Reference

Civil Engineering Formula Reference

The formulas behind every QSCivilCalc tool, written out in full — what each symbol means, the units it must be in, a worked example with real numbers, and the mistakes that most often produce a wrong answer.

Maintained by the QSCivilCalc editorial team Last updated 25 September 2026 Free to cite and link

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QSCivilCalc (2026). Civil Engineering Formula Reference. https://qscivilcalc.com/engineering-formulas/ (accessed [date]).

Concrete & materials

Quantity take-off for concrete and masonry. These are volumetric relationships, not code provisions — they hold in any country, with local practice entering only through the mix ratio and the wastage allowance you choose.

Dry volume from wet volume

The first step in every concrete take-off, and the one most often skipped.

Vdry = Vwet × 1.54
SymbolMeaningUnit
VwetFinished volume of concrete in place — the volume you measure off the drawingm³
VdryCombined loose volume of cement, sand and aggregate needed to produce itm³
1.54Bulking factor: dry ingredients occupy roughly 54% more loose volume than the compacted concrete they make, because water fills the voids between particlesdimensionless
Worked example A slab 6 m × 4 m × 0.125 m gives V_wet = 3.0 m³. Dry volume = 3.0 × 1.54 = 4.62 m³. That 4.62 m³ is what you divide among cement, sand and aggregate by the mix ratio — not the 3.0 m³.
Common mistakes
  • Applying the mix ratio to wet volume, which under-orders every material by about a third.
  • Using 1.54 for mortar and plaster. Mortar has no coarse aggregate, so the accepted factor is about 1.27–1.33 — see the plaster formula below.
  • Treating 1.54 as exact. It is an industry convention covering typical voids and bulking; values from 1.50 to 1.57 are all defensible.
Run this in the Concrete Calculator →

Cement, sand and aggregate from a mix ratio

Splitting the dry volume between the three constituents.

Vcement = Vdry × ( a ÷ (a + b + c) ) · Bags = Vcement ÷ 0.0347
SymbolMeaningUnit
a : b : cMix ratio by volume — cement : sand : coarse aggregate (e.g. 1 : 1.5 : 3 for M20 nominal mix)dimensionless
VdryDry volume from the formula abovem³
0.0347Loose volume of one 50 kg cement bag, taking bulk density as 1440 kg/m³ (50 ÷ 1440)m³/bag
Worked example M20 nominal mix 1 : 1.5 : 3, dry volume 4.62 m³. Sum of ratio = 5.5.
Cement = 4.62 × 1/5.5 = 0.840 m³ → 0.840 ÷ 0.0347 ≈ 24.2 bags, order 25.
Sand = 4.62 × 1.5/5.5 = 1.26 m³. Aggregate = 4.62 × 3/5.5 = 2.52 m³.
Common mistakes
  • Using a 50 kg bag volume where the local bag is 40 kg or 42.6 kg (94 lb). Recalculate 0.0347 as bag mass ÷ 1440.
  • Ordering sand by the calculated volume without allowing for bulking — damp sand can occupy 20–30% more volume than dry.
  • Applying nominal-mix ratios to grades above M20. Higher grades require design mixes, not volumetric ratios.
Run this in the Concrete Calculator →

Plaster and mortar volume

Vwet = A × t · Vdry = Vwet × 1.27 (to 1.33)
SymbolMeaningUnit
ANet plastered area, with openings deductedm²
tCoat thickness — commonly 12 mm internal, 15–20 mm externalm
1.27–1.33Bulking factor for mortar. Lower than concrete's 1.54 because there is no coarse aggregatedimensionless
Worked example A wall 10 m × 3 m with a 2 m × 1.2 m window: net area = 30 − 2.4 = 27.6 m². At 12 mm: V_wet = 27.6 × 0.012 = 0.331 m³; V_dry = 0.331 × 1.27 = 0.421 m³. In 1:4 mortar, cement = 0.421 × 1/5 = 0.084 m³ ≈ 2.4 bags.
Common mistakes
  • Forgetting that both faces of a wall are usually plastered — double the area if so.
  • Deducting openings smaller than the local measurement convention allows. Many standard methods of measurement say openings below a stated area are not deducted.
  • Using concrete's 1.54 factor, which over-orders mortar materials by roughly 20%.
Run this in the Plaster Calculator →

