Soil Mechanics Chapter-wise MCQs with PYQs & MEQs | Civil Engineering

Soil Mechanics – Chapter-wise MCQs

PYQs • MEQs • Conceptual & Numerical Questions | Civil Engineering

Note: Har question ke niche “Answer Dekho” button dabao. Questions mein variety hai – theoretical, numerical, PYQ-style aur expected (MEQ). Major chapters cover kiye gaye hain. Practice ke liye best!

1. Phase Relations 2. Index Properties 3. Classification 4. Compaction 5. Permeability 6. Effective Stress 7. Consolidation 8. Shear Strength 9. Earth Pressure 10. Slope Stability

Chapter 1: Basic Definitions & Phase Relationships

1Soil mass in its natural state is a ________ system. MEQ

a) One-phase b) Two-phase c) Three-phase d) Four-phase

Answer: c) Three-phase (solids + water + air). Fully saturated soil behaves as two-phase.

2Void ratio (e) is defined as the ratio of: PYQ

a) Volume of voids to total volume b) Volume of voids to volume of solids c) Volume of solids to volume of voids d) Volume of water to volume of voids

Answer: b) e = V_v / V_s

3Porosity (n) of a soil with void ratio e = 0.35 is approximately: Numerical

a) 19.5% b) 25.9% c) 35% d) 40.6%

Answer: b) n = e/(1+e) = 0.35/1.35 ≈ 0.259 → 25.9%

4Relationship between porosity n and void ratio e is: MEQ

a) n = e/(1+e) b) e = n/(1+n) c) Both a and b d) n = 1 + e

Answer: c) Both relations are valid and interchangeable.

5Degree of saturation S for a fully saturated soil is: PYQ

a) 0 b) 0.5 c) 1.0 d) > 1.0

Answer: c) S = 1 (or 100%)

6A soil sample has G = 2.7, e = 0.66, w = 20%. Degree of saturation is: Numerical

a) 70% b) 81.8% c) 90% d) 100%

Answer: b) S = (wG)/e = (0.2 × 2.7)/0.66 ≈ 0.818 → 81.8%

7Bulk unit weight γ is related to dry unit weight γ_d by: MEQ

a) γ = γ_d (1 + w) b) γ = γ_d / (1 + w) c) γ = γ_d × w d) γ = γ_d + w

Answer: a) γ = γ_d (1 + w)

8If volume of voids equals volume of solids, then: PYQ

a) e = 1, n = 0.5 b) e = 0.5, n = 1 c) e = 1, n = 1 d) e = 0, n = 0

Answer: a) e = V_v/V_s = 1; n = e/(1+e) = 0.5

9Air content a_c is the ratio of: MEQ

a) Volume of air to total volume b) Volume of air to volume of voids c) Volume of air to volume of water d) Volume of voids to total volume

Answer: b) a_c = V_a / V_v = 1 − S

10Specific gravity G of soil solids is usually determined by: PYQ

a) Oven drying method b) Pycnometer method c) Sand bath method d) Core cutter method

Answer: b) Pycnometer (or density bottle) method is standard for G.

11A soil weighs 190 kN. After oven drying it weighs 150 kN. Weight of water is: Numerical

a) 30 kN b) 40 kN c) 45 kN d) 50 kN

Answer: b) W_w = 190 − 150 = 40 kN

12Relative density is used for: MEQ

a) Cohesive soils only b) Cohesionless soils c) Both d) Organic soils

Answer: b) Relative density (Density Index) is primarily for sands/gravels.

Chapter 2: Index Properties & Consistency Limits

1Liquid Limit is the water content at which soil: PYQ

a) Starts to shrink b) Passes from liquid to plastic state c) Passes from plastic to semi-solid state d) Has maximum dry density

Answer: b) LL is the boundary between liquid and plastic states.

2Plasticity Index (PI) is equal to: MEQ

a) LL − PL b) LL − SL c) PL − SL d) LL + PL

Answer: a) PI = Liquid Limit − Plastic Limit

3Shrinkage Limit is the water content below which: PYQ

a) Soil starts cracking b) Further reduction in water content does not cause volume reduction c) Soil becomes plastic d) Soil loses strength completely

Answer: b) Below SL, volume remains constant even if water content decreases.

4Atterberg limits are determined for: MEQ

a) Coarse-grained soils b) Fine-grained soils c) Gravels only d) All soils

Answer: b) Atterberg limits are meaningful for fine-grained (cohesive) soils.

5Liquidity Index (LI) is given by: Numerical

a) (w − PL)/PI b) (LL − w)/PI c) (w − LL)/PI d) PI/(w − PL)

Answer: a) LI = (Natural water content − PL) / PI

6Casagrande’s liquid limit device uses how many blows for LL determination? PYQ

a) 10 b) 25 c) 30 d) 50

Answer: b) Standard is 25 blows for groove closure of 12.5 mm.

7Activity of clay is defined as: MEQ

a) PI / % clay fraction b) LL / % clay fraction c) PI × % clay d) % clay / PI

Answer: a) Activity = Plasticity Index / (% of clay-sized particles)

8Sensitivity of a clay is the ratio of: PYQ

a) Undisturbed strength to remoulded strength b) Remoulded strength to undisturbed strength c) Peak strength to residual strength d) Cohesion to friction

Answer: a) S_t = q_u (undisturbed) / q_u (remoulded)

9Which method is most accurate for water content determination? MEQ

a) Sand bath method b) Oven drying method c) Calcium carbide method d) Alcohol method

Answer: b) Oven drying at 105–110°C is the standard accurate method.

10A soil has LL = 50%, PL = 25%. Its plasticity index is: Numerical

a) 25% b) 50% c) 75% d) 12.5%

Answer: a) PI = 50 − 25 = 25%

Chapter 3: Soil Classification

1Unified Soil Classification System (USCS) was developed by: PYQ

a) Rankine b) Casagrande c) Terzaghi d) Coulomb

Answer: b) Arthur Casagrande developed USCS.

2According to IS classification, a soil is coarse-grained if more than ____ % is retained on 75 micron sieve. MEQ

a) 30% b) 50% c) 70% d) 90%

Answer: b) > 50% retained on 75 μ sieve → coarse-grained.

