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Question 1Structural engineering models and applications

A simply reinforced concrete rectangular beam has a cross section of b × h = 250 mm × 500 mm with an effective depth h₀ = 460 mm. The concrete grade is C30 (design axial compressive strength f_c = 14.3 N/mm², α₁ = 1.0, β₁ = 0.80) and the longitudinal reinforcement consists of 3Φ20 HRB400 bars (f_y = 360 N/mm², total area A_s = 942 mm²). According to GB 50010-2010 (2015 Edition), what is the design flexural moment capacity M_u of the beam section?

  1. 121.4 kN·m
  2. 139.9 kN·m Correct answer.
  3. 155.8 kN·m
  4. 168.2 kN·m

Explanation

Per GB 50010-2010 (Section 6.2.10): 1. Calculate the concrete compression zone depth x: x = (f_y · A_s) / (α₁ · f_c · b) = (360 N/mm² × 942 mm²) / (1.0 × 14.3 N/mm² × 250 mm) = 339,120 / 3,575 = 94.86 mm. 2. Check the relative compression depth limit for HRB400 (ξ_b = 0.518): x_max = ξ_b · h₀ = 0.518 × 460 mm = 238.28 mm. Since x = 94.86 mm ≤ 238.28 mm (and x ≥ 2a'_s = 80 mm), the section is ductile under-reinforced. 3. Compute the design flexural moment capacity M_u: M_u = α₁ · f_c · b · x · (h₀ - x / 2) = 1.0 × 14.3 × 250 × 94.86 × (460 - 94.86 / 2) = 339,120 × 412.57 N·mm = 139.91 × 10⁶ N·mm = 139.9 kN·m.

Question 2Structural engineering models and applications

According to the Code for Design of Concrete Structures (GB 50010-2010, 2015 Edition), what is the relative limit compression zone depth (界限相对受压区高度 ξ_b) for a flexural member with C30 concrete (ultimate compressive strain ε_cu = 0.0033, β₁ = 0.80) reinforced with HRB400 longitudinal steel bars (design yield strength f_y = 360 N/mm², modulus of elasticity E_s = 2.0 × 10⁵ N/mm²)?

  1. 0.485
  2. 0.518 Correct answer.
  3. 0.550
  4. 0.614

Explanation

Per GB 50010-2010 Section 6.2.7, the relative limit compression zone depth ξ_b is determined from strain compatibility at the balanced failure state: ξ_b = β₁ / [1 + f_y / (ε_cu · E_s)] = 0.80 / [1 + 360 / (0.0033 × 200,000)] = 0.80 / [1 + 360 / 660] = 0.80 / (1 + 0.54545) = 0.80 / 1.54545 ≈ 0.518. For HRB500 it is 0.482, and for HRB335 it is 0.550.

Question 3Structural engineering models and applications

A T-shaped reinforced concrete beam has flange width b'_f = 600 mm, flange thickness h'_f = 100 mm, web width b = 250 mm, and effective depth h₀ = 550 mm. The materials are C30 concrete (f_c = 14.3 N/mm², α₁ = 1.0) and HRB400 rebar (f_y = 360 N/mm²). If the design bending moment is M = 380 kN·m, how is this section classified per GB 50010-2010 Section 6.2.11?

  1. Class I T-beam (第一类T形截面), because M ≤ α₁ · f_c · b'_f · h'_f · (h₀ - h'_f / 2) Correct answer.
  2. Class II T-beam (第二类T形截面), because M > α₁ · f_c · b'_f · h'_f · (h₀ - h'_f / 2)
  3. Rectangular beam with width b = 250 mm, because the flange is in the tension zone.
  4. Over-reinforced section requiring revision of cross-sectional dimensions.

Explanation

Per GB 50010-2010 Section 6.2.11, the boundary moment capacity when the compression zone just fills the flange (x = h'_f = 100 mm) is: M_f = α₁ · f_c · b'_f · h'_f · (h₀ - h'_f / 2) = 1.0 × 14.3 N/mm² × 600 mm × 100 mm × (550 mm - 100 mm / 2) = 858,000 × 500 N·mm = 429.0 × 10⁶ N·mm = 429.0 kN·m. Since the applied design moment M = 380 kN·m ≤ M_f = 429.0 kN·m, the neutral axis lies within the flange (x < h'_f), which is classified as a Class I T-beam (第一类T形截面).

Question 4Structural engineering models and applications

In reinforced concrete eccentric compression member analysis (GB 50010-2010), which condition distinguishes large eccentric compression (大偏心受压, ductile tension failure) from small eccentric compression (小偏心受压, brittle compression failure)?

  1. e₀ > 0.5 h
  2. ξ = x / h₀ ≤ ξ_b Correct answer.
  3. N ≤ 0.5 f_c · b · h
  4. l₀ / h > 8

Explanation

Per GB 50010-2010 Section 6.2.17 and 6.2.18, the fundamental mechanical distinction between large and small eccentric compression is the relative compression zone depth ξ = x / h₀ compared to the balanced relative compression depth limit ξ_b: - When ξ ≤ ξ_b: Large eccentricity failure (tension steel yields before concrete reaches ultimate compressive strain); - When ξ > ξ_b: Small eccentricity failure (concrete crushes while tension/less compressed steel does not reach yield).

Question 5Structural engineering models and applications

A column with a square section of 400 mm × 400 mm transmits a design punching axial load F_l = 480 kN to a concrete flat slab of effective depth h₀ = 180 mm (total thickness h = 200 mm). The concrete grade is C30 (f_t = 1.43 N/mm², β_h = 1.0). According to GB 50010-2010 Section 6.5.1, what is the design punching shear capacity F_lu of the unreinforced concrete slab?

  1. 375.5 kN
  2. 418.0 kN Correct answer.
  3. 462.6 kN
  4. 512.4 kN

Explanation

Per GB 50010-2010 Section 6.5.1: 1. The critical punching perimeter $u_m$ is located at a distance of $h_0 / 2$ from the column face: Side of critical square perimeter $b_m = 400\text{ mm} + h_0 = 400\text{ mm} + 180\text{ mm} = 580\text{ mm}$. Critical perimeter $u_m = 4 \times 580\text{ mm} = 2,320\text{ mm}$. 2. Slab thickness factor $\beta_h = 1.0$ (for $h \le 800\text{ mm}$). 3. Punching shear capacity of concrete: $F_{\text{lu}} = 0.7 \beta_h f_t u_m h_0 = 0.7 \times 1.0 \times 1.43\text{ N/mm}^2 \times 2,320\text{ mm} \times 180\text{ mm} = 418,018\text{ N} \approx 418.0\text{ kN}$.

