Solving Static Equilibrium at Point A: 15 kip/ft & 11 ft

[Jim_Earle]

I am really stuck on this and am looking for any type of direction anyone can give. I have an 11 foot long rigid body that supports a 15 kip/ft uniformly distributed load. I need to find the reacting moment in (lb-in) and force in (lb) at point A.

Point a is at the pined connection (see diagram below)

|
| w=15 kip/ft
| vvvvvvvvvvvvv
A ------------------
| L=11 feet
|
|
|

 

[Yafa]

what are you confused about? What have you done so far?

 

[Jim_Earle]

I'm confused on where to start. I can't find any example of a problem that resembles this situation anywhere.

 

[Jim_Earle]

This is what seems odd. What I think I have to do seems way to simple which is why I think I am heading in the wrong direction.

For my force I am thinking the sum of my forces need to equal 0, thus my force at point A will be negative 165 kips. right?

For my moment I think I have to take the weight times the length and divide by 2 or use WL/8

so it would be 165 kips X 11 ft / 8 = 226.875 kips

I realize i need to get these in the proper units but am I correct in my approach?

 

[p21bass]

If the system is in equilibrium, joint A cannot be pinned - it must be fixed.

Therefore, you are tackling A_y correctly - that it is the length of the distributed load * load distribution. In short, yes - 165K is correct.

For the M_A, though, you need to convert the distributed load to a concentrated load at its centroid. I have no idea where the 8 comes from in your equation.

Also, don't forget to solve for A_x. Of course, that should be very easy.

 

[Jim_Earle]

so this is what I came up with as I am running out of time.
w = 15 kip/ft X 1000 lb/kip = 15,000 lb/ft


Mmax = wL²/ 2 = 15,000 lb/ft (11ft x 11ft) / 2 = 907,500 lb-ft X 12 in/ft = 10,890,000 lb-in

Thus my Mmax =10,890,000 lb-in

For my forces I assume the sum of all forces must equal 0

∑Fx = 0
∑Fy = 0
∑M = 0

So if I have a positive applied force of 165,000 lb (15 kip x 11 ft = 165 kip X 1,000 lbs / kip = 165,000 lb)

My reacting force at my point A should be -165,000 lb

Is this correct?

 

[p21bass]

Well, yes - but you did it in a weird sort of way for the Moment.

 

[Jim_Earle]

thats what my formula tells me to do. I have a cantilevered beam with a fixed connection @ point A, and it caries a uniformly distributed load of 15 kips/ ft for 11 ft.

my formula that I found for this situation says my Mmax = wL2/2

 

[Jim_Earle]

I think i screwed up my conversions though. does my answer seem high. I times by 12 and I am thinking I should have divided by 12.

Out of curiosity what's another way I could have solved for the moment?

 

[p21bass]

Since you converted things from K to lbs, and from ft to in, your answers are going to be high. But, as long as the units cancel correctly (which they did) - it'll be fine.

And using the formula you had, while it works, is simply a bad way of doing things. It seems like a formula for this particular type of problem was given, but the "why" wasn't taught. That equation wouldn't work, had the distributed load been over only the first 9 feet of the beam.

In short, formulas are nice, but aren't always appropriate.

 

[p21bass]

Ah, just saw the last line of your last post...

You'd have to find the centroid of the distributed load. In this case, as the load is equally distributed over the entire beam, it's just the length of the beam/2 => 5.5ft. Were the load just over the first 9 feet, then 4.5ft. However, if the load didn't start until 3 feet from point A - so that the load was over the last 8ft - then the centroid for the load would be located at 3ft + 8ft/2 => 7ft. And then it's simply the total applied load * the centroid location.

 

[Jim_Earle]

I see what you are saying. I have had problems such as that and I did have to use a different approach for that problem. Thanks for your help and thanks for looking at my problem. I think I was making it way harder than what I needed to and was really second guessing myself :)