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Grade-2 Registered Structural Engineer (China) - independent engineering foundations study materials
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Grade-2 Registered Structural Engineer (China) - independent engineering foundations practice questions

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Question 1Structural 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 2Structural 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 3Structural 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 4Structural 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 5Structural 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 6Structural 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.

Question 7Structural engineering models and applications

A welded I-section steel girder in a workshop floor is subjected to a maximum design bending moment $M_x = 350\text{ kN}\cdot\text{m}$ (about its strong axis). The steel is Q235B ($f = 215\text{ N/mm}^2, t \le 16\text{ mm}$) and the section plastic development coefficient is $\gamma_x = 1.05$. Assuming lateral-torsional stability is fully ensured by a concrete floor slab, what is the minimum required net section modulus $W_{nx}$ to satisfy the flexural strength criterion per GB 50017-2017?

  1. $1.750 \times 10^6\text{ mm}^3$ ($1750\text{ cm}^3$)
  2. $1.628 \times 10^6\text{ mm}^3$ ($1628\text{ cm}^3$)
  3. $1.489 \times 10^6\text{ mm}^3$ ($1489\text{ cm}^3$)
  4. $1.550 \times 10^6\text{ mm}^3$ ($1550\text{ cm}^3$) Correct answer.

Explanation

According to GB 50017-2017 (Section 6.1.1): $$\sigma = \frac{M_x}{\gamma_x W_{nx}} \le f$$ Rearranging for the required net section modulus $W_{nx}$: $$W_{nx} \ge \frac{M_x}{\gamma_x f} = \frac{350 \times 10^6\text{ N}\cdot\text{mm}}{1.05 \times 215\text{ N/mm}^2} = \frac{350,000,000}{225.75} = 1,550,387\text{ mm}^3 \approx 1550\text{ cm}^3$$.

Question 8Structural engineering models and applications

A welded I-section steel beam has web dimensions $h_w = 600\text{ mm}$ and $t_w = 8\text{ mm}$. At the support, the design shear force is $V = 400\text{ kN}$. The steel is Q235B ($f_v = 125\text{ N/mm}^2$). What is the calculated average shear stress $\tau$ in the web and does it satisfy the shear strength check per GB 50017-2017?

  1. $\tau = 66.7\text{ N/mm}^2 \le 125\text{ N/mm}^2$; Satisfies shear strength
  2. $\tau = 133.3\text{ N/mm}^2 > 125\text{ N/mm}^2$; Fails shear strength
  3. $\tau = 83.3\text{ N/mm}^2 \le 125\text{ N/mm}^2$; Satisfies shear strength Correct answer.
  4. $\tau = 104.2\text{ N/mm}^2 \le 125\text{ N/mm}^2$; Satisfies shear strength

Explanation

According to GB 50017-2017 (Section 6.1.2), the shear stress in the web of an I-section beam can be checked using the average shear formula: $$\tau = \frac{V}{I t_w} S \approx \frac{V}{h_w t_w} = \frac{400 \times 10^3\text{ N}}{600\text{ mm} \times 8\text{ mm}} = \frac{400,000}{4800} = 83.33\text{ N/mm}^2$$ Since $\tau = 83.33\text{ N/mm}^2 \le f_v = 125\text{ N/mm}^2$, the shear strength condition is fully satisfied.

Question 9Structural engineering models and applications

According to GB 50017-2017 (Section 6.2.1), under which of the following conditions is it permitted to OMIT the lateral-torsional overall stability check (整体稳定验算) for a steel flexural beam?

  1. Whenever the steel beam span does not exceed 12 meters
  2. When a cast-in-place reinforced concrete slab or composite deck is rigidly connected to the compression flange with shear studs along the entire span Correct answer.
  3. Whenever the steel yield grade is Q355 or higher
  4. When the beam web height-to-thickness ratio $h_w/t_w \le 80$

Explanation

According to GB 50017-2017 (Section 6.2.1), when a beam has a reinforced concrete slab or profiled steel deck firmly attached to its compression flange (e.g. via welded shear stud connectors or continuous welding) such that lateral displacement and twist of the compression flange are completely restrained, overall lateral-torsional buckling will not occur, and overall stability calculation may be omitted.

Question 10Structural engineering models and applications

An axially compressed steel column has an effective buckling length $l_0 = 6.0\text{ m}$. The cross-section has a gross area $A = 120\text{ cm}^2$ ($12,000\text{ mm}^2$) and a moment of inertia about the governing buckling axis $I = 1.08 \times 10^8\text{ mm}^4$. What is the radius of gyration $i$ and the slenderness ratio $\lambda$ of this column?

  1. $i = 85.2\text{ mm}, \lambda = 70.4$
  2. $i = 90.0\text{ mm}, \lambda = 66.7$
  3. $i = 104.0\text{ mm}, \lambda = 57.7$
  4. $i = 94.9\text{ mm}, \lambda = 63.2$ Correct answer.

