Nepal Engineering Council · Civil Engineering · Chapter 5
Design of Structures
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184 questions in 6 syllabus topics.
5.1 Loads and load combinations
30 questions · ACiE0501
1. In the IS 875 series, the Part that deals with dead loads (unit weights of building materials) is
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Part numbering starts with the most permanent load.
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Answer: A. Part 1
IS 875 (Part 1) gives unit weights of building materials and stored materials for dead load calculation; Part 2 is imposed load, Part 3 wind, Part 4 snow and Part 5 special loads and combinations.
2. Load combinations for design are specified in which part of IS 875?
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It is the last part of the series.
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Answer: D. Part 5
IS 875 (Part 5) covers special loads and load combinations; the other parts give individual loads (dead, imposed, wind, snow).
3. In Nepal, seismic design of buildings is covered by which Nepal Building Code?
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Think of the 100-series (requirements/design loads), not 200-series (rules of thumb).
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Answer: D. NBC 105
NBC 105 is the code for seismic design of buildings in Nepal; NBC 102 gives unit weights, NBC 104 imposed loads and NBC 205 rules of thumb for RC buildings.
4. The unit weight of plain cement concrete and reinforced cement concrete as per IS 875 (Part 1) are respectively
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Reinforced concrete is heavier than plain.
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Answer: C. 24 kN/m³ and 25 kN/m³
IS 875 (Part 1) gives 24 kN/m³ for plain concrete and 25 kN/m³ for reinforced concrete, the steel making it heavier.
5. A 150 mm thick RCC slab (unit weight 25 kN/m³) carries a floor finish of 1.0 kN/m². The total dead load per m² of slab is
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Convert thickness to metres before multiplying.
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Answer: C. 4.75 kN/m²
Self weight = 0.15 × 25 = 3.75 kN/m²; adding finish gives 4.75 kN/m².
6. A 230 mm thick brick masonry wall 3.0 m high (unit weight of masonry 19 kN/m³, plaster ignored) acting on a beam produces a line load of
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Line load = area of wall cross-section per metre length × unit weight.
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Answer: B. 13.11 kN/m
w = thickness × height × unit weight = 0.23 × 3.0 × 19 = 13.11 kN/m.
7. Which of the following is correctly classed as a dead load?
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Dead loads do not change with time or occupancy.
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Answer: B. Weight of a permanent partition wall and floor finish
Dead loads are permanent loads including self weight, finishes and fixed partitions; occupants and movable furniture are imposed loads.
8. As per IS 875 (Part 2), the minimum imposed load on floors of residential building rooms is
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Small value used also for hostel rooms.
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Answer: B. 2.0 kN/m²
IS 875 (Part 2) prescribes 2.0 kN/m² for dwelling-house rooms; balconies, corridors and stairs carry 3.0 kN/m².
9. Imposed load on a flat roof not accessible except for maintenance, with slope ≤ 10°, as per IS 875 (Part 2) is
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Lower than an accessible roof.
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Answer: A. 0.75 kN/m²
For access provided roofs 1.5 kN/m² applies; where no access is provided other than maintenance and slope ≤ 10°, IS 875 (Part 2) takes 0.75 kN/m².
10. A column supports three floors including the roof, each carrying a total imposed load of 50 kN. IS 875 (Part 2) allows a reduction of 20% for three floors. The design imposed load is
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Reduce the total load, not each floor differently.
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Answer: D. 120 kN
Total = 3 × 50 = 150 kN; reduced by 20% gives 150 × 0.8 = 120 kN.
11. Basic wind speed is 47 m/s, k1 = 1.0, k2 = 1.07, k3 = 1.0 and k4 = 1.0. The design wind speed Vz is
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Design speed is the product of basic speed and four factors.
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Answer: A. 50.29 m/s
Vz = Vb k1 k2 k3 k4 = 47 × 1.0 × 1.07 × 1.0 × 1.0 = 50.29 m/s.
12. The design wind speed at a building height is 36 m/s. The wind pressure pz = 0.6 Vz² is
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Square the speed first; the unit is N/m².
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Answer: C. 777.6 N/m²
pz = 0.6 × 36² = 777.6 N/m² (about 0.78 kN/m²).
13. In the IS 875 (Part 3) formula Vz = Vb k1 k2 k3 k4, the factor k3 accounts for
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It deals with the ground profile.
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Answer: B. local topography such as hills and escarpments
k1 is probability (risk) factor, k2 terrain, height and structure size factor, and k3 topography factor.
14. As per IS 875 (Part 3), the internal pressure coefficient Cpi for a building with large openings (more than 20% of wall area) is taken as
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Larger openings give larger internal pressure.
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Answer: D. ± 0.7
Cpi = ±0.2 for low permeability (<5%), ±0.5 for medium (5–20%) and ±0.7 for large openings (>20%).
15. A wall of area 60 m² has design wind pressure pd = 1.2 kN/m², Cpe = +0.7 (windward) and Cpi = +0.2. The net wind force on the wall is
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Net coefficient is the difference of external and internal values.
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Answer: B. 36 kN
F = (Cpe − Cpi) pd A = (0.7 − 0.2) × 1.2 × 60 = 36 kN.
16. As per IS 875 (Part 3), a building is considered wind-sensitive (dynamic effects must be considered) if its height-to-least-width ratio exceeds
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A slender ratio, between 3 and 10.
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Answer: C. 5
Buildings with h/b greater than 5 or natural frequency less than 1 Hz are treated as flexible and need dynamic wind analysis.
17. Ground snow load is 1.5 kN/m² and the shape coefficient for a roof of slope 25° is 0.8. The design snow load on the roof is
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Snow load on roof is shape coefficient times ground snow load.
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Answer: D. 1.20 kN/m²
s = μ s0 = 0.8 × 1.5 = 1.20 kN/m².
18. The snow load shape coefficient μ in IS 875 (Part 4) for roof slopes steeper than 60° is
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Snow cannot rest on very steep roofs.
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Answer: D. zero
Snow slides off steep roofs, so μ = 0.8 for slope ≤ 30° and reduces to 0 for slopes above 60°.
19. The number of seismic zones in the map of IS 1893 (Part 1):2016 is
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Zone I was removed in the latest revision.
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Answer: C. 4 (Zones II to V)
The 2016 revision merged Zone I into II, leaving Zones II, III, IV and V.
20. The zone factor Z in IS 1893 (Part 1) for Seismic Zone V is
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Highest value in the table.
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Answer: C. 0.36
Z = 0.10, 0.16, 0.24 and 0.36 for Zones II, III, IV and V respectively.
21. For seismic weight calculation of a building floor, IS 1893 considers imposed load of up to 3 kN/m² as
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Only a fraction of live load is present at shaking.
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Answer: A. 25% of the imposed load
Seismic weight = full dead load plus 25% of imposed load (50% when imposed load exceeds 3 kN/m²), since full live load is unlikely during an earthquake.
22. An RC moment-resisting frame building without infill has height 16 m. The approximate fundamental period Ta = 0.075 h^0.75 is
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16 raised to 0.75 is a whole number.
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Answer: A. 0.60 s
16^0.75 = 8, so Ta = 0.075 × 8 = 0.60 s.
23. A building in Zone IV (Z = 0.24) on medium soil has I = 1.0, R = 5, T = 1.0 s (Sa/g = 1.36/T) and seismic weight 8000 kN. The design base shear is
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Ah = (Z/2)(I/R)(Sa/g), then multiply by seismic weight.
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Answer: A. 261.1 kN
Ah = (Z/2)(I/R)(Sa/g) = 0.12 × 0.2 × 1.36 = 0.03264; Vb = Ah W = 261.1 kN.
