Question

A metal beam of length 1 m, breadth 2.5 cm and thickness 5 mm supported at its ends is loaded at its center by a weight of 25 N. The metal beam sags at the middle by an amount of (Young's modulus of the metal $= 2\times10^{11} Nm^{-2}$)

• A. 1 cm
• B. 1 mm
• C. 0.5 cm
• D. 3 mm

Hint:

Be very careful with units. Convert everything to SI units (meters) before plugging into the formula. Remember $d$ is cubed, so small thickness has a huge effect on sagging.

Answer 1

The Correct Option is A

Step 1: Formula for Depression of a Beam:
The depression (sag) $\delta$ at the center of a beam supported at ends and loaded at the center is given by: \[ \delta = \frac{WL^3}{4Ybd^3} \] Where: $W =$ Load ($25$ N) $L =$ Length ($1$ m) $Y =$ Young's Modulus ($2 \times 10^{11} \, Nm^{-2}$) $b =$ Breadth ($2.5$ cm $= 2.5 \times 10^{-2}$ m) $d =$ Thickness/Depth ($5$ mm $= 5 \times 10^{-3}$ m)
Step 2: Substitution and Calculation:
\[ \delta = \frac{25 \times (1)^3}{4 \times (2 \times 10^{11}) \times (2.5 \times 10^{-2}) \times (5 \times 10^{-3})^3} \] Calculate the denominator terms:

• $d^3 = (5 \times 10^{-3})^3 = 125 \times 10^{-9}$
• $4 \times 2 \times 10^{11} = 8 \times 10^{11}$
• Denom $= (8 \times 10^{11}) \times (2.5 \times 10^{-2}) \times (125 \times 10^{-9})$
• Combine numbers: $8 \times 2.5 \times 125 = 20 \times 125 = 2500$
• Combine powers of 10: $10^{11} \times 10^{-2} \times 10^{-9} = 10^0 = 1$
• Denom $= 2500$

\[ \delta = \frac{25}{2500} = \frac{1}{100} \, \text{m} \] \[ \delta = 0.01 \, \text{m} = 1 \, \text{cm} \]
Step 4: Final Answer:
The sag is 1 cm.