1892 United States elections

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1892 United States elections
Senate elections
Overall controlDemocratic gain
Seats contested29 of 88 seats[1]
Net seat changeDemocratic +4[2]
House elections
Overall controlDemocratic hold
Seats contestedAll 356 voting members
Net seat changeRepublican +38[2]

The 1892 United States elections was held on November 8, electing member to the

1858 elections
.

In the presidential election,

Populist James B. Weaver also carried five Western states and won a little over eight percent of the vote.[3] At the 1892 Republican National Convention, Harrison fended off a challenge from supporters of former Secretary of State James G. Blaine and Governor William McKinley of Ohio. At the 1892 Democratic National Convention, Cleveland defeated Senator David B. Hill from New York and Governor Horace Boies of Iowa on the first ballot. Harrison had previously defeated Cleveland in 1888, and Cleveland's win made him the first President to serve non-consecutive terms. Cleveland's win in the popular vote also made him the second person, after Andrew Jackson
, to win the popular vote in three presidential elections.

1890 census added twenty four seats to the House. Republicans picked up several seats in the House, but Democrats continued to command a large majority in the chamber.[4]

In the

See also

References

  1. ^ Not counting special elections.
  2. ^ a b Congressional seat gain figures only reflect the results of the regularly-scheduled elections, and do not take special elections into account.
  3. ^ a b "1892 Presidential Election". The American Presidency Project. Retrieved 25 June 2014.
  4. ^ "Party Divisions of the House of Representatives". United States House of Representatives. Retrieved 25 June 2014.
  5. ^ "Party Division in the Senate, 1789-Present". United States Senate. Retrieved 25 June 2014.

Further reading