24. A three-storey building has equal floor weights at heights 3 m, 6 m and 9 m. For design base shear 300 kN, the lateral force at the roof level using Qi = Vb Wi hi²/ΣWj hj² is
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Distribution is proportional to weight times height squared.
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Answer: A. 192.9 kN
Σ W h² ∝ 9 + 36 + 81 = 126; roof force = 300 × 81/126 = 192.9 kN.
25. The response reduction factor R in seismic design is larger for a structural system that
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Ductile detailing earns reduction in force.
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Answer: B. has greater ductility and overstrength
A ductile system dissipates energy by inelastic action, so the elastic force may be reduced by a larger R (e.g. SMRF has R = 5 versus OMRF 3).
26. The importance factor I in IS 1893 is taken as 1.5 for
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Must remain functional after an earthquake.
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Answer: B. hospitals and other important service buildings
Important buildings such as hospitals, schools and fire stations carry I = 1.5 so that they stay operational after a quake.
27. In a design load combination, earthquake load and wind load are
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Both are rare short-duration events.
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Answer: B. not considered to act simultaneously
Per IS 875 (Part 5) and IS 1893, wind and earthquake are not assumed to occur together; the larger effect governs.
28. A slab has dead load 4.75 kN/m² (including finish) and imposed load 2.0 kN/m². The factored design load for the limit state of collapse using 1.5(DL + IL) is
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Both loads get the same partial factor of 1.5.
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Answer: A. 10.125 kN/m²
wu = 1.5 × (4.75 + 2.0) = 10.125 kN/m².
29. To check overturning or uplift when dead load is beneficial, the load combination used in IS 456 / IS 875 (Part 5) is
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Dead load helps resist overturning, so it is reduced.
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Answer: C. 0.9 DL + 1.5 WL
A reduced dead load factor 0.9 is used with 1.5 times the wind or seismic load when stability against overturning is critical.
30. Earthquake force on a structure is essentially
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Newton's second law.
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Answer: D. an inertia force proportional to mass and ground acceleration
Ground shaking accelerates the structure, giving inertia force F = m a, so lighter and more ductile structures attract smaller loads.
5.2 Concrete technology
32 questions · ACiE0502
31. Among the main compounds of Portland cement, the one chiefly responsible for strength gain in the first seven days is
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It is the fast-hydrating silicate.
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Answer: C. tricalcium silicate (C3S)
C3S hydrates quickly and gives early strength; C2S hydrates slowly and adds strength after about 28 days, while C3A mainly causes flash set and heat.
32. Gypsum is added in small quantity while grinding cement clinker mainly to
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It retards a very fast reacting compound.
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Answer: C. control the flash set caused by C3A
C3A reacts instantly with water; gypsum forms ettringite on its surface and delays this reaction, regulating the setting time.
33. As per IS code for Ordinary Portland Cement, the minimum initial setting time and the maximum final setting time are respectively
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Final is 10 hours.
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Answer: D. 30 minutes and 600 minutes
OPC must not set before 30 min (initial) and must be hard within 10 hours = 600 min (final), measured with the Vicat apparatus.
34. Unsoundness of cement caused by excess free lime or magnesia is tested by the Le Chatelier apparatus, where the permissible expansion for OPC is not more than
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Values between 5 and 20 mm; it is the middle round figure.
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Answer: A. 10 mm
IS specifications limit Le Chatelier expansion to 10 mm (0.8% in the autoclave test).
35. The grade number in 43 grade OPC indicates
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It is a strength index in MPa.
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Answer: A. minimum 28-day compressive strength of standard mortar cubes in N/mm²
43 grade cement gives at least 43 N/mm² compressive strength at 28 days when tested on standard mortar cubes.
36. Cumulative percentage retained on sieves 4.75, 2.36, 1.18 mm, 600, 300 and 150 µm are 5, 15, 35, 60, 85 and 97. The fineness modulus of the sand is
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Add the cumulative values and divide by 100.
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Answer: B. 2.97
FM = Σ cumulative % retained/100 = 297/100 = 2.97.
37. As per IS 456, the nominal maximum size of coarse aggregate should not be greater than
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A small fraction of the thinnest dimension.
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Answer: C. one-fourth of the minimum dimension of the member
IS 456 limits aggregate size to 1/4 of the smallest member dimension and 5 mm less than the minimum clear spacing between bars.
38. As per IS 383 and IS 2386 tests, the maximum permissible aggregate crushing value for aggregates used in concrete for wearing surfaces such as roads and pavements is
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Stricter than the 45% limit for ordinary concrete.
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Answer: D. 30%
Aggregate crushing value is limited to 45% for concrete other than wearing surfaces and 30% for wearing surfaces.
39. As per IS 456, the permissible limit of chlorides (as Cl) in mixing water for reinforced concrete work is
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RCC limit is stricter than for PCC.
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Answer: C. 500 mg/litre
Chlorides are limited to 2000 mg/L for plain concrete and 500 mg/L for reinforced concrete because chloride induces corrosion of steel.
40. The slump test of fresh concrete is used mainly to measure
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It checks how easily the concrete flows.
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Answer: C. workability or consistency
The slump of a moulded concrete cone indicates the consistency and workability of a mix (IS 1199).
41. In the compaction factor test, the compaction factor is the ratio of
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The value is always less than 1.
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Answer: D. weight of partially compacted concrete to weight of fully compacted concrete
The cylinder is filled by free fall (partial compaction) and weighed; the same concrete is fully compacted and weighed. The ratio is below 1.
42. Upward movement of mixing water to the surface of freshly placed concrete is called
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The water seeps out as if from a sponge.
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Answer: D. bleeding
Bleeding is the rise of water to the top surface, leaving laitance and a weak, porous layer.
43. According to Abrams' law, for a given workable and fully compacted mix, the strength of concrete
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Excess water leaves voids.
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Answer: D. decreases as the water–cement ratio increases
Strength varies inversely with the w/c ratio: f = A/B^(w/c). A greater w/c increases porosity and decreases strength.
44. The water required for complete chemical hydration of cement is about 23% by weight, and the water held in gel pores about 15%. So the minimum w/c ratio for complete hydration is about
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Add the two percentages.
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Answer: C. 0.38
Total water needed = 23% + 15% = 38% of cement weight, i.e. w/c = 0.38; extra water is needed only for workability.
45. A 150 mm concrete cube fails at a load of 675 kN in the compression test. The compressive strength is
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Divide load in N by area in mm².
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Answer: A. 30.0 N/mm²
Strength = P/A = 675000/(150×150) = 30.0 N/mm².
46. As per IS 456, the short-term modulus of elasticity of M30 concrete (Ec = 5000√fck) is
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Square root of fck in N/mm² times 5000.
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Answer: B. 27386 N/mm²
Ec = 5000 √30 = 5000 × 5.477 = 27386 N/mm².
47. The flexural (modulus of rupture) strength of M25 concrete from IS 456, fcr = 0.7√fck, is
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Take the square root first.
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Answer: A. 3.5 N/mm²
fcr = 0.7 × √25 = 0.7 × 5 = 3.5 N/mm².
48. The characteristic strength of concrete is defined as the strength of 150 mm cubes at 28 days below which not more than
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The same fractile used for characteristic loads.
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Answer: C. 5% of test results are expected to fall
Characteristic strength is the 5th percentile value: 95% of results are expected to exceed it.
49. Using IS 10262, the target mean strength for M25 concrete with standard deviation 4 N/mm² is
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Factor 1.65 corresponds to 5% defective.
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Answer: B. 31.6 N/mm²
f't = fck + 1.65 s = 25 + 1.65 × 4 = 31.6 N/mm².
