RCC & Prestressed Concrete – Complete Study Notes for GATE, ESE & SSC JE

🧱 RCC & Prestressed Concrete 🔥 Day 1

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RCC & Prestressed Concrete — Complete Study Notes

16 chapters · GATE, ESE (IES), SSC JE, State PSC, RRB JE

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RCC & Prestressed Concrete is the largest and highest-weightage design subject in civil engineering exams — from concrete/steel material behaviour and cement testing, through the Working Stress and Limit State design philosophies, singly/doubly reinforced beams, shear, bond, torsion, flanged beams, serviceability, two-way slabs, columns, foundations, prestressed concrete (IS 1343), earthquake-resistant design (IS 1893/IS 13920), and a masonry design appendix (IS 1905). Every formula, IS code clause, diagram, solved example and exam-pattern table is included.

Learning Objectives

After studying this chapter you will be able to:

Explain concrete and steel material behaviour (stress-strain, creep, modulus of elasticity) and the tests used to verify cement quality (IS 4031)
Apply both the Working Stress Method (transformed section) and the Limit State Method (IS 456 stress block) to design singly and doubly reinforced beam sections
Design RCC members for shear, bond/development length, torsion, and T/L flanged sections per IS 456
Check serviceability (deflection, crack width) and design one-way and two-way slabs
Design short and slender columns (axial, uniaxial and biaxial bending) and isolated/combined footings
Analyse prestressed concrete beams (pre-tensioning vs post-tensioning, losses, stress limits) per IS 1343
Apply earthquake-resistant design principles and IS 13920 ductile detailing requirements, and design load-bearing masonry per IS 1905

Exam Relevance

ESE★★★★★

GATE★★★★★

SSC JE★★★★★

State PSC★★★★★

💡 RCC design is the single highest-weightage subject across GATE Civil, ESE and SSC JE — LSM flexure design, shear design, and column design appear in nearly every paper, often worth 15–20+ marks combined.

Concept Map

Material Behaviour (Concrete, Steel, Cement Tests) → Design Philosophies (WSM, LSM)

↓

Singly/Doubly Reinforced Beams ↔ Shear ↔ Bond ↔ Torsion ↔ Flanged Beams

↓

Serviceability (Deflection, Cracking) → Two-Way Slabs

↓

Columns ↔ Foundations

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Prestressed Concrete → Earthquake-Resistant Design → Masonry Design (Appendix)

Where This Fits

Prerequisite: Theory of Structures (analysis results — BM, SF, axial force — are the direct inputs to every design calculation here) and Steel Structures (parallel design philosophy for the other major structural material). Leads to: Soil Mechanics, since foundation design (Ch 13) depends directly on soil bearing capacity.

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Ch 1 · Introduction to RCC

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1.1 Introduction

Reinforced Cement Concrete (RCC) is a composite material in which the relatively low tensile strength and ductility of concrete are compensated by the inclusion of reinforcement — steel bars, fibres, or meshes — which carry tensile forces. The two materials act together due to the excellent bond between them and a near-identical coefficient of thermal expansion (~12×10⁻⁶/°C for both).

⭐ Key premise: Concrete is strong in compression ($f_{ck}$) but weak in tension (≈10–15% of compressive strength). Steel is strong in both tension and compression. RCC combines them to carry complex loading.

1.2 Types of Concrete

Type	$f_{ck}$ Range	Key Characteristic
Plain Cement Concrete (PCC)	M5–M20	No reinforcement; good in compression only
Reinforced Concrete (RCC)	M20–M50	Steel bars embedded; resists all types of forces
Prestressed Concrete (PSC)	M40+	Pre-compressive stress applied via high-tensile steel
Fibre Reinforced Concrete	M30–M60	Random discrete fibres for ductility
High Performance Concrete	M60–M100+	Low w/c, SCMs, superior durability

1.3 Importance of Design Codes

IS 456:2000 — Plain and Reinforced Concrete, Code of Practice (primary RCC design code)
IS 1343:2012 — Prestressed Concrete Code of Practice
IS 875 (Parts 1–5) — Code of Practice for Design Loads
IS 1893:2016 — Earthquake Resistant Design
IS 13920:2016 — Ductile Detailing of RC Structures
Codes ensure safety, economy, serviceability, and uniformity across the profession

1.4 & 1.5 Characteristic Strength and Grades of Concrete

Characteristic Strength $f_{ck}$: strength below which not more than 5% of test results are expected to fall.

$$ f_{ck} = f_{mean} - 1.65\sigma $$

Target Mean Strength (Mix Design): $f_{target} = f_{ck} + 1.65\sigma$.

Grade	$f_{ck}$ (MPa)	Min. Use
M15	15	PCC, lean concrete
M20	20	Min. for RCC (mild exposure)
M25	25	Min. for moderate exposure
M30	30	Min. for severe exposure
M35	35	Min. for very severe exposure
M40	40	Min. for extreme exposure; PSC
M50–M100	50–100	HPC, PSC, special structures

1.8 Behaviour under Uniaxial Compression

The IS 456 idealized design stress-strain block for concrete rises parabolically to a strain of 0.002, then stays flat until the ultimate strain $\varepsilon_{cu} = 0.0035$ (see diagram tab). The actual curve softens beyond the peak stress instead of staying flat.

1.9 Behaviour in Tension

Concrete tensile strength ≈ 10–15% of compressive strength
Modulus of Rupture (flexural tension): $f_{r} = 0.7\sqrt{f_{ck}}$ MPa (IS 456)
Direct tensile strength: $f_t \approx 0.4$–$0.5\sqrt{f_{ck}}$ MPa
Split tensile strength: $f_{sp} = 2P/(\pi dL)$
Concrete cracks in tension at very small strains (~0.0001 to 0.0002)

1.10 Modulus of Elasticity & Poisson's Ratio

Short-term static $E_c = 5000\sqrt{f_{ck}}$ MPa (IS 456 Cl. 6.2.3.1). Long-term (with creep): $E_{c,eff} = E_c/(1+\theta)$. Poisson's ratio $\mu = 0.1$ to $0.2$ (IS 456 uses $\mu = 0.2$). Shear modulus $G = E_c/[2(1+\mu)] = E_c/2.4 \approx 0.4E_c$.

1.12 Creep of Concrete

Creep Coefficient $\theta$ = Creep Strain / Elastic Strain. IS 456 values: $\theta = 2.2$ (loading at 7 days), 1.6 (at 28 days), 1.1 (at 1 year). Effective Modulus: $E_{ce} = E_c/(1+\theta)$ — used for long-term deflection.

Factors increasing creep:

Higher w/c ratio, early loading, high temperature, low humidity, thin sections, higher stress/strength ratio, higher cement content, fine aggregate

1.13 Compressive Strength in Structural Design

IS 456 LSM: Design compressive strength $f_{cd} = 0.67f_{ck}/\gamma_m$. With $\gamma_m$ (concrete) = 1.5: $f_{cd} = 0.67f_{ck}/1.5 = 0.446f_{ck} \approx 0.45f_{ck}$. With $\gamma_m$ (steel Fe415) = 1.15: $f_{yd} = 0.87f_y = 0.87\times415 = 361$ MPa.

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Ch 1 · Formulas

Characteristic & Target Mean Strength

$$ f_{ck}=f_{mean}-1.65\sigma,\quad f_{target}=f_{ck}+1.65\sigma $$

σ = standard deviation from trial mixes

MPa

Modulus of Elasticity of Concrete

$$ E_c = 5000\sqrt{f_{ck}} $$

IS 456 Cl. 6.2.3.1; long-term: $E_{ce}=E_c/(1+\theta)$

MPa

Modulus of Rupture

$$ f_r = 0.7\sqrt{f_{ck}} $$

Flexural tensile strength used for cracking moment

MPa

Design Strengths (LSM)

$$ f_{cd}=0.45f_{ck},\quad f_{yd}=0.87f_y $$

$\gamma_m$: concrete = 1.5, steel = 1.15

MPa

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Ch 1 · Solved Example

GATE CE / ESE level

Find the modulus of elasticity, modulus of rupture, and design compressive strength $f_{cd}$ for M25 grade concrete.

Approach hint

Apply the three IS 456 formulas directly with $f_{ck}=25$ MPa.

$E_c = 5000\sqrt{25} = 5000\times5 = 25{,}000$ MPa
$f_r = 0.7\sqrt{25} = 0.7\times5 = 3.5$ MPa
$f_{cd} = 0.45\times25 = 11.25$ MPa

✓ $E_c = 25{,}000$ MPa; $f_r = 3.5$ MPa; $f_{cd} = 11.25$ MPa

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Ch 1 · Diagrams

Click any diagram to zoom

Fig 1.1 – Concrete stress-strain curve under uniaxial compression. IS 456 idealized design block: parabolic rise to 0.002 strain, then flat to ultimate strain $\varepsilon_{cu}$ = 0.0035.

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Ch 1 · Quick Quiz

3 questions · tap an option to reveal answer

Question 1

The ultimate compressive strain of concrete used in IS 456 design is:

The IS 456 design stress block reaches its flat plateau at strain 0.002 and terminates at the ultimate strain $\varepsilon_{cu}=0.0035$ ✓

Question 2

The modulus of elasticity of M30 concrete (IS 456) is closest to:

$E_c=5000\sqrt{30}=5000\times5.477\approx27{,}386$ MPa ✓

Question 3

Which factor does NOT increase creep in concrete?

Low humidity increases creep (faster moisture loss); high humidity actually reduces creep — the other three options all increase it ✓

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Ch 2 · Basic Design Concepts

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2.1 Introduction

Structural design is the process of proportioning structural members so that they safely and economically resist applied loads throughout their intended service life. In RCC design, two materials — concrete and steel — act compositely; the designer must understand the properties of both, the nature of applied loads, and the design philosophy adopted by the governing code (IS 456:2000 for India).

Objective of design: Achieve adequate safety (no collapse), serviceability (no excessive deflection or cracking), and durability (service life ≥ design life, typically 50–100 years)
Design process: Determine loads → structural analysis (BM, SF, axial forces) → design of sections → detailing → check serviceability
IS 456:2000 is the primary Indian code; it follows the Limit State Method (LSM) based on probabilistic concepts and partial safety factors
Design must comply with IS 456, IS 875 (loads), IS 1893 (seismic), IS 13920 (ductile detailing) and relevant material codes

💡 The designer's primary responsibility is to ensure that the structure performs its intended function safely and economically — not just to pass the numbers. Sound engineering judgement, knowledge of material behaviour, construction practice and failure modes is essential.

2.2 Necessity of Designing RCC Structures

RCC structures must be designed — not guessed or over-built arbitrarily — for the following reasons:

Reason	Explanation
Safety	Undesigned structures may collapse under loads they were never checked against — axial, flexural, shear, torsion, earthquake, wind
Economy	Over-design wastes material and cost; under-design causes failure. Rational design finds the optimum
Serviceability	Even if a member doesn't collapse, excessive deflection, cracking or vibration renders it unusable. Design controls these
Durability	Design specifies minimum cover, w/c ratio, cement content — without which steel corrodes and concrete deteriorates prematurely
Legal & Code Compliance	Building bye-laws and IS codes make structural design mandatory. A structure without a design basis cannot receive a completion certificate
Predictable Behaviour	Design provides a known, predictable failure mode (ductile, not sudden). Under-reinforced beams give warning before collapse
Sustainability	Properly designed structures last their full design life (50–100 years), reducing material consumption and environmental impact over time

Why RCC specifically needs design (unlike timber or masonry by rule-of-thumb):

RCC is a composite material — the interaction of concrete and steel must be designed, not assumed
Concrete is brittle in tension; steel is ductile — a wrongly proportioned section may fail without warning
The quantity and arrangement of steel reinforcement directly controls the mode and load of failure
High variability in concrete strength (site conditions, w/c ratio, curing) makes statistical design (LSM with partial factors) essential

2.3 Hydraulic & Non-Hydraulic Cements

Non-Hydraulic Cements

Non-hydraulic cements do not harden by chemical reaction with water alone — they require air (CO₂ or O₂) to set and harden. They cannot be used underwater and lose strength if kept wet.

Type	Main Constituent	Setting Mechanism	Use
Lime (Fat/Rich Lime)	Calcium oxide CaO (from pure limestone)	Carbonation: Ca(OH)₂ + CO₂ → CaCO₃ (slow, needs air)	Plastering, lime-wash, pointing in non-structural work
Gypsum Plaster	CaSO₄·½H₂O (hemihydrate)	Rehydration: CaSO₄·½H₂O + 1½H₂O → CaSO₄·2H₂O (sets rapidly in air)	Internal plastering, partition boards, ornamental work
Oxychloride Cement (Sorel)	MgO + MgCl₂ solution	Chemical reaction forming magnesium oxychloride	Industrial flooring, grinding wheels (not for wet areas)

Hydraulic Cements

Hydraulic cements harden by chemical reaction with water (hydration) — independent of air. They set and gain strength even underwater. Portland cement is the most important hydraulic cement.

Type	IS Code	Key Feature	Structural Use
OPC 33 Grade	IS 269:2015	28-day strength ≥ 33 MPa; moderate early strength	General construction, plastering, mortar
OPC 43 Grade	IS 8112:2013	28-day strength ≥ 43 MPa; most widely used	RCC slabs, beams, columns — standard grade
OPC 53 Grade	IS 12269:2013	28-day strength ≥ 53 MPa; high early strength	Prestressed concrete, precast, high-strength mixes
PPC	IS 1489 Pt 1	Fly ash blended (15–35%); lower heat; better long-term durability	Mass concrete, foundations, marine, canals
PSC (PBFSC)	IS 455:2015	GGBS blended (25–70%); high sulphate resistance	Marine, sewage, sulphate-bearing soils
SRC	IS 12330	Very low C₃A (<5%); resists sulphate attack	Foundations in aggressive sulphate soils
Low Heat Cement	IS 12600	Low C₃S + C₃A content; max heat 271 kJ/kg at 7d	Gravity dams, massive raft foundations
Rapid Hardening (RHPC)	IS 8041	Higher C₃S; finer grinding; fast early strength	Cold weather concreting, emergency repairs, precast
High Alumina Cement (HAC)	IS 6452	Al₂O₃ > 32%; rapid strength, acid & heat resistant	Refractory linings, chemical plants (NOT structural RCC)
White Portland Cement	IS 8042	Very low Fe₂O₃ (<0.5%); white colour	Architectural, decorative, terrazzo flooring

⚠️ Hydraulic Lime occupies an intermediate position — it contains silica/clay impurities that react with water, giving it partial hydraulic properties. Used in lime-pozzolana mixes and traditional construction. It is not suitable as a sole binder for structural RCC.

2.4 Tests on Cement

All tests on cement are governed by IS 4031 (Parts 1–15). The following tests are conducted to verify the quality of cement before use in structural concrete.

① Fineness Test (IS 4031 Pt 1 & Pt 2)

Method	Apparatus	What it Measures	IS Requirement
Sieve Test	90 µm IS sieve	% residue on 90 µm sieve	OPC: ≤ 10% residue (IS 269); ≤ 5% on 45 µm for OPC 53
Blaine's Air Permeability	Blaine apparatus	Specific surface area (m²/kg)	OPC 43/53: ≥ 225 m²/kg (typically 300–350 m²/kg in practice)

💡 Finer cement → faster hydration → higher early strength, more heat, more shrinkage. Coarser cement → slower hydration → better for mass concrete (low heat). Fineness directly controls rate of strength gain, not the ultimate strength.

