February 5th, 2018, 03:13 AM  #21 
Newbie Joined: Feb 2018 From: Alabama Posts: 3 Thanks: 0 
Yes I mean the constants of integration, I have read the thread but am unsure of how to determine the values of a and b, I think the method is by simultaneous equation but I donâ€™t know how to do this

February 5th, 2018, 08:11 AM  #22 
Senior Member Joined: Jun 2015 From: England Posts: 887 Thanks: 266 
OK from post#11 $\displaystyle EI\frac{{{d^2}y}}{{d{x^2}}} = M = {\rm{first}}\;{\rm{expression}}$ 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. $\displaystyle EIy = {\rm{second}}\;{\rm{integral}} + Ax + B$ So the second integration leads us to an exquation for the vertical deflection, y. But it introduces two constants A and B. However we know the value of the deflection at two points, (the supports). This is because the beam neither lifts off the supports nor do the supports themselves sink. So the deflection is zero at these points on the beam. so y(x=1.4) = y(x=5.6) = 0 So putting these values into the deflection equation we find $\displaystyle EIy =  \frac{{720 < x{ > ^3}}}{6} + 1198\frac{{ < x  1.4{ > ^3}}}{6}  450\frac{{ < x  2.5{ > ^3}}}{6}  100\frac{{ < x  3.6{ > ^4}}}{{24}} + 1022\frac{{ < x  5.6{ > ^3}}}{6} + 100\frac{{ < x  5.6{ > ^4}}}{{24}} + Ax + B$ $\displaystyle 0 =  \frac{{720 < 5.6{ > ^3}}}{6} + 1198\frac{{ < 5.6  1.4{ > ^3}}}{6}  450\frac{{ < 5.6  2.5{ > ^3}}}{6}  100\frac{{ < 5.6  3.6{ > ^4}}}{{24}} + 1022\frac{{ < 5.6  5.6{ > ^3}}}{6} + 100\frac{{ < 5.6  5.6{ > ^4}}}{{24}} + A*5.6 + B$ I will do the second value for you Since these are macaulay brackets, we ignore all terms for which the bracket is negative (or zero) reducing the equation to $\displaystyle 0 =  \frac{{720\left( {175.6} \right)}}{6} + 1198\frac{{74.1}}{6}  450\frac{{29.8}}{6}  100\frac{{16}}{{24}} + 5.6A + B$ $\displaystyle 0 =  21072 + 14795  2235  67 + 5.6A + B$ $\displaystyle 8579 = 5.6A + B$ Can you do the first one You should have $\displaystyle 0 =  \frac{{720 < 1.4{ > ^3}}}{6} + 1198\frac{{ < 1.4  1.4{ > ^3}}}{6}  450\frac{{ < 1.4  2.5{ > ^3}}}{6}  100\frac{{ < 1.4  3.6{ > ^4}}}{{24}} + 1022\frac{{ < 1.4  5.6{ > ^3}}}{6} + 100\frac{{ < 1.4  5.6{ > ^4}}}{{24}} + 1.4A + B$ reducing to $\displaystyle 0 =  \frac{{720 < 1.4{ > ^3}}}{6} + 1.4A + B$ Last edited by studiot; February 5th, 2018 at 08:35 AM. 
February 5th, 2018, 10:13 AM  #23 
Newbie Joined: Feb 2018 From: Alabama Posts: 3 Thanks: 0 
So a = 1963.86 and b = 2420.6? Does that sound about right?

February 5th, 2018, 10:20 AM  #24 
Member Joined: Jan 2017 From: Huddersfield Posts: 36 Thanks: 0 
Studiot I would like to thank you for your help on this question, I was able to work my way through it and complete it on time and correctly. Once I had got the values for a and b it was a doddle Seems like thereâ€™s quiet a few people having the same sort of problems as I did on this kind of question, I noticed vincot and Aaronmooy47 also asking for some help. I do find that the resources available when looking at solving beams is sometimes a bit scarce and takes a while to dig out exactly what youâ€™re looking for. But then again I suppose thatâ€™s why people use websites like this so that people like you can pass on your knowledge and give them better guidance than a textbook ever could. 

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