01

What is Reinforced Concrete?

Civil View

Why engineers love it

Concrete is cheap, fireproof, moldable into any shape, and excellent in compression. But you can't build a beam from plain concrete — it'll crack the moment it bends. Steel bars embedded in the tension zone fix that. The result is a material that's strong everywhere, economical, and durable for 100+ year service lives.

Physics View

What's happening inside

Under bending, a beam develops a neutral axis. Above it, the material is squeezed (compression). Below it, the material is stretched (tension). Concrete can handle the squeezing — its aggregate particles interlock and resist crushing. But tension pulls the brittle cement matrix apart. Steel, being ductile, handles the pulling with ease.

Math View

The numbers that matter

Composite action requires compatibility — both materials must strain together. This means:

\[ \varepsilon_s = \varepsilon_c \quad \text{(strain compatibility)} \] \[ f_c \approx 3{,}000 \text{–} 6{,}000 \text{ psi (compression)} \] \[ f_t \approx 300 \text{–} 600 \text{ psi (tension — ignored in design)} \]
Core Idea

Reinforced concrete is a composite: concrete resists compression, steel resists tension — together they handle everything a structure can throw at them.

Why it matters: Every RC design decision — bar placement, cover, section depth — traces back to this fundamental division of labor between the two materials.

Animation: RC Beam Under Load

Watch how the neutral axis stays at mid-height for a symmetric section. Compression zone (top) resists crushing; tension zone (bottom) relies on steel bars.

Calculator: Concrete Cover & Effective Depth

Find the effective depth d from overall beam depth, cover, stirrup size, and bar diameter.

Worked Example 1.1

Identifying Tension & Compression Zones

Problem: A simply-supported RC beam with span L = 20 ft carries a uniform load w = 2 k/ft. The section is 12 in × 24 in overall, with 3 × #8 bars at the bottom. Identify the stress regions and explain where each material is working.

1
Loading condition: Simply supported → sagging moment at midspan. Midspan moment: \( M = \frac{wL^2}{8} = \frac{2(20)^2}{8} = 100 \text{ kip·ft} \)
2
Compression zone: The top of the beam is compressed. Concrete handles this with its high f'c capacity (say 4000 psi). The compression block depth 'a' will be calculated in Chapter 2.
3
Tension zone: The bottom is in tension. Plain concrete (ft ≈ 400–500 psi) would crack immediately. The 3 × #8 bars (As = 3 × 0.79 = 2.37 in²) carry the entire tensile force.
4
Neutral axis: The boundary between compression and tension. For this section it sits above the bar centroid. Finding its exact position is the core task of flexural analysis.
Conclusion: Top 40–50% of section → concrete in compression. Bottom → steel in tension, concrete ignored. This is the fundamental RC assumption.

Quick Reference — RC Material Properties

Concrete compressive strengthf'c = 3000–6000 psi
Concrete tensile strengthfr ≈ 7.5√f'c psi (ignored in flexure)
Steel yield strengthfy = 40, 60, or 75 ksi
Thermal expansion (steel)6.5 × 10⁻⁶ per °F
Thermal expansion (concrete)5.5 × 10⁻⁶ per °F (compatible!)
Unit weight of RCwc = 150 pcf (normal weight)
02

Concrete Constituents & Cement

Civil View

What goes into the mix

Concrete = cement + coarse aggregate + fine aggregate + water + (optional) admixtures. In Bangladesh, coarse aggregate is typically brick chips or stone chips. Fine aggregate is sylhet sand or river sand. Each ingredient has a job: cement binds, aggregates provide bulk and economy, water triggers hydration, admixtures tune properties.

Physics View

Hydration: the chemistry that hardens concrete

Cement (CaO, SiO₂, Al₂O₃) reacts with water in a process called hydration. This produces calcium silicate hydrate (C-S-H) gel — the glue that binds everything together. The gel fills voids and continues to gain strength for years, though 28 days is the standard design benchmark. More water = more pores = weaker matrix.

Math View

Mix proportioning

Mix ratios are expressed as cement:sand:coarse aggregate by weight or volume. A common 1:2:4 mix means:

\[ \text{Ratio} = 1 \text{ (cement)} : 2 \text{ (sand)} : 4 \text{ (gravel)} \] \[ V_\text{total} = V_c + V_{fa} + V_{ca} + V_w + V_\text{air} \] \[ \text{For 1 m}^3\text{: } M_c = \frac{\rho_c \cdot 1}{1 + 2 + 4} \approx 280 \text{ kg/m}^3 \]
Core Idea

Every ingredient in a concrete mix has a specific purpose — change one and you change the balance between strength, workability, durability, and cost.