Question 6Structural engineering models and applications

A Grade HRB400 ribbed steel bar with diameter d = 20 mm (design tensile strength f_y = 360 N/mm²) is anchored into C30 concrete (design tensile strength f_t = 1.43 N/mm²). The rebar shape coefficient is α = 0.14. According to GB 50010-2010 Section 8.3.1, what is the basic anchorage length l_ab of the rebar?

  1. 503 mm
  2. 705 mm Correct answer.
  3. 840 mm
  4. 1,000 mm

Explanation

Per GB 50010-2010 Section 8.3.1: Basic anchorage length formula: l_ab = α · (f_y / f_t) · d = 0.14 × (360 N/mm² / 1.43 N/mm²) × 20 mm l_ab = 0.14 × 251.75 × 20 mm = 704.9 mm ≈ 705 mm (approx. 35.25d).

Question 7Structural engineering models and applications

A concrete frame column in a Seismic Grade 2 structure has C35 concrete (design axial compressive strength f_c = 16.7 N/mm²) and HPB300 stirrups (design tensile strength f_yv = 270 N/mm²). The stirrup characteristic value required by GB 50011-2010 Table 6.3.9 is λ_v = 0.12. What is the minimum volumetric stirrup ratio ρ_v required in the column plastic hinge confinement zone?

  1. 0.55%
  2. 0.74% Correct answer.
  3. 0.96%
  4. 1.15%

Explanation

Per GB 50011-2010 Section 6.3.9 and GB 50010-2010 Section 11.4.17: The volumetric stirrup ratio ρ_v is calculated as: ρ_v = λ_v · (f_c / f_yv) = 0.12 × (16.7 N/mm² / 270 N/mm²) = 0.12 × 0.06185 = 0.00742 = 0.742% (or 0.74%). Per code, ρ_v must also satisfy absolute minimum limits (≥ 0.60% for Grade 2). Thus, ρ_v = 0.74%.

Question 8Structural engineering models and applications

A steel tension plate fabricated from Q235B steel (design tensile strength f = 215 N/mm²) has a gross cross section b × t = 200 mm × 12 mm. Two bolt holes of diameter d₀ = 22 mm are aligned in a single transverse section. According to GB 50017-2017 Section 5.1.1, what is the design axial tensile capacity N_t of this steel plate?

  1. 386.4 kN
  2. 402.5 kN Correct answer.
  3. 442.8 kN
  4. 516.0 kN

Explanation

Per GB 50017-2017 Section 5.1.1: 1. Gross area A = 200 mm × 12 mm = 2,400 mm². 2. Net area A_n deducting 2 bolt holes: A_n = A - 2 · d₀ · t = 2,400 mm² - 2 × 22 mm × 12 mm = 2,400 - 528 = 1,872 mm². 3. Design tensile load capacity: N_t = A_n · f = 1,872 mm² × 215 N/mm² = 402,480 N = 402.48 kN ≈ 402.5 kN.

Question 9Structural engineering models and applications

In the net section area calculation of a steel tension member with staggered bolt holes per GB 50017-2017 Section 5.1.1, what is the formula for the net width b_n along a zig-zag failure path passing through holes with longitudinal pitch s and transverse gauge g?

  1. b_n = b - n · d₀ + Σ (s² / 4g) Correct answer.
  2. b_n = b - n · d₀ + Σ (s · g / 2)
  3. b_n = b - n · d₀ - Σ (s² / 2g)
  4. b_n = b - (n - 1) · d₀ + Σ (s / g)

Explanation

Per GB 50017-2017 Section 5.1.1, when bolt holes are staggered along the tension path, the net section width is calculated as b_n = b - n · d₀ + Σ (s² / 4g), where b is the gross width, n is the number of hole centers along the failure path, d₀ is the bolt hole diameter, s is the staggered pitch parallel to force, and g is the gauge perpendicular to force.

Question 10Structural engineering models and applications

An axially loaded pinned-pinned steel column of length L = 6.0 m (effective length factor μ = 1.0) is constructed from Q355 steel (design strength f = 305 N/mm² for t ≤ 16 mm) with gross cross-sectional area A = 8,000 mm² and minimum radius of gyration i_min = 75 mm. The member belongs to Buckling Curve b, for which the stability factor at slenderness ratio λ = 80 is φ = 0.688. What is the design axial compressive capacity N_u of the column per GB 50017-2017?

  1. 1,425 kN
  2. 1,678 kN Correct answer.
  3. 1,952 kN
  4. 2,440 kN

Explanation

Per GB 50017-2017 Section 7.1.1: 1. Slenderness ratio λ = l₀ / i_min = (1.0 × 6,000 mm) / 75 mm = 80. 2. Overall stability factor φ = 0.688 (given from Curve b lookup). 3. Design compressive capacity: N_u = φ · A · f = 0.688 × 8,000 mm² × 305 N/mm² = 1,678,720 N ≈ 1,678.7 kN (or 1,678 kN).

Question 11Structural engineering models and applications

A welded I-section steel beam carries a design bending moment M_x = 420 kN·m about its strong axis. The cross-section properties are: gross elastic section modulus W_nx = 1.85 × 10⁶ mm³, section class allows partial plastic development with γ_x = 1.05, and steel grade is Q355 (design bending strength f = 305 N/mm²). Lateral buckling is fully prevented by a continuous cast-in-place concrete slab. According to GB 50017-2017 Section 6.1.1, what is the maximum flexural bending stress σ in the beam?

  1. 185.6 N/mm²
  2. 216.2 N/mm² Correct answer.
  3. 227.0 N/mm²
  4. 248.5 N/mm²

Explanation

Per GB 50017-2017 Section 6.1.1, the flexural bending stress is calculated as: σ = M_x / (γ_x · W_nx) σ = (420 × 10⁶ N·mm) / (1.05 × 1.85 × 10⁶ mm³) = 420 × 10⁶ / 1,942,500 mm³ = 216.22 N/mm² ≤ f = 305 N/mm².

Question 12Structural engineering models and applications

A connection utilizes Grade 10.9 M20 high-strength bolts (pretension design value P = 155 kN per bolt) in a friction-type connection (摩擦型连接). The contact surfaces are blast-cleaned with shot (slip coefficient μ = 0.45), and the connection has a single friction shear plane (n_f = 1). According to GB 50017-2017 Section 11.2.1, what is the design shear capacity N_v^b of a single high-strength bolt?

