1940 United States Senate elections

1940 United States Senate elections

← 1938 November 5, 1940 1942 →
1941 (MS, SC, TX) →
34 of the 96 seats in the United States Senate 49 seats needed for a majority
Majority party Minority party   Leader Alben Barkley Warren Austin (acting) Party Democratic Republican Leader since July 22, 1937 January 3, 1940 Leader's seat Kentucky Vermont Seats before 69 24 Seats after 66 28 Seat change 3 4 Popular vote 19,715,831 19,831,926 Percentage 47.5% 47.8% Seats up 25 9 Races won 22 13   Third party Fourth party   Party Farmer–Labor Progressive Seats before 1 1 Seats won 0 1 Seat change 1 Popular vote 310,875[1] 605,609 Percentage 2.57%[1] 1.5% Seats up 1 1 Races won 0 1   Fifth party   Party Independent Seats before 1 Seats after 1 Seat change Seats up 0 Races won 0
Results of the elections:      Democratic gain      Democratic hold      Republican gain      Republican hold      Progressive hold      No election
Majority Leader before election Alben Barkley Democratic Elected Majority Leader Alben Barkley Democratic

The 1940 United States Senate elections coincided with the election of Franklin D. Roosevelt to his third term as president. The 32 seats of Class 1 were contested in regular elections, and special elections were held to fill vacancies.

Although Roosevelt was re-elected, support for his administration had dropped somewhat after eight years, and the Republican opposition gained three seats from the Democrats. However, the New Deal Democrats regained firm control of both the House and Senate because Progressives dominated the election.^[2] The Minnesota Farmer–Labor Party also disappeared from the Senate, as Henrik Shipstead joined the Republican party and Ernest Lundeen had died during the preceding term. Senator Harry S. Truman of Missouri was elected to his final term in the Senate in 1940. Truman resigned in 1945 to serve as President Roosevelt's third Vice President.

Republicans later gained an additional seat through an appointment in Colorado.