Chapter 13 Problems

A series of cold rolling operations are to be used to reduce the thickness of a plate from 50 mm down to 25 mm in a reversing two-high mill. Roll diameter = 700 mm and coefficient of friction between rolls and work = 0.15. The specification is that the draft is to be equal on each pass. Determine (a) minimum number of passes required, and (b) draft for each pass?

(a) Maximum draft dmax = μ2 R = (0.15)2 (350) = 7.875 mm Minimum number of passes = (to - tf)/dmax = (50 - 25)/7.875 = 3.17 → 4 passes (b) Draft per pass d = (50 - 25)/4 = 6.25 mm

A cold heading operation is performed to produce the head on a steel nail. The strength coefficient for this steel is 600 MPa, and the strain hardening exponent is 0.22. Coefficient of friction at the die-work interface is 0.14. The wire stock out of which the nail is made is 5.00 mm in diameter. The head is to have a diameter of 9.5 mm and a thickness of 1.6 mm. The final length of the nail is 120 mm. (a) What length of stock must project out of the die in order to provide sufficient volume of material for this upsetting operation? (b) Compute the maximum force that the punch must apply to form the head in this open-die operation.

(a) Volume of nail head V = πDf hf/4 = π(9.5) (1.6)/4 = 113.4 mm . Ao = πDo2/4 = π(4.75)2/4 = 19.6 mm2 ho = V/Ao = 113.4/19.6 = 5.78 mm (b) ε = ln(5.78/1.6) = ln 3.61= 1.2837 Yf = 600(1.2837)0.22 = 634 MPaAf = π(9.5)2/4 = 70.9 mm2Kf = 1 + 0.4(0.14)(9.5/1.6) = 1.33 F = 1.33(634)(70.9) = 59,886 N

Rod stock that has an initial diameter of 0.50 in is drawn through a draw die with an entrance angle of 13°. The final diameter of the rod is = 0.375 in. The metal has a strength coefficient of 40,000 lb/in2 and a strain hardening exponent of 0.20. Coefficient of friction at the work-die interface = 0.1. Determine (a) area reduction, (b) draw force for the operation, and (c) horsepower to perform the operation if the exit velocity of the stock = 2 ft/sec.

(a) r = (Ao - Af)/AoAo = 0.25π(0.50)2 = 0.1964 in2Af = 0.25π(0.35)2 = 0.1104 in2r = (0.1964 - 0.1104)/0.1964 = 0.4375 (b) Draw force F:ε = ln(0.1964/0.1104) = ln 1.778 = 0.5754 Y f = 40,000(0.5754)0.20/1.20 = 29,845 lb/in2 φ = 0.88 + 0.12(D/Lc)D = 0.5(.50 + 0.375) = 0.438Lc = 0.5(0.50 - 0.375)/sin 13 = 0.2778 φ = 0.88 + 0.12(0.438/0.2778) = 1.069 F=Af Yf (1+μ/tanα)φ(lnAo/Af)F = 0.1104(29,845)(1 + 0.1/tan 13)(1.069)(0.5754) = 2907 lb (c) P = 2907(2 ft/sec x 60) = 348,800 ft/lb/min HP = 348800/33,000 = 10.57 hp

A cylindrical billet that is 100 mm long and 50 mm in diameter is reduced by indirect (backward) extrusion to a 20 mm diameter. The die angle is 90°. The Johnson equation has a = 0.8 and b = 1.4, and the flow curve for the work metal has a strength coefficient of 800 MPa and strain hardening exponent of 0.13. Determine (a) extrusion ratio, (b) true strain (homogeneous deformation), (c) extrusion strain, (d) ram pressure, and (e) ram force.

(a) rx = Ao/Af = Do2/Df2 = (50)2/(20)2 = 6.25 (b)ε=lnrx =ln6.25=1.833(c)εx =a+blnrx =0.8+1.4(1.833)=3.366 (d) Y f = 800(1.833)0.13/1.13 = 766.0 MPa p = 766.0(3.366) = 2578 MPa (e) Ao = πDo2/4 = π(50)2/4 = 1963.5 mm2 F = 2578(1963.5) = 5,062,000 N

A part is designed to be hot forged in an impression die. The projected area of the part, including 22flash, is 16 in . After trimming, the part has a projected area of 10 in . Part geometry is complex. As 2heated the work material yields at 10,000 lb/in , and has no tendency to strain harden. At room temperature, the material yields at 25,000 lb/in2 Determine the maximum force required to perform the forging operation.

Since the work material has no tendency to work harden, n = 0. From Table 19.1, choose Kf = 8.0.F = 8.0(10,000)(16) = 1,280,000 lb.

An indirect extrusion process starts with an aluminum billet with diameter = 2.0 in and length = 3.0 in. Final cross section after extrusion is a square with 1.0 in on a side. The die angle = 90°. The operation is performed cold and the strength coefficient of the metal K = 26,000 lb/in2 and strain- hardening exponent n = 0.20. In the Johnson extrusion strain equation, a = 0.8 and b = 1.2. (a) Compute the extrusion ratio, true strain, and extrusion strain. (b) What is the shape factor of the product? (c) If the butt left in the container at the end of the stroke is 0.5 in thick, what is the length of the extruded section? (d) Determine the ram pressure in the process.

(a) rx = Ao/AfAo = πDo2/4 = π(2)2/4 = 3.142 in2 Af =1.0x1.0=1.0in2rx = 3.142/1.0 = 3.142ε = ln 3.142 = 1.145εx = 0.8 + 1.3(1.145) = 2.174 (b) To determine the die shape factor we need to determine the perimeter of a circle whose area is 2 0.5 equal to that of the extruded cross section, A = 1.0 in . The radius of the circle is R = (1.0/π) = 0.5642 in, Cc = 2π(0.5642) = 3.545 inThe perimeter of the extruded cross section Cx = 4(1.0) = 4.0 inKx = 0.98 + 0.02(4.0/3.545)2.25 = 1.006 (c) Given that the butt thickness = 0.5 in Original volume V = (3.0)(π x 22/4) = 9.426 in3 The final volume consists of two sections: (1) butt, and (2) extrudate. The butt volume V1 = 232 (0.5)(π2 /4) = 1.571 in . The extrudate has a cross-sectional area Af = 1.0 in . Its volume V2 = LAf = 3 9.426 - 1.571 = 7.855 in . Thus, length L = 7.855/1.0 = 7.855 in (d) Y f = 26,000(1.145)0.2/1.2 = 22,261 lb/in3 p = 1.006(22,261)(2.174) = 48,698 lb/in2

A 42.0-mm-thick plate made of low carbon steel is to be reduced to 34.0 mm in one pass in a rolling operation. As the thickness is reduced, the plate widens by 4%. The yield strength of the steel plate is 174 MPa and the tensile strength is 290 MPa. The entrance speed of the plate is 15.0 m/min. The roll radius is 325 mm and the rotational speed is 49.0 rev/min. Determine (a) the minimum required coefficient of friction that would make this rolling operation possible, (b) exit velocity of the plate, and (c) forward slip.

(a)Maximumdraftdmax =μ2R Giventhatd=to -tf =42-34=8.0mm, μ2 = 8/325 = 0.0246μ = (0.0246)0.5 = 0.157 (b) Plate widens by 4%.towovo = tfwfvfwf = 1.04 wo42(wo)(15) = 34(1.04wo)vfvf = 42(wo)(15)/ 34(1.04wo) = 630/35.4 = 17.8 m/min (c) vr = π r2N= π(0.325)2(49.0) = 16.26 m/min s = (vf - vr)/vr = (17.8 - 16.26)/16.26 = 0.0947