  1. 55.8 kN
  2. 62.8 kN Correct answer.
  3. 69.8 kN
  4. 78.5 kN

Explanation

Per GB 50017-2017 Section 11.2.1, the design shear capacity of a single friction-type high-strength bolt is: N_v^b = 0.9 · n_f · μ · P N_v^b = 0.9 × 1 × 0.45 × 155 kN = 0.405 × 155 kN = 62.775 kN ≈ 62.8 kN.

Question 13Structural engineering models and applications

A steel tie plate connection transmits an axial tensile load N = 360 kN using side fillet welds (侧面角焊缝). The weld leg size is h_f = 8 mm (effective throat h_e = 0.7 h_f = 5.6 mm) on both sides of the plate (2 weld lines). The design strength of the fillet weld is f_f^w = 160 N/mm². According to GB 50017-2017 Section 11.1.2, what is the minimum required calculated design length l_w of each side fillet weld?

  1. 160 mm
  2. 201 mm Correct answer.
  3. 250 mm
  4. 402 mm

Explanation

Per GB 50017-2017 Section 11.1.2: 1. Total effective weld throat area required: A_we = N / f_f^w = 360,000 N / 160 N/mm² = 2,250 mm². 2. Total weld length for 2 symmetric side welds: Σ l_w = A_we / h_e = 2,250 mm² / (0.7 × 8 mm) = 2,250 / 5.6 = 401.79 mm. 3. Calculated length per side weld: l_w = 401.79 mm / 2 = 200.89 mm ≈ 201 mm (actual detailing length L = l_w + 2h_f = 201 + 16 = 217 mm).

Question 14Structural engineering models and applications

A bracket connected to a steel column by a bolt group of n = 6 high-strength bolts arranged in two vertical columns (3 bolts per column, vertical spacing s = 100 mm, horizontal column gauge g = 120 mm). The connection carries a design vertical shear load V = 120 kN acting at an eccentricity e = 250 mm from the centroid of the bolt group. What is the maximum resultant shear force N_max on the most critical corner bolt?

  1. 20.0 kN
  2. 48.7 kN
  3. 69.2 kN Correct answer.
  4. 78.1 kN

Explanation

1. Direct vertical shear force per bolt: $V_i = V / n = 120\text{ kN} / 6 = 20.0\text{ kN}$. 2. Applied eccentric torsional moment: $M = V \cdot e = 120\text{ kN} \times 250\text{ mm} = 30,000\text{ kN}\cdot\text{mm}$. 3. Polar second moment of bolt coordinates: $\sum r_i^2 = 6 \times (60\text{ mm})^2 + 4 \times (100\text{ mm})^2 = 61,600\text{ mm}^2$. 4. Torsional components on critical corner bolt ($x = 60\text{ mm}, y = 100\text{ mm}$): $N_{Mx} = \frac{M \cdot y}{\sum r_i^2} = \frac{30,000 \times 100}{61,600} = 48.70\text{ kN}$, and $N_{My} = \frac{M \cdot x}{\sum r_i^2} = \frac{30,000 \times 60}{61,600} = 29.22\text{ kN}$. 5. Combined vertical shear: $N_y = V_i + N_{My} = 20.0 + 29.22 = 49.22\text{ kN}$. 6. Total resultant shear force: $N_{\max} = \sqrt{N_{Mx}^2 + N_y^2} = \sqrt{48.70^2 + 49.22^2} = 69.24\text{ kN} \approx 69.2\text{ kN}$.

Question 15Structural engineering models and applications

In the design of an exposed column base plate (外露式柱脚底板) under axial compression N and bending moment M per GB 50017-2017 Section 12.7, if the eccentricity e = M / N ≤ L / 6 (where L is the base plate length parallel to bending), how is the maximum concrete bearing pressure σ_max under the base plate verified?

  1. σ_max = N / (B · L) · (1 + 6e / L) ≤ f_c Correct answer.
  2. σ_max = 2 N / [3 B · (L / 2 - e)] ≤ f_c
  3. σ_max = N / (B · L) ≤ 0.5 f_c
  4. σ_max = M / (B · L²) ≤ f_c

Explanation

Per GB 50017-2017 Section 12.7.2, when e ≤ L / 6, the resultant compression force falls within the middle third (kern) of the base plate, meaning the entire contact area remains under compression. The linear trapezoidal bearing stress distribution gives σ_max = N / (B · L) · (1 + 6e / L) ≤ f_c (or modified design bearing strength β_cc · f_c).

Question 16Structural engineering models and applications

A bearing-type high-strength bolt connection (承压型高强螺栓连接) uses Grade 8.8 M20 bolts (shear strength f_v^b = 250 N/mm², nominal diameter d = 20 mm, shank shear area A_v = 314 mm²). The bolt passes through a Q235 steel plate of thickness t = 10 mm (bearing strength f_c^b = 405 N/mm²). According to GB 50017-2017 Section 11.2.2, what is the design shear capacity N_b of a single bolt in single shear?

  1. 78.5 kN Correct answer.
  2. 81.0 kN
  3. 94.2 kN
  4. 102.5 kN

Explanation

Per GB 50017-2017 Section 11.2.2, the design capacity of a bearing-type bolt is the minimum of its shear capacity N_v^b and plate bearing capacity N_c^b: 1. Bolt shear capacity (single shear n_v = 1): N_v^b = n_v · (π · d² / 4) · f_v^b = 1 × 314.16 mm² × 250 N/mm² = 78,540 N = 78.54 kN. 2. Plate bearing capacity: N_c^b = d · Σ t · f_c^b = 20 mm × 10 mm × 405 N/mm² = 81,000 N = 81.0 kN. 3. Governing capacity: N_b = min(78.54 kN, 81.0 kN) = 78.5 kN (governed by bolt shear).

Question 17Structural engineering models and applications

In a rigid beam-to-column steel moment connection per GB 50017-2017 and JGJ 99-2015, how are the design bending moment M and design shear force V transmitted through the beam flanges and web into the column?

  1. Flanges transmit the entire shear force V, and the web transmits the entire bending moment M.
  2. Flanges transmit the bending moment M via axial flange forces N_f = M / (h - t_f), and the web transmits the entire shear force V. Correct answer.
  3. Flanges and web transmit equal proportions (50% each) of moment and shear.
  4. The entire moment and shear are transmitted purely through friction in the web bolt group.

