A vertical force Facts on the cross-seaction of a cylinder body of radius .r. and length L. If Y is young.s modulus and `sigma` = possoing ratio, find the change in volume of the body?

Answer

To find the change in volume of a cylindrical body subjected to a vertical force, we can follow these steps:

### Step 1: Understand the Geometry of the Cylinder
The volume \( V \) of a cylinder is given by the formula:
\[ V = \pi r^2 L \]
where \( r \) is the radius and \( L \) is the length of the cylinder.

### Step 2: Relate Change in Volume to Strain
The change in volume \( \Delta V \) can be expressed in terms of the original volume \( V \) and the strains in the radial and longitudinal directions:
\[ \frac{\Delta V}{V} = 2 \frac{\Delta r}{r} + \frac{\Delta L}{L} \]
where \( \Delta r \) is the change in radius and \( \Delta L \) is the change in length.

### Step 3: Use Poisson's Ratio
Poisson's ratio \( \sigma \) relates the change in radius to the change in length:
\[ \sigma = -\frac{\Delta r / r}{\Delta L / L} \]
From this, we can express \( \frac{\Delta r}{r} \) in terms of \( \Delta L \):
\[ \frac{\Delta r}{r} = -\sigma \frac{\Delta L}{L} \]

### Step 4: Substitute into the Volume Change Equation
Substituting \( \frac{\Delta r}{r} \) into the volume change equation gives:
\[ \frac{\Delta V}{V} = 2 \left(-\sigma \frac{\Delta L}{L}\right) + \frac{\Delta L}{L} \]
This simplifies to:
\[ \frac{\Delta V}{V} = -2\sigma \frac{\Delta L}{L} + \frac{\Delta L}{L} \]
\[ \frac{\Delta V}{V} = \left(1 - 2\sigma\right) \frac{\Delta L}{L} \]

### Step 5: Relate Change in Length to Stress
The change in length \( \Delta L \) can be related to the applied force \( F \) using the formula for stress and Young's modulus \( Y \):
\[ \text{Stress} = \frac{F}{A} = \frac{F}{\pi r^2} \]
\[ \text{Strain} = \frac{\Delta L}{L} = \frac{\text{Stress}}{Y} = \frac{F}{Y \pi r^2} \]
Thus,
\[ \frac{\Delta L}{L} = \frac{F}{Y \pi r^2} \]

### Step 6: Substitute Back to Find Change in Volume
Now substituting \( \frac{\Delta L}{L} \) back into the volume change equation:
\[ \frac{\Delta V}{V} = (1 - 2\sigma) \left(\frac{F}{Y \pi r^2}\right) \]
Thus, the change in volume \( \Delta V \) can be expressed as:
\[ \Delta V = V \cdot \frac{\Delta V}{V} = \pi r^2 L \cdot (1 - 2\sigma) \left(\frac{F}{Y \pi r^2}\right) \]
This simplifies to:
\[ \Delta V = L \cdot (1 - 2\sigma) \left(\frac{F}{Y}\right) \]

### Final Answer
The change in volume of the body is:
\[ \Delta V = \frac{L(1 - 2\sigma)F}{Y} \]