Franklin, Nebraska

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Franklin is a city in Franklin County, Nebraska, United States. The population was 1,026 at the 2000 census. It is the county seat of Franklin County[3].



Franklin is located at 40°5′43″N 98°57′11″W / 40.09528°N 98.95306°W / 40.09528; -98.95306 (40.095357, -98.953083)[4].

According to the United States Census Bureau, the city has a total area of 1.0 square miles (2.6 km²), all of it land.


As of the census[1] of 2000, there were 1,026 people, 440 households, and 272 families residing in the city. The population density was 1,028.4 people per square mile (396.1/km²). There were 498 housing units at an average density of 499.2/sq mi (192.3/km²). The racial makeup of the city was 99.42% White, 0.39% Native American, 0.10% Asian, and 0.10% from two or more races. Hispanic or Latino of any race were 0.97% of the population.

There were 440 households out of which 25.2% had children under the age of 18 living with them, 51.4% were married couples living together, 8.4% had a female householder with no husband present, and 38.0% were non-families. 36.8% of all households were made up of individuals and 24.3% had someone living alone who was 65 years of age or older. The average household size was 2.18 and the average family size was 2.85.

In the city the population was spread out with 22.8% under the age of 18, 4.1% from 18 to 24, 20.3% from 25 to 44, 20.1% from 45 to 64, and 32.7% who were 65 years of age or older. The median age was 48 years. For every 100 females there were 76.3 males. For every 100 females age 18 and over, there were 73.3 males.

The median income for a household in the city was $27,315, and the median income for a family was $32,639. Males had a median income of $25,156 versus $16,739 for females. The per capita income for the city was $15,181. About 7.8% of families and 10.8% of the population were below the poverty line, including 13.7% of those under age 18 and 8.4% of those age 65 or over.

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