Concrete Calculator

A concrete calculator is an online tool to estimate how much concrete you need, like volume, weight, and bag counts for common jobs like slabs, footings, columns, tubes, curbs, and stairs. You can use it to avoid under-ordering, reduce waste, and translate building dimensions into real-world quantities you can buy or mix on site.

Modules (inputs the calculator should offer):

• Slab / Wall / Rectangular Footing: Length, Width, Thickness, Units, Quantity.
• Round Footing / Column (solid): Diameter, Height (Depth), Units, Quantity.
• Circular Slab / Tube (hollow cylinder): Outer diameter, Inner diameter, Height (or thickness), Units, Quantity.
• Curb & Gutter: Length, Curb depth, Gutter width, Curb height, Flag thickness (if any), Units, Quantity.
• Stairs: Run (tread depth), Rise (riser height), Width, Platform depth (if any), Number of steps, Units, Quantity.

How to Use the Concrete Calculator

Below are short, plain-language steps for each module and one concrete example input for each. Follow the steps exactly, and the calculator will show volume, mass, and bag estimates.

Slabs / Walls / Footings (Rectangular prism)

What to enter: Length, width, thickness (all in the same units), quantity.

How it works: The calculator multiplies length × width × thickness for each slab and sums across quantity.

Example input: Length = 5.00 m, Width = 2.50 m, Thickness = 0.10 m, Quantity = 1.

What you get: Volume in cubic meters and cubic feet, mass using density, and bag estimates.

Round Footings / Columns (Solid cylinder)

What to enter: Diameter, height (depth), quantity.

How it works: The calculator uses the cylinder formula π × (d/2)² × height.

Example input: Diameter = 0.60 m, Height = 1.00 m, Quantity = 2 (two columns).

What you get: Volume per column × quantity, mass, bag counts.

Circular Slab or Tube (Hollow cylinder)

What to enter: Outer diameter (d1), inner diameter (d2), height (thickness), quantity.

How it works: Calculates the volume of the outer cylinder minus the inner cylinder: π × ( (d1/2)² – (d2/2)² ) × height.

Example input: Outer diameter = 2.00 m, Inner diameter = 1.50 m, Height = 0.15 m, Quantity = 1.

What you get: Hollow volume, mass, and bag estimate.

Curbs & Gutters

What to enter: Length, curb depth, gutter width, curb height, flag (top) thickness if applicable, quantity.

How it works: Breaks the curb+gutter into rectangular pieces (or trapezoids) and sums their volumes.

Example input: Length = 10.0 m, Curb depth = 0.04 m, Gutter width = 0.10 m, Curb height = 0.04 m, Flag thickness = 0.05 m, Quantity = 1.

What you get: Total volume, mass, and bags.

Stairs

What to enter: Run (tread depth), rise (riser height), width, platform depth (optional), number of steps, quantity.

How it works: Approximates each step as a rectangular prism (tread × width × thickness/height) or sums up prism volumes for more complex profiles.

Example input: Run = 0.12 m, Rise = 0.06 m, Width = 0.50 m, Platform depth = 0.05 m, Number of steps = 5, Quantity = 1.

What you get: Total stepped volume, mass, and bag estimate.

Concrete Volume Formula: the math (plain language + formulas)

Below are the exact formulas used for each shape, explained simply and followed by a worked example.

1. Rectangular prism (Slab / Wall / Rectangular footing)

Formula:

Volume=Length×Width×Thickness\text{Volume} = \text{Length} \times \text{Width} \times \text{Thickness}Volume=Length×Width×Thickness

Why this works: A slab (or rectangular wall/footing) is a rectangular box. Multiply its three orthogonal dimensions to get cubic units.

Worked example (step-by-step):
Inputs: Length L=2.50 mL=2.50\ \text{m}L=2.50 m, Width W=5.00 mW=5.00\ \text{m}W=5.00 m, Thickness T=0.10 mT=0.10\ \text{m}T=0.10 m.

1. Multiply length × width: 2.50×5.00=12.50 m22.50 \times 5.00 = 12.50\ \text{m}^22.50×5.00=12.50 m2.

    

2. Multiply by thickness: 12.50×0.10=1.25 m312.50 \times 0.10 = 1.25\ \text{m}^312.50×0.10=1.25 m3.
   Result: Volume = 1.25 m³.

