Question

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4. Two previously undeformed specimens of the same metal are to be plastically deformed by reducing their cross-sectional areas. One has a circular cross section, and the other is rectangular during deformation the circular cross section is to remain circular, and the rectangular is to remain as such. Their original and deformed dimensions are as follows: Rectangular (mm) Circular (diameter, mm) 125x 175 15.2 Original dimensions 7.5x 200 Deformed dimensions 11.4 Which of these specimens will be the hardest after plastic deformation, and why? 5. A cylindrical rod of 1040 steel originally 15.2 mm (0.60 in.) in diameter is to be cold worked by drawing; the circular cross section will be maintained during deformation. A cold-worked tensile strength in excess of 840 MPa (122,000 psi) and a ductility of at least 12 %EL are desired. Furthermore, the final diameter must be 10 mm (0.40 in.). Explain how this may be accomplished. 140 70 120 900 800 1040 Steel 60 120 1040 Steel 800 700 100 so 700 600 100 8C 4C 600 500 Brass Brass 30- 60 400- Brass 500 Copper 40 20 300 60 400 Copper 1040 Steel 10 200 300 40 20 Copper 100 200 30 50 60 10 20 40 60 10 40 50 70 20 40 50 60 70 10 30 Percent cold work Percent cold work Percent cold work fel

Answer 1

4th Question Solution:

Calculate the area of the circular specimen for the original dimension. Aocirdar d Here, original diameter of the circular specimen is do. Substitute 15.2 mm for do. Aoiralar = x (15.2mm)2 4 A0ciralar = 181.49mm2 Calculate the area of the circular specimen for the deformed dimension. Adcredar d Here, deformed diameter of the circular specimen is da Substitute 11.4 mm for da (11.4mm)2 Adciralar = X Adirdar 102.07mm2 From Equation (1), calculate the percentage of cold work for the circular specimen. AocirularAdcircular A0irular %CWCircular x 100 Substitute 181.49mm2 for Aocirular and 102.07mm2 for Adeirular 2)-(102.07mm2 ) (181.49mm 181.49mm %CW x 100 circular 43.76% %CW circular

Calculate the area of the rectangle specimen for the original dimensions. rectangle Here, original length of the rectangle specimen is lo and original breadth of the rectangle specimen is bo Substitute 125 mm for lo and 175 mm for bo (125mm) x (175mm) Aoctgle 21875mm2 _ rectangle Calculate the area of the rectangle specimen for the deformed dimensions. Adectsngle = l/4X ba Here, deformed length of the rectangle specimen is la and deformed breadth of the rectangle specimen is bd. Substitute 75 mm for l4 and 200 mm forba Aetang (75mm) x (200mm) = 15000mm2 Ao 'rectangle From Equation (1), calculate the percentage of cold work for the rectangular specimen. Aorectangle-Adrectangle Aogectangle %CW.rectangle x 100 Substitute 21875mm2 for Ao and 15000mm2 for Ad (21875mm2)-(15000mm2) (21875mm*) %CWrectangle x 100 %CWrectangle = 31.43%

The comparison of percentages of cold work for the circular and rectangular specimens is expressed as, %CW circular > %CWrectangle 43.75% 31.42% The one which has greater cold work percentage will be the hardest after plastic deformation. Therefore, the deformed circular specimen will be harder.

5 th Question Solution:

The percent cold work and attendant tensile strength and ductility if the drawing is carried out without interruption is given by the equation 2 %CW x 100 d 15.2 mm 10 mm 2 100 56%CW x 15.2 mm 1040 130 700 600 s00 20 400 Ceppe i040 S Cpper 200 20 50 20 20 Percent cold work Percent cold work (2) (1) At 56%CW, the steel will have a tensile strength on the order of 920 MPa (133,000 psi) [Figure-(1) which is adequate. And, the dictility will be less than 10%EL [Figure-(2), which is insufficient Instead of performing the drawing in a single operation, initially draw some fraction of the total deformation, then anneal to recrystallize, and, finally, cold-work the material a second time in order to achieve the final diameter, tensile strength, and ductility Reference to Figure-(1) indicates that 20 %CW is necess ary to yield a tensile strength of 840 MPa (122,000 psi) Similarly, a maximum of 21%CW is possible for 12%EL [Figure (2)] The average of these extremes is 20.5%CW. Again using the above we equation, if the final diameter after the first drawing is d, then 10 mm 2 20.5%CW x 100 10 mm 20.5%CW 100 d 11.2 mm 0.45in Tensile strength (MPa) Tensile strength (MPa) Ductility (%EL)