2 KN 225 mm BF 550 mm 75 mm 100 mm Each of rods BD and...

Question

Question

2 KN 225 mm BF 550 mm 75 mm 100 mm Each of rods BD and CE is made of brass (E = 105 GPa) and has a cross-sectional area of 20

Answer 1

Answer #1

Solution:

Draw the free body diagram of the member ABC. FBD 2 KN 550 mm 75 mm | 100 mm Е СЕ

Take moment about point F. EM, = 0 (2000)(0.625) - Fx (0.075) - Fc (0.1) = 0 .... (1) FxD (0.075)+Ft (0.1) = 1.25x103

Draw the geometry of the member after applying the load. O

Use the similar triangles method and determine the compatibility equations. From the triangle FAA, FBB and FCC, Substitute

Write the deformation equation in the member CE. &c FELCE ......(4) AcgEce Here, Fc is the force in the member CE, Ece is you

Take the deflection at point B and C. Substitute the equations (4) and (5) in equation (2). 85 = 0.758 (EL) EL Since the leng

Calculate the force in the member BD. Substitute 6kN for Fp in the equation (6). FCE = 1.333F8D = 1.333(6) = 8kN Substitute 2

Answer 2

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