50. As per IS 456, the minimum grade of concrete for reinforced concrete exposed to severe environment is
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Third exposure class in order of severity.
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Answer: B. M30
Minimum grades: mild M20, moderate M25, severe M30, very severe M35 and extreme M40.
51. IS 456 specifies maximum water–cement ratio and minimum cement content for RCC in severe exposure as
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Between moderate (0.50) and extreme (0.40).
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Answer: A. 0.45 and 320 kg/m³
Severe exposure requires w/c ≤ 0.45 and cement ≥ 320 kg/m³; mild is 0.55/300 and extreme 0.40/360.
52. Minimum period of moist curing for OPC concrete and for concrete with mineral admixtures or blended cement, as per IS 456, are respectively
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The slower-hydrating cements need more days.
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Answer: B. 7 days and 10 days
IS 456 requires at least 7 days of moist curing for OPC and 10 days where mineral admixtures or blended cements are used.
53. The nominal mix for M20 concrete as per IS 456 (cement : sand : coarse aggregate by volume) is
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Richer than M15.
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Answer: A. 1 : 1.5 : 3
IS 456 nominal mixes are M5 1:5:10, M7.5 1:4:8, M10 1:3:6, M15 1:2:4 and M20 1:1.5:3.
54. For 1 m³ of nominal concrete 1:2:4, take dry volume of materials 1.54 m³ and bulk density of cement 1440 kg/m³. The cement required is about
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Cement is 1 part in 7 of the dry volume.
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Answer: C. 6.3 bags
Cement volume = 1.54 × 1/(1+2+4) = 0.22 m³; mass = 0.22 × 1440 = 317 kg = 6.3 bags of 50 kg.
55. A mix design starts with 186 kg/m³ water for a 50 mm slump (20 mm aggregate), increased by 3% for every additional 25 mm slump. For 100 mm slump and w/c = 0.45, the cement content is about
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Two 25 mm steps give a 6% increase.
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Answer: B. 438 kg/m³
Water = 186 × 1.06 = 197.2 kg/m³; cement = 197.2/0.45 = 438 kg/m³.
56. An admixture used to produce flowing concrete at a low w/c ratio, without loss of strength, is a
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Gives high slump with less water.
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Answer: D. superplasticizer
Superplasticizers disperse cement grains, allowing high workability at low water content; retarders delay setting, accelerators speed it, and air-entraining agents improve freeze–thaw resistance.
57. IS 456 acceptance: mean of four consecutive M25 results must be not less than fck + 0.825 s (s = 4 N/mm²) or fck + 3 N/mm², whichever is greater. The required mean is
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Evaluate both and take the larger.
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Answer: D. 28.3 N/mm²
fck + 0.825 s = 25 + 3.3 = 28.3; fck + 3 = 28; greater is 28.3 N/mm².
58. As per IS 456, the minimum number of samples for concrete quantity of 12 m³ in a day's work is
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Quantity lies between 6 and 15 m³.
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Answer: A. 2
IS 456 requires 1 sample for 1–5 m³, 2 for 6–15 m³, 3 for 16–30 m³, 4 for 31–50 m³ and 4 plus one per extra 50 m³ above that.
59. In the ultrasonic pulse velocity test, concrete is generally rated as excellent quality when the pulse velocity is
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Dense concrete transmits pulses faster.
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Answer: D. above 4.5 km/s
UPV above 4.5 km/s indicates excellent, 3.5–4.5 good, 3.0–3.5 medium and below 3.0 km/s doubtful quality concrete.
60. Cube strengths of five samples are 28, 30, 32, 34 and 36 N/mm². The standard deviation (n − 1 method) is
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Divide by (n − 1) = 4 before the square root.
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Answer: A. 3.16 N/mm²
Mean = 32; Σ(x−mean)² = 40; s = √(40/4) = 3.16 N/mm².
61. A mix needs 0.5 m³ of dry sand. If the sand is damp and bulks by 25%, the volume of damp sand to be measured is
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Bulked sand occupies more volume for the same dry mass.
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Answer: B. 0.625 m³
Bulked volume = 0.5 × 1.25 = 0.625 m³ so that dry sand content is correct.
62. Increase in strain with time under a constant sustained stress in concrete is known as
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Needs a sustained load to occur.
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Answer: B. creep
Creep is time-dependent deformation under sustained load; shrinkage occurs without load, and relaxation is stress loss at constant strain.
5.3 RCC structures-1
31 questions · ACiE0503
63. In working stress design to IS 456 (Annex B), the modular ratio m = 280/(3σcbc) for M20 concrete (σcbc = 7 N/mm²) is
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Denominator is three times the bending compressive stress.
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Answer: D. 13.33
m = 280/(3 × 7) = 13.33.
64. Permissible tensile stress σst in HYSD Fe415 reinforcement (bars up to 20 mm) in the working stress method of IS 456 is
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Roughly 0.55 times fy.
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Answer: B. 230 N/mm²
IS 456 Annex B gives σst = 140 N/mm² for Fe250 and 230 N/mm² for Fe415 (275 for Fe500).
65. For a balanced section (M20, Fe415, σcbc = 7, σst = 230, m = 13.33) the critical neutral axis depth xc for d = 450 mm is about
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k = 1/(1 + σst/(m σcbc)).
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Answer: A. 130 mm
k = mσcbc/(mσcbc + σst) = 93.3/323.3 = 0.289; xc = 0.289 × 450 = 130 mm.
66. The moment of resistance of a balanced working stress section, b = 250 mm, d = 450 mm, M20 and Fe415 (k = 0.289, j = 0.904, Q = 0.5σcbc k j), is about
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M = Q b d², Q in N/mm², convert to kN·m.
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Answer: A. 46.2 kN·m
Q = 0.5 × 7 × 0.289 × 0.904 = 0.913 N/mm²; M = Q b d² = 0.913 × 250 × 450² = 46.2 kN·m.
67. In the limit state method of IS 456, the limiting value of xu,max/d for Fe415 steel is
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Between Fe250 and Fe500 values.
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Answer: B. 0.48
IS 456 gives xu,max/d = 0.53 (Fe250), 0.48 (Fe415), 0.46 (Fe500) and 0.44 (Fe550); higher steel grades have smaller limits.
68. The maximum strain in concrete at the extreme compression fibre in flexure at the limit state of collapse as per IS 456 is
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Value used in the strain compatibility for beams.
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Answer: D. 0.0035
IS 456 adopts 0.0035 for flexure; 0.002 applies to pure axial compression.
69. The equivalent rectangular stress block of IS 456 for a beam has an average compressive stress and depth, measured from the extreme fibre, of
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Force coefficient 0.36 and lever position 0.42.
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Answer: B. 0.36 fck and 0.42 xu
Compressive force C = 0.36 fck b xu acting at 0.42 xu from the extreme compression fibre; 0.446 fck is the design stress.
70. The partial safety factors γm for concrete and for reinforcing steel in the limit state design of IS 456 are respectively
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Concrete's material factor is larger than steel's.
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Answer: B. 1.5 and 1.15
Design strengths are 0.45 fck (= 0.67 fck/1.5) for concrete and 0.87 fy (= fy/1.15) for steel.
71. The limiting moment of resistance of a singly reinforced rectangular beam (b = 250 mm, d = 450 mm, M20, Fe415, Mu,lim = 0.138 fck b d²) is
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Convert N·mm to kN·m by dividing by 10⁶.
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Answer: C. 139.7 kN·m
Mu,lim = 0.138 × 20 × 250 × 450² = 139.7 kN·m.