Explanation

1. Radius of gyration: $$i = \sqrt{\frac{I}{A}} = \sqrt{\frac{1.08 \times 10^8\text{ mm}^4}{12,000\text{ mm}^2}} = \sqrt{9000} = 94.87\text{ mm}$$ 2. Slenderness ratio $\lambda$: $$\lambda = \frac{l_0}{i} = \frac{6000\text{ mm}}{94.87\text{ mm}} = 63.24 \approx 63.2$$.

Question 11Structural engineering models and applications

When checking the Out-of-Plane Stability (平面外整体稳定) of a steel beam-column under axial compression $N$ and bending moment $M_x$ per GB 50017-2017 (Section 8.2.5), which factor represents the lateral-torsional stability factor for the beam under pure bending?

  1. $\varphi_x$ (Major-axis column stability factor)
  2. $\varphi_b$ (Beam lateral-torsional buckling coefficient) Correct answer.
  3. $\beta_{mx}$ (Equivalent moment factor for in-plane stability)
  4. $\alpha_1$ (Concrete stress block coefficient)

Explanation

According to GB 50017-2017 (Section 8.2.5), out-of-plane stability of a beam-column member is verified by $\frac{N}{\varphi_y A} + \frac{\eta \beta_{tx} M_x}{\varphi_b W_{1x}} \le f$, where $\varphi_b$ is the uniform bending lateral-torsional buckling factor of the steel beam, and $\varphi_y$ is the column buckling factor about the weak axis.

Question 12Structural engineering models and applications

A lap splice between two Q235 steel plates ($t_1 = 10\text{ mm}, t_2 = 10\text{ mm}$) is connected using Grade 4.6 ordinary bolts of diameter $d = 20\text{ mm}$ (single shear plane, $n_v = 1$). For Grade 4.6 bolts, the design shear strength is $f_v^b = 140\text{ N/mm}^2$ and design bearing strength is $f_c^b = 305\text{ N/mm}^2$. What is the design shear bearing capacity $N_b$ per bolt?

  1. 44.0 kN Correct answer.
  2. 61.0 kN
  3. 52.5 kN
  4. 38.2 kN

Explanation

According to GB 50017-2017 (Section 11.2.1): 1. Shear capacity of single bolt: $$N_v^b = n_v \frac{\pi d^2}{4} f_v^b = 1 \times \frac{\pi \times 20^2}{4} \times 140 = 314.16 \times 140 = 43,982\text{ N} \approx 44.0\text{ kN}$$ 2. Bearing capacity of single bolt on connected plates: $$N_c^b = d \sum t f_c^b = 20\text{ mm} \times 10\text{ mm} \times 305\text{ N/mm}^2 = 61,000\text{ N} = 61.0\text{ kN}$$ 3. The design capacity is governed by the smaller value: $$N_b = \min(N_v^b, N_c^b) = \min(44.0\text{ kN}, 61.0\text{ kN}) = 44.0\text{ kN}$$.

Question 13Structural engineering models and applications

A friction-type high-strength bolt connection uses M20 Grade 10.9s bolts. The pretension force per bolt is $P = 155\text{ kN}$. The friction surface treatment is blast-cleaned (喷砂), providing a slip coefficient $\mu = 0.45$. The joint is configured in single shear ($n_f = 1$ friction surface). According to GB 50017-2017 (Section 11.2.2), what is the design slip shear capacity $N_v^b$ per bolt?

  1. 55.8 kN
  2. 69.8 kN
  3. 139.5 kN
  4. 62.8 kN Correct answer.

Explanation

According to GB 50017-2017 (Section 11.2.2, Equation 11.2.2-1): $$N_v^b = 0.9 n_f \mu P$$ 1. Substitute values ($n_f = 1, \mu = 0.45, P = 155\text{ kN}$): $$N_v^b = 0.9 \times 1 \times 0.45 \times 155\text{ kN} = 62.775\text{ kN} \approx 62.8\text{ kN}$$ The $0.9$ factor is the statutory coefficient accounting for pretension relaxation and multi-bolt load distribution.

Question 14Structural engineering models and applications

Two side fillet welds have actual length200mm each, leg8mm, stipulated effective throat0.7*8mm and design shear stress160N/mm^2. For this exercise define effective length of each as200-2*8mm. What total resistance follows the stated model?

  1. 358.4 kN
  2. 329.7 kN Correct answer.
  3. 460.8 kN
  4. 230.4 kN

Explanation

According to GB 50017-2017 (Section 11.1.2): 1. Effective throat thickness: $$h_e = 0.7 h_f = 0.7 \times 8\text{ mm} = 5.6\text{ mm}$$ 2. Effective length per weld (deducting $2 h_f$ for start/stop arc craters): $$l_w = l - 2 h_f = 200 - 2 \times 8 = 184\text{ mm}$$ 3. Total design capacity for 2 welds: $$N = 2 \times (h_e l_w f_f^w) = 2 \times (5.6\text{ mm} \times 184\text{ mm} \times 160\text{ N/mm}^2) = 2 \times 164,864\text{ N} = 329,728\text{ N} \approx 329.7\text{ kN}$$.

Question 15Structural engineering models and applications

In single-story light steel portal frames designed in accordance with the Technical Code for Steel Portal Frame Light Buildings (GB 51022-2015), what structural mechanism is routinely utilized in tapered I-section webs to achieve high material efficiency?