2. Cylinder (Solid column / round footing)

Formula:

Volume=π×(d2)2×h\text{Volume} = \pi \times \left(\frac{d}{2}\right)^2 \times hVolume=π×(2d​)2×h

where ddd = diameter and hhh = height.

Why this works: A column or round footing is a circular cross-section extruded along its height, the area of the circle times height.

Worked example:

Inputs: Diameter d=0.60 md = 0.60\ \text{m}d=0.60 m, Height h=1.00 mh = 1.00\ \text{m}h=1.00 m, Quantity = 2 columns.

1. Radius r=d/2=0.60/2=0.30 mr = d/2 = 0.60/2 = 0.30\ \text{m}r=d/2=0.60/2=0.30 m.

    

2. Circle area =πr2=π×0.302= \pi r^2 = \pi \times 0.30^2=πr2=π×0.302. Compute 0.302=0.090.30^2 = 0.090.302=0.09. So area =π×0.09≈3.14159265×0.09=0.282743338 m2= \pi \times 0.09 \approx 3.14159265 \times 0.09 = 0.282743338\ \text{m}^2=π×0.09≈3.14159265×0.09=0.282743338 m2.

    

3. Volume per column =0.282743338×1.00=0.282743338 m3= 0.282743338 \times 1.00 = 0.282743338\ \text{m}^3=0.282743338×1.00=0.282743338 m3.

    

4. For 2 columns, total volume =0.282743338×2=0.565486676 m3= 0.282743338 \times 2 = 0.565486676\ \text{m}^3=0.282743338×2=0.565486676 m3.

Result (rounded): 0.5655 m³ total.

3. Hollow cylinder (Tube or circular slab with hole)

Formula:

Volume=π×((d12)2−(d22)2)×h\text{Volume} = \pi \times \bigg(\left(\frac{d_1}{2}\right)^2 – \left(\frac{d_2}{2}\right)^2\bigg) \times hVolume=π×((2d1​​)2−(2d2​​)2)×h

where d1d_1d1​ = outer diameter, d2d_2d2​ = inner diameter, hhh = height (or thickness).

Why this works: Take the volume of the outer cylinder and subtract the inner (empty) cylinder.

Worked example:
Inputs: Outer diameter d1=2.00 md_1 = 2.00\ \text{m}d1​=2.00 m, Inner diameter d2=1.50 md_2 = 1.50\ \text{m}d2​=1.50 m, Height h=0.15 mh = 0.15\ \text{m}h=0.15 m.

1. Outer radius r1=2.00/2=1.00 mr_1 = 2.00/2 = 1.00\ \text{m}r1​=2.00/2=1.00 m. Inner radius r2=1.50/2=0.75 mr_2 = 1.50/2 = 0.75\ \text{m}r2​=1.50/2=0.75 m.

    

2. Outer area =πr12=π×1.002=π×1.00=3.14159265 m2= \pi r_1^2 = \pi \times 1.00^2 = \pi \times 1.00 = 3.14159265\ \text{m}^2=πr12​=π×1.002=π×1.00=3.14159265 m2.

    

3. Inner area =πr22=π×0.752= \pi r_2^2 = \pi \times 0.75^2=πr22​=π×0.752. Compute 0.752=0.56250.75^2 = 0.56250.752=0.5625. So inner area =π×0.5625≈3.14159265×0.5625=1.7671458676 m2= \pi \times 0.5625 \approx 3.14159265 \times 0.5625 = 1.7671458676\ \text{m}^2=π×0.5625≈3.14159265×0.5625=1.7671458676 m2.

    

4. Annular area =3.14159265−1.7671458676=1.3744467824 m2= 3.14159265 – 1.7671458676 = 1.3744467824\ \text{m}^2=3.14159265−1.7671458676=1.3744467824 m2.

    

5. Volume =1.3744467824×0.15=0.20616701736 m3= 1.3744467824 \times 0.15 = 0.20616701736\ \text{m}^3=1.3744467824×0.15=0.20616701736 m3.

Result (rounded): 0.2062 m³.

4. Stairs (sum of rectangular prisms)

Approach / Formula:

Treat each step as a rectangular block (tread depth × width × step height or thickness) and sum across all steps. For steps that are not full prisms (e.g., triangular noses), adjust accordingly, but for simplicity, most concrete stairs poured monolithically are approximated by the sum of rectangular volumes.