② Consistency Test — Normal / Standard Consistency (IS 4031 Pt 4)

Purpose: Determine the water content (%) needed to produce a cement paste of standard consistency — required before conducting setting time and soundness tests
Apparatus: Vicat apparatus with plunger (10 mm dia, flat end)
Procedure: Prepare paste with trial water content; fill Vicat mould; lower plunger gently; record penetration depth
Standard consistency (P): When plunger penetrates to 5–7 mm from bottom (33–35 mm from top of 40 mm mould)
Typical P value: 26–33% by mass of cement for OPC
All subsequent tests (IST, FST, soundness autoclave) use paste made at this normal consistency

③ Setting Time Test (IS 4031 Pt 5)

Parameter	Apparatus / Needle	End-Point	IS Requirement (OPC)
Initial Setting Time (IST)	Vicat needle: 1 mm sq., 50 mm long	Needle penetrates to 5–7 mm from bottom (33–35 mm mark)	IST ≥ 30 minutes
Final Setting Time (FST)	Vicat needle with annular attachment (5 mm dia, 0.5 mm projection)	Needle makes impression but annular attachment does not	FST ≤ 600 minutes (10 hours)

⚠️ Flash Set (without gypsum): immediate, irreversible stiffening due to rapid C₃A hydration — no workability regained. False Set (excess gypsum): premature stiffening that reverses on re-mixing — workability is regained. IST must be ≥ 30 min to allow sufficient time for transport, placing and compaction.

④ Soundness Test (IS 4031 Pt 3)

Tests for expansion due to excess free lime (CaO) or magnesia (MgO) which cause delayed, disruptive expansion after concrete hardens — making the cement unsound.

Test	Detects	Apparatus	IS Limit
Le Chatelier Test	Excess free CaO	Le Chatelier mould; immerse 24h at 27±2°C, boil 3h, measure needle separation	Expansion ≤ 10 mm (OPC)
Autoclave Test	Excess MgO (periclase)	Autoclave (steam pressure vessel, 2.1 MPa) for 3h	Expansion ≤ 0.8%

⑤ Strength Test (IS 4031 Pt 6 & IS 10080)

Test	Specimen	Mix	Strength Requirement
Compressive Strength	70.6 mm cube (or 50 mm)	Cement : Standard sand = 1:3; w/c = 0.40	OPC 43: 3d ≥ 23 MPa; 7d ≥ 33 MPa; 28d ≥ 43 MPa
Tensile (old method)	Briquette / cylinder	Neat cement paste	Largely replaced by compressive strength test

⑥ Heat of Hydration Test (IS 4031 Pt 9)

Measures total heat released during hydration (J/g)
Method: Solution calorimeter or conduction calorimeter
Significance: High heat → thermal cracking in mass concrete; Low Heat Cement preferred for dams and massive structures
Limits (IS 12600 — Low Heat Cement): ≤ 271 kJ/kg at 7 days; ≤ 314 kJ/kg at 28 days

⑦ Specific Gravity of Cement (IS 4031 Pt 11)

Apparatus: Le Chatelier flask (kerosene as liquid — does not react with cement). Specific Gravity of OPC = 3.10–3.15 (typically 3.12–3.15). PSC (with GGBS): ~2.90 | PPC (with fly ash): ~2.85–2.90. HAC: ~3.20 | White cement: ~3.05.

Summary Table – All Cement Tests

Test	IS Code	Property Measured	Key Limit (OPC)
Fineness (Sieve)	IS 4031 Pt 1	% residue on 90 µm sieve	≤ 10%
Fineness (Blaine)	IS 4031 Pt 2	Specific surface (m²/kg)	≥ 225 m²/kg
Normal Consistency	IS 4031 Pt 4	Water for std. paste (P%)	Penetration 5–7 mm from bottom
Initial Setting Time	IS 4031 Pt 5	Time to IST	≥ 30 minutes
Final Setting Time	IS 4031 Pt 5	Time to FST	≤ 600 minutes
Soundness (Le Chatelier)	IS 4031 Pt 3	Expansion due to free CaO	≤ 10 mm
Soundness (Autoclave)	IS 4031 Pt 3	Expansion due to MgO	≤ 0.8%
Compressive Strength	IS 4031 Pt 6	28-day mortar cube strength	OPC 43: ≥ 43 MPa; OPC 53: ≥ 53 MPa
Heat of Hydration	IS 4031 Pt 9	Heat released (kJ/kg)	LHC: ≤ 271 kJ/kg at 7d
Specific Gravity	IS 4031 Pt 11	Density ratio	OPC: 3.10–3.15

⭐ Exam Tip: Most frequently asked values — IST ≥ 30 min, FST ≤ 600 min, Le Chatelier expansion ≤ 10 mm, Autoclave ≤ 0.8%, Specific gravity 3.15, Blaine ≥ 225 m²/kg. These appear in GATE, ESE and SSC JE every year.

2.5 Durability – Chemical Attack

Exposure Class	Min. Grade	Max. w/c	Min. Cement (kg/m³)	Min. Cover (mm)
Mild	M20	0.55	300	20
Moderate	M25	0.50	300	30
Severe	M30	0.45	320	45
Very Severe	M35	0.45	340	50
Extreme	M40	0.40	360	75

2.6 Design Philosophies

Method	Full Name	Basis	IS Code Era
WSM	Working Stress Method	Elastic theory; linear stress-strain; stresses must not exceed permissible values under working loads	Older; IS 456:1978 & earlier
LSM	Limit State Method	Statistical; partial safety factors applied to loads and material strengths; considers actual failure modes	IS 456:2000 (current)
ULM	Ultimate Load Method	Plastic theory; find ultimate load; divide by FOS for design load	Intermediate; not in IS 456:2000

2.7 & 2.8 Limit State Method – Partial Safety Factors (IS 456:2000)

Partial Safety Factors for Materials ($\gamma_m$)

Material	LSM $\gamma_m$	Design Strength
Concrete	1.5	$f_{cd}$ = 0.67 $f_{ck}$/1.5 = 0.45 $f_{ck}$
Steel (Fe250, Fe415, Fe500)	1.15	$f_{yd} = f_y$/1.15 = 0.87 $f_y$

Partial Safety Factors for Loads ($\gamma_f$) – IS 456 Table 18

Load Combination	DL	LL	Wind/Seismic
DL + LL	1.5	1.5	–
DL + LL + Wind	1.2	1.2	1.2
DL + Wind only	1.5 (or 0.9)	–	1.5
DL + LL + Seismic (IS 1893)	1.2	1.2	1.2

2.10 Properties of Reinforcement

Grade	$f_y$ (MPa)	Type	Ductility	Use
Fe250 (Mild Steel)	250	Plain round bars	Very High	Links, stirrups; seismic zones
Fe415 (TMT/HYSD)	415	Deformed (ribbed)	High	Most common for RCC
Fe500	500	Deformed (TMT)	Moderate-High	Beams, columns, slabs
Fe550	550	Deformed (TMT)	Moderate	Heavy structures
Fe500D / Fe550D	500/550	D = enhanced ductility	High	Seismic zones (IS 13920)

$E_s = 2\times10^5$ MPa (200 GPa) for all grades of steel (IS 456 Cl. 6.2.4). Yield strain (Fe415): $\varepsilon_y = f_y/E_s + 0.002 = 415/200000 + 0.002 = 0.00208 + 0.002 = 0.00383$. Design yield strength: $f_{yd} = 0.87\times415 = 361.05$ MPa.

2.11 Cover Requirements (IS 456:2000 Cl. 26.4)

Member	Min. Nominal Cover (mild exposure)
Slabs	20 mm
Beams	25 mm
Columns	40 mm
Footings	50 mm (75 mm against earth)
Prestressed members	20 mm (min.)

⚠️ Nominal cover ≥ bar diameter. For severe/very severe/extreme exposure, follow IS 456 Table 16A (increase cover by 10–15 mm). Fire resistance also governs minimum cover (IS 456 Table 16).

2.12 Spacing of Reinforcement (IS 456 Cl. 26.3)

Min. horizontal clear distance: max (bar dia, MSA + 5 mm, 25 mm)
Min. vertical clear distance: 15 mm, or 2/3 × max aggregate size, whichever is greater
Maximum bar spacing in slabs: 3d or 300 mm (whichever less) for main bars; 5d or 450 mm for distribution bars

2.14 Major Causes of Structural Failure

Design errors: incorrect load assumptions, wrong design philosophy, arithmetical mistakes
Construction defects: inadequate cover, poor concrete quality, insufficient curing, wrong bar placement
Material failures: substandard reinforcement (low ductility), poor aggregate quality
Overloading beyond design assumptions
Environmental attack: corrosion, sulphate attack, freeze-thaw, carbonation
Foundation settlement, seismic forces

2.15 List of Key IS Codes for RCC

IS Code	Subject
IS 456:2000	Plain & Reinforced Concrete (main design code)
IS 1343:2012	Prestressed Concrete
IS 875 Pt 1/2/3/5	Dead / Imposed / Wind Loads / Special Load Combinations
IS 1893:2016	Earthquake Resistant Design (Criteria)
IS 13920:2016	Ductile Detailing of RC Structures
IS 2911	Design & Construction of Pile Foundations
IS 1080	Design of Shallow Foundations
SP 16:1980	Design Aids for Reinforced Concrete
SP 34:1987	Handbook on Concrete Reinforcement & Detailing

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Ch 2 · Formulas

Partial Safety Factors — LSM

$$ \gamma_m(\text{concrete})=1.5,\quad \gamma_m(\text{steel})=1.15 $$

DL+LL load factor = 1.5 each (IS 456 Table 18)

—

Reinforcement Modulus & Yield Strain

$$ E_s=2\times10^5\ \text{MPa},\quad \varepsilon_y=\frac{f_y}{E_s}+0.002 $$

Fe415: $\varepsilon_y=0.00383$; $f_{yd}=0.87f_y=361$ MPa

—

OPC Strength Requirements

$$ \text{OPC43: 28d} \ge 43\ \text{MPa} $$

OPC33: 33 MPa; OPC53: 53 MPa (28-day mortar cube)

MPa

Minimum Nominal Cover

$$ \text{Slab:20, Beam:25, Column:40, Footing:50 mm} $$

Nominal cover ≥ bar diameter always

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Ch 2 · Solved Example

GATE CE / ESE level

In a Le Chatelier soundness test, the indicator needle separation before boiling is 6 mm, and after 3 hours of boiling it is 15 mm. Is the cement sound per IS 4031 Pt 3?

Approach hint

Compute the expansion as the difference between the two readings, then compare against the IS limit of 10 mm.

Expansion $= d_2 - d_1 = 15 - 6 = 9$ mm
IS 4031 Pt 3 limit for OPC: expansion ≤ 10 mm
9 mm ≤ 10 mm → within limit

✓ The cement is sound (expansion 9 mm ≤ 10 mm limit)

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Ch 2 · Diagrams

Click any diagram to zoom

Fig 2.1 – Classification of cements. Hydraulic cements harden with water (even underwater); non-hydraulic require air. Portland cement is the most important hydraulic cement for structural use.

Fig 2.2 – Vicat apparatus for setting time test. IST: needle penetrates to 5–7 mm from bottom. FST: annular attachment (1 mm projection) makes mark but does not pierce surface.

Fig 2.3 – Le Chatelier test: indicator needles spread apart as free lime expands on boiling. Expansion (d₂ − d₁) ≤ 10 mm → sound cement. > 10 mm → unsound, reject.

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Ch 2 · Quick Quiz

4 questions · tap an option to reveal answer

Question 1

The minimum Initial Setting Time (IST) required for OPC per IS 4031 is:

IST ≥ 30 minutes ensures enough time for transport, placing and compaction before the cement begins to stiffen ✓

Question 2

Which cement type does NOT harden by reaction with water, requiring air (CO₂) instead?

Fat (rich) lime is a non-hydraulic binder — it hardens by carbonation (reaction with atmospheric CO₂), not by hydration, and cannot be used underwater ✓

Question 3

The minimum nominal cover for a column under mild exposure (IS 456) is:

Columns require 40 mm minimum cover under mild exposure — more than slabs (20 mm) or beams (25 mm) since columns carry critical axial loads and are more exposed ✓

Question 4

The partial safety factor $\gamma_m$ for steel in the Limit State Method is:

$\gamma_m$(steel) = 1.15, giving $f_{yd}=f_y/1.15=0.87f_y$; concrete uses $\gamma_m=1.5$ ✓

⚖️

Ch 3 · Working Stress Method (WSM)

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3.1 Introduction

WSM (also called Elastic Method) assumes both concrete and steel remain within elastic limits under working loads. The allowable stresses are fractions of ultimate strengths, providing an implicit factor of safety.

Concrete and steel both linear-elastic; plane sections remain plane (Bernoulli's hypothesis)
Tensile stress in concrete is neglected (concrete below NA is cracked)
Perfect bond between steel and concrete (no slip)
Strain compatibility: steel strain = concrete strain at same level

3.5 Transformed Section & 3.6 Modular Ratio

Modular Ratio $m = E_s/E_c$. IS 456 (WSM): $m = 280/(3\sigma_{cbc})$, where $\sigma_{cbc}$ = permissible bending stress in concrete. For M20: $m = 280/(3\times7) = 13.33$. For M25: $m = 280/(3\times8.5) = 10.98 \approx 11$.

Transformed area of tension steel = $m \times A_{st}$. Transformed area of compression steel = $(1.5m-1)\times A_{sc}$ [subtract concrete already accounted for].

3.6 Permissible Stresses in WSM (IS 456 Table B-1 & B-2)

Grade	$\sigma_{cbc}$ (MPa) bending	$\sigma_{cc}$ (MPa) direct compression
M15	5.0	4.0
M20	7.0	5.0
M25	8.5	6.0
M30	10.0	8.0
M35	11.5	9.0
M40	13.0	10.0

Steel Grade	$\sigma_{st}$ tension (MPa)	$\sigma_{sc}$ compression (MPa)
Fe250 (mild)	140	130
Fe415 (HYSD/deformed)	230	190
Fe500 (HYSD)	275	190

3.3 Transformed Section Analysis (Key Equations)

NA depth (x) is found by equating moments of areas about the NA: $b\cdot x\cdot(x/2) = m\cdot A_{st}\cdot(d-x)$, i.e. $bx^2/2 = m\cdot A_{st}\cdot(d-x)$.

Moment of Inertia (cracked): $I_{cr} = bx^3/3 + m\cdot A_{st}\cdot(d-x)^2$. Moment of resistance: $M = \sigma_{cbc}\cdot I_{cr}/x = \sigma_{st}\cdot I_{cr}/(m\cdot(d-x))$. Lever arm: $z = d - x/3$. Critical NA depth (balanced): $x_c/d = m\sigma_{cbc}/(m\sigma_{cbc}+\sigma_{st})$.

3.9 Cracking Moment

$M_{cr} = f_r\cdot I_g/y_t$, where $f_r = 0.7\sqrt{f_{ck}}$ MPa (modulus of rupture, IS 456), $I_g$ = gross moment of inertia (uncracked section), and $y_t$ = distance from NA to tension fibre (= D/2 for a symmetric section).

