joints E and F are fixed right ? ('cause A and B are pinned)

why isn't the moment 25.8/2*5 = 64.5 kNm ? (without the swaybrace)

"Wir hoffen, dass dieses Mal alles gut gehen wird!"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

@rb1957 without the sway brace it would most definitely be PL/2 however with the brace I’m stuck.

Depending on the stiffness of the beams and columns, the column moment diagram is going to look something like the red shaded areas in the sketch below. It is clear that the maximum beam moment will be at points C and D and will be greater than 22.6 because of the contribution from the reactions at A and B.

Review your textbook to find out how to handle sidesway in frames.

so the ground (vertical) reactions at A nd B are what they are ... 25.8*5/7.9, and this is the load in AC and BD.
But the load in CE and DF is much smaller because the moment is taken out by CD.
CD's loading is two moments (and the ends) with the same sign, reacted by a couple.

And the moment taken at C is dependent on the stiffness of CD ... the stiffness of CD is not only keeping the angle between CD and AE as 90 deg (though this is more a function of the joint design), but it is also stopping CD adopting an "S" shape.

"Wir hoffen, dass dieses Mal alles gut gehen wird!"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

@BAretired Thank you for your valuable reply.
I have been using Hibbeler for my analysis, it works be great if you could point me to a specific book to use.

I do understand the moments at Joints where the bracing after would be greater.
I’m unable to wrap my head around how to break this frame into Free body diagrams set points of Contraflexure to be analysed.

@rb1958
From my understanding the stiffness K = EI/ L
assuming E and E are constant won’t that mean that K = 1/ L, so if the beam shown is 7.7 m
won’t that make it less stiffer?

BAretired although it seems reasonable as an assumption that the horizontal reaction at the pinned joints is split 50/50, I can't quite justify it.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376: Eng-Tips.com Forum Policies http://eng-tips.com/market.cfm?

if the two vertical legs are the same section, if the joints are equally fixed .. then why not.
this being a problem, the assumptions should be stated.
In my "real world", we might often assume a 60/40 split, and conservatively either split (40/60 or 60/40)

I believe the analysis assumes that joints C and D are not rotating (more than assuming the CD is perpendicular to AE).

"Wir hoffen, dass dieses Mal alles gut gehen wird!"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

Quote (Bogusbogedi )

I have been using Hibbeler for my analysis, it works be great if you could point me to a specific book to use.

I hesitate to recommend specific books, as I am not familiar with most of them. The Hibbeler text may be perfectly adequate for your purposes. The book I used prior to retirement is "Statically Indeterminate Structures" by Chua-Kia Wang, PhD, published by MGraw-Hill in 1953. If that is available, I recommend that.

Quote (GregLocock)

BAretired although it seems reasonable as an assumption that the horizontal reaction at the pinned joints is split 50/50, I can't quite justify it.

You may be correct, but I would have to review Wang to see how it would be affected. If I remember correctly, in cases of sidesway, there are some corrections to be made to account for movement of joints. My guess is that the horizontal reactions would be pretty close to 50/50.

My general question remains as to how to proceed with the analysis, do i break the frame into smaller members by introducing points of contraflexure between the rigid joints.
The moments in member EF and EC are around 30 KNm and in the CD member slightly higher at 32 KNm(from software analysis)

Regarding the behaviour of the frame under the lateral load, I understnd that the joints at C tend to open more (as compared to E) due to the pin support at A.
Hence to resist the joints opening a larger moment would be required at that joint.

How would we proceed with analysing the frame by hand?

Review the moment distribution method for sway frames in the Hibbeler text, section 12.5 in the 8th edition. Slope-deflection method would be another hand method as well.

That will get you closer to the software answer, the additional difference will be due to the axial deformation effects of the members. Another possible source of difference might be shear deformation if the software considers it. For textbook problems like this where the member properties are not intended to influence the results use very stout sections and make sure the results you are looking at do not include member self weights.

Here it is using the MDM-Sway procedure:


MDM for Joint C displacement (Mq1 model)
Assumes a moment for the upper column of 6EI delta,E / (3.5)^2 = 100 kN*m, fixed end moment for the lower column is scaled accordingly:
Note I only did 5 cycles of MDM so that will impact the final answers a bit.


