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 requires finding the elongation of a square bar subjected to an axial tensile load using the relationship between load, dimensions, and Young's modulus.

Step 2: Key Formula or Approach:
The extension of a prismatic bar under axial loading is calculated using the expression \(\delta=\frac{PL}{AE}\), where \(P\) is the applied load, \(L\) is the original length, \(A\) is the cross-sectional area, and \(E\) is the modulus of elasticity.

Step 3: Detailed Explanation:
The applied tensile load is 60 kN, which is equal to \(60\times10^3\) N. The original length of the bar is 400 mm. Since the cross-section is square with each side measuring 10 mm, the area becomes \(10\times10=100\) mm². The modulus of elasticity is 200 GPa, equivalent to \(200\times10^3\) N/mm². Substituting these values into the elongation formula gives \(\delta=\frac{60\times10^3\times400}{100\times200\times10^3}=1.2\) mm. Hence, the bar stretches by 1.2 mm under the given load.

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