Determine the force in members AB, BH, GH, EF, CE, and CD of the truss shown in Figure Q2. Indicate whether the members are in tension or compression.
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Answer:
The reaction at the supports can be determined by considering equilibrium of the entire truss. Since both the external loads are vertical, only the vertical component of the reaction at the hinged ends A need to be considered. Since the triangle AEC is a right angle triangle, with angle AEC = 90º. Then, AE = AC cos 60º = 5 X 0.5 = 2.5m CE = AC sin60º = 5 X 0.866 = 4.33m Since triangle ABE is an equilateral triangle and therefore, AB = BC = AE = 2.5m Distance of line of action of force 10KN from joint A, AF = AE cos 60º = 2.5 X 0.5 = 1.25m Again, the triangle BDC is a right angle triangle with angle BDC = 90º Also, BC = AC – AB = 5 – 2.5 = 2.5m BD = BC cos 60º = 2.5 X 0.5 = 1.25m Distance of line of action of force 12KN from joint A, AG = AB + BG = AB + BD cos 60º = 2.5 + 1.25 X 0.5 = 3.125m Taking moment about end A, We get RC X 5 = 12 X 3.125 + 10 X 1.25 = 50 RC = 10KN ...(i) ∑V = 0, RC + RA = 10 + 12 = 22KN RA = 12KN ...(ii)Read more on Sarthaks.com - https://www.sarthaks.com/512182/determine-the-forces-in-all-the-members-of-the-truss-loaded-and-supported-as-shown-in-fig