Cement type matters: OPC (Grade 43/53) for general structures. PPC for marine/mass concrete (lower heat, better sulfate resistance). Rapid hardening for urgent repairs. Low heat for dams. Choosing wrong adds cost or risks premature failure.

Cement Types at a Glance

TypeKey PropertyUse Case
OPC (33/43/53 grade)Standard strength gainBuildings, bridges, pavements
Portland Pozzolana (PPC)Low heat, sulfate resistantMarine works, hydraulic structures
Rapid HardeningEarly high strengthPrecast, urgent repairs, road work
Low HeatMinimal hydration heatMass concrete (dams, retaining walls)
Sulphate ResistingResists sulfate attackFoundations in sulfate-rich soil
White / ColoredAesthetic finishDecorative, architectural

Animation: Concrete Mix — Aggregate Packing & Hydration

Left: dry mix (cement, coarse & fine aggregate). Right: after water addition — hydration gel forms and fills voids, binding everything into an artificial stone.

Calculator: Mix Proportioning (Gravimetric)

Given a target volume and mix ratio, find the mass of each ingredient.

Worked Example 2.1

Materials for 1 m³ of 1:2:4 Concrete

Problem: Calculate the mass of cement, sand, and coarse aggregate required for 1 m³ of nominal mix concrete using a 1:2:4 ratio by weight. Assume bulk unit weight of dry concrete ingredients = 1500 kg/m³. W/C = 0.50.

1
Sum of ratios: \( 1 + 2 + 4 = 7 \)
2
Total dry mass for 1 m³: \( M_\text{total} = 1500 \times 1 = 1500 \text{ kg} \)
3
Cement: \( M_c = \frac{1}{7} \times 1500 = 214 \text{ kg} \approx 4.3 \text{ bags (50 kg each)} \)
4
Fine aggregate (sand): \( M_{fa} = \frac{2}{7} \times 1500 = 429 \text{ kg} \)
5
Coarse aggregate: \( M_{ca} = \frac{4}{7} \times 1500 = 857 \text{ kg} \)
6
Water: \( M_w = 0.50 \times 214 = 107 \text{ kg} = 107 \text{ liters} \)
Answer: Cement = 214 kg, Sand = 429 kg, Coarse Agg = 857 kg, Water = 107 L per 1 m³ of concrete.

Quick Reference — Concrete Constituents

Coarse aggregate (CA)Size > ¼ in (6 mm); stone/brick chips
Fine aggregate (FA)Size ≤ ¼ in; sylhet/river sand
Cement hydration time~28 days for full design strength
Cement raw materialsCaO, SiO₂, Al₂O₃ (limestone, shale)
Portland cement patentJoseph Aspdin, 1824 (England)
RC patentJoseph Monier, 1867 (France)
03

Water-Cement Ratio

Civil View

The most important single number in mix design

Every spec sheet for structural concrete includes a maximum W/C ratio. ACI 318 requires W/C ≤ 0.45 for exposure to freezing/thawing, ≤ 0.50 for moderate sulfate exposure. Why? Because W/C controls durability and strength simultaneously. Reducing it is the single most effective way to produce better concrete.

Physics View

Water creates pores — pores kill strength

Only about W/C = 0.25 of water is needed for complete hydration. Any water beyond that creates capillary pores that remain empty after evaporation. These voids are stress concentrators — cracks initiate there under load. More excess water → more pores → lower strength. Think of it like Swiss cheese: more holes = weaker cheese.

Math View

Abrams' Law

\[ w/c = \frac{W_w}{W_c} \] \[ f'_c \approx \frac{A}{B^{w/c}} \quad \text{(Abrams, 1918)} \] \[ \text{General range: } w/c = 0.40 \text{ to } 0.60 \] \[ \text{High strength: } w/c \leq 0.35 \]

Where A and B are empirical constants depending on cement type and curing.

Core Idea

Lower W/C = more strength and durability, but less workability. Mix design is about finding the sweet spot for your specific job conditions.

The workability tradeoff: A concrete mix with W/C = 0.40 might give you f'c = 6000 psi but will be stiff and hard to place in congested rebar cages. Superplasticizers (admixtures) break this tradeoff — they fluidize the mix without adding water.

