January 16th, 2018, 07:45 AM  #1 
Member Joined: Jan 2017 From: Huddersfield Posts: 36 Thanks: 0  Complicated beam
I have a question on a complicated beam that I can't get my head around (see attatched picture) I have worked out the forces at both supports however I am stuck on finding the moment for the beam So far I have m=720(x) + 1198.095(x  1.4) I know that it's needs integrating to find the slope and deflection but I cannot get the correct equation for the moment Any help much appreciated 
January 16th, 2018, 01:08 PM  #2 
Senior Member Joined: Jun 2015 From: England Posts: 740 Thanks: 208 
OK I agree your reactions, though the numebr of decimal places are ridiculous. 1198N and 1022N are sufficient. As regards the bending moment, there are two methods depending upon what you have been taught. If you have been taught McCauly brackets or the unit impulse method then you can write down a single equation. Otherwise you need to break the beam into sections between points where the loading conditions change. There will then be a separate equation for each of these sections. In your case there are five sections I have relabelled the beam points ABCDE and F. You will then have five equations for sections AB, BC, CD, DE and EF. Looking at your equation attempt you may have been taught McCauly or unit impulse, do you think your brackets have special significance? It is usual to indicate this by using different parentheses < > instead of ( ). 
January 16th, 2018, 01:28 PM  #3 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,956 Thanks: 800 
What you have so far is a little off. First the "100 N/m" 2 meter length can be treated as 200 N at the center point of that length. So you have a downward total force of 720+ 420+ 200+ 850= 2190 N. The two upward forces add to 1198.095+ 1021.90= 2219.995. That's not in equilibrium! The linear force equation should be $\displaystyle R_a+ R_b= 2190$ unless you are asking about a position and size of a third downward force so this beam will be in equilibrium. Now, for the moment, about the left end, each of the downward forces is multiplied by the distance from the left end and added then that is set equal to the sum of each of the upward forces times the distance from the left end: $\displaystyle 450(2.5)+ 200(4.6)+ 850(6.30)= 1.4R_a+ 5.6R_b$. Again that is not satisfied by your values of $\displaystyle R_a$ and $\displaystyle R_b$. If you are asking where and what strength a third downward force must be added to get equilibrium, then writing $\displaystyle R_x$ as the strength of the force and x as the distance from the left end it is applied then you must have $\displaystyle R_a+ R_b+ R_x= 2219.995= 2190+ R_x$ and $\displaystyle 450(2.5)+ 200(4.6)+ 850(6.30) + R_x x= 1.4R_a+ 5.6R_b= 1.4(1198.095)+ 5.6(1021.90)$. Solve those equations for x and $\displaystyle R_x$. 
January 16th, 2018, 03:23 PM  #4  
Senior Member Joined: Jun 2015 From: England Posts: 740 Thanks: 208 
I made the load moment equation about Rb 720(5.6)  Ra(4.2) + 450(3.1) + 200(1)  850(0.7) = 0 4032  Ra(4.2) +1395 + 200 595 = 0 Ra = 5032/4.2 = 1198 @Country boy Quote:
Last edited by studiot; January 16th, 2018 at 03:41 PM.  
January 17th, 2018, 12:08 PM  #5 
Member Joined: Jan 2017 From: Huddersfield Posts: 36 Thanks: 0 
I realise the rounding is a tad excessive now, it was just what was calculated. Also I'm aware that the Udl will turn into a point load The McCauly method sounds most familiar, the thing I'm struggling to get my head round is where the dimensions are struck from, are they all taken from the left edge of the beam? Also do they become positive and negative dependant on where abouts they are in relation to the supports? 
January 17th, 2018, 02:04 PM  #6  
Senior Member Joined: Jun 2015 From: England Posts: 740 Thanks: 208  Quote:
Yes the coordinate system starts from x = 0 at the left hand end and works its way rightwards. And yes you have started correctly with Macaulay brackets (poor fellow I missed a couple of a's out of his name  hope he didn't mind. The value of x in the special bracket is zero if the bracket is negative so < x1.4 > is negative up to x = 1.4 and is therefore taken as zero until you get to the first reaction. Does this ring a bell? You continue this way to the othr end of the beam. Remember the distributed load has a square function of x. Try it and see. I have to go now but I will look in again and set out the complete equation.  
January 18th, 2018, 03:56 PM  #7 
Senior Member Joined: Jun 2015 From: England Posts: 740 Thanks: 208 
Well I did say I would post some working today, sorry it's a bit near the end. I have worked the question for each of the two methods I suggested. You can see they are similar except for the last bit (E to F). You should convince yourself that although the formulae there are different, they lead to the same numerical solution. You really do need to say more about your needs and ask questions about the formulae (or anyone else interested as I note lots of views). Also you did not answer my question. I asked because you mentioned slope/deflection, which suggests you had further use for the moment equation? 
January 21st, 2018, 03:48 AM  #8 
Member Joined: Jan 2017 From: Huddersfield Posts: 36 Thanks: 0 
Thank you very much for looking at this question for me and sorry for the late reply The Macauley method you've shown was very similar to the example I had In my notes, however that was for a much simpler beam and I'd gotten a bit muddled up with the addition of a Udl and an overhang, however your working is very clear and I can now understand where I went wrong Also the moment equation will then be integrated twice to determine the slope and deflection, this shouldn't be a struggle for me It was just determining the actual moment equation that was difficult 
January 21st, 2018, 06:02 AM  #9 
Senior Member Joined: Jun 2015 From: England Posts: 740 Thanks: 208 
Glad it was some use. Do you understand how to obtain the boundary conditions for your integration? The slopes must match at each change of load and the deflections are zero at two points. 
January 21st, 2018, 01:01 PM  #10 
Member Joined: Jan 2017 From: Huddersfield Posts: 36 Thanks: 0 
This is what ive got for the integration, does it look correct to you?


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