  1. Concrete encasement composite action
  2. Prestressed post-tensioned tendon anchoring
  3. Full plastic hinge redistribution across multiple spans
  4. Post-buckling shear strength (tension field action / 屈曲后强度) of thin webs Correct answer.

Explanation

According to GB 51022-2015 (Section 6.1.1 and 6.4), light portal frame members typically feature slender webs with $h_w / t_w$ exceeding elastic shear buckling limits. The design standard explicitly permits utilizing the post-buckling shear resistance (tension field action / 屈曲后强度) provided by the web and intermediate transverse stiffeners.

Question 16Structural engineering models and applications

In light portal frame roof systems per GB 51022-2015, what is the primary structural function of cold-formed steel purlin sag rods (拉条) and stay rods (隅撑)?

  1. Sag rods prevent purlin thermal expansion; stay rods support suspended MEP piping
  2. Sag rods carry main roof vertical live loads; stay rods carry longitudinal crane brake forces
  3. Sag rods reduce the out-of-plane weak-axis bending span of purlins; stay rods prevent lateral-torsional buckling of the portal rafter compression flange Correct answer.
  4. Both sag rods and stay rods serve solely as temporary erection bracing during assembly

Explanation

According to GB 51022-2015 (Sections 7.1 and 7.2): 1. Sag rods (拉条) are installed between purlins to provide intermediate lateral support, reducing the span for bending about the weak axis under roof slope gravity and wind uplift. 2. Stay rods / knee braces (隅撑) connect the purlin/girt to the inner flange of the portal frame rafter or column, bracing the compression flange against lateral-torsional buckling under gravity or wind suction load reversals.

Question 17Structural engineering models and applications

A nominally pinned column base plate (铰接柱脚底板) transmits a factored axial compressive load $N = 900\text{ kN}$ uniformly onto a concrete pedestal. The base plate dimensions are $B = 300\text{ mm}$ and $L = 400\text{ mm}$ ($A = 120,000\text{ mm}^2$). The steel is Q235B ($f = 215\text{ N/mm}^2$). The cantilever projection of the base plate beyond the column flange is $c = 60\text{ mm}$. What is the required minimum base plate thickness $t$ based on cantilever bending per GB 50017-2017?

  1. 24.5 mm (Select 25 mm plate)
  2. 19.4 mm (Select 20 mm plate) Correct answer.
  3. 10.2 mm (Select 12 mm plate)
  4. 32.0 mm (Select 35 mm plate)

Explanation

According to GB 50017-2017 (Section 12.7): 1. Uniform compressive bearing pressure on the base plate: $$q = \frac{N}{B L} = \frac{900 \times 10^3\text{ N}}{300\text{ mm} \times 400\text{ mm}} = 7.5\text{ N/mm}^2$$ 2. Maximum cantilever bending moment per unit width: $$M = \frac{1}{2} q c^2 = \frac{1}{2} \times 7.5 \times 60^2 = 13,500\text{ N}\cdot\text{mm/mm}$$ 3. Required minimum plate thickness from $M / W \le f$ (where $W = t^2 / 6$): $$t = \sqrt{\frac{6 M}{f}} = \sqrt{\frac{6 \times 13,500}{215}} = \sqrt{376.74} = 19.41\text{ mm} \approx 19.4\text{ mm}$$. Selecting a standard $20\text{ mm}$ thick steel plate satisfies design requirements.

Question 18Structural engineering models and applications

For a steel tension plate whose governing net-section path is explicitly a straight transverse line through aligned bolt holes, how is the hole deduction made from gross area? No staggered path is considered in this exercise.

  1. By adding the bolt hole diameters to the flange width
  2. By multiplying the gross area by a constant $0.85$
  3. By subtracting the total area of bolt holes located in the governing critical fracture path from the gross cross-sectional area ($A_n = A - \sum d_0 t$) Correct answer.
  4. Net area check is not required for tension members

Explanation

According to GB 50017-2017 (Section 5.1.1), for tension members with bolted connections, net section tensile fracture must be verified by subtracting the area of all bolt holes along the critical transverse or staggered zig-zag failure section: $A_n = A - \sum d_0 t + \sum \frac{s^2}{4g} t$.

Question 19Structural engineering models and applications

A 240 mm thick unreinforced sintered common brick wall ($h = 240\text{ mm}$) in a multi-story residential building has a clear story height $H = 3.3\text{ m}$. The floor and roof structural scheme is classified as a Rigid Scheme (刚性方案), for which the wall calculation height is $H_0 = 1.0 H = 3.3\text{ m}$. What is the calculated wall slenderness / height-to-thickness ratio $\beta$ (高厚比)?

  1. 13.75 Correct answer.
  2. 16.50
  3. 11.00
  4. 22.00

Explanation

According to the Code for Design of Masonry Structures (GB 50003-2011, Section 5.1.1): $$\beta = \frac{H_0}{h} = \frac{3.30\text{ m}}{0.24\text{ m}} = 13.75$$ For sintered common brick masonry with mortar grade $\ge \text{M5}$, the allowable slenderness ratio $[\beta]_0 = 24$. Since $\beta = 13.75 \le [\beta] = 24$, the wall satisfies the basic height-to-thickness stability check.