72. A beam b = 250 mm, d = 450 mm has Ast = 942 mm² (3–20 mm bars), M20 and Fe415. The depth of neutral axis xu is about
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Equate tension 0.87 fy Ast to compression 0.36 fck b xu.
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Answer: B. 189 mm
xu = 0.87 fy Ast/(0.36 fck b) = 0.87×415×942/(0.36×20×250) = 189 mm. Since xu/d = 0.42 < 0.48, the section is under-reinforced.
73. For the same beam (xu = 189 mm), the ultimate moment of resistance Mu = 0.87 fy Ast (d − 0.42 xu) is about
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Lever arm is d − 0.42 xu.
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Answer: C. 126 kN·m
Lever arm = 450 − 0.42 × 189 = 370.6 mm; Mu = 0.87×415×942×370.6/10⁶ = 126 kN·m.
74. A simply supported beam of span 6 m carries dead load 20 kN/m (including self weight) and imposed load 15 kN/m. The design factored moment at mid span is
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Factor the load first, then use wl²/8.
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Answer: C. 236.25 kN·m
wu = 1.5 × 35 = 52.5 kN/m; Mu = wu l²/8 = 52.5×36/8 = 236.25 kN·m.
75. As per IS 456, the minimum area of tension reinforcement in a beam is As/(bd) ≥ 0.85/fy. For b = 230 mm, d = 400 mm and Fe415, As,min is about
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Use fy = 415 N/mm².
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Answer: A. 188 mm²
As,min = 0.85 × 230 × 400/415 = 188 mm².
76. The minimum reinforcement in each direction of a RC slab with HYSD bars, as a percentage of gross cross-sectional area, as per IS 456 is
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HYSD needs less than mild steel.
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Answer: C. 0.12%
IS 456 requires 0.15% of gross area for mild steel and 0.12% for HYSD bars (Fe415/Fe500) in slabs.
77. A rectangular RC slab supported on all four edges is designed as a one-way slab when the ratio of long span to short span is
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Spans differ by more than a factor of two.
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Answer: D. greater than 2
If ly/lx > 2, nearly all load is carried in the short direction, so the slab is treated as one-way.
78. As per IS 456, the basic span to effective depth ratios for cantilever, simply supported and continuous beams are respectively
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Cantilevers have the smallest value.
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Answer: A. 7, 20 and 26
IS 456 Cl. 23.2.1 gives basic ratios 7 (cantilever), 20 (simply supported) and 26 (continuous), modified for steel stress and flanges.
79. Final deflection of a beam due to all loads, including creep and shrinkage, should not normally exceed
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The overall limit is larger than the 'after construction' limit.
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Answer: B. span/250
IS 456 limits total deflection to span/250 and deflection after construction of partitions to span/350 or 20 mm, whichever is less.
80. Design shear force at a section is 150 kN, b = 250 mm and d = 450 mm. The nominal shear stress τv is
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Use force in N and area in mm².
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Answer: B. 1.33 N/mm²
τv = Vu/(b d) = 150000/(250 × 450) = 1.33 N/mm².
81. For Vu = 150 kN, b = 250 mm, d = 450 mm and τc = 0.62 N/mm² (pt = 1%), two-legged 8 mm stirrups (Asv = 100.5 mm², fy = 415) are used. Spacing required from Vus = 0.87 fy Asv d/sv is about
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Concrete takes τc b d; stirrups take the rest.
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Answer: A. 204 mm
Vus = 150 − 0.62×250×450/1000 = 80.25 kN; sv = 0.87×415×100.5×450/80250 = 204 mm.
82. As per IS 456, the maximum spacing of vertical stirrups in a beam of effective depth d is the smaller of
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Fraction of d under 1.
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Answer: C. 0.75 d and 300 mm
To ensure every potential diagonal crack is crossed by a stirrup, spacing ≤ 0.75 d or 300 mm, whichever is less.
83. For M20 concrete, the design bond stress τbd for plain bars in tension is 1.2 N/mm². For HYSD (deformed) bars it is increased by
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Deformed bars grip better than plain bars.
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Answer: D. 60%, giving 1.92 N/mm²
IS 456 increases the bond stress by 60% for deformed bars; for bars in compression a further 25% increase is permitted.
84. Development length Ld = φσs/(4τbd) of a 20 mm HYSD bar of Fe415 in M20 concrete (σs = 0.87 fy, τbd = 1.92 N/mm²) is about
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σs = 0.87 × 415 = 361 N/mm².
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Answer: B. 940 mm
Ld = 20 × 0.87 × 415/(4 × 1.92) = 940 mm (about 47 φ).
85. As per IS 456, the anchorage value of a standard U-type hook is
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Largest of the three hook types in the bend list.
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Answer: A. 16 times the bar diameter
A 180° (U-type) hook has an anchorage value of 16φ, a 90° bend 8φ and each 45° bend 4φ.
86. The minimum lap length for flexural tension bars as per IS 456 is the greater of
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Flexural tension laps use the larger of the two stated values.
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Answer: D. Ld and 30 φ
Lap length in flexural tension is not less than Ld or 30φ; in direct tension 2Ld or 30φ; in compression Ld or 24φ.
87. A simply supported T-beam has span 6 m, web width 300 mm and flange thickness 120 mm. Effective flange width bf = lo/6 + bw + 6Df is
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Take lo = 6000 mm for a simply supported beam.
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Answer: C. 2020 mm
bf = 6000/6 + 300 + 6 × 120 = 2020 mm (limited to the actual centre-to-centre beam spacing).
88. A doubly reinforced rectangular beam becomes necessary when
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The section is too small for the moment.
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Answer: A. the design moment exceeds Mu,lim of the singly reinforced section and depth cannot be increased
When Mu > Mu,lim, steel in the compression zone provides the extra moment so that xu stays within xu,max.
89. For a working stress section b = 250 mm, d = 450 mm, Ast = 942 mm² and m = 13.33, the neutral axis depth x from b x²/2 = m Ast (d − x) is about
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Solve the quadratic from equating first moments.
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Answer: C. 168 mm
125x² + 12557x − 5650587 = 0 gives x = 168 mm, which exceeds the balanced xc = 130 mm, so the section is over-reinforced in the working stress sense (concrete reaches its permissible stress first).
90. An RC section in which the steel reaches yield before the concrete reaches its crushing strain is called under-reinforced. This is preferred in design because it
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Think of warning before collapse.
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Answer: C. gives ductile failure with warning
Under-reinforced beams yield and deflect visibly before crushing, giving warning; over-reinforced beams fail suddenly by concrete crushing.
91. The effective span of a simply supported beam, as per IS 456, is the lesser of
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Two candidate lengths; smaller governs.
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Answer: A. centre-to-centre of supports and clear span plus effective depth
IS 456 Cl. 22.2: effective span = clear span + d, or centre to centre of bearings, whichever is smaller.
92. In Nepal, the code NBC 205 gives mandatory rules of thumb for
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A 200-series code for RC frames.
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Answer: D. reinforced concrete buildings without masonry infill
NBC 205 provides rules of thumb for the design of low-rise RC buildings without masonry infill; NBC 201 applies to RC buildings with infill and NBC 202 to load-bearing masonry.
93. The partial safety factor on load for dead and imposed loads combined in the basic limit state of collapse combination of IS 456 is
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Used for ordinary gravity design.
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Answer: D. 1.5
The basic combination is 1.5 (DL + IL); the 1.2 factor appears when wind/earthquake is combined, and 0.9 for overturning checks.
5.4 RCC structures-2
30 questions · ACiE0504
94. As per IS 456, a braced compression member is classified as a short column when the ratio of effective length to least lateral dimension is
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Value between 6 and 20.
