Assignment [3]  forming solutions
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  • 1. ASSIENMENT [2] METAL FORMING REVIEW QUESTIONS1- What are the differences between bulk deformation processes and sheet metal processes?Answer. In bulk deformation, the shape changes are significant, and the work parts have alow area-to-volume ratio. In sheet metal processes, the area-to-volume ratio is high.2- Extrusion is a fundamental shaping process. Describe it. Answer. Extrusion is a compression process in which the work material is forced to flow through a die orifice, thereby forcing its cross section to assume the profile of the orifice.3- Why is the term press working often used for sheet metal processes? Answer. The term press working is used because most sheet metal operations are performed on presses.4- What is the difference between deep drawing and bar drawing? Answer. Deep drawing is a sheet metal forming process used to fabricate cup-shaped parts; bar drawing is a bulk deformation process used to reduce the diameter of a cylindrical work part.5- Indicate the mathematical equation for the flow curve. Answer. The flow curve is defined in Eq. (18.1) as Yf = Kεn.6- List some of the products produced on a rolling mill. Answer. Rolled products include flatsheet and plate stock, round bar and rod stock, rails, structural shapes such as I-beams andchannels.7- What is draft in a rolling operation? Answer. Draft is the difference between the starting thickness and the final thickness as the workpiece passes between the two opposing rolls.8- What is sticking in a hot rolling operation? Answer. Sticking is a condition in hot rolling in which the surface of the workpiece adheres to the rolls as the piece passes between the rolls, causing severe deformation of the metal below the surface in order to allow passage through the roll gap.9- Identify some of the ways in which force in flat rolling can be reduced. Answer. Ways to reduce force in flat rolling include (1) use hot rolling, (2) reduce draft in each pass, and (3) use smaller diameter rolls. 19.19 Distinguish between direct and indirect extrusion. Answer. In direct extrusion, also known as forward extrusion, a metal billet is loaded into a container, and a ram compresses the material, forcing it to flow through a die opening at the opposite end of the container. In indirect extrusion, also known as backward extrusion, the die is incorporated into the ram, and as the ram compresses into the metal billet, the metal is forced to flow through the die opening in a direction that is opposite (backwards) of the ram motion.10- Name some products that are produced by extrusion.
  • 2. Answer. Products produced by continuous extrusion include structural shapes (window frames, shower stalls, channels), tubes and pipes, and rods of various cross sections. Products made by discrete extrusion include toothpaste tubes, aluminium beverage cans, and battery cases.11- In blanking of a circular sheet metal part, is the clearance applied to the punch diameter orthe die diameter? Answer. The die diameter equals the blank diameter, and the punch diameter is smaller by twice the clearance. Multiple Choice QUESTIONS 1- Which of the following is typical of the starting work geometry in sheet metal processes: (a) high volume-to-area ratio or (b) low volume-to-area ratio? Answer. (b).2- The flow curve expresses the behaviour of a metal in which of the following regions of the stress-strain curve: (a) elastic region or (b) plastic region? Answer. (b).3- The average flow stress is the flow stress multiplied by which of the following factors: (a) n, (b) (1+n), (c) 1/n, or (d) 1/(1+n), where n is the strain hardening exponent? Answer. (d).4- Hot working of metals refers to which one of the following temperature regions relative to the melting point of the given metal on an absolute temperature scale: (a) room temperature, (b) 0.2Tm, (c) 0.4Tm, or (d) 0.6Tm? Answer. (d).5- The starting workpiece in steel hot rolling of plate and sheet stock is which of the following(one best answer): (a) bar stock, (b) billet, (c) bloom, (d) slab, or (e) wire stock? Answer. (d).6- The maximum possible draft in a rolling operation depends on which of the following parameters (two correct answers): (a) coefficient of friction between roll and work, (b) roll diameter, (c) roll velocity, (d) stock thickness, (e) strain, and (f) strength coefficient of the work metal? Answer. (a) and (b).7- Which of the following are alternative names for indirect extrusion (two correct answers):(a) backward extrusion, (b) direct extrusion, (c) forward extrusion, (d) impact extrusion, and(e) reverse extrusion?Answer. (a) and (e).8- The production of tubing is possible in indirect extrusion but not in direct extrusion: (a) true or (b) false?
