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Solution Manual for Finite Element Analysis: Theory and Application with ANSYS, 5th Edition

Solution Manual for Finite Element Analysis: Theory and Application with ANSYS, 5th Edition is the ultimate guide for understanding and solving textbook problems.

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    Chapter 1 1.1 (a) Using two elements ( ) 0.25 0.0125 A y y   2 1 0.25 in A  2 2 0.1875 in A  2 3 0.125 in A  1 ( ) 2 i i eq A A E K    6 1 (0.25 0.1875)(10.4 10 ) lb 455,000 2(5) in K     6 2 (0.1875 0.125) (10.4 10 ) lb 325,000 2(5) in K     3 455 10 1 455  0 455  1 2 0 3 1 2 3 0 455 325 325 0 1000 325 325 0, 0.002197 in 0.005274 in u u u u u u                                        (b) 2 1 0.25 in A  2 2 0.234375 in A  2 3 0.21875 in A  2 4 0.203125 in A  2 5 0.1875 in A  2 6 0.171875 in A  2 7 0.15625 in A  2 8 0.140625 in A  2 9 0.125 in A 

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    1.1 Cont. 1 lb 2015000 in K  2 lb 1885000 in K  3 lb 1755000 in K  4 lb 1625000 in K  5 lb 1495000 in K  6 lb 1365000 in K  7 lb 1235000 in K  8 lb 1105000 in K  2015000 1 2015000  0 2015000  0 2015000 1885000 1885000 1885000 1885000 1755000 1755000 1755000 1755000 1625000 1625000 1625000 1625000 1495000 1495000 1495000 1495000 1365000 1365000 1365000 1365000 1235000 1235000 1235000 1235000 11050                    1 2 3 4 5 6 7 8 9 0 0 0 0 0 0 0 0 1000 1105000 00 1105000 1105000 u u u u u u u u u                                                                                                                      1 0, u  2 0.00049628, u  3 0.0010268, u  4 0.0015965, u  5 0.0022119, u  6 0.0028808, u  7 0.0036134, u  8 0.0044232, u  9 0.0053281in. u  y (in) exact defl. Two - element eight element 0 0 0 0 1.25 0.00049645 0.00049628 2.5 0.0010272 0.0010268 3.75 0.0015972 0.0015965 5.0 0.0022129 0.002197 0.0022119 6.25 0.0028822 0.0028808 7.5 0.0036154 0.0036134 8.75 0.0044259 0.0044232 10.0 0.0053319 0.005274 0.0053281

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    1.2 avg. A E K  6 1 5 10.5 12 (3)(3.27 10 ) 2 6 K K           1 5 lb 18,393,750 in K K   6 2 4 10.5 9 (3)(3.27 10 ) lb 2 15,941,250 6 in K K            6 3 (9)(3)(3.27 10 ) lb 22, 072,500 4 in K    18393750 1 18393750  0 18393750  1 0 2 3 4 5 6 18393750 15941250 15941250 15941250 15941250 22072500 22072500 22072500 22072500 15941250 15941250 15941250 15941250 18393750 18393750 18393750 18393750 u u u u u u                                                    0 0 0 0 0 500                                      5 5 5 4 4 0 2.71831 10 5.85483 10 { } in 8.12009 10 1.12566 10 1.39749 10 u                                       5 5 6 (3) ( 8.12009 10 5.85483) 10 (3.27 10 ) 4           (3) 18.5 Psi C  

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    1.2 C ont. as a check: (3) 500 18.5 Psi (9)(3) F A     1 ( ) i i u u E     5 6 (1) ( 2.71831 10 0)(3.27 10 ) 14.8 Psi C 6         4 6 (5) ( 1.39749 1.12566) 10 (3.27 10 ) 14.8 Psi C 6          as a check: (1) (5) 500 14.8 Psi C 10.5 12 (3) 2 F A                  5 6 2 5.85483 2.71831 10 (3.27 10 ) 17.1 Psi C 6          4 5 6 (4) ( 1.12566 10 8.12009 10 ) (3.27 10 ) 17.1 Psi C 6           as a check: (2) (4) 500 17.1 Psi 10.5 9 (3) 2 F A             