Explanation

In standard structural steel rigid connections (flange-welded, web-bolted/welded), the classic structural mechanics distribution assumes the tension and compression flange couples carry the entire design moment M as axial forces N_f = M / (h - t_f), while the beam web plate carries the entire design transverse shear force V.

Question 18Structural engineering models and applications

In the design of cold-formed thin-walled steel members (GB 50018-2002 / GB 50017-2017), how is the post-buckling strength of slender plate elements utilized when width-to-thickness ratios exceed elastic buckling thresholds?

  1. By applying the Effective Width Method (有效宽度法), which replaces the gross width with an effective width b_e that carries the maximum edge yield stress. Correct answer.
  2. By reducing the steel yield strength f_y to an equivalent reduced stress across the entire gross area.
  3. By adding diagonal internal stiffener plates inside every section.
  4. Post-buckling strength is prohibited from being utilized in Chinese structural standards.

Explanation

GB 50018-2002 and GB 50017-2017 Section 7.3 adopt the Effective Width Method (有效宽度法 / Effective Width Criterion) derived from Winter's equation. When a thin plate buckles elastically, the center sheds load while the supported edges continue to resist stress up to yield. The calculation replaces gross width b with effective width b_e.

Question 19Structural engineering models and applications

A single-story unreinforced brick masonry wall has a height H = 4.2 m and a thickness h = 240 mm (sintered common brick with M7.5 mortar, allowable high-thickness ratio [β] = 24). The roof is a rigid diaphragm, giving an effective height factor μ = 1.0 (effective height H₀ = 1.0 H = 4,200 mm). The wall is solid with no openings (γ₁ = 1.0, γ₂ = 1.0). According to GB 50003-2011 Section 5.1.1, what is the wall's high-thickness ratio β and is it compliant?

  1. β = 14.5 ≤ [β] = 24, compliant
  2. β = 17.5 ≤ [β] = 24, compliant Correct answer.
  3. β = 21.0 ≤ [β] = 24, compliant
  4. β = 26.5 > [β] = 24, non-compliant

Explanation

For this question use the supplied numerical assumptions and the stated model only. Per GB 50003-2011 Section 5.1.1: 1. High-thickness ratio: β = H₀ / h = 4,200 mm / 240 mm = 17.5. 2. Allowable high-thickness ratio: [β] · γ₁ · γ₂ = 24 × 1.0 × 1.0 = 24. 3. Since β = 17.5 ≤ 24, the wall satisfies the code stability and slenderness limit. This is not independent confirmation of a current local code or complete design approval.

Question 20Structural engineering models and applications

An unreinforced rectangular brick masonry pier of cross section b × h = 370 mm × 490 mm (area A = 0.1813 m² < 0.3 m²) has design compressive strength f = 1.69 N/mm² (γ_a = A + 0.7 = 0.8813). Under eccentric axial load, the calculated high-thickness ratio is β = 12 and the relative eccentricity is e / h = 0.15, giving a stability coefficient φ = 0.65. With importance factor γ₀ = 1.0, what is the design axial compressive capacity N_u of this pier per GB 50003-2011 Section 5.1.1?

  1. 152.6 kN
  2. 175.4 kN Correct answer.
  3. 199.2 kN
  4. 235.8 kN

Explanation

Per GB 50003-2011 Section 5.1.1 and Section 3.2.3: 1. Cross-sectional area A = 0.37 m × 0.49 m = 0.1813 m² < 0.3 m². 2. Area strength reduction factor γ_a = A + 0.7 = 0.1813 + 0.7 = 0.8813. 3. Modified design compressive strength: f_design = γ_a · f = 0.8813 × 1.69 N/mm² = 1.4894 N/mm². 4. Design compressive capacity N_u: N_u = φ · f_design · A = 0.65 × 1.4894 N/mm² × (370 × 490 mm²) = 0.65 × 1.4894 × 181,300 N = 175,518 N ≈ 175.4 kN (or 175.5 kN).

Question 21Structural engineering models and applications

A reinforced concrete floor beam has a support reaction N_l = 85 kN bearing directly onto a 240 mm thick unreinforced brick wall (masonry design compressive strength f = 1.50 N/mm²). The beam width is b = 250 mm and bearing length is a = 190 mm (bearing area A_l = 250 mm × 190 mm = 47,500 mm²). The calculated local bearing strength enhancement factor is η = 1.45. According to GB 50003-2011 Section 5.2.1, what is the local compressive bearing capacity N_lu of the masonry?

  1. 71.3 kN
  2. 85.0 kN
  3. 103.3 kN Correct answer.
  4. 125.6 kN

Explanation

Per GB 50003-2011 Section 5.2.1, the local compressive bearing capacity of unreinforced masonry directly under concentrated load is: N_lu = η · f · A_l N_lu = 1.45 × 1.50 N/mm² × 47,500 mm² = 2.175 N/mm² × 47,500 mm² = 103,312.5 N ≈ 103.3 kN. Since N_lu = 103.3 kN ≥ N_l = 85 kN, the local bearing capacity is satisfied.

Question 22Structural engineering models and applications

In mesh-reinforced brick masonry columns (网状配筋砖砌体柱) per GB 50003-2011 Section 8.1.1, how does the transverse steel wire mesh enhance the column axial compressive capacity N_u?

  1. By resisting longitudinal tension directly like beam rebar
  2. By providing lateral confinement (triaxial compressive stress state in mortar and brick), formulated as N_u = φ_n · (f + 2 · ρ_v · f_y) · A Correct answer.
  3. By preventing temperature cracking without increasing compressive strength
  4. By doubling the modulus of elasticity E of masonry

Explanation

GB 50003-2011 Section 8.1.1 states that horizontal steel wire mesh placed in horizontal bed joints restrains the transverse lateral expansion of brick and mortar, placing masonry under triaxial compression and significantly increasing compressive capacity: N_u = φ_n · [f + 2 ρ_v f_y (1 - 2 e/y)] · A.

Question 23Structural engineering models and applications

A glued laminated timber (Glulam 胶合木) beam of rectangular cross section b × h = 150 mm × 400 mm has a design flexural strength f_m = 16.0 N/mm². Lateral buckling is fully restrained. According to GB 50005-2017 Section 5.1.1, what is the design flexural bending moment capacity M_u of the glulam beam?

