Question

The bean shown in Figure -2 force and monent Loadings wich are located at the center of The beam. Assuming The baan is discrétized in to the two elenests, determine the dispiaconest and rotation along with latemal Forces at each node using FEM . The bounding Conditions are shown figners and the applied moment 70.000 Num 1 and element Length is 7m The young's modules is E = 210 Gpa and the second moment of inertia is I= 4x104 throughat the beam leagth. lokn L-XXnd L-xxm MEXX KNM na find The local matrix Determine The global matrix usig super position Use the reccess matrix To calculat The displacmest and yotation ase local matrix to calculate internal forces and nonest.

Answer 1

WHILE MIT HAURRAIN the elemental According to Euler Bernoulli's Beam theory Stefness matrix 1 given by, 12 6L -12 6L [k] = El 6L 422 -62 2 L2 -12 -6L 12 -6L 6L 2L2 -bL 4L] Elemental stiffness matrix of element , [kjo). 210x 109 x 4x10-4 12 73 42 42 196 - 42 98 -12 42 -42 98 12 - 42 - 42 196 -12 42 244897.96 12 42 -12 42 196 – 42 -12 -42 42 98 -42 42 98 -42 196 12 2 12 42 -12 42 Elemental stiffness matrix of element @ Dxy(2) - 210 x 10°x 4x10-4 42 196 -42 98 -12 -42 12 -42 42 98 -42 196 73 244897.96 12 42 -12 42 42 196 - 42 98 -12 -42 12 -42 42 98 -42 196

The global stiffness matt fe geven [kg] = 244897.96 12 42 -12 42 42 196 -42 98 o 42 -12 -12 -42 24 42 98 392 -42 98 O -12 -42 1242 42 98 o -42 196. The equation is given by dig di [ka] [day © F2 d3y JP IN 03 M3 Boundary conditions: 20 § föred supports Affex applying boundary conditions eqo & reduced to diy = 0,20 S-10,000) -12 2 244891.96 70,000 24 42 "day 02 392 -42 98 -12 – 42 12 -42 1 day 42 98 -42 1961.03 After saving the above system of equation we get.

0.0068 day day 2 = 0.002917 $3 = 0·002917 0·0272284 Local forces for element s f 2 mi = 244897.96 42 f2 12 42 -12 42 196 -42 98 -12 - 42 12 -42 42 98 -42 196 0.0068 0.002917 maz NS 07 = 244897.96 ( -12 X0·0068 + 42K0-002917) fi= 10019.755 N fi- 10 KN mi = 244897.96 (-42X0.0068 + 9 8 x 0·002917) mi 65.142897 Nom 2 244897.96 ( 12X0.0066 12X0.0068 - 42X0.002917) f2 (1) =-10019.755 N or f2" 2 - 10 kn m₂ (1) 244897-96 (-42 - 42x0.0068 +196 0-002917) M₂ W) = 700 73.143 Nm.

ette Local forces for element 7212) 12 42 -12 42 0.0068 (2) m2 +3 ( 244 897. 96) 42 196-42 98 0.002917 m3 0.0272224 -12 -42 12 42 42 98 -42 196 0.002917) fzce)244897.96 ( 128 12X0.0068 +42x0.002917-12X0.0272224 +42X0.002917) -9.9918 N f2(2) = mace) 244897.96 ( 42 42x0.0068 + 196 x0.002917 - 42X0-0272224 +98x0.002917) m₂ (2 = -34.97 Nom. PA to • 244897-96 (-12 -12X0.0068 -42X0-002917 + 12x00272224 42 X 0.002917) PP f = 9.9918 N L.PL M3 2 24489 7.96 ( 4260-0068 +98 X0-002917 -42X000272224 + 1964 0.002917) M3 = -34-97 Nom,