1 CHAPTER 31 31 1 The equation to be solved is d 2T 15 dx 2 Assume a solution of the form T ax2 bx c which can be differentiated twice to give T 2a Substituting this result into the differential equation gives a 7 5 The boundary conditions can be used to evaluate the remaining coefficients For the first condition at x 0 75 7 5 0 2 b 0 c or c 75 Similarly for the second condition 150 7 5 10 2 b 10 75 which can be solved for b 82 5 Therefore the final solution is T 7 5 x 2 82 5 x 75 The results are plotted as 300 200 100 0 0 2 4 6 8 10 31 2 The heat source term in the first row of Eq 31 26 can be evaluated by substituting Eq 31 3 and integrating to give 2 5 15 0 2 5 x dx 18 75 2 5 Similarly Eq 31 4 can be substituted into the heat source term of the second row of Eq 31 26 which can also be integrated to yield 2 5 15 0 x 0 dx 18 75 2 5 These results along with the other parameter values can be substituted into Eq 31 26 to give 0 4T1 0 4T2 dT x1 18 75 dx and PROPRIETARY MATERIAL The McGraw Hill Companies Inc All rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by McGraw Hill for their individual course preparation If you are a student using this Manual you are using it without permission 2 0 4T1 0 4T2 dT x2 18 75 dx 31 3 In a manner similar to Fig 31 7 the equations can be assembled for the total system 0 4 0 4 T1 dT x1 dx 18 75 37 5 0 4 0 8 0 4 T2 T3 0 4 0 8 0 4 37 5 37 5 0 4 0 8 0 4 T4 0 4 0 4 T5 dT x5 dx 18 75 The known end temperatures can be substituted to give 1 0 4 dT x1 dx 11 25 67 5 T2 0 8 0 4 0 4 0 8 0 4 T3 37 5 0 4 0 8 T4 97 5 0 4 1 dT x5 dx 41 25 These equations can be solved for dT x1 dx 82 5 234 375 T2 T3 300 T4 271 875 dT x5 dx 67 5 The solution along with the analytical solution dashed line is shown below 300 200 100 0 0 2 4 6 8 10 31 4 0 D R D x2 x1 d 2c dx 2 d 2 c 2 U dc kc dx U dc kc dx dx d 2 c dc kc N i dx D 2 U dx dx PROPRIETARY MATERIAL The McGraw Hill Companies Inc All rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by McGraw Hill for their individual course preparation If you are a student using this Manual you are using it without permission 3 d 2 c x2 1 Ni x dx dx 2 x2 dc U Ni x dx x1 dx D x1 k x2 x1 2 i x dx cN 3 Term 1 D c1 c2 dc dx x1 x x 2 1 Ni x dx D c c dc dx 2 x2 2 1 x2 x1 dx d c 2 x2 x1 Term 2 dc Ni x dx dx x2 x1 x2 x2 x1 c2 c1 N i x dx x2 x1 x2 x1 2 c c dc N i x dx 2 1 2 dx Ni x dx x1 x2 x1 c2 c1 dc U N i x dx U 2 c2 c1 x1 dx 2 Term 3 x2 k x2 x1 i x dx cN k x2 x1 c1 c2 2 Total element equation 1 2 3 a11 a 21 a12 c1 b1 a 22 c 2 b2 where a11 D U k x2 x1 x2 x1 2 2 a22 D U k x2 x1 x2 x1 2 2 b1 D dc x1 dx a12 b2 D D U x2 x1 2 a21 D U x2 x1 2 dc x2 dx 31 5 First we can develop an analytical function for comparison Substituting parameters gives PROPRIETARY MATERIAL The McGraw Hill Companies Inc All rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by McGraw Hill for their individual course preparation If you are a student using this Manual you are using it without permission 4 0 1 200 109 ddxu 1000 2 2 Assume a solution of the form u ax 2 bx c This can be differentiated twice to yield d2u dx2 2a This can be substituted into the ODE which can then be solved for a 2 5 10 8 The boundary conditions can then be used to evaluate the remaining coefficients At the left side u 0 0 and 0 2 5 10 8 0 2 b 0 c and therefore c 0 At the right side of the bar u 10 0 and 0 2 5 10 8 10 2 b 10 and therefore b 2 5 10 7 and the solution is u 2 5 10 8 x 2 2 5 10 7 x which can be displayed as 0 2 4 6 8 10 0 E 00 2 E 07 4 E 07 6 E 07 8 E 07 The element equation can be written as du dx x1 Ac E 1 1 u1 Ac E du x 2 x1 1 1 u 2 x2 dx x2 x1 x2 x1 P x N 1 x dx P x N 2 x dx The distributed load can be evaluated as 2 0 1000 2 x dx 1000 2 2 0 1000 x 0 dx 1000 2 Thus the final element equation is PROPRIETARY MATERIAL The McGraw Hill Companies Inc All rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by McGraw Hill for their individual course preparation If you are a student using this Manual you are using it without permission 5 1 10 1 1010 10 du x 1 10 u1 A E dx 1 1000 1000 c du 1 1010 u 2 x2 dx 10 Assembly yields 2 1010 1 1010 2 1010 1 1010 1 1010 2 1010 1 1010 1 1010 2 1010 1 1010 1 1010 2 1010 1 1010 du x1 dx 1000 …
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