  1. 48.0 kN·m
  2. 64.0 kN·m Correct answer.
  3. 85.3 kN·m
  4. 96.0 kN·m

Explanation

Per GB 50005-2017 Section 5.1.1: 1. Elastic section modulus W_n: W_n = (b · h²) / 6 = [150 mm × (400 mm)²] / 6 = (150 × 160,000) / 6 = 4.0 × 10⁶ mm³. 2. Design flexural moment capacity M_u: M_u = f_m · W_n = 16.0 N/mm² × 4.0 × 10⁶ mm³ = 64.0 × 10⁶ N·mm = 64.0 kN·m.

Question 24Structural engineering models and applications

In a ductile dowel-type timber connection model, which two yielding mechanisms can combine when evaluating lateral resistance per shear plane? This question concerns the modeled yielding modes, not every possible brittle timber failure.

  1. Timber wood embedment crushing (承压屈服) and dowel steel plastic bending (销杆屈服弯曲) Correct answer.
  2. Adhesive delamination and longitudinal shear
  3. Timber tensile rupture and weld fracture
  4. Thermal expansion and shrinkage split

Explanation

Per GB 50005-2017 Section 6.2 and the European Yield Model (Johansen yield theory), the load-carrying capacity of dowel-type fasteners (bolts, steel dowels) is governed by the interaction between timber embedment crushing (木材承压屈服) and the plastic bending hinge formation of the steel dowel pin (销杆受弯屈服).

Question 25Structural engineering models and applications

A reinforced concrete structure with fundamental period T = 1.0 s has structural damping ratio ζ = 0.05 (damping adjustment factor η₂ = 1.0, decay index γ = 0.9). The site characteristic period is T_g = 0.40 s and the maximum horizontal seismic influence coefficient is α_max = 0.16 (Intensity 8 degrees frequent earthquake). According to GB 50011-2010 Section 5.1.5, what is the horizontal seismic influence coefficient α(T)?

  1. 0.048
  2. 0.070 Correct answer.
  3. 0.092
  4. 0.160

Explanation

Per GB 50011-2010 Section 5.1.5, in the period range T_g ≤ T ≤ 5 T_g (0.40 s ≤ 1.0 s ≤ 2.0 s): α(T) = (T_g / T)^γ · η₂ · α_max α(1.0) = (0.40 / 1.0)⁰·⁹ × 1.0 × 0.16 = (0.40)⁰·⁹ × 0.16 Calculating (0.40)⁰·⁹: ln(0.40) = -0.9163; 0.9 × (-0.9163) = -0.8247; e^(-0.8247) = 0.4384. α(1.0) = 0.4384 × 0.16 = 0.07014 ≈ 0.070.

Question 26Structural engineering models and applications

For a stipulated simplified seismic exercise, total representative gravity load is60000kN and equivalent participating weight is0.85times that value. The supplied lateral coefficient is0.08, with no separate roof concentration. What modeled total base shear follows?

  1. 3,600 kN
  2. 4,080 kN Correct answer.
  3. 4,800 kN
  4. 5,100 kN

Explanation

Per GB 50011-2010 Section 5.2.1: 1. Effective equivalent total gravity load G_eq: For multi-story structures modeled as single-degree-of-freedom equivalent systems, G_eq = 0.85 G_E = 0.85 × 60,000 kN = 51,000 kN. 2. Total horizontal seismic base shear F_Ek: F_Ek = α₁ · G_eq = 0.08 × 51,000 kN = 4,080 kN.

Question 27Structural engineering models and applications

According to GB 50011-2010 Section 5.2.1, after deducting the roof concentrated force ΔF_n, how is the remaining base shear distributed to each floor level i as the story horizontal seismic force F_i?

  1. F_i = [G_i · H_i / Σ (G_j · H_j)] · F_Ek · (1 - δ_n) Correct answer.
  2. F_i = (G_i / Σ G_j) · F_Ek
  3. F_i = (H_i / Σ H_j) · F_Ek
  4. F_i = [G_i · H_i² / Σ (G_j · H_j²)] · F_Ek

Explanation

Per GB 50011-2010 Section 5.2.1, the horizontal seismic force at floor level i is formulated as an inverted triangular distribution proportional to floor mass G_i and height H_i above ground: F_i = [G_i · H_i / Σ (G_j · H_j)] · F_Ek · (1 - δ_n).

Question 28Geotechnical engineering models and applications

A shallow foundation has base width b = 4.0 m and embedment depth d = 2.0 m. The subgrade soil has characteristic bearing capacity f_ak = 180 kPa, unit weight γ = 18 kN/m³ (below base), weighted average unit weight above base γ_m = 17 kN/m³, and bearing capacity correction coefficients η_b = 0.3 and η_d = 1.5. According to GB 50007-2011 Section 5.2.4, what is the corrected characteristic bearing capacity f_a of the subgrade?

  1. 198.5 kPa
  2. 223.7 kPa Correct answer.
  3. 245.2 kPa
  4. 268.0 kPa

Explanation

For this question use the supplied numerical assumptions and the stated model only. Per GB 50007-2011 Section 5.2.4: f_a = f_ak + η_b · γ · (b - 3) + η_d · γ_m · (d - 0.5) 1. Width correction term (b = 4.0 m > 3.0 m): Δf_b = 0.3 × 18 kN/m³ × (4.0 - 3.0) m = 5.4 kPa. 2. Depth correction term (d = 2.0 m > 0.5 m): Δf_d = 1.5 × 17 kN/m³ × (2.0 - 0.5) m = 1.5 × 17 × 1.5 = 38.25 kPa. 3. Corrected characteristic bearing capacity: f_a = 180 kPa + 5.4 kPa + 38.25 kPa = 223.65 kPa ≈ 223.7 kPa. This is not independent confirmation of a current local code or complete design approval.

Question 29Geotechnical engineering models and applications

A rectangular spread footing of dimensions l × b = 3.0 m × 2.0 m (area A = 6.0 m², section modulus W = b · l² / 6 = 2.0 × 9 / 6 = 3.0 m³) carries a characteristic vertical load F_k = 900 kN and bending moment M_k = 180 kN·m acting along the length direction l. The foundation embedment depth is d = 1.5 m (average unit weight of footing and backfill soil γ_G = 20 kN/m³). What are the average contact pressure p_k and maximum edge contact pressure p_kmax at the base?

