Question

A bar of square cross section of side 10 mm and length 400 mm is subjected to a tensile load of 60 kN. If the modulus of elasticity of the material is 200 GPa, the elongation of the bar is

• A. 12 mm
• B. 6 mm
• C. 2.4 mm
• D. 1.2 mm

Hint:

The formula $\delta L = PL/AE$ is fundamental in mechanics of materials.
A common source of error is inconsistent units.
It's often easiest to convert everything to N and mm:
- 1 kN = $10^3$ N
- 1 GPa = $10^3$ N/mm$^2$
- 1 MPa = 1 N/mm$^2$

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question asks to calculate the total elongation (change in length) of a bar under a given tensile load.

Step 2: Key Formula or Approach:
The elongation ($\delta L$) of an axially loaded bar is given by the formula:
\[ \delta L = \frac{PL}{AE} \] where:
$P$ = Axial load
$L$ = Original length of the bar
$A$ = Cross-sectional area of the bar
$E$ = Modulus of Elasticity (Young's Modulus) of the material

Step 3: Detailed Explanation:
First, identify and convert all the given values to a consistent set of units (e.g., Newtons and millimeters).

• Load ($P$) = 60 kN = $60 \times 10^3$ N

• Length ($L$) = 400 mm

• Side of square cross section = 10 mm

• Cross-sectional area ($A$) = side $\times$ side = $10 \text{ mm} \times 10 \text{ mm} = 100 \text{ mm}^2$

• Modulus of Elasticity ($E$) = 200 GPa = $200 \times 10^9$ Pa = $200 \times 10^9$ N/m$^2$.
To convert GPa to N/mm$^2$: 1 GPa = $10^3$ N/mm$^2$.
So, $E = 200 \times 10^3$ N/mm$^2$.


Now, substitute these values into the elongation formula:
\[ \delta L = \frac{(60 \times 10^3 \text{ N}) \times (400 \text{ mm})}{(100 \text{ mm}^2) \times (200 \times 10^3 \text{ N/mm}^2)} \] \[ \delta L = \frac{24 \times 10^6}{20 \times 10^6} \text{ mm} \] \[ \delta L = \frac{24}{20} \text{ mm} = 1.2 \text{ mm} \]

Step 4: Final Answer:
The elongation of the bar is 1.2 mm.