Question

A rectangular steel alloy A-36 (structural steel) plate is hanging vertically and supporting a hanging weight of 90 kN. The plate has a length of 0.5 m, a width of 75 mm, and a thickness of 6 mm. Assuming that the bar deforms elastically under this loading condition and given the elastic properties for the steel: E = 207 GPa and u= 0.30

a.) What is the final length of the bar?
b.) What are the final dimensions (width and thickness) of the plate?
c.) What is the % reduction in area?

Answer 1

a) Final length of bar = 0.5 + 0.4838 *10^-3 = 0.5004838 M

b)Final Thickness = 6- -1.739 * 10^-3 mm = 5.998260mm\

c) % Reduction in area = (450-449.7391/450 ) = 0.58 %.

Step-by-step explanation:

a) Change in length = Pl /AE = 90*1000*0.5*1000/75*207*6*10^3

= 0.4838mm Expansion.

Final length of bar = 0.5 + 0.4838 *10^-3 = 0.5004838 M

b )Change in width = - μpt/AE = -(0.3*90*1000*6/ 75*207*6*10^3)

= -1.739 * 10^-3 mm

Final Thickness = 6- -1.739 * 10^-3 mm = 5.998260mm

c )

New C/s area = 74.97827 *5.998260 = 449.7391 mm^2

% Reduction in area = (450-449.7391/450 ) = 0.58 %.