I posted this as a part of a thread on the classical physics forums, but the doubt wasn't resolved, so I thought this might be a more appropriate place... A fixed beam is considered equivalent to a Simply supported beam (SSB) with a point load at its mid point(and thus having reactions Ra* and Rb* at its ends) and another beam with no external load (as the end reactions are both R in magnitude but opposite in direction), but the forces R on this beam are what cause the end moments Ma and Mb......Now, I know that the function of the end moments Ma and Mb is to neutralise the deflection produced on the beam and thus their directions are shown as in the diagram... ...the problem is that from the direction of the forces 'R' at either end of the beam (which are supposed to cause the end moments), the direction of the moments Ma and Mb do not neutralise the deflection of the SSB......Any idea what might be wrong?(I've picked this up from my book exactly as its given) FOR A DIAGRAM, PLEASE SEE POST#15 ON https://www.physicsforums.com/showthread.php?t=371640
Looking through your previous thread I have picked out the above misconceptions from your last posts. Whether a beam is simply supported or fixed, the beam ends do not deflect. Think about this - if they deflected what you are saying is that the supports move. (Yes there is a theory of elastic foundations but this is not the place for it) The downwards centre load in a beam causes sagging. The end fixing moments causes hogging. So the end moments do indeed work in the opposite direction to reduce the deflection. They do not eliminate it completely. Now let us return to the ends of the beams. In a simply supported beam there is rotation at the ends. The end moments cause counter rotation, not counter deflection. Perhaps this is what your book was describing, but did not make clear.
"The downwards centre load in a beam causes sagging. The end fixing moments causes hogging. So the end moments do indeed work in the opposite direction to reduce the deflection. They do not eliminate it completely." Thanks for stating this clearly,Studiot.....but then could you elaborate more on my original doubt as to why the 'R' forces that are the forces causing end moments act in the directions as shown in my diagram,please? If required, please tell me in detail as to what you infer from the diagram on the other thread...- I'm referring to post#15 on page 1.
OK I don't like the diagrams from your book so I have drawn some of my own. Further your discussion with Tiny-Tim in the other thread about moments and couples needs clarifying. A couple acts in a plane and has no (zero) resultant force associated with it. The value and direction of the couple is the same about every point in its plane of action. You can demonstrate this with a sheet of plywood. Set a few screws into (not through) the plywood at random points so the screw heads are projecting above the surface of the ply. Lay the ply on a flat surface so that it is free to rotate. Then insert a torque measuring screwdriver and turn the assembly by each screw. You will find the torque to overcome friction is the same whichever screw you turn. A single force (F) has a point of application, say A. This force has a moment about any other point in the plane, say B, which varies according to the distance between points A and B. If we move F to apply it at another point on a body, say C, it will, in general, exert a different moment about B, although its effect on vertical and horizontal force equilibrium will be nil. We can account for this by considering the effect of F applied at A is the same as the effect of F applied at C along with an additional couple to allow for the difference in moment about B. This additional couple is the same for moments about every point in the plane from the new location, C An understanding of this is needed to follow the answer to your question. Ask again if you cannot show this. Before tackling your fixed end beam I have shown a simpler situation in the first 5 sketches. Take note of this as you will meet it again when studying buckling. sketch1 Shows a slender strut under axial compression. The dashed line shows the strut bowing or buckling sideways under the action of this compression. sketch2 Shows why. The load C is never applied perfectly along the centreline. The sketch shows an exaggerated loading eccentricity. sketch3 If we now introduce two equal and opposite forces on the centreline, we have not changed the the situation overall. Note we are just considering force C. We are not showing or considering the reaction. None of my arrows are reactions. sketch4 The upward pointing C on the centreline and the eccentric downward pointing C form a couple. That is they are in vertical force equilibrium, but between them they exert a constant moment everywhere in the plane of the section. If we cancel these two forces out and replace them by a couple as I have shown, we can place the couple anywhere, including the centreline. sketch5 So we are left with the other C acting downwards at the centeline plus a couple, which we conveniently ascribe to the centreline as well. Hopefully these sketches make clear the development of replacing an eccentric force with an inline force plus a couple. sketch6 Is the centre loading force, P, on your beam. Note no reactions are shown. sketch7 For convenience I have split P into two halves, still acting at the centre. sketch8 I have added two equal and opposite pairs of forces (p/2) acting at each end. Note again these are not reactions. sketch9 As in the previous example cancel each of the centre (p/2) forces with one of the end (p/2) forces and introduce a compensating couple. Since I do this twice there are two couples. sketch10 As before I move the couples to the ends, where it can be seen that they are sagging moments. The original central load has now been replaced by a downward end load and a sagging end moment at each end of the beam. sketch11 Finally I get to the reactions. These are equal and opposite to the end loads and end moments so provide upward support reaction forces and hogging reaction moments. go well
Firstly, thankyou Studiot for helping me out in this way....After having read, re-read and re-reread your post, there are still some lingering doubts... To be honest, I still don't understand the bolded part...I tried proving it on the lines provided by TinyTim(post#5 on the other thread)....my attempt is shown on the diagram attached to this post .....but you see, following this method(of just adding pairs of cancelling forces), we could even create a couple that'd be double in magnitude to the original couple(please see diagram, where I've explained my point better). Besides, the concept of the moment applied to one point being equal to the moment caused by the same set of (equal and opposite) forces at another point is still vague.... Your diagram and explanation is intuitively clear to me....but it is contradictory to what is in my book (so my book may be wrong)....as you will recall, my original question was as to why the forces causing end moments are in the opposite directions (and not in the same direction as seen in your diagram....please see post #18 on the other thread if it is not clear what I'm referring to.
