SA31: Slope-Deflection Equations (Part 3)

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Solution for Exercise Problem 1: • Slope-Deflection Equat...
Solution for Exercise Problem 2: • Slope-Deflection Equat...

[@tauhawani4294]

Phenomenal work

[@importantmcqhub6030]

thanks for such a wonderful work....

[@nitin5777]

u ppl are awesome

[@christopherpaladio2683]

really appreciate the explanation

[@pineyshelden4903]

I understand how you got Mcd=wl^2/2 but how did you get -5wl^2/12 for the fem?

[@DrStructure]

The equation of interest is Mcd - wL^2/2 = 0. The fixed-end moment for Mcd, given by the slope-deflection equation, is wL^2/12. So, when we expand the above equation, we end up with wL^2/12 - wL^2/2 which equals to -5 wL^2/12

[@pineyshelden4903]

Thank you

[@nirbhaytiwary1776]

great work........will you please provide lectures on strength of materials??

[@omidhassani]

At the example give, there are 3 joints , where is the 4th one? the cantilever end can be assumed as a joint?

[@DrStructure]

Yes. The two ends, and the interior support points, are considered joints for the purpose of writing the equilibrium equations.

[@mohammedhisham5682]

What does it mean when we get the values of the member end moments (Mab, Mba and Mbc) as zero values noting that already there is rotations happened for the joints where they exist? And can I know the deformation response for this case and how to relate it with the Member end moments and joint rotations? Thanking you in advance

[@DrStructure]

A bending moment in a beam (at point X) causes rotation in the beam (at point Y). That is to say, a bending moment does not have to be present at a specific point for the point to rotate or deflect. A good example would be a simply supported beam subjected to a load at its midpoint. The load causes the beam to deflection downward, resulting in a rotation at the pin and roller supports. Yet, we know that bending moment at the pin/roller is zero. The response of a beam to loads manifests itself in (1) internal forces (i.e., moment and shear), and (2) deformation and rotation. Both these responses can be determined using the slope-deflection method. The method yields a set of equations the collectively describe the relationships among the joint rotations, displacements, and moments. Of significance here is to keep in mind that there is no one-to-one relationship that tells us a zero moment at point X mean zero a rotation/displacement at X, or vice versa.

[@mohammedhisham5682]

That was truly helping.. thank you

[@ahmedhyd867]

Thanks..for this very nice explanation about slope deflection method. If possible could you please explain how a fixed beam with uniformly varying load reaction derivation in steps

[@DrStructure]

Unfortunately, this kind of derivation can not be explained effectively using text only in the comment section. We'll try to accommodate your request in a future lecture, but that video most probably will not be available online anytime soon.

[@raheelmalook5884]

Dr structures, should we not delta negative in our equation as it deflects downward( clockwise)? Or the answer of delta is negative which means that it would have been positive if we had taken negative in equation?

[@DrStructure]

The negative sign in front of delta in the slope-deflection equations simply indicates that a positive delta value reduces (subtracts from) the positive (counterclockwise) member-end moments. That negative sign is not telling us anything about the sign of delta itself. After solving for delta, if we end up with a negative value, as we do in the example problem, then we know that the delta (relative displacement) is caused by a clockwise rotation of the beam segment. This does not necessarily mean a downward deflection however. It would be a downward deflection if the right end moves down relative to the left end. But it could also happen if the left end of the segment moves up relative to the right end. In both cases, the segment’s rotation would be clockwise, meaning a negative delta.

[@soaibdiwan]

Hi Dr.Structure, is there a separate link to the last two exercise? thank you

[@DrStructure]

Here are the links: Solution for Exercise Problem 1: kzbin.info/www/bejne/eZPFdItrfa6chJY Solution for Exercise Problem 2: kzbin.info/www/bejne/qafMgaajmch2qKM

[@musttobewild8355]

Thanks for the videos. Very helpful. I have a question. What will happen when we have just nodes in case of simple supports? Here in the example @ 2:00, say the mid node becomes just a node, not a support. Can we use these formulas directly? Since the mid node will deflect downwards, can we say that this is the same 'delta' specified as support deflection? In no where i saw a solution like that.