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    1.3 AE K  9 1 3 (0.08)(0.006)(68.9 10 ) 0.025 K K    9 1 3 N 1.32288 10 m K K    9 2 (0.02)(0.006)(68.9 10 ) 0.1 K   9 2 N 0.08268 10 m K   9 1.32288 10 1 1.32288  0 1.32288  1 0 2 3 4 0 0 1.32288 0.08268 0.08268 0 0.08268 0.08268 1.32288 1.32288 1800 1.32288 1.32288 u u u u                                                    2 3 9 4 6 5 5 2 3 4 1.40556 0.08268 0 0 0.08268 1.40556 1.32288 0 0 1.32288 1.32288 1800 10 1.360668 10 m 2.313135 10 m 2.449202 10 m u u u u u u                                               6 (1) 9 2 1 (1.360668 10 0) (68.9 10 ) 3750000 3.75 MPa 0.025 u u E           as a check: (1) 1800 N 3.75 MPa (0.08)(0.006) F A     5 6 (2) 9 3 2 (2.313135 10 1.360668 10 ) (68.9 10 ) 15 MPa 0.1 u u E            as a check: (2) 1800 N 15 MPa (0.02)(0.006)    5 (3) 9 4 3 (2.449202 2.313135) 10 (68.9 10 ) 3.75 MPa 0.025 u u E         

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    1.4 6 1 6 (4)(0.125)(28 10 ) 2 K K    6 1 6 lb 7 10 in K K    6 2 3 4 5 (0.625)(0.125)(28 10 ) 8 K K K K      2 lb 273438 in K  2 new lb ( ) (4)(273438) 1093750 in K   6 7 10  1 6 7 10   0 6 7 10   1 0 2 6 3 6 6 4 6 6 0 0 7 10 1093750 1093750 0 1093750 1093750 7 10 7 10 500 7 10 7 10 u u u u                                                           2 6 3 6 6 4 5 4 4 2 3 4 8093750 1093750 0 0 1093750 8093750 7 10 0 500 0 7 10 7 10 7.142857 10 in 5.285714 10 in 6 10 in u u u u u u                                                5 (1) 6 2 1 2 7.142857 10 in 1b 28 10 1000 2 in u u E                       as a check: (1) 2 500 lb 1000 (4)(0.125) in F A     4 5 (2) 6 3 2 2 5.285714 10 7.142857 10 lb 28 10 1600 8 in u u E                          as a check: (2) 2 500 lb 1600 (2.5)(0.125) in    4 4 (3) 6 4 3 2 6 10 5.285714 10 lb 28 10 1000 2 in u u E                         

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    1.5 1 2 1 ( ) 0 R K u u    1 2 1 2 3 2 ( ) ( ) 0 K u u K u u     2 3 2 3 4 3 ( ) ( ) 0 K u u K u u     ` 3 4 3 ( ) 0 K u u P    In matrix fo r m: 1 1 1 1 1 2 2 2 2 2 3 3 3 3 3 4 0 0 K K u R K K K K u K K K K u K K u P                                                 

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    1.6 Size of the global matrix: 5 × 5   1 [ ] 5 5 5 5 K          ① ② ① ②   2 [ ] 8 8 8 8 K          ② ③ ② ③   3 [ ] 5 5 5 5 K          ② ③ ② ③   4 [ ] 20 20 20 20 K          ② ④ ② ④   5 [ ] 10 10 10 10 K          ③ ④ ③ ④   6 [ ] 20 20 20 20 K          ④ ⑤ ④ ⑤ (6) 5 5 5 5 8 5 20 8 5 20 [ ] 8 5 8 5 10 10 20 10 20 10 20 20 20 20 K                                     applying B.Cs and loads; 1 5 0 u u   2 10 lb, F  4 101b F  2 3 4 2 3 4 38 13 20 10 13 23 10 0 20 10 50 10 0.962 in 0.874 in 0.760 in u u u u u u                                         1 5(0.962 0) 4.8 lb R    5 20(0.760 0) 15.2 lb R   

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    1.6 Cont. 4 5 ( ) K u u  note also as a check, 1 5 external R R F    4.8 + 15.2 = 10 +10

