Question

A cylindrical car component with 7 mm diameter is to be manufactured. The component is required to possess a yield strength of at least 700 MPa and an elongation to fracture of at least 11%. For this purpose, you use a cylindrical steel 1040 rod with an initial diameter of 9 mm. The percent cold work cannot exceed 40% during any single cold deformation step in order to avoid cracking. Using the diagram below, describe your manufacturing procedure for this component! (10 pts) 70 800 60 1040 Steel 700 50 600 500 Brass 30 400 Brass Copper 300 20 200 1040 Steel 10 Capper 100 10 20 30 40 50 60 70 0 0 40 60 70 10 20 30 Percent cold work Percent cold wcrk 50 40 n3%) ong Yield strength (MPal

Answer 1

sol. Given, df = 7mm, TS 2700 MPa, &>, 11% S do= amm %.cw required = | fo- Af) x100 I don- I df X100 dode "pico - (1-64.95/100 = (1-f4ye ] x100 %cw = 89.5% - At 39.5% of cw for 1040 steel, IS = 760 MPa, E = 85%. But le gequired, ductility for elongation) is 114. . There fore we need to do it in stages. From the diagram : for 1040 sleel, at TS= 100 MPa, % (W = 12% - at EL = 11%, % cw = 30% So, aff for our objective we need a cold work between it and 30%. Therefore, we will use too 291. cw. (Average of 126 4 306)

sleps: ① cold working of 20% (first drawing) © Annealing ② cold drawing to final diameter © Diameter after the 1st celd draw % (w = (1 - (de Jx100 0.21= 1- Country 0.21 = 1(a, y = 0.888= d. di = 7.8156 mm ♡ In annegling, diameter is constant - Therefore, de = d, = 1.8156 km © 2nd cold draw to fried diarreter to Tomos (de) r. cw of 1- Polygr ) X100 = (- Grenada ]xoco. %0w = 214 At 21% CW, TS = 750 MPa (( EL = 12:57 - There fore, all Conditions are met: X100 8756)