Sheffield, Iowa

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Sheffield is a city in Franklin County, Iowa, United States. The population was 930 at the 2000 census.



Sheffield is located at 42°53′37″N 93°13′1″W / 42.89361°N 93.21694°W / 42.89361; -93.21694 (42.893649, -93.217067)[1].

According to the United States Census Bureau, the city has a total area of 5.6 square miles (14.5 km²), of which, 5.6 square miles (14.4 km²) of it is land and 0.04 square miles (0.1 km²) of it (0.54%) is water.


As of the census[2] of 2000, there were 930 people, 369 households, and 259 families residing in the city. The population density was 167.4 people per square mile (64.7/km²). There were 397 housing units at an average density of 71.5/sq mi (27.6/km²). The racial makeup of the city was 99.14% White, 0.32% Asian, and 0.54% from two or more races. Hispanic or Latino of any race were 0.65% of the population.

There were 369 households out of which 30.4% had children under the age of 18 living with them, 58.5% were married couples living together, 8.4% had a female householder with no husband present, and 29.8% were non-families. 27.6% of all households were made up of individuals and 20.9% had someone living alone who was 65 years of age or older. The average household size was 2.37 and the average family size was 2.86.

In the city the population was spread out with 23.4% under the age of 18, 6.8% from 18 to 24, 21.3% from 25 to 44, 22.9% from 45 to 64, and 25.6% who were 65 years of age or older. The median age was 44 years. For every 100 females there were 84.9 males. For every 100 females age 18 and over, there were 74.9 males.

The median income for a household in the city was $38,594, and the median income for a family was $48,472. Males had a median income of $31,544 versus $22,237 for females. The per capita income for the city was $16,980. About 3.6% of families and 5.3% of the population were below the poverty line, including 0.9% of those under age 18 and 15.1% of those age 65 or over.


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