3In USCS, symbol ‘M’ stands for: PYQ

a) Clay b) Silt c) Sand d) Organic

Answer: b) M = Silt (from Swedish word “Mo”)

4A well-graded soil has: MEQ

a) Excess of one size particles b) Good representation of all particle sizes c) Gap in particle sizes d) Only fine particles

Answer: b) Well-graded soils have particles of all sizes in good proportion.

5Coefficient of uniformity C_u = D₆₀/D₁₀. For well-graded gravel, C_u should be: Numerical

a) > 4 b) > 6 c) < 4 d) = 1

Answer: a) For gravel C_u > 4; for sand C_u > 6 (along with C_c between 1–3).

6Plasticity chart is used for classification of: PYQ

a) Coarse-grained soils b) Fine-grained soils c) Organic soils only d) Gravels

Answer: b) A-line on plasticity chart separates clays from silts.

7IS classification system is based on: MEQ

a) AASHTO b) USCS with some modifications c) MIT system only d) USDA system

Answer: b) IS 1498 is largely based on USCS with Indian modifications.

8Group symbol ‘CH’ means: PYQ

a) Clay of high plasticity b) Clay of low plasticity c) Silt of high plasticity d) Organic clay

Answer: a) C = Clay, H = High plasticity (LL > 50%)

Chapter 4: Soil Compaction

1Compaction is a process of: MEQ

a) Expelling air by mechanical means b) Expelling water by drainage c) Reducing volume by consolidation d) Increasing water content

Answer: a) Compaction densifies soil by expelling air (not water).

2Standard Proctor test uses: PYQ

a) 2.5 kg rammer, 30 cm drop, 3 layers b) 4.5 kg rammer, 45 cm drop, 5 layers c) 2.5 kg rammer, 45 cm drop, 5 layers d) 4.5 kg rammer, 30 cm drop, 3 layers

Answer: a) Standard Proctor: 2.5 kg, 305 mm drop, 3 layers, 25 blows each.

3Optimum Moisture Content (OMC) corresponds to: MEQ

a) Maximum dry density b) Minimum dry density c) Zero air voids d) Maximum bulk density

Answer: a) At OMC, dry density is maximum for given compactive effort.

4Modified Proctor test has higher energy than Standard Proctor because: PYQ

a) Heavier rammer and more layers b) Lighter rammer c) Less number of blows d) Smaller mould

Answer: a) 4.54 kg rammer, 457 mm drop, 5 layers → about 4.5 times energy.

5Zero air voids line represents: MEQ

a) Fully saturated condition b) Dry soil c) Optimum condition d) Partially saturated

Answer: a) Theoretical maximum dry density for given water content (S=100%).

6Relative compaction is: Numerical

a) (γ_{d field} / γ_{d max}) × 100% b) (γ_{d max} / γ_{d field}) × 100% c) e_max − e / e_max − e_min d) Both a and c

Answer: a) Relative compaction = field dry density / maximum dry density × 100%

7Which of the following increases with increase in compactive effort? PYQ

a) OMC b) MDD c) Both d) Neither

Answer: b) Higher energy → higher MDD and lower OMC.

Chapter 5: Permeability & Seepage

1Darcy’s law is valid for: PYQ

a) Laminar flow through soils b) Turbulent flow c) Both laminar and turbulent d) Only coarse gravels

Answer: a) Darcy’s law: v = ki holds for laminar flow (Reynolds number < 1 approximately).

2Coefficient of permeability k has units of: MEQ

a) m/s or cm/s b) m²/s c) N/m² d) Dimensionless

Answer: a) Velocity units (length/time).

3Which soil has the highest permeability? PYQ

a) Clay b) Silt c) Sand d) Gravel

Answer: d) Gravel > Sand > Silt > Clay

4Constant head permeability test is suitable for: MEQ

a) Fine-grained soils b) Coarse-grained soils c) Both d) Organic soils

Answer: b) Constant head for permeable soils (sands/gravels); falling head for clays/silts.

5Seepage velocity is related to discharge velocity by: Numerical

a) v_s = v / n b) v_s = v × n c) v_s = v × e d) v_s = v

Answer: a) Seepage velocity = discharge velocity / porosity

6Quick sand condition occurs when effective stress becomes: PYQ

a) Zero b) Maximum c) Equal to total stress d) Negative

Answer: a) When seepage force makes σ′ = 0, soil loses strength (boiling/quick condition).

7Flow net is used to determine: MEQ

a) Seepage quantity and exit gradient b) Only permeability c) Only shear strength d) Compaction characteristics

Answer: a) Flow nets give seepage discharge, uplift pressure and exit gradient.

Chapter 6: Effective Stress Principle

1Effective stress principle was given by: PYQ

a) Rankine b) Terzaghi c) Coulomb d) Darcy

Answer: b) Karl Terzaghi (σ′ = σ − u)

2Effective stress σ′ is equal to: MEQ

a) Total stress − pore water pressure b) Total stress + pore water pressure c) Pore water pressure only d) Total stress only

Answer: a) σ′ = σ − u

3Increase in pore water pressure causes: PYQ

a) Increase in effective stress b) Decrease in effective stress c) No change d) Increase in shear strength

Answer: b) Higher u → lower σ′ → lower strength.

4In a submerged soil, effective unit weight is: Numerical

a) γ_sat − γ_w b) γ_sat + γ_w c) γ_d d) γ_sat

Answer: a) γ′ = γ_sat − γ_w (submerged/buoyant unit weight)

5Capillary rise in soil causes: MEQ

a) Increase in effective stress above water table b) Decrease in effective stress c) Zero effective stress d) Negative pore pressure only without stress change

Answer: a) Negative pore pressure (suction) increases effective stress in capillary zone.

Chapter 7: Consolidation

1Consolidation is a process of: PYQ

a) Expelling air b) Expelling water under sustained load c) Mechanical densification d) Increasing void ratio

Answer: b) Time-dependent volume reduction due to expulsion of water.

2Terzaghi’s one-dimensional consolidation theory assumes: MEQ

a) Soil is fully saturated b) Flow is one-dimensional c) Darcy’s law is valid d) All of the above

Answer: d) All are assumptions of Terzaghi’s theory.