Animation: W/C Ratio vs. Strength & Workability

Drag the slider to change W/C ratio. Watch how compressive strength (blueprint bars) and workability (amber bars) respond inversely.

Calculator: Water-Cement Ratio & Water Content

Worked Example 3.1

W/C Ratio for Target Strength

Problem: A structural engineer specifies f'c = 4000 psi for a column. Using Abrams' law approximation and ACI guidance, estimate the required W/C ratio and calculate the water content if cement = 380 kg/m³. Also check if workability will be acceptable for placement.

1
Target strength with margin: Required average \( f'_{cr} = f'_c + 1200 = 5200 \text{ psi} \) (ACI 301 margin for unknown variability)
2
Estimate W/C from ACI Table 6.3.4.3: For f'cr ≈ 5200 psi, W/C ≈ 0.44 (interpolating between 5000 psi → 0.48 and 6000 psi → 0.41)
3
Water content: \( W_w = 0.44 \times 380 = 167 \text{ kg/m}^3 = 167 \text{ liters} \)
4
Workability check: For 167 L/m³ with moderate cement content → slump likely 50–75 mm. Acceptable for beam/column placement. If rebar is congested, consider adding superplasticizer, not more water.
Answer: W/C = 0.44, Water = 167 kg/m³. Workability is acceptable; if not, use superplasticizer — never increase water.

Quick Reference — W/C Ratio

FormulaW/C = Weight of water ÷ Weight of cement
General structural range0.40 – 0.60
High-strength concreteW/C ≤ 0.35 (with HRWR)
Freeze-thaw exposure (ACI)W/C ≤ 0.45
Min. water for hydration onlyW/C ≈ 0.25 (not workable)
Workability measureSlump test (ASTM C143)
04

Compressive Strength f'c

Civil View

The master design variable

f'c is the 28-day compressive strength of a standard 6×12 in cylinder, tested in a compression machine. Every formula in ACI 318 — for flexure, shear, development length, deflection — references f'c. It's the number your concrete mix design must hit, and what your QC program must verify batch by batch.

Physics View

What happens when concrete crushes

Initially the concrete is elastic — stress and strain increase proportionally. Around 30–40% of f'c, microcracks begin forming at aggregate-paste interfaces. By 70–80% of f'c, cracks propagate and link up. At f'c, the specimen suddenly fractures in a cone-and-shear pattern (brittle failure). The post-peak behavior is very sensitive to loading rate and confinement.

Math View

Cylinder test and acceptance criteria

\[ f'_c = \frac{P}{A} = \frac{P}{\pi r^2} \quad \text{(6×12 in cylinder)} \] \[ \varepsilon_{cu} \approx 0.003 \quad \text{(ACI ultimate strain)} \] \[ \text{Acceptance: } \bar{f}_{c,3} \geq f'_c \text{ AND min} \geq f'_c - 500 \text{ psi} \]
Core Idea

f'c is the fundamental material property of concrete — everything in RC design is calibrated around it. Know it, specify it, test it, and never guess it.

Animation: Concrete Stress-Strain Curve in Compression

The curve climbs linearly (elastic range), curves over near peak (microcracking), reaches f'c at ε ≈ 0.002, then drops. ACI assumes the design limit at εcu = 0.003.

Calculator: Cylinder Compressive Strength

Given cylinder dimensions and failure load, compute f'c. Also check ACI acceptance.

Worked Example 4.1

Cylinder Test Result & ACI Acceptance Check

Problem: Three consecutive 6×12 in cylinders from a batch were tested at 28 days: P₁ = 122 k, P₂ = 135 k, P₃ = 128 k. The specified f'c = 4000 psi. Check ACI acceptance criteria.

1
Cylinder area: \( A = \pi(3)^2 = 28.27 \text{ in}^2 \)
2
Individual strengths: \( f_1 = \frac{122{,}000}{28.27} = 4315 \text{ psi}, \quad f_2 = \frac{135{,}000}{28.27} = 4776 \text{ psi}, \quad f_3 = \frac{128{,}000}{28.27} = 4529 \text{ psi} \)
3
Average of 3: \( \bar{f}_c = \frac{4315 + 4776 + 4529}{3} = 4540 \text{ psi} \geq 4000 \text{ psi} \checkmark \)
4
Min individual check: Min = 4315 psi. Required min = 4000 − 500 = 3500 psi. 4315 ≥ 3500 ✓
Answer: Both ACI criteria satisfied. Batch accepted. f'c ≈ 4315–4776 psi, well above the 4000 psi spec.