MDM for Joint E displacement (Mq2 model):


Solve for the c1 and c2 constants in the joint load scaling functions:


Final Joint Moments:

From here you can see the top column moments aren't symmetric so the mid-height inflection point assumption of the portal method is not accurate for this case.

Analysis Program with large members to limit secondary effects:

hummm ... so shear reaction is not 50/50 ... hummm ... because the two verticals have the same stiffness, but the far one sees the axial stiffness on the frame member (EF) ... therefore is slightly more flexible than the direct loadpath ?

"Wir hoffen, dass dieses Mal alles gut gehen wird!"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

Good one, Celt83. I am surprised that the axial stiffness made even that much difference in FX.
If the height is 5m and the span is 7.9m, FZ should be 25.8*5/7.9 = 16.3291; you found it to be 16.7532.
I don't know why there is a difference. (7.9 should be 7.7)

Using hand methods, the axial deformation was ignored in my day, but perhaps in some cases it should be considered.

"..and the span is 7.9m, .."
Oh I read the span to be 7.7m.

"I am surprised that the axial stiffness made even that much difference in FX."

To clarify my hand calc. did not include axial stiffness and in the software I used large members to make the impact of the axial deformation negligible I also disabled shear deformations.

Quote (BAretired)

The span is 7.7 m I apologise for the poor handwriting and photo.

@Celt83
Is there a reference book I can use as a guide to understand the sway model and the moments at the joints as you have mentioned, I will use this as a guide to develop my own personal excel sheet.

I am slowly understanding the frame behaviour thanks to all of y’all.

Which Hibbeler textbook do you have?

Celt83, I got a problem.

Consider the question as given, and its mirror image. Now add the two together.
We hav e net compression of 25.8 in the top member, and no forces anywhere else.

yet if we add the reactions for each pinjoint, at the left we have -12.46--13.33=0.7 kN

same at the other side -0.7 kN

Is this due to axial compression of the top member springing the posts apart at the bottom?

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376: Eng-Tips.com Forum Policies http://eng-tips.com/market.cfm?

Incidentally this answers my own previous question to BAretired, in the absence of compressive effects the reaction at each pin will be the same size.

The reason is that in the absence of compression, we are agnostic about where the force is applied to the top member, so split it into two and apply half at each end of the top member. Now cut the structure down the middle and we have two essentially identical structures taking into account various sign changes, hence the reaction force must be equal.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376: Eng-Tips.com Forum Policies http://eng-tips.com/market.cfm?

Greg:

Interestingly enough it seems this structure is sensitive to the axial rigidity of the members when the load is applied to the corner nodes only.
Models:

25.8kN applied to upper left node                        25.8kN applied as opposing forces to upper corner nodes
25.8kN applied to upper beam mid span                    25.8kN/7.7m applied as a uniform axial load to the upper beam 

@infinite/nearly infinite axial rigidity (W44x335):



@Soft axial rigidity (W8x10):

hummm, maybe people who harp on about "validate your model" are onto something ?

"Wir hoffen, dass dieses Mal alles gut gehen wird!"
General Paulus, Nov 1942, outside Stalingrad after the launch of Operation Uranus.

So with the top left one, if we add that to its mirror image we should get the top right one.
So for the LH pin joint we'd get -12.2-(-13.6)=+1.4, and for the right -13.6-(-12.2)=-1.4

and top right agrees.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376: Eng-Tips.com Forum Policies http://eng-tips.com/market.cfm?

Thank you all that have helped me.
@celt83 I was able to understand most of it following the Hibbeler book and resource material found here :http://www.facweb.iitkgp.ac.in/~baidurya/CE21004/o...

Some final thoughts before I close the thread would be regarding the behaviour of elements under loadings and how these rotation or displacement formula were derived in the first place.

Thank you all.

Quote (Celt83)

Assumes a moment for the upper column of 6EI delta,E / (3.5)^2 = 100 kN*m

I can't catch this, can you elaborate more on this? Shouldn't it be read as 6EI big_delta/L^2