Reaction forces in simple beam involving applied moments

[nick.falconer]

Homework Statement

Homework Equations

The Attempt at a Solution

The problem I'm having is that I'm not sure how to work out the reaction forces because of the applied moment of 50kN.m. I know that the formulas are the sum of forces in a single direction equals 0 and the sum of moments about a point equal 0. But does the applied moment count as a force? And as a moment do I multiply it by the distance from the point I'm taking?

 

[tiny-tim]

welcome to pf!

hi nick! welcome to pf!

… does the applied moment count as a force? And as a moment do I multiply it by the distance from the point I'm taking?

an applied moment counts as a pair of equal and opposite (but not in-line) forces …

so no, it doesn't "count" as a linear force, and its moment is the same about any point

 

[nick.falconer]

So if my working is correct is the reaction force at A=4.3kN and B=11.7kN

 

[tiny-tim]

So if my working is correct is the reaction force at A=4.3kN and B=11.7kN

if your working is correct, then your answer will be correct also

(what working? )

 

[nick.falconer]

Oops left out the working.

Taking down as positive and and clockwise as positive

sum of forces = 0 = -RA + 4x4 - RB

sum of moments about A = 0 = 16x2 + 50 - RBx7

So RB = 11.7kN and RA = 4.3kN

 

[nvn]

nick.falconer: Your answers in post 3 were correct, except generally always maintain at least four significant digits throughout all your intermediate calculations, then round only the final answer to three (or four) significant digits. RB in your post should instead be RD. Also, always leave a space between a numeric value and its following unit symbol. E.g., 4.29 kN, not 4.29kN. See the international standard for writing units (ISO 31-0).