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    1.7 5.88 150 1 150 5.88 0 5.88 5.88 2.27 2.27 2.27 2.27 10 10 10 10 0.581 0.581 0.581 0.581 0.781 0.781 0.781 0.781 2.22 2.22 2.22 2.22 1.47 1.47 1.47                    0 1.47 1 2 3 4 5 6 7 8 1 150 10 0 0 0 0 0 0 68 T T T T T T T T                                                                                        10 12.04 17.31 18.51 { } 39.12 54.46 59. F 85 68 T                             q UA T   Btu (1.47)(150)(68 59.85) 1800 hr q    or as another example: Btu (0.781)(150)(54.46 39.12) 1800 hr q    or in out 1 (150)(68 10) Btu ( ) 1800 Resistance 0.17 0.44 0.1 1.72 1.28 0.45 0.68 hr q A T T            

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    1.8 5.88 150 1 150 5.88  0 5.88 5.88 1.23 1.23 1.23 1.23 0.76 0.76 0.76 0.76 0.053 0.053 0.053 0.053 2.22 2.22 2.22 2.22 1.47 1.47 1.47                 0 1.47 1 2 3 4 5 6 7 1 150 20 0 0 0 0 0 68 T T T T T T T                                                                            20 20.37 22.12 { } 24.95 65.57 66. F 54 68 T                         q UA T   as an example: Btu (1.47)(150)(68 66.54) 320 hr q    or in out 1 (150)(68 20) Btu ( ) 320 Resistance 0.17 0.81 1.32 19 0.45 0.68 hr q A T T           

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    1.9 with the help of a psychometric chart, using a dry bulb T emp . of 68 F  and 40%,   we identify condensation temperature to be 42 F.  Thus condensation will occur between surfaces 4 and 5.

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    1.10 1.47 1000 1 1000 1.47  0 1.47 1.47 0.053 0.053 0.053 0.053 2.22 2.22 2.22 2.22 1.47 1.47 1.47           0 1.47 1 2 3 4 5 1 1000 15 0 0 0 70 T T T T T                                                              15 16.81 { } F 66.99 68.19 70 T                     q UA T   as an example: Btu (1.47)(1000)(70 68.19) 2660 hr q    or Btu (1.47)(1000)(16.81 15) 2660 hr q   

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    1.11 5.88 22.5 1 22.5 5.88  0 5.88 5.88 2.56 2.56 2.56 2.56 1.47 1.47 1.47        0 1.47 1 2 3 4 1 22.5 20 0 0 70 T T T T                                                  20 26.85 { } F 42.59 70 T                 1 Btu (5.88)(22.5)(26.85 20) 910 hr q U A T      or as an o ther example: 2 Btu (2.56)(22.5)(42.59 26.85) 910 hr q U A T      o r in out 1 1 ( ) (22.5)(70 20) Resistance (0.17 0.39 0.68) q A T T        Btu 910 hr q 

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    1.12 2 2 1 5 12 10.5 ( ) ( ) (3) (3) (0.5) 33.161 in 2 4 c c A A             2 2 2 4 (10.5 9) ( ) ( ) (3) 3 (0.5) 28.661 in 2 4 c c A A       2 2 3 ( ) (9)(3) 3 (0.5) 26.411 in 4 c A     6 1 5 (33.161)(3.27 10 ) lb ( ) ( ) 18,072,745 6 in c c K K     6 2 4 (28.661)(3.27 10 ) 1b ( ) ( ) 15,620,245 6 in c c K K     6 3 (26.411)(3.27 10 ) lb ( ) 21,590,992 4 in c K    2 6 1 2 4 5 (3) (0.5) (29 10 ) lb 4 ( ) ( ) ( ) ( ) 2,847,068 6 in s s s s K K K K              2 6 3 (3) (0.5) (29 10 ) lb 4 ( ) 4,270,602 4 in s K           The Combined stiffnesses: 1 5 18,072,745 2,847,068 20,919,813 K K     2 4 15,620,245 2,847,068 18,467,313 K K     3 21,590,992 4,270,602 25,861,594 K    20,919,813 1 20,919,813  0 20,919,813  0 20,919,813 18, 467,313 18, 467,313 18, 467,313 18, 467,313 25,861,594 25,861,594 25,861,594 25,861,594 18, 467,313 18, 467,313 18, 467,313 18, 467,313 20,919,813 20,919,813 20,919,813 20,919,813                              1 2 3 4 5 6 0 0 0 0 0 1000 u u u u u u                                                                5 4 1 2 3 4 4 4 4 5 6 0 4.78016 10 in 1.01951 10 in 1.40618 10 1.94768 10 in 2.42570 10 in u u u u u u                     

