What is the reaction at point A for a loaded beam?

[portofino]

Homework Statement

a beam is loaded and supported as shown, determine the reaction at point A

see attachment --> the small circle in the beam is not a hole, it is supposed to represent the point at which the moment is about

Homework Equations

torque tau = r*F*sin(theta)

The Attempt at a Solution

i do know there is a torque being applied by the 2kN force, such that torque = (4m)(-2kN)sin(90) = -8kN, assuming the positive is upwards

the moment at 2m is given to be 3kN m

i know i did something wrong, because the reaction at A is +2kN (upward), opposite to the applied load at 4m, and the given moment about A, M_A = 11kN m CCW

i can understand the reasoning behind the reaction at A, but how did they calculate the moment about A?

i tried using a free body diagram, but i feel i am missing something and thus not moving in the correct direction with this problem, did i do the FBD correct? see attachment

 

[portofino]

sorry to double post, but after some thought i have somewhat figured out how to get 11 kN m.

using the torque i calculated earlier --> the 8kN, summed with the given moment = 8kN + 3 kN m = 11, but my units are off?

any help here?

 

[PhanthomJay]

sorry to double post, but after some thought i have somewhat figured out how to get 11 kN m.

using the torque i calculated earlier --> the 8kN, summed with the given moment = 8kN + 3 kN m = 11, but my units are off?

any help here?

You just got a little careless with your units of the moment caused by the 2kN force. Moments always have units of force times length. 2kN*4m = 8____?

 

[portofino]

ohhh, i should of seen that, thanks for clearing it up. as for the reaction at point A, not the moment, there is no calculations to be done, other than knowing the reaction is the same magnitude but opposite direction of the load, correct?

could you/someone have a look at my free body diagram i attached in the original post, is it correct?

 

[PhanthomJay]

sorry to double post, but after some thought i have somewhat figured out how to get 11 kN m.

using the torque i calculated earlier --> the 8kN, summed with the given moment = 8kN + 3 kN m = 11, but my units are off?

any help here?

ohhh, i should of seen that, thanks for clearing it up. as for the reaction at point A, not the moment, there is no calculations to be done, other than knowing the reaction is the same magnitude but opposite direction of the load, correct?

could you/someone have a look at my free body diagram i attached in the original post, is it correct?

No, that is not correct, and your FBD is therefore not correct. Your FBD forces and moments must satisfy all equilibrium equations, namely, sum of F_y = 0, and sum of Moments about any point =0. What happened to the ccw moment of 11kN-m?

 

[portofino]

i always thought a free body diagram was supposed to represent the given loads, not the reactions i am supposed to solve for.

is my free body diagram supposed to have both such that the sum is zero in each axis?

i attached the correct free body diagram, now each axis sums to zero.

how is it? could you explain why the reaction at A is +2kN other than making the assumption is an equal but opposite reaction.

 

[PhanthomJay]

i always thought a free body diagram was supposed to represent the given loads, not the reactions i am supposed to solve for.

In a free body diagram, whether of the entire beam or any portion of it, you must show ALL forces and moments acting on the body. This includes the load at the supports, if the support is part of your FBD.

is my free body diagram supposed to have both such that the sum is zero in each axis?

yes, all forces in the y direction, and in the x direction, and all moments about any point, must sum to zero.

i attached the correct free body diagram, now each axis sums to zero.

how is it? could you explain why the reaction at A is +2kN other than making the assumption is an equal but opposite reaction.

I haven't seen your new FBD yet, but i hope you are not confusing Newton's 3rd law with his 1st. The vertical force at support A, acts up on the beam, in order for Newton 1 to be satisfied in the y direction. There is also a momemt or couple at support A that acts on the beam.