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Agricultural and Biosystems Engineers Licensure Examination Building Materials, Theory of Structures & Construction: practice questions

Agricultural and Biosystems Engineers Licensure Examination (ABELE) · Philippines · Professional Regulatory Board of Agricultural and Biosystems Engineering, PRC

Original practice questions written to the published syllabus · not past papers
10worked here, free
60written to this exam in this subject
9syllabus units

What this subject covers

Materials, structural members, foundations and estimates for farm buildings and small works.

In the exams: NCEES PE Facilities & Structures (materials, foundations, structural analysis); PRC ABE structures; TNPSC Unit V

Where the 60 questions fall

How our questions for Agricultural and Biosystems Engineers Licensure Examination spread across the units of Building Materials, Theory of Structures & Construction. Difficulty: 5 easy, 30 medium, 25 hard.

Reinforced concrete beams and slabs11
Concrete and concrete mix design8
Reinforced concrete columns7
Bricks, stone and cement6
Steel and timber6
Trusses6
Foundations6
Estimation, costing and rate analysis5
Specifications and contracts5

10 practice questions, worked

Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.

1 · Bricks, stone and cement · medium

Brick earth that contains too much lime is used to make bricks. What is the most likely result?

  1. The bricks melt and lose shape in burning, and lime lumps may later split them
  2. The bricks turn bright red because lime reacts with the iron oxide during firing
  3. The bricks become very porous and light because the lime burns out completely
  4. The bricks crack during drying because lime can make the clay too plastic to shape
Answer and working

A. The bricks melt and lose shape in burning, and lime lumps may later split them

Excess lime acts as a flux, so bricks fuse and distort in the kiln; lumps of lime become quicklime that slakes and expands when wet, splitting the brick. Plasticity comes from alumina, not lime, and the red colour comes from iron oxide; lime tends to make bricks lighter in colour.

Further reading: Duggal, Building Materials

2 · Bricks, stone and cement · medium

How does a pozzolana such as fly ash or rice-husk ash contribute to the strength of a lime or blended-cement binder?

  1. Its reactive silica combines with calcium hydroxide and water to form calcium silicate hydrate
  2. It reacts with atmospheric carbon dioxide to form calcium carbonate, as fat lime does
  3. It hydrates on its own like Portland cement clinker, producing calcium hydroxide and heat
  4. It fills voids as an inert powder only, so any strength gain is due to better particle packing
Answer and working

A. Its reactive silica combines with calcium hydroxide and water to form calcium silicate hydrate

A pozzolana has little or no cementing value alone; it reacts with calcium hydroxide (from lime or from cement hydration) to form extra C–S–H. It does not hydrate on its own like clinker, and the carbonation route is how fat lime hardens, not how a pozzolana works. Packing (filler) effects exist but are not the main mechanism.

Further reading: Mehta and Monteiro, Concrete

3 · Reinforced concrete beams and slabs · hard

A singly reinforced beam is 250 mm wide with effective depth 450 mm, M25 concrete and Fe415 steel. What is its limiting moment of resistance Mu,lim by IS 456?

  1. 139.7 kN·m
  2. 218.7 kN·m
  3. 169.1 kN·m
  4. 174.6 kN·m
Answer and working

D. 174.6 kN·m

xu,max = 0.48 × 450 = 216 mm. Mu,lim = 0.36 fck b xu,max (d − 0.42 xu,max) = 0.36 × 25 × 250 × 216 × (450 − 90.72) = 174.6 kN·m (≈ 0.138 fck b d²). Using 0.46 (the Fe500 limit) gives 169.1; omitting the 0.42 xu lever-arm reduction gives 218.7. 139.7 kN·m uses fck = 20 (M20) instead of 25.

Further reading: IS 456:2000, Annex G

4 · Reinforced concrete beams and slabs · hard

A beam 300 mm wide with effective depth 500 mm has four 16 mm Fe500 bars in tension and M25 concrete. Using IS 456, what is its moment of resistance?

  1. 250.5 kN·m
  2. 155.9 kN·m
  3. 174.9 kN·m
  4. 175.9 kN·m
Answer and working

B. 155.9 kN·m

Ast = 4 × π × 16²/4 = 804.2 mm². xu = 0.87 × 500 × 804.2/(0.36 × 25 × 300) = 129.6 mm < xu,max = 0.46 × 500 = 230 mm, so the section is under-reinforced. Mu = 0.87 fy Ast (d − 0.42 xu) = 155.9 kN·m. Ignoring the lever-arm reduction gives 174.9; dropping 0.87 gives 175.9; 250.5 is Mu,lim, the capacity only if the section were reinforced up to the limit.