⚖️

Ch 3 · Formulas

Modular Ratio

$$ m = \frac{E_s}{E_c} = \frac{280}{3\sigma_{cbc}} $$

M20: m=13.33; M25: m≈11

—

Neutral Axis Depth (Cracked Elastic)

$$ \frac{bx^2}{2} = mA_{st}(d-x) $$

Equates moments of transformed areas about NA

Cracked Moment of Inertia & Lever Arm

$$ I_{cr}=\frac{bx^3}{3}+mA_{st}(d-x)^2,\quad z=d-\frac{x}{3} $$

Used to find moment of resistance

mm⁴, mm

Cracking Moment

$$ M_{cr}=\frac{f_r I_g}{y_t},\quad f_r=0.7\sqrt{f_{ck}} $$

$y_t=D/2$ for symmetric section

N·mm

⚖️

Ch 3 · Solved Example

GATE CE / ESE level

A singly reinforced beam of width b = 250 mm and effective depth d = 450 mm uses M20 concrete and Fe415 steel with $A_{st} = 1500$ mm². Using WSM with m = 13.33, find the neutral axis depth x (solve the quadratic).

Approach hint

Set up $bx^2/2 = mA_{st}(d-x)$, substitute values, and solve the resulting quadratic in x.

$250x^2/2 = 13.33\times1500\times(450-x)$
$125x^2 = 19{,}995(450-x) = 8{,}997{,}750 - 19{,}995x$
$125x^2 + 19{,}995x - 8{,}997{,}750 = 0$
Solving the quadratic: $x \approx 199$ mm

✓ Neutral axis depth x ≈ 199 mm

⚖️

Ch 3 · Diagrams

Click any diagram to zoom

Fig 3.1 – WSM singly reinforced beam: linear strain distribution; triangular compression stress block; tensile concrete stress neglected; steel carries all tension.

⚖️

Ch 3 · Quick Quiz

3 questions · tap an option to reveal answer

Question 1

In WSM analysis, the tensile stress in concrete below the neutral axis is:

WSM assumes concrete below the NA is cracked and carries no tension — all tensile force is carried by the steel reinforcement ✓

Question 2

The modular ratio m for M20 concrete under WSM (IS 456) is approximately:

$m = 280/(3\sigma_{cbc}) = 280/(3\times7) = 13.33$ for M20 ($\sigma_{cbc}=7$ MPa) ✓

Question 3

The cracking moment $M_{cr}$ marks the transition between:

Below $M_{cr}$ the full gross (uncracked) section resists the load; above $M_{cr}$, concrete cracks in tension and the cracked transformed section must be used ✓

📐

Ch 4 · Limit State Method (LSM)

○ Not Started

4.1 Introduction – Limit States

Limit State	Type	Concerns
Ultimate Limit State (ULS)	Safety	Collapse, yielding, buckling, overturning, fatigue; design loads = factored loads (1.5 DL + 1.5 LL)
Serviceability Limit State (SLS)	Functional	Excessive deflection (span/250 or span/350 after construction), cracking ($w_{max}$ = 0.3 mm severe), vibration
Durability Limit State	Longevity	Minimum cover, w/c ratio, cement content for given exposure

4.3 Assumptions in LSM (IS 456 Cl. 38.1)

Plane sections remain plane after bending (Bernoulli hypothesis)
Maximum compressive strain in concrete at ultimate = 0.0035
Tensile strength of concrete is ignored
Stress distribution in compression zone: IS 456 parabolic-rectangular block (peak = 0.67 $f_{ck}$/1.5 = 0.446 $f_{ck}$)
Maximum strain in tension steel ≥ $f_y$/(1.15 $E_s$) + 0.002 (ensures yielding before failure)
Perfect bond; no slip between steel and concrete

4.4 Singly Reinforced Sections – Key Formulae

Design Compressive Force: $C = 0.36f_{ck}\cdot b\cdot x_u$. Design Tensile Force: $T = 0.87f_y\cdot A_{st}$. Equilibrium (C = T): $x_u = 0.87f_yA_{st}/(0.36f_ckb)$. Lever arm: $z = d - 0.42x_u$. Moment of Resistance: $M_u = 0.36f_{ck}bx_u(d-0.42x_u)$, or equivalently $M_u = 0.87f_yA_{st}(d-0.42x_u)$.

4.5 Limiting (Balanced) Neutral Axis Depth (IS 456 Cl. G-1.1)

$x_{u,max}/d$ depends on steel grade (from strain compatibility, $\varepsilon_{cu}=0.0035$): Fe250: 0.53. Fe415: 0.48. Fe500: 0.46. Fe550: 0.44.

If $x_u < x_{u,max}$ → Under-reinforced (preferred — steel yields first → ductile failure). If $x_u = x_{u,max}$ → Balanced section. If $x_u > x_{u,max}$ → Over-reinforced (brittle failure — NOT permitted per IS 456).

4.5 Limiting Moment of Resistance ($M_{u,lim}$)

$M_{u,lim} = 0.36(x_{u,max}/d)[1-0.42(x_{u,max}/d)]f_{ck}bd^2$. Fe415: $M_{u,lim} = 0.138f_{ck}bd^2$ [very frequently asked in GATE]. Fe250: $0.148f_{ck}bd^2$. Fe500: $0.133f_{ck}bd^2$. Fe550: $0.128f_{ck}bd^2$.

⭐ Most frequently asked formula in GATE: $M_{u,lim}$ = 0.138 $f_{ck}$ b d² for Fe415. Memorise this.

4.6 Deflection Control – Span/Depth Ratio (IS 456 Cl. 23.2)

Support Condition	Basic l/d Ratio
Simply Supported	20
Continuous	26
Cantilever	7

Modified l/d = Basic l/d × Modification Factor ($MF_{tension}$) × Modification Factor ($MF_{comp}$). MF for tension steel: from IS 456 Fig. 4 (function of $f_s = 0.58f_y \times (A_{st,req}/A_{st,prov})$ and $p_t$). For cantilevers > 10 m: deflection calculation is mandatory (not the l/d approach).

4.7–4.8 Minimum & Maximum Reinforcement Limits (IS 456)

Parameter	Beams	Slabs
Min. $A_{st}$	0.85 bd / $f_y$ (IS 456 Cl. 26.5.1.1)	0.12% of total cross-section (HYSD); 0.15% (mild steel)
Max. $A_{st}$	4% of gross cross-sectional area (tension + compression combined)	As for beams

4.12 One-Way Slabs

When $l_y/l_x > 2$: load carried primarily in shorter direction (one-way action)
Main reinforcement in short span direction; distribution reinforcement in long span direction
Design as singly reinforced beam of unit width (b = 1000 mm)
Distribution steel: ≥ 0.12% (HYSD) or 0.15% (mild steel) of cross-sectional area
Max. bar spacing: main = 3d or 300 mm (lesser); distribution = 5d or 450 mm (lesser)

📐

Ch 4 · Formulas

Moment of Resistance (Singly Reinforced)

$$ M_u = 0.36f_{ck}\,b\,x_u\,(d-0.42x_u) $$

Equivalently $M_u=0.87f_yA_{st}(d-0.42x_u)$

N·mm

Neutral Axis Depth (LSM)

$$ x_u=\frac{0.87f_yA_{st}}{0.36f_{ck}b} $$

From C = T equilibrium

Limiting Moment of Resistance (Fe415)

$$ M_{u,lim} = 0.138\,f_{ck}\,b\,d^2 $$

Fe250: 0.148; Fe500: 0.133; Fe550: 0.128 (× $f_{ck}bd^2$)

N·mm

Min/Max Reinforcement

$$ A_{st,min}=\frac{0.85bd}{f_y},\quad A_{st,max}=0.04A_g $$

Beams (IS 456 Cl. 26.5.1.1)

mm²

📐

Ch 4 · Solved Example

GATE CE / ESE level

A rectangular beam of width b = 300 mm and effective depth d = 500 mm uses M20 concrete and Fe415 steel. Find the limiting moment of resistance $M_{u,lim}$.

Approach hint

Apply the Fe415 coefficient directly: $M_{u,lim}=0.138f_{ck}bd^2$.

$M_{u,lim} = 0.138\times20\times300\times500^2$
$= 0.138\times20\times300\times250{,}000$
$= 207{,}000{,}000$ N·mm $= 207$ kN·m

✓ $M_{u,lim} = 207$ kN·m — if the applied $M_u$ exceeds this, doubly reinforced design is required

📐

Ch 4 · Diagrams

Click any diagram to zoom

Fig 4.1 – LSM stress block at ULS. Compression = 0.36 $f_{ck}$·b·$x_u$ at 0.42$x_u$ from top; Tension = 0.87 $f_y·A_{st}$. Moment of resistance = C × z = T × z.

📐

Ch 4 · Quick Quiz

3 questions · tap an option to reveal answer

Question 1

If $x_u > x_{u,max}$ for a beam section, the section is classified as:

$x_u>x_{u,max}$ means concrete would crush before steel yields — a sudden, brittle failure with no warning, which IS 456 does not permit ✓

Question 2

The limiting NA depth ratio $x_{u,max}/d$ for Fe415 steel is:

Fe415 has $x_{u,max}/d=0.48$ — higher-grade steel has a progressively lower limiting NA depth ratio (0.53 → 0.44 as grade increases) ✓

Question 3

The basic span/effective-depth (l/d) ratio for a cantilever beam (IS 456 Cl. 23.2) is:

Cantilever = 7 (most restrictive since deflection at the free end is largest); simply supported = 20; continuous = 26 ✓

🔗

Ch 5 · Doubly Reinforced Beams (LSM)

○ Not Started

5.1 Introduction

A doubly reinforced beam has steel in both the compression zone ($A_{sc}$) and tension zone ($A_{st}$). Required when $M_u > M_{u,lim}$ and section size cannot be increased, or when the beam is subject to reversal of bending.

5.4 Stress in Compression Steel ($f_{sc}$)

Strain in compression steel: $\varepsilon_{sc} = 0.0035\times(x_{u,max}-d')/x_{u,max}$. The d'/d ratio determines $f_{sc}$ (IS 456 / SP-16 Table F): d'/d = 0.05 → $f_{sc}$ = 354.8 MPa (Fe415); d'/d = 0.10 → 351.8; d'/d = 0.15 → 342.4; d'/d = 0.20 → 329.4.

5.7–5.8 Design Steps for Doubly Reinforced Beam

Compute $M_{u,lim}$ for the given section (using the limiting NA depth for the steel grade)
If $M_u > M_{u,lim}$: $M_{u2} = M_u - M_{u,lim}$
Determine the d'/d ratio; find $f_{sc}$ from the IS 456 table or strain compatibility
$A_{sc} = M_{u2}/[f_{sc}\times(d-d')]$
$A_{st,lim} = 0.36f_{ck}bx_{u,max}/(0.87f_y)$
$A_{st2} = f_{sc}\times A_{sc}/(0.87f_y)$
$A_{st} = A_{st,lim} + A_{st2}$

🔗

Ch 5 · Formulas

Additional Moment Carried by Comp. Steel

$$ M_{u2} = M_u - M_{u,lim} $$

Only when $M_u > M_{u,lim}$ (else singly reinforced suffices)

N·mm

Compression Steel Area

$$ A_{sc} = \frac{M_{u2}}{f_{sc}(d-d')} $$

$f_{sc}$ from strain compatibility or SP-16 Table F

mm²

Additional Tension Steel

$$ A_{st2} = \frac{f_{sc}A_{sc}}{0.87f_y} $$

$A_{st}=A_{st,lim}+A_{st2}$

mm²

🔗

Ch 5 · Solved Example

GATE CE / ESE level

A beam needs $M_u = 260$ kN·m but $M_{u,lim} = 207$ kN·m for the given section (d − d' = 450 mm, $f_{sc} = 353$ MPa). Find the required compression steel area $A_{sc}$.

Approach hint

First find $M_{u2}=M_u-M_{u,lim}$, then apply $A_{sc}=M_{u2}/[f_{sc}(d-d')]$.

$M_{u2} = 260 - 207 = 53$ kN·m $= 53\times10^6$ N·mm
$A_{sc} = 53\times10^6/(353\times450) = 53\times10^6/158{,}850 \approx 334$ mm²

✓ $A_{sc} \approx 334$ mm² of compression steel required

🔗

Ch 5 · Diagrams

Click any diagram to zoom

Fig 5.1 – Doubly reinforced beam design. Total $M_u = M_{u,lim}$ (singly) + $M_{u2}$ (from comp. steel). $A_{sc}$ and additional $A_{st2}$ carry $M_{u2}$.

🔗

Ch 5 · Quick Quiz

3 questions · tap an option to reveal answer

Question 1

A doubly reinforced beam is needed when:

When the applied moment exceeds the singly-reinforced section's limiting capacity and the section cannot be enlarged, compression steel is added to carry the excess moment $M_{u2}$ ✓

Question 2

In a doubly reinforced beam, the additional tension steel $A_{st2}$ is required to:

For force equilibrium, the extra compression force $f_{sc}A_{sc}$ provided by the compression steel must be balanced by an equal additional tension force, requiring $A_{st2}=f_{sc}A_{sc}/(0.87f_y)$ ✓

Question 3

As the d'/d ratio (cover-to-compression-steel ratio) increases, the stress $f_{sc}$ in the compression steel:

As d'/d increases from 0.05 to 0.20, $f_{sc}$ decreases from 354.8 to 329.4 MPa (Fe415) — the compression steel sits closer to the neutral axis, reducing its strain and hence its stress ✓

✂️

Ch 6 · Design for Shear in RCC

○ Not Started

6.6 Nominal Shear Stress (IS 456 Cl. 40.1)

$\tau_v = V_u/(b\cdot d)$ — nominal shear stress at ULS, where $V_u$ = factored shear force at the critical section. Critical section for shear: 'd' from the face of the support (for beams with vertical loads).

6.8 Shear Strength without Reinforcement ($\tau_c$) – IS 456 Table 19

Design shear strength of concrete $\tau_c$ depends on grade of concrete and % tension reinforcement ($p_t = 100A_{st}/(bd)$).

$p_t$ (%)	M20 $\tau_c$ (MPa)	M25 $\tau_c$ (MPa)	M30 $\tau_c$ (MPa)
0.15	0.28	0.29	0.29
0.25	0.36	0.36	0.37
0.50	0.48	0.49	0.50
0.75	0.56	0.57	0.59
1.00	0.62	0.64	0.66
1.25	0.67	0.70	0.71
1.50	0.72	0.74	0.76
2.00	0.79	0.82	0.84
≥ 3.00	0.82	0.85	0.88

Maximum shear stress $\tau_{c,max}$ (IS 456 Table 20): M20 = 2.8, M25 = 3.1, M30 = 3.5, M35 = 3.7, M40 = 4.0 MPa.

6.9 Design of Shear Reinforcement

Case 1: $\tau_v \le \tau_c$ → No shear reinforcement required (but provide minimum). Case 2: $\tau_c < \tau_v \le \tau_{c,max}$ → Design shear reinforcement. Case 3: $\tau_v > \tau_{c,max}$ → Redesign section (increase b or d).

$Vus = V_u - \tau_c\cdot b\cdot d$ — shear to be carried by reinforcement. For vertical stirrups: $Vus = 0.87f_yA_{sv}d/S_v$, so $S_v = 0.87f_yA_{sv}d/Vus$.

6.10 Minimum Shear Reinforcement (IS 456 Cl. 26.5.1.6)

$A_{sv}/(b\cdot S_v) \ge 0.4/(0.87f_y)$. For Fe415: min. $A_{sv}/(b\cdot S_v) \ge 0.4/(0.87\times415) = 0.001109$.