3Coefficient of consolidation C_v has units of: Numerical

a) m²/year or cm²/s b) m/s c) kN/m² d) Dimensionless

Answer: a) C_v = k / (m_v γ_w) → length²/time

4Primary consolidation is due to: PYQ

a) Expulsion of air b) Expulsion of water c) Plastic readjustment of particles d) Creep

Answer: b) Primary = hydrodynamic (pore pressure dissipation). Secondary = creep.

5Compression index C_c is the slope of: MEQ

a) e–log σ′ curve in virgin compression b) e–σ′ curve c) Void ratio vs time d) Settlement vs time

Answer: a) C_c = Δe / Δlog σ′ (virgin compression line)

6Time factor T_v for 50% consolidation (U=50%) is approximately: Numerical

a) 0.197 b) 0.848 c) 1.0 d) 0.05

Answer: a) T_v ≈ 0.197 for U = 50%

Chapter 8: Shear Strength of Soil

1Mohr-Coulomb failure criterion is: PYQ

a) τ = c + σ tan φ b) τ = c − σ tan φ c) τ = σ tan φ only d) τ = c only

Answer: a) τ_f = c + σ′ tan φ

2For saturated clay under undrained condition, φ_u is approximately: MEQ

a) 0° b) 30° c) 45° d) Equal to φ′

Answer: a) φ_u ≈ 0 for saturated clay in undrained loading (c_u only).

3Direct shear test is suitable for: PYQ

a) Determining both c and φ of sands and clays b) Only clays c) Only residual strength d) Pore pressure measurement

Answer: a) Widely used for both, though has some limitations (failure plane forced).

4Unconfined compression test is a special case of: MEQ

a) Triaxial UU test with σ₃ = 0 b) Direct shear c) Consolidated drained test d) Vane shear

Answer: a) UC test → q_u = 2c_u (σ₃ = 0)

5In triaxial test, the major principal stress at failure is: Numerical

a) Cell pressure + deviator stress b) Cell pressure only c) Deviator stress only d) Pore pressure

Answer: a) σ₁ = σ₃ + (σ₁ − σ₃)

6Shear strength of soil increases with: PYQ

a) Increase in effective normal stress b) Decrease in effective normal stress c) Increase in pore pressure d) None

Answer: a) Higher σ′ → higher shear strength (Coulomb).

Chapter 9: Earth Pressure

1Rankine’s theory of earth pressure assumes: PYQ

a) Smooth vertical wall and horizontal backfill b) Rough wall c) Inclined wall only d) Cohesive soil only

Answer: a) Rankine assumes no wall friction, vertical wall, horizontal backfill.

2Active earth pressure coefficient K_a for φ = 30° is: Numerical

a) 1/3 b) 1/2 c) 1 d) 3

Answer: a) K_a = (1−sinφ)/(1+sinφ) = 1/3 for φ=30°

3Passive earth pressure is greater than active because: MEQ

a) Wall moves towards the soil b) Wall moves away from soil c) Soil is denser d) Water pressure is higher

Answer: a) Passive state develops when wall pushes into the soil (higher resistance).

4Earth pressure at rest K₀ is approximately: PYQ

a) 1 − sin φ b) (1 − sin φ)/(1 + sin φ) c) tan²(45 − φ/2) d) 1

Answer: a) Jaky’s formula: K₀ ≈ 1 − sin φ

5Coulomb’s theory considers: MEQ

a) Wall friction and inclined backfill b) Only horizontal backfill c) No wall friction d) Only cohesive soils

Answer: a) Coulomb’s wedge theory includes wall friction (δ) and backfill inclination.

Chapter 10: Stability of Slopes

1Factor of safety against sliding for infinite slope in cohesionless soil is: PYQ

a) tan φ / tan β b) tan β / tan φ c) c / (γ H) d) 1

Answer: a) FOS = tan φ / tan i (where i = slope angle)

2Swedish circle method is used for: MEQ

a) Finite slopes (mainly cohesive) b) Infinite slopes only c) Retaining walls d) Bearing capacity

Answer: a) Method of slices / Swedish circle for finite slopes.

3Critical height of a vertical cut in pure clay (φ=0) is: Numerical

a) 4c / γ b) 2c / γ c) c / γ d) 8c / γ

Answer: a) H_c = 4c_u / γ (for φ_u = 0)

4Taylor’s stability number is: PYQ

a) c / (γ H F) b) γ H / c c) c / γ d) F / (γ H)

Answer: a) S_n = c / (γ H F_c)

5For a slope, the most critical failure surface is the one with: MEQ

a) Minimum factor of safety b) Maximum factor of safety c) Maximum length d) Steepest angle

Answer: a) The surface giving the lowest FOS is critical.

Tip: Yeh page offline bhi kaam karega. Bookmark kar lo. Agar kisi particular chapter mein aur questions chahiye ya PDF version chahiye to batao!

Soil Mechanics MCQ Blog • Prepared for Civil Engineering students (GATE / ESE / SSC-JE / University exams)
Questions based on standard concepts & previous year patterns

MP Sub Engineer Civil Important IS Codes List | Soil, Survey, BMC, RCC & Estimation

MP Sub Engineer Civil
Important IS Codes List
Soil + Survey + BMC + CT/RCC + Estimation

Legend:
● Very Important   ● Important   Normal = Useful / Occasional

Exam Tip: MP Sub Engineer / JE level में IS 456, IS 383, IS 1200, IS 2720, IS 1498, IS 1077, IS 269 सबसे ज्यादा पूछे जाते हैं। इन्हें जरूर याद रखें।

1. Soil Mechanics & Foundation – IS Codes

IS Code	Title / Subject	Importance
IS 1498 : 1970	Classification and Identification of Soils	Very Important
IS 2720 (All Parts)	Methods of Test for Soils (Water content, Specific gravity, Grain size, Atterberg limits, Compaction, Shear, CBR etc.)	Very Important
IS 1892 : 1979 / 2021	Code of Practice for Subsurface Investigation for Foundations	Important
IS 1904 : 1986	Code of Practice for Design & Construction of Foundations in Soils (Shallow Foundations)	Important
IS 6403 : 1981	Code of Practice for Determination of Bearing Capacity of Shallow Foundations	Important
IS 2131 : 1981	Method for Standard Penetration Test (SPT) for Soils	Useful
IS 1888 : 1982	Method of Load Test on Soils (Plate Load Test)	Useful
IS 2911 (Parts 1-4)	Design and Construction of Pile Foundations	Useful
IS 1080 : 1985	Code of Practice for Design & Construction of Shallow Foundations in Soils	Useful
IS 2950	Code of Practice for Raft Foundations	Occasional