Worked example:
Inputs: Run (tread depth) = 0.12 m, Rise (riser height) = 0.06 m, Width = 0.50 m, Number of steps = 5, Platform depth = 0.05 m (1 platform). Assume each step has a thickness equal to the tread depth for the volume estimate (conservative).

1. Volume per step =Run×Width×Rise= \text{Run} \times \text{Width} \times \text{Rise}=Run×Width×Rise. Compute: 0.12×0.50=0.060.12 \times 0.50 = 0.060.12×0.50=0.06. Then 0.06×0.06=0.0036 m30.06 \times 0.06 = 0.0036\ \text{m}^30.06×0.06=0.0036 m3 per step.

    

2. For 5 steps: 0.0036×5=0.018 m30.0036 \times 5 = 0.018\ \text{m}^30.0036×5=0.018 m3.

    

3. Platform volume =Platform depth×Width×thickness (assume 0.05 m)=0.05×0.50×0.05=0.00125 m3= \text{Platform depth} \times \text{Width} \times \text{thickness (assume 0.05 m)} = 0.05 \times 0.50 \times 0.05 = 0.00125\ \text{m}^3=Platform depth×Width×thickness (assume 0.05 m)=0.05×0.50×0.05=0.00125 m3.

    

4. Total stair volume =0.018+0.00125=0.01925 m3= 0.018 + 0.00125 = 0.01925\ \text{m}^3=0.018+0.00125=0.01925 m3.

Result (rounded): 0.0193 m³.

Converting Volume to Concrete Bags or Weight

Once you have volume (m³ or ft³), you usually want two practical outputs:

1. Mass (kilograms and pounds): useful when ordering by tonnage.

    

2. Bag counts: how many pre-mixed bags (60-lb or 80-lb) are needed if you plan to mix on site.

    

We present two commonly used conversion methods and show how they differ: mass-based (using density) and bag-yield (volume-based) (using the manufacturer’s yield per bag). In practice, bag-yield is often the construction preference for estimating how many bags to buy; mass-based is useful for ordering ready-mix by weight or volume.

Standard reference values used

• Nominal density of normal-weight concrete: 2400 kg/m³ (this is a common average, many mixes vary 2300–2500 kg/m³).

   

• 1 cubic meter = 35.3146667 cubic feet.

   

• 1 kilogram = 2.2046226218 pounds (lb).

   

• Typical bag yields (approximate, manufacturer-specific):

   

  • 60-lb bag (premoistened dry mix) yields ~0.45 ft³ of concrete.

     

  • 80-lb bag yields ~0.60 ft³ (not universal, check bag label).

     

Step-by-step: Convert the earlier slab example (1.25 m³) to weight and bags

We’ll compute both mass-based and volume-yield bag estimates, showing all arithmetic step-by-step.

Starting volume: V=1.25 m3V = 1.25\ \text{m}^3V=1.25 m3.

1) Mass (kg) using density 2400 kg/m³

Mass (kg)=V×density=1.25×2400.\text{Mass (kg)} = V \times \text{density} = 1.25 \times 2400.Mass (kg)=V×density=1.25×2400.

Compute: 1.25×2400=(1.25×2000)+(1.25×400)=2500+500=3000 kg.1.25 \times 2400 = (1.25 \times 2000) + (1.25 \times 400) = 2500 + 500 = 3000\ \text{kg}.1.25×2400=(1.25×2000)+(1.25×400)=2500+500=3000 kg.

Mass in kilograms: 3000 kg.

2) Mass (lb)

Mass (lb)=3000 kg×2.2046226218 lb/kg.\text{Mass (lb)} = 3000\ \text{kg} \times 2.2046226218\ \text{lb/kg}.Mass (lb)=3000 kg×2.2046226218 lb/kg.

Compute: 3000×2.2046226218=6613.8678654 lb.3000 \times 2.2046226218 = 6613.8678654\ \text{lb}.3000×2.2046226218=6613.8678654 lb.

Mass in pounds (rounded): 6613.87 lb.

To express in tons:

• Metric tons (tonne): 3000 kg=3.000 tonnes.3000\ \text{kg} = 3.000\ \text{tonnes}.3000 kg=3.000 tonnes.