6.11 Maximum Spacing of Shear Reinforcement

Vertical stirrups: min (0.75d, 300 mm) — IS 456 Cl. 26.5.1.5
Inclined bars: min (d, 300 mm)
For high shear zone ($\tau_v > 0.5\tau_{c,max}$): max spacing = 0.5d or 200 mm (whichever less)

✂️

Ch 6 · Formulas

Nominal Shear Stress

$$ \tau_v = \frac{V_u}{bd} $$

Critical section at 'd' from face of support

MPa

Stirrup Spacing

$$ S_v = \frac{0.87f_y A_{sv}d}{V_{us}} $$

$V_{us}=V_u-\tau_c bd$; max spacing 0.75d or 300mm

Minimum Shear Reinforcement

$$ \frac{A_{sv}}{bS_v} \ge \frac{0.4}{0.87f_y} $$

Fe415: ≥ 0.001109

—

Maximum Shear Stress

$$ \tau_{c,max}:\ \text{M20=2.8, M25=3.1, M30=3.5 MPa} $$

If $\tau_v>\tau_{c,max}$: redesign section

MPa

✂️

Ch 6 · Solved Example

GATE CE / ESE level

A beam (b = 300 mm, d = 450 mm) carries a factored shear $V_u = 120$ kN, with $\tau_c = 0.5$ MPa and Fe415 2-legged 8 mm stirrups ($A_{sv} = 100.5$ mm²). Find the required stirrup spacing.

Approach hint

Find $V_{us}=V_u-\tau_c bd$ first, then apply $S_v=0.87f_yA_{sv}d/V_{us}$.

$\tau_c\cdot b\cdot d = 0.5\times300\times450 = 67{,}500$ N = 67.5 kN
$V_{us} = 120 - 67.5 = 52.5$ kN $= 52{,}500$ N
$S_v = 0.87\times415\times100.5\times450/52{,}500 \approx 310$ mm
Check max spacing: min(0.75×450, 300) = 300 mm → governs

✓ Provide stirrups at 300 mm c/c (governed by the maximum spacing limit, not the calculated 310 mm)

✂️

Ch 6 · Diagrams

Click any diagram to zoom

Fig 6.1 – Crack patterns in RC beam. Flexural (vertical) cracks at mid-span; diagonal shear-flexure cracks near supports. Critical section for shear at 'd' from face of support.

✂️

Ch 6 · Quick Quiz

3 questions · tap an option to reveal answer

Question 1

The critical section for shear design in an RC beam is located at:

IS 456 permits the critical shear section to be taken at 'd' from the face of the support, since compression-strut action carries load directly into the support within that region ✓

Question 2

If $\tau_v > \tau_{c,max}$ for a beam section, the correct design action is to:

$\tau_{c,max}$ is an absolute ceiling — beyond it, diagonal compression (web-crushing) failure occurs regardless of how much shear reinforcement is added, so the section itself must be enlarged ✓

Question 3

The maximum permissible spacing of vertical stirrups (IS 456 Cl. 26.5.1.5) is:

Vertical stirrups: min(0.75d, 300mm); this tightens to min(0.5d, 200mm) only in high-shear zones where $\tau_v>0.5\tau_{c,max}$ ✓

🪢

Ch 7 · Bond in Reinforced Concrete

○ Not Started

7.4 Bond Stress & 7.8 Development Length

Bond stress $\tau_{bd}$ = Force in bar / (perimeter × embedment length). Development Length $L_d$: length of bar needed to develop the full design stress $0.87f_y$.

$$ L_d = \frac{0.87f_y\cdot\phi}{4\tau_{bd}} $$

IS 456 Design Bond Stress $\tau_{bd}$ (plain bars): M20=1.2, M25=1.4, M30=1.5, M35=1.7, M40=1.9 MPa. For deformed (HYSD) bars: multiply $\tau_{bd}$ by 1.6. For bars in compression: multiply $\tau_{bd}$ by 1.25.

$L_d$ for Fe415 HYSD bar in M20 concrete (tension): $\tau_{bd} = 1.2\times1.6 = 1.92$ MPa; $L_d = (0.87\times415\times\phi)/(4\times1.92) = 47\phi$ (approx.). General approximate values: $L_d \approx 47\phi$ (Fe415, M20); $\approx 55\phi$ (Fe415, M15); $\approx 40\phi$ (Fe415, M25).

7.5 Types of Bond

Type	Definition	Significance
Flexural Bond	Bond stress arising from change in bar force along length due to varying bending moment	Controls cracking; distributes cracks
Anchorage / Development Bond	Bond needed to develop full bar capacity (yield) within available embedment	Controls bar pullout failure; governs $L_d$

7.11 Hooks, Bends & Standard Anchorages (IS 456 Cl. 26.2)

Type	Equivalent Anchorage Length	Bend Angle
Standard 180° hook (U-hook)	16φ (equivalent anchorage)	180° + 4φ tail
Standard 90° bend	8φ (equivalent anchorage)	90° + 12φ tail (min.)
Standard 45° bend	4φ	45° bend

⚠️ Hooks are mandatory for plain (mild steel) bars; not normally required for deformed bars if $L_d$ is available straight. Deformed bars have better mechanical interlock.

7.16 Splicing of Reinforcement (IS 456 Cl. 26.2.5)

Lap splice length: 1.3 $L_d$ in tension; 1.0 $L_d$ in compression
Max. 50% of bars spliced at one cross-section for tension; 100% in compression
Bars > 36 mm dia: prefer mechanical couplers or butt-weld, not laps
Stagger laps at least 1.3 × lap length apart

🪢

Ch 7 · Formulas

Development Length

$$ L_d = \frac{0.87f_y\phi}{4\tau_{bd}} $$

Deformed bars: ×1.6; compression bars: ×1.25

Approximate L_d (Fe415, HYSD)

$$ L_d \approx 47\phi\ \text{(M20, tension)} $$

≈55φ (M15); ≈40φ (M25)

Lap Splice Length

$$ L_{lap,tension}=1.3L_d,\quad L_{lap,comp}=1.0L_d $$

Max 50% bars spliced at one section (tension)

🪢

Ch 7 · Solved Example

GATE CE / ESE level

Find the development length $L_d$ for a 20 mm dia Fe415 HYSD bar in M20 concrete under tension.

Approach hint

Use the design bond stress for M20 plain bars (1.2 MPa), scale by 1.6 for deformed bars, then apply $L_d=0.87f_y\phi/(4\tau_{bd})$.

$\tau_{bd} = 1.2\times1.6 = 1.92$ MPa (HYSD bar, M20)
$L_d = (0.87\times415\times20)/(4\times1.92) = 7221/7.68 \approx 940$ mm
Check: $L_d = 47\phi = 47\times20 = 940$ mm ✓ matches the approximate rule

✓ $L_d \approx 940$ mm (= 47φ) for a 20 mm Fe415 HYSD bar in M20 concrete

🪢

Ch 7 · Diagrams

Click any diagram to zoom

Fig 7.1 – Development length: bond stress distributed along the embedded length transfers the full bar tension $0.87f_y\cdot A_{bar}$ into the surrounding concrete.

🪢

Ch 7 · Quick Quiz

3 questions · tap an option to reveal answer

Question 1

Compared to plain bars, the design bond stress $\tau_{bd}$ for deformed (HYSD) bars is:

Deformed bars have mechanical interlock (ribs) in addition to adhesion/friction, so IS 456 allows a 60% increase (×1.6) in design bond stress over plain bars ✓

Question 2

The lap splice length for reinforcement in tension is:

Tension laps require 1.3 $L_d$ (longer, since tension is more critical); compression laps only need 1.0 $L_d$ ✓

Question 3

A standard 180° hook provides an equivalent anchorage length of:

A standard 180° (U) hook is equivalent to 16φ of anchorage length; a 90° bend gives 8φ, and a 45° bend gives 4φ ✓

🌀

Ch 8 · Design for Torsion in RCC

○ Not Started

8.1 Introduction – When Does Torsion Occur?

Equilibrium torsion: Must be resisted for structural stability (e.g., curved beams, L-beams with eccentric loading). Must be fully designed for.
Compatibility torsion: Arises to maintain compatibility with adjacent members (e.g., edge beams of flat slabs). IS 456 allows redistribution; minimum design required.

8.8 IS 456 LSM Torsion Design (Cl. 41)

IS 456 uses an equivalent shear method: torsion is converted to equivalent shear and equivalent moment, then designed using standard shear and flexure procedures.

Equivalent Shear: $V_e = V_u + 1.6T_u/b$. Equivalent Moment: $M_{e1} = M_u + M_t$ (for the face with sagging moment), where $M_t = T_u(1+D/b)/1.7$. $M_{e2} = M_t - M_u$ (for the opposite face, if $M_t > M_u$).

Nominal shear stress: $\tau_{ve} = V_e/(b\cdot d) \le \tau_{c,max}$. If $\tau_{ve} > \tau_c$: design transverse + longitudinal torsion reinforcement.

8.9 Torsional Reinforcement Requirements (IS 456 Cl. 41.4)

Closed stirrups (rectangular links) mandatory — open stirrups cannot resist torsion
Longitudinal bars at corners of stirrups; max 300 mm spacing or b (width)
Stirrup spacing ≤ $x_1$ (shorter dimension of stirrup), $y_1$/4, or 300 mm (whichever least)

Area of transverse reinforcement per unit length: $A_{sv}/s_v = T_u/(b_1\cdot d_1\cdot0.87f_y) + V_u/(2.5\cdot d_1\cdot0.87f_y)$, where $b_1, d_1$ = distance between corner bars (centre to centre).

🌀

Ch 8 · Formulas

Equivalent Shear

$$ V_e = V_u + \frac{1.6T_u}{b} $$

Converts torsion into an equivalent shear demand

Equivalent Moment

$$ M_t = \frac{T_u(1+D/b)}{1.7} $$

$M_{e1}=M_u+M_t$; $M_{e2}=M_t-M_u$ (opposite face)

N·mm

Transverse Torsion Reinforcement

$$ \frac{A_{sv}}{s_v}=\frac{T_u}{b_1d_1\cdot0.87f_y}+\frac{V_u}{2.5d_1\cdot0.87f_y} $$

Closed stirrups mandatory; $b_1,d_1$ = corner-bar spacing

mm²/mm

🌀

Ch 8 · Solved Example

GATE CE / ESE level

A beam of width b = 300 mm carries a factored shear $V_u = 80$ kN and a factored torsion $T_u = 15$ kN·m. Find the equivalent shear $V_e$.

Approach hint

Apply $V_e=V_u+1.6T_u/b$ directly, keeping units consistent (convert T_u to N·mm and b to mm, or keep kN and m together carefully).

$1.6T_u/b = 1.6\times15/0.3 = 80$ kN (using T_u in kN·m, b in m)
$V_e = V_u + 1.6T_u/b = 80 + 80 = 160$ kN

✓ Equivalent shear $V_e = 160$ kN — the beam must now be checked/designed for this equivalent shear instead of the actual 80 kN

🌀

Ch 8 · Diagrams

Click any diagram to zoom

Fig 8.1 – Closed rectangular stirrup with corner longitudinal bars — the only detailing that can resist torsional shear flow around the section perimeter.

🌀

Ch 8 · Quick Quiz

3 questions · tap an option to reveal answer

Question 1

Which type of stirrup MUST be used to resist torsion in an RCC beam?

Torsional shear flow circulates around the full perimeter of the section, so only a closed stirrup (with corner longitudinal bars) can resist it — open stirrups cannot ✓

Question 2

Torsion that arises purely to maintain compatibility with an adjacent member (e.g., an edge beam supporting a flat slab), where redistribution is possible, is called:

Compatibility torsion can redistribute through cracking without endangering overall stability, unlike equilibrium torsion which must be fully designed for structural stability ✓

Question 3

IS 456's approach to torsion design (Cl. 41) works by:

The equivalent shear method ($V_e=V_u+1.6T_u/b$) and equivalent moment method ($M_t=T_u(1+D/b)/1.7$) let designers reuse standard shear and flexure procedures instead of a separate torsion theory ✓

🏗

Ch 9 · Flanged Beams (T-beams & L-beams)

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9.3 Effective Width of Flange (IS 456 Cl. 23.1.2)

For T-beams (intermediate beam): $b_f = l_o/6 + b_w + 6D_f$. For L-beams (edge beam): $b_f = l_o/12 + b_w + 3D_f$. Where $l_o$ = distance between points of zero moment (~0.7 × span for continuous; = span for SS), $b_w$ = web width, $D_f$ = flange (slab) thickness.

$b_f$ ≤ clear distance between beams + $b_w$ (T-beam); ≤ $b_w$ + (clear distance)/2 (L-beam).

9.5 Analysis of Flanged Sections

Case 1: Neutral Axis within flange ($x_u \le D_f$) — most common

Treat as a rectangular beam of width $b_f$ and depth d: $C = 0.36f_{ck}\cdot b_f\cdot x_u$, $T = 0.87f_y\cdot A_{st}$, $M_u = 0.87f_yA_{st}(d-0.42x_u)$.

Case 2: Neutral Axis in web ($x_u > D_f$)

Total compression = Flange compression + Web compression. $C_f = 0.36f_{ck}(b_f-b_w)D_f$ [flange portion beyond web]. $C_w = 0.36f_{ck}\cdot b_w\cdot x_u$ [web compression]. $M_u = C_f(d-D_f/2) + C_w(d-0.42x_u)$.

🏗

Ch 9 · Formulas

Effective Flange Width — T-Beam

$$ b_f = \frac{l_o}{6} + b_w + 6D_f $$

Also ≤ clear distance between beams + $b_w$

Effective Flange Width — L-Beam

$$ b_f = \frac{l_o}{12} + b_w + 3D_f $$

Edge beam (flange on one side only)

Moment of Resistance — NA in Flange

$$ M_u = 0.87f_yA_{st}(d-0.42x_u) $$

Valid only when $x_u \le D_f$ (most common case)

N·mm

Moment of Resistance — NA in Web

$$ M_u = C_f\left(d-\frac{D_f}{2}\right) + C_w(d-0.42x_u) $$

$C_f=0.36f_{ck}(b_f-b_w)D_f$; $C_w=0.36f_{ck}b_wx_u$

N·mm

🏗

Ch 9 · Solved Example

GATE CE / ESE level

A simply supported T-beam has span $l_o = 6$ m, web width $b_w = 300$ mm, and flange (slab) thickness $D_f = 120$ mm. Find the effective flange width $b_f$.

Approach hint

Apply $b_f = l_o/6 + b_w + 6D_f$ directly with consistent units (mm).

$l_o/6 = 6000/6 = 1000$ mm
$6D_f = 6\times120 = 720$ mm
$b_f = 1000 + 300 + 720 = 2020$ mm

✓ $b_f = 2020$ mm (subject to the clear-distance-between-beams check)

🏗

Ch 9 · Diagrams

Click any diagram to zoom

Fig 9.1 – T-beam (both sides) and L-beam (one side) cross-sections. Effective flange width $b_f$ per IS 456 Cl. 23.1.2. NA usually falls within flange ($D_f$ zone) for normal loading.

🏗

Ch 9 · Quick Quiz

3 questions · tap an option to reveal answer

Question 1

For most practical loading, the neutral axis of a T-beam falls:

Since the flange is wide, the NA usually falls within the thin flange depth, allowing the T-beam to be analysed as an equivalent rectangular beam of width $b_f$ — the common Case 1 ✓

Question 2

An L-beam differs from a T-beam in that:

A T-beam has flange (slab) on both sides of the web (interior beam); an L-beam only has flange on one side, since it sits at the edge of the slab ✓

Question 3

When the neutral axis falls in the web ($x_u > D_f$), the total compression force is computed as:

Case 2 splits the compression zone into the flange overhang ($C_f$, beyond the web) and the web portion down to $x_u$ ($C_w$), each contributing to the total moment of resistance at its own lever arm ✓

📏

Ch 10 · Serviceability Limit State – Deflection & Cracking

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10.3 Deflection Limits (IS 456 Cl. 23.2)

Criterion	Limit
Total final deflection (affecting appearance & comfort)	Span/250
Final deflection after construction of partitions & finishes	Span/350 or 20 mm (whichever less)
For flat roofs (ponding)	Span/480 + camber may be needed

10.3 Deflection Calculation – Components

Total deflection δ = Short-term $\delta_i$ + Long-term additional (creep + shrinkage). Short-term $\delta_i = k\cdot WL^3/(E_{ce}\cdot I_{eff})$, where k = 5/384 (SS UDL); 1/48 (SS point load at mid); 1/8 (cantilever tip). $E_{ce} = E_c$ (short term) or $E_c/(1+\theta)$ (long term).