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Answer: D. less than 12
IS 456 treats columns with le/D < 12 as short (failure by crushing); greater values are slender and need additional moments for buckling.
95. Minimum eccentricity for a column of unsupported length 3.5 m and lateral dimension 400 mm is emin = l/500 + D/30 (not less than 20 mm), which equals
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Add both terms and compare with 20 mm.
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Answer: A. 20.33 mm
emin = 3500/500 + 400/30 = 7.00 + 13.33 = 20.33 mm, which exceeds the 20 mm minimum.
96. A short tied column 400 mm × 400 mm, M25, Fe415, reinforced with 8–20 mm bars (Asc = 2513 mm²) has axial capacity Pu = 0.4 fck Ac + 0.67 fy Asc of about
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Use net concrete area (gross minus steel).
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Answer: D. 2274 kN
Ac = 160000 − 2513 = 157487 mm²; Pu = 0.4×25×157487 + 0.67×415×2513 = 2274 kN.
97. The percentage of longitudinal reinforcement in an RC column, as per IS 456, should lie between
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Minimum is below 1%.
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Answer: D. 0.8% and 6% of the gross cross-sectional area
IS 456 prescribes minimum 0.8% and maximum 6% of gross area (practically 4% to allow lap splices).
98. As per IS 456, the minimum number of longitudinal bars in a rectangular column and in a circular column are respectively
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A circle needs more bars than a rectangle.
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Answer: D. 4 and 6
Rectangular columns need at least 4 bars (one at each corner) and circular columns at least 6.
99. In a 400 mm × 400 mm column with 20 mm longitudinal bars, the maximum pitch of lateral ties as per IS 456 (least of least lateral dimension, 16φ and 300 mm) is
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Compute all three limits and pick the smallest.
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Answer: C. 300 mm
The limits are 400 mm, 16×20 = 320 mm and 300 mm; the least is 300 mm.
100. A helically reinforced circular column compared with a tied column of the same section, as per IS 456, may be assigned a strength
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A modest increase.
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Answer: A. 5% higher
Helical confinement improves ductility and strength; IS 456 allows 1.05 times the tied column strength if the helix satisfies the pitch and area conditions.
101. A slender column has Pu = 1200 kN, D = 400 mm and le = 6 m. The additional moment Ma = (Pu D/2000)(le/D)² is
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le/D = 15, squared.
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Answer: B. 54 kN·m
Ma = 1200×10³×400/2000 × 15² = 5.40e+07 N·mm = 54 kN·m.
102. For biaxial bending of columns (IS 456), the exponent αn in (Mux/Mux1)^αn + (Muy/Muy1)^αn ≤ 1 equals 1 at Pu/Puz ≤ 0.2, 2 at Pu/Puz ≥ 0.8 and varies linearly between. For Pu/Puz = 0.5, αn is
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Mid-way between the end ratios.
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Answer: B. 1.5
αn = 1 + (0.5 − 0.2)/(0.8 − 0.2) × (2 − 1) = 1.5.
103. In the limit state of collapse for members under axial compression, the maximum compressive strain in concrete is limited to
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Smaller than the flexural value.
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Answer: B. 0.002
IS 456 uses 0.0035 minus 0.75 times the strain at the least compressed face for eccentric loads, which reduces to 0.002 for pure axial compression.
104. A column transmits 1000 kN to an isolated footing; safe bearing capacity of soil is 150 kN/m² and 10% of the load is allowed for footing self weight. The required square footing side is about
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Increase the load by 10% before dividing by SBC.
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Answer: B. 2.71 m
A = 1.1 × 1000/150 = 7.33 m²; side = √7.33 = 2.71 m.
105. In a rectangular footing with long side/short side ratio β = 2, the fraction of the short-direction steel to be placed in the central band of width equal to the short side is 2/(β + 1), i.e.
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Substitute β = 2 in the formula.
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Answer: A. 2/3
IS 456 Cl. 34.3.1: central band reinforcement = 2/(β+1) × total short-direction steel = 2/3 for β = 2.
106. For an isolated RC footing, the critical sections for one-way (beam) shear and for two-way (punching) shear are respectively at
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Punching perimeter is closer to the column.
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Answer: C. d from the column face and d/2 from the column face
One-way shear is checked at a distance d from the face of the column; punching shear on a perimeter at d/2 from the face.
107. A footing has a net factored upward soil pressure of 200 kN/m² and a projection of 1.0 m beyond the column face. The bending moment per metre width at the column face is
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Cantilever under uniform pressure.
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Answer: B. 100 kN·m/m
The projection acts as a cantilever: M = w l²/2 = 200 × 1²/2 = 100 kN·m per metre.
108. A footing 3 m × 2 m carries a vertical load of 900 kN at 0.3 m eccentricity along the 3 m side. The maximum soil pressure is
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Eccentricity is measured along the length L.
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Answer: A. 240 kN/m²
p = (P/A)(1 ± 6e/L) = 150(1 ± 0.6); pmax = 240 kN/m², pmin = 60 kN/m².
109. Columns carrying 800 kN and 1200 kN are 4.0 m apart. For a rectangular combined footing giving uniform pressure, with the 800 kN column 0.5 m from the end, the footing length is
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The footing centroid coincides with the load resultant.
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Answer: A. 5.8 m
Resultant from the 800 kN column = 1200×4/2000 = 2.4 m. The centroid must lie there: L = 2 × (2.4 + 0.5) = 5.8 m.
110. Minimum factor of safety against overturning and sliding of a footing under lateral loads as per IS 456 is
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Slightly above unity.
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Answer: C. 1.4
IS 456 Cl. 34.2.4: factor of safety of 1.4 against sliding and overturning (dead load plus wind/earthquake, with 0.9 DL taken for stabilising).
111. In pretensioned concrete members, the prestress is transferred to concrete mainly through
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Think of tendons that are cast in without ducts.
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Answer: C. bond between tendons and concrete
Tendons are stressed before casting; after the concrete hardens, the tendons are released and prestress is transferred by bond.
112. A pretensioned member has stress at the level of steel fc = 10 N/mm², Es = 2×10⁵ N/mm² and Ec = 35000 N/mm². The loss of stress due to elastic shortening, m fc, is
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Modular ratio times concrete stress.
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Answer: C. 57.1 N/mm²
m = Es/Ec = 5.71; loss = m fc = 5.71 × 10 = 57.1 N/mm².
113. A post-tensioned tendon has P0 = 1000 kN at the jack. For μ = 0.2, α = 0.1 rad, k = 0.0015 per m and x = 20 m, the force at the far end, P = P0 e^−(μα + kx), is
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Add the curvature and wobble terms.
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Answer: C. 951 kN
Exponent = 0.2×0.1 + 0.0015×20 = 0.05; P = 1000 e^−0.05 = 951 kN (about 4.9% loss).
114. A prestressed rectangular beam 300 mm × 600 mm carries P = 1500 kN at eccentricity 150 mm below the centroid (self weight ignored). The stresses at the top and bottom fibres (compression positive, tension negative) are respectively
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Z = bh²/6; the eccentric moment causes tension at the top.
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Answer: D. −4.17 and 20.83 N/mm²
P/A = 8.33; Pe/Z = 12.50 N/mm². Top = 8.33 − 12.50 = -4.17 (tension); bottom = 20.83 N/mm² (compression).
115. A parabolic tendon with P = 1500 kN, central dip e = 0.15 m in a simply supported span of 10 m balances a uniformly distributed load wb = 8Pe/L² equal to
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Upward load from the parabolic tendon: 8Pe/L².
