A stepped circular bar 150 mm long with diameters 20, 15, and 10 mm along length AB = 40 mm, BC - 45 mm, and CD = 65 mm, respectively is subjected to various forces as shown in Fig. P-14.12(a). Determine the change in its length if E = 200 \(kN/m{m^2}\). (AMIE Winter 2023) Solution Considering portion AB, a tensile force of 20 kN is to act to maintain equilibrium. For portion BC, if it is considered to be subjected to a tensile force of 5 kN, then the net force at section B will be 15 kN as shown. Similarly, for the portion CD, a compressive force of 10 kN would make a net force of 15 kN at section C. Area of cross sections \(\begin{array}{l}{A_1} = \frac{\pi }{4}{(20)^2} = 314.16\,m{m^2}\\{A_2} = \frac{\pi }{4}{(15)^2} = 176.7\,m{m^2}\\{A_3} = \frac{\pi }{4}{(10)^2} = 78.54\,m{m^2}\end{array}\) Extension of AB \(\begin{array}{l}\delta {L_1} = \frac{{20x40}}{{314.16E}}\\ = \frac{{20x40}}{{314.16x200}}\\ = 1.273x{10^{ - 2}}\,mm\end{array}\) Extension of BC \(\begin{array}{l}\delta {L_2}...
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