7/30/2019 TA201N_Deformation Processing_Self Assessment Questions Exam_20.pdf

1/1

TA201N: Introduction to Manufacturing Processes2005-2006 Semester II

Bulk and Sheet Deformation Processing

Self-Assessment Questions

1. A metal has a flow curve with parameters: K = 850 MPa and strain hardening

exponent n = 0.30. A tensile specimen of the metal with guage length 100 mm isstrectched to a length 157 mm. Determine the flow stress at the new length and the

average flow stress that the metal has been subjected to during the deformation.

(Ans: flow stress Yf= 669.4 MPa and avg. flow stress fY = 514.9 MPa)

2. Determine the value of the strain hardening exponent for a metal that will cause the

average flow stress to be 3/4th

of the final flow stress after deformation.

(Ans: n = 0.333)

3. A 300 mm wide strip 25 mm thick is fed through a rolling mill with two powered rollseach of radius 250 mm. The work thickness is to be reduced to 22 mm in one pass at a

roll speed of 50 rev/min. The work material has a flow curve defined by K = 275 MPa

and n = 0.15 and the coefficient of friction between the rolls and the work is assumed tobe 0.12. Determine if the friction is sufficient to permit the rolling operation. If so,

calculate the roll force, torque and the power.{Ans: Yes, since dmax = 3.6 mm; F= 1,444,786 N ; T = 19,786 N-m; Power: 207,201N-m/s (W) }

4. A cylindrical billet that is 100 mm long and 40 mm in diameter is reduced by indirectextrusion to 15 mm diameter. Die angle is 90 deg. If the Johnson equation has a = 0.8 and

b = 1.5 and the flow curve fot he work metal has K = 750 MPa and n = 0.15, determine:

(a) extrusion ratio; (b) true strain (assuming homogeneous deformation); (c) extrusion

strain, (d) ram pressure, and (e) ram force.{Ans: (a) rx = 7.111; (b) = ln rx = 1.962; (c) Johnson equation used to estimate extrusion strain x = a + b

ln (rx) therefore ans of (c) 3.742; (d) fY = 721.5 MPa and p = x fY = 2700 MPa; (e) F = 3,392,920 N}

5. Derive an expression for the reduction r in drawing as function of the drawing ratio

DR.{Ans: r = (D-DP)/D; DR = D/DP; therefore r = 1- (DP/D) = 1-(1/DR). Note that for drawing to occur DR

2.0; r 1%}

6. A round disk of 150 mm diameter is to be blanked from a 3.2 mm thick strip of hard

cold-rolled steel (a=0.075) whose shear strength = 310 MPa. Determine (a) theappropriate punch and die diameters, and (b) blanking force{Ans: c = 0.24; die opening dia= 150 mm & punch dia = 149.52 mm; L=471.2mm & F= 467,469 N (shear

strength*stock thickness*length of cut edge=StL; if only the ultimate tensile strength TS of the material is

known then S= 0.7*TS)}