I hope I didn't say that. A moment is the turning effect of one force and is applied at one point. It varies from point to point. It posseses both a single force and a lever arm. A couple is a turning effect that applies to the whole plane simultaneously. So we can consider its seat or location at any point. It possesses neither a force nor a lever arm. It is possible but not necessary to make a couple from two forces. I have made one in sketch 4. In sketch 5 I have used the property that I can ascribe the couple to any point I like to say that it operates at the centreline. If this was a moment, not a couple, then I could not do this as per note above. Since both are turning effects they can be set to operate against or with each other and the terms moment and couple are often mixed up or used for both mechanical effects.
Not quite. 1)cancel FR and -GL. This leaves no couple behind as they are in the same line 2)cancel GR and GL. This leaves an clockwise couple = +c 3)There is now an anticlockwise couple of -2c between FL and GR (as you say) 4)Find the resultant couple by adding (2) and (3) = c-2c = -c, the same anticlockwise couple as the original between FR and FL. I don't see that the circles add anything useful to the analysis? Incidentally I am only trying to make the method in your book work, it would not be my first choice of approach to fixed (end) moments. It does have the advantage of being an engineer's graphical method.
hi Studiot! As you can see, it took me a while, but I think I understand why the couple at any point can be shiftes to any other point....your last post-#7 really helped, because now I can apply your logic to any situation and realise how the whole thing works :-)
Btw, while we're still doing this, can I ask another quick question? (I won't ask too many more, promise!) Well, the question is- Do the external shear forces applied on a beam (say a simply supported beam) in the form of point load or Uniformly distributed load play any role in determining the magnitude of the bending stresses? (My thoughts- the internal stresses i.e shear stresses in a beam (that has external shear forces acting on it) act to cancel the residual bending stresses on the beam....and the same shear stresses act to cancel the external shear forces(as a kind of reactional force) ....so the external forces must also have an effect on the bending stresses....)
I don't think it is a good idea to classify loads as shear. I use the terms axial, transverse and oblique to describe them. I prefer to reserve term shear for the internal response of the structure. As your knowledge of the mechanics of materials develops it is a good idea to keep mentally separate the loads and their effects. This will really help when you come to the detail. Remember loads are forces. The effects are stresses. The loads only have to obey the three conditions of equilibrium. The effects have further conditions to obey which are determined by the geometry of the situation. These extra conditions are known as the conditions of compatibility and it is these that allow us to evaluate structural responses in terms of stresses and strains and deflections.
Hmm..meaning that what I call the shear stresses satisfy the two situations as a necessity of one of these conditions of compatibility....but there is no direct link between the two things that the shear stress separately relates to...right?
Compatibility (geometric) conditions might be The deflection at a support is zero The slope of the deflected beam shape is zero at midspan The sum of the angles of a deflected triangle = 180 degrees The compression shortening in the steel reinforcement = the compression shortening in the concrete There are many more possibilities depending upon each individual circumstance.
Just to consolidate I have attached two sketches to emphasise the difference between the turning effects we call moments and couples It is most unfortunate that people who should know better often mix them up. Sketch1 Shows any three colinear points, R, S and T distances a and b apart with a>b. I have shown a force, F applied at T (perpendicular to this line for convenience). T induces clockwise moments at R and S, as shown. The moments are clearly different and will vary with the locations of S and R. Sketch2 The same three points now have a couple applied at S, equal in magnitude to the moment described in sketch1. To apply this couple I have removed the original force at T and replaced it with a two forces as shown. The addition of the separate moments induced by each of the two forces show quite clearly that the turning effect is the same at R, S and T (or indeed any other point). In particular it is the same even at the point of application of one of the forces themselves (ie at T). Of course in sketch1 the turning effect of F at T is zero. This thead was a spin off from your earlier question thread. I suggest this would be a good endpoint for this subject and that we start another about shear and moment in beams rather than mix it up here. go well
You know, what really struck me while looking at your sketch is that the Moment of F about R is F(a+b) while the turning effect of F about R is Fb......and the moment of F about different points is different....but its turning effect is the same.....turning effect is thus when you make a couple from a force(like you did in your sketch) whereas moment is when you consider the force alone...so if I had a action-reaction pair(like in a beam), there would be no moment but a turning effect.....did I just get it?? :-)
The points are the same and the turning effect is the same about the chosen point (which is S), but the forces are different in the two sketches. So there is nothing to suggest I am saying that mechanics shows us that In the top sketch - The turning effect about S of all forces acting is F(b) In the bottom sketch - The turning effect about S of all forces acting is F(b) BUT In the top sketch - The turning effect about R of all forces acting is F(a+b) In the bottom sketch - The turning effect about R of all forces acting is F(b) So the situations are clearly different. That is why we need two names for turning effects - to account for the different situations. I like to call the situation in the bottom sketch a couple and the situation in the top sketch a moment - and not to mix them up. Before posting a new thread about beams, you might like to look through this one and comment on the maths involved. It is about the fixed end moments in more complicated beams than yours. https://www.physicsforums.com/showthread.php?t=388754
Okay...so whenever we have a pair of equal and opposite forces that are not on the same line of action, the turning effect of this pair is the same about every point(forms a couple) whereas when there's one force, without another force to cancel it, it has different turning effects about different points(moment)....
