CELE Strength of Materials Reviewer 2026

Answer: A. σ = My / I

Step 1 — Recall the derivation: plane sections remain plane under bending, so strain (and stress for elastic material) varies linearly with distance y from the neutral axis. Step 2 — Equilibrium of the cross-section gives σ = My/I, where M is the bending moment and I is the second moment of area about the neutral axis. Step 3 — Option B (VI/Q) is not a real formula. Option C (VQ/Ib) is the shear stress formula τ, not bending stress. Option D (M/A) has no distance term and gives the wrong units (N·mm/mm² = MPa only accidentally — but it ignores depth variation). Step 4 — The correct formula is σ = My/I.

Answer: A. S = I / c

Step 1 — The maximum bending stress formula is σ_max = Mc/I. Step 2 — Rearranging, σ_max = M / (I/c). The quantity I/c is defined as the section modulus S, so σ_max = M/S. Step 3 — S = I/c is a property of the cross-section shape alone, making it a convenient design parameter. Step 4 — Option B (I × c) would have units of mm⁵ — nonsensical for stress calculations. Option C (Q/b) is related to shear stress, not the section modulus. Option D (A × c) is not a standard section property. Step 5 — Remember: bigger S → lower bending stress for the same moment.

Answer: A. 2.25 × 10⁶ mm³

Step 1 — For a rectangle: S = bh²/6. Step 2 — Substitute b = 150 mm and h = 300 mm: S = 150 × (300)² / 6. Step 3 — (300)² = 90,000 mm². Step 4 — S = 150 × 90,000 / 6 = 13,500,000 / 6 = 2,250,000 mm³ = 2.25 × 10⁶ mm³. Step 5 — Option B (4.50 × 10⁶) is twice the answer — a common error of forgetting the '6' in the denominator and dividing by 3 instead. Option C is half the answer. Option D has units of mm⁴ — that is the moment of inertia I, not S.

Answer: A. 5.33 MPa

Step 1 — Maximum moment for a simply supported beam with UDL: M_max = wL²/8 = 6(4)²/8 = 12 kN·m = 12 × 10⁶ N·mm. Step 2 — Section modulus: S = bh²/6 = 150(300)²/6 = 2.25 × 10⁶ mm³. Step 3 — Maximum bending stress: σ_max = M/S = (12 × 10⁶) / (2.25 × 10⁶) = 5.33 MPa. Step 4 — Option B (2.67 MPa) results from incorrectly computing M = wL²/16 instead of /8. Option C (10.67 MPa) comes from doubling the moment. Option D (0.40 MPa) is the maximum shear stress for this beam — a common mix-up.

Answer: A. τ = VQ / Ib

Step 1 — The horizontal shear stress formula is derived by considering the equilibrium of a longitudinal slice of the beam. Step 2 — τ = VQ/(Ib), where V is shear force, Q is the first moment of area of the section above (or below) the cut level about the neutral axis, I is the moment of inertia of the whole section, and b is the width at the cut level. Step 3 — Option B (My/I) is the bending stress formula — a very common mix-up in the board exam. Option C (V/A) gives only an average shear stress, not the distribution. Option D replaces I with A — dimensionally incorrect (gives N/mm³ instead of N/mm² = MPa).

Answer: A. τ_max = 3V / 2A

Step 1 — Applying τ = VQ/(Ib) at the neutral axis of a rectangle (where shear is maximum): Q at NA = b(h/2)(h/4) = bh²/8. Step 2 — I = bh³/12 and b is the width. So τ_max = V(bh²/8) / [(bh³/12)(b)] = (3V)/(2bh) = 3V/(2A). Step 3 — The factor 1.5 (= 3/2) is the key number to memorise for rectangular sections. Step 4 — Option B (V/A) is the average shear stress — the maximum is 50% higher. Option C (4V/3A) is the formula for a solid circular section. Option D (2V/A) is too high and does not correspond to any standard section.

Answer: A. 0.40 MPa

Step 1 — Cross-sectional area: A = 150 × 300 = 45,000 mm². Step 2 — Apply the rectangular maximum shear formula: τ_max = 3V/(2A). Step 3 — Convert V to Newtons: V = 12 kN = 12,000 N. Step 4 — τ_max = 3(12,000) / [2(45,000)] = 36,000 / 90,000 = 0.40 MPa. Step 5 — Option B (0.27 MPa) comes from using τ = V/A (average shear stress without the 1.5 factor). Option C (0.60 MPa) uses a factor of 2 instead of 1.5. Option D (5.33 MPa) is the maximum bending stress for this same beam under UDL — a common mix-up.

Answer: A. Maximum bending stress occurs at the extreme fibers; maximum shear stress occurs at the neutral axis.

Step 1 — Bending stress: σ = My/I. This is directly proportional to y, the distance from the neutral axis. At y = 0 (neutral axis), σ = 0. At y = c (extreme fiber), σ is maximum. Step 2 — Shear stress: τ = VQ/(Ib). Q = 0 at the top and bottom fibers (no area beyond the extreme fiber), so τ = 0 there. Q is maximum at the neutral axis, so τ is maximum at the NA. Step 3 — These two distributions are opposite to each other — a key conceptual point that frequently appears in board examinations. Step 4 — Options B, C, and D all incorrectly pair the locations.

Answer: A. q = VQ / I

Step 1 — Shear flow q is defined as the shear force per unit length acting along the interface between connected pieces. Step 2 — q = VQ/I (units: N/mm), where Q is the first moment of the attached area about the neutral axis. Step 3 — Note the difference from the shear stress formula: τ = VQ/(Ib) divides by the width b as well, giving stress in MPa. Shear flow q does NOT divide by b. Step 4 — The connector spacing is then found from s = F/q, where F is the shear capacity of one connector (nail, bolt, or weld per unit length). Step 5 — Option D (M/S) is the maximum bending stress formula — completely different.

Answer: A. σ = P/A ± Mc/I

Step 1 — When a member carries both an axial force P and a bending moment M, superposition applies (for linearly elastic material). Step 2 — The axial load produces a uniform stress P/A on the cross-section (compressive if the load is compressive). Step 3 — The bending moment produces ±Mc/I at the extreme fibers (tension on one side, compression on the other). Step 4 — Adding both: σ = P/A ± Mc/I, where + and − apply to the two opposite extreme fibers. Step 5 — Option B incorrectly restricts M/S to only one fiber. Option C ignores the axial load entirely. Option D incorrectly uses the shear stress formula in a combined-stress context.

Answer: A. Yes, because t/r = 0.08, which is less than or equal to 0.10.

Step 1 – Apply the thin-wall criterion: t/r ≤ 1/10 = 0.10. Step 2 – Compute the ratio: t/r = 40 mm / 500 mm = 0.08. Step 3 – Compare: 0.08 ≤ 0.10, so the thin-wall assumption is valid. Option B is wrong because no 0.05 limit exists in standard practice. Option C states an incorrect ratio of 0.125. Option D is wrong because the thin-wall criterion is purely a ratio — absolute size does not matter.

Answer: B. 40 MPa

Step 1 – Identify the hoop stress formula: σ_h = pD / (2t). Step 2 – Substitute values: p = 2 MPa, D = 400 mm, t = 10 mm. Step 3 – Compute: σ_h = (2 × 400) / (2 × 10) = 800 / 20 = 40 MPa. Option A (20 MPa) is what you get if you mistakenly use the longitudinal formula but with D in the numerator. Option C (80 MPa) results from using 2t in the numerator or dropping the factor of 2 in the denominator. Option D (10 MPa) comes from using 4 × 2t in the denominator.