Application of Copper Slag in Ultra-high Performance Concrete

Mechanical Strength and Microstructure Properties of Concrete Incorporating Copper Slag as Fine Aggregate: A State-of-the-Art Review

“…Table 4 summarizes the CS chemical compositions from various research. The primary element present is iron (Fe), which indicates that the increased iron (Fe) content is responsible for the greater density and hardness of CS when compared with other industrial waste materials [43], followed by silica, alumina, and calcium oxide. As a result, the existence of silica and alumina in CS makes it a viable choice for use as a starting material in alkaliactivated substances.…”

“…The decrease in mechanical strength of the specimens can be attributed to the relatively low absorption rate of CS, resulting in an increased presence of free water within the mixture. Consequently, this promotes the creation of pores/voids in the hardened concrete, ultimately leading to ultimately reduction in its overall strength [2,43,54]. The inclusion of heavy metals within CS, which has the potential to hinder the hydration process in concrete mixtures, may provide insights into the observed reduction, as reported by Sharma and Khan [56].…”

“…Its water absorption capacity is notably low; concrete requires less water when mixed with CS, compared to quartz sand. Therefore, an increase in content of CS is anticipated to reduce the quantity of water in the mixture [43,44]. The current implementation of copper smelting slag treatment lacks a cost-effective and efficient approach.…”

Properties and Microstructure of an Interfacial Transition Zone Enhanced by Silica Fume in Concrete Prepared with Coal Gangue as an Aggregate

“…For the compressive strength testing, the load was uniformly applied until the specimen was destroyed at a rate of 2400 ± 200 N/s, and the compressive strength R c (MPa) of the specimen was calculated according to formula . Particularly worth mentioning was that the compressive strength of the specimen was the average of those of the six testing blocks R f = 1.5 × F f × L / b 3 R c = F c / b 2 where F f (N) is the damage load to apply on the middle of the specimen, L (mm) is the distance between the brackets of the tester, b (mm) is the side length of the square cross section of the specimen (40 mm), F c (N) is the load applied to the specimen when broke.…”