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    1.12 Cont. (1) 6 5 2 1 Concrete 2 ( ) (3.27 10 )( 4.78016 10 ) lb 26 C 6 in c u u E          (11 2 1 Steel 2 ( ) lb 231 C in s u u E     as a check: (1) (1) 2 Concrete 1 1 ( ) ( ) (26)(33.161) (231)(3) (0.5) 1000 lb 4 c s s A A              6 4 5 (2) 3 2 2 (3.27 10 )( 1.01951 10 4.78016 10 ) 1b 30 C 6 in c c u u E                   (2) 2 lb 231 C in s   C heck: (2) (2) 2 2 2 ( ) ( ) (30)(28.661) (231)(3) (0.5) 1000 lb 4 c c s s A A              6 4 4 (3) 4 3 2 ( ) (3.27 10 )( 1.40618 10 1.01951 10 ) lb 32 C 4 in c c u u E             6 4 4 (3) 4 3 2 ( ) (29 10 )( 1.40618 10 1.01951 10 ) lb 280 C 4 in s s u u E             C heck: (3) (3) 2 3 3 ( ) ( ) (32)(26.411) (280)(3) (0.5) 1000 lb 4 c c s s A A              6 4 4 (4) 5 4 2 ( ) (3.27 10 )( 1.94768 10 1.40618 10 ) lb 30 C 6 in c c u u E             (4) 2 lb 231 C in s   6 4 (5) 6 5 2 ( ) (3.27 10 )( 2.42570 1.94768) 10 lb 26 C 6 in c c u u E           (5) 2 lb 231 C in s   note (1) (5) c c    and (2) (4) c c    as expected.

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    1.13   2 avg ( ) 1 2 e i i A E u u     (1) (2) (3) (4) (5) total            (1) (5) 2 2 1 1 1 1 2 1 1 2 1 ( ) ( ) ( ) ( ) 2 2 ( ) ( ) ( ) 2 c c s s i i i i c c s s A E A E u u u u A E A E u u             6 2 6 (1) 5 2 (33.161)(3.27 10 ) 3 (0.5) (29 10 ) 4 ( 4.78016 10 ) 0.024 lb in (2)(6)                              or 6 2 6 (5) 4 2 (33.161)(3.27 10 ) 3 (0.5) (29 10 ) 4 [( 2.4257 1.94768) 10 ] 0.024 (2)(6)                                (2) 2 2 3 2 6 2 6 4 5 2 ( ) ( ) ( ) 2 (28.661)(3.27 10 ) 3 (0.5) (29 10 ) 4 [( 1.01951 10 4.78016 10 )] 0.027 2(6) c c s s A E A E u u                           or 6 2 6 (4) 4 2 (28.661)(3.27 10 ) 3 (0.5) (29 10 ) 4 [( 1.94768 1.40618) 10 ] 0.027 2(6)                        (3) 3 3 4 3 ( ) ( ) ( ) 2 c c s s A E A E u u     6 2 6 (3) 4 2 (26.411)(3.27 10 ) 3 (0.5) (29 10 ) 4 [( 1.40618 1.01951 10 ] 2(4) 0.019 lb in                       total (2)(0.024) 0.019 2(0.027) 0.121 lb in      

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    1.14 (a) 0 0 M   ( 6 lb)(5 in) (35 lb)(6 in) (8 in) 0 s F     30 lb s F  30 lb 60 0.5 in s F KX X X    (b) ( ) 1 1 0 n m e i i i i i i i F u u u u                2 1 3 5 35 (6) 0 2 4 8 KX X X X X                                3 5 (35) (6) 0 4 8 KX                3 5 (35) (6) 4 8 60 X               0.5 in X 

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    1.15 ( or ) Because of the fact that charge is conserved in a c ircuit (Kirchhoff’s c urrent law), at any time, the algebraic s um of the currents entering any node must be zero. Thus we can write 1 1 2 1 ( ) I V V R   2 2 1 1 ( ) I V V R   Note 1 2 0 I I   In matrix form: 1 1 2 2 1 1 1 1 1 V I V I R                     

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    1.16 (1) 1 1 1 [ ] 1 1 5 K          (2) 1 1 1 [ ] 1 1 10 K          (3) 1 1 1 [ ] ; 1 1 15 K          Apply 3 0 V  as a boundary c ondition 1 1 0 5 5 1 1 1 1 1 1 5 5 10 15 10 15 1 1 0 10 15         0 1 1 10 15  1 2 3 1 1 2 0.01 0 0 0.11 0.06 V V V V V                                              1 2 0.05 Volts V V   2 3 0.06 Volts V V  

