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: 844 Thanks: 252 
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. 

Tags 
beam, complicated 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Infinite beam under moving mass  geenatrombetta  Differential Equations  0  June 24th, 2017 06:39 AM 
Question in CANTILEVER BEAM  rsoy  Physics  0  December 29th, 2011 03:07 AM 
A double beam oscilloscope  rsoy  Physics  9  July 7th, 2011 07:50 AM 
Beam Equation  LariRudi  Algebra  0  March 18th, 2010 02:09 AM 
Most efficient choice of beam length  fb2003  Applied Math  3  June 22nd, 2009 02:03 PM 