A mild Steel bar 20cm long and 8cm x 6cm in cross section is subjected to a long longitudinal axial compression of 72 kN. Determine the value of the lateral force necesary to prevent any transverse strain in length and poisson ratio.

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: A mild steel bar

• Step 1: Identify given values and calculate cross-sectional area and axial stress. The length of the bar is $L = 20 \, \text{cm}$. The cross-section dimensions are $8 \, \text{cm}$ (height, let's call this $h$) and $6 \, \text{cm}$ (width, let's call this $b$). The cross-sectional area $A = h \times b = 8 \, \text{cm} \times 6 \, \text{cm} = 48 \, \text{cm}^2$. The axial compression force is $P = 72 \, \text{KN} = 72 \times 10^3 \, \text{N}$. Young's modulus is $E = 2 \times 10^5 \, \text{N/cm}^2$. Poisson's ratio is $\nu = 0.25$. The axial stress $\sigma_x$ (in the longitudinal direction, along the 20 cm length) is:

  $$\sigma_x = -\frac{P}{A} = -\frac{72 \times 10^3 \, \text{N}}{48 \, \text{cm}^2} = -1500 \, \text{N/cm}^2$$

  The negative sign indicates a compressive stress.

• Step 2: Determine the lateral forces necessary to prevent any transverse strain. To prevent transverse strain, we set $\epsilon_y = 0$ and $\epsilon_z = 0$. Using the generalized Hooke's Law for strains:

  $$\epsilon_y = \frac{1}{E} [\sigma_y - \nu(\sigma_x + \sigma_z)] = 0$$

  This simplifies to:

  $$\sigma_y - \nu\sigma_x - \nu\sigma_z = 0 \implies \sigma_y - \nu\sigma_z = \nu\sigma_x \quad (1)$$

  Similarly for $\epsilon_z$:

  $$\epsilon_z = \frac{1}{E} [\sigma_z - \nu(\sigma_x + \sigma_y)] = 0$$

  This simplifies to:

  $$\sigma_z - \nu\sigma_x - \nu\sigma_y = 0 \implies \sigma_z - \nu\sigma_y = \nu\sigma_x \quad (2)$$

  From equations (1) and (2), we can deduce that $\sigma_y = \sigma_z$. Substitute $\sigma_z = \sigma_y$ into equation (1):

  $$\sigma_y - \nu\sigma_y = \nu\sigma_x$$ $$\sigma_y(1-\nu) = \nu\sigma_x$$ $$\sigma_y = \frac{\nu}{1-\nu} \sigma_x$$

  Now, substitute the given values $\nu = 0.25$ and $\sigma_x = -1500 \, \text{N/cm}^2$:

  $$\sigma_y = \frac{0.25}{1-0.25} (-1500 \, \text{N/cm}^2) = \frac{0.25}{0.75} (-1500 \, \text{N/cm}^2) = \frac{1}{3} (-1500 \, \text{N/cm}^2) = -500 \, \text{N/cm}^2$$

  Therefore, $\sigma_y = \sigma_z = -500 \, \text{N/cm}^2$. Now, we calculate the lateral forces. The force in the y-direction ($F_y$) acts on the face with area $L \times b = 20 \, \text{cm} \times 6 \, \text{cm} = 120 \, \text{cm}^2$.

  $$F_y = \sigma_y \times (L \times b) = (-500 \, \text{N/cm}^2) \times (120 \, \text{cm}^2) = -60000 \, \text{N} = -60 \, \text{KN}$$

  The force in the z-direction ($F_z$) acts on the face with area $L \times h = 20 \, \text{cm} \times 8 \, \text{cm} = 160 \, \text{cm}^2$.

  $$F_z = \sigma_z \times (L \times h) = (-500 \, \text{N/cm}^2) \times (160 \, \text{cm}^2) = -80000 \, \text{N} = -80 \, \text{KN}$$

  The negative signs indicate that these are compressive forces.

• Step 3: Evaluate the resultant alteration in length. The alteration in length $\Delta L$ is given by $\Delta L = L \epsilon_x$. We need to find the longitudinal strain $\epsilon_x$ under the condition that $\epsilon_y = \epsilon_z = 0$. Using the generalized Hooke's Law for longitudinal strain:

  $$\epsilon_x = \frac{1}{E} [\sigma_x - \nu(\sigma_y + \sigma_z)]$$

  Substitute $\sigma_y = \sigma_z = \frac{\nu}{1-\nu} \sigma_x$:

  $$\epsilon_x = \frac{1}{E} \left[ \sigma_x - \nu \left( \frac{\nu}{1-\nu} \sigma_x + \frac{\nu}{1-\nu} \sigma_x \right) \right]$$ $$\epsilon_x = \frac{1}{E} \left[ \sigma_x - \frac{2\nu^2}{1-\nu} \sigma_x \right] = \frac{\sigma_x}{E} \left[ 1 - \frac{2\nu^2}{1-\nu} \right]$$ $$\epsilon_x = \frac{\sigma_x}{E} \left[ \frac{1-\nu-2\nu^2}{1-\nu} \right]$$

  Substitute $\sigma_x = -1500 \, \text{N/cm}^2$, $E = 2 \times 10^5 \, \text{N/cm}^2$, and $\nu = 0.25$:

  $$\epsilon_x = \frac{-1500 \, \text{N/cm}^2}{2 \times 10^5 \, \text{N/cm}^2} \left[ \frac{1-0.25-2(0.25)^2}{1-0.25} \right]$$ $$\epsilon_x = -0.0075 \left[ \frac{0.75 - 2(0.0625)}{0.75} \right] = -0.0075 \left[ \frac{0.75 - 0.125}{0.75} \right]$$ $$\epsilon_x = -0.0075 \left[ \frac{0.625}{0.75} \right] = -0.0075 \times \frac{5}{6} = -0.00625$$

  Now, calculate the alteration in length:

  $$\Delta L = L \epsilon_x = 20 \, \text{cm} \times (-0.00625) = -0.125 \, \text{cm}$$

  The negative sign indicates a reduction in length.

The lateral forces necessary to prevent any transverse strain are $\boxed{\text{60 KN (on 20cm x 6cm face) and 80 KN (on 20cm x 8cm face)}}$.
The resultant alteration in length is $\boxed{\text{-0.125 cm (reduction)}}$.