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    1.17 2 2 ( ) ( ) 2 d Y M X wX L X EI EI dX    2 3 1 1 2 2 3 dY wX L wX C dX EI             3 4 1 2 1 2 6 12 wX L wX Y C X C EI             Applying Boundary Conditions: 2 3 1 0 @ 0 0 0 @ 24 Y X C wL Y X L C EI        3 2 3 exact ( 2 ) 24 wX Y X LX L EI     (a) Note t hat the assumed Solution s atisfies the boundary c onditions. 2 1 X X Y C L L                       1 2 2 1 dY X C dX L L         2 1 2 2 2 d Y C dX L          1 2 2 2 wX L X C EI L    R We may force the error function to equal z ero at 2 L X  4 1 1 2 2 2 2 0 2 16 L L w L C wL C EI EI L            2 4 16 wL X X Y EI L L                      

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    1.17 Cont. (b) 0 0 L R dX   1 2 0 2 ( ) 0 2 L c wX L X dX EI L           3 3 4 1 1 2 2 0 2 2 3 24 c w L L wL L c EI EI L              2 4 24 wL X X Y EI L L                       For W 2 4 × 104 Beam 4 3100 in ; I  6 2 lb 29 10 in E   Exact   4 4 max 6 4 2 lb 1 ft 12 in 5 5000 20 ft ft 12 in ft 5 0.20 in lb 384 (384) 29 10 3100 in in wL Y EI                           Collocation 4 4 max 6 4 2 lb 1 ft 12 in 5000 20 ft ft 12 in ft 0.24 in lb 64 (64) 29 10 (3100 in ) in wL Y EI                           Subdomain   4 4 max 6 4 2 lb 1 ft 12 in 5000 20 ft ft 12 in ft 0.16 in lb 96 (96) 29 10 3100 in in wL Y EI                          

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    1.18 S ee S ection 1.7 1.19 2 2 d u dP dX dy   1 1 du dP y c dy dX    2 1 2 1 ( ) 2 dP y u y c y c dX     applying B.Cs (0) 0 u  and ( ) 0 u H  2 2 1 2 1 1 0 0 0 and 0 2 0 and 2 dP H c c H dX H dP c c dX           2 1 ( ) ( ) 2 dP u y Hy y dX     note because P ressure drops in the direction of flow, 0 dP dX  in velocity equation. (a) 2 assumed 1 2 sin d u dP y u c dX H dy           note the assumed solution S atisfies the boundary c onditions. 1 1 sin cos du d y y c c dy dy H H H                        2 2 1 1 2 cos sin d u d y y c c dy H H H H dy                                2 1 sin y dP c H H dX                           R We may force the error function to equal z ero at 2 H y  2 1 sin 0 2 H dP c H H dX                           2 2 1 2 2 then, ( ) sin H dP H dP y c u y dX dX H                   

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    1.19 Cont. (b) 0 0 H R dy   2 1 0 sin 0 H y dP c dy H H dX                              2 1 0 1 cos 0 H y dP c H H H dX H                       1 ( 2) dP c H H dX     2 2 1 then, ( ) sin 2 2 H dP H dP y c u y dX dX H                    For C omparison, let’s use: 5 8 2 N S Pa 0.02 ; 0.01 mm 1 10 m; 1 10 m m dP H dX           evaluating max vel. @ 2 H y  Exact 2 2 8 5 2 max 1 1 ( 1 10 )(1 10 ) m 0.06 2 2 2 2 4 (8)(0.02) s dP H H dP H u H dX dX                                   Collocation method 2 2 5 2 8 max 2 2 2 (1 10 ) ( 1 10 ) m 2 sin 0.05 s (0.02) H H dP H dP u dX H dX                                      Subdomain method 2 2 5 2 8 max (1 10 ) ( 1 10 ) m 2 sin 0.08 2 2 (2)(0.02) s H H dP H dP u dX H dX                                     