Reinforced concrete

Detailing limits and rebar quantities. The clause numbers below are quoted from IS 456:2000, which is published openly by the Bureau of Indian Standards; each was checked against the standard text rather than a secondary source. Where an equivalent provision exists in ACI 318 or Eurocode 2 the numerical limit differs, so check the code that governs your project.

Rebar unit weight

The d²/162 rule, and where the constant comes from.

w = d² ÷ 162.2 (kg per metre, d in mm)
SymbolMeaningUnit
dNominal bar diametermm
wMass per metre length of barkg/m
162.2Derived, not arbitrary: w = (π/4)d² × 7850 / 10⁶, and 4 × 10⁶ / (π × 7850) = 162.2, taking steel density as 7850 kg/m³—
Worked example A 16 mm bar: 16² ÷ 162.2 = 256 ÷ 162.2 = 1.578 kg/m. Forty bars of 12 m: 1.578 × 12 × 40 = 757.6 kg, about 0.76 tonne.
Common mistakes
  • Entering diameter in centimetres or inches. The constant 162.2 is valid only for millimetres.
  • Using nominal length instead of cutting length — hooks, bends and laps add materially to the total.
  • Rounding 162.2 to 162 and then treating the result as exact. The difference is about 0.1%, immaterial for ordering but not for reconciliation.
Run this in the Steel Weight Calculator →

Minimum and maximum tension reinforcement in beams

As,min ÷ (b·d) = 0.85 ÷ fy · As,max = 0.04 · b · D
SymbolMeaningUnit
As,minMinimum area of tension reinforcementmm²
bBreadth of beam, or breadth of the web for a T-beammm
dEffective depthmm
DOverall depthmm
fyCharacteristic strength of reinforcementN/mm²
Worked example A 230 × 450 mm beam, effective depth 415 mm, Fe 415 steel.
A_s,min = 0.85 × 230 × 415 / 415 = 195.5 mm² — two 12 mm bars give 226 mm², sufficient.
A_s,max = 0.04 × 230 × 450 = 4140 mm², a limit you would reach only in a heavily loaded transfer beam.
Common mistakes
  • Using overall depth D in the minimum formula. The minimum uses effective depth d; only the maximum uses D.
  • Assuming the same limit applies under ACI 318 or Eurocode 2 — both express minimum steel differently and give different areas.
  • Treating the minimum as optional on lightly loaded beams. Its purpose is to prevent sudden failure at first cracking, so it governs precisely when the calculated area is small.

Source: IS 456:2000, clause 26.5.1.1 (a) and (b), Plain and Reinforced Concrete — Code of Practice. Text verified against the standard as published at law.resource.org.

Size the beam in the Beam Load Calculator →

Span-to-effective-depth ratios for deflection control

The quickest sanity check on whether a member is deep enough.

Cantilever 7 · Simply supported 20 · Continuous 26
ItemMeaningUnit
span/dClear span divided by effective depth. Vertical deflection limits may generally be assumed satisfied if the ratio does not exceed the basic valuedimensionless
≤ 10 mBasic values apply directly for spans up to 10 mm
> 10 mMultiply the basic value by 10/span in metres — except cantilevers, where a deflection calculation must be done insteadm
Worked example A simply supported beam of 5 m clear span: minimum effective depth ≈ 5000 ÷ 20 = 250 mm. With 25 mm cover and a 16 mm bar, overall depth ≈ 250 + 25 + 8 = 283 mm, so a 300 mm deep beam clears the check before any modification factors.
Common mistakes
  • Applying the basic value unmodified. The standard then modifies it for tension steel, compression steel and flanged sections — the basic figure is a starting point, not the final limit.
  • Using overall depth instead of effective depth.
  • Carrying the 10/span reduction over to cantilevers, where the code requires a full deflection calculation instead.