Effective Moment of Inertia $I_{eff}$ (IS 456 Cl. 23.2.1 / Branson's formula):

$$ I_{eff} = I_{cr} + (I_g - I_{cr})\times\left(\frac{M_{cr}}{M_{max}}\right)^3 $$

$I_g$ = gross MoI, $I_{cr}$ = cracked MoI, $M_{cr}$ = cracking moment. $I_{eff}$ transitions smoothly between the uncracked ($I_g$) and fully cracked ($I_{cr}$) stiffness as the applied moment $M_{max}$ grows relative to $M_{cr}$.

10.4 Cracking Limits (IS 456 Cl. 35.3)

Exposure Condition	Max. Crack Width $w_{max}$
Mild / Moderate	0.3 mm
Severe / Very Severe / Extreme	0.2 mm
Prestressed concrete members	0.1 mm (or no cracking in Type I/II)

Crack width (IS 456 Annex F): $w_{cr} = 3a_{cr}\varepsilon_m/[1+2(a_{cr}-c_{min})/(D-x)]$, where $a_{cr}$ = distance from crack to nearest bar surface, and $\varepsilon_m$ = average strain at the level of tension steel.

10.5 Other Serviceability Limits

Vibration: Natural frequency of floors ≥ 8 Hz for residential buildings (to avoid perception of vibration)
Fire resistance: Member size + cover requirements from IS 456 Table 16A
Fatigue: For bridges and crane girders; limit stress range in reinforcement

📏

Ch 10 · Formulas

Short-Term Deflection

$$ \delta_i = k\cdot\frac{WL^3}{E_{ce}I_{eff}} $$

k = 5/384 (SS UDL); 1/8 (cantilever tip)

Effective Moment of Inertia (Branson)

$$ I_{eff}=I_{cr}+(I_g-I_{cr})\left(\frac{M_{cr}}{M_{max}}\right)^3 $$

Bridges between gross ($I_g$) and cracked ($I_{cr}$) stiffness

mm⁴

Deflection Limits

$$ \delta \le \frac{L}{250}\ \text{(total)},\ \frac{L}{350}\ \text{(after finishes)} $$

L/350 also capped at 20 mm absolute

Crack Width

$$ w_{cr}=\frac{3a_{cr}\varepsilon_m}{1+2(a_{cr}-c_{min})/(D-x)} $$

IS 456 Annex F; limit 0.3mm mild, 0.2mm severe

📏

Ch 10 · Solved Example

GATE CE / ESE level

A simply supported beam has span L = 6 m. What is the maximum permissible total final deflection, and the maximum permissible deflection after construction of partitions?

Approach hint

Apply the two IS 456 Cl. 23.2 limits directly: span/250 for total, and the lesser of span/350 or 20 mm for the post-construction case.

Total final deflection limit = $L/250 = 6000/250 = 24$ mm
Post-construction limit = $\min(L/350, 20) = \min(6000/350, 20) = \min(17.1, 20) = 17.1$ mm

✓ Total deflection ≤ 24 mm; deflection after partitions/finishes ≤ 17.1 mm

📏

Ch 10 · Diagrams

Click any diagram to zoom

Fig 10.1 – Branson's effective moment of inertia: $I_{eff}$ starts at $I_g$ below the cracking moment, then transitions toward $I_cr$ as $M_{max}$ grows, following a cube-power weighting.

📏

Ch 10 · Quick Quiz

3 questions · tap an option to reveal answer

Question 1

The total final deflection limit for a beam (IS 456 Cl. 23.2), affecting appearance and general comfort, is:

Total final deflection ≤ span/250; the stricter span/350 (or 20 mm) limit applies only to deflection occurring after partitions/finishes are constructed ✓

Question 2

In Branson's effective moment of inertia formula, as $M_{max}$ increases well beyond $M_{cr}$, $I_{eff}$ approaches:

As $M_{max}/M_{cr}$ grows large, $(M_{cr}/M_{max})^3\to0$, so $I_{eff}\to I_{cr}$ — the section behaves as fully cracked under heavy loading ✓

Question 3

The maximum allowable crack width for RCC members under severe/very severe/extreme exposure is:

Severe/very severe/extreme exposure requires the tighter 0.2 mm limit (vs 0.3 mm for mild/moderate) since wider cracks let in more moisture and chlorides, accelerating corrosion ✓

🔲

Ch 11 · Two-Way Slab Design

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11.2 One-Way vs Two-Way Slabs

Criterion	One-Way Slab	Two-Way Slab
Aspect ratio $l_y/l_x$	> 2	≤ 2
Load transfer	Primarily in shorter span (x)	In both directions
Main reinforcement	In short span only	In both spans
Examples	Verandah slabs, one-way spanning	Rooms, two-way panels

11.5 IS 456 Method for Two-Way Slabs (Cl. D-1)

Design BM per unit width: $M_x = \alpha_x\cdot w\cdot l_x^2$ (short span direction), $M_y = \alpha_y\cdot w\cdot l_x^2$ (long span direction). $\alpha_x, \alpha_y$ = BM coefficients from IS 456 Table 26 (function of $l_y/l_x$ and support conditions). $w$ = total factored load per unit area = 1.5(DL + LL) kN/m².

11.6 Thickness of Two-Way Slabs

l/d (short span) for two-way slabs = 28 (two short edges continuous). Minimum thickness: 120–150 mm for general floors; 200 mm for parking decks. IS 456 Table 10 (l/d for deflection) applies; modify for $p_t$.

11.4 Support Conditions & Edge Discontinuity

IS 456 provides α coefficients for 9 different edge conditions (Cases 1–9)
Free (discontinuous) edges: torsional reinforcement required at corners
Corner reinforcement: ≥ 75% of area of short span mid-span steel; in both directions; extend 1/5 of short span from corner
For corner held down: 3 layers (top + bottom in 2 directions) at 45° or orthogonal

11.8 Shear in Two-Way Slabs

Shear check: $\tau_v = V_u/(b\cdot d) \le k_s\cdot\tau_c$, where $k_s = (0.5+\beta_c)$ but ≤ 1.0, and $\beta_c$ = short side / long side of loaded area. For flat slabs: punching shear is critical at 'd/2' from the column face (two-way action).

🔲

Ch 11 · Formulas

Design BM per Unit Width

$$ M_x=\alpha_xwl_x^2,\quad M_y=\alpha_ywl_x^2 $$

α coefficients from IS 456 Table 26

N·mm/mm

Two-Way Slab l/d

$$ l/d = 28\ \text{(two short edges continuous)} $$

Min thickness 120–150 mm general floors

—

Shear in Two-Way Slabs

$$ \tau_v = \frac{V_u}{bd} \le k_s\tau_c $$

$k_s=0.5+\beta_c\le1.0$

MPa

Corner Reinforcement

$$ A_{corner} \ge 0.75\times A_{mid-span,short} $$

Both directions; extends 1/5 of short span from corner

mm²

🔲

Ch 11 · Solved Example

GATE CE / ESE level

A square two-way slab panel ($l_y/l_x = 1.0$) has short span $l_x = 4$ m and factored load $w = 12$ kN/m². Using $\alpha_x = 0.062$, find the design moment $M_x$ per metre width.

Approach hint

Apply $M_x=\alpha_xwl_x^2$ directly.

$M_x = \alpha_x\times w\times l_x^2 = 0.062\times12\times4^2$
$= 0.062\times12\times16 = 11.9$ kN·m/m

✓ $M_x \approx 11.9$ kN·m per metre width in the short-span direction

🔲

Ch 11 · Diagrams

Click any diagram to zoom

Fig 11.1 – Two-way slab panel. Dashed lines show diagonal load distribution. Short span carries more load. BM coefficients $\alpha_x, \alpha_y$ from IS 456 Table 26.

🔲

Ch 11 · Quick Quiz

3 questions · tap an option to reveal answer

Question 1

A slab panel is classified as two-way (per IS 456) when the aspect ratio $l_y/l_x$ is:

When $l_y/l_x \le 2$, the slab spans significantly in both directions and must be designed as two-way; beyond 2, one-way action dominates ✓

Question 2

Torsional (corner) reinforcement is required in a two-way slab panel when:

At corners where edges are free to lift (discontinuous), the slab tends to curl up — torsional corner reinforcement in both directions restrains this uplift ✓

Question 3

In a rectangular two-way slab, which span direction generally carries the LARGER share of the load?

The short span is stiffer (deflects less for the same load), so it attracts a larger share of the load — this is why $\alpha_x > \alpha_y$ as the aspect ratio increases ✓

🏛

Ch 12 · Design of Compression Members (Columns)

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12.3 Classification of Columns

Basis	Type	Definition
Slenderness	Short Column	$l_{eff}$/D ≤ 12 (both directions); failure by material
Slenderness	Slender/Long Column	$l_{eff}$/D > 12; additional moments due to deflection (IS 456 Cl. 39.7)
Loading	Axially Loaded	Concentric axial force only (rare in practice)
Loading	Eccentrically Loaded	Axial force + bending moment (uniaxial or biaxial)
Lateral Ties	Tied Column	Rectangular links confine concrete; most common
Lateral Ties	Spiral Column	Helical reinforcement; better ductility; circular section

12.5 IS 456 Recommendations for Columns (Cl. 26.5.3)

Parameter	IS 456 Requirement
Min. $A_{sc}$	0.8% of gross cross-sectional area
Max. $A_{sc}$	4% (general); 6% at lapping (local maximum)
Min. bar dia (longitudinal)	12 mm
Min. no. of bars: rectangular / circular	4 / 6
Lateral ties (links): dia	Max ($\phi_{main}$/4, 6 mm)
Lateral ties: spacing	Min ($b_{least}$, 16$\phi_{main}$, 300 mm)
Cover to main bars	40 mm (mild exposure)

12.6 Short Columns – Axial Load Only (IS 456 Cl. 39.3)

$P_u = 0.4f_{ck}A_c + 0.67f_yA_{sc}$, where $A_c$ = net concrete area = $A_g - A_{sc}$, giving $P_u = 0.4f_{ck}(A_g-A_{sc}) + 0.67f_yA_{sc}$. Minimum eccentricity $e_{min} = \max(l/500+D/30,\ 20\text{ mm})$ must always be considered.

12.7 Short Column with Uniaxial Bending

The P-M interaction diagram (see Diagrams tab) plots all combinations of axial load $P_u$ and moment $M_u$ that the section can safely carry. Any point inside the curve is safe.

12.8 Failure Modes in Eccentric Compression

Large eccentricity (e > $e_{balanced}$): Tension-controlled failure — steel yields first, then concrete crushes (ductile)
Small eccentricity (e < $e_{balanced}$): Compression-controlled failure — concrete crushes before steel yields (less ductile)
Balanced failure: Both occur simultaneously at specific e = $e_{balanced}$

12.13 Biaxial Bending (IS 456 Cl. 39.6 / SP-16)

Interaction equation for biaxial bending: $(M_{ux}/M_{ux1})^{\alpha_n} + (M_{uy}/M_{uy1})^{\alpha_n} \le 1.0$.

$\alpha_n = 1$ when $P_u/P_{uz} \le 0.2$; $\alpha_n = 2$ when $P_u/P_{uz} \ge 0.8$ (interpolate for intermediate values). $P_{uz} = 0.45f_{ck}A_g + 0.75f_yA_{sc}$.

12.17–12.18 Slender Columns (IS 456 Cl. 39.7)

Additional moment due to slenderness (P-delta effect): $M_{ax} = P_u\cdot e_{ax}$ where $e_{ax} = (l_{ex}/D)^2\cdot D/2000$. $M_{ay} = P_u\cdot e_{ay}$ where $e_{ay} = (l_{ey}/b)^2\cdot b/2000$. Design moment: $M_{u,design} = M_u + M_{ax}$ (or $M_{ay}$). $l_{ex}, l_{ey}$ = effective lengths in respective directions, from IS 456 Table 28 (based on end condition: fixed, pinned, free).

🏛

Ch 12 · Formulas

Short Column — Axial Capacity

$$ P_u = 0.4f_{ck}A_c + 0.67f_yA_{sc} $$

$A_c=A_g-A_{sc}$; always check $e_{min}$

Minimum Eccentricity

$$ e_{min}=\max\left(\frac{l}{500}+\frac{D}{30},\ 20\text{mm}\right) $$

Must be considered even for "axial" load cases

Biaxial Bending Interaction

$$ \left(\frac{M_{ux}}{M_{ux1}}\right)^{\alpha_n}+\left(\frac{M_{uy}}{M_{uy1}}\right)^{\alpha_n}\le1.0 $$

$\alpha_n=1$ at $P_u/P_{uz}\le0.2$; =2 at ≥0.8

—

Slender Column — Additional Moment

$$ M_{ax}=P_u\cdot e_{ax},\quad e_{ax}=\left(\frac{l_{ex}}{D}\right)^2\frac{D}{2000} $$

P-delta effect; $M_{u,design}=M_u+M_{ax}$

N·mm

🏛

Ch 12 · Solved Example

GATE CE / ESE level

A short square RCC column (400 mm × 400 mm), M20 concrete, Fe415 steel, carries $A_{sc} = 3600$ mm² (2.25% of gross area). Find its axial load capacity $P_u$.

Approach hint

Compute $A_g$, then $A_c = A_g - A_{sc}$, and apply $P_u = 0.4f_{ck}A_c + 0.67f_yA_{sc}$.

$A_g = 400\times400 = 160{,}000$ mm²
$A_c = 160{,}000 - 3600 = 156{,}400$ mm²
$P_u = 0.4\times20\times156{,}400 + 0.67\times415\times3600$
$= 1{,}251{,}200 + 1{,}001{,}\!880 = 2{,}253{,}080$ N ≈ 2253 kN

✓ $P_u \approx 2253$ kN (before applying the minimum-eccentricity check)

🏛

Ch 12 · Diagrams

Click any diagram to zoom

Fig 12.1 – Column P-M interaction diagram. Any ($P_u, M_u$) point inside the curve is safe. Balanced point corresponds to simultaneous concrete crushing ($\varepsilon_{cu}$=0.0035) and steel yielding.

