Calculating support reactions in a beam

[Studious_stud]

Homework Statement

Trying to calculate the support reactions here.

Homework Equations

ƩFy = 0
ƩMA = 0

The Attempt at a Solution

Well I've come up with the first equation (I assume two are needed to solve for both support reactions from the above equations).

ƩFy = 0

RA + RB (up) = 10kN + 10kN + 4(10kN) (down)

RA + RB = 60kN

Not too sure how to go about calculating the sum of the moments in this case. But here's an attempt, I may be completely wrong! But then again that's why I'm here.

ƩMA = 0

Taking moments about A in a positive clockwise direction:

(4+2+2)(10kN) + (4+2)(10kN) (clockwise+) = 4(10kN)(4 + 4/2) + 4RB (counter clockwise-)

(8)(10kN) + (6)(10kN) - 4(10kN)(4 + 4/2) - 4RB = 0

4/2 is the centroid of the uniformly distributed load 10kN

Any help would be greatly appreciated!

Thanks all.

 

[Blugga]

Is A the pin support and B the roller support? I don't see it on picture

 

[Studious_stud]

Is A the pin support and B the roller support? I don't see it on picture

Yep!

 

[Blugga]

Your mistake is where you take the moment of the uniformly distributed load.

You have "4(10kN)(4 + 4/2)" for the moment of the uniformly distributed load.
You're right that the resultant force from the 10kN/m load will be 4*10 because the load is acting on the 4 meter length. This resultant force is now 40KN at the center of that length (which is 2m). So as you know, the moment will be the resultant force multiplied by the distance to point A. So you don't need to add the original distance, just where the resultant force is acting.

Edit: Also, the moment from the resultant force acts clockwise(+) because it points down and is to the right of A

 

[Studious_stud]

Your mistake is where you take the moment of the uniformly distributed load.

You have "4(10kN)(4 + 4/2)" for the moment of the uniformly distributed load.
You're right that the resultant force from the 10kN/m load will be 4*10 because the load is acting on the 4 meter length. This resultant force is now 40KN at the center of that length (which is 2m). So as you know, the moment will be the resultant force multiplied by the distance to point A. So you don't need to add the original distance, just where the resultant force is acting.

Edit: Also, the moment from the resultant force acts clockwise(+) because it points down and is to the right of A

Thanks Blugga!

So the moment for the UDL is 40kN(2m), 2m being the centroid of that section? I understand now.

Which is the moment from the resultant force?

 

[Blugga]

Thanks Blugga!

So the moment for the UDL is 40kN(2m), 2m being the centroid of that section? I understand now.

Yes it is, and it's clockwise.

 

[Studious_stud]

Yes it is, and it's clockwise.

Great. So just to clarify, the correct summing of moments about A is:

(8)(10kN) + (6)(10kN) + 40kN(2) - 4RB = 0

Thank you very much

 

[Blugga]

Great. So just to clarify, the correct summing of moments about A is:

(8)(10kN) + (6)(10kN) + 40kN(2) - 4RB = 0

Thank you very much

That's what I got.