Topic: Strength of Materials
Member CD of the assembly shown is an aluminium bar 0.80 meter long. Cross-sectional dimension of the bar is such that, its width is three times its thickness. For functional design of the bar CD, the allowable axial stress is 70 MPa, and the total elongation is not to exceed 0.72 mm. E value for aluminium is 70 GPa. Assume the connection at C does not reduce the net area of the bar. The thickness of the bar in metre is
- A. 2/$\sqrt{21*10^5}$ m
- B. 5/$\sqrt{21*10^5}$ m
- C. 2/$\sqrt{21*10^3}$ m
- D. 3/$\sqrt{21*10^3}$ m
Answer: C. 2/$\sqrt{21*10^3}$ m
Explanation: First, take the moment about pivot B to find the force P acting on member CD: Sum of moments M_B = 0 gives 6000 * 0.6 = P * 0.1, which yields P = 36000 N. The elongation is restricted to delta_CD = 0.72 mm = 0.72 * 10^-3 m. The elongation formula is delta = (P * L) / (A * E). Substituting the known values: A = (P * L) / (E * delta) = (36000 * 0.8) / (70 * 10^9 * 0.72 * 10^-3) = 28800 / (50.4 * 10^6) = 5.714 * 10^-4 m². The area is width * thickness = (3t) * t = 3t^2. Setting 3t^2 = 4 / (7 * 10^3) gives t² = 4 / (21 * 10^3). Taking the square root, t = 2 / $\sqrt{21 * 10^3}$ m.