Question

5) A bar made of steel (E = 200 GPa, Poisson ratio = 0.32) has the dimensions shown in Fig. 5. If an axial force of P= 80 kN is applied to the bar, determine the change in its length and the change in the dimensions of its cross-section. The material behaves elastically Selastically NOKN 100 mm Fig. 5

Answer 1

Ans P=80 KN = 80x103N E = 200 GPa = 200 x103 MPa = 0.32 , des = Area of cross-section of bar Li-initial length = 1.5m = 1500 mm bi - initial width = 100 mm ti-initial thicknen - 50 mm Axial stress on bar a J = 1 - P Ausb ixti 100x50 52= 16 N or MPa Using Hooke's law. (As the material behaves elastically) J = Ex & E= strain in z-direction. - AL 16= 200 x 10²X OL 80x103 imp AL- Change in its length ALF 16 x 1500 200x103 OL - 0.12 mm Poisson Ratio v= – lateral strain longitudinal strain V ve - Ex

0.32 - - (at) 0.32x 0.12 1500 x 50 = – at At= change in thickness -- 0.00128 mm At=tf-ti = – 0.00128 te = final thicknen = 50 - 0.00128 th= 49.998 mm Strain in a direction is given by formulas) Ex- en 97 / 1 - 5 / as no stresses in x and y direction ie Tx= Tyco - Ex = – cute = – 0.34х 16 - 200 x103) bi ob = -2.72x10-5 Negative sign shows contraction Ab=- 2.72x10-5x 100 Ab= – 0.00272 mm. ben bi = – 0.00272 min. be a final width = 100 – 0.00272 be- 99.997 mm