Question 20Structural engineering models and applications

What is the primary structural purpose of providing continuous Reinforced Concrete Ring Beams (现浇混凝土圈梁) in multi-story brick masonry buildings per GB 50003-2011 and GB 50011-2010?

  1. To replace standard window and door lintels throughout the building
  2. To tie exterior and interior masonry walls together into a rigid three-dimensional box, mitigate differential settlement cracking, and enhance seismic shear capacity Correct answer.
  3. To allow complete removal of transverse load-bearing walls
  4. To eliminate the need for foundation footings

Explanation

According to GB 50003-2011 (Section 7.1.5) and GB 50011-2010 (Section 7.3.3), cast-in-place RC ring beams encircle the building at floor and roof levels, acting as horizontal tension belts. They tie transverse and longitudinal walls together, increase spatial diaphragm stiffness, resist uneven foundation settlement, and prevent out-of-plane wall collapse during earthquakes.

Question 21Structural engineering models and applications

When a brick masonry wall is reinforced with concrete constructional tie-columns (构造柱), GB 50011-2010 (Section 7.2.4) calculates the combined seismic shear capacity as $V \le \frac{1}{\gamma_{RE}} (f_{vE} A + \zeta_c f_t A_c + 0.08 f_y A_s)$. What do the terms $\zeta_c f_t A_c$ and $0.08 f_y A_s$ represent?

  1. The shear contribution of the concrete tie-column core and the dowel/truss action of its longitudinal reinforcement Correct answer.
  2. The compressive capacity of the foundation soil and the tensile strength of the roof slab
  3. The thermal expansion coefficient and wind drag resistance
  4. The dynamic impact resistance of window lintels

Explanation

According to GB 50011-2010 (Section 7.2.4, Equation 7.2.4-1), in a masonry wall confined by constructional tie-columns: - $f_{vE} A$ is the seismic shear resistance of the masonry pier. - $\zeta_c f_t A_c$ is the shear contribution of the concrete tie-column core ($A_c$ is column concrete area, $f_t$ is tensile strength, $\zeta_c$ is participation factor). - $0.08 f_y A_s$ is the shear resistance contributed by dowel and truss action of the longitudinal rebar ($A_s$).

Question 22Structural engineering models and applications

When unreinforced masonry walls are subjected to bending tension, under which failure plane mode is the design tensile strength of the masonry lowest per GB 50003-2011 (Table 3.2.1-3)?

  1. Bending tension through the stepped zigzag joint (沿齿缝截面弯曲受拉, $f_{tm,z}$)
  2. Bending tension along the horizontal bed joint (沿通缝截面弯曲受拉, $f_{tm}$) Correct answer.
  3. Direct shear across the brick units
  4. Concentric compression

Explanation

According to GB 50003-2011 (Table 3.2.1-3), when bending tension acts perpendicular to the bed joints (沿通缝), failure occurs purely along the plane mortar-brick interface where bond strength is lowest (typically $f_{tm} \approx 0.08\text{--}0.18\text{ N/mm}^2$), which is approximately half the strength along stepped joints (沿齿缝, $f_{tm,z} \approx 0.17\text{--}0.33\text{ N/mm}^2$).

Question 23Structural engineering models and applications

What is the core philosophical framework governing seismic structural design in China, officially known as the 'Three-Level Seismic Fortification Objective' (三水准设防目标) in GB 50011-2010?

  1. Zero horizontal lateral displacement under maximum considered earthquakes
  2. Complete elastic response under all possible earthquake intensities without any cracking
  3. Structure is allowed to collapse as long as emergency alarms sound
  4. No damage under frequent/minor earthquakes; repairable under moderate earthquakes; no collapse under rare/major earthquakes (小震不坏，中震可修，大震不倒) Correct answer.

Explanation

According to GB 50011-2010 (Section 1.0.1): 1. Frequent Minor Earthquakes (小震, 63% exceedance in 50 years): structural elements remain in the elastic range without structural damage (小震不坏); 2. Design-Basis Moderate Earthquakes (中震, 10% exceedance in 50 years): moderate non-structural and repairable structural damage (中震可修); 3. Rare Major Earthquakes (大震, 2%–3% exceedance in 50 years): severe structural damage and yielding, but ductile collapse is strictly prevented to safeguard human life (大震不倒).

Question 24Structural engineering models and applications

Under GB 50011-2010 (Section 4.1.6, Table 4.1.6), what two primary geotechnical parameters are used to classify a construction site into Class I0, I1, II, III, or IV (场地类别划分)?

