Question

Figure 3b() shows a step beam with different moment of inertia in member 1 and 2. Assemble the structure stiffness matrix, Ke. Then, calculate the reactions at both supports by using matrik stifness method. Assuming the elastic modulus of beam, E 200 GPa. 150 kN 3 5m 2 10 m 1 = 500 x 106 mm4 I = 250 x 106 mm 4 Rajah 3b(@)/Figure 3b() Given: Stiffness relations for a beam element 12 6 12 6 z12 12 6 12 If the beam is modified as shown in Figure 3b(), show how the modification is taken into consideration into the structure stifness matrix KBAM A- 4000 mm2 E = 200 GPa 5m C 150 kN 5m 10 m I-500 x 10mm I-250 x 10 mm Rajah 3b(i)/Figure 3b(i)

Answer 1

The solution is provided below:

PRINT REACTION SOLUTIONS PER NODE o POST1 TOTAL REACTION SOLUTION LISTING LOAD STEP 1 SUBSTEP-1 TIHE 1.0000 LOAD CASE- THE FOLLOUING,Y,Z SOLUTIONS ARE IN THE GLOBAL COORDINATE SYSTEH NODE FY H2 1 0.00000.10010E+006 0.0000 0.32106E-010-0.42918E-011 0.25078E+006 0.0000 0.16003E-010 0.42648E-011-0.24922E+006 FX FZ HK HY 52 0.0000 49896 TOTAL VALUES VALUE 0.0000 0.15000E+006 0.0000 0.48109E-010-0.27031E-013 1563.6

ANSYS H19.2 NODAL SOLUTION STEP-1 Acacemi TIME 1 USUM(AVG) MAY 31 2019 RSYS 0 21:56:04 DMX.595E-03 .132E-03 .264E-03 3 .3318-0 F-n3 :3975-03 .529E-03 661E-04 .595E-03

ANSYS H19,2 LINE STRESS STEP-1 Acacemi亡 SUB =1 TIME-1 MAY 31 2019 SMIS3 SMIS16 21:59:33 MIN =-245777 ELEM-1 MAX =244750 ELEM=51 27764.8 190247 -245777 -136771 81241 82267.8 244750 -191274 26738.1 135744

ANSYS 19,21 LINE STRESS STEP-1 Acacemi SUB 1 TIME 1 SMIS6 SMIS19 MIN =-100104 MAY 31 2019 22:00:11 MAX -49895.8 ELEM-52 -66770.9 -33437.6 33229.1 -100104 61z0 a -104.24 83437.6 16770.9 16562.4

ANSYS H19,2 Academic HAY 31, 2111.9 22:09:36

ANSYS LINE STRESS F19.2 STEP=1 Acacemi SUB-1 TIME-1 MAY 31 2019 SMIS3 SMIS16 22:15:27 MIN =-188959 ELEM-1 MAX-163993 יו-- 32091.6 110525 46342.1 124776 -188959 -149742 '11.3nn .. 얹 7125.21 85558.9 163993

ANSYS H19,2 LINE STRESS STEP-1 Acacemi亡 SUB =1 TIME-1 MAY 31 2019 SMIS6 SMIS19 22:15:59 MIN 56724.8 ELEM=4 9 MAX-283625 ELEM=51 -56724.8 -18906 -37815.4 3.43241 18912.8 -47270.1 -28360.7 -9451.27 9458.14 28367.5

PRINT REACTION SOLUTIONS PER NODE POST1 TOTAL REACTION SOLUTION LISTING k LOAD STEP 1SUBSTEP 1 TIHE- 1.0000 LOAD CASE0 THE FOLLOHING X,Y,Z SOLUTIONS ARE IN THE GLOBAL COORDINATE SYSTEH NODE Fr FY 56725 28368 64908 FZ 0.0000 0.0000 0.0000 0.88497E-011-0.53064E-011 0.14569E+006 HX 0.18193E-010-0.32997E-011 0.19180E+006 0.90982E-011 0.19909E-011-0.11685E+006 HY H2 1 13847. 52 6923.3 102 -20770 TOTAL VALUES ALUE 0.25939E-008 0.15000E+006 0.0000 0.36141E-010-0.66151E-011 0.22064E+006