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Answer: D. 18 kN/m
wb = 8 × 1500 × 0.15/10² = 18 kN/m.
116. As per IS 1343, minimum characteristic strength grades of concrete for post-tensioned and pretensioned members are respectively
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Pretensioned needs the higher grade.
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Answer: A. M30 and M40
IS 1343 specifies at least M30 for post-tensioned and M40 for pretensioned work; high-strength concrete resists high bearing and bond stresses and reduces creep losses.
117. A loss of prestress that occurs only in post-tensioned members and not in pretensioned members is due to
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Pretensioned tendons have no duct.
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Answer: A. friction between tendon and duct
Shrinkage, creep and steel relaxation occur in both; friction (curvature and wobble) arises only where the tendon is stressed inside ducts.
118. In prestressed concrete, high-strength steel (about 1860 N/mm²) is necessary mainly because
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Fixed strain losses matter more at low stress.
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Answer: D. low-strength steel would lose a large part of its prestress through shrinkage and creep
Concrete shrinkage and creep shorten the member by a fixed strain; with ordinary steel this strain would wipe out the initial prestress, whereas a high initial strain keeps a net stress.
119. In IS 1343, the type of prestressed member in which no tensile stress is permitted in concrete under service loads is called
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The most conservative of the types.
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Answer: B. Type 1
Type 1: no tension; Type 2: tension but no visible cracking; Type 3: tension with cracks of limited width.
120. The Lee–McCall system of post-tensioning is mainly used with
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Think threaded bars.
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Answer: B. high-tensile alloy steel bars anchored by nuts
Lee–McCall uses threaded high-strength bars with nuts and bearing plates; Freyssinet uses wedge anchors and Magnel–Blaton uses sandwich plates.
121. A combined footing is provided for two columns mainly when
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Think of limited space near a boundary.
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Answer: C. an isolated footing for the edge column would project beyond the property line or the footings would overlap
When columns are close or an edge column cannot be centred on its footing because of a boundary, a common footing whose centroid coincides with the load resultant is used.
122. A strap (cantilever) footing consists of
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Two footings plus a connecting member.
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Answer: A. two isolated footings connected by a strap beam
The strap beam transfers the moment due to the eccentric edge column load to the interior footing, giving uniform pressure under both.
123. The minimum clear cover to main reinforcement in footings (bottom steel cast against earth) as per IS 456 is
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Greater than for columns.
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Answer: B. 50 mm
IS 456 requires at least 50 mm cover for footings; 40 mm for columns and 20 mm for beams and slabs in mild exposure.
5.5 Steel structures
30 questions · ACiE0505
124. The present Indian code IS 800:2007 for general construction in steel adopts the
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It replaced the permissible-stress approach of 1984.
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Answer: D. limit state method
IS 800:2007 is based on limit state design with partial safety factors; the older IS 800:1984 used the working stress method.
125. For structural steel of grade E250 (Fe410) as per IS 2062, the minimum yield stress and ultimate tensile stress are respectively
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Fe410 is the older name.
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Answer: C. 250 N/mm² and 410 N/mm²
E250 steel has fy = 250 N/mm² and fu = 410 N/mm² (hence the name Fe410).
126. In IS 800:2007, the partial safety factors for material γm0 (yielding and buckling) and γm1 (ultimate stress, rupture) are respectively
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Ultimate-stress checks use the larger factor.
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Answer: A. 1.10 and 1.25
Resistance governed by yielding uses γm0 = 1.10; that governed by ultimate stress (net section, bolts) uses γm1 = 1.25.
127. The yield stress of a bolt of property class 4.6 is
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Multiply the two digits and by 10.
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Answer: C. 240 N/mm²
In '4.6' the first number × 100 = fub = 400 N/mm² and the product of the two numbers × 10 = fyb = 4×6×10 = 240 N/mm².
128. In the designation ISMC 300, the number 300 indicates
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Same convention as ISMB 300.
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Answer: D. the depth of the channel section in mm
ISMC means Indian Standard Medium-weight Channel, and the number is its nominal depth (300 mm).
129. A steel cross-section that can form a plastic hinge with the rotation capacity required for plastic analysis is classified as
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The most ductile class.
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Answer: B. plastic (Class 1)
Class 1 plastic sections reach full plastic moment with sufficient rotation capacity; compact sections reach Mp but with limited rotation.
130. In IS 800:2007, the limiting width-to-thickness ratio b/tf for the outstand of the compression flange of a rolled I-section to be of plastic class is
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The smallest of the three class limits.
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Answer: B. 9.4ε, where ε = √(250/fy)
Limits for rolled outstand flanges: plastic 9.4ε, compact 10.5ε, semi-compact 15.7ε.
131. The standard hole diameter for an M20 bolt (clearance 2 mm for 16–24 mm bolts) is
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Add the clearance to nominal diameter.
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Answer: D. 22 mm
IS 800 gives d0 = d + 1 mm for bolts up to 14 mm, d + 2 for 16–24 mm and d + 3 for larger bolts, so d0 = 22 mm for M20.
132. Design shear strength of one M20 bolt of grade 4.6 (fub = 400 N/mm²) in single shear with the threaded portion in the shear plane (net tensile area Anb = 245 mm², γmb = 1.25) is about
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Shear strength involves fub/√3 and the net area.
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Answer: A. 45.3 kN
Vnsb = fub Anb/√3 = 400×245/1.732 = 56.6 kN; Vdsb = Vnsb/1.25 = 45.3 kN.
133. Bearing strength of M20 bolt (fub = 400) on a 10 mm plate (fu = 410 N/mm²), end distance 40 mm, pitch 60 mm, hole 22 mm, is Vdpb = 2.5 kb d t fu/γmb with kb = least of e/3d0, p/3d0 − 0.25, fub/fu and 1. The value is about
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Select the smallest of the four kb terms.
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Answer: C. 99.4 kN
kb = min(0.606, 0.659, 0.976, 1) = 0.606; Vnpb = 2.5×0.606×20×10×410 = 124.2 kN; Vdpb = 99.4 kN.
134. The proof load (minimum bolt tension) Fo = 0.7 fub Anb for an M20 HSFG bolt of grade 8.8 (Anb = 245 mm²) is
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Grade 8.8 has fub = 800 N/mm².
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Answer: B. 137.2 kN
Fo = 0.7 × 800 × 245 = 137.2 kN.
135. For a fillet weld of size s joining plates meeting at 90°, the effective throat thickness is
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Cos 45° ≈ 0.707.
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Answer: D. 0.7 s
For equal legs at 90° the throat = s cos 45° = 0.707 s, taken as 0.7 s.
136. Design strength of a 6 mm shop fillet weld, overall length 200 mm, fu = 410 N/mm², γmw = 1.25 (effective length = overall − 2s) is about
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Use throat thickness and reduced length.
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Answer: C. 150 kN
Throat = 4.2 mm; Leff = 200 − 12 = 188 mm; fwd = 410/(√3×1.25) = 189.4 N/mm²; strength = 189.4×4.2×188 = 150 kN.
137. As per IS 800:2007, the minimum size of a fillet weld joining plates whose thicker part is 12 mm is
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Thickness range 10–20 mm.
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Answer: A. 5 mm
Minimum fillet size: 3 mm (thicker part up to 10 mm), 5 mm (10–20 mm), 6 mm (20–32 mm) and 8 mm (32–50 mm).
138. The minimum effective length of a fillet weld in IS 800:2007 should not be less than
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Multiple of weld size, less than 5.
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Answer: A. four times the size of the weld
A fillet weld shorter than 4s cannot be considered to carry load reliably because of end effects.