2. Surveying – IS Codes

IS Code	Title / Subject	Importance
IS 962 : 1989	Code of Practice for Architectural and Building Drawings (Symbols, Lettering, Dimensioning)	Important
IS 1191	Hydrometric Determinations – Vocabulary	Useful
IS 2405	Specification for Industrial Screens (for sieves used in testing)	Useful
IS 1498	Also used for soil classification related to survey data	Useful
SP 46	Engineering Drawing Practice for Schools & Colleges (related to drawing standards)	Useful

Note: Surveying में ज्यादा IS codes नहीं पूछे जाते। ज्यादातर instruments, methods और curve formulas आते हैं। IS 962 drawing symbols के लिए याद रखें।

3. Building Materials & Construction (BMC) – IS Codes

IS Code	Title / Subject	Importance
IS 269 : 2015	Ordinary Portland Cement (OPC) – Specification	Very Important
IS 1489 (Part 1 & 2)	Portland Pozzolana Cement (PPC) – Specification	Very Important
IS 383 : 2016	Coarse and Fine Aggregate for Concrete – Specification	Very Important
IS 1077 : 1992	Common Burnt Clay Building Bricks – Specification	Very Important
IS 8112 / IS 12269	43 Grade & 53 Grade OPC	Important
IS 4031 (All Parts)	Methods of Physical Tests for Hydraulic Cement	Important
IS 2386 (Parts 1-8)	Methods of Test for Aggregates for Concrete	Important
IS 3495	Methods of Tests of Burnt Clay Building Bricks	Important
IS 2116	Sand for Masonry Mortars	Useful
IS 2250	Code of Practice for Preparation and Use of Masonry Mortars	Useful
IS 2212	Code of Practice for Brickwork	Useful
IS 4082	Stacking and Storage of Construction Materials	Useful
IS 1542	Sand for Plaster	Useful
IS 712	Building Limes	Occasional

4. Concrete Technology + RCC – IS Codes

IS Code	Title / Subject	Importance
IS 456 : 2000	Plain and Reinforced Concrete – Code of Practice	Very Important (Most Asked)
IS 1786 : 2008	High Strength Deformed Steel Bars and Wires for Concrete Reinforcement	Very Important
IS 10262 : 2019	Concrete Mix Proportioning – Guidelines	Important
IS 516	Method of Tests for Strength of Concrete	Important
IS 1199	Methods of Sampling and Analysis of Concrete	Important
IS 2502	Code of Practice for Bending and Fixing of Bars for Concrete Reinforcement	Important
IS 13920 : 2016	Ductile Design and Detailing of Reinforced Concrete Structures	Useful (Earthquake)
IS 3370	Concrete Structures for Storage of Liquids	Useful
IS 1343	Prestressed Concrete – Code of Practice	Occasional
IS 4926	Ready Mixed Concrete	Useful

5. Estimation & Costing – IS Codes

IS Code	Title / Subject	Importance
IS 1200 (All Parts)	Method of Measurement of Building and Civil Engineering Works	Very Important
IS 3861	Method of Measurement of Plinth, Carpet and Rentable Areas of Buildings	Important
IS 10067	Guidelines for Classification of Civil Engineering Works	Useful
SP 27	Handbook of Method of Measurement of Building Works	Useful

Most Important Codes to Remember (Must Revise):
1. IS 456 : 2000 – RCC
2. IS 383 – Aggregates
3. IS 1200 – Measurement (Estimation)
4. IS 269 / 1489 – Cement
5. IS 1077 – Bricks
6. IS 2720 – Soil Testing
7. IS 1498 – Soil Classification
8. IS 1786 – Reinforcement Steel
9. IS 2502 – Bar Bending
10. IS 10262 – Mix Design

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MP Sub Engineer Civil – Important IS Codes List
Soil + Survey + BMC + CT/RCC + Estimation
Very Important codes को लाल/गुलाबी और Important को नारंगी बैकग्राउंड से mark किया गया है।

MP Sub Engineer Civil Formulas – Soil, Survey, BMC, RCC & Estimation | Complete Formula Sheet

MP Sub Engineer Civil
Complete Important Formulas
Soil + Survey + BMC + CT/RCC + Estimation

Last 5 Years Average Weightage (Technical 100 Qs)

Subject	Approx. Questions	Priority
Soil Mechanics + Foundation	12 – 20	Highest
Surveying	10 – 15	Highest
Building Materials + Construction	10 – 18	Highest
Concrete Technology + RCC	10 – 15	High
Estimation & Costing	8 – 13	High
Total from these 5	55 – 70	Most Important

