You have a 40 mm x 8 mm cross-section low carbon steel billet available (the length can be cut to size as needed) and your company possesses equipment with the following capabilities: 1. Hot open die forging (MF = 2) 2. Hot closed die forging (MF = 5) 3. Hot impression die forging (MF = 9) 4. Cold open die forging (MF = 1.5) 5. Cold closed die forging (MF-4) 6. Cold impression die forging (MF = 7) a) Select a process from the list above for the manufacture of the component shown in Figure 1 on page 2! There is no requirement for the surface finish and the design allows relatively large tolerances. b) To what length would you cut the blanks for forging? c) Calculate the required forging force! Material parameters and a list of equations are provided in the appendix d) Indicate 1 method to reduce the forging force requirement! 10 15 15 10 - All dimensions are in mm -Length in third direction = 10 mm Figure 1 17 8 8 Material: Low carbon steel Following process parameters apply: Coefficient of friction = 0.6 (hot) = 0.15 (cold) K = 600 MPa; C = 300 MPa; n = 0.27; m = 0.13 Radius of rolls in rolling stand = 0.6 m Shear yield strength of low carbon steel = 110 MPa (hot) = 320 MPa (cold) Tensile yield strength of low carbon steel = 210 MPa (hot) = 620 MPa (cold) Ultimate tensile strength of low carbon steel = 410 MPa (hot) = 820 MPa (cold) Forging speed = 0.25 m/sec Roll surface speed = 1.2 m/sec Extrusion speed = 0.12 m/sec. Max. length of extruder = 5 m. a = 14 (for drawing) Taverage q=k& O = CET ET = In &T= DEPARTMENT OF MECHANICAL ENGINEERING Close book Equations KET n+1 (1)| L = R(ho-h) ts= Fa average = m V hf A = LxW O yield 2 F Do F = OTaverage * HR *w* MF F = o*A*MF [6Vin (Aorginal) Afinal ET = In +- ( )| hfaverage ho ET = In average = Aoriginal Afinal 1.7 * A * Oraverage hfaverage - n(A) V A projected haverage ho L V & Taverage m ]] Aoriginal [0.8+1.2/n (Aoriginal]