139. The effective throat thickness of a full-penetration butt weld between two plates of unequal thickness is taken as
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The weld cannot be thicker than what it joins at the weaker side.
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Answer: A. the thickness of the thinner plate
A complete-penetration butt weld has throat equal to the thinner part, and its strength equals the strength of the parent metal there.
140. A 180 mm × 10 mm flat (fy = 250, fu = 410 N/mm²) has two 22 mm bolt holes in a line across the width. Considering only yielding of the gross section and rupture of the net section, the design tensile strength is about
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Compute both limit states and take the lesser.
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Answer: A. 401.5 kN
Tdg = 1800×250/1.10 = 409.1 kN; Tdn = 0.9×1360×410/1.25 = 401.5 kN; the smaller governs.
141. As per IS 800:2007, the maximum slenderness ratio permitted for a compression member carrying dead and imposed loads, and for a tension member that is always in tension, are respectively
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Compression members must be stockier.
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Answer: B. 180 and 400
Compression member (dead + live) λ ≤ 180; wind/earthquake-only compression member 250; member always in tension 400; reversal members 350.
142. Two bolt holes of 22 mm diameter are staggered in a 180 mm × 10 mm plate with pitch p = 50 mm and gauge g = 60 mm. Considering the zig-zag failure path through both holes, the net area An = (b − Σd0 + Σp²/4g) t is about
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The stagger adds back p²/4g.
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Answer: D. 1464 mm²
Net width = 180 − 44 + 2500/240 = 146.4 mm; An = 1464 mm².
143. A column of length 4 m with effective length factor 0.8 has least radius of gyration 32 mm. The Euler buckling stress fcc = π²E/λ² with E = 2×10⁵ N/mm² is about
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First compute the slenderness ratio KL/r.
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Answer: D. 197.4 N/mm²
λ = 0.8×4000/32 = 100; fcc = π²×200000/100² = 197.4 N/mm².
144. For fcc = 197.4 N/mm², fy = 250 N/mm² and buckling class c (α = 0.49), IS 800 gives non-dimensional slenderness λ = √(fy/fcc), φ = 0.5[1 + α(λ − 0.2) + λ²] and χ = 1/(φ + √(φ² − λ²)). The design compressive stress fcd = χ fy/γm0 is about
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χ is below 1, so fcd < fy/1.1.
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Answer: A. 107 N/mm²
λ = 1.125; φ = 1.360; χ = 0.471; fcd = 0.471 × 250/1.10 = 107 N/mm².
145. As per IS 800:2007, the recommended design effective length of a column fixed in position and direction at both ends is
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Smaller than 0.8L.
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Answer: B. 0.65 L
Recommended values: 0.65 L (both ends fixed), 0.80 L (one fixed, one hinged), 1.0 L (both hinged), 2.0 L (cantilever).
146. IS 800 takes the bearing strength of concrete as 0.45 fck. For a column load of 1500 kN on M20 concrete, the minimum plan area of the base plate is
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Allowable bearing stress is 9 N/mm² for M20.
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Answer: B. 166667 mm²
A = P/(0.45 fck) = 1500×10³/9 = 166667 mm² (e.g. about 410 mm × 410 mm).
147. A slab base plate 450 mm × 450 mm carries 1500 kN. The projections are a = 100 mm (larger) and b = 40 mm (smaller). With ts = √[2.5 w (a² − 0.3 b²) γm0/fy], fy = 250 N/mm² and γm0 = 1.10, the plate thickness is about
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First find the bearing pressure w under the plate.
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Answer: C. 28 mm
w = 1500×10³/450² = 7.41 N/mm²; ts = √[2.5×7.41×(10000 − 480)×1.1/250] = 28 mm.
148. In a laced built-up column, the lacing bars should be inclined to the axis of the column at an angle of
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Neither too flat nor near vertical.
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Answer: B. 40° to 70°
IS 800 requires lacing inclination between 40° and 70° (single or double lacing); steeper or flatter angles are inefficient.
149. In working stress design of steel as per IS 800:1984, the permissible axial tensile stress is 0.6 fy. For Fe410 (fy = 250 N/mm²) it equals
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Fraction 0.6 of yield.
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Answer: B. 150 N/mm²
σat = 0.6 fy = 0.6 × 250 = 150 N/mm²; permissible bending stress is 0.66 fy = 165 N/mm² and shear 0.4 fy = 100 N/mm².
150. A single angle connected through one leg and loaded in tension has lower strength than its gross area suggests because of
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Think of eccentric load transfer.
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Answer: D. shear lag, which makes the outstanding leg only partly effective
Load transfer through one leg causes non-uniform stress; IS 800 accounts for shear lag with the β factor reducing the net section resistance.
151. The shape factor (ratio of plastic moment to yield moment) of a rectangular steel section is
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Plastic modulus over elastic modulus.
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Answer: A. 1.5
Zp/Ze = (bd²/4)/(bd²/6) = 1.5 for a rectangle; rolled I-sections have about 1.12–1.15.
152. For structural steel, IS 800:2007 takes the modulus of elasticity and Poisson's ratio as
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200 GPa.
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Answer: C. 2 × 10⁵ N/mm² and 0.3
E = 2.0 × 10⁵ N/mm² (200 GPa) and ν = 0.3 for structural steel.
153. In a friction-grip (HSFG) bolted connection under service loads, the force is transferred mainly by
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The bolts are pre-tensioned to clamp the plates.
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Answer: C. friction between the clamped plates
Highly tightened HSFG bolts clamp plates so tightly that slip resistance arises from friction; bearing-type bolts rely on shank shear and bearing.
5.6 Timber and masonry structures
31 questions · ACiE0506
154. Because wood is an anisotropic material, its weakest strength is in
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Think of splitting wood with a wedge.
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Answer: B. tension perpendicular to the grain
Fibres are strong along the grain, but across the grain only the weak lignin bonds resist tension, so splitting occurs easily.
155. The moisture content at which cell walls are saturated but cell cavities are free of water, called the fibre saturation point, is about
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Around one-quarter to one-third of dry weight.
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Answer: A. 25–30%
Below the fibre saturation point (about 25–30%) timber shrinks and gains strength as it dries; above it, properties remain nearly constant.
156. Which of the following timbers is classed as a softwood?
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Softwoods are conifers.
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Answer: C. Chir pine (Pinus roxburghii)
Softwoods come from conifers such as pine; sal, sissoo and teak are broad-leaved hardwoods.
157. Permissible compressive stress in timber perpendicular to the grain is generally
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Wood fibres are like bundled tubes.
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Answer: B. much lower than that parallel to the grain
Crushing of the cell walls across the grain occurs at much lower stress than along it, so IS 883 gives lower permissible values for bearing across the grain.
158. A timber beam 100 mm wide and 200 mm deep has permissible bending stress 10 N/mm². The permissible bending moment is
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Z = bd²/6 with d = 200 mm.
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Answer: C. 6.67 kN·m
Z = bd²/6 = 100×200²/6 = 666667 mm³; M = σ Z = 10 × 666667 = 6.67 kN·m.
159. A rectangular timber beam 100 mm × 200 mm carries a maximum shear force of 10 kN. The maximum shear stress is
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Peak shear stress in a rectangle is 1.5 times the average.
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Answer: B. 0.75 N/mm²
τmax = 1.5 V/(b d) = 1.5×10000/(100×200) = 0.75 N/mm².
160. A simply supported timber beam of span 3 m, section 100 mm × 200 mm, E = 10000 N/mm², carries a UDL of 2 kN/m. The mid-span deflection 5wL⁴/384EI is
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Use the stronger axis for I and w in N/mm.
