Problem 3 Using the engineering strcssstrain plot shown in t

Problem 3 Using the engineering strcss-strain plot shown in the next page, determine the following quantities, for a reduction in cross-section area at fracture equal to 63.3%. (1) ultimate tensile strength (2) 0.2% offset yield strength (3) clastic modulus (4) true plastic strain at necking (5) true stress at necking (6) true fracture strength (7) true fracture strain (8) strength coefficient (9) clongation at fracture

Solution

1)

Ultimate tensile strength = Stress at the point of fracture.

From the plot, stress at the end of the curve is = 75 ksi

Hence, Ultimate Tensile strength = 75 ksi

2)

In the graph locate the point where strain is 0.002. Draw a line parallel to the modulus line (linear portion of curve). Notice that the line intersects the stress-strain curve at a certain point. The ordinate of this point is the Yield Strength at 0.2% Offset.

We get it close to 85 ksi.

3)

At the end of linear portion of the curve, we can read from the graph that stress = 90 ksi and corresponding strain = 0.01

Hence, Elastic Modulus = Stress / Strain

= 90 / 0.01

= 9000 ksi

4)

Necking occurs when slope of curve becomes negative. In the given plot, it occurs at the highest stress point.

Engineering strain at necking = 0.2

Assuming original length of specimen is L,

Change in length = 0.2*L

New length = L + 0.2*L = 1.2*L

True plastic strain at necking = ln (New length / original length)

= ln (1.2*L / L)

= ln 1.2

= 0.182