Creating moment function for a 2D frame

[Tygra_1983]

Hi there,

I am trying to solve a rigid frame that is supported by two pin supports. I am doing this using the unit load (or virtual work) method. To do this I am trying to create moment functions for each segement of the frame.

This is what my frame looks like:

I have replaced the pin support at node 1 to a roller (which is what my text book said to do).

The virtual structure looks like the following:

For segment 2:3 I am attempting to create a moment function. I can do this easily with a beam in one dimension. However, because this is a frame I need to take the momenmt function in two directions.

The formula for virtual work across sgement 2:3 is:

Where 'M' is the moment fuction of the real structure and 'm' is the moment function for the virtual structure.

Thus, I have created the moment function to be:

From here, I have tried solving this using the double inegral procedure, but I am not getting the correct result. Am I on the right path here, or am I wqay off?

 

[rb1957]

I don't understand the term "moment function". Draw a free body for each load, and superimpose.

Your vertical reactions are wrong ... you need to include the overturning moments from the horizontal loads,
and this'll show you the additional moments in member 2.

Re-read your post. So this is a redundant structure, pinned at 1 and 3. So you're using virtual work to final the horizontal reaction at 1 (the redundancy).
So yes, release the constraint (as you have) and apply a unit load. But you need to consider all three elements of the frame.

Draw the FBDs, it'll give you some clarity on the problem.

If still in doubt, look for a worked solution ... there are many around.

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[Tygra_1983]

Hi rb1857,

By moment function I mean the equation you construct to find the moment at anywhere along the length of the member 'x'.

If I were just constructing the equation along member 2:3 the equation would be 5.1667*x - 24*(x-3). However, this is a frame with 2 dimensions, so I have added the dimension 'y'. For the virtual work procedure you solve the integral of the moment equation of the real structure 'M' multiplied by the virtual moment equation 'm'. Thus, 'M' equals 5.1667*x - 24*(x-3), and 'm' equals 1*y. The emphasis of my question here is I have I dont this correctly? Because I have an inegral with two independant variables which I have never tackled before till now.

I know you have to consider all three elements of the frame. Member 1:2 equals zeros. I am on member 2:3 and will next solve member 3:4

Sorry for the confusion.

Many thanks

 

[driftLimiter]

You need to write these functions for each member alone in 2d space. or just for the entire frame in 2D space as if it were a line.

 

[rb1957]

ok, m = y, but "y" is a constant, = L[sub]1-2[/sub]; but your M(x) is not correct, as your ground reactions are not correct ... unless you are only solving for the vertical load.

I'm not sure what solving for member 2-3 in isolation gets you ? Sure, in member 1-2 there is no M ('cause there is no transverse shear, at 1 or anywhere) ... 1-2 is axially loaded only. But 3-4 does have a moment, both M and m, so it'll be part of the solution. But then you say you'll move onto 3-4 once you've done 2-3, so ok.



"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[BAretired]

I believe the reaction at point 1 is correct. I checked it with scaled dimensions, assuming your sketch is drawn to some scale, and found it to be 5.19 kN ( compared to your 5.1667).

The equation for Member 2 is not correct. It should be M = V1*x - 24(x>3)(x-3). The term (x>3) is a Boolean expression; when x is less than 3, it is zero. If you don't include the boolean expression you won't get the right answer.

And by the way, it would be helpful if you put some dimensions on your sketch. We are good, but we're not mind readers.

 

[rb1957]

that's for the vertical ... what about the couple ... (12*H+15*H/2)/W ?

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[Tygra_1983]

I have resolved this problem! I wasn't considering the discontinuity at the 24 kN point of member 2-3. My moment equation was correct, it was just for the section of the member 3 < x < 9, Another moment equation of 5.1667x between 0 < x < 3 was required. I applied the same to member 3-4 and got the correct answer there too.

 

[BAretired]

Well, that is good news. Congratulations, Tygra 1983.

 

[Tygra_1983]

Thanks BAretired.

 

[rb1957]

yeah, I guess we should've seen that (that your moment is to the right of the load) ...

but your reactions aren't correct ? The posted drawing is not a free body. So the moment in 2-3 will be different ...

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[BAretired]

The moment equation for Member #2 was correct for 0<x<9 if it had been written M = 5.1667x-24(x>3)(x-3). The boolean term (x>3) prevents the term 24(x-3) from being calculated for 0<x>3.

The reaction at the left support appears to be correct, i.e. V[sub]1[/sub]=5.1667, H[sub]1[/sub]=0 for the determinate structure where the pinned support is changed to a roller.

 

[rb1957]

what about the overturning moment ?

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[BAretired]

The dimensions were not known at first,so I checked for V[sub]1[/sub] using scaled dimensions. They are shown on the sketch below:

Dimensions a, b and L are 3, 6 and 9m respectively. We still don't know dimensions c and d, although it appears they are probably 5 and 2.5m.

what about the overturning moment ?

Overturning moment = 12*5+15*2.5 = 97.5kN-m
Stabilizing moment = 24*6 = 144kN-m

V1 = (144-97.5)/9 = 5.1667kN

 

[rb1957]

ok, I'm a dumba$$ ... I took the ground reactions as reacting the vertical load ... but of course the near reaction would be higher ... like V1v = 16 and V4v = 8. and then v1m is like -10 and V4m +10 so all is right.


sorry

"Hoffen wir mal, dass alles gut geht !"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

 

[BAretired]

Okay, no problem rb1957.