ETABS6 用戶準備資料
ETABS 6 用戶如何準備檔案
2. 載重組合之處理流程(loading combination)：
2.1 自*.FRM，讀取1~6個載重即， I、II、…D2等8個類型：
| sample file: ETX.FRM |
| LOAD CASE DEFINITION DATA LOAD LTYP I II III A B C D1 D2 1 0 1.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 2 0 0.000 1.000 0.000 0.000 0.000 0.000 0.000 0.000 3 0 0.000 0.000 0.000 1.000 0.000 0.000 0.000 0.000 4 0 0.000 0.000 0.000 0.000 1.000 0.000 0.000 0.000 5 0 0.000 0.000 0.000 -1.000 0.000 0.000 0.000 0.000 6 0 0.000 0.000 0.000 0.000 -1.000 0.000 0.000 0.000 |
2.2 由此解析出對應關係為 I=1、 II=2、 A=3、B=4、至於III、C、D1、D2等則無對應值
2.3 自*.INP，讀取載重組合公式：
| sample file: CNKXD6.INP |
| $ HEADING DATA CONKER 6.2 SAMPLE EXAMPLE FOR CONKER MANUAL SPECIAL MOMENT RESISTING CONCRETE FRAME $ $CONTROL DATA $ ICODE NFR NLC LDC LLC NRMP NRCP NRBP NCRV NPTS IPRI IPHI 2 1 33 1 2 3 3 8 7 21 1 0 $ MBB MBV MCI MCV MJV MJR 1 1 1 1 1 1 $ $LOAD COMBINATION DEFINITION DATA $L LTYP XI XII XIII XA XB XC XD1 XD2 1 0 1.400 1.700 0 0 0 0 0 0 2 0 0.048+1.050 1.275 0 1.403 0 0 0 0 3 0 0.048+1.050 1.275 0 -1.403 0 0 0 0 4 0 -0.048+1.050 1.275 0 1.403 0 0 0 0 5 0 -0.048+1.050 1.275 0 -1.403 0 0 0 0 6 0 0.048+1.050 1.275 0 0 1.403 0 0 0 7 0 0.048+1.050 1.275 0 0 -1.403 0 0 0 8 0 -0.048+1.050 1.275 0 0 1.403 0 0 0 9 0 -0.048+1.050 1.275 0 0 -1.403 0 0 0 10 0 0.049+0.900 0 0 1.430 0 0 0 0 11 0 0.049+0.900 0 0 -1.430 0 0 0 0 12 0 -0.049+0.900 0 0 1.430 0 0 0 0 13 0 -0.049+0.900 0 0 -1.430 0 0 0 0 14 0 0.049+0.900 0 0 0 1.430 0 0 0 15 0 0.049+0.900 0 0 0 -1.430 0 0 0 16 0 -0.049+0.900 0 0 0 1.430 0 0 0 17 0 -0.049+0.900 0 0 0 -1.430 0 0 0 18 0 0.161+1.050 1.275 0 0.421 0 0 0 0 19 0 0.161+1.050 1.275 0 -0.421 0 0 0 0 20 0 -0.161+1.050 1.275 0 0.421 0 0 0 0 21 0 -0.161+1.050 1.275 0 -0.421 0 0 0 0 22 0 0.161+1.050 1.275 0 0 0.421 0 0 0 23 0 0.161+1.050 1.275 0 0 -0.421 0 0 0 24 0 -0.161+1.050 1.275 0 0 0.421 0 0 0 25 0 -0.161+1.050 1.275 0 0 -0.421 0 0 0 26 0 0.165+0.900 0 0 0.429 0 0 0 0 27 0 0.165+0.900 0 0 -0.429 0 0 0 0 28 0 -0.165+0.900 0 0 0.429 0 0 0 0 29 0 -0.165+0.900 0 0 -0.429 0 0 0 0 30 0 0.165+0.900 0 0 0 0.429 0 0 0 31 0 0.165+0.900 0 0 0 -0.429 0 0 0 32 0 -0.165+0.900 0 0 0 0.429 0 0 0 33 0 -0.165+0.900 0 0 0 -0.429 0 0 0 $MATERIAL PROPERTY REDEFINITION DATA 注意 以下各行單位必與EKO檔一致，此例為 ton-m $MID MTYPE E U W M ALPHA FY FC FYS FCS 1 C 2.17E+06 0.17 2.400 0 0 4.2E4 0.21E4 2.8E4 0.21E4 2 C 2.17E+06 0.17 1.800 0 0 4.2E4 0.21E4 2.8E4 0.21E4 3 C 2.17E+06 0.17 0 0 0 4.2E4 0.21E4 2.8E4 0.21E4 |
2.4 算例：求Beam 12之loading-33之內力 = ?
2.4.1 已知loading-33= 0.735 x I – 0.429 x B = 0.735 x (FRM之1) – 0.429 x (FRM之4)
2.4.1.1 (FRM之1) 即某桿件於3165.FRM 之第12 CASE 1內力值
2.4.1.2 (FRM之4) 即某桿件於3165.FRM 之第12 CASE 4內力值
| BEAM FORCES AT LEVEL PHFL IN FRAME FRAME ALL BAY OUTPUT OUTPUT MAJOR MAJOR MINOR MINOR AXIAL TORSIONAL ID ID POINT MOMENT SHEAR MOMENT SHEAR FORCE MOMENT 12 CASE 1 END-I -2.74 -4.90 0.00 0.00 0.00 0.00 1/4-PT 2.00 -2.53 0.00 1/2-PT 3.71 -0.16 0.00 3/4-PT 2.39 2.22 0.00 END-J -1.94 4.59 0.00 12 CASE 2 END-I -0.80 -1.44 0.00 0.00 0.00 0.00 1/4-PT 0.59 -0.74 0.00 1/2-PT 1.09 -0.04 0.00 3/4-PT 0.69 0.66 0.00 END-J -0.60 1.36 0.00 12 CASE 3 END-I 0.00 0.00 0.00 0.00 0.00 0.02 1/4-PT 0.00 0.00 0.00 1/2-PT 0.00 0.00 0.00 3/4-PT 0.00 0.00 0.00 END-J 0.00 0.00 0.00 12 CASE 4 END-I 0.00 0.00 0.00 0.00 0.00 0.00 1/4-PT 0.00 0.00 0.00 1/2-PT 0.00 0.00 0.00 3/4-PT 0.00 0.00 0.00 END-J 0.00 0.00 0.00 12 CASE 5 END-I 0.00 0.00 0.00 0.00 0.00 -0.02 1/4-PT 0.00 0.00 0.00 1/2-PT 0.00 0.00 0.00 3/4-PT 0.00 0.00 0.00 END-J 0.00 0.00 0.00 12 CASE 6 END-I 0.00 0.00 0.00 0.00 0.00 0.00 1/4-PT 0.00 0.00 0.00 1/2-PT 0.00 0.00 0.00 3/4-PT 0.00 0.00 0.00 END-J 0.00 0.00 0.00 |