Source: IS 456:2000, clause 23.2.1 (a)–(c). Text verified against the published standard.

Nominal cover for durability

Mild 20 · Moderate 30 · Severe 45 · Very severe 50 · Extreme 75 (mm, minimum)
ExposureTypical conditionNominal cover
MildProtected against weather or aggressive conditions20 mm
ModerateSheltered from severe rain; buried concrete; permanently under water30 mm
SevereAlternate wetting and drying; exposed to coastal air45 mm
Very severeSea-water spray; corrosive fumes; freezing while wet50 mm
ExtremeTidal zone; direct contact with liquid or solid aggressive chemicals75 mm
Common mistakes
  • Measuring cover to the centre of the bar. Nominal cover is to the outermost surface of all steel, including links.
  • Ignoring the concessions and penalties attached to the table — for mild exposure with bars up to 12 mm the cover may be reduced by 5 mm, and for severe and very severe exposure a 5 mm reduction is permitted at M35 and above.
  • Using durability cover where fire resistance governs and demands more.

Source: IS 456:2000, Table 16 (clause 26.4.2), with the notes to that table. Values verified against the published standard.

Structural analysis

Statics for the standard load cases. These are derivations from equilibrium, not code provisions, so they are the same everywhere; only the load factors applied to them change between codes.

Simply supported beam under uniform load

R = wL ÷ 2 · Mmax = wL² ÷ 8 · δmax = 5wL⁴ ÷ (384·E·I)
SymbolMeaningUnit
wUniformly distributed load per unit lengthkN/m
LEffective spanm
RReaction at each supportkN
MmaxMaximum bending moment, at mid-spankN·m
EModulus of elasticity of the materialN/mm²
ISecond moment of area about the bending axismm⁴
Worked example A 6 m beam carrying 20 kN/m: R = 20 × 6 / 2 = 60 kN at each support; M_max = 20 × 6² / 8 = 90 kN·m; shear at the support equals the reaction, 60 kN.
Common mistakes
  • Omitting the beam's own self-weight from w. For a 230 × 450 mm RCC beam that is 0.23 × 0.45 × 25 = 2.59 kN/m before anything else is applied.
  • Mixing units in the deflection formula — L in metres with I in mm⁴ gives an answer wrong by 10¹².
  • Using wL²/8 for a cantilever, where the moment is wL²/2 — four times larger, at the support rather than mid-span.
Run this in the Beam Load Calculator →

Section properties of a rectangle

I = b·d³ ÷ 12 · Z = b·d² ÷ 6 · σ = M ÷ Z
SymbolMeaningUnit
bBreadth, measured perpendicular to the bending axismm
dDepth, measured parallel to the bending axismm
ISecond moment of areamm⁴
ZElastic section modulusmm³
σExtreme fibre bending stressN/mm²
Worked example A 230 × 450 mm section bending about its strong axis: I = 230 × 450³ / 12 = 1.747 × 10⁹ mm⁴; Z = 230 × 450² / 6 = 7.763 × 10⁶ mm³. Under 90 kN·m: σ = 90 × 10⁶ / 7.763 × 10⁶ = 11.6 N/mm².
Common mistakes
  • Swapping b and d. Depth is cubed, so the error is large — a 230 × 450 section is 3.8 times stiffer upright than flat.
  • Applying the gross rectangular I to a cracked reinforced section, which is considerably less stiff.
  • Forgetting to convert kN·m to N·mm (× 10⁶) before dividing by Z in mm³.