  1. p_k = 150 kPa, p_kmax = 210 kPa
  2. p_k = 180 kPa, p_kmax = 240 kPa Correct answer.
  3. p_k = 200 kPa, p_kmax = 260 kPa
  4. p_k = 220 kPa, p_kmax = 310 kPa

Explanation

Per GB 50007-2011 Section 5.2.2: 1. Total weight of footing and soil overburden G_k: G_k = γ_G · A · d = 20 kN/m³ × 6.0 m² × 1.5 m = 180 kN. 2. Average base contact pressure p_k: p_k = (F_k + G_k) / A = (900 + 180) kN / 6.0 m² = 1,080 / 6 = 180.0 kPa. 3. Maximum edge contact pressure p_kmax: p_kmax = (F_k + G_k) / A + M_k / W = 180 kPa + (180 kN·m / 3.0 m³) = 180 + 60 = 240.0 kPa. 4. Eccentricity check: e = M_k / (F_k + G_k) = 180 / 1080 = 0.1667 m ≤ l / 6 = 0.50 m (no base detachment).

Question 30Geotechnical engineering models and applications

In the bearing capacity check of a soft underlying soil layer (软弱下卧层承载力验算) beneath a shallow foundation per GB 50007-2011 Section 5.2.7, what is the verification condition at the top of the soft layer at depth z below the foundation base?

  1. p_z + p_cz ≤ f_az Correct answer.
  2. p_z ≤ f_ak
  3. p_z + p_cz ≤ 1.2 f_az
  4. p_cz ≤ f_az

Explanation

Per GB 50007-2011 Section 5.2.7, the soft underlayer check requires that the sum of the diffused vertical net additional stress p_z and the effective self-weight stress of soil p_cz at the top surface of the soft layer must not exceed the corrected characteristic bearing capacity f_az of the soft layer at that depth: p_z + p_cz ≤ f_az.

Question 31Geotechnical engineering models and applications

In the structural design of a reinforced concrete spread footing per GB 50007-2011 Section 8.2.11, where is the critical cross section located for calculating the maximum flexural bending moment M and sizing the bottom tension reinforcement?

  1. At the center of the footing
  2. At the face of the column (or column pedestal / stepped interface) Correct answer.
  3. At a distance of h₀ / 2 from the column face
  4. At the outer edge of the footing

Explanation

Per GB 50007-2011 Section 8.2.11, the critical cross section for calculating flexural bending moments in a spread footing is located at the face of the column (or the edge of the reinforced concrete column pedestal / stepped footing step). The upward net soil pressure cantilevered from this section determines the required tension reinforcement.

Question 32Geotechnical engineering models and applications

In the design of an uplift tension pile (抗拔桩) per JGJ 94-2008 Section 5.2.11, how is the ultimate vertical tensile capacity T_uk of a single pile calculated from soil shaft resistance and pile self-weight G_p?

  1. T_uk = Σ (λ_i · q_sik · u_i · l_i) + G_p (where λ_i is the uplift skin friction coefficient) Correct answer.
  2. T_uk = u · Σ (q_sik · l_i) + 2 G_p
  3. T_uk = Q_uk - q_pk · A_p
  4. T_uk = G_p only

Explanation

Per JGJ 94-2008 Section 5.2.11, the ultimate uplift capacity of a single pile is T_uk = Σ (λ_i · q_sik · u_i · l_i) + G_p, where λ_i is the uplift resistance coefficient (typically 0.5 to 0.8 depending on soil type and pile slenderness l/d), reflecting the reduction in lateral effective stress on the pile shaft during upward tensile pullout.

Question 33Geotechnical engineering models and applications

When deep soft soil layers settle under self-weight consolidation or ground surcharge, Negative Skin Friction (负摩阻力) acts on the pile shaft above a specific depth. According to JGJ 94-2008 Section 5.4.2, what is this boundary location called and what occurs at this depth?

  1. Plastic Hinge Point, where pile bending moment is maximum
  2. Neutral Plane / Neutral Point (中性点), where pile settlement equals surrounding soil settlement (relative displacement is zero) Correct answer.
  3. Bedrock Interface, where tip resistance reaches 100%
  4. Critical Liquefaction Plane

Explanation

Per JGJ 94-2008 Section 5.4.2, the Neutral Point (中性点) is the depth along the pile shaft where the downward settlement of the surrounding consolidating soil exactly equals the downward settlement of the pile shaft under load. Above the neutral point, soil settles faster than the pile, creating downward drag (negative skin friction); below the neutral point, positive skin friction resists the load.

Question 34Geotechnical engineering models and applications

A gravity retaining wall has vertical back (α = 0), smooth backface (δ = 0), and retains cohesionless sand (unit weight γ = 19 kN/m³, internal friction angle φ = 30°, cohesion c = 0) of height H = 6.0 m with horizontal backfill surface. According to Rankine earth pressure theory per GB 50007-2011 and GB 50330-2013, what is the total active earth pressure resultant E_a per linear meter of wall?

  1. 76.0 kN/m
  2. 114.0 kN/m Correct answer.
  3. 152.0 kN/m
  4. 228.0 kN/m

Explanation

Per Rankine earth pressure theory: 1. Rankine active earth pressure coefficient K_a: K_a = tan²(45° - φ / 2) = tan²(45° - 15°) = tan²(30°) = (1 / √3)² = 1 / 3 ≈ 0.3333. 2. Total active earth pressure resultant E_a: E_a = 0.5 · γ · H² · K_a E_a = 0.5 × 19 kN/m³ × (6.0 m)² × (1 / 3) = 0.5 × 19 × 36 × 0.3333 = 19 × 6 = 114.0 kN/m.

Question 35Construction engineering models and applications

In Critical Path Method (CPM) network scheduling, how is Total Float (TF) of an activity calculated from Early Start (ES), Late Start (LS), Early Finish (EF), and Late Finish (LF)?

  1. TF = LS - ES = LF - EF Correct answer.
  2. TF = ES - LS
  3. TF = EF - ES
  4. TF = LF - LS

Explanation

Total Float is the maximum duration an activity can be delayed from its early start without delaying the project completion date: TF = LS - ES = LF - EF. Activities with zero total float form the Critical Path.

Question 36Construction engineering models and applications

Under an NEC4 Option C Target Cost Contract, the initial Target Price is HK$ 100M. Due to efficiency, the actual final Defined Cost + Fee incurred by the Contractor is HK$ 88M. If the contract pain/gain share arrangement specifies a 50/50 split for cost savings below target, what is the Contractor's Gain Share bonus payout?

  1. HK$ 6M Correct answer.
  2. HK$ 12M
  3. HK$ 44M
  4. HK$ 88M

Explanation

Total cost saving below target = Target Price - Actual Defined Cost = HK$ 100M - HK$ 88M = HK$ 12M. Under the 50/50 gain share rule, the savings are shared equally between Employer and Contractor: Contractor Gain Share = 50% * HK$ 12M = HK$ 6M. Total payout to Contractor = Defined Cost (HK$ 88M) + Gain Share (HK$ 6M) = HK$ 94M.

