A mild steel bar of length 450 mm tapers uniformly. The diameters at the ends are 36 mm and 18 mm, respectively. An axial load of 12kN is applied on the bar. E = 2 × 105 N/mm2. The elongation of the bar will be

Concept:Deflection (\(\Delta\)) of circular Tapering BAR:Extension of the rod under the axial pull, \(\delta l = \frac{{PL}}{{\left( {\frac{\PI }{4}} \right) \times \left( {{d_1}{d_2}} \right) \times E}}\)whereP = load appliedL = LENGTH of the barE = MODULUS of elasticityd1 and d2 are the tapered diameters\(\delta = \frac{{4PL}}{{\pi {d_1}{d_2}E}}\)Calculation: Elongation in bar, using standard formula is given by,\({\rm{\Delta }}L = \frac{{4PL}}{{\mu {D_1}{D_2}E}}\)P = 12 kNL = 450 mmD1 = 36 mmD2 = 18 mmE = 2 × 105 N/mm2\({\rm{\Delta }}L = \frac{{4\; \times \;\left( {12\; \times \;{{10}^3}} \right)\; \times \;450}}{{\pi\; \times \;36\; \times \;18\; \times \;2\; \times \;105}}\)\({\rm{\Delta }}L = \frac{1}{{6\pi }}\;mm\)