ME 303 MANUFACTURING ENGINEERING

ME 303 MANUFACTURING ENGINEERING

PROBLEM SETS

Prepared by: Kamil ?zden Ural Uluer Salih Alan Mehmet Bilal Atar

PROBLEM SET FOR CHAPTER 2

Q2.1 A copper bar is to be cold rolled into a section which must have a min. tensile strength of 390 MPa. If the final cross-sectional area is 20.13 mm2 and assuming the flow properties of the workpiece material are given as K = 450 MPa, n = 0.33. Calculate the followings: a) The tensile strength of the annealed material. b) The initial diameter of the copper bar.

Solution It is important to understand the question well. In this question, the copper bar is annealed first then a cold rolling operation is done on it. Initial diameter of the copper bar is same before and after the annealing operation.

a) Flow stress of the plastically deformed bar could be shown as

. As

shown in Course Slides at UTS

Then UTS of the original bar (no annealing, no cold working) is found as below:

u = 450 (0.33)0.33

u = 312.12 MPa

Since before necking occurs ( < n) area stays constant at the whole bar, the formula below could be used:

By putting the area ratio into the formula (i) UTS of the annealed material is found as:

b) For this question d0 can be found by using the ratio: if < n (which means necking is not started yet) with the UTScw value given in the question as UTScw = 390 MPa, to check the value

So > n which means that necking started, then UTScw becomes equal to flow stress of the deformed material,

Q2.2 A steel tensile specimen with an initial diameter of 20mm and gage length of 120mm is subjected to a load of 125 kN and a gage length of 135mm is observed. Assuming uniform deformation at this point, calculate the followings: a) calculate the true stress, strain and the instantaneous diameter b) supposing the plastic behavior of the specimen material is expressed as

(MPa) estimate the yield strength of the deformed specimen. Solution

a)

from volume constancy

b)

Q2.3

A tensile bar was machined with a stepped gage section consisting of two regions of

different diameters. The initial diameters of the two regions were 2.0 cm and 1.9 cm.

After a certain amount of stretching in tension, the diameters of the two regions were

measured as 1.893 cm and 1.698 cm, respectively. Assuming the tensile strain hardening

is described by

, find n for the material.

Solution

Before: 2.0 cm ccm cm

After: 1.893 cm rrrrr

1.9 cm

cm

1.698 cm

Strains at larger and smaller diameter sections are respectively:

()

()

The load carrying capacities at two sections must be the same:

/

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