  • 3. Answer. (b). Tube and pipe cross sections can be produced by either direct or indirectextrusion.9- Which of the following stress or strength parameters is used in the computation of the force in an extrusion operation (one best answer): (a) average flow stress, (b) compression strength, (c) final flow stress, (d) tensile strength, (e) yield strength? Answer. (a).10- In which of the following extrusion operation is friction a factor in determining theextrusion force (one best answer): (a) direct extrusion or (b) indirect extrusion? Answer. (a).11- A circular sheet metal slug produced in a hole punching operation will have the samediameter as which of the following: (a) the die opening or (b) the punch?Answer. (a).12- The cutting force in a sheet metal blanking operation depends on which mechanicalproperty of the metal (one correct answer): (a) compressive strength, (b) modulus of elasticity,(c) shear strength, (d) strain rate, (e) tensile strength, or (f) yield strength?Answer. (c).13- The holding force in drawing is most likely to be (a) greater than, (b) equal to, or (c) lessthan the maximum drawing force?Answer. (c). PROBLEMS1- The strength coefficient = 550 MPa and strain-hardening exponent = 0.22 for a certainmetal. During a forming operation, the final true strain that the metal experiences = 0.85.Determine the flow stress at this strain and the average flow stress that the metal experiencedduring the operation. Solution: Flow stress Yf = 550(0.85)0.22 = 531 MPa. Average flow stress fY= 550(0.85)0.22/1.22 = 435 MPa. 2- A metal has a flow curve with parameters: strength coefficient = 850 MPa and strain-hardening exponent = 0.30. A tensile specimen of the metal with gage length = 100 mm isstretched to a length = 157 mm. Determine the flow stress at the new length and the averageflow stress that the metal has been subjected to during the deformation. Solution: ε = ln (157/100) = ln 1.57 = 0.451 Flow stress Yf = 850(0.451)0.30 = 669.4 MPa. Average flow stress fY= 850(0.451)0.30/1.30 = 514.9 MPa.3- For a certain metal, the strength coefficient = 700 MPa and strain-hardening exponent= 0.27. Determine the average flow stress that the metal experiences if it is subjected to astress that is equal to its strength coefficient K. Solution: Yf = K = 700 = Kεn = 700ε.27 ε must be equal to 1.0.
  • 4. fY= 700(1.0).27/1.27 = 700/1.27 = 551.2 MPa4- Determine the value of the strain-hardening exponent for a metal that will cause theaverage flow stress to be 3/4 of the final flow stress after deformation. Solution: fY= 0.75 Yf Kεn/(1+n) = 0.75 Kεn 1/(1+n) = 0.75 1 = 0.75(1+n) = 0.75 + 0.75n 0.25 = 0.75n n = 0.3335- A 2.0 in thick slab is 10.0 in wide and 12.0 ft long. Thickness is to be reduced in threesteps in a hot rolling operation. Each step will reduce the slab to 75% of its previousthickness. It is expected that for this metal and reduction, the slab will widen by 3% in eachstep. If the entry speed of the slab in the first step is 40 ft/min, and roll speed is the same forthe three steps, determine: (a) length and (b) exit velocity of the slab after the final reduction. Solution: (a) After three passes, tf = (0.75)(0.75)(0.75)(2.0) = 0.844 in wf = (1.03)(1.03)(1.03)(10.0) = 10.927 in towoLo = tfwfLf (2.0)(10.0)(12 x 12) = (0.844)(10.927)Lf Lf = (2.0)(10.0)(12 x 12)/(0.844)(10.927) = 312.3 in = 26.025 ft (b) Given that roll speed is the same at all three stands and that towovo = tfwfvf ,6- Determine the blanking force required in Problem 20.2, if the steel has shear strength = 325MPa and the tensile strength is 450 MPa. Solution: F = StL t = 2.0 mm from Problem 20.2. L = πD = 75π = 235.65 mm F = 325(2.0)(235.65) = 153,200 N7- Determine the tonnage requirement for the blanking operation in Problem 20.4, given thatthe stainless steel has a shear strength = 600 MPa. Solution: F = StL t = 4 mm from Problem 20.4. L = 85 + 50 + 25 + 25 + 35 + 25 + 25 + 50 = 320 mm F = 600(4.0)(320) = 768,000 N. This is about 86.3 tons