1869 United States Senate election in Massachusetts explained

Election Name:1869 United States Senate election in Massachusetts
Type:presidential
Ongoing:no
Previous Election:1863 United States Senate election in Massachusetts
Previous Year:1863
Next Election:1874 United States Senate special election in Massachusetts
Next Year:1874 (special)
Election Date:January 19, 1869
Votes For Election:40 Members of the Massachusetts Senate
231 Members of the Massachusetts House
Majority vote of each house needed to win
1Blank:Senate
2Blank:Percentage
3Blank:House
4Blank:Percentage
Image1:Charles Sumner Brady-Handy.jpg
Nominee1:Charles Sumner
Party1:Republican Party (United States)
1Data1:37
2Data1:92.5%
3Data1:216
4Data1:93.51%
Nominee2:Josiah Abbott
Party2:Democratic Party (United States)
1Data2:2
2Data2:5%
3Data2:14
4Data2:6.06%
Senator
Before Election:Charles Sumner
Before Party:Republican Party (United States)
After Election:Charles Sumner
After Party:Republican Party (United States)

The 1869 United States Senate election in Massachusetts was held on January 19, 1869. Incumbent Charles Sumner was re-elected to a fourth term in office.

At the time, Massachusetts elected United States senators by a majority vote of each separate house of the Massachusetts General Court: the House and the Senate.

Background

In the 1868 state legislative elections, Republicans maintained an overwhelming majority in both houses. Only 20 Democratic Representatives and two Democratic Senators were elected. This ensured Sumner's re-election in the January session, though there was some speculation that Sumner would vacate his seat to accept a Cabinet appointment in the newly elected Grant administration.[1]

Election

Election in the Senate

Notes and References

  1. News: THE STATE OF MASSACHUSETTS AND "THE NATION" OF NEW-YORK . New York Tribune . 13 Nov 1868 . 2 .