  1. Groundwater depth ($d_w$) and soil plastic index ($I_p$)
  2. Standard Penetration Test blow count ($N$) and soil moisture content ($w$)
  3. Equivalent shear wave velocity of soil layers ($v_{se}$) and thickness of the soft/overburden soil deposit ($d_{ov}$) Correct answer.
  4. Total building weight ($G$) and foundation area ($A$)

Explanation

According to GB 50011-2010 (Section 4.1.6), site categorization is determined based on: 1. The equivalent shear wave velocity of the upper $20\text{ m}$ of soil: $v_{se} = \frac{d_0}{\sum (d_i / v_{si})}$; 2. The total thickness of the overburden soil stratum ($d_{ov}$) down to bedrock ($v_s \ge 500\text{ m/s}$).

Question 25Structural engineering models and applications

In GB 50011-2010 (Section 5.1.4), when the fundamental natural period of a building satisfies $T_g \le T_1 \le 5 T_g$, what is the equation for the horizontal seismic influence coefficient $\alpha_1$ (水平地震影响系数)?

  1. $\alpha_1 = \left(\frac{T_g}{T_1}\right)^\gamma \eta_2 \alpha_{\max}$ Correct answer.
  2. $\alpha_1 = \alpha_{\max}$
  3. $\alpha_1 = \eta_2 \alpha_{\max} \left(1 - \frac{T_1}{T_g}\right)$
  4. $\alpha_1 = 0.2 \alpha_{\max}$

Explanation

According to GB 50011-2010 (Section 5.1.4, Equation 5.1.4-3), in the descending curve region ($T_g \le T \le 5 T_g$), the seismic influence coefficient is given by: $$\alpha = \left(\frac{T_g}{T}\right)^\gamma \eta_2 \alpha_{\max}$$ where $\gamma = 0.9 + \frac{0.05 - \zeta}{0.3 + 6 \zeta}$ (for standard $5\%$ damping, $\gamma = 0.9, \eta_2 = 1.0$), $\alpha_{\max}$ is the maximum seismic influence coefficient, and $T_g$ is the characteristic period.

Question 26Structural engineering models and applications

In the Equivalent Base Shear Method per GB 50011-2010 (Section 5.2.1), how is the total base shear $F_{Ek}$ distributed to the individual story levels ($F_i$) when the fundamental period $T_1 \le 1.4 T_g$ (additional top concentrated force $\delta_n = 0$)?

  1. $F_i = \frac{G_i H_i}{\sum_{j=1}^n G_j H_j} F_{Ek}$ (Inverted triangular distribution proportional to story height and weight) Correct answer.
  2. $F_i = \frac{F_{Ek}}{n}$ (Uniform distribution)
  3. $F_i = \frac{G_i}{\sum G_j} F_{Ek}$ (Proportional to story weight only)
  4. $F_i = \frac{H_i^2}{\sum H_j^2} F_{Ek}$ (Parabolic distribution)

Explanation

According to GB 50011-2010 (Section 5.2.1, Equation 5.2.1-2), when $T_1 \le 1.4 T_g$, higher mode effects at the roof are negligible ($\\delta_n = 0$), and horizontal seismic story forces are distributed along the building height according to an inverted triangle: $$F_i = \frac{G_i H_i}{\sum_{j=1}^n G_j H_j} F_{Ek}$$ where $G_i$ is the representative gravity load at level $i$ and $H_i$ is the height of level $i$ from the base.

Question 27Structural engineering models and applications

According to GB 50011-2010 (Section 5.2.5, Table 5.2.5), what is the purpose of the Minimum Seismic Shear-to-Weight Ratio (剪重比 / 楼层最小地震剪力系数 $\lambda = V_{Ei} / G_E$)?

  1. To verify wind resistance during construction
  2. To limit total structural weight to prevent foundation settlement
  3. To eliminate the need for seismic stirrups in columns
  4. To ensure that long-period structures designed by dynamic analysis do not have their seismic story shears under-designed below statutory minimum safety baselines due to period lengthening or higher-mode truncation Correct answer.

Explanation

According to GB 50011-2010 (Section 5.2.5), response spectrum and time-history analysis can yield overly low story shears for flexible, long-period buildings. To safeguard against long-period ground motion resonance and uncertainty, the code mandates a minimum shear-to-weight ratio (e.g. $\lambda \ge 1.6\%$ for 7-degree, $3.2\%$ for 8-degree zones); if calculated shear is lower, story shear forces must be scaled up to this baseline.

Question 28Geotechnical engineering models and applications

In the Code for Design of Building Foundation (GB 50007-2011, Section 5.2.4), what is the statutory equation for correcting the characteristic value of subgrade soil bearing capacity ($f_a$) for footing width $b$ and embedment depth $d$?

  1. $f_a = f_{ak} + \eta_b \gamma (b - 3) + \eta_d \gamma_m (d - 0.5)$ Correct answer.
  2. $f_a = f_{ak} + \eta_b \gamma b + \eta_d \gamma_m d$
  3. $f_a = f_{ak} \times \left(1 + \frac{b}{3}\right) \times \left(1 + \frac{d}{0.5}\right)$
  4. $f_a = f_{ak} + \gamma d$

Explanation

According to GB 50007-2011 (Section 5.2.4, Equation 5.2.4): $$f_a = f_{ak} + \eta_b \gamma (b - 3) + \eta_d \gamma_m (d - 0.5)$$ where $f_{ak}$ is the uncorrected bearing capacity from plate load tests (evaluated at standard $b=3\text{ m}, d=0.5\text{ m}$), $\eta_b, \eta_d$ are correction coefficients for width and depth, $\gamma$ is the weighted soil unit weight below the footing base, and $\gamma_m$ is the weighted soil unit weight above the base.