Quick Reference — Compressive Strength f'c

Standard cylinder6 × 12 in or 4 × 8 in (ASTM C39)
Test age28 days (standard)
ACI: average of 3≥ f'c required
ACI: min individual≥ f'c − 500 psi (if f'c ≤ 5000 psi)
Ultimate strain (ACI)εcu = 0.003
Strain at peak stressε₀ ≈ 0.002
05

Modulus of Elasticity, Ec

Civil View

Why Ec matters for deflection

Ec controls how much a beam deflects under service loads. ACI 318 limits live-load deflection to L/360 for floors, L/480 for floors supporting brittle partitions. If you calculate the wrong Ec, your deflection prediction is off. In Bangladesh, where most concrete is normal-weight with modest f'c, Ec typically runs 2.5–3.5 million psi.

Physics View

What Ec represents physically

Ec is the slope of the initial linear portion of the stress-strain curve. A stiffer concrete (higher f'c, denser aggregates) resists deformation more — higher Ec. Normal-weight concrete (wc = 145–150 pcf) follows ACI's empirical formula closely. Lightweight concrete (wc = 90–115 pcf) has significantly lower Ec and needs a modified expression.

Math View

ACI Formula

\[ E_c = 57{,}000\sqrt{f'_c} \quad \text{(psi, normal-weight)} \] \[ E_c = 33 w_c^{1.5} \sqrt{f'_c} \quad \text{(general, } w_c \text{ in pcf)} \] \[ E_c = 4700\sqrt{f'_c} \quad \text{(MPa, SI)} \] \[ n = \frac{E_s}{E_c} = \frac{29{,}000{,}000}{E_c} \quad \text{(modular ratio)} \]
Core Idea

Ec links f'c to stiffness: stronger concrete is stiffer concrete. Ec grows only as the square root of f'c — doubling f'c only increases Ec by ~41%.

The modular ratio n = Es/Ec is used in transformed-section analysis (WSD method). For f'c = 3000 psi, n ≈ 9. For f'c = 4000 psi, n ≈ 8. ACI permits rounding n to the nearest whole number.

Ec & n Values for Common Concrete Grades

f'c (psi)Ec (psi)Ec (ksi)n = Es/Ec
3,0003,122,0003,1229.3 → use 9
4,0003,605,0003,6058.0 → use 8
5,0004,031,0004,0317.2 → use 7
6,0004,415,0004,4156.6 → use 7

Animation: Ec as the Slope of the Stress-Strain Curve

Ec = 3,605 ksi

The dashed blueprint line is the chord from origin — its slope is Ec. The amber lines mark f'c and εcu = 0.003.

Calculator: Modulus of Elasticity Ec & Modular Ratio n

Worked Example 5.1

Computing Ec and n for a Column Design

Problem: A normal-weight concrete column uses f'c = 4000 psi. Find Ec using both the simplified and general ACI formulas, then compute the modular ratio n. Es = 29,000,000 psi.

1
Simplified ACI formula (normal-weight, wc ≈ 145 pcf): \( E_c = 57{,}000\sqrt{4000} = 57{,}000 \times 63.25 = 3{,}605{,}000 \text{ psi} \approx 3605 \text{ ksi} \)
2
General ACI formula check: \( E_c = 33(145)^{1.5}\sqrt{4000} = 33 \times 1746.0 \times 63.25 = 3{,}645{,}000 \text{ psi} \) (≈ 1% difference — acceptable)
3
Modular ratio: \( n = \frac{E_s}{E_c} = \frac{29{,}000{,}000}{3{,}605{,}000} = 8.04 \approx 8 \)
Answer: Ec = 3,605 ksi, n = 8. Use n = 8 in transformed-section calculations for this column.

Quick Reference — Ec

ACI simplified (normal-weight)Ec = 57,000√f'c (psi)
ACI generalEc = 33 wc^1.5 √f'c (pcf, psi)
SI versionEc = 4700√f'c (MPa)
Modular ratio nn = Es/Ec = 29,000,000 / Ec
Es (steel)29,000 ksi (constant, all grades)
Normal-weight unit weightwc = 145–150 pcf
06

Steel Reinforcement

Civil View

Choosing bars for the job

ACI recognizes 11 standard bar sizes: #3 through #11, plus #14 and #18 (number = 1/8" increments of diameter). Grade 60 (fy = 60 ksi) dominates structural work. Grade 40 is used for light stirrups. Grade 75 for high-strength applications. For seismic design, ASTM A706 (controlled chemistry) is specified instead of A615 to ensure ductility.