🏛

Ch 12 · Quick Quiz

3 questions · tap an option to reveal answer

Question 1

A column is classified as "short" (IS 456) when its slenderness ratio $l_{eff}/D$ is:

$l_{eff}/D \le 12$ (in both directions) classifies a column as short — failure governed by material strength, not buckling ✓

Question 2

The minimum longitudinal steel percentage required in an RCC column (IS 456 Cl. 26.5.3) is:

Minimum $A_{sc}$ = 0.8% of gross cross-sectional area, ensuring the column has ductile behaviour and can resist accidental moments ✓

Question 3

In a column P-M interaction diagram, a design point that falls OUTSIDE the curve indicates:

The interaction curve is the boundary of safe (P,M) combinations — any point outside it means the section will fail under that combined loading and must be redesigned ✓

⛰

Ch 13 · Shallow Foundations

○ Not Started

13.4 Types of Footings

Type	Description	Use
Isolated Column Footing	Square / rectangular / circular footing under single column	Most common; widely spaced columns
Wall Footing / Strip Footing	Continuous footing under load-bearing wall	Masonry walls, load-bearing construction
Combined Footing	Single footing under two or more columns	Adjacent columns close together / near property line
Mat / Raft Footing	Single slab under entire structure	Weak soil; closely spaced columns; differential settlement control
Strap (Cantilever) Footing	Two isolated footings connected by strap beam	Eccentric column near property boundary

13.6 Design of Isolated Column Footing (IS 456 Cl. 34)

Design Procedure

Area of footing: A = $P_{service}$ / SBC (use service loads for geotechnical design)
Net upward factored pressure: $q_u = P_u / A_{footing}$
Bending moment at face of column: $M = q_u\cdot a^2/2$ (a = cantilever projection)
Depth from shear: One-way: $\tau_v = q_u\cdot a'/d \le \tau_c$; Two-way punching: $\tau_v = V_u/(b_o\cdot d) \le k_s\cdot\tau_c$
Steel: $M_u = 0.87f_yA_{st}d(1-A_{st}f_y/(bdf_{ck}))$; solve for $A_{st}$
Development length: $L_d$ available ≥ $L_d$ required

One-way shear: critical at 'd' from face of column. Punching shear (two-way): critical perimeter at 'd/2' from each face. $b_o$ = perimeter at d/2 from column = $2(l_c+b_c+2d)$, where $l_c, b_c$ = column dimensions. Permissible punching: $\tau_c = k_s\times0.25\sqrt{f_{ck}}$ (MPa); $k_s = 0.5+\beta_c \le 1.0$.

13.12 Combined Footings

Used when two columns are too close for separate footings, or one column is near property boundary
Design so resultant of column loads passes through centroid of footing (for uniform pressure)
Rectangular combined footing: design as inverted beam/slab under uniform pressure
Trapezoidal footing: when centroid cannot be aligned with rectangular shape
Strap footing: strap beam connects eccentric outer footing to inner footing; strap beam designed for bending and shear

⛰

Ch 13 · Formulas

Soil Pressure — Eccentric Load

$$ q_{max/min} = \frac{P}{A} \pm \frac{6M}{BL^2} $$

Uplift possible if e > L/6

kN/m²

Bending Moment at Column Face

$$ M = \frac{q_u a^2}{2} $$

a = cantilever projection beyond column face

N·mm/mm

Punching Shear Perimeter & Strength

$$ b_o=2(l_c+b_c+2d),\quad \tau_c=k_s\times0.25\sqrt{f_{ck}} $$

Critical at d/2 from column face; $k_s=0.5+\beta_c\le1.0$

mm, MPa

Footing Area

$$ A = \frac{P_{service}}{SBC} $$

Use unfactored (service) loads — geotechnical design

m²

⛰

Ch 13 · Solved Example

GATE CE / ESE level

A square isolated footing (2.5 m × 2.5 m) carries a factored load $P_u = 900$ kN with a square column of side 400 mm. Find the net factored upward pressure $q_u$, and the bending moment per metre width at the face of the column.

Approach hint

Compute $q_u = P_u/A$, find the cantilever projection $a$ = (footing width − column width)/2, then apply $M = q_ua^2/2$.

$A = 2.5\times2.5 = 6.25$ m²
$q_u = 900/6.25 = 144$ kN/m²
Cantilever projection $a = (2.5-0.4)/2 = 1.05$ m
$M = q_u a^2/2 = 144\times1.05^2/2 = 79.4$ kN·m per metre width

✓ $q_u = 144$ kN/m²; $M \approx 79.4$ kN·m/m at the face of the column

⛰

Ch 13 · Diagrams

Click any diagram to zoom

Fig 13.1 – Soil pressure under isolated footing. Concentric load → uniform. Eccentric load → trapezoidal. Uplift (tension) occurs when eccentricity e > L/6.

⛰

Ch 13 · Quick Quiz

3 questions · tap an option to reveal answer

Question 1

Punching (two-way) shear in an isolated footing is checked at a critical section located:

Two-way (punching) shear is checked on a perimeter at d/2 from each face of the column, unlike one-way shear which is checked at 'd' from the face ✓

Question 2

Which footing type is used when a column is located right at the property boundary and cannot have a symmetric isolated footing?

A strap footing connects an eccentric outer (boundary) column's footing to an interior footing via a strap beam, balancing the eccentric moment without extending the footing past the property line ✓

Question 3

For an eccentrically loaded footing, soil pressure uplift (tension) occurs when the eccentricity e exceeds:

When e > L/6 (outside the "middle third"), the formula $q_{min}=P/A-6M/(BL^2)$ becomes negative, implying tension — soil cannot take tension, so part of the footing lifts off ✓

🔩

Ch 14 · Prestressed Concrete (PSC) – IS 1343:2012

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14.4 Comparison: RCC vs PSC

Feature	RCC	PSC
Steel type	Mild steel / HYSD ($f_y$ 250–550 MPa)	High-tensile (HT) wires/strands ($f_{pu}$ 1600–2000 MPa)
Concrete grade	M20–M50	Min. M40 (pre-tensioned); M30 (post-tensioned)
Cracking	Cracked in tension (design accepts cracks)	No cracking (Type I) or limited cracking (Type II, III)
Deflection	Higher; affected by cracking	Lower; camber by prestress; load-balancing possible
Shear resistance	Stirrups needed	Better inherent shear resistance; fewer stirrups
Span range	Up to ~20 m economical	20–100+ m bridges, long-span structures
Initial cost	Lower	Higher (high-tensile steel + equipment)

14.11–14.13 Pre-tensioning vs Post-tensioning

Pre-tensioning: Steel is tensioned before casting (against fixed abutments in a factory bed); concrete is cast around the tensioned steel and cured; on release, the prestress transfers to the concrete by bond, developing camber. Post-tensioning: Concrete is cast with ducts already in place; HT steel is threaded through the ducts after the concrete hardens, tensioned with a jack, anchored, and the duct is grouted. See the Diagrams tab for the step-by-step comparison.

14.14 Assumptions in PSC Analysis (IS 1343 Cl. 21.1)

Plane sections remain plane (Bernoulli hypothesis)
Elastic behaviour assumed for serviceability analysis; inelastic for ultimate limit state
For bonded tendons: perfect bond after grouting (strain compatibility)
For unbonded tendons: tendon strain based on overall elongation, not local strain

14.16–14.17 Stress Distribution in PSC Beam

Stress at any fibre of a prestressed beam:

$$ f = \frac{P}{A} \pm \frac{P\cdot e\cdot y}{I} \pm \frac{M\cdot y}{I} $$

Top fibre: $f_{top} = P/A - Pe\,y_t/I + My_t/I$. Bottom fibre: $f_{bot} = P/A + Pe\,y_b/I - My_b/I$. Where P = effective prestress force, e = eccentricity of tendon, A = cross-sectional area, I = moment of inertia, $y_t, y_b$ = distances from centroid to top and bottom fibres.

14.22 Losses in Prestress

Loss	Occurs in	Magnitude (approx.)	Cause
Elastic Shortening	Pre-tensioned (immediate); Post-tensioned (during tensioning)	1–5%	Concrete shortens on application of prestress
Creep of Concrete	Both	5–10%	Long-term shortening under sustained prestress
Shrinkage of Concrete	Both	2–6%	Drying shrinkage causes tendon shortening
Steel Relaxation	Both	2–5%	HT steel loses stress under constant strain
Friction	Post-tensioned only	5–15%	Curvature friction (μ) + wobble (kx) along duct
Anchorage Slip	Post-tensioned only	1–3%	Slip at wedge anchors during lock-off
Total loss	—	15–25%	Typical range for well-designed PSC

Friction loss: $P_x = P_0\cdot e^{-(\mu\alpha+kx)}$, where $\mu$ = coefficient of friction (0.2–0.5), $\alpha$ = cumulative angle of curvature (radians), k = wobble coefficient (0.0015–0.005/m), x = length from jacking end. Elastic shortening loss (pre-tensioning): $\Delta f_p = m_c\cdot f_c$, where $m_c = E_s/E_c$ (modular ratio for steel and concrete).

14.19–14.20 Pressure Line (C-Line) & Load Balancing

Equivalent load (UDL) from a parabolic tendon profile: $w_{eq} = 8Pe/L^2$ (upward UDL balancing downward gravity load). When $P\times8e/L^2 = w_{self-weight+DL}$: net deflection = 0. Pressure line shift: $e_p = M/P$ (distance of the pressure line from the centroidal axis).

14.23–14.24 Design of PSC Beams – Stress Limits (IS 1343)

Stage	Stress Limit in Concrete
Transfer (initial prestress, no live load): Compression	0.44 $f_{ci}$ (IS 1343)
Transfer: Tension	0 (no tension in Type I) or 1.0 MPa (Type II pre-tensioned)
Service (effective prestress + DL + LL): Compression	0.33 $f_{ck}$
Service: Tension (Type I — fully prestressed)	0 (no tension permitted)
Service: Tension (Type II — limited prestress)	≤ 0.5 MPa (or 1.0 MPa in some cases)
Service: Tension (Type III — partial prestress)	Limited crack width ≤ 0.1 mm

🔩

Ch 14 · Formulas

Fibre Stress (Combined Prestress + Bending)

$$ f = \frac{P}{A} \pm \frac{Pey}{I} \pm \frac{My}{I} $$

e = tendon eccentricity; sign per fibre location

MPa

Equivalent Balancing Load

$$ w_{eq} = \frac{8Pe}{L^2} $$

Parabolic tendon; balances downward DL when equal

kN/m

Friction Loss (Post-tensioning)

$$ P_x = P_0\,e^{-(\mu\alpha+kx)} $$

μ=0.2–0.5; k=0.0015–0.005/m

Elastic Shortening Loss

$$ \Delta f_p = m_c\cdot f_c,\quad m_c=\frac{E_s}{E_c} $$

Pre-tensioning; total losses typically 15–25%

MPa

🔩

Ch 14 · Solved Example

GATE CE / ESE level

A PSC beam of span L = 10 m has a parabolic tendon with effective prestress P = 800 kN and eccentricity e = 150 mm at midspan. Find the equivalent upward UDL $w_{eq}$ that balances the self-weight.

Approach hint

Apply $w_{eq}=8Pe/L^2$ directly, keeping consistent units.

$w_{eq} = 8\times800\times0.15/10^2$
$= 960/100 = 9.6$ kN/m

✓ $w_{eq} = 9.6$ kN/m — if this equals the beam's actual self-weight + superimposed dead load, net deflection at that load stage is zero (full load balancing)

🔩

Ch 14 · Diagrams

Click any diagram to zoom

Fig 14.1 – Pre-tensioning (factory; steel tensioned before casting; bonded) vs Post-tensioning (site; steel tensioned after concrete hardens; anchorage required).

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Ch 14 · Quick Quiz

3 questions · tap an option to reveal answer

Question 1

In post-tensioning, the high-tensile steel is stressed:

Post-tensioning threads HT steel through pre-cast ducts and tensions it against the already-hardened concrete using a jack, then anchors and grouts — the reverse order of pre-tensioning ✓

Question 2

Which prestress loss occurs ONLY in post-tensioned members, not pre-tensioned?

Friction loss (curvature + wobble) only occurs where a tendon is pulled through a duct against the sides — this is unique to post-tensioning; pre-tensioned tendons are straight and external to the concrete during tensioning ✓

Question 3

The equivalent upward UDL from a parabolic tendon profile is given by $w_{eq} = 8Pe/L^2$. If the sag/eccentricity e is doubled (P, L constant), $w_{eq}$:

$w_{eq}$ is directly proportional to e, so doubling e doubles the equivalent balancing load ✓

🌍

Ch 15 · Earthquake Resistant Design (IS 1893:2016 & IS 13920:2016)

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15.1–15.6 Seismology Basics

Term	Definition
Focus / Hypocenter	Point within earth where fault rupture originates
Epicenter	Point on earth's surface directly above the focus
Magnitude (Richter)	$\log_{10}(A/A_0)$; measures energy released; each unit = 10× more shaking, 31.6× more energy
Intensity (MMI)	Subjective measure of shaking at a location (I–XII scale)
P-waves	Primary (compressional) waves; fastest; travel through all media
S-waves	Secondary (shear) waves; ~0.6× P-speed; travel through solid only; cause more damage
Surface waves (R & L)	Rayleigh and Love waves; slowest; largest amplitude; most destructive

15.7 Seismic Zoning (IS 1893:2016)

Zone	Zone Factor Z	Seismicity	Examples
Zone II	0.10	Low	Southern peninsula (most of Kerala, Tamil Nadu stable areas)
Zone III	0.16	Moderate	Parts of MP, UP, Karnataka, Rajasthan
Zone IV	0.24	High	Delhi, parts of J&K, HP, Sikkim
Zone V	0.36	Very High (most severe)	North-east India, Uttarakhand, Andaman & Nicobar

⚠️ Zone I has been removed in IS 1893:2016. India now has 4 seismic zones (II to V). Z = 0.36 for Zone V means peak ground acceleration = 0.36g/2 = 0.18g (at 5% damping reference).

15.24 Design Base Shear (IS 1893:2016 Cl. 7.6)

Design Horizontal Seismic Coefficient: $A_h = Z\cdot I\cdot S_a/(2\cdot R\cdot g)$, where Z = Zone factor, I = Importance factor, R = Response reduction factor, $S_a/g$ = Average response acceleration coefficient (from the response spectrum).

Design Base Shear: $V_B = A_h\cdot W$, where W = Seismic weight = DL + % of LL (IS 1893 Table 8). For LL ≤ 3 kN/m²: 25% of LL; for LL > 3 kN/m²: 50% of LL; roof: 0%.

15.23 Important Seismic Design Parameters

Parameter	Symbol	Values
Importance Factor	I	1.0 (normal); 1.2 (schools, hospitals >300 persons); 1.5 (critical infra)
Response Reduction Factor	R	1.5 (unreinforced masonry); 3.0 (ordinary RC shear wall); 5.0 (special SMRF)
Damping ratio (RC structures)	ξ	5% (design default)
$T_a$ – Empirical period (moment frame)	$T_a$	0.075$h^{0.75}$ (RC frame); 0.085$h^{0.75}$ (steel); $0.09h/\sqrt d$ (other)
$S_a/g$ for medium soil		T ≤ 0.1: 2.5 \| 0.1 < T ≤ 0.4: 2.5 \| T > 0.4: 1.0/T

15.11–15.16 Structural Concepts for Earthquake Resistance

Symmetry: Regular plan and elevation; no torsional irregularity
Continuous load path: Forces must transfer continuously to foundation
Ductility: Ability to deform beyond yield without collapse; key for energy dissipation
Redundancy: Multiple load paths; structure tolerates local failure
Soft storey: Storey stiffness < 70% of adjacent storey — must be avoided or specially designed
Short column effect: Partially infilled frames; columns with window openings fail in shear — avoid

Ductile Detailing (IS 13920:2016) – Key Requirements

Member	Key Requirement
Beams	$p_{t,min} \ge 0.24\sqrt{f_{ck}}/f_y$; $p_{t,max}$ ≤ 2.5%; confinement hoops; top steel ≥ bottom steel/2 at all sections
Columns	Min 1% steel; max 4%; confinement hoops in plastic hinge zones (both ends, full height for short columns)
Joints (Beam-Column)	Shear check per IS 13920; closed hoops in joint; no bar splicing in joint
Shear walls	Min. thickness 150 mm; min. vertical + horizontal steel 0.25% each; boundary elements at ends
Confinement hoops	Spacing ≤ min (D/4, 8$\phi_{long}$, 100 mm) in critical zones
Strong column – Weak beam	$\Sigma M_{columns} \ge 1.2\Sigma M_{beams}$ at joint (IS 13920 Cl. 9.2)

Storey Drift & Irregularities (IS 1893:2016)

Storey drift = $(\Delta_i - \Delta_{i-1})/\text{storey height}$. Max allowable storey drift under design earthquake: 0.004 × storey height (IS 1893 Cl. 7.11.1).