Question 37Construction engineering models and applications

A Chartered Civil Engineer reviewing site drawings discovers a critical calculation error in a temporary works cofferdam design prepared by another engineering firm that poses an imminent collapse risk. What is the engineer's ethical duty under HKIE Rules of Conduct?

  1. Ignore the error because it was prepared by a competing firm
  2. Immediately notify the site project team, contractor, and designer to rectify the hazard before work proceeds Correct answer.
  3. Wait until the cofferdam fails before reporting
  4. Sell the information to local news media anonymously

Explanation

HKIE Code of Ethics mandates that if a member identifies a structural hazard or error that threatens life safety, they must take immediate action to notify responsible parties (contractor, designer, and relevant authorities) to stop unsafe work and prevent collapse. Public safety overrides commercial courtesies.

Question 38Construction engineering models and applications

An employer explicitly instructs a HKIE member engineer to sign and certify a structural submission containing unverified calculation assumptions that violate the HK Concrete Code. What is the engineer's proper ethical course of action?

  1. Sign the submission immediately to keep their job
  2. Refuse to certify the non-compliant submission, explain the safety non-compliance in writing, and escalate if pressure continues Correct answer.
  3. Sign the submission but add a hidden footnote in fine print
  4. Delegate the signing to an unqualified junior intern

Explanation

Under HKIE Code of Ethics, a member must not sign or seal documents that do not comply with engineering standards or statutory safety codes. If pressured by an employer or client, the engineer must refuse certification, document the technical non-compliance in writing, and if necessary report the matter to HKIE or relevant regulatory authorities.

Question 39Construction engineering models and applications

A construction project requires excavating 5,000 cubic yards of soil in its bank (in-place) state. The soil has a swell factor of 1.4 (1 bank cy expands to 1.4 loose cy when excavated). How many cubic yards of loose soil must be hauled away?

  1. 3,571 cy
  2. 5,000 cy
  3. 6,000 cy
  4. 7,000 cy Correct answer.

Explanation

Loose volume = bank volume × swell factor = 5,000 × 1.4 = 7,000 cubic yards. Soil expands (swells) when excavated from its compacted in-place (bank) state to a loose hauled state; the swell factor (loose/bank) is typically 1.1–1.4 for most soils.

Question 40Construction engineering models and applications

In a critical path method (CPM) schedule, Activity A has a duration of 5 days, Activity B has a duration of 8 days, and Activity C has a duration of 4 days. Activity C can only start after both A and B are complete. What is the early start (ES) for Activity C?

  1. 5 days
  2. 8 days Correct answer.
  3. 9 days
  4. 13 days

Explanation

Activity C requires both A and B to be complete. Activity A finishes at day 5, Activity B finishes at day 8. Activity C can start when the longer predecessor (B) finishes, so ES = 8 days. This is determined by the maximum of all predecessor early finish times.

Question 41Construction engineering models and applications

A concrete mix design requires 350 lbs of cement per cubic yard. For a 100 cubic yard pour with 8% waste factor, how many 94-lb sacks of cement are required?

  1. 371 sacks
  2. 372 sacks
  3. 402 sacks
  4. 403 sacks Correct answer.

Explanation

Total concrete needed = 100 × 1.08 = 108 cy. Total cement = 350 × 108 = 37,800 lbs. Number of sacks = 37,800 ÷ 94 = 402.13 sacks. Round up to 403 sacks (next whole sack). Always round up cement quantities for construction ordering.

Question 42Construction engineering models and applications

What is the primary purpose of a slump test for fresh concrete?

  1. To determine compressive strength
  2. To measure workability and consistency Correct answer.
  3. To test for air content
  4. To verify cement content

Explanation

The slump test (ASTM C143) measures the workability and consistency of fresh concrete. A cone is filled with concrete, lifted, and the settlement (slump) is measured. Typical specifications require 2-4 inches for most structural work. Compressive strength is tested with cylinders cured for 28 days.

Question 43Construction engineering models and applications

A temporary excavation support system uses soldier piles with timber lagging. The lagging spans 8 feet horizontally between soldier piles. If the lateral earth pressure is 600 psf and the lagging is 3-inch thick Douglas Fir-Larch (Fb = 1,500 psi), is the lagging adequate?

  1. Yes, stress is 768 psi
  2. Yes, stress is 1,200 psi
  3. No, stress is 3,200 psi Correct answer.
  4. No, stress is 2,400 psi

Explanation

For a uniformly loaded simple span (1-ft vertical strip of lagging): M = wL²/8 = (600 psf × 1 ft) × (8 ft)² / 8 = 4,800 ft-lb = 57,600 in-lb. For 3" lagging (S = bd²/6 = 12 × 3² / 6 = 18 in³), f = M/S = 57,600/18 = 3,200 psi. This exceeds 1,500 psi allowable, so the lagging is not adequate.

Question 44Construction engineering models and applications

A horizontal curve on a construction site has a radius of 500 feet and a central angle of 30°. What is the length of the curve?

  1. 131 ft
  2. 262 ft Correct answer.
  3. 524 ft
  4. 785 ft

Explanation

Curve length L = R × Δ × (π/180) = 500 × 30 × (π/180) = 500 × 0.5236 = 261.8 ft ≈ 262 ft. This is the arc length along the curve centerline.

Question 45Structural engineering models and applications

For a reinforced concrete flexural member with a rectangular section, what is the fundamental condition required by GB 50010-2010 to prevent over-reinforced brittle compression failure (超筋破坏) at the ultimate limit state?

  1. $\xi \le \xi_b$ (Relative depth of compressive zone $\xi = x / h_0$ must not exceed the balanced relative depth $\xi_b$) Correct answer.
  2. $\xi \ge \xi_b$ (Relative depth of compressive zone $\xi = x / h_0$ must exceed the balanced relative depth $\xi_b$)
  3. $\rho \le 0.20\%$ (Tensile reinforcement ratio must be less than the minimum limit)
  4. $x \le 2 a'_s$ (Depth of compressive zone must be less than twice the compression steel cover depth)

Explanation

According to GB 50010-2010 (Section 6.2.10), to ensure ductile flexural failure where tension steel yields before concrete crushes (preventing brittle over-reinforced failure), the relative compressive zone depth $\xi = x / h_0$ must satisfy $\xi \le \xi_b$, where $\xi_b = \frac{\beta_1}{1 + \frac{f_y}{E_s \varepsilon_{cu}}}$.