1. Soil Mechanics – Complete Important Formulas

Basic Properties

1. Void Ratio (e) = V_v / V_s

2. Porosity (n) = (V_v / V) × 100   |   n = e/(1+e)

3. Degree of Saturation (S) = (V_w / V_v) × 100

4. Water Content (w) = (W_w / W_s) × 100

5. Specific Gravity (G) = γ_s / γ_w

6. Bulk Unit Weight γ = (G + S e)/(1 + e) × γ_w

7. Dry Unit Weight γ_d = G/(1 + e) × γ_w = γ/(1 + w)

8. Saturated Unit Weight γ_sat = (G + e)/(1 + e) × γ_w

9. Submerged Unit Weight γ' = (G – 1)/(1 + e) × γ_w

10. Fundamental Relation: S e = G w

11. Relative Density I_D = (e_max – e)/(e_max – e_min) × 100

12. Density Index I_d = (γ_d – γ_{d min})/(γ_{d max} – γ_{d min}) × 100

Index Properties & Classification

13. Liquidity Index I_L = (w – PL)/(LL – PL)

14. Consistency Index I_C = (LL – w)/(LL – PL)

15. Plasticity Index PI = LL – PL

16. Shrinkage Index = PL – SL

17. Activity = PI / (% clay fraction)

18. Coefficient of Uniformity C_u = D₆₀ / D₁₀

19. Coefficient of Curvature C_c = (D₃₀)² / (D₆₀ × D₁₀)

20. Group Index GI = 0.2a + 0.005ac + 0.01bd

Permeability

21. Darcy’s Law: Q = k i A   |   v = k i

22. Constant Head: k = (QL)/(A h t)

23. Falling Head: k = (2.303 a L / A t) log₁₀(h₁/h₂)

24. Hazen’s Formula: k = C D₁₀² (C ≈ 100–150)

Compaction

25. Dry Density γ_d = γ / (1 + w)

26. Zero Air Void Line: γ_d = (G γ_w)/(1 + w G)

27. Relative Compaction = (γ_{d field} / γ_{d max}) × 100

Consolidation

28. Compression Index C_c = (e₁ – e₂)/log(σ₂/σ₁)

29. Settlement S_c = [C_c H /(1 + e₀)] log₁₀[(σ₀ + Δσ)/σ₀]

30. Time Factor T_v = (C_v t)/d²

31. Coefficient of Consolidation C_v = (T_v d²)/t

32. Coefficient of Volume Compressibility m_v = a_v/(1 + e₀)

Shear Strength

33. Coulomb’s Equation: τ = c + σ tan φ

34. Unconfined Compressive Strength q_u = 2c (for φ=0)

35. Sensitivity = q_u undisturbed / q_u remoulded

Earth Pressure

36. Rankine Active: K_a = (1 – sinφ)/(1 + sinφ) = tan²(45° – φ/2)

37. Rankine Passive: K_p = (1 + sinφ)/(1 – sinφ) = tan²(45° + φ/2)

38. Active Pressure P_a = ½ γ H² K_a

39. Passive Pressure P_p = ½ γ H² K_p

Bearing Capacity

40. Terzaghi Ultimate: q_u = c N_c + γ D_f N_q + 0.5 γ B N_γ

41. Net Ultimate q_nu = q_u – γ D_f

42. Safe Bearing Capacity = q_nu / FOS + γ D_f

43. Plate Load Test: Settlement relation S_f/S_p = [(B_f(B_p+0.3))/(B_p(B_f+0.3))]²

2. Surveying – Complete Important Formulas

Chain Surveying & Corrections

1. True Length L = L' × (l / l')   (l = true length of chain, l' = incorrect)

2. Temperature Correction C_t = α (T_m – T₀) L

3. Pull Correction C_p = (P – P₀) L / (A E)

4. Sag Correction C_s = – (W² L)/(24 P²)

5. Slope Correction C_slope = – (h²)/(2L)

6. Absolute Error = Difference between true & measured

7. Relative Error = Absolute Error / True Value

Compass Surveying

8. Fore Bearing – Back Bearing = ± 180°

9. Included Angle = Difference of bearings of two lines

10. Declination: True Bearing = Magnetic Bearing ± Declination

Levelling

11. HI = RL of BS station + BS reading

12. RL = HI – FS (or IS)

13. Rise & Fall: New RL = Old RL + Rise – Fall

14. Check: Σ BS – Σ FS = Last RL – First RL

15. Curvature Correction C_c = –0.0785 D² (D in km)

16. Refraction Correction C_r = +0.0112 D²

17. Combined Correction = –0.0673 D²

Contouring & Area/Volume

18. Horizontal Equivalent = Contour Interval / tan(slope)

19. Trapezoidal Rule: A = (d/2)(y₁ + 2y₂ + … + 2y_n-1 + y_n)

20. Simpson’s Rule: A = (d/3)(y₁ + 4y₂ + 2y₃ + 4y₄ + … + y_n)

21. Volume by Trapezoidal: V = (d/2)(A₁ + 2A₂ + … + 2A_n-1 + A_n)

22. Volume by Prismoidal: V = (d/3)(A₁ + 4A_m + A₂)

Theodolite & Curves

23. Latitude = L cos θ   |   Departure = L sin θ

24. Radius of Curve R = 1720 / D (D = degree of curve for 30m chain)

25. Length of Curve L = (π R Δ)/180

26. Tangent Length T = R tan(Δ/2)

27. Chainage of T₁ = Chainage of Intersection – T

28. Chainage of T₂ = Chainage of T₁ + L

29. Mid Ordinate M = R (1 – cos(Δ/2))

30. External Distance E = R (sec(Δ/2) – 1)

31. Length of Long Chord = 2 R sin(Δ/2)

32. Apex Distance = R (sec(Δ/2) – 1)

3. Building Materials & Construction – Complete Formulas & Values

Cement

1. Specific Gravity of Cement ≈ 3.15

2. Standard Consistency = 25% – 35%

3. Initial Setting Time ≥ 30 min (OPC)

4. Final Setting Time ≤ 600 min (10 hrs)

5. Fineness (by sieve) ≤ 10% retained on 90 micron

6. Compressive Strength (28 days): 33/43/53 Grade

Aggregates & Concrete

7. Fineness Modulus = Σ Cumulative % retained / 100

8. Bulking of Sand = (V_moist – V_dry)/V_dry × 100

9. Water-Cement Ratio w/c = Weight of water / Weight of cement

10. Nominal Mixes (by volume):

• M10 → 1 : 3 : 6
• M15 → 1 : 2 : 4
• M20 → 1 : 1.5 : 3
• M25 → 1 : 1 : 2

11. Slump Values:

• Mass concrete / Roads → 25–50 mm
• Beams & Slabs → 50–100 mm
• Columns → 75–150 mm

12. Aggregate Impact Value ≤ 30% (for wearing surface)

13. Aggregate Crushing Value ≤ 30%

14. Los Angeles Abrasion Value ≤ 30% (for pavement)

Bricks & Masonry

15. Modular Brick Size = 190 × 90 × 90 mm

16. Brick with mortar = 200 × 100 × 100 mm

17. Number of modular bricks per m³ ≈ 500

18. Water Absorption of brick ≤ 20% (1st class)

19. Compressive Strength of 1st class brick ≥ 10.5 N/mm²

Timber & Others

20. Moisture Content of Timber = (W_wet – W_dry)/W_dry × 100

21. Seasoning reduces moisture to 10–12%

4. Concrete Technology + RCC – Complete Formulas

Working Stress Method (Most Asked)

