Question

3. (25%) The horizontal rigid beam ABCD is supported by vertical bars BE and CF and is loaded by vertical forces P - 400 kN and P, -360 kN acting at points A and D respectively (see figure). Bars BE and CF are made of steel (E 25 cross-sectional areas ABE 11,000 mm and AcF 9,000 mm2. The distances betwee various points on the bars are shown in the figure. (a) Determine the reactions at points B and C. (b) Determ (b) Calculate the total strain energy of the two vertical bars. 0 GPa) and have ine the vertical displacements ?, and ?D of points A and D, respectively. -1.5m-1.5 m 2.1 m P 400kN 2.4 m P2 360kN 0.6 m

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Answer 1

Given data: P_1 = 400kN P_2 = 360kN E = 250Gpa A_BE = 11000 mm^2 A_cF = 9000 mm^2 From force equilibrium (on member ABCD) sum F = 0 R_B + R_C = 400 + 360 R_B + R_C = 760 KN----(A) from moment equilibrium sum M = 0 (on member ABCD) Moment about B P_2 times (1.5 + 2.2) - R_C times 1.5 - 400 times 1.5 = 0 36. Times 3.6 - 1.5 R_c - 600 = 0 R_c = (696/1.5) = 464 kN R_c = 464 kN (A) longrightarrow R_B = 296 KN FBD R_B = 296 KN R_C = 464 KN on the member BE and CF b) delta_(BE) = (R_B times L/AE)_BE (296 times 1000 times 3000)/(11000 times 250 times 10^3) = 0.323 mm delta_(CF) = (R_CL/AE)_CF = (464 times 1000 times 2.4 times 1000)/(9000 times 250 times 10^3) = 0.495 mm from similar triangle delta_BE - delta_A/1.5