CELE Steel & Timber Design Reviewer 2026

Answer: A. Pn = Fy × Ag

Step 1 – Identify the two limit states: (1) tensile yielding on gross area, (2) tensile rupture on effective net area. Step 2 – For yielding, NSCP 2015 Section 504.2 (AISC 360 D2-1) states Pn = Fy·Ag, where Ag is the full, unreduced gross cross-sectional area. Step 3 – The resistance factor for yielding is φt = 0.90, so the design strength is φtPn = 0.90·Fy·Ag. Step 4 – Option B (Fu·Ae) is the rupture formula, not the yield formula. Option C incorrectly mixes φ into the nominal strength. Option D uses the net area An, which applies to rupture, not yielding. Yielding uses the gross area because, at early loading, the full section resists the force before any fracture occurs.

Answer: C. 0.75

Step 1 – NSCP 2015 (AISC 360 D2-2) distinguishes two φ values for tension members. Step 2 – Tensile yielding (ductile, gradual): φt = 0.90. Step 3 – Tensile rupture (sudden, brittle fracture through holes): φt = 0.75. Step 4 – The lower φ for rupture reflects greater uncertainty and the sudden, non-ductile nature of fracture failure. Step 5 – Option A (0.90) is for yielding. Option B (0.85) is used for bearing in some provisions. Option D (0.70) is not a standard NSCP tension φ. Always remember: rupture is more dangerous → lower φ.

Answer: C. 22 mm

Step 1 – The bolt hole is made slightly larger than the bolt diameter to allow insertion. Step 2 – NSCP 2015 / AISC 360 specifies that the standard punched-hole diameter = bolt diameter + 1.6 mm (1/16 in); then for net-area deduction an additional 1.6 mm is added for hole damage, totalling bolt diameter + 3.2 mm ≈ bolt + 3 mm in common Philippine practice, or simply bolt + 2 mm in many review references. Step 3 – Most local review sources use dh = db + 2 mm as the clearance allowance for net area: dh = 20 + 2 = 22 mm. Step 4 – Option A (20 mm) ignores the clearance allowance entirely. Option B (21 mm) uses only +1 mm. Option D (24 mm) overstates the hole. Always add the clearance to the bolt diameter when computing An.

Answer: A. 1 320 mm²

Step 1 – Gross area: Ag = 150 × 10 = 1 500 mm². Step 2 – Hole deduction for one bolt: dh × t = 18 × 10 = 180 mm². Step 3 – Net area: An = Ag − (dh × t) = 1 500 − 180 = 1 320 mm². Step 4 – Only one bolt in the critical cross-section, so only one hole is deducted. Step 5 – Common mistake: using the bolt diameter (16 mm) instead of the design hole diameter (18 mm). Always use dh, not db, for the net-area calculation.

Answer: D. 1.00

Step 1 – Shear lag occurs when only part of a cross-section is connected; the unconnected elements carry less load, reducing effectiveness. Step 2 – The effective net area is Ae = U·An, where U is the shear-lag factor (0 < U ≤ 1.0). Step 3 – When the load is transmitted uniformly through ALL cross-sectional elements (e.g., a flat plate bolted across its full width), there is no shear lag, so U = 1.0 and Ae = An. Step 4 – For angles connected by only one leg, U < 1.0 (typically 0.80–0.85 depending on bolt count). Step 5 – Options A, B, C are values applicable to specific shear-lag cases for angles and channels, not the ideal full-connection case.

Answer: B. The smaller of 0.90·Fy·Ag and 0.75·Fu·Ae

Step 1 – Two limit states must each be checked: yielding (φtPn = 0.90·Fy·Ag) and rupture (φtPn = 0.75·Fu·Ae). Step 2 – Design is governed by the LOWER (more critical) of the two computed values — the member fails at the weaker condition. Step 3 – Taking the larger value would be unconservative and could lead to failure. Step 4 – Neither limit state can be ignored; both must always be computed and compared. Step 5 – In Example 1 of the reference, yielding gave 535.7 kN and rupture gave 561.6 kN → yielding governed because it was lower.