Question 29Geotechnical engineering models and applications

A column pad footing has width $b = 4.0\text{ m}$ and embedment depth $d = 2.5\text{ m}$. The subgrade soil is silty clay with characteristic bearing capacity $f_{ak} = 160\text{ kPa}$, soil unit weight below base $\gamma = 18.0\text{ kN/m}^3$, weighted unit weight above base $\gamma_m = 17.5\text{ kN/m}^3$, and correction coefficients $\eta_b = 0.3, \eta_d = 1.6$. According to GB 50007-2011, what is the corrected characteristic bearing capacity $f_a$?

  1. 198.5 kPa
  2. 160.0 kPa
  3. 277.4 kPa
  4. 221.4 kPa Correct answer.

Explanation

For this question use the supplied numerical assumptions and the stated model only. According to GB 50007-2011 (Section 5.2.4): $$f_a = f_{ak} + \eta_b \gamma (b - 3) + \eta_d \gamma_m (d - 0.5)$$ 1. Width correction term: $$\Delta f_b = 0.3 \times 18.0 \times (4.0 - 3.0) = 5.4\text{ kPa}$$ 2. Depth correction term: $$\Delta f_d = 1.6 \times 17.5 \times (2.5 - 0.5) = 1.6 \times 17.5 \times 2.0 = 56.0\text{ kPa}$$ 3. Corrected bearing capacity: $$f_a = 160.0 + 5.4 + 56.0 = 221.4\text{ kPa}$$. This is not independent confirmation of a current local code or complete design approval.

Question 30Geotechnical engineering models and applications

A rectangular column footing has length $l = 3.0\text{ m}$ and width $b = 2.0\text{ m}$. The total vertical standard load including foundation self-weight is $F_k + G_k = 1200\text{ kN}$, acting with an eccentricity $e = 0.60\text{ m}$ along the length direction ($l$). Since $e = 0.60\text{ m} > l/6 = 0.50\text{ m}$, the base experiences tensile liftoff. According to GB 50007-2011 (Section 5.2.2, Equation 5.2.2-4), what is the maximum edge compressive pressure $p_{k\max}$?

  1. 200.0 kPa
  2. 444.4 kPa Correct answer.
  3. 266.7 kPa
  4. 600.0 kPa

Explanation

According to GB 50007-2011 (Section 5.2.2, Equation 5.2.2-4): When $e > l/6$, the contact pressure distribution is triangular over an effective contact length $3 a$ (where $a = l/2 - e$ is the distance from the resultant force to the near edge): 1. Distance $a = \frac{l}{2} - e = \frac{3.0}{2} - 0.60 = 1.50 - 0.60 = 0.90\text{ m}$. 2. Effective contact length: $L_{eff} = 3 a = 3 \times 0.90 = 2.70\text{ m}$. 3. Maximum edge pressure: $$p_{k\max} = \frac{2 (F_k + G_k)}{3 a b} = \frac{2 \times 1200\text{ kN}}{3 \times 0.90\text{ m} \times 2.0\text{ m}} = \frac{2400}{5.40} = 444.44\text{ kPa} \approx 444.4\text{ kPa}$$.

Question 31Geotechnical engineering models and applications

In GB 50007-2011 (Section 5.2.7), when checking the bearing capacity of a soft underlying layer (软弱下卧层承载力验算) beneath a shallow footing, what is the governing verification criterion at the top of the soft layer?

  1. $p_z + p_{cz} \le f_{az}$ (Diffused additional stress $p_z$ plus effective overburden self-weight stress $p_{cz}$ must not exceed the corrected bearing capacity of the soft layer $f_{az}$) Correct answer.
  2. $p_z \le f_{az}$ (Additional stress alone is checked)
  3. $p_z + p_{cz} \le f_a$ (Checked against the upper layer bearing capacity)
  4. $p_{cz} \le f_{az}$

Explanation

According to GB 50007-2011 (Section 5.2.7, Equation 5.2.7-1), the total vertical stress at the top of the soft underlayer equals the diffused foundation additional stress $p_z = \frac{b (p_k - p_c)}{b + 2 z \tan\theta}$ (for strip footing) plus the self-weight stress of the soil overburden $p_{cz}$. The verification condition is: $$p_z + p_{cz} \le f_{az}$$ where $f_{az}$ is the characteristic bearing capacity of the soft layer corrected for depth to the top of the soft layer.

Question 32Geotechnical engineering models and applications

When calculating the bending moment $M$ for sizing flexural bottom reinforcement in an isolated reinforced concrete column pad footing under concentric axial load per GB 50007-2011 (Section 8.2.11), which critical cross-section must be checked?

  1. A section located at a distance $h_0$ away from the column face
  2. The section at the centerline of the column
  3. The outermost edge of the footing
  4. The section at the column face (截面位于柱边 / 变阶处) Correct answer.