Column axial load by tributary area

P = Σfloors ( Atrib × wslab + Lbeam × wbeam + Lwall × wwall ) + self-weight
SymbolMeaningUnit
AtribTributary area — half the span to each adjacent column in both directionsm²
wslabSlab load per unit area, dead plus imposedkN/m²
wbeam, wwallSelf-weight per metre run of the beams and walls framing into the columnkN/m
PTotal axial load at the column basekN
Worked example An internal column on a 5 m × 6 m grid: A_trib = (5/2 + 5/2) × (6/2 + 6/2) = 30 m². At 10 kN/m² over four floors: 30 × 10 × 4 = 1200 kN, before beam, wall and column self-weight.
Common mistakes
  • Using the full bay area instead of half the span each way — this doubles the load on internal columns.
  • Applying internal-column tributary areas to edge and corner columns, which carry roughly half and a quarter respectively.
  • Omitting the column's own self-weight, which accumulates over every storey below.
Run this in the Column Load Calculator →

Levelling

Reducing staff readings to levels. Both methods below must give the same answer — that is the point of running them together.

Height of instrument (collimation) method

HI = RLknown + BS · RL = HI − IS or FS
SymbolMeaningUnit
BSBacksight — first reading after setting up, always onto a point of known levelm
ISIntersight — any reading between the backsight and foresightm
FSForesight — last reading before moving the instrumentm
HIHeight of instrument: the reduced level of the line of collimationm
RLReduced level of the pointm
Worked example Benchmark RL 100.000 m, backsight 1.425 m → HI = 101.425 m. A foresight of 2.310 m gives RL = 101.425 − 2.310 = 99.115 m, so that point is 885 mm below the benchmark.
Common mistakes
  • Treating "height of instrument" as the telescope's height above the ground. It is a reduced level, usually a number like 101.425, not 1.5.
  • Recording a reading as a foresight when the instrument does not move afterwards — it is an intersight.
  • Forgetting that a rising staff reading means falling ground.
Run this in the Levelling Calculator →

Arithmetic checks

These prove the arithmetic, not the fieldwork.

ΣBS − ΣFS = ΣRise − ΣFall = RLlast − RLfirst
Worked example A run with ΣBS = 6.815 m and ΣFS = 8.230 m gives −1.415 m. The last reduced level must be exactly 1.415 m below the first. If it is not, the reduction contains an arithmetic error.
Common mistakes
  • Believing a passing check means the survey is correct. It only proves the reductions are self-consistent — a misread staff that was booked as read will pass every check.
  • Including intersights in the backsight or foresight totals.
  • Not closing back onto a known benchmark, which is the only check that catches field error.

Traversing & coordinates

Turning bearings and distances into coordinates, then proving the loop closes.

Latitude and departure

Latitude = L · cos θ · Departure = L · sin θ
SymbolMeaningUnit
LHorizontal length of the linem
θWhole circle bearing, measured clockwise from northdegrees
LatitudeNorthing component — positive north, negative southm
DepartureEasting component — positive east, negative westm
Worked example A line of 125.40 m on a bearing of 142° 30′: 142°30′ = 142.5°. Latitude = 125.40 × cos 142.5° = −99.49 m (south); Departure = 125.40 × sin 142.5° = +76.34 m (east).
Common mistakes
  • Swapping the functions. Latitude takes cosine because it is the north component and bearings are measured from north.
  • Leaving the calculator in radians, or converting 142° 30′ as 142.30 instead of 142.5.
  • Using slope distance instead of horizontal distance.
Run this in the Traverse Calculator →

Closing error and relative precision

e = √( (ΣLat)² + (ΣDep)² ) · Relative precision = 1 : (P ÷ e)
SymbolMeaningUnit
ΣLat, ΣDepAlgebraic sums of latitudes and departures. Both are zero for a perfectly closed loopm
eLinear misclosure — the gap between the computed and true start pointm
PTotal perimeter of the traversem
Worked example ΣLat = +0.042 m, ΣDep = −0.031 m over a 1 240 m perimeter.
e = √(0.042² + 0.031²) = 0.052 m; relative precision = 1 : (1240 / 0.052) = 1 : 23 750, which meets the 1:5 000 commonly required for ordinary engineering traverses.
Common mistakes
  • Adding latitudes without regard to sign. Norths and souths must cancel; taking absolute values makes any traverse look catastrophic.
  • Quoting relative precision the wrong way round — it is perimeter divided by error, so a bigger second number is better.
  • Adjusting a traverse before checking that the misclosure is acceptable. A gross error must be found and fixed, not distributed.
Run this in the Traverse Calculator →

