A horizontal rod is supported at both ends and loaded at the middle. If L and Y are length and Young's modulus repectively, then depression at the middle is directly proportional to

Options

A. L
B. `L^(2)`
C. Y
D. `(1)/(Y)`

Answer

To solve the problem of finding the relationship between the depression (deflection) at the middle of a horizontal rod supported at both ends and loaded at the middle, we can use the formula for deflection in a beam. Here’s a step-by-step solution:

### Step 1: Understand the Setup
We have a horizontal rod (beam) of length \( L \) that is supported at both ends. A load \( W \) is applied at the midpoint of the rod. This setup causes the rod to sag or deflect in the middle.

**Hint:** Visualize the rod and the load applied at its center to understand how it will sag.

### Step 2: Identify the Relevant Formula
The deflection \( \delta \) (depression at the middle) for a beam loaded at the center can be expressed using the following formula:
\[ \delta = \frac{W L^3}{48 Y I} \]
where:
- \( W \) is the load applied at the center,
- \( L \) is the length of the beam,
- \( Y \) is Young's modulus of the material,
- \( I \) is the moment of inertia of the beam's cross-section.

**Hint:** Remember that the moment of inertia \( I \) depends on the shape of the beam's cross-section.

### Step 3: Simplify the Formula
For a rectangular cross-section, the moment of inertia \( I \) can be expressed as:
\[ I = \frac{b t^3}{12} \]
where \( b \) is the breadth and \( t \) is the thickness of the beam. Substituting this into the deflection formula gives:
\[ \delta = \frac{W L^3}{48 Y \left(\frac{b t^3}{12}\right)} = \frac{W L^3 \cdot 12}{48 Y b t^3} = \frac{W L^3}{4 Y b t^3} \]

**Hint:** Focus on how the parameters \( W \), \( L \), \( Y \), \( b \), and \( t \) affect the deflection.

### Step 4: Analyze the Proportionality
From the derived formula, we can see that the deflection \( \delta \) is directly proportional to \( L^3 \) and \( W \), and inversely proportional to \( Y \), \( b \), and \( t^3 \):
\[ \delta \propto \frac{W L^3}{Y b t^3} \]

**Hint:** Identify which variables are directly proportional and which are inversely proportional.

### Step 5: Conclusion
Since the question specifically asks for the relationship of depression at the middle with respect to Young's modulus \( Y \), we conclude:
\[ \delta \propto \frac{1}{Y} \]
This means that the depression at the middle is directly proportional to \( \frac{1}{Y} \).

**Final Answer:** The depression at the middle is directly proportional to \( \frac{1}{Y} \).