Explanation

According to GB 50007-2011 (Section 8.2.11), the cantilever upward soil reaction produces maximum bending moment in the footing at the face of the column (or at stepped section transitions). The flexural reinforcement in both orthogonal directions is calculated based on the moments at these column-face critical sections.

Question 33Geotechnical engineering models and applications

Under GB 50007-2011 (Section 8.2.7), what is the difference between the Punching Shear Failure Mode (冲切破坏) and the One-Way Diagonal Shear Failure Mode (剪切破坏) in shallow pad footings?

  1. Punching shear is caused by wind loads; one-way shear is caused by gravity loads
  2. Punching shear applies only to circular columns; one-way shear applies only to square columns
  3. Punching shear involves a truncated 45-degree pyramidal failure surface around the column perimeter; one-way shear involves an inclined shear plane extending across the full footing width Correct answer.
  4. There is no difference; they share the identical design equation

Explanation

According to GB 50007-2011 (Sections 8.2.7 and 8.2.9): - Punching shear (受冲切) is a two-way shear mechanism where the column punches through the slab forming a truncated $45^\circ$ pyramid ($F_l \le 0.7 \beta_{hp} f_t a_m h_0$). - One-way diagonal shear (斜截面受剪) occurs in elongated rectangular footings or wall strip footings where an inclined shear crack traverses across the full footing width ($V \le 0.7 \beta_{hs} f_t b h_0$).

Question 34Geotechnical engineering models and applications

A continuous brick masonry wall transmits a characteristic vertical load $F_k = 220\text{ kN/m}$ to a concrete strip footing. The footing embedment depth is $d = 1.5\text{ m}$. The corrected subgrade bearing capacity is $f_a = 150\text{ kPa}$. The average unit weight of the foundation footing and overburden soil is $\gamma_G = 20\text{ kN/m}^3$. According to GB 50007-2011 (Section 5.2.2), what is the required strip footing base width $b$?

  1. 1.47 m (Select 1.5 m)
  2. 1.83 m (Select 1.9 m) Correct answer.
  3. 2.25 m (Select 2.3 m)
  4. 1.20 m (Select 1.2 m)

Explanation

For this question use the supplied numerical assumptions and the stated model only. According to GB 50007-2011 (Section 5.2.2, Equation 5.2.2-1): $$p_k = \frac{F_k + G_k}{A} = \frac{F_k + \gamma_G b d}{b \times 1.0} \le f_a$$ Rearranging for base width $b$ per unit run ($1.0\text{ m}$): $$b \ge \frac{F_k}{f_a - \gamma_G d} = \frac{220\text{ kN/m}}{150\text{ kPa} - (20\text{ kN/m}^3 \times 1.5\text{ m})} = \frac{220}{150 - 30} = \frac{220}{120} = 1.833\text{ m} \approx 1.83\text{ m}$$ Thus, a strip footing width of $1.9\text{ m}$ is selected. This is not independent confirmation of a current local code or complete design approval.

Question 35Geotechnical engineering models and applications

In the calculation of foundation settlement using the Layer Summation Method (分层总和法) per GB 50007-2011 (Section 5.3.5), what is the function of the empirical settlement calculation coefficient $\psi_s$ (沉降计算经验系数)?

  1. It calibrates the theoretical 1D elastic-consolidation settlement to account for 3D stress state, soil lateral yield, disturbance, and regional regional soil calibration Correct answer.
  2. It accounts for seismic inertial liquefaction settlement
  3. It converts short-term concrete elastic shrinkage to long-term creep
  4. It reduces the calculated settlement to zero if groundwater is present

Explanation

According to GB 50007-2011 (Section 5.3.5), the calculated settlement $s = \psi_s s'$ uses the empirical coefficient $\psi_s$ (obtained from Table 5.3.5 based on subgrade characteristic pressure and compression modulus) to calibrate the 1D layer summation theory against extensive field plate/building settlement observation data, correcting for 3D lateral strain, sample disturbance, and soil non-linearity.

Question 36Geotechnical engineering models and applications

A cast-in-place reinforced concrete basement exterior wall is restrained by the basement ground floor slab at the top and the foundation raft at the bottom. When calculating the lateral earth pressure acting on this wall per GB 50007-2011 and JGJ 120-2012, which earth pressure state must be adopted?

  1. Hydrostatic pressure only, neglecting all soil particles
  2. Active earth pressure (主动土压力, $E_a$), because any small strain mobilizes active state
  3. Passive earth pressure (被动土压力, $E_p$), because soil pushes into the wall
  4. At-rest earth pressure (静止土压力, $E_0 = \frac{1}{2} K_0 \gamma H^2$), because the rigid floor diaphragms prevent wall lateral outward displacement Correct answer.

Explanation

According to GB 50007-2011 and technical standards for basement retaining walls, because the basement perimeter wall is rigidly propped by multiple floor slabs and monolithic floor diaphragms, lateral displacement $\Delta$ is practically zero. Therefore, active earth pressure cannot fully mobilize, and the wall must be designed for At-Rest Earth Pressure (静止土压力, $K_0 \approx 0.5$ for normal clay/sand) plus hydrostatic water pressure.