Physics View

Why steel works so well with concrete

Three reasons: (1) Bond — deformed bar ribs mechanically interlock with concrete, preventing slip. (2) Thermal compatibility — steel expands at 6.5×10⁻⁶/°F, concrete at 5.5×10⁻⁶/°F — close enough to avoid differential cracking. (3) Protection — the alkaline concrete environment (pH ≈ 12.5) passivates the steel surface, preventing corrosion for decades.

Math View

Key stress-strain points

\[ E_s = 29{,}000{,}000 \text{ psi} = 29{,}000 \text{ ksi} \] \[ \varepsilon_y = \frac{f_y}{E_s} = \frac{60{,}000}{29{,}000{,}000} = 0.00207 \] \[ f_s = E_s \cdot \varepsilon_s \quad (\text{for } \varepsilon_s \leq \varepsilon_y) \] \[ f_s = f_y \quad (\text{for } \varepsilon_s > \varepsilon_y, \text{ perfect plasticity}) \]
Core Idea

Steel's perfectly plastic yield behavior is what makes RC design predictable and safe — a bar at yield gives the structure warning (large deflections) before it fails, unlike brittle concrete.

Tension-controlled failure: ACI requires RC beams to fail in a ductile tension-controlled mode (steel yields first, large deflection warns occupants) rather than a brittle compression-controlled mode (sudden concrete crushing, no warning). This is why there's a maximum reinforcement ratio ρmax.

Standard Rebar Sizes (ASTM A615 / A706)

Bar #Diameter (in)Area (in²)Weight (lb/ft)Common Use
#30.3750.110.376Stirrups, ties, light slabs
#40.5000.200.668Slabs, footings
#50.6250.311.043Beams, columns (light)
#60.7500.441.502Beams, columns
#70.8750.602.044Beams, columns
#81.0000.792.670Beams, columns (heavy)
#91.1281.003.400Heavy beams, columns
#101.2701.274.303Large columns, foundations
#111.4101.565.313Heavy members

Animation: Steel Stress-Strain Curve (Grade Comparison)

Notice the well-defined yield plateau in all grades. After yielding, stress stays at fy while strain keeps increasing — this is the ductility that makes RC safe.

Calculator: Rebar Area & Yield Force

Select bar size and quantity to get total steel area and the maximum tensile force T = As × fy.

Worked Example 6.1

Rebar Selection for a Required Tensile Force

Problem: A beam tensile force at midspan requires As ≥ 2.15 in². Using Grade 60 steel, select a suitable bar arrangement. Check that bars fit within a 12-in wide beam with adequate spacing.

1
Required As = 2.15 in². Try 3 × #8: As = 3 × 0.79 = 2.37 in² ≥ 2.15 ✓ (10% over — acceptable)
2
Alternate: 4 × #7: As = 4 × 0.60 = 2.40 in² ✓. More bars, smaller diameter.
3
Spacing check for 3 × #8 in 12-in beam: Cover = 1.5 in, stirrup = 0.375 in, bar dia = 1.0 in. Available width = 12 − 2(1.5) − 2(0.375) − 3(1.0) = 12 − 3 − 0.75 − 3 = 5.25 in for 2 gaps → spacing = 5.25/2 = 2.63 in ≥ 1.5 × 1.0 = 1.5 in ✓
4
Tensile force provided: \( T = A_s \cdot f_y = 2.37 \times 60 = 142.2 \text{ kips} \)
Answer: Use 3 × #8 bars (As = 2.37 in²). Spacing = 2.63 in > minimum. T = 142 kips available.

Quick Reference — Steel Reinforcement

Es (all grades)29,000 ksi (constant)
Grade 40fy = 40 ksi, εy = 0.00138
Grade 60fy = 60 ksi, εy = 0.00207
Grade 75fy = 75 ksi, εy = 0.00259
Bar # = diameter inN × (1/8) in (e.g. #8 = 1.0 in dia)
Min clear spacingmax(1.5db, 1.5 in, 4/3 × max aggregate)