Question 46Structural engineering models and applications

A reinforced concrete T-shaped beam has flange width $b'_f = 600\text{ mm}$, flange thickness $h'_f = 100\text{ mm}$, web width $b = 200\text{ mm}$, and effective depth $h_0 = 500\text{ mm}$. The concrete grade is C30 ($f_c = 14.3\text{ N/mm}^2, \alpha_1 = 1.0$) and the tension rebar is HRB400 ($f_y = 360\text{ N/mm}^2, A_s = 2000\text{ mm}^2$). Which type of T-beam is this section, and does the equivalent compressive zone fall inside the flange or extend into the web?

  1. Type II T-beam ($x > h'_f$); the neutral axis extends into the web ($x = 142.5\text{ mm} > 100\text{ mm}$)
  2. Type I T-beam ($x \le h'_f$); the neutral axis falls within the flange ($x = 83.9\text{ mm} \le 100\text{ mm}$) Correct answer.
  3. Type I T-beam ($x > h'_f$); the neutral axis extends into the web
  4. Over-reinforced T-beam ($x > \xi_b h_0$); brittle failure governs

Explanation

According to GB 50010-2010 (Section 6.2.11): 1. Check the maximum tension force that the entire flange can resist in compression: $$f_y A_s = 360 \times 2000 = 720,000\text{ N} = 720\text{ kN}$$ $$\alpha_1 f_c b'_f h'_f = 1.0 \times 14.3 \times 600 \times 100 = 858,000\text{ N} = 858\text{ kN}$$ 2. Since $f_y A_s \le \alpha_1 f_c b'_f h'_f$ ($720\text{ kN} \le 858\text{ kN}$), the depth of the equivalent compressive stress block $x$ is less than or equal to the flange thickness $h'_f = 100\text{ mm}$. 3. Calculate $x$ as a rectangular beam with width $b'_f$: $$x = \frac{f_y A_s}{\alpha_1 f_c b'_f} = \frac{720,000}{1.0 \times 14.3 \times 600} = 83.92\text{ mm} \le 100\text{ mm}$$ 4. Therefore, it is a Type I T-beam (第一类T形截面) with its neutral axis within the compression flange.

Question 47Structural engineering models and applications

A simply supported reinforced concrete rectangular beam carrying uniform load has section dimensions $b = 250\text{ mm}, h = 600\text{ mm}$, and effective depth $h_0 = 550\text{ mm}$. It is cast with C30 concrete ($f_t = 1.43\text{ N/mm}^2, f_c = 14.3\text{ N/mm}^2$) and fitted with two-legged HRB400 stirrups $\Phi 8@150$ ($A_{sv} = 101\text{ mm}^2, f_{yv} = 360\text{ N/mm}^2$). According to GB 50010-2010 (Section 6.3.4), what is the design diagonal shear capacity $V_{cs}$ of this beam?

  1. 270.8 kN Correct answer.
  2. 137.6 kN
  3. 133.2 kN
  4. 315.0 kN

Explanation

According to GB 50010-2010 (Equation 6.3.4-1) for members under uniform load: $$V_{cs} = V_c + V_s = 0.7 f_t b h_0 + 1.25 f_{yv} \frac{A_{sv}}{s} h_0$$ 1. Concrete contribution $V_c$: $$V_c = 0.7 \times 1.43 \times 250 \times 550 \times 10^{-3} = 137.64\text{ kN}$$ 2. Stirrups contribution $V_s$ (with $1.25$ factor): $$V_{cs} = 137.64 + 133.32 = 270.96\text{ kN} \approx 270.8\text{ kN}$$.

Question 48Structural engineering models and applications

When checking an RC column for eccentric compression, under which condition is the failure mode classified as Large Eccentricity Compression Failure (大偏心受压破坏 / Tension Failure)?

  1. The section remains completely in compression with no tension zone
  2. $\xi = \frac{x}{h_0} > \xi_b$ (Relative compressive zone depth exceeds the balanced relative depth $\xi_b$)
  3. $e_0 \le 0.5 h$ (Initial eccentricity is smaller than half the column depth)
  4. $\xi = \frac{x}{h_0} \le \xi_b$ (Relative compressive zone depth is less than or equal to the balanced relative depth $\xi_b$) Correct answer.

Explanation

According to GB 50010-2010 (Section 6.2.17), large eccentricity compression failure (tension-controlled failure) occurs when $\xi = x / h_0 \le \xi_b$. In this ductile mode, the tension reinforcement on the far side yields ($f_y$) before the concrete in the compression zone reaches its ultimate crushing strain $\varepsilon_{cu}$.

Question 49Structural engineering models and applications

In the structural design of reinforced concrete corbels (牛腿) supporting crane girders per GB 50010-2010 (Section 9.3), what is the primary strut-and-tie failure mechanism that dictates the inclined concrete compression strut dimensioning?

  1. Diagonal compressive crushing along the inclined strut between the load pad and column face ($F_{vk} \le \beta_c f_c b_w s_n$) Correct answer.
  2. Pure flexural tensile yielding of vertical stirrups
  3. Horizontal sliding along the column construction joint
  4. Punching shear through the corbel tip

Explanation

According to GB 50010-2010 (Section 9.3.1), corbels behave as internal strut-and-tie trusses. The vertical load from the crane beam is transferred through an inclined concrete compression strut into the column. The section dimensions must ensure the inclined compressive stress does not cause diagonal crushing failure of the concrete strut.

Question 50Structural engineering models and applications

In the design of embedded steel plates with welded anchor bars (预埋件锚筋) under shear force $V$ and tensile force $N$ per GB 50010-2010 (Section 9.7), why must anchor bars use deformed steel bars (HRB400/HRB500) rather than plain round bars (HPB300)?

  1. Because plain bars have higher carbon content and cause weld embrittlement
  2. Because plain bars are prohibited from all structural welding in China
  3. To ensure high bond-slip anchorage resistance and prevent pullout slip failure under cyclic and dynamic loads Correct answer.
  4. To reduce the anchor plate thickness by 50%

Explanation

According to GB 50010-2010 (Section 9.7.1), anchor bars for embedded steel parts carrying significant shear and tension must consist of deformed bars (HRB400 or HRB500). Deformed ribs provide mechanical interlocking with the surrounding concrete, preventing premature anchorage pullout slip failure under high tensile and shear stress combinations.