Question

Problem 7: Use element matrices potential energy formulations to solve the beam problem shown in Figure.6. Find nodal displacement values. Make sure to construct the grid information table and one-dimensional elements' line/mesh with appropriate labeling, numbering and displacement, loads nodal points. Steel: E = 20(105) N/cm2 A = 20 cm2 A = 12 cm? 15000N 8T=15°C 22500 a = 11(10-6)/C -85 cm- 85 -70 Figure.7: Beam under axial loads

Answer 1

From the given data:

Answer: The given beam is: steel. En 200106) w/cm? A = 200m2 22500N A = 12cm² SODU - Element O clement © Elemento nodal y Displacement or ou -sen-me AU o us .85 —* -70 -- un > Beam under anial loads steel (E) = 20x100 w/em? 67 3.15°C d = 16100) lc A = 20 cm Element :(() Element (2) Element (3) 1 = 12000 Az = leema Az = 20cm? Lewyth. 350M 22 = 85cm dg = 70 cm olange temperature (41) (67) = 15 c 150 Thermal GIL () o 3 il mio per nxiolc

find the Nodal Displacement values: Let the nodal displacement u bly, Uz and le at the nodes. » Let the fifo ifs, and fu are load es at nodes. And R, and Ru are the reaction forces. for, fod for are thermal loads for elements 1, 2, 3 Respectively. condition: Boundary P, το Pa = 15000 N Pzz'-22500 N Thermal load for elements for = A, E, DAT 112.xQ0 xiló x 12 x 16°215 12x20 X11 X15 For . 39,60 food - Aata CAT 512-x20 x 10" xlx 66 x 15

2 12 X 20 X11 X 15 Fox = -39,000 N fos = Az E ₂ KAT 20 xao xide xul x18" x15 = 20 x 20 x 1 X15 for a 66,000 N The Nodal forces are a F = P, R,+ P, E, -R, +0 -39600 f = R, -39600 Pa = Pa P, + fol. - Foz - 15000 + 39660 -39600 F < 15000 W f or P₂ = P₂ t Fox for - 29500 +39600 - 66000

= -88500 +39600 F = -48900 N "The stiffness values. 12x20 x 10 . 85 240 o duo xlo Ks = 2.8 xlo mlem . we know that Ki = K2 So K = 9.6 mowiem H3 = AB : 20X20 X1o. 4x106 = 40 x100 Ks - 5.4.2106 N.com

Assembled Global stiffness matäit and forces Lo O nline megtan JL Now putting the values we get I-216 (88+28) e o lo ins Kits -5.7 toll Bats -5.7 5.7 So, we gelo ras - 2800 R-39600 6. 2.3 5.6 2.8 15000 10 0 -2.8 8.5 -5.7 --48900 0 -5.7 5.71 uu l L Rut 66000

where uifuu 20 rol / R = 39660 2.6 2.8 15000 -2.8 56 250 lo-28 45 -48900 Rut 66000 0 -5.7 5.7 Elimination method r 150007 -48900 10. (15.664 -2.us] = 15000 (5.600- 2843) 10 = 15X102 5.642 -2.84; 15 x103 lor " : 5.642 - 2.849 - 15X10 3 ind 10€ (-2.50 +6.5 43) = -48100

(-2.8 68+8,543) 106 = -48.9 x103 -2.647 +8,5uz 48.9 x 10 -2.6418 +5 Uz = 48.9 xio? Solving the equation (1) 4 (2) (5.6.4. – 2.845 EIS 403) X22.6 (-2.842 +1.5.43 = -48.9 x108) *5.6 we get ,15.680 - 7.8 4 uz = 4a xio 3 - -15.664 + 47.643 = 27-3.84 2105 Now adding the equations (3) flu) 15.by - 7.& 403 -42 X10 -15.68ua + 47.64₂ -273. 84 x 10 29.7613 -231.84 2165 -231.84 X1003 39. 76

4 . -5.8 X10 uz 7 - 6x183 cm put in ea ① SO! 5.6 Wy2.8 (-6 x16 3 ) = 15 x16+3 5.642 + 16.8 810 3 - 158103 5.6(la = 15x163 - 46.5810? 5.600 (15-1618) xro? -led = (15-16.5) xio? 5.6 1 U S -0.32x10 cm The Nodal displacement at Nodes: EU,=0 Ugaro. 32x10 3 en = -6x10² in cor) a 5.8 x 103 cm Uu=0