Torsional irregularity: max displacement > 1.5 × avg displacement at that level. Mass irregularity: seismic weight of a storey > 200% of adjacent storey.

🌍

Ch 15 · Formulas

Design Horizontal Seismic Coefficient

$$ A_h = \frac{Z\cdot I\cdot S_a}{2\cdot R\cdot g} $$

Design Base Shear: $V_B=A_h\cdot W$

—

Seismic Force Distribution over Height

$$ Q_i = V_B\cdot\frac{W_ih_i^2}{\Sigma(W_jh_j^2)} $$

Parabolic; largest at roof level

Empirical Fundamental Period

$$ T_a = 0.075h^{0.75}\ \text{(RC moment frame)} $$

0.085$h^{0.75}$ for steel frames

Maximum Storey Drift

$$ \Delta_{max} = 0.004 \times \text{storey height} $$

IS 1893 Cl. 7.11.1

—

🌍

Ch 15 · Solved Example

GATE CE / ESE level

A building in Zone IV (Z = 0.24) has an ordinary RC moment-resisting frame (R = 3.0), importance factor I = 1.0, and $S_a/g = 2.5$. Find the design horizontal seismic coefficient $A_h$.

Approach hint

Substitute directly into $A_h = ZIS_a/(2Rg)$, treating $S_a/g$ as the given ratio.

$A_h = (Z/2)\times(I/R)\times(S_a/g)$
$= (0.24/2)\times(1.0/3.0)\times2.5$
$= 0.12\times0.333\times2.5 = 0.10$

✓ $A_h = 0.10$ — the design base shear $V_B = 0.10\times W$ (10% of the seismic weight)

🌍

Ch 15 · Diagrams

Click any diagram to zoom

Fig 15.1 – Seismic lateral force distribution over building height. Forces increase parabolically (proportional to $W_ih_i^2$), largest at roof. Sum of all $Q_i = V_B$.

🌍

Ch 15 · Quick Quiz

3 questions · tap an option to reveal answer

Question 1

The seismic zone factor Z for Zone V (the most severe zone in IS 1893:2016) is:

Zone V (North-east India, Uttarakhand, Andaman & Nicobar) has Z = 0.36, the highest of the four zones (II=0.10, III=0.16, IV=0.24, V=0.36) ✓

Question 2

A "soft storey" irregularity (a major cause of building collapse in earthquakes) occurs when a storey's lateral stiffness is:

A soft storey (stiffness < 70% of the storey above, often the open ground floor of a building) concentrates drift and damage in that one storey, frequently causing pancake-type collapse ✓

Question 3

IS 13920's "strong column–weak beam" philosophy requires that at any joint:

Ensuring column capacity exceeds beam capacity by at least 20% forces plastic hinges to form in the beams (not columns), preserving the vertical load path and preventing storey-collapse mechanisms ✓

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Appendix · Masonry Design (IS 1905:1987)

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Masonry Reinforcement

Reinforced masonry: steel bars embedded in mortar joints or grouted cavity to resist tension and shear
Used in seismic zones; lintel bands, plinth bands, gable bands — mandatory in IS 4326 (seismic zones III–V)
Vertical bars in corners, junctions, around openings for resistance to out-of-plane forces

Effective Height of Walls (IS 1905 Cl. 5.2.2)

End Condition	Effective Height $H_{eff}$
Both ends restrained (horizontal AND rotational)	0.75 H
Both ends restrained horizontally only (pinned)	1.0 H
One end free, one end restrained horizontally + rotationally	1.5 H
One end free, one end fully restrained	2.0 H

Effective Length of Walls (IS 1905 Cl. 5.2.3)

End Condition	Effective Length $L_{eff}$
Wall with returns at both ends	0.8 L
Wall restrained at one end, free at other	1.5 L
Wall free at both ends	2.0 L
Wall restrained at both ends	1.0 L (pinned)

Slenderness Ratio of Masonry (IS 1905)

SR = $H_{eff}/t$ OR SR = $L_{eff}/t$ (use the smaller of the two), where t = thickness of wall. Max. SR = 27 (unreinforced masonry, IS 1905 Cl. 5.4.1). For superimposed loads, the allowable stress is reduced by a stress reduction factor ($k_s$) based on SR.

Permissible Compressive Stresses in Masonry (IS 1905 Table 8)

Mortar Grade	Brick Strength (MPa)	Basic Compressive Stress (MPa)
M1 (1:0:3)	5.0	0.35
M2 (1:0.5:4.5)	7.5	0.50
M3 (1:1:6)	10.0	0.75
M4 (1:2:9)	12.5	0.95
H1 (1:0:3 with hydraulic lime)	5.0	0.35

Masonry in Seismic Zones – IS 4326 Requirements

Zone II: Bricks in cement mortar; horizontal steel band (lintel band) mandatory
Zone III: All of Zone II + vertical bars at corners and junctions + gable band + plinth band
Zone IV and V: All of Zone III + seismic bands at all floor levels; max opening area 11% of wall area; max opening width ≤ B/2 (B = wall length)
Reinforcement in seismic bands: 2 bars minimum of 8–12 mm dia; stirrups at 150 mm c/c

🧰

Appendix · Formulas

Slenderness Ratio

$$ SR = \frac{H_{eff}}{t}\ \text{or}\ \frac{L_{eff}}{t} $$

Use whichever is smaller; max SR = 27

—

Effective Height — Both Ends Restrained

$$ H_{eff} = 0.75H $$

Stiffest case; 2.0H for one end free

Effective Length — Returns Both Ends

$$ L_{eff} = 0.8L $$

2.0L if wall is free at both ends

🧰

Appendix · Solved Example

GATE CE / ESE level

A masonry wall has height H = 3.0 m, thickness t = 230 mm, and is restrained horizontally and rotationally at both ends. Check its slenderness ratio against the IS 1905 limit.

Approach hint

Find $H_{eff}=0.75H$ for the given end condition, then compute SR = $H_{eff}/t$ and compare to the max SR = 27.

$H_{eff} = 0.75\times3.0 = 2.25$ m = 2250 mm
$SR = H_{eff}/t = 2250/230 = 9.8$
Compare to max SR = 27

✓ SR = 9.8 ≤ 27 — well within the IS 1905 slenderness limit

🧰

Appendix · Diagrams

Click any diagram to zoom

Fig 16.1 – Effective height factors for masonry walls under four end-restraint conditions (IS 1905 Cl. 5.2.2) — stiffer restraint gives a shorter effective (buckling) height.

🧰

Appendix · Quick Quiz

3 questions · tap an option to reveal answer

Question 1

The maximum permissible slenderness ratio for unreinforced masonry (IS 1905 Cl. 5.4.1) is:

Max SR = 27 for unreinforced masonry walls/columns — well below the RCC column short/slender threshold of 12 in a different ratio definition ✓

Question 2

In seismic Zone IV and V, per IS 4326, the maximum opening area permitted in a load-bearing masonry wall is:

IS 4326 restricts openings to 11% of wall area in Zones IV–V (with additional width limits ≤ B/2) since large openings weaken the wall's in-plane shear capacity during shaking ✓

Question 3

A masonry wall fully restrained (both horizontally and rotationally) at both ends has an effective height of:

Full restraint at both ends gives the shortest effective height, 0.75H — the same "stiffer ends → shorter effective length" principle seen in RCC and steel columns ✓

🧱

Ch 16B · Retaining Walls, Water Tanks & Staircases

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16B.1 RCC Retaining Walls

A retaining wall holds back earth (or other material) where there is an abrupt change in ground level. RCC cantilever and counterfort types resist the lateral earth pressure computed by Rankine/Coulomb theory.

Type	Description	Economic height
Gravity wall	Plain concrete/masonry; resists overturning by self-weight only	up to ~3 m
Cantilever wall	RCC stem + base slab (heel + toe) acting as cantilevers	~3–6 m
Counterfort wall	Stem & heel tied by counterforts (tension ties) at intervals	> 6 m
Buttress wall	Like counterfort but ribs on the front (compression) face	> 6 m

Stability checks (per unit length): (i) factor of safety against overturning $\ge 1.55$ about the toe; (ii) against sliding $\ge 1.55$; (iii) base pressure $\le$ SBC with no tension (resultant in the middle third). The active earth pressure varies linearly, giving a triangular thrust $P_a$ acting at $H/3$ from the base.

16B.2 RCC Water Tanks (Liquid-Retaining Structures, IS 3370)

Liquid-retaining structures are designed on the no-crack (uncracked) philosophy so that steel stress is kept low and concrete stays in the permissible tensile range — crack width is limited to 0.2 mm to prevent leakage. Working-stress method with reduced permissible stresses is used even where LSM governs elsewhere.

Classification: resting on ground, underground, and elevated (overhead) tanks; shapes — rectangular, circular, Intze.
Circular tank (walls in hoop tension): hoop tension $T = \gamma H \cdot D/2$ per unit height; $A_{st} = T/\sigma_{st}$ with $\sigma_{st}$ = 115–150 MPa (HYSD, IS 3370).
Rectangular tank: walls act in combined bending + direct tension; analysed by moment-coefficient (IS 3370 Part IV) methods for the wall spanning conditions.
Base slab & roof: designed for the water load, self weight and, for underground tanks, uplift when empty.

16B.3 RCC Staircases (IS 456 Cl. 33)

A stair slab is designed as a one-way slab spanning either along the going (longitudinal) or between supporting walls/stringer beams (transverse), depending on how it is supported.

Effective span (IS 456 Cl. 33.1): depends on how the landing is supported — includes going + landing(s) when the stair spans longitudinally.
Loading: self weight of waist slab + steps + finishes + live load (3–5 kN/m²); the waist-slab self weight acts along the inclined length, so it is converted to the horizontal (plan) span.
Dog-legged & open-well arrangements; the flight is designed as a slab, main bars along the span, distribution bars transverse.

🧱

Ch 16B · Formulas

Active Earth Pressure (Rankine)

$$ P_a = \tfrac{1}{2}\,K_a\,\gamma\,H^2,\qquad K_a = \frac{1-\sin\phi}{1+\sin\phi} $$

Thrust acts at H/3 above base

kN/m

Retaining Wall — Stability

$$ \text{FoS}_{overturn}=\frac{\sum M_R}{\sum M_O}\ge 1.55,\quad \text{FoS}_{slide}=\frac{\mu\sum W}{P_a}\ge 1.55 $$

No tension at base ⇒ resultant in middle third

—

Circular Tank — Hoop Tension

$$ T = \frac{\gamma\,H\,D}{2},\qquad A_{st} = \frac{T}{\sigma_{st}} $$

IS 3370: crack width ≤ 0.2 mm; σ_st ≈ 115–150 MPa

kN/m, mm²

📚

Ch 17 · Master Formulae & Quick Revision — RCC & PSC

Cross-chapter recall sheet, mnemonics, critical numerical values, mini quiz and practice questions

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Critical Formulae — One-Page Reference

Concrete Properties

$$ E_c=5000\sqrt{f_{ck}},\quad f_r=0.7\sqrt{f_{ck}} $$

Creep θ=1.6 (28d); $\varepsilon_{cu}=0.0035$; $f_{cd}=0.45f_{ck}$; $f_{yd}=0.87f_y$

—

WSM Section

$$ m=\frac{280}{3\sigma_{cbc}},\quad \frac{bx^2}{2}=mA_{st}(d-x) $$

Lever arm z=d−x/3; $x_c/d=m\sigma_{cbc}/(m\sigma_{cbc}+\sigma_{st})$

—

LSM Flexure (Singly Reinforced)

$$ M_{u,lim}(Fe415)=0.138f_{ck}bd^2 $$

$x_{u,max}/d=0.48$ (Fe415); C=0.36$f_{ck}$bx_u; T=0.87$f_y A_{st}$; z=d−0.42x_u; $A_{st,min}=0.85bd/f_y$; $A_{st,max}$=4%$A_g$

—

LSM Flexure (Doubly Reinforced)

$$ M_{u2}=M_u-M_{u,lim},\quad A_{sc}=\frac{M_{u2}}{f_{sc}(d-d')} $$

$A_{st2}=f_{sc}A_{sc}/(0.87f_y)$; $A_{st}=A_{st,lim}+A_{st2}$

—

Shear & Bond

$$ \tau_v=\frac{V_u}{bd},\quad S_v=\frac{0.87f_yA_{sv}d}{V_{us}} $$

Min $A_{sv}/(bS_v)\ge0.4/(0.87f_y)$; max Sv=0.75d or 300mm. $L_d=0.87f_y\phi/(4\tau_{bd})\approx47\phi$ (Fe415,M20). Lap: 1.3$L_d$ tension, 1.0$L_d$ compression

—

Torsion

$$ V_e=V_u+\frac{1.6T_u}{b},\quad M_t=\frac{T_u(1+D/b)}{1.7} $$

Equivalent shear/moment method (IS 456 Cl. 41)

—

Columns

$$ P_u=0.4f_{ck}A_c+0.67f_yA_{sc} $$

$e_{min}=\max(L/500+D/30,20mm)$; additional moment $M_a=P_u(l_{eff}/D)^2\cdot D/2000$ (slender column)

—

Prestressed Concrete

$$ f=\frac{P}{A}\pm\frac{Pey}{I}\pm\frac{My}{I},\quad w_{eq}=\frac{8Pe}{L^2} $$

Friction: $P_x=P_0e^{-\mu\alpha-kx}$; total losses ≈ 15–25%

—

Earthquake

$$ A_h=\frac{ZIS_a}{2Rg},\quad V_B=A_h\cdot W $$

$Q_i=V_B W_ih_i^2/\Sigma(W_jh_j^2)$; $T_a=0.075h^{0.75}$ (RC frame); max drift=0.004×storey height

—

Critical Numerical Values Table

Parameter	Value
Max strain in concrete at ULS	0.0035
Max strain in steel (Fe415) for yielding	$0.87\times415/200000+0.002=0.00383$
$x_{u,max}/d$: Fe250 / Fe415 / Fe500	0.53 / 0.48 / 0.46
$M_{u,lim}$ coefficient: Fe415	0.138 $f_{ck}$ b d²
Design compressive strength $f_{cd}$	0.45 $f_{ck}$ (= 0.67$f_{ck}$/1.5)
Design tensile strength in steel $f_{yd}$	0.87 $f_y$ (= $f_y$/1.15)
Partial safety factor: concrete ($\gamma_m$)	1.5
Partial safety factor: steel ($\gamma_m$)	1.15
Partial safety factor: DL + LL	1.5 DL + 1.5 LL
Min. cover: beams / columns / slabs / footings	25 / 40 / 20 / 50 mm (mild exp.)
Min. $A_{sc}$ in column	0.8% of $A_g$
Max. $A_{sc}$ in column	4% of $A_g$ (6% at lap)
Min. bars: rectangular column / circular	4 / 6
Dev. length $L_d$: Fe415, M20, tension (HYSD)	≈ 47φ
Lap splice tension / compression	1.3 $L_d$ / 1.0 $L_d$
Max stirrup spacing (shear)	0.75d or 300 mm
Max storey drift (IS 1893)	0.004 × storey height
Seismic Zone V factor Z	0.36
PSC min. concrete grade (pre-tensioned)	M40
PSC total prestress losses (typical)	15–25%
Punching shear perimeter (footing)	at d/2 from column face
Modulus of elasticity of steel $E_s$	2 × 10⁵ MPa