Question 37Geotechnical engineering models and applications

A vertical, smooth retaining wall has height $H = 6.0\text{ m}$. The backfill is a cohesionless sand with unit weight $\gamma = 18.0\text{ kN/m}^3$ and internal friction angle $\varphi = 30^\circ$. The ground surface behind the wall is horizontal with no surcharge. Using Rankine's earth pressure theory per GB 50007-2011, what is the total resultant active earth pressure $E_a$ acting on a unit length of the wall?

  1. 162.0 kN/m
  2. 108.0 kN/m Correct answer.
  3. 324.0 kN/m
  4. 81.0 kN/m

Explanation

According to Rankine's earth pressure theory in GB 50007-2011: 1. Rankine active earth pressure coefficient: $$K_a = \tan^2\left(45^\circ - \frac{\varphi}{2}\right) = \tan^2\left(45^\circ - 15^\circ\right) = \tan^2(30^\circ) = \frac{1}{3} \approx 0.3333$$ 2. Active earth pressure intensity at base ($z = 6.0\text{ m}$): $$p_a = \gamma H K_a = 18.0 \times 6.0 \times \frac{1}{3} = 36.0\text{ kN/m}^2$$ 3. Total active earth pressure resultant: $$E_a = \frac{1}{2} \gamma H^2 K_a = \frac{1}{2} \times 18.0 \times 6.0^2 \times \frac{1}{3} = \frac{1}{2} \times 18.0 \times 36.0 \times \frac{1}{3} = 108.0\text{ kN/m}$$.

Question 38Geotechnical engineering models and applications

A bored cast-in-place concrete pile has a circular diameter $d = 0.8\text{ m}$ (perimeter $u = 2.513\text{ m}$, base area $A_p = 0.5027\text{ m}^2$) and length $L = 15.0\text{ m}$. The ultimate shaft resistance is $Q_{sk} = 900\text{ kN}$ and the ultimate base resistance is $Q_{pk} = 600\text{ kN}$. According to the Technical Code for Building Pile Foundations (JGJ 94-2008, Section 5.2.2), what is the characteristic vertical compressive bearing capacity $R_a$ (单桩竖向承载力特征值) of this single pile?

  1. 600.0 kN
  2. 1500.0 kN
  3. 1200.0 kN
  4. 750.0 kN Correct answer.

Explanation

For this question use the supplied numerical assumptions and the stated model only. According to JGJ 94-2008 (Section 5.2.2): 1. Total ultimate vertical compressive bearing capacity: $$Q_{uk} = Q_{sk} + Q_{pk} = 900\text{ kN} + 600\text{ kN} = 1500.0\text{ kN}$$ 2. The characteristic value of vertical compressive bearing capacity $R_a$ is defined as the ultimate capacity divided by a global safety factor of $K = 2.0$: $$R_a = \frac{Q_{uk}}{2} = \frac{1500.0\text{ kN}}{2} = 750.0\text{ kN}$$. This is not independent confirmation of a current local code or complete design approval.

Question 39Geotechnical engineering models and applications

In pile foundation engineering per JGJ 94-2008 (Section 5.4.1), what is Negative Skin Friction (负摩阻力) and at what critical point along the pile shaft does the shaft shear stress reverse from downward drag to upward supporting resistance?

  1. Tensile friction generated during pile dynamic driving rebound
  2. Upward friction caused by groundwater buoyancy acting at the pile tip
  3. Downward drag caused when surrounding soft consolidating soil settles more than the pile shaft; the transition occurs at the Neutral Point (中性点) Correct answer.
  4. Frictional resistance along the top pile cap interface

Explanation

According to JGJ 94-2008 (Section 5.4.1), when compressible soft soil surrounding a pile undergoes self-weight consolidation or settles due to surcharge/dewatering faster than the pile, the soil drags downward on the pile shaft. This downward shear is negative skin friction. The location along the shaft where the settlement of the soil exactly equals the settlement of the pile is the Neutral Point (中性点), above which friction is downward and below which friction is upward.

Question 40Geotechnical engineering models and applications

In seasonally frozen ground regions per GB 50007-2011 (Section 5.1.7), what is the primary structural reason for setting the minimum foundation embedment depth below the design frost penetration depth $z_d$?

  1. To increase the soil thermal conductivity
  2. To prevent frost heave uplift forces (冻胀力) and thawing settlement (融沉) from cracking footings and masonry walls Correct answer.
  3. To ensure concrete hydration does not exceed 50 degrees Celsius
  4. To reduce the structural wind load

Explanation

According to GB 50007-2011 (Section 5.1.7), water in frost-susceptible soils expands upon freezing, generating massive upward normal and tangential frost heave forces. Upon spring thaw, the subgrade collapses in thawing settlement (融沉). Placing the footing base below the frost depth $z_d$ ensures the subgrade bearing layer remains unfrozen.

Question 41Construction 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 42Construction 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 43Construction 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 44Construction 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 45Construction 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 46Construction 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 47Construction 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 48Construction 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 49Construction 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 50Construction 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.