1940 Iowa Senate election

1940 Iowa Senate election

← 1938 November 5, 1940 1942 →

29 out of 50 seats in the Iowa State Senate
26 seats needed for a majority
  Majority party Minority party
 
Party Republican Democratic
Last election 38 12
Seats after 45 5
Seat change Increase7 Decrease7

Results
     Democratic gain      Republican gain
     Democratic hold      Republican hold

The 1940 Iowa State Senate elections took place as part of the biennial 1940 United States elections. Iowa voters elected state senators in 29 of the state senate's 50 districts. State senators serve four-year terms in the Iowa State Senate.

A statewide map of the 50 state Senate districts in the 1940 elections is provided by the Iowa General Assembly here.

The primary election on June 3, 1940, determined which candidates appeared on the November 5, 1940 general election ballot.[1][2]

Following the previous election, Republicans had control of the Iowa state Senate with 38 seats to Democrats' 12 seats.

To claim control of the chamber from Republicans, the Democrats needed to net 14 Senate seats.

Republicans maintained control of the Iowa State Senate following the 1940 general election with the balance of power shifting to Republicans holding 45 seats and Democrats having 5 seats (a net gain of 7 seats for Republicans).

  1. ^ "Primary Election 1940 For State Senator" (PDF). Iowa General Assembly. Retrieved June 21, 2020.
  2. ^ "General Election 1940 For State Senator" (PDF). Iowa General Assembly. Retrieved June 21, 2020.