   

• Short tons (US): 6613.8678654 lb÷2000=3.3069339327 short tons≈3.307 tons.6613.8678654\ \text{lb} \div 2000 = 3.3069339327\ \text{short tons} \approx 3.307\ \text{tons}.6613.8678654 lb÷2000=3.3069339327 short tons≈3.307 tons.

   

3) Bag estimate (Volume-yield method)

Why use this: Bag yields (ft³ per bag) come from the manufacturer and reflect mix proportions; they are how many bags you’ll physically buy to achieve the volume.

Conversion constants used:

• 1 m3=35.3146667 ft3.1\ \text{m}^3 = 35.3146667\ \text{ft}^3.1 m3=35.3146667 ft3.

   

• 60-lb bag yields ≈ 0.45 ft³.

   

• An 80-lb bag yields ≈ 0.60 ft³.

   

Procedure:

1. Convert volume to cubic feet: Vft3=1.25×35.3146667.V_{ft^3} = 1.25 \times 35.3146667.Vft3​=1.25×35.3146667.

Compute: 35.3146667×1.25=35.3146667+8.828666675=44.143333375 ft3.35.3146667 \times 1.25 = 35.3146667 + 8.828666675 = 44.143333375\ \text{ft}^3.35.3146667×1.25=35.3146667+8.828666675=44.143333375 ft3. (alternate: 35.3146667×1=35.314666735.3146667 \times 1 = 35.314666735.3146667×1=35.3146667; times 0.25 = 8.828666675; sum = 44.143333375.)

1. Number of 60-lb bags =Vft3÷0.45= V_{ft^3} ÷ 0.45=Vft3​÷0.45.

Compute: 44.143333375÷0.45=98.096296388944.143333375 ÷ 0.45 = 98.096296388944.143333375÷0.45=98.0962963889.
Round up because you cannot buy fractional bags: 99 bags (or order 102–109 to include waste).

1. Number of 80-lb bags =Vft3÷0.60= V_{ft^3} ÷ 0.60=Vft3​÷0.60.

Compute: 44.143333375÷0.60=73.572222291744.143333375 ÷ 0.60 = 73.572222291744.143333375÷0.60=73.5722222917.
Round up: 74 bags.

Result (volume-yield): ~98 (60-lb) bags → round up to 99, or ~74 (80-lb) bags for this 1.25 m³ slab.

Note: This method typically gives fewer bags than a straight mass-based division by bag mass because bag yield accounts for the voids and actual mix proportions.

4) Bag estimate (Mass-based method)

Procedure: Divide total mass (kg) by mass per bag (kg).

• Mass per 60-lb bag: 60 lb=60÷2.2046226218=27.215542 kg.60\ \text{lb} = 60 \div 2.2046226218 = 27.215542\ \text{kg}.60 lb=60÷2.2046226218=27.215542 kg.
  Compute: 60÷2.2046226218≈27.215542 kg.60 ÷ 2.2046226218 \approx 27.215542\ \text{kg}.60÷2.2046226218≈27.215542 kg.

   

• Mass per 80-lb bag: 80 lb=80÷2.2046226218≈36.287389 kg.80\ \text{lb} = 80 ÷ 2.2046226218 \approx 36.287389\ \text{kg}.80 lb=80÷2.2046226218≈36.287389 kg.

Then:

• 60-lb bag count = 3000 kg÷27.215542≈110.231131093000\ \text{kg} ÷ 27.215542 \approx 110.231131093000 kg÷27.215542≈110.23113109 → round up 111 bags.

   

• 80-lb bag count = 3000÷36.287389≈82.722348313000 ÷ 36.287389 \approx 82.722348313000÷36.287389≈82.72234831 → round up 83 bags.

   

Observation: Mass-based numbers (111, 83) differ from volume-yield numbers (98, 74). This difference occurs because bag manufacturers quote a yield volume per bag (which factors in mix proportions and packing) that does not equal simply mass ÷ density. For purchase planning, use the bag-yield method (volume per bag) because it aligns with manufacturer claims. Use the mass-based check when ordering by tonnage or for ready-mix trucks.

Practical ordering recommendation (waste and rounding)

• After selecting a bag count, add 5–10% extra to cover spillage, uneven subgrade, slope, over-excavation, and depth variation. For complex forms or thickened edges, consider 10–15% extra.
• Example for our 60-lb bag estimate of 98 bags: add 10% → 98×1.10=107.898 \times 1.10 = 107.898×1.10=107.8 → order 108 bags.