[@DrStructure]

In the particular beam that you are referring to, if we replace the middle roller with a hinge, the beam becomes unstable. So that beam is not a good example to use here. Here is a better example. Suppose we have a continuous beam with five supports where the middle support is a roller. We can then ask the question: "What happens if we replace the roller with a hinge?" Several things would happen: 1. The beam is going to deflect downward at that joint. This means the Delta term would no longer be zero in the slope-deflection equations associated with the two members connected to the joint. This introduces an additional unknown (Delta) into the equilibrium equations. 2. Bending moment at the joint would be zero since it is an internal hinge. 3. The beam segment to the left of the hinge would have different rotation (theta) than the beam segment to the right of the hinge. This means at the hinge, we will have two distinct rotations, one for the left side of the hinge and one for the right side of the hinge. So, how do figure out these additional unknowns? We need more equilibrium equations. Where do we get these additional equations from? Generally, when Delta is present in the slope-deflection equations, we need to write an equilibrium equation using a shear force. Here, we can do that using either the left or the right beam segment attached to the hinge. Further, for the hinge we can write two moment equations: The end moment (at the hinge) in the left beam segment must be zero. The end moment (at the hinge) in the right beam segment must be zero. Given these additional equations, we now should be able to analyze the beam and find the unknowns. I'll include this problem as a subject of a future video lecture. Thanks for brining it up.

[@musttobewild8355]

Thanks very much for the detailed info. How about replacing the roller on the midspan of the example on 2:00 with just a node (carries both moment and shear) , not an internal hinge? So that the beam will deflect at that joint under the given loading. When we write moment - slope equations according to the beam part - x=0 to the mid node- we have to consider the deflection at the node. Does this vertical deflection can be counted as the support settlement, Delta as usual? (All the examples in text books are with internal pins or rollers. So when we write the Mij equations delta cancells out.)

[@DrStructure]

In that case, we will end up with a simply supported beam which can be easily analyzed without needing to use the slope-deflection method. But, if we insisted on using the slope-deflection method to analyze the determinate beam, we need to divide it into two segments: AB and BC, B being the mid point of the beam. Then, we would get four unknowns: rotation at A, rotation at B, rotation at C and vertical deflection at B. We would need four equilibrium equations: Sum of Moments at A = 0 ==> (Mab - 0) Sum of Moments at B = 0 ==> (Mba + Mbc = 0) Sum of Moments at C = 0 ==> (Mcb = 0) And, a shear force Equation, Sum of vertical (shear) forces at B = 0 ==> (Vba + Vbc = 0)

[@tingtong6224]

at 11:06, shouldn't it gives us -Mcd-wL^2/2=0?

[@DrStructure]

Mcd is drawn in the counterclockwise direction, but the distributed load creates a clockwise moment about the left end of the segment. So, if we assume counterclockwise is positive, we get: Mcd - wL^2/2 = 0

[@sintowndec]

Thanks. You offer very clear concepts. by the way, which textbook of Structural Analysis do you refer ?

[@DrStructure]

Thanks! There is no specific textbook in use here.

[@rikdevghosh7324]

@@DrStructure sir if the beam has a overhang part as shown in the example but does not have load on it.. Will we still conider non zero del in the equation ?

[@DrStructure]

@@rikdevghosh7324 Yes! In that case, we can remove the overhang part from the beam before starting the analysis. This means the delta terms vanishes from the equations since the other joints of the beam are resting on pin/rollers.

[@mariostudio7]

What about the moment diagrams?

[@kimquocanh6946]

I am a bit confused: When is the bending moment at A clockwise and when is it counter-clockwise ? Thanks

[@DrStructure]

In the slope-deflection method, we assume the member-end moments to be in the counterclockwise direction. That is the same as saying, counterclockwise is the assumed positive direction. But this is completely arbitrary. We could have assumed them to be acting the clockwise direction which would have resulted in a similar, but not identical, set of slope-deflection equations. In general, one can assume any direction for an unknown moment, or force. We can assume clockwise, or counterclockwise, to be positive,. We then set up our equations and solve for the unknown moment(s). If a moment value turns out to be negative, then we know that the moment acts in the opposite direction to what was assumed initially.

[@kimquocanh6946]

@@DrStructure Thank you so much

[@DrStructure]

You're welcome!

[@greenjellymc6850]

Is it possible to unprivate the solution video? Thanks

[@DrStructure]

The updated lectures and exercise problem solutions are available in the (free) online course referenced in the video description field.

[@samibaloch277]

AND plz add some videos eplaining moment distribution method...as u have not yet uploaded any video about MDM