1. Modular Ratio m = 280 / (3 σ_cbc)

2. Neutral Axis factor k = m / (m + σ_st/σ_cbc)

3. Lever Arm factor j = 1 – k/3

4. Moment of Resistance Factor Q = ½ σ_cbc k j

5. Moment of Resistance M = Q b d²

6. Area of Tension Steel A_st = M / (σ_st × j × d)

7. Percentage Steel p_t = (A_st /(b d)) × 100

8. Critical Neutral Axis depth n_c = (m σ_cbc /(m σ_cbc + σ_st)) × d

Shear & Bond

9. Nominal Shear Stress τ_v = V / (b d)

10. Development Length L_d = (φ σ_s) / (4 τ_bd)

11. Minimum Shear Reinforcement: A_sv / (b s_v) ≥ 0.4 / (0.87 f_y)

12. Maximum Spacing of Stirrups = 0.75 d (vertical) or 300 mm (whichever less)

Limit State Basics (IS 456)

13. Design Strength of Concrete = 0.67 f_ck / 1.5 = 0.446 f_ck

14. Design Strength of Steel = 0.87 f_y

15. Xu max / d = 0.53 (Fe250), 0.48 (Fe415), 0.46 (Fe500)

16. Limiting Moment of Resistance M_{u lim} = 0.149 f_ck b d² (for Fe415)

17. pt lim = 41.61 (f_ck/f_y) × (Xu max/d) (approx values used in exam)

Cover & Deflection

18. Nominal Cover (IS 456):

• Mild → 20 mm
• Moderate → 30 mm
• Severe → 45 mm
• Very Severe → 50 mm
• Extreme → 75 mm

19. Basic L/d ratio: Simply Supported = 20, Continuous = 26, Cantilever = 7

20. Effective Depth d = Overall depth D – Effective cover

5. Estimation & Costing – Complete Formulas

Types of Estimates & Areas

1. Plinth Area Estimate = Plinth Area × Plinth Area Rate

2. Cubic Content Estimate = Plinth Area × Height × Rate per m³

3. Carpet Area ≈ 50–65% of Plinth Area (residential)

4. Floor Area Ratio (FAR) = Total Floor Area / Plot Area

5. Built-up Area = Carpet Area + Wall Area + Wet Areas

Quantity of Materials

6. Number of Bricks = Volume of brickwork / Volume of 1 brick with mortar

7. Cement Bags (approx for 1 m³ concrete):

• 1:3:6 (M10) → 4.4 bags
• 1:2:4 (M15) → 6.4 bags
• 1:1.5:3 (M20) → 8.0 bags
• 1:1:2 (M25) → 10.2 bags

8. Dry Volume of Concrete = 1.54 × Wet Volume

9. Steel Weight = (D² / 162) × Length (m)   → kg

10. Thumb Rule Steel Quantity:

• Slab → 0.7% – 1.0% of concrete volume
• Beam → 1.0% – 2.0%
• Column → 1.0% – 5.0%
• Foundation → 0.5% – 0.8%

Valuation

11. Scrap Value = Original Cost – Total Depreciation

12. Salvage Value = Value at the end of utility period without being dismantled

13. Straight Line Depreciation = (C – S)/n

14. Constant Percentage Method: Annual Dep. = 1 – (S/C)^1/n

15. Sinking Fund A = S i / [(1+i)ⁿ – 1]

16. Year’s Purchase = 100 / Rate of Interest (%)

17. Capitalized Value = Net Annual Income × Year’s Purchase

18. Obsolescence = Loss in value due to change in fashion/design

Exam Tip: D²/162, Cement bags, Nominal mixes, Modular ratio, Se = Gw, Rankine Ka/Kp, Curve formulas (T, L, M, E) और Terzaghi bearing capacity सबसे ज्यादा पूछे जाते हैं। इन्हें रोज revise करें।

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MP Sub Engineer Civil – Complete Formula Sheet
Soil (43) + Survey (32) + BMC (21) + CT/RCC (20) + Estimation (18) = 134 Important Formulas
Practice daily for maximum score in Technical Section.

RCC Short Notes & 50 MCQs for MP Sub Engineer Civil | LSM, WSM & RCC Beam

RCC Short Notes + 50 MCQs (English)

Basic Concepts • LSM • WSM • Singly & Doubly Reinforced Beams • IS 456:2000

Based on recurring PYQ patterns of SSC JE / RRB JE / MP Sub Engineer Civil

What is RCC?

RCC stands for Reinforced Cement Concrete. Concrete is strong in compression but weak in tension. Steel reinforcement is provided to resist tensile stresses.

Exam Point: Concrete mainly resists compression; steel mainly resists tension.

1. Basic Concepts

• Cement, Fine Aggregate, Coarse Aggregate, Water, Steel Reinforcement

Material	Main Function
Concrete	Resists compression
Steel	Resists tension
Concrete Cover	Protects steel from corrosion & fire
Stirrups	Shear resistance + holds longitudinal bars

Effective Depth (d) = Overall depth − Effective cover

2. Limit State Method (LSM)

Two main limit states: Collapse and Serviceability.

• Collapse → Flexure, Compression, Shear, Torsion
• Serviceability → Deflection, Cracking, Vibration

Design Load = Characteristic Load × Partial Safety Factor

3. Working Stress Method (WSM)

Traditional elastic method using permissible stresses and Factor of Safety.

Permissible Stress = Strength / Factor of Safety

4. Singly Reinforced Beam

Main flexural reinforcement only in tension zone.

Mu = T × z

5. Doubly Reinforced Beam

Reinforcement in both tension and compression zones. Used when depth is restricted or required moment > limiting moment of singly reinforced section.

RCC 50 MCQs (English) – SSC JE / JE Pattern

Instructions: Attempt all 50 questions. Click Submit Quiz to see score and explanations.
Questions based on recurring concepts from SSC JE, RRB JE and State JE exams (IS 456:2000).

Q1. As per IS 456:2000, the minimum grade of concrete for reinforced concrete is: (SSC JE)

A. M15 B. M20 C. M25 D. M30

As per IS 456:2000 Clause 6.1.2 – minimum grade for RCC is M20.