Further reading: IS 456:2000, Annex G

5 · Reinforced concrete columns · hard

A short, axially loaded 350 × 350 mm column has four 20 mm Fe500 bars and M25 concrete. Using Pu = 0.4 fck Ac + 0.67 fy Asc (IS 456), what is its design capacity?

  1. 1823 kN
  2. 1759 kN
  3. 1633 kN
  4. 1646 kN
Answer and working

C. 1633 kN

Asc = 4 × π × 20²/4 = 1256.6 mm²; Ac = 122,500 − 1256.6 = 121243.4 mm². Pu = 0.4 × 25 × 121243 + 0.67 × 500 × 1256.6 = 1633 kN. Using the gross area for concrete gives 1646; using 0.87 fy gives 1759; 1823 uses the pure-axial coefficients without the code's allowance for minimum eccentricity.

Further reading: IS 456:2000, clause 39.3

6 · Foundations · medium

How does a rise of the water table to the base of a shallow footing in sand affect its bearing capacity?

  1. It reduces it only for clays, because sands drain freely and their strength is not affected
  2. It reduces it, because the submerged unit weight, close to half the bulk value, enters the γ terms
  3. It has no effect, because the bearing capacity factors depend only on the angle of shearing resistance
  4. It increases it, because water pressure supports part of the footing load and reduces settlement
Answer and working

B. It reduces it, because the submerged unit weight, close to half the bulk value, enters the γ terms

Effective stresses fall when soil is submerged; the γ in the Nγ term (and in the Nq term if the water rises above the base) is replaced by the submerged unit weight, cutting capacity, especially in cohesionless soils whose strength is purely frictional.

Further reading: Das, Principles of Foundation Engineering

7 · Foundations · medium

Why is differential settlement more damaging to a building than uniform settlement of the same maximum magnitude?

  1. It reduces the soil's bearing capacity, while uniform settlement increases it by densifying the soil
  2. It affects only the floor slab, while uniform settlement cracks the roof trusses and walls
  3. It distorts the building, causing cracks, tilt and jammed doors; uniform settlement does not
  4. It happens instantly, while uniform settlement always takes decades to develop in any soil
Answer and working

C. It distorts the building, causing cracks, tilt and jammed doors; uniform settlement does not

Angular distortion between adjacent supports induces stresses and cracks in walls and frames; a building that settles uniformly may only need service connections adjusted. Both kinds can be immediate or time-dependent.

Further reading: Das, Principles of Foundation Engineering

8 · Estimation, costing and rate analysis · hard

One cubic metre of brickwork uses 500 bricks of actual size 190 × 90 × 90 mm. If dry mortar volume is taken as 1.27 times the wet volume of joints, what dry mortar volume is needed (ignoring frog filling and wastage)?

  1. 0.293 m³
  2. 0.355 m³
  3. 0.231 m³
  4. 0.769 m³
Answer and working

A. 0.293 m³

Volume of bricks = 500 × 0.19 × 0.09 × 0.09 = 0.7695 m³; wet mortar = 1 − 0.7695 = 0.2305 m³; dry = 0.2305 × 1.27 = 0.293 m³. 0.231 is the wet volume; 0.769 is the brick volume; 0.355 uses the concrete dry-volume factor.

Further reading: Dutta, Estimating and Costing

9 · Estimation, costing and rate analysis · hard

A godown cost 800,000 currency units, has a scrap value of 80,000 and a useful life of 40 years. By the straight-line method, what is its book value after 15 years?

  1. 610,000
  2. 530,000
  3. 270,000
  4. 500,000
Answer and working

B. 530,000

Annual depreciation = (800,000 − 80,000)/40 = 18,000. After 15 years: 800,000 − 15 × 18,000 = 530,000. 500,000 ignores the scrap value; 270,000 is the accumulated depreciation, not the book value. 610,000 adds the scrap value on top of the book value.

Further reading: Dutta, Estimating and Costing

10 · Specifications and contracts · medium

What are liquidated damages in a construction contract?

  1. Payments to the contractor for delays caused by the owner's late drawings
  2. A pre-agreed sum per day or week of delay, payable by the contractor for late completion
  3. The actual losses the owner proves in court after the work is completed late
  4. A penalty for poor workmanship assessed by the engineer at final inspection
Answer and working

B. A pre-agreed sum per day or week of delay, payable by the contractor for late completion

Liquidated damages are a genuine pre-estimate of loss, written into the contract so that the owner does not need to prove actual loss. Delays caused by the owner entitle the contractor to an extension of time, not to pay damages. Defective work is handled by rectification and the defects liability period.

Further reading: FIDIC Conditions of Contract

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Building Materials, Theory of Structures & ConstructionThe full study manual, free Agricultural and Biosystems Engineers Licensure ExaminationFormat, sections and how it has changed

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