = 239.5 MPa, below σ_y = 435 MPa, so the behaviour is elastic. Axial strain ε = 1.28/275 = 4.655 × 10⁻³.

E = σ/ε = 239.5/(4.655 × 10⁻³) = 51.46 × 10³ MPa, about 51.5 GPa.

From E = 2G(1 + ν): ν = E/(2G) − 1 = 51.46/44 − 1 = 0.17.

Lateral strain = −νε = −0.1695 × 4.655 × 10⁻³ = −7.89 × 10⁻⁴, so Δd = −7.89 × 10⁻⁴ × 30.50 = −0.024 mm. The diameter contracts by about 0.024 mm.

(b) Maximum permissible torque on the 65 mm shaft

J = πd⁴/32 = π × 65⁴/32 = 1.7525 × 10⁶ mm⁴.

Strength: T₁ = τJ/r = 60 × 1.7525 × 10⁶/32.5 = 3.235 × 10⁶ N·mm = 3.24 kN·m.

Stiffness: θ = 1.05° per m = 0.018326 rad/m = 1.8326 × 10⁻⁵ rad/mm. From θ = T/(GJ), T₂ = 80 000 × 1.7525 × 10⁶ × 1.8326 × 10⁻⁵ = 2.569 × 10⁶ N·mm = 2.57 kN·m.

The smaller value governs, so T_max = 2.57 kN·m, controlled by the twist limit. At this torque the maximum shear stress is 2.569 × 10⁶ × 32.5/1.7525 × 10⁶ = 47.6 MPa, below 60 MPa.

(c) Strong-box: largest force F and highest position h

W = 85 × 9.81 = 833.9 N, acting at the centre, 0.35 m from each edge (width 0.70 m).

No sliding: N = W, so F_max = μN = 0.25 × 833.9 = 208.5 N.

No tipping: at the point of tipping the box pivots about its front bottom edge, and N and friction act at that edge. Moments about it give F·h = W × 0.35, so h = 833.9 × 0.35/208.5 = 0.35/μ = 1.40 m.

A force of 208.5 N applied at any height up to 1.40 m neither slides nor tips the box. At exactly 1.40 m both become imminent together. Applied above 1.40 m, the same force tips the box before it slides.

(d) Prestressed I-beam: balancing force and pressure-line shift

Section: flanges 250 × 80 mm, web 80 mm wide and 290 mm high, depth 450 mm. A = 2 × 250 × 80 + 80 × 290 = 63 200 mm². Self-weight g = 0.0632 × 24 = 1.517 kN/m. Total load w = 1.517 + 2.5 = 4.017 kN/m.

(i) A parabolic cable with mid-span eccentricity e exerts an upward load 8Pe/L². Equating it to w: P = wL²/(8e) = 4.017 × 8²/(8 × 0.150) = 214.2 kN.

(ii) Mid-span moment M = wL²/8 = 4.017 × 64/8 = 32.13 kN·m. The pressure line lies M/P from the tendon line: 32.13/214.2 = 0.150 m = 150 mm, equal to the eccentricity. The pressure line therefore coincides with the centroidal axis all along the span, since both M and the eccentricity vary parabolically, and the beam carries uniform compression P/A = 214.2 × 10³/63 200 = 3.39 MPa with no bending.

(e) Site-welded ISA 75 × 50 × 8 tie

Design strength of the member: T_dg = A_g f_y/γ_m0 = 938 × 250/1.10 = 213.2 kN.

Weld: throat t_t = K s = 0.7 × 6 = 4.2 mm. f_wd = f_u/(√3 γ_mw) = 410/(√3 × 1.5) = 157.8 MPa. Strength per mm = 157.8 × 4.2 = 662.8 N/mm.

End weld (75 mm): P₃ = 75 × 662.8 = 49.7 kN, acting 37.5 mm from the L_w1 line. The centroid is 25.2 mm from the L_w1 side. Moments about the L_w1 line: T × 25.2 = P₃ × 37.5 + P₂ × 75, so P₂ = (213.2 × 25.2 − 49.7 × 37.5)/75 = 46.8 kN. Then P₁ = 213.2 − 49.7 − 46.8 = 116.7 kN.

L_w1 = 116 700/662.8 = 176 mm, say 180 mm. L_w2 = 46 800/662.8 = 70.6 mm, say 75 mm. These are effective lengths; the total 180 + 75 + 75 = 330 mm exceeds the 321.6 mm needed overall.