Sheffield, Ohio

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Sheffield is a village in Lorain County, Ohio, United States, along the Black River. The population was 2,949 at the 2000 census.



Sheffield is located at 41°26′53″N 82°5′0″W / 41.44806°N 82.083333°W / 41.44806; -82.083333 (41.448009, -82.083397)[3].

According to the United States Census Bureau, the village has a total area of 10.8 square miles (28.0 km²), all of it land.


As of the census[2] of 2000, there were 2,949 people, 1,089 households, and 866 families residing in the village. The population density was 272.6 people per square mile (105.2/km²). There were 1,147 housing units at an average density of 106.0/sq mi (40.9/km²). The racial makeup of the village was 90.88% White, 4.27% African American, 0.10% Native American, 0.81% Asian, 0.10% Pacific Islander, 2.61% from other races, and 1.22% from two or more races. Hispanic or Latino of any race were 5.93% of the population.

There were 1,089 households out of which 34.5% had children under the age of 18 living with them, 67.4% were married couples living together, 9.1% had a female householder with no husband present, and 20.4% were non-families. 16.5% of all households were made up of individuals and 7.3% had someone living alone who was 65 years of age or older. The average household size was 2.71 and the average family size was 3.06.

In the village the population was spread out with 26.3% under the age of 18, 6.6% from 18 to 24, 29.3% from 25 to 44, 26.2% from 45 to 64, and 11.6% who were 65 years of age or older. The median age was 39 years. For every 100 females there were 94.4 males. For every 100 females age 18 and over, there were 95.2 males.

The median income for a household in the village was $59,816, and the median income for a family was $66,136. Males had a median income of $43,313 versus $35,174 for females. The per capita income for the village was $25,218. About 2.5% of families and 3.2% of the population were below the poverty line, including 5.1% of those under age 18 and none of those age 65 or over.

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