Demonstration Problem: 5.1. This example illustrates the procedure for a relatively simple case when the three-force member is horizontal and the two-force member is an inclined strut. Calculate the force in member “BD” and reaction at the pinned connection “A”. The force “F” = 50 kN, “x1” = 1.5 m “x2” = 0.8 m and “y1” = 1.05 m. Figure 5.1 – Demonstration Problem 5.1 Step (1):Inspection of the above diagram reveals that “BD” is a two-force member and the line of action of the force that it applies to “ABC” passes along the member through points “D” and “B”. Since “ABC” tends to rotate downwards about “A”, under the action of the applied force, “BD” is likely to be in compression. Step (2):The free body diagram of the three-force member is shown below with an assumed compression force in the two-force member “BD”. Horizontal and vertical components of the reaction at “A” are included, with assumed positive directions.
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