Mnemonics & Memory Aids

💡 $x_{u,max}/d$ for steel grades (easy recall):
Fe250 → 0.53 | Fe415 → 0.48 | Fe500 → 0.46 | Fe550 → 0.44
Higher grade steel → lower limiting NA depth (more ductile behaviour enforced)

$M_{u,lim}$ key values:
Fe250: 0.148 | Fe415: 0.138 | Fe500: 0.133 | Fe550: 0.128 → all × $f_{ck}bd^2$
Higher strength steel → smaller $M_{u,lim}$ coefficient (NA moves up)

Cover (mm) – mild exposure:
Slab: 20 | Beam: 25 | Column: 40 | Footing: 50 | PSC: 20
"Slabs Barely Cause Footing Problems"

Seismic Zone factors:
Zone II: 0.10 | Zone III: 0.16 | Zone IV: 0.24 | Zone V: 0.36
Zone II → 10; each zone ×1.5 roughly; Zone V worst

PSC Loss Sources (mnemonic FESCAR):
F-riction | E-lastic shortening | S-hrinkage | C-reep | A-nchorage slip | R-elaxation

Design stages for a beam:
Size → Effective depth → Neutral axis check → Reinforcement → Shear → Bond → Deflection check

Exam-Angle Comparison Table

Topic	GATE Focus	ESE Focus	SSC JE Focus
Design Philosophy	LSM computations; partial safety factors	WSM vs LSM comparison; historical context	Basic definitions
Flexure (Singly/Doubly)	$M_{u,lim}$, $A_{st}$ design	Full doubly-reinforced design from scratch	$M_{u,lim}$ formula recall
Shear & Bond	Stirrup spacing design; $\tau_c$ table lookup	Full shear + bond design; detailing checks	$\tau_v$ formula; basic spacing rules
Slabs	Two-way BM coefficients; l/d ratios	Complete slab design incl. corners	One-way vs two-way identification
Columns	Axial capacity; interaction diagram reading	Biaxial bending; slender column design	Min/max steel; basic axial formula
Foundations	Footing area, one/two-way shear	Combined/strap footing design	Footing types identification
PSC	Losses; fibre stress formula	Complete PSC beam design; load balancing	Pre vs post-tensioning concept
Earthquake	Base shear calculation; zone factors	Ductile detailing; irregularity checks	Zone factor values only

Mini Quiz

A singly reinforced beam (b=300mm, d=550mm) uses M25 concrete and Fe415 steel. Find $M_{u,lim}$.

$M_{u,lim}=0.138\times25\times300\times550^2=0.138\times25\times300\times302{,}500\approx313.5\times10^6$ N·mm = 313.5 kN·m ✓

Which of the following is NOT one of the six recognised sources of prestress loss (FESCAR)?

FESCAR = Friction, Elastic shortening, Shrinkage, Creep, Anchorage slip, Relaxation — fatigue is a separate serviceability concern, not a prestress loss source ✓

A column in seismic Zone V (Z=0.36) with I=1.0, R=5.0 (special SMRF), and $S_a/g=2.5$: find $A_h$.

$A_h=(Z/2)(I/R)(S_a/g)=(0.36/2)\times(1.0/5.0)\times2.5=0.18\times0.2\times2.5=0.09$ ✓

Which reinforcement limit is correct for a beam per IS 456?

For beams: $A_{st,min}=0.85bd/f_y$ and $A_{st,max}=4\%$ of gross area — the 0.12%/0.15% rule applies to slabs, not beams ✓

A short RCC column has $l_{eff}/D = 10$. Per IS 456, this column is classified as:

Since $l_{eff}/D=10 \le 12$, this is a short column — buckling (P-delta) effects are negligible and no additional slenderness moment is required ✓

Practice Questions

EasyState the IS 456 assumption regarding the maximum compressive strain in concrete and the corresponding stress block shape at the ultimate limit state.

The maximum compressive strain in concrete at the extreme fibre is taken as 0.0035. The IS 456 design stress block is parabolic from zero strain up to 0.002 strain (peak stress 0.446$f_{ck}$), then constant (flat) from 0.002 to the ultimate strain 0.0035.

EasyWhy must the minimum eccentricity $e_{min}$ always be considered in column design, even for a load that is theoretically applied exactly at the centroid?

In practice, no column is perfectly straight, no load is perfectly centred, and construction tolerances (formwork misalignment, bar placement error) always introduce some unintended eccentricity. IS 456 accounts for this by mandating a minimum design eccentricity $e_{min}=\max(L/500+D/30,\ 20\text{mm})$ even for nominally "axial" columns, ensuring the design is never unconservative due to idealised assumptions.

MediumA doubly reinforced beam requires $M_u=350$ kN·m with $M_{u,lim}=280$ kN·m, $f_{sc}=353$ MPa, and (d−d')=460 mm. Find the required compression steel area $A_{sc}$.

$M_{u2}=M_u-M_{u,lim}=350-280=70$ kN·m = 70×10⁶ N·mm. $A_{sc}=M_{u2}/[f_{sc}(d-d')]=70\times10^6/(353\times460)=70\times10^6/162{,}380\approx431$ mm².

MediumExplain why the punching (two-way) shear check for a footing is performed at a critical perimeter d/2 from the column face, while the one-way (beam-type) shear check uses a section at d from the face — why the difference?

One-way shear behaves like beam shear along a straight critical section, and experimental/theoretical studies place the critical section at 'd' from the support face (accounting for the shear-transfer zone through the compression strut near the support). Punching shear, by contrast, is a fundamentally different failure mode — a truncated-pyramid-shaped cone punches through the slab around the column — and the critical perimeter for this cone geometry, calibrated against test data, is located at d/2 (half the effective depth) from each face of the loaded area, following the typical ~45° punching cone angle.

HardA structural engineer proposes reducing prestress losses in a post-tensioned bridge girder by using a straight (rather than parabolic) tendon profile. Explain the trade-off this creates and why parabolic profiles are almost always preferred for simply supported PSC beams under UDL.

A straight tendon does reduce friction losses somewhat (no curvature angle α, only wobble), but it provides constant eccentricity along the span, meaning the balancing upward force is concentrated only at the anchorages (as end moments) rather than distributed as a UDL — it cannot counteract a distributed gravity load (self-weight + UDL) efficiently, since $w_{eq}=8Pe/L^2$ requires a parabolic profile to generate an equivalent *distributed* upward load matching the shape of the applied UDL. A straight tendon would leave significant net sagging moment un-balanced at midspan even though the end regions might be over-compressed, requiring a much higher prestress force (and correspondingly larger losses and larger anchorage zones) to achieve the same service-stress limits — so despite the friction penalty, parabolic profiles are almost universally preferred for simply-supported beams under uniformly distributed load.

Material	LSM \(\gamma_m\)	Design Strength
Concrete	1.5	\(f_{cd}\) = 0.67 \(f_{ck}\)/1.5 = 0.45 \(f_{ck}\)
Steel (Fe250, Fe415, Fe500)	1.15	\(f_{yd} = f_y\)/1.15 = 0.87 \(f_y\)

Parameter	Beams	Slabs
Min. \(A_{st}\)	0.85 bd / \(f_y\) (IS 456 Cl. 26.5.1.1)	0.12% of total cross-section (HYSD); 0.15% (mild steel)
Max. \(A_{st}\)	4% of gross cross-sectional area (tension + compression combined)	As for beams

Criterion	One-Way Slab	Two-Way Slab
Aspect ratio \(l_y/l_x\)	> 2	≤ 2
Load transfer	Primarily in shorter span (x)	In both directions
Main reinforcement	In short span only	In both spans
Examples	Verandah slabs, one-way spanning	Rooms, two-way panels

Basis	Type	Definition
Slenderness	Short Column	\(l_{eff}\)/D ≤ 12 (both directions); failure by material
Slenderness	Slender/Long Column	\(l_{eff}\)/D > 12; additional moments due to deflection (IS 456 Cl. 39.7)
Loading	Axially Loaded	Concentric axial force only (rare in practice)
Loading	Eccentrically Loaded	Axial force + bending moment (uniaxial or biaxial)
Lateral Ties	Tied Column	Rectangular links confine concrete; most common
Lateral Ties	Spiral Column	Helical reinforcement; better ductility; circular section

Parameter	IS 456 Requirement
Min. \(A_{sc}\)	0.8% of gross cross-sectional area
Max. \(A_{sc}\)	4% (general); 6% at lapping (local maximum)
Min. bar dia (longitudinal)	12 mm
Min. no. of bars: rectangular / circular	4 / 6
Lateral ties (links): dia	Max (\(\phi_{main}\)/4, 6 mm)
Lateral ties: spacing	Min (\(b_{least}\), 16\(\phi_{main}\), 300 mm)
Cover to main bars	40 mm (mild exposure)

Feature	RCC	PSC
Steel type	Mild steel / HYSD (\(f_y\) 250–550 MPa)	High-tensile (HT) wires/strands (\(f_{pu}\) 1600–2000 MPa)
Concrete grade	M20–M50	Min. M40 (pre-tensioned); M30 (post-tensioned)
Cracking	Cracked in tension (design accepts cracks)	No cracking (Type I) or limited cracking (Type II, III)
Deflection	Higher; affected by cracking	Lower; camber by prestress; load-balancing possible
Shear resistance	Stirrups needed	Better inherent shear resistance; fewer stirrups
Span range	Up to ~20 m economical	20–100+ m bridges, long-span structures
Initial cost	Lower	Higher (high-tensile steel + equipment)

Stage	Stress Limit in Concrete
Transfer (initial prestress, no live load): Compression	0.44 \(f_{ci}\) (IS 1343)
Transfer: Tension	0 (no tension in Type I) or 1.0 MPa (Type II pre-tensioned)
Service (effective prestress + DL + LL): Compression	0.33 \(f_{ck}\)
Service: Tension (Type I — fully prestressed)	0 (no tension permitted)
Service: Tension (Type II — limited prestress)	≤ 0.5 MPa (or 1.0 MPa in some cases)
Service: Tension (Type III — partial prestress)	Limited crack width ≤ 0.1 mm

Member	Key Requirement
Beams	\(p_{t,min} \ge 0.24\sqrt{f_{ck}}/f_y\); \(p_{t,max}\) ≤ 2.5%; confinement hoops; top steel ≥ bottom steel/2 at all sections
Columns	Min 1% steel; max 4%; confinement hoops in plastic hinge zones (both ends, full height for short columns)
Joints (Beam-Column)	Shear check per IS 13920; closed hoops in joint; no bar splicing in joint
Shear walls	Min. thickness 150 mm; min. vertical + horizontal steel 0.25% each; boundary elements at ends
Confinement hoops	Spacing ≤ min (D/4, 8\(\phi_{long}\), 100 mm) in critical zones
Strong column – Weak beam	\(\Sigma M_{columns} \ge 1.2\Sigma M_{beams}\) at joint (IS 13920 Cl. 9.2)

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Learning Objectives

Exam Relevance

Concept Map

Where This Fits

1.1 Introduction

1.2 Types of Concrete

1.3 Importance of Design Codes

1.4 & 1.5 Characteristic Strength and Grades of Concrete

1.8 Behaviour under Uniaxial Compression

1.9 Behaviour in Tension

1.10 Modulus of Elasticity & Poisson's Ratio

1.12 Creep of Concrete

Factors increasing creep:

1.13 Compressive Strength in Structural Design

2.1 Introduction

2.2 Necessity of Designing RCC Structures

Why RCC specifically needs design (unlike timber or masonry by rule-of-thumb):

2.3 Hydraulic & Non-Hydraulic Cements

Non-Hydraulic Cements

Hydraulic Cements

2.4 Tests on Cement

① Fineness Test (IS 4031 Pt 1 & Pt 2)

② Consistency Test — Normal / Standard Consistency (IS 4031 Pt 4)

③ Setting Time Test (IS 4031 Pt 5)

④ Soundness Test (IS 4031 Pt 3)

⑤ Strength Test (IS 4031 Pt 6 & IS 10080)

⑥ Heat of Hydration Test (IS 4031 Pt 9)

⑦ Specific Gravity of Cement (IS 4031 Pt 11)

Summary Table – All Cement Tests

2.5 Durability – Chemical Attack

2.6 Design Philosophies

2.7 & 2.8 Limit State Method – Partial Safety Factors (IS 456:2000)

Partial Safety Factors for Materials (\(\gamma_m\))

Partial Safety Factors for Loads (\(\gamma_f\)) – IS 456 Table 18

2.10 Properties of Reinforcement

2.11 Cover Requirements (IS 456:2000 Cl. 26.4)

2.12 Spacing of Reinforcement (IS 456 Cl. 26.3)

2.14 Major Causes of Structural Failure

2.15 List of Key IS Codes for RCC

3.1 Introduction

3.5 Transformed Section & 3.6 Modular Ratio

3.6 Permissible Stresses in WSM (IS 456 Table B-1 & B-2)

3.3 Transformed Section Analysis (Key Equations)

3.9 Cracking Moment

4.1 Introduction – Limit States

4.3 Assumptions in LSM (IS 456 Cl. 38.1)

4.4 Singly Reinforced Sections – Key Formulae

4.5 Limiting (Balanced) Neutral Axis Depth (IS 456 Cl. G-1.1)

4.5 Limiting Moment of Resistance (\(M_{u,lim}\))

4.6 Deflection Control – Span/Depth Ratio (IS 456 Cl. 23.2)

4.7–4.8 Minimum & Maximum Reinforcement Limits (IS 456)

4.12 One-Way Slabs

5.1 Introduction

5.4 Stress in Compression Steel (\(f_{sc}\))

5.7–5.8 Design Steps for Doubly Reinforced Beam

6.6 Nominal Shear Stress (IS 456 Cl. 40.1)

6.8 Shear Strength without Reinforcement (\(\tau_c\)) – IS 456 Table 19

6.9 Design of Shear Reinforcement

6.10 Minimum Shear Reinforcement (IS 456 Cl. 26.5.1.6)

6.11 Maximum Spacing of Shear Reinforcement

7.4 Bond Stress & 7.8 Development Length

7.5 Types of Bond

7.11 Hooks, Bends & Standard Anchorages (IS 456 Cl. 26.2)

7.16 Splicing of Reinforcement (IS 456 Cl. 26.2.5)

8.1 Introduction – When Does Torsion Occur?

8.8 IS 456 LSM Torsion Design (Cl. 41)

8.9 Torsional Reinforcement Requirements (IS 456 Cl. 41.4)

9.3 Effective Width of Flange (IS 456 Cl. 23.1.2)

9.5 Analysis of Flanged Sections

Case 1: Neutral Axis within flange (\(x_u \le D_f\)) — most common

Case 2: Neutral Axis in web (\(x_u > D_f\))

10.3 Deflection Limits (IS 456 Cl. 23.2)

10.3 Deflection Calculation – Components

10.4 Cracking Limits (IS 456 Cl. 35.3)

10.5 Other Serviceability Limits

11.2 One-Way vs Two-Way Slabs

11.5 IS 456 Method for Two-Way Slabs (Cl. D-1)

11.6 Thickness of Two-Way Slabs

11.4 Support Conditions & Edge Discontinuity