Q2. The modulus of elasticity of concrete as per IS 456:2000 is: (SSC JE)

A. 5000 √fck B. 5700 √fck C. 570 √fck D. 500 √fck

Ec = 5000 √fck (N/mm²) as per IS 456:2000.

Q3. For Fe 500 steel, the limiting xu/d for a singly reinforced section is: (SSC JE)

A. 0.53 B. 0.48 C. 0.46 D. 0.42

xu,max/d = 0.46 for Fe 500 (IS 456 Annex G / Table G).

Q4. Design bond stress for M20 concrete (plain bars in tension) as per IS 456 is: (SSC JE)

A. 1.0 N/mm² B. 1.2 N/mm² C. 1.4 N/mm² D. 1.5 N/mm²

Table 26.2.1.1 of IS 456 → τbd for M20 = 1.2 N/mm².

Q5. Minimum cement content for RCC under moderate exposure (20 mm aggregate) as per IS 456 is: (SSC JE)

A. 280 kg/m³ B. 300 kg/m³ C. 320 kg/m³ D. 350 kg/m³

Table 5 of IS 456:2000 – Moderate exposure, RCC → 300 kg/m³.

Q6. Partial safety factor for concrete in limit state design is: (SSC JE)

A. 1.15 B. 1.5 C. 1.0 D. 1.2

γm for concrete = 1.5; for steel = 1.15.

Q7. Maximum percentage of tension reinforcement in a beam as per IS 456 is: (SSC JE)

A. 2% B. 3% C. 4% D. 6%

Maximum tension reinforcement = 4% of gross sectional area (Cl. 26.5.1.1).

Q8. Nominal cover for moderate exposure as per IS 456 is: (SSC JE)

A. 20 mm B. 30 mm C. 45 mm D. 50 mm

Table 16 – Moderate exposure → 30 mm.

Q9. Development length Ld for a bar in tension is given by: (SSC JE)

A. (ϕ × σs)/(4 τbd) B. (ϕ × 0.87 fy)/(4 τbd) C. (ϕ × fy)/(4 τbd) D. (ϕ × 0.87 fy)/(2 τbd)

Ld = (ϕ × 0.87 fy)/(4 τbd) as per IS 456.

Q10. Doubly reinforced beams are mainly provided when: (SSC JE)

A. Depth is restricted B. Span is very large C. Load is very small D. Shear force is high

Used when available depth is restricted and required Mu > Mulim of singly reinforced section.

Q11. Characteristic strength of concrete is the strength below which not more than ___ % of results are expected to fall: (SSC JE)

A. 5% B. 10% C. 15% D. 20%

Definition of characteristic strength – not more than 5% results fall below it.

Q12. Target mean strength of concrete is given by: (SSC JE)

A. fck + 1.65σ B. fck + 1.5σ C. fck + σ D. fck + 2σ

fm = fck + 1.65σ (IS 456).

Q13. Maximum spacing of main reinforcement in a slab shall not exceed: (SSC JE)

A. 3d or 300 mm B. 2d or 300 mm C. 3d or 450 mm D. 2d or 450 mm

Main bars: lesser of 3d or 300 mm (Cl. 26.3.3).

Q14. Minimum percentage of tension reinforcement for Fe 415 in beams is approximately: (SSC JE)

A. 0.12% B. 0.15% C. 0.20% D. 0.85%

Ast,min = 0.85 bd/fy → for Fe 415 ≈ 0.205%.

Q15. Approximate lever arm for a balanced singly reinforced section is: (SSC JE)

A. 0.87d B. 0.90d C. 0.95d D. d

z ≈ 0.90d is commonly taken for balanced section.

Q16. In an under-reinforced section: (SSC JE)

A. Steel fails first B. Concrete fails first C. Both fail simultaneously D. Failure is sudden

Under-reinforced → steel yields first → ductile failure.

Q17. In an over-reinforced section: (SSC JE)

A. Steel fails first B. Concrete fails first C. Both fail together D. Large deflection occurs

Over-reinforced → concrete crushes first → brittle failure.

Q18. Minimum nominal cover for footing as per IS 456 is: (SSC JE)

A. 40 mm B. 50 mm C. 60 mm D. 75 mm

Minimum nominal cover for footing = 50 mm.

Q19. For Fe 415 steel, limiting xu/d is: (SSC JE)

A. 0.53 B. 0.48 C. 0.46 D. 0.42

xu,max/d = 0.48 for Fe 415.

Q20. Maximum shear stress in beams (even with shear reinforcement) shall not exceed: (SSC JE)

A. 0.63√fck B. 0.16√fck C. 0.36√fck D. 0.45√fck

τc,max = 0.63√fck (Cl. 40.2.3 of IS 456).

Q21. Minimum diameter of longitudinal bar in a column is: (SSC JE)

A. 8 mm B. 10 mm C. 12 mm D. 16 mm

Minimum diameter of longitudinal bars in column = 12 mm.

Q22. Minimum number of longitudinal bars in a rectangular column is: (SSC JE)

A. 3 B. 4 C. 6 D. 8

Minimum 4 bars in rectangular column.

Q23. Pitch of lateral ties in a column shall not exceed the least of: (SSC JE)

A. Least lateral dimension, 16 × dia of smallest longitudinal bar, 300 mm B. Least lateral dimension, 12 × dia, 300 mm C. 300 mm only D. 16 × dia only

Pitch ≤ least of: least lateral dimension, 16 × smallest longitudinal bar dia, 300 mm.

Q24. In limit state of collapse – flexure, the maximum compressive strain in concrete is taken as: (SSC JE)

A. 0.002 B. 0.0035 C. 0.003 D. 0.004

Maximum compressive strain in concrete = 0.0035.

Q25. Design stress-strain curve for concrete in limit state method is: (SSC JE)

A. Linear B. Parabolic up to 0.002 then constant C. Rectangular D. Triangular

Parabolic up to strain 0.002 and then constant till 0.0035.

Q26. The factor of safety for steel in working stress method is generally: (SSC JE)

A. 1.5 B. 1.78 C. 2.0 D. 3.0

For mild steel ≈ 1.78 (based on yield stress).