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    1.20 (a) Galerkin method 2 1 0 sin sin 0 H y y dP C dy H H H dX                                     2 2 1 0 0 sin sin H H y dP y C dy dy H H dX H                           2 1 0 0 1 1 2 1 sin cos 2 4 H H dP C y y y H H dX H H H                                                         2 1 1 ( 2) 2 H dP C H H dX                      2 2 1 3 3 4 4 y then, sin H dP H dP C u y dX dX H              (b) least - squares method 1 2 2 1 1 0 0 sin sin sin 0 R c R H H R y y dP y R dy C dy C H H H dX H H                                                             result s in: 2 2 1 3 3 4 4 and ( ) sin H dP H dP y C u y dX dX H              using values from Problem 1.19 for c omparison 2 5 2 8 max 3 3 4 4(1 10 ) ( 1 10 ) m 0.06 s (0.02) H dP u dX           

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    1.21 2 2 d y PX EI dX   2 1 2 dy p X c dX EI    3 1 2 6 p X y c X c EI     B.Cs ① ( ) 0, y L  ② 0 X L dy dX   ② 2 2 1 1 0 2 2 P L P L c c EI EI     ① 3 2 3 2 2 1 0 6 2 3 L P L P P L C C L EI EI             3 2 3 exact ( 3 2 ) 6 P y X L X L EI     In terms of x : Substitute for X L x   and s implify 3 2 3 exact ( ( ) 3 ( ) 2 ) 6 P y L x L L x L EI         3 2 exact 3 6 P y x L x EI     2 2 0 d y EI P L x dx    B.Cs   0 0 y  and 0 0 x dy dx   Let us assume: 2 3 assumed 1 2 y c x c x   note, the assumed s olution s atisfies the boundary c onditions.

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    1.21 Cont. 2 1 2 2 3 dy c x c x dx   2 1 2 2 2 6 d y c c x dx   then Residual R becomes: 1 2 (2 6 ) ( ) R EI c c x P L x     Subdomain method: 2 2 1 2 0 0 0 [ (2 6 ) ( )] 0 L L R dx EI c c x P L x dx        (1) 2 2 1 2 0 [ (2 6 ) ( )] 0 L L L L R dx EI c c x P L x dx        (2) Integrating (1) : 2 2 1 2 2 2 2 6 0 2 2 2 2 L L L L EI c c P L                                                         Simplifying: 1 2 3 3 2 2 4 PL c c L EI    (3) Integrating (2) and Simplifying: 1 2 9 4 8 PL c c L EI    (4) Solving (3) and (4) Simultaneously: 1 2 and 2 6 PL P c c EI EI    3 2 ( 3 ) 6 P y x Lx EI   note, because the assumed s olution has the s ame functional form as the exact s olution, the c oefficients are exact.

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    1.21 Cont. Galerkin method: 2 2 1 2 0 0 0 [ (2 6 ) ( )] 0 L L x R dx x EI c c P L x dx        (1) 3 3 1 2 0 0 0 [ (2 6 ) ( )] 0 L L x R dx x EI c c P L x dx        (2) Integrating and Simplifying (1) and (2), we get: 1 2 1 2 2 3 3 2 12 1 6 2 5 20 PL c c L EI PL c c L EI              Solving (3) and (4) Simultaneously 1 2 and 2 6 PL P c c EI EI    3 2 ( 3 ) 6 P y x Lx EI   (3) ( 4 )

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    1.22 4 1 3 1 2 J J r    4 4 1 3 1 1.5 0.497 in 2 2 J J           4 4 2 1 1 0.0982 in 2 2 J          6 (1) (3) 1 1 1 1 202942 202942 (0.497)(9.8 10 ) [ ] [ ] 1 1 1 1 202942 202942 (2)(12) JG K K                              6 (2) 1 1 61102 61102 (0.0982)(11.2 10 ) [ ] 1 1 61102 61102 (1.5)(12) K                    202942 1 202942  0 202942  0 202942 61102 61102 61102 61102 202942 202942      0 202942  0 202942 1 2 3 4 1 1 2 3 4 0 0 2400 0 0 0.002222 rad 0.009604 rad 0                                                              4 3 3 ( 0) 202942 0.009604 1949 lb in R K        1 1 2 ( 0) 202942 0.002222 451 lb in R K        note they add up to 2400 lb ⋅ in

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    1.23 2 3 264044 61102 1200 61102 264044 1200                        2 3 0.005913 rad     1 4 1 2 3 3 ( 0) ( 0) 202942 0.005913 1200 lb in R R K K           

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