Bowditch (compass) rule adjustment

Correction to a line = − ΣLat (or ΣDep) × ( Lline ÷ P )
Worked example With ΣLat = +0.042 m, P = 1 240 m and a line of 186 m: latitude correction = −0.042 × 186 / 1240 = −0.0063 m. Every line's corrections sum to −0.042 m, closing the traverse exactly.
Common mistakes
  • Dropping the minus sign — the correction opposes the misclosure.
  • Using the Bowditch rule when angles are much more reliable than distances; the Transit rule is the appropriate choice there.
  • Rounding corrections before summing, so the adjusted traverse still does not close.

Area from coordinates (shoelace formula)

A = ½ · | Σ ( xi · yi+1 − xi+1 · yi ) |
SymbolMeaningUnit
xi, yiCoordinates of corner i, taken in order around the boundarym
AEnclosed aream²
Worked example Corners (0,0), (40,0), (40,25), (0,25): Σ = (0×0 − 40×0) + (40×25 − 40×0) + (40×25 − 0×25) + (0×0 − 0×25) = 0 + 1000 + 1000 + 0 = 2000; A = ½ × 2000 = 1000 m², which matches 40 × 25 as it must.
Common mistakes
  • Not returning to the first point — the last term must pair corner n with corner 1.
  • Listing corners out of order, or mixing clockwise and anticlockwise, which produces a meaningless area.
  • Using unadjusted coordinates. Compute the area only after the traverse has been balanced.
Run this in the Traverse Calculator →

Highway geometry

Circular and parabolic curve elements. The geometry is universal; the design limits that feed into it — maximum superelevation, minimum radius, design speed — come from AASHTO, IRC, BS or your national authority, and differ between them. The formulas below are stated without clause citation because those documents are not openly published; check the governing standard for limits.

Circular curve elements

T = R·tan(Δ/2) · L = πRΔ ÷ 180 · E = R·(sec(Δ/2) − 1) · M = R·(1 − cos(Δ/2))
SymbolMeaningUnit
RRadius of the circular curvem
ΔDeflection angle between the two tangentsdegrees
TTangent length, from PI to PC or PTm
LLength of curve along the arcm
EExternal distance, PI to the mid-point of the curvem
MMid-ordinate, from the long chord to the curvem
Worked example R = 300 m, Δ = 40°.
T = 300 × tan 20° = 109.19 m; L = π × 300 × 40 / 180 = 209.44 m; E = 300 × (1/cos 20° − 1) = 19.25 m; M = 300 × (1 − cos 20°) = 18.09 m.
If the PI is at chainage 1 450.00, then PC = 1450 − 109.19 = 1340.81 and PT = 1340.81 + 209.44 = 1550.25.
Common mistakes
  • Computing PT as PI + T. The curve is shorter than the two tangents, so PT = PC + L.
  • Using the full deflection angle where the formula calls for Δ/2.
  • Mixing degree-of-curve and radius definitions midway through a calculation.
Run this in the Horizontal Curve Calculator →

Degree of curve

Two conventions are in use, and they give different numbers for the same curve.

