U.S. presidential election, 1940
| Presidential Candidate | Electoral Vote | Popular Vote | Pct | Party | Running Mate (Electoral Votes) |
|---|---|---|---|---|---|
| Franklin D. Roosevelt of New York (W) | 449 | 27,244,160 | 54.9 | Democrat | Henry A. Wallace of Iowa (449) |
| Wendell Wilkie of New York | 82 | 22,305,198 | 44.9 | Republican | Charles L. McNary of Oregon (82) | Norman Thomas of New York | Socialist | Maynard C. Krueger of Illinois |
| Total | 100.0% | ||||
| Other elections: 1928, 1932, 1936, 1940, 1944, 1948, 1952 | |||||
| Source: U.S. Office of the Federal Register | |||||
Wendell Wilkie, a former Democrat, switched to the Republican party in 1939 in opposition to incumbant president Franklin D. Roosevelt's New Deal. Wilkie campaigned against the New Deal and the government's lack of military preparedness. During the election, Roosevelt preempted the military issue by expanding military contracts. Wilkie then reversed his approach and accused Roosevelt of warmongering. On election day Roosevelt received 27 million votes to Wilkie's 22 million, and in the Electoral College, Roosevelt defeated Wilkie 449 to 82.
Norman Thomas of the Socialist Party received 116,410 popular votes (0.2%).
The election was held on November 5, 1940.
See also: President of the United States, U.S. presidential election, 1940, History of the United States (1918-1945)
