January 22nd, 2018, 09:23 AM  #11 
Senior Member Joined: Jun 2015 From: England Posts: 820 Thanks: 243 
Ok so first problem You seem to have forgotten the physical characteristics of the beam. We have been loosely talking baout the moment = expression but the full equation is $\displaystyle EI\frac{{{d^2}y}}{{d{x^2}}} = M = {\rm{first}}\;{\rm{expression}}$ We get the slope from the first integration $\displaystyle EI\frac{{dy}}{{dx}} = {\rm{first}}\;{\rm{integral}} + A$ and the deflection from the second integration. $\displaystyle EIy = {\rm{second}}\;{\rm{integral}} + Ax + B$ You have nearly got the integrations right , although you said that B was 1.4m from A and your first attempt reflected this, a figure of 1.9m seems to have slipped into your subsequent equations. I apologise if my hand written script which shows a different type of 4 from yours was responsible. However you will need to learn to recognise 4 written either way in your future career. The other point is that when you integrate macaulay brackets it is not a normal integration, there's lots here or ask again in this thread. https://www.google.co.uk/search?sour....0.WkdEPxBj9qc The point is that the expressions within the macaulay brackets are effectively new variables, so Ax term in the deflection equation has limits that runs the full length of the beam, but the expressions within the macaulay brackets do not. If you think about it this is the way what is happening to the right of B to reduce the deflection in the section AB from what it would be as a cantilever, fixed at B. It is this Ax term which does this. This is why I (and everyone else) recommend distinguishing the macaulay brackets by not using the ordinary curved ones. I have tried to put all this into type for the deflection equation. You may have to wait a bit for the page here to load Mathjax or evenr efresh the page a couple of times. $\displaystyle EIy =  \frac{{720 < x{ > ^3}}}{6} + 1198\frac{{ < x  1.4{ > ^3}}}{6}  450\frac{{ < x  2.5{ > ^3}}}{6}  100\frac{{ < x  3.6{ > ^4}}}{{24}} + 1022\frac{{ < x  5.6{ > ^3}}}{6} + 100\frac{{ < x  5.6{ > ^4}}}{{24}} + Ax + B$ As a matter of interest, this is applied maths, why did you not put the question in that section? Last edited by studiot; January 22nd, 2018 at 09:28 AM. 
January 22nd, 2018, 12:12 PM  #12 
Member Joined: Jan 2017 From: Huddersfield Posts: 36 Thanks: 0 
Once again thanks for taking the time to look in depth at this question for me, it is hugely appreciated. You seem to have a very good knowledge of beams. As for the 1.9 instead of 1.4 I can only put that down to a sloppy error, possibly following previous examples and getting the numbers mixed up. And as for the placing of this question I was torn between real analysis and applied maths, wasn't 100% sure which category it fell under so I opted for apparently the wrong one After the bending moment and slope is determined the question then asks to determine the stress in the beam at 3.8m from the left end if the beam has a rectangular section 50mm wide by 100 deep I am not overly too concerned about this part of the question as I'm sure as the more I do of this kind it will become apparent but as a quick summary how would I go about tackling this? 
January 22nd, 2018, 02:00 PM  #13 
Senior Member Joined: Jun 2015 From: England Posts: 820 Thanks: 243 
So what do you know about the relationship between the bending moment and the vertical shear force?

January 23rd, 2018, 03:05 AM  #14 
Senior Member Joined: Jun 2015 From: England Posts: 820 Thanks: 243 
If you look first at constants A and B, normally the left support is at left hand end so making B zero. This is not the case here. However, as you observed you have 2 conditions for the deflection equations where you know the deflection so you can write 2 simultaneous equations to solve for A and B. Back to the stress part of your question. In general the normal stress induced by the bending moment varies from point to point across any section. (Have you covered the variation of stress due to bending over a section yet?) So asking for the stress at any general section has no meaning unless it is a section of contraflexure. I have not calculated this, but you should check your BM equation. That leaves the shear stress which is even over the section and brings us back to the question I asked in my previous post. 
January 28th, 2018, 06:32 AM  #15 
Newbie Joined: Jan 2017 From: brussels Posts: 28 Thanks: 0 
I am looking at a very similar question to this, in fact I think it's the exact same one with the values altered, please can you show how to determine the stress at 3.8m from the left end and also the rectangular section part, I understand the question up to this point

January 28th, 2018, 01:19 PM  #16  
Senior Member Joined: Jun 2015 From: England Posts: 820 Thanks: 243  Quote:
Then what happens when substitute your values of x?  
January 29th, 2018, 02:30 AM  #17 
Newbie Joined: Jan 2017 From: brussels Posts: 28 Thanks: 0 
Yes I used the Macauley method used as you showed in your example, to determine the stress 3.8m from the left end do I have to just substitute 3.8 in instead of x on one of the integrations? I honestly have no idea what it's asking in terms of the rectangular section part of the question 
January 29th, 2018, 04:13 AM  #18 
Senior Member Joined: Jun 2015 From: England Posts: 820 Thanks: 243 
I'm not sure if your difficulty is due to a language problem? In your last thread you seemed to understand stress. Presumably you understand that you need to differentiate the moment equation to get the shear (force) equation and thus the shear force at any section? In a beam of rectangular section, the shear force is distributed over the section in a parabola according to the equation $\displaystyle {S_y} = \frac{V}{{2I}}\left( {\frac{{{h^2}}}{4}  {y^2}} \right)$ Where Sy is the shear stress at any distance y from the neutral axis, h is the height of the beam V is the shear force I is the moment of inertial of that section. This leads to a maximum shear force at the neutral axis of 50% of the average shear stress over the section (=foce divided by area, A) $\displaystyle Max{S_y} = \frac{3}{2}\frac{V}{A}$ Finally the Maximum bending stress occurs at the outer edge, furtherst from the neutral axis and is related to the moment at a section by the equation $\displaystyle \left( {Max{S_{bending}}} \right) = \frac{{6M}}{{b{h^2}}}$ 
February 4th, 2018, 10:47 AM  #19 
Newbie Joined: Feb 2018 From: Alabama Posts: 3 Thanks: 0 
Can someone show how to get a and b please

February 5th, 2018, 12:33 AM  #20 
Senior Member Joined: Jun 2015 From: England Posts: 820 Thanks: 243  

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