D = (s × 180) ÷ (π · R) → s = 30 m: D = 1718.87/R · s = 100 ft: D = 5729.58/R
SymbolMeaningUnit
DDegree of curve — the central angle subtended by the standard arcdegrees
sStandard arc length: 30 m in metric practice, 100 ft in US practicem or ft
RRadius, in the same length unit as sm or ft
Worked example R = 300 m with the 30 m arc convention: D = 1718.87 / 300 = 5.73°. The same curve in US practice with R = 984.25 ft: D = 5729.58 / 984.25 = 5.82° — close but not identical, because 100 ft is not 30 m.
Common mistakes
  • Using 5729.58 with a radius in metres, which understates the degree of curve by a factor of about 3.3.
  • Confusing the arc definition above with the chord definition, which uses sin(D/2) = s/(2R) and differs slightly on sharp curves.
  • Assuming 1746.38 and 1718.87 are interchangeable — the first is for a 30.48 m (100 ft) arc expressed in metres.

Superelevation and side friction

e + f = V² ÷ (127 · R) (V in km/h, R in m)
SymbolMeaningUnit
eSuperelevation rate, as a decimal (0.07 = 7%)dimensionless
fSide friction factor assumed for design — typically 0.10–0.16, falling as speed risesdimensionless
VDesign speedkm/h
RCurve radiusm
127Derived from g × 3.6² — converts km/h and metres into consistent units (9.81 × 12.96 ≈ 127.1)—
Worked example V = 80 km/h, R = 300 m: e + f = 80² / (127 × 300) = 6400 / 38100 = 0.168. With f taken as 0.14, e = 0.028, about 2.8% — comfortably below a 7% maximum, so the curve is generous for the speed.
Common mistakes
  • Entering V in m/s. The 127 constant is specific to km/h; for m/s the relation is e + f = V²/(g·R).
  • Using the full e + f as the superelevation. Friction carries part of the lateral force; only the remainder is built into the road.
  • Exceeding the maximum e permitted by the governing standard, which depends on climate — snow and ice regions use lower maxima.
Run this in the Superelevation Calculator →

Vertical curve offsets and K value

y = A·x² ÷ (200·L) · K = L ÷ |A| · A = g₂ − g₁
SymbolMeaningUnit
g₁, g₂Entry and exit grades, as percentages — upgrade positive, downgrade negative%
AAlgebraic grade difference. Negative gives a crest, positive a sag%
LLength of vertical curvem
xDistance from the start of the curvem
yOffset from the tangent grade line at distance xm
KLength per 1% of grade change — the usual measure of how gentle a curve ism/%
Worked example g₁ = +3%, g₂ = −2%, L = 150 m. A = −2 − 3 = −5% (a crest); K = 150 / 5 = 30 m/%.
At x = 75 m (mid-curve): y = −5 × 75² / (200 × 150) = −0.9375 m, so the road surface sits 0.94 m below the projected tangent.
Common mistakes
  • Getting the sign of A wrong by subtracting the wrong way round. A is always exit minus entry.
  • Entering grades as decimals. The 200 in the denominator assumes percentages.
  • Choosing K from the wrong table — crest and sag curves have different sight-distance criteria, and headlight sight distance usually governs sags.
Run this in the Vertical Curve Calculator →

Earthwork & hydraulics

Volumes between cross-sections, and open channel flow.

Average end area and prismoidal volume

Vend area = L · (A₁ + A₂) ÷ 2 · Vprismoidal = L · (A₁ + 4Am + A₂) ÷ 6
SymbolMeaningUnit
A₁, A₂Cross-sectional areas at the two endsm²
AmArea at the mid-section — measured, not averaged from A₁ and A₂m²
LDistance between the end sectionsm
Worked example A₁ = 18.4 m², A₂ = 25.6 m², L = 30 m. End area: 30 × (18.4 + 25.6)/2 = 660 m³. With a measured mid-area of 21.5 m²: 30 × (18.4 + 4×21.5 + 25.6)/6 = 650 m³ — the end-area method overstates by 10 m³ here, as it usually does.
Common mistakes
  • Computing Am as the mean of A₁ and A₂. That collapses the prismoidal formula back into the end-area one and gains nothing.
  • Adding cut and fill volumes together. They are separate quantities and usually separate pay items.
  • Ignoring bulking and shrinkage. Excavated soil swells by roughly 10–30% loose, and compacted fill shrinks — neither equals the in-situ volume.
Run this in the Excavation Calculator →