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Answer: A. 3.16 mm
I = 100×200³/12 = 6.667e+07 mm⁴; δ = 5×2×3000⁴/(384×10000×6.667e+07) = 3.16 mm.
161. Bridging or herringbone strutting is provided between timber floor joists mainly to
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Deep and narrow sections twist sideways.
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Answer: C. prevent lateral buckling and help distribute loads between joists
Deep, narrow joists tend to buckle sideways; cross-bracing between them keeps them upright and shares concentrated loads with neighbouring joists.
162. A timber strut 100 mm × 100 mm, pin-ended, length 2 m, E = 9000 N/mm². Euler's critical load π²EI/L² is about
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Use L in mm and I = bd³/12.
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Answer: C. 185 kN
I = 100⁴/12 = 8.333e+06 mm⁴; Pcr = π²×9000×8.333e+06/2000² = 185055 N = 185 kN.
163. A scarf joint in timber is used to
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Think of lengthening a member.
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Answer: A. join two members end to end in a line to increase the length
A scarf joint splices timbers in length, with the overlapping inclined faces held by bolts or straps.
164. Nepal Building Code NBC 202 is titled
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200-series are rules of thumb; this one concerns walls that carry load.
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Answer: B. Mandatory Rules of Thumb: Load Bearing Masonry
NBC 202:1994 gives mandatory rules of thumb for load-bearing masonry; NBC 201 covers RC buildings with masonry infill and NBC 105 seismic design.
165. NBC 203 and NBC 204 of Nepal give guidelines for earthquake-resistant construction of
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Both are masonry codes distinguished by strength.
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Answer: C. low-strength and medium-strength masonry buildings respectively
NBC 203:1994 is for low-strength masonry (e.g. stone or brick in mud mortar) and NBC 204:1994 for medium-strength masonry.
166. Which of the following mortars is weakest and loses strength rapidly when wetted?
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Plain earth with water.
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Answer: B. mud mortar
Mud mortar has no hydraulic binder and softens in water; it is common in traditional low-strength masonry but unsuitable in wet conditions.
167. Ordinary (fat) lime mortar hardens mainly by
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The slow carbonation reaction.
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Answer: D. absorbing carbon dioxide from the air
Slaked lime Ca(OH)₂ slowly carbonates to CaCO₃ by absorbing CO₂, so lime mortar gains strength slowly; cement mortar sets by hydration.
168. Hydraulic lime differs from fat lime because it
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Clay impurities give hydraulicity.
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Answer: D. contains some clay and can set even under water
Hydraulic lime, made from limestone with 5–30% clay, forms calcium silicates and aluminates and sets in damp conditions; fat lime is pure and sets only by carbonation.
169. A 1:6 cement–sand mortar is made taking dry volume as 1.33 m³ per m³ of mortar and cement bulk density 1440 kg/m³. The cement needed per m³ of mortar is about
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Cement is one part in seven parts.
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Answer: A. 274 kg
Cement volume = 1.33 × 1/(1+6) = 0.19 m³; mass = 0.19 × 1440 = 274 kg.
170. In brick masonry, the bond with alternate courses of headers and stretchers, giving strong and commonly used walls, is called
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Courses alternate, not bricks within a course.
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Answer: C. English bond
English bond has alternate header and stretcher courses with queen closers; Flemish bond has headers and stretchers alternately in each course.
171. Under vertical compression, masonry prisms commonly fail by vertical splitting cracks in bricks because
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Compare lateral deformation of mortar and brick.
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Answer: C. the mortar expands laterally more than the bricks, putting bricks in lateral tension
Mortar is softer with a larger lateral strain; bond forces it to restrain the bricks, producing lateral tension and splitting in the units.
172. X-shaped (diagonal) cracks in a masonry wall under in-plane earthquake loading are characteristic of
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Two diagonals in opposite directions.
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Answer: D. diagonal shear (tension) failure
Reversing in-plane shear creates principal tension along both diagonals, giving crossed cracks through the mortar joints and units.
173. A slender masonry pier that fails by cracking at its base and pivoting about the toe under in-plane lateral load shows
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Like a rigid block tilting.
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Answer: D. rocking failure
Rocking occurs in slender piers when horizontal bed joints crack in tension and the wall rotates about the compressed toe.
174. Separation of walls at corners during an earthquake, causing the walls to fall outward, mainly results from
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Walls need to act like a box.
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Answer: D. inadequate bonding between orthogonal walls and absence of horizontal ties
Out-of-plane vibration pulls the walls apart at the corners unless they are bonded and tied by bands and corner reinforcement.
175. Horizontal seismic bands (plinth, lintel and roof bands) in masonry buildings are provided mainly to
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Think of tying a box with a strap.
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Answer: B. tie the walls together so that the building acts as a box
Continuous RC or timber bands hold walls together, preventing out-of-plane collapse and distributing lateral loads to shear walls.
176. As per IS 1905, the maximum slenderness ratio (effective height/effective thickness) of a load-bearing masonry wall is
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Between 12 and 50.
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Answer: A. 27
IS 1905 limits the slenderness ratio of load-bearing walls to 27; walls exceeding this ratio need to be thickened or stiffened with buttresses or cross walls.
177. A 230 mm thick masonry wall of height 3.2 m is laterally supported with rotational restraint at top and bottom by RC slabs (effective height = 0.75 H). Its slenderness ratio is
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Hef divided by thickness, both in metres.
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Answer: B. 10.4
Hef = 0.75×3.2 = 2.4 m; SR = 2.4/0.23 = 10.4.
178. The area reduction factor Ka for a masonry column of cross-sectional area A (A < 0.2 m²) is Ka = 0.7 + 1.5A. For a 0.1 m² column section, Ka is
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Substitute A in m².
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Answer: A. 0.850
Ka = 0.7 + 1.5 × 0.1 = 0.85; small sections are reduced since they are more sensitive to defects.
179. Permissible stress in masonry is fd = fb·ks·ka·kp. For basic stress 1.2 N/mm², ks = 0.8, ka = 1.0, kp = 1.0, the capacity of 1 m length of a 230 mm thick wall is
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Multiply allowable stress by area per metre length.
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Answer: A. 220.8 kN
fd = 1.2×0.8 = 0.96 N/mm²; load = 0.96 × 230 × 1000 N = 220.8 kN per metre.
180. A masonry wall 0.23 m thick carries 100 kN per metre at eccentricity 0.02 m. The maximum compressive stress at the base (per metre length) is
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Axial plus bending stress with no tension.
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Answer: D. 662 kN/m²
σmax = (P/A)(1 + 6e/t) = (100/0.23)(1 + 6×0.02/0.23) = 662 kN/m².
181. Through stones (bonders) are provided in stone masonry walls mainly to
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They go through the full thickness.
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Answer: A. tie the inner and outer faces of the wall together
Long stones extending across the wall thickness prevent the faces from separating (delamination) under vertical and lateral loads.
182. Unreinforced masonry is considered a brittle material in earthquakes because it
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Strength in compression is much greater than in tension.
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Answer: B. has low tensile and shear strength compared with its compressive strength
Masonry resists compression well, but tension and shear strengths are small and mortar joints crack suddenly with little warning.
183. The stress at the base of a 3.2 m high masonry wall due to its own weight (unit weight 19 kN/m³) is
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Weight of a prism divided by its base area.
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Answer: D. 60.8 kN/m²
σ = γ h = 19 × 3.2 = 60.8 kN/m², independent of wall thickness.
184. A 230 mm × 230 mm brick masonry column carries an axial load of 300 kN. The average compressive stress is
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Force in N over area in mm².
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Answer: C. 5.67 N/mm²
σ = P/A = 300000/(230×230) = 5.67 N/mm².