As we saw in this discussion, a couple applied to a beam has the same turning effect about any point and thus we can shift it to any point along the beam.... ....then why is it that while using Macaulay's method for a fixed beam (considering a section of the beam at a time like we're supposed to do) we consider only one of the fixed end moments,say Ma and not Mb?? Doesn't Mb have a turning effect on the considered part of the beam??
I did say , a couple of posts back, that we should start a new thread about beams and their loadings as this one has too many questions for anyone else to follow. Further I don't know where you are in your study of beam loadings, so I don't know where to start. Did you understand how to write the equation I started with in the attachment in the link I gave? In particular do you understand the process of dividing the beam into two free bodies for the purpose of analysis? You should note that I haven't included V_{B} either - for the same reason.
Could we finish off this question on this thread and then if I have any further queries, I'll start off a new thread,please?.....Actually, I have only one chapter left in my syllabus, so I might not have too many other queries :-) We've done theory of pure bending (with problems involving calculation of bending stresses and moment of resistance), shear stresses in beams, and now I'm doing deflection of beams(of which the last topic is fixed beams)...there's one more unit to go-- involving columns, struts and cylinders(both thick and thin). Well...I've done derivations of formulae of this nature while doing calculation of deflection by double integration method... Yes....I've seen this done in books like Beer and Johnson...(for e.g I've seen it used to show that to an externally applied point load,there is an equal and opposite reaction at each point in the beam,provided there are no other loads in between,ofcourse) Thanks for your help,Studiot :-)
OK here is a quick tour of beams. It is important to fully understand what is happening in Figs 1 to 6 as this forms the basis for much of mechanics of meaterials Fig1 I've started with a simple beam supporting a single point load somewhere towards the right hand end. Fig2 I have broken the beam into two free bodies a C and D. In terms of external loads the right hand FB carries the load and one of the support reactions. The left hand FB only carries the left hand support reaction. Fig3 I'd like to concentrate only on the left hand free body. As shown this only has one force acting on it and so is not in equilibrium. Fig4 In order to maintain vertical force equilibrium something must be exerting a downward force on the free body. The only candidate is the right hand free body so it must be exerting a downward force I have shown as V_{C} at the break. Fig5 However the left hand free body is still not in full equilibrium because the reaction R_{A} exerts a clockwise moment at C so to maintain full equilibrium the right hand FB must also exert an anticlockwise moment at C. I have shown this as M_{C}. Fig 6 It should be realised that we have not used the loading on the right hand free body in any way to determine V_{C} and M_{C}. They are completely determined by the external forces acting on the left hand free body. This is useful as we could replace the right hand end of the beam with a totally differnt one, so long as it provides the required vertical force and moment at the junction. Note also that I have not at this juncture specified a sign convention. Due to the laws of mechanics, the forces and moments shown on the left hand free body must act in the directions shown, regardless of whether we call them positive or negative. In what follows I have chose to call them positive. I do not know your testbook and I apologise if they use the opposite sign convention. Fig 7 Having established the idea of working our way from the left hand end of the beam and not considering any forces until we encounter them I illustrate further with a more complicated beam ABCD. This time I have put some numbers in to make the presentation simpler. This beam has three distinct sections, AB, BC and CD. If we wrote ordinary expressions for bending moment as before different ordinary expressions would be needed for the different parts of the beam. I have shown these expressions in the attachment. Note the first equation is like out first example, it contains only the left hand reaction R_{A} As we move right R_{A} still affects our left hand free body but the point load at B must also be considered so the second equation contains two terms. Similarly the effect of the distributed load comes into effect as we pass C, still moving rightwards with our exploratory section and the left hand free body. We never include the moment effect of the right hand reaction R_{B} because when the section reaches the right hand support R_{B} passes through it and has no moment about itself. Macaulay looked over those three equations and noted that the third one incorporates the other two so long as we only consider appropriate values of x for each term. Each term in the equation is a product of a force and a distance. He realised that we could do this if we disregarded the terms when the values made the distance part of each term negative. So if we disregard the second term when x<3 and the third term when x<6 we effectively have one equation we can integrate for the whole length of the beam. Finally Fig8 This is your question about the end moments revisited in the light of the foregoing. Starting at the left hand end we have R_{A} and M_{A} only acting. So at section C we will not encounter any terms including the centre load L. However at section D will will have a term for the left hand supports and the centre load L. There are no further loads on the beam so this two term equation will extend to the right hand end. As before we do not add the right hand moment. go well