Question

The sag developed in a bar of length $l$, breadth $b$, and thickness $d$ when subjected to a load $w$ at the nter is (Y -- Young's modulus of the material of the bar)

• A. $\frac{w^2l^2}{4bd^2Y}$
• B. $\frac{wl^3}{bd^2Y}$
• C. $\frac{wl^3}{4bd^3Y}$
• D. $\frac{wl}{4bd^3}$

Hint:

Beam Deflection Formula:

• Use $\delta = \fracwl^348EI$ for center-loaded beam.
• Moment of inertia for rectangular section: $I = \fracbd^312$.
• Larger $d$ gives smaller sag due to cubic power in denominator.

Answer 1

The Correct Option is C

This is a simply supported beam with a point load at the center. Standard formula for central sag (deflection): $\delta = \frac{wl^3}{48EI}$.
For a rectangular cross-section, moment of inertia $I = \frac{bd^3}{12}$. Substituting into the formula:
$\delta = \frac{wl^3}{48Y \cdot \frac{bd^3}{12}} = \frac{wl^3}{4Ybd^3}$.
Thus, sag is $\frac{wl^3}{4bd^3Y}$ which matches option (3).

Answer 2

Step 1: Understand the problem
We need to find the sag (deflection) developed in a bar subjected to a load at its center.

Step 2: Given parameters
- Length of the bar = l
- Breadth of the bar = b
- Thickness of the bar = d
- Load applied at the center = w
- Young's modulus of the material = Y

Step 3: Relevant theory
For a bar supported at both ends and loaded at the center, the bending deflection (sag) is given by the formula:
\[ \delta = \frac{wl^3}{4bd^3Y} \]
where the moment of inertia I = \(\frac{bd^3}{12}\) and the formula accounts for the bending stiffness.

Step 4: Explanation
- The sag increases with the cube of the length l, indicating longer bars bend more.
- It decreases with increasing breadth b and thickness d (cubed), showing thicker and wider bars resist bending better.
- Higher Young's modulus Y means stiffer material and less sag.

Step 5: Conclusion
The sag developed in the bar is \(\displaystyle \frac{wl^3}{4bd^3Y}\).