【文章內容簡介】 7 8 9 .000000 10 11 12 13 .000000 14 15 16 17 .000000 18 19 20 21 .000000 22 23 24 The Information of Supports IS VS 11 .000000 12 .000000 13 .000000 21 .000000 22 .000000 23 .000000 31 .000000 32 .000000 33 .000000 41 .000000 42 .000000 43 .000000 ( NA= 864 ) ( NW= 2612 ) Loading Case 1 The Loadings at Joints NLJ= 10 ILJ PX PY PM 5 .0000 6 .0000 .00000 9 .0000 10 .0000 .00000 13 .0000 14 .0000 .00000 17 .0000 18 .0000 .00000 21 .0000 22 .0000 .00000 The Loadings at Members NLM= 5 ILM ITL PV DST 5 4 12 4 19 4 26 4 33 4 The Results of Calculation The Joint Displacementsjoint u v phi 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 The Terminal Forcesmember N(st) Q(st) M(st) N(en) Q(en) M(en) 1 2 3 .422 .161 4 .070 .307 5 6 .238 7 .946 .718 8 9 10 11 12 .392 13 .175 14 .337 15 16 17 18 19 .150 20 .064 21 .328 22 23 24 25 26 .106 27 .038 28 .145 29 30 31 32 33 34 .739 35 ( NA= 864 ) ( NW= 2632 )************************************** * * * 中布荷載內力計算 * * * ************************************** The Input Data The General Information E NM NJ NS NLC +07 35 24 12 1 The Information of Membersmember start end A I 1 1 5 2 2 6 3 3 7 4 4 8 5 5 6 6 6 7 7 7 8 8 5 9 9 6 10 10 7 11 11 8 12 12 9 10 13 10 11 14 11 12 15 9 13 16 10 14 17 11 15 18 12 16 19 13 14 20 14 15 21 15 16 22 13 17 23 14 18 24 15 19 25 16 20 26 17 18 27 18 19 28 19 20 29 17 21 30 18 22 31 19 23 32 20 24 33 21 22 34 22 23 35 23 24 The Joint Coordinatesjoint X Y 1 .000000 .000000 2 .000000 3 .000000 4 .000000 5 .000000 6 7 8 9 .000000 10 11 12 13 .