Minimum Diameter Calculation Using True Stress-Strain

This question involves calculating the minimum diameter a metal rod can achieve when subjected to tensile loading, based on its true stress-strain relationship. We need to understand the concept of necking and how it relates to the material's properties.

Parameters Analysis

The problem provides the following information:

• Initial diameter of the metal rod, \(d_0 = 8\) mm.
• True stress-strain relationship: \(\sigma = 900 \times \epsilon^{0.6}\). This relationship follows the general form \(\sigma = K \epsilon^n\), where \(K = 900\) MPa represents the strength coefficient and \(n = 0.6\) is the strain hardening exponent.
• The rod is under tensile loading, which means it is being stretched.

Understanding Necking and Minimum Diameter

When a ductile metal rod is pulled (tensile loading) beyond its yield point, it experiences plastic deformation. Initially, this deformation is uniform across the rod's cross-section. However, after the stress reaches the Ultimate Tensile Strength (UTS), the deformation becomes localized, leading to a rapid decrease in the cross-sectional area in a specific region. This phenomenon is called "necking".

The minimum diameter of the rod is achieved precisely at the point where necking begins. This critical point corresponds to the maximum value of the *engineering stress* (\(\sigma_e\)) on the engineering stress-strain curve.

The relationship between true stress (\(\sigma\)) and engineering stress (\(\sigma_e\)) is given by:

$$ \sigma_e = \sigma \times \frac{A}{A_0} $$

Where \(A\) is the instantaneous cross-sectional area and \(A_0\) is the original cross-sectional area.

Due to volume conservation during plastic deformation, the product of cross-sectional area and length remains constant: \(A \times L = A_0 \times L_0\). This relationship implies \(\frac{A}{A_0} = \frac{L_0}{L}\).

True strain (\(\epsilon\)) is defined as \(\epsilon = \ln(\frac{L}{L_0})\). From this definition, we get \(\frac{L}{L_0} = e^{\epsilon}\), and consequently, \(\frac{L_0}{L} = e^{-\epsilon}\).

Substituting \(\frac{A}{A_0} = e^{-\epsilon}\) into the engineering stress equation gives:

$$ \sigma_e = \sigma \times e^{-\epsilon} $$

Now, using the given true stress-strain relationship, \(\sigma = K \epsilon^n\), we can express engineering stress as:

$$ \sigma_e = K \epsilon^n e^{-\epsilon} $$

To determine the condition for maximum engineering stress (\(\sigma_e\)), we need to find the derivative of \(\sigma_e\) with respect to true strain (\(\epsilon\)) and set it to zero:

$$ \frac{d\sigma_e}{d\epsilon} = \frac{d}{d\epsilon} (K \epsilon^n e^{-\epsilon}) $$

Using the product rule for differentiation:

$$ \frac{d\sigma_e}{d\epsilon} = K \left[ \left( n \epsilon^{n-1} \right) e^{-\epsilon} + \epsilon^n \left( -e^{-\epsilon} \right) \right] $$

$$ \frac{d\sigma_e}{d\epsilon} = K e^{-\epsilon} [n \epsilon^{n-1} - \epsilon^n] $$

Setting the derivative to zero to find the maximum:

$$ K e^{-\epsilon} [n \epsilon^{n-1} - \epsilon^n] = 0 $$

Since \(K\) and \(e^{-\epsilon}\) are always positive, the term in the brackets must be zero:

$$ n \epsilon^{n-1} - \epsilon^n = 0 $$

Assuming \(\epsilon \neq 0\), we can divide by \(\epsilon^{n-1}\):

$$ n - \epsilon = 0 $$

$$ \epsilon = n $$

This result is fundamental: necking initiates, and consequently the minimum diameter is reached, when the true strain (\(\epsilon\)) is equal to the strain hardening exponent (\(n\)).

Calculating Minimum Diameter Step-by-Step

For this problem, the strain hardening exponent is given as \(n = 0.6\). Therefore, the true strain at the point of minimum diameter is \(\epsilon = 0.6\).

We use the relationship derived from volume conservation, which connects the initial diameter (\(d_0\)) to the diameter (\(d\)) at any given true strain (\(\epsilon\)):

$$ d = d_0 e^{-\epsilon/2} $$

Let's substitute the given values into this formula:

• Initial diameter \(d_0 = 8\) mm
• True strain \(\epsilon = 0.6\)

Plugging these values in:

$$ d = 8 \times e^{-0.6 / 2} $$

$$ d = 8 \times e^{-0.3} $$

Now, calculate the numerical value:

$$ e^{-0.3} \approx 0.740818 $$

$$ d = 8 \times 0.740818 $$

$$ d \approx 5.9265 \text{ mm} $$

Result Comparison

The calculated minimum diameter is approximately 5.9265 mm. We compare this result with the options provided:

The calculated value \(5.9265\) mm is closest to option 3, which is 5.92 mm. Thus, 5.92 mm is the minimum diameter that can be achieved.