To calculate the change in volume ($\Delta V$) of the segment, we need to use the concept of dilatation, which is the relative change in volume of a material under stress.

Given:

• Tensile load: $F = 4 \, \text{kN} = 4000 \, \text{N}$
• Thickness of the steel plate: $t = 1.6 \, \text{mm}$
• Modulus of elasticity: $E = 200 \, \text{GPa} = 200 \times 10^9 \, \text{Pa}$
• Poisson’s ratio: $v = 0.30$
• Gage length: $L_0 = 50 \, \text{mm}$
• Width of the steel plate: Unknown (assuming as $w$)

Step 1: Calculate the stress ($\sigma$)

Stress is defined as the force per unit area. The cross-sectional area of the test coupon is the product of the width ($w$) and thickness ($t$):

$A = w \times t$

Thus, the stress is:

$\sigma = \frac{F}{A} = \frac{4000}{w \times 1.6 \times 10^{-3}} \, \text{N/m}^2$

Step 2: Calculate the longitudinal strain ($\varepsilon_L$)

The strain in the longitudinal direction (the direction of the tensile force) is given by:

$\varepsilon_L = \frac{\sigma}{E} = \frac{4000}{w \times 1.6 \times 10^{-3} \times 200 \times 10^9}$

Step 3: Calculate the lateral strain ($\varepsilon_T$)

Poisson's effect dictates that as the material stretches longitudinally, it will contract laterally. The lateral strain is given by:

$\varepsilon_T = -v \varepsilon_L$

Step 4: Calculate the volumetric strain (dilatation)

Volumetric strain (dilatation) is the sum of the strains in all three directions. Since the strain in two lateral directions are the same, we can write:

$\Delta V = \varepsilon_L + 2 \varepsilon_T = \varepsilon_L (1 - 2v)$

Finally, the change in volume for the segment can be found by multiplying the initial volume $V_0 = A \times L_0$ by the dilatation:

$\Delta V = V_0 \times \Delta V$

Would you like to proceed with a width estimate for final volume change, or provide more context on width $w$?