Calculate The Moments about the Point A, [Guidance needed]

[Dellis]

The exercise was done in post #24...The last post just showed you how to pick the answers from it...

CS

What?, I though you were just showing mostly how to calculate B in there

my previous post you quoted is updated.

 

[Dellis]

You said the reaction of AB= 7, so I tried it out and got this.

\Sigma{M_A} = 5(6) - 5.57(7) - 2(24) = 0

 

[stewartcs]

What?, I though you were just showing mostly how to calculate B in there

my previous post I updated it.

Your original question was to find the moments about point A right?

In order to find them, you needed to know the reaction at B. This was found in post #24.

While finding the reaction at B (in post #24) you had to sum the moments about A as shown below:

\Sigma{M_A} = 5(6) - 5.57(14) + 2(24) = 0

From that part (summing the moments about A), all you needed to do was pick out the moments and list them as I did in post #29.

5(6)      =  30 ft-lb
-5.57(14) = -78 ft-lb
2(24)     =  48 ft-lb
---------------------
Sum       =   0 ft-lb

Hence the moments about A are listed above. This is your answer based on the question you first asked in this thread.

Does this help?

CS

 

[stewartcs]

You said the reaction of AB= 7, so I tried it out and got this.

\Sigma{M_A} = 5(6) - 5.57(7) - 2(24) = 0

That does not sum to zero.

The negative sign in front of 2(24) should not be there based on your sign convention.

CS

 

[Dellis]

Your original question was to find the moments about point A right?

In order to find them, you needed to know the reaction at B. This was found in post #24.

While finding the reaction at B (in post #24) you had to sum the moments about A as shown below:

\Sigma{M_A} = 5(6) - 5.57(14) + 2(24) = 0

From that part (summing the moments about A), all you needed to do was pick out the moments and list them as I did in post #29.

5(6)      =  30 ft-lb
-5.57(14) = -78 ft-lb
2(24)     =  48 ft-lb
---------------------
Sum       =   0 ft-lb

Hence the moments about A are listed above. This is your answer based on the question you first asked in this thread.

Does this help?

CS

So that is the answer to the problem?, showing all the calculated moments, justthat?

 

[stewartcs]

So that is the answer to the problem?, showing all the calculated moments, you don't need to add them or anything like that?, just show that?

Based on the question you asked...yes. Those are the moments about A.

CS

 

[Dellis]

That does not sum to zero.

The negative sign in front of 2(24) should not be there based on your sign convention.

CS

Ok so this below and the "Moments about A" calculation equal to= 0 as stated above?, that is the answer to all of this?

\Sigma{M_A} = 5(6) - 5.57(7) + 2(24) = 72.43 kips/ft


5(6) = 30 ft-lb
-5.57(14) = -78 ft-lb
2(24) = 48 ft-lb
---------------------
Sum = 0 ft-lb

 

[stewartcs]

Ok so this is? and the "Moments about A" equal to 0 as stated above?, that is the answer to all of this?

\Sigma{M_A} = 5(6) - 5.57(7) + 2(24) = 72.43 kips/ft


5(6) = 30 ft-lb
-5.57(14) = -78 ft-lb
2(24) = 48 ft-lb
---------------------
Sum = 0 ft-lb

No.

-5.57(7) is not correct...it should be, as stated numerous times above, -5.57(14). The 14 comes from 6 + 8 which is the distance between point A and reaction B.

CS

 

[stewartcs]

The answer is in post #34.

Answer: 30 ft-lb, -78 ft-lb, and 48 ft-lb

CS

 

[Dellis]

No.

-5.57(7) is not correct...it should be, as stated numerous times above, -5.57(14). The 14 comes from 6 + 8 which is the distance between point A and reaction B.

CS

Ok sorry about that\Sigma{M_A} = 5(6) - 5.57(14) + 2(24) = .025(6) = 30 ft-lb
-5.57(14) = -78 ft-lb
2(24) = 48 ft-lb
---------------------
Sum = 0 ft-lb

 

[Dellis]

The answer is in post #34.

Answer: 30 ft-lb, -78 ft-lb, and 48 ft-lb

CS

I know but I need to show how I got them though, I need to know how to properly display the answer to the whole exercise.

Does that calculation I did in my previous post even part of it?, I don't see it asking for that right. It is just asking to calculate the Moments about A, I can just show how I get those and that's it right?.

 

[stewartcs]

Ok sorry about that

\Sigma{M_A} = 5(6) - 5.57(14) + 2(24) = 72.43 kips/ft

I want you to multiply each term in this equation and right it down without adding them up and post the results.

CS

 

[stewartcs]

I know but I need to show how I got them though, I need to know how to properly display the answer to the whole exercise.

Does that calculation I did in my previous post even part of it?, I don't see it asking for that right, I can just show how I get those and that's it?.

Combine post #24 and #34 and that is the "work" that you should show.

CS

 

[stewartcs]

Ok sorry about that


\Sigma{M_A} = 5(6) - 5.57(14) + 2(24) = .02


5(6) = 30 ft-lb
-5.57(14) = -78 ft-lb
2(24) = 48 ft-lb
---------------------
Sum = 0 ft-lb

The 0.02 is due to rounding errors...it should be 0.

CS

 

[Dellis]

The 0.02 is due to rounding errors...it should be 0.

CS

Ah ok, so other then that(which I will fix below) that is the end of this exercise?

\Sigma{M_A} = 5(6) - 5.57(14) + 2(24) = 05(6) = 30 ft-lb
-5.57(14) = -78 ft-lb
2(24) = 48 ft-lb
---------------------
Sum = 0 ft-lb

 

[stewartcs]

Ah ok, so other then that(which I will fix below) that is the end of this exercise?




\Sigma{M_A} = 5(6) - 5.57(14) + 2(24) = 0


5(6) = 30 ft-lb
-5.57(14) = -78 ft-lb
2(24) = 48 ft-lb
---------------------
Sum = 0 ft-lb

Yes.

CS

 

[Dellis]

Yes.

CS

Thank you 1 million times, I really appreciate all the help and guidance.


As you seen on the forum I got a new thread, dealing with Free Body Diagram though and I got it done in a picture, just want to see what forces I missed, can you help me with that please?.