(a) A T-section beam is constructed by gluing two pieces of wood together as shown in the figure. The maximum stress in the glue…

(a) Glued T-section beam

Section properties. The section is an inverted T. The web (50 × 150 mm) sits on top of the flange (150 × 50 mm), and the glue line is at the web–flange interface, 50 mm above the base.

• Area: A = 7500 + 7500 = 15 000 mm²
• Neutral axis from the base: ȳ = (7500 × 125 + 7500 × 25)/15 000 = 75 mm. The glue line is therefore 25 mm below the NA.
• I = [50 × 150³/12 + 7500 × 50²] + [150 × 50³/12 + 7500 × 50²] = 32.81 × 10⁶ + 20.31 × 10⁶ = 53.125 × 10⁶ mm⁴
• First moment of the flange about the NA: Q = 7500 × 50 = 375 000 mm³. The glued width is b = 50 mm.

Beam actions (w in N/m, anticlockwise couple 4w at A). ΣM_A gives 2R_B − 2w + 4w = 0, so R_B = −w (downward) and R_A = 3w. M(x) = 3wx − wx²/2 − 4w and V(x) = 3w − wx. Over 0 ≤ x ≤ 2 m, M rises from −4w (hogging) at A to 0 at B, and V falls from 3w to w. Both are largest just right of A. The critical element P is taken there: M = −4w N·m and V = 3w N.

(i) Stress components at P (x along the beam, y upward).

• σₓ = −My/I = −(−4000w)(−25)/(53.125 × 10⁶) = −1.882 × 10⁻³ w MPa (compression; the hogging moment compresses the bottom).
• τ = VQ/(Ib) = 3w × 375 000/(53.125 × 10⁶ × 50) = 0.4235 × 10⁻³ w MPa. It acts downward on the +x face, so τₓᵧ = −0.4235 × 10⁻³ w.
• σᵧ = 0.

(ii) Principal stresses. Centre = σₓ/2 = −0.941 × 10⁻³ w; radius R = √[(0.941)² + (0.4235)²] × 10⁻³ w = 1.032 × 10⁻³ w.

• σ₁ = +0.0909 × 10⁻³ w, σ₂ = −1.973 × 10⁻³ w, τ_max = R = 1.032 × 10⁻³ w
• tan 2θₚ = 2τₓᵧ/σₓ = 0.450, so θₚ = 12.1°. σ₂ acts at 12.1° anticlockwise from the beam axis and σ₁ at 102.1°. τ_max acts on planes at 57.1° and −32.9°, with a normal stress of σₓ/2.

(iv) Maximum w.

• Tension: 0.0909 × 10⁻³ w ≤ 2, so w ≤ 22 000 N/m.
• Shear: 1.032 × 10⁻³ w ≤ 1.7, so w ≤ 1647 N/m. (On the glue plane itself, τ = 0.4235 × 10⁻³ w ≤ 1.7 gives 4014 N/m, which is less critical.)
• w_max ≈ 1647 N/m ≈ 1.65 kN/m. Shear governs.

At this load: σₓ = −3.10 MPa, τₓᵧ = −0.70 MPa; σ₁ = +0.15 MPa, σ₂ = −3.25 MPa; τ_max = 1.70 MPa with σ = −1.55 MPa.

(iii) Oriented elements (w = 1647 N/m):

A third element rotated 45° from the principal one (at 57.1°) carries τ_max = 1.70 MPa on all faces, with σ = −1.55 MPa on each face.

(b) Fixed–hinge–fixed beam: reaction at C and BMD

The two cantilevers AC and CB are joined by a hinge at C, and only AC carries w. Let the hinge transmit a vertical force R_C, acting upward on AC and downward on CB. The one extra unknown is found by compatibility: both tips must deflect equally at C.

• Tip of AC: δ = wL⁴/(8EI) − R_C L³/(3EI)
• Tip of CB: δ = R_C L³/(3EI)

Equating the two: wL⁴/8 = 2R_C L³/3, so R_C = 3wL/16.

Support reactions.

• A: V_A = wL − 3wL/16 = 13wL/16 (up) and M_A = wL²/2 − (3wL/16)L = 5wL²/16 (hogging).
• B: V_B = 3wL/16 (up) and M_B = 3wL²/16 (hogging).

Moments. Measure x from C towards A: M(x) = (3wL/16)x − wx²/2.

• M = 0 at C and at x = 3L/8 (point of contraflexure).
• Maximum sagging moment occurs at x = 3L/16: M = 9wL²/512 ≈ 0.0176wL².
• In CB, M varies linearly from 0 at C to −3wL²/16 at B.

Equilibrium check. Vertical forces: 13wL/16 + 3wL/16 = wL ✓.

(c) Balanced-failure strength of the 500 × 300 column (major axis)

When bending is about the major axis, D = 500 mm and b = 300 mm. The six bars then form three rows of 2–25 φ (982 mm² per row) along the depth.

• Effective cover: d′ = 40 + 8 + 12.5 = 60.5 mm
• Effective depth: d = 439.5 mm
• Row distances from the centroid: yᵢ = +189.5, 0 and −189.5 mm

Neutral axis. ε_y = 0.87 × 500/(2 × 10⁵) + 0.002 = 0.004175. At balanced failure, ε_cu = 0.0035 at the compression face and ε_y is reached in the extreme tension row. x_ub = 0.0035/(0.0035 + 0.004175) × 439.5 = 200.4 mm

Concrete.

• C_c = 0.36 f_ck b x_u = 0.36 × 25 × 300 × 200.4 = 541.1 kN
• x̄ = 0.42x_u = 84.2 mm
• M_c = 541.1 × (250 − 84.2) = 89.73 kN·m

Steel (strains from the linear strain diagram, stresses from the given table):

• 1 (compression, 60.5): Strain — +0.002443; fₛ (MPa) — 399.1; f_c (MPa) — 11.15; Force (kN) — +380.9; yᵢ (mm) — +189.5; Moment (kN·m) — +72.18
• 2 (centre, 250): Strain — −0.000866; fₛ (MPa) — −173.1; f_c (MPa) — 0; Force (kN) — −169.9; yᵢ (mm) — 0; Moment (kN·m) — 0
• 3 (tension, 439.5): Strain — −0.004175; fₛ (MPa) — −434.8; f_c (MPa) — 0; Force (kN) — −426.9; yᵢ (mm) — −189.5; Moment (kN·m) — +80.89

The row 1 stress is interpolated between 0.00226 (391.3 MPa) and 0.00277 (413.0 MPa), and f_c = 0.446 f_ck because ε exceeds 0.002. The totals are C_s = −215.9 kN and M_s = 153.07 kN·m.

Balanced design strength components.

• P_ub = C_c + C_s = 541.1 − 215.9 = 325.3 kN
• M_ub = M_c + M_s = 89.73 + 153.07 = 242.8 kN·m
• e_b = M_ub/P_ub ≈ 746 mm

A load above 325 kN, or an eccentricity below 746 mm, gives compression failure; the converse gives tension failure.