Answer: C. s²/(4g)

Step 1 – When bolt holes are staggered, the diagonal failure path is longer than a straight horizontal cut, so the effective net width is slightly larger. Step 2 – The Cochrane formula adds s²/(4g) for each diagonal (zig-zag) segment in the failure path. Step 3 – Here, s = longitudinal pitch (spacing along the member axis) and g = transverse gage (spacing perpendicular to the member axis). Step 4 – Net width = Wg − ΣdH + Σ(s²/4g), where the summation covers all diagonal segments in the chosen failure path. Step 5 – Option A uses 2g in the denominator (wrong). Option B inverts s and g incorrectly. Option D confuses which variable is squared.

Answer: C. 300

Step 1 – Unlike compression members, tension members do not buckle, so there is no mandatory slenderness limit based on strength. Step 2 – However, very slender tension members can sag excessively under self-weight or vibrate under dynamic loads, causing serviceability problems. Step 3 – NSCP 2015 (AISC 360 D1) therefore recommends L/r ≤ 300 for tension members (rods are excluded from this recommendation). Step 4 – This is a recommendation, not a mandatory code requirement — the designer should use engineering judgment. Step 5 – Option A (200) is the limit for compression members (columns). Options B and D are not standard NSCP limits.

Answer: B. Sudden, brittle fracture through the net section at the bolt holes

Step 1 – Tensile yielding is the gradual, ductile elongation of the gross section as steel reaches Fy across the full area. Step 2 – Tensile rupture is fracture — a sudden, non-ductile failure — that occurs at the weakest cross-section, which is the net section through the bolt holes where the area is smallest. Step 3 – Because rupture is sudden (no warning), it is assigned a lower resistance factor (φ = 0.75) than yielding (φ = 0.90). Step 4 – Option A describes yielding, not rupture. Option C (lateral buckling) is a compression/bending failure mode, irrelevant to pure tension. Option D describes block shear or bolt bearing, which is a different limit state.

Answer: C. Ae < An because U < 1.0 reduces the effectiveness of the unconnected leg

Step 1 – When an angle is bolted through only one leg, the force must 'lag' (transfer by shear) from the connected leg into the unconnected leg. Step 2 – This unequal load distribution means the unconnected leg is not fully effective at the connection zone. Step 3 – The shear-lag factor U (< 1.0 for one-leg connections) accounts for this: Ae = U·An, so Ae < An. Step 4 – Option A is wrong: the unconnected element reduces, not increases, the effective area. Option B ignores shear lag entirely. Option D is overly conservative — the unconnected leg does contribute, just at a reduced level quantified by U.

Answer: A. Fe = π²E / (KL/r)²

Step 1 — Recall Euler's column formula: the elastic critical stress equals π²E divided by the square of the effective slenderness ratio (KL/r)². Step 2 — Option A matches this exactly; this is the standard NSCP 2015 / AISC 360 expression for Fe. Step 3 — Option B is wrong because the slenderness is not squared. Step 4 — Option C incorrectly incorporates the 0.877 imperfection factor inside Fe; 0.877 is applied to Fe when computing Fcr in the elastic range, not to Fe itself. Step 5 — Option D has the wrong constant and missing π².

Answer: B. 133.7

Step 1 — The transition slenderness is given by 4.71√(E/Fy). Step 2 — Substitute: 4.71 × √(200,000 / 248) = 4.71 × √806.45. Step 3 — √806.45 ≈ 28.40. Step 4 — 4.71 × 28.40 ≈ 133.7. Step 5 — Columns with KL/r ≤ 133.7 buckle inelastically; those with KL/r > 133.7 buckle elastically. Option 113.4 corresponds to Fy = 345 MPa, a common mix-up.