Q27. Modular ratio ‘m’ in working stress method is taken as: (SSC JE)

A. 280 / (3σcbc) B. 280 / σcbc C. 3σcbc / 280 D. Es / Ec

m = 280 / (3σcbc) as per IS 456 (WSM).

Q28. Minimum clear cover for beams as per IS 456 (mild exposure) is: (SSC JE)

A. 20 mm B. 25 mm C. 30 mm D. 40 mm

Minimum nominal cover for beam = 25 mm (mild exposure).

Q29. Side face reinforcement is provided in beams when depth exceeds: (SSC JE)

A. 450 mm B. 750 mm C. 900 mm D. 1000 mm

Side face reinforcement required when overall depth > 750 mm.

Q30. The maximum spacing of shear reinforcement (vertical stirrups) shall not exceed: (SSC JE)

A. 0.75d B. d C. 300 mm D. Both A and C (lesser of the two)

Maximum spacing of vertical stirrups = 0.75d or 300 mm, whichever is less.

Q31. In a balanced section (LSM): (SSC JE)

A. xu = xu,max B. xu < xu,max C. xu > xu,max D. Steel stress is less than 0.87fy

Balanced section → xu = xu,max.

Q32. The design compressive strength of concrete in limit state is taken as: (SSC JE)

A. 0.36 fck B. 0.45 fck C. 0.67 fck D. 0.87 fck

Average design stress in compression block = 0.36 fck.

Q33. The depth of rectangular stress block in limit state is taken as: (SSC JE)

A. 0.36 xu B. 0.42 xu C. 0.45 xu D. 0.48 xu

Centroid of stress block is at 0.42 xu from extreme compression fibre.

Q34. Minimum reinforcement in a slab (HYSD bars) is: (SSC JE)

A. 0.12% B. 0.15% C. 0.20% D. 0.25%

For HYSD bars / welded wire fabric → 0.12% of gross sectional area.

Q35. Lap length in tension for HYSD bars is generally taken as: (SSC JE)

A. Ld B. 1.3 Ld C. 2 Ld D. 40 ϕ

Lap length in tension = 1.3 Ld (or as per development length requirements).

Q36. The basic value of span to effective depth ratio for simply supported beam (span ≤ 10 m) is: (SSC JE)

A. 7 B. 20 C. 26 D. 35

Simply supported → 20; continuous → 26; cantilever → 7.

Q37. Effective length of a column fixed at both ends is: (SSC JE)

A. 0.65 L B. 0.80 L C. 1.0 L D. 1.2 L

Both ends fixed → recommended effective length = 0.65 L (IS 456 Table 28).

Q38. The minimum eccentricity for a column as per IS 456 is: (SSC JE)

A. L/300 + D/30 B. L/500 + D/30 C. L/300 + D/50 D. L/500 + D/50

emin = L/500 + D/30 but not less than 20 mm.

Q39. In working stress method, the permissible tensile stress in HYSD bars (Fe 415) is: (SSC JE)

A. 140 N/mm² B. 230 N/mm² C. 0.55 fy D. 0.87 fy

Half of the guaranteed yield stress subject to maximum of 230 N/mm² for HYSD bars.

Q40. The percentage of steel for a balanced singly reinforced section (Fe 415, M20) is approximately: (SSC JE)

A. 0.96% B. 1.2% C. 1.5% D. 2.0%

pt,lim ≈ 0.96% for Fe 415 & M20.

Q41. If T = 100 kN and lever arm z = 0.4 m, the moment of resistance is: (SSC JE pattern)

A. 25 kNm B. 40 kNm C. 100 kNm D. 250 kNm

M = T × z = 100 × 0.4 = 40 kNm.

Q42. If overall depth D = 500 mm and effective cover = 50 mm, effective depth is: (SSC JE pattern)

A. 450 mm B. 500 mm C. 550 mm D. 400 mm

d = D − effective cover = 500 − 50 = 450 mm.

Q43. Design load = Characteristic load × : (SSC JE)

A. Partial safety factor B. Factor of safety C. Load factor only D. 1.0

Design load = Characteristic load × Partial safety factor for load.

Q44. In limit state method, the design strength of steel in tension is taken as: (SSC JE)

A. 0.87 fy B. 0.67 fy C. fy D. 0.55 fy

Design stress in steel = 0.87 fy.

Q45. The ratio of the depth of neutral axis to the effective depth for a balanced section with Fe 250 is: (SSC JE)

A. 0.53 B. 0.48 C. 0.46 D. 0.42

xu,max/d = 0.53 for Fe 250.

Q46. Bond between steel and concrete is mainly due to: (SSC JE)

A. Adhesion, friction and mechanical interlock B. Only adhesion C. Only friction D. Chemical reaction

Bond is due to adhesion, friction and mechanical interlock (especially for deformed bars).

Q47. A doubly reinforced beam has reinforcement in: (SSC JE)

A. Tension zone only B. Compression zone only C. Both tension and compression zones D. Neutral axis only

Doubly reinforced = steel in both tension and compression zones.

Q48. Effective depth is measured from: (SSC JE)

A. Compression face to centroid of tension steel B. Top of beam to bottom of beam C. Neutral axis to tension steel D. Cover to cover

Effective depth d = distance from extreme compression fibre to centroid of tension reinforcement.

Q49. Limit state of collapse primarily ensures: (SSC JE)

A. Structural safety against failure B. Comfort under service loads C. Aesthetic appearance D. Economy only

Collapse limit state → strength and safety against failure.

Q50. For MP Sub Engineer / SSC JE Civil, the most important combination for RCC preparation is: (Exam oriented)

A. LSM + Beam design + IS 456 provisions B. Only painting & finishing C. Only architectural drawings D. Only landscaping

Core topics: Limit State Method, singly/doubly reinforced beams, and important codal provisions of IS 456:2000.

Your Score

Last Minute Revision

• Minimum grade for RCC → M20
• Ec = 5000 √fck
• xu,max/d → 0.53 (Fe250), 0.48 (Fe415), 0.46 (Fe500)
• γm (concrete) = 1.5 | γm (steel) = 1.15
• Design stress in steel = 0.87 fy
• Min cover (moderate) = 30 mm | Footing = 50 mm
• Ld = ϕ × 0.87 fy / (4 τbd)
• Max Ast in beam = 4%
• Under-reinforced → steel yields first (ductile)