Question

(15 pts) 6. A cylindrical part is forged in an open die. The starting diameter = 50 mm and height = 40 mm. Height after forging = 30 mm. Coefficient of friction at the die-work interface = 0.20. Yield strength of the low-carbon steel work material = 105 MPa; its strength coefficient = 500 MPa and strain hardening exponent = 0.25. Determine (a) the romarse force for the viviu point is reacha ... tegenkor Lici........ .he longer tha n (a) Use a spread sheet calculator (such as Excel or MATLAB) to calculate and plot the compressive force - forging height curve. The height interval of the calculation should be 1 mm. Hence, forces at the height of 40 mm, 39 mm, 38 mm, ..., 30 mm should be obtained. (b) To obtain the forging work, we can integrate the equation dW=Fd disp = Fd -h =Fd Vhvi from h=40 mm to h=30 mm. Integration can be approximated by the summation W F , 1 mm. Please use the data you obtained in (a) to evaluate the forging work and compare your result with the one you get in problem 5 (c). Explain the discrepancy if there is any and provide possible suggestions to reduce the discrepancy.

Answer 1

bi - Yorun Di z omen initial diarneter bir corrien initial height hf Born a Height after forging + Di — To calculate, final diametse we belance the volume Initial volume (Vi) z final. Volume (VA) & Di²x hi 7 + w2x hf ! (5038 x 40 z 1 (06)? x 30 Df z 57.7335 mm Olfa 28.86 75 min 2 -0.2876 G = True Strain = cm (that) 2 in (20) Chi . 60 2 flow Stress z K ET? where, Kz strength Coefficient 500 mpa nz Strain boordening exponent z Oslo 60 z 500 X (0.28 768) Too z 366.183 MACE forging force in open die Jorging (f) f, 60 А; ГL+ ще a) L 3hf Where Goz flow Stress . As a final area If a final vadius of cylinder hf, final heights Me Coffident of friction

winds are for height, hz Homm. D 250 man – 3661e3x. Ty neog [ +220.2. f = 778.9151 KN D257.73 35 mm. for height, ha 30mm 9XO2X 29.8675 3x 30 2 366763* * 563 > 366-193 * I *57-2335)** | = 1081.6056 kN. Rest of value are plotted by the help of excel and Shown below! 06) to calculate forging work. - dwa fah Saw of cox Af> [ 1 2 ] ah We Go As (San + 24 a San coes Lendero + 2x0.20-288675*fenn] ) - 3661123* 100 * TX 66.05773357* [6-03-0amer => 12.7705 kJ. + 0.0115cmo 2]

? Book1 - Excel FORMULAS DATA - . Sign in REVIEW CHART TOOLS DESIGN FORMAT VIEW X GS FILE HOME INSERT PAGE LAYOUT Cut Calibri (Body) - 10 6 Copy Paste Format Painter Painter BI U - - Clipboard Font -A A === D X AutoSumĄ # - A E On Wrap Text Merge & Center - Alignment General , % , Number 809 Insert Delete Format Conditional Format as Cell Formatting Table Styles Styles Clear Sort & Find & Filter Select Editing Cells Chart 1 J K L M N O P Q R S T U V W X Y Z AA 1 2 Compressive force-forging height curve 1200.00 4 5 1000.00 H I height | force in kN 40 779.02 39 801.37 38 825.06 37 850.22 876.97 905.49 935.95 968.56 1003.55 1041.19 30 1081.80 800.00 600.00 32 400.00 31 200.00 0.00 - 30 32 34 36 38 40 Sheet1 + READY + 100% @ - -- 6 o Ę 18:01 Search the web and Windows SWP w x] PA ENG 17-03-2020

The question told us about the comparison with problem 5(c), but there is no trace of it. Rest of the question is solved above.