High Energy Absorbent Sustainable Concrete Element From Used Tin Cans

Abstract

Large quantities of solid wastes are being generated worldwide which lead to environmental concerns. One such major component is tin cans. Utilization of tin cans wastes in making building construction materials can reduce the magnitude of many associated problems. “Energy Block” is a type of block which has high energy absorption when kept under load. It is mainly composed of cement, sand, fibers, and tin cans. This paper shows the energy block is an environment friendly and lightweight building block from cement mortar, waste tin cans and fibers, without the use of coarse aggregate providing high energy absorption to the concrete. The paper concludes the results of laboratory tests and the actual behavior and performance of such building materials in practical application providing lightweight and the usage of tin cans can be economical for conservation of environment.

Appendix

• Quantity Estimate

• The volume of a single block is πr²h = π × 75² × 300

• =5301.4 × 10³ mm³

• =5301.4 mL

• Therefore volume of mortar in cylinder with plain mortar only

• =5301.4 mL = 0.005301 m³

• Fiber content added in one block is 0.75% of the volume of the mortar filled in cylindrical block.

• 0.75% × 5301.4 cm³ = 40 g for cylinder with plain mortar and fiber only

• Therefore volume of fiber used

• =40/8.05 (since density of steel = 8.05 g/cm³) = 4.97 cm³

• =4.97 mL

• Volume of mortar in cylinder with mortar + fiber

• =5301.4 − 4.97

• =5296.43 mL = 0.005296 m³

• Volume of mortar in cylinder with mortar + tin

• =5301.4 − 500

• =4801.4 mL = 0.004801 m³

• For the cylinder having plain mortar, tin and fiber, mass of fiber used is calculated in following steps

• Volume of plain mortar

• =[volume of cylinder − volume of tin cans]

• =5301.4 − 500

• =4801.4 mL (reduced volume)

• Mass of fiber used

• =0.75% of 4801.4

• =(0.75/100) × 4801.4

• =36 g

• Volume of fiber

• =36/8.05 = 4.47 cm³ = 4.47 mL

• Therefore volume of total mortar is used in cylinder

• =(volume of cylinder − volume of tin cans − volume of fiber)

• =5301.4 − 500 − 4.47

• =4796.92 mL = 0.004797 m³

• Details of Cost Estimation

• Total volume of cylinder = 0.0053 m³

• Volume of mortar in plain mortar cylinder

• =0.0053 m³

• As cement:sand = 1:3

• Therefore quantity of cement

• =[0.0053014/(1 + 3)] = 0.001325 m³

• Density of cement = 1440 kg/m³

• 1 bag of cement = 50 kg of cement

• Volume of cement = 50/1440 = 0.0347 m³

• Therefore for volume of 0.001325 m³

• =[(50/0.0347) * 0.001325]

• =1.9 kg of cement for 1 cylinder

• Therefore for the plain mortar cylinder the volume of mortar used is 1.9 kg

• 1 bag = 50 kg of cement = Rs. 375

• 1.9 kg = [(375/50) * 1.9]

• =Rs. 14.25 (cement price)

• Volume of sand = 3 * 0.001325 = 0.003975 m³

• 0.0283 m³ = Rs. 43

• Therefore for 0.14037 m³ = [(43/1) * 0.14037]

• =Rs. 6.036 (sand price)

• Total price = 14.25 + 6.036

• =Rs. 20.286 (for plain mortar cylinder)

• For 1 m³ = [20.286/0.0053014]

• =Rs. 3826.54

• Similarly,

• Volume of mortar used in cylinder filled with plain mortar + fiber only

• =0.005296 m³

• Therefore volume of cement used

• =[0.005296/(1 + 3)]

• =0.001324 m³

• Quantity of cement used in cylinder

• =0.001324/0.0347

• =0.0382 bags

• 1 bag = 50 kg

• Therefore 0.0382 bag = 1.91 kg for 1 cylinder

• 1 bag = 50 kg of cement = Rs. 375

• Therefore for 1.91 bag = Rs. 14.325

• Volume of sand = 3 * 0.001324

• =0.00397 m³

• 0.0283 m³ = Rs. 43

• Therefore for 0.00397 m³ = Rs. 6.032

• Quantity of fiber used = 4.97 mL = 0.04 kg

• 1 kg of fiber = Rs. 150

• Therefore for 0.04 kg = Rs. 6

• Total cost of 1 cylinder

• =[14.325 + 6.032 + 6]

• =Rs. 26.357

• For 1 m³ = [26.357/0.0053014]

• =Rs. 4971.71

• Now estimation for cylinder with plain mortar + tin cans only

• Volume mortar used in cylinder

• =0.0048 m³

• Quantity of cement used

• =[0.0048/(1 + 3)]

• =0.0012 m³

• Quantity of cement in bag

• =[0.0012/0.0347]

• =0.0346 bags

• 1 bag of cement = 1.73 kg

• For 0.0346 bags = 1.73 kg

• 1 bag = 50 kg = Rs. 375

• For 1.73 kg = Rs. 12.975

• Quantity of sand used

• =3 * 0.0012

• =0.0036 m³

• 0.0283 m³ = Rs. 43

• For 0.0036 m³ = Rs. 5.4667

• Total cost of a cylinder

• =[5.4667 + 12.975]

• =Rs. 18.4417

• For 1 m³ = [18.447/0.0053014]

• Now estimation for cylinder with all mortar, fiber and tin cans

• Volume of mortar used

• =0.004797 m³

• Quantity of cement used

• =[0.004797/(1 + 3)]

• =0.001199 m³

• Quantity of cement in term of bag

• =0.001199/0.0347

• =0.03455 bags

• 1 bag = 50 kg

• Therefore in 0.03455 bags = 1.7277 kg

• 1 bag = 50 kg = Rs. 375

• For 1.7277 kg = Rs. 12.957

• Quantity of sand used in cylinder = 0.001199 * 3

• =0.003597 m³

• 0.0283 m³ = Rs. 43

• Therefore in 0.003597 m³ = Rs. 5.46

• Now quantity of fiber used = 4.97 mL = 0.04 kg

• 1 kg = Rs. 150

• Therefore in 0.04 kg = Rs. 6

• Total cost of a cylinder = [12.957 + 5.46 + 6] = Rs. 24.417

• For 1 m³ = [24.417/0.0053014]

• =Rs. 4605.76