Manning's equation for open channel flow

V = (1 ÷ n) · Rh2/3 · S1/2 · Q = A · V · Rh = A ÷ P
SymbolMeaningUnit
VMean flow velocitym/s
nManning's roughness coefficient — about 0.013 for concrete, 0.025 for an earth channels/m1/3
RhHydraulic radius: flow area divided by wetted perimeterm
SChannel bed slope, as a decimalm/m
A, PFlow cross-sectional area and wetted perimeterm², m
QDischargem³/s
Worked example A rectangular concrete channel 2 m wide flowing 0.8 m deep on a 1 in 500 slope.
A = 1.6 m²; P = 2 + 2(0.8) = 3.6 m; R_h = 0.444 m; S = 0.002.
V = (1/0.013) × 0.444^(2/3) × 0.002^(1/2) = 76.9 × 0.581 × 0.0447 = 2.00 m/s; Q = 1.6 × 2.00 = 3.20 m³/s.
Common mistakes
  • Using the full channel depth instead of the flow depth when computing A and P.
  • Including the free surface in the wetted perimeter. Only the boundary in contact with water counts.
  • Entering slope as a ratio like "1 in 500" rather than 0.002.

Quantity surveying

Take-off relationships used in bills of quantities. Measurement conventions — what is deducted, how items are grouped — come from the standard method of measurement governing your contract, and differ by country.

Bricks per cubic metre of brickwork

N = 1 ÷ ( (l + t) × (h + t) × w )
SymbolMeaningUnit
l, h, wBrick length, height and widthm
tMortar joint thickness, typically 0.010 mm
NNumber of bricks per cubic metre of finished brickworkbricks/m³
Worked example A 0.190 × 0.090 × 0.090 m brick with 10 mm joints: (0.190 + 0.010) × (0.090 + 0.010) × 0.090 = 0.0018 m³ per brick with its mortar; N = 1 / 0.0018 = 556 bricks/m³. Mortar volume is the remainder: 1 − 556 × (0.190 × 0.090 × 0.090) = 1 − 0.855 = 0.145 m³, about 14.5%.
Common mistakes
  • Adding the joint to all three dimensions. A brick has mortar on two faces in the plane of the wall, not on its width.
  • Using a nominal brick size where the actual size differs — brick dimensions vary widely between countries and even between kilns.
  • Applying wastage to cement and sand as well as to bricks. Mortar is normally calculated on net volume.
Run this in the Brick Calculator →

Paint quantity from coverage

Litres = ( A × n ) ÷ c
SymbolMeaningUnit
ANet area to be painted, openings deductedm²
nNumber of coats—
cSpreading rate from the manufacturer's data sheet — typically 10–14 m²/litre per coat for emulsionm²/litre
Worked example A room with 62 m² of net wall area, two coats, spreading rate 12 m²/litre: (62 × 2) / 12 = 10.3 litres. Primer at 10 m²/litre for one coat adds 6.2 litres.
Common mistakes
  • Using the spreading rate for a smooth surface on fresh plaster or masonry, which absorbs considerably more on the first coat.
  • Applying the same rate to primer, putty and finish, which all differ.
  • Forgetting the ceiling, or including it when it takes a different product.
Run this in the Paint Calculator →

Sources and how they were checked

IS 456:2000 — every clause cited on this page (26.5.1.1, 23.2.1, Table 16 / clause 26.4.2) was read from the standard as published openly by the Bureau of Indian Standards, not from a secondary source.

AASHTO, IRC, BS and ACI documents are not openly published, so this page cites them at document level only and never quotes a clause number for them. Where a design limit comes from one of those standards — maximum superelevation, minimum radius, K values, minimum reinforcement under ACI 318 — check the governing document rather than relying on a figure quoted here.

Everything else — statics, curve geometry, the shoelace formula, Manning's equation, volumetric take